Uploaded by alberto_pita23

(Wiley survival guides in engineering and science) Venkatarama Krishnan - Probability and random processes-Wiley-Interscience (2006) este

advertisement
B
PROBABILITY AND
RANDOM PROCESSES
WILEY SURVIVAL GUIDES IN ENGINEERING
AND SCIENCE
Emmanuel Desurvire, Editor
Wiley Survival Guide in Global Telecommunications: Signaling Principles,
Network Protocols, and Wireless Systems Emmanuel Desurvire
Wiley Survival Guide in Global Telecommunications: Broadband Access,
Optical Components and Networks, and Cryptography
Emmanuel Desurvire
Fiber to the Home: The New Empowerment Paul E. Green, Jr.
Probability and Random Processes Venkatarama Krishnan
B
PROBABILITY AND
RANDOM PROCESSES
Venkatarama Krishnan
Professor Emeritus of Electrical Engineering
University of Massachusetts Lowell
A John Wiley & Sons, Inc., Publication
Copyright # 2006 by John Wiley & Sons, Inc. All rights reserved
Published by John Wiley & Sons, Inc., Hoboken, New Jersey
Published simultaneously in Canada
No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form
or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as
permitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the prior
written permission of the Publisher, or authorization through payment of the appropriate per-copy
fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400,
fax (978) 750-4470, or on the web at www.copyright.com. Requests to the Publisher for permission
should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken,
NJ 07030, (201) 748-6011, fax (201) 748-6008, or online at http://www.wiley.com/go/permission.
Limit of Liability/Disclaimer of Warranty: While the publisher and author have used their best efforts
in preparing this book, they make no representations or warranties with respect to the accuracy or
completeness of the contents of this book and specifically disclaim any implied warranties of
merchantability or fitness for a particular purpose. No warranty may be created or extended by sales
representatives or written sales materials. The advice and strategies contained herein may not be
suitable for your situation. You should consult with a professional where appropriate. Neither the
publisher nor author shall be liable for any loss of profit or any other commercial damages, including
but not limited to special, incidental, consequential, or other damages.
For general information on our other products and services or for technical support, please contact our Customer
Care Department within the United States at (800) 762-2974, outside the United States at (317) 572-3993
or fax (317) 572-4002.
Wiley also publishes its books in a variety of electronic formats. Some content that appears in print
may not be available in electronic formats. For more information about Wiley products, visit our web site at
www.wiley.com.
Library of Congress Cataloging-in-Publication Data:
Krishnan, Venkatarama,
Probability and random processes / by Venkatarama Krishnan.
p. cm. — (Wiley survival guides in engineering and science)
Includes bibliographical references and index.
ISBN-13: 978-0-471-70354-9 (acid-free paper)
ISBN-10: 0-471-70354-0 (acid-free paper)
1. Probabilities. 2. Stochastic processes. 3. Engineering—Statistical methods.
4. Science—Statistical methods. I. Title. II. Series
QA273.K74 2006
519.2–dc22
2005057726
Printed in the United States of America
10 9 8 7
6 5 4 3
2 1
Contents
Preface, xi
4.6 Hypergeometric Distribution, 46
4.7 Poisson Distribution, 48
CHAPTER 1
Sets, Fields, and Events, 1
1.1
Set Definitions, 1
1.2
Set Operations, 3
1.3
Set Algebras, Fields, and Events, 8
4.8 Logarithmic Distribution, 55
4.9 Summary of Discrete Distributions, 62
CHAPTER 5
Random Variables, 64
5.1 Definition of Random Variables, 64
CHAPTER 2
Probability Space and Axioms, 10
2.1
Probability Space, 10
2.2
Conditional Probability, 14
2.3
Independence, 18
2.4
Total Probability and Bayes’
Theorem, 20
5.2 Determination of Distribution and
Density Functions, 66
5.3 Properties of Distribution and Density
Functions, 73
5.4 Distribution Functions from Density
Functions, 75
CHAPTER 6
CHAPTER 3
Basic Combinatorics, 25
Continuous Random Variables and Basic
Distributions, 79
3.1
Basic Counting Principles, 25
6.1 Introduction, 79
3.2
Permutations, 26
6.2 Uniform Distribution, 79
3.3
Combinations, 28
6.3 Exponential Distribution, 80
CHAPTER 4
Discrete Distributions, 37
6.4 Normal or Gaussian Distribution, 84
CHAPTER 7
4.1
Bernoulli Trials, 37
Other Continuous Distributions, 95
4.2
Binomial Distribution, 38
7.1 Introduction, 95
4.3
Multinomial Distribution, 41
7.2 Triangular Distribution, 95
4.4
Geometric Distribution, 42
7.3 Laplace Distribution, 96
4.5
Negative Binomial Distribution, 44
7.4 Erlang Distribution, 97
vii
viii
Contents
7.5 Gamma Distribution, 99
10.4
Higher-Order Moments, 153
7.6 Weibull Distribution, 101
10.5
Bivariate Gaussian, 154
7.7 Chi-Square Distribution, 102
7.8 Chi and Other Allied Distributions,
104
7.9 Student-t Density, 110
7.10 Snedecor F Distribution, 111
CHAPTER 11
Characteristic Functions and Generating
Functions, 155
11.1
Characteristic Functions, 155
11.2
Examples of Characteristic
Functions, 157
11.3
Generating Functions, 161
11.4
Examples of Generating
Functions, 162
11.5
Moment Generating Functions, 164
11.6
Cumulant Generating Functions, 167
11.7
Table of Means and Variances, 170
7.11 Lognormal Distribution, 112
7.12 Beta Distribution, 114
7.13 Cauchy Distribution, 115
7.14 Pareto Distribution, 117
7.15 Gibbs Distribution, 118
7.16 Mixed Distributions, 118
7.17 Summary of Distributions of
Continuous Random Variables, 119
CHAPTER 12
Functions of a Single Random Variable, 173
CHAPTER 8
12.1
Random Variable g(X ), 173
12.2
Distribution of Y ¼ g(X ), 174
12.3
Direct Determination of Density
fY ( y) from fX (x), 194
12.4
Inverse Problem: Finding g(x) Given
fX(x) and fY ( y), 200
12.5
Moments of a Function of
a Random Variable, 202
Conditional Densities and Distributions, 122
8.1 Conditional Distribution and Density
for P(A)= 0, 122
8.2 Conditional Distribution and Density
for P(A) ¼ 0, 126
8.3 Total Probability and Bayes’ Theorem
for Densities, 131
CHAPTER 9
Joint Densities and Distributions, 135
CHAPTER 13
9.1 Joint Discrete Distribution
Functions of Multiple Random
Variables, 206
Functions, 135
13.1
Function of Two Random Variables,
Z ¼ g(X,Y ), 206
13.2
Two Functions of Two Random
Variables, Z ¼ g(X,Y ),
W ¼ h(X,Y ), 222
13.3
Direct Determination of Joint Density
fZW(z,w ) from fXY (x,y ), 227
13.4
Solving Z ¼ g(X,Y ) Using an Auxiliary
Random Variable, 233
13.5
Multiple Functions of
Random Variables, 238
9.2 Joint Continuous Distribution
Functions, 136
9.3 Bivariate Gaussian Distributions, 144
CHAPTER 10
Moments and Conditional Moments, 146
10.1 Expectations, 146
10.2 Variance, 149
10.3 Means and Variances of Some
Distributions, 150
Contents
CHAPTER 14
Inequalities, Convergences, and Limit
Theorems, 241
14.1
Degenerate Random Variables, 241
14.2
Chebyshev and Allied
Inequalities, 242
14.3
Markov Inequality, 246
14.4
Chernoff Bound, 248
14.5
Cauchy–Schwartz Inequality, 251
14.6
Jensen’s Inequality, 254
14.7
Convergence Concepts, 256
14.8
Limit Theorems, 259
17.4 Diagonalization of
Covariance Matrices, 330
17.5 Simultaneous Diagonalization of
Covariance Matrices, 334
17.6 Linear Estimation of Vector
Variables, 337
CHAPTER 18
Estimation Theory, 340
18.1 Criteria of Estimators, 340
18.2 Estimation of Random
Variables, 342
18.3 Estimation of Parameters (Point
Estimation), 350
CHAPTER 15
Computer Methods for Generating
Random Variates, 264
18.4 Interval Estimation (Confidence
Intervals), 364
18.5 Hypothesis Testing (Binary), 373
15.1
Uniform-Distribution Random
Variates, 264
15.2
Histograms, 266
15.3
Inverse Transformation
Techniques, 269
Random Processes, 406
15.4
Convolution Techniques, 279
19.1 Basic Definitions, 406
15.5
Acceptance –Rejection
Techniques, 280
19.2 Stationary Random Processes, 420
CHAPTER 16
Elements of Matrix Algebra, 284
16.1
Basic Theory of Matrices, 284
16.2
Eigenvalues and
Eigenvectors of Matrices, 293
16.3
16.4
Vectors and Matrix
Differentiations, 301
Block Matrices, 308
18.6 Bayesian Estimation, 384
CHAPTER 19
19.3 Ergodic Processes, 439
19.4 Estimation of Parameters of
Random Processes, 445
19.5 Power Spectral Density, 472
CHAPTER 20
Classification of Random Processes, 490
20.1 Specifications of Random
Processes, 490
20.2 Poisson Process, 492
CHAPTER 17
Random Vectors and
Mean-Square Estimation, 311
20.3 Binomial Process, 505
20.4 Independent Increment Process, 507
17.1
Distributions and Densities, 311
20.5 Random-Walk Process, 512
17.2
Moments of Random Vectors, 319
20.6 Gaussian Process, 521
17.3
Vector Gaussian Random
Variables, 323
20.7 Wiener Process (Brownian
Motion), 523
ix
x
Contents
20.8 Markov Process, 527
CHAPTER 23
20.9 Markov Chain, 536
Probabilistic Methods in Transmission
Tomography, 666
20.10
Martingale Process, 551
23.1
Introduction, 666
20.11
Periodic Random Process, 557
23.2
Stochastic Model, 667
20.12
Aperiodic Random Process
(Karhunen – Loeve Expansion), 563
23.3
Stochastic Estimation Algorithm, 671
23.4
Prior Distribution PfMg, 674
23.5
Computer Simulation, 676
23.6
Results and Conclusions, 678
23.7
Discussion of Results, 681
23.8
References for Chapter 23, 681
CHAPTER 21
Random Processes and Linear
Systems, 574
21.1 Review of Linear Systems, 574
21.2 Random Processes through
Linear Systems, 578
21.3 Linear Filters, 592
21.4 Bandpass Stationary Random
Processes, 606
APPENDIXES
A
A Fourier Transform Tables, 683
B Cumulative Gaussian Tables, 687
C Inverse Cumulative Gaussian Tables, 692
D
Inverse Chi-Square Tables, 694
Weiner and Kalman Filters, 625
E
Inverse Student-t Tables, 701
22.1 Review of Orthogonality
F Cumulative Poisson Distribution, 704
CHAPTER 22
Principle, 625
G
Cumulative Binomial Distribution, 708
22.2 Wiener Filtering, 627
22.3 Discrete Kalman Filter, 647
22.4 Continuous Kalman Filter, 660
References, 714
Index, 716
Preface
Many good textbooks exist on probability and random processes written at the undergraduate level to the research level. However, there is no one handy and ready book
that explains most of the essential topics, such as random variables and most of their
frequently used discrete and continuous probability distribution functions; moments,
transformation, and convergences of random variables; characteristic and generating
functions; estimation theory and the associated orthogonality principle; vector random
variables; random processes and their autocovariance and cross-covariance functions; stationarity concepts; and random processes through linear systems and the associated
Wiener and Kalman filters. Engineering practitioners and students alike have to delve
through several books to get the required formulas or tables either to complete a project
or to finish a homework assignment. This book may alleviate this difficulty to some
extent and provide access to a compendium of most distribution functions used by
communication engineers, queuing theory specialists, signal processing engineers, biomedical engineers, and physicists. Probability tables with accuracy up to nine decimal
places are given in the appendixes to enhance the utility of this book. A particular
feature is the presentation of commonly occurring Fourier transforms where both the
time and frequency functions are drawn to scale.
Most of the theory has been explained with figures drawn to scale. To understand the
theory better, more than 300 examples are given with every step explained clearly. Following the adage that a figure is worth more than a thousand words, most of the examples are
also illustrated with figures drawn to scale, resulting in more than 400 diagrams. This book
will be of particular value to graduate and undergraduate students in electrical, computer,
and civil engineering as well as students in physics and applied mathematics for solving
homework assignments and projects. It will certainly be useful to communication and
signal processing engineers, and computer scientists in an industrial setting. It will also
serve as a good reference for research workers in biostatistics and financial market
analysis.
The salient features of this book are
.
Functional and statistical independence of random variables are explained.
.
Ready reference to commonly occurring density and distribution functions and their
means and variances is provided.
.
A section on Benford’s logarithmic law, which is used in detecting tax fraud, is
included.
.
More than 300 examples, many of them solved in different ways to illustrate the
theory and various applications of probability, are presented.
Most examples have been substantiated with graphs drawn to scale.
.
xi
xii
Preface
.
More than 400 figures have been drawn to scale to improve the clarity of understanding the theory and examples.
.
Bounds on the tails of Gaussian probability have been carefully developed.
A chapter has been devoted to computer generation of random variates.
.
.
.
Another chapter has been devoted to matrix algebra so that some estimation
problems such as the Kalman filter can be cast in matrix framework.
Estimation problems have been given considerable exposure.
.
Random processes defined and classified.
A section on martingale processes with examples has been included.
.
Markov chains with interesting examples have been discussed.
.
Wiener and Kalman filtering have been discussed with a number of examples.
Important properties are summarized in tables.
.
.
.
.
.
The final chapter is on applications of probability to tomographic imaging.
The appendixes consist of probability tables with nine-place decimal accuracy for
the most common probability distributions.
An extremely useful feature are the Fourier transform tables with both time and
frequency functions graphed carefully to scale.
After the introduction of functional independence in Section 1.2, the differences
between functional and statistical independences are discussed in Section 2.3 and clarified
in Example 2.3.2. The foundation for permutations and combinations are laid out in
Chapter 3, followed by discrete distributions in Chapter 4 with an end-of-chapter
summary of discrete distributions. Random variables are defined in Chapter 5, and most
of the frequently occurring continuous distributions are explained in Chapters 6 and
7. Section 6.4 discusses the new bounds on Gaussian tails. A comprehensive table is
given at the end of Chapter 7 for all the continuous distributions and densities. After
discussion of conditional densities, joint densities, and moments in Chapters 8 –10,
characteristic and other allied functions are explained in Chapter 11 with a presentation
of an end-of-chapter table of means and variances of all discrete and continuous distributions. Functions of random variables are discussed in Chapters 12 and 13 with numerous
examples. Chapter 14 discusses the various bounds on probabilities along with some commonly occurring inequalities. Computer generation of random variates is presented in
Chapter 15. Elements of matrix algebra along with vector and matrix differentiations
are considered in Chapter 16. Vector random variables and diagonalization of covariance
matrices and simultaneous diagonalization of two covariance matrices are taken up in
Chapter 17. Estimation and allied hypothesis testing are discussed in Chapter 18. After
random processes are explained in Chapter 19, they are carefully classified in Chapter
20, with Section 20.10 presenting martingale processes that find wide use in financial
engineering. Chapter 21 discusses the effects of passing random processes through
linear systems. A number of examples illustrate the basic ideas of Wiener and Kalman
filters in Chapter 22. The final chapter, Chapter 23, presents a practical application of
probability to tomographic imaging. The appendixes include probability tables up to
nine decimal places for Gaussian, chi-square, Student-t, Poisson, and binomial distributions, and Fourier transform tables with graphs for time and frequency functions carefully drawn to scale.
This book is suitable for students and practicing engineers who need a quick reference
to any probability distribution or table. It is also useful for self-study where a number of
Preface
xiii
carefully solved examples illustrate the applications of probability. Almost every topic has
solved examples. It can be used as an adjunct to any textbook on probability and may also
be prescribed as a textbook with homework problems drawn from other sources.
During the more than four decades that I have been continuously teaching probability
and random processes, many authors who have written excellent textbooks have
influenced me to a considerable extent. Many of the ideas and concepts that are expanded
in this book are the direct result of this influence. I owe a scientific debt of gratitude to all
the authors who have contributed to this ubiquitous and exciting field of probability. Some
of these authors are listed in the reference, and the list is by no means exhaustive.
Most of the graphs and all the probability tables in this book were created with
Mathcad software. All algebraic calculations were verified with Mathcad software.
Mathcad and Mathsoft are registered trademarks of Mathsoft Engineering and Education,
Inc., http://www.mathcad. com.
While facing extreme personal difficulties in writing this book, the unequivocal
support of my entire family have been a source of inspiration in finishing it.
I acknowledge with great pleasure the support given to me by the Wiley staff. George
Telecki welcomed the idea of this book; Rachel Witmer did more than her share of
keeping me in good humor with her cheerful disposition in spite of the ever-expanding
deadlines; and Kellsee Chu who did an excellent job in managing the production of this
book. Finally, the enthusiastic support given to me by the Series Editor Emmanuel
Desurvire was very refreshing. This is my second book with Wiley, and the skills of
their copyeditors and staff who transform highly mathematical manuscript into a
finished book continue to amaze me.
I have revised this book several times and corrected errors. Nevertheless, I cannot
claim that it is error-free since correcting errors in any book is a convergent process.
I sincerely hope that readers will bring to my attention any errors of commission or
omission that they may come across.
VENKATARAMA KRISHNAN
Chelmsford, Massachusetts
March 2006
CHAPTER
1
Sets, Fields, and Events
B 1.1
SET DEFINITIONS
The concept of sets play an important role in probability. We will define a set in the
following paragraph.
Definition of Set
A set is a collection of objects called elements. The elements of a set can also be sets.
Sets are usually represented by uppercase letters A, and elements are usually represented
by lowercase letters a. Thus
A ¼ {a1 ;a2 ; . . . , an }
(1:1:1)
will mean that the set A contains the elements a1,a2, . . . , an. Conversely, we can write that
ak is an element of A as
ak [ A
(1:1:2)
ak A
(1:1:3)
and ak is not an element of A as
A finite set contains a finite number of elements, for example, S ¼ f2,4,6g. Infinite sets will
have either countably infinite elements such as A ¼ fx : x is all positive integersg or
uncountably infinite elements such as B ¼ fx : x is real number 20g.
Example 1.1.1
The set A of all positive integers less than 7 is written as
A ¼ {x : x is a positive integer ,7}: finite set:
Example 1.1.2
The set N of all positive integers is written as
N ¼ {x : x is all positive integers}: countably infinite set:
Example 1.1.3
The set R of all real numbers is written as
R ¼ {x : x is real}: uncountably infinite set:
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
1
2
Sets, Fields, and Events
Example 1.1.4
The set R 2 of real numbers x, y is written as
R2 ¼ {(x,y) : x is real, y is real}
Example 1.1.5
The set C of all real numbers x,y such that x þ y 10 is written as
C ¼ {(x,y) : x þ y 10}: uncountably infinite set:
Venn Diagram
Sets can be represented graphically by means of a Venn diagram. In this case we
assume tacitly that S is a universal set under consideration. In Example 1.1.5, the universal
set S ¼ fx : x is all positive integersg. We shall represent the set A in Example 1.1.1 by
means of a Venn diagram of Fig. 1.1.1.
Empty Set
An empty is a set that contains no element. It plays an important role in set theory and
is denoted by 1. The set A ¼ f0g is not an empty set since it contains the element 0.
Cardinality
The number of elements in the set A is called the cardinality of set A, and is denoted
by jAj. If it is an infinite set, then the cardinality is 1.
Example 1.1.6 The cardinality of the set A ¼ f2,4,6g is 3, or jAj ¼ 3. The cardinality of
set R ¼ fx : x is realg is 1.
Example 1.1.7
The cardinality of the set A ¼ fx : x is positive integer ,7g is jAj ¼ 6.
Example 1.1.8 The cardinality of the set B ¼ fx : x is a real number ,10g is infinity
since there are infinite real numbers ,10.
Subset
A set B is a subset of A if every element in B is an element of A and is written as
B # A. B is a proper subset of A if every element of A is not in B and is written
as B , A.
FIGURE 1.1.1
1.2 Set Operations
3
FIGURE 1.1.2
Equality of Sets
Two sets A and B are equal if B # A and A # B, that is, if every element of A is contained in B and every element of B is contained in A. In other words, sets A and B contain
exactly the same elements. Note that this is different from having the same cardinality, that
is, containing the same number of elements.
Example 1.1.9 The set B ¼ f1,3,5g is a proper subset of A ¼ f1,2,3,4,5,6g, whereas the
set C ¼ fx : x is a positive even integer 6g and the set D ¼ f2,4,6g are the same since they
contain the same elements. The cardinalities of B, C, and D are 3 and C ¼ D.
We shall now represent the sets A and B and the sets C and D in Example 1.1.9 by
means of the Venn diagram of Fig. 1.1.2 on a suitably defined universal set S.
Power Set
The power set of any set A is the set of all possible subsets of A and is denoted by
PS(A). Every power set of any set A must contain the set A itself and the empty set 1.
If n is the cardinality of the set A, then the cardinality of the power set jPS(A)j ¼ 2n.
Example 1.1.10 If the set A ¼ f1,2,3g then PS(A) ¼ f1, (1,2,3), (1,2), (2,3), (3,1), (1),
(2), (3)g. The cardinality jPS(A)j ¼ 8 ¼ 23.
B 1.2
SET OPERATIONS
Union
Let A and B be sets belonging to the universal set S. The union of sets A and B is
another set C whose elements are those that are in either A or B, and is denoted by
A < B. Where there is no confusion, it will also be represented as A þ B:
A < B ¼ A þ B ¼ {x : x [ A or x [ B}
(1:2:1)
4
Sets, Fields, and Events
FIGURE 1.2.1
Example 1.2.1 The union of sets A ¼ f1,2,3g and B ¼ f2,3,4,5g is the set
C ¼ A < B ¼ f1,2,3,4,5g.
Intersection
The intersection of the sets A and B is another set C whose elements are the same as
those in both A and B and is denoted by A > B. Where there is no confusion, it will also be
represented by AB.
A > B ¼ AB ¼ {x : x [ A and x [ B}
(1:2:2)
Example 1.2.2 The intersection of the sets A and B in Example 1.2.1 is the set C ¼ f2,3g
Examples 1.2.1 and 1.2.2 are shown in the Venn diagram of Fig. 1.2.1.
Mutually Exclusive Sets
Two sets A and B are called mutually exclusive if their intersection is empty. Mutually
exclusive sets are also called disjoint.
A>B¼1
(1:2:3)
One way to determine whether two sets A and B are mutually exclusive is to check whether
set B can occur when set A has already occurred and vice versa. If it cannot, then A and B
are mutually exclusive. For example, if a single coin is tossed, the two sets, fheadsg and
ftailsg, are mutually exclusive since ftailsg cannot occur when fheadsg has already
occurred and vice versa.
Independence
We will consider two types of independence. The first is known as functional
independence [42]. Two sets A and B can be called functionally independent if the occurrence of B does not in any way influence the occurrence of A and vice versa. The second
one is statistical independence, which is a different concept that will be defined later. As
an example, the tossing of a coin is functionally independent of the tossing of a die because
they do not depend on each other. However, the tossing of a coin and a die are not mutually
exclusive since any one can be tossed irrespective of the other. By the same token, pressure
and temperature are not functionally independent because the physics of the problem,
namely, Boyle’s law, connects these quantities. They are certainly not mutually exclusive.
Numbers in brackets refer to bibliographic entries in the References section at the end of the book.
1.2 Set Operations
5
Cardinality of Unions and Intersections
We can now ascertain the cardinality of the union of two sets A and B that are not
mutually exclusive. The cardinality of the union C ¼ A < B can be determined as
follows. If we add the cardinality jAj to the cardinality jBj, we have added the cardinality
of the intersection jA > Bj twice. Hence we have to subtract once the cardinality jA > Bj
as shown in Fig. 1.2.2. Or, in other words
jA < Bj ¼ jAj þ jBj jA > Bj
(1.2.4a)
In Fig. 1.2.2 the cardinality jAj ¼ 9 and the cardinality jBj ¼ 11 and the cardinality
jA < Bj is 11 þ 9 2 4 ¼ 16.
As a corollary, if sets A and B are mutually exclusive, then the cardinality of the union
is the sum of the cardinalities; or
jA < Bj ¼ jAj þ jBj
(1.2.4b)
The generalization of this result to an arbitrary union of n sets is called the inclusion–
exclusion principle, given by
n
n
n n X
[
X
X
Ai > A j þ
A i > A j > Ak jA i j Ai ¼
i¼1 i¼1
i, j¼1
i, j, k¼1
i=j
(+1)n
i=j=k
n
X
Ai > A j > A k > An (1:2:5a)
i, j, k¼1
i=j=k...=n
If the sets fAig are mutually exclusive, that is, Ai > Aj ¼ 1 for i = j, then we have
n
n
X
[
jA j
(1:2:5b)
A i ¼
i¼1 i¼1 i
This equation is illustrated in the Venn diagram for n ¼ 3 in Fig. 1.2.3, where if
jA < B < Cj equals, jAj þ jBj þ jCj then we have added twice the cardinalites of
jA > Bj, jB > Cj and jC > Aj. However, if we subtract once jA > Bj, jB > Cj, and
jC > Aj and write
jA < B < Cj ¼ jAj þ jBj þ jCj jA > Bj jB > Cj jC > Aj
then we have subtracted jA > B > Cj thrice instead of twice. Hence, adding jA > B > Cj
we get the final result:
jA < B < Cj ¼ jAj þ jBj þ jCj jA > Bj jB > Cj jC > Aj jA > B > Cj (1:2:6)
FIGURE 1.2.2
6
Sets, Fields, and Events
FIGURE 1.2.3
Example 1.2.3 In a junior class, the number of students in electrical engineering (EE) is
100, in math (MA) 50, and in computer science (CS) 150. Among these 20 are taking both
EE and MA, 25 are taking EE and CS, and 10 are taking MA and CS. Five of them are
taking EE, CS and MA. Find the total number of students in the junior class.
From the problem we can identify the sets A ¼ EE students, B ¼ MA students, and
C ¼ CS students, and the corresponding intersections are A > B, B > C, C > A, and
A > B > C. Here, jAj ¼ 100, jBj ¼ 50, and jCj ¼ 150. We are also given
jA > Bj ¼ 20, jB > Cj ¼ 10, and jC > Aj ¼ 25 and finally jA > B > Cj ¼ 5. Using
Eq. (1.2.6) the total number of students are 100 þ 50 þ 150 2 20 2 10 2 25 þ 5 ¼ 250.
Complement
If A is a subset of the universal set S, then the complement of A denoted by A is the
elements in S not contained in A; or
A ¼ S A ¼ {x : x A , S and x [ S}
(1:2:7)
Example 1.2.4 If the universal set S ¼ f1,2,3,4,5,6g and A ¼ f2,4,5g, then the complement of A is given by A ¼ f1,3,6g.
Difference
The complement of a subset A , S as given by Eq. (1.2.7) is the difference between
the universal set S and the subset A. The difference between any two sets A and B denoted
by A 2 B is the set C containing those elements that are in A but not in B and, B 2 A is the
set D containing those elements in B but not in A:
C ¼ A B ¼ {x : x [ A and x B}
D ¼ B A ¼ {x : x [ B and x A}
(1:2:8)
A 2 B is also called the relative complement of B with respect to A and B 2 A is called the
relative complement of A with respect to B. It can be verified that A 2 B = B 2 A.
Example 1.2.5 If the set A is given as A ¼ f2,4,5,6g and B is given as B ¼ f5,6,7,8g, then
the difference A 2 B ¼ f2,4g and the difference B 2 A ¼ f7,8g. Clearly, A 2 B = B 2 A.
1.2 Set Operations
7
FIGURE 1.2.4
Symmetric Difference
The symmetric difference between sets A and B is written as ADB and defined by
ADB ¼ (A B) < (B A) ¼ {x : x [ A and x B) or {x : x [ B and x A) (1:2:9)
Example 1.2.6 The symmetric difference between the set A given by A ¼ f2,4,5,6g and
B given by B ¼ f5,6,7,8g in Example 1.2.5 is ADB ¼ f2,4g < f7,8g ¼ f2,4,7,8g.
The difference and symmetric difference for Examples 1.2.5 and 1.2.6 are shown in
Fig. 1.2.4.
Cartesian Product
This is a useful concept in set theory. If A and B are two sets, the Cartesian product
A B is the set of ordered pairs (x,y) such that x [ A and y [ B and is defined by
A B ¼ {(x,y) : x [ A, y [ B}
(1:2:10)
When ordering is considered the cartesian product A B = B A. The cardinality of a
Cartesian product is the product of the individual cardinalities, or jA Bj ¼ jAj . jBj.
Example 1.2.7 If A ¼ f2,3,4g and B ¼ f5,6g, then C ¼ A B ¼ f(2,5), (2,6), (3,5),
(3,6), (4,5), (4,6)g with cardinality 3 2 ¼ 6. Similarly, D ¼ B A ¼ f(5,2), (5,3),
(5,4), (6,2), (6,3), (6,4)g with the same cardinality of 6. However, C and D do not
contain the same ordered pairs.
Partition
S
A partition is a collection
T of disjoint sets fAi, i ¼ 1, . . . , ng of a set S such that Ai
over all i equals S and Ai
Aj , with i = j empty. Partitions play a useful role in conditional probability and Bayes’ theorem. Figure 1.2.5 shows the concept of a partition.
FIGURE 1.2.5
8
Sets, Fields, and Events
FIGURE 1.2.6
TABLE 1.2.1
Table of set properties
Property
Equation
Identity
A<1¼A
A>S¼A
Domination
A>1¼1
A<S¼S
Idempotent
A>A¼A
A<A¼A
¼
Complementation
A ¼A
Commutative
A<B¼B<A
A>B¼B>A
Associative
A < (B < C ) ¼ (A < B) < C
A > (B > C ) ¼ (A > B) > C
Distributive
A > (B < C ) ¼ (A > B) < (A > C )
A < (B > C ) ¼ (A < B) > (A < C )
Noncommutative
A2B = B2A
De Morgan’s law
A<B¼A>B
A>B¼A<B
Example 1.2.8 If set S ¼ f1,2,3,4,5,6,7,8,9,10g, then the collections of sets fA1, A2, A3,
A4g where A1 ¼ f1,3,5g, A2 ¼ f7,9g, A3 ¼ f2,4,6g, A4 ¼ f8,10g is a partition of S as shown
in Fig. 1.2.6.
A tabulation of set properties is shown on Table 1.2.1. Among the identities,
De Morgan’s laws are quite useful in simplifying set algebra.
B 1.3
SET ALGEBRAS, FIELDS, AND EVENTS
Boolean Field
We shall now consider a universal set S and a collection F of subsets of S consisting
of the sets fAi: i ¼ 1,2, . . . , n,n þ 1, . . .g. The collection F is called a Boolean field if the
1.3 Set Algebras, Fields, and Events
9
following two conditions are satisfied:
1. If Ai [ F, then Ai [ F.
2. If A1 [ F and A2 [ F, then A1 < A2 [ F.
3. If fAi , i ¼ 1, . . . , ng [ F, then < ni¼1 Ai [ F.
Let us see the consequences of these conditions being satisfied. If A1 [ F and A1 F, then
A1 < A1 ¼ S [ F. If S [ F, then S ¼ 1 [ F. If A1 < A2 [ F, then by condition 1
A1 < A2 [ F, and by De Morgan’s law, A1 < A2 [ F. Hence A1 > A2 [ F. Also
A1 2 A2 [ F and (A1 2 A2) < (A2 2 A1) [ F. Thus the conditions listed above are
sufficient for any field F to be closed under all set operations.
Sigma Field
The preceding definition of a field covers only finite set operations. Condition 3 for
finite additivity may not hold for infinite additivity. For example, the sum of
1 þ 2 þ 22 þ 23 þ 24 ¼ (25 2 1)/(2 2 1) ¼ 31, but the infinite sum 1 þ 2 þ 22 þ 23 þ
24 þ . . . diverges. Hence there is a need to extend finite additivity concept to infinite set
operations. Thus, we have a s field if the following extra condition is imposed for infinite
additivity:
4. If A1, A2, . . . An, . . . [ F then <1
i¼1 Ai [ F
Many s fields may contain the subsets fAig of S, but the smallest s field containing the
subsets fAig is called the Borel s field. The smallest s field for S by itself is F ¼ fS, 1g.
Example 1.3.1 We shall assume that S ¼ f1,2,3,4,5,6g. Then the collection of the following subsets of S, F ¼ fS, 1, (1,2,3), (4,5,6)g is a field since it satisfies all the set operations such as (1,2,3) < (4,5,6) ¼ S, ð 1,2,3 Þ ¼ (4,5,6). However, the collection of the
following subsets of S, F1 ¼ {S, 1, (1,2,3), (4,5,6), ð2Þ} will not constitute a field
because (2) < (4,5,6) ¼ (2,4,5,6) is not in the field. But we can adjoin the missing sets
and make F1 into a field. This is known as completion. In the example above, if we
adjoin the collection of sets F2 ¼ {(2,4,5,6), (1,3)} to F1 , then the resulting collection
F ¼ F1 < F2 ¼ {S, 1, (1,2,3), (4,5,6), {2}, (2,4,5,6), (1,3)} is a field.
Event
Given the universal set S and the associated s field F, all subsets of S belonging to the
field F are called events. All events are subsets of S and can be assigned a probability, but
not all subsets of S are events unless they belong to an associated s field.
CHAPTER
2
Probability Space and Axioms
B 2.1
PROBABILITY SPACE
A probability space is defined as the triplet fS, F, Pg, where S is the sample space, F is the
field of subsets of S, and P is a probability measure. We shall now define these elements.
Sample Space S
Finest-grain, mutually exclusive, and collectively exhaustive listing of all possible
outcomes of a mathematical experiment is called the sample space. The outcomes can
be either finite, countably infinite, or uncountably infinite. If S is countable, we call it a
discrete sample space; if it is uncountable, we call it continuous sample space.
Examples of Discrete Sample Space
Example 2.1.1 When a coin is tossed, the outcomes are heads {H} or tails {T}. Thus the
sample space S ¼ {H,T}.
Example 2.1.2 When a die is tossed, the outcomes are 1,2,3,4,5,6. Thus the sample
space is S ¼ {1,2,3,4,5,6}.
Example 2.1.3 The set of all positive integers is countable, and the sample space is
S ¼ fx : x is a positive integerg
Examples of Continuous Sample Space
Example 2.1.4 The set of all real numbers between 0 and 2 is the sample space
S ¼ fx : 0 x 2g.
Example 2.1.5 The set of all real numbers x and y such that x is between 0 and 1 and y is
between 1 and 2 is given by S ¼ fx, y : 0 x 1, 1 y 2g.
The continuous sample space given in Examples 2.1.4 and 2.1.5 is shown in Fig. 2.1.1.
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
10
2.1 Probability Space
11
FIGURE 2.1.1
Field F
A field has already been defined in Section 1.3. We shall assume that a proper Borel s
field can be defined for all discrete and continuous sample spaces.
Probability Measure P
A probability measure is defined on the field F of events of the sample space S. It is a
mapping of the sample space to the real line between 0 and 1 such that it satisfies the
following three Kolmogorov axioms for all events belonging to the field F:
Axiom 1: If A belongs to F; then P(A) 0: nonnegativity:
(2:1:1)
Axiom 2: The probability of the sample space equals 1.
P(S) ¼ 1: normalization:
(2:1:2)
Axiom 3a: If A and B are disjoint sets belonging to F, that is, A > B ¼ 1, then
P{A < B} ¼ P(A) þ P(B): Boolen additivity:
(2:1:3)
Axiom 3b: If fA1, A2, . . . g is a sequence of sets belonging to the field 1 such that
Ai > Aj ¼ 1 for all i = j, then
P fA 1 < A 2 < < A i < g ¼
1
X
PfA j g: sigma additivity:
(2:1:4)
i¼1
From these axioms we can write the following corollaries:
Corollary 1:
P{1}
(2:1:5)
Note that the probability of the empty event is zero, but if the probability of an
event A is zero, we cannot conclude that A ¼ 1. All sets with zero probability
are defined as null sets. An empty set is also a null set, but null sets need not be
empty.
Corollary 2. Since A < A ¼ S and A > A ¼ 1, we can write
P(A) ¼ 1 P(A)
(2:1:6)
12
Probability Space and Axioms
Corollary 3. For any A and B that are subsets of S, not mutually exclusive, we can
write
Pf A < Bg ¼ P{A < A > B} ¼ P{A} þ P{A > B}
P{B} ¼ P{(A > B) < (A > B)} ¼ P{A > B} þ P{A > B}
Eliminating P{A < B} between the two equations, we obtain
P{A < B} ¼ P{A} þ P{B} P{A > B}
(2:1:7)
Clearly, if P{A > B} ¼ 0 implying A and B are mutually exclusive, then Eq. (2.1.7)
reduces to Kolmogorov’s axiom 3a. If the probabilities P{A} and P{B} are
equal, then sets A and B are equal in probability. This does not mean that A is
equal to B.
The probability of the union of three sets can be calculating from the Venn diagram
for the three sets shown in Fig. 1.2.3 and redrawn in Fig. 2.1.2.
The results of the analysis for the cardinalities given in Eq. (1.2.6), can be extended to
finding the P(A < B < C). We subtract {P(A > B) þ P(B > C) þ P(C > A)} from
{P(A) þ P(B) þ P(C)} and add P(A > B > C), resulting in
P{A < B < C} ¼ P{A} þ P{B} þ P{C}
P{A > B} P{B > C} P{C > A}
þ P{A > B > C}
FIGURE 2.1.2
(2:1:8)
2.1 Probability Space
13
Using the inclusion– exclusion principle of Eq. (1.2.5a), we can also generalize Eq. (2.1.8)
to the probabilities of the union of n sets A1, A2, . . . , An as follows:
Pfni¼1 Ai g ¼
n
X
P{Ai } i¼1
n
X
P{Ai > A j } þ
i, j¼1
i=j
(+1)n
n
X
n
X
P{Ai > A j > Ak }
i, j, k¼1
i=j=k
P{Ai > A j > Ak > An }
(2:1:9)
i, j, k¼1
i=j=k...=n
Example 2.1.6 A fair coin is tossed 3 times. Since each toss is a functionally independent
trial and forms a Cartesian product, the cardinality of the three tosses is 2 2 2 ¼ 8
and the sample space consists of the following 8 events:
S ¼ {½HHH, ½HHT, ½HTT, ½HTH, ½THH, ½TTH, ½THT, ½TTT}
In this sample space
P{2 heads} ¼ P{½HHH, ½HHT, ½HTH, ½THH} ¼
4
8
P{both heads and tails} ¼ {½HHT, ½HTT, ½HTH, ½THH, ½TTH, ½THT} ¼
6
8
Example 2.1.7 A telephone call occurs at random between 9:00 a.m. and 11:00 a.m.
Hence the probability that the telephone call occurs in the interval 0 , t t0 is given
by t0 =120. Thus the probability that the call occurs between 9:30 a.m. and 10:45 a.m. is
given by:
P{30 , t 105} ¼
105 30
75
5
¼
¼
120
120 8
Example 2.1.8 Mortality tables give the probability density that a person who is born
today assuming that the lifespan is 100 years as
f (t) ¼ 3 109 t2 (100 t2 )u(t)
and the distribution function that the person will die at the age t0 years as
ð t0
F(t0 ) ¼ P{t t0 } ¼
3 109 t2 (100 t2 ) dt
0
These probabilities are shown in Fig. 2.1.3. The probability that this person will die
between the ages 70 to 80 is given by P{70 , t 80} ¼ 0:1052, obtained from the
following integral:
ð 80
P{70 , t 80} ¼ F(80) F(70) ¼
3 109 t2 (100 t2 ) dt ¼ 0:1052
70
The density and distribution functions are shown in Fig. 2.1.3.
14
Probability Space and Axioms
FIGURE 2.1.3
B 2.2
CONDITIONAL PROBABILITY
The conditional probability of an event B conditioned on the occurrence of another event B
is defined by
P{B j A} ¼
P{B > A}
P{A}
P{B > A} ¼ P{B j A}P{A}
9
>
=
>
;
if P{A} = 0
(2:2:1)
If P{A} ¼ 0, then the conditional probability P{B j A} is not defined. Since P{B j A} is a
probability measure, it also satisfies Kolmogorov axioms.
Conditional probability P{B j A} can also be interpreted as the probability of the
event B in the reduced sample space S1 ¼ A. Sometimes it is simpler to use the reduced
sample space to solve conditional probability problems. We shall use both these interpretations in obtaining conditional probabilities as shown in the following examples.
2.2
Conditional Probability
15
FIGURE 2.2.1
Example 2.2.1 In the tossing of a fair die, we shall calculate the probability that 3 has
occurred conditioned on the toss being odd. We shall define the following events:
A ¼ {odd number}: B ¼ 3
1
2
1
6
P{A} ¼ and P{B} ¼ and the event A > B ¼ {3} and hence
1
p{3}
1
¼6¼
P{3 j odd} ¼
1
P{odd}
3
2
Note that P{2 j odd} ¼ 0 since 2 is an impossible event given that an odd number has
occurred. Also note that P{3 j 3} ¼ 1. The conditional probability of 3 occurring, given
that 3 has occurred is obviously equal to 1.
We can also find the solution using the concept of reduced sample space. Since
we are given that an odd number has occurred, we can reduce the sample space from
S ¼ {1,2,3,4,5,6} to S1 ¼ {1,3,5} as shown in Fig. 2.2.1.
The probability of 3 in this reduced sample space is 13, agreeing with the previous result.
Example 2.2.2 Consider a family of exactly two children. We will find the probabilities
(1) that both are girls, (2) both are girls given that one of them is a girl, (3) both are girls
given that the elder child is a girl, and (4) both are girls given that both are girls.
The sample space S has 4 points {gg, gb, bg, bb}, where g is girl and b is boy. The
event B ¼ fboth children are girlsg. The conditioning event A will depend on the following
four cases:
1. The event A ¼ S. Hence P{B} ¼ 14.
2. The event A ¼ fone child is a girlg. The reduced sample space is S1 ¼ fgg, gb, bgg
and the conditional probability P{B j A} ¼ 13.
3. The event A ¼ felder child is a girlg. In this case, the reduced sample space is
S2 ¼ {gg, gb} and P{B j A} ¼ 12.
4. The event ¼ fboth are girlsg. The reduced sample space has only one point,
namely, S3 ¼ {gg} and P{B j A} ¼ 1.
As our knowledge of the experiment increases by the conditioning event, the uncertainty decreases and the probability increases from 0.25, through 0333, 0.5, and finally to
1, meaning that there is no uncertainty.
16
Probability Space and Axioms
Example 2.2.3 The game of craps as played in Las Vegas has the following rules.
A player rolls two dice. He wins on the first roll if he throws a 7 or a 11. He loses if
the first throw is a 2, 3, or 12. If the first throw is a 4, 5, 6, 8, 9, or 10, it is called a
point and the game continues. He goes on rolling until he throws the point for a win or
a 7 for a loss. We have to find the probability of the player winning.
We will solve this problem both from the definition of conditional probability and the
reduced sample space. Figure 2.2.2 shows the number of ways the sums of the pips on the
two dice can equal 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, and their probabilities.
Solution Using Definition of Conditional Probability. The probability of winning in
the first throw is
P{7} þ P{11} ¼
6
2
8
þ
¼
¼ 0:22222
36 36 36
The probability of losing in the first throw is
P{2} þ P{3} þ P{12} ¼
1
2
1
4
þ þ
¼
¼ 0:11111
36 36 36 36
To calculate the probability of winning in the second throw, we shall assume that i is
the point with probability p. The probability of not winning in any given throw after
the first is given by r ¼ P{not i and not 7} ¼ 1 p 16. We compute the conditional probability
P{win j i in first throw} ¼ p þ rp þ r 2 p þ ¼
p
p
¼
1 r p þ 16
as an infinite geometric series:
P{win in second or subsequent throws} ¼
p2
p þ 16
The probability p of making the point i ¼ 4,5,6,8,9,10 is obtained from Fig. 2.2.2.
FIGURE 2.2.2
2.2
Conditional Probability
17
Thus, for i ¼ 4,5,6, we obtain
2 2 2 3
P{win after point i ¼ 4} ¼
36
P{win after point i ¼ 5} ¼
P{win after point i ¼ 6} ¼
4
36
5
36
3 1
1
þ
¼
36 6
36
4 1
2
þ
¼
36 6
45
5 1
25
þ
¼
36 6
396
with similar probabilities for 8,9,10. Thus the probability of winning in craps is
P{win} ¼ P{win in roll 1} þ P{win in roll 2,3, . . .}
8
1
2
25
244
¼
þ2
þ þ
¼ 0:4929
¼
36
36 45 396
495
Solution Using Reduced Sample Space. We can now use the reduced sample space to
arrive at the same result in a simpler manner. After the point i has been rolled in the
first throw, the reduced sample space consists of (i 2 1 þ 6) points for i ¼ 4,5,6 and
(13 2 i þ 6) points for i ¼ 8,9,10 in which the game will terminate. We can see
from the figure that there are 6 ways of rolling 7 for a loss, (i 2 1, i ¼ 4,5,6)
ways of rolling the point for a win, and (13 2 i, i ¼ 8,9,10) ways of rolling the
point for a win. Thus
P{win in roll 2,3, . . . j i in roll 1} ¼ P{W j i}
8 i1
>
,
<
¼ i1þ6
>
: 13 i ,
13 i þ 6
i ¼ 4,5,6
i ¼ 8,9,10
4
5
5
4
Thus, P{W j 4} ¼ 39 , P{W j 5} ¼ 10
, P{W j 6} ¼ 11
, P{W j 8} ¼ 11
, P{W j 9} ¼ 10
,
3
P{W j 10} ¼ 9. Hence probability of a win after the first roll is P{W j i} P{i}:
Hence
3 3
4 4
5 5
134
þ þ ¼
9 36 10 36 11 36
495
3 3
4 4
5 5
8 134 244
þ
¼ 0:4929
P{win} ¼ 2 þ þ ¼ þ
9 36 10 36 11 36
36 495 495
P{win after first roll} ¼ 2
a result obtained with comparative ease.
Example 2.2.4 We shall continue Example 2.1.8 and find the probability that a person
will die between the ages 70 and 80 given that this individual has lived upto the age 70. We
will define the events A and B as follows:
A ¼ {person has lived upto 70}
B ¼ {person dies between ages 70 and 80}
18
Probability Space and Axioms
The required conditional probability equals the number of persons who die between the ages 70 and 80 divided by the number of persons who have lived upto the age 70:
P{B j A} ¼
P{70 , t 80 and t 70} P{70 , t 80}
¼
P{t 70}
P{t 70}
Ð 80
70
¼ Ð 100
70
3 109 t2 (100 t2 )dt
3
109 t2 (100
t2 )dt
¼
0:105
¼ 0:644
0:163
This type of analysis is useful in setting up actuarial tables for life insurance purposes.
B 2.3
INDEPENDENCE
We have already seen the concept of functional independence of sets in Chapter 1. We will
now define statistical independence of events. If A and B are two events in a field of events
F, then A and B are statistically independent if and only if
P{A > B} ¼ P{A}P{B}
(2:3:1)
Statistical independence neither implies nor is implied by functional independence.
Unless otherwise indicated, we will call statistically independent events “independent”
without any qualifiers.
Three events A1 ,A2 ,A3 are independent if they satisfy, in addition to
P{A1 > A2 > A3 } þ P{A1 }P{A2 }P{A3 }
(2:3:2)
the pairwise independence
P{Ai > A j } ¼ P{Ai }P{A j },
i, j ¼ 1,2,3, i = j
(2:3:3)
Note that if A and B are mutually exclusive, then, from Kolmogorov’s axiom 3a, we obtain
P{A < B} ¼ P{A} þ P{B}
(2:3:4)
The concepts of independence and mutual exclusivity are fundamental to probability
theory. For any two sets A and B, we recall from Eq. (2.1.7) that
P{A < B} ¼ P{A} þ P{B} P{A > B}
(2:1:7)
If A and B are mutually exclusive, then P{A > B} ¼ 0, but if they are independent,
then P{A > B} ¼ P{A} P{B}. If this has to be zero for mutual exclusivity, then one of
the probabilities must be zero. Therefore, we conclude that mutual exclusivity and independence are incompatible unless we are dealing with null events.
We shall now give examples to clarify the concepts of functional independence and
statistical independence.
Example 2.3.1 The probability of an ace on the throw of a die is 16. The probability of a
head on the throw of a coin is 12. Since the throw of a die is not dependent on the throw of a
coin, we have functional independence and the combined sample space is the Cartesian
product with 12 elements as shown in Table 2.3.1.
2.3 Independence
19
TABLE 2.3.1
1, H
2, H
3, H
4, H
5, H
6, H
1, T
2, T
3, T
4, T
5, T
6, T
We assume as usual that all the 12 elements are equiprobable events and the probability of
1
ace and head is 12
since there is only one occurrence of f1, Hg among the 12 points as seen
in the first row and first column in Table 2.3.1. Using Eq. (2.3.1), we have
1
P{1, H} ¼ P{1} P{H} ¼ 16 12 ¼ 12
, illustrating that in this example, functional independence coincides with statistical independence.
Example 2.3.2 Two dice, one red and the other blue, are tossed. These tosses are functionally independent, and we have the Cartesian product of 6 6 ¼ 36 elementary events
in the combined sample space, where each event is equiprobable. We seek the probability
that an event B defined by the sum of the numbers showing on the dice equals 9. There are
4
four points {(6,3), (5,4), (4,5), (3,6)}, and hence P{B} ¼ 36
¼ 19. We now condition the
event B with an event A defined as the red die shows odd numbers. The probability of
1
the event A is P{A} ¼ 18
36 ¼ 2. We want to determine whether the events A and B are statistically independent. These events are shown in Fig. 2.3.1, where the first number is for
the red die and the second number is for the blue die.
2
1
From Fig. 2.3.1 P{A > B} ¼ P{(3,6), (5,4)} ¼ 36
¼ 18
and ¼ P{A} P{B} ¼ 12 19 ¼
1
18 showing statistical independence. We compute P{B j A} from the point of view of
reduced sample space. The conditioning event reduces the sample space from 36 points
to 18 equiprobable points and the event {B j A} ¼ {(3,6), (5,4)}. Hence
2
P{B j A} ¼ 18
¼ 19 ¼ P{B}, or, the conditioning event has no influence on B. Here, even
though the events A and B are functionally dependent, they are statistically independent.
However, if another set C is defined by the sum being equal to 8 as shown in
5
Fig. 2.3.2, then P{C} ¼ 36
.
Here the events C and A are not statistically independent because P{C} P{A} ¼
5
1
5
2
4
36 2 ¼ 72 = P{C > A} ¼ P{(3,5), (5,3)} ¼ 18 ¼ 72.
In this example, we have the case where the events A and C are neither statistically
independent nor functionally independent.
FIGURE 2.3.1
20
Probability Space and Axioms
FIGURE 2.3.2
B 2.4
TOTAL PROBABILITY AND BAYES’ THEOREM
Let {Ai , i ¼ 1, . . . , n} be a partition of the sample space and let B an arbitrary event in the
sample space S as shown in Fig. 2.4.1.
We will determine the total probability PfBg, given the conditional probabilities
P{B j A1 }, P{B j A2 }, . . . , P{B j An }. The event B can be given in terms of the partition as
B ¼ {B > S} ¼ B > {A1 < A1 < < An }
¼ {B > A1 } < {B > A2 } < < {B > An }
(2:4:1)
The events {B > A1 }, {B > A2 }, . . . , {B > An } in Eq. (2.4.1) are mutually exclusive since
A1, A2, . . . , An are mutually exclusive and we have from Kolmogorov’s axiom 3a:
P{B} ¼ P{B > A1 } þ P{B > A2 } þ þ P{B > An }
(2:4:2)
Using the definition of conditional probability from Eq. (2.2.1), we can rewrite Eq. (2.4.2) as
P{B} ¼ P{B j A1 }P{A1 } þ P{B j A2 }P{A2 } þ þ P{B j An }P{An }
¼
n
X
P{B j Ai }P{Ai }
(2:4:3)
i¼1
Equation (2.4.3) is the expression for the total probability P{B} in terms of the conditional
probabilities P{B j Ai , i ¼ 1, . . . , n}. We shall give an example to illustrate this concept.
FIGURE 2.4.1
2.4 Total Probability and Bayes’ Theorem
21
Example 2.4.1 Four computer companies A, B, C, D supply microprocessors to a
company. From previous experience it is known that the probability of the processor
being bad if it comes from A is 4%, B is 2%, C is 5%, and D is 1%. The probabilities
of picking supplier A is 20%, B is 30%, C is 10%, and D are 40%. We want to find the
probability that a processor chosen at random is bad.
Let us define X as the event that the processor is bad. We need to find P{X} from
the conditional probabilities P{X j A} ¼ 0:04, P{X j B} ¼ 0:02, P{X j C} ¼ 0:05, and
P{X j D} ¼ 0:01. We are also given that P{A} ¼ 0:2, P{B} ¼ 0:3, P{C} ¼ 0:1, and
P{D} ¼ 0:4. Substituting these quantities in Eq. (2.4.3), we obtain
P{X} ¼ P{X j A}P{A} þ P{X j B}P{B} þ P{X j C}P{C} þ P{X j D}P{D}
¼ 0:04 0:2 þ 0:02 0:3 þ 0:05 0:1 þ 0:01 0:4 ¼ 0:023
Thus the probability that the processor is bad is 2.3%.
Bayes’ Theorem
The next logical question to ask is, “Given that the event B has occurred, what are
the probabilities that A1, A2, . . . , An are involved in it?” In other words, given the conditional probabilities P{B j A1 }, P{B j A2 }, . . . , P{B j An }, we have to find the reverse
conditional probabilities P{A1 j B}, P{A2 j B}, . . . , P{An j B}. Using the total probability
P{B} from Eq. (2.4.3), we can write
P{Ai j B} ¼
P{Ai B} P{B j Ai }P{Ai }
P{B j Ai }P{Ai }
¼
¼ Pn
P{B}
P{B}
j¼1 P{B j A j }P{A j }
(2:4:4)
This is the celebrated Bayes’ theorem. This theorem connects the a priori probability
P{Ai } to the a posteriori probability P{Ai j B}. This useful theorem has wide applicability
in communications and medical imaging, as we will see in Chapter 23.
Example 2.4.2 We will now continue Example 2.4.1 and find the probability that the
processor came from companies A, B, C, or D given that the processor is bad. Using
the result from Example 2.4.1 that P{X} ¼ 0:023, we can write from Eq. (2.4.4)
P{X j A} ¼
P{X j A}P{A} 0:04 0:2
¼
¼ 0:348
P{X}
0:023
P{X j B} ¼
P{X j B}P{B} 0:02 0:3
¼
¼ 0:261
P{X}
0:023
P{X j C} ¼
P{X j C}P{C} 0:05 0:1
¼
¼ 0:217
P{X}
0:023
P{X j D} ¼
P{X j D}P{D} 0:01 0:4
¼
¼ 0:174
P{X}
0:023
Note that the sum of the resulting probabilities, namely, 0.348 þ 0.261 þ 0.217 þ
0.174 ¼ 1, as it should be because the bad processor has to come only from any one of
the four companies A, B, C, or D resulting in the probability being 1.
Example 2.4.3 A communication receiver sends numbers 1,2,3. As a result of inherent
noise in the channel, the conditional probabilities of receiving 1,2,3 under the condition
that 1,2,3 were sent are shown in Fig. 2.4.2.
22
Probability Space and Axioms
FIGURE 2.4.2
Transition Probabilities:
P{1R j 1S } ¼ 0:60
P{2R j 1S } ¼ 0:25
P{3R j 1S } ¼ 0:15
P{1R j 2S } ¼ 0:35
P{2R j 2S } ¼ 0:55
P{3R j 2S } ¼ 0:10
P{1R j 3S } ¼ 0:05
P{2R j 3S } ¼ 0:30
P{3R j 3S } ¼ 0:65
The total probability that 1 was received is obtained from Eq. (2.4.3):
P{1R } ¼ P{1R j 1S }P{1S } þ P{1R j 2S }P{2S } þ P{1R j 3S }P{3S }
¼ 0:6 0:25 þ 0:35 0:3 þ 0:05 0:45 ¼ 0:2775
From Bayes’ theorem, Eq. (2.4.4), the a posteriori probabilities of 1, 2, and 3 being
sent conditioned on 1 received are found as follows:
P{1S j 1R } ¼
P{1R j 1S }P{1S } 0:6 0:25
¼ 0:5405:
¼
P{1R }
0:2775
P{1S } ¼ 0:25
P{2S j 1R } ¼
P{1R j 2S }P{2S } 0:35 0:3
¼ 0:3784:
¼
P{1R }
0:2775
P{2S } ¼ 0:3
P{3S j 1R } ¼
P{1R j 3S }P{3S } 0:05 0:45
¼ 0:0811:
¼
P{1R }
0:2775
P{3S } ¼ 0:45
The sum of these probabilities is equal to 1, as it should be. Note that the a priori
probability that 3 was sent is 0.45 and after 1 has been received, the a posteriori
probability is 0.0811, a very small probability. Similar posterior probabilities of
1, 2, and 3 being sent conditioned on 2 and 3 received can be found.
Example 2.4.4 (Monty Hall Problem) This classic problem is called the “Monty Hall
problem” because of the game show host Monty Hall who designed this game. There are
three doors A, B, and C, and behind one of them is a car and behind the other two are goats.
The contestant is asked to select any door, and the host Monty Hall opens one of the other
two doors and reveals a goat. He then offers the choice to the contestant of switching to
the other unopened door or keep the original door. The question now is whether the
2.4 Total Probability and Bayes’ Theorem
23
probability of winning the car is improved if she switches or it is immaterial whether she
switches or not.
The answer is counterintuitive in the sense that one may be misled into thinking that it is
immaterial whether one switches or not. We will analyze this problem in two different ways.
Mathematical Analysis. Let us analyze this problem from the point of view of Bayes’
theorem. We shall first assume that the host has prior knowledge of the door behind
which the car is located. Let us also assume that the host opens door B given that the
contestant’s choice is door A. The a priori probabilities of a car behind the doors A,
B, and C are
1
1
1
P{A} ¼ ; P{B} ¼ ; P{C} ¼
3
3
3
We can make the following observations. If the contestant’s choice is door A and the
car is behind door A, then the host will open the door B with probability of 12 (since
she has a choice between B and C ). On the other hand, if the car is behind door B,
then there is zero probability that she will open door B because of her prior knowledge. If the car is behind door C, then she will open door B with probability 1. Thus
we can write the following conditional probabilities for the host opening door B:
1
P{B j A} ¼ ; P{B j B} ¼ 0; P{B j C} ¼ 1
2
We can now calculate the total probability P(B) of the host opening door B.
P{B} ¼ P{B j A}P{A} þ P{B j B}P{B} þ P{B j C}P{C}
¼
1 1
1
1 1
þ0 þ1 ¼
2 3
3
3 2
Using Bayes’ theorem [Eq. (2.4.4)] we can now find the a posteriori probabilities of the
car behind the doors A or C (B has already been opened by the host) conditioned on B:
1 1
P{BA} P{B j A}P{A} 2 3 1
P{A j B} ¼
¼
¼
¼
1
P{B}
P{B}
3
2
1
P{BC} P{B j C}P{C} 1 3 2
P{C j B} ¼
¼
¼
¼
1
P{B}
P{B}
3
2
Similar analysis holds for other cases of opening the doors A or C. We can see from
the result above that the contestant must switch if she wants to double her probability of winning the car.
TABLE 2.4.1
First Choice
Host Opens
Second Choice
Probability
A
B
A
P{B} ¼ 12 13 ¼ 16
A
C
A
P{C} ¼ 12 13 ¼ 16
B
C
B
P{C} ¼ 1 13 ¼ 13
C
B
B
P{B} ¼ 1 13 ¼ 13
24
Probability Space and Axioms
TABLE 2.4.2
First Choice
A
A
B
C
TABLE 2.4.3
First Choice
A
A
B
C
For not switching
Host Opens
Second Choice
B
C
C
B
A
A
B
B
Win or Loss
Probability
P{B} ¼ 12 13 ¼ 16
P{C} ¼ 12 13 ¼ 16
P{C} ¼ 1 13 ¼ 13
P{B} ¼ 1 13 ¼ 13
Win
Win
Loss
Loss
For switching
Host Opens
Second Choice
B
C
C
B
C
B
A
A
Win or Loss
Probability
P{B} ¼ 12 13 ¼ 16
P{C} ¼ 12 13 ¼ 16
P{C} ¼ 1 13 ¼ 13
P{B} ¼ 1 13 ¼ 13
Loss
Loss
Win
Win
Heuristic Analysis. In this case we shall enumerate all the possibilities and determine
which possibility gives a better chance. Assume that the car is behind door A. The
probabilities of A,B,C are once again
1
P{A} ¼ ;
3
1
P{B} ¼ ;
3
P{C} ¼
1
3
Conditioned on the contestant choosing door A, the host has two equiprobable choices
of opening either door B or door C. However, conditioned on the contestant choosing B or C, the host has only one choice of either opening C or B. Thus
1
P{B j A} ¼ ;
2
1
P{C j A} ¼ ;
2
P{C j B} ¼ 1;
P{B j C} ¼ 1
From these conditional probabilities we can evaluate the unconditional probabilities
as shown in Table 2.4.1.
We can now enumerate the various win or loss possibilities for not switching and
for switching in Tables 2.4.2 and 2.4.3. From the Table 2.4.2 we can see that the
probability of winning the car by not switching is 16 þ 16 ¼ 13.
From Table 2.4.3 we can see that the probability of winning the car by switching
is 13 þ 13 ¼ 23. Thus the probability of winning the car doubles with switching.
CHAPTER
3
Basic Combinatorics
B 3.1
BASIC COUNTING PRINCIPLES
Combinatorics is the study of arrangement and manipulation of mathematical elements in
sets and essentially involves counting. We will discuss two basic counting principles.
The Addition Rule
A task A consists of m subtasks A1, A2, . . . , Am that can be done in n(A1), n(A2), . . . ,
n(Am) ways, respectively. If no two of them can be done at the same time, then the total
number of ways n(A) of doing the task A is
n(A) ¼ n(A1 ) þ n(A2 ) þ þ n(Am )
(3:1:1)
This rule is the same as in set theory [Eq. (1.2.5a)] where the cardinality of the union of
mutually exclusive sets is the sum of the cardinalities of the individual sets. Thus, the
addition rule is a restatement of one of the properties of mutually exclusive sets.
Example 3.1.1 An engineering student undertaking a capstone project T has a choice
of 4 professors A,B,C,D who can give him a list of projects from which he can
choose one. A has 5 projects, B has 7 projects, C has 6 projects, and D has 2 projects.
Since the student can not take any other project once he has chosen a project, the
number of choices he has is
n(T) ¼ n(A) þ n(B) þ n(C) þ n(D) ¼ 5 þ 7 þ 6 þ 2 ¼ 20
The Multiplication Rule
A task A consists of m sequential subtasks A1, A2, . . . , Am that can be performed only
after the previous one is completed. The numbers of ways n(A1), n(A2), . . . , n(Am) of
performing the subtasks are functionally independent; that is, n(A2) ways are needed
for the second step regardless of n(A1) of the first step. The number of ways n(A) of
performing the task A is
n(A) ¼ n(A1 ) n(A2 ) n(Am )
(3:1:2)
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
25
26
Basic Combinatorics
This is very similar to the Cartesian product in set theory, where the cardinality of the
product is the product of the cardinalities of the individual sets. These sets are also functionally independent. They may or may not be statistically independent.
Example 3.1.2 A professor assigns a capstone project T to senior engineering students
consisting of four parts A,B,C,D that must be done in sequence independent of the previous
ones. Part A can be done in 2 ways, part B in 3 ways, part C in 5 ways, and part D in 4 ways.
Since A,B,C,D are functionally independent, the project can be done in
n(T) ¼ n(A) n(B) n(C) n(D) ¼ 2 3 5 4 ¼ 120 ways
B 3.2
PERMUTATIONS
Many problems in combinatorics can be modeled as placing r balls into n cells. The two
ways of considering permutations are (1) sampling with replacement and with ordering
and (2) sampling without replacement and with ordering.
Sampling with Replacement and with Ordering
We consider placing r balls into n cells with replacement and with ordering. The first
ball can be placed in n cells in n ways. The second ball can be placed also in n ways. Since
the placing of the second ball is functionally independent of that of the first ball, there are
now n 2 ordered 2-permutations. Thus, for placing r balls into n cells, there will be nr ways
of placing r balls into n cells with replacement and ordering:
Number of distinct ordered r-tuples ¼ nr
(3:2:1)
The probability of each one of these r-tuples assuming that they are equally likely is n – r.
Clearly we can place 100 balls into one cell or 1 ball in 100 cells. Hence in Eq. (3.2.1) it
does not matter whether r . n, r ¼ n, or r , n.
Example 3.2.1 The number of distinct ordered ways we can draw 2 balls from an urn
containing 4 balls numbered from 1 to 4 is obtained as follows. The cells correspond to
n ¼ 4, the total number of balls. The balls correspond to r ¼ 2. The number of ways 2
balls can be placed into 4 cells is, from Eq. (3.2.1), 42 ¼ 16 ways. These 16 ordered
ways are shown in Table 3.2.1 for clarity.
Sampling without Replacement and with Ordering
We now consider placing r balls into n cells without replacement and with ordering.
This situation is different from the previous case. The first ball can be placed in n cells in
TABLE 3.2.1
1,1
3,1
1,2
3,2
1,3
3,3
1,4
3,4
2,1
4,1
2,2
4,2
2,3
4,3
2,4
4,4
3.2 Permutations
27
n different ways. Since one of the cells is occupied, the second ball can be placed only
in (n – 1) ways. The rth ball can be placed in (n – r þ 1) ways. Since placements of
the balls are again functionally independent, we have
Number of distinct r-tuples ¼ n(n 1) (n r þ 1)
(3:2:2)
These will be called r-permutations and will be defined by (n)r or nPr. Using the factorial
notation, we can write
(n)r ¼
n!
(n r)!
(3:2:3)
We cannot place 5 balls into 2 cells without replacement because at least some cells will
have more than 1 ball, violating the nonreplacement condition. Hence we define (n)r ¼ 0
whenever r . n. If r ¼ n, then (n)r ¼ (n)n ¼ n!.
Example 3.2.2 We continue Example 3.2.1 and now draw 2 balls from 4 balls without
replacement and with ordering. The number of ways this can be done is obtained by
applying Eq. (3.2.3) with n ¼ 4 and r ¼ 2 which is equal to 4.3 ¼ 12 ways
instead of the 16 ways that we obtained in Table 3.2.1. The 12 ways are shown in
Table 3.2.2.
In Table 3.2.2, the arrangement with ordering, (1,2), (1,3), (1,4), (2,3), (2,4), (3,4)
represents an ordering different from (2,1), (3,1), (4,1), (3,2), (4,2), (4,3).
We can now summarize the preceding results as follows. The number of different
ways in which r balls can be placed into n cells with ordering and replacement is n r,
and with ordering and without replacement is (n)r.
Example 3.2.3 In the Olympic 100-m dash, there are 9 runners. In how many ways can
the gold, silver, and bronze can be given to these runners?
Here the 9 runners are the cells n and the 3 medals are the balls r. Hence, from
Eq. (3.2.3), the number of ways that the medals can be distributed is (9)3 ¼
9 8 7 ¼ 504 ways.
Example 3.2.4 (Birthdays Problem) We want to determine the probability that in a
class size of r two or more birth dates match. We shall assume that the year consists of
365 days and the year of birth is inconsequential.
Here the 365 days in a year correspond to n cells and the class size, to r. If the class
size is over 365, then two or more birth dates will match with probability 1. The
number of ways r students can be placed into 365 birth dates with replacement is
365r. The number of ways they can be placed without replacement is (365)r,
meaning that all the birth dates are different. Thus, the probability of no two birth
dates match is
P{no birthdays match} ¼
(365)r
365r
TABLE 3.2.2
1,2
3,1
1,3
3,2
1,4
3,4
2,1
4,1
2,3
4,2
2,4
4,3
28
Basic Combinatorics
FIGURE 3.2.1
TABLE 3.2.3
#
Probability
#
Probability
#
Probability
20
21
22
23
24
25
26
0.411438
0.443668
0.475695
0.507297
0.538344
0.5687
0.598241
30
31
32
33
34
35
36
0.706316
0.730455
0.753348
0.774972
0.795317
0.814383
0.832182
44
45
46
46
48
49
50
0.932885
0.940976
0.948253
0.954774
0.960598
0.96578
0.970374
The probability P(r) that two or more birth dates match is the complement of the
preceding equation and is given by
P{r} ¼ 1 (365)r
365r
P(r) is graphed for r ¼ 0– 50 in Fig. 3.2.1 and tabulated for r ¼ 20– 26, 30– 36, and
44 – 50 in Table 3.2.3.
Note that for two or more birthdays to match, there is about 50% chance for
23 students, 71% chance for 30 students, and 95% chance for 46 students. In other
words, we need only 46 students for 2 or more birthdays to match with 95% probability,
but to get to 100% probability, we need to go to 366 students!
B 3.3
COMBINATIONS
Similar to the case of permutations we have two ways of considering combinations:
1. Sampling without replacement and without ordering
2. Sampling with replacement and without ordering
3.3 Combinations
29
Sampling without Replacement and without Ordering
We now discuss the case where we want to obtain a sample of r balls from n balls
without replacement and without ordering. We have already seen that the number of
ways to get r balls from n balls with ordering is (n)r. If we disregard ordering, then the
number of ways is called the r-combination from the n balls. We can define this quantity as
n
{r-combination from n balls} ¼
(3:3:1)
r
n
By ordering the elements in each r-combination, we can obtain from
n r the rpermutations (n)r. This can be done in r! ways. Hence we can determine r from either
of the following formulas:
n
r! ¼ (n)r
r
(3:3:2)
n
(n)r
n!
¼
¼
r!(n r)!
r!
r
The quantity nr is also called the binomial coefficient since it occurs as the coefficient of
the binomial expansion, given by
n n
n n1
n nr r
n n
n
(a þ b) ¼
a þ
a b þ þ
a b þ þ
b
0
1
r
n
Properties of Binomial Coefficients
n
n
¼
nr
r
1.
(3:3:3)
Expanding the righthand side of Eq. (3.3.3), we obtain
n!
n!
n
n
¼
¼
¼
r
nr
(n r)!½n (n r)! (n r)!r!
2. Pascal’s Identity:
nþ1
r
¼
n
r1
þ
n
r
(3:3:4)
Expanding the righthand side of Eq. (3.3.4), we obtain
n
r1
þ
n
r
n!
n!
þ
(r 1)!(n r þ 1)! r!(n r)!
n!
r
þ1
¼
r!(n r)! n r þ 1
nþ1
n!
nþ1
(n þ 1)!
¼
¼
¼
r!(n r)! n r þ 1
r!(n r þ 1)!
r
¼
Pascal’s identity shows that if two adjacent binomial coefficients are added, the
binomial coefficient in the next row in between these two coefficients results, as
shown in Fig. 3.3.1. It is called the Pascal triangle.
30
Basic Combinatorics
FIGURE 3.3.1
3. Sum of Binomial Coefficients:
n X
n
r¼0
r
¼ 2n
(3:3:5)
Substituting in the binomial expansion
(1 þ x)n ¼ 1 þ
n 2
n
n n
x þ þ
xþ
x
2
1
n
x ¼ 1, we obtain Eq. (3.3.5).
4. Vandermonde’s Identity:
nþm
r
r X
n
m
¼
k
rk
k¼0
(3:3:6)
Let us assume that there are two urns with n balls in the first and m balls in the
second.
nþm The total number of ways of picking r balls from both of these urns is
r . We can also choose these r balls by first picking k balls from the first urn
and (r 2 k) balls from the second urn. Since thechoice
m is functionally independent,
the number of ways of accomplishing this is nk rk
. The realizations for every
0 k r are mutually exclusive, and hence these numbers add, resulting in
Eq. (3.3.6). This is shown in Fig. 3.3.2.
FIGURE 3.3.2
3.3 Combinations
31
Example 3.3.1 We continue Example 3.2.2 with sampling 2 balls from 4 balls without
replacement and without ordering. Thus means that there is no distinction between (12)
and (21), (23) and (32), (34), and (43). Thus we have only 6 ways and this is obtained
by applying Eq. (3.3.1) with n ¼ 4 and r ¼ 2, yielding (4 3)/(2 1) ¼ 6. These are
shown in Table 3.3.1.
TABLE 3.3.1
1,2
2,3
1,3
2,4
1,4
3,4
Example 3.3.2
1. A committee of 3 women and 4 men must be formed from a group of 6 women
and 8 men. The number of ways 3 women can be chosen from 6 women is 63
from
Eq. (3.2.2) and the number of ways 4 men can be chosen from 8 men is
8
4 . Since these choices are functionally
independent, the total number of ways
the committee can be formed is 63 84 ¼ 1400.
2. We modify the problem above by having 2 particular men be always on the
committee. In this case we have a choice of only
2 men from the 6 men after
taking out the 2 required men. Thus, there are 63 62 ¼ 300 ways.
3. We again modify the problem by requiring that two particular women should not
be on the committee. In this case, we calculate the number of ways that the two
women will be together and subtract it from the total number of ways. There
are 51 84 ¼ 350
ways
that the two women will be together, and hence there
are 63 84 51 84 ¼ 1400 350 ¼ 1050 ways in which two given women
will not be together.
Example 3.3.3 (Catalan Numbers) There are n objects to be placed in computer
stacks in the following way. These objects are first stored and deleted later with the
result that when all the n objects are deleted, the stack is empty. Hence, the operation
store (S ) has to precede the operation delete (D), and at any point along the stack the
number of stores should always be more than or equal to the number of deletes.
Finally, the stack should begin with store and end with delete. We need to find the
number C(n) of operations required for n objects. For example, if there are n ¼ 3
objects, the number of ways of ordering the store and delete operations are
2 3 ¼ 6 given by SSSDDD, SDSDSD, SSDDSD, SDSSDD, SSDSDD, SDSDSSD.
Thus, the operation starts with an S and ends in a D, and at no point along the
chain can the number of D operations exceed the S operations. We will show that
the number C(n) is given by
C(n) ¼
1
(2n)!
n þ 1 n!n!
Since there are a total
of 2n operations, the total number of ways that we can obtain an
n-combination is 2n
n . Out of these combinations there will be many realizations in
which the number of Ds will exceed the number of Ss. We will now find the
number of ways in which the Ds will exceed the Ss. If we are at the point (n þ 1)
32
Basic Combinatorics
TABLE 3.3.2
#
Stacks
#
Stacks
#
Stacks
1
2
3
4
5
6
7
1
2
5
14
42
132
429
8
9
10
11
12
13
14
1430
4862
16796
58786
208012
742900
2674440
15
16
17
18
19
20
9694845
35357670
129644790
477638700
1767263190
6564120420
2n
2n
along the stack, the number of ways that
D
will
exceed
S
will
be
nþ1 ¼ n1 . Hence
2n 2n
1 2n
the required number of stacks is n n1 ¼ nþ1 n , which is the desired result.
The Catalan numbers are tabulated in Table 3.3.2 for n ¼ 1 to 20.
Sampling with Replacement and without Ordering
We shall now consider forming r-combinations with replacement and without ordering. In this case we pick r balls with replacement from balls in n distinct categories such as
red, blue, green, and brown. We do not distinguish among balls of the same color, and the
distinction is among different colors. For example, if we are picking 6 balls from the 4
color categories, then it can be arranged as shown in Figure 3.3.3.
There are now 2 red balls, 1 blue ball, no green balls, and 3 brown balls. The bar
symbol | separates the four categories. The arrangement cannot start with a bar or end
with a bar; hence we have 4 2 1 ¼ 3 bars. We now have to choose 6 balls from 6 balls
and 3 bars without replacement and without ordering. There are then
987
9
9
¼
¼
¼ 84
3
6
321
ways of picking 6 balls from 4 different categories.
Generalizing this result, if there are n distinct categories of balls, and if we have to
pick r balls with replacement and without regard to order, then
Number of r-combinations ¼
n1þr
r
(3:3:7)
This is the celebrated Bose –Einstein statistics, where the elementary particles in the same
energy state are not distinguishable.
FIGURE 3.3.3
3.3 Combinations
33
Example 3.3.4 An urn contains red, blue, green, yellow, white, and black balls. In how
many ways can we choose
1. 10 balls? This is a direct application of Eq. (3.3.7) with n ¼ 6, r ¼ 10 and the
number of ways is
6 1 þ 10
15
15
¼
¼
¼ 3003
10
10
5
2. 20 balls with least 2 colors each? If we draw 2 balls of each color, then we
have drawn 12 balls, leaving us to draw 20 – 12 ¼ 8 balls to be drawn without
restriction. Hence, applying Eq. (3.2.7) with n ¼ 6 and r ¼ 8, we obtain the
number of ways as
13
13
61þ8
¼ 1287
¼
¼
5
8
8
3. 20 balls with no more than 3 blue balls? We first calculate the number of unrestricted ways of drawing 20 balls as
6 1 þ 20
25
¼
¼ 53,130
20
5
We next draw at least 4 blue balls out of 20 balls. Having drawn the 4 blue balls,
the number of ways to draw the remaining 16 balls is
21
21
6 1 þ 16
¼ 20,349
¼
¼
5
16
26
Thus the number of ways of drawing 20 balls with no more than 3 blue balls is
53,130 2 20,349 ¼ 32,781.
4. 20 balls with at least 4 red balls, at least 2 green balls, at least 6 white balls,
and no more than 3 brown balls? We first draw the lower bounded balls,
namely, 4 þ 2 þ 6 ¼ 12, before we go to the upper-bounded brown balls. We
then draw the remaining 8 balls in an unrestricted manner, and the number
of ways is
61þ8
13
¼
¼ 1287
8
5
We have to subtract from this number the number of ways 4 brown balls can be
picked from the 6 colors. This number is
61þ4
9
¼
¼ 126
4
4
Hence the number of ways in which all the restrictions will be satisfied is
1287 2 126 ¼ 1161.
Example 3.3.5
the equation
We tackle a different kind of a problem. How many solutions are there to
x1 þ x2 þ x3 þ x4 þ x5 þ x6 ¼ 24
where fxk, k ¼ 1, . . . , 6g are positive integers such that
34
Basic Combinatorics
1. xk . 2 for all k? We need xk 3 for all k. This takes up 6 3 ¼ 18 of the required
total of 24, leaving us to solve the equation
x1 þ x2 þ x3 þ x4 þ x5 þ x6 ¼ 6
in an unrestricted manner. Here we identify, in Eq. (3.2.7), n ¼ 6 corresponding to
the number of xk values and r ¼ 6 corresponding to the righthand side of the
equation. Hence the number of solutions is
11
11
61þ6
¼ 462
¼
¼
5
6
6
2. x1 . 2, x2 1, x3 . 3, x4 4, x5 . 2, and x6 2? Using x1 3, x3 4, x5 3,
the number of restrictions add upto 3 þ 1 þ 4 þ 4 þ 3 þ 2 ¼ 17. This leaves
24 2 17 ¼ 7 unrestricted choices. Hence the number of solutions is
61þ7
12
12
¼
¼
¼ 792
7
7
5
3. x2 , 5, x3 . 10? The number of balls that can be drawn with the restriction
x3 11 corresponding to n ¼ 6 and r ¼ (24 2 11) ¼ 13 is given by
18
18
6 1 þ 13
¼ 8568
¼
¼
5
13
13
The number of draws that violate the additional restriction x2 , 5 is obtained by
first finding the number of draws with the additional restriction x2 6. The
number of ways corresponding to n ¼ 6 and r ¼ (24 2 11 2 6) ¼ 7 is given by
61þ7
12
12
¼
¼
¼ 792
7
7
5
Hence the required result is 8568 2 792 ¼ 7776 ways.
Example 3.3.6 Here is an interesting example to illustrate the versatility of the
formula in Eq. (3.2.7). Four dice, each with only 4 faces marked 1,2,3,4, are tossed.
We have to calculate the number of ways in which the tosses will result in exactly 1
number, 2 numbers, 3 numbers, and 4 numbers showing on the faces of the dice
without regard to order and with replacement. The total number of ways is tabulated
in Table 3.3.3.
TABLE 3.3.3
1,1,1,1
1,1,1,2
1,1,2,2
1,2,2,2
1,1,1,3
1,1,3,3
2,2,2,2
2,2,2,3
2,2,3,3
2,3,3,3
2,2,2,4
2,2,4,4
3,3,3,3
3,3,3,4
3,3,4,4
3,4,4,4
3,3,3,1
4,4,4,4
4,4,4,1
4,4,1,1
4,1,1,1
4,4,4,2
1,1,2,3
1,1,3,4
1,1,4,2
2,2,3,4
2,2,4,1
2,2,1,3
3,3,4,1
3,3,1,2
3,3,2,4
4,4,1,2
4,4,2,3
4,4,3,1
1,2,3,4
)
4 single digits
18 double digits
12 triple digits
1 quadruple digit
3.3 Combinations
35
We can see from Table 3.3.3 that there are 4 single digits, 18 double digits, 12 triple
digits, and 1 quadruple digit totaling 35 arrangements without regard to order and with
replacement. The problem can be stated as follows. Four dice each with 4 faces are
tossed without regard to order and with replacement. Find the number of ways in which
1. This can be achieved. With n ¼ 4 and r ¼ 4 in Eq. (3.2.7), the number of ways is
41þ4
4
¼
7
7
¼ 35
¼
3
4
which agrees with the total number in the table.
2. Exactly one digit is present. With n ¼ 4 and r ¼ 1, the number of ways is
41þ1
1
¼
4
¼4
1
which agrees with the table.
3. Exactly 2 digits
are present. The number of ways 2 digits can be drawn from 4
digits is 42 . The number of ways any particular 2 digits, for example f1,2g, will
occupy a 4-digit number is a restricted case as in Example 3.2.4 with both 1 and
2 occupying two slots and the rest of the two remaining unrestricted. With
n ¼ 2 and r ¼ 4 2
2 ¼ 2 in Eq.
21þ2
(3.2.7), the number of ways for this particular
3
sequence
f1,2g
is
¼
2
2 ¼ 3. The total number using all the combinations
are 42 32 ¼ 6 3 ¼ 18 ways. This agrees with the table.
4. Exactly 3 digits
are present. The number of ways 3 digits can be drawn from 4
digits is 43 ¼ 4 ways. The number of ways 3 digits can be arranged in 4 slots
such that all the three digits
3 are present is again given by the restriction
4formula
3
Eq. (3.2.7) 31þ(43)
¼
¼
3.
Hence
the
total
number
of
ways
is
2
1
3 1 ¼ 12
ways, agreeing with the table.
5. Exactly 4 digits are present. The number of ways 4digits will
be present is
obtained from the argument in item 3 above, giving 44 41þ(4þ4)
¼1
0
Generalization
The result in Example 3.3.6 can be generalized to the tossing of n n-sided
dice. The total number of sample points without regard to order and with replacement
is, from Eq. (3.2.7)
n1þn
n
¼
2n
n
(3:3:8)
Out of these, the number of ways exactly r digits will
occur can be derived as follows:
The number of ways r digits can be chosen from n is nr , and if any particular sequence of r
digits has to fill n slots, then the number of ways this can be done is obtained by
substituting r ¼ (n 2 r) in Eq. (3.2.7), yielding
r 1 þ (n r)
nr
¼
n1
nr
(3:3:9)
36
Basic Combinatorics
FIGURE 3.3.4
Since all the nr sequences are functionally independent, the number of ways N(n,r) is as
follows, from Eq. (3.1.2):
n1
n
(3:3:10)
¼
N(n,r) ¼
nr
r
Using Eq. (3.3.10), N(n,r) tabulated for n ¼ 1, . . . , 8 and r ¼ 1, . . . , 8 in Fig. 3.3.4.
In addition, we have the following interesting identity. Combining Eqs. (3.3.8) and
(3.3.9), we have
X
n 2n 1
n
n1
¼
(3:3:11)
n
r
nr
r¼0
Equation (3.3.11) can also be expressed by substituting m ¼ n – 1, r ¼ n, and k ¼ r in
the Vandermonde identity of Eq. (3.3.6)
X
X
r n nþm
n
m
nþn1
n
n1
¼
¼
¼
:
(3:3:12)
r
k
rk
n
k
nk
k¼0
k¼0
CHAPTER
4
Discrete Distributions
B 4.1
BERNOULLI TRIALS
We will elaborate on the Cartesian product already discussed in Chapter 1. Consider the
tossing of an unfair coin with success given by fhead ¼ sg with probability p and failure
given by ftail ¼ f g with probability q, and p þ q ¼ 1. The sample space is S1 ¼ fs, f g with
field F ¼ ½S1 , 1, {s}, { f }. If the coin is tossed twice, then the sample space S2 ¼ fss, sf,
fs, ffg is an ordered set from the Cartesian product S1 S1. The cardinality of S2 is
2 2 ¼ 4. The probability of two successes is p 2 and two failures is q 2. If we toss the
coin n times, then the resulting sample space is Sn ¼ S1 S1 S1 S1 and the
cardinality of Sn is 2n. The probability of n successes is p n.
Bernoulli Trials
Bernoulli trials are repeated functionally independent trials with only two events s and
f for each trial. The trials are also statistically independent with the two events s and f in
each trial having probabilities p and (1 2 p), respectively. These are called independent
identically distributed (i.i.d.) trials.
Example 4.1.1 Tossing an unfair coin 4 times constitutes 4 Bernoulli trials with
heads ¼ s and tails ¼ f with Pfsg ¼ p and Pf fg ¼ 1 2 p. The sample space S is the
Cartesian product with ordered 4-tuples. The cardinality of S is 24 ¼ 16 points.
Example 4.1.2 Tossing a die 5 times can be considered Bernoulli trials if we define
getting an ace as success, s with Pfsg ¼ 16 and not getting an ace as failure, f with
Pf fg ¼ 56. The resulting sample space S is the Cartesian product with ordered quintuples
with number of points equaling the cardinality 25 ¼ 32.
Bernoulli Distribution
Bernoulli distribution is the simplest of a discrete distribution. A single event is
Bernoulli distributed if the probability of success is p and probability of failure is q
with p þ q ¼ 1.
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
37
38
Discrete Distributions
Example 4.1.3 A single die is thrown. The probability of getting an ace is p ¼ 16 and the
probability of not getting an ace is q ¼ 56. Hence this experiment is Bernoulli distributed.
B 4.2
BINOMIAL DISTRIBUTION
We are given an experiment S with an event A , S with PfAg ¼ Pfsuccessg ¼ p and
PfAg ¼ Pffailureg ¼ q with p þ q ¼ 1. We repeat this experiment n times, and the resulting sample space Sn is the Cartesian product
Sn ¼ {S S S}
n times
containing 2n elementary events. We will now determine the probability b(k; n, p) that the
event A occurs exactly k times in n Bernoulli trials. We shall address this problem in two
parts. In the first part we will find the probability of k successes in a particular sequence in
n trials. For example, one such sequence can be
{s, s, f , s, f , f , . . . , s, f }
exactly k successes
For the k independent successes the probability is p k and for the (n 2 k) independent failures the probability is q n – k. Since both successes and failures are independent, we have
Probability{k successes in a given sequence} ¼ pk qnk
(4:2:1)
In the second part we address the situation of k successes in any sequence in n trials. The
number of ways k successes can be obtained from n trials is by sampling
without replacement and without regard to order. This number from Eq. (3.3.2) is kn . These realizations
are mutually exclusive since the occurrence of one realization precludes the occurrence of
another. Hence the probabilities will add, giving rise to
n k nk
Probability{k successes in any sequence in n trials} ¼ b(k; n,p) ¼
pq
(4:2:2)
k
This is called the binomial distribution, and b(k;n,p) is called a probability mass function.
over all k. Since
We
will now show that this is a probability distribution by summing
n
n k nk
q
is
the
kth
term
in
the
binomial
expansion
of
(
p
þ
q)
,
we
can
write
p
k
n
n
X
X
n k nk
p q ¼ ( p þ q)n ¼ 1
b(k; n,p) ¼
(4:2:3)
k
k¼0
k¼0
The probability of getting utmost r successes is given by
P{number of successes r} ¼
r
X
k¼0
b(k; n,p) ¼
r X
n k nk
pq
k
k¼0
(4:2:4)
and the probability of getting at least r successes is given by
n X
n k nk
P{number of successes r} ¼
pq
b(k; n,p) ¼
k
k¼r
k¼r
n
X
(4:2:5)
4.2 Binomial Distribution
39
Since n! grows very rapidly with n, computational difficulties in calculating b(k;n,p)
can be avoided by using the recursive formula shown below with starting value b(0;n,p).
n
n k p n k nk
kþ1 nk1
p q
b(k þ 1; n,p) ¼
¼
pq
kþ1 q k
kþ1
¼
Example 4.2.1
nk p
b(k; n,p)
kþ1 q
(4:2:6)
A fair die is rolled 10 times, and we will calculate
1. The probability that an ace comes up exactly 4 times. Substituting p ¼ 16 and q ¼ 56,
n ¼ 10, and k ¼ 4 in Eq. (4.2.3), we obtain
10! 1 4 5 6
P{6 aces in 10 trials} ¼
¼ 0:05427
6!4! 6
6
2. The probability that utmost 4 aces show up. In probability 1 (above) we wanted
exactly 4 aces coming up. Now we are looking into the possibility of 4 or fewer
aces showing up. This probability is obtained as follows:
k 10k
10!
1
5
P{number of aces 4} ¼
¼ 0:98454
k!(10 k)! 6
6
k¼0
4
X
3. The probability that the number of aces showing is between 2 and 6 inclusive. Here
we have to sum b(k: 10, 16) between 2 and 6. Performing the indicated summation,
we have
k 10k
10!
1
5
¼ 0:22451
P{2 number of aces 6} ¼
k!(10
k)!
6
6
k¼2
k 5 10k
The graph of the probability mass function 10i ( 16 P
) ( 10 ) is
i shown
10iin Fig. 4.2.1,
and that of the cumulative distribution function ki¼0 10i 16 56
is shown in
Fig. 4.2.2 for k ¼ 0, . . . , 10.
6
X
FIGURE 4.2.1
40
Discrete Distributions
FIGURE 4.2.2
Example 4.2.2 We shall use the recursive form [Eq. (4.2.6)] to compute the binomial
coefficients in Example 4.2.1 with initial condition b(0;n,p) ¼ 0.161506. These values
are substituted in Eq. (4.2.4), and the first four steps are shown:
1
b 0; 10,
¼ 0:161506
6
1
10 1
0:161506 ¼ 0:323011
¼
b 1; 10,
6
1 5
1
9 1
b 2; 10,
¼ 0:323011 ¼ 0:29071
6
2 5
1
8 1
b 3; 10,
¼ 0:29071 ¼ 0:155045
6
3 5
The true values and the computed values are shown in Table 4.2.1, and they are the same.
Example 4.2.3 Two experiments are performed. In the first experiment 6 fair dice are
tossed and the person wins if she gets 1 or more aces. In the second experiment 12 dice
are tossed and the person wins if she gets 2 or more aces. We want to find in which experiment the probability of winning is higher.
TABLE 4.2.1
b(k;n,p)
k
Computed
Calculated
k
Computed
Calculated
0
1
2
3
4
5
0.161506
0.323012
0.290711
0.155046
0.054266
0.013024
0.161506
0.323011
0.29071
0.155045
0.054266
0.013024
6
7
8
9
10
0.002171
0.000248
0.000019
0.000001
0
0.002171
0.000248
0.000019
0.000001
0
4.3 Multinomial Distribution
41
The probability of getting one or more aces out of 6 tosses is the same as the complement of the probability of getting no aces in 6 throws. This is obtained from Eq. (4.2.5)
with n ¼ 6, p ¼ 16 and q ¼ 56:
0 6
1
5
6
1 P{no aces in 6 throws} ¼ 1 ¼ 0:6651
0
6
6
The probability of getting 2 or more aces in 12 throws is the same as the complement of the
sum of the probabilities of getting no aces and one ace in 12 throws. With n ¼ 12, we
obtain from Eq. (4.2.5)
1 P{no aces in 12 throws} P{one ace in 12 throws}
0 12
1 11
12
12
1
5
1
5
¼1
1 6
6
6
6
0
1
¼ 1 0:1122 0:2692 ¼ 0:6186
Hence the probability of getting one or more aces in 6 throws is better than getting two or
more aces in 12 throws.
Example 4.2.4 An urn consists of 40 red balls and 60 green balls. What is the probability
of getting exactly k red balls in a sample size of 10 if the sampling is done
1. With replacement? The meaning of “with replacement” is to pick a ball from the
100 in the urn note its color and replace it and pick again, and this process is continued for 10 times. Thus Pfred ballg ¼ p ¼ 25 and Pfgreen ballg ¼ q ¼ 35 and with
n ¼ 10, the probability of k red balls in 10 trials is given by
k 10k
2
2
3
10
b k; n,
¼
k
5
5
5
2. Without replacement? Here we sample the balls and do not replace it in the urn.
100
Thus, the number of ways of getting 10 balls from 100 balls
40 is 10 . Out of
these 10 balls, the number of ways of getting
kred balls is k and the number
of ways of getting (10 2 k) blue balls is 1060 k . Since getting k red balls and
40
(10
602k) blue balls are functionally independent, the total number of ways are k 10 k using the product rule. Hence the probability P(k) of k red balls without
replacement is given by
40
60
k
10 k
P(k) ¼
100
10
B 4.3
MULTINOMIAL DISTRIBUTION
The binomial distribution of the previous section can be generalized to n repeated independent trials where each trial has m outcomes fEi, i ¼ 1, . . . , mg with probabilities PfEig ¼
fpi, i ¼ 1, . . . , mg and
p1 þ p2 þ þ pm ¼ 1
(4:3:1)
42
Discrete Distributions
Each of the outcomes fEi, i ¼ 1, . . . , mg occurs fki, i ¼ 1, . . . , mg times with
k1 þ k2 þ þ km ¼ n
(4:3:2)
The probability Pfk1, k2, . . . , kmg is given by the multinomial probability distribution,
n
pk1 pk2 pkmm
P{k1 , k2 , . . . , km } ¼
k1 , k2 , . . . , km 1 2
n!
pk1 pk2 pkmm
k1 !k2 !; km ! 1 2
¼
(4:3:3)
If m ¼ 2, the multinomial distribution reduces to the binomial distribution.
Example 4.3.1 A pair of dice is rolled 8 times. We will find the probability that 7 or 11
occurs thrice; 2, 3, or 12 twice; any pair twice; and any other combination once. For each
rolling of a pair there are four events E1, E2, E3, E4 and their probabilities are obtained
from Table 2.2.1 and are shown below:
E1 ¼ {7 or 11}
8
36
4
P{E2 } ¼ p2 ¼
36
6
P{E3 } ¼ p3 ¼
36
18
P{E4 } ¼ p4 ¼
36
P{E1 } ¼ p1 ¼
E2 ¼ {2, 3 or 12}
E3 ¼ {any pair}
E4 ¼ {other combination}
In Eq. (4.3.3) we substitute k1 ¼ 3, k2 ¼ 2, k3 ¼ 2, k4¼1 with p1, p2, p3, p4 as given above,
yielding
3 2 2 1
8!
8
4
6
18
P{3, 2, 2, 1} ¼
¼ 0:003161
3!2!2!1! 36
36
36
36
B 4.4
GEOMETRIC DISTRIBUTION
Closely allied to the binomial distribution is the geometric distribution. A sequence of
independent Bernoulli trials with p as the probability of success can be modeled in two
different ways. The distribution of the number of successes k is binomial. The waiting
time until the first success occurs is geometrically distributed. If the first success is to
occur at the kth trial, then we must have (k 2 1) failures preceding the first success.
Since the trials are independent, this probability is given by
g(k,p) ¼ (1 p)k1 p, k ¼ 1,2, . . .
(4:4:1)
This is the geometric distribution with parameter p. Further, the waiting times between
each success and the next are independent and are geometrically distributed.
We can show that the sum to infinity of g(k,p) is indeed 1 as shown below:
1
X
k¼1
g(k,p) ¼ p
1
X
k¼1
(1 p)k1 ¼
p
¼1
1 (1 p)
4.4 Geometric Distribution
43
Example 4.4.1 In an urn there are r red balls and b blue balls. Balls are selected at
random with replacement so that the first blue ball is obtained. We have to find the
probability that
1. Exactly k draws are needed for the first blue ball. The probability of a blue ball is
p ¼ b/(b þ r). Substituting this value p in Eq. (4.4.1), the required probability is
b
b k1 b
g k,
¼ 1
bþr
bþr
bþr
k1
r
b
br k1
¼
¼
bþr
b þ r (b þ r)k
2. At least m draws needed for the first blue ball. The first success occurs at the mth
draw after (m 2 1) failures. Since the failure probability is q ¼ [r/(b þ r)], we
have from the definition of the geometric distribution this probability is
[r/(b þ r)]m21.
The same result can also be obtained from the cumulative summation as
follows:
k1
1
1 X
b
b X
r
g k,
¼
bþr
b þ r k¼m b þ r
k¼m
#
m1 "
2
b
r
r
r
þ
¼
1þ
þ
bþr bþr
bþr
bþr
2
3
m1
m1
b
r
1
r
6
7
¼
¼
4
r 5
bþr bþr
bþr
1
bþr
This geometric distribution is shown in Fig. 4.4.1 for r ¼ 100, b ¼ 20, and k ¼ 1, . . . , 10
and the cumulative distribution in Fig. 4.4.2.
FIGURE 4.4.1
44
Discrete Distributions
FIGURE 4.4.2
B 4.5
NEGATIVE BINOMIAL DISTRIBUTION
The binomial distribution finds the probability of exactly k successes in n independent
trials. The geometric distribution found the number of independent trials needed for the
first success. We can generalize these two results and find the number of independent
trials required for k successes. This distribution is known as the negative binomial or
the Pascal distribution. Let us now define the following events to find this distribution:
E ¼ n trials for exactly k successes
A ¼ exactly (k 1) successes occur in (n 1) trials
B ¼ nth trial results in kth success
Since E ¼ A > B and A and B are independent, we have
P{E} ¼ P{A} P{B}
(4:5:1)
We will now compute PfAg. In a sequence of (n 2 1) trials we have exactly (k 2 1) successes and [n 2 1 2 (k 2 1)] ¼ (n 2 k) failures. Since these trials
are independent, the
probability of that particular sequence is p k21q n21. There are n1
k1 mutually exclusive
sequences, and hence
n 1 k1 nk
p q
P{A} ¼ b(k 1; n 1, p) ¼
(4:5:2)
k1
Since the probability PfBg ¼ p, we can substitute this and Eq. (4.5.2) in Eq. (4.5.1) and
write
n 1 k1 nk
n 1 k nk
p q
nb(k; n,p) ¼ P{E} ¼ p
p q , n ¼ k, k þ 1, . . . (4:5:3)
¼
k1
k1
We have to show that the sum of nb(k;n,p) from n ¼ k to 1 equals 1. Or
1 X
n 1 k nk
pq
nb(k; n,p) ¼
¼1
k1
n¼k
n¼k
1
X
(4:5:4)
4.5
TABLE 4.5.1
Negative Binomial Distribution
45
Negative binomial
n
Probability
n
Probability
n
Probability
5
6
7
8
0.16807
0.252105
0.226894
0.158826
9
10
11
12
0.095296
0.05146
0.02573
0.01213
13
14
15
0.005458
0.002365
0.000993
By substituting (n 2 k) ¼ m, we can rewrite Eq. (4.5.4) as
1
1 1 X
X
kþm1 k m X kþm1 k m
pq ¼
p q ¼1
nb(k; k þ m,p) ¼
k1
m
m¼0
m¼0
m¼0
(4:5:5)
The Taylor series expansion of (1 2 x)2k yields
(k)( k 1)
(k)(k 1)(k 2)
(x)2 þ
(x)3 þ 2!
3!
1 X
kþm1 m
¼
x
m
m¼0
(1 x)k ¼ 1 þ kx þ
(4:5:6)
Substituting x ¼ q in Eq. (4.5.6) and (1 2 q) ¼ p, we obtain
1 1 X
X
kþm1 m
kþm1 k m
k
p ¼
q ,
p q , which is Eq. (4:5:5)
or 1 ¼
m
m
m¼0
m¼0
In Eq. (4.5.3), if we substitute k ¼ 1, then we get the geometric distribution.
Example 4.5.1 A computer software company wants to recruit 5 new engineers. The
probability that a person interviewed is given a job offer is 0.7. We want to find the distribution of the number of people interviewed for the 5 positions.
We use Eq. (4.5.3) with p ¼ 0.7 and k ¼ 5. Thus
n1
P{E ¼ n} ¼
(0:7)5 (0:3)n5 , n ¼ 5, 6; . . .
51
PfE ¼ ng is shown for n ¼ 5,6, . . . ,15 in Table 4.5.1, and the probability mass function is
shown in Fig. 4.5.1.
FIGURE 4.5.1
46
Discrete Distributions
From the table and the figure we see that the maximum probability of 0.2521 occurs at
k ¼ 6 and it is most likely that company will interview exactly 6 persons for the 5 jobs.
B 4.6
HYPERGEOMETRIC DISTRIBUTION
The binomial distribution is obtained while sampling with replacement. Let us select a
sample of n microprocessor chips from a bin of N chips. It is known that K of these
chips are defective. If we sample with replacement then the probability of any defective
chip is K/N. The probability that k of them will be defective in a sample size of n is
given by the binomial distribution since each drawing of a chip constitutes a Bernoulli
trial. On the other hand, if we sample without replacement, then the probability for
the first chip is K/N, the probability for the second chip is (K 2 1)/(N 2 1), and for
each succeeding sample the probability changes. Hence the drawings of the chips
do not constitute Bernoulli trials and binomial distribution does not hold. Hence we
have to use alternate methods to arrive at the probability of finding k defective chips
in n trials.
The number of sample points in this sample space
is choosing n chips out of the total
N chips without regard to order. This can be done in Nn ways. Out of this sample size of n,
there are k defective chips and (n 2 k) good chips.The
number of ways of picking k defective chips out of the total of K defective chips is Kk and the number
of ways of drawing
(n 2 k) good chips out of the remaining (N 2 K) good chips is NK
nk . We denote the total
number of ways of drawing a sample of n chips with k bad chips as the
event
E.
Since these
drawings are functionally independent the number of ways is Kk NK
nk . Hence the
probability of the event E is given by
K
NK
k
nk
(4:6:1)
P{E ¼ k} ¼
N
n
This is known as the hypergeometric distribution. We came across this distribution in
Example 4.2.4. Since the summation over all k must be one, we must have the following
in Eq. (4.6.1):
n X
K
N K
N
¼
(4:6:2)
k
n
k
n
k¼0
This result follows directly from Vandermonde’s identity of Eq. (3.3.6) given by
n X
n
m
nþm
¼
(3:3:6)
k
rk
r
k¼0
with n ¼ K, m ¼ N 2 K, and r ¼ n.
Example 4.6.1 (Mass Megabucks) Massachusetts megabucks lottery is played as
follows. A player has to choose 6 numbers from a slate of 42 numbers. Twice a week
the lottery commission draws 6 numbers. If all the 6 numbers of the player match all
the 6 numbers drawn by the commission, the player wins megabucks. There are smaller
prizes for matching 5, 4, or 3 numbers. We have to calculate the probability of winning
megabucks or the smaller prizes.
4.6 Hypergeometric Distribution
TABLE 4.6.1
47
Mass megabucks
k
N(k)
C(k)
P(k)
0
1
2
3
4
5
6
1947792
2261952
883575
142800
9450
216
1
3
2
6
37
555
24286
5,245,786
0.371306035
0.431194105
0.168435197
0.02722185
0.001801446
0.000041176
0.000000191
Let k ¼ 0,1, . . . , 6. Substituting N ¼ 42, n ¼ 6, and K ¼ 6, we can find from
Eq. (4.6.1) PfE ¼ kg:
6
42 K
k
6k
N(k)
, k ¼ 0,1, . . . , 6
¼
P{E ¼ k} ¼
D(k)
42
6
The results
for k ¼ 0, . . . , 6 in Table 4.6.1. In this table, the second column
are tabulated
N(k) ¼ 6k 42K
is
the
number
of ways k digits can be formed from 6 digits. In the third
6k
column C(k) ¼ D(k)/N(k) represents the chance of winning with k digits rounded to the
nearest digit and in the fourth column P(k) is the probability of winning.
Example 4.6.2 Hypergeometric distribution is used to estimate an unknown population
from the data obtained. To estimate the population N of tigers in a wildlife sanctuary, K
tigers are caught, tagged, and released. After the lapse of a few months, a new batch of
n tigers is caught, and k of them are found to be tagged. It is assumed that the population
of tigers does not change between the two catches. The total number of tigers N in the
sanctuary is to be estimated.
In Eq. (4.6.1), the known quantities are K, k, and n. The total number N has to be
estimated in the following equation:
K
NK
k
nk
Pk {N} ¼
N
n
We will find the value of N that maximizes the probability PkfNg. Such an estimate is
called the maximum-likelihood estimate, which will be discussed in Chapter 18. Since
the numbers are very large, we investigate the following ratio:
K
NK , K
N1K
k
nk
k
nk
Pk {N}
¼
N
N1
Pk {N 1}
n
n
N K
N1
nk
n
Nn
NK
¼
¼ N
N1K
N NKnþk
n
nk
48
Discrete Distributions
The ratio (Pk {N})=(Pk {N 1}) is greater than 1 if (N 2 n)(N 2 K ) . N (N 2 K 2 n þ k),
Nk , nK, or N , nK/k.
Conclusion. If N is less than nK/k, then Pk(N) is increasing, and if N is greater than
nK/k, then Pk(N ) is decreasing. Hence the maximum value of Pk(N ) occurs when N
is equal to the nearest integer not exceeding nK/k. Thus, if K ¼ 100, n ¼ 50, and
k ¼ 20, then the population of tigers is 100.50/20 ¼ 250.
Approximation to Binomial Distribution
If N and K are very large with respect to n and k, then the hypergeometric distribution
can be approximated by the binomial distribution as shown below:
P{E ¼ k} ¼
K
k
NK
nk
N
n
¼
k!
(N K)!
n!(N n)!
k!(K k)! (n k)!(N K n þ k)!
N!
¼
n!
K! (N n)!
(N K)!
k!(n k)! (K k)! N! (N K n þ k)!
¼
n K(K 1) (K k þ 1)
k N(N 1) (N k þ 1)
(N K)(N K 1) (N K n þ k þ 1)
(N k)(N K 1) (N k n þ k þ 1)
(4:6:3)
Since K k and N n, Eq. (3.3.7) can be approximated as
n K K K (N K)(N K) (N K)
P{E ¼ k} ¼
N N N
k N N N
k times
(4:6:4)
Nk times
Approximating p ffi K/N and q ffi (N 2 K )/n in Eq. (3.3.8), we obtain
n k nk
P{E ¼ k} ffi
pq
k
(4:6:5)
which is a binomial distribution.
B 4.7
POISSON DISTRIBUTION
Perhaps three of the most important distributions in probability theory are (1) Gaussian,
(2) Poisson, and (3) binomial. We have already discussed the binomial and other allied
distributions that are characterized by discrete time (n) and discrete state, that is, the cumulative distribution function has jumps occurring only at discrete points. In the regular
Poisson distribution the cumulative distribution has unit jumps at random times and not
at discrete times. Thus, it is discrete in state but continuous in time.
4.7 Poisson Distribution
49
The Poisson distribution takes values 0,1, . . . k, . . . at random points with a parameter
l and is given by
p{k ; l} ¼ el
lk
,
k!
k0
(4:7:1)
We will show that this is a probability function by summing k from 0 to 1 as shown below:
1
X
p{k ; l} ¼
k¼0
1
X
el
k¼0
1
X
lk
lk
¼ el
¼ el el ¼ 1
k!
k!
k¼0
(4:7:2)
Similar to Eq. (4.2.6) we can derive a recursive form for evaluating the Poisson distribution as shown below:
1
X
k¼0
p{k þ 1 ; l} ¼
1
X
k¼0
el
1
lkþ1
l X
lk
l
p{k ; l}
¼
el ¼
(k þ 1)! k þ 1 k¼0
k! k þ 1
(4:7:3)
This last equation is solved from the starting value p(0, l) ¼ e 2l.
Example 4.7.1 During a certain experiment the number of a particles passing through a
sensor in a 1-ms interval is 3.2. The Poisson distribution from Eq. (4.7.1) with l ¼ 3.2 can
be written as
p{k ; 3:2} ¼ e3:2
(3:2)k
,
k!
k0
This is graphed in Fig. 4.7.1.
We want to find the probability that
1. Six a particles pass through the sensor in a given millisecond. Substituting l ¼ 3.2
and k ¼ 6, we obtain
p{6 ; 3:2} ¼ e3:2
ð3:2Þ6
¼ 0:06079
6!
FIGURE 4.7.1
50
Discrete Distributions
2. No more than three a particles pass through the sensor. We have to find the
cumulative distribution given by
3
X
p{k ; 3:2} ¼
k¼0
3
X
k¼0
e3:2
(3:2)k
¼ 0:6025
k!
Example 4.7.2 The average number of customers arriving at the service counter in a
supermarket is l, and the probability of k customers arriving in the interval (0,t)
minutes is given by the Poisson distribution pfk ; lg ¼ e 2l[(lt)k/k!]. In the nonoverlapping intervals (0,t1) and (t1,t), we can write the following Poisson law
P{n1 customers in (0,t1 ) and n2 in (t1 ,t)}
¼ P{n1 customers in (0,t1 )} P{n2 customers in (t1 ,t)}
and the two events are independent in disjoint intervals. We will find the conditional probability, Pfn1 customers in (0,t1) j (n1 þ n2) customers in (0,t)g. From the definition of conditional probability, we can write
P{n1 ; (0,t1 ) j (n1 þ n2 ); (0,t)} ¼
¼
P{½n1 ; (0,t1 )½(n1 þ n2 ); (0,t)}
P{(n1 þ n2 ); (0, t)}
P{½n1 ; (0,t1 )½n2 ; (t1 ,t)}
P{(n1 þ n2 ); (0, t)}
elt1 (lt1 )n1 el(tt1 ) ½l(t t1 )n2
¼
n1 !
n2 !
¼
(lt1 )n1 ½l(t t1 )n2 (n1 þ n2 )!
(lt)n1 þn2 n1 !n2 !
¼
t1n1 (t t1 )n2 (n1 þ n2 )!
tn1 þn2 n1 !n2 !
,
elt (lt)n1 þn2
(n1 þ n2 )!
which yields a surprising result independent of l!
Poisson distribution has been used in a number of applications such as the
1. Number of telephone calls coming into an exchange
2. Number of cars entering a highway ramp
3. Number of packets transmitted along wireless lines
4. Number of customers at the checkout counter
5. Number of misprints in a book
6. Number of photons emitted
Approximation to Binomial Distribution
Poisson distribution is also used for approximating a binomial distribution with
parameters (n,p) under the condition that n is large compared to k and p is small
enough that the product np ¼ l is of moderate magnitude. As a rule of thumb, l may
4.7 Poisson Distribution
51
be a single digit. We shall now derive this approximation:
n k nk
n!
pk (1 p)nk
b(k;n,p) ¼
pq
¼
k!(n k)!
k
n(n 1) (n k þ 1) l k
l nk
¼
1
k!
n
n
¼
n(n 1) (n k þ 1) lk (1 l=n)n
nk
k! (1 l=n)k
(4:7:4)
As n becomes very large and for moderate l, the quantity (1 l=n)n can be approximated
by e 2l and the quantity (1 l=n)k can be approximated as 1. Finally, as k is much smaller
than n, the quantity ½n(n 1) (n k þ 1)=nk can be approximated by 1. Thus the binomial distribution b(k;n,p) can be approximated as
b(k; n,p)
lk l
e ,
k!
k0
(4:7:5)
Example 4.7.3 A computer system consists of 1000 microchips, and the probability of
failure of a microchip is 0.005. We have to find the probability of k microchips failing
using both the binomial distribution and its Poisson approximation for k ¼ 0,1, . . . ,15.
We will also find the percentage error between the two.
Using binomial distribution, k chips failing in 1000 trials given by b(k;
1000,0.005) ¼ b(k) is
b(k; 1000,0:005) ¼ b(k) ¼
1000!
0:005k 0:9951000k
k!(1000 k)!
The value b(k) can be solved using the recursive formula developed in Eq. (4.2.6), and the
steps are
1000 k 0:005
b(k)
k þ 1 0:995
1000 k 1
¼
b(k)
k þ 1 188
b(k þ 1) ¼
Using the Poisson approximation we have np ¼ l ¼ 5, and substituting in the Poisson distribution, we have
k
5 5
p(k ; l) ¼ p(k ; 5) ¼ p(k) ¼ e
k!
Both b(k) and p(k) are shown for k ¼ 0,1, . . . , 15 in Table 4.7.1 along with the percentage
error.
Table 4.7.1 is also graphed in Fig. 4.7.2 for the binomial distribution and the Poisson
approximation. The figure shows that the approximation is very close. The error in the
approximation is also graphed in Fig. 4.7.3.
From the table and the graphs we see that the approximation becomes worse as
k is increased and the approximation is excellent in the vicinity of k ¼ 5 ¼ l, about
0.05 –0.25%.
52
Discrete Distributions
TABLE 4.7.1
Comparison of binomial with Poisson approximation
k
Binomial
Poisson
%Error
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0.006654
0.033437
0.083929
0.140303
0.175731
0.175908
0.14659
0.104602
0.065245
0.036138
0.017996
0.008139
0.003371
0.001287
0.000456
0.000151
0.006738
0.03369
0.084224
0.140374
0.175467
0.175467
0.146223
0.104445
0.065278
0.036266
0.018133
0.008242
0.003434
0.001321
0.000472
0.000157
21.26208
20.75577
20.352343
20.050683
0.150021
0.250272
0.250272
0.14992
20.051188
20.353762
20.758823
21.267706
21.88207
22.603907
23.435549
24.379687
FIGURE 4.7.2
FIGURE 4.7.3
4.7 Poisson Distribution
53
Example 4.7.4 (See Ref. 47) We will revisit the two or more birthdays matching
problem posed in Example 3.2.4, and solve it using the Poisson approximation and
compare the results. In addition, we will use the Poisson approximation to find probability
of matching three or more birthdays that would have been very difficult using the exact
method.
Given the total numbers of persons n 365, we will find, using the Poisson approximation, the probability of two or more birthdays matching. The probability of two birth1
days matching is 365
, and this we will set equal to p. The number
n of ways we can obtain a
sample of two without replacement and without ordering
is
2 , each of which has the
1
of matching. Hence the 2n ways constitute Bernoulli trials and
same probability of 365
1
is very small and
since p ¼ 365
1
n(n 1)
n
¼
¼l
np ¼
2 365
730
is less than 10, all the conditions for approximation by a Poisson distribution are satisfied.
Hence
P{2 or more people share the same birthday}
¼ 1 P{no birthdays match}
n(n 1) l0
¼ 1 p(0; l) ¼ 1 exp 730
0!
and [1 2 p(0 ; l)] is shown in Table 4.7.2 for the same values of n as given in Table 3.2.3
with the corresponding percentage error shown in Table 4.7.3.
The percent error curve between the true value and the Poisson approximation for
values of n ¼ 1, . . . ,100 is shown in Fig. 4.7.4.
An examination of the error curve reveals that the Poisson approximation underestimates the true values, and for values of n higher than 82, the error is almost zero. Even then
the maximum error is only about 1.46% occurring at n ¼ 26.
As mentioned before, this approximation can be extended to the case
of three or more
matches. Approaching, as before, the number of 3-combinations is n3 ¼ n(n1)(n2)
, and
6
TABLE 4.7.2
Poisson approximation for two matches
#
Probability
#
Probability
#
Probability
20
21
22
23
24
25
26
0.0405805
0.437488
0.468938
0.500002
0.530536
0.560412
0.589513
30
31
32
33
34
35
36
0.69632
0.720282
0.743058
0.764625
0.784972
0.804097
0.82201
44
45
46
47
48
49
50
0.925113
0.933618
0.941318
0.948266
0.954517
0.960121
0.965131
54
Discrete Distributions
TABLE 4.7.3
Error in Poisson approximation
#
%Error
#
%Error
#
%Error
20
21
22
23
24
25
26
1.369162
1.397497
1.420488
1.438108
1.450358
1.457273
1.458918
30
31
32
33
34
35
36
1.415262
1.392672
1.365868
1.335125
1.300742
1.263039
1.222352
44
45
46
47
48
49
50
0.833107
0.781948
0.731364
0.681649
0.633067
0.58586
0.540238
FIGURE 4.7.4
the probability p of three matches is
1 2
365 .
Substituting in the Poisson formula we obtain
P{3 or more share the same birthday}
¼ 1 P{no 3 birthdays match}
n(n 1)(n 2) l0
¼ 1 p(0; l) ¼ 1 exp 6 3652
0!
n(n 1)(n 2)
¼ 1 exp 799350
This probability is graphed in Fig. 4.7.5 for values of n ¼ 1, . . . , 150 and tabulated in
Table 4.7.4 for values of n ¼ 80 – 86, 90– 96, and 104 –110.
Examination of Fig. 4.7.3 and Table 4.7.4 reveals that for an even chance for at least
three matches, there must be about 84 people and for a 70% chance there must be about
100 people compared to 23 and 30 for at least two matches. This approximation technique
can be extended to any number of matches from 2 to 365, whereas exact computation
becomes extremely cumbersome, if not impossible, for anything more than two matches.
4.8 Logarithmic Distribution (Benford’s Law)
55
FIGURE 4.7.5
TABLE 4.7.4
B 4.8
4.8.1
Poisson approximation for three matches
#
Probability
#
Probability
#
Probability
80
81
82
83
84
85
86
0.460278
0.472929
0.485593
0.498257
0.510911
0.523543
0.536141
90
91
92
93
94
95
96
0.58597
0.598231
0.610393
0.622444
0.634375
0.646176
0.657838
104
105
106
107
108
109
110
0.745102
0.755146
0.764978
0.774593
0.783987
0.793155
0.802095
LOGARITHMIC DISTRIBUTION
Finite Law (Benford’s Law)
First Significant Digit
Most physical data depend on units. Examples are rainfall in inches, land area in
acres, blood pressure in millimeters of mercury, stock prices in dollars, distances in
miles, and volume in cubic feet. If clusters of these quantities can be described by a probability distribution P(x), then such a probability distribution must be invariant under a
scale change, so that
P(k x) ¼ KP(x)
(4:8:1)
Ð
Ð
where k and K are scale factors. Since P(x)dx ¼ 1, we have P(k x)dx ¼ 1=k, and hence
K ¼ 1=k is the normalization factor. Thus
1
P(kx) ¼ P(x) or kP(kx) ¼ P(x)
k
Differentiating Eq. (4.8.2) with respect to k, we obtain
P(kx) þ k
d
d(kx)
P(kx)
¼0
d(kx)
dk
(4:8:2)
56
Discrete Distributions
Substituting k ¼ 1 in the equation above, we have
d
x P(x) ¼ P(x)
dx
The solution for Eq. (4.8.3) is
P(x) ¼
1
x
(4:8:4)
Equation (4.8.4) can be considered as a probability distribution only for finite values
of x since the integral for infinite values of x diverges. We shall assume that the data are
defined to some arbitrary base B rather than base 10. If X1 is the random variable corresponding to the occurrence of the first digit d to the base B, then the probability of occurrence of PX1(d) is given by
Ð dþ1 1
dx
d
PX1 (d) ¼ dÐ B
¼ Ð x ; d ¼ 1,2, . . . , B 1
B1
1 P(x)dx
1 x dx
1
ln 1 þ
ln (d þ 1) ln (d)
d
¼
¼
ln (B) ln (1)
ln (B)
Ð dþ1
P(x)dx
(4:8:5)
We can show that indeed Eq. (4.8.5) is a probability function as follows:
B1 X
1
1
1
1
ln 1 þ
¼ ln 1 þ
þ ln 1 þ
þ ln 1 þ
d
1
2
B1
1
¼ ln
B
1
Y
1
1
234
B
1þ
¼ ln
¼ ln (B)
d
123 B 1
(4:8:6)
Thus PX1(d ) over all digits 1 to B21 sums to 1.
If the base B is 10, the generic ln becomes log, and the probability of the first significant digit d from 1 to 9 is,
1
PX1 (d) ¼ log 1 þ ; d ¼ 1,2, . . . , 9
(4:8:7)
d
This equation is called Benford’s law, first stated in 1938 by Benford [5]. It gives the probability of occurrence of the first significant digit in a cluster of distributions. Incidentally, 0
cannot be the first significant digit. It can be a significant digit only after the first one.
Example 4.8.1 In a cluster of numbers from different scale factors, we want to find the
occurrence of 1,2, . . . , 9 as the first significant digit. Using Eq. (4.8.7), we have
1
PX1 (1) ¼ log 1 þ
¼ 0:30103
1
The probability of 1 as the first significant digit is 30.1% and not a uniform distribution of
11.1% as one would expect! The probability of 2 is 17.6%. The probabilities of other significant digits are tabulated in Table 4.8.1 and plotted in Fig. 4.8.1. They show that 9 as the
first significant digit has the least probability of occurrence of 4.6%.
Example 4.8.2 As a realistic example we have taken the values of physical constants (73
in number), in Chapter 17 of Ref. 18 and tabulated the significant digits from 1 to 9 in
4.8 Logarithmic Distribution (Benford’s Law)
TABLE 4.8.1
57
Benford distribution
d
fD(d )
1
2
3
4
5
6
7
8
9
0.30103
0.176091
0.124939
0.09691
0.079181
0.066947
0.057992
0.051153
0.045757
FIGURE 4.8.1
Table 4.8.2. The physical constants have totally different units. The probabilities of the
significant digits are shown in Fig. 4.8.2 along with the graph of Benford’s law.
The table and the graphs show that even for a small sample of 73, the probabilities
follow the logarithmic distribution reasonably closely except for the number 3.
In conclusion, in a mixed population of different scale factors, the probabilities of the
digits 1 to 9 are logarithmically distributed and are not uniform—a surprising result!
TABLE 4.8.2
k
p(k)
1
2
3
4
5
6
7
8
9
0.342
0.205
0.055
0.096
0.096
0.055
0.027
0.041
0.082
58
Discrete Distributions
FIGURE 4.8.2
Second Significant Digit
We shall now compute the occurrence of 0,1, . . . , B 2 1 as the second significant
digit. Note that 0 can occur as a subsequent significant digit after the first. We will only
consider numbers of only first-order magnitude since the pattern will be the same for
any other order. With base B and the total number of digits being B 2 1, the occurrences
of 2 as the second significant digit is 1.2,2.2, . . . (B 2 1) . 2. Figure 4.8.3 shows in logarithmic scale the arrangement of numbers for base 10.
Thus, the probability of 2 as the second significant digit occurring after the first digit 1
is given by an equation similar to Eq. (4.8.5):
13
ln
ln (1:3) ln (1:2)
12
P{2 as second digit after first digit 1} ¼
¼
ln (B) ln (1)
ln (B)
(4:8:8)
If X2 is the random variable corresponding to the second digit, then the probability of 2
occurring after all the B 2 1 first digits is given by
13
23
(B 1)3
ln
þ ln
þ þ ln
12
22
(B 1)2
PX2 (2) ¼
ln (B)
(4:8:9)
Summing Eq. (4.8.10) over all the first digits d ¼ 0, . . . , B 2 1, we obtain the probability
FIGURE 4.8.3
4.8 Logarithmic Distribution (Benford’s Law)
of occurrence of the second significant digit, PX2(d ), as
B1 1 X
1
PX2 (d) ¼
ln 1 þ
; d ¼ 0,1, . . . ,B 1
ln (B) k¼1
kB þ d
!
B
1
Y
1
1
ln
1þ
¼
ln (B)
kB þ d
k¼1
It can be shown that
" B1
B
1
Y
X
ln
1þ
d¼0
k¼1
1
kB þ d
#
"
¼ ln
B
1 B
1
Y
Y
d¼0 k¼1
1
1þ
kB þ d
59
(4:8:10)
#
¼ ln (B)
(4:8:11)
and hence, PX2(d ) is a probability function.
We can extend the preceding analysis to obtain a general probability function for the
(n þ 1)st significant digit after the first one digit n ¼ 0,1, . . . , B21 as shown below.
The third digit after the first one can be obtained from Eq. (4.8.10). The occurrence of
3 as the third significant digit is 103, 113, . . . , 1(B 2 1)3; 203, 213, . . . , 2(B 2 1)3; . . . ;
(B 2 1)03,(B 2 1)13, . . . , (B 2 1)(B 2 1)3. Thus, there are B(B 2 1) possible ways in
which the digit 3 or any other digit d ¼ 0,1, . . . , B21 can occur as the third significant
number. An equation analogous to Eq. (4.8.9) can be written for the probability of 3 occurring as the third significant digit:
2
1 1 BX
1
PX3 (3) ¼
ln 1 þ
(4:8:12)
ln (B) k¼B
kB þ 3
Similar to Eq. (4.8.10), we can write the probability of any digit n occurring as the third
significant digit:
2
1 1 BX
1
ln 1 þ
PX3 (d) ¼
ln (B) k¼B
kB þ d
!
2
BY
1
1
1
ln
1þ
¼
;
ln (B)
kB þ d
k¼B
d ¼ 0, . . . , B 1
(4:8:13)
We can extend Eq. (4.8.13) to all (n þ 1) significant digits after the first with
n ¼ 1,2, . . . , B 2 1:
Bn1
1 X
1
ln 1 þ
PX(nþ1) (d) ¼
ln (B) k¼Bn1
kB þ d
n ¼ 1,2, . . . , B 1;
d ¼ 0, . . . , B 1
(4:8:14)
Example 4.8.3 Using Eq. (4.8.14) with the base B ¼ 10, we will find the occurrence of the
digits 0,1, . . . , 9 occurring as the first, second, . . . , tenth significant digit, noting that the
probability of occurrence of 0 as the first significant digit is 0. Equation (4.8.14) becomes
n
10
1
X
1
1
PX(nþ1) (d) ¼
ln 1 þ
ln (10) k¼10n1
k 10 þ d
n ¼ 0,1, . . . , 9;
d ¼ 0,1, . . . , 9
60
Discrete Distributions
TABLE 4.8.3
Benford distribution for 1 – 9 significant digits
d
PX1(d )
PX2(d )
PX3(d)
PX4(d )
PX5(d )
PX6(d )
PX7(d )
0
1
2
3
4
5
6
7
8
9
0
0.30103
0.176091
0.124939
0.09691
0.079181
0.066947
0.057992
0.051153
0.045757
0.119679
0.11389
0.108821
0.10433
0.100308
0.096677
0.093375
0.090352
0.08757
0.084997
0.101784
0.101376
0.100972
0.100573
0.100178
0.099788
0.099401
0.099019
0.098641
0.098267
0.100176
0.100137
0.100098
0.100059
0.100019
0.09998
0.099941
0.099902
0.099863
0.099824
0.100018
0.100014
0.10001
0.100006
0.100002
0.099998
0.099994
0.09999
0.099986
0.099982
0.100002
0.100001
0.100001
0.100001
0.1
0.1
0.099999
0.099999
0.099999
0.099998
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
0.1
FIGURE 4.8.4
The values of PX (nþ1)(d) have been tabulated in Table 4.8.3 for values of d ¼ 0, . . . , 9 and
n ¼ 1, . . . , 6.
It is interesting to note that the first significant digit is considerably different from
uniform distribution. As the level of significance increases, the distribution tends to be
more and more uniform and the significant digit is practically uniform. In fact, the third
significant digit is almost a uniform distribution as shown in Fig. 4.8.4.
Benford’s law is being used widely by the Internal Revenue Service for detecting
fraudulent tax returns.
4.8.2
Infinite Law
We will also present two versions of a logarithmic distribution where the number of outcomes are infinite:
Version 1. The first version is defined by
p(k,u) ¼
uk
;
k ln (1 u)
0 , u , 1, k 1
(4:8:15)
4.8 Logarithmic Distribution (Benford’s Law)
61
TABLE 4.8.4
k
P(u,k)
1
2
3
4
5
6
7
8
9
10
0.390865
0.175889
0.105534
0.071235
0.051289
0.038467
0.029675
0.023369
0.018695
0.015143
where u is the shape parameter and k are the infinite outcomes. Here u is between 0
and 1, and k is greater than or equal to 1. We will show that this is a probability
distribution by summing over all k:
1
X
p(k,u) ¼
k¼1
1
X
uk
1
uk ln (1 u)
¼1
¼
¼
ln (1 u)
k ln (1 u) ln (1 u) k¼1 k
k¼1
1
X
(4:8:16)
In Eq. (4.8.15), as u tends to zero, p(k,u) tends to 1 as shown below:
uk
kuk1 (1 u)
¼1
¼ lim
u!0 k ln (1 u)
u!0
k
lim p(k,u) ¼ lim
u!0
(4:8:17)
The probability mass function (pmf) given by Eq. (4.8.15) is tabulated for
k ¼ 1, . . . ,10 and u ¼ 0.9 in Table 4.8.4 and shown in Fig. 4.8.5. This logarithmic
distribution is used to model the number of items purchased by a customer in a
supermarket in a given amount of time.
FIGURE 4.8.5
62
Discrete Distributions
Version 2. The second version of this logarithmic distribution is obtained by substituting l ¼ (1 2 u) in Eq. (4.8.15) to obtain
p(k,l) ¼
(1 l)k
: 0 , l , 1, k 1
k ln (l)
(4:8:18)
Here, as l tends to 1, p(k,l) tends to 1 as shown below:
(1 l)k
klk1 l
¼1
¼ lim
u!0 k ln (l)
u!0
k
(4:8:19)
lim p(k,l) ¼ lim
u!0
B 4.9
SUMMARY OF DISCRETE DISTRIBUTIONS
Assumptions: Pfsuccessg ¼ p; Pffailureg ¼ q; p þ q ¼ 1.
Distribution Type
Definition
Bernoulli
8
< p;
b(p,q) ¼ q;
:
0;
Binomial
8 < n k nk
pq ;
k
b(k; n,p) ¼
:
0;
k¼1
k¼0
otherwise
Multinomial
p(k1 ,k2 , . . . , km ) ¼
m
X
ki ¼ n;
i¼1
m
X
k ¼ 0,1, . . . , n
otherwise
n
k1 ,k2 , . . . , km
pi ¼ 1
i¼1
qk1 ;
0;
k ¼ 1, 2, . . .
k0
Geometric
g(k,p) ¼
Negative binomial: Pascal
8
< n 1 k nk
pq ;
k1
nb( p) ¼
:
0;
Hypergeometric
pk11 pk22 pkmm
8 K
NK
>
>
> k
>
< n k
;
p(n,k) ¼
N
>
>
>
>
: n
0;
k ¼ k, k þ 1, . . .
otherwise
k ¼ 0,1, . . . , n
9
>
>
>
>
=
otherwise
>
>
>
>
;
(Continued)
4.9
Summary of Discrete Distributions
Distribution Type
Poisson
Logarithmic – Benford
63
Definition
8
<
9
k
=
l
l
p(k,l) ¼ e k! ; k ¼ 0,1, . . .
:
;
0;
otherwise
1
1
p(d) ¼ log 1 þ ; d ¼ 1, . . . , 10N 1
N
d
Logarithmic 1—semiinfinite
p(k,u) ¼
uk
;
k ln (1 u)
Logarithmic 2—semiinfinite
p(k,u) ¼
(1 u)k
;
k ln (u)
0 , u , 1, k 1
0 , u , 1, k 1
CHAPTER
5
Random Variables
B 5.1
DEFINITION OF RANDOM VARIABLES
What is a random variable? An outlandish definition would be that it is neither random nor
a variable! The real definition of a random variable is that it is a function X that assigns a
rule of correspondence for every point j in the sample space S called the domain, a unique
value X(j) on the real line R called the range. Let F be the field associated with the sample
space and FX be the field associated with the real line. The random variable X induces a
probability measure PX in R and hence X is a mapping of the probability space {S, F, P}
to the probability space {R, FX , PX } as shown below:
X: {S,F,P} ! {R, FX , PX }
(5:1:1)
Consider an event D , S [ F. The random variable X maps every point ji in the event D
to points xi in the event Ix , called the image of D under X, where Ix , R [ FX . X can be a
random variable only if the inverse image X 1 (Ix ) belongs to the field F of subsets of S,
and hence it must be an event. The mapping and the inverse mapping are shown in
Fig. 5.1.1. With this restriction we should be able to find the induced probability
measure PX in terms of the probability measure P as follows:
PX {Ix } ¼ P{X 1 (Ix )} ¼ P{D} ¼ P{j : X(j) [ Ix }
(5:1:2)
Since Ix belongs to the field FX , on the real line it may consist of sets of the form
{1 , h x}, {x , h , 1}, {x1 , h x2 }, {h ¼ x}. Out of these sets we define Ix by
Ix ¼ {1 , h x}
(5:1:3)
All the other sets {x , h , 1}, {x1 , h x2 }, {h ¼ x} can be expressed in terms of the
defined Ix as follows:
I x ¼ {x , h 1}
Ix2 Ix1 ¼ {1 , h x2 } {1 , h x1 } ¼ {x1 , h x2 }
lim {IxþDx Ix } ¼ {h ¼ x}
Dx!0þ
With the definition of Ix given by Eq. (5.1.3), we can write Eq. (5.1.2) as follows:
PX {Ix } ¼ P{j : X(j) [ Ix } ¼ P{j : X(j) x}
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
64
(5:1:4)
5.1 Definition of Random Variables
65
FIGURE 5.1.1
We now define the quantity P{j : X(j) x} as the cumulative distribution function (cdf)
FX (x), and rewrite Eq. (5.1.4) in abbreviated notation as
P{j : X(j) [ Ix } ¼ P{X x} ¼ FX (x)
(5:1:5)
Equation (5.1.5) converts from a cumbersome set function P{X x} into a convenient point
function FX(x). The value x scans the entire real line, that is, 21 , x , 1, and FX(x) must be
defined for all x on the real line.
Note that {X x} in Eq. (5.1.5) is meaningless unless it is defined carefully, because
X is a function and a function without an argument cannot be less than a number!
For example, we cannot have an exponential less than 2, unless we specify the argument
of the exponential. The abbreviated notation in Eq. (5.1.5) is to be defined as the probability of the set of all points j [ S such that the number X(j) is less than or equal to
the number x. The random variable X should not confused with the value x that the
random variable takes. The random variable X can take any value, for example,
FX (y) ¼ P{X y}.
Having defined the cumulative distribution function (cdf), we now define the probability density function (pdf) as the derivative of the cdf FX(x) with respect to x:
P(X x þ Dx) P(X x)
Dx!1
Dx
fX ðxÞ ¼ lim
FX (x þ Dx) FX (x)
d
¼ FX (x)
Dx!1
Dx
dx
¼ lim
(5:1:6)
The cdf FX(x) can be given in terms of the pdf fX(x) as follows:
ðx
FX (x) ¼
fX (j)dj
(5:1:7)
1
Whereas the distribution function FX(x) is a probability measure, the density function fX(x)
is not a probability measure unless it is multiplied by the infinitesimal Dx to yield
fX (x)Dx ¼ P(x , X x þ Dx)
(5:1:8)
66
Random Variables
B 5.2 DETERMINATION OF DISTRIBUTION AND
DENSITY FUNCTIONS
Most of the difficulty in determining distribution functions for random variables can be
avoided if the following step by step procedure is followed.
Step 1. Since the random variable is a mapping from the sample space S to the real line,
the mapping diagram is drawn connecting every point j in the sample space to the
points on the real line R.
Step 2. The regions for x on the entire real line from the mapped values are determined.
The regions may correspond to the different subsets of S that map to the real line R. The
set of points Ix will be equal to the right closed set ð1, x or, Ix ¼ {1 , h x}.
Note that whereas Ix is a right closed subset, the region of definition of x is a subset
of the real line R.
Step 3. On the sample space S, the probability of the set of all points that map into Ix for
every region of x in step 2 is determined.
Step 4. The cdf FX (x) ¼ P{X x} is graphed for all the regions of x in step 2 on the entire
real line.
Step 5. The derivative of FX (x) determines the density function fX (x).
Example 5.2.1 A die is tossed, and the random variable X is defined by the amount won
(þ) or lost (2) on the face of the die as shown in Table 5.2.1. We have to find the cdf FX(x)
and the pdf fX(x).
We will follow the steps to find the probability functions.
Step 1. The mapping diagram (Fig. 5.2.1) is drawn with positive numbers indicating win
and negative numbers indicating loss.
Step 2. From the mapping diagram the regions of x are (a) x 29, (b) 29 , x 24,
(c) 24 , x 5, (d) 5 , x 8, and (e) x . 8. The corresponding set Ix on the real
line for all the 5 regions is Ix ¼ {1 , h x}.
Step 3. We will find the all the points in the sample space that map into Ix for every region
of x:
(a) x 29: Since no points in S map into Ix (Fig. 5.2.1a), we have FX(x) ¼ 0.
(b) 29 , x 24: In this region (Fig. 5.2.1b), only one point f3g maps into Ix. Since
P{3} ¼ 16 , FX (x) ¼ 16.
(c) 24 , x 5: Here the points f3,2,6g from S map into Ix (Fig. 5.2.1c). Hence
FX (x) ¼ 16 þ 16 þ 16 ¼ 12.
(d) 5 , x 8: In this region (Fig. 5.2.1d) the points f3,2,6,5g in S map in to Ix.
Hence FX (x) ¼ 12 þ 16 ¼ 23
TABLE 5.2.1
Pips on the Die
1,4
2,6
3
5
Win or Loss $
þ8
24
29
þ5
5.2 Determination of Distribution and Density Functions
67
FIGURE 5.2.1
(e) x . 8: In this region (Fig. 5.2.1e) all six points, that is, the entire sample space S,
map into Ix. Hence FX (x) ¼ 1.
In terms of unit step functions we can write the cdf FX(x) as follows:
1
1
1
1
FX (x) ¼ u(x þ 9) þ u(x þ 4) þ u(x 5) þ u(x 8)
6
3
6
3
Step 4. The cdf FX(x) can be graphed as shown in Fig. 5.2.2.
68
Random Variables
FIGURE 5.2.2
Step 5. We can now find the density function fX(x) by differentiating the distribution
function FX(x), bearing in mind that the differentiation of the unit step is the Dirac delta
function. Performing the indicated differentiation, we obtain
1
1
1
1
fX (x) ¼ d(x þ 9) þ d(x þ 4) þ d(x 5) þ d(x 8)
6
3
6
3
Note that the sum of the strengths of the impulse function adds to 1. The pdf fX(x) is shown
in Fig. 5.2.3.
From Fig. 5.2.2 we can write the following probabilities:
(a) P{X ¼ 4} ¼ FX (4þ) FX (4) ¼ 12 16 ¼ 13
1
1
(b) P{X 4} ¼ FX (4þ) FX (1) ¼ 2 0 ¼ 2
(c) P{5 , X 4} ¼ FX ( 4þ) FX (5) ¼ 12 16 ¼ 13
1
1
1
(d) P{4 , X 4} ¼ FX (4þ) FX (4) ¼ 2 6 ¼ 3
(e) P{5 X , 8} ¼ FX (8) FX (5þ)(Note closure on the left)
¼ 12=3 2=3 ¼ 0
The example above shows that the mapping is from a discrete sample space S to discrete
points on the real line R.
FIGURE 5.2.3
5.2 Determination of Distribution and Density Functions
69
TABLE 5.2.2
Time Interval
Win or Loss ($)
6:00– 6:15
6:15– 6:45
6:45– 7:15
7:15– 8:00
þ2
24
24
þ6
Example 5.2.2 A telephone call occurs at random between 6:00 p.m. and 8:00 p.m.
Depending on the interval of arrival, the win/loss table is shown in Table 5.2.2. We
have to find the cdf FX(x) and the pdf fX(x).
Since the phone call is uniformly distributed in 120 min, the probability of a phone call
in the region (0,x] is given by x/120. We follow the steps as outlined in Example 5.2.1:
Step 1. The mapping diagram is shown in Fig. 5.2.4. Note that the mapping is from the
real line R to the real line R.
Step 2. From the mapping diagram the regions of x are (a) x 24, (b) 24 , x 2,
(c) 2 , x 6, (d) x . 6.
Step 3.
(a) x 24: No points map FX(x) ¼ 0.
60
¼ 12.
(b) 24 , x 2: The interval that maps is 30 min; hence FX (x) ¼ 120
75
(c) 2 , x 6: The interval that maps is 45 min; hence FX (x) ¼ 120 ¼ 58.
(d) x . 6: The entire sample space of 120 min maps; hence FX (x) ¼ 1.
In terms of unit step functions, the cdf FX(x) is
1
1
3
FX (x) ¼ u(x þ 4) þ u(x 2) þ u(x 6)
2
8
8
Step 4. The cdf FX(x) is graphed in Fig. 5.2.5
Step 5. The pdf fX(x) is obtained by differentiating FX(x), and the result is
1
1
3
fX (x) ¼ d(x þ 4) þ d(x 2) þ d(x 6)
2
8
8
fX(x) is graphed in Fig. 5.2.6.
FIGURE 5.2.4
70
Random Variables
FIGURE 5.2.5
FIGURE 5.2.6
In the preceding example, the sample space is continuous whereas the range space is discrete. Since the random variable takes discrete values, we have a discrete random variable.
Example 5.2.3 A telephone call occurs at random between 6:00 p.m. and 8:00 p.m. We
want to find the cdf FX(x) and the pdf fX(x). Here we have a continuous mapping from the
real line to the real line and X is a continuous random variable.
Step 1. See the mapping diagram in Fig. 5.2.7.
Step 2. From the mapping diagram the regions of x are (a) x 0, (b) 0 , x 120,
(c) x . 120, and Ix ¼ {1 , h x).
Step 3
(a) x 0: No points in S map into Ix. Hence FX(x) ¼ 0.
(b) 0 , x 120: The call is uniformly distributed in the interval (0,120], and since
(0,x] is an interval mapping into Ix, we have FX(x) ¼ x/120.
(c) x . 120: Here the entire sample space maps into Ix, and hence FX (x) ¼ 1.
Step 4. The cdf FX (x) is graphed in Fig. 5.2.8.
Step 5. The density function fX(x) is obtained by taking the derivative of FX (x), or
1
fX (x) ¼ 120
, 0 x 120. This is sketched in Fig. 5.2.9.
We found the cdf and pdf for discrete random variables in Examples 5.2.1 and 5.2.2
and for continuous random variables in Example 5.2.3. We will now give an example of a
mixed random variable that has both continuous and discrete components.
5.2 Determination of Distribution and Density Functions
71
FIGURE 5.2.7
FIGURE 5.2.8
FIGURE 5.2.9
Example 5.2.4 A telephone call occurs at random between 6:00 p.m. and 8:00 p.m.
However, between 6:20 and to 6:40 p.m. calls accumulate and are released exactly at
6:40 p.m. with a similar situation between 7:00 and 7:20 p.m. and are released at
exactly 7:20 p.m. We want to find the cdf FX(x) and the pdf fX(x).
Step 1. The mapping diagram shown in Fig. 5.2.10 is similar to Fig. 5.2.7 except that there
are deadspots between 6:20 and 6:40 p.m. and 7:00 and 7:20 p.m.
Step 2. From the mapping diagram the regions of x are (a) x 0, (b) 0 , x 20, (c)
20 , x , 40, (d) x ¼ 40, (e) 40 , x 60, (f) 60 , x , 80, (g) x ¼ 80, (h)
80 , x 120, (i) x . 120.
Step 3.
(a) x 0: No points in S map into Ix; hence FX(x) ¼ 0.
72
Random Variables
FIGURE 5.2.10
(b) 0 , x 20: The call is uniformly distributed in the interval (0,120]; hence FX
(x) ¼ x/120.
20
¼ 16.
(c) 20 , x , 40: There is a deadspot in this region; hence FX (x) ¼ FX (20) ¼ 120
1
(d) x ¼ 40: There is jump at this point equal to P{X ¼ 40} ¼ 6; hence
FX (40) ¼ 16 þ 16 ¼ 13.
(e) 40 , x 60: Here also (0,x] is an interval in Ix and we have FX (x) ¼ x=120 and
FX (40) , 13.
(f) 60 , x , 80: Here again we have a deadspot; and hence FX (x) ¼ FX (60) ¼
60
1
120 ¼ 2.
(g) x ¼ 80: There is again a jump at this point equal to P{X ¼ 80} ¼ 16; hence
FX (80) ¼ 12 þ 16 ¼ 23.
(h) 80 , x 120: Here (0,x] is an interval in Ix and we have FX (x) ¼ x=120 and
FX (80) ¼ 23.
(i) x . 120: The entire sample space maps into Ix; hence FX(x) ¼ 1
Step 4. The cdf FX(x) is graphed in Fig. 5.2.11.
FIGURE 5.2.11
5.3 Properties of Distribution and Density Functions
73
FIGURE 5.2.12
Step 5. The pdf fX(x) is the derivative of FX(x) and is graphed in Fig. 5.2.12.
From the graph of the cdf FX(x) of Fig. 5.2.11, we can obtain the following
probabilities:
1. P{X ¼ 40} ¼ P{X ¼ 80} ¼ 16.
2. P{X , 80} ¼ 12.
3. P{X 80} ¼ 23 :
4. P{40 X 80} ¼ 13 :
5. P{X ¼ 100} ¼ 0:
5
5
6. P{50 , X 100} ¼ 56 12
¼ 12
:
B 5.3 PROPERTIES OF DISTRIBUTION AND DENSITY
FUNCTIONS
Cumulative Distribution Function FX(x)
1. From the three examples given in Section 5.2 the cdf of a discrete random variable
is a staircase function with jumps.
2. FX(x) 0, that is, nonnegative.
3. FX(21) ¼ 0, FX(1) ¼ 1.
4. If x1 , x2, then FX(x1) FX(x2), that is, FX(x) is a nondecreasing function (monotone increasing). For a continuous random variable, if x1 , x2, then FX(x1) ,
FX(x2), that is, FX(x) is an increasing function (strict monotone increasing).
5. FX(x) is right-continuous:
FX (x) ¼ lim FX (x þ 1)
1!0
for 1 . 0
From this definition, the value that FX (x) takes at the point of a jump, is to the right
of the jump, and if there is an event A in the interval {x1 , X x2 }, then the interval is closed on the righthand side so that we can write P{x1 , X x2 } ¼
P{X x2 } P{X x1 } ¼ FX (x2 ) FX (x1 ). This ensures that the point x1 is not
counted twice. Figure 5.3.1 will clarify this point.
ðx
FX (x) ¼
fX (j)dj ¼ P{X x}
(5:3:1)
6.
1
74
Random Variables
FIGURE 5.3.1
7. At the point x ¼ a of the jump
ð aþ
FX (x) ¼
fX (j)dj ¼ FX (aþ) FX (a) ¼ P{X ¼ a}
(5:3:2)
a
Probability Density Function fX(x)
1. fX (x) 0, nonnegative.
2. fX (x) consists of the sum of Dirac delta functions for a discrete random variable.
3. fX (x) is not a probability measure, but fX (x)Dx is a probability measure as shown:
fX (x)Dx P{x , X x þ Dx}
4.
(5:3:3)
P{x , X x þ Dx}
d
¼ FX (x)
Dx
dx
ð5:3:4Þ
fX (j)dj FX (b) FX (a) ¼ P{a , X b}
(5:3:5)
fX (x) lim
Dx!0
ðb
5.
a
Probability Mass Functions (pmf’s)
We have already come across probability mass functions in connection with discrete
random variables. We will now formalize this concept. The probability of a random
variable equal to a number is called the probability mass function (pmf). In Eq. (5.3.2),
P{X ¼ a} is the pmf.
1. P{X ¼ a} is the probability of the jump at the point x ¼ a in the cdf FX (x). Or,
P{X ¼ a} ¼ FX (a) FX (a).
2. The pmf P{X ¼ a} is nonzero for discrete random variables at x ¼ a, where they
jump.
3. P{X ¼ a} ¼ 0 for continuous random variables, since at any point of continuity a,
FX (a) FX (a) ¼ 0.
4. The density function fX(x) for discrete random variables involves summation
of the pmf’s with Dirac delta functions multiplying the pmf at every point in
the range. For example, the pdf of the binomial distribution given by the
5.4 Distribution Functions from Density Functions
n
pk qnk is
n X
n k nk
p q d(x k)
fX (x) ¼
k
k¼0
probability mass function
75
k
(5:3:6)
The pdf of the Poisson distribution given by the pmf el (lk =k!) is
fX (x) ¼
1
X
el
k¼0
lk
d(x k)
k!
(5:3:7)
B 5.4 DISTRIBUTION FUNCTIONS FROM DENSITY
FUNCTIONS
Given any density function (pdf), the distribution function (cdf) can be found by integrating the pdf over all x on the real line as given in Eq. (5.3.1). Care must be taken in
integrating over various regions along the x axis.
Example 5.4.1
expressed as
We will start with a simple example. The density function fX (x) is
fX (x) ¼
8
k
>
>
<4,
0,x2
>
>
:1,
2
2,x3
and is shown in Fig. 5.4.1. We have to first find k and then the distribution function FX (x)
for all x.
Integrating fX (x) over the range of definition of x, we have
ð3
ð3
ð2
k
1
k 1
dx þ
dx ¼ þ ¼ 1, [ k ¼ 1
fX (x) ¼
2 2
0
04
22
1
1
fX (x) ¼ u(x) þ u(x 2)
4
4
FIGURE 5.4.1
76
Random Variables
FIGURE 5.4.2
We note that there are four natural ranges for x given by (1) x 0, (2) 0 , x 2,
(3) 2 , x 3, (4) x . 0. We will evaluate FX(x) in all these ranges:
1. x 0: Since fX (x) is zero, FX (x) ¼ 0.
2. 0 , x 2: Using Eq. (5.3.5), we can write
ðx
FX (x) ¼
1
x
dj ¼
4
04
3. 2 , x 3: Again using Eq. (5.3.5) and taking care to carry over FX (2), we have
ð2
FX (x) ¼
1
dj þ
04
ðx
1
1 1
1
dj ¼ þ (x 2) ¼ (x 1)
2
2
2
2
2
4. x . 3: Since fX(x) is zero in this range, FX(x) ¼ FX(3) ¼ 1.
FX(x) is graphed in Fig. 5.4.2.
FIGURE 5.4.3
5.4 Distribution Functions from Density Functions
77
Example 5.4.2 This is a more involved example that characterizes a mixed random variable. The density function is given by
fX (x) ¼ K{ex ½u(x 1) u(x 5) þ ex ½u(x 1) u(x 5)g
1
þ ½d(x þ 2) þ d(x) þ d(x 2)
8
(see Fig. 5.4.3). We will find K and graph the distribution function FX(x) for all x.
We integrate fX(x) over the range (25,5] to evaluate the constant K. Thus
ð 5
K
ex dx þ
5
1 1
ex dx þ þ ¼ 1
8 8
1
ð5
2K½e1 e5 þ
3
¼1
8
[K¼
5
e5 )
16(e1
The ranges for x for integrating fX(x) are given by (1) x 25, (2) 25 , x 22,
(3) 22 , x 21, (4) 21 , x 0, (5) 0 , x 1, (6) 1 , x 2, (7) 2 , x 5, (8)
x . 5. We shall now evaluate FX(x) in each of these ranges:
1. x 25: Since fX(x) ¼ 0, we have FX(x) ¼ 0
2. 5 , x 2:
3. 2 , x 1:
4. 1 , x 0:
5. 0 , x 1:
FX (x) ¼
FX (x) ¼
FX (x) ¼
FX (x) ¼
5
16(e1 e5 Þ
ðx
ej dj ¼
5
5(ex e5 )
1
þ
1
5
16(e e Þ 8
5(e1 e5 ) 1
7
þ ¼
1
5
16(e e Þ 8 16
7 1
9
þ ¼
16 8 16
FIGURE 5.4.4
5(ex e5 )
16(e1 e5 Þ
78
Random Variables
6. 1 , x 2:
7. 2 , x 5:
8. x . 5:
FX (x) ¼
FX (x) ¼
FX (x) ¼
9
þ
16
ðx
ej dj ¼
1
9
5(e1 ex )
þ
16 16(e1 e5 )
9
5(e1 ex ) 1
þ
þ
16 16(e1 e5 ) 8
9
5(e1 e5 ) 1
þ
þ ¼1
16 16(e1 e5 ) 8
The graph of FX (x) is shown in Fig. 5.4.4.
In the next chapter we will discuss the various continuous distributions.
CHAPTER
6
Continuous Random Variables
and Basic Distributions
B 6.1
INTRODUCTION
In the last chapter we defined a continuous random variable as the mapping from a continuous sample space S to continuous points on the real line R. We also discussed some
of the properties of continuous pdf fX(x) and the continuous cdf, FX(x). In this chapter
we will discuss three of the basic continuous distributions: uniform, exponential, and
Gaussian. Other distributions will be discussed in the next chapter.
B 6.2
UNIFORM DISTRIBUTION
The basic continuous distributions are uniform, exponential, and Gaussian. We have
already seen (in Example 5.2.3) the distribution of a telephone call occurring at random
in a given interval. The pdf fX(x) of a random variable uniformly distributed in the interval
(a, b], (notation X U(a, b]), is given by
8
0
xa
>
>
>
<
1
(6:2:1)
fX (x) ¼
a,xb
>
b
a
>
>
:
0
x.b
and the corresponding distribution function FX(x) is
8
0
xa
>
>
>
<xa
a,xb
FX (x) ¼
ba
>
>
>
:
1
x.b
(6:2:2)
Such distributions occur in transmission of sinusoids where the phase angle will be uniformly distributed in (0,2p). The density function is shown in Fig. 6.2.1, and the
distribution function is shown in Fig. 6.2.2.
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
79
80
Continuous Random Variables and Basic Distributions
FIGURE 6.2.1
FIGURE 6.2.2
Example 6.2.1 A sinusoidal signal with random phase angle Q, uniformly distributed
between (0,2p] is sent along a communications channel. We want to find the probability
that the phase angle is between (p/4, 3p/4].
Since it is uniformly distributed, the pdf is 1/2p in (0,2p], and hence the Pfp/4, 3p/
4g is given by
p
3p
1 3p p 1
P
,u
¼
¼
4
4
2p 4
4
B 6.3
EXPONENTIAL DISTRIBUTION
Exponential distributions find wide applicability in queuing theory, reliability theory, and
communication theory. It is the continuous analog of the geometric distribution discussed
in Chapter 4. The pdf of an exponential distribution is given by
0
x0
fX (x) ¼
(6:3:1)
lelx x . 0
where the parameter l is the average rate at which events occur. The corresponding cdf
FX(x) is given by
0
x0
FX (x) ¼
(6:3:2)
1 elx x . 0
The pdf is shown in Fig. 6.3.1, and the cdf is shown in Fig. 6.3.2 for values of
l ¼ 0.5,1,2,3,4.
The exponential distribution belongs to a class of distributions possessing memoryless
(Markov) properties that we will presently show. A random variable X is called memoryless if it satisfies the conditional probability
P{X . x þ x0 j X . x0 } ¼ P{X . x}
(6:3:3)
6.3 Exponential Distribution
81
FIGURE 6.3.1
FIGURE 6.3.2
The condition given by Eq. (6.3.3) is equivalent to
P{X . x þ x0 , X . x0 }
¼ P{X . x}
P{X . x0 }
(6:3:4)
or
P{X . x þ x0 } ¼ P{X . x}P{X . x0 }
To demonstrate the memoryless property, let X be a random variable representing the
time to failure of a computer with cdf given by Eq. (6.3.2). We are given that the computer
has lasted x0 hours. We want to find the conditional probability that the computer will fail
at time x beyond x0 given that it has lasted for x0 hours. We proceed as follows:
P{X x þ x0 j X . xo } ¼ FX (x þ x0 j X . x0 )
¼
P{X x þ x0 , X . x0 } P{x0 , X x þ x0 }
¼
P{X . x0 }
1 FX (x0 )
¼
FX (x þ x0 ) FX (x0 ) 1 el(xþx0 ) 1 þ elx0
¼
1 FX (x0 )
1 1 þ elx0
¼
elx0 (1 elx )
¼ 1 elx , x . x0
elx0
(6:3:5)
82
Continuous Random Variables and Basic Distributions
The conditional distribution FX(x þ x0 j X . x0) is the same as FX(x)! The computer
“forgets” that it has been operating for x0 hours and has no memory of previous hours
of operation. Hence the exponential distribution is called a memoryless distribution.
Poisson Arrival Process
The number k of events occurring in an interval (0,t] is Poisson-distributed with
p(k,l) ¼ e 2lt[(lt)k/k!], and the waiting times for the occurrence from one event to the
next are independent and are exponentially distributed as e 2lt.
Example 6.3.1 A computer company leases its computers. The lifetime of one of the
1
computers is exponentially distributed with l ¼ 50;000
h. If a person leases the computer
for 10,000 h, we have to find the probability that the computer company will have to
replace the computer before the lease runs out.
By the memoryless property derived in Eq. (6.3.5), it is immaterial how long it has
been leased before, and hence we have
P{computer fails in t 10,000} ¼ 1 e10000l ¼ 1 e(1=5) ¼ 0:1813
and the probability that the computer will not be replaced before the lease runs out is
P{lifetime . 10,000} ¼ 1 0:1813 ¼ 0:8187
However, if the failure time FX(x) is not exponentially distributed, then the probability
of replacement has to be computed, taking into account the prior use of x0 hours of the
computer. This is given by the conditional probability
P{lifetime . x0 þ 10,000 j lifetime . 10000} ¼
1 FX (x0 þ 1000)
1 FX (x0 )
which means that we have to know x0, the number of hours the computer was operational
before the lease.
Hazard Rate
Given that an item has not failed upto time t, the conditional probability that it will fail
in the next small interval of time Dt is determined as follows
P{t , X t þ Dt j X . t} ¼
¼
P{t , X t þ Dt, X . t}
P{X . t}
P{t , X t þ Dt}
fX (t)Dt
¼
¼ b(t)Dt
P{X . t}
1 FX (t)
(6:3:6)
where FX(t) ¼ PfX tg represents the failure distribution and the quantity b(t) ¼ fx(t)/
[1 2 FX(t)] is called the instantaneous hazard rate. It is not a probability function as
determined by the following analysis:
ðt
ðt
ðt
fX (j)
dfX (x)
dj ¼ ¼ ln½1 FX (t)
b(j)dj ¼
1
F
1
FX (j)
(j)
X
0
0
0
6.3 Exponential Distribution
83
Ð
Since t0b(j)dj ! 1 as t ! 1, we conclude that b(t) is not a probability function.
However, solving for FX(t), we obtain
ðt
ln½1 FX (t) ¼ b(j)dj
(6:3:7)
0
or
ðt
1 FX (t) ¼ exp b(j)dj ¼ 0 as t ! 1
0
From Eq. (6.3.7) we can obtain the cdf FX(t), given by
ðt
FX (t) ¼ 1 exp b(j)dj
(6:3:8)
0
Thus, b(t) uniquely determines the failure probability distribution function FX(t). While
b(t) is not a probability density function, it is to be noted that the function b(x), given by
8
< fX (x) , x . t
(6:3:9)
b(x) ¼ 1 FX (t)
:
0,
xt
is the conditional failure density conditioned on x . t and b(t) is the value of b(x) at
x ¼ t.
Example 6.3.2 Physicians are of the opinion that the death rate of a person E exercising
regularly is half that of a person S who has sedentary habits. We want to compare the probabilities of survival of E and S given that the death rate of an exercising person E is bE(t)
and the rate for a sedentary person S is bS(t) with bE(t) ¼ 12bS(t).
Since the rates are conditional failure rates, we will assume that both persons have
lived upto the age T1, and we want to compare the probabilities of their reaching the
age T2 with T2 . T1. Thus we need to find
P{Xi . T2 j Xi . T1 } ¼
P{Xi . T2 } 1 FXi (T2 )
¼
P{Xi . T1 } 1 FXi (T1 )
where Xi ¼ E or S. Substituting from Eq. (6.3.8), we have
ÐT
ð T2
1 FXi (T2 ) exp 0 2 bXi (j)dj
¼
Ð
¼ exp bXi (j)dj ,
1 FXi (T1 ) exp T1 bX (j)dj
T1
0
i ¼ E or S
i
Simplifying, we obtain
ð
1 T2
P{E . T2 j E . T1 } ¼ exp bS (j)dj
2 T1
ð T2
P{S . T2 j S . T1 } ¼ exp bS (j)dj
T1
Thus, the probability of an exercising person who has lived upto T1 years living to T2 years
is the square root of the probability of the sedentary person. If the death rate of a sedentary
person, bS(t) ¼ 0.001t, T1 ¼ 60 years and T2 ¼ 70 years, then the following probabilities
84
Continuous Random Variables and Basic Distributions
can be evaluated:
ð 70
P{S . 70 j S . 60} ¼ exp 0:001j dj ¼ 0:522
60
ð
pffiffiffiffiffiffiffiffiffiffiffi
1 70
P{E . 70 j E . 60} ¼ exp 0:001j dj ¼ 0:522 ¼ 0:7225
2 60
The probability of a 60 year old sedentary person reaching 70 years is 0.522, whereas the
same person exercising regularly will reach 70 years with a probability of 0.7225.
B 6.4
NORMAL OR GAUSSIAN DISTRIBUTION
The most important distribution in probability theory is the Gaussian distribution. Under
some general conditions, a random variable X consisting of a number of components, each
with a general distribution tends to a normal distribution as the number becomes very
large.
The probability density fX(x) of a Gaussian distribution is given by
2
1
fX (x) ¼ pffiffiffiffiffiffiffiffiffi eð1=2Þ½(xm)=s
2ps
(6:4:1)
where m is called the mean and s2 is called the variance. These quantities will be defined
later. The cumulative distribution function FX(x) is given by
ðx
1
pffiffiffiffiffiffiffiffiffi e(1=2)½(jm)=s dj
FX (x) ¼
(6:4:2)
2ps
1
The Gaussian density function fX(x) given by Eq. (6.4.1) is shown in Fig. 6.4.1 for m ¼ 3
and s2 ¼ 4. It also shows that the area under the Gaussian pdf between (m 2 s, m þ s] is
68.3%, between (m 2 2s, m þ 2s] is 95.5%, and between (m 2 3s, m þ 3s] is 99.7%.
Thus, almost the entire pdf (99.7%) is between the m + 3s points of the Gaussian curve.
The Gaussian distribution function, FX(x) is shown in Fig. 6.4.2 along with (m 2 s,
m þ s], (m 2 2s, m þ 2s], and (m – 3s, m þ 3s] points.
FIGURE 6.4.1
6.4 Normal or Gaussian Distribution
85
FIGURE 6.4.2
A standard Gaussian is defined as the random variable with m ¼ 0 and s2 ¼ 1,
given by
1
2
fX (x) ¼ pffiffiffiffiffiffi e(1=2)x
2p
(6:4:3)
and the corresponding cumulative distribution function is denoted by FX(x), given by
ðx
FX (x) ¼
2
1
pffiffiffiffiffiffi e(1=2)j dj
2p
1
(6:4:4)
There is no closed-form solution for Eq. (6.4.4). Hence, FX(x) is tabulated for values of
x between 25 and 5 in Appendix B. The standard Gaussian pdf fX(x) is shown in
Fig. 6.4.3 with the 68.3% points between (21,1], 95.5% points between (22,2], and
the 99.7% points between (23,3].
FIGURE 6.4.3
86
Continuous Random Variables and Basic Distributions
Given a random variable X with any m and s2, the random variable
Z¼
X mX
sX
is a standard Gaussian. Hence, the distribution function FX(x) can be given in terms of the
standard Gaussian FZ (x) as follows:
FX (x0 ) ¼ P{X x0 ¼} ¼ P
X mX x0 mX
x0 mX
¼ FX
sX
sX
sX
(6:4:5)
We shall now prove that the fX(x) is a probability density function by showing that if
ð1
I¼
1
2
pffiffiffiffiffiffi e(1=2)x dx
2p
1
then I ¼ 1. Direct evaluation of the integral is difficult. Hence we square I and write
ð1
1 (1=2)x2
1
2
pffiffiffiffiffiffi e(1=2)y dy
e
dx 2p
1 2p
1
ð ð
1 1 1 ð1=2Þ(x2 þy2 )
¼
e
dx dy
2p 1 1
I2 ¼
ð1
(6:4:6)
Substituting x ¼ r cos(u) and y ¼ r sin(u) in Eq. (6.4.5) we obtain,
I2 ¼
1
2p
ð 2p ð 1
0
2
re(1=2)r dr du ¼ 1
(6:4:7)
0
and the result is proven.
Example 6.4.1 The number of malfunctioning computers is Gaussian-distributed with
m ¼ 4 and s ¼ 3. We want to find the probability that the number of bad computers is
between 3 and 5.
Denoting the malfunctioning computers by the random variable X, we have to find
Pf3 , X 5g. Using Eq. (6.4.5) and referring to the Gaussian tables, we have
34 X4 54
1
1
P{3 , X 5} ¼ P
,
¼P ,Z
3
3
3
3
3
1
1
¼ FX
FX ¼ 0:2611
3
3
Example 6.4.2 We shall study what grading on the curve in examinations means. If the
average of the class in a particular examination is m, then one method of grading is to
assign A to those receiving over m þ 2s, AB to those receiving between m þ s and
m þ 2s, B to those receiving between m and m þ s, C to those receiving between
m 2 s and m, D to those receiving between m 2 s and m 2 2s, and F to those receiving
less than m 2 2s. We want to calculate the percentage of students receiving the grades A,
AB, B, C, D, and F.
6.4 Normal or Gaussian Distribution
We can calculate these probabilities from Eq. (6.4.5) as follows:
Xm
. 2 ¼ 1 F(2) ¼ 0:0228 ¼ 2:28%
A: P{X . m þ s} ¼ P
s
Xm
2 ¼ F(2) F(1)
AB: P{m þ s , X m þ 2s} ¼ P 1 ,
s
¼ 0:1359 ¼ 13:59%
B:
Xm
1 ¼ F(1) F(0)
P{m , X m þ s} ¼ P 0 ,
s
¼ 0:3413 ¼ 34:13%
C:
Xm
0 ¼ F(0) F(1)
P{m s , X m} ¼ P 1 ,
s
¼ 0:3413 ¼ 34:13%
D:
F:
Xm
1 ¼ F(1) F(2)
P{m 2s , X m s} ¼ P 2 ,
s
¼ 0:1359 ¼ 13:59%
Xm
2 ¼ F(2) ¼ 0:0228 ¼ 2:2:8%
P{X m 2s} ¼ P
s
These grades are shown in Fig. 6.4.4.
Properties of Standard Gaussian
Density and Distribution
1. The pdf fX(x) is an even function.
pffiffiffiffiffiffi
2. fX(x) attains its maximum value of 1= 2p at x ¼ 0.
Ð
3. The cdf FX(x) ¼ x21fX(j)dj.
FIGURE 6.4.4
87
88
Continuous Random Variables and Basic Distributions
4. FX (0) ¼ 1 :
2
5. The error function erf(x) is defined as
2
erf(x) ¼ pffiffiffiffi
p
ðx
2
ej dj
(6:4:8)
0
6. The complement of the error function erfc(x) is defined as
ð
2 1 j2
e dj ¼ 1 erf(x)
erfc(x) ¼ pffiffiffiffi
p x
(6:4:9)
7. The tails of the distribution Q(x) is used in describing the probability of error in
transmission of communication signals. It is defined as
ð
1 1 (j2 =2)
Q(x) ¼ pffiffiffiffiffiffi
e
dj
(6:4:10)
2p x
We will discuss this function in greater detail in the next section.
8. Q(x) in terms of error function complement erfc(x) is
1
x
Q(x) ¼ erfc pffiffiffi
2
2
(6:4:11)
9. The general distribution function FX(x) in terms of erf(x) and erfc(x) is
pffiffiffiffi
ðx
ð
1
1 2 ðxmÞ= 2s j2
(1=2)½(jm)=s2
FX (x) ¼ pffiffiffiffiffiffiffiffiffi
e
dj ¼ pffiffiffiffi
e dj
2 p 1
2ps 1
1 1
xm
1
xm
¼ þ erf pffiffiffiffiffiffi ¼ 1 erfc pffiffiffiffiffiffi
(6:4:12)
2 2
2
2s
2s
1
10.
(6:4:13)
FX (m) ¼
2
Gaussian Tails Q(x)
As mentioned in property 7 [Eq. (6.4.10)] Q(x) is used to express the bit error probability in transmission of communication signals. Since Q(x) does not have a closed-form
solution, bounds can be established.
The distribution Q(x) can expanded into an asymptotic series [1] given by
2
ex 1
1 1:3
n 1:3 (2n 1)
p
ffiffiffiffiffiffi
1 2 þ 4 þ þ (1)
Q(x) ¼
(6:4:14)
x
x
x2n
2p x
and for large x, Q(x) can be approximated by
2
ex 1
Q(x) pffiffiffiffiffiffi
2p x
(6:4:15)
For small x, Q(x) diverges, and hence this is not a very good bound. A better bound is
2
ex
1
Q(x) QA(x) ¼ pffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi
2p 1 þ x2
(6:4:16)
6.4 Normal or Gaussian Distribution
89
Even this bound, although better than Eq. (6.4.15), is not very good for small values of x.
However, a tighter bound has been established by Börjesson and Sundberg [8] by first integrating Q(x) by parts as shown below:
ð
pffiffiffiffiffiffi ð 1 1 (j2 =2)
1 1 (j2 =2)
je
Q(x) ¼ pffiffiffiffiffiffi
e
dj ¼ 1 2p
dj
2p x
x j
ð1
1
1
1 (j2 =2)
2
¼ pffiffiffiffiffiffi e(x =2) e
dj
2
2p x
x j
and the expression for Q(x) can be manipulated to obtain
Q(x) ¼
1
1
2
pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi e(x =2)
2
(1 a)x þ a x þ b 2p
(6:4:17)
Substituting the values a ¼ 1 and b ¼ 1 in Eq. (6.4.17), we obtain Eq. (6.4.16). Better
values for a and b can be obtained by minimizing the absolute relative error function
1(x), given by
1(x) ¼
Q0 (x) Q(x)
Q0 (x)
(6:4:18)
and constraining 1(0) ¼ 0. In Eq. (6.4.18) Q0(x) is the true value given by Eq. (6.4.10). As
a consequence, the upper bound QU(x) is obtained for values of a ¼ 0.33919 and
b ¼ 5.533425, and the lower bound QL(x) is obtained for values of a ¼ 1/p and
b ¼ 2p. Thus the upper bound is
QU(x) ¼
1
1
2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi e(x =2)
2
0:66081x þ 0:33919 x þ 5:33425 2p
(6:4:19)
and the lower bound is,
QL(x) ¼
1
1
2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi eðx =2Þ
(1 1=p) þ 1=p x2 þ 2p 2p
(6:4:20)
The functions Q(x), QU(x), QL(x), and QA(x) are shown in Fig. 6.4.5 and tabulated in
Table 6.4.1 showing the upper and lower bounds are for all x . 0. The bound QA(x) is
not good for low values of x, whereas QU(x) and QL(x) practically lie on Q(x).
FIGURE 6.4.5
90
Continuous Random Variables and Basic Distributions
TABLE 6.4.1
x
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
Lower Bound
QL(x)
True Value
Q(x)
Upper Bound
QU(x)
Approx. Value
QA(x)
0.5
0.30496416
0.15704991
0.06633866
0.0226461
0.00619131
0.00134729
0.00023233
0.00003164
0.0000034
0.00000029
0.5
0.30853754
0.15865525
0.0668072
0.02275013
0.00620967
0.0013499
0.00023263
0.00003167
0.0000034
0.00000029
0.5
0.3071816
0.15837866
0.06684731
0.02279147
0.00622374
0.00135302
0.00023314
0.00003173
0.0000034
0.00000029
0.39894228
0.3148968
0.17109914
0.07184344
0.02414549
0.00650985
0.00140147
0.00023974
0.00003246
0.00000347
0.00000029
The absolute percentage errors for the upper QU(x) and lower bounds QL(x) are
shown in Fig. 6.4.6. Table 6.4.2 shows in addition to the upper- and lower-bound errors
the absolute percentage error for the approximate value QA(x). Since these errors are
large, they are not shown in Fig. 6.4.6.
The errors for the approximate value are high (20.2%). The upper-bound errors are
better for smaller values of x, whereas the lower-bound is better for higher values of x.
Even then, these bounds clearly illustrate that the approximation for all x is excellent
with the maximum error of about 1.16% occurring at low values of x. This is much
better than the usual approximation QA(x), where the maximum error is 20.2% for low
values. For high values of x, QU(x), QL(x), and QA(x) provide reasonably good
approximation.
Properties of Q Function
1. The limit as x ! 1 is zero:
lim Q(x) ¼ 0
x!1
FIGURE 6.4.6
(6:4:21)
6.4 Normal or Gaussian Distribution
TABLE 6.4.2
x
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
91
Error percentages
Lower Bound
Upper Bound
Approximate Value
0
1.15816594
1.01184414
0.70133395
0.45727327
0.29559943
0.19344425
0.12922798
0.08836646
0.06185728
0.04427826
0.00000176
0.43947331
0.17433835
0.06003046
0.1816945
0.226622
0.23152986
0.21848937
0.19884525
0.1779578
0.15815582
20.21154392
2.06109849
7.8433496
7.53846047
6.13342513
4.83410996
3.82069533
3.05864395
2.48602699
2.05102577
1.71571131
2. The limit as x ! 21 is 1:
lim Q(x) ¼ 1
(6:4:22)
x!1
3. The value of Q(x) at x ¼ 0 is half:
Q(0) ¼
1
2
(6:4:23)
4. Q(x) is even:
Q(x) ¼ Q(x)
(6:4:24)
5. Q(x) can be expressed in terms of error function as
1
x
1
x
Q(x) ¼ 1 erf pffiffiffi ¼ erfc pffiffiffi ,
2
2
2
2
x0
(6:4:25)
6. If F(x) is a standard Gaussian, then
Q(x) ¼ 1 F(x)
(6:4:26)
7. For any Gaussian distributed random variable X with mean m and variance s2
P{X . a} ¼ 1 FX (a) ¼ Q
am
s
(6:4:27)
Gaussian Approximation to Binomial Distribution
In Example 4.7.3 we saw that the binomial distribution can be approximated by a
Poisson distribution under the condition that n is large compared to k and p is small
enough that the product np ¼ l is of moderate magnitude. On the other hand, if npq is
quite large, the Poisson approximation does not fare very well. In this case we use the
Gaussian approximation, which is called the DeMoivre –Laplace limit theorem, which
states that, if Sn is the total number of successes in n Bernoulli trials, each with probability
of success p and failure q with ( p þ q ¼ 1) with pmf b(k;n,p), then, as n ! 1, the
pffiffiffiffiffiffiffiffi
standardized random variable Z ¼ (Sn 2 np)/ npq converges to a standard normal
distribution function where m ¼ np is the mean and s2 ¼ npq is the variance. Or, if
92
Continuous Random Variables and Basic Distributions
n1 , n2, then
Sn np
lim Z ¼ P n1 , pffiffiffiffiffiffiffiffi n2 ! F(n2 ) F(n1 )
n!1
npq
(6:4:28)
and the probability mass function b(k;n,p) converges to the density function fZ (z). Or
lim
2
n!
1
pk qnk ! pffiffiffiffiffiffiffiffiffiffiffiffiffi e(1=2)½(xnp)=npq
k)!
2pnpq
n!1 k!(n
(6:4:29)
The Demoivre –Laplace limit theorem is a special case of the central limit theorem
presented in a later chapter.
Example 6.4.3 (Gaussian and Poisson Approximation) We will find the Gaussian
approximation for the binomial distribution given in Example 4.7.3, where the Poisson
approximation has already been found. In the binomial distribution b(k;n,p) of that
example, n ¼ 1000, p ¼ 0.005 with np ¼ 5, and npq ¼ 4.975.
The approximate Gaussian pdf will be given by
2
1
fX (k) ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi eð1=2Þ½(k5) =4:975 b(k; 1000,0:005)
2p:4:975
For the Poisson approximation np ¼ l ¼ 5, and this is given in Example 4.7.3 as
k
5
pX (k;l) ¼ pX (k;5) ¼ pX (k) ¼ e5
k!
Since l ¼ 5 is a single digit and npq ¼ 4.975 is not very large, we would reasonably infer
that the Poisson approximation will be better than the Gaussian. The binomial distribution
with the Poisson and Gaussian approximations are shown in Fig. 6.4.7 for k ¼ 0, . . . , 15.
The true value of b(5; 1000, 0.005) ¼ 0.17591, the Gaussian approximation
fX(5) ¼ 0.17886 with an absolute error of 1.678% and the Poisson approximation pX(5;
5) ¼ 0.17547 with an error of 0.25%, confirming our inference that the Poisson approximation is better. Figure 6.4.7 clearly shows that the Poisson approximation follows the
binomial more closely than the Gaussian.
FIGURE 6.4.7
6.4 Normal or Gaussian Distribution
93
FIGURE 6.4.8
Example 6.4.4 We shall now modify Example 6.4.3 by keeping n ¼ 1000 and changing
p ¼ 0.5, yielding np ¼ l ¼ 500, and npq ¼ 250. Here both l and npq are quite large, so
we can infer that the Gaussian will be a better approximation than the Poisson. The
approximate Gaussian pdf is given by
2
1
fX (k) ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e(1=2)½(k500) =250 b(k; 1000,0:5)
2p:250
and the corresponding Poisson approximation is
PX (k;l) ¼ PX (k;500) ¼ PX (k) ¼ e500
500k
k!
The binomial distribution with the Poisson and Gaussian approximations for this case are
shown in Fig. 6.4.8.
TABLE 6.4.3
Comparison of binomial with Gaussian and Poisson approximations
k
Binomial
Gaussian
Error 1BG
Error 1BP
2
3
4
5
6
7
8
0.083929
0.140303
0.175731
0.175908
0.14659
0.104602
0.035245
n ¼ 1000, p ¼ 0.005, np ¼ 5, npq ¼ 4.975
0.072391
0.084224
13.74699
0.119653
0.140374
14.718176
0.161758
0.175467
7.951344
0.17886
0.175467
1.678441
0.161758
0.146223
10.347485
0.119653
0.104445
14.38896
0.072391
0.065278
10.953113
0.352343
0.50683
0.150021
0.250272
0.250272
0.14992
0.051188
497
498
499
500
501
502
503
0.024775
0.025024
0.25175
0.25225
0.025175
0.025024
0.024775
n ¼ 1000, p ¼ 0.5, np ¼ 500, npq ¼ 250
0.024781
0.017731
0.023389
0.02503
0.017803
0.02438
0.025181
0.017838
0.024978
0.025231
0.017838
0.025177
0.025181
0.017803
0.024978
0.02503
0.017732
0.02438
0.024781
0.017626
0.023389
28.431591
28.85844
29.141872
29.283306
29.283306
29.141589
28.857017
Poisson
94
Continuous Random Variables and Basic Distributions
The true value for k ¼ 500 for binomial b(500,1000,0.5) ¼ 0.025225, the Gaussian
approximation fX(k) ¼ 0.025231 with an absolute error of 0.025% and the Poisson approximation pX(k) ¼ 0.017838 with an error of 29.28%. In this case, the Gaussian approximation, as expected, fares very much better than the Poisson one. The results graphed in
Fig. 6.4.8 for k ¼ 400, . . . , 600 clearly show that the Gaussian approximation follows
the binomial much more closely than does the Poisson, unlike the scenario described in
Example 6.4.3.
The percentage absolute errors for the Gaussian and Poisson approximations to the
binomial are defined by
Gaussian:
1BG ¼
b(k;n,p) fX (k;np)
100
b(k;n,p)
Poisson:
1BP ¼
b(k;n,p) pX (k;l)
100
b(k;n,p)
These errors are tabulated for p ¼ 0.005 and 0.5 in Table 6.4.3 in the neighborhoods of
np ¼ l ¼ 5 and 500, respectively. The absolute error percentages also show that for
large n and for values of moderate np and npq, the Poisson approximation is better and
for large values of np and npq, the Gaussian approximation is better.
CHAPTER
7
Other Continuous Distributions
B 7.1
INTRODUCTION
We will now discuss some of the other important continuous distributions that occur in
communications, signal processing, queuing, statistical inference, and other probabilistic
situations. This list is by no means exhaustive. For a more complete list of distributions,
refer to Web references W1 and W2 (last two entries in References list at end of the book).
Most of these distributions are derivable from the three basic distributions discussed in
Chapter 6.
B 7.2
TRIANGULAR DISTRIBUTION
If two independent random variables X and Y are uniformly distributed in the interval
(a, b], then Z ¼ X þ Y has a triangular density defined by
8 za
>
a,zb
>
>
2
>
< (b a)
(7:2:1)
fZ (z) ¼ (z 2b þ a)
b , z 2b a
>
2
>
>
(b
a)
>
:
0
otherwise
The density fZ (z) is shown in Fig. 7.2.1 with the corresponding uniform densities of X and
Y for a ¼ 1 and b ¼ 5.
The distribution function FZ (z) is given by
8
0
za
>
>
>
> 1 z a 2
>
>
a,zb
<2 b a
(7:2:2)
FZ (x) ¼
1 z þ a 2b 2
>
>
>1 b , z 2b a
>
>
2
ba
>
:
1
z . 2b a
The distribution function FZ (z) is shown for a ¼ 1 and b ¼ 5 in Fig. 7.2.2.
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
95
96
Other Continuous Distributions
FIGURE 7.2.1
FIGURE 7.2.2
B 7.3
LAPLACE DISTRIBUTION
The Laplace density, also called a double-exponential density, is used to model navigation
and pilot errors, speech, and image processing. As the term suggests, double-exponential
density is defined by
fX (x) ¼
l ljxj
e
2
l.0
(7:3:1)
The cdf FX(x) is given by
8
1
>
< elx
FX (x) ¼ 2
>
: 1 1 elx
2
x0
(7:3:2)
x.0
It is an even density function, as shown for l ¼ 0.5 in Fig. 7.3.1.
The cdf FX(x) is shown in Fig. 7.3.2.
An alternate form of Laplace density in terms of the variance s2 is defined by
pffiffi
1
fX (x) ¼ pffiffiffi e 2(jxj=s)
2s
(7:3:3)
7.4 Erlang Distribution
97
FIGURE 7.3.1
FIGURE 7.3.2
with the corresponding cdf FX(x) given by
8 pffiffi
1
>
< e 2(x=s)
FX (x) 2
pffiffi
>
: 1 1 e 2(x=s)
2
x0
(7:3:4)
x.0
where s2 is the variance of the Laplace distribution.
B 7.4
ERLANG DISTRIBUTION
We already saw that the waiting time between successive events of a Poisson arrival
process is exponentially distributed with parameter l. The distribution of the waiting
time for n successive independent events of a Poisson arrival process is given by the
Erlang distribution with n degrees of freedom and is defined for positive integral values
of n by
fX (x) ¼
l(lx)n1 elx
,
(n 1)!
where n is an integer
(7:4:1)
98
Other Continuous Distributions
The Erlang distribution is obtained by the n-fold convolution of n independent exponential
distributions. The cumulative distribution corresponding to Eq. (7.4.1) can be derived as
follows. Let Xn denote the time at which the nth event occurs. Let N(x) be the number of
events in the interval (0,x]. Hence the event fXn x) can occur only if the number of
events occurring in (0,x] is at least n. Since N(x) is Poisson-distributed, the cdf FXn(x)
can be given by
P{Xn x} ¼ FXn (x) ¼ P{N(x) . n}
¼
1
X
el x
k¼n
n1
X
(l x)k
(l x)k
¼1
el x
k!
k!
k¼0
(7:4:2)
We can obtain the pdf fXn(x) by differentiating Eq. (7.4.2) with respect to x. Performing the
indicated differentiation, we can write
(
)
n1
k
X
d
d
l x (l x)
{FXn (x)} ¼
1
e
dx
dx
k!
k¼0
¼ el x
n1
X
k(l x)k1 l
k¼0
(
¼ lel x
k!
n1
X
(l x)k
n1
X
(l x)k
k¼0
n1
X
(l x)k1
k!
)
(k 1)!
(
)
n1
n2
X
(l x)k X
(l x)k
l x
¼ le
k!
k!
k¼0
k¼0
k¼0
k!
þ lel x
k¼1
or
d
lel x (l x)n1
{FXn (x)} ¼ fXn (x) ¼
dx
(n 1)!
(7:4:3)
and Eq. (7.4.3) is the Erlang distribution. This distribution is shown in Fig. 7.4.1 for
n ¼ 10 and l ¼ 2,2.5,3,3.5,4,5.
Another family of Erlang distributions is shown for l ¼ 4 and n ¼ 2,3,4,5,6 in
Fig. 7.4.2.
FIGURE 7.4.1
7.5 Gamma Distribution
99
FIGURE 7.4.2
In both these cases the Erlang distributions reduce to an exponential distribution for
n ¼ 1. In cases where the data may not be modeled as a simple exponential, the Erlang
family provides greater flexibility.
B 7.5
GAMMA DISTRIBUTION
A gamma distribution is a generalization of the Erlang distribution where n ¼ a, which
may not be an integer. It finds wide use in statistics. Gamma density is a two-parameter
family with parameters a . 0, termed the shape parameter, and l . 0, called the scale
parameter, and is defined as
fX (x) ¼
l(lx)a1 elx
,
G(a)
a, l, x . 0
(7:5:1)
where G(a) is the gamma function defined by the integral
ð1
G(a) ¼ xa1 ex dx, a . 0
(7:5:2)
0
Integrating Eq. (7.5.2) by parts with u ¼ x a21 and dv ¼ e 2x dx, we have
1
ð1
ð1
a1 x
a1 x G(a) ¼ x e dx, a . 0 ¼ x e þ (a 1)
xa2 ex dx
0
0
¼ (a 1)G(a 1)
0
(7:5:3)
If a is an integer n, then G(n) ¼ (n –1)G(n –1) and continued expansion yields
G(n) ¼ (n– 1)! Hence the gamma function can be regarded as the generalization of the factorial function for all positive real numbers. The most useful fractional argument for the
gamma function is 12 with
pffiffiffiffi
1
G
¼ p
(7:5:4)
2
100
Other Continuous Distributions
The incomplete gamma function Gx(a) is defined by
ðx
Gx (a) ¼ ja1 ej dj
(7:5:5)
0
We will use Eq. (7.5.5) later for the x2 distribution function FX(x) in Section 7.7. We shall
now show that the density function given by Eq. (7.5.1) indeed integrates to 1. If we define
the integral Ix by
ðx
l(lh)a1 elh
dh
(7:5:6)
Ix ¼
G(a)
0
Then, substituting lh ¼ j in Eq. (7.5.6), we can rewrite
ð
1 lx a1 j
Ix ¼
j e dj
G(a) 0
(7:5:7)
Applying Eq. (7.5.5) to Eq. (7.5.7), we have
Ix ¼
Glx (a)
G(a)
or
lim Ix ¼
x!1
G(a)
¼1
G(a)
thus showing the validity of Eq. (7.5.1) as a proper density function. If a is an integer,
then we have the Erlang distribution.
The gamma density family is shown in Fig. 7.5.1 with the parameter l ¼ 2 and
a ¼ 0.5,1,1.5,2.5,3.5,4.5,5.5.
The cumulative distribution function FX(x) is given by
ð
1 x
Gx (a)
(7:5:8)
FX (x) ¼
l(lj)a1 elj dj ¼
G(a) 0
G(a)
This integral in Eq. (7.5.8) has no closed-form solution unless a is an integer, in which
case it is the Erlang distribution given by Eq. (7.4.2). However, the integral given by
Eq. (7.5.8) has been extensively tabulated.
Reproductive Property of Gamma Distribution
If X1, X2, . . . , Xn are n independent gamma-distributed random variables with shape
parameters a1, a2, . . . , an and having the same scale parameter l, then random variable
FIGURE 7.5.1
7.6
Weibull Distribution
101
Z ¼ X1 þ X2 þ þ Xn is also gamma-distributed with shape parameter b ¼ a1 þ
a2 þ þ an. As a consequence of this property, since an exponential random variable
is gamma-distributed with parameters (1,l), the sum of n exponential random variables
is gamma-distributed with parameters (n,l), which is an Erlang distribution given by
Eq. (7.4.1).
B 7.6
WEIBULL DISTRIBUTION
We will define the Weibull distribution as a three-parameter family of distribution functions, in which the parameter l . 0 is the scale parameter, a . 0 is the shape parameter,
and m is the location parameter. It is defined by
a
a
a l (x m)a1 e½l(xm) x . m
(7:6:1)
fX (x) ¼
0
xm
We will now find the cdf FX (x) of the Weibull distribution. We can write
ðx
a
FX (x) ¼ a la (j m)a1 e½l(jm) dj
(7:6:2)
m
Substituting h ¼ ½l(j m)a and hence dh ¼ ½l(j m)a1 al dh, we can rewrite
Eq. (7.6.2) as follows:
½l(xm)a
ð ½l(xm) a
eh dh ¼ eh , x.m
(7:6:3)
FX (x) ¼
0
0
Substitution of the limits in Eq. (7.6.3) yields
a
1 e½l(xm) x . m
FX (x) ¼
0
xm
and
FX (1) ¼ 1
(7:6:4)
Two families of Weibull distributions are shown for m ¼ 0 and a ¼ 2, l ¼ 0.5,1,2,3,4
(Fig. 7.6.1) and for l ¼ 2 and a ¼ 0.5,1,2,3,4,5 (Fig. 7.6.2).
When m ¼ 0 and l ¼ 1 in Eqs. (7.6.1) and (7.6.4), then the resulting distribution is
called the standard Weibull.
FIGURE 7.6.1
102
Other Continuous Distributions
FIGURE 7.6.2
The Weibull distribution was originally used for modeling fatigue data and is at
present used extensively in engineering problems for modeling the distribution of the lifetime of an object that consists of several parts and that fails if any component part fails.
B 7.7
CHI-SQUARE DISTRIBUTION
The Laplace, Erlang, gamma, and Weibull distributions are all based on exponential
distributions. The chi-square distribution straddles both the exponential and Gaussian
distributions. The general gamma distribution has parameters (a, l). In the gamma distribution, if we substitute a ¼ n/2 and l ¼ 12, we obtain the standard chi-square distribution,
defined by
8 (n=2)1 (x=2)
x
e
>
<
, x.0
n=2
2 G(n=2)
(7:7:1)
fX (x) ¼
>
:
0,
x0
where n is the degrees of freedom for the x2 distribution. We can introduce another
parameter s2 and rewrite Eq. (7.7.1) as
8
(n=2)1 (x=2s2 )
>
e
>
<x
, x.0
n=2
2
(7:7:2)
fX (x) ¼ (2s ) G(n=2)
>
>
:
0
, x0
If we substitute x ¼ x2 in Eq. (7.7.2), we obtain the familiar form of the chi-square distribution, given by
2
fX (x2 ) ¼
2
x2½(n=2)1 e(x =2s )
(2s2 )n=2 G(n=2)
(7:7:3)
where G(x) is the gamma function defined in Eq. (7.5.2). If n is even in Eq. (7.7.2), then
G(n=2) ¼ (n=2 1)!, and if n is odd, then G(n=2) in Eqs. (7.7.1)
pffiffiffiffi –(7.7.3) is obtained by
iterating (n=2 1)G(n=2 1) with the final iterant Gð12Þ ¼ p.
Equation (7.7.3) can also be obtained from the following. If X1, X2, P
. . . , Xn are
n i.i.d. Gaussian random variables with zero mean and variance s2, then X ¼ nk¼1 X k2 is
x2 -distributed with n degrees of freedom.
7.7 Chi-Square Distribution
103
The cumulative chi-square distribution function FX(x) is given by integrating
Eq. (7.7.2):
8 ð x n=21 (j=2s2 )
e
< j
dj x . 0
n=2
2
(7:7:4)
FX (x) ¼
: 0 (2s ) G(n=2)
0
x0
This equation can be given in terms of the incomplete gamma function Gx(a) given by
Eq. (7.5.5). Substituting h ¼ j/2s2 and dj ¼ 2s2 dh in Eq. (7.7.4), we obtain
ðx
FX (x) ¼
2
jn=21 e(j=2s )
2s2 dj
2 n=21 G(n=2)
0 (2s )
ð x=2s2
¼
0
Gx=2s2 (n=2)
hn=21 e(h)
dh ¼
(7:7:5)
G(n=2)
G(n=2)
When x ! 1 then Gx=2s2 (n=2) ! G(n=2) and hence FX(x) ¼ 1. The cdf FX(x) in
Eq. (7.7.5) has been extensively tabulated.
The chi-square distribution is shown in Fig. (7.7.1) for the number of degrees of
freedom n ¼ 2,3,4,5,6,10.
The chi-square distribution is used directly or indirectly in many tests of hypotheses.
The most common use of the chi-square distribution is to test the goodness of a hypothesized fit. The goodness-of-fit test is performed to determine whether an observed
value of a statistic differs enough from a hypothesized value of a distribution to draw the
inference whether the hypothesized quantity is the true distribution. Its utility is shown
in Chapter 15.
Example 7.7.1 The density function of fZ (z) of Z ¼ X 2 þ Y 2, where X and Y are independent random variables distributed as N(0, s2), can be found by substituting n ¼ 2 in
Eq. (7.7.2), yielding
8
2
< e(z=2s )
z.0
FZ (z) ¼
2
: 2s
0
z0
FIGURE 7.7.1
104
Other Continuous Distributions
which is a x2 density with 2 degrees of freedom. The corresponding cdf, FZ (z), is given by
(
2
1 e(z=2s ) z . 0
FZ (z) ¼
0
z0
B 7.8
CHI AND OTHER ALLIED DISTRIBUTIONS
The Rayleigh and Maxwell distributions are special cases of the chi distribution. Just as the
sum of the squares of n zero mean Gaussian random variables with variance s2 is chisquare-distributed with n degrees of freedom, we can consider that the sum of the
square roots of the sum of the squares of n zero mean Gaussian random variables with
variance s2 is chi-distributed with n degrees of freedom. Accordingly, let Zn be the
random variable given by
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
(7:8:1)
Zn ¼ X1 þ X2 þ þ Xn2
where fXi . i ¼ 1, . . . , ng are zero mean Gaussian random variables with variance s2. Then
the x-density fZ (z) is given by
9
8
1
zn1 (1=2)(z=s)2
=
<
e
z.0
fZ (z) ¼ 2n=21 G(n=2) sn
(7:8:2)
;
:
0
z0
where G(n/2) is the gamma function defined in Eq. (7.5.2).
A family of chi distributions is shown in Fig. 7.8.1 for s ¼ 1.5 and n ¼ 1,2,3,5,7,9,
11,13,15.
The distribution function FZ (z) is given by integrating Eq. (7.8.2):
ðz
1
z n1 (1=2)(z=s)2
e
dz z . 0
(7:8:3)
Fz (z) ¼
n=21
G(n=2) sn
02
There is no closed-form solution for Eq. (7.8.3), but it can be expressed in terms of the
incomplete gamma function GZ (z) defined in Eq. (7.5.5). By substituting y ¼ z2 =2s2 and
FIGURE 7.8.1
7.8 Chi and Other Allied Distributions
105
dz ¼ (s2 dy)=z in the in Eq. (7.8.3), we can write
2n=21
FZ (z) ¼ n=21
2
G(n=2)
¼
2n=21
2n=21 G(n=2)
ð z2 =2s2
y(n=21) ey dy
0
G(z2 =2s2 )
n G 2 2 (n=2)
(z =2s )
¼
2
G(n=2)
(7:8:4)
where G(z2 =2s2 ) (n=2) is the incomplete gamma function defined in Eq. (7.5.5). Clearly, as
z ! 1, G(z2 =2s2 ) (n=2) ! G(n=2), thus showing that FZ (1) ¼ 1.
There are two very important special cases of the chi distribution when n ¼ 2 and
n ¼ 3.
Rayleigh Distribution (n 5 2)
When n ¼ 2 in the chi distribution of Eq. (7.8.2), we have
( z
)
(1=2)(z=s)2
e
z
.
0
fZ (z) ¼ s2
0
z0
(7:8:5)
This is called the Rayleigh density and finds wide application in communications fading
channels. If we have two independent zero mean Gaussian random variables X and Y
with variance
s2, and if we make the transformation X ¼ Z cos(Q) and Y ¼ Z sin(Q),
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
then Z ¼ X þ Y 2 is Rayleigh-distributed, which is a chi distribution with 2 degrees
of freedom. Q is uniformly p
distributed
in the interval (0,2p]. The Rayleigh density is
ffiffiffiffiffi
shown in Fig. 7.8.2 for s ¼ 10,4,5.
The cdf corresponding to the Rayleigh density can be expressed in a closed form as
follows:
ðz
j j2 =2s2
FZ (z) ¼
e
dj or,
2
0s
(
2
2
1 ez =2s z . 0
¼
(7:8:6)
0
z0
FIGURE 7.8.2
106
Other Continuous Distributions
Maxwell Density (n 5 3)
This is one of the earliest distributions derived by Maxwell and Boltzmann from statistical mechanics considerations. They derived that the velocity vi of gas molecules in any
direction i, based on the assumptions that the molecules are identical and distinguishable,
and can occupy any energy state, follow the probability density function
fV (vi ) ¼
e(1i =kT)
Z
(7:8:7)
where 1i is the energy function given by 1i ¼ mv2i =2 with m as the mass of the particle, k is
the Boltzmann constant, and T is the absolute temperature. Z is the normalization constant
found from integrating Eq. (7.8.7) over all vi, or
rffiffiffiffiffiffiffiffiffiffiffi
ð
1 1 (mv2i =2kT)
2pkT
e
dvi ¼ 1 and Z ¼
Z 1
m
so that the velocity density function along any direction follows a Gaussian distribution
given by
rffiffiffiffiffiffiffiffiffiffiffi
m (mv2i =2kT)
, i ¼ x,y,z
(7:8:8)
e
fV (vi ) ¼
2pkT
From Eq. (7.8.8) the velocity vector V ¼ fvx,vy,vzg will be distributed with density
function,
"
#
m 3=2
m(v2x þ v2y þ v2z )
exp fV (vx ,vy ,vz ) ¼
(7:8:9)
2pkT
2kT
Eq. (7.8.9) is a chi-square density with 3 degrees of freedom. The speed x of gas molecules
will be defined by
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x ¼ v2x þ v2y þ v2z
Hence x is a distributed as a chi-density with three degrees of freedom. Its probability
density can be derived as
h m i3=2
2
f (x) ¼ 4p
x2 emx =2kT , x . 0
(7:8:10)
2pkT
If we substitute kT=m ¼ s2 in the Maxwell – Boltzmann distribution of Eq. (7.8.10), we
obtain
rffiffiffiffi
2 x2 x2 =2s2
e
, x.0
(7:8:11)
fX (x) ¼
p s3
Equation (7.8.11) is a special case of the chi density of Eq. (7.8.2) with n ¼ 3 degrees of
freedom. The pdf values fX(x)
for the Maxwell and Rayleigh densities are compared in
pffiffiffiffiffi
Fig. 7.8.3 for values of s ¼ 10, 4, and 5.
The cdf FX(x) corresponding to the Maxwell density fX(x) is obtained by integrating
Eq. (7.8.11) and is given by
rffiffiffiffi ð x 2
2 j j2 =2s2
e
dj
(7:8:12)
FX (x) ¼
p 0 s3
7.8 Chi and Other Allied Distributions
107
FIGURE 7.8.3
There is no closed form solution for Eq. (7.8.12). However, integrating Eq. (7.8.12) by
parts, it can be expressed as
rffiffiffiffi
pffiffiffi ð x
2
2 x x2 =2s2
j2 =2s2
e
FX (x) ¼ pffiffiffiffi e
dj ps
s p 0
rffiffiffiffi
x
2 x x2 =2s2
e
¼ erf pffiffiffi
x.0
(7:8:13)
ps
2s
where erf(x) is as defined in Eq. (6.4.6).
Rice’s Density
In the Rayleigh distribution discussed earlier, the random variables X and Y must be
zero mean with the same variances s2. In many communication problems involving fading
channels, X and Y will not be zero mean p
and
will have mean values equal to mX and mY. If
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
the random variable Z is given by Z p
¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
X 2 þffi Y 2 , then fZ (z) has a Rice density [49,50]
with a noncentrality parameter m ¼ m2X þ m2Y , given by
2
z
(z þ m2 ) mz fZ (z) ¼ 2 exp
I0 2
(7:8:14)
2s2
s
s
where I0(z) is the zeroth-order modified Bessel function of the first kind. The general
nth-order modified Bessel function is given by the series
In (z) ¼
1
z n X
2
(z2 =4)k
k! G(n þ k þ 1)
k¼0
If n is an integer, then In(z) can also be given by
ð
1 p z cos u
In (z) ¼
e
cos (nu) du
p 0
and I0(z) the zeroth-order modified Bessel function by
ð
1
X
1 p z cos u
(z2 =4)k
e
du ¼
I0 (z) ¼
2
p 0
k¼0 (k!)
and I0(z), I1(z), I2(z), I3(z) are shown in Fig. 7.8.4.
(7:8:15)
(7:8:16)
(7:8:17)
108
Other Continuous Distributions
FIGURE 7.8.4
From Fig. 7.8.4 we see that I0(0) ¼ 1, and hence when m ¼ 0 (zero mean), then fZ (z)
is Rayleigh-distributed. Rice’s density is shown in Fig. 7.8.5 for m ¼ 0,2,4,6,8,10,12,14,
and m ¼ 0 corresponds to the Rayleigh density.
The distribution function FZ(z) corresponding to Eq. (7.8.14) is given by integrating
fZ (z):
ðz
j
(j2 þ m2 )
mj
exp
I0 2 dj
(7:8:18)
FZ (z) ¼
2
2
s
2s
s
0
There is no closed-form solution for this integral.
Nakagami Density
The Nakagami distribution [49], like the Rayleigh density and unlike Rice’s distribution, also belongs to the class of central chi-square distributions. It is used for modeling
data from multipath fading channels and has been shown to fit empirical results more generally than the Rayleigh distributions. Sometimes, the pdf of the amplitude of a mobile
signal can also be described by the Nakagami m distribution.
FIGURE 7.8.5
7.8 Chi and Other Allied Distributions
109
The Nakagami distribution may be a reasonable means to characterize the back scattered echo from breast tissues to automate a scheme for separating benign and malignant
breast masses. The shape parameter m has been shown to be useful in tissue characterization. Chi-square tests showed that this distribution is a better fit to the envelope than the
Rayleigh distribution. Two parameters, m (effective number) and V (effective cross
section), associated with the Nakagami distribution are used for the effective classification
of breast masses.
The Nakagami m density is defined by
fX (x) ¼
2 h m im 2m1 mx2 =V
x
e
G(m) V
(7:8:19)
where m is the shape parameter and V controls the spread of the distribution. The interesting feature of this density is that, unlike the Rice distribution, it is not dependent on
the modified Bessel function. The Nakagami family is shown in Fig. 7.8.6 for V ¼ 1
and for m ¼ 3,2,1.5,1,0.75,0.5.
The cdf FX(x) is given by
ðx
2 h m im 2m1 mj2 =V
FX (x)
j
e
dj
(7:8:20)
0 G(m) V
This has no closed form solution. However, it can be given in terms of the incomplete
gamma function by substituting y ¼ mj2/V and dj ¼ (V=2mj) dy in Eq. (7.8.20). This
results in
m
2 mj2 1 mj2 =V
e
dj
FX (x) ¼
j
V
0 G(m)
ð mx2 =V
m
2
2 mj2
1
emj =V
¼
dy
G(m) V
2(mj2 =V)
0
ð mx2 =V
2 m y 1
y e
dy
¼
G(m)
2y
0
ð mx2 =V
Gmx2 =V (m)
1 m1 y
y
¼
e dy ¼
G(m)
G(m)
0
ðx
Since Gmx2 =V (m) ! G(m) as x ! 1 we have FX (1) ¼ 1:
FIGURE 7.8.6
(7:8:21)
110
Other Continuous Distributions
B 7.9
STUDENT-t DENSITY
The t distribution first developed by William Gosset under the pseudonym “Student” is
used to estimate the confidence intervals for the unknown population mean where the
variance is also unknown. In this case the statistic used for determining the confidence
interval is given by
T¼
m^ X mX
pffiffiffi
s=
^ n
(7:9:1)
and the distribution of T is governed by the Student-t distribution with density
given by
nþ1
½(nþ1)=2
G
t2
2
1þ
fT (t) ¼
pffiffiffiffiffiffi
n
n
np G
2
1 , t , 1
(7:9:2)
where n is the number of degrees of freedom of T. It will be shown in Example
13.4.5 that if Z is a standard Gaussian random variable, Wn is a chi-square random
variable with n degreespffiffiffiffiffiffiffiffiffiffiffi
of freedom, and Z and Wn are independent, then the
random variable T ¼ Z= Wn =n is t-distributed with n degrees of freedom.
The Student-t distribution with n ¼ 2 is compared to a standard Gaussian in
Fig. 7.9.1.
After integrating the density function, we can write the distribution function FT (t) as
1
½(nþ1)=2
ðt G n þ
t2
2
1þ
FT (t) ¼
dt
n
n
1 pffiffiffiffiffiffi
np G
2
This is shown in Fig. 7.9.2.
FIGURE 7.9.1
(7:9:3)
7.10
Snedecor F Distribution
111
FIGURE 7.9.2
Limit of Student-t as n ! 1
We shall find the limit as n ! 1 in Eq. 7.9.2. We have
½(nþ1)=2
t2
2
lim 1 þ
¼ et =2
n!1
n
1
G nþ
1
2
¼ pffiffiffiffiffiffi
lim
n!1 pffiffiffiffiffiffi
n
2p
np G
2
(7:9:4)
and hence
½vþ(1=2)
G( nþ1 )
t2
1
2
lim pffiffiffiffiffiffi 2 n 1 þ
¼ pffiffiffiffiffiffi et =2
n!1 np G( )
n
2p
2
(7:9:5)
that is, as the number of degrees of freedom n goes to infinity, the Student-t tends to a
standard Gaussian.
If X1, X2, . . . , Xn are n independent identically distributed P
Gaussian random variables
with mean mX and standard deviation sX, then m^ X ¼ (1=n) ni¼1 Xi is also Gaussiandistributed with mean mX and variance s2X =n.PSince we do not know s2X , we use the
sample variance given by s^ 2X ¼ 1=(n 1) ni¼1 (Xi m^ X )2 and s^ 2X is chi-squaredistributed with n – 1 degrees of freedom. Under these conditions the random variable
T¼
(m^ X mX )
pffiffiffi
s^ X = n
(7:9:6)
has a t distribution with n 2 1 degrees of freedom.
B 7.10
SNEDECOR F DISTRIBUTION
If X1 and X2 are two independent chi-squared-distributed random variables with n1 and n2
degrees of freedom, then the density of the random variable F ¼ (X1 =n1 )=(X2 =n2 ) is
n þ n n n1 =2
1
2
1
G
2
n2
x(n1 2)=2 , x . 0
(7:10:1)
fn1 ,n2 (x) ¼ n1
n2 1 þ n1 (n1 þn2 )=2
G
x
G
2
2
n2
112
Other Continuous Distributions
FIGURE 7.10.1
FIGURE 7.10.2
This density is shown in Fig. 7.10.1 for three cases: n1 ¼ 15, n2 ¼ 25: n1 ¼ 5, n2 ¼ 25:
n1 ¼ 10, n2 ¼ 5. Note that in general fn1 ,n2 (x) is not equal to fn2 ,n1 (x).
The corresponding distribution function is given by
n þ n n n1 =2
1
2
1
ðx
G
2
n2
(n1 2)=2
Fn2 , n1 (x) ¼
dj
(7:10:2)
n n 1 þ n (n1 þn2 )=2) j
0
1
2
1
G
j
G
2
2
n2
Equation (7.10.2) is shown in Fig. 7.10.2 for the same values of n1 and n2 as in Fig. 7.10.1.
As n1 and n2 tend to infinity, then the density function becomes more sharply spiked
around 1. The F distribution is used to draw inferences about the population variances
in the case of two-sample situations.
B 7.11
LOGNORMAL DISTRIBUTION
A random variable X has lognormal density with parameters a, m, and s2 if the transformed random variable Y ¼ ln(X ) has a normal distribution with mean m and variance
7.11
Lognormal Distribution
113
FIGURE 7.11.1
s2. This density, defined only for positive values of x, is given by
(
)
1
1 ln (x a) m 2
exp fX (x) ¼ pffiffiffiffiffiffi
,
2
s
2p s (x a)
x . a; m, s . 0
(7:11:1)
where a is the location parameter, m is the scale parameter, and s is the shape parameter.
When a and m are zero, fX(x) is called the standard lognormal distribution with
parameter s. The case where a ¼ 0 and m ¼ 1 is shown in Fig. 7.11.1 for
s ¼ 0.4,0.5,1.0,1.5,2 for values of x . 0.1.
The cumulative distribution function FX(x) is obtained by integrating Eq. (7.11.1) and
is given by
ðx
2
1
2
pffiffiffiffiffiffi
e( ln (ja)m) =2s dj
2p
s(j
a)
0
ð ln (xa)m
s
1
1
ln (x a) m
2
pffiffiffiffiffiffi eh =2 dh ¼ erf
pffiffiffi
¼
2
2p
2s
0
FX (x) ¼
(7:11:2)
where
1
1
x
2
pffiffiffi eh =2 dh ¼ erf pffiffiffi
2
2p
2
0
ðx
The cdf FX(x) is shown in Fig. 7.11.2 for a ¼ 0, m ¼ 1 and for the same values of s as the
density function.
In finance, prices for primary instruments such as stocks and bonds are often assumed
to be lognormally distributed. A standard assumption for the evolution of a stock price is
that in an interval Dt the fractional increase in the stock price, DS/S, is normally distributed with mean mDt and variance s2Dt. The lognormal distribution is also used to model
the lifetime of units whose failures are of a fatigue– stress nature. Consequently, the
lognormal distribution is an adjunct to the Weibull distribution when attempting to
model these types of units.
114
Other Continuous Distributions
FIGURE 7.11.2
B 7.12
BETA DISTRIBUTION
The beta distribution is a two-parameter a and b family of density functions fX(x) for
0 x 1. It is often used to represent proportions and percentages. It has the following
probability density function:
fX (x) ¼
G(a þ b) a1
x (1 x)b1 ,
G(a)G(b)
0x1
(7:12:1)
The choices of the two parameters a and b makes it amenable to a variety of shapes. For
example, for a ¼ b ¼ 1, fX(x) becomes the rectangular distribution. The triangular distribution arises when one parameter has the value 2 and the other is unity. If both parameters
are less than unity and positive, the density becomes U-shaped, and if only one of the parameters is less than unity, it becomes J-shaped. The distribution is unimodal when both
parameters are positive, and symmetric when they are equal. The pdf curves are shown
in Fig. 7.12.1 for the cases described above. The beta distribution is related to the
Student-t distribution [Eq. (7.9.2)] by the following transformation. If Z is a t-distributed
FIGURE 7.12.1
7.13
Cauchy Distribution
115
FIGURE 7.12.2
random variable with n degrees of freedom, then the transformation
1
Z
X ¼ þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 2 n þ Z2
is beta-distributed with a ¼ n/2 and b ¼ n/2.
Example 7.12.1 We will give an example of how the beta distribution is used in stockmarket analysis. The analyst from his model of stock performances predicts before the
start of each day that the proportion of the listed stocks has a beta distribution with parameters a ¼ 7.3 and b ¼ 5.2. The average value is a=(a þ b) ¼ 0:584. We want to
find the probability that more than 70% of the stocks increase in value and this is given by
ð 0:7
P{X . 0:7} ¼ 1 fX (x) dx ¼ 1 0:794 ¼ 0:206
0
This probability is 20.6%, as illustrated in Fig. 7.12.2.
B 7.13
CAUCHY DISTRIBUTION
Cauchy distribution resembles a normal distribution in appearance but has much heavier
tails. However, in studying hypothesis tests that assume normality, seeing how the tests
perform on data from a Cauchy distribution is a good indicator of how sensitive the
tests are to heavy-tail departures from normality. The general Cauchy density is given by
fX (x) ¼
b
p½(x a)2 þ b2 (7:13:1)
where a is the location parameter and b is the shape parameter. It is an example of a pathological case in the sense the moments do not exist even though the first moment can be
evaluated to be equal to 0 using the Cauchy principle value method. The Cauchy
density is plotted for a ¼ 5 and b ¼ 2,3,4,5 in Fig. 7.13.1.
Since the moments do not exist for the Cauchy distribution, it is characterized by the
width at half the maximum value equal to 2b, called full width half maximum (FWHM).
For b ¼ 2 the maximum is 0.159154 and the half-maximum is 0.079577. This is shown in
Fig. 7.13.2. For comparison, a Gaussian distribution is superposed on the same figure for
the same FWHM of 4 that is, a Gaussian with mean 5 and s ¼ 4/2.35482 ¼ 1.69864.
116
Other Continuous Distributions
FIGURE 7.13.1
FIGURE 7.13.2
FIGURE 7.13.3
7.14
Pareto Distribution
The distribution function FX(x) can be determined as follows:
ðx
2
1
2
e( ln (ja)m) =2s dj
FX (x) ¼ pffiffiffiffiffiffi
2p s(j a)
0
Substituting (j a)=b ¼ tan(u) in Eq. (7.13.2), we have
ð tan1 {(xa)=b}
sec2 (u)du
FX (x) ¼
p sec2 (u)
p=2
tan1 {(xa)=b}
u 1
xa
1
¼ ¼ tan1
þ
p p=2
p
b
2
117
(7:13:2)
(7:13:3)
The term FX(x) is shown in Fig. 7.13.3 for the same values of a and b as in Fig. 7.13.1.
B 7.14
PARETO DISTRIBUTION
Many synthetic and naturally occurring phenomena, including city sizes, incomes, word
frequencies, and earthquake magnitudes, are distributed according to a power-law distribution. A power law implies that small occurrences are extremely common, whereas large
instances are extremely rare. These can be modeled as a Pareto distribution. In addition,
traffic measurements reveal features such as nonstationarity, long-range dependence,
and significant fluctuations in different timescales (e.g., self-similarity). The essential features of the burstiness observed in measurements can be modeled as a heavy-tailed on-time
and off-time distributions such as the Pareto distribution.
The Pareto density is given by
a
b aþ1
fX (x) ¼
, x.0
(7:14:1)
b xþb
and the distribution function can be obtained from integrating Eq. (7.14.1):
ðx a
b aþ1
b a
FX (x) ¼
dj ¼ 1 xþb
0b jþb
FIGURE 7.14.1
(7:14:2)
118
Other Continuous Distributions
The Pareto density and the distribution functions are plotted in Fig. 7.14.1 for a ¼ 5 and
b ¼ 10,15,20. Note that the tails are long compared to other distributions. One of the uses
of the Pareto distribution is to model the high end of the distribution of incomes.
B 7.15
GIBBS DISTRIBUTION
The Gibbs distribution has been widely used in image processing and pattern recognition
problems. An example using this distribution in tomographic imaging is presented in
Chapter 23. It is a generalization of the Maxwell – Boltzmann distribution and has
origins in statistical mechanics. It is defined on a random field N as
P(v) ¼
where,
1 ð1=TÞU(v)
e
Z
(7:15:1)
P
U(v) ¼
wjk V(vj vk )
j,keN
V(v)
Total energy function
Potential function, where v is some vector field
wjk
j, k
Weighting functions
Pixels in neighboring field N
T
Absolute temperature that controls sharpness of
distribution
P
Normalization constant given by Z ¼ v e(1=T)U(v) ,
also called the partition function
Z
The potential function V(v) depends on the local configuration of the field. Some of the
properties of potential functions [38] are given below:
1. V(0) ¼ 0
2. V(v) 0: nonnegative
3. V(v) ¼ V(v): even function
d
4.
V(v) . 0 for v . 0
dv
5. lim V(v) ! 1
v!1
6.
sup
0v,1
2
7.
B 7.16
d
dv V(v)
,1
d
V(v) . 1
dv2
MIXED DISTRIBUTIONS
There are any number of derived and mixed distributions, depending on the application,
and new distributions are also constantly being formulated. For example, the doubledouble-exponential or double-Laplace distribution, given by
f (x) ¼
p ðjxjÞ 1 p e a þ
e
2a
2b
jxj
b
(7:16:1)
7.17
Summary of Distributions of Continuous Random Variables
119
FIGURE 7.16.1
with
f1 (x) ¼
p ðjxjÞ
1p e a : f2 (x) ¼
e
2a
2b
jxj
b
is shown in Fig. 7.16.1 with p ¼ 0.3, a ¼ 1, b ¼ 4.
This distribution has been specially constructed for modeling navigation errors over
transatlantic flights consisting of both the pilot and instrument errors [51]. The pilot errors
are modeled by f2(x) and the instrument errors, by f1(x). The instrument errors have a
sharper peak than do pilot errors, which extend over a greater distance. Whereas the instrument are easier to eliminate, the pilot errors are more difficult. The composite error curve
is f(x) shown in Fig. 7.16.1.
The distributions given here are by no means exhaustive, and only some of the
more important distributions are discussed. More comprehensive sets of distributions
can be obtained from Web references W1 and W2 (see References at end of book). A
complete list of all the densities and distributions discussed in this chapter is tabulated in
Section 7.17.
B 7.17 SUMMARY OF DISTRIBUTIONS OF
CONTINUOUS RANDOM VARIABLES
Density
Function
Name
Uniform
Triangular
1
ba
xa
; a,xb
(b a)2
(x 2b þ a)
(b a)2
b , x 2b a
Closed-Form
Distribution Function
xa
ba
1 x a 2
a,xb
2 b a
1 x þ a 2b 2
1
2
ba
b , x 2b a
(continued)
120
Other Continuous Distributions
Density
Function
Name
Exponential
Laplace
Closed-Form
Distribution Function
1 lx
e u(x), l . 0
l
l ljxj
e
, l0
2
(1 elx )u(x),
1 lx
e ,
2
l.0
x0
1 12 elx , x . 0
Erlang
l(lx)n1 elx
,
(n 1)!
1
n integer
n integer; x . 0
l(lx)a1 elx
,
G(a)
a,l,x . 0
ala (x m)a1 e½l(xm)
x.m
1 ðx2 =2Þ
Standard Gaussian pffiffiffiffiffiffi e
2p
Weibull
Chi-square
k!
k¼0
x.0
Gamma
n1
X
(lx)k elx
2
(x2 )(n=2)1 ex =2s
n
(2s2 )n=2 G
2
Gx (a)
G(a)
Incomplete gamma
a
a
1 e½l(xm) ,
1 1
x
þ erf pffiffiffi
2 2
2
n
Gx2 =2s2
n 2 ,
G
2
2
x.m
x ¼ x2
Incomplete gamma
1
Chi
2(n=2)1 G
n
xn1 (1=2)(x=s)2
e
s2
2
x.0
n
Gx2 =2s2
n2
G
2
Incomplete gamma
Rayleigh
z (z2 =s2 )
e
,
s2
Maxwell
rffiffiffiffi
2 x3 ðx2 =2s2 Þ
e
,
p s3
Rice
x ½(x2 þm2 )=2s2 mx
e
I0 2 ,
s2
s
2
2
1 eðz =2s Þ , z . 0
z.0
x pffiffiffi
s 2 erf
2s2
x.0
rffiffiffiffi
2 ðx2 =2s2 Þ
e
p
x.0
x.0
No closed form
(continued)
7.17
Summary of Distributions of Continuous Random Variables
Density
Function
Name
Nakagami
Student-t(n)
Snedecor
F(n1 ,n2 )
Closed-Form
Distribution Function
2 m m 2m1 mx2 =V
x
e
G(m) V
Gmx2 =V (m)
G(m)
x.0
Incomplete gamma
vþ1
½(vþ1)=2
G
t2
2
1
þ
pffiffiffiffiffiffi v
v
vp G
2
n þ n n n1 =2
1
2
1
x(n1 2)=2
2
n2
n n n1 (n1 þn2 )=2
1
2
G
G
1þ x
2
2
n2
G
121
No closed form
No closed form
x.0
Lognormal
Beta
2
1
2
pffiffiffiffiffiffi
e½ln (xa)m =2s
2p s(x a)
G(a þ b) a1
x (1 x)b1
G(a)G(b)
1
ln (x a) m
pffiffiffi
erf
2
s 2
No closed form
0x1
Cauchy
b
p½(x a)2 þ b2 1 1 1 x a
þ tan
2 p
b
Pareto
aþ1
a
b
, x.0
b xþb
1
Definitions
Zero-order modified
Bessel function
ð
1 p x cos (u)
I0 (x) ¼
e
du
p 0
b
xþb
a
,
x.0
Error function
ð
2 x j2
erf (x) ¼ pffiffiffiffi e dj
p 0
Incomplete gamma function
ðx
Gx (a) ¼ ja1 ej dj
0
CHAPTER
8
Conditional Densities and
Distributions
B 8.1 CONDITIONAL DISTRIBUTION AND DENSITY
FOR P{A}= 0
In Section 2.2 we defined conditional probability for events A and B that are subsets of the
sample space S as
9
P{A > B}
=
P{B j A} ¼
if P{A} = 0
(8:1:1)
P{A}
;
P{A > B} ¼ P{B j A}P{A}
We will extend this to distributions FX(x) and densities fX(x). The conditional distribution
FX(x j A), given that an event A has occurred, is defined as
FX (x j A) ¼ P{X x j A} ¼
P{X x, A}
P{A}
if
P{A} = 0
(8:1:2)
and if the event fX xg is independent of A, then PfX x j Ag ¼ PfX xg. The corresponding conditional density fX(x j A) is defined by
dFX (x j A)
dx
P{x , X x þ Dx j A}
¼ lim
Dx!0
Dx
fX (x j A) ¼
Properties of Conditional Distributions
1. FX (1 j A) ¼ 1; FX (1 j A) ¼ 0
2. FX (x j A) is right-continuous
3. FX (x2 j A) FX (x1 j A) ¼ P{x1 , X x2 j A}
Ðx
4. FX (x j A) ¼ 1 fX (j j A) dj
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
122
(8:1:3)
8.1 Conditional Distribution and Density for P{A}= 0
123
Example 8.1.1 A die is tossed and, corresponding to the faces of the die, a random variable is defined by X ¼ 10k, k ¼ 1,2,3,4,5,6. The conditioning event A is given by A ¼ fthe
toss resulted in an even numberg.
We will first find the unconditional distribution FX(x) and then the conditional
distribution FX(x j A). Using the same mapping techniques discussed in Chapter 2, the
unconditional cdf is given by
FX (x) ¼
6
1X
u(x 10k)
6 k¼1
Since the sample space has been reduced to 3 points by the conditioning event A, the
conditional cdf FX(x j A) is given by
Fx (x j A) ¼
¼
P{X 10k, X even} FX (10k), k ¼ 2,4,6
¼
P{X even}
1=2
3
1X
u(x 20k)
3 k¼1
Both these distributions are shown in Fig. 8.1.1.
Conditioning Event A Depends on x. We shall now consider the case when the event A
is dependent on x. Three cases arise:
1. A ¼ {X x1 }: Eq. (8.1.2) takes the form:
FX (x j A) ¼
P{X x, X x1 }
P{X x1 }
(8:1:4)
Two situations arise:
(a) x . x1 : Since x . x1, the joint event fX x, X x1g ¼ fX x1g, and hence
Eq. (8.1.4) can be rewritten as
FX (x j A) ¼
P{X x1 }
¼1
P{X x1 }
(8:1:5)
(b) x x1: In this case the joint event fX x, X x1g ¼ fX xg, and hence
Eq. (8.1.4) becomes
FX (x j A) ¼
P{X x}
FX (x)
¼
P{X x1 } FX (x1 )
FIGURE 8.1.1
(8:1:6)
124
Conditional Densities and Distributions
The corresponding density function is given by
8
< fX (x)
fX (x j A) ¼ FX (x1 )
:
0
x x1
(8:1:7)
x . x1
2. A ¼ fx1 , X x2g with x1 , x2
Analogous to Eq. (8.1.4), we have the following:
FX (x j A) ¼
P{X x, x1 , X x2 }
P{x1 , X x2 }
(8:1:8)
Here three situations arise:
(a) x . x2:
FX (x j A) ¼
P{x1 , X x2 }
¼1
P{x1 , X x2 }
(8:1:9)
(b) x1 . x x2:
FX (x j A) ¼
P{x1 , X x}
FX (x) FX (x1 )
¼
P{x1 , X x2 } FX (x2 ) FX (x1 )
(8:1:10)
FX (x j A) ¼
P{1}
¼0
P{x1 , X x2 }
(8:1:11)
(c) x x1:
The corresponding density function is given by
8
0
x x1
>
>
<
fX (x)
x1 , x x2
fX (x j A) ¼
>
F (x ) FX (x1 )
>
: X 2
0
x . x1
Example 8.1.2
(8:1:9)
A random variable has a Gaussian distribution, given by
( " #)
1
1 x4 2
fX (x) ¼ pffiffiffiffiffiffi exp 2
2
2 2p
We will find the conditional distribution FX(x j A) and the conditional density fX(x j A)
given that (1) A ¼ fX 4.5g and (2) A ¼ f1.5 , X 4.5g. Using Eq. (8.1.6), we obtain
1. A ¼ fX 4.5g:
FX (x j A) ¼
FX (x)
1
1
x4
¼ 1 þ erf pffiffiffi
9
9 2
2 2
FX
FX
2
2
and using Eq. (8.1.10), we obtain
2. A ¼ f 1.5 , X 4.5 g:
1
x 4)
3
pffiffiffi
1 þ erf
FX
FX (x) FX (x1 )
2
2
(2
2
FX (x j A) ¼
¼
9
3
FX (x2 ) FX (x1 )
FX
FX
2
2
8.1 Conditional Distribution and Density for P{A}= 0
125
FIGURE 8.1.2
where
2
erf(x) ¼ pffiffiffiffi
p
ðx
2
ej dj
0
The corresponding density functions are as follows for (1)
( " 2 #)
1
1 x4
pffiffiffiffiffiffi exp 2
2
2 2p
fX (x j A) ¼
9
FX
2
and for (2)
( " #)
1
1 x4 2
pffiffiffiffiffiffi exp 2
2
2 2p
fX (x j A) ¼
9
3
FX
FX
2
2
The conditional distribution and densities FX(x j A) and fX(x j A) along with the
unconditional distributions and densities FX(x) and fX(x) are shown in Fig. 8.1.2
for (1) A ¼ fX 4.5g and in Fig. 8.1.3 for (2) A ¼ f1.5 , X 4.5g.
FIGURE 8.1.3
126
Conditional Densities and Distributions
B 8.2 CONDITIONAL DISTRIBUTION AND DENSITY
FOR P{A} 5 0
In Section 8.1 we assumed that the probability of the conditioning event A is not zero. We
will now determine the conditional probabilities when PfAg ¼ 0. We need to find
PfB j X ¼ x1g. Since PfX ¼ x1g ¼ 0 for a continuous random variable, we have to use limiting forms to determine this probability. We will write PfB j X ¼ x1g as
P{B j X ¼ x1 } ¼ lim P{B j x1 , X x1 þ Dx}
Dx!0
(8:2:1)
The conditional probability PfB j x1 , x x1 þ Dxg in Eq. (8.2.1) can be written as
follows:
P{B j x1 , X x1 þ Dx} ¼
P{B, x1 , X x1 þ Dx}
P{x1 , X x1 þ Dx}
¼
P{x1 , X x1 þ Dx j B}P{B}
FX (x1 þ Dx) FX (x1 )
¼
FX {x1 þ Dx j B} FX {x1 j B}P{B}
FX (x1 þ Dx) FX (x1 )
(8:2:2)
Dividing numerator and denominator of Eq. (8.2.2) by Dx and taking the limit as Dx ! 1
results in
P{B j X ¼ x1 } ¼ lim
Dx!0
¼
{FX {x1 þ Dx j B} FX {x1 j B}P{B}g=Dx
{FX (x1 þ Dx) FX (x1 )}=Dx
fX (x1 j B)P{B}
fX (x1 )
(8:2:3)
Case of B: fY 5 yg
In an analogous manner we can find an expression for the conditional probability
PfB j X ¼ xg when B is the event B:fY ¼ y). In this case we can write
P{Y ¼ y j X ¼ x} ¼ lim P{y , Y y þ Dy j x , X x þ Dx}
Dx!0
Dy!0
¼ lim
Dx!0
Dy!0
P{y , Y y þ Dy, x , X x þ Dx}
P{x , X x þ Dx}
Dividing the numerator and denominator of Eq. (8.2.4) by Dx Dy, we have,
lim P{y , Y y þ Dy j x , X x þ Dx}
Dx!0
Dy!0
½P{y , Y y þ Dy j x , X x þ Dx}=DxDy
Dx!0
½P{x , X x þ Dx=DxDy
¼ lim
Dy!0
¼
fXY (x,y)
Dy
fX (x)
(8:2:4)
8.2 Conditional Distribution and Density for P{A} 5 0
127
Hence
lim
Dx!0
Dy!0
P{y , Y y þ Dy j x , X x þ Dx}
¼ fYjX ( y j X ¼ x)
Dy
¼ fYjx ( y j x) ¼
fXY (x,y)
fX (x)
(8:2:5)
The conditional distribution FYjX( y j X xg ¼ PfY y j X xg can be obtained from the
basic definition of conditional probability as follows:
FYjX ( y j X x) ¼ P{Y y j X x} ¼
P{Y y, X xg FXY (x, y)
¼
P{X x}
FX (x)
(8:2:6)
Even though Eq. (8.2.5) is of exactly the same form as Eq. (8.2.6), it cannot be obtained
from the definition of conditional probability but must be obtained from limiting
considerations.
Point of Clarity
Since fYjX( y j x) has been defined as fYjX( y j X ¼ x), it is not the derivative with respect
to y of the distribution FYjX ( y j x), which has been defined as FYjX( y j X x). Bearing this
in mind, we can write the following equations for clarity:
dFYjX ( y j X x) dFyjX ( y j x)
¼
¼ fYjX ( y j X x)
dy
dy
(8:2:7)
dFYjX ( y j X ¼ x)
¼ fYjX ( y j X ¼ x) ¼ fYjX ( y j x)
dy
Example 8.2.1 The probability that a machine fails in x months is given by (1 2 e 2x/d)
u(x). If the machine failed sometime before c months, it is desired to find the conditional
distribution FX(x j X c) and the corresponding conditional density. Here we can use the
definition of conditional distribution and write for FX(x j X c):
FX (x j X c) ¼ P{X x j X c} ¼
¼
P{X x, X c}
P{X c}
8
1 ex=d
>
>
>
< 1 ec=d
0,xc
>
>
>
:
x.c
x0
1
0
The corresponding density function for the event fX cg is
dex=d
fX (x j X c) ¼ 1 ec=d
0
0,xc
otherwise
128
Conditional Densities and Distributions
FIGURE 8.2.1
We can now find the conditional distribution FX(x j X ¼ c) if it failed at x ¼ c
months. Since PfX ¼ cg ¼ 0, this has to be solved using limiting considerations:
FX (x j X ¼ c) ¼ lim P{X x j c , X c þ Dc}
Dc!0
¼ lim
Dc!0
P{X x, c , X c þ Dc}
P{c , X c þ Dc}
P{c , X c þ Dc j X x}P{X x}=Dc
Dc!0
P{c , X c þ D c}=Dc
¼ lim
¼ fX (c j X x)
FX (x) fX (c, X x) FX (x)
¼
fX (c)
FX (x)
fX (c)
Hence
FX (x j X ¼ c) ¼
1 x.c
0 xc
and the corresponding density function for the event fX ¼ cg is given by
fX (x j X ¼ c) ¼ d(x c)
These distribution and density functions are shown in Fig. 8.2.1 for c ¼ 2, d ¼ 2.
Example 8.2.2
The joint density function of random variables X and Y is
(1
fXY (x, y) ¼
p
0
x2 þ y2 1
otherwise
The marginal density function fX(x) is obtained by integrating fXY(x,y) over all y:
ffi
ð pffiffiffiffiffiffiffi
1x2
pffiffiffiffiffiffiffiffiffiffiffiffiffi
dy 2 1 x2
fX (x) ¼ pffiffiffiffiffiffiffiffi ¼
p
1x2 p
8.2 Conditional Distribution and Density for P{A} 5 0
129
FIGURE 8.2.2
From Eq. (8.2.5), the conditional density fYjX (y j X ¼ x) is given by
8
1
< pffiffiffiffiffiffiffiffiffiffiffiffi
fXY (x, y)
1=p
ffi x2 þ y2 1
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 2 1 x2
fYjX ( y j X ¼ x) ¼
fX (x)
2 1 x2 =p : 0
otherwise
Note that for each value of x the pdf fYjX (y j X ¼ x) is a constant and not a function of y.
The corresponding conditional
pffiffiffiffiffiffiffiffiffiffiffiffiffi distribution function FYjX (y j X ¼ x) is obtained by integrating the pdf from 1 x2 to y as shown below:
ðy
1
y
1
FYjX (y j X ¼ x) ¼ pffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi dh ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi þ
2
2
2
2
2 1x
1x 2 1 x
In Fig. 8.2.2, the conditional pdf fYjX (y j X ¼ x) is shown as a function of x and the conditional cdf FYjX (y j X ¼ x) is shown as a function of y for x = 0.5.
Example 8.2.3 This is an interesting example that again illustrates finding the conditional probability when the probability of the conditioning event is zero. Two friends
A and B plan to meet between 7:00 and 8:00 p.m. However, they agree that each of
them will wait only 15 min for the other. Each of them arrives independently at a
random time between 7 and 8 p.m. We have to compute the probability that they will
meet given that A arrives at 7:20 p.m.
Let us first define these random variables.
X ¼ {time of arrival of A after 7:00 p.m. in minutes}
Y ¼ {time of arrival of B after 7:00 p.m. in minutes}
The pdf fX(x) and fY( y) are both uniformly distributed in (0, 60] as shown in Fig. 8.2.3.
Since X and Y are independent, the joint density fXY(x,y) ¼ fX(x) . fY( y) and is given by
1
1
1
fXY(x,y) ¼ 60
60
¼ 3600
: The 60 60 Cartesian product space is shown in Fig. 8.2.4.
FIGURE 8.2.3
130
Conditional Densities and Distributions
FIGURE 8.2.4
If A comes at time x minutes after 7:00 p.m., then for the meeting to take place, we
have to find the probability of the event j Y 2 X j 15 under the condition X ¼ x.
Hence, we can write PfY 2 X j 15 j X ¼ xg ¼ Pf215 þ X Y 15 þ X j X ¼ xg, and
we have to find the area under the shaded portion in Fig. 8.2.4. In essence, we have to integrate fY( y) in three different regions: (1) 0 , y x þ 15, 0 , x ¼ 15, (2)
x 2 15 , y x þ 15, 15 , x 45, and (3) x þ 15 , y 60, 45 , x 60. Performing
the indicated integration, we obtain
8 ð
15þx
>
dx
>
>
0 , x , 15
>
>
60
>
0
>
>
< ð 15þx
dx
P{A and B meeting j X ¼ x} ¼
15 , x 45
>
60
>
15þx
>
>
ð
60
>
>
dx
>
>
45 , x 60
:
60
15þx
8
15 þ x
>
>
0 , x 15
>
>
60
>
>
<
1
¼
15 , x 45
>
2
>
>
>
>
>
: 75 x 45 , x 60
60
The probability that A and B meet if A arrives exactly x minutes after 7:00 p.m. is shown in
Fig. 8.2.5.
From the curve, the Pfmeet j X ¼ 20g ¼ 0.5. The total probability of A and B meeting
is given by
ð 60
P{meet} ¼
P{meet j x ¼ x}fX (x)
0
¼
1 60 30 15
105
7
þ þ
¼
¼ 0:4375
¼
60 4
4
4
4:60 16
8.3 Total Probability and Bayes’ Theorem for Densities
131
FIGURE 8.2.5
B 8.3 TOTAL PROBABILITY AND BAYES’ THEOREM
FOR DENSITIES
We will use Eq. (8.2.3), shown in Eq. (8.3.1) as the starting point to derive the total
probability law for density functions:
P{B j X ¼ x1 } ¼
fX (x1 j B)P{B}
fX (x1 )
(8:3:1)
Cross-multiplying and integrating both sides of the resulting equation, we obtain
ð1
ð1
P{B j X ¼ x1 }fX (x1 )dx1 ¼
1
fX (x1 j B) P{B}dx1
(8:3:2)
1
In Eq. (8.3.2) we obtain the term
ð1
ð1
fX (x1 j B)P{B}dx1 ¼ P{B}
1
fX (x1 j B)dx1 ¼ P{B}
1
and hence Eq. (8.3.2) becomes
ð1
P{B j X ¼ x}fX (x)dx ¼ P{B}
(8:3:3)
1
where the subscript on x has been omitted without loss of generality. Equation (8.3.3) is
the total probability law for density functions. We can now formulate Bayes’ theorem
for density functions. From Eq. (8.3.1) we can write
fX (x j B) ¼
P{B j X ¼ x}fX (x)
P{B}
P{B j X ¼ x}fX (x)
1 P{B j X ¼ x}fX (x)dx
¼ Ð1
(8:3:4)
132
Conditional Densities and Distributions
where fX(x j B) is the a posteriori pdf after observing the event B given the a priori pdf
fX(x). If B is the event B ¼ {Y ¼ y}, then Eq. (8.3.4) can be written as
fYjX (y j x)fX (x)
1 fYjX (y j x)fX (x)dx
fXjY (x j y) ¼ Ð 1
(8:3:5)
Equation (8.3.5) is Bayes’ theorem for connecting the a priori density fX(x) with the a
posteriori density fXj Y (x j y) after observing the event fY ¼ yg.
Example 8.3.1 Let X be the random variable that a navigation system fails before
x hours. Three suppliers A, B, and C have supplied the system to airlines with probabilities
PfAg ¼ 0.35, PfBg ¼ 0.25, PfCg ¼ 0.4. The conditional probabilities of failure for the
suppliers A, B, C are as follows:
FX (x j A) ¼ (1 eax )u(x), a ¼ 1:625 104
FX (x j B) ¼ (1 ebx )u(x), b ¼ 5:108 104
FX (x j C) ¼ (1 ecx )u(x),
c ¼ 2:877 104
1. From the conditional probabilities of failure, we can find the probability density
fX(x) as follows:
fX (x j A) ¼ aeax u(x),
a ¼ 1:625 104 : P{A} ¼ 0:35
fX (x j B) ¼ bebx u(x),
b ¼ 5:108 104 : P{B} ¼ 0:25
fX (x j C) ¼ cebx u(x), c ¼ 2:877 104 : P{C} ¼ 0:40
and
fX (x) ¼ fX (x j A)P{A} þ fX (x j B)P{B} þ fX (x j C)P{C}
¼ (0:56875eax þ 1:277ebx þ 1:1508ecx )104 u(x)
2. If the navigation system fails after x hours, we will find the probabilities PfA j X . xg,
PfB j X . xg, PfC j X . xg.
P{X . x} is given by
P{X . x} ¼ (1 FX (x j A)ÞP{A} þ (1 FX (x j B))P{B} þ (1 FX (x j C))P{C}
¼ eax 0:35 þ ebx 0:25 þ ecx 0:40
The probability PfA j X . xg can be found from the equation
P{A j X . x} ¼
P{X . x j A}P{A} (1 FX (x j A))P{A}
¼
P{X . x}
P{X . x}
8.3 Total Probability and Bayes’ Theorem for Densities
133
with similar equations for PfB j X . xg and PfC j X . xg. Hence these probabilities
can be given by
P{A j X . x} ¼
P{B j X . x} ¼
P{C j X . x} ¼
eax 0:35
eax 0:35
þ ebx 0:25 þ ecx 0:40
eax 0:35
ebx 0:25
þ ebx 0:25 þ ecx 0:40
ecx 0:40
eax 0:35 þ ebx 0:25 þ ecx 0:40
3. If the navigation system fails exactly at time X ¼ x, we will find PfA j X ¼ xg,
P{B j X ¼ x}; and P{C j X ¼ x}. From Eq. (8.2.3) we have
P{A j X ¼ x} ¼
fX (x j A)P{A} 0:56875 104 eax
¼
fX (x)
fX (x)
P{B j X ¼ x} ¼
fX (x j A)P{B} 1:277 104 ebx
¼
fX (x)
fX (x)
P{C j X ¼ x} ¼
fX (x j A)P{C} 1:1508 104 ecx
¼
fX (x)
fX (x)
4. We will find the most likely supplier in part (3) if x ranges from 1000 to 10,000 and
the total probability of a wrong decision Pe.
The most likely supplier if the system failed at exactly x hours can be obtained by drawing
the graphs of the probabilities PfA j X ¼ xg, PfB j X ¼ xg, and PfC j X ¼ xg as shown in
Fig. 8.3.1.
From the graph in Fig. 8.3.1 we can see that if the system fails between x ¼ 1000 and
x ¼ 5629.17 h, the most likely supplier is C, and if it fails between x ¼ 5629.17 and
FIGURE 8.3.1
134
Conditional Densities and Distributions
10,000 h, the most likely supplier is A. The conditional probability PfB j X ¼ xg is smaller
than the other two conditional probabilities in the two ranges.
The probability of wrong decision Pe is given by
Pe ¼ P{X 5629:17 j A}P{A} þ P{X . 5629:17 j C}P{C} þ P{B}
Since supplier B does not figure in the wrong decision, PfBg appears without any multiplier. Hence
P{wrong decision} ¼ Pe ¼ ½1 e5629:17a 0:35 þ ½1 (1 e5629:17c ) 0:4 þ 0:25
¼ 0:539
CHAPTER
9
Joint Densities and Distributions
B 9.1
JOINT DISCRETE DISTRIBUTION FUNCTIONS
If the random variables fX ¼ X1, X2, . . . , Xng, are discrete, then we can define a joint
probability mass function as
pX1 Xn (x1 , . . . , xn ) ¼ P{X1 ¼ x1 , . . . , Xn ¼ xn }
(9:1:1)
and the total sum
n
X
m
n X
n
X
j
pX1 Xn (xi , . . . , xm ) ¼ 1
i
In addition, the marginal distributions are defined by
n
X
pX1 Xn (x1 , . . . , xk ) ¼ pX1 Xn1 (x1 , . . . , xn1 )
(9:1:2)
k
Example 9.1.1 (Discrete) An urn contains 4 red, 5 green, and 6 blue balls, and 3 balls
are taken at random. Given the number of red balls represented by the random variable X
and the number of green balls by the random variable Y, we want to find the joint probability pXY (k,m) of PfX ¼ k, Y ¼ mg, where k ranges from 0 to 4 and m ranges from 0
to 5. The total number of points in the sample space is
n!
15!
n
C(n,r) ¼
¼
¼
¼ 455
r
r!(n r)! 3!12!
The number of ways in which 3 balls can be picked from 15 balls is to choose k balls from
4 red balls, m balls from 5 green balls, and 3 2 k – m balls from 6 blue balls. Hence, the
probability pXY (k,m) is given by
C(6, 3 k m)C(4, k)C(5, m) k þ m 3
pXY (k,m) ¼
0
otherwise
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
135
136
Joint Densities and Distributions
FIGURE 9.1.1
These probabilities are shown in Fig. 9.1.1
2
p(0,0)
1 6
p(1,0)
6
PXY ¼
C(15,3) 4 p(2,0)
p(3,0)
for all values of k and m.
3
p(0,1) p(0,2) p(0,3)
p(1,1) p(1,2) p(1,3) 7
7
p(2,1) p(2,2) p(2,3) 5
p(3,1) p(3,2) p(3,3)
As the probability mass functions PfX ¼ k) and PfY ¼ m) appear as sums in the margin,
they are referred to as marginal probability mass functions. Thus, the probability of 2 red
balls PfX ¼ 2g ¼ 66/455 and the probability of 3 green balls PfY ¼ 3g ¼ 10/455.
B 9.2
JOINT CONTINUOUS DISTRIBUTION FUNCTIONS
The joint distribution of n random variables fX1, X2, . . . , Xng, is the probability of the joint
event fX1 x1g, . . . , Xn xn and is given by
FX1 Xn (x1 , . . . , xn ) ¼ P{X1 x1 , . . . , Xn xn }
(9:2:1)
In particular, if X and Y are two random variables, the joint distribution function FXY (x,y)
is given by
FXY (x,y) ¼ P{X x, Y y}
(9:2:2)
The corresponding density function is given by
fXY (x,y) ¼
@2 FXY (x,y)
@[email protected]
(9:2:3)
Using the definition of the second derivative, Eq. (9.2.3) can also be given in terms of the
distribution function as follows:
FXY (x þ D x, y þ Dy) þ FXY (x,y)
FXY (x þ D x,y) þ FXY (x,y þ Dy)
lim
fXY (x,y) ¼ D x!0
Dy!0
D xDy
(9:2:4)
Similarly, the distribution function FXY (x,y) can also be given in terms of the density
function by
ðx ðy
FXY (x,y) ¼
fXY (j,h)dhdj
(9:2:5)
1 1
9.2
Joint Continuous Distribution Functions
137
The marginal density functions fX(x) and fY( y) are obtained by integrating the joint density
function FXY (x,y) over all y and x, respectively, giving
ð1
fX (x) ¼
ð1
fXY (x,y)dy;
fY ( y) ¼
1
fXY (x,y)dx
(9:2:6)
1
Properties of Joint Densities and Distributions
1. FXY (1, 1) ¼ 0; FXY (1,1) ¼ 1
2. FXY (1,y) ¼ 0; FXY (x,1) ¼ 0
3. FXY (x,1) ¼ FX (x); FXY (1,y) ¼ FY (y)
4. P{x1 , X x2 , Y y} ¼ FXY (x2 ,y) FXY (x1 ,y)
P{X x, y1 , Y y2 } ¼ FXY (x,y2 ) FXY (x,y1 )
5. P{x1 , X x2 , y1 , Y y2 }
¼ FXY (x2 ,y2 ) FXY (x2 ,y1 ) FXY (x1 ,y2 ) þ FXY (x1 ,y1 )
6. @FXY (x,y) ¼
@x
@FXY (x,y)
¼
@y
(9:2:7)
(9:2:8)
ðy
fXY (x,h)dh
1
ðx
(9:2:9)
fXY (j,y)dj
1
In many problems we are given fXY (x,y) in the region {(x1,x2], ( y1,y2]} and the distribution function FXY (x,y) is desired in the entire XY plane. In these cases, we can visualize
the XY plane and delineate the various regions as shown in Fig. 9.2.1 for which FXY (x,y)
can be determined.
Given FXY (x,y) in region I, we find that using the properties enunciated earlier:
FXY (x,y) ¼ 0 in region V, FXY (x,y) ¼ FX (x) in region II, FXY (x,y) ¼ FY ( y) in region
III, and FXY (x,y) ¼ 1 in region IV. We shall give three different examples to illustrate
how the joint distribution function can be determined.
FIGURE 9.2.1
138
Joint Densities and Distributions
Example 9.2.1 (Regions Bounded by Constants) The joint pdf fXY (x,y) is given by
K(8 x y) 1 , x 3, 1 , y 2
fXY (x,y) ¼
0
otherwise
The constant K can be found by integrating over the range, yielding
ð3 ð2
1
K(8 x y)dy dx ¼ 9K ¼ 1 or K ¼
9
1 1
We plot the regions in Fig. 9.2.2 as shown in Fig. 9.2.1.
Region V: x 1 or y 1
FXY (x,y) ¼ 0
Region IV: 2x . 3, y . 2
FXY (x,y) ¼ 1
Region I: 1 , x 3, 1 , y 2
y
ðx 1
1
h2 (8 j h)dh dj ¼
8h jh 2 1
1 19
19
ðx 1
y2
15
8y jy þ j ¼
dj
2
2
19
x
1
j2 y jy2 j2 15j 8yj þ ¼
9
2
2 1
2
2
1
j2 y jy2 j2 15j
8yj þ ¼
9
2
2
2
2
2
2
2
2
1
x y xy
x
y
15x 15y
8xy þ7
þ þ ¼
9
2
2
2
2
2
2
ðx ðy
FIGURE 9.2.2
9.2
Joint Continuous Distribution Functions
139
Region II: 1 , x 3, y . 2 [marginal distribution FX(x)]
1 x2 13x
6
FXY (x,2) ¼ FX (x) ¼
þ
9 2
2
Region III: x . 3, 1 , y 2 [marginal distribution FY ( y)]
FXY (3,y) ¼ FY ( y) ¼
1 2
y þ 12y 11
9
By differentiating the distribution function, we can write the density function for the
various regions as follows:
8
0
>
>
>
>1
>
>
>
> 9 (8 x y)
>
>
<1
(12 2y)
fXY (x,y) ¼
9
>
>
>
>
1 13
>
>
x
>
>
>
9 2
>
:
0
x1
or y 1
1 , x 3, 1 y 2
x . 3, 1 y 2
1 , x 3, y . 2
x . 3, y . 2
We can also conclude that since fXY (x,y) = fX (x) fY ( y), the random variables X and Y
are not independent. Having determined the joint probability FXY (x,y), we can find the
probability for other regions such as f1.5 , X 2.5, 1 , Y 1.5g as shown in
Fig. 9.2.3.
ð 2:5 ð 1:5
P{1:5 , X 2:5, 1 , Y 1:5} ¼
1:5
1
1
(8 x y)dy dx
9
¼ FXY (2:5, 1:5) FXY (1:5, 1:5)
FXY (2:5, 1) þ FXY (1:5, 1)
¼ 0:4167 0:1528 0 0 ¼ 0:2639
FIGURE 9.2.3
140
Joint Densities and Distributions
We can also find the conditional probability of Y y given that X x as follows
FXY (x,y)
FX (x)
1
x2 y xy2 x2 y2 15x 15y
8xy þ7
þ þ 9
2
2
2
2
2
2
2
¼
1 x
13x
6
þ
9 2
2
FY j X (y j X x) ¼
¼
y2 þ xy x 15y þ 14
x 12
At the endpoints x ¼ 3 and y ¼ 2, we have FY j X ( y j X 3) ¼ FY ( y) and FY j X
(2 j X 3) ¼ FY ( y) ¼ 1.
The
corresponding
conditional
density
function
fY j X( y j X x) is given by
fY j X (y j X x) ¼
x þ 2y 15
,
x 12
1,x3
1,y2
We will now find the conditional density fY j X ( y j X ¼ x) from Eq. (8.2.5)
1
fXY (x,y) 9 (8 x y) 2(8 x y)
¼
¼ fY j X (y j X ¼ x) ¼
1 13
fX (x)
13 2x
x
9 2
in the region 1 , x 3, 1 , y 2. The corresponding distribution function FY j X
( y j X ¼ x) in the region 1 , x 3, 1 , y 2 is obtained by integrating the density
function and is given by
ðy
2(8 x h)
dh
FY j X (y j X ¼ x) ¼
13 2x
1
¼
y2 þ 2xy 2x 16y þ 15
2x 13
We see clearly from this example that FY j X ( y j X x) is different from FY j X ( y j X ¼ x)
and the corresponding conditional density functions fY j X ( y j X x) and fY j X ( y j X ¼x)
are also different. Note that fY j X ( y j X ¼ x) has been defined as fY j X ( y j x) in Chapter 8.
Example 9.2.2 (One Region Bounded by a Function) In this example one of the
bounds on the region is a functional inequality. The joint pdf fXY (x,y) of two random variables X and Y is given by
(8
(1 þ 3xy) 0 , x y 1
fXY (x,y) ¼ 7
0
otherwise
We have to find the joint distribution function FXY (x,y) in the entire xy plane. Integrating
first along the h axis and then along the j axis (Fig. 9.2.4), we have
ðx ðy
8
FXY (x,y) ¼
(1 þ 3jh)dh dj
0 j7
¼
1
3x4 þ 6x2 y2 4x2 þ 8xy ,
7
0,xy1
9.2
Joint Continuous Distribution Functions
141
FIGURE 9.2.4
Integrating along the j axis first and then along the h axis (Fig. 9.2.4), we have
ðx ðh
8
(1 þ 3jh)dj dh
FXY (x,y) ¼
0 0 7
I
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl
ffl}
ðy ðx
8
(1 þ 3jh)dj dh
þ
x 07
II
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
1
1
¼ ð3x4 þ 4x2 Þ þ ð6x4 þ 6x2 y2 8x2 þ 8xyÞ
7
7
1
¼ ð3x4 þ 6x2 y2 4x2 þ 8xyÞ; 0 , x y 1
7
Example 9.2.3 (Both Regions Bounded by Functions) The joint probability density
function of two random variables X and Y is given by
8
2
>
0,x1
>
>
<
6
k(1 þ 2xy) 4
1
fXY (x,y) ¼
0 , x2 , y x þ 1
>
>
2
>
:
0
otherwise
The two boundaries are shown in Fig. 9.2.5.
FIGURE 9.2.5
142
Joint Densities and Distributions
The constant k is found by integrating fXY (x,y) over the ranges of definition as shown
below:
ð 1=2 ð xþ1=2
ð1 ð1
k(1 þ 2xy)dy dx þ
k(1 þ 2xy)dy dx ¼ 1
0
x2
1=2 x2
Solving the above equation we obtain,
k
161
¼1
192
or k ¼
192
161
Method 1: Marginal Distributions by Integrating fXY (x,y). We will now find the
marginal density fX (x) by integrating fXY (x,y) over all y. In the region 0 , x 12,
we have
ð xþ1=2
192
(1 þ 2xy)dy
fX (x) ¼
161
x2
48 1
¼
2 þ 5x þ 4x3 4x 5 , 0 , x 161
2
In the region 12 , x 1, we have
ð1
192
(1 þ 2xy)dy
x2 161
192 ¼
1 þ x x2 x5 ,
161
fX (x) ¼
1
,x1
2
By integrating fX (x) in these two regions, we can write the distribution function
FX(x) as
8
8
1
>
>
(12x þ 15x2 þ 16x4 4x6 )
0,x
>
>
161
2
>
<
31 32
1
2
3
6
FX (x) ¼
þ
(6x þ 3x 2x x )
,x1
> 161 161
2
>
>
>
x.1
>
:1
0
otherwise
Similarly, we integrate fXY (x,y) over all x to obtain the marginal density fY ( y) as
shown below:
8 ð pyffiffi
192
1
>
>
>
(1 þ 2xy)dx,
0,y
<
161
2
fY ( y) ¼ ð 0pyffiffi
>
192
1
>
>
(1 þ 2xy)dx,
,y1
:
161
2
y1=2
or
8
192 pffiffiffi
>
>
y y2
>
< 161
pffiffiffi
fY ( y) ¼ 16 6 þ 12 y 15y þ 24y2 12y3
>
>
> 161
:
0
0,y
1
2
1
,y1
2
otherwise
9.2
Joint Continuous Distribution Functions
143
Integrating fY ( y) in the above regions we obtain the distribution function FY ( y) as
shown below:
8
64 3=2
>
>
2y þ y3
>
> 161
>
<
1 16
FY ( y) ¼
þ
(6y þ 8y3=2 15y2 þ 8y3 3y4 )
>
7
161
>
>
>
>
:1
0
0,y
1
2
1
,y1
2
y.1
otherwise
fX(x), FX (x), fY ( y) and FY ( y) are shown in Fig. 9.2.6.
Method 2: Marginal Distributions from FXY (x,y). We will now find the marginal distributions from the joint distribution FXY (x,y) obtained by integrating the joint
density fXY (x,y) over the regions of definition.
The joint distribution FXY (x,y) in the region 0 , x 12, 0 , y 12 is obtained from
the following integration:
ðx ðy
FXY (x,y) ¼
0
x2
192
(1 þ 2jh)dh dj
161
32 ¼
6xy 2x3 þ 3x2 y2 x6 ,
161
0,x1
0 , y 1=2
and in the region 0 , x 1, 12 , y 1, we have
ð y(1=2) ð jþ(1=2)
FXY (x,y) ¼
0
ðx
j2
ðy
þ
y(1=2) j
2
192
(1 þ 2jh)dh dj
161
192
(1 þ 2jh)dh dj
161
or
FXY (x,y) ¼
1
8 þ
12y 15y2 þ 8y3 6y4
7
161
2 2
3
þ 24xy þ 12x y 8x 4x
FIGURE 9.2.6
6
(
0,x,1
1
,y1
2
144
Joint Densities and Distributions
The marginal distributions FX (x) and FY ( y) can be obtained by substituting suitable
boundary values in the joint distribution FXY (x,y) as shown below:
FX (x) ¼
8
1
>
>
F
x,x
þ
,
>
XY
>
>
2
>
>
>
>
>
< FXY (x,1),
>
>
>
FXY (1,1),
>
>
>
>
>
>
>
: FXY (0,0),
1
1
0,x , y.
2
2
1
1
, x 1, y .
2
2
1
x . 1, y .
2
1
x 0, y .
2
Similarly
8
pffiffiffi
>
>
FXY ( y,y),
>
>
>
<
pffiffiffi
FY (y) ¼ FXY ( y,y),
>
>
>
>
>
: FXY (1,1),
FXY (0,0),
1
0,y , 0,x1
2
1
, x 1, 0 , x 1
2
y . 1, x . 1
y 0, x . 1
and these are exactly the same as derived previously.
B 9.3
BIVARIATE GAUSSIAN DISTRIBUTIONS
If X and Y are jointly distributed random variables with mean value mX and mY and variances s2X and s2Y, then the joint density function fXY (x,y) is given by
(
"
1
1
x mX 2
pffiffiffiffiffiffiffiffiffiffiffiffiffi exp fXY (x,y) ¼
2(1 r2 )
sX
2psX sY 1 r2
#)
2r(x mX )(y mY )
y mY 2
þ
(9:3:1)
sX sY
sY
where r is the correlation coefficient that determines the degree of similarity between X
and Y, given by
sXY
r¼
(9:3:2)
sX sY
We have to show that
ðð 1
I¼
FXY (x,y)dy dx ¼ 1
(9:3:3)
1
Substituting j ¼ (x 2 mX)/sx and h ¼ ( y 2 mY)/sy in Eq. (9.3.1), we can write the
integral I as follows:
1
1
pffiffiffiffiffiffiffiffiffiffiffiffiffi exp
½j2 2rjh þ h2 dh dj
2
2(1
r2 )
1 2p 1 r
ðð 1
I¼
(9:3:4)
145
9.3 Bivariate Gaussian Distributions
Completing the squares in j in the integral (9.3.4) above, we have
ðð 1
1
1 (j rh)2
pffiffiffiffiffiffiffiffiffiffiffiffiffi exp
I¼
þ h dh dj
2
1 r2
1 2p 1 r2
(9:3:5)
Substituting
j rh
w ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi
1 r2
and
dj
dw ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi
1 r2
the integral (9.3.5) can be written as follows:
2
2
ð1
ð
1
w
1 1
h
pffiffiffiffiffiffi exp
exp
I¼
dw
dh ¼ 1
2p 1
2
2
2p
1
(9:3:6)
If r ¼ 0 in Eq. (9.3.3), then X and Y are independent because fXY (x,y) ¼ fX (x) fY ( y). Thus,
if Gaussian-distributed random variables X and Y are uncorrelated, they are also
independent.
We shall find the conditional pdf fY j X (X ¼ x) given X ¼ x. From Eq. (8.2.5) we have
fXY (x,y)
fX (x)
8
9
1
1
x mX >
>
>
>
pffiffiffiffiffiffiffiffiffiffiffiffiffi exp >
>
>
>
2
>
>
2(1 r2 )
sX
1
r
2ps
s
>
>
X
Y
>
>
>
>
>
>
2
>
>
>
>
(2r(x
m
)(
y
m
)
y
m
X
Y
Y
>
>
>
>
þ
=
<
sX sY
sY
¼
2
>
>
1
1 x mX )
>
>
>
>
pffiffiffiffiffiffi exp >
>
>
>
>
>
2
s
2p
s
X
>
>
X
>
>
>
>
>
>
>
>
>
>
>
>
;
:
fY j X (y j X ¼ x) ¼
(
"
2 )
1
1
rsY
¼ pffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffi exp
(x mX Þ
y mY
sX
2s2Y (1 r2 )
2psY 1 r2
Thus the conditional density has a conditional mean value of
rsY
mY j X ¼ mY þ
(x mX )
sX
and a conditional variance of s2Y j X ¼ s2Y (1 r2 ):
CHAPTER
10
Moments and Conditional
Moments
B 10.1
EXPECTATIONS
The expected value or the mean value of any random variable X is defined by
ð1
E ½X ¼ mX ¼
x fX (x) dx
(10:1:1)
1
This definition is valid regardless of whether the random variable is continuous or discrete.
Since the pdf of a discrete random variable is always given by a sum of delta functions,
we have
)
ð 1 (X
1
E½X ¼ mX ¼
x
P{X ¼ xi }d(x xi ) dx
1
¼
1
X
i¼0
xi P{X ¼ xi }
(10:1:2)
i¼0
The expected value is the first moment of the random variable X about the origin.
Example 10.1.1
For a random variable uniformly distributed in the interval (a,b], we have
fX (x) ¼
1
ba
and hence the mean value is given by
E½X ¼
1
ba
ðb
x dx ¼
a
b
1 x2 1 b2 a2 b þ a
¼
¼
b a 2 a 2 b a
2
Example 10.1.2 On a toss of a die, a person wins $10 if {1} or {3} results, loses $5 if
{4} or {6} results, wins $5 if {2} results, and loses $10 if {5} results. From Eq. (10.1.2),
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
146
10.1 Expectations
147
the expected value of the winnings is
6
X
E{X} ¼ mX ¼
xi P{X ¼ xi }
i¼1
1
5
¼ {10 þ 10 5 5 þ 5 10} ¼
6
6
The expected value of any random variable is a constant. In other words, it smooths the
variations of a random variable to a constant. It is a one-way operator, in the sense that
once the random variable X is smoothed, there is no way of getting back the density function.
Thus, the expected value of a constant a is the same constant a.
Properties of Expectations
1. It is a linear operator irrespective of the nature of random variables, in the sense
that the expected value of the sum equals the sum of expected values
E½X þ Y ¼ E½X þ E½Y
(10:1:3)
as shown below:
ð1 ð1
E½X þ Y ¼
(x þ y)fXY (xy)dx dy
1 1
ð1 ð1
ð1 ð1
xfXY (xy)dx dy þ
¼
1 1
ð1
ð1
xfX (x)dx þ
¼
1
yfXY (xy)dx dy
1 1
yfY (y)dy ¼ E{X} þ E{Y}
1
2. On the contrary, the expected value of the product given by
ð1 ð1
E½XY ¼
xyfXY (xy)dx dy
(10:1:4)
1 1
is not, in general, equal to the product of expected values. However, Equation
(10.1.4) can be split only if X and Y are independent, in which case we have
ð1 ð1
E½XY ¼
xy fXY (xy)dx dy
1 1
ð1
ð1
xfX (x)dx
¼
1
yfY (y)dy
1
¼ E½X E½Y
and the expected value of the product is equal to the product of expected values. The
converse is not true. If E[XY ] ¼ E[X ] E[Y ], then X and Y are uncorrelated and
need not be independent. Analogously, if E[XY ] ¼ 0, then X and Y are orthogonal.
We can summarize the following definitions:
Uncorrelated: E[XY ] ¼ E[X ] E[Y ]
Orthogonal: E[XY ] ¼ 0
Independence: fXY (xy) ¼ fX (x) fY (y)
148
Moments and Conditional Moments
Conditional Expectation
Given that the random variable X has occurred, the expected value of Y is defined by
ð1
E½Y j X ¼
yfYjX ( y j x)dy
(10:1:5)
1
The conditional expectation operator does not completely smooth the random variable Y to
a constant but smooths it to random variable dependent on X. Thus, if we take the expectation of the conditional expectation, we get the expected value of Y:
ð 1 ð 1
E{E½Y j X} ¼
y fY j X ( y j x) dy fX (x)dx
1
1
ð1
y fXY (xy) dy dx
¼
1
ð1
y fY ( y)dy ¼ E½Y
¼
(10:1:6)
1
If X and Y are independent, then E[Y j X ] ¼ E[Y ] since fY j X ( y j x) ¼ fY ( y).
Example 10.1.3 A random variable Y is uniformly distributed in the interval (0,5}.
Hence the distribution function FY( y) is given by y/5 for 0 , y 5. We will find the
conditional expectation of Y given that Y . x and 0 , x 5. We first find the conditional
distribution FY ( y j Y . x):
FY ( y j Y . x) ¼
P(x , Y y) FY ( y) FY (x)
¼
P(Y . x)
1 FY (x)
Differentiating this equation with respect to y, the conditional density fY( y j Y . x) is
obtained as follows:
fY (y) ¼
fY ( y)
1
¼
1 FY (x) 5 x
Hence the conditional expectation E[Y j Y . x] is given by
5
y
y2 E½Y j Y . x ¼
dy ¼
2(5 x)x
x5 x
ð5
¼
25 x2
5þx
¼
2
2(5 x)
Joint Moment
We can now define the joint moment between two random variables X and Y as follows:
ð1 ð1
E½XY ¼ mXY ¼
xy fXY (x,y) dx dy
(10:1:7)
1
1
The joint moments determine the correlation between X and Y.
10.2
B 10.2
Variance
149
VARIANCE
The variance is the second moment about the mean value also known as the second central
moment. It is defined by
ð1
var½X ¼ E½X mX 2 ¼ s2X ¼
(x mX )2 fX (x) dx
1
2
¼ E½X 2mX E½X þ
m2X
¼ E½X 2 m2X
(10:2:1)
The variance expresses the variability of the random variable about the mean value. If this
variability is zero, then the random variable will be no longer random but a constant. This
will be discussed further in Chapter 14.
The preceding definition applies for any random variable. However, for a discrete
random variable, analogous to Eq. (10.1.2), we have
E½X mX 2 ¼ sX2 ¼
1
X
(xi mX )2 P{X ¼ xi )
(10:2:2)
i¼0
Example 10.2.1 For a random variable, X uniformly distributed between (a,a, the
mean value mX ¼ 0. Hence the variance is given by
ð
1 a 2
a2
2
2
sX ¼ E½X ¼
x dx ¼
2a a
3
Example 10.2.2 From Eq. (10.2.2), the variance of the random variable given in
Example 10.1.2 is
6 X
5 21
s2X ¼
xi 6 6
i¼1
2
1 55
552 252 252 252 552
10,950
¼
þ
þ
þ
þ
þ
¼
6 62
216
62
62
62
62
62
Properties of Variance
1.
2.
var½aX ¼ a2 var½X ¼ a2 s2X
(10:2:3)
var½X + Y ¼ E½X + Y2 {E½X þ E½Y}2
¼ E½X 2 {E½X}2 þ E½Y 2 {E½Y}2
+ 2{E½XY E½XE½Y}
¼ sX2 þ s2Y + 2{E½XY E½XE½Y}
(10:2:4)
3. If X and Y are independent, then
var½X + Y ¼ s2X þ s2Y
(10:2:5)
Conditional Variance. We can also define a conditional variance of Y given X has
occurred as
var½Y j X ¼ E½(Y mYjX )2 j X
(10:2:6)
150
Moments and Conditional Moments
Covariance. The covariance between two random variables X and Y is defined by
cov½XY ¼ sXY ¼ E ½X mX E ½Y mY ¼ E ½XY mX mY
(10:2:7)
If X and Y are independent, then cov[XY ] ¼ 0. The converse is not true. If
cov[XY ] ¼ 0, then X and Y are uncorrelated but not necessarily independent.
Correlation Coefficient. The correlation coefficient is the normalized covariance and
is defined by
cov½XY
sXY
rXY ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼
var½X var½Y sX sY
(10:2:8)
The correlation coefficient satisfies 1 rXY 1 and shows the degree of selfsimilarity between X and Y.
B 10.3 MEANS AND VARIANCES OF SOME
DISTRIBUTIONS
Discrete Random Variables
Binomial Distribution
Mean:
n k nk
b(k; n, p) ¼
pq
k
E½X ¼
n
X
k¼0
¼
k
(10:3:1)
n!
pk qnk
k!(n k)!
n
X
n(n 1)!
p pk1 qnk
(k
1)!(n k)!
k¼0
¼ np
n
X
(n 1)!
pk1 qnk
(k
1)!(n
k)!
k¼1
¼ np( p þ q)n1 ¼ np
(10:3:2)
Variance:
var½X ¼ E½X 2 m2X
E½X 2 ¼
n
X
k¼0
k2
n!
pk qnk
k!(n k)!
(10:3:3)
(10:3:4)
Substituting k 2 ¼ k (k 2 1) þ k in Eq. (10.3.4), we have
E½X 2 ¼
n
X
k¼0
k(k 1)
n
X
n!
n!
pk qnk þ
pk qnk
k
k!(n k)!
k!(n
k)!
k¼0
(10:3:5)
10.3
Means and Variances of Some Distributions
151
The second summation in Eq. (10.3.5) is the mean value, already calculated as np in
Eq. (10.3.2). The first summation can be written as follows:
n
X
k(k 1)
k¼0
n
X
n!
(n 2)!
pk qnk ¼ n(n 1) p2
pk2 qnk
k!(n k)!
(k
2)!(n k)!
k¼2
¼ n(n 1) p2 (p þ q)n2 ¼ n(n 1)p2
(10:3:6)
Thus
E½X 2 ¼ n (n 1) p2 þ np
(10:3:7)
and the variance from Eq. (10.3.3) is given by
var½X ¼ n (n 1) p2 þ np n2 p2 ¼ npq
(10:3:8)
Poisson Distribution
Mean:
P(k; l) ¼ el
E½X ¼ el
lk
k!
1
X
klk
k¼0
k!
(10:3:9)
¼ lel
1
X
lk 1
¼ l el el ¼ l
(k
1)!
k¼1
(10:3:10)
Variance. We use the same technique as in the binomial distribution by substituting
k 2 ¼ k(k 2 1) þ k. Hence
E½X 2 ¼ el
1
X
k(k 1)lk
k¼0
¼ l2 el
k!
þ el
1
X
klk
k¼0
k!
1
X
lk 2
þl
(k 2)!
k¼2
¼ l2 el el þ l ¼ l2 þ l
(10:3:11)
Substituting in var½X ¼ E½X 2 m2X from Eqs. (10.3.11) and (10.3.10), we have
var ½X ¼ l2 þ l l ¼ l
(10:3:12)
For a Poisson distribution, the mean and the variance are the same and equal
to l.
Continuous Random Variables
Exponential Distribution
Mean:
fX (x) ¼ l elx u(x)
ð1
E½X ¼ l
xelx dx
0
(10:3:13)
(10:3:14)
152
Moments and Conditional Moments
Integrating Eq. (10.3.14) by parts, we obtain
ð1
ð1
delx
¼
E½X ¼ l
x
x delx
l
0
0
1 ð 1
1
lx ¼ xe þ
delx ¼
l
0
0
Variance. We first calculate the second moment:
ð1
E½X 2 ¼ l
x2 elx dx
(10:3:15)
(10:3:16)
0
Integrating by parts [Eq. (10.3.16)] twice, we get the result as
ð1
l
x2 elx dx ¼
0
lx
1
e
2 2
¼ 2
(l
x
þ
2lx
þ
2)
2
l2
l
0
(10:3:17)
2
1
1
¼
l2 l2 l2
(10:3:18)
and the variance is given by
var½X ¼
Gaussian Distribution
Mean:
1
1 (x mX )2
pffiffiffiffiffiffi exp 2
s2X
sX 2p
ð1
1
1 (x mX )2
x exp dx
E½X ¼ pffiffiffiffiffiffi
2
s2X
sX 2p 1
ð1
1
1 (x mX )2
¼ pffiffiffiffiffiffi
(x mX ) exp dx
2
s2X
sX 2p 1
ð1
m
1 (x mX )2
exp dx
þ pXffiffiffiffiffiffi
2
s2X
sX 2p 1
fX (x) ¼
(10:3:19)
(10:3:20)
In Eq. (10.3.20), the first integral is zero because it is an odd function and the second
integral equals mX. Hence
E½X ¼ mX
(10:3:21)
Variance:
1
var½X ¼ pffiffiffiffiffiffi
sX 2p
1 (x mX )2
(x mX ) exp dx
2
s2X
1
ð1
2
Substituting y ¼ (x mX )=sX in Eq. (10.3.22), we have
ð
1 1 2 2 (1=2)y2
sX y e
dy ¼ s2X
var½X ¼ pffiffiffiffiffiffi
2p 1
pffiffiffiffiffiffi Ð 1
2
where it can be shown that 1= 2p 1 y2 e(1=2)y dy ¼ 1
(10:3:22)
(10:3:23)
10.4
B 10.4
Higher-Order Moments
153
HIGHER-ORDER MOMENTS
We can now define higher-order moments. The kth-order moment is defined by
E½X k ¼ mk ¼
ð1
xk fX (x) dx
(10:4:1)
1
Clearly, m0 ¼ 1 and m1 ¼ mX. Similarly, the kth-order central moments are defined by
E½X mkX ð1
(x mkX ) fX (x) dx
¼ vk ¼
(10:4:2)
1
Here v0 ¼ 1, v1¼0, and v2 ¼ s2X.
We can find relationships between the kth-order moments and the kth-order central
moments. We can write these relationships as follows:
v0 ¼ 1
v1 ¼ 0
v 2 ¼ s2
v3 ¼ m3 3m1 m2 þ 2m31
m0 ¼ 1
m1 ¼ m
m2 ¼ v2 þ m2
m3 ¼ v3 þ 3m1 v2 þ m31
(10:4:3)
The moments of random variables are not arbitrary numbers but must satisfy certain
inequalities. For example, since E½X mX 2 must be nonnegative, we have
s2 ¼ m2 m21 0
and
m2k m2k 0
for all k
(10:4:4)
A random variable is completely determined if the density function is known.
However, knowledge of only two moments gives only a limited information of a
random variable. But under certain conditions, knowledge of moments mk for all k will
uniquely determine the random variable. However, because only two of the moments of
a Gaussian random variable are independent, it can be completely determined with two
moments. All other moments are dependent on two moments. We will show that for a
Gaussian random variable with density function
1
2
2
fX (x) ¼ pffiffiffiffiffiffi ex =2s
s 2p
satisfies the following moment equations:
0
k
E½X ¼
1 3 5 (n 1)sn
n ¼ 2k þ 1 for all k
n ¼ 2k for all k
(10:4:5)
(10:4:6)
function.
To show the result for
Clearly the odd moments are 0 because fXÐ(x) is an even p
ffiffiffiffiffiffiffiffi
2
1
the even moments, we take the equation 1 ebx dx ¼ p=b and differentiate it k time
with respect to b, yielding
rffiffiffiffiffiffiffiffiffiffiffi
ð1
1 3 5 (2k 1)
p
2k bx2
x e
dx ¼
(10:4:7)
k
2kþ1
2
b
1
and the result is obtained by substituting b ¼ 1/2s2 in Eq. (10.4.7). Thus, the Gaussian
random variable is completely determined if only two moments are known.
154
Moments and Conditional Moments
Higher-Order Cross-Moments
The concept of joint moments can be extended to higher-order cases. The k, jth
higher-order cross-moment is defined by,
ð1 ð1
k
j
E½X , Y ¼ mkj ¼
xk y j fXY (x,y) dx dy
(10:4:8)
1
1
and the k, jth higher order central moment is defined by
ð1 ð1
k
j
E½(X mX ) , (Y mY ) ¼ hkj ¼
(x mX )k ( y mY ) j fXY (x,y) dx dy
1
(10:4:9)
1
and m11 ¼ E [XY ] the joint moment and h11 ¼ cov[XY ] ¼ sXY is the covariance.
B 10.5
BIVARIATE GAUSSIAN
Two random variables are jointly Gaussian-distributed if they satisfy the following density
function with r as the correlation coefficient:
(
"
1
1
x mX 2
pffiffiffiffiffiffiffiffiffiffiffiffiffi exp
fXY (x,y) ¼
2(1 r2 )
sX
2psX sY 1 r2
#)
(x mX )(y my )
y mY 2
þ
(10:5:1)
2r
sY
s X sY
Ordinarily, uncorrelatedness does not imply independence, even though independence
implies uncorrelatedness. However, in the case of Gaussian-distributed random variables
if r ¼ 0, then Eq. (10.5.1) becomes
( "
#)
1
1 x mX 2 y mY 2
fXY (x,y) ¼
exp
þ
(10:5:2)
2psX sY
2
sX
sY
and clearly
fXY (x,y) ¼ fX (x) fY (y)
( ( )
)
1
1 x mX 2
1
1 x mY 2
pffiffiffiffiffiffi exp
¼ pffiffiffiffiffiffi exp
2
2
sX
sY
2psX
2psY
and hence X and Y are independent.
(10:5:3)
CHAPTER
11
Characteristic Functions and
Generating Functions
B 11.1
CHARACTERISTIC FUNCTIONS
The characteristic function (CF) of any random variable X (discrete or continuous) is
defined by
ð1
FX (v) ¼ E{e jvX} ¼
e jvx fX (x) dx
(11:1:1)
1
From this definition it is seen that the characteristic function is the Fourier transform of the
pdf fX (x) where the sign of v is positive instead of negative. However, all the properties of
the Fourier transform hold good. From the definition of FX (v), we can see that
FX (0) ¼ 1
(11:1:2)
This equation is used to check the correctness of the CF.
Existence of CF
The CF is bounded as shown below. Applying the fact that the absolute value of an
integral is less than or equal to the integral of its absolute value, we have
ð1
ð 1
jvx
e fX (x) dx je jvx jj fX (x)j dx
1
and since je
jvx
1
j ¼ 1 we have
ð1
ð 1
jvx
e fX (x) dx je jvx jj fX (x)j dx
1
1
ð1
¼
j fX (x)jdx ¼ 1
1
Thus, the CF always exists.
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
155
156
Characteristic Functions and Generating Functions
We can also obtain the pdf fX(x) from the CF from the inversion formula for the
Fourier transform given by
ð
1 1
FX (v)ejvx dv
(11:1:3)
fX (x) ¼
2p 1
[Comment: Since FX(v) is the Fourier transformation of the pdf fX(x), the random variable X is also completely determined if FX(v) is known. A Fourier transform of a signal characterizes the frequency content of the signal. No such characterization for the CF can be given; it
is just a convenient form of representing the pdf and has other very useful properties.]
Example 11.1.1
We will find the CF of an exponential distribution given by
fX (x) ¼ lelx u(x)
Using Eq. (11.1.1), we have
ð1
FX (v) ¼
lelx e jvx dx ¼
0
l
l jv
Moment Generating Properties of CF
In Chapter 10 we found the moments of a random variable using integration techniques. Once the CF is known, the moments can be obtained by differentiation, which
is a simpler process than integration. Expanding the CF as a Taylor series about the
origin, we can write
ð1
jvx ( jvx)2 (jvx)3
( jvx)k
þ
FX (v) ¼
þ
þ þ
þ dv
(11:1:4)
fX (x) 1 þ
1!
2!
3!
k!
1
Substituting
ð1
mk ¼
xk fX (x) dx,
k ¼ 0,1; . . .
(11:1:5)
1
in Eq. (11.1.2), we have
jvm1 ( jv)2 m2 ( jv)3 m3
( jv)k mk
FX (v) ¼ 1 þ
þ
þ
þ þ
þ 1!
2!
3!
k!
We now differentiate FX(v) term by term to yield
d
jm1 2( j)2 vm2 3( j)3 v2 m3
k( j)k vk1 mk
FX (v) ¼
þ
þ
þ þ
þ dv
1!
2!
3!
k!
(11:1:6)
Substituting v ¼ 0 in Eq. (11.1.6), we get
F0X (v)v¼0 ¼ jm1
and m1 ¼
F0X ð0Þ
j
(11:1:7)
Differentiating FX(v) successively k times and substituting in the resulting equations
v ¼ 0 we can obtain all the moments of the random variables as shown below:
(k)
2
k
F(2)
(11:1:8)
X (v) v¼0 ¼ ( j) m2 , . . . , FX (v) v¼0 ¼ ( j) mk , . . . ,
We will give more examples in a later section.
11.2
Examples of Characteristic Functions
157
Joint Characteristic Functions
We can also define a joint characteristic function for two random variables X and
Y as
FXY (v1 , v2 ) ¼ E e j(v1 Xþv2 Y)
ð1 ð1
¼
fXY (x, y)e j(v1 xþv2 y) dx dy
(11:1:9)
1 1
B 11.2
EXAMPLES OF CHARACTERISTIC FUNCTIONS
We will now give several examples of characteristic functions for both discrete and
continuous random variables and determining their means and variances.
Discrete Random Variables
Example 11.2.1 (Binomial Distribution)
n k nk
pq
b(k: n,p) ¼
k
(11:2:1)
The CF is given by
n X
n k nk jvk
pq e
k¼0 k
n X
n
¼
pe jvk qnk ¼ ( pe jv þ q)n
k
k¼0
FX (v) ¼
(11:2:2)
We find that FX(0) ¼ ( p þ q) ¼ 1
Mean:
F0X (v) ¼ n( pe jv þ q)n1 jpe jv
F0X (0) ¼ jnp
or E{X ¼ k} ¼ m1 ¼ np
(11:2:3)
Variance. We first find EfX 2 ¼ k 2g ¼ m2:
F00X (v) ¼ n(n 1)( pe jv þ q)n2 ( jpe jv )2 þ n( pe jv þ q)n1 j2 pe jv
F00X (0) ¼ n(n 1)p2 np and
m2 ¼ (np)2 þ npq
Hence
var½X ¼ m2 m21 ¼ npq
(11:2:4)
158
Characteristic Functions and Generating Functions
Example 11.2.2 (Poisson Distribution)
p(k; l) ¼ el
FX (v) ¼
1
X
lk
k!
el
lk jvk
e
k!
el
le jv
jv
jv
¼ el ele ¼ el(1e Þ
k!
k¼0
¼
1
X
(11:2:5)
k
k¼0
(11:2:6)
and FX(0) ¼ 1.
Mean:
jv
F0X (v) ¼ el(1e ) l(e jv )j
(11:2:7)
F0X (0) ¼ lj or E{X ¼ k} ¼ m1 ¼ l
Variance. We first find EfX 2 ¼ k2g ¼ m2:
jv
jv
F00X (v) ¼ el(1e ) l(e jv )j l(e jv )j el(1e ) lj(e jv )j
F00X (0) ¼ l2 l or
m2 ¼ l2 þ l
Hence
var{X} ¼ m2 m21 ¼ l
(11:2:8)
Example 11.2.3 (Geometric Distribution)
p{k : p} ¼ P{k failures followed by a success} ¼ qk p
FX (v) ¼
1
X
pqk e jvk ¼ p
k¼0
1
X
(11:2:9)
(qe jv )k
k¼0
¼
p
1 qe jv
(11:2:10)
and FX(0) ¼ 1.
Mean:
F0X (v) ¼
p
jpqe jv
jv
(jqe
)
¼
(1 qe jv )2
(1 qe jv )2
F0X (0) ¼
jpq
p2
or E{x ¼ k} ¼ m1 ¼
q
p
(11:2:11)
Variance. We first find EfX 2 ¼ k 2g ¼ m2.
F00X (v) ¼
F00X (0)
2jpqe jv
j2 pqe jv
jv
(jqe
)
þ
(1 qe jv )3
(1 qe jv )2
2pq2 pq
¼
2
p
p3
2q2 q
or m2 ¼ 2 þ
p
p
(11:2:12)
11.2
Examples of Characteristic Functions
159
Hence variance is
m2 m2X ¼
2q2 q q2 q(p þ q)
q
þ 2¼
¼ 2
p p
p2
p
p2
(11:2:13)
Continuous Random Variables
Example 11.2.4 (Uniform Distribution)
1
½u(x b) u(x a)
ba
ð b jvx
e
e jvb e jva
FX (v) ¼
dx ¼
jv(b a)
aba
fX (x) ¼
(11:2:14)
(11:2:15)
This is one of the more difficult CFs to obtain the mean and the variance by differentiating
Eq. (11.2.15) because of the limiting processes involved. Instead, we expand Eq. (11.2.15)
using Taylor series and differentiate the resulting expression. Expansion of Eq. (11.2.15)
results in
1
( jv)2 2
( jv)3 3
FX (v) ¼
jv(b a) þ
(b a2 ) þ
(b a3 ) þ jv(b a)
2!
3!
jv
( jv)2 2
(b þ a) þ
(b þ ab þ a2 ) þ ¼ 1þ
(11:2:16)
2!
3!
Mean:
j
2j2 v 2
(b þ a) þ
(b þ ab þ a2 ) þ 2!
3!
F0X (v)
¼
F0X (0)
j(b þ a)
¼
2
(11:2:17)
(b þ a)
or m1 ¼
2
Variance:
F00X (v) ¼
F00X (0) ¼
2j2 2
(b þ ab þ a2 ) þ 3!
(b2 þ ab þ a2 )
3
or
m2 ¼
b2 þ ab þ a2
3
Hence
var½X ¼ m2 m21 ¼
(b a)2
12
(11:2:18)
In most cases it is easier to find the mean and variance using CF. However, in this
case it is easier to obtain mean and variance by integration techniques using their
definitions.
Example 11.2.5 (Exponential Distribution)
fX (x) ¼ lelx u(x)
(11:2:19)
160
Characteristic Functions and Generating Functions
From Example 11.1.1 the CF is
ð1
FX (v) ¼
lelx e jvx dx ¼
0
l
l jv
(11:2:20)
Mean:
F0X (v) ¼
F0X (0)
jl
(l jv)2
j
¼
l
(11:2:21)
1
m1 ¼
l
or
Variance:
F00X (v) ¼
2l
(l jv)3
F00X (0) ¼
2
l2
or
m2 ¼
(11:2:22)
2
l2
Hence
var½X ¼ m2 m21 ¼
1
l2
(11:2:23)
Example 11.2.6 (Gaussian Distribution)
2
1
2
fX (x) ¼ pffiffiffiffiffiffi e(xm) =2s
2p s
ð1
2
1
2
FX (v) ¼ pffiffiffiffiffiffi
e(xm) =2s þjvx dx
2p s 1
(11:2:24)
(11:2:25)
Completing the squares in the exponent of Eq. (11.2.25), we obtain the resulting
expression as follows:
ð1
1
(x m jvs2 )2
jvm½(s2 v2 )=2
p
ffiffiffiffiffiffi
FX (v) ¼ e
exp dx
2s2
2p s 1
The value of the integral within braces is 1 and hence the CF of a Gaussian is
2
FX (v) ¼ e jvm½(s v
2
)=2
(11:2:26)
Mean:
2
2
F0X (v) ¼ e jvm½(s v )=2 ( jm s2 v)
F0X (0) ¼ jm or m1 ¼ m
(11:2:27)
Variance:
2
F00X (v) ¼ e jvm½ðs v
2
F00X (0) ¼ m2 s2
Þ=2
( jm s2 v)2 e jvm½(s
or m2 ¼ m2 þ s2
2
v2 )=2 2
s
11.3 Generating Functions
161
Hence
var½X ¼ m2 m21 ¼ s2
B 11.3
(11:2:28)
GENERATING FUNCTIONS
Certain simplifications are possible in the definition of the characteristic functions for
discrete random variables taking positive integral values. This is called the generating
function (GF) and is defined for a discrete random variable X taking positive discrete
values as
PX (z) ¼ E{zk } ¼
1
X
zk P{X ¼ k}
(11:3:1)
k¼0
with PX(1) ¼ 1.
The infinite series of Eq. (11.3.1) will converge for values of jzj , 1. Equation
(11.3.1) can be recognized as the z transform of discrete data with the sign reversal for z.
Example 11.3.1 Six dice are tossed. Using the generating function, we want to find the
probability that the sum of the faces of the dice add to 24.
The GF for the toss of a single die is
1
PX (z) ¼ ½z þ z2 þ z3 þ z4 þ z5 þ z6 6
and in closed form this can be written as
z 1 z6
PX (z) ¼
6 1z
Since the tosses are independent, the GF for the toss of 6 dice is [PX(z)]6 given by
QX (z) ¼ ½PX (z)6 ¼
z
6
6
1 z6
1z
6
To find Pffaces add up to 24g, we have to first find the coefficient of z 24 in the power
series expansion of Q(z). Alternately, we have to find the coefficient of z 18 in the
expansion of
RX (z) ¼
(1 z6 )6
(1 z)6
The expansion of the numerator of RX(z) is
(1 z)6 ¼ 1 6z6 þ 15z12 20z18 þ 15z24 6z30 þ z36
and the expansion of the denominator of RX(z) is given by the negative exponential
(1 z)6 ¼
1
X
(5 þ k 1)!
k¼1
5!(k 1)!
zk1
162
Characteristic Functions and Generating Functions
We now multiply the two expansions above and find the coefficient z 18 in the resulting
product. The terms containing z 18 are
(5 þ 7 1)
(5 þ 13 1)!
(5 þ 19 1)!
6 þ1 z18 20 þ 15 5!6!
5!12!
5!18!
¼ 3451z18
Hence the probability that the sum of the faces add to 24 is
P{sum ¼ 24} ¼
3451
¼ 0:073967
66
a result that would have been very difficult to obtain without generating functions.
Moment Generating Properties of GF
It is easier to use the GF in finding the moments of a discrete random variable taking
positive integral values than finding it through CF.
We can differentiate Eq. (11.3.1) and obtain
PX (z) ¼
1
X
zk P{X ¼ k}
k¼0
P0X (z) ¼
1
X
kzk1 P{X ¼ k}
(11:3:2)
k¼0
P0X (1) ¼
1
X
kP{X ¼ k} ¼ m1
k¼0
Differentiating PX0 (z) again, we obtain
P00X (z) ¼
1
X
k(k 1)P{X ¼ k}
k¼0
¼
1
X
k2 P{X ¼ k} k¼0
1
X
kP{X ¼ k}
(11:3:3)
2
(11:3:4)
k¼0
P00X ð1Þ ¼ m2 m1
Hence the variance is given by
var½X ¼ P00X (1) þ P0X (1) P0X (1)
Unlike the CF, there is no general formula for the kth moment in the case of GF. In particular
P000
X (1) ¼ m3 3m2 þ 2m1
B 11.4
(11:3:5)
EXAMPLES OF GENERATING FUNCTIONS
We will now derive generating functions for all discrete random variables discussed in
Section 11.2 and determine their means and variances.
11.4 Examples of Generating Functions
163
Example 11.4.1 (Binomial Distribution)
b(k: n,p) ¼
n k nk
pq
k
(11:4:1)
The GF is given by
N X
n
PX (z) ¼
pzk qnk ¼ (pz þ q)N
k
k¼0
(11:4:2)
Clearly PX(1) ¼ 1.
Mean:
P0X (z) ¼ N(pz þ q)N1 p
P0X (1) ¼ Np
or m1 ¼ Np
(11:4:3)
Variance:
P00X (z) ¼ N(N 1)(pz þ q)N2 p2
P00X (1) ¼ N(N 1)p
or m2 ¼ N(N 1)p2 Np
Hence
var½X ¼ m2 m21 ¼ Npq
(11:4:4)
This result has been obtained with comparative ease.
Example 11.4.2 (Poisson Distribution)
p(k; l) ¼ el
lk
k!
(11:4:5)
The GF is given by
PX (z) ¼
1
X
el
k¼0
lzk
¼ el(1z)
k!
(11:4:6)
Note that PX(1) ¼ 1.
Mean:
P0X (z) ¼ el(z1) ¼ el(z1) l
P0X (1) ¼ l or m1 ¼ l
(11:4:7)
a result that has been trivially obtained compared to Eq. (11.2.7).
Variance:
P00X (z) ¼ el(z1) l2
P00X (1) ¼ l2
or
m2 ¼ l2 l
(11:4:8)
Hence
var½X ¼ m2 m21 ¼ l
A result that has been obtained much more easily than using Eq. (11.2.8).
(11:4:9)
164
Characteristic Functions and Generating Functions
Example 11.4.3 (Geometric Distribution)
p{k: p} ¼ qk p
(11:4:10)
The GF is
PX (z) ¼
1
X
qzk p ¼
k¼0
p
1 qz
(11:4:11)
Here also PX(1) ¼ 1.
Mean:
pq
(1 qz)2
q
P0X (1) ¼
p
P0X (z) ¼
(11:4:12)
Variance:
P00X (z) ¼ 2
P00X (1)
pq2
(1 qz)3
pq2
2q2
¼2
¼
p2
(1 qz)3
(11:4:13)
or
m2 ¼
2q2 q
þ
p
p2
Hence
var½X ¼ m2 m21 ¼
B 11.5
q
p2
(11:4:14)
MOMENT GENERATING FUNCTIONS
The moment generating function (MGF) MX(t) if it exists is defined by
ð1
MX (t) ¼ E(etX ) ¼
fX (x)etx dx
(11:5:1)
1
where t is a real variable. This is similar to the characteristic function (CF) [Eq. (11.1.1)],
except that jv has been replaced by the real variable t. Unlike the CF, which always exists,
the MGF may or may not exist. Expanding E(e tX) in a power series about the origin and
taking expectations, if they exist, we can write
t2 X 2
tn X n
tX
þ þ
þ MX (t) ¼ E(e ) ¼ E 1 þ tX þ
2!
n!
¼ 1 þ tm þ
t2 m2
t n mn
þ þ
þ 2!
n!
(11:5:2)
11.5
Moment Generating Functions
165
If MX(t) exists, Eq. (11.5.2) can be differentiated term by term to obtain
d
tm2
tn1 mn
MX (t) ¼ m þ
þ þ
þ dx
2!
n!
Substituting t ¼ 0 in Eq. (11.5.3), we obtain
MX0 (t)t¼0 ¼ m
(11:5:3)
(11:5:4)
Differentiating Eq. (11.5.2) successively with respect to t and setting t ¼ 0 in the resulting
equation, we obtain the moments as follows:
MX(2) (t)t¼0 ¼ m2 , . . . , m(n)
(11:5:5)
X (t)jt¼0 ¼ mn , . . .
These
results
are very similar to those in Eq. (11.1.5) except for the absence of
pffiffiffiffiffiffi
ffi
j ¼ 1.
Since the methodology of finding the moments using MGF is very similar to that of
using CF, we will give only three examples.
Example 11.5.1 (Negative Binomal—Pascal)
pmf is
If N is the random variable, then the
nb{n; k,p} ¼ P{n trials produce k successes} ¼
n 1 k nk
p q : n ¼ k, k þ 1, . . .
k1
Or substituting m ¼ n 2 k, we can write the pmf with M as the random variable:
nb{m; k, p} ¼ P{(m þ k) trials produce k successes}
kþm1 k m
p q , m ¼ 0, 1, . . .
¼
k1
Since
kþm1
k1
¼
kþm1
m
we can write the MGF for the random variable M:
1 X
kþm1
MM (t) ¼
(qet )m pk
m
m¼0
The negative binomial expansion of (1 2 qe t)2k can be written as follows:
(1 qet )k ¼
1 X
kþm1
(qet )m
m
m¼0
Substituting this equation in the previous one, we obtain the MGF:
MM (t) ¼ pk (1 qet )k
Clearly MM(0) ¼ 1.
166
Characteristic Functions and Generating Functions
Mean:
0
MM
ðtÞ ¼ pk qk(1 qet )k1 et
0
MM
(0) ¼ E(M) ¼
kq
p
and
E(N) ¼
kq
k
þk ¼
p
p
Variance. We first find EfM 2g ¼ m2:
00
MM
(t) ¼ pk qk(1 qet )k2 (kqet þ 1)et
00
MM
(0) ¼ E(M 2 ) ¼
qk(1 þ qk)
p2
Hence the variance of M is
var½M ¼
qk(1 þ qk) (qk)2 qk
2 ¼ 2
p2
p
p
The variance of N can be calculated by using N ¼ M þ k in the equations above.
The MGF of the Gaussian distribution is
Example 11.5.2 (Gaussian Distribution)
given by
1
MX (t) ¼ pffiffiffiffiffiffi
2p s
ð1
e½(xm)
2
=2s2 þtx
dx
1
Completing the squares, we obtain
2 2
MX (t) ¼ etmþ½ðs t
Þ=2
ð1
1
(x m ts2 )2
pffiffiffiffiffiffi
exp dx
2s2
2p s 1
The value of the integral within braces is 1, and hence the MGF of a Gaussian is
2 2
MX (t) ¼ etmþ½(s t
)=2
Mean:
MX0 (t) ¼ etm½(s
2 2
t )=2
(m þ s2 t)
MX0 (0) ¼ m
Variance:
MX00 (t) ¼ etmþ½(s
2 2
t )=2
MX00 (0) ¼ m þ s2
2 2
(m þ s2 t)2 þ etmþ½(s t
and
var½X ¼ s2
)=2 2
s
11.6
Example 11.5.3 (Gamma Distribution)
given by
167
We will find the MGF of a gamma density
l(lx)a1 elx
G(a)
fX (x) ¼
Cumulant Generating Functions
a, l; x . 0
From Eq. (11.5.1), we have
ð1
l(lx)a1 elx tx
e dx, a, l . 0
G(a)
0
ð
la 1 a1 (lt)x
x e
dx
¼
G(a) 0
MX (t) ¼
Substituting (l 2 t)x ¼ y and using the definition of a gamma function [Eq. (7.5.2)],
we can write this equation as
ð1
la
MX (t) ¼
ya1 ey dy
G(a)(l t)a 0
MX (t) ¼
la
la
a G(a) ¼
G(a)(l t)
(l t)a
Mean:
la (a)(1)
ala
¼
(l t)aþ1
(l t)aþ1
a
MX0 (0) ¼
l
MX0 (t) ¼
Variance. We first find EfX 2g ¼ m2:
MX00 (t) ¼
a(a þ 1)la
(l t)aþ2
MX00 ð0Þ ¼ m2 ¼
a(a þ 1)la a2 þ a
¼
l2
laþ2
Hence the variance is
m2 m2X ¼
a2 þ a a2
a
2¼ 2
l2
l
l
The means and variances of most of the commonly occurring discrete (Chapter 4)
and continuous (Chapters 6 and 7) distributions are shown in Section 11.7.
B 11.6
CUMULANT GENERATING FUNCTIONS
The cumulant generating function (CGF) KX(t) is defined as the logarithm of the moment
generating function MX(t), or KX (t) ¼ ln½MX (t). The cumulants also called semiinvariants
{ki } are the coefficients of CGF:
KX (t) ¼ ln½MX (t) ¼ tk1 þ
1 i
X
t2
tn
t
k2 þ þ kn þ ¼
ki
2!
n!
i!
i¼0
(11:6:1)
168
Characteristic Functions and Generating Functions
We will now find the cumulants in terms of the moments mk. From Eq. (11.5.2), we can
expand ln[MX(t)] in terms of a Taylor series as follows:
t 2 m2
t n mn
þ þ
þ KX (t) ¼ ln½MX (t) ¼ ln 1 þ tm1 þ
2!
n!
t2 m2
t n mn
¼ tm1 þ
þ þ
þ 2!
n!
2
1
t 2 m2
tn mn
tm1 þ
þ þ
þ 2
2!
n!
3
1
t 2 m2
tn mn
tm1 þ
þ þ
þ +
þ
3
2!
n!
(11:6:2)
Collecting terms in t, we have
KX (t) ¼ tm1 þ
t2
(m2 m21 )
2!
þ
t3
(m3 3m1 m2 þ 2m31 )
3!
þ
t4
(m4 3m22 4m1 m3 þ 12m21 m2 6m41 )
4!
þ
t5
(m5 5m4 þ 30m1 m22 þ 40m21 m3 60m31 m2 24m51 )
5!
þ (11:6:3)
Equating coefficients of ti in Eqs. (11.6.1) and (11.6.3), the cumulants have been expressed
in terms of the moments fmig in Eq. (11.6.4), where they are also expressed in terms of the
central moments vi ¼ E½X mX i :
k1 ¼ m1 ¼ mX
k2 ¼ m2 m21 ¼ s2X ¼ v2
k3 ¼ m3 3m1 m2 þ 2m31 ¼ v3
(11:6:4)
k4 ¼ m4 3m22 4m1 m3 þ 12m21 m2 6m41 ¼ v4 3v2
k5 ¼ m5 5m4 þ 30m1 m22 þ 40m21 m3 60m31 m2 24m51
¼ v5 10v2 v3 þ Skewness measures the lack of symmetry of the tails of a density. A density is symmetric
if it looks the same on both the left and the right of the centerpoint. Skewness is
defined as
v3
g1 ¼ 3=2
(11:6:5)
v2
Kurtosis is defined as a normalized form of the fourth central moment of a distribution. It
is a measure of whether the density is peaked or flat relative to a normal distribution.
High-kurtosis density functions will have a high peak near the mean, and low-kurtosis
11.6
Cumulant Generating Functions
functions will have a flat top. Kurtosis is defined as
v4
g2 ¼ 2
v2
169
(11:6:6)
For Gaussian distributions the kurtosis is 3.
Example 11.6.1 We will find the cumulants of the Poisson distribution given by
p(k; l) ¼ el (lk =k!).
t
The MGF of the Poisson distribution is MX (t) ¼ el(1e ) , and the CGF is given by
t
KX (t) ¼ ln el(1e ) ¼ l þ let
Expanding KX (t) in Taylor series, we have
t2 t3 t4 t5
l þ let ¼ l t þ þ þ þ þ 2! 3! 4! 5!
Thus, all the cumulants are equal to l, or ki ¼ l, i ¼ 1,2, . . . :
Example 11.6.2
We will find the cumulants of a Gaussian distribution
"
#
1
1 x mX 2
fX (x) ¼ pffiffiffiffiffiffi exp 2
sX
2psX
2 2
The MGF is given by MX (t) ¼ etmX þ½(t sX )=2 and the CGF, by
h
i
t2 s2X
2 2
KX (t) ¼ ln etmX þ½(t sX )=2 ¼ tmX þ
2
and the cumulants are k1 ¼ mX , k2 ¼ s2X and ki ¼ 0 for i . 2. Thus, for the Gaussian
distribution, there are only two cumulants.
Joint Cumulants
By analogy to Eq. (11.6.4), we can define joint cumulants for one to four random
variables, X1, X2, X3, X4, as follows:
k1 ¼ E½X1 ¼ mX
k12 ¼ E½X1 X2 E½X1 E½X2 ¼ sX1 X2
k123 ¼ E½X1 X2 X3 E½X1 E½X2 X3 E½X2 E½X1 X3 E½X3 E½X1 X2 þ 2E½X1 E½X2 E½X3 k1234 ¼ E½X1 X2 X3 X4 E½X1 X2 E½X3 X4 E½X1 X3 E½X2 X4 E½X1 X4 E½X2 X3 E½X1 E½X2 X3 X3 E½X2 E½X1 X3 X4 E½X3 E½X1 X2 X4 E½X4 E½X1 X2 X3 þ 3E½X1 E½X2 E½X3 X4 þ 3E½X2 E½X3 E½X1 X4 þ 3E½X3 E½X4 E½X1 X2 þ 3E½X1 E½X4 E½X2 X3 6E½X1 E½X2 E½X3 E½X4 (11:6:7)
170
Characteristic Functions and Generating Functions
If the mean values of random variables Z1, Z2, Z3, Z4 are zero, then Eq. (11.6.7) can be
considerably simplified as follows:
k1 ¼ E½Z1 ¼ 0
k12 ¼ E½Z1 Z2 k123 ¼ E½Z1 Z2 Z3 k1234 ¼ E½Z1 Z2 Z3 Z4 E½Z1 Z2 E½Z3 Z4 (11:6:8)
E½Z1 Z3 E½Z2 Z4 E½Z1 Z4 E½Z2 Z3 If X1, X2, Y1, Y2 are four random variables, not necessarily independent, we will
express the cov[X1 X2, Y1 Y2] in terms of the joint cumulant k1234 :
cov½X1 X2 , Y1 Y2 ¼ E{½X1 X2 E(X1 X2 )½Y1 Y2 E(Y1 Y2 )}
¼ E½X1 X2 Y1 Y2 E½X1 X2 E½Y1 Y2 (11:6:9)
Expressing E½X1 X2 Y1 Y2 in terms of the fourth cumulant k1234 given in Eq. (11.6.8), we
can rewrite Eq. (11.6.9) as follows:
cov½X1 X2 ; Y1 Y2 ¼ k1234 þ E½X1 Y1 E½X2 Y2 þ E½X1 Y2 E½X2 Y1 B 11.7
(11:6:10)
TABLE OF MEANS AND VARIANCES
pþq¼1
Random Variable
Mean
Variance
Discrete Distributions
Bernoulli
p
pq
Binomial
Np
Npq or Np(1 2 p)
Multinomial
npi ,
q
p
Geometric
i ¼ 1, . . . , m
Negative binomial E(M) ¼
Hypergeometric
Kn
N
Poisson
l
Logarithmic—
Benford
kq
p
npi (1 pi ), i ¼ 1, . . . , m
q
p2
kq
for M ¼ N k var(M) ¼ 2 for M ¼ N k
p
nK(N K)(N n)
N 2 (n 1)
l
2 8
2 :5
log
ffi 3:44
7 : 34
(d ¼ 1, . . . , 9)
2 8 2
28 : 572
2 :5
ffi 6:06
log 13 50 log
7 :3
7 : 34
(d ¼ 1, . . . , 9)
Continuous Distributions
Uniform
bþa
2
(b a)2
12
(Continued )
11.7
Random Variable
Table of Means and Variances
Mean
1
l
Variance
1
l2
Gaussian
m
Triangular
b
s2
(b a)2
6
2
l2
n
, n integer
l2
a
l2
1
aþ2
2 aþ1
G
G
a
a
l2
Exponential
Laplace—double- 0
exponential
n
, n integer
Erlang
l
a
Gamma
l
1 aþ1
Weibull
mþ G
l
a
Chi-square
Chi
Rayleigh
Maxwell
ns2
2ns4
pffiffiffi
nþ1
2sG
2
n
G
2
rffiffiffiffi
p
s
2
rffiffiffiffi
2
2s
p
3
nþ1
2G
6
7
2
7
n
s2 6
n
4
5
G2
2
p
s2 2 2
8
2
s 3
p
mR ¼ s2
Rice
2
2
171
rffiffiffiffi
2
2 p (m2 =4s2 )
m2
m
m2
m
1 þ 2 I0
I
þ
e
1
2
2s
4s2
2s2
4s2
s2R ¼ m2 þ 2s2 m2R ; m: noncentrality parameter
Nakagami
1
1=2 G m þ
V
2
m
G(m)
3
2 1 2
G mþ
V6
2 7
7
V 6
4
m
G(m) 5
n
, n.2
v2
Student-t
0
Snedecor F
n2
,
n2 2
Lognormal
e½mþ(s =2)
e(2mþs
Beta
a
aþb
ab
(a þ b þ 1)(a þ b)2
2
n2 . 2
2v22 (n1 þ n2 2)
,
n1 (n2 2)2 (n2 4)
2
)
v2 . 4
2
es 1
(Continued )
172
Characteristic Functions and Generating Functions
Random Variable
Mean
Variance
Cauchy
Not defined
Not defined
Pareto
b
,
a1
ab2
, a.2
(a 1)2 (a 2)
a.1
CHAPTER
12
Functions of a Single
Random Variable
B 12.1
RANDOM VARIABLE g(X )
In many probability problems, we have to find the pdf of a function of random variables.
We have already defined (in Chapter 5) a random variable X as the mapping from the
sample space S to space of real numbers Rx; that is, it is the set of all outcomes j belonging
to the sample space S such that the unique number X(j) belongs to the real line Rx. Thus,
the random variable X is a mapping of the probability space {S, F, P} to the probability
space on the real line {Rx , FX , PX }:
X: {S, F, P} ! {Rx , FX , PX }
(12:1:1)
We can now define a random variable Y as a function of another random variable X, or
Y ¼ g(X ), as the mapping from the real line Rx to another real line Ry, such that it
induces a probability space {Ry , FY , PY }. In other words, the random variable Y is a composite function of the random variable X such that Y(j) ¼ g(X ) ¼ g[X (j)]. The mapping is
given by
Y: {S, F, P} ! {Rx , FX , PX } ! {Ry , FY , PY }
(12:1:2)
and shown in Fig. 12.1.1.
Conditions on Y 5 g(X ) to be a Random Variable
1. The domain of g(x) must include the range of the random variable (RV) X.
2. For each y [ Iy, fg(x) yg must form a field FY ; that is, it must consist of the
unions and intersections of a countable number of intervals of the real line Ry.
3. The probability of the event fg(x) ¼ +1g is 0.
In essence, the transformation Y ¼ g(X ) induces a probability measure PY on FY , and
we have to find this probability measure. From Fig. 12.1.1 we can write
PY {Y} ¼ PX {g1 ( Y)} ¼ PX {x : g1 ( Y) , Ix }
(12:1:3)
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
173
174
Functions of a Single Random Variable
FIGURE 12.1.1
The set Ix consists of all the countable unions and intersections of the type a , x b.
Without loss in generality, we will assume that Ix ¼ {X y}. Thus, Eq. (12.1.3) becomes
PY {Y} ¼ P{Y y} ¼ FY ( y) ¼ PX {x : g1 ( Y) y} ¼ PX {x : g(x) y}
(12:1:4)
In conclusion, we can say that the probability of Y y is given by the probability of
the set of all values of x such that g(x) y.
B 12.2
DISTRIBUTION OF Y 5 g(X )
We will find the distribution FY (y) of the RV Y ¼ g(X) in terms of the distribution FX (x) of
the RV X and the function g(x). We shall do this with an arbitrary function g(x) bounded
between a and b as shown in Fig. 12.2.1. As y traverses from 21 to þ1 along the g(x)
axis, we have to find the values of x along the x axis that satisfy g(x) y. In Fig. 12.2.1
there are five distinct regions for y corresponding to the intersections of y with g(x):
1. y a, no points of intersection
2. a , y y1, one point of intersection x1
3. y1 , y y2, three points of intersection x2, x20 , x002
4. y2 , y b, one point of intersection x3
5. y . b, no points of intersection
FIGURE 12.2.1
12.2 Distribution of Y 5 g(X )
175
We have to find FY ( y) ¼ probability of g(x) y in each of these five regions.
1. y a: There are no points along the x axis for which g(x) y, or IX ¼ {1}; hence
FY ( y) ¼ 0.
2. a , y y1: The set of all points along the x axis for which g(x) y, or Ix ¼
(x1, 1); hence FY ( y) ¼ 1 2 FX(x1).
3. y1 , y y2: The set of all points along the x axis for which g(x) y, or Ix ¼
(x2, x20 ] < (x002, 1); hence FY ( y) ¼ FX (x20 ) 2 FX (x2) þ 1 2 FX (x002).
4. y2 , y b: The set of all points along the x axis for which g(x) y or Ix ¼ (x3, 1);
hence FY ( y) ¼ 1 2 FX(x3).
5. y . b: Here the entire g(x) is below y; and hence Ix ¼ (21, 1); hence FY ( y) ¼ 1.
Thus, we have used Eq. (12.1.4) to find PfY y) along the entire g(x) axis.
Special Case
We now consider a situation where g(x) ¼ y1 for x1 , x x2. In this case we have
P{Y ¼ y1 } ¼ P{x1 , X , x2 } ¼ FX (x2 ) FX (x1 )
(12:2:1)
Hence FY ( y) is discontinuous at y ¼ y1, and the discontinuity is equal to FX (x2 )
FX (x1 ).
Summary of Steps for Finding FY( y) 5 PfY y)
1. Graph the function g(x) along the x and g(x) axes. Do not label the g(x) axis as y.
2. Draw y parallel to the x axis.
3. Find the distinct regions along the g(x) axis. The regions are determined by the
number of intersections of y with g(x) as y goes from 21 to 1 along the g(x)
axis. Let these be y1, y2, . . . .
4. In each region solve for all values of x for which y ¼ g(x). Let these be x, x0 , x00 , . . . .
5. For each of these regions, find Ix such that g(x) y.
6. Find FY( y) ¼ PfY yg ¼ Pfx [ Ix) ¼ Pfx : g(x) yg.
We will take a number of examples of increasingly difficult transformations g(x) and
solve for PfY yg using the steps listed above. We shall assume that in all the following
examples FX(x) or fX(x) is given and we have to find FY( y).
Example 12.2.1 Given that Y ¼ aX þ b with a . 0, we have to find FY( y).
We follow the steps listed above:
1. The function g(x) ¼ ax þ b has been graphed.
2. y parallel to the x axis has been drawn in Fig. 12.2.2.
3. In this simple example there is only one region 21 , y , 1.
4. In this region, solving for y ¼ ax þ b, we have x ¼ ( y 2 b)/a.
5. The region Ix such that g(x) y is given by
yb
Ix ¼ 1 , x a
y
b
yb
6.
FY ( y) ¼ P{x : g(x) y} ¼ P X ¼ FX
a
a
176
Functions of a Single Random Variable
FIGURE 12.2.2
We can now find the density fY ( y) by differentiating FY ( y) with respect to y. The result is
fY ( y) ¼
Example 12.2.2
before.
dFX (ð y bÞ=a) 1
yb
¼ fX
dy
a
a
We repeat Example 12.2.1 with a 0. We follow the same steps as
1. The function g(x) is graphed in Fig. 12.2.3.
2. y has been drawn parallel to x.
3. In this example also, there is only one region 21 , y ,1.
4. In this region, solving for y ¼ ax þ b, we have x ¼ ( y 2 b)/a or x ¼ [( y 2 b)/jaj].
5. The region IX such that g(x) y is given by
6.
yb
,x1
IX ¼ jaj
yb
FY ( y) ¼ P{x : g(x) y} ¼ P X . jaj
yb
yb
¼ 1 FX ¼1P X .
jaj
jaj
The corresponding density function fY ( y) is given by differentiating FY ( y) with respect to
y, and the result is
dFX
fY ( y) ¼
yb
1
yb
a
fX
¼
jaj
a
dy
The result is very similar to that obtained in Example 12.2.1 except for the absolute
value of a.
12.2 Distribution of Y 5 g(X )
177
FIGURE 12.2.3
Example 12.2.3 Let g(x) ¼ (x 2 a)2. This example is different from the previous
two examples because there are multiple regions for y. Following similar steps, we
obtain
1. The function g(x) is graphed in Fig. 12.2.4 and
2. y has been drawn parallel to x.
3. Here the two regions of y are fy 0g, no points of intersection and fy . 0g, two
points of intersection.
4. In the region fy 0g, Ix ¼ 1. In the region fy.0g, we solve for y ¼ g(x) and
pffiffiffi
pffiffiffi
obtain x1 ¼ a þ y and x2 ¼ a y.
pffiffiffi
pffiffiffi
5. Thus, Ix ¼ (a y, a þ y , y . 0.
6. In the first region FY ( y) ¼ 0 : y 0. In the second region
pffiffiffi
pffiffiffi
y , X a þ y)
pffiffiffi
pffiffiffi
¼ FX (a þ y ) FX (a y ) : y . 0
FY ( y) ¼ P(a FIGURE 12.2.4
178
Functions of a Single Random Variable
The corresponding density function is
fY ( y) ¼ 0 : y 0
fY ( y) ¼
d
pffiffiffi
pffiffiffi
(FX (a þ y ) FX (a y ))
dy
1
pffiffiffi
pffiffiffi
¼ pffiffiffi (fX (a þ y ) þ fX (a y )) : y . 0
2 y
If X is a zero mean Gaussian distributed random variable given by
1
2
2
fX (x) ¼ pffiffiffiffiffiffi e(x =2s )
s 2p
and if Y ¼ X 2, then fY ( y) is given by
1
1 1
2
2
2
fY ( y) ¼ pffiffiffi pffiffiffiffiffiffi e( y=2s ) þ e( y=2s ) ¼ pffiffiffiffiffiffiffiffi e( y=2s )
2 y s 2p
s 2py
which is a chi-square distribution with one degree of freedom [Eq. (7.7.1)].
Example 12.2.4 This example is a complement of Example 12.2.3. Here g(x) ¼ 1=(x a)2
and we have to find FY ( y). We follow the same procedure laid out earlier.
1. The function g(x) is graphed in Fig. 12.2.5.
2. y has been drawn parallel to the x axis.
3. Here the two regions of y are fy 0g, no points of intersection and fy . 0g, two
points of intersection.
4. In the region f y 0g, Ix ¼ {1} and FY( y) ¼ 0. In the region f y . 0g, we solve for
pffiffiffi
pffiffiffi
y ¼ g(x) and obtain x1 ¼ a þ ð1= y Þ and x2 ¼ a (1= y ).
5. Thus
1
1
Ix ¼ 1, a pffiffiffi < a þ pffiffiffi , 1
y
y
FIGURE 12.2.5
12.2 Distribution of Y 5 g(X )
179
6. In the region indicated in item 5 (above), we obtain
1
1
FY ( y) ¼ P X a pffiffiffi þ 1 P X a þ pffiffiffi
y
y
1
1
¼ FX a pffiffiffi þ 1 FX a þ pffiffiffiffi : y . 0
y
Y
The corresponding density function is
fY ( y) ¼ 0 : y 0
d
1
1
FX a pffiffiffi þ 1 FX a þ pffiffiffi
fY ( y) ¼
dy
y
y
1
1
1
¼ 3=2 fX a pffiffiffi þ fX a þ pffiffiffi : y . 0
2y
y
y
Example 12.2.5 The previous examples illustrated continuous functions of g(x). In this
example g(x) is a discrete function as shown in Fig. 12.2.6.
Note that g(x) is not a true function because there are many points of y at x ¼ +a. We
can still find FY ( y) given FX (x). The procedure is no different from that in the previous
examples.
1. The function g(x) has already been graphed in Fig. 12.2.6.
2. y has been drawn parallel to the x axis.
3. There are four region of y, given by (1) y 2b, (2) 2b , y 0, (3) 0 , y b,
(4) y . b.
4. The intersection points are (1) none, (2) x1 ¼ 2a, (3) x2 ¼ a, (4) none.
5. The corresponding four regions along the x axis are (1) Ix ¼ {1}, (2)
Ix ¼ {1, a}, (3) Ix ¼ f1, a}, (4) Ix ¼ {1, 1}.
6. The distribution function FY( y) in the various regions are (1) y 2 b : FY ( y) ¼ 0,
(2) 2b , y 0 : FY ( y) ¼ FX (2a), (3) 0 , y b : FY ( y) ¼ FX (a), (4) y . b :
FY ( y) ¼ 1. The distribution function FY( y) is a “staircase” function, indicating
that Y is a discrete random variable and the density function fY ( y)
FIGURE 12.2.6
180
Functions of a Single Random Variable
FIGURE 12.2.7
is (a) fY( y) ¼ 0, (b) fY( y) ¼ FX (2a)d(x þ b), (c) fY( y) ¼ [FX(a) 2 FX(2a)]d(x),
(d) [1 2 FX(a)]d(x 2 b).
The distribution and density functions are shown in Fig. 12.2.7.
Example 12.2.6 We shall now take an example of a piecewise-continuous g(x) with
a specified fX(x) and FX(x). The functions g(x), fX(x) and FX(x) are shown in
Fig. 12.2.8.
FIGURE 12.2.8
12.2 Distribution of Y 5 g(X )
181
We have to find FY( y) given that fX(x) is defined by Fig. 12.2.8 as
2
3
1
: 3 , x 2
5
fX (x) ¼ 4 5
0: otherwise
Since fX(x) is specified, FX(x) is determined by integrating fX(x). The result is
2
0:
6x 3
FX (x) ¼ 4 þ :
5 5
1:
x 3
3
7
3 , x 2 5
x.2
and is shown in Fig. 12.2.8. In this example, the range of fX(x), x [ f23, 2g has to be
factored in determining FY ( y). Other than that, we will still follow the procedure laid
out earlier.
1. The function g(x) has already been graphed in Fig. 12.2.8.
2. y will scan from 21 to 1 along the g(x) axis.
3. The regions of y are given by (a) y 24, (b) 24 , y 22, (c) 22 , y21,
(d) 21 , y 0, (e) 0 , y 1, (f) 1 , y 2, (g) 2 , y 4, (h) y . 4.
4. The points of intersection of y with g(x) are (a) none, (b) x ¼ 22, (c) x ¼ y,
(d) x ¼ 0, (5) x ¼ 1, (6) x ¼ y, (7) x ¼ 2, (8) none.
5. The corresponding regions Ix are (a) Ix ¼ f1g, (b) Ix ¼ f21, 22g, (c) Ix ¼ f21,
yg, (d) Ix ¼ f21, 0g, (e) Ix ¼ f21, 1g, (f) Ix ¼ f21, yg, (g) Ix ¼ f21, 2g, (h)
Ix ¼ f21, 1g.
6. The distributions functions in the six regions are
(a) y 4 : FY ( y) ¼ 0
1
(b) 4 , y 2 : FY ( y) ¼ FX (2) FX (4) ¼
5
y 3
(c) 2 , y 1 : FY ( y) ¼ FX ( y) ¼ þ
5 5
3
(d) 1 , y 0 : FY ( y) ¼ FX (0) ¼
5
4
(e) 0 , y 1 : FY ( y) ¼ FX (1) ¼
5
y 3
(f) 1 , y 2 : FY ( y) ¼ FX ( y) ¼ þ
5 5
(g) 2 , y . 4 : FY ( y) ¼ FX (2) ¼ 1 since FX (x) ¼ 1 for all values of x . 2
(h) y . 4 : FY ( y) ¼ 1
By differentiating FY( y) in every region, we can express the density function fY ( y) as
1
1
1
1
fY ( y) ¼ d( y þ 3) þ ½u( y þ 1) u( y þ 2) þ d( y þ 1) þ d( y)
5
5
5
5
1
þ ½u( y 2) u( y 1)
5
where u( y) is the usual step function. FY ( y) and fY ( y) are graphed in Fig. 12.2.9.
182
Functions of a Single Random Variable
FIGURE 12.2.9
Example 12.2.7 This example is similar to the previous one with a piecewisecontinuous function g(x). The functions g(x), fX(x) and FX(x) are given by
2
3
0:
x 2
"1
#
6 1: 2 , x 0 7
6
7
:
2
,
x
2
7
;
g(x) ¼ 6
6 2 x: 0 , x 3 7; fX (x) ¼ 4
4
5
0:
otherwise
3
2:
x.3
2
3
0:
x 2
6x 1
7
FX (x) ¼ 4 þ : 2 , x 2 5
4 2
1:
x,2
The functions g(x), fX(x), and FX(x) are shown in Fig. 12.2.10. We have to find FY( y).
Following the same procedure as before
1. The function g(x) has been graphed in Fig. 12.2.10.
2. y has been drawn, and the intersection of y with g(x) is given by x ¼ 32y.
3. The regions of y are given by (a) y 21, (b) 21 , y 0, (c) 0 , y 43,
(d) 43 , y 2, (e) y . 2. Even though g(x) extends upto x ¼ 3, the function
fX(x) terminates at x ¼ 2 and FX(x) ¼ 1 beyond x ¼ 2.
4. The points of intersection of y with g(x) are (a) none, (b) x ¼ 22,0, (c) x ¼ 32y,
(d) x ¼ 32y, (e) none.
5. The corresponding regions Ix are (a) Ix ¼ f1g, (b) Ix ¼ (– 2, 0], (c) Ix ¼ (21, 23 x],
(d) Ix ¼ (21, 23x], (e) Ix ¼ (21, 1].
6. The distribution functions in the four regions are (a) y 21: FY ( y) ¼ 0,
(b) 21 , y 0: FY( y) ¼ FX(0) 2 FX(22) ¼ 12, (c) 0 , y 43: FY( y) ¼ FXð32 yÞ ¼
3
1
4
8 y þ 2, (d) 3 , y 2: FY ( y) ¼ 1, (e) y 2: FY( y) ¼ 1.
12.2 Distribution of Y 5 g(X )
FIGURE 12.2.10
The corresponding density function fY ( y) is given by
1
3
4
fY ( y) ¼ d( y þ 1) þ u y u( y)
2
8
3
The functions FY ( y) and fY ( y) are shown in Fig. 12.2.11.
Example 12.2.8 We shall now take a more complicated example and find FY ( y).
In this example g(x) is defined by
g(x) ¼
1
x2 1
for all x
FIGURE 12.2.11
183
184
Functions of a Single Random Variable
FIGURE 12.2.12
The probability density function fX(x) and the distribution function FX(x) are given by
2
2
fX (x) ¼ 4 9
0
3
if 1:5 , x 3 5
;
otherwise
2
0
1
62
FX (x) ¼ 4 x þ
9
3
1
3
if
x 1:5
if
7
1:5 x 3 5
if
x.3
The function g(x) is shown in Fig. 12.2.12 and fX(x) and FX(x) are shown in Fig. 12.2.13.
1. g(x) has been graphed in Fig. 12.2.12.
2. y scans from 21 to þ1 along the g(x) axis.
3. The regions of y are given by (a) y 21, two points of intersection in the range
of fX(x), (21.5, 3]; (b) 21 , y 0, no points of intersection; (c) 0 , y 18,
two points of intersection beyond the range of fX(x), (21.5, 3]; (d) 18 , y 45, one
point of intersection in the range of fX(x), (21.5, 3], (e) y . 45, two points of
intersection in the same range.
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
4. The p
points
of ffiintersection of y with
g(x) areffi (a) x1 ¼pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ (1=y)
and
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
x2 ¼ 1 þ (1=y), (b) none, (c) x1 ¼ 1 þ (1=y) and x2 p
¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ (1=y)
beyond
ffi
the range of (21.5, 3], (d) one point of intersection x2 ¼ 1 þ (1=y) within the
FIGURE 12.2.13
12.2 Distribution of Y 5 g(X )
185
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
rangep
and
anotherffi x1 ¼ 1 þ (1=y) beyond the range, (e) x1 ¼ 1 þ (1=y) and
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 ¼ 1 þ (1=y).
5. The corresponding regions Ix are
sffiffiffiffiffiffiffiffiffiffiffi) (sffiffiffiffiffiffiffiffiffiffiffi )
(
1
1
ðaÞ Ix ¼ 1, 1 þ
<
1þ , 1
y
y
ðbÞ Ix ¼ (1, 1
sffiffiffiffiffiffiffiffiffiffiffi)
(sffiffiffiffiffiffiffiffiffiffiffi
)
1
1
Ix ¼ 1, 1 þ
< {1, 1}
1þ , 1
y
y
(
ðcÞ
sffiffiffiffiffiffiffiffiffiffiffi)
(sffiffiffiffiffiffiffiffiffiffiffi
)
1
1
ðdÞ Ix ¼ 1, 1 þ
< {1, 1}
1þ , 1
y
y
(
sffiffiffiffiffiffiffiffiffiffiffi)
(sffiffiffiffiffiffiffiffiffiffiffi
)
1
1
Ix ¼ 1, 1 þ
< {1, 1}
1þ , 1
y
y
(
ðeÞ
6. The distribution functions in the various regions are
sffiffiffiffiffiffiffiffiffiffiffi!
sffiffiffiffiffiffiffiffiffiffiffi!
1
1
1þ
ðaÞ y 1 : FY ( y) ¼ FX 1 þ
FX (1) þ FX (1) FX
y
y
sffiffiffiffiffiffiffiffiffiffiffi
2
1 1
2 1
2 1
þ
1þ þ þ
¼
þ
9
y 3
9 3
9 3
sffiffiffiffiffiffiffiffiffiffiffi
!
2
1 1
1þ þ
9
y 3
sffiffiffiffiffiffiffiffiffiffiffi
4 4
1
1þ
¼ 9 9
y
ðbÞ
ðcÞ
1 , y 0 : FY ( y) ¼ FX (1) þ FX (1) ¼
2 1
2 1
4
þ
þ
¼
9 3
9 3
9
sffiffiffiffiffiffiffiffiffiffiffi!
1
1
0 , y : FY ( y) ¼ FX 1 þ
FX (1) þ FX (1) FX (1)
8
y
sffiffiffiffiffiffiffiffiffiffiffi!
1
þ FX (1) FX
1þ
y
4
4
¼00þ þ11¼
9
9
186
Functions of a Single Random Variable
sffiffiffiffiffiffiffiffiffiffiffi!
1
4
1
, y : FY ( y) ¼ FX 1 þ
ðdÞ
FX (1) þ FX (1) FX (1)
8
5
y
sffiffiffiffiffiffiffiffiffiffiffi!
1
þ FX (1) FX
1þ
y
sffiffiffiffiffiffiffiffiffiffiffi
sffiffiffiffiffiffiffiffiffiffiffi
!
4
2
1 1
10 2
1
1þ þ
1þ
¼00þ þ1
¼
9
9
y 3
9 9
y
sffiffiffiffiffiffiffiffiffiffiffi!
4
1
ðeÞ y . : FY ( y) ¼ FX 1 þ
FX (1) þ FX (1) FX (1)
5
y
sffiffiffiffiffiffiffiffiffiffiffi!
1
þ FX (1) FX
1þ
y
sffiffiffiffiffiffiffiffiffiffiffi
sffiffiffiffiffiffiffiffiffiffiffi
!
!
2
1 1
4
2
1 1
1þ þ
1þ þ
¼ 0þ þ1
9
y 3
9
9
y 3
sffiffiffiffiffiffiffiffiffiffiffi
13 4
1
¼
1þ
9 9
y
The density function fY ( y) can be obtained by differentiating FY ( y). Functions FY ( y) and
fY ( y) are shown in the following equation. FY ( y) is graphed in Fig. 12.2.14a, and fY ( y) is
graphed in Fig. 12.2.14b.
sffiffiffiffiffiffiffiffiffiffiffi
3
2
4 4
1
1þ
if
y 1 7
6 7
6 9 9
y
7
6
64
17
6
if 1 , y 7
69
87
7
6
sffiffiffiffiffiffiffiffiffiffiffi
7;
FY ( y) ¼ 6
6 10 2
1
1
4 7
7
6 6 9 9 1 þ y if 8 , y 5 7
7
6
7
6
sffiffiffiffiffiffiffiffiffiffiffi
7
6
5
4 13 4
1
4
1þ
if
y.
9 9
y
5
3
22
1
rffiffiffiffiffiffiffiffiffiffiffi if
y 1
7
69 2
7
6 y 1þ1
7
6
y
7
6
6
17
60
if 1 , y 7
6
87
7
6
61
1
1
4 7
fY ( y) ¼ 6
7
6 9 rffiffiffiffiffiffiffiffiffiffiffi1 if 8 , y 5 7
7
6 2
7
6 y 1þ
7
6
y
7
6
62
1
4 7
7
6
y.
4 9 rffiffiffiffiffiffiffiffiffiffiffi1 if
5 5
2
y 1þ
y
12.2 Distribution of Y 5 g(X )
187
FIGURE 12.2.14
Example 12.2.9 In Example 12.2.8 we will find the conditional probability of Y conditioned on the event A ¼ jXj . 1. Since jXj . 1, the part of the function g(x) that lies
between 21 and 1 has no effect on the conditional distribution. Thus, g(x) is given by
g(x) ¼
1
x2 1
for
jxj , 1
The conditional density function fX(x j A) with A ¼ jXj . 1 is given by
fX (x j A) ¼
fX (x)
fX (x)
¼
p(jXj . 1) FX (1) þ 1 FX (1)
f (x)
9
X
¼ fX (x)
¼
2 1
2 1
5
þ
þ
þ1
9 3
9 3
for
jxj . 1
188
Functions of a Single Random Variable
Hence, the conditional density fX(x j A) and the conditional distribution FX(x j A)
can be written as
22
3
if 1:5 , x 1
65
7
6
7
fX (x j A) ¼ 6 2
7;
if
1,x3 5
4
5
0 otherwise
3
2
0
if
x , 1:5
7
6 2x þ 3
if 1:5 , x 1 7
6
7
6 5
7
6 1
7
6
if
1 , x 1 7
FX (x j A) ¼ 6
7
6 5
7
6 2x 1
6
if
1,x3 7
5
4
5
1
if
x.3
The function g(x j A) is shown in Fig. 12.2.15. f (x j A) and FX(x j A) are shown in
Fig. 12.2.16.
Using exactly the same procedure as in Example 12.3.9, we can find the conditional
distribution FY ( y j A) and the conditional density fY ( y j A) for the regions (a) y 18,
(b) 18 , y 45, and (c) y . 45. The results are shown below:
3
2
1
if y 7
60
8
7
6
sffiffiffiffiffiffiffiffiffiffiffi!
7
6
1
1
47
6 2
7
6
3
1
þ
,
x
if
FY ( y j A) ¼ 6 5
y
8
57
7
6
sffiffiffiffiffiffiffiffiffiffiffi!
7
6
7
61
1
4
5
4
94 1þ
if y .
5
y
5
2
0
6
1
6 0
6
6
C
61B
1
6 B rffiffiffiffiffiffiffiffiffiffiffiC
[email protected]
1A
6
y2 1 þ
fY ( y j A) ¼ 6
y
6 0
1
6
6
6 B
C
1
62B
6 @ rffiffiffiffiffiffiffiffiffiffiffiC
45
1A
y2 1 þ
y
if
if
if
y
1
8
3
7
7
7
7
1
47
,x 7
8
57
7
7
7
7
7
7
4
7
7
y.
5
5
FY ( y j A) and fY (Y j A) are also graphed in Fig. 12.2.17.
Example 12.2.10 In the previous examples g(x) was explicitly given, and we had to find
FY ( y) given fX(x). In this example g(x) is not explicitly given, but we will have to determine it from the problem. Points are uniformly distributed along the circumference of a
unit circle. From each point a tangent is drawn that intersects the x axis (Fig. 12.2.18).
We have to find the distribution of the points of intersection along the x axis. Since the
12.2 Distribution of Y 5 g(X )
189
FIGURE 12.2.15
points are uniformly distributed, the angle Q that the point subtends at the center of the
circle is a random variable that is uniformly distributed as U(0,2p).
The transformation mapping X ¼ g(Q) between Q and X is found as follows. Since
the radius of the circle is 1, the cosine of the angle u subtended by any point on the
unit circle at the center of the circle is given by cos(u) ¼ (1/x). Hence, the transformation
between the random variable Q and the random variable X is given by
X ¼ g(Q) ¼ sec(Q)
The function g(u) is graphed in Fig. 12.2.19.
The density function fQ(u) and the distribution function FQ(u) of the random variable
Q are given by
2
3
0 if x 0
" 1
#
for 0 , u 2p
6 u
7
; FQ (u) ¼ 4
fQ (u) ¼ 2p
if 0 , u 2p 5
2p
0 otherwise
1 if x . 2p
They are shown in Fig. 12.2.20.
FIGURE 12.2.16
190
Functions of a Single Random Variable
FIGURE 12.2.17
The usual procedure can now be followed to determine the distribution function
FX(x).
1. The function g(u) has been graphed.
2. The line x has been drawn in Fig. 12.2.19.
FIGURE 12.2.18
12.2 Distribution of Y 5 g(X )
191
FIGURE 12.2.19
3. The regions along the x axis are given by (a) x21, two points of intersection;
(b) 21 , x 1, no points of intersection; (c) x . 1, two points of intersection.
4. Intersection points of x with g(u) are (a) u1 ¼ cos21 (1/x) and u10 ¼ 2p 2 cos21(1/x),
(b) no points of intersection, (c) u2 ¼ cos21(1/x) and u20 ¼ 2p 2 cos21(1/x).
5. The corresponding regions Iu are
p
1 S
1
3p
, cos1
(a) Iu ¼
2p cos1
,
2
x
x
2
p 3p
,
(b) Iu ¼
2 2
1 S p 3p S
1
,
(c) Iu ¼ 0, cos1
2p cos1
, 2p
x
2 2
x
6. The distribution functions in the various regions are
1
1 1 3
1
1
1
1 1
cos1 þ 1 cos1
(a) FX (x) ¼
¼ cos1 2p
x 4 4
2p
x
x 2
p
3 1 1
(b) FX (x) ¼ ¼
4 4 2
FIGURE 12.2.20
192
Functions of a Single Random Variable
FIGURE 12.2.21
(c) FX (x) ¼
1
1 1
1
1
1
1 1
cos1 þ þ 1 1 cos1
¼ cos1 þ
2p
x 2
2p
x
x 2
p
This distribution is called the arcsine law. The density function fX(x) is obtained by differentiating the distribution function FX(x). They are both shown below, and the graphs are
shown in Fig. 12.2.21:
2
3
1
1 1
cos1 if
x 1
6p
7
x 2
6
7
61
7
FX (x) ¼ 6
if 1 , x 1 7;
62
7
41
5
1
1 1
cos
þ
if
x.1
p
x 2
2
3
1
p
ffiffiffiffiffiffiffiffiffiffiffiffi
ffi
if
x
1
6 px x2 1
7
6
7
6
fX (x) ¼ 6 0
if 1 , x 1 7
7
4
5
1
pffiffiffiffiffiffiffiffiffiffiffiffiffi if
x.1
px x2 1
Example 12.2.11 As a final example, we will take the case when g(u) ¼ cos(u) in the
region 2p , u p. In the previous example g(x) ¼ sec(u), and here g(x) ¼ cos(u).
We will also assume as in the previous example that fQ(u) is uniformly distributed in
193
12.2 Distribution of Y 5 g(X )
FIGURE 12.2.22
(2p, p]. Thus, fQ(u) and FQ(u) are given by
" 1
fQ (u) ¼ 2p
0
for
p , u p
otherwise
2
#
;
0
1
6 u
FQ (u) ¼ 4
þ
2p 2
1
if
if
if
x0
3
7
p , u p 5
x.p
The function g(u) ¼ cos(u) is shown in Fig. 12.2.22. The density and distribution
functions are graphed in Fig. 12.2.23.
Following the usual procedure
1. g(u) has been graphed.
2. The line x has been drawn (Fig. 12.2.22).
3. The regions along the x axis are given by (a) x 21, no points of intersection;
(b) 21 , x 1, two points of intersection; (c) x . 1, no points of intersection.
4. Intersection points of x with g(u) are (a) no points of intersection, (b) u1 ¼
2cos21(x) and u10 ¼ cos21(x), (c) no points of intersection.
5. The corresponding regions Iu are (a) Iu ¼ {1}, (b) Iu ¼ { p, cos1 (x)}
{cos1 (x), p}, (c) Iu ¼ f.
FIGURE 12.2.23
194
Functions of a Single Random Variable
FIGURE 12.2.24
6. The distribution functions in the various regions are
(a) FX (x) ¼ 0
1
1
p 1
(b)
1
cos (x) þ þ
FX (x) ¼ 2p
2
2p 2
p 1
1
1
1
1
þ cos (x) þ
þ
¼ 1 cos1 (x)
2p 2
2p
2
p
(c) FX (x) ¼ 1
Here also the distribution is the arcsine law.
The density function fX(x) is obtained by differentiating the distribution function FX(x).
The distribution and density functions are given below and graphed in Fig. 12.2.24.
2
0
if
1
6
FX (x) ¼ 4 1 cos (x) if
p
1
if
2
0
if
6
1
ffi if
fX (x) ¼ 6
4 ppffiffiffiffiffiffiffiffiffiffiffiffi
1 x2
0
if
x 1
3
7
1 , x 1 5;
x.1
x 1
3
7
1 , x 1 7
5
x.1
B 12.3 DIRECT DETERMINATION OF DENSITY fY ( y)
FROM fX (x)
In the previous section, for any given Y ¼ g(X ), we determined fY ( y) from fX (x) through a
circuitous route by converting fX (x) to FX(x), using the transformation g(X ) to find FY ( y),
and then differentiating FY ( y) to obtain fY ( y) as shown in the schematic representation in
Fig. 12.3.1.
Consider the function g(x) given in Fig. 12.3.2. The probability of the random variable
Y lying between y and y þ dy is given by
fY ( y)dy ¼ P{y , Y y þ dy}
(12:3:1)
12.3 Direct Determination of Density fY ( y) from fX (x)
195
FIGURE 12.3.1
In Fig. 12.3.2 y ¼ g(x) has three real solutions: x1, x2, x3. Hence the set of values for which
{y , g(x) y þ dy} is given by the union of the set of values corresponding to the three
points of intersection, {x1 , X x1 þ dx1 } {x2 þ dx2 , X x2 } {x3 , X x3 þ dx3 },
where dx1 . 0, dx3 . 0, and dx2 , 0. By the conservation of probability measure,
we have
fY ( y)dy ¼ P{y , Y y þ dy}
¼ P{x1 , X x1 þ dx1 } þ P{x2 þ dx2 , X x2 } þ P{x3 , X x3 þ dx3 }
¼ fX (x1 )dx1 þ fX (x2 ) j dx2 j þ fX (x3 )dx3
(12:3:2)
We can now write
fY ( y) ¼ fX (x1 )
dx1
jdx2 j
dx3
þ fX (x3 )
þ fX (x2 )
dy
dy
dy
(12:3:3)
jdx2 j
1
¼ 0
,
dy
jg (x2 )j
(12:3:4)
But
dx1
1
¼ 0
,
jg (x1 )j
dy
FIGURE 12.3.2
dx3
1
¼ 0
jg (x3 )j
dy
196
Functions of a Single Random Variable
Substituting Eq. (12.3.4) into Eq. (12.3.3), we obtain the result for fY ( y) directly
from fX(x):
fY ( y) ¼
fX (x1 )
fX (x2 )
fX (x3 )
þ 0
þ 0
0
jg (x1 )j jg (x2 )j jg (x3 )j
(12:3:5)
Equation (12.3.5) can be generalized to the case when there are n real solutions to the
equation y ¼ g(x), given by {x1 , x2 , . . . , xn }:
fY ( y) ¼
n
X
fX (xi )
0 (x )j
jg
i
i¼1
(12:3:6)
From the derivation it is clear that certain restrictions have to be satisfied if we have to
determine fY ( y) directly from fX(X ).
Conditions for Direct Determination of fX(X)
1. The derivative of g(x) must exist.
2. There can be no straight-line solutions to g(x) ¼ y.
We now formulate the following steps for the direct determination of fY ( y) from fX(x):
Steps for Direct Determination of fY ( y)
1. Check whether g(x) satisfies restrictions 1 and 2.
2. Solve for y ¼ g(x). Let the n real roots be fx1, x2, . . . , xng.
3. Determine
dg(x)
dx
,
i ¼ 1, 2, . . . , n
x¼xi
4. Calculate
fY ( y) ¼
n
X
fX (xi )
jg0 (xi )j
i¼1
Thus, if the restrictions on g(x) are satisfied, it is easier to calculate fY ( y) directly from
fX(x). We will take some of the examples in the previous section and solve for fX(x) directly.
Example 12.3.1
We will find fY ( y) for Y ¼ aX þ b with a . 0 or, a 0.
1. g(x) satisfies the restrictions.
2. We solve for y ¼ ax þ b. The solution yields x ¼ ( y 2 b)/a.
dg(x)
d(ax þ b)
¼
¼ jaj
dx
dx
4. fY ( y) ¼ 1 fX y b
jaj
a
3.
Example 12.3.2
(12:3:7)
Y ¼ g(X ) ¼ (X 2 a)2:
1. g(x) satisfies the restrictions.
pffiffiffi
pffiffiffi
2. Solving for x in y ¼ (x 2 a)2, we obtain two roots: x1 ¼ a y, x2 ¼ a þ y
12.3 Direct Determination of Density fY ( y) from fX (x)
3.
dg(x)
d(x a)2
pffiffiffi
¼
¼ j2(x a)j ¼ 2 y
dx
dx
4.
1
pffiffiffi
pffiffiffi
fY ( y) ¼ pffiffiffi ½ fX (a y ) þ fX (a þ y );
2 y
Example 12.3.3
y.0
197
(12:3:8)
Y ¼ g(X) ¼ 1=(X a)2 .
1. g(x) satisfies the restrictions.
pffiffiffi
2. Solving for x in y ¼ 1=(x a)2 , we obtain two roots x1 ¼ a (1= y),
pffiffiffi
x2 ¼ a þ (1= y).
d
1
1
3. dg(x)
¼
¼ 2
¼ j2y3=2 j
dx
dx (x a)2
(x a)3
1
1
1
4.
fY ( y) ¼ 3=2 fX a pffiffiffi þ fX a þ pffiffiffi ; y . 0
2y
y
y
(12:3:9)
Example 12.3.4 (See Example 12.2.8) The probability density functions fX(x) and g(x)
and are given by
2
3
2
1
if 1:5 , x 3 5
g(x) ¼ 2
for all x
fX (x) ¼ 4 9
x
1
0 otherwise
Even though g(x) is defined from 21 to 1, the range of fX(x) plays a critical role in
determining fY ( y). Figure 12.2.12 is redrawn in Fig. 12.3.3 to indicate the ranges.
1. g(x) satisfies the restrictions.
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2. Solving
for x in
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi y ¼ 1/(x 2 1), we obtain two roots: x1 ¼ 1 þ (1=y),
x2 ¼ 1 þ (1=y).
sffiffiffiffiffiffiffiffiffiffiffi
3. dg(x)
d
1
x
1
2
¼
¼ 2 2
¼ 2y 1 þ
dx
dx (x2 1)
y
(x 1)2
4. fY ( y) have to be calculated in the four ranges of y as shown in Fig. 12.3.3.
The density function fY ( y) is calculated as follows:
Region I: y 21, two points of intersection:
1
2 2
2
1
rffiffiffiffiffiffiffiffiffiffiffi
rffiffiffiffiffiffiffiffiffiffiffi
þ
fY ( y) ¼
¼
9 2
1 9 9
1
1þ
2y2 1 þ
y
y
y
1
Region II: 1 , y , no points of intersection:
8
fY ( y) ¼ 0
1
4
Region III: , y , one point of intersection:
8
5
1
2
1
1
rffiffiffiffiffiffiffiffiffiffiffi
rffiffiffiffiffiffiffiffiffiffiffi
fY ( y) ¼
¼
9 2
1 9
1
2y2 1 þ
y
1þ
y
y
198
Functions of a Single Random Variable
FIGURE 12.3.3
4
Region IV: y . , two points of intersection:
5
1
2 2
2
1
rffiffiffiffiffiffiffiffiffiffiffi
rffiffiffiffiffiffiffiffiffiffiffi
þ
fY ( y) ¼
¼
9
9
9
1
1
2y2 1 þ
y2 1 þ
y
y
Here we have found the result far more easily than in Example 12.2.8.
Example 12.3.5 (See Example 12.2.10)
by g(u) ¼ sec(u) and fQ(u) is given by
" 1
fQ (u) ¼
2p
0
In Example 12.2.10, the function g(u) is given
for 0 , u 2p
#
otherwise
We will now find fX(x) directly from fQ(u).
1. g(u) satisfies the restrictions.
2. Solving for u in x ¼ sec(u), we have u1 ¼ sec21(x) and u10 ¼ 2p 2 sec21(x).
pffiffiffiffiffiffiffiffiffiffiffiffiffi
3. dg(u)
d
¼
sec (u) ¼ j sec (u) tan (u)j ¼ x x2 1
du
du
4. From Fig. 12.2.19, the regions of x are (a) x 21, (b) 2 1 , x 1, (c) x . 1.
(a) x 21, two points of intersection:
1
1
1
1
fX (x) ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi
þ
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi
x x2 1 2p 2p
px x2 1
(b) 21 , x 1, no points of intersection, fX(x) ¼ 0.
199
12.3 Direct Determination of Density fY ( y) from fX (x)
(c) x . 1, two points of intersection:
1
1
1
1
þ
fX (x) ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2p
2p
x x 1
px x2 1
These are the same results obtained in Example 12.2.10.
Example 12.3.6 (See Example 12.2.11)
" 1
fQ (u) ¼ 2p
0
Here g(u) ¼ cos(u) and fQ(u) is given by
for p , u p
#
otherwise
The three regions of x are (a) x 2 1, (b) 21 , x 1, (c) x . 1.
1. g(u) satisfies the restrictions.
2. The solutions for x ¼ cos(u) are u1 ¼ 2cos21(x) and u10 ¼ cos21(x).
3. dg(u)
pffiffiffiffiffiffiffiffiffiffiffiffiffi
d
¼
cos (u) ¼ j sin (u)j ¼ 1 x2
du
du
4. fX(x) can now be found in the three regions.
(a) x 21, no points of intersection: fX (x) ¼ 0:
(b) 21 , x 1, two points of intersection:
1
1
1
1
þ
fX (x) ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2p
2p
1x
p 1 x2
(12:3:10)
(c) x . 1, no points of intersection: fX (x) ¼ 0:
These are the same results obtained in Example 12.2.11.
Example 12.3.7 This example will have relevance in the next section. Here g(x) and
fX(x) are given by
2
0
6
4
2
þ
ln
g(x) ¼ 6
4
5x
0
3
if
x1
if
7
1 , x , 57
5
if
x5
"1
fX (x) ¼
if
1,x5
#
4
0 otherwise
The functions g(x) and fX(x) are shown in Fig. 12.3.4.
1. g(x) satisfies the restrictions.
2. There are no points of intersection in the region y 2. Solving for y ¼ g(x) in the
region y . 2, we obtain
y ¼ 2 þ ln
4
4
; y 2 ¼ ln
5x
5x
or 5 x ¼ 4e( y2)
200
Functions of a Single Random Variable
FIGURE 12.3.4
or
x ¼ 5 4e( y2)
for
y.2
4
1
3. dg(x) ¼ d 2 þ ln
¼
¼ 4e( y2) ;
dx
dx
5x
5x
y.2
4. Since there are no points of intersection of y with g(x) in the region y 2, it
follows that fY ( y) ¼ 0. In the region y . 2,
1
fY ( y) ¼ 4e( y2) ¼ e( y2)
4
for y . 2
B 12.4 INVERSE PROBLEM: FINDING g(x) GIVEN fX(x)
AND fY ( y)
In the previous sections we studied finding fY ( y) given fX(x) and g(x). In this section we
will find g(x) given fX(x) and fY ( y). We will use the conservation of probability measure
given by an equation similar to Eq. (12.3.3):
fY (h)Dh ¼ fX (j1 )Dj1 þ fX (j2 )Dj2 þ þ fX (jn )Djn
(12:4:1)
Taking the limit as Dh, Dj1, Dj2, . . . , Djn ! 0, all the jk coalesce to j, and integrating h
between 21 to y and j between 21 to x, we obtain
ðy
ðx
fY (h)dh ¼
fX (j)dj
(12:4:2)
1
1
The lefthand side of Eq. (12.4.2) is a function of y, the righthand side is a function of x, and
the equation can be solved for y, which yields g(x) as the transforming function.
Example 12.4.1 Given fX (x) ¼ 2e2(x2) u(x 2) and fY ( y) ¼ 14 ½u( y 2) u( y 6),
we have to find the transforming function g(x). The functions fX(x) and fY ( y) are shown
in Fig. 12.4.1.
12.4
Inverse Problem: Finding g(x) Given fX(x) and fY ( y)
201
FIGURE 12.4.1
Integrating both fX(x) and fY ( y), we obtain the following:
ðy
ðy
1
y2
dh ¼
for
fY (h)dh ¼
4
4
2
2
ðx
ðx
fX (j)dj ¼ 2e2(j2) dj ¼ 1 e(x2) for
2
2,y6
x.2
2
Thus
y2
¼ 1 e(x2)
4
for
2 , y 6 and x . 2
or
y ¼ g(x) ¼ 6 4e(x2)
for
x.2
The function g(x) is graphed in Fig. 12.4.2. Using the same techniques as in the previous
section, we can show that the transformation function g(x) indeed gives a uniform distribution in (2,6], given that fX(x) is an exponentially distributed function starting at y ¼ 2.
Example 12.4.2 (See Example 12.3.7) We will revisit Example 12.3.7, where fX(x) is
uniformly distributed in the interval (1,5] and fY ( y) is exponentially distributed as
FIGURE 12.4.2
202
Functions of a Single Random Variable
e 2( y22) starting at y ¼ 2. We have to find the transforming functions. Proceeding as in the
previous example, we have
ðy
ðx
1
x1
(h2)
dj or 1 e( y2) ¼
, y . 2, x . 1
e
dh ¼
4
4
2
1
or
e( y2) ¼
5x
,
4
y . 2, x . 1
Taking logarithm on both sides, we obtain
5x
4
( y 2) ¼ ln
or y ¼ g(x) ¼ 2 þ ln
,
4
5x
x.1
g(x) is exactly the same function that we used in Example 12.3.7 and is shown in Fig. 12.3.4.
B 12.5 MOMENTS OF A FUNCTION OF
A RANDOM VARIABLE
Expectation
We will now find the expectation of the random variable Y ¼ g(X ). We have already
found the density function fY ( y). We can use the usual definition of expectations and write
ð1
mY ¼ E½Y ¼
y fY ( y)dy
(12:5:1)
1
However, if only the expected value of Y, and not the density function fY ( y) is required, we
can find the result from the knowledge of fX(x) and g(x). Using the law of conservation of
probability measure [Eq. (12.3.2)], we have
ð1
ð1
E½Y ¼
y fY ( y)dy ¼
y fX (x)dx
(12:5:2)
1
1
Substituting y ¼ g(x) in Eq. (12.5.2), we obtain the equation for E[Y ] in terms of the
density function fX(x):
ð1
E½Y ¼
g(x)fX (x)dx
(12:5:3)
1
If the random variable is of the discrete type, then
X
X
E½Y ¼
yi P( y ¼ yi ) ¼
g(xi ) P(x ¼ xi )
i
(12:5:4)
i
Variance
In a similar manner, the variance of Y ¼ g(X ) can be defined as
ð1
2
2
sY ¼ E½Y mY ¼
( y mY )2 fY ( y) dy
ð1
¼
1
1
( y mY )2 fX (x)dx ¼
ð1
1
½g(x) mY 2 fx (x)dx
(12:5:5)
12.5
Moments of a Function of a Random Variable
203
Higher-Order Moments
The rth moment of Y ¼ g(X ) can be defined as follows:
mr ¼ E½Yr ¼
ð1
ð1
( y)r fY ( y)dy
1
r
ð1
( y) fX (x)dx ¼
¼
1
½g(x)r fX (x)dx
(12:5:6)
1
Example 12.5.1 We will take Example 12.2.7 and calculate the expected values by
using the density functions fY ( y) and fX(x). Function fX(x) is uniformly distributed in
(22, 2] as shown in Fig. 12.2.10. The density function fY ( y) for that example is given by
1
3
4
fY ( y) ¼ d( y þ 1) þ u y u( y)
2
8
3
and g(x), fX(x), and fY ( y) are shown in Fig. 12.5.1.
Using Eq. (12.5.1) to obtain E[Y ], we have
1
3
4
d( y þ 1) þ u y u( y) dy
2
8
3
1
ð 4=3
1
3
1 3 16
1
y dy ¼ þ ¼
¼ þ
2
8
2
8
9:2
6
0
ð1
E½Y ¼
y
FIGURE 12.5.1
204
Functions of a Single Random Variable
Using Eq. (12.5.3) to obtain E[Y ], we have
ð
ð2
1 0
2
x dx
E½Y ¼
1 dx þ
4 2
03
2 1 2 4
1 1
1
¼ þ ¼ þ ¼
4 4 3 2
2 3
6
This result could just as well have been obtained by inspection of g(x) by adding the areas
of the rectangle (22) and the triangle 43 and dividing them by 14, yielding 16. All three of
these methods yield exactly the same result.
We will now compute the variance of Y from fY ( y) using Eq. (12.5.5):
ð1 1 2 1
3
4
d( y þ 1) þ u y yþ
var½Y ¼
u( y)
dy
6
2
8
3
1
2 ð 4=3 1
5
3
1 2
25 91
83
yþ
¼
¼
þ
dy ¼ þ þ
2
6
6
72 216 108
0 8
Computing var[Y ] from fX(x) and using var[Y ] ¼ E[Y ]2 2 m2Y, we have
ð
ð 1 0
1 2 2 2
1
2
x dx var½Y ¼
(1) dx þ
4 2
4 0 3
36
¼
2 1 32 1
83
þ ¼
4 4 27 36 108
and the two results are the same.
Example 12.5.2 We are given fX(x) and fY ( y), and we have to find g(x) and the expectation and variance of Y using fY ( y) and fX(x):
2 2
3
"1
#
3y
2,x6
;
fY ( y) ¼ 4 8 0 , y 2 5
fX (x) ¼ 4
0 otherwise
0 otherwise
Integrating fX(j) from 2 to x and fY (h) from 0 to y, we have
ðy 2
ðx
1
3h
dj ¼
dh
4
2
0 8
1
y3
(x 2) ¼ ; 2 , x 6, 0 , y 2
4
8
Hence, y ¼ g(x) ¼ ½2(x 2)1=3 . We also note the following:
Expectation with fY ( y):
ð2
y
E½y ¼
0
3y2
3
dy ¼
2
8
Expectation with fX(x):
ð6
E½ y ¼
1
21=3
3
½2(x 2)1=3 dx ¼
3 41=3 ¼
4
2
4
2
12.5
Moments of a Function of a Random Variable
205
FIGURE 12.5.2
Variance with fY ( y):
ð2
var½ y ¼
0
3
y
2
2
3y2
3
dy ¼
20
8
Variance with fX(x), using var[Y ] ¼ E[Y ]2 2 m2Y:
ð6
1
9 3
9
3
var½ y ¼ ½2(x 2)2=3 dx ¼ 82=3 ¼
4
4
5
4
20
2
Both methods give the same results.
Indicator Functions
An indicator function IA(j) is a real-valued function defined on the sample space S,
having a value 0 or 1 depending on whether the j point is in the event A , S. Or
1 j[A
(12:5:7)
IA (j) ¼
0 jA
The indicator function is a unit step for sets. It is shown in Fig. 12.5.2.
The indicator function is a zero – one random variable, and if
P{j [ A} ¼ p
P{j A} ¼ 1 p ¼ q
(12:5:8)
then the expected value and the variance of the indicator function IA are
E½IA ¼ 1 p þ 0 q ¼ p
var½IA ¼ 12 p p2 ¼ pq
(12:5:9)
Properties of Indicator Functions
1. A , B () IA IB
2. IA ¼ IA2 ¼ ¼ IAn
3. Is ¼ 1: sample space
4. IA<B ¼ max (IA , IB )
5. IA>B ¼ min (IA , IB )
6. IA ¼ 1 IA : A ¼ complement of A
(12.5.10)
CHAPTER
13
Functions of Multiple
Random Variables
B 13.1 FUNCTION OF TWO RANDOM VARIABLES,
Z 5 g(X,Y )
In the previous chapter we discussed a single function of a single random variable. Here
we will find the probability fZ(z) of a single function Z of two random variables X and Y
defined by the joint probability density function fXY(x, y). The analysis is very similar to
that in the previous chapter.
Following similar analysis given in Chapter 12, we have Z ¼ g(X, Y ) as a
function that maps the domain space of Z given by fCxy,FXY,PXYg into the range space
fRz,FZ ,PZg, which is a real line, as shown in Fig. 13.1.1
{Z ¼ g(X,Y)} : {Cxy ,FXY ,PXY } ! {Rz ,FZ ,PZ }
(13:1:1)
and the mapping of fX,Y g is from the probability space fS, F, Pg (where, S is the product
space S ¼ SX SY) into the domain space of Z given by fCxy,FXY,PXYg
X,Y: {S,F,P} ! {Cxy ,FXY ,PXY }
(13:1:2)
Alternately, if Dxy , Cxy and Iz , Rz, then Iz is the image of Dxy under the transformation g
and Dxy is the inverse image of Iz under the transformation g – 1. Hence Iz is the set of all
points (x,y) belonging to Dxy such that g(x,y) belongs to Iz. This is represented by
Dxy ¼ {x,y : g(x,y) [ Iz }
(13:1:3)
We also have conditions similar to those in the last chapter for Z to be a random
variable.
Conditions on Z 5 g(X,Y) to Be a Random Variable
1. The domain of g(x,y) must include the range of the RV Z.
2. For each z [ Iz, fg(x,y) zg must form a field FZ; that is, it must consist of the
unions and intersections of a countable number of intervals.
3. The probability of the event fg(x,y) ¼ +1g is 0.
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
206
Function of Two Random Variables, Z 5 g(X,Y )
13.1
207
FIGURE 13.1.1
In essence, the transformation Z ¼ g(X,Y ) induces a probability measure PZ on FZ , and
we have to find this probability measure. From Fig. 13.1.1, we can write
PZ (Z) ¼ P(Z [ Iz ) ¼ PXY (g1 (Z) [ Dxy ) ¼ PXY (X,Y [ Dxy )
(13:1:4)
The set Iz consists of all the countable unions and intersections of the type a , Z b.
Without loss in generality, we will assume that Iz ¼ fZ zg, and substituting for Dxy
from Eq. (13.1.3) in Eq. (13.1.4), we have
PZ (Z) ¼ PXY ½(X,Y) [ Dxy ¼ PXY ½x,y:g(x,y) [ Iz ¼ PXY ½x, y:g(x,y) z ¼ FXY (z) ¼ P(Z z) ¼ FZ (z)
(13:1:5)
Equation (13.1.5) can also be written as
ð
ð
PZ (Z) ¼ PXY ½(X,Y) [ Dxy ¼ FZ (z) ¼
fXY (j,h)dj dh
(13:1:6)
Dxy
where fXY (x,y) is the joint density of the random variables X and Y.
In conclusion, we can say that the probability of Z z is given by the probability
of the set of all values of x and y such that g(x,y) z, or by integrating the joint density
fXY (x,y) over the domain Dxy. We will use both these interpretations in solving problems.
Summary of Steps for Finding FZ(z) 5 P(Z z)
1. Graph the function g(x,y) in the x– y plane.
2. Equate g(x,y) ¼ z
3. Find Dxy, the distinct regions in the x– y plane such that g(x,y) z.
4. Find FZ(z) from Eq. (13.1.5) or (13.1.6).
Example 13.1.1 Given Z ¼ g(X,Y ) ¼ min(X,Y ), we will find FZ(z) given the joint
distribution FXY(x,y). We follow the scheme outlined above.
1. g(x,y) is shown in Fig. 13.1.2.
2. g(x,y) ¼ z is marked on the diagram.
3. The region Dxy is the set of all points in the x –y plane such that either x z or
y z. It is shown by the region ABHCDF in Fig. 13.1.2.
208
Functions of Multiple Random Variables
FIGURE 13.1.2
4. The probability of Dxy is given by Pf(X z) or (Y z)g, which can be written as
Pfrectangle ABEFg þ Pfrectangle GCDFg 2 Pfrectangle GHEFg. We subtract
Pfrectangle GHEFg since it occurs twice in the addition of the first two terms.
Hence
FZ (z) ¼ FY (z) þ FX (z) FXY (z,z)
(13:1:7)
If X and Y are independent, then Eq. (13.1.7) becomes
FZ (z) ¼ FY (z) þ FX (z) FX (z)FY (z)
(13:1:8)
and the density function fZ(z) is given by
fZ (z) ¼ fY (z) þ fX (z) fX (z)FY (z) FX (z)fY (z)
(13:1:9)
We will define the joint density function fXY (x,y) in the region f0 , x 4g, f0 , y 2g as
shown in Fig. 13.1.3 and find the distribution FZ(z).
FIGURE 13.1.3
13.1
Function of Two Random Variables, Z 5 g(X,Y )
209
The joint density fXY(x,y) and the distribution FXY(x,y) are shown below:
2
0 for x 0, y 0
61
fXY (x,y) ¼ 4
for 0 , x 4, 0 , y 2 ;
8
0 for x . 0, y . 0
2
0 for x 0, y 0
6 xy for 0 , x 4, 0 , y 2
6
68
6x
FXY (x,y) ¼ 6
6 4 for 0 , x 4, y . 2
6y
6
for x . 4, 0 , y 2
4
2
1 for x . 0, y . 0
1. In region I, for fz 0g, FZ(z) ¼ 0.
2. In region II, f0 , z 2g or f0 , x 2g, f0 , y 2g we have the following
formula from the equation derived earlier:
FZ (z) ¼
z z z2 z
þ ¼ (6 z)
4 2 8 8
for
0,z2
3. In region III, f2 , z 4g or f2 , x 4g, f2 , y 4g
FZ (z) ¼ FX (z) þ FY (2) FXY (z,2) ¼
z
z
þ1 1¼1
4
4
for
2,z4
The same result can also be obtained from the fact that the random variable Y is
always less than 2. Since the region Dxy is either X z or Y z, and z . 2, we
have FZ(z) ¼ 1.
4. In region IV, fz . 4g or fx . 4g, fy . 4g, FZ(z) ¼ 1.
The density function fZ(z) obtained by differentiating FZ(z) is shown below.
2
0
for z 0
63 z
6 for 0 , z 2
fZ (z) ¼ 6
64 4
for 2 , z , 4
40
0
for z . 4
Example 13.1.2 Given Z ¼ g(X,Y ) ¼ max(X,Y ), we will find FZ(z) given the joint
distribution FXY(x,y).
We follow the scheme outlined above.
1. g(x,y) is shown in Fig. 13.1.4.
2. g(x,y) ¼ max(x,y) ¼ z is marked on the diagram.
3. The region Dxy is the set of all points in the x –y plane such that x z and y z.
This region is shown in Fig. 13.1.4.
4. The probability of Dxy is given by Pf(X,Y z)g, which can be written as
FZ (z) ¼ FXY (z,z)
(13:1:10)
210
Functions of Multiple Random Variables
FIGURE 13.1.4
If X and Y are independent, then Eq. (13.1.10) becomes
FZ (z) ¼ FX (z)FY (z)
(13:1:11)
and the density function fZ(z) is given by
fZ (z) ¼ fX (z)FY (z) þ FX (z)fY (z)
(13:1:12)
We will use the same joint density function fXY(x,y) in the region f0 , x 4g, f0 , y 2g
as shown in Fig. 13.1.3 and find the distribution FZ(z).
1. In region I, for fz 0g, FZ(z) ¼ 0.
2. In region II, f0 , z 2g or f0 , x 2g, f0 , y 2g, FZ(z) is given by
FZ (z) ¼
z2
8
for
0 , x 2, 0 , y 2
3. In the region III, f2 , z 4g or f2 , x 4g, f2 , y 4g we have
z 2 z
FZ (z) ¼ FXY (z, 2) ¼ ¼
4 2 4
for
2 , x 4, 2 , y 4
We can also integrate the joint density function fXY(x,y) in the ranges f0 , x zg
and f0 , y 2g:
ðz ð2
FZ (z) ¼
0
1
dy dx ¼
08
ðz
2
z
dx ¼
8
4
0
for
2 , x 4, 2 , y 4
4. In region IV, fz . 4g or fx . 4g, fy . 4g, FZ(z) ¼ 1.
13.1
Function of Two Random Variables, Z 5 g(X,Y )
211
The density function fZ(z) obtained by differentiating FZ(z) is shown below.
2
0 for z 0
6 z for 0 , z 2
6
64
fZ (z) ¼ 6 1
6
for 2 , z , 4
4
4
0 for z . 4
Example 13.1.3
edge of fXY(x,y).
Given that Z ¼ g(x,y) ¼ X þ Y, we have to find FZ(z) from the knowl-
1. The function g(x,y) is shown in Fig. 13.1.5.
2. The equation x þ y ¼ z is shown in Fig. 13.1.5.
3. Dxy is shown as the shaded region in the figure.
4. The distribution function FZ(z) is
fZ (z) ¼ fX (z)FY (z) þ FX (z)fY (z)
(13:1:13)
The density function fZ(z) is obtained by differentiating FZ(z), and the result is
ð 1 ð zx
ð1
ð zx
d
d
fZ (z) ¼
fXY (x,y)dy dx ¼
f (x,y)dy dx
dz 1 1
1 dz
1
(13:1:14)
ð1
¼
fXY (x,z x)dx
1
If X and Y are independent in Eq. (13.1.14), then
ð1
fX (x)fY (z x)dx
fZ (z) ¼
(13:1:15)
1
Equation (13.1.15) is a convolution integral, and we can use characteristic functions to
write
FZ (v) ¼ FX (v)FY (v)
fZ(z) can obtained from FZ(v).
FIGURE 13.1.5
(13:1:16)
212
Functions of Multiple Random Variables
Example 13.1.4 We are given X and Y as independent random variables with
Z ¼ g(X,Y ) ¼ 2X 2 12Y. The density functions fX(x) and fY( y) are given by
fX (x) ¼ 2e2x u(x);
1
fY (y) ¼ ey=2 u(y)
2
We have to find fZ(z).
Substituting U ¼ 2X and V ¼ 12Y, we can obtain the corresponding densities of U and
V from Eq. (12.3.7):
1 u
fU (u) ¼ fX
¼ eu u(u); fV (v) ¼ 2fY (2v) ¼ ev u(v)
2
2
With this substitution, Z can be given by Z ¼ U þ (2V ). Since U and V are also
independent, we can use Eq. (13.1.16) to find fZ(z). The characteristic functions of U
and (2V ) are
FU (v) ¼
1
1
; FV (v) ¼
1 jv
(1 þ jv)
Using the independence property, we obtain
1
1
1
1
¼
þ
FZ (v) ¼ FU (v)FV (v) ¼
(1 jv)(1 þ jv) 2 1 jv 1 þ jv
Taking the inverse transform, we obtain
1
1
fZ (z) ¼ ½ez u(z) þ ez u(z) ¼ ejzj
2
2
The functions fX(x), fY( y), and fZ(z) are shown in Fig. 13.1.6.
FIGURE 13.1.6
13.1
Example 13.1.5
Function of Two Random Variables, Z 5 g(X,Y )
213
Two independent random variables X and Y are distributed normally as
2
1
fX (x) ¼ pffiffiffiffiffiffiffiffiffiffiffiffi e(1=2)½(x2)=3 ;
2p 3
2
1
fY (y) ¼ pffiffiffiffiffiffiffiffiffiffiffiffi e(1=2)½(y3)=4
2p 4
Another random variable Z is given by Z ¼ X 2 Y. We have to find fZ(z).
The characteristic functions of X and (2Y ) are given by
FX (v) ¼ e jv
2
½ðv2 9Þ=2
; FY (v) ¼ ejv3½ðv
2
16Þ=2
Using the independence of X and Y, we have
2 2
3 2
2
FZ (v) ¼ e jv ½ðv 9Þ=2 e jv ½ðv 16Þ=2 ¼ ejv½ðv 25Þ=2
and fZ(z) is also normally distributed with mean 21 and variance 25, and is given by
2
1
fZ (z) ¼ pffiffiffiffiffiffiffiffiffiffiffiffi e(1=2)½(x1)=5
2p 5
Hence, the probability of a sum of independent Gaussian random variables is also a
Gaussian random variable with mean equaling the sum of means, and variance equaling
the sum of variances.
Example 13.1.6 This example illustrates when the two random variables X and Y are not
independent but are given by a joint Gaussian density function:
(
)
1
1
2
2
fXY (x,y) ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi exp pffiffiffiffiffiffiffiffiffiffiffiffiffi ½x 2rxy þ y (13:1:17)
2p 1 r2
2 1 r2
where the correlation coefficient is r and the random variable Z ¼ X þ Y. Here we can use
Eq. (13.1.14) only. Substituting fXY (x,y) into Eq. (13.1.14), we have
ð1
fZ (z) ¼
fXY (x,z x)dx
1
1
2
2
½x 2rx(z x) þ (z x) dx
exp 2(1 r2 )
1
ð1
1
1
2
2
2
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi
½2x
exp 2rxz
þ
2rx
þ
z
2zx
dx
2(1 r2 )
2p 1 r2 1
ð1
1
1
2
2
½2x
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi
exp
(1
þ
r)
2xz(1
þ
r)
þ
z
dx
2(1 r2 )
2p 1 r2 1
ð1
1
2
2
z2 =2(1r2 )
p
ffiffiffiffiffiffiffiffiffiffiffiffiffi
¼
e½(1þr)=(1r )½x xz dx
e
2
2p 1 r
1
ð1
1
2
2
2
e½1=(1r)½x xz dx
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ez =2(1r )
2
2p 1 r
1
1
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi
2p 1 r2
ð1
(13:1:18)
214
Functions of Multiple Random Variables
Completing squares under the integral sign in Eq. (13.1.18), we have
ð1
1
1 h
z i 2 z2
z2 =2(1r2 )
x
fZ (z) ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi e
exp
dx
1r
2
4
2p 1 r2
1
ð1
1
1 h
z i2
2
2
2
x
exp
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi e½z =2(1r )þ½z =4(1r)
dx
1r
2
2p 1 r2
1
( )
ð1
1 z2 =4(1þr)
1
1 x (z=2) 2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ pffiffiffiffiffiffi e
exp
dx
2 (1 r)=2)
2p
2pð1 r2 Þ 1
( )
ð1
1 z2 =4(1þr)
1
1
1 x (z=2) 2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
exp
dx
¼ pffiffiffiffiffiffi e
2 (1 r)=2)
2(1 þ r)
2p
1 r 1
2p
2
|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}
¼1
1
1
2
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffi ez =4(1þr)
2(1 þ r)
2p
(13:1:19)
Thus, we see that the sum of jointly Gaussian-distributed zero mean, unit variance
random variables with correlation coefficient r is also a zero mean Gaussian with variance 2(1þ r). If r ¼0, we have independent Gaussian random variables and variance
equals 2, which is the sum of unit variances. If r ¼ 1, we have essentially Z ¼ 2X and
the variance is 4.
Example 13.1.7 Two independent Gaussian random variables X and Y are with zero
mean and variance s2. The probability density functions of X and Y are
1
1
2
2
(13:1:20)
fX (x) ¼ pffiffiffiffiffiffi e(x=2s ) ; fY (x) ¼ pffiffiffiffiffiffi e(y=2s )
s 2p
s 2p
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Another random variable Z is given by Z ¼ g(X,Y) ¼ X 2 þ Y 2 . We have to find the pdf
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
fZ(z). The region Dxy ¼ x2 þ y2 z is shown in Fig. 13.1.7.
Using Eq. (13.1.6), we have
ðð
FZ (z) ¼
1 (j2 þh2 Þ=2s2
e
dj dh
2 2p
s
pffiffiffiffiffiffiffiffiffi
2
2
x þy z
We can solve this integral by using polar coordinates using the following substitution:
j ¼ r cos u; h ¼ r sin u; dj dh ¼ r dr du;
u ¼ tan1
h
j
Thus
ð 2p ð z
1 (r2 =2s2 )
e
r dr du ¼
2
0
0 s 2p
h
i
2
2
¼ 1 e(z =2s ) u(z)
FZ (z) ¼
ðz
r (r2 =2s2 )
e
dr
2
0s
(13:1:21)
13.1
Function of Two Random Variables, Z 5 g(X,Y )
215
FIGURE 13.1.7
and
fZ (z) ¼
z (z2 =2s2 )
e
u(z)
s2
(13:1:22)
where u(z) is the usual unit step function. This is a Rayleigh distribution as given in
Eq. (7.8.5) or a chi-square distribution with 2 degrees of freedom.
Example 13.1.8 This example is similar to the previous example except that
Z ¼ g(X,Y) ¼ X 2+Y 2.
The region Dxy ¼ x 2+y 2 z. We can now find the pdf fZ(z) using polar coordinates as
in the previous example. Thus
ð 2p ð pffiz
ð pffiz
1 (r2 =2s2 )
r (r2 =2s2 )
e
r dr du ¼
e
dr
FZ (z) ¼
2
2
0 s 2p
0 s
0
h
i
2
¼ 1 e(z=2s ) u(z)
(13:1:23)
and the density function fZ(z) is given by
1 z=2s2
e
fZ (z) ¼
u(z)
2s2
(13:1:24)
This is an exponential distribution or a chi-square distribution given by Eq. (7.7.3) with 2
degrees of freedom.
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
The density functions for Z ¼ X 2 þ Y 2 and Z ¼ X 2 þ Y 2 given by Eqs. (13.1.22)
and (13.1.24) are shown in Fig. 13.1.8 for comparison.
Example 13.1.9 In Example 13.1.7 we assumed that the independent random variables
X and Y are of zero mean. In this example, we will assume that X and Y have the same
variance s2 and that their mean values are mX and mY respectively. The density
functions are
1
2
fX (x) ¼ pffiffiffiffiffiffi e½(xmx )=2s ;
s 2p
1
2
fY (y) ¼ pffiffiffiffiffiffi e½(ymy )=2s s 2p
(13:1:25)
216
Functions of Multiple Random Variables
FIGURE 13.1.8
We will now find fZ(z) for Z ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
X 2 þ Y 2 . Function FZ(z) is now given by
1
(j mX )2 þ (h mY )2
exp
dj dh;
2
s 2p
2s2
pffiffiffiffiffiffiffiffiffi
2
2
ðð
FZ (z) ¼
z.0
(13:1:26)
x þy z
Expanding Eq. (13.1.26) we have,
1
j2 þ h2 2(mX j þ mY h) þ m2X þ m2Y
exp
FZ (z) ¼
dj dh;
s2 2p
2s2
pffiffiffiffiffiffiffiffiffi
2
2
ðð
z.0
x þy z
(13:1:27)
Substituting polar coordinates in Eq. (13.1.27), we obtain the following:
2
1
r 2r(mX cos u þ mY sin u) þ m2X þ m2Y
exp
r dr du
2
2s2
0
0 s 2p
2
2
2 ð
e(mX þmY )=s z ð1=2Þðr=sÞ2
e
¼
s2
0
ð 2p
1
r(mX cos u þ mY sin u)
exp
du r dr; z . 0
2p 0
s2
ð 2p ð z
FZ (z) ¼
(13:1:28)
We now define
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
m2X þ m2Y cos (c);
mX ¼
c ¼ tan
1
mY
mX
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
m2X þ m2Y sin (c)
mY ¼
ð13:1:29Þ
Function of Two Random Variables, Z 5 g(X,Y )
13.1
and a noncentrality parameter m ¼
(13.1.29), we have
217
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
m2X þ m2Y . Substituting m in Eqs. (13.1.28) and
rm( cos u cos c þ mY sin u sin c)
e
exp
du r dr
s2
0
0
ð 2p
2
2 ð
em =s z (1=2)ðr=sÞ2 1
rm( cos u c)
e
exp
du r dr; z . 0
(13:1:30)
¼
2p
s2
s2
0
0
2
em =s
FZ (z) ¼
s2
2
ðz
(1=2)ðr=sÞ2
1
2p
ð 2p
The integral under u is the modified Bessel function of zero order with argument rm=s2
given by
I0
ð 2p
rm
1
2
erm( cos uc)=s du
¼
s2
2p 0
Substituting this value in Eq. (13.1.30) and differentiating with respect to z, we obtain
zm
z
2
2
2
fZ (z) ¼ 2 e½(z þm )=s I0 2 ; z . 0
(13:1:31)
s
s
This is Rice’s distribution as given in Eq. (7.8.11). With m ¼ 0, we get the Rayleigh
distribution.
Example 13.1.10 In this example we have Z ¼ g(X,Y ) ¼ XY. The function
g(x,y) ¼ xy ¼ z is shown in Fig. 13.1.9 for z . 0. We have to find fZ(z) given the joint
density fXY(x,y).
The distribution function FZ(z) is obtained by integrating fXY(x,y) in the shaded region
given by g(x,y) z. Since the region Dxy ¼ fxy zg is bounded by two disjoint curves, we
have to integrate fXY (x,y) along the two regions f –1 , x 0g and f0 , x 1g. Hence
ðð
FZ (z) ¼
fXY (x,y)dy dx
xyz
ð0 ð1
ð 1 ð z=x
fXY (x,y)dy dx þ
¼
1
z=x
fXY (x,y)dy dx
0
FIGURE 13.1.9
1
(13:1:32)
218
Functions of Multiple Random Variables
We differentiate Eq. (13.1.32) using the chain rule of differentiation and obtain fZ(z):
ð1 z 1
z 1
fXY x, dx þ
fXY x, dx
x
x
x
x
1
0
ð1
1
z
fXY x, dx; z . 0
¼
x
1 jxj
ð0
fZ (z) ¼ (13:1:33)
It can be shown by similar techniques that Eq. (13.1.33) is valid for z 0. If X and Y are
independent, then
ð1
z
1
fX (x)fY
dx
x
1 jxj
fZ (z) ¼
for all z
(13:1:34)
We note that Eq. (13.1.34) is not a convolution integral even if the product of independent
random variables is involved.
The problem in Example 13.1.10 can also be solved using the techniques developed in
Chapter 12. We will first find the conditional density fZ(z j X ¼ x), given the conditional
density fY( y j X ¼ x) with Z ¼ xY. Since x is a scaling factor for the random variable Y,
we can use Eq. (12.3.7) and obtain
fZ (z j X ¼ x) ¼
1 z
fY j X ¼ x
jxj x
Hence, the unconditional density fZ(z) is given by
ð1
1 z
fY j X ¼ x fX (x)dx ¼
fZ (z) ¼
x
1 jxj
ð1
z
1
fXY x, dx
x
1 jxj
which is the same as before.
Example 13.1.11 We will now assume that X and Y are independent random variables
with identical Cauchy distributions given by
fX (x) ¼
2
2
; fY (y) ¼
2
p ð y þ 4Þ
þ 4Þ
pðx2
Substituting these values in Eq. (13.1.34), we have
ð1
fZ (z) ¼
1
2
2
dx
2
2 þ 4Þ
jxj
p
x
ð
z
1
p 2þ4
x
Since fZ(z) is an even function, we can rewrite this equation as
fZ (z) ¼
42
p2
ð1
0
x
dx
ðx2 þ 4Þðz2 þ 4x2 Þ
13.1
219
Function of Two Random Variables, Z 5 g(X,Y )
Substituting w ¼ x 2 and dw ¼ 2x dx and expanding into partial fractions, we have
4
p2
ð1
1
dw
2
0 ðw þ 4Þðz þ 4wÞ
ð
4 1
1
1
1
¼ 2
dw
p 0 ðz2 16Þ ðw þ 4Þ ðz2 þ 4wÞ
fZ (z) ¼
4
4
wþ4
1
2
ln(w þ 4) ln(z þ 4w) 0 ¼ 2 2
ln 2
¼ 2 2
p ðz 16Þ
p ðz 16Þ
z þ 4w
1
4
wþ4
4
1
4
ln 2
ln
¼ 2 2
¼ 2 2
ln 2
p ðz 16Þ
z þ 4w
p
4
z
ð
z
16
Þ
0
1
0
or
fZ (z) ¼
2
4
z
ln
p2 ðz2 16Þ
16
and fZ(z) is shown in Fig. 13.1.10.
Example 13.1.12 Given Z ¼ g(X,Y ) ¼ Y/X, we have to find fZ(z) from the knowledge of
the joint density function fXY(x,y). The region Dxy(x,y) ¼ fx,y: y xzg is shown in
Fig. 13.1.11. In the two regions of Dxy given by (1) x 0 and (2) x . 0, FZ(z) can be
written as follows:
ðð
FZ (z) ¼
fXY (x,y)dy dx
yxz
ð0 ð1
ð 1 ð xz
fXY (x,y)dy dx þ
¼
1 xz
fXY (x,y)dy dx
0
FIGURE 13.1.10
1
(13:1:35)
220
Functions of Multiple Random Variables
FIGURE 13.1.11
Using the chain rule of differentiation in Eq. (13.1.35), we can obtain fZ(z) as
ð0
ð1
fZ (z) ¼ ð1
fXY (x,xz)x dx þ
fXY (x,xz)x dx
0
1
jxjfXY (x,xz)dx
¼
(13:1:36)
1
As an illustration, we will find fZ(z) given that X and Y are jointly Gaussian random variables with means 0 and variances s2X and s2Y with correlation coefficient r, given by
fXY (x,y) ¼
2
1
1
x
xy
y2
pffiffiffiffiffiffiffiffiffiffiffiffiffi exp 2r
þ
2(1 r2 ) s2X
sX sY s2Y
2psX sY 1 r2
Substituting fXY(x,y) into Eq. (13.1.36) and using the fact that fXY(x,y) is an even function
we have,
2
pffiffiffiffiffiffiffiffiffiffiffiffiffi
fZ (z) ¼
2psX sY 1 r2
ð1
0
x2
1
z
z2
x exp 2r
þ
dx
sX sY s2Y
2(1 r2 ) s2x
1
s2 s2 (1 r2 )
pffiffiffiffiffiffiffiffiffiffiffiffiffi 2 X Y
2
psX sY 1 r2 sY 2rsX sY þ z2 sX
pffiffiffiffiffiffiffiffiffiffiffiffiffi
sX sY 1 r2
¼
sY s2Y
þ 2
ps2X z2 2rz
s X sX
¼
(13:1:37)
By completing the squares in the denominator of Eq. (13.1.37), we can write
pffiffiffiffiffiffiffiffiffiffiffiffiffi
sX s Y 1 r2
fZ (z) ¼ 2
p s2X z r ssXY þ s2Y ð1 r2 Þ
(13:1:38)
13.1
Function of Two Random Variables, Z 5 g(X,Y )
221
FIGURE 13.1.12
This is a Cauchy distribution centered at z ¼ r(sY/sX) and is shown in Fig. 13.1.12 with
s2X ¼ 4, s2Y ¼ 9, and r ¼ 35 :
Integrating Eq. (13.1.38) with respect to z, we obtain the distribution function FZ(z) as
"
#
1 1 1 sX z rsY
pffiffiffiffiffiffiffiffiffiffiffiffiffi
FZ (z) ¼ þ tan
2 p
sY 1 r 2
(13:1:39)
Example 13.1.13 We are given two independent random variables, X and Y, which are
exponentially distributed as
fX (x) ¼ 2e2x u(x);
fY (x)ey u( y)
We have to find the density function fZ(z) of the random variable given by Z ¼ (Y/X ).
From Eq. (13.1.36), we have
ð1
fZ (z) ¼
jxjfXY (x,xz)dx
1
ð1
x 2e
¼
2x
e
xz
ð1
dx ¼
0
¼
2xex(2þz) dx
0
2
(2 þ z)2
Integrating fZ (z), we obtain the distribution FZ (z) as
FZ (z) ¼ 1 These functions are shown in Fig. 13.1.13.
2
u(z)
2þz
222
Functions of Multiple Random Variables
FIGURE 13.1.13
B 13.2 TWO FUNCTIONS OF TWO RANDOM
VARIABLES, Z 5 g(X,Y ), W 5 h(X,Y )
We are given two random variables, X and Y, and their joint density function fXY(x,y). We
form the functions Z ¼ g(X,Y ) and W ¼ h(X,Y ) and determine the joint density fZW(z,w)
from knowledge of fXY (x,y). The analysis is very similar to the one presented in the
previous section.
The functions Z ¼ g(X,Y ) and W ¼ h(X,Y ) map the domain space of fZ,Wg given by
fCxy, FXY, PXYg into the range space fCzw, FZW, PZWg as shown in Fig. 13.2.1.
From Fig. 13.2.1, we obtain
Z ¼ g(X,Y), W ¼ h(X,Y);
{Cxy , FXY ,PXY } ! {Czw , FZW , PZW }
(13:2:1)
and the mapping of fX, Y g is from the probability space fS, F, Pg (where S is the product
space S ¼ SX SY) into the domain space of fZ,Wg given by fCxy, FXY, PXYg:
X,Y : {S, F, P} ! {Cxy , FXY , PXY }
(13:2:2)
Alternately, if Dxy , Cxy and Dzw , Czw, then Dzw is the image of Dxy under the transformation fg,hg and Dxy is the inverse image of Dzw under the transformation {g – 1, h – 1}.
Hence, Dzw is the set of all points (x,y) belonging to Dxy such that g(x,y), h(x,y) belongs
FIGURE 13.2.1
13.2
Two Functions of Two Random Variables, Z 5 g(X,Y ), W 5 h(X,Y )
223
to Dzw. This is represented by
Dxy ¼ {x,y:g(x,y),h(x,y) [ Dzw }
(13:2:3)
We also have conditions similar to the last section for Z and W to be a random variables.
Conditions on Z 5 g(X,Y ) and W 5 h(X,Y ) to Be Random Variables
1. The domain of fg(x,y), h(x,y)g must include the range of the RV Z and W.
2. For each z,w , Dzw, the set fg(x,y) z, h(x,y) wg must form a field FZW; that
is, it must consist of the unions and intersections of a countable number of
intervals.
3. The probability of the events fg(x,y) ¼ +1g and fh(x,y) ¼ +1g is 0.
In essence, the transformation fZ ¼ g(X,Y ), W ¼ h(X,Y )g induces a probability measure
PZW on FZW, and we have to find this probability measure. From Fig. 13.2.1 we can write
PZW (Z,W) ¼ P(Z,W [ Dzw )
¼ PXY g1 (Z,W),h1 (Z,W) [ Dxy ) ¼ PXY X,Y [ Dxy
(13:2:4)
The set Dzw consists of all the countable unions and intersections of the type
fa , Z bg, fc , W dg. Without loss of generality, we will assume that
Dzw ¼ fZ z,W wg. Thus, substituting for Dxy from Eq. (13.2.3) in Eq. (13.2.4),
we have
PZW (Z,W) ¼ PXY X,Y [ Dxy ¼ PXY ½ x,y:g(x,y),h(x,y) [ Dzw ¼ PXY ½ x,y:g(x,y) z,h(x,y) w ¼ FXY (z,w)
¼ P(Z z,W w) ¼ FZW (z,w)
(13:2:5)
Equation (13.2.5) can also be written as
ðð
PZW (Z,W) ¼ PXY (X,Y) [ Dxy ¼ FZW (z,w) ¼
fXY (j,h)dj dh
(13:2:6)
Dxy
where fXY(x,y) is the joint density of the random variables X and Y.
In conclusion, we can say that the probability of fZ z, W wg is given by the probability of the set of all values of x and y such that fg(x,y) z, h(x,y) wg or by integrating
the joint density fXY(x,y) over the domain Dxy. We will use both these interpretations in
solving problems.
Example 13.2.1 Here we are given that Z ¼ max(X,Y ), W ¼ min(X,Y ) and we have to
find FZW(z,w) from the knowledge of the joint density FXY(x,y). Both Z and W are shown in
Fig. 13.2.2.
The regions min(X,Y ) w ¼ fX wg < fY zg and max(X,Y ) z ¼ fX zg >
X,Y ) z ¼ fX zg > fY zg are shown in Fig. 13.2.2. There are two regions of interest:
w . z and w z. In the first region w . z, Dxy equals the semiinfinite square defined by
the point B ¼ (z, z). Thus
FZ,W (z,w) ¼ FXY (z,z) ¼ FZ (z):
w.z
(13:2:7)
224
Functions of Multiple Random Variables
FIGURE 13.2.2
In the second region w z, Dxy equals the semiinfinite square defined by the point
B ¼ (z,z) minus the square ABCD. Hence
FZ,W (z,w) ¼ FXY (z,z) P(ABCD)
¼ FXY (z,z) ½FXY (z,z) FXY (z,w) FXY (w,z) þ FXY (w,w)
¼ FXY (z,w) þ FXY (w,z) FXY (w,w); w z
(13:2:8)
When w ¼ z, Eq. (13.2.8) is identical to Eq. (13.2.7).
We can obtain FW(w) by letting z tend to 1 in Eq. (13.2.8), yielding
FZ,W (z,w) ¼ FXY (z,w) þ FXY (w,z) FXY (w,w)
¼ FX (w) þ FY (w) FXY (w,w) ¼ FW (w)
which is similar to Eq. (13.1.7) derived in the previous section.
If X and Y are independent, then Eqs. (13.2.7) and (13.2.8) become
w.z
FX (z)FY (z):
FZ,W (z,w) ¼
FX (z)FY (w) þ FX (w)FY (z) FX (w)FY (w): w z
(13:2:9)
(13:2:10)
Example 13.2.2 To continue the previous example, the density functions of two independent random variables X and Y are given by
fX (x) ¼ 2e2x u(x);
fY (y) ¼ ey u(y)
Hence the joint density function fXY(x,y) is
fXY (x;y) ¼ 2eð2xþyÞ u(x)u(y)
Given random variables Z ¼ max(X,Y ) and W ¼ min(X,Y ), we have to find the joint
distribution FZW(z,w) and the marginal distributions FZ(z) and FW(w).
The joint distribution FXY(x,y) can be obtained by integrating fX(x) and fY( y):
FXY (x,y) ¼ 1 e2x ð1 ey Þu(x)u(y)
¼ 1 e2x ey þ e(2xþy) u(x)u(y)
13.2
Two Functions of Two Random Variables, Z 5 g(X,Y ), W 5 h(X,Y )
Using Eq. (13.2.10), we obtain
8
1 e2z ð1 ez Þ;
z . 0,
>
<
2z
w
2w
z
ð1 e Þ þ 1 e
ð1 e Þ
1e
FZW (z,w) ¼
>
:
1 e2w ð1 ew Þ;
z . 0,
z . 0,
1 e2z ez þ e3z ;
¼
2z
z
(2zþw)
(zþ2w)
3w
1e e þe
þe
e ; z . 0,
225
w.z
wz
w.z
wz
as the joint distribution function FZW(z,w). The marginal distribution FZ(z) is obtained by
letting w ¼ z in the second equation of FZW(z,w) since w cannot exceed z. Thus
FZ (z) ¼ 1 e2z ez þ e3z ,
z.0
which is the same as the first equation. To obtain FW(w), we let z tend to 1 in the second
equation of FZW(z,w) and obtain
FW (w) ¼ 1 e3w ,
w.0
Example 13.2.3 We are given Z ¼ X 2 þ Y 2 and W ¼ Y/X, where X and Y are zero mean
independent Gaussian random variables with joint density function
fXY (x,y) ¼
1 ½(x2 þy2 )=2s2 e
2ps2
The region Dxy is the set of all values of x,y such that x 2 þ y 2z and y/x w and is shown
in Fig. 13.2.3. The joint distribution FZW(z,w) is found by integrating FZW(z,w) over the
shaded region of the figure.
ðð
FZW (z,w) ¼
1 ½(x2 þy2 )=2s2 e
dx dy
2
Dxy 2ps
FIGURE 13.2.3
226
Functions of Multiple Random Variables
pffiffi
pffiffi
pffiffi pffiffi
Using polar coordinates, x ¼ z cos (u), y ¼ z sin (u), dx dy ¼ z d z du, we can
rewrite the preceding equation as
ð pffiz
ðu
FZW (z,u) ¼ 2
(p=2) 0
1 (z=2s2 ) pffiffi pffiffi
e
z d z da
2ps2
with the factor 2 appearing since there are two shaded regions. This last equation can be
simplified as follows:
ð
1
p z 1 (z=2s2 ) pffiffi dz
uþ
z pffiffi
e
p
2 0 s2
2 z
1 u
p
p
2
þ
¼
1 eðz=2s Þ ; z . 0, , u 2 p
2
2
FZW (z,u) ¼
Since w ¼ (y/w) ¼ tan(u), we can substitute u ¼ tan – 1(w) and obtain the marginal
distributions FZ(z) and FW(w) as follows:
2
FZ (z) ¼ 1 e(z=2s ) ;
z . 0;
FW (w) ¼
1 tan1 (w)
þ
; 1 , w , 1
2
p
The corresponding density functions are obtained by differentiation:
fZ (z) ¼
1 (z=2s2 )
e
; z . 0;
2s2
fW (w) ¼
1
; 1 , w , 1
p(1 þ w2 )
The functions FW(w) and fW(w) are Cauchy-distributed and are shown in Figures 13.2.4
and 13.2.5.
The examples given above are some of the simple problems to solve with transformations of multiple functions of multiple random variables. In general, this method of
solving for distribution functions of two functions of two random variables is more
difficult. We now analyze alternate methods of finding the joint density functions.
FIGURE 13.2.4
13.3
Direct Determination of Joint Density fZW(z,w ) from fXY (x,y )
227
FIGURE 13.2.5
B 13.3 DIRECT DETERMINATION OF JOINT DENSITY
fZW(z,w ) FROM fXY (x,y )
It is easier to obtain fZW(z,w) directly from the joint density function fXY(x,y) exactly as
we did in the case of a single function of a single random variable. We are given the transformation Z ¼ g(X,Y ) and W ¼ h(X,Y ), and the inverse images g – 1(Z,W ) and h – 1(Z,W )
exist. However, corresponding to the differential parallelogram DDzw in the Z –W
P plane,
there may be several differential unconnected parallelograms DDxy ¼ dxkdyk,
k ¼ 1, . . . , n in the X– Y plane as shown in Fig. 13.3.1. These differential regions are connected by the relation
dz dw ¼ dx1 dy1 jJ(x1 ,y1 )j þ þ dxn dyn jJ(xn ,yn )j
where jJ(x,y)j is the determinant of the Jacobian matrix J(x,y), defined by
2
3
@z(x,y) @z(x,y)
6 @x
@y 7
7
J(x,y) ¼ 6
4 @w(x,y) @w(x,y) 5
@x
@x
FIGURE 13.3.1
(13:3:1)
(13:3:2)
228
Functions of Multiple Random Variables
The joint density fZW(z,w) can be obtained from fXY(x,y) as follows:
fZW (z,w)dz dw ¼ fXY (x1 ,y1 )dx1 dy1 þ þ fXY (xn ,yn )dxn dyn
(13:3:3)
Substitution of Eq. (13.3.1) in Eq. (13.3.3) results in
fZW (z,w)dz dw ¼
¼
fXY (x1 ,y1 )dz dw
fXY (xn ,yn )dz dw
þ þ
jJ(x1 ,y1 )j
jJ(xn ,yn )j
n
X
fXY (xk ,yk )dz dw
jJ(xk ,yk )j
k¼1
(13:3:4)
n
X
fXY (xk ,yk )
jJ(x
k ,yk )j
k¼1
(13:3:5)
or
fZW (z,w) ¼
From this derivation it is clear that certain restrictions have to be satisfied for g(x,y) and
h(x,y) if we have to determine fZW(z,w) directly from fXY(x,y):
1. All the partial derivative of g(x,y) and h(x,y) must exist.
2. There cannot be any planar solutions to g(x,y) ¼ z and h(x,y) ¼ w.
We can now formulate the methodology of finding fZW(z,w) from fXY(x,y).
Steps for Direct Determination of fZW(z, w)
1. Check whether the restrictions on g(x,y) and h(x,y) are satisfied.
2. Find the n real solutions to the system of equations
g(x,y) ¼ z;
h(x,y) ¼ w
and express them in terms of z and w. Let the n solutions be fxk, yk, k ¼ 1, . . . , ng.
3. Find the Jacobian determinant jJj for each xk,yk, and express this also in terms of z
and w.
4. Find the density function fZW(z,w) from Eq. (13.3.5):
fZW (z,w) ¼
Example 13.3.1
n
X
fXY (xk ,yk )
jJ(x
k ,yk )j
k¼1
The functions Z ¼ g(X,Y ) and W ¼ h(X,Y ) are defined by
g(X,Y) ¼ X þ Y; h(X,Y) ¼ X Y
where X and Y are independent positive random variables with joint density fXY (x,y) given
by 6e – (3xþ2y) . u(x)u( y). We have to find fZW(z,w).
1. All partials exist for g(x,y) ¼ x þ y and h(x,y) ¼ x y.
2. In the equations x þ y ¼ z and x y ¼ w, we note that z . 0 and w can have negative values. Solving for x and y results in one real solution:
1 1 1
zþw
zw
z
x
or x ¼
; y¼
¼
1 w
y
2 1
2
2
Since x and y are positive, we must have z . jwj and z . w.
3. The Jacobian determinant is jJj ¼ 2.
13.3
Direct Determination of Joint Density fZW(z,w ) from fXY (x,y )
229
FIGURE 13.3.2
4. The joint density function fZW(z,w) from Eq. (13.3.5) is given by
h zþw
1
z wi
þ2
fZW (z,w) ¼ 6 exp 3
¼ 3e½(5zþw)=2 ; z . 0, z , w z
2
2
2
Finding the marginal densities fZ(z) and fW(w) are a little tricky because of finding the
limits of integrations with respect to z and w.
ðz
3e½(5zþw)=2 dw; z . 0, z , w z
fZ (z) ¼
z
¼ 6 e2z e3z u(z)
Function fZ(z) is shown in Fig. 13.3.2.
Since w can take positive and negative values, we have the following integrals for
fW(w) for the two regions w . 0 and w 0. In the region w . 0, we have
ð1
fW (w) ¼
3e½(5zþw)=2 dz; w . 0
w
6
¼ e3w u(z)
5
and in the region w 0, we have
ð1
fW (w) ¼
3e½(5zþw)=2 dz;
w0
w
¼
6 2w e u(w)
5
The function fW(w) is shown in Fig. 13.3.3.
The distribution function FW(w) is obtained by integrating fW(w) and is given by
8
3
>
< (e2w )
w0
FW (w) ¼ 5
>
: 3 þ 2 (1 e3w ) w . 0
5 5
It is shown in Fig. 13.3.4.
230
Functions of Multiple Random Variables
FIGURE 13.3.3
Example 13.3.2 Two independent positive random variables X and Y are jointly exponentially distributed with density function fXY(x,y) ¼ 6e 2(3x þ 2y) u(x)u( y), exactly as in
the previous example. Two other random variables are Z and W, given by, Z ¼
2X=(X þ Y) and W ¼ X þ Y. We have to find the joint density fZW(z,w) and the marginal
densities fZ(z) and fW(w).
1. All partials exist for g(x,y) ¼ 2x=(x þ y) and h(x,y) ¼ x þ y.
2. Solving for 2x=(x þ y) ¼ z and x þ y ¼ w, we have x ¼ zw=2
y ¼ ½1 (z=2)w. Since x and y are positive, w . 0 and 0 , z 2.
3. The Jacobian matrix is
2
2y
4
J ¼ (x þ y)2
1
3
2x
(x þ y)2 5
1
and the determinant is
j Jj ¼
2
2
¼
xþy w
FIGURE 13.3.4
and
13.3
Direct Determination of Joint Density fZW(z,w ) from fXY (x,y )
231
4. The joint density function fZW(z,w) from Eq. (13.3.5) is given by
nh zw
w
z io
fZW (z,w) ¼ 6 exp 3 þ 2(1 )w ¼ 3wew½2þ(z=2) ; w . 0, 0 , z 2
2
2
2
Integrating over all w, we obtain fZ(z)
ð1
fZ (z) ¼
3wew½2þðz=2Þ dw ¼
0
12
;
(z þ 4)2
0,z2
and the corresponding distribution function FZ(z) is
8
0:
>
>
z0
< 4
: 0,z2
FZ (z) ¼ 3 1 >
zþ4
>
:
1:
z.2
The functions fZ(z) and FZ(z) are shown in Fig. 13.3.5.
Integrating over all z, we obtain fW(w)
ð2
fW (w) ¼ 3w ew½2þðz=2Þ dz ¼ 6(e2w e3w );
w.0
0
and this result is the same one obtained in the previous example and shown in Fig. 13.3.2.
Integrating fW(w), we obtain FW(w):
FW (w) ¼ 1 þ 2e3w 3e2w ;
w.0
Example 13.3.3 We will find fZW(z,w) from fXY(x,y) for Example 13.2.3, where Z ¼
2
2
2
X 2 þ Y 2 and W ¼ Y=X and the joint density fXY (x,y) ¼ (1=2ps2 )e½(x þy )=2s .
1. All partials exist for g(x,y) ¼ x2 þ y2 and h(x,y) ¼ y=x.
2. We solve x2 þ y2 ¼ z and y=x ¼ w. Substituting y ¼ wx in x2 þ y2 ¼ z, we obtain
rffiffiffiffiffiffiffiffiffiffiffiffiffiffi
rffiffiffiffiffiffiffiffiffiffiffiffiffiffi
z
w2 z
2
2
and
y
¼
+
x 1 þ w ¼ z or x ¼ +
1 þ w2
1 þ w2
FIGURE 13.3.5
232
Functions of Multiple Random Variables
or
2 0rffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffi1
3
2
z
w
z
A; z . 0, 1 , w , 1
[email protected]
7
,
6
7
1 þ w2 1 þ w2
(x1 , y1 )
6
7
¼6
7
rffiffiffiffiffiffiffiffiffiffiffiffiffiffi!
rffiffiffiffiffiffiffiffiffiffiffiffiffiffi
(x2 , y2 )
6
7
2
z
w z
4
5
,
;
z
.
0,
1
,
w
,
1
2
2
1þw
1þw
3. The Jacobian matrix is given by
2
2x
J¼4 y
2
x
3
2y
15
x
and the Jacobian determinant
y2
jJj ¼ 2 1 þ 2 ¼ 2(1 þ w2 )
x
4. The joint density fZW(z,w) is given by
1
1 (z=2s2 )
1 ðz=2s2 Þ
e
þ
e
fZW (z,w) ¼
2(1 þ w2 ) 2ps2
2ps2
or
1
1 (z=2s2 )
e
;
fZW (z,w) ¼
p(1 þ w2 ) 2s2
z . 0, 1 , w 1
Integrating this equation with respect to w, the marginal density fZ(z) is
ð1
1
dw
2
fZ (z) ¼ 2 e(z=2s )
2
2s
1 p(1 þ w )
¼
1 (z=2s2 ) tan1 (w)
e
2s2
p
1
¼
1
1 1(z=2s2 )
e
;
2s2
z.0
and by a similar integration with respect to z we obtain fW(w) as
1
fW (w) ¼
p(1 þ w2 )
which is a Cauchy density. These results correspond to the results obtained in Example
13.2.3.
Example 13.3.4 Two independent zero mean Gaussian random variables X and Y has
joint density function fXY(x,y) given by
1 (1=2)½(x2 þy2 Þ=s2 fXY (x,y) ¼
e
2ps2
Two other random variables Z and W are defined by
8
1 Y
>
>
< tan
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
X
Z ¼ X2 þ Y 2; W ¼
>
Y
>
: p þ tan1
X
if
X.0
if
X0
13.4
Solving Z 5 g(X,Y ) Using an Auxiliary Random Variable
233
The angle W has been defined in this manner so that it covers the region ((p=2), (3p=2)
since tan1 (Y=X) covers only the region ((p=2),(p=2)]. We have to find the joint density
fZW(z,w) and the marginal densities fZ(z) and fW(w). We apply the usual four steps to find a
solution to this problem.
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1. All partials exist for g(x,y) ¼ x2 þ y2 and h(x,y) ¼ tan1 (y=x).
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2. We solve for x2 þ y2 ¼ z and tan1 (y=x) ¼ w for x . 0 and p þ tan1 (y=x) ¼
w ffiffiffiffiffiffiffiffiffiffiffiffiffiffi
for x 0. We note that tan(w) ¼ y/x for both x 0 and x . 0. Squaring
p
x2 þ y2 ¼ z and substituting y=x ¼ tan (w), we can write
y2
2
x 1 þ 2 ¼ z2 or x2 1 þ tan2 (w) ¼ z2 or x2 ¼ z2 cos2 (w)
x
However, for x . 0; (p=2) , w (p=2) and cos (w) . 0 and for x 0; (p=2) , w (3p=2) and cos(w) 0. Hence, the only solutions are x ¼
z cos(w) and y ¼ z sin(w) in the range (p=2) , w (3p=2).
3. The Jacobian matrix and its determinant are given by
2
3
x
y
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
6 x2 þ y2
1
1
x2 þ y2 7
7
J¼6
¼
4 y
y 5 and jJj ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
z
x þy
x2 þ y2
x2 þ y2
4. The joint density fZW(z,w) is given by
fZW (z,w) ¼
1 z (1=2)(z2 =s2 )
e
; z.0
2p s2
Integrating with respect to w in the range (p=2) , w (3p=2), we obtain
z
2
2
fZ (z) ¼ 2 e(1=2)(z =s ) ; z . 0
s
which is a Rayleigh distribution, and the angle w is uniformly distributed in
(p=2) , w (3p=2),
or
fW (w) ¼
1
p
3p
; ,w
2p
2
2
and Z and W are also independent random variables.
B 13.4 SOLVING Z 5 g(X,Y ) USING AN AUXILIARY
RANDOM VARIABLE
The problem of finding the density function fZ(z) of a function of two random variables
Z ¼ g(X,Y ) can be considerably simplified by converting to two functions of two
random variables with the use of an auxiliary random variable W ¼ X or Y, provided
g(x,y) satisfies the two restrictions of differentiability and no straight-line solutions.
Example 13.4.1 We will now solve Example 13.1.10 using an auxiliary random variable. Here g(X,Y ) ¼ Z ¼ XY and we introduce an auxiliary random variable
W ¼ h(X,Y ) ¼ X.
234
Functions of Multiple Random Variables
1. The restrictions on g(x,y) ¼ xy and h(x,y) ¼ x are satisfied.
2. Solving for xy ¼ z and x ¼ w, we obtain the real solution, x ¼ w and y ¼ z=w.
3. The Jacobian matrix and its absolute determinant are
y x
or kJk ¼ jxj ¼ jwj
J¼
1 0
4. The joint density fZW(z,w) is given by
fZW (z,w) ¼
z
1
fXY w,
jwj
w
The function fZ(z) is obtained by integrating this equation with respect to w
ð1
z
1
fXY w, dw
FZ (z) ¼
w
1 jwj
which is the same as Eq. (13.1.33).
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Example 13.4.2 This is the same as Example 13.1.7, where Z ¼ g(X,Y) ¼ X 2 þ Y 2 .
We introduce the auxiliary random variable W ¼ h(X,Y ) ¼ X and proceed to solve it
using the techniques of the previous section. The joint density fXY(x,y) is given by
fXY (x,y) ¼
1 (1=2)½(x2 þy2 )=s2 e
2ps2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1. The restrictions on g(x,y) ¼ x2 þ y2 and h(x,y) ¼ x are satisfied.
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2. Solving for x2 þ y2 ¼ z and x ¼ w,pwe
obtain x2 þ y2 ¼ z2 or y2 p
¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
z2 x2 , and
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
the two solutions are x1 ¼ w, y1 ¼ z w and x2 ¼ w, y2 ¼ z2 w2 with
2z , w z for real y1 and y2.
3. The Jacobian matrix and its absolute determinant are
2
3
x
y
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
z2 w2
y
2
2
2
2
4
5
;
x þy
x þy
J¼
or kJk ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼
z
x2 þ y2
1
0
4. The joint density function fZW(z,w) is given by
z
2 (1=2)(z2 =s2 )
fZW (z,w) ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
e
; z . 0, z , w z
z2 w2 2ps2
Integrating with respect to the variable w, we obtain
ðz
z
dw
2
2 1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; z . 0
fZ (z) ¼ 2 e(1=2)(z =s )
s
p z z2 w2
Since the integrand is even, we can write this equation as
ð
z (1=2)(z2 =s2 ) 2 z
dw
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; z . 0
fZ (z) ¼ 2 e
s
p 0 z2 w2
z
z
2
2 2 p
2
2
¼ 2 e(1=2)(z =s ) ; z . 0
¼ 2 e(1=2)(z =s )
s
p2 s
which is the same as the result obtained in Eq. (13.1.22).
jwj z
13.4
Solving Z 5 g(X,Y ) Using an Auxiliary Random Variable
235
Example 13.4.3 This is the same as Example 13.1.8, where Z ¼ g(X,Y) ¼ X 2 þ Y 2 . We
introduce the auxiliary random variable W ¼ h(X,Y ) ¼ X and proceed to solve it using the
techniques of the previous section. The joint density fXY (x,y) is given by
fXY (x,y) ¼
1 (1=2)½(x2 þy2 )=s2 e
2ps2
1. The restrictions on g(x,y) ¼ x2 þ y2 and h(x,y) ¼ x are satisfied.
2. Solving for x2pþffiffiffiffiffiffiffiffiffiffiffiffiffi
y2 ¼ z and x ¼ w, we obtain
from Y 2 ¼ z x2 , the two solutions
pffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffi
pffiffi
x1 ¼ w, y1 ¼ z w2 and x2 ¼ w, y2 ¼ z w2 , with z , w z for real
y1 and y2.
3. The Jacobian matrix and its absolute determinant are
pffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffi
2x 2y
or kJk ¼ 2y ¼ 2 z w2 ; z . 0, jwj z
J¼
1 0
4. The joint density function fZW(z,w) corresponding to the two solutions is given by,
1
2 (1=2)(z=s2 )
e
;
fZW (z,w) ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
2ps
2 zw
pffiffi
pffiffi
z . 0, z , w z
Integrating with respect to the variable w, we obtain
ð pffi
1 (1=2)(z=s2 ) z
dw
pffiffiffiffiffiffiffiffiffiffiffiffiffi ; z . 0
fZ (z) ¼ 2 e
pffi
2
2s
zp zw
pffiffi
pffiffi
Substituting w ¼ z sin (u) and dw ¼ z cos (u)du, we obtain
ð p pffiffi
2
1
z cos (u)du
2 1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ; z . 0
fZ (z) ¼ 2 e(1=2)(z=s )
2s
p p2
z cos2 (u)
¼
1 (1=2)(z=s2 )
e
; z.0
2s2
which is the same result obtained in Eq. (13.1.24).
Example 13.4.4 The random variable Z is given by Z ¼ (X þ Y )2, where X and Y are
zero mean independent Gaussian random variables given by the joint density function
fXY (x,y) ¼
1 (1=2)½(x2 þy2 )=s2 e
2ps2
We have to find the density function fZ(z) using auxiliary random variable W ¼ X.
1. The restrictions on g(x,y) ¼ (x þ y)2 and h(x,y) ¼ x are satisfied.
pffiffi
2. Solving for (x þ y)2 ¼ z and x ¼ w, we obtain two solutions: x1 ¼ w, y1 ¼ z w
pffiffi
and x2 ¼ w, y2 ¼ z w.
3. The Jacobian matrix and its absolute determinant are
pffiffi
2(x þ y) 2(x þ y)
J¼
or kJk ¼ 2 z;
1
0
z.0
236
Functions of Multiple Random Variables
4. The joint density fZW(z,w) corresponding to the two solutions is
1
1 (1=2s2 )½w2 þ(pffizw)2 fZW (z,w) ¼ pffiffi
e
2 z 2ps2
1
1 (1=2s2 )½w2 þ(pffizw)2 þ pffiffi
e
; z.0
2 z 2ps2
We will now simplify this expression. Completing the squares in the first and the
second terms on the righthand side of the equation, we obtain
pffiffi
1
1
1
2
2
exp
½w
þ
(
z
w)
pffiffi
2 z 2ps2
2s2
(
"
pffiffi2 #)
1
1
2
z
z
¼ pffiffi
exp 2 w þ
2 z 2ps2
2s
4
2
pffiffi
1
1
1
2
2
exp
½w
þ
(
z
w)
pffiffi
2 z 2ps2
2s2
(
"
pffiffi2 #)
1
1
2
z
z
¼ pffiffi
exp 2 w þ
þ
2
2 z 2ps2
2s
4
Hence the joint density fZW(z,w) is given by
1
1 (z=4s2 ) n (1=s2 )½w(pffiz=2)2 o
fZW (z,w) ¼ pffiffi
e
e
2 z 2ps2
pffi
(1=s2 )½wþ( z=2)2
þe
; z . 0, 1 , w , 1
We integrate this equation over all w to obtain fZ(z):
ð
o
pffi
1
1 (z=4s2 ) 1 n (1=s2 )½w(pffiz=2)2
(1=s2 )½wþ( z=2)2
fZ (z) ¼ pffiffi
e
e
þ
e
dw
2 z 2ps2
1
2
pffiffiffiffi
1
1 (z=4s2 ) pffiffiffiffi
1 e(z=4s )
pffiffiffiffi pffiffi ; z . 0
¼ pffiffi
e
½s
p
þ
s
p
¼
2 z 2ps2
2s p
z
The distribution function FZ(z) is obtained by integrating fZ(z) over z:
pffiffi
ð z (z=4s2 )
1
e
z
pffiffiffi dz ¼ erf
FZ (z) ¼ pffiffiffiffi
2s
2s p 0
z
Ðz
2
where erf(z) is defined by erf(z) ¼ p2ffiffipffi 0 ej dj and
ð
pffiffi
1 z ej
erf( z) ¼ pffiffiffiffi pffiffiffi dj
p 0 j
Functions fZ(z) and FZ(z) are shown in Fig.13.4.1.
Example 13.4.5 In this example X is a zero mean unit variance random variable given by
the density function
1
2
fX (x) ¼ pffiffiffiffiffiffi e(x =2)
2p
13.4
Solving Z 5 g(X,Y ) Using an Auxiliary Random Variable
237
FIGURE 13.4.1
and Y is an independent chi-square random variable with n degrees of freedom whose
density is given by Eq. (7.7.2):
y(v=2)1 e(y=2)
v u(y)
2v=2 G
2
pffiffiffiffiffiffiffiffi
The density function of the random variable Z ¼ X Y=v is called the Student-t distribution (Section 7.9) and finds extensive application in statistical analyses. We will use
an auxiliary variable W ¼ Y:
fY (y) ¼
pffiffiffiffiffiffiffi
1. The restrictions on gðx;yÞ ¼ x= y=v and h(x,y) ¼ y are satisfied.
pffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffi
2. Solving for x= y=v ¼ z and y ¼ w, we obtain the solution x ¼ z w=v and y ¼ w.
3. The Jacobian matrix and its absolute determinant are
2
3
1
x
p
ffiffiffiffiffiffiffi
J ¼ 4 y=v 2v(y=v3=2 ) 5
0
1
1
1
or kJk ¼ pffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffi ;
y=v
w=v
w.0
4. The density function fZW(z,w) is given by
pffiffiffiffiffiffiffiffi
w(v=2)1 e(w=2)
w=v
2
v u(w)
fZW (z,w) ¼ pffiffiffiffiffiffi e(z w=v2)
2p
2v=2 G
2
w(v1)=2
2
e(w=2)½(z =vÞþ1
2
u(w)
¼
pffiffiffiffiffiffi v
2 pv G
2
The density function fZ(z) is found by integrating this equation over w,
1
fZ (z) ¼ pffiffiffiffiffiffi v
2 pv G
2
ð 1 (v1)=2
w
2
e(w=2)½(z =v)þ1 dw
0 2
238
Functions of Multiple Random Variables
This equation can be rewritten by substituting j ¼ ðw=2Þ½(z2 =nÞ þ 1, resulting in
ð1
1
fZ (z) ¼
j(v1)=2 ej dj
(vþ1)=2
pffiffiffiffiffiffi v z2
0
þ1
pv G
2 v
The integral in this equation is a gamma function given by G½(v þ 1)=2. Thus
vþ1
(vþ1)=2
G
z2
2
fZ (z) ¼ pffiffiffiffiffiffi v 1 þ
v
pv G
2
which is a Student-t density [Eq. (7.9.2)].
B 13.5 MULTIPLE FUNCTIONS OF
RANDOM VARIABLES
We can now generalize the results of the previous section to n functions of n random variables. Let {X1 , . . . , Xn } be the n random variables defined by the joint density function
fX1 , . . . , Xn (x1 , . . . , xn ). Let the n functions be
Z1 ¼ g1 (X1 , . . . , Xn ), Z2 ¼ g2 (X1 , . . . , Xn ), . . . , Zn ¼ gn (X1 , . . . , Xn )
(13:5:1)
We assume that these functions satisfy the restrictions of existence of partial derivatives
and no planar solutions. We now solve for
g1 (x1 , . . . , xn ) ¼ z1 , g2 (x1 , . . . , xn ) ¼ z2 , . . . , gn (x1 , . . . , xn ) ¼ zn
(13:5:2)
Let the solutions be {x1 , . . . , xn } be expressed as functions of {z1 , . . . , zn }:
x1 ¼ h1 (z1 , . . . , zn ), x2 ¼ h2 (z1 , . . . , zn ), . . . , xn ¼ hn (z1 , . . . , zn )
(13:5:3)
Let the Jacobian matrix be given by
2
@g1
6 @x1
J(x1 , . . . , xn ) ¼ 6
4 @gn
@x1
...
...
3
@g1
@xn 7
7
@gn 5
(13:5:4)
@xn
and its determinant be expressed in terms of {z1 , . . . , zn }:
kJk ¼ kJ(z1 , . . . , zn )k
(13:5:5)
The joint density function fZ1 , . . . , Zn (z1 , . . . , zn ) can now be given in terms of Eqs. (13.5.3)
and (13.5.5):
fZ1 , . . . , Zn (z1 , . . . , zn ) ¼
fX1 , . . . , Xn (x1 , . . . , xn )
kJ(z1 , . . . , zn )k
¼
fX1 , . . . , Xn ½h1 (z1 , . . . , zn )1 , . . . , hn (z1 , . . . , zn )
kJ(z1 , . . . , zn )k
(13:5:6)
For general functions {gk (x1 , . . . , xn , k ¼ 1, . . . , n}, {hk (z1 , . . . , zn , k ¼ 1, . . . , n} and
kJ(z1 , . . . , zn )k are not that easy to solve. However, for linear functions
13.5
Multiple Functions of Random Variables
239
{gk (x1 , . . . , xn , k ¼ 1, . . . , n}, the solutions are not that difficult, as shown by the following
example.
Example 13.5.1 Three independent zero mean unit variance Gaussian random variables
X, Y, and Z are given by the joint density function
fXYZ (x,y,z) ¼
1
2
2
2
pffiffiffiffiffiffi e(1=2)(x þy þz )
2p 2p
Three other random variables U, V, and W are given by
2 3 2
32 3
U
2 2 1
X
4 V 5 ¼ 4 5 3 3 54 Y 5
W
3 2 2
Z
We have to find the joint density fUVW (u, v, w) and the marginal densities fU(u), fV(v),
and fW(w).
The solution to
2 3 2
32 3
u
2 2 1
x
4 v 5 ¼ 4 5 3 3 54 y 5
w
3 2 2
z
is
2 3 2
x
0
4y5 ¼ 4 1
z
1
2
1
2
32 3 2
3
3
u
2v 3w
1 54 v 5 ¼ 4 u v þ w 5
4
w
u 2v þ 4w
The Jacobian matrix and its determinant are given by
2
3
2 2 1
J ¼ 4 5 3 3 5; kJk ¼ 1
3 2 2
Hence, using Eq. (13.5.6), we obtain the joint density function fUVW(u, v, w)
fUVW (u, v, w) ¼
2
2
2
1
1
pffiffiffiffiffiffi e(1=2)½(2v3w) þ(uvþw) þ(u2vþ4w) 1 2p 2p
1
2
2
2
pffiffiffiffiffiffi e(1=2)½(2u 9v 426w þ2uv30vw6wu)
2p 2p
8
2 319
0
u
>
>
=
<
1
1B
1 6 7C
½
u
v
w
½C
v
¼
exp
@
A
4
5
>
>
(2p)3=2 jCj1=2
;
: 2
w
2
3
9 19 12
where the matrix C ¼ 4 19 43 27 5and jCj ¼ 1.
12 27 17
¼
The random variables X, Y, Z are independent, whereas the transformed random
variables U, V, W are not independent. Their correlation coefficients can be calculated
as follows:
19
rUV ¼ pffiffiffiffiffiffiffiffiffiffiffi ¼ 0:966,
9 43
27
rVW ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:999,
43 17
12
rWU ¼ pffiffiffiffiffiffiffiffiffiffiffi ¼ 0:970
17 9
240
Functions of Multiple Random Variables
By successively integrating with respect to (v,w), (w,u) and (u,v), we obtain the marginal
density functions fU(u), fV(v), and fW(w) as
ð1 ð1
1
2
2
2
fU (u) ¼ pffiffiffiffiffiffi
e½(1=2)(2u þ9v 426w þ2uv30vw6wu) dv dw
2p 2p 1 1
1
2
¼ pffiffiffiffiffiffi e½(1=2)(u =9)
3 2p
ð1 ð1
1
2
2
2
fV (v) ¼ pffiffiffiffiffiffi
e½(1=2)(2u þ9v þ26w þ2uv30vw6wu) dw du
2p 2p 1 1
1
2
¼ pffiffiffiffiffiffi e½(1=2)(v =43)
43 2p
ð1 ð1
1
2
2
2
fW (w) ¼ pffiffiffiffiffiffi
e½(1=2)(2u þ9v 426w þ2uv30vw6wu) du dv
2p 2p 1 1
1
2
¼ pffiffiffiffiffiffi e½(1=2)(w =17)
17 2p
These results can be verified from the results of Example 13.1.5. Since var(ax) ¼ a 2
var(X ), we have var(U ) ¼ 22 þ 22 þ 12 ¼ 9, var(V ) ¼ 52 þ 32 þ 32 ¼ 43, and
var(W ) ¼ 32 þ 22 þ 22 ¼ 17, which checks with the results above.
CHAPTER
14
Inequalities, Convergences, and
Limit Theorems
B 14.1
DEGENERATE RANDOM VARIABLES
In Section 10.2 it was stated that if the variance is zero, then the random variable will no
longer be random. We can state this concept more fully as follows.
If E(X 2 a)2¼0, where a is a constant, then the random variable X equals a with probability 1. We shall show this result as follows. We construct a small hole of nonzero width
21 with 1 . 0 about the value a in the region of integration of x from 21 to 1 as shown in
Fig. 14.1.1.
We integrate (x2a)2 fX(x) from 21 to 1 except for the hole about a and write
ð1
ð
E(X a)2 ¼
(x a)2 fX (x)dx (x a)2 fX (x)dx
(14:1:1)
jxaj.1
1
By substituting the lowest value of 1 for jx –aj in Eq. (14.1.1), we can write
ð
ð
2
2
2
E(X a) 1 fX (x)dx ¼ 1
fX (x)dx ¼ 12 P(jX aj . 1)
jxaj.1
jxaj.1
2
Since E(X a) ¼ 0 by assumption, we must have 12 P(jX aj . 1) ¼ 0: Since 1 is a
nonzero constant, we must have P(jX aj . 1 ¼ 0: As a consequence
P(jX aj ¼ 0) ¼ 1
and
X ¼ a with probability 1:
(14:1:2)
2
What this result conveys is that if E(X2a) ¼ 0, then almost all realizations of
X will be equal to a constant a. The random variable X is called a degenerate random
variable. This method of analysis will be useful in showing some of the inequalities
that will follow.
Example 14.1.1 If the variance s2X ¼ 0, then the random variable X ¼ mX almost
certainly. Since variance of X equals E(X mX )2 ¼ 0, then, by Eq. (14.1.2) X ¼ mX
with probability 1.
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
241
242
Inequalities, Convergences, and Limit Theorems
FIGURE 14.1.1
B 14.2
CHEBYSHEV AND ALLIED INEQUALITIES
In many problems in communications we are interested in finding bounds on certain performance measures. The idea in using a bound is to substitute a simpler expression for
something more complicated, thus gaining some insight into the problem at the expense
of accuracy. This process usually requires a fair amount of experience with the results
that are to be expected.
One of the bounds that we are interested in is the tails probabilities of the type
P(X . 1) or P(jXj . 1) (Fig. 14.2.1). Even if we cannot find these probabilities exactly,
we should be able to find bounds on these quantities. These bounds are based on partial
information about the random variable, such as mean and variance.
Chebyshev Bound
The tails probability bound of the random variable X of the form P(jX –mXj , 1) is
based on the knowledge of only the mean and the variance of X and is known as the
Chebyshev inequality, given by
P{jX mX j 1} s2X
12
or
FIGURE 14.2.1
CB(1) ¼
s2X
12
(14:2:1)
14.2
Chebyshev and Allied Inequalities
243
where 1 is a positive constant .0 and CB(1) is the Chebyshev bound (this result can be
shown using an analysis similar to that in the previous section):
ð1
ð
s2X ¼
(x mX )2 fX (x)dx (x mX )2 fX (x)dx
(14:2:2)
jxmX j.1
1
By substituting the lowest value of 1 in Eq. (14.2.2), we obtain
ð
s2X 12
fX (x)dx ¼ 12 P(jx mX j 1)
(14:2:3)
jxmX j.1
and hence Eq. (14.2.1).
Geometric Derivation of the Chebyshev Bound We will now derive the Chebyshev
bound from a different perspective. We will define a random variable Y ¼ X – mX. We will
make use of the indicator functions defined in Eq. (12.5.7). Let A be a set on the real line
such that A ¼ f –1, – 1g < f1, 1g. We define an indicator function IA as follows:
1 if Y [ A
(14:2:4)
IA ¼
0 if Y A
The indicator function IA( y) is shown in Fig. 14.2.2.
Taking the expected value of IA, we have
E½IA ¼ 1 P(Y [ A) þ 0 P(Y A) ¼ P(Y [ A)
¼ P(jYj 1) ¼ P(jX mX j 1)
(14:2:5)
We will now bound IA by another random variable Y 2/12 as shown in Fig. 14.2.2.
From the figure we can see that
IA Y 2 jX mX j2
¼
12
12
FIGURE 14.2.2
(14:2:6)
244
Inequalities, Convergences, and Limit Theorems
Taking expectation on both sides of Eq. (14.2.6), we have
E½IA ¼ P(jX mX j 1) E½Y 2 EjX mX j2 s2X
¼
¼ 2
12
12
1
(14:2:7)
which is Chebyshev inequality. The bound is E½Y 2 =12 is called the Chebyshev bound. We
will use this technique for deriving the Chernoff bound.
Alternate Forms of the Chebyshev Bound By rearranging the terms in Eq. (14.2.1),
we can rewrite Chebyshev’s inequality in terms of the center probability:
P{jX mX j , 1} 1 s2X
12
(14:2:8)
By substituting 1 ¼ nsX, we can write Eq. (14.2.1) as follows:
P{jX mX j nsX } 1
n2
(14:2:9)
Thus, for any random variable X, the probability of a deviation from the mean mX of more
than n standard deviations is 1/n 2. For example, in a standard Gaussian if
n ¼ 3, 1=n2 ¼ 19 ¼ 0:1111. Chebyshev inequality tells us that the probability of deviation
by more than 3 from 0 in a standard Gaussian is 0.1111. In actual cases this probability
is 0.0027.
We can also rewrite Eq. (14.2.5) analogous to Eq. (14.2.4) as
P{jX mX j , nsX } 1 1
n2
(14:2:10)
Generalized Chebyshev Inequality
If f(x) is a convex function (Section 14.6) in the sense that its second derivative is
positive for all x, then the inequality in Eq. (14.2.1) can be generalized to
P{jXj . 1} E½f(X)
f(1)
(14:2:11)
This can be shown by using techniques similar to those used in demonstrating Chebyshev
inequality, applying the fact that f(x) is convex:
ð1
ð
E½f(X) ¼
f(x)fX (x)dx f(x)fX (x)dx
(14:2:12)
1
jxj.1
Using the least value for f(x), namely, f(1) in Eq. (14.2.12), we have
ð
E½f(X) f(1)fX (x)dx ¼ f(1)P{jXj . 1}
(14:2:9)
jxj.1
and hence Eq. (14.2.11). If f(X ) ¼ (X– mX)2 in Eq. (14.2.11), we get Chebyshev
inequality.
Example 14.2.1 We will apply the Chebyshev bounds to two different distributions, a
Gaussian distribution and a Laplace distribution with zero means and unit variances.
14.2
Chebyshev and Allied Inequalities
245
FIGURE 14.2.3
They are given by
1
2
fN (x) ¼ pffiffiffiffiffiffi eðx =2Þ ;
2p
pffiffi
1
fL (x) ¼ pffiffiffi e 2jxj
2
and shown in Fig. 14.2.3.
We will find the tails probabilities P(jXj 1) for both the distributions and the
Chebyshev bound, which will be the same for both since they have the same means and
variances. Applying Eq. (14.2.1), the Chebyshev bounds are
P{jXj 1} 1
12
Thus the Chebyshev bound CB(1) ¼ 1/12. Since we know both the density functions, we
can calculate the exact tails probabilities P (jXj 1), which are given by
ð
ð
1 1 x2 =2
1 1 x2 =2
1
e
dx þ pffiffiffiffiffiffi
e
dx ¼ 1 erf pffiffiffi
FN (1) ¼ pffiffiffiffiffiffi
2p 1
2p 1
2
ð 1 pffiffiffiffi
ð 1 pffiffiffiffi
p
ffiffi
1
1
e 2x dx þ pffiffiffi
e 2x dx ¼ e 2j1j
FL (1) ¼ pffiffiffi
2 1
2 1
where
rffiffiffiffi ð 1
1
2
2
erf pffiffiffi ¼
ex =2 dx
p
2
0
The one-sided tails probabilities FN (1) and FL(1) for the Gaussian density and the Laplace
density, and the Chebyshev bound CB(1) are shown in Fig. 14.2.4 and tabulated in
Table 14.2.1 for 1 ranging from 1.5 to 5.
Conclusions
It is clear from Fig. 14.2.4 and Table 14.2.1 that the Chebyshev bound is a very loose
one. As 1 increases, the bound becomes somewhat tighter (Fig. 14.2.4). Table 14.2.1 also
shows that the Chebyshev bound is the same for both density functions, which have the
same mean and variance. In fact, the Chebyshev bound is applicable to all density
246
Inequalities, Convergences, and Limit Theorems
FIGURE 14.2.4
TABLE 14.2.1
1
1.5
1.75
2
2.25
2.5
2.75
3
3.25
3.5
3.75
4
4.25
4.5
4.75
5
Normal
FN(1)
Laplace
FL(1)
Chebyshev
CB(1) ¼ 1/12
0.133614
0.080118
0.0455
0.024449
0.012419
0.00596
0.0027
0.001154
0.000465
0.000177
0.000063
0.000021
0.000007
0.000002
0.000001
0.119873
0.084174
0.059106
0.041503
0.029143
0.020464
0.01437
0.01009
0.007085
0.004975
0.003493
0.002453
0.001723
0.00121
0.000849
0.444444
0.326531
0.25
0.197531
0.16
0.132231
0.111111
0.094675
0.081633
0.071111
0.0625
0.055363
0.049383
0.044321
0.04
functions having the same means and variances. It is useful as a theoretical tool to determine the bounds of the tails probabilities, where only means and variances are known
about the density functions.
B 14.3
MARKOV INEQUALITY
There are several different versions of Markov inequality. The usual version is defined for
a positive random variable X with a finite mean mX. It states that for a positive 1 . 0
m
P(X 1) X
(14:3:1)
1
14.3
Markov Inequality
247
This result can be shown by the usual techniques developed in the previous section
ð1
ð1
ð1
mX ¼
xfX (x)dx xfX (x)dx 1
fX (x)dx ¼ 1P(X 1)
(14:3:2)
0
1
1
and hence Eq. (14.3.1). In Eq. (14.3.1), the quantity mX =1 is called the Markov bound.
Equation (14.3.1) can be slightly generalized to include both positive and negative
random variables
P(jX aj 1) E(jX aj)
1
(14:3:3)
where a is an arbitrary constant and 1 . 0. Equation (14.3.3) can be further generalized to
P(jX aj 1) E(jX ajk )
1k
(14:3:4)
where k is a positive integer. This inequality is called the Bienayme inequality.
We will derive the value of a for which P(jX– aj 1) is a minimum in the equation
P(jX aj 1) E½jX aj2 12
(14:3:5)
for which we have to minimize E[jX – aj]2 with respect to a:
E½jX aj2 ¼ E½X 2 2Xa þ a2 ¼ E½X 2 2aE½X þ a2
(14:3:6)
Differentiating Eq. (14.3.6) with respect to a and setting equal to 0, we have
d
{E½X 2 2aE½X þ a2 } ¼ 2a 2E½X ¼ 0
da
from which we obtain a ¼ E[X ]. Thus we obtain Chebyshev inequality as follows:
P(jX mX j 1) E½jX mX j2 12
(14:2:1)
Example 14.3.1 A materials manager purchases 10,000 memory chips from a company
where 3% of the chips are faulty. We have to find the probability that the number of good
chips lies between 9675 and 9775. Without loss in generality, we will assume that the
chips are independent.
Let Xk be the zero – one random variable
1 if the kth chip is good
Xk ¼
0 if the kth chip is bad
with the probability of Xk ¼ 1 is 0.97. Thus E[Xk] ¼ 1 0.97 þ 0 0.03 ¼ 0.97 and
E[X2k ] ¼ 1 0.97 þ 0 0.03 ¼ 0.97. Hence var[Xk] ¼ 0.97 2 (0.97)2 ¼ 0.0291.
If Y is the sum of the random variables fXk, k ¼ 1, . . . , 10,000), then the mean of Y is
"
#
10,000
10,000
X
X
Xk ¼
E½Xk ¼ 9700
E½Y ¼ mY ¼ E
k¼1
k¼1
With the assumed independence of Xk, the variance of Y is
"
#
10,000
10,000
X
X
2
Xk ¼
var½Xk ¼ 291
var½Y ¼ sY ¼ var
k¼1
k¼1
248
Inequalities, Convergences, and Limit Theorems
With a ¼ ð9675 þ 9775Þ=2 ¼ 9725, 1 ¼ 50, and E½Y a2 given by
E½Y a2 ¼ E½Y 2 2Ya þ a2 ¼ EY 2 2mY a þ a2
¼ s2Y þ m2Y 2mY a þ a2 ¼ 291 þ 97002 2 9700 9725 þ 97252
¼ 916
we can use Eq. (14.3.5) and write
P(jY 9725j 50 916
¼ 0:3664
502
What we want is the complementary event P(jY 9725j 50), and this is given by
1 2 0.3664 ¼ 0.6336. Thus, the probability that the number of good chips will lie
between 9675 and 9775 is 63.36%. On the other hand, if we want the number of good
chips to lie in the region 9700 + 50, namely, 9650 and 9750, then the probability using
Chebyshev inequality is
P(jY 9700j 50) ¼ 1 P(jY 9700j 50) ¼ 1 291
¼ 0:8836 ¼ 88:36%
2500
Example 14.3.2 In a binomially distributed random variable b(k; n, p) ¼ nk pk qnk , we
have n ¼ 50 and p ¼ 0.1. Using Markov inequality, we want to find the probability of
X . 10 and compare it with the true value. The mean value is np ¼ 5 and 1 ¼ 10.
Thus, from Eq. (14.3.1), we have the Markov bound
P(X 10) 5
¼ 0:5
10
and the true value is
P(X 10) ¼
50
X
50!
0:1k 0:950k ¼ 0:02454
k!(50
k)!
k¼10
This Markov bound is very loose!
We will now find the Chebyshev bound from Eq. (14.2.7) with f(x) ¼ x 2 and
compare the results. Equation (14.2.7) becomes P{jXj . 1} E½X 2 =12 and E[X 2] ¼
npq þ (np)2 ¼ 29.5. The Chebyshev bound is {X . 10} (29:5=100) ¼ 0:295, which
is better than the Markov bound of 0.5, but nowhere close to the true value of 0.02454.
B 14.4
CHERNOFF BOUND
In determining the Chebyshev bound, the indicator function was bounded by a quadratic.
In an effort to improve the tightness of the bound, we will use a positive exponential bound
the indicator function as shown in Fig. 14.4.1.
The expectation of the indicator function yields
E½IA ¼ P(jYj 1) ¼ P(jX mX j 1)
l(Y – 1)
(14:4:1)
, l . 0, 1 . 0, as shown in Fig. 14.4.1. From
The indicator function is bounded by e
the figure, IA e l(Y21) and taking expectations on both sides, we obtain
E½IA ¼ P(Y 1) E½el(Y1) 14.4
Chernoff Bound
249
FIGURE 14.4.1
or
P(Y 1) el1 E½elY ¼ el1 MY (l)
(14:4:2)
where jYj has been replaced with Y since we are considering only the positive bound and
MY(l) is the moment generating function of Y.
The bound given above will hold for any l . 0. However, we would like to choose l
so that e 2l1E[e lY] is a minimum. Differentiating e 2l1E[e lY] with respect to l and setting
it equal to zero, we obtain
d
d l(Y1) E½el(Y1) e
¼E
dl
dl
l¼l0
l¼l0
or
E (Y 1)el0 (Y1) ¼ el0 1 E (Y 1)el0 Y ¼ 0
(14:4:3)
Rearranging terms in Eq. (14.4.3), we obtain an implicit relation for the minimum l0 in
terms of the moment generating functions (MGFs):
1¼
E½Yel0 Y MY0 (l0 )
¼
E½el0 Y MY (l0 )
(14:4:4)
The Chernoff bound is given by
P(Y 1) el0 1 MY (l0 )
(14:4:5)
where l0 is given by Eq. (14.4.4). A similar result holds for the bound on the negative side.
Example 14.4.1 We will find the bounds on tails probability for the standard Gaussian
random variable Y using Chernoff’s formula and compare it to the Chebyshev bound found
in Example 14.2.1. The MGF for any Gaussian random variable N is given by
MN (t) ¼ etmþ(t
MY (t) ¼ et
2
=2
2 2
s =2)
and for a standard Gaussian, Y N(0,1)
250
Inequalities, Convergences, and Limit Theorems
FIGURE 14.4.2
We will now find the optimum l from Eq. (14.4.4):
2
M 0 (l0 ) l0 e(l0 =2)
¼
1¼ Y
2
MY (l0 )
e(l0 =2)
or l0 ¼ 1
The Chernoff bound from Eq. (14.4.5) is
2
2
P(Y 1) el0 1 el0 =2 ¼ e1 e1
2
=2
¼ e 1
2
=2
¼ ChG(1)
The Chernoff bound, ChG(1), the one-sided
Ð 1 x2 =2 Chebyshev bound, CBO(1), and the one-sided
normal distribution, FNo (1) ¼ p1ffiffiffiffi
e
dx are shown in Fig. 14.4.2 and in Table 14.4.1
2p e
for 1 ¼ 1.5, . . . , 3.
TABLE 14.4.1
1
1.5
1.6
1.7
1.8
1.9
2
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3
Normal
FNo(1)
Chernoff
ChG(1)
Chebyshev
CBO(1) ¼ 1/212
0.066807
0.0548
0.044566
0.035931
0.028717
0.02275
0.017864
0.013903
0.010724
0.008197
0.00621
0.004661
0.003467
0.002555
0.001866
0.00135
0.324652
0.278037
0.235746
0.197899
0.164474
0.135335
0.110251
0.088922
0.071005
0.056135
0.043937
0.034047
0.026121
0.019841
0.014921
0.011109
0.222222
0.195312
0.17301
0.154321
0.138504
0.125
0.113379
0.103306
0.094518
0.086806
0.08
0.073964
0.068587
0.063776
0.059453
0.055556
14.5
Cauchy–Schwartz Inequality
251
From the figure and the table we see that for 1 . 2 the Chernoff bound is better than the
Chebyshev bound. The crossover point occurs when 1 ¼ 2.075.
B 14.5
CAUCHY–SCHWARTZ INEQUALITY
An important inequality used in communication problems is the Cauchy – Schwartz
inequality. Let g(x) and h(x) be two complex functions with finite norm defined by
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð
1
g(x)g (x)dx , 1
kg(x), g(x)l ¼ kg(x)k ¼
1
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð
(14:5:1)
1
h(x)h (x)dx , 1
kh(x), h(x)l ¼ kh(x)k ¼
1
The Cauchy–Schwartz inequality is given by
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð 1
ð1
ð1
2
g(x)h (x)dx jg(x)j dx jh(x)j2 dx
1
1
(14:5:2)
1
We can also state this inequality in terms of the inner product defined by
ð1
kg(x), h(x)l ¼
g(x)h (x)dx ¼ kh(x), g(x)l
(14:5:3)
1
and the inequality becomes
kg(x), h(x)l kg(x)k kh(x)k
(14:5:4)
If we define two real-valued functions, a(x) and b(x), as
a(x) ¼ jg(x)j,
b(x) ¼ jh(x)j
then we have to show that
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð
ð
ð1
1
1
1
a2 (x)dx
a(x)b(x)dx 1
b2 (x)dx
(14:5:5)
1
Let us represent a(x) and b(x) in a function space spanned by the basis vectors f1(x) and
f2(x). These vectors have the following orthonormal property:
ð1
ð1
ð1
2
2
f1 (x)dx ¼
f2 (x)dx ¼ 1;
f1 (x)f2 (x)dx ¼ 0
(14:5:6)
1
1
1
Thus
a(x) ¼ a1 f1 (x) þ a2 f2 (x);
b(x) ¼ b1 f1 (x) þ b2 f2 (x)
(14:5:7)
where a1,a2,b1,b2 are coordinates in the function space of ff1(x), f2(x)g and the vectors,
a ¼ ½a1 ,a2 ; b ¼ ½b1 ,b2 as shown in Fig. 14.5.1.
252
Inequalities, Convergences, and Limit Theorems
FIGURE 14.5.1
We will find the inner product, ka(x), b(x)l ¼
can write
ð1
Ð1
1
a(x)b(x)dx. From Eq. (14.5.7) we
ð1
½a1 f1 (x) þ a2 f2 (x)½b1 f1 (x) þ b2 f2 (x)dx
a(x)b(x)dx ¼
1
1
½using orthonormal properties of f1 (x) and f2 (x)
ð1
½a1 b1 f21 (x) þ a2 b2 f22 (x) dx ¼ a1 b1 þ a2 b2
¼
(14:5:8)
1
From Fig. 14.5.1 we can write
cos(u) ¼ cos(ua ub ) ¼ cos(ua ) cos(ub ) þ sin(ua ) sin(ub )
¼
a1 b1 a2 b2
ab
þ
¼
jaj jbj jaj jbj jajjbj
(14:5:9)
Since jcos(u)j 1, we have
ja bj
1 and ja bj jajjbj
(14:5:10)
jajjbj
ffi
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
Ð1
Ð1
Substituting for jaj ¼ 1 a2 (x)dx; jbj ¼ 1 b2 (x)dx in Eq. (14.5.10), we obtain
Eq. (14.5.5) and hence Eq. (14.5.2). The equality is obtained when we substitute
a ¼ Kb in Eq. (14.5.10) or a(x) ¼ Kb(x) and hence g(x) ¼ Kh(x).
If g(X ) and h(X ) are real random variables, then we define the norms and the inner
product given by Eqs. (14.5.1) and (14.5.3) as
kg(X), g(X)l ¼ E½g2 (X) ¼
2
ð1
1
ð1
kh(X), h(X)l ¼ E½h (X) ¼
g2 (x)f (x)dx , 1
(14:5:11)
2
h (x)f (x)dx , 1
1
ð1
g(x)h(x)f (x)dx ¼ kh(X), g(X)l
kg(X), h(X)l ¼ E½g(X)h(X) ¼
1
(14:5:12)
14.5
Cauchy–Schwartz Inequality
253
FIGURE 14.5.2
and Cauchy–Schwartz inequality of Eq. (14.5.2) becomes
ffi sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ð 1
ð1
ð1
2
2
g(x)h(x)f (x)dx g (x)f (x)dx h (x)f (x)dx
1
1
(14:5:13)
1
or
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
E½g(X)h(X) E½g2 (X)E½h2 (X)
Example 14.5.1 A communication system is shown in Fig. 14.5.2, where s(t) is the input
signal and n(t) is a random variable representing input noise, so(t) and no(t) are the output
signal and output noise, H( f ) is the transfer function. gi(t) the input signal-to-noise ratio
(SNR), and go(t) the output signal-to-noise ratio are given by
gi (t) ¼
s2 (t)
;
E½n2 (t)
go (t) ¼
s2o (t)
E½n2o (t)
The output so(t) is given by the inverse Fourier transform of S( f )H( f ), or
ð1
ð1
so (t) ¼
s(t)h(t t)dt ¼
S( f )H( f )e j2pf df
1
(14:5:14)
(14:5:15)
1
and the output noise is characterized by
E½n2o (t) ¼
ð1
jH( f )j2 N( f )df
(14:5:16)
1
where N( f ) is the frequency-domain characterization of the input noise n(t). With these
characterizations of Eqs. (14.5.15) and (14.5.16), the output SNR go(t) can be given by
Ð 1
2
j2pf
df s2o (t)
1 S( f )H( f )e
¼ Ð1
go (t) ¼
2
E½n2o (t)
1 jH( f )j N( f )df
(14:5:17)
The Cauchy–Schwartz inequality of Eq. (14.5.2) is modified to suit the frequency-domain
characterization as follows
2 ð 1
ð1
ð1
2
A(
f
)B
(
f
)df
jA(
f
)j
df
jB( f )j2 df
1
1
1
with the equality holding for A( f ) ¼ KB ( f ). We substitute
pffiffiffiffiffiffiffiffiffiffi
A( f ) ¼ H( f ) N( f )
(14:5:18)
254
Inequalities, Convergences, and Limit Theorems
and
S( f )e j2pft
B( f ) ¼ pffiffiffiffiffiffiffiffiffiffi
N( f )
(14:5:19)
A( f )B( f ) ¼ S( f )H( f )e j2pft
(14:5:20)
Thus
Applying Eqs. (14.5.19) and (14.5.20) in Eq. (14.5.18), the Cauchy–Schwartz inequality
becomes
2 ð 1
ð 1
ð1
jS( f )j2
j2pft 2
df
(14:5:21)
S(
f
)H(
f
)e
df
jH(
f
)j
N(
f
)df
1
1
1 N( f )
Substituting Eq. (14.5.21) in Eq. (14.5.17), the output SNR is given by
ð1
ð1
jS( f )j2
2
df
jH( f )j N( f )df 2
so (t)
1
1 N( f )
Ð
¼
g
(t)
o
1
2
E½n2o (t)
1 jH( f )j N( f )df
or
ð1
go (t) jS( f )j2
df
1 N( f )
(14:5:22)
The equality will hold for A( f ) ¼ KB ( f ), or
pffiffiffiffiffiffiffiffiffiffi
S ( f )ej2pfT
H( f ) N( f ) ¼ K pffiffiffiffiffiffiffiffiffiffi
N( f )
or Hopt ( f ) ¼ K
S ( f )ej2pfT
jN( f )j
(14:5:23)
where we have substituted T the time of observation for t. The maximum output SNR is
ð1
jS( f )j2
go, max df
(14:5:24)
1 jN( f )j
B 14.6
JENSEN’S INEQUALITY
Convex Functions
Before we enunciate Jensen’s inequality, we want to make clear the idea of a convex
function. We are interested in a convex function because convexity guarantees a unique
minimum over the interval of convexity. A function h(x) is convex in the interval (a, b)
if for every x1 belonging to the interval (a,b), there exists a constant m such that the following inequality is satisfied for all x:
h(x) h(x1 ) þ m(x x1 )
(14:6:1)
As seen in Fig. 14.6.1, h(x1 ) þ m(x x1 ) is the equation to a straight line passing tangentially to the point h(x1) and m is the slope of the straight line at the point x1. It is clear from
the figure that the inequality is satisfied geometrically. For the sake of completeness, we
will give a more formal definition of convexity.
14.6
Jensen’s Inequality
255
FIGURE 14.6.1
A function h(x) is convex over an interval (a,b), if every x1 and x2 belonging to (a,b)
and for 0 l 1, the following equation holds.
h½lx1 þ (1 l)x2 lh(x1 ) þ (1 l)h(x2 )
(14:6:2)
If this equation is a strict inequality, then h(x) is strictly convex. Geometrically, this
equation conveys that if the function h(x) lies below the line segment connecting two
points in the interval (a,b), then it is convex.
It can also be shown that it is necessary and sufficient that the second derivative
d2 h(x)=dx2 0 for all x in the interval (a,b), for Eq. (14.6.2) to be satisfied, in which
case h(x) is convex.
Example 14.6.1
is nonnegative:
The following functions h(x) are convex because the second derivative
1. x2 , since h00 (x) ¼ 2, positive for all x
2. jxj, since h00 (x) ¼ 0, nonnegative for all x
3. ex , since h00 (x) ¼ ex , positive for all x
4. ln(x), for 0 , x , 1, since h00 (x) ¼ 1=x2 , positive in the range of x
5. x ln(x), for 0 , x , 1, since h00 (x) ¼ 1=x, positive in the range of x
Jensen’s Inequality
Jensen’s inequality states that if h(x) is convex and if X is a random variable with
EjXj , 1, then
E½h(X) h E½X
(14:6:3)
Substituting E[X ] ¼ x1 in the convexity [Eq. (14.6.1)], we obtain
h(x) h E½X þ m x E½X
Taking expectations on both sides of Eq. (14.6.4) gives
E½h(X) h E½X þ m E½X E½X
¼0
and hence Jensen’s inequality [Eq. (14.6.3)].
(14:6:4)
256
Inequalities, Convergences, and Limit Theorems
Example 14.6.2 Since E[X 2n]¼ E[(X n)2] is a convex function from Example 14.6.1, it
follows from Jensen’s inequality that
2
E½X 2n E½X n (14:6:5)
and as a consequence
E½X 2 E½X
2
(14:6:6)
which is a restatement of the fact that the variance of a random variable is nonnegative.
B 14.7
CONVERGENCE CONCEPTS
We saw in Section 14.1 that if E(X a)2 ¼ 0, then X ¼ a with probability 1 or X
converges to a with probability 1. There are a few other convergence concepts, and we
shall make these a little clearer.
A machinist wants to make piston rings of diameter a inches to a tolerance limit
of +1. Let us assume that the variability of the measurement is s2X. According to the
Chebyshev inequality of Eq. (14.2.8), we have
P{jX aj , 1} 1 s2X
12
(14:7:1)
If sX is very much smaller than 1, then the observed random variable X is between a 2 1
and a þ 1 is almost certain, and one measurement of X is sufficient. However, if sX is not
sufficiently small compared to 1, then the estimate a will not give results of sufficient accuracy. To improve the accuracy of the estimate a, we take n measurements corresponding to
n random variables fXi, i ¼ 1, . . . , ng with mean a and noise random variables Wi given by
Xi ¼ a þ W i ,
i ¼ 1, . . . , n
where we will assume that the random variables Wi are independent with zero mean and
variance s2. We will represent the average of these n random variables by
X 1 þ X2 þ þ Xn
b
X¼
(14:7:2)
n
where the mean of b
X is a and the variance is ns2X and the corresponding Chebyshev
inequality is given by
P{jX aj , 1} 1 s2X
n12
(14:7:3)
We can see from Eq. (14.7.3) that as the number of samples n becomes very large, the
probability that the jX aj is less than 1 becomes almost certain even though s2X may
not be small compared to 1.
Pointwise Convergence
A discrete sequence of random variables
{X1 , X2 , . . . , Xn , . . .}
(14:7:4)
converges to a limiting random variable X if and only if for any 1 . 0, however small, we
can find a number n0 such that
jXn (j) X(j)j , 1
(14:7:5)
14.7
Convergence Concepts
257
for every n . n0 and every j. This type of convergence is also called pointwise or everywhere convergent. This may be all right for a real variable, but it is highly restrictive for a
random variable. It is possible that the sequence fXng may converge for j points belonging
to a subset of the sample space S. Thus, convergences using probability measures are more
useful than this type of convergence.
Almost Sure Convergence (a.s.)
A sequence of random variables fXng converges almost surely (a.s.), or almost everywhere (a.e.), or strongly to the random variable X, if for every j point in the sample space S
satisfies the following criterion:
lim jXn (j) X(j)j ! 0 with probability 1
n!1
(14:7:6)
The criterion can also be written as
P(Xn ! X) ¼ 1
as
n ! 1
(14:7:7)
In Eq. (14.7.7) the quantity Xn ! X is to be interpreted as the set of all outcomes j such
that the number Xn(j) tends to the number X(j). In some cases the limit X may not be
known a priori. In this case the Cauchy criterion of convergence comes handy to define
mutual convergence.
Cauchy Convergence Criterion
A sequence of random variables fXng on the probability space {S, F, P} is convergent almost surely if the set of points j for which the real sequence {Xn (j)} is convergent
with probability 1. Or, in other words
P{j [ S : Xnþm (j) Xn (j) ! 0}
as
n ! 1
(14:7:8)
for every m . 0. Using this criterion, the sequence fXng converges mutually almost
surely if
P{Xnþm Xn } ! 0
as
n ! 1
(14:7:9)
Convergence in Probability (p)
A sequence of random variables fXng converges in probability or weakly converges to
the random variable X if for every 1 . 0, however small
lim P{jXn Xj 1} ¼ 0
(14:7:10)
lim P{jXn Xj 1} ¼ 1
(14:7:11)
n!1
or equivalently
n!1
Using the Cauchy criterion, the sequence fXng converges mutually in probability to X if
lim P{jXnþm Xn j 1} ! 0
n!1
(14:7:12)
Convergence in probability is widely used in probability theory. In particular, it is used to
determine the weak law of large numbers and the consistency of estimators.
258
Inequalities, Convergences, and Limit Theorems
Convergence in Mean Square (m.s.)
A sequence of random variables fXng converges in the mean square or limit in the
mean to the random variable X if
lim E{jXnþm Xj2 } ! 0
n!1
(14:7:13)
This is also called quadratic mean convergence.
Mutual convergence in the mean square is defined by
lim E{jXnþm Xn j2 } ! 0
n!1
(14:7:14)
for all m . 0.
Convergence in Distribution
The widest form of convergence is the convergence in distribution. If we denote the
distribution functions of Xn and X by Fn(x) and F(x), respectively, then the sequence fXng
converges in distribution to the random variable X if
lim Fn (x) ! F(x)
n!1
(14:7:15)
for every continuity point x of F(x). However, the sequence fXn(j)g need not converge for
any j.
Properties of Convergences
1. Almost-sure convergence implies convergence in probability. The converse is not
true.
2. Mean-square convergence also implies convergence in probability. Again the converse is not true.
3. If fXng converges in probability, then there exists a subsequence fXnkg of fXng that
converges almost surely.
4. Convergences almost surely, probability, and mean square all imply convergence
in distribution. Of course, the converse is not true.
All the four convergences are shown schematically in Fig. 14.7.1.
FIGURE 14.7.1
14.8
B 14.8
Limit Theorems
259
LIMIT THEOREMS
Weak Law of Large Numbers (WLLN)
Armed with the convergence concepts, we will now explore some limit theorems. Let
fXng be a sequence of independent identically distributed random variables with finite
means m and finite variances s2. Let Sn be the sum of the n random variables given by
Sn ¼ X 1 þ X2 þ þ Xn
(14:8:1)
Let the random variable Zn, called the sample mean, be defined by
X1 þ X2 þ þ X n
n
Zn ¼
Clearly the expected value of Zn is
X1 þ X2 þ þ Xn
nm
¼m
E½Zn ¼ E
¼
n
n
and because of the independence of the sequence fXng, the variance of Zn is
X 1 þ X2 þ þ Xn
ns2 s2
var½Zn ¼ var
¼ 2 ¼
n
n
n
(14:8:2)
(14:8:3)
(14:8:4)
The weak law of large numbers states that for every 1 . 0
lim P½jZn mj . 1 ! 0
n!1
(14:8:5)
To show this result, we apply Chebyshev’s inequality [Eq. (14.2.1)]
P½jZn mj . 1 s2
n12
(14:8:6)
and since 1 . 0, the limit of Eq. (14.8.6) as n tends to 1 is indeed 0. In other words, the
probability of the sample mean Zn for a specific n of a sequence fXng of random variables
deviating from the true mean m by an arbitrary 1 . 0 becomes vanishingly small as
n tends to 1. Or the sample mean Zn converges in probability to the true mean.
Example 14.8.1 In the design of a gyroscope for a navigation system we want the probability of the sample drift rate Zn calculated for n production gyros differing from the
unknown true drift rate m by more than 0.1 be less than 2%. We want to find
the number n of the gyros needed to establish this criterion expressed in terms of the
unknown variance s2. Applying Chebyshev inequality given by Eq. (14.2.8), we can write
1
s2 100
. 0:98
P jZn mj ,
1
10
n
From 1 ð100s2 =nÞ . 0:98 we have
1 0:98 n
100s2
n
100s2
¼ 5000s2
0:02
If the quality control is very good, s will be small, in which case the number n will be
small. However, since we are using Chebyshev inequality that gives very loose bounds,
260
Inequalities, Convergences, and Limit Theorems
we expect n to be overly conservative. We will revisit this example after discussing the
central limit theorem in the next section.
Example 14.8.2 We will assume that there are a sequence of independent events fAig
that will be denoted by the sequence of indicator functions IAi (j), given by
IAi (j) ¼
Zn ¼
1
n
1
if
j [ Ai
0
if
j [ Ai
n
X
IAi
i¼1
We will assume that P(Ai ) ¼ p and P(Ai ) ¼ 1 p for i ¼ 1, . . . , n. Thus
E½IAi ¼ p
and
E½IA2i ¼ p
and
E½Zn ¼ p
and
var½Zn ¼
p(1 p)
n
var½IAi ¼ p(1 p), i ¼ 1, . . . , n
We can now apply WLLN and write
P{jZn pj , 1} 1 p(1 p)
n12
Assuming p ¼ 0.4, we will estimate p within 0.1. If n ¼ 500, then this equation becomes
P{jZn pj , 0:1} 1 0:4 0:6
8
117
¼1
¼
500 0:01
125 125
What this equation tells us is that if we perform 500 trials, then 468 of these will result in
an error of fZn 2 pg of less than 0.1.
Strong Law of Large Numbers
The weak law of large numbers given by Eq. (14.8.5) states that the sample mean Zn
converges in probability to the true mean m as n tends to 1. The strong law of large
numbers states that this convergence is not only in probability but also with probability
1, which is almost surely (a.s.). Or
n
o
(14:8:7)
P lim (Zn m) ¼ 0 ¼ 1 (a.s.)
n!1
Unlike the weak law of large numbers, which is easy to show with Chebyshev’s inequality,
the strong law of large numbers is difficult to show and will not be attempted in this book.
Even though Eqs. (14.8.5) and (14.8.7) look similar, they are dramatically different.
Equation (14.8.7) states that the sequence of the sample means Zn for every n will tend
to the true mean m and will stay close to it. The strong law gives us confidence that
sample means calculated from a number of realizations will always tend to the true
mean as the number increases and isolated discrepancies become unimportant.
Central Limit Theorem (CLT)
Perhaps the most important theorem in statistics is the central limit theorem, which
in its most general form states that if we sum a sufficient number of random variables,
the resulting probability is a Gaussian. Let fXng be a sequence of n independent
14.8
Limit Theorems
261
identically distributed random variables with mean mX and variance s2X. Their sum Sn is
given by
Sn ¼ X 1 þ X2 þ þ Xn
(14:8:8)
The mean m and the variance s2 of Sn are
E½Sn ¼ m ¼ nmX
var½Sn ¼ s2 ¼ ns2X
(14:8:9)
We will now define a normalized random variable Zn with mean 0 and variance 1 as
follows:
Zn ¼
Sn m Sn nmX
pffiffiffi
¼
s
sX n
(14:8:10)
The central limit theorem states that as n tends to 1, Zn tends to a standard normal
distribution:
ðz
1
2
pffiffiffiffiffiffi e(x =2) dx
lim P(Zn z) ¼
(14:8:11)
n!1
2p
1
We can show this result heuristically by using moment generating functions (MGFs) in
Zn ¼
n
Sn nmX
1 X
pffiffiffi ¼ pffiffiffi
(Xi mX )
sX n i¼1
sX n
(14:8:12)
Taking MGF on both sides, we obtain
pffiffi Pn
(t=sX n)
(X
m
)
tZn
i
X
i¼1
MZn (t) ¼ E(e ) ¼ E e
"
¼E
n
Y
e
pffiffi
(t=sX n)(Xi mX )
#
i¼1
( )n
n
Y
pffiffi
pffiffi
¼
E e(t=sX n)(Xi mX ) ¼ E e(t=sX n)(XmX )
(14:8:13)
i¼1
In Eq. (14.8.13) we have used the independence property of fXng and X has the same pdf as
Xn. We can now expand the last exponential in the above equation and write
h
i
pffiffi
1
t2
(t=sX n)(XmX )
2
E e
(X mX ) þ hn (X; t)
¼ E 1 þ pffiffiffi (X mX ) þ
sX n
2ns2X
¼1þ0þ
t2
þ E½hn (X; t)
2n
where hn(X; t) represents the higher-order terms. Hence
MZn (t) ¼ 1 þ 0 þ
t2
lim MZn (t) ¼ 1 þ
n!1
2n
n
t2
þ E½hn (X; t)
2n
n
¼e
t2 =2
(14:8:14)
262
Inequalities, Convergences, and Limit Theorems
2
Since E[hn(X; t)] is of the order of 1/n 2, it is zero in the limit as n tends to 1. The term et =2
is the MGF of a standard Gaussian, and we have shown the central limit theorem. The
number n is usually assumed to be 30 for CLT to hold, but as we shall see in the following
example, it holds for n much less than 30.
Example 14.8.3 Random variables Xi, i ¼ 1,2,3,4 are independent and identically distributed with a uniform distribution in the interval (0,1] with fXi(x) ¼ u(x) 2 u(x 2 1).
1
The mean values are mXi ¼ 12 and the variances are sXi2 ¼ 12
. We form the random
variables Y1 ¼ X1 , Y2 ¼ X1 þ X2 , Y3 ¼ X1 þ X2 þ X3 , Y4 ¼ X1 þ X2 þ X3 þ X4 , and their
density functions are fY1 (y) ¼ fX1 (y), fY2 (y) ¼ fX1 (y) fX2 (y), fY3 (y) ¼ fY2 (y) fX1 (y), fY4 (y) ¼
fY3 (y) fX1 (y). These densities are given by
fY1 ( y) ¼
1
0
if 0 , y 1
otherwise
8
if 0 , y 1
>
<y
fY2 (y) ¼ 2 y if 1 , y 2
>
:
0
otherwise
8 2
y
>
>
>
if 0 , y 1
>
>
2
>
>
>
3
< 2
y þ 3y if 1 , y 2
fY3 ( y) ¼
2
>
> y2
>
9
>
>
if 2 , y 3
3y þ
>
>
2
>
:2
0
otherwise
8 3
y
>
>
if 0 , y 1
>
>
6
>
>
>
>
y3
2
>
>
>
þ 2y2 2y þ
if 1 , y 2
>
< 2
3
3
fY2 ( y) ¼ y
22
>
if 2 , y 3
4y2 þ 10y >
>
3
>2
>
>
>
y3
32
>
>
>
þ 2y2 8y þ
if 3 , y 4
>
> 6
3
:
0
otherwise
These densities are compared to the corresponding normal densities given by
1 12
fNi ( y) N(mNi , s2Ni ) ¼ N i , i
, i ¼ 1,2,3,4
2 12
or
2
1
fN1 ( y) ¼ rffiffiffiffi e½x(1=2) =(1=6)
p
6
2
1
fN2 ( y) ¼ rffiffiffiffi e(x1) =(1=3)
p
3
2
1
fN3 ( y) ¼ rffiffiffiffi e½x(3=2) =(1=2)
p
2
2
1
fN4 ( y) ¼ rffiffiffiffiffiffi e(x2) =(2=3)
2p
3
These densities are shown in all four panels of Fig. 14.8.1.
We see that the approximation progressively improves as n increases from 1 to 4, and
it is very good when n ¼ 4 (Fig. 14.8.1d).
14.8
Limit Theorems
263
FIGURE 14.8.1
Example 14.8.4 We will revisit Example 14.8.1 in light of the CLT and determine
whether we can get a better bound on the number n of gyros needed. In order to use the
Gaussian tables, we have to convert the random variable Zn into the standard Gaussian
form by substituting
pffiffiffi
n(Zn m)
Zn m
Wn ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼
s
var(Zn )
1
0:98 and arrive at
We can substitute Wn in P½jZn mj , 10
pffiffiffi n
sjWn j
1
P pffiffiffi ,
0:98
0:98 or P jWn j ,
10s
10
n
Referring
pffiffiffi to Gaussian tables, we have the value for a probability of 0.98 is x ¼ 2.3263.
Thus, n=10s 2:3263 or n 2:32632 100s2 ¼ 541:17s2 and n 542 s2. This is a
much better number than the 5000s2 that we obtained in Example 14.8.1.
CHAPTER
15
Computer Methods for
Generating Random Variates
If we have an actual probabilistic system, we can obtain the pdf by analyzing the data from
the system. On the other hand, if we are designing a system such as transmission of packets
through a communication channel, we must get some idea as to how the system will
behave before it is actually implemented. We resort to simulation of the system on a computer. The probability distributions may be known a priori, and we have to generate
random data for the given probability distributions. We call such generated values
random variates. It is not feasible to develop techniques for generating random variates
for every possible distribution function. As it turns out, generating a uniform distribution
in the interval (0,1] is fast on a computer. Starting with the uniform distribution, we can
generate random variates for any given distribution function by a number of techniques
that will be discussed.
B 15.1
UNIFORM-DISTRIBUTION RANDOM VARIATES
Almost every computer will have a uniform number generator built into the system using
some form of the modulus function. One such generator uses the linear congruential or
power residue method that produces a sequence of integers fxi, i ¼ 0, . . . , m 2 2g. It is
given by the recursive relationship
xiþ1 ¼ (a xi þ c)mod m
(15:1:1)
where a is a constant multiplier, c is the increment, and m is the modulus function. The
initial value x0 is called the “seed.” Equation (15.1.1) can be recursively solved to yield
n
a 1
x n ¼ an x 0 þ c
mod m
(15:1:2)
a1
All these parameters affect the mean, variance, and cycle length of the sequences. It is
called a mixed congruential generator if c = 0 and multiplicative congruential generator
if c ¼ 0. The sequence repeats after xm22 and the period of repetition is less than or equal
to m 2 1 depending on the parameters.
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
264
15.1
265
Uniform-Distribution Random Variates
FIGURE 15.1.1
We will form two sequences with
Example 15.1.1
1. fxig : x0 ¼ 1, a ¼ 7, c ¼ 0, m ¼ 13
2. fyig : y0 ¼ 1, a ¼ 7, c ¼ 3, m ¼ 13
Using the recursive relationship of Eq. (15.1.1), these two sequences fxig and fyig are
shown in Fig. 15.1.1 and Table 15.1.1.
As we can see from Fig. 15.1.1 and Table 15.1.1, the repetition period is 13 2 1 ¼ 12.
Also changing c from 0 to 3 has radically changed the sequence. We can conclude that
almost all random-number generators are not truly random but pseudorandom sequences
that have the appearance in the statistical sense of being random. Thus, when we refer to
random numbers we really mean pseudorandom sequences where the period is very high.
A random-number generator must have the following properties:
1. The numbers generated must have good statistical properties. They must be independent and identically distributed. In other words, the values may not be
correlated.
2. The repetition period must be long for any simulation to be useful.
3. Random numbers must be repeatable. For each specified x0, a, c, and m, the
generator must produce the same sequence of numbers.
4. Many simulations will require thousands of random numbers, and hence the
generation of these numbers must be fast and thus must also have computational
efficiency.
5. It must be easy to generate separate sequences of random numbers.
TABLE 15.1.1
i
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
2
1
7
10
5
9
1
10
8
7
Series 1: x0 ¼ 1, a ¼ 7, c ¼ 0, m ¼ 13
xi
1
7
10
5
9
11
12
6
3
8
4
Series 2: x0 ¼ 1, a ¼ 7, c ¼ 3, m ¼ 13
yi
1
10
8
7
0
3
11
2
4
5
12
266
Computer Methods for Generating Random Variates
The linear congruential generator may satisfy all these conditions provided a and m
are carefully chosen. Good statistical properties have been obtained with a ¼ 75 ¼ 16,807
and m ¼ 231 2 1 ¼ 2,147,483,647. Under these conditions the period of the sequence is
231 2 2 ¼ 2,147,483,646. Sometimes a ¼ 742,938,285 and m ¼ 248 are also used.
B 15.2
HISTOGRAMS
Having generated random variates, we have to see how well they approximate the desired
probability density. From the definition of the density function fX (x), we have
fX (x) P(X x þ Dx) P(X x)
Dx
or
fX (x)Dx P(x , X x þ Dx)
(15:2:1)
which gives the value of the probability mass at the point x and width Dx. Hence with the
generated random variates fxig we will be able to construct a histogram that will approximate the given mass function fX (x)Dx. The purpose of the histogram is to construct the
model of the desired distribution from the generated data. The usual form of the histogram
is to divide the range of data points N into equal-size bins or subintervals Dx. Or the necessary number of bins K needed will be given by
Number of bins K ¼
range of data N
bin width Dx
(15:2:2)
Next, the number of data points fni, i ¼ 1, . . . , Kg falling into each bin is counted. The
vertical axis (ordinate, y) represents the counts in each bin, and the horizontal axis
(abscissa, x ) represents K . Dx, the product of the number of bins K and the bin interval
Dx. If fX (x) is finite, then the product K . Dx must cover the entire range of x. If it is infinite
or semiinfinite, then the product can cover a reasonable range. Finally, the histogram must
be normalized so that it can be compared to the desired density function. The normalization proceeds as follows. The normalized count is the count in each bin fni, i ¼ 1, . . . , Kg,
divided by the total number N of observations multiplied by the bin width Dx so that the
area under the histogram equals one. This type of normalization results in a histogram that
is similar to the probability density function. Even though this may be less intuitive, it is
the one to be used if the histogram is to model the desired probability density function.
We will now model a uniform distribution U(0,1) from the random variates generated
by the linear congruential generator of a computer. For this model, N ¼ 10,000 points
were generated and K ¼ 100 bins were chosen with each bin interval Dx ¼ 0.01. Note
that K . Dx ¼ 1 covers the x axis of the U(0,1) distribution. The histogram subroutine is
called that places the counts fni, i ¼ 1, . . . , Kg in each of the K bins. The number of
points falling in the first 10 bins is shown in Table 15.2.1. The expected number in
each bin is 100.
The resulting histogram is divided by N . Dx ¼ 10,000 0.01 ¼ 100 to yield the
model for U(0,1) distribution. The normalized histogram is shown in Fig. 15.2.1.
Superimposed over the histogram is the actual uniform distribution. By “eyeballing”
the two plots in Fig. 15.2.1, we see that the modeling is indeed good. More formally, we
can also test the goodness of fit by a number of mathematical tests, and we will use the chisquare test to ascertain whether the histogram fits the true pdf sufficiently well. We will
now enumerate the procedure for the chi-square test as applied to this case.
15.2 Histograms
267
TABLE 15.2.1
Bins
Count
Expected
0
1
2
3
4
5
6
7
8
9
102
100
100
100
109
100
100
100
117
100
107
100
86
100
93
100
86
100
81
100
Chi-Square Test
1. From the histogram find the number of observations ni in each of the i ¼ 1, . . . , K
bins.
2. From the probability density function fX (x) calculate the occupancy
mi ¼ fX(xi)NDx in each of the i ¼ 1, . . . , K bins.
3. Calculate the deviation Di from Di ¼ ni mi ; i ¼ 0; . . . ; K 1:
PK1
4. Form the chi-square statistic from x ¼ i¼0
(Di =mi )2 . This will have K22
degrees of freedom.
5. Set the probability level pg or the significance level a ¼ 1 2 pg. Usually pg is set at
95% or 99% or a is set at 5% or 1%. The higher the significance level, the tighter
will be the bounds.
6. From the chi-square tables find the threshold value ta for the given significance
level and K22 degrees of freedom.
7. If the x value is less than ta, the hypothesis that the fit is good is accepted. Otherwise it is rejected.
We can use the x2 test to determine the goodness of the fit to the uniform distribution
that has been achieved. The x value is found to be x ¼ 91.39 at the 5% significance level.
The corresponding ta value for 100 2 2 ¼ 98 degrees of freedom at the 5% significance
level is obtained from Table 15.2.2 as ta ¼ 122.10773. The chi-square density and distribution for 98 degrees of freedom is shown in Figs. 15.2.2 and 15.2.3 with the probability
and significance levels. The hypothesis that the histogram is a good fit to the true normal
distribution is accepted since 91.39 , 122.10773.
FIGURE 15.2.1
268
Computer Methods for Generating Random Variates
TABLE 15.2.2
K
pg 95%
pg 97.5%
pg 99%
K
pg 95%
pg 97.5%
pg 99%
1
2
3
4
5
6
7
8
9
10
12
14
16
18
20
22
24
26
28
30
3.84146
5.99146
7.81473
9.48773
11.07050
12.59159
14.06714
15.50731
16.91898
18.30704
21.02601
23.68475
26.29620
28.86928
31.41042
33.92443
36.41502
38.88513
41.33713
43.77297
5.02389
7.37776
9.34840
11.14329
12.83250
14.44938
16.01276
17.53455
19.02277
20.48306
23.33660
26.11891
28.84532
31.52636
34.16959
36.78070
39.36407
41.92316
44.46079
46.97924
6.63490
9.21034
11.34487
13.27670
15.08627
16.81189
18.47531
20.09024
21.66599
23.20918
26.21693
29.14122
31.99992
34.80530
37.56623
40.28936
42.97982
45.64168
48.27823
50.89218
35
40
45
50
55
60
65
70
75
80
82
84
86
88
90
92
94
96
98
100
49.80185
55.75848
61.65623
67.50481
73.31149
79.08194
84.82064
90.53122
96.21667
101.87947
104.13874
106.39484
108.64789
110.89800
113.14527
115.38979
117.63165
119.87094
122.10773
124.34211
53.20335
59.34170
65.41016
71.42019
77.38046
83.29767
89.17714
95.02318
100.83934
106.62857
108.93729
111.24226
113.54360
115.84143
118.13589
120.42708
122.71511
125.00007
127.28207
129.56120
57.34207
63.69074
69.95683
76.15389
82.29212
88.37942
94.42208
100.42518
106.39292
112.32879
114.69489
117.05654
119.41390
121.76711
124.11632
126.46165
128.80325
131.14122
133.47567
135.80672
Starting from uniformly distributed random variates, we can generate other random
variates given the desired probability distribution. We will discuss three widely used techniques for generating these variates:
1. Inverse transformation
2. Convolution
3. Acceptance –rejection
In all the simulations that follow we have used ui U(0,1) distribution shown in Section
15.2 with N ¼ 10,001 points.
FIGURE 15.2.2
15.3
Inverse Transformation Techniques
269
FIGURE 15.2.3
B 15.3
INVERSE TRANSFORMATION TECHNIQUES
One of the more useful ways of generating random variates is through the inverse transformation technique, which is based on the following result. If U is a random variable uniformly distributed in (0,1] with distribution function FU (u) ¼ u,u 1(0,1], and X is a
random variable with distribution function FX(x), then the random variates for X can be
21
generated from X ¼ F21
X (U ). We will show that FX (U ) is a random variable with distri21
bution FX(x). Assuming that X ¼ FX (U ), we have
P{X x} ¼ P{FX1 (U) x}
(15:3:1)
Since FX (x) is monotonically increasing, we can write Eq. (15.3.1) as follows:
P{FX1 (U) x} ¼ P{U FX (x)} ¼ FU {FX (x)}
(15:3:2)
However, U is uniformly distributed in (0,1] and hence FU(u) ¼ u, 0 , u 1. Applying
this fact in Eq. (15.3.2), we have
FU {FX (x)} ¼ FX (x)
(15:3:3)
This result can be used to obtain the random variates for any distribution function FX(x)
only if F21
X can be solved in closed form or FX can be accurately and efficiently inverted
numerically. The methodology of finding random variates fXig by the inversion technique
can be enumerated as follows:
1. FX(x) is known a priori.
2. Uniform random variates fui, i ¼ 1, . . . , Ng in the interval (0,1] are generated.
3. The inverse F21
X (u) of the cdf FX(x) can be analytically or numerically determined.
4. The variates fxig are generated from xi ¼ F21
X (ui), i ¼ 1, . . . , N.
5. The histogram for fxig is constructed and tested using chi-square goodness of fit.
We will use this procedure to find random variates for a number of discrete and continuous
distributions starting from 10,001 points of the uniform random variates in (0,1] generated
in the previous section.
270
Computer Methods for Generating Random Variates
Discrete Distributions
Since FX(x) is a monotonically increasing function, the inverse for a finitedimensional integer-valued discrete distribution functions can be obtained in a simple
manner as shown in the following two examples.
Example 15.3.1
given by
The distribution of packets in bytes in a communication system is
8
0,
>
>
>
>
0:2,
>
>
<
0:4,
FX (x) ¼
0:6,
>
>
>
>
0:8,
>
>
:
1,
0 , x 64
64 , x 128
128 , x 256
256 , x 320
320 , x 384
x . 384
and is shown in Fig. 15.3.1.
It is not difficult to find the inverse function F21
X (u). From Fig. 15.3.1, for values of the
uniform random variate ui given by 0.4 , ui 0.6, F21
X (u) ¼ 256 bytes. Similarly, we can
obtain the complete inverse function as
8
64,
0 , ui 0:2
>
>
>
>
< 128, 0:2 , x 0:4
xi ¼ FX1 (ui ) ¼ 256, 0:4 , x 0:6
>
>
320, 0:6 , x 0:8
>
>
:
384, 0:8 , x 1
and the sequence fxig represents the random variates for this discrete distribution. The generated uniform random variates ui are assigned values 64, 128, 256, 320, and 384 depending on the ranges in which they fall along the vertical axis. They are then sorted in
ascending order and plotted with x ¼ F21
X (u) along the horizontal axis and i scaled by
1/10,000 along the vertical axis so that the range is from 0 to 1. The plot is shown in
Fig. 15.3.2. As an example, the uniform random variates the falling in the shaded
region in the figure between 0.4 and 0.6 represent 256 bytes.
FIGURE 15.3.1
15.3
Inverse Transformation Techniques
271
FIGURE 15.3.2
Out of the 10,001 uniform random variates generated, 2017 fall in the range 0– 0.2,
1912 fall in the range 0.2– 0.4, 1976 fall in the range 0.4– 0.6, 2017 fall in the range
0.6 –0.8, and 2079 fall in the range 0.8 –1.0 as shown in Fig. 15.3.2, which is exactly
the same as Fig. 15.3.1.
Example 15.3.2 In this example, we will simulate a binomial distribution FB(x) with
n ¼ 5 and p ¼ 0.4 given by
FB (x) ¼
x
X
5!
(0:4)k (0:6)nk
k!(5
k)!
k¼0
and shown below and graphed in Fig. 15.3.3:
FIGURE 15.3.3
272
Computer Methods for Generating Random Variates
FIGURE 15.3.4
8
0:07776,
>
>
>
>
0:33696,
>
>
<
0:68256,
FB (x) ¼
0:91296,
>
>
>
>
0:98976,
>
>
:
1,
0,x1
1,x2
2,x3
3,x4
4,x5
x.5
The inverse function F 21
B (U) can be found in a manner similar to that in the previous
example. In Fig. 15.3.3, for values of the uniform random variate ui between 0.33696
and 0.68256, F21
B (u) ¼ 2. Similarly, we can obtain the complete inverse function as
8
0,
0 , ui 0:07776
>
>
>
>
1,
0:07776
, ui 0:33696
>
>
<
2,
0:33696
,
ui 0:68256
FB1 (ui ) ¼
3,
0:68256
,
ui 0:91296
>
>
>
>
4,
0:91296
,
ui 0:98976
>
>
:
1, 0:98976 , ui 1
which is shown in Fig. 15.3.4.
1
The vertical axis has been scaled by 10,000
for the range to correspond to (0,1].
Out of the 10,001 uniform random variates generated, 797 fall in the range
0 – 0.07776, 2555 fall in the range 0.07776 –0.33696, 3404 fall in the range 0.33696 –
0.68256, 2312 fall in the range 0.68256 –0.91296, 801 fall in the range 0.91296 –
0.98976, and 132 fall in the range 0.98976 –1. Figure 15.3.4 is the same as Fig. 15.3.3.
Continuous Distributions
If the cumulative distribution function of a random variable can be expressed in a
closed form, then it is not difficult to find the inverse in a closed form. We will give
several examples to clarify the inverse transformation method.
Example 15.3.3 (Exponential Distribution) In this example we will obtain random
variates for an exponential distribution given by
FX (x) ¼ 1 elx ,
x0
(15:3:4)
15.3
Inverse Transformation Techniques
273
FIGURE 15.3.5
The inverse is given by
1 elx ¼ u
or elx ¼ 1 u
and
1
x ¼ FX1 (u) ¼ ln (1 u),
l
x0
The exponential random variates are generated from this equation with l ¼ 13, resulting in
the following equation:
xi ¼ 3 ln (1 ui ), x 0, i ¼ 0, 1, . . . , 10,000
The histogram subroutine hist(K,Dx,x) is called with K ¼ 100 bins and bin width
Dx ¼ 0.1. The resulting histogram is divided by N . Dx ¼ 1000 to yield the normalized
histogram fE(xk), k ¼ 1, . . . , 100. This is shown in Fig. 15.3.5. Superimposed on top of
the histogram is the true density function, fX (x) ¼ 13 e(x=3) , x . 0. We can see from
figure that the fit is very good. We can also apply the chi-square goodness of fit, which
it easily passes.
Example 15.3.4 (Weibull Distribution)
The cdf of a Weibull distribution is given by
a
FX (x) ¼ 1 e½l(xm) ,
x.m
(15:3:5)
This distribution, as mentioned in Chapter 7, is a generalization of the exponential with
parameters, scale l, shape a, and location m. Since this is in a closed form, the inverse
is given by
a
a
1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 e½l(xm) ¼ u or e½l(xm) ¼ 1 u and FX1 (u) ¼ m þ a ln(1 u)
l
with u , 1. The Weibull random variates are generated with l ¼ 13 , a ¼ 3 and m ¼ 2
from the following equation:
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
xi ¼ 2 þ 3 3 ln(1 mi ), i ¼ 0, . . . , 10,000
The histogram subroutine hist(K, Dx, x) is called with K ¼ 100 bins and bin width
Dx ¼ 0.1. The resulting histogram is divided by N . Dx ¼1000 to yield the normalized histogram fW (xk), k ¼ 1, . . . , 100. This is shown in Fig. 15.3.6.
3Superimposed on 3top of the
histogram is the true Weibull density function, fX (x) ¼ 3 13 (x 2)2 e½(x2)=3 , x . 2.
274
Computer Methods for Generating Random Variates
FIGURE 15.3.6
We can see from the figure that the fit is very good. We can also apply the chi-square
goodness of fit, which it readily passes.
Example 15.3.5 (Rayleigh Distribution) The random variates corresponding to a
Rayleigh distribution can be generated by taking (1) the direct inverse of the distribution
or (2) the square root of the sum of the squares of two independent Gaussian distributions.
The cdf of a Rayleigh distribution is given by
FZ (z) ¼ 1 e(z
2
=2s2 )
, z.0
(15:3:6)
Since this is in a closed form, the inverse is found as follows:
2
2
1 e(z =2s ) ¼ u
or e(z
2
=2s2 )
¼ 1 u and
z ¼ FZ1 (u) ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2s2 ln(1 u)
The Rayleigh random variates are generated from this equation with s2 ¼ 1, resulting in
the following equation:
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
zi ¼ 2s2 ln(1 ui ), z . 0, i ¼ 0, 1, . . . , 10,000
The histogram subroutine hist(K, Dx, x) is called with K ¼ 200 bins and bin width
Dx ¼ 0.05. The resulting histogram is divided by N . Dx ¼ 500 to yield the normalized histogram fR(xk), k ¼ 1, . . . , 200. This is shown in Fig. 15.3.7. Superimposed on top of the
2
histogram is the true density function, fZ (z) ¼ ze(z =2) , z . 0. We can see from figure
that the fit is very good. We can also apply the chi-square goodness of fit, which it
easily passes.
Special Methods
Even though some distributions may not have a closed form, random variates can be
obtained by special transformations. The very important Gaussian random variates can be
generated by these special transformation techniques. Two independent random variables
U and V are both uniformly distributed in (0,1] and whose joint density is given by
1, 0 , u 1, 0 , v 1
fUV (u,v) ¼
(15:3:7)
0, otherwise
15.3
Inverse Transformation Techniques
275
FIGURE 15.3.7
The transformation used to generate standard Gaussian random variables is
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
X ¼ 2 ln(U) cos(2pV)
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Y ¼ 2 ln(U) sin(2pV)
(15:3:8)
We can find the joint density fXY (x,y) from the transformation. Solving for u and v in
Eq. (15.3.8), we obtain
u ¼ e½(x
2
þy2 )=2
The Jacobian matrix of transformation is
2
3 2 cos (2pv)
dx dx
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
u 2 ln(u)
6 du dv 7 6
6
¼
J¼4
dy dy 5 4 sin (2pv)
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
du dv
u 2 ln(u)
: v¼
1
y
tan1
2p
x
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2p sin(2pv) 2 ln(u)
(15:3:9)
3
7
7
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 5
2p cos(2pv) 2 ln(u)
(15:3:10)
and the absolute value of the Jacobian determinant is
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
cos (2pv)
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2p sin (2pv) 2 ln(u)
2p
2
2
u 2 ln(u)
¼
¼ 2pe(x þy )=2
kJk ¼
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
sin (2pv)
u
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2p cos (2pv) 2 ln(u)
u 2 ln(u)
Hence, from Eqs. (15.3.7) and (15.3.11) we have
fUV (u,v)
1 (x2 þy2 )=2
1 (x2 =2)
1 (y2 =2)
pffiffiffiffiffiffi e
¼
e
¼ pffiffiffiffiffiffi e
fXY (x,y) ¼
kJk
2p
2p
2p
(15:3:11)
(15:3:12)
which yields two independent standard Gaussian random variables. This transformation is
called the Box– Mueller transformation.
Example 15.3.6 (Gaussian Distribution) We will now use the Box – Mueller transformation to generate two independent standard Gaussian random variates. In this example,
10,001 points of two independent uniformly distributed random variates fuig and fvig
are generated. They are passed through the transformation of Eq. (15.3.8) to form the
276
Computer Methods for Generating Random Variates
Gaussian random variates fzig and fwig. Theoretically, these must be zero mean and unit
variance. If they are not, they can be transformed into exactly zero mean and unit variance
by another transformation
wi w
; i ¼ 0; . . . , 10;000
sw
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
P
P
where z ¼ (1=N) Ni¼0 zi and sz ¼ (1=N) Ni¼0 (zi z)2 , N ¼ 10,000, with similar definitions for w and sw . The histogram subroutines are called with K ¼ 100 bins and bin
width Dx ¼ Dy ¼ 0.08. The resulting histogram is normalized by NDx ¼ NDy ¼ 800 to
yield histograms fXk and fYk, k ¼ 1, . . . , 100 for the Gaussian random variates.
plots
pffiffiffiffiffiffi Both
2
are shown p
inffiffiffiffiffiffi
Fig. 15.3.8 with the true standard Gaussian fX (x) ¼ (1= 2p)e(x =2) and
2
fY (y) ¼ ð1= 2pÞe(y =2) superimposed on them. From these figures we find that the fit
is very good. These random variates are stored for later use in simulating stochastic
systems.
xi ¼
zi z
;
sz
yi ¼
FIGURE 15.3.8
15.3
Inverse Transformation Techniques
277
Example 15.3.7 (Rayleigh Distribution with Phase) We have already found random
variates for a Rayleigh distribution using a direct transformation. We will now apply
the relationship to two independent standard Gaussian random variables by
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Z ¼ X2 þ Y 2
to generate random variates for the Rayleigh distribution. The Gaussian random variates
are generated by the Box – Mueller transformation as explained in the previous section.
The Rayleigh random variates and the corresponding phase variates are given by
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 9
zi ¼ x2i þ y2i >
=
, i ¼ 0, . . . , N
y
i
>
wi ¼ tan1
;
xi
The sequences fzig and fwig are both passed through histogram subroutines with the
number of bins K ¼ 200 and the bin width Dz ¼ Dw ¼ 0.05. Both these histograms are
normalized with division by NDz ¼ NDw ¼ 500. The resulting histograms fZk and fWk
2
are shown in Fig. 15.3.9 with the true Rayleigh density fZ (z) ¼ ze(z =2) , z . 0 and
uniform density U(– p/2, p/2] superimposed on them. We can see from both plots that
the fit is very good.
Table 15.3.1 shows the densities with their closed-form distribution functions and
their inverses.
FIGURE 15.3.9
278
a
b aþ1
, x.0
b xþb
b
p½(x a)2 þ b2 1 (x2 þy2 )=2
e
2p
Pareto
Cauchy
Gaussian
z.0
z (z2 =s2 )
e
,
s2
ala (x m)a1 e½l(xm)
Rayleigh
Weibull
a
x.m
=2s2 )
,
b a
,
xþb
2
1
2p
1 1
ðx ðy
e(j
2
þh2 )=2
dhdj
x.0
z.0
x.m
1 1 1 x a
þ tan
2 p
b
1
1 e(z
a
1 e½l(xm) ,
1 lx
1
e , x 0 1 elx ,
2
2
l ljxj
e
,
2
Laplace
l0
(1 elx )u(x)
x.0
1 xa 2
a,xb
2 ba
1 x þ a 2b 2
b , x 2b a
1
ba
2
xa
ba
cdf F(x)
1 lx
e u(x)
l
1
ba
pdf f (x)
Exponential
Triangular
Uniform (a,b)
Type
TABLE 15.3.1
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1p
a
ln (1 u),
l
u,1
u,1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 ln (u1 ) cos (2pu2 )
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
y ¼ 2 ln (u1 ) sin (2pu2 )
x¼
1
a þ b tan p u 2
1
ffiffiffiffiffiffiffiffiffiffiffi
b p
1
,
a
1u
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2s2 ln (1 u), u , 1
mþ
1
1 1
ln (2u), u ln½2(1 u),
l
2 l
1
ln (1 u), u , 1
l
u.
pffiffiffiffiffi
1
a þ (b a) 2u, u 2
h
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffii
1
b þ (b a) 1 2(1 u) u .
2
a þ u(b a)
Inverse F 21(u)
1
2
15.4
B 15.4
Convolution Techniques
279
CONVOLUTION TECHNIQUES
In cases where it may not be possible to obtain inverses, we can use the convolution technique. From Eq. (13.1.15) we recall that the density of the sum Z of two independent
random variables X and Y is the convolution of the densities of X and Y:
ð1
fZ (z) ¼
fX (x)fY (z x)dx
(15:4:1)
1
We can generate random variates from independent random variates that have already
been generated. We will give two examples.
Example 15.4.1 (Triangular Distribution) We want to generate random variates for a
triangular distribution. Two independent uniformly distributed random variates, faig and
fbig, are generated. We form the sum
x i ¼ ai þ bi ,
i ¼ 0, . . . , 10,000
The sequence fxig is passed through the histogram subroutine with the number of bins
K ¼ 100 and the bin width Dx ¼ 0.02. The resulting histogram is normalized with
NDx ¼ 10,000 0.02 ¼ 200. The histogram fXk is graphed in Fig. 15.4.1. The true triangular density given by
fX (x) ¼
x,
2 x,
0,x1
1,x2
is superimposed on the histogram. The fit is very close and passes the chi-square
test easily.
Example 15.4.2 (Erlang Distribution) The Erlang distribution is obtained by the n-fold
convolution of n independent exponential distributions as given in Eq. (7.4.3). Hence we
can generate Erlang random variates with n degrees of freedom by summing n independent
exponential random variates. We will now generate Erlang random variates with 2 degrees
FIGURE 15.4.1
280
Computer Methods for Generating Random Variates
FIGURE 15.4.2
of freedom. From two sets of uniform random variates faig and fbig, we generate two identically distributed exponential random variates fxig and fyig, with l ¼ 13 given by
xi ¼ 3 ln (ai );
yi ¼ 3 ln (bi )
Erlang random variates fzig are formed by summing fxig and fyig. The variates fzig are then
passed through the histogram subroutine with number of bins K ¼ 200 and bin width
Dz ¼ 0.1. The resulting histogram is normalized by division with NDx ¼ 1000. The histogram fZk is graphed for k ¼ 0, . . . , 200 and shown in Fig. 15.4.2. The true Erlang density
for 2 degrees of freedom given by
2
1
fZ (z) ¼
ze(z=3) ,
3
z0
is superimposed on the histogram as shown in the figure. The fit is very good as seen in
the diagram.
B 15.5
ACCEPTANCE – REJECTION TECHNIQUES
The rejection method does not require that the cumulative distribution function [indefinite
integral of p(x)] be readily computable, much less the inverse of that function—which was
required for the transformation method in the previous section.
Usually the inverse transformation technique uses the cumulative distribution function in closed form, in which case it is the preferred method of generating random variates.
However, in many cases the cdf may not be expressible in closed form or it may be very
inefficient to calculate it numerically. On the other hand, the density functions may be
available in closed form such as the beta or the gamma density. The rejection method is
a technique for generating random deviates fxig whose density function fX(x), if known
in closed form, is amenable directly to determine these variates. We will now outline
the acceptance– rejection method of generating variates.
Consider a function g(x) that upper-bounds the density fX(x) or g(x) fX(x) for all x.
This is called the bounding or comparison function g(x) which, is not a density function
15.5 Acceptance – Rejection Techniques
281
Ð1
Ð1
since
1 g(x)dx 1 fX (x)dx ¼ 1. We define a normalization constant c ¼
Ð1
1 g(x)dx 1, and another function wX (x) ¼ g(x)=c for all x, thus rendering wX(x) a
proper density function.
We can now state the acceptance – rejection algorithm:
1. Generate random variates fyi, i ¼ 0, . . . , Ng having the density function wY( y).
2. Generate random variates fui, i ¼ 0, . . . , Ng uniformly distributed in (0,1] and
independent of fyig.
3. If ui ½ fX ( yi )=½g( yi ) 1, accept the point yi and set fxi ¼ yig. Otherwise, reject
yi. The number of accepted points fxig will be equal to N/c.
4. The sequence fxig are the desired random variates with density function fX(x).
We will now show from conditional probability considerations that the generated
variates fxig do indeed have the distribution FX(x). From the definition of conditional probability, we obtain
P(generated X x) ¼ P(Y x j accept) ¼
P(Y x, acceptÞ
P(accept)
(15:5:1)
We need the following result for any y:
fX (y)
P(accept j Y ¼ y) ¼ P U g(y)
(15:5:2)
Since U is uniformly distributed in (0,1], we have FU(u) ¼ u, and it is independent of Y.
Hence Eq. (15.5.2) can be written as follows:
fX (y)
fX (y)
fX (y)
P U
¼ FU
¼
(15:5:3)
g(y)
g(y)
g(y)
Incorporating Eq. (15.5.3) into Eq. (15.5.2), we obtain
P(accept j Y ¼ y) ¼
fX (y)
g(y)
(15:5:4)
We can evaluate the denominator in Eq. (15.5.1) from Eq. (15.5.4) with substitution of
wY (y) ¼ g(y)=c.
ð1
P(accept) ¼
P(accept j Y ¼ y)wY (y)dy
1
ð1
¼
1
fX (y) g(y)
1
dy ¼
g(y) c
c
We will now evaluate the numerator of Eq. (15.5.1):
ð1
P(Y x, accept) ¼
P(accept, Y x j Y ¼ y)wY (y)dy
(15:5:5)
(15:5:6)
1
Since Y ¼ y in Eq. (15.5.6) we can rewrite it as
ðx
P(Y x, accept) ¼
P(accept, y x j Y ¼ y)wY (y)dy
1
(15:5:7)
282
Computer Methods for Generating Random Variates
where the integration is terminated at x because y x. Applying Eq. (15.5.4) and substituting wY (y) ¼ g(y)=c in Eq. (15.5.7), we obtain
ðx
fX (y) g(y)
P(Y x, accept) ¼
dy
1 g(y) c
¼
FX (x)
c
(15:5:8)
Combining Eqs. (15.5.5) and (15.5.8), we obtain the result that
P(generated X x) ¼
FX (x)
c ¼ FX (x)
c
(15:5:9)
thus showing that the acceptance –rejection method yields the desired probability
distribution.
Example 15.5.1 We will give an example of generating beta density variates by the
method described above. The general beta density is given by
fb (x) ¼
G(a þ b) a1
x (1 x)b1 ,
G(a)G(b)
0x1
and there is no closed-form solution for the cdf. For a and b integers, we have
fb (x) ¼
(a þ b 1)! a1
x (1 x)b1 , 0 x 1
(a 1)!(b 1)!
Random variates will be generated for a beta(3,6) density (Fig. 15.5.1) given by
fb (x) ¼ 168x2 (1 x)5 ,
0x1
The maximum of the density function occurs at x ¼ 27 ¼ 0.2857 and fb(27) ¼ 2.55. We can
bound this density with a rectangle of width 2.55 so that g(x) ¼ 2.55, 0 ,x 1. Hence the
normalizing constant c ¼ 2.55 and wX (x) ¼ g(x)=c is a U(0,1] density function.
Algorithm for Acceptance – Rejection
1. Generate a sequence fyi, i ¼ 0, . . . , Ng with uniform density wY (y) ¼ g(y)=c.
2. Generate uniform random variates fui, i ¼ 0, . . . , Ng independent of fyig.
3. If ui ½168y2i (1 yi )5 =2:55; return xi ¼ yi, i ¼ 0, . . . , N.
Sequences fyi, i ¼ 0, . . . , 10,000g and fui, i ¼ 0, . . . , 10,000g are generated, and if
the condition 3 (above) is satisfied, xi was set equal to yi. The sequence fxig has 3936
points out of the original total of 10,001 points. This corresponds to the expected
10,001/2.55 ¼ 3922 points. The sequence fxig consisting of 3936 points was passed
through a histogram subroutine with K ¼ 100 bins and bin width Dx ¼ 0.01. The
histogram was normalized by division with 3936 0.01 ¼ 39.36. The normalized
histogram is graphed in Fig. 15.5.1 with the true beta(3,6) density superimposed for
comparison. The bounding function g(x) is also shown. From the figure it is seen that
the fit is good.
Discussion
We have discussed three methods for generation of random variates for a given distribution function. The inversion method is used where the cumulative distribution
15.5 Acceptance – Rejection Techniques
283
FIGURE 15.5.1
function can be expressed in a closed form or can be computed efficiently. The convolution method can be used when the density function is expressible as the resultant of a sum
of independent random variates. The acceptance –rejection method is used directly on the
probability density function. In generating histograms the number of bins K and the bin
width Dx have to be manipulated first by “eyeballing” and later confirming that the x
value is less than the ta value in the chi-square test at the significance level a. Note
also that higher significance levels yield tighter bounds.
CHAPTER
16
Elements of Matrix Algebra
A number of results depend on an understanding of matrices and manipulations of
matrices. We will now discuss relevant aspects of matrix algebra. We will assume familiarity with linear vectors spaces [56].
B 16.1
BASIC THEORY OF MATRICES
Definitions
A matrix X is defined as an array of m column vectors {xj , j ¼ 1, . . . , m} with each
column vector x.j having n real or complex elements x.j ¼ fx1j, x2j, . . . , xi j, . . . , xn jgT. Thus
the matrix X is an n m rectangular array of real or complex numbers and written as
3
2
x1 x2 xj . . . xm
2 36 x11 x12 x1j x1m 7
7
x1 6
7
6
6 x2 76 x21 x22 x2j x2m 7
7
76
(16:1:1)
X¼6
7
4 xi 56
6 xi1 xi2 xij xim 7
7
xn 6
5
4
xn1 xn2 xnj xnm
Notation: In the double subscript (i j ) the first subscript denotes the row, and the second
subscript denotes the column. Thus, an n m array consists of n rows and m columns.
Sometimes the matrix X is also written as fxijg. A square matrix is a matrix with equal
number of rows and columns. It is denoted as n n square matrix or an n matrix. If the
column vectors {xj ¼ ½x1j , x2j , . . . , xnj T , j ¼ 1, 2, . . . , m} are linearly independent, then
the matrix X is column-independent. On the other hand, if the row vectors denoted by
{xi ¼ ½xi1 , xi2 , . . . , xim , i ¼ 1, 2, . . . , n} are linearly independent, then the matrix X is
row-independent. The dots in the column-vector x j and the row-vector x i represent the
integers 1, . . . , n. An arbitrary rectangular matrix cannot be both column- and rowindependent unless it is a square matrix. However, not all square matrices need be
row-independent and column-independent.
Parts of this chapter are based, with permission, on Chapters 30, 31, and 33 from the book Linear Systems
Properties—a Quick Reference, by Venkatarama Krishnan, published by CRC Press, 1988.
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
284
16.1
Basic Theory of Matrices
285
Matrix Addition
Addition is defined for two matrices X and Y of similar dimensions:
X þ Y ¼ {xij } þ {yij } ¼ {xij þ yij }
(16:1:2)
The addition rule is commutative: X þ Y ¼ Y þ X.
Example 16.1.1
"
#
"
1
5 2
3
4 7
3
8
6
11
10
16
,
X¼
"
3 4
8
6 9
#
"
,
Y¼
4 6
#
Z¼
3 2
#
,
XþY¼
2
X þ Z and Y þ Z are not defined
Matrix Multiplication
Matrix multiplication is a little complex. If we denote X Y ¼ Z, then the number of
columns of X must be equal to the number of rows of Y. Otherwise, multiplication is not
defined. If X is ? n, then Y has to be n ? In this case the dimension of Z is ? ?. The
product X Y is defined by
Z¼XY¼
n
X
xik ykj ¼ {zij}
(16:1:3)
k¼1
Example 16.1.2
"
3
1 2
2
#
,
X¼
4
"
5 1
4
6
6
Y ¼ 63
4
1
1
3
7
7
27
5
5
34þ13þ21 31þ12þ25
#
XY¼
"
17 15
#
¼
44þ53þ11 41þ52þ15
2
32 19
3
2
3
43þ14
41þ15
42þ11
16
9
9
6
6
Y X ¼ 63 3 þ 2 4
4
13þ54
31þ25
7 6
7 6
3 2 þ 2 1 7 ¼ 6 17
5 4
12þ51
23
13
7
7
18 7
5
7
11þ55
26
In general, matrix multiplication is not commutative, that is, XY = YX as shown above.
As an example, consider the following symmetric matrices:
e f
a b
,
Y¼
X¼
f h
b d
ae þ bf af þ bh
ae þ bf be þ df
XY ¼
,
YX ¼
be þ df bf þ dh
af þ bh bf þ dh
286
Elements of Matrix Algebra
The matrices X and Y commute only if af þ bh ¼ be þ df.
We usually denote in the product XY that Y is premultiplied by X or X is postmultiplied by Y. Unlike real numbers, division of one matrix by another matrix is not
defined.
Transpose of Matrix
The transpose of a matrix X ¼ fxijg is obtained by interchanging the rows and
columns and is denoted by
XT ¼ {xij }T ¼ {x ji }
(16:1:4)
The transpose of a matrix product (XY) T is given by Y TX T whereas (X þ Y)T ¼
X T þ Y T. If X ¼ X T, then we say that the matrix X is symmetric.
Example 16.1.3
2
X¼
1
5
3
27
(XY) ¼
21
T
4
6
2
3
6
Y ¼ 41
4
30
;
23
3
2
7
1 5;
2
6
XT ¼ 4 5
3
4
27 30
Y X ¼
21 23
T
T
2
3
1
3
7
3 5 YT ¼
2
6
1
1
4
3
Determinant of a Matrix
The nonzero determinant of a square matrix is one of the two most important
invariants of a matrix. It is usually denoted by det X or jXj. It is computed as
det X ¼
n
X
xij Dij ¼ x1j D1j þ x2j D2j þ þ xnj Dnj
(16:1:5)
i¼1
with no implied summation over j and Dij is the cofactor of the matrix X defined by
Dij ¼ (1)iþj Mij
(16:1:6)
where Mij is the minor of the matrix X obtained by taking the determinant after deleting
the ith row, i ¼ 1, . . . , n, and the fixed jth column. Equation (16.1.5) is called a cofactor
expansion about the jth column. The difference between the cofactor and the minor is the
sign as shown in Eq. (16.1.6).
Example 16.1.4
The determinant of the following matrix is evaluated as follows:
2
3
2
5
3
2
6 1
2 4
57
7
X¼6
4 4
1
3 2 5
3 4
2
1
16.1
Basic Theory of Matrices
287
Expanding in terms of minors about the second column, the expression for the determinant
jXj is
2 3
1 4
2 5 jXj ¼ 5 4
3 2 þ 2 4 3 2 3
3 2
1
2
1
2
2
3
2 3 2 5
þ 1 1 4 5 þ 4 1 4
3
4
3 2 2 1
Each determinants in the above equation can again be expanded in terms of minors about
the second column as shown by the expansion of the first term:
1 4
5 1 5
4 2 5 4
3 2 ¼ 5 4 5 3 3
3 1
1
3
2
1
1
5
5 (2) ¼ 350
4 2 Evaluating the determinant
42 þ 376 ¼ 632.
by
this
expansion,
we
have
jXj ¼ 350 252 2
If only the determinant of the matrix is zero, we call that it a singular or simply degenerate matrix. In addition to the determinant, if the first- and higher-order minors are also
zero, then we call the matrix multiply degenerate.
Properties of Determinants (X and Y Nonsingular)
1. det X ¼ det X T: The following properties (2,3,4,5,6) are very useful in the
manipulation of matrices.
2. If any row (or column) of X is multiplied by a constant a to yield a new matrix Y,
then det X ¼ a . det Y.
3. If any row (or column) of X is proportional to the other one, then the rows (or
columns) are linearly dependent and det X ¼ 0.
4. If any two rows (or columns) of X are interchanged an odd number of times to
yield Y, then det Y ¼ 2det X. If they are interchanged an even number of times
to yield Z, then det Z ¼ det X.
5. If any row (or column) of X is modified by adding to it a times, the corresponding
elements of another row (or column) to yield Y, then det Y ¼ det X. (The determinant remains unchanged.)
6. The determinant of a triangular matrix X is the product of its diagonal terms
det X ¼ x11.x22 . . . . .xnn.
7. If the matrices X and Y are both n n, then det ( XY) ¼ (det X)(det Y).
8. det (aX) ¼ an det X.
9. Even if X and Y are n n, det (X þ Y) = det X þ det Y in general.
288
Elements of Matrix Algebra
Trace of a Matrix
The trace of a square matrix Tr(X) is the second most important invariant of a matrix.
The trace is the sum of the diagonal terms of a matrix. The trace of the matrix given in
Example 16.1.4 is 2 þ 2 þ 3 þ 1 ¼ 8.
Rank of a Matrix
The column rank of a rectangular matrix is the number of linearly independent
column vectors. Similarly, the row rank of a matrix is the number of linearly independent
rows. If the column rank and the row rank are the same, then we call that number the rank
of the matrix. Thus in a nonsingular square matrix the column rank equals the row rank and
the rank is the dimension of the matrix. In general, rank of any n m matrix can be
defined as the number of elements 2, 3, . . . , n or m required to form a nonzero determinant.
Forming all possible combinations of determinants from any given matrix is a difficult job.
Finding a rank is not always easy, but some methods are easier than others. We show one
method for finding the rank and the determinant of any matrix using pivotal 2 2
determinants.
Example 16.1.5
2
a11
6
6 a21
6
6
X¼6
6 a31
6
6 a41
4
a51
Find the rank and determinant of the following nonsingular matrix:
a12
a13
a14
a22
a23
a24
a32
a33
a34
a42
a43
a44
a52
a53
a54
a15
2
3
1
2
3
2
7 6
6
a25 7
7 6 1
7 6
6
a35 7
7¼6 2
7 6
6
a45 7
5 4 1
1
1
2
1
2
2
2
2
1
2
1
1
2
a55
3
3
7
87
7
7
1 7
7
7
07
5
(16:1:7)
5
Step 1. Remove a11 ¼ 1 outside the matrix in Eq. (16.1.7) using property 2 as shown in
the previous section. We form a 4 4 matrix X1 from 2 2 determinants of the
form fa11 . ajk 2 aj1 . a1k, j ¼ 2, . . . , 5 : k ¼ 2, . . . , 5) as shown below:
2
3
6
6 3
6
X1 ¼ 1 6
6 0
4
5
4
4
8
6
1
1
5
2
11
3
7
7 7
7
7
3 7
5
1
(16:1:8)
Step 2. Normalize the first element ¼ 3 as in step 1, and the resultant matrix is shown
below.
2
1
6
6 3
6
X1 ¼ 1:3 6
6 0
4
5
4
3
8
4
3
6
1
1
5
2
11 3
3 7
7 7
7
7
3 7
5
1
(16:1:9)
16.1
Basic Theory of Matrices
289
Step 3. Form a 3 3 matrix X2 in an exactly analogous fashion by forming 2 2 determinants as in step 1 and normalize the first element in X2 as shown in Eq. (16.1.10)
2
3
2
3
1
4
2
4
1
1
6
7
2
6
7
7
1
1
3 7 ¼ 1:3 4 6
(16:1:10)
X2 ¼ 1:3 6
1
1
3
6
7
4 5 14
4 5 14
52 5
52 5
3
3
3
3
3
3
Step 4. Form a 2 2 matrix X3 in an exactly analogous fashion by forming 2 2 determinants as in step 1 and normalize the first element in X3 as shown in Eq. (16.1.11):
2
3
1
"
#
4
1
8
1
6 2
7
23
(16:1:11)
X3 ¼ 1:3 4 4
5 ¼ 1:3 4 23
19
2
19
6
6
Step 5. Continue this process of reducing the order of the matrix until we end up with either
a 1 1 matrix or a matrix whose elements are all zero. Finally we form a 1 1 matrix
X4 in an exactly analogous fashion by forming 2 2 determinants as in step 1:
X4 ¼ 1:3 4 1 35
¼ 70
2
3
(16:1:12)
Since we have come to a scalar, the operations stop at this point. The number of cycles is
5 and hence the rank of this matrix is 5. The determinant is given by X4 ¼ 70. Thus, we
can obtain both the rank and the determinant by this method.
Example 16.1.6 In this example we will determine the rank and the maximum possible
determinant of a singular matrix given by
2
3
6
11
5
4
1
6 5
3 2 3
47
6
7
7
X¼6
8
12
4
2
5
6
7
4 6
7
1
0
35
3
0 3 1
2
We form the various matrices as in the previous example, and the results are shown below:
3
2
2
3
11
5
4
1
6
11
5
4
1
1
6
6
6
6
67
6
7
7
6
3 2 3
47
6 5
6 5
3 2 3
47
6
7
7
6
X¼6
2 5 7
6 8 12 4
7¼66
8 12 4
2 5 7
7
6
6
7
7
6
7
1
0
35
4 6
4 6
7
1
0
35
3
0 3 1
2
3
0 3 1
2
2
37
6 6
6
6 8
6
6
X1 ¼ 6 6 3
6
6 4
6
4 11
2
37
6
8
3
4
19
3
22
3
4
11
2
3
3
2
19
1
6
6 7
7
6
6 8
11 7
7
37 6
7
6
6 3
3 7¼6
7
6 6
6 4
2 7
7
6
5
4 11
3
2
2
1
8
3
4
11
2
38
37
22
3
4
3
3
19
37 7
7
11 7
7
7
3 7
7
27
7
35
2
290
Elements of Matrix Algebra
We now form a 3 3 matrix using 2 2 pivotal determinants. This will yield the first
column to be all zeros. Interchanging two columns of the resulting matrix is also shown
below:
2
3
2
170 85
170
6
7
6
37
37 7
6
6 37
37
37
4
2 7
6
6 4
16 0
16
X2 ¼ 6 7¼6 6
7
6 37
6
6
37
37
4
4 98
98 49 5
0
37
37
37
2
3
1
0
1
6
7
2
7
37 170 6
2
6 4
7
¼6 07
6
7
6
37 6 37 37
4 98 49
5
0
37 37
0
3
85
0
7
37
7
2
7
07
7
37
5
49
0
37
We now form a 2 2 matrix
"
37 170 0
X3 ¼ 6 6
37 0
0
#
"
0
0
0
0
#
¼ 170
0
After three cycles we have a premature termination with a 2 2 zero matrix. Hence
170
the rank of this matrix is 3 and the maximum determinant is 6 37
6 37 ¼ 170:
Inverse of a Matrix
Since division by a matrix is not defined, we define the inverse of a nonsingular matrix
as the matrix X 21 such that XX 21 ¼ X 21X ¼ I, an identity matrix. Thus inverse matrices
commute with their direct counterparts. The cofactor expansion of the determinant of the
matrix X from Eq. (16.1.5) is
det X ¼
n
X
xij Dij ¼ x1j D1j þ x2j D2j þ þ xnj Dnj
(16:1:5)
i¼1
where the expansion is carried about the n elements in any column j. Equation (16.1.5)
can be rewritten as
(
n
X
det X, k ¼ j
det X ¼
xij Dik ¼
(16:1:13)
0,
k=j
i¼1
In Eq. (16.1.13), when k ¼ j, we obtain det X by Eq. (16.1.5), but when k = j, then
we have cofactor expansion about the n elements in the jth column but with cofactors
obtained from the kth column. This is equivalent to finding the determinant by replacing the kth column with the jth column in the matrix X. Thus X will contain two
identical columns and hence from property 3 in the previous section, det X ¼ 0.
We divide Eq. (16.1.13) by det X and obtain
(
n X
1kn
Dik
(16:1:14)
xij ¼ dkj ,
det X
1jn
i¼1
16.1
Basic Theory of Matrices
291
where dkj is the Krönecker delta, which equals 1 if k ¼ j and 0 if k = j. Equation (16.1.14)
is a scalar statement of the matrix equation X21X ¼ I, and the inverse is given by
2
3
D11 D21 Dn1
7
Dik
1 6
6 D12 D22 Dn2 7
(16:1:15)
X1 ¼ {aki } ¼
xij ¼
6 ..
.. 7
..
det X 4 .
det X
. 5
.
D1n D1n
Dnn
Note that the cofactors are in reverse order of the sequence of elements of the matrix X.
The matrix fDjig of the cofactors is sometimes called the adjoint matrix. Finding the
inverse using Eq. (16.1.15) is very inefficient and is of only theoretical interest since it
gives an explicit expression for the inverse.
Solution of Linear Equations
We can use matrix analysis to solve linear equations of the form Ax ¼ y where x and
y are n-vectors:
32 3 2 3
2
y1
a11 a12 a1n
x1
7
7
7
6a
6
6
6 21 a22 a2n 7 6 x2 7 6 y2 7
7
7
6 .
6
6
(16:1:16)
¼
.. 7 6 .. 7 6 .. 7
..
7
6 .
. 54 . 5 4 . 5
.
4 .
an1
an2
ann
A
xn
x
yn
y
We will assume that A is a nonsingular n n matrix and x and y are n 1 vectors.
We want to solve for x in terms of y. Since A is invertible, we can write the solution as
x ¼ A1 y
This equation can be expanded as
2 3
2
2
32 3
3
D j1
y1
D11 D21 Dn1
x1
n 6 D j2 7
6 x2 7
76 7
1 6
1 X
6 7
6 D12 D22 Dn2 7 6 y2 7
6
7
¼
6 .. 7 ¼
6 ..
6
6 . 7yj
7
7
..
..
.. 5 det A
.
4 . 5 det A 4 .
4
4
5
.
.
.
. 5
j¼1
D1n D1n
xn
Dnn
yn
D jn
x
A1
y
(16:1:17)
(16:1:18)
Note again that the order of the cofactors is transposed from that of the original matrix A,
that is, if aij is an element in the ith row and jth column, then Dji is in the jth row and ith
column.
Example 16.1.7
x2, x3, x4:
We shall now solve the following set of linear equations for x1,
x1 3x3 2x4 ¼ 4
5
1
15
23
x1 þ x2 þ x3 þ 6x4 ¼
2
2
2
2
2x1 7x3 6x4 ¼ 11
7
1
25
39
x1 þ x2 þ x3 þ 10x4 ¼
2
2
2
2
292
Elements of Matrix Algebra
or in matrix form
2
1
0
3
6 5
6
6 2
6
6
6 2
6
4
7
2
1
2
0
15
2
7
1
2
25
2
3
2
2 3
4
72 x 3 6 23 7
7
6
1
6 7
76 7 6 2 7
76 x2 7 6
7
74 5 ¼ 6
7
x3
6 11 7
6 7
7
7
6
5 x4
5
4
39
10
2
The determinant of the matrix A is found as follows:
2 1 0 3
2 3
2
3
2
1
15
5
7
6 5 1
6 7
15
7
6
6
6
2
2
6 7
6
7
6 2
6 2 2
2
7
6
7
6
6
jAj ¼ 6
6 7 3 6 2
7 ¼ 1 6 0 7
7
6
7
6
6 2 0 7
6
7
41
5
4 7
6
25
5
4
10
7 1
25
2
2
2
10
2 2
2
2
3
5 1
15
6 2 2
2 7
6
7
6
7
þ 2 6 2 0 7 7
6
7
4 7 1
25 5
2 2
2
1
2
0
1
2
Or jAj ¼ 1 þ 3 3 ¼ 1. The inverse of matrix A is
2
A1
1
0
3
6 5
6
6 2
6
¼6
6 2
6
4
7
2
1
2
0
15
2
7
1
2
25
2
2 31
2
1
7
7
6 7
6
16 2
7
7 ¼ 6 1
14
6 7
3
7
5
2
10
2
1
1
1
2
1
2
1
1
The solution to the original set of equations is given by
3
2
2
3 4
2 3
1 2
1
2 6 23 7 2 3
1
x1
7
6
6 2
1
2
17
7 6 7
6 x2 7 6
76
2
17
7
6
6 7 ¼ 6 1 1
1
17
7¼6
4
4 x3 5 6
76
15
7
6
4 3
1
1 56 11 7
1
x4
5
4
1
39
2
2
2
2
Properties of Inverses (X, Y, Z Same Order and Nonsingular)
1. X X21 ¼ X21X matrix commutes with its inverse
2. Xm Xn ¼ X m1n
3. (XY)21 ¼ Y21X21
4. XY ¼ XZ ) Y ¼ Z
5. (X T)21 ¼ (X21)T
3
2
17
7
17
5
1
2
3
6 7
7
7
6 7
7
5
10
16.2 Eigenvalues and Eigenvectors of Matrices
1
det X
(1)iþj Mij
, where Mij is the minor of X
¼
det X
293
6. det X21 ¼
7. X1
(16.1.19)
Orthogonal Matrices
The matrix X is orthonormal if X TX ¼ I. Since X21X is also equal to the identity
matrix I, we have
XT X ¼ I ¼ X1 X
or XT ¼ X1
(16:1:20)
It immediately follows from Eq. (16.1.20) that the determinant of an orthogonal matrix is
equal to +1. Note that the converse is not true. In any orthogonal matrix X, real or
complex, individual row or column vectors x are orthonormal:
1, j ¼ i
H
xi x j ¼
(16:1:21)
0, j = i
where H stands for conjugate transpose.
An example of a 4 4 orthogonal matrix used in discrete Fourier transform analysis
is shown below:
2
3
1
1
1
1
6 1 ej(2p=4) ej(4p=4) ej(6p=4) 7
7
W¼6
(16:1:22)
4 1 ej(4p=4) ej(8p=4) ej(12p=4) 5
1 ej(6p=4) ej(12p=4) ej(18p=4)
It can be verified that the rows and columns of W are orthonormal and det W ¼ 1.
B 16.2 EIGENVALUES AND
EIGENVECTORS OF MATRICES
Eigenvalues
The concept of eigenvalues and eigenvectors plays an important role. We motivate
this concept by investigating solutions to a linear system of equations given by
Ax ¼ lIx ¼ lx
where A is an n n matrix and x is an n 1 vector, l is a scalar constant, and I is an
n n identity matrix. Simplification of the equation above yields
(lI A)x ¼ 0
(16:2:1)
A trivial solution to Eq. (16.2.1) is x ¼ 0. Nontrivial solutions can exist only if
det(lI A) ¼ jlI Aj ¼ 0
(16:2:2)
Equation (16.2.2) is called the characteristic equation of the matrix A. Expansion of the
determinant in Eq. (16.2.2) yields a polynomial in l of degree n called the characteristic
polynomial of the matrix A, given by
ln þ an1 ln1 þ an2 ln2 þ þ a0 ¼ 0
(16:2:3)
294
Elements of Matrix Algebra
The characteristic equation has a fundamental significance in linear systems.
The n roots of the characteristic equation are called the eigenvalues of the matrix
A, and in linear systems they are also known as the “poles” of the transfer function.
The eigenvalues can be real, repeated, or complex. From the well-known result
for the roots of polynomials, we also have the sum of the eigenvalues as 2an21, and
the product of the eigenvalues, which is also the determinant of the matrix A, is
given by (21)n a0.
Example 16.2.1
determined:
The eigenvalues of the following nonsingular matrix have to be
2
3
1:5625
4:5
2:0625
5:25
6 0:4375
6
6
A¼6
6 0:4375
6
4 0:0625
2:5
1:9375
6:25
2:5
1:9375
7:25
3:5
0:4375
6:25
1:25 7
7
7
1:25 7
7
7
0:25 5
1:5
3
1
0:5
0
0:75
The characteristic equation of A is det (lI 2 A)¼0 is given by
2
l þ 1:5625
4:5
6
6 0:4375 l þ 2:5
6
6
jlI Aj ¼ 6
0:4375
2:5
6
6
6
3:5
4 0:0625
0:5
0
5
4
3
2:0625
5:25
1:9375
6:25
l þ 1:9375
7:25
0:4375
1:5
l 6:25
3
3
0:75 7
1:25 7
7
7
1:25 7
7
7
0:25 7
5
l1 2
¼ l 1:25l 2:375l þ l þ 1:375l þ 0:25 ¼ 0
The solutions of the fifth-order polynomial of this equation are the eigenvalues of A,
given by
l1 ¼ 2, l2 ¼ 1, l3 ¼ 1, l4 ¼ 0:5, l5 ¼ 0:25
The product of the eigenvalues is the determinant given by (21)5 . (a0) ¼ 20.25 and the
sum of the eigenvalues is a4 ¼ 21.25.
We can also find the eigenvalues of a square matrix that is singular. In this case the
number of nonzero eigenvalues represents the rank of the matrix and the number of zero
eigenvalues, the degree of singularity.
Cayley– Hamilton Theorem
The Cayley– Hamilton theorem states that every square matrix satisfies its own
characteristic equation. In Eq. (16.2.3), if we substitute the matrix A for l, we obtain
the Cayley –Hamilton equation
An þ an1 An1 þ an2 An2 þ þ a0 A0 ¼ 0
where 0 represents a zero matrix.
(16:2:4)
16.2 Eigenvalues and Eigenvectors of Matrices
Example 16.2.2
295
Given the matrix
2
0
A¼4 6
6
1
0
11
3
0
15
6
equation is l3 þ 6l2 þ 11l þ 6 ¼ 0. According to the Cayley–
we should have
33
2
2
32
31
0
0
1
0
0
1
0
7
6
6
7
7
1 5 þ 64 0
0
1 5 þ 114 0
0
15
6
6 11 6
6 11 6
30 2
3 2
3
1
0
6
11 6
0
0 6
7 6
7 6
7
0
1 5 ¼ 4 36
60
25 5 þ 4 36 66 36 5
11 6
150 239 90
216 360 150
3 2
3
11
0
6 0 0
7 6
7
0
11 5 þ 4 0 6 0 5
121 66
0 0 6
the characteristic
Hamilton theorem,
2
0
1
6
0
4 0
6
11
2
0
6
þ 64 0
6
2
0
6
þ4 0
66
or
2
0
¼ 40
0
0
0
0
3
0
05
0
showing the validity of the theorem.
Definiteness of Matrices
A matrix A is positive definite if all its eigenvalues are either positive or have positive real
parts. For example, the matrix A in Example 16.2.2 is positive definite, whereas the matrix A
in Example 16.2.1 is nondefinite. Similarly, matrices having eigenvalues that are negative or
have negative real parts are known as negative definite. If some of the eigenvalues are zero,
then the matrix is called positive semidefinite or negative semidefinite accordingly.
Since the determinants are (21)n times the product of eigenvalues, definiteness can
also be defined by the sign of the determinants of all principal minors of a matrix.
Principal minors are those minors obtained by successively deleting the rows and
columns of a matrix starting from the first row and first column. If all the determinants
of principal minors are positive, then the matrix is positive-definite: if they are negative
then they are negative definite.
Example 16.2.3
eigenvalues:
We will check the definiteness of the following matrix by finding the
2
1
3
7
7
5
6
9
11
4
87
7
7
87
7
7
75
63
6
6
A¼6
63
6
4 2:5
9
11
3
8
9:5
4
0:5
0
1:5
3
1
296
Elements of Matrix Algebra
As in the previous example, setting det (lI 2 A) ¼ 0 gives
3
2
7
5
6 l 1 7
6 3
l9
11
4
8 7
6
7
6
7
2 3
2
9 l þ 11
3
8 7
jlI Aj ¼ 6
6 3
7 ¼ l (l þ 4l þ l 6) ¼ 0
6
7
4 2:5
8
9:5 l þ 4 7 5
0:5
0
1:5 3
l1
Thus the rank of the matrix A is 3 since there are two zero eigenvalues. The eigenvalues
are given by l1 ¼ 23, l2 ¼ 22, l3 ¼ 1, l4 ¼ 0, l5 ¼ 0. Here the matrix A is nondefinite
since it has positive, negative, and zero eigenvalues.
Eigenvectors
Eigenvectors are solutions to Eq. (16.2.1) for any given eigenvalue. We shall assume
that all eigenvalues are distinct and there are no repeated eigenvalues. If we substitute any
eigenvalue li in det(lI 2 A) ¼ 0, then, by the very definition, we have det(liI– A) ¼ 0
and hence solutions to Eq. (16.2.1) are possible, and these solutions are the unnormalized
eigenvectors ci. However, these solutions are unique only upto a multiplicative constant.
To make the solutions unique we can normalize these eigenvectors and the normalized
eigenvectors are denoted by fi ¼ ci/jcij.
Example 16.2.4
repeated below:
Find the eigenvectors of the nonsingular matrix of Example 16.2.1,
2
2:0625
6 0:4375 2:5
6
6
A¼6
6 0:4375 2:5
6
4 0:0625 3:5
1:9375
6:25
1:9375
7:25
0:4375
6:25
1:25 7
7
7
1:25 7
7
7
0:25 5
1:5
3
1
0:5
0
5:25 0:75
3
1:5625 4:5
Substituting l1 ¼ 2 in this equation, we can solve for the unnormalized eigenvectors:
2
32
3
c11
3:5625 4:5 2:0625 5:25
0:75
6 0:4375 4:5
6
7
1:9375 6:25 1:25 7
6
76 c12 7
6
76
7
6
7
3:9375 7:25 1:25 7
ðl1 I AÞC1 ¼ 6
6 0:4375 2:5
76 c13 7 ¼ 0
6
76
7
4 0:0625 3:5
0:4375 4:25 0:25 54 c14 5
0:5
0
1:5
3
1
c15
As mentioned earlier, the equation above can be solved for only four of the unknowns in
terms of the fifth. Let us assume that c11 ¼ 1 and solve for c12,c13,c14,c15 in terms of c11.
The preceding equation can be rewritten by substituting c11 ¼ 1:
3
32
2
3 2
c12
0:4375
4:5 1:9375 6:25 1:25
7
76
6
7 6
6 2:5 3:9375 7:25 1:25 76 c13 7 6 0:4375 7
7
76
6
7¼6
6 3:5 0:4375 4:25 0:25 76 c 7 6 0:0625 7
5
54 14 5 4
4
0:5
0
1:5
3
1
c15
16.2 Eigenvalues and Eigenvectors of Matrices
297
Solving for c12 ,c13 ,c14 ,c15 and adjoining c11 ¼ 1, we obtain the unnormalized eigenvector corresponding to the eigenvalue l1 ¼ 2:
3
2
3 2
c11
1
3
2
3 2
1:333333
c12
7
6
7 6
6 c12 7 6 1:333333 7
6 c 7 6 1:952381 7
6
7
6
7
7
6 13 7 6
¼6
1:952381 7
c13 7
7 and 6
6
7¼6
6
7
6
7
4 c14 5 4 1:238095 5
7
6
7 6
4 c14 5 4 1:238095 5
1:285714
c15
1:285714
c15
Normalization of this equation yields the eigenvector f1:
3
2
3 2
2
3
0:319838
1
c11
7
6
7 6
6
7
6 1:333333 7 6 0:426451 7
6 c12 7
7
6
7 6
7
c1 6
1
6 1:952381 7 ¼ 6 0:624446 7
6 c13 7 ¼
f1 ¼
6
7
6
7
6
7
jc1 j 6
7
7 6
7 3:126581 6
4 1:238095 5 4 0:395990 5
4 c14 5
0:411220
1:285714
c15
l1 ¼2
In a similar manner, the other eigenvectors corresponding to l2 ¼ 1, l3 ¼ 21, l4 ¼ 20.5,
and l5 ¼ 20.25 are shown below:
2
2
2
3
3
3
0:275010
0:774597
0:679189
6
6
6
7
7
7
6 0:366679 7
6 0:516398 7
6 0:549819 7
6
6
6
7
7
7
6
6
7
7
7
f2 ¼ 6
6 0:641681 7 , f3 ¼ 6 0:258199 7 , f4 ¼ 6 0:032342 7 ,
6
6
6
7
7
7
4 0:275010 5
4 0:258199 5
4 0:291081 5
0:550019
0
0:388108
2
l2 ¼1
3
l3 ¼1
l4 ¼0:5
0:679104
6
7
6 0:325871 7
6
7
7
f5 ¼ 6
6 0:340796 7
6
7
4 0:166667 5
0:531373
l5 ¼0:25
Modal Matrix
A matrix F consisting of the linearly independent eigenvectors of a matrix A is called
the modal matrix of A. In the previous example the modal matrix of A was
2
3
0:319838 0:319838 0:774597
0:679189 0:679104
6
7
0:516398 0:549819
0:325871 7
6 0:426451 0:426451
6
7
F¼6
0:258199
0:032342 0:340796 7
6 0:624446 0:624446
7
6
7
0:258199 0:291081
0:166667 5
4 0:395990 0:395990
0:411220 0:411220
0
0:388108 0:531373
The determinant of F ¼ 20.00141.
298
Elements of Matrix Algebra
Diagonalization of Matrices
An n n square matrix A is diagonalizable if there exists another nonsingular n n
matrix T such that T21AT ¼ D, where D is a diagonal matrix. The conditions of diagonalizability reduce to the existence of n linearly independent eigenvectors of A or the modal
matrix F of A is invertible, in which case the jth diagonal element of D is the jth eigenvalue
of A. In this case the modal matrix F is the desired matrix T. The modal matrix F is not the
only matrix that will diagonalize A. There are other matrices that may also diagonalize A,
but all the diagonal elements obtained from the modal matrix only are the eigenvalues of A.
As a consequence of the preceding discussion, we can write the matrix equivalent of
the eigenequation (16.2.1) as follows:
or F1 AF ¼ L
AF ¼ FL
Example 16.2.5
Diagonalize the following matrix:
2
0 1 0
6 4
0 0
A¼6
4 0
2 0
5
1 1
(16:2:5)
3
0
07
7
15
0
The eigenvalues of A are obtained by solving the determinant equation jlI 2 Aj ¼ 0, or
2
3
l
1
0
0 6 4
l
0
07
7 ¼ l4 5l2 þ 4 ¼ 0
jlI Aj ¼ 6
4 0 2
5
l 1 5 1 1
l The characteristic equation can be factored as (l 2 2)(l þ 2)(l 2 1)(l þ 1) ¼ 0, yielding
the eigenvalues as l1 ¼ 2, l2 ¼ 22, l3 ¼ 1, l4 ¼ 21. We find the eigenvectors corresponding to these eigenvalues. The eigenvectors following the Example 16.2.4 are
3
3
2
2
3
3
3
3
2
2
p
ffiffiffiffiffi
p
ffiffiffiffiffiffiffi
ffi
6 74 7
6 146 7
0
0
7
7
6
6
6 6 7
6 6 7
7
7
6
6
6 pffiffiffiffiffi 7
6 pffiffiffiffiffiffiffiffi 7
6 0 7
6 0 7
7
7
6
6
7
7
6
6
6 74 7
6 146 7
6 1 7
6 1 7
f1 ¼ 6
7, f2 ¼ 6 1 7, f3 ¼ 6 pffiffiffi 7, f4 ¼ 6 pffiffiffi 7
7
6 5 7
6
6 27
6 27
6 pffiffiffiffiffi 7
6 pffiffiffiffiffiffiffiffi 7
7
7
6
6
6 74 7
6 146 7
4 1 5
4 1 5
7
7
6
6
pffiffiffi
pffiffiffi
4 2 5
4 10 5
2
2
pffiffiffiffiffiffiffiffi
pffiffiffiffiffi
146
74
or
2
0:348743
3
2
0:248282
3
6 0:697486 7
6 0:496564 7
6
7
6
7
f1 ¼ 6
7 , f2 ¼ 6
7,
4 0:581238 5
4 0:082761 5
0:827606
0:232495
l1 ¼2
2
l2 ¼2
3
0
6
7
0
6
7
f4 ¼ 6
7
4 0:707107 5
0:707107
l4 ¼1
2
0
3
6
7
0
6
7
f3 ¼ 6
7,
4 0:707107 5
0:707107
l3 ¼1
16.2 Eigenvalues and Eigenvectors of Matrices
299
The modal matrix F is given by
2
F ¼ ½f1
f2
f3
3
6 pffiffiffiffiffi
6 74
6 6
6 pffiffiffiffiffi
6
6 74
f4 ¼ 6
6 5
6 pffiffiffiffiffi
6 74
6
4 2
pffiffiffiffiffi
74
3
pffiffiffiffiffiffiffiffi 0
146
6
pffiffiffiffiffiffiffiffi 0
146
1
1
pffiffiffiffiffiffiffiffi pffiffiffi
146
2
10
1
pffiffiffiffiffiffiffiffi pffiffiffi
146
2
3
0 7
7
7
0 7
7
7
7
1 7
pffiffiffi 7
27
7
1 5
pffiffiffi
2
and
2
2
6
0
F1 AF ¼ 6
40
0
0
2
0
0
0
0
1
0
3
0
07
7
05
1
Symmetric Matrices
Real symmetric matrices are those matrices that satisfy the condition
AT ¼ A
(16:2:6)
Complex symmetric matrices satisfy the Hermitian symmetry condition
AH ¼ (A )T ¼ A
(16:2:7)
where the symbol H defined earlier stands for the conjugate transpose.
Symmetric matrices play a very important role in probability; for example, the covariance matrices are all real symmetric matrices. They have some important properties that
come in handy in simplifying algebraic manipulations. Some of the more important
properties of symmetric matrices are
1. All eigenvalues are real.
2. The eigenvectors corresponding to distinct eigenvalues are orthogonal. Hence the
eigenvectors can form the basis vectors of an n-dimensional coordinate system.
3. The modal matrix F is an orthogonal matrix, and hence F21 ¼ F T. The diagonalization of a symmetric matrix can be accomplished by the transformation
F TAF ¼ L.
Example 16.2.6 The eigenvectors and eigenvalues of the symmetric matrix given below
are to be determined and their properties studied:
2
3
1 2
1 1
6 2
1
1 1 7
7
A¼6
4 1
1
2 2 5
1 1 2
0
The characteristic equation is det [lI 2 A] ¼ 0:
l 1
2
1
1
2 l 1 1
1 jlI Aj ¼ ¼ l4 4l3 7l2 þ 22l þ 24 ¼ 0
1
1
l
2
2
1
1
2 l
300
Elements of Matrix Algebra
The eigenvalues are l1 ¼ 3, l2 ¼ 22, l3 ¼ 21, l4 ¼ 4, obtained by factoring the characteristic equation. The four normalized eigenvectors are obtained as outlined in the
previous examples and they are given by
2
2 3
2
2
3
3
3
1
1
1
1
7
7
7
7
1 6
1 6
1 6
1 6
6 17
617
6 1 7
6 17
f1 ¼ pffiffiffi 6
f2 ¼ pffiffiffi 6 7 , f3 ¼ pffiffiffiffiffi 6
7,
7 , f4 ¼ pffiffiffiffiffi 6
7
3405
10 4 2 5
24 05
15 4 3 5
0
1
l1 ¼3
2
l2 ¼2
2
l3 ¼1
l4 ¼4
The modal matrix F is
2
1
1
0
1
1
1 1 1
1 6
1
6
F ¼ pffiffiffi pffiffiffi pffiffiffiffiffi pffiffiffiffiffi 4
0
2 3 10 15
0
It can be verified that
fi fj ¼
1,
0,
1
1
2
2
3
1
17
7,
35
2
jFj ¼
30
¼1
30
j¼i
j=i
showing that the modal matrix is orthogonal and jFj ¼ 1.
The original matrix A can be diagonalized by the modal matrix as shown below:
2
1
6
1 6 1
FT AF ¼ 6
30 4 1
1
1
0
0
1
2
1
1
3
2
1
6 1
6
6
4 0
0
1 1
1 1
0
1
2
2
32
1
0
6
7
1 76 2
76
2 54 1
2
1
1
1
1
2
3
1
1 7
7
7
2 5
1 1 2
0
2
3
2
3
1
3
0
0 0
6 0 2
17
0 07
7 1
6
7
¼6
7
7
3 5 30 4 0
0 1 0 5
2
0
0
0 4
Similarity Transformations
In the previous section we obtained the diagonal matrix L by transforming the matrix
A by the modal matrix F. The resulting diagonal matrix L has all the fundamental properties of the original matrix A. For example these two matrices have the same determinants
and the same traces. Such matrices are called similar matrices, and the transformation is
known as a similarity transformation. Any nonsingular matrix A can be transformed into
another similar matrix B by the following similarity transformation
T1 AT ¼ B
where T is another nonsingular transforming matrix.
Example 16.2.7 With
2
0 1
6 1 0
T¼6
4 1 1
1 1
1
1
1
1
3
0
17
7
05
1
2
and
1 2
6 2 1
A¼6
4 1
1
2
1
3
1 2
2
17
7
1
05
1
2
16.3
Vector and Matrix Differentiation
301
we have
2
7
6 2
1
T AT ¼ B ¼ 6
4 7
2
0 7
2
2
4
6
3
1
3
5
37
7
75
2
In this example the determinants of A and B are 1 and the traces of A and B are 21.
B 16.3
VECTOR AND MATRIX DIFFERENTIATION
The definitions for differentiation with respect to a real vector and a complex vector are
slightly different. In the case of differentiation of a complex vector, the question of
whether the derivative is a total or partial also makes a difference. We shall first discuss
differentiation with respect to real vectors.
Differentiation with Respect to Real Vectors
Let x be an n-vector variable defined by x ¼ fx1, x2, . . . , xngT. The total and partial
derivative operators with respect to x are defined by
2
2
3
3
d
@
6 dx1 7
6 @x1 7
6
6
7
7
6 d 7
6 @ 7
d
@
6
7,
7
¼6
¼
(16:3:1)
6
7
7
dx 6
6 dx2 7 @x 6 @x2 7
4 d 5
4 @ 5
@xn
dxn
Just as the (d=dx)x ¼ 1, we want the derivative with respect to the vector x to be equal to
the identity matrix I, taking care that the dimensional compatibility conditions are satisfied. Thus, (d=dx)x cannot be defined since the compatibility conditions are not met.
We can, however, define (d=dx)xT ¼ I as shown below:
2
3
d
6 dx1 7
2
3
6
7
1 0 0
7
d T 6
d
7
4
5
x ¼6
(16:3:2)
6 dx2 7 x1 x2 xn ¼ 0 1 0 ¼ I
dx
6
7
0
0
0
1
4 d 5
dxn
In an analogous fashion we can define the following operation:
2
3
d
6 dx1 7
2
3
6
7
2x1
6
7
d T
d 7 2
2
2
4
5
x x¼6
6 dx2 7 x1 þ x2 þ þ xn ¼ 2x2 ¼ 2x
dx
6
7
2x
n
4 d 5
dxn
(16:3:3)
302
Elements of Matrix Algebra
We can also write the following differentiation formulas as follows
d T
x Ax ¼ (A þ AT )x
dx
d T
d
y Ax ¼ xT AT y ¼ AT y
dx
dx
(16:3:4)
Note that in Eq. (16.3.4) x TAx is a scalar, known as the quadratic form, and y TAx is
also a scalar, known as the bilinear form. A scalar is its own transpose, that is,
x TAx ¼ x TA Tx and y TAx ¼ x TA Ty.
Corresponding to Eq. (16.3.2), we can also define the following operations:
2 3
x1
6 x 7
T
27
d
d
d
d 6
6 . 7¼n
x¼
7
.
dx
dx1 dx2
dxn 6
4 . 5
xn
d
dx
T
xxT ¼
d
dx1
2
d
dx2
x21
x1 x2
6x x
2
6 2 1 x2
6
6 .
..
4 ..
.
xn x1 xn x2
T
d
d
d
xyT ¼
dx
dx1 dx2
2
x1 y1 x1 y2
6x y x y
6 2 1 2 2
6
..
6 ..
4 .
.
xn y1 xn y2
d
dxn
3
x1 xn
x2 xn 7
7
T
.. 7
7 ¼ (n þ 1)x
. 5
d
dxn
(16:3:5)
x2n
x1 yn
3
x2 yn 7
7
T
.. 7
7 ¼ ny
. 5
xn yn
Differentiate f(x) with respect to the vector x:
"
#" #
3 2
x1
T
f (x) ¼ x Ax ¼ x1 x2
¼ 3x21 þ 6x1 x2 þ 5x22
x2
4 5
Example 16.3.1
Differentiating f(x) with respect to x1 and x2, we obtain
9
d
"
>
(3x21 þ 6x1 x2 þ 5x22 ) ¼ 6x1 þ 6x2 >
=
6
dx1
¼
d
>
6
>
(3x21 þ 6x1 x2 þ 5x22 ) ¼ 6x1 þ 10x2 ;
dx2
Using the formula ½d(xT Ax)=dx ¼ ½A þ AT x, we
("
# "
3 2
3
d
T
f (x) ¼ ½A þ A x ¼
þ
dx
4 5
2
obtain
# )" #
4
x1
5
x2
6
#"
10
"
x1
#
x2
6
6
6
10
¼
#"
x1
x2
#
16.3
Vector and Matrix Differentiation
303
We differentiate f (x, y) with respect to the vector x in the following
Example 16.3.2
example:
T
f (x, y) ¼ y Ax ¼ y1
y2
3
4
2
5
x1
x2
¼ 3x1 y1 þ 4x1 y2 þ 2x2 y1 þ 5x2 y2
Using the formula ½d(yT Ax)=dx ¼ AT y, we can write
3 4 y1
3y1 þ 4y2
¼
AT y ¼
2 5 y2
2y1 þ 5y2
Complex Differentiation
Total Derivatives Differentiation with respect to a complex variable z ¼ x þ jy is a
bit more involved. If we have a function of a complex variable
w ¼ f (z) ¼ u(x, y) þ jv(x, y)
(16:3:6)
then the total derivative of f(z) with respect to z is defined by
d
Dw
f (z þ Dz) f (z)
w ¼ lim
¼ lim
Dz!0 Dz
Dz!0
dz
Dz
(16:3:7)
If the derivative df (z)=dz is to exist, then the limit must exist independent of the way in
which Dz approaches zero. This condition imposes restrictions on the existence of the
total derivative. With Dw ¼ Du þ j Dv and Dz ¼ Dx þ j Dy, we have the following from
Eq. (16.3.7):
lim Dz ¼ lim
Dz!0
Dx!0
Dy!0
Du þ jDv
Dx þ jDy
(16:3:8)
If we first let Dy ! 0 and then let Dx ! 0 in Eq. (16.3.8), we have
df (z) @u
@v
¼ þj
dz
@x
@x
(16:3:9)
Similarly, if we first let Dx ! 0 and then let Dy ! 0, we obtain
df (z) @v
@u
¼ j
dz
@y
@y
(16:3:10)
If the derivative df (z)=dz is to be unique then Eqs. (16.3.9) and (16.3.10) must match.
Hence
@u @v
¼
@x @y
@v
@u
¼
@x
@y
(16:3:11)
These conditions are known as the Cauchy –Riemann (C– R) conditions. If C – R
conditions are satisfied, then the derivative of f (z) is given by either Eq. (16.3.9) or
Eq. (16.3.10).
304
Elements of Matrix Algebra
A complex function f (z) is said to be analytic in the neighborhood of z ¼ z0, if C –R
conditions are satisfied in that neighborhood. In addition, if it is also analytic at z ¼ z0,
then the point z0 is a regular point. However, if it is not analytic at z ¼ z0, then z0 is
called a singular point.
Example 16.3.3 We will find the derivative of f (z) ¼ z 2 ¼ (x þ jy)2 ¼ x 2 2 y 2 þ 2jxy.
We will determine whether f (z) satisfies the Cauchy – Riemann conditions of
Eq. (16.3.9).
z2 ¼ (x þ jy)2 ¼ x2 y2 þ 2jxy
u(x, y) ¼ x2 y2
:
v(x, y) ¼ 2xy
@u
@v
¼ 2x ¼ ¼ 2x
@x
@y
and hence
@v
@u
¼ 2y ¼ ¼ 2y
@x
@y
The C – R conditions are satisfied and df (z)=dz is given by
d 2 @u
@v
z ¼ þ j ¼ 2(x þ jy) ¼ 2z
dz
@x
@x
Example 16.3.4 On the other hand, we will find the derivative of f (z) ¼ z 2 ¼
(x 2 jy)2 ¼ x 2 2 y 2 2 2jxy. We will determine whether f (z) satisfies the Cauchy –
Riemann conditions of Eq. (16.3.9):
z2 ¼ (x jy)2 ¼ x2 y2 2jxy
u(x, y) ¼ x2 y2
: v(x, y) ¼ 2xy
@u
@v
¼ 2x = ¼ 2x
@x
@y
@v
@u
¼ 2y = ¼ 2y
@x
@y
Since the C –R conditions are not satisfied at points other than at z ¼ 0, z 2 does not have a
total derivative dz2 =dz at points other than at z ¼ 0.
Partial Derivatives Many functions occurring in engineering practice are not analytic, and we are interested in optimization problems involving partial derivatives with
respect to a complex variable z. With z ¼ x þ jy, we shall define the partial derivative
operator @[email protected] as
@
@
@
¼ þj
@z @x
@y
(16:3:12)
16.3
Vector and Matrix Differentiation
305
From the definition given in Eq. (16.3.12), we have the following:
@
@
@
z¼
þj
(x þ jy) ¼ 1 1 ¼ 0
@z
@x
@y
@ @
@
z ¼
þj
(x jy) ¼ 1 þ 1 ¼ 2
@z
@x
@y
(16:3:13)
Note that from Example 16.3.3 the total derivative (d/dz)z does not exist since z is not
analytic except at z ¼ 0.
From Eq. (16.3.12) we can also write
@ @
@
(z z) ¼
þj
(x2 þ y2 )
@z
@x
@y
¼ 2(x þ jy) ¼ 2z
@ 2
@
@
z ¼
þj
(x2 þ 2jxy y2 )
@z
@x
@y
¼ 2(x þ jy) 2(x þ jy) ¼ 0
@ @
@
(z w) ¼
þj
(xu þ yv þ j(xv yu)Þ
@z
@x
@y
(16:3:14)
¼ 2(u þ jv) ¼ 2w
@
@
@
(zw) ¼
þj
(xu yv þ j(xv þ yu)Þ
@z
@x
@y
¼ (u þ jv) (u þ jv) ¼ 0
Note that if we used the chain rule of differentiation in Eq. (16.3.14) and apply
Eq. (16.3.13), we obtain the same results. This fact gives credibility to the definition of
the partial derivative @[email protected] Note also that if z and w were real, then we would get
(@[email protected])z2 ¼ 2z and (@[email protected])zw ¼ w. These differences also manifest themselves in defining
vector differentiation with respect to a complex variable.
Vector Partial Differentiation Analogous to Eq. (16.3.12), partial derivatives with
respect to a vector complex variable can be defined. With z ¼ [x1 þ jy1, x2 þ jy2, . . . ,
xn þ jyn]T, partial derivatives @[email protected] and @[email protected] are defined as follows:
3
@
@
þj
6 @x1
@y1 7
7
6
6 @
@ 7
7
6
þj
@
6
@y2 7
¼ 6 @x2
7;
7
@z 6
..
7
6
.
7
6
4 @
@ 5
þj
@xn
@yn
2
2
x1 þ jy1
3
6 x þ jy 7
27
6 2
7;
z¼6
..
6
7
4
5
.
xn þ jyn
3
@
@
j
6 @x1
@y1 7
7
6
6 @
@ 7
7
6
@
6 @x2 j @y2 7
¼
7
6
7
@z 6
..
7
6
.
7
6
4 @
@ 5
j
@xn
@yn
2
(16:3:15)
306
Elements of Matrix Algebra
With this definition of @[email protected] and @[email protected] , we can define the following operations:
3
@
@
þj
6 @x1
@y1 7
7
6
6 @
@ 7
7
6
þ
j
@ H 6 @x
@y2 7
z ¼6 2
7 x1 jy1
7
6
@z
..
7
6
.
7
6
4 @
@ 5
þj
@xn
@yn
2
3
2 0 0 0
60 2 0 07
6
7
¼6
7 ¼ 2In
40 0 2 05
2
0
0 0
x2 jy2
xn jyn
(16:3:16)
2
where z H is the conjugate transpose of z and In is an nth-order identity matrix. In a similar
manner, we obtain
3
@
@
j
6 @x1
@y1 7
7
6
6 @
@ 7
7
6
j
@ T 6 @x
@y2 7
2
z
¼
7 x1 þ jy1
6
7
6
@z
..
7
6
.
7
6
4 @
@ 5
j
@xn
@yn
2
3
2 0 0 0
60 2 0 07
6
7
¼6
7 ¼ 2In
40 0 2 05
2
0
0
0
x2 þ jy2
xn þ jyn
(16:3:17)
2
In addition, we also have
3
@
@
þj
6 @x1
@y1 7
7
6
6 @
@ 7
7
6
þ
j
@ T 6 @x
@y2 7
z ¼6 2
7 x1 þ jy1
7
6
@z
..
7
6
.
7
6
4 @
@ 5
þj
@xn
@yn
2
¼ 0n ¼
x2 þ jy2
xn þ jyn
(16:3:18)
@ H
z
@z
where 0n is an nth-order zero matrix.
Equations (16.3.16) – (16.3.18) are very useful in complex vector minimization of
scalar quadratic error performance criteria, as we will show in the following examples.
Example 16.3.5 The terms z and w are two complex 2-vectors as shown below. It is
desired to minimize the complex matrix scalar criteria 11 ¼ zH w and 12 ¼ wH z with
16.3
respect to the complex vector z. With
a þ jb
;
z¼
c þ jd
Vector and Matrix Differentiation
w¼
e þ jf
g þ jh
307
and the corresponding error criteria
11 ¼ zH w ¼ ae þ bf þ cg þ dh þ j(af be þ ch dg)
12 ¼ wH z ¼ ae þ bf þ cg þ dh j(af be þ ch dg)
we can differentiate 11 and 12 partially with respect to z, and using Eqs. (16.3.16) and
(16.3.18), we obtain
e þ e þ j( f þ f )
@
@ H
¼ 2w
11 ¼
z w¼
@z
@z
g þ g þ j(h þ h)
e e þ j( f f )
@
@
@ T z w ¼
¼0
12 ¼ w H z ¼
@z
@z
@z
g g þ j(h h)
The same result can be obtained by the chain rule of differentiation and Eqs. (16.3.16) –
(16.3.18).
Example 16.3.6 We take an example of minimizing a scalar quadratic error surface with
respect to a complex weight vector a p called adaptive coefficients and finding the optimum
value of ap. The quadratic error surface is given by
H
H
E p ¼ s2 þ a H
p R p ap þ r p ap þ a p r p
(16:3:19)
We differentiate with respect to a p and set the result to zero and solve for a po using
Eqs. (16.3.16) – (16.3.18):
@
@
H
H
Ep ¼
½s2 þ aH
p Rp ap þ rp ap þ ap rp ¼ 0
@ap
@ap
¼
@ 2
@ H
@ H
@ H
s þ
a R p ap þ
r ap þ
a rp ¼ 0
@a p
@a p p
@ap p
@ap p
(16:3:20)
¼ 0 þ 2Rp ap þ 2rp ¼ 0,
or
apo ¼ R1
p rp
The solution a po ¼ R1
p rp is called the Wiener solution for the adaptive coefficients a p.
Summary of Some Vector Partial Derivatives of Real and Complex Quadratic Forms
@
@
@ T
@ T
J(x) ¼ ½xT Ax ¼
x Ax þ
x
AT x ¼ ½A þ AT x : x real
@x
@x
@x
@x
@
@
@ H
@ T
J(x) ¼ ½xH Ax ¼
x Ax þ
x
AT x ¼ 2Ax þ 0 : x complex
@x
@x
@x
@x
(16:3:21)
@
@
@ T
J(x, y) ¼ ½xT Ay ¼
x Ay ¼ Ay : x real
@x
@x
@x
@
@ H
@ H
J(x, y) ¼ ½x Ay ¼
x Ay ¼ 2Ay : x complex
@x
@x
@x
308
Elements of Matrix Algebra
A final point to note in taking the vector partial derivatives is that if x is real, then
(@[email protected])xT ¼ I, and if x is complex, then (@[email protected])xH ¼ 2I and (@[email protected])xT ¼ 0.
Differentiation with Respect to Real Matrices
In some minimization problems there is a need to differentiate a scalar function of a
real matrix A with repect to the real matrix A. We will use the following result in Section
22.3. If B is any square matrix and A is any other matrix, then
@
Tr½ABAT ¼ A½B þ BT @A
(16:3:22)
@
Tr½ABAT ¼ 2AB
@A
(16:3:23)
If B is symmetric, then
Example 16.3.7
We are given the matrices X and B as
x11 x12
b11 b12
X¼
; B¼
x21 x22
b21 b22
We will show that Eq. (16.3.22) is true. The trace of XBX T is given by
Tr½XBXT ¼ x211 b11 þ x222 b22 þ x212 b11 þ x221 b22
þ x11 x21 b12 þ x11 x21 b21 þ x22 x12 b12 þ x22 x12 b21
Taking partial derivatives with respect to x11, x22, x12, and x21, we obtain
@
Tr½XBXT ¼ 2x11 b11 þ x21 (b21 þ b12 )
@x11
@
Tr½XBXT ¼ 2x22 b22 þ x12 (b21 þ b12 )
@x22
@
Tr½XBXT ¼ 2x12 b11 þ x22 (b21 þ b12 )
@x12
@
Tr½XBXT ¼ 2x21 b22 þ x11 (b21 þ b12 )
@x21
b11
x11 x12
b11 b12
þ
¼
x21 x22
b21 b22
b12
B 16.4
b21
b22
¼ X{B þ BT }
BLOCK MATRICES
In some applications we will encounter block matrices whose elements are also matrices.
An (n m) (n m) matrix Z can be formulated as a block matrix consisting of
16.4
individual blocks A,B,C,D, as
2
x11
6
6
6
6 xn1
6
Z¼6
6
6 xnþ1,1
6
6
4
xnþm,1
2
A
6
6nn
¼6
6 C
4
mn
Block Matrices
309
shown below:
x1n
x1, nþ1
xn,n
xn, nþ1
B
xnþ1,n
xnþ1, nþ1
A
C
x1, nþm
3
7
7
7
xn, nþm 7
7
7
7
xnþ1, nþm 7
7
7
D
5
xnþm, nþm
xnþm,n xnþm, nþ1
3
B
7
nm7
7
D 7
5
mm
(16:4:1)
We can enumerate some properties of block matrices that may be useful in later chapters:
1. Multiplication. The compatibility of multiplying blocks must be satisfied as
shown:
2
3
2
3
F
E
A
B
6
7
6
6nn
nm7
nm7
6
76 n n
7
4 C
D 56
H 7
4 G
5
mm
mn
mm
mn
2
3
AE þ BG
AF þ BH
6 nn
nm 7
6
7
6
7
(16:4:2)
¼6
CF þ DH 7
4 CE þ DG
5
mm
mn
2. Determinant. If A is nonsingular and D is square, then
"
#
A
B
det
¼ det A det(D CA1 B)
C
D
(16:4:3)
3. Inverse.
a. Matrix Inversion Lemma. If A and D are nonsingular n n and m m matrices
respectively, then,
(A þ BD1 C)1 ¼ A1 A1 B(D þ CA1 B)1 CA1
(16:4:4)
b. If inverses for A and D exist then,
A
C
0
D
A
0
B
D
"
1
¼
"
1
¼
A1
0
D1 CA1
D1
A1
0
A1 BD1
D1
#
;
#
(16:4:5)
310
Elements of Matrix Algebra
c. If inverses for A and D exist then,
1 "
D1
A B
A
¼
1
C D
D1
D CA
A1 BD1
D
D1
D
#
(16:4:6)
where D A and D D are the Schur components of the matrices A and D defined by
DA ¼ A BD1 C
DD ¼ D CA1 B
(16:4:7)
d. Using property 3a, the inverses of the Schur component D A and D D are given
by
1
1
þ A1 BD1
D1
A ¼A
D CA
1
1
D1
þ D1 CD1
D ¼D
A BD
(16:4:8)
1
We can substitute for either D1
A or DD from Eq. (16.4.8) and solve Eq.
(16.4.6).
CHAPTER
17
Random Vectors and
Mean-Square Estimation
B 17.1
DISTRIBUTIONS AND DENSITIES
Joint Distributions and Densities
We have come across sequences of random variables in Chapter 14, and these can be
represented by a random n-vectors. We can define two random vectors X and Y as
2 3
2 3
Y1
X1
6X 7
6Y 7
6 27
6 27
7
6 7
(17:1:1)
X¼6
6 .. 7; Y ¼ 6 .. 7
4 . 5
4 . 5
Xn
Ym
The joint distribution of the random vector X is given by
FX (X) ¼ P½X x ¼ P½X1 x1 , . . . , Xn xn (17:1:2)
with FX(1) ¼ 1 and FX(21) ¼ 0. The corresponding joint density function fX(x) can be
defined as
P(x , X x þ DxÞ
DX!0
Dx
fX (x) ¼ lim
¼ lim
Dx1 !0
...
Dxn !0
Pðx1 , X1 x1 þ Dx1 , . . . , xn , Xn xn þ Dxn Þ
Dx1 Dxn
(17:1:3)
which is the nth partial derivative of the distribution function FX(x):
fX (x) ¼
@n FX (x)
@x1 @xn
(17:1:4)
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
311
312
Random Vectors and Mean-Square Estimation
Given the density function fX(x), the distribution function FX(x) can be obtained by an
n-fold integration:
ð xn
ð x1
FX (x) ¼
fX (x0 )dx01 dx0n
(17:1:5)
1
1
T
Given an n-vector X ¼ X1 Xn
T
, the joint distribution of X and Y is given by
Joint Distribution of Two Vectors
X and Y
and an m-vector Y ¼ Y1 Ym
FXY (x,y) ¼ PðX x, Y yÞ
(17:1:6)
and the joint density, by
Pðx , X x þ Dx, y , Y y þ DyÞ
Dx Dy
fXY (x,y) ¼ lim
Dx!0
Dy!0
x1 , X1 x1 þ Dx1 , . . . , xn , Xn xn þ Dxn
!
P
¼ lim
Dx1 !0
..
.
y1 , Y1 y1 þ Dy1 , . . . , ym , Ym ym þ Dym
Dx1 Dxn Dy1 Dym
Dxn !0
..
.
Dy1 !0
..
.
Dym !0
or
fXY (x,y) ¼
@n @m FXY (x,y)
@x1 @xn @y1 @ym
(17:1:7)
The distribution function FXY(x,y) can also be obtained by integrating Eq. (17.1.7):
ðx ðy
fxy (x0 ,y0 )dy0 dx0
(17:1:8)
FXY (x,y) ¼
1
1
Since Eq. (17.1.8) is a distribution function, we infer the following relationships:
FXY (x,1) ¼ FXY (1, y) ¼ 0
FXY (x,þ1) ¼ FX (x), marginal distribution of X
FXY (þ1,y) ¼ FY (y), marginal distribution of Y
(17:1:9)
FXY (þ1,þ1) ¼ 1
The marginal density functions are obtained from Eq. (17.1.7) by integration as
follows:
ð1
fX (x) ¼
ð1
fXY (x,y)dy1 dym
1
ð1
fY (y) ¼
1
ð1
fXY (x,y)dx1 dxn
1
(17:1:10)
1
17.1
Distributions and Densities
313
T
Example 17.1.1 Thejoint density
T function of a random 2-vector X ¼ X1 X2 and a
random 2-vector Y ¼ Y1 Y2 is given by
8
< 0 , x1 1, 0 , x2 2
1
1 , y1 2, 1 , y2 3
fXY (x,y) ¼
(x1 þ 2x2 )(4y1 þ y2 ) ,
:
80
0
otherwise
We have to determine (1) marginal density fX(x), (2) marginal density fY(y), (3) marginal
distribution FX(x), (4) marginal distribution FY(y), (5) the distribution FX(x1), and (6) the
distribution FX(x2).
1. The marginal density fX(x) is obtained by integrating fXY(x,y) over all y:
ð3 ð2
1
1
fX (x) ¼
½(x1 þ 2x2 )(4y1 þ y2 )dy1 dy2 ¼ (x1 þ 2x2 )
80
5
1 1
2. The marginal density fY(y) is obtained by integrating fXY(x,y) over all x:
ð2 ð1
1
1
fY (y) ¼
½(x1 þ 2x2 )(4y1 þ y2 )dx1 dx2 ¼ (4y1 þ y2 )
16
0 0 80
3. The marginal distribution FX(x) is obtained by integrating fX(x) as shown below:
ð x2 ð x 1
1
1 2
(j1 þ 2j2 )dj1 dj2 ¼
x1 x2 þ 2x1 x22
FX (x) ¼
10
0 0 5
4. The marginal distribution FY(y) is obtained by integrating fY(y) as shown below:
ð y2 ð y1
1
(4h1 þ h2 )dh1 dh2
FY (y) ¼
1 1 16
1 4( y2 1Þy21 þ ( y22 1)y1 y22 4y2 þ 5
¼
32
5. The distribution FX(x1) is obtained by substituting the end value x2 ¼ 2 in FX(x):
1 2
1
2 x1 x2 þ 2x1 x2 FX (x1 ) ¼
¼ x21 þ 4x1
10
5
x2 ¼2
6. The distribution FX(x2) is obtained by substituting the end value x1 ¼ 1 in FX(x):
1 2
1 2
x1 x2 þ 2x1 x22 2x2 þ x2
FX (x2 ) ¼
¼
10
10
x1 ¼1
The random vectors X and Y are independent since fXY(x,y) ¼ fX(x) . fY(y), but the
random variables X1 and X2 are not independent since FX(x1, x2) = FX(x1) . FX(x2).
Conditional Distributions and Densities
The conditional distribution function FXjA (xjA) given that an event A has occurred is
defined by
FX j A (x j A) ¼ P(X x j A) ¼
P(X x, A)
, P(A) = 0
P(A)
(17:1:11)
314
Random Vectors and Mean-Square Estimation
Analogous to the one-dimensional case, we can also define a total probability. If there are
disjoint events fAi, i ¼ 1, . . . , ng such that
n
[
Ai ¼ S, and
\
Ai Aj ¼ 1
i=j
i¼0
then the total probability is given by
FX (x) ¼ P(X x) ¼
n
X
P(X x j Ai )P(Ai ) ¼
i¼1
n
X
FX j Ai (x j Ai )P(Ai )
(17:1:12)
i¼1
The density functions corresponding to Eqs. (17.1.11) and (17.1.12) are given by
@n FX j A (x j A)
, P(A) = 0
@x1 @xn
n
X
fX j Ai (x j Ai )P(Ai )
fX (x) ¼
fX j A (x j A) ¼
(17:1:13)
(17:1:14)
i¼1
If there are events A and B defined by A: x , X x þ Dx and B : y , Y y þ Dy,
then the probability of B occurring conditioned on A is given by
P(B j A) ¼ P(y , Y y þ Dy j x , X x þ Dx) ¼ fY j X (y j X)Dy
¼
P(y , Y y þ Dy, x , X x þ Dx)
P(x , X x þ Dx)
¼
fXY (x,y)Dx Dy fXY (x,y)Dy
¼
fX (x)Dx
fX (x)
(17:1:15)
From Eq. (17.1.15) we can conclude that the conditional density fY j X (y j X) is given by
fY j X (y j X) ¼
fXY (x,y)
fX (x)
(17:1:16)
The corresponding expression for the conditional distribution FY j X (Y j X x) is
FY j X (y j X x) ¼
P(Y y, X x) FYX (y,x)
¼
P(X x)
FX (x)
(17:1:17)
Example 17.1.2 In Example 17.1.1 we established that the random variables X and Y
are independent and the random variables X1 and X2 are not independent. We will now
find the conditional density fX2 j X1 (x2 j X1 ¼ x1 ). By Eq. (17.1.16), we have
fX2 j X1 (x2 j X1 ¼ x1 ) ¼
fX2 X1 (x2 , x1 )
fX1 (x1 )
(x1 þ 2x2 )
x1 þ 2x2
5
¼
¼
2(x1 þ 2)
2(x1 þ 2)
5
and the corresponding conditional distribution FX2 j X1 (x2 j X1 ¼ x1 ) is obtained by integrating the conditional density and is given by
ð x2
x1 þ 2j2
x2 (x1 þ x2 )
FX2 j X1 (x2 j X1 ¼ x1 ) ¼
dj2 ¼
2(x1 þ 2)
2(x
þ
2)
1
0
17.1
Distributions and Densities
315
Note that when the end value of x2 ¼ 2 is substituted in the distribution shown above, we
obtain FX2 j X1 (2 j X1 ¼ x1 ) ¼ 1, as we should.
We can also find the conditional distribution FX2 j X1 (x2 j X1 x1 ) as follows:
FX2 j X1 (x2 j X1 x1 ) ¼
P(X2 x2 , X1 x1 ) FX2 X1 (x2 , x1 )
¼
P(X1 x1 )
FX1 (x1 )
1 2
(x1 x2 þ 2x1 x22 ) x (x þ 2x )
2 1
2
10
¼
¼
1 2
2(x1 þ 4)
(x þ 4x1 )
5 1
Here also if the end value of x2 ¼ 2 is substituted in the preceding distribution, we obtain
FX2 j X1 (2 j X1 x1 ) ¼ 1, as we should.
By differentiating the preceding equation with respect to x2, we can find the corresponding conditional density function fX2 j X1 (x2 j X1 x1 ).
fX2 j X1 (x2 j X1 x1 ) ¼
d x2 (x1 þ 2x2 )
x1 þ 4x2
¼
dx2 2(x1 þ 4)
2(x1 þ 4)
The difference between these two conditional distributions is to be particularly noted. In
the first case the conditioning event is fX1 ¼ x1g, and in the second case it is fX1 x1g.
Points of Clarity
In Eq. (8.2.7) we noted some points of clarity regarding conditional densities and distributions. We will elaborate on this point. The joint density function fXY(x,y) is defined in
Eq. (9.2.3) but can also be defined as an infinitesimal probability
fXY (x,y)Dx Dy ¼ P(x , X x þ Dx, y , Y y þ Dy)
(17:1:18)
and the corresponding distribution function FXY(x,y), by
FXY (x,y) ¼ P(X x, Y y)
which can also be obtained by integrating fXY(x,y) over x and y:
ðx ðy
FXY (x,y) ¼
fXY (j,h)dh dj
(9:2:2)
(17:1:19)
1 1
The function fXY(x,y) can also be obtained by differentiating FXY(x,y) as given in
Eq. (9.2.4):
fXY (x,y) ¼
@2 FXY (x,y)
@x @y
(9:2:4)
There is a need for defining another density – distribution function fXY (X x,y) that can be
used in formulating directly conditional density problems involving the conditioning event
such as fX x):
fXY (X x,y) ¼ lim
Dy!0
P(X x,y , Y y þ Dy)
Dy
(17:1:20)
316
Random Vectors and Mean-Square Estimation
Equation (17.1.20) can be obtained by integrating fXY(x,y) over x as shown below:
ðx
fXY (j,y)dj
(17:1:21)
fXY (X x,y) ¼
1
We can integrate Eq. (17.1.21) over y and obtain FXY (x,y):
ðy
FXY (x,y) ¼
fXY (X x,h)dh
(17:1:22)
1
The conditonal density fY j X (y j X x) can be determined as follows:
fY j X (y j X x)Dy ¼ P(y , Y y þ Dy) j X x)
¼
P(y , Y y þ Dy, X x)
P(X x)
¼
fYX (y, X x)Dy
FX (x)
Hence
fY j X (y j X x) ¼
fYX (y, X x)
FX (x)
(17:1:23)
Example 17.1.3 We will continue the previous example (Example 17.1.2) and find the
conditional density fX2 j X1 (x2 j X1 x1 ). From Eq. (17.1.23) we have
fX2 j X1 (x2 j X1 x1 ) ¼
fX2 X1 (x2 , X1 x1 )
FX1 (x1 )
From Eq. (17.1.21) we have
ð x1
j1 þ 2x2
x1 (x1 þ 4x2 )
dj1 ¼
fX2 X1 (x2 ,X1 x1 ) ¼
10
5
0
and
FX1 (x1 ) ¼
x21 þ 4x1
5
Hence
x1 (x1 þ 4x2 )
x1 þ 4x2
fX2 j X1 (x2 j X1 x1 ) ¼ 2 10
¼
2(x1 þ 4)
x1 þ 4x1
5
By integrating fX2 j X1 (x2 j X1 x1 ) with respect to x2, we obtain the conditional
distribution
FX2 j X1 (x2 j X1 x1 ) ¼
x2 (x1 þ 2x2 )
2(x1 þ 4)
Bayes’ Theorem
Similar to the one-dimensional case, we can formulate a vector form of Bayes’
theorem given that the event Ai has occurred. We want to find the a posteriori probability
P(Ai j X x) given that the event fX xg has occurred. This probability is given by
P(Ai j X x) ¼
P(X x, Ai ) P(X x j Ai )P(Ai )
¼
P(X x)
P(X x)
17.1
Distributions and Densities
317
Substituting for P(X x) from Eq. (17.1.12), we obtain the vector form of Bayes’
theorem:
P(X x j Ai )P(Ai )
P(Ai j X x) ¼ Pn
k¼1 P(X x j Ak )P(Ak )
(17:1:24)
We can also formulate Bayes’ rule for random vector densities. If fXY(x,y) represents the
joint density associated with random vectors X and Y, then, from Eq. (17.1.16), we
obtain
fXY (x,y) ¼ fY j X (y j X ¼ x)fX (x)
and Bayes’ theorem can be formulated for fX j Y(x j Y ¼ y) as follows:
fX j Y (x j Y ¼ y) ¼
fY j X (y j X ¼ x)fX (x)
fY (y)
fY j X (y j X ¼ x)fX (x)
0
0
0
1 fY j X (y j X ¼ x )fX (x )dx
¼ Ð1
(17:1:25)
In many engineering estimation problems the random vector X cannot be observed but
the random vector Y connected to X can be observed. In this connection fX(x) is
called the a priori density and fX j Y(x j Y ¼ y) is called the a posteriori density after
observing fY ¼ yg.
The density fX(x) of a two-dimensional random vector X is given by
Example 17.1.4
fX (x) ¼
4x1 þ 7x2
, 0 , x1 1, 0 , x2 2
18
The conditional density fY j X(y j x) of another two-dimensional random vector Y is
fY j X (y j x) ¼
4x1 y1 þ 2x1 y2 þ 5x2 y1 þ 5x2 y2
,
5(4x1 þ 7x2 )
0 , x1 1, 0 , x2 2
1 , y1 2, 1 , y2 3
Given this information, we have to find the a posteriori density fX j Y(x j y) using Bayes’
theorem of Eq. (17.1.24). The joint density function fYX(x,y) is given by
fYX (y,x) ¼ fY j X (y j x)fX (x)
4x1 y1 þ 2x1 y2 þ 5x2 y1 þ 5x2 y2 4x1 þ 7x2
5(4x1 þ 7x2 )
18
0 , x1 1, 0 , x2 2
1
¼ (4x1 y1 þ 2x1 y2 þ 5x2 y1 þ 5x2 y2 ),
90
1 , y1 2, 1 , y2 3
¼
The marginal density function fY(y) is obtained by integrating the preceding equation with
respect to x:
ð1 ð2
fY (y) ¼
0
1
7y1 þ 6y2
4x1 y1 þ 2x1 y2 þ 5x2 y1 þ 5x2 y2 dx2 dx1 ¼
45
0 90
318
Random Vectors and Mean-Square Estimation
The a posteriori density can be obtained from Eq. (17.1.25) as
fX j Y (x j y) ¼
fY j X (y j x)fX (x)
fY (y)
4x1 y1 þ 2x1 y2 þ 5x2 y1 þ 5x2 y2 4x1 þ 7x2
5(4x1 þ 7x2 )
18
¼
7y1 þ 6y2
45
0 , x1 , 1, 0 , x2 2
4x1 y1 þ 2x1 y2 þ 5x2 y1 þ 5x2 y2
,
¼
2(7y1 þ 6y2 )
1 , y1 2, 1 , y2 3
which, when integrated over all x, indeed gives the result 1.
Example 17.1.5
The density fX(x) of a two-dimensional vector X is given by
fX (x) ¼ a1 d(x1 m1 , x2 m2 ) þ a2 d(x1 m1 , x2 m2 )
where a1 þ a2 ¼ 1, d(x1,x2) is a two-dimensional Dirac delta function and m1,m2 are
constants. The conditional density of a two-dimensional random vector Y correlated with
X is
1 (y1 x1 )2 þ (y2 x2 )2
e
fY j X (y j x) ¼
p
From Eq. (17.1.6) the joint density function fYX(x,y) is given by
fYX (y,x) ¼ fY j X (y j x)fX (x)
1 (y1 x1 )2 þ(y2 x2 )2
e
a1 d(x1 m1 , x2 m2 ) þ a2 d(x1 m1 , x2 m2 )
¼
p
1 (y1 m1 )2 þ(y2 m2 )2
¼
e
a1 d(x1 m1 , x2 m2 ) þ a2 d(x1 m1 , x2 m2 )
p
The marginal density function fY(y) is obtained by integrating this equation with respect to x:
ð1 ð1
a1 d(x1 m1 , x2 m2 )
1 (y1 m1 )2 þ(y2 m2 )2
e
fY (y) ¼
dx1 dx2
þa2 d(x1 m1 , x2 m2 )
1 1 p
2
2
2
2
1
a1 e (y1 m1 ) þ(y2 m2 ) þ a2 e (y1 m1 ) þ(y2 m2 )
¼
p
The a posteriori density can be obtained from Eq. (17.1.25) as
fy j x (y j x)fX (x)
fY (y)
(y1 m1 )2 þ(y2 m2 )2
a1 d(x1 m1 , x2 m2 ) þ a2 d(x1 m1 , x2 m2 )
e
¼
2
2
2
2
a1 e (y1 m1 ) þ(y2 m2 ) þ a2 e (y1 m1 ) þ(y2 m2 )
fx j y (x j y) ¼
¼ b1 (y)d(x1 m1 , x2 m2 ) þ b2 (y)d(x1 m1 , x2 m2 )
where
2
2
ai e½(y1 m1 ) þ(y2 m2 ) ,
bi (y) ¼
2
2
2
2
a1 e (y1 m1 ) þ(y2 m2 ) þ a2 e (y1 m1 ) þ(y2 m2 )
i ¼ 1,2
17.2
B 17.2
Moments of Random Vectors
319
MOMENTS OF RANDOM VECTORS
Expectation Vector
Analogous to the single-dimensional case, the expected value of a random n-vector X
is another n-vector mX defined by
ð1
ð1
x f X (x)dx1 dxn
(17:2:1)
E½X ¼ mX ¼
1
1
and
ð1
E½Xi ¼ mXi ¼
ð1
ð1
...
1
xi f X (x)dx1 dxi dxn
1
xi f Xi (xi )dxi
¼
(17:2:2)
1
since
ð1
ð1
fX (x)dx1 dxi1 dxiþ1 dxn ¼ 1
1
1
Hence the mean value n-vector mX can be written as
T
mX ¼ mX1 mXi mXn
(17:2:3)
The term X is a 3-vector whose pdf fX(x) is given by
2 2x 3
2e 1
4
fX (x) ¼ 3e3x25, x1 ,x2 ,x3 . 0
4e4x3
Example 17.2.1
The mean value vector mX is:
2 3
1
2
3 627
E(X1 )
6 7
617
mX ¼ 4E(X2 )5 ¼ 6 7
637
E(X3 )
415
4
Correlation Matrix
The correlation matrix is a square matrix RX representing the set of all second
moments of the components of a real random vector X of the form
ð1
xi xj fXi Xj (xi , xj )dxi dxj , i = j, i, j ¼ 1, . . . , n
(17:2:4)
E½Xi X j ¼
1
and is defined by
2
E½X12 E½X1 X2 6
6 E½X2 X1 E½X22 E½XX ¼ RX ¼ 6
6
..
..
4
.
.
E½Xn X1 E½Xn X2 T
E½X1 Xn 3
7
E½X2 Xn 7
7
7
..
5
.
2
E½Xn (17:2:5)
320
Random Vectors and Mean-Square Estimation
The quantity E[XX T] is defined as the outer product of the vector X, which
is equal to the
P
correlation matrix RX. The inner product RX is defined by E½XT X ¼ ni¼1 E½Xi2 , which is
equal to the trace of RX or E[X TX] ¼ Tr[RX].
The expected value of a scalar function g(X) of a vector random variable X can be
defined as follows:
ð1
ð1
ð1
E½g(X) ¼
g(X)fX (x)dx ¼
g(X)fX (x)dx1 dxn
(17:2:6)
1
1
1
Example 17.2.2 The correlation matrix of Example 17.2.1 can be obtained from
Eq. 17.2.4 assuming that X1, X2,X3 are independent and hence the joint densities
fXi Xj (xi , xj ) are given by the product of individual densities as shown below:
fXi Xj (xi , xj ) ¼ fXi (xi )fXj (xj ), i = j, i, j ¼ 1,2,3
or
fX1 X2 (x1 , x2 ) ¼ 6e(2x1 þ3x2 ) u(x1 )u(x2 )
fX2 X3 (x2 , x3 ) ¼ 12e(3x2 þ4x3 ) u(x2 )u(x3 )
fX1 X3 (x1 , x3 ) ¼ 8e(2x1 þ4x3 ) u(x1 )u(x3 )
Hence RX is given by
2
E½X12 6
RX ¼ 4 E½X2 X1 E½X3 X1 E½X1 X2 E½X22 E½X3 X2 2
1
6
E½X1 X3 62
7 61
E½X2 X3 5 ¼ 6
66
41
E½X32 8
3
3
1
1
6
8 7
7
2
1 7
7
9 12 7
1
1 5
12 8
The trace of the matrix Tr(Rx Þ ¼ 12 þ 29 þ 18 ¼ 61
72. We note that the correlation matrix is
symmetric.
Covariance Matrix
The covariance matrix CX associated with a real random
value of the outer product of the vector (X 2 mX) and is given
2 2
sX1 sX1 X2 sX1 Xj
6
6 sX2 X1 s2X2 sX2 Xj
6
6 .
..
..
6 ..
.
.
T
CX ¼ E½X mX ½X mX ¼ 6
6s
6 Xi X1 sXi X2 sXi Xj
6
..
..
6 ..
4 .
.
.
sXn X1 sXn X2 sXn Xj
vector X is the expected
by
3
sX 1 X n
7
sX 2 X n 7
7
.. 7
. 7
7 (17:2:7)
sXi Xn 7
7
7
.. 7
. 5
s2Xn
where sXi Xj is the covariance between the random variables Xi and Xj given by
sXi Xj ¼ E(Xi mXi )(Xj mXj )
(17:2:8)
Equation (17.2.7) can be expanded to express the covariance matrix CX in terms of the
correlation matrix RX:
CX ¼ E XXT E X mTX mX E XT þ mX mTX
¼ E XXT mX mTX ¼ RX mX mTX
(17:2:9)
17.2
Moments of Random Vectors
321
Example 17.2.3 We will find the covariance matrix for Examples 17.2.1 and 17.2.2. The
correlation matrix RX has already been found. The matrix mX mTX is given by
2 3
1
2
3
E(X1 )
6 2 7
7
6
7
6
617 1 1 1
6
7
T
mX mX ¼ 6 E(X2 ) 7 E(X1 ) E(X2 ) E(X3 ) ¼ 6 7
637 2 3 4
4
5
415
E(X3 )
4
2
3
1 1
1
64 6
8 7
6
7
1 7
61 1
¼6
7
6 6 9 12 7
41 1
1 5
8 12 16
From Eq. (17.2.9) the covariance matrix can be given by
2
3 2
1 1
1
1 1
62 6
6
7
8 7 64 6
6
1 7 61 1
61 2
T
CX ¼ RX mX mX ¼ 6
76
6 6 9 12 7 6 6 9
41 1
1 5 41 1
8 12 8
8 12
3 2
1
1
6
7
8 7 64
1 7 6
7 ¼ 60
12 7 6
1 5 4
0
16
0
1
9
0
0
3
7
7
7
0 7
7
1 5
16
Since the cross-terms are zero, the random variables X1,X2, and X3 are pairwise
uncorrelated.
Joint Moments of Two Random Vectors
Given a random n-vector X and m-vector Y, with joint density fXY(x,y), the expectation of a scalar function g(X,Y) of the random vectors X,Y is defined by
ð1 ð1
g(x,y)fXY (x,y)dx dy
E½g(X,Y) ¼
1
1
1
1
ð1 ð1
g(x,y)fXY (x,y)dx1 dxn dy1 dym
¼
(17:2:10)
The cross-correlation matrix RXY represents the set of all joint moments of the components of real random vectors X and Y of the form
ð1
xi yj fXi Yj (xi ,yj )dxi dyj , i ¼ 1, . . . , n, j ¼ 1, . . . ,m
(17:2:11)
E½Xi Yj ¼
1
and is defined by the outer product E[XY T] as follows:
2
3
E½X1 Y1 E½X1 Y2 E½X1 Ym 6
7
6 E½X2 Y1 E½X2 Y2 E½X2 Ym 7
6
7
E½XYT ¼ RXY ¼ 6
7
..
..
..
6
7
.
.
.
4
5
E½Xn Y1 E½Xn Y2 E½Xn Ym (17:2:12)
It is not a square matrix unless mP¼ n. The inner product E[X TY] is not defined unless
m ¼ n, in which case E½XT Y ¼ ni¼1 E½Xi Yi .
322
Random Vectors and Mean-Square Estimation
Cross-Covariance Matrix
The cross-covariance matrix CXY associated with random n-vector X and a
random m-vector Y is expected value of the outer product (X 2 mX) and (Y 2 mY) and
is given by
2
CXY
sX 1 Y1
6 sX 2 Y1
6
6 .
6 ..
6
T
¼ E½X mX ½Y mY ¼ 6
6 sXi Y1
6
6 ..
4 .
sX n Y1
sX 1 Y2
sX 2 Y2
..
.
sX i Y2
..
.
sX n Y2
sX1 Yj
sX2 Yj
..
.
sX i Yj
..
.
sXn Yj
3
sX 1 Ym
sX 2 Ym 7
7
.. 7
. 7
7
(17:2:13)
sXi Ym 7
7
7
.. 7
. 5
sX n Ym
Again CXY is not a square matrix unless m ¼ n. If m ¼ n then Tr½CXY ¼ Sni¼1 E½Xi Yi .
Similar to Eq. (17.2.9), Eq. (17.2.13) can be expanded to express the cross-covariance
matrix CXY in terms of the cross-correlation matrix RXY:
CXY ¼ E XYT E X mTY mX E YT þ mX mTY ¼ E XYT mX mTY
¼ RXY mX mTY
(17:2:14)
Example 17.2.4 We will now find the cross-covariance matrix of the two-dimensional
vectors X and Y given the joint density
fXY (x,y) ¼
1
(4x1 y1 þ 2x1 y2 þ 5x2 y1 þ 5x2 y2 ),
90
0 , x1 1, 0 , x2 2
1 , y1 2, 1 , y2 3
We will first find the four densities, fX1 (x1 ), fX2 (x2 ), fY1 (y1 ), fY2 (y2 ) as follows:
ð2 ð2 ð3
1
1
4x1 þ 7
(4x1 y1 þ 2x1 y2 þ 5x2 y1 þ 5x2 y2 )dy2 dy1 dx2 ¼
9
1 90
1
1
7x2 þ 2
(4x1 y1 þ 2x1 y2 þ 5x2 y1 þ 5x2 y2 )dy2 dy1 dx1 ¼
90
18
1
0
1
14y1 þ 24
(4x1 y1 þ 2x1 y2 þ 5x2 y1 þ 5x2 y2 )dy2 dx2 dx1 ¼
90
45
1
0
1
4y1 þ 7
(4x1 y1 þ 2x1 y2 þ 5x2 y1 þ 5x2 y2 )dy1 dx2 dx1 ¼
90
30
1
fX1 (x1 ) ¼
0
ð1 ð2 ð3
fX2 (x2 ) ¼
0
ð1 ð2 ð3
fY1 (y1 ) ¼
0
ð1 ð2 ð2
fY2 (y2 ) ¼
0
The mean vectors mX,mY and the outer product mXmTY can be obtained from the densities
as follows:
2
3
29
6 54 7
7
mX ¼ 6
4 5 5,
27
2
3
134
6 45 7
7
mY ¼ 6
4 119 5,
180
2
3
1943 3451
6 1215 9720 7
7
mX mTY ¼ 6
4 134
119 5
243
972
17.3
Vector Gaussian Random Variables
323
To find the cross-correlation matrix RXY, we need the joint densities fX1 Y1 (x1 ,y1 ),
fX1 Y2 (x1 ,y2 ), fX2 Y1 (x2 ,y1 ), fX2 Y2 (x2 ,y2 ). They can be determined from fXY(x,y) and are
shown below:
8x1 y1 þ 8x1 þ 10y1 þ 20
45
4x1 y2 þ 12x1 þ 10y2 þ 15
fX1 Y2 (x1 ,y2 ) ¼
90
5x2 y1 þ 10x2 þ 2y1 þ 2
fX2 Y1 (x2 ,y1 ) ¼
45
10x2 y2 þ 15x2 þ 2y2 þ 6
fX2 Y2 (x2 ,y2 ) ¼
180
fX1 Y1 (x1 ,y1 ) ¼
We can now find the various joint expectations as follows:
332
405
778
E½X2 Y1 ¼
405
E½X1 Y1 ¼
454
405 ,
1066
E½X2 Y2 ¼
405
2
E½X1 Y2 ¼
hence
RXY
3
454
405 7
7
1066 5
405
332
6 405
¼6
4 778
405
From Eq. (17.2.14) we can evaluate CXY ¼ RXY mX mTY :
2
CXY
3 2
332 454
1943
6 405 405 7 6 1215
7 6
¼6
4 778 1066 5 4 134
243
405 405
1:516 0:766
¼
1:370 2:510
3 2
163,963
3451
6 108,135
9720 7
7¼6
119 5 4 1664
972
1215
3
1489
1944 7
7
12,197 5
4860
Definitions
1. Two random vectors X and Y are independent if fXY(x,y) ¼ fX(x) fY(y).
2. Two random vectors X and Y are uncorrelated if RXY ¼ mXmTY, or CXY ¼ 0.
3. Two random vectors X and Y are orthogonal if RXY ¼ 0.
The definitions and properties of correlation and covariance matrices for two real random
vectors X and Y are shown in Table 17.2.1.
B 17.3
VECTOR GAUSSIAN RANDOM VARIABLES
Gaussian random variables play an important role in signal processing and communication
problems. For a real n-vector random variable, the multivariate Gaussian density takes
the form
fX (X) ¼
1
e(1=2) (x mx )T C1
X (x mx )
(2p)n=2 j CX j1=2
(17:3:1)
324
Random Vectors and Mean-Square Estimation
TABLE 17.2.1
Property
Correlation functions
Covariance functions
Autocorrelation and Autocovariance
RX ¼ E[XX T]
RX ¼ RTX
Positive definite
Greater than 0
Orthogonal
Greater than 0
RX ¼ CX þ mXmTX
Definition
Symmetry
Definiteness
Eigenvalues
Eigenvectors
Principal diagonals
Relationship
CX ¼ E[(X 2 mX) (X 2 mX)T]
CX ¼ CTX
Positive definite
Greater than 0
Orthogonal
Greater than 0
CX ¼ RX 2 mXmTX
Cross-Correlation and Cross-Covariance
RXY ¼ E[XY T]
RXY ¼ RTYX
Nondefinite
RXY ¼ CXY þ mXmTY
RXY ¼ mXmTY
RXY ¼ 0 ¼ RYX
RXþY ¼ RX þ RXY þ RYX þ RY
RXþY ¼ RX þ RY if X, Y are
orthogonal
Definition
Symmetry
Definiteness
Relationship
Uncorrelated
Orthogonal
Sum of X and Y
CXY ¼ E[(X 2 mX) (Y 2 mY)T]
CXY ¼ CTYX
Nondefinite
CXY ¼ RXY 2 mXmTY
CXY ¼ 0 ¼ CYX
CXþY ¼ CX þ CXY þ CYX þ CY
CXþY ¼ CX þ CY if X, Y are
uncorrelated
The joint density of two random variables X and Y is given by a bivariate Gaussian as in
Eq. (10.5.1)
(
"
1
1
x mx
pffiffiffiffiffiffiffiffiffiffiffiffiffi exp fXY (x,y) ¼
2)
2
2(1
r
sx
2psx sy 1 r
#)
(x mx )(y my )
y my 2
2r
þ
sy
s x sy
2
(10:5:1)
where r is the correlation coefficient between random variables X and Y given by
r ¼ sxy =(sx sy ). This can also be expressed in the matrix form of Eq. (17.3.1) with n ¼ 2
(
1
1
½x mx
fXY (x,y) ¼ pffiffiffiffiffiffiffiffiffi exp 2
2p jCX j
s2
y my X
rsX sY
rsX sY
s2Y
1 x mx
y my
)!
(17:3:2)
where the covariance matrix CX, its inverse CX 21, and the mean vector m are given by
"
CX ¼
s2X
rsX sY
rsX sY
s2Y
2
#
,
3
1
r
6 s2X (1 r2 ) sX sY (1 r2 ) 7
6
7,
C1
X ¼4
5
r
1
sX sY (1 r2 ) s2Y (1 r2 )
mX
m¼
mY
(17:3:3)
17.3
Vector Gaussian Random Variables
325
2
In a similar fashion, when n ¼ 1, CX ¼ s2X, C21
X ¼ 1/sX, r ¼ 0, and the density for a
single-dimensional random variable X from Eq. (17.3.1) is
2
1
fX (x) ¼ pffiffiffiffiffiffiffiffiffiffiffi e(1=2)½(xmX )=sX 2psX
(17:3:4)
a result that has already been stated in Eq. (6.4.1).
If random n-vector X and m-vector Y are Gaussian-distributed, then the joint density
fXY(x,y) can be obtained by forming an (n þ m) vector
X
Z¼
Y
(17:3:5)
with the corresponding mean vector and the covariance matrix given by
mZ ¼ E
X
¼
mX
mY
X mX CX
(X mX )T (Y mY )T ¼
CZ ¼ E
Y mY
CYX
Y
CXY
(17:3:6)
CY
Since CZ is block symmetric, we note that CTYX ¼ CXY.
The joint density fZ(z) ¼ fXY(x,y) can be obtained from Eq. (17.3.1) as
fXY (x,y) ¼
T 1
1
e(1=2)(zmZ ) CZ (zmZ )
(2p)ðnþmÞ=2 j CZ j1=2
(17:3:7)
where the determinant jCZj is obtained from Eq. (16.4.3)
jCZ j ¼ jCX j j CY CYX C1
X CXY j
(17:3:8)
and the inverse C1
Z , from Eq. (16.4.6)
CX
CXY 1
CYX CY
2 1
1
1
CX þ C1
X CXY DCY CYX CX
4
¼
1
D1
CY CYX CX
C1
Z ¼
1
C1
X CXY DCY
D1
CY
3
5
(17:3:9)
where the inverse of the Schur component D1
CY of CY is obtained from Eq. (16.4.7):
1
1
D1
CY ¼ (CY CYX CX CXY )
It can be seen from Eq. (17.3.7) that when the densities fX(x) and fY(y) are jointly
Gaussian, the density fXY(x,y) is also jointly Gaussian.
326
Random Vectors and Mean-Square Estimation
The conditional density fY j X(y j x) is obtained from fY j X(y j x) ¼ fXY(x,y)/ fX(x).
Performing the indicated division and after some involved matrix manipulation, we
obtain
fY j X (y j x) ¼
m=2
(2p)
T 1
1
e(1=2)(ymY j X ) CY j X (ymY j X )
1=2
j CY j X j
(17:3:10)
1
where mY j X ¼ mY þ CYX C1
X (x mX ) and CY j X ¼ CY CYX CX CXY .
We note that the conditional density is also Gaussian.
Example 17.3.1 Two-dimensional zero mean Gaussian random vectors X and Y have
covariance matrices equal to
1
CX ¼
0
0
9
4
CY ¼
0
and
0
16
The cross-covariance matrix between X and Y is given by
CXY ¼
CX1 Y1
CX1 Y2
CX2 Y1
CX2 Y2
2
6
¼4
1
18
5
3
6
5 7 ¼ CT
YX
12 5
5
We have to find the joint density fXY(x,y) and the conditional density fY j X(y j x) using
Eqs. (17.3.7) and (17.3.10).
Joint Density
From the data given, we can form the matrix
CX
CZ ¼
CYX
CXY
CY
1 0
,
CX ¼
0 9
with
4
CY ¼
0
0
16
and the cross-covariance matrices CXY and CYX, given by
2
1
6
CXY ¼ 6
4 18
5
3
6
5 7
7
12 5
5
2
and
1
6
CYX ¼ CTXY ¼ 6
46
5
3
18
5 7
7
12 5
5
The
p
ffiffiffiffiffiffiffiffiffimean vectors mX and mY are zero vectors. From Eq. (17.3.8), jCZj and
jCZ j are
jCZ j ¼
95904
¼ 153:446
625
and
pffiffiffiffiffiffiffiffiffi 36 pffiffiffiffiffi
jCZ j ¼
74 ¼ 12:387
25
17.3
Vector Gaussian Random Variables
327
The inverse matrix C1
Z is obtained from Eq. (17.3.9):
"
C1
Z
¼
1
1
1
C1
X þ CX CXY DCY CYX CX
1
C1
X CXY DCY
1
D1
CY CYX CX
D1
CY
#
3
250
50
215 35
7
6
111
222
148 7
6 111
7
6
6 50
275
40 25 7
7
6
7
6 111
999
111
333
7
6
¼6
7
725
75 7
6 215 40
7
6
6 222
111
888
592 7
7
6
4 35 25
75
325 5
148
333
592 3552
3
2
2:252
0:45
0:968 0:236
7
6
7
6
7
6
6 0:45
0:275 0:36
0:075 7
7
6
7
6
¼6
7
7
6
7
6 0:968 0:36
0:816
0:127
7
6
7
6
5
4
2
0:236
0:075
0:127
0:091
With this information, the joint density fXY(x,y) is given by
fXY (x,y) ¼
1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
(2p) 153:446
8
2
0
2:252
>
>
>
B 1<
6
B
6 0:45
½x j y6
expB
@ 2>
4 0:968
>
>
:
0:236
2
0:45
0:275
0:36
0:075
91
>
>
=C
7 >
0:075 7 x C
C
7
A
0:127 5 y >
>
>
;
0:091
0:968 0:236
0:36
0:816
0:127
3
or
fXY (x,y) ¼
1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
(2p) 153:446
2
3
0:408y21 þ (0:127y2 þ 0:968x1 þ 0:36x2 )y1
6
7
6
7
exp6 0:045y22 þ (0:236x1 þ 0:075x2 )y2 1:126x21 7
4
5
0:45x1 x2 0:1375x22
2
328
Random Vectors and Mean-Square Estimation
Conditional Density. We will now calculate the conditional density fY j X(y j x) from
Eq. (17.3.10). The conditional expectation vector mY j X is given by
0
mY j X ¼ mY þ CTXY C1
(x
m
)
¼
X
X
0
2
3
2
3
18 2
2x2
3
1 0 x1 þ
1
6
6
x1
5 7
5 7
74
7
¼6
þ6
15
4 6 12 5
4
x2
6x1 4x2 5
0
þ
9
5 5
5
15
The conditional covariance matrix CY j X is obtained as follows:
2
3
2
3
18 2
1
0
61 5 7
6 1
4 0
56
74
CY j X ¼ CY CTXY C1
6
1
X CXY ¼
4 6 12 5
4 18
0 16
0
9
5 5
5
3
2
39 54
6 25
25 7
7
¼6
4 54 348 5
25
3
6
5 7
7
12 5
5
25
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
10656
The determinant jCY j X j ¼
¼ 17:05 and jCY j X j ¼ 4:129 and
625
2
2
3
3
725
75
39 54 1
6 25
6 888 592 7
0:816 0:127
25 7
6
6
7
7
¼
¼
¼
C1
YjX
4 54 348 5
4 75
0:127 0:091
325 5
592 3552
25
25
Substituting these values in Eq. (17.3.10), we obtain
8
0
>
<
1
2x2
B 1
fY j X (y j x) ¼ pffiffiffiffiffiffiffiffiffiffiffi [email protected]
y1 x1 2>
5
2p 17:05
:
0:816
0:127
y1 6x1 4x2
5
15
2
391
y1 x1 2x2 >
=
0:127 6
C
5 7
A
4
5
6x
4x
>
1
2
0:091
;
y1 5 15
Or expanding this equation results in the conditional density fY j X(y j x) as follows:
fY j X (y j x) ¼
1
pffiffiffiffiffiffiffiffiffiffiffi
2p 17:05
"
#
0:408y21 þ (0:127y2 þ 0:968x1 þ 0:36x2 )y1 0:045y22
exp
þ(0:236x1 þ 0:075x2 )y2 0:626x21 0:45x1 x2 0:082x22
Linear Transformations
In Section 13.5 we discussed the general case of the transformation Z ¼ g(X), where
Z and X are random n-vectors and g is a function n-vector. We will now discuss a linear
17.3
Vector Gaussian Random Variables
329
transformation of the form
2
3 2
y1
a11
6 y2 7 6 a21
6 7 6
Y¼6 . 7¼6 .
4 .. 5 4 ..
an1
yn
a12
a22
..
.
an2
32 3
x1
a1n
6 x2 7
a2n 7
76 7
.. 76 .. 7 ¼ AX
. 54 . 5
xn
ann
(17:3:11)
where X is a Gaussian random n-vector and A is a square nonsingular n-matrix faijg. The
density function for X is given in Eq. (17.3.1):
fX (x) ¼
T 1
1
e(1=2)(xmX ) CX (xmx )
(2p)n=2 jCX j1=2
(17:3:1)
Using the transformation techniques developed in Chapter 13, the solution for X is
given by X ¼ A21Y. The Jacobian determinant kJk ¼ jAj. The density function fY(y) is
given by
fY (y) ¼
(2p)
n=2
(2p)
n=2
(2p)
n=2
¼
¼
1
1
T 1
1
eð1=2Þ(A ymx ) CX (A ymx )
1=2
jAjjCX j
1
ð1=2Þ(yAmX )T (AT CX1 A1 )(yAmX )
e
T 1=2
jAjjCX jjA j
1
ð1=2Þ(yAmX )T (ACX AT )1 (yAmX )
e
T 1=2
jAjjCX jjA j
(17:3:12)
Taking the expectation and outer product of Eq. (17.3.11), we obtain
mY ¼ AmX
E½YYT ¼ CY ¼ E½AXXT AT ¼ AE½XXT AT ¼ ACX AT
(17:3:13)
Substituting Eqs. (17.3.13) into Eq. (17.3.12), we can write fY(y) as
fY (y) ¼
(2p)
n=2
1
ð1=2Þ(ymY )T C1
Y (ymY )
1=2 e
jCY j
ð17:3:14Þ
Example 17.3.2 Starting with the same example (Example 13.5.1), we will solve for
fY(y) using matrix methods. A zero mean Gaussian random 3-vector X has the density
function fX(x)
fX (x) ¼
1
T 1
e(1=2)x CX x
3=2
(2p)
where
2
1
CX ¼ 4 0
0
0
1
0
3
0
0 5,
1
2
jCX j ¼ 1,
C1
X
1 0
¼ 40 1
0 0
3
0
05
1
330
Random Vectors and Mean-Square Estimation
The transformation Y ¼ AX is given by
2 3 2
y1
2
Y ¼ 4 y2 5 ¼ 4 5
3
y3
2
3
2
32 3
1
x1
3 54 x2 5 ¼ AX
2
x3
Taking expectation on both sides of the preceding
mX ¼ 0. The covariance matrix CY is given by
2
32
32
2 2 1
1 0 0
2
CY ¼ ACX AT ¼ 4 5 3 3 54 0 1 0 54 2
3 2 2
0 0 1
1
equation we have mY ¼ 0, since
3 2
5 3
9 19
3 2 5 ¼ 4 19 43
3 2
12 27
3
12
27 5
17
The inverse of CY is
2
C1
Y
2
1
¼4 1
9
13 15
3
3
15 5
26
and the determinant of CY ¼ 1. Hence fY(y) is obtained from Eq. (17.3.14)
8
2
2
< 1
1
T4
y
1
fY (y) ¼
exp
: 2
(2p)3=2 1
13
1
9
15
3 9
3
=
15 5y
;
26
or
fY (y) ¼
1
e(1=2)
(2p)3=2
2y21 þ9y22 þ26y23 þ2y1 y2 30y2 y3 6y3 y1
which is exactly the same as obtained in Example 13.5.1.
B 17.4 DIAGONALIZATION OF
COVARIANCE MATRICES
Diagonalization Principles
In Section 16.2 we discussed diagonalization of a nonsingular square matrix by
another transformation matrix of eigenvectors called the modal matrix. Since covariance
matrices are symmetric and positive definite, the modal matrix is orthogonal. This property makes diagonalization of covariance matrices somewhat simpler. Diagonalization
is needed to make a problem simpler, and correlated Gaussian random variables
become independent.
Given any positive definite covariance n-matrix C with distinct eigenvalues, we find
the eigenvalue n-matrix L by solving the characteristic equation jlI 2 Cj ¼ 0 as outlined
in Section 16.2.
2
3
l1 0 0
6 0 l2 0 7
6
7
L¼6 .
(17:4:1)
.. 7
..
4 ..
. 5
.
0 0 ln
17.4
Diagonalization of Covariance Matrices
331
We find the eigenvectors ffi, i ¼ 1, . . . , ng corresponding to each of the eigenvalues fli,
i ¼ 1, . . . , ng and form the modal matrix M of eigenvectors:
2
3
f11 f21 fn1
7
6
6 f12 f22 fn2 7
(17:4:2)
M ¼ f1 f2 fn ¼ 6 .
7
.
.
.. 5
..
4 ..
f1n f2n fnn
Since C is symmetric, M will be orthogonal:
1, i ¼ j
fTi fj ¼
and MT ¼ M1
0, i = j
(17:4:3)
With these definitions, the following eigenequation can be formed:
CM ¼ ML
and MT CM ¼ L
or M1 CM ¼ L
(17:4:4)
The problem now is to find a transformation T between two random n-vectors X and
Y such that the covariance matrix CX is diagonalized to the covariance matrix CY. We take
the expected value and the covariance of Y in the transformation
(17:4:5)
(17:4:6)
Y ¼ TX
mY ¼ TmX
Thus
E½(Y mY )(Y mY )T ¼ E½T(X mX )(X mX )T TT ,
or CY ¼ TCX TT (17:4:7)
Or, identifying terms in Eqs. (17.4.4) and (17.4.7), we obtain
CY L, CX C,
T MT
(17:4:8)
T
with the required transformation matrix T M .
Diagonalizing to an Identity Matrix
We can carry the diagonalization process one step further and find a transformation
matrix D such that CY is an identityp
matrix.
ffiffiffiffiT Since
pffiffiffiffi L is a diagonal matrix, we can take the
positive square root and write L ¼
L In L ,where
identity
n-matrix. Incorpffiffiffiffi TIn isan
pffiffiffiffi
porating this result in Eq. (17.4.4) and noting that
L ¼
L , we obtain
pffiffiffiffi1
pffiffiffiffi pffiffiffiffi
pffiffiffiffi1
MT CM ¼
or
L MT CM L ¼ In
L In L
(17:4:9)
Thus the required transformation matrix D in the transformation D TCD ¼ In is
pffiffiffiffi1
D¼M L
(17:4:10)
The transformation T is orthogonal, whereas D is not.
Example 17.4.1
given by
The covariance matrix CX of a zero mean Gaussian random 3-vector is
2
3
6
CX ¼ 4 1
0
1 0
3
7
2 15
1 3
332
Random Vectors and Mean-Square Estimation
We have to determine
1. The joint density fX(x)
2. The correlation coefficient matrix r, given by
3
2
1
rX1 X2 rX1 X3
7
6
1
rX2 X3 7
r¼6
5
4 rX 2 X 1
rX3 X1 rX3 X2
1
3. The modal matrix M and show that it is orthogonal
4. The transformation matrix T in Y ¼ TX that will make CY diagonal
5. The transformation matrix D that will make CX an identity matrix
6. The joint density fY(y)
1. With jCXj ¼ 12 and
2
C1
X
5 1
6 12
4
6
6
1
3
6
¼6
6 4
4
6
4 1 1
12
4
3
1
12 7
7
7
1 7
7
4 7
7
5 5
12
in the equation
fX (x) ¼
T 1
1
pffiffiffiffiffiffiffiffiffi e(1=2)X CX X
(2p)3=2 jCX j
the joint density is given by
fX (x) ¼
1 5x21 x3 x2 x2 x3 3x22 5x23
p
ffiffiffiffiffi
þ
þ
þ
exp
x
1
2 12
6
2
2
4
12
(2p)3=2 12
1
2. From the given covariance matrix CX, we have
s2X1 ¼ 3,
s2X2 ¼ 2, s2X3 ¼ 3 and
sX1 X2 ¼ 1, sX1 X2 ¼ 0,
Hence the correlation matrix r is given by
2
3
1
1 pffiffiffi 0
6
7
6
6
7
6 1
7
1
7
p
ffiffi
ffi
p
ffiffi
ffi
1
r¼6
6 6
67
6
7
4
5
1
p
ffiffi
ffi
0
1
6
3. From the determinant of the matrix
2
l3
jlI CX j ¼ 4 1
0
1
l2
1
3
0 1 5 ¼ 0
l3 sX2 X3 ¼ 1
17.4
Diagonalization of Covariance Matrices
333
we obtain the characteristic equation, l3 2 8l2 þ 19l 2 12 ¼ (l 2 1)(l 2 3)
(l 2 4) ¼ 0, and the eigenvalues are l1 ¼ 1, l2 ¼ 3, l4 ¼ 4. The eigenvectors
corresponding to these eigenvalues are obtained by the methods described in
Example 16.2.5 and are shown below with the modal matrix M:
3
1
pffiffiffi
6 67
7
6
6 2 7
6 pffiffiffi 7
f1 ¼ 6
7,
6 67
7
6
4 1 5
pffiffiffi
6
2
1
1
pffiffiffi pffiffiffi
6 6
2
6
6 2
6
M ¼ 6 pffiffiffi 0
6 6
6
4 1
1
pffiffiffi pffiffiffi
6
2
2
2
3
1
pffiffiffi
6 27
6
7
6
7
f2 ¼ 6 0 7,
6
7
4 1 5
pffiffiffi
2
3
1
pffiffiffi
6 37
7
6
6 1 7
6 pffiffiffi 7
f3 ¼ 6
7,
6 37
7
6
4 1 5
pffiffiffi
3
2
3
1
pffiffiffi
37
7
1 7
pffiffiffi 7
7
37
7
1 5
pffiffiffi
3
By direct multiplication we can see that fT1 f2 ¼ 0, fT1 f3 ¼ 0, fT2 f3 ¼ 0, and
fT1 f1 ¼ fT2 f2 ¼ fT3 f3 ¼ 1, thus showing orthogonality. Further, jMj ¼ 1.
4. From Eq. (14.7.8) the transformation matrix T is
2 1
pffiffiffi
6 6
6
6 1
T ¼ MT ¼ 6
6 pffiffi2ffi
6
4 1
pffiffiffi
3
2 1 3
pffiffiffi pffiffiffi
6
67
7
1 7
0 pffiffiffi 7
27
7
1
1 5
pffiffiffi pffiffiffi
3
3
2
and
1
TCX T ¼ CY ¼ 4 0
0
T
0
3
0
3
0
05
4
5. From Eq. (17.4.10) the transformation matrix D that takes CX to an identity matrix I
is given by
pffiffiffiffi1
D¼M L
2 1
1
1 32
pffiffiffi pffiffiffi pffiffiffi
1 0
6 6
37
2
1
76
6
6 2
6 0 pffiffiffi
1 7
7
6
6
¼ 6 pffiffiffi 0 pffiffiffi 76
3
3 74
6 6
4 1
1
1 5 0 0
pffiffiffi pffiffiffi pffiffiffi
3
6
2
3
2
1
1
1
p
ffiffi
ffi
p
ffiffi
ffi
p
ffiffi
ffi
6 6
6 2 37
7
6 pffiffiffi
7
6
2
1 7
6
¼ 6 pffiffiffi 0
pffiffiffi 7
6
3
2 37
7
6
4 1
1
1 5
pffiffiffi pffiffiffi pffiffiffi
6
6 2 3
0
3
7
0 7
7
7
1 5
pffiffiffi
4
334
Random Vectors and Mean-Square Estimation
6. Since CY is a diagonal matrix, the random variables Y1, Y2, Y3 are independent,
Gaussian, and zero mean. Hence the joint density is
fY (y) ¼
1
(2p)
3=2
pffiffiffiffiffi e(1=2)½(y1 =1)þ(y2 =3)þ(y3 =4)
12
2
2
2
B 17.5 SIMULTANEOUS DIAGONALIZATION OF
COVARIANCE MATRICES
We are given two positive definite covariance matrices A and B with distinct eigenvalues.
In pattern recognition problems it will be very useful to diagonalize both matrices by
means of the same transformation matrix T. The transforming matrix T will take A to
an identity matrix while at the same time diagonalizing B. We shall now analyze the procedure that will accomplish this. The two covariance matrices A and B with distinct eigenvalues are
0
1
0
1
a11 a12 a1n
b11 b12 b1n
B a21 a22 a2n C
B b21 b22 b2n C
B
C
B
C
A¼B .
B¼B .
(17:5:1)
C
.
.. C
.
..
.. A
..
@ ..
@ ..
. A
.
an1 an2 ann
bn1 bn2 bnn
We have seen in Section 16.2 how the modal matrix of eigenvectors F ¼ ff1,f2, . . . ,
fng of A will transform A to a diagonal matrix of its eigenvalues fl1,l2,l3, . . . , lng as
shown:
0
1
l1 0 0
B 0 l2 0 C
B
C
FT AF ¼ L ¼ B .
(17:5:2)
.. C
..
@ ..
. A
.
0 0 ln
From Eq. (17.4.9) we can write
pffiffiffiffi pffiffiffiffi
pffiffiffiffi1
pffiffiffiffi1
FT AF ¼
L In L
or
L FT AF L ¼ In
pffiffiffiffi1
By defining U ¼ F L , we can write Eq. (17.5.3) as follows:
pffiffiffiffi1
pffiffiffiffi1
L FT AF L ¼ UT AU ¼ In
(17:5:3)
(17:5:4)
This process is known as “whitening” the covariance matrix A. Since B is a covariance
matrix, we can show that C ¼ U TBU is also a symmetric matrix as follows:
pffiffiffiffi
pffiffiffiffi1 T pffiffiffiffi1
pffiffiffiffi1 1
T
T T
T
L F BF L
¼
L F B F L
C ¼
¼
pffiffiffiffi1
pffiffiffiffi1
L FT BF L ¼ C
(17:5:5)
The symmetric matrix C ¼ U TBU can therefore be diagonalized by its orthogonal
modal matrix M, that is, M TCM ¼ M TU TBUM ¼ D ¼ diagfd1,d2, . . . , dng. The
This section is adapted, with permission, from Chapter 31 of the book Linear Systems Properties—a Quick
Reference, by Venkatarama Krishnan, published by CRC Press, 1988.
17.5
Simultaneous Diagonalization of Covariance Matrices
335
transformation UM has diagonalized B, and now we will show that UM applied to A
retains the whitening transformation. Using Eq. (17.5.4), we can write
pffiffiffiffi1
pffiffiffiffi1
MT UT AUM ¼ MT L FT AF L M ¼ MT In M ¼ MT M
(17:5:6)
¼ In
since M is an orthogonal matrix M TM ¼ In.
Thus, the required transformation matrix T that takes A to an identity matrix and B to
a diagonal matrix is given by
pffiffiffiffi1
T ¼ F L M ¼ UM
(17:5:7)
With this transformation, we have
TT AT ¼ In , TT BT ¼ D
(17:5:8)
For computational purposes the procedure described above can be simplified as
follows. From Eq. (17.5.8) we can write
AT ¼ TT
and
BT ¼ TT D
(17:5:9)
Hence BT ¼ ATD, and we have the generalized eigenequation
A1 BT ¼ TD
ð17:5:10Þ
where the matrix T serves as the modal matrix for the product matrix A21B with D as its
diagonal eigenvalue matrix. From Eq. (17.5.10) we can now formulate a procedure for
finding the transformation matrix T that diagonalizes simultaneously the covariance
matrices A and B.
Summary of Procedure for Simultaneous Diagonalization
1. Calculate the n distinct eigenvalues fd1, d2, . . . , dng of the matrix R ¼ A21B from
det (lI –R) ¼ 0.
2. Calculate the eigenvectors ti of R from the eigenequation (lI –R)ti ¼ 0.
3. Form the modal matrix N ¼ ft1, t2, . . . , tng.
4. Form the matrix N TAN ¼ W (normalization with respect to A).
5. Since
with positive values, step 4 can be rewritten,
as
pffiffiffiffiffi W is diagonal
pffiffiffiffiffi
pffiffiffiffiffi
( W)1 NT AN( W)1 ¼ I and the desired transformation is T ¼ N( W)1 .
Example 17.5.1
Two covariance matrices A and B are
0
1
0
2 1 1
4
A ¼ @ 1 3 1 A,
B ¼ @2
1 1 4
2
given by
1
2 2
5 2A
2 6
There is no need to find the eigenvalues of A, and for comparison, the eigenvalues of B are
2
3
2:37
0
0
LB ¼ 4 0
3:52
0 5
0
0
9:11
336
Random Vectors and Mean-Square Estimation
1. The matrix R ¼ A – 1B is given by
2
2
R ¼ A1 B ¼ 4 0
0
3
0:176 0:235
1:588 0:118 5
0:059 1:412
2. The characteristic equation of R is (l 2 2)(l2 –3l þ 2.235) ¼ 0, and the eigenvalues are
2
3
0
0
1:621
0 5
0
1:379
2
D ¼ 40
0
3. The modal matrix N of eigenvectors of R is
2
3
0:525 0:188
0:82
0:481 5
0:23
0:856
1
N ¼ 40
0
4. The matrix W ¼ N TAN is given by
2
2
0
W ¼ 4 0 2:053
0
0
3
0
0 5
2:733
and
2
1:414
pffiffiffiffiffi
W¼4 0
0
3
0
0
1:433
0 5
0
1:653
pffiffiffiffiffi 5. The desired transfromation T ¼ N W 21 is given by
2
0:707 0:366
pffiffiffiffiffi1
0:572
T¼N W ¼4 0
0
0:161
3
0:114
0:291 5
0:518
Comments
1. The matrix T is not orthogonal since
2
0:707
0
0
3
6
7
TT ¼ 4 0:366
0:572 0:616 5 = T1
0:114 0:291
0:518
2
3
1:414 0:707 0:707
6
7
¼4 0
1:51 0:848 5 and jTj ¼ 0:243
0
0:161 0:518
17.6
Linear Estimation of Vector Variables
2. We can check the diagonalization process:
2
2
3
1 0 0
2
6
6
7
TT AT ¼ 4 0 1 0 5, TT BT ¼ 4 0
0 0
1
0
1:621
0
0
0
1:379
3
0
2
2:37
6
whereas LB ¼ 4 0
0
0
3:52
0
337
3
7
5
0
7
0 5
9:11
3. The diagonal elements in the transformation of B are the eigenvalues of R and not
the eigenvalues LB of B.
B 17.6
LINEAR ESTIMATION OF VECTOR VARIABLES
We will take up more detailed estimation problems in the next chapter. We are interested
in estimating the vector X given the equation
Y ¼ H
X þ N
m1 mn n1 m1
(17:6:1)
where Y is an m 1 observation vector, H is an m n system vector, X is an n 1 vector
to be estimated, and N is an m 1 noise vector with zero mean and covariance matrix,
CN ¼ E[NN T] ¼ s2I. We need a linear estimator of the form
^ ¼ AY
X
(17:6:2)
where A is an n m matrix to be determined. We have to choose a scalar criterion
function J, which is some measure of the estimation error 1 ¼ Y 2 HX̂, which can be
minimized with respect to the vector X̂. It is reasonable to choose a sum-squared error
function given by
T ^ ¼ Y HX
^
^
J(X)
Y HX
(17:6:3)
This is known as the linear least-squares estimation. Using Eq. (16.3.21), we can minimize Eq. (17.6.3) and write
T @
^ ¼ @
^
^
J(X)
Y HX
Y HX
^
^
@X
@X
T @ T
T T
T ^
T T ^
^
^
¼
Y Y X H Y Y HX þ X H HX
^
@X
h
i
^ ¼0
¼ 0 ¼ 2HT Y þ 2HT HX
ð17:6:4Þ
or
^ ¼ HT H 1 HT Y
X
Thus, the desired n m matrix A in Eq. (17.6.2) is given by
1
A ¼ HT H HT
(17:6:5)
338
Random Vectors and Mean-Square Estimation
The quantity (H TH)21H T is called the pseudoinverse. We have essentially derived the
estimator X̂ deterministically without involving the noise covariance CN ¼ s2I. In this
case the estimation error 1 ¼ Y 2 HX̂ is not random. We ask the question as to the
meaning of this deterministic estimator X̂. If m . n, then we have an overdetermined
system where the number of observations are more than the unknowns and the n n
matrix H TH is invertible. In this case among the many possible solutions, X̂ is the one
that gives a minimum mean-square solution. If m ¼ n, then H is a square matrix that
may be invertible, in which case we have (H TH)21H T ¼ H 21H 2TH T ¼ H 21 and
X ¼ H 21Y giving an exact solution. If m , n, then we have an underdetermined
system where the number of observations is less than the number of unknowns and the
n n matrix H TH will be singular. The m-vector estimator X̂ can be given only in
terms of the n 2 m unknown quantities of the vector X.
The presence of the noise vector N makes the estimator X̂ random that estimates the
deterministic parameter vector X, and the criterion function J is
J¼E
T ^
^
Y HX
Y HX
(17:6:6)
and the estimator X̂ ¼ (H TH)21H TY is in the minimum mean-square error sense.
By substituting X̂ ¼ (H TH)21H TY in Eq. (17.6.3), we can find the least-square
value as
T Jmin ¼ Y HðHT HÞ1 HT Y Y HðHT HÞ1 HT Y
¼ YT I HðHT HÞ1 HT Y
(17:6:7)
since
I HðHT HÞ1 HT I HðHT HÞ1 HT ¼ I HðHT HÞ1 HT
or the matrix ðI HðHT HÞ1 HT Þ is idempotent.
Example 17.6.1
An observation system, Y ¼ HX þ N, is given by
3 2
1
3
4 5 5 ¼ 42
1
10
2
Y
H
3
2 3
1 n1
x
4 5 1 þ 4 n2 5
x2
2
n3
X
N
We have to estimate in the least-squares sense the 2-vector X. We also will find the noise
sequence and the least squared error. From Eq. (17.6.4), we have
2
3
1 1
1 2 1 6
6 11
7
HT H ¼
42 45 ¼
1 4 2
11 6
1 2
2
3
2
3
21 11
2 1
2
6
T 1 T 6 5
1 2 1
5 7
5
5 7
7
7
H H H ¼6
¼6
4
4 11
5
1 4 2
2
1 5
6
1
5
5
5
5
17.6
Linear Estimation of Vector Variables
Hence the estimator X̂ is given by
^ ¼ HT H
X
1
6
HT Y ¼
11
11
6
1
1
2
4
2 3
3
1 4 5
2
5 ¼
2
1
10
The error vector N is given by
2
3 2 3 2
3
3
3
0
^ ¼ 4 5 5 4 8 5 ¼ 4 3 5
N ¼ Y HX
10
4
6
and the least-squared error Jmin is obtained from Eq. (17.6.7) as
Jmin ¼ YT I H(HT H)1 HT Y ¼ 45
339
CHAPTER
18
Estimation Theory
B 18.1
CRITERIA OF ESTIMATORS
Estimation is a process where we draw inferences about a set, random or otherwise, from
knowledge of a suitably selected finite observation set. This inference is based on the minimization of some criterion function. The collection of elements in the entire set is called
the population. The suitably selected observation set is called the sample. Any function of
the elements of the sample is called an estimator or a statistic. Generally, estimators are
random variables, but they can be constants under some circumstances. The criterion used
for estimation in the previous chapter is the least-squares error. There are several other
criteria for judging the proximity to the true value and the reasonableness of estimators.
We shall enumerate some of these criteria assuming that u is a variable, random or otherwise, to be estimated by u^ from a set of finite observations:
^ of an estimator (u)
^ is defined by
1. Bias. The bias B(u)
^ ¼ E½u^ u
B(u)
(18:1:1)
^ ¼ 0 and the pdf of the estimator is cenAn estimator is said to be unbiased if B(u)
tered around u. Generally we would like to select an unbiased estimator. In some
cases a biased estimator may be preferable to an unbiased one.
2. Variance. The variance of an estimator is defined by
^ ¼ s2^ ¼ E{u^ E½u}
^ 2
var(u)
u
(18:1:2)
Generally we would choose an estimator with minimum variance. However, this
may not be compatible with a minimum bias.
3. Mean-Square Error. The mean-square error is defined by
^ ¼ E½u^ u2
MSE (u)
(18:1:3)
^ ¼ E{u^ E½u
^ þ E½u
^ u}2
MSE(u)
n
o
^ 2 þ ðE½(u
^ u)2 þ (u^ E½u)(E½
^
^ u)
¼ E (u^ E½u)
u
n
o
^
¼ s2u^ þ E ðE½u^ uÞ2 þ ðu^ E½uÞðE½
u^ uÞ
if u is a constant
n
o
^
¼ s2u^ þ B2 if u^ is a constant
(18:1:4)
¼ s2u^ þ B2 þ E ðu^ E½uÞB
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
340
18.1
Criteria of Estimators
341
If the estimator is unbiased, then the minimum mean square is the same as
minimum variance. However, in the presence of a bias we will be able to obtain
a mean-square error smaller than the variance.
4. Consistency. Let u^ n be an estimator obtained from n observations. Then u^ n is a
consistent estimator if
lim P½(u^ n u) . 1 ! 0
for every 1 . 0
n!1
(18:1:5)
or u^ n converges in probability to u as n ! 1: We would prefer to have a consistent estimator; otherwise we will have a bias for infinite samples that may not be
desirable.
5. Maximum Likelihood. In Example 4.6.2 we estimated the population N of tigers in
a wildlife sanctuary by maximizing the probability of occurrence of N using
the maximum-likelihood criterion. We will expand on this concept further. The
maximum-likelihood estimate of the parameter u from a probability density
function f (x,u) of a random variable X is obtained by maximizing the likelihood
function formed from n independent observations of the population fXi ¼ xi,
i ¼ 1, . . . , ng. The likelihood function L(X,u) is the product of the density functions
f(xi,u), i ¼ 1, . . . , n:
L(X,u) ¼ f (x1 ,u) f (xn ,u)
(18:1:6)
Under certain regularity conditions the value of u^ that maximizes Eq. (18.1.6) is
called the maximum-likelihood (ML) estimator. The monotonicity of the logarithm
allows us to maximize the log likelihood function lnL(X,u). Thus the ML estimator
is obtained from
n
@
@
@ X
ln½L(X,u) ¼ ln ½ f (x1 ,u) . . . f (xn ,u) ¼
ln f (xi , u)
@u
@u
@u i¼1
¼
n
X
i¼1
1 @f (xi ,u)
f (xi ,u) @u
(18:1:7)
This method works well for large samples, but for small samples it may be
biased.
^ and if it
6. Efficiency. The basic problem in estimation is to find the best estimator u,
exists, it will be unique. However, in some problems finding the best estimator may
not be simple. In such cases we are satisfied if we can find an estimator with the
greatest lower bound. This bound is called the Rao –Cramer bound of an estimator,
given by
^ (
var(u)
E
1
@
ln f (X,u)
@u
2 )
(18:1:8)
An unbiased estimate that satisfies the Rao – Cramer bound is called an efficient
estimator.
342
Estimation Theory
B 18.2
ESTIMATION OF RANDOM VARIABLES
It is not possible for any estimator to satisfy all the criteria mentioned above. We will
discuss unbiased minimum variance estimators under three broad categories:
1. Estimation of a random variable by a constant
2. Estimation of a random variable by a function of another random variable
3. Estimation of a constant by a random variable called parameter estimation
We will discuss the first two in this section and the third one in the next section.
Estimation of a Random Variable by a Constant
This is the simplest of the estimation problems. A random variable X is to be estimated
by a constant a such that the mean-square error J ¼ E[X 2 a]2 is minimized.
J ¼ E½X a2 ¼ E½X 2 2Xa þ a2 dJ
d
¼ ½E½X 2 2mX a þ a2 ¼ 2mX þ 2a ¼ 0
dx dx
or
a ¼ mX :
(18:2:1)
Thus, the constant that minimizes the mean-square error is the expected value E[X ]. The
minimum mean-square error Jmin ¼ E[X – mX]2 ¼ s2X is the variance. Equation (18.2.1)
can be interpreted in a different way. Since
ð1
E½X a2 ¼
(x a)2 fX (x)dx
(18:2:2)
1
the value of a that minimizes the mean-square error is
ð1
xfX (x)dx ¼ mX
(18:2:3)
1
Estimation of a Random Variable Y by a Function of Another Random
Variable g(X)
Here we address the problem of estimating a random variable Y by a function of
another random variable g(X ) using the mean-square error criterion J ¼ E[Y – g(X )]2:
ð1 ð1
J ¼ E½Y g(X)2 ¼
½y g(x)2 fXY (x,y)dy dx
(18:2:4)
1 1
Inserting fXY (x,y) ¼ fY j X ( y j x)fX (x) in Eq. (18.2.4), we obtain
ð 1 ð 1
2
½ y g(x) fY j X ( y j x)dy fX (x)dx
J¼
1
(18:2:5)
1
Minimizing Eq. (18.2.5) amounts to minimizing the inner integral
ð1
½ y g(x)2 fY j X ( y j x)dy
1
(18:2:6)
18.2
Estimation of Random Variables
343
From Eqs. (18.2.2) and (18.2.3), the value of g(x) that minimizes Eq. (18.2.6) is
ð1
g(x) ¼
yfY j X (y j x)dy
1
or
g(X) ¼ Y^ ¼ E½Y j X
(18:2:7)
Thus, the estimator Ŷ is the conditional expectation of Y given X. This estimator is
unbiased because
^ ¼ E½Y j X ¼ E½Y
E½Y
It is also minimum variance, and the minimum mean-square error is the conditional
variance:
s2Y j X ¼ E{Y E½Y j X}2
In general, the conditional expectation E[Y j X ] is a complicated nonlinear function of
X that may not be expressible in an explicit form. We confine ourselves to a linear function
of the form
E½Y j x ¼ g(x) ¼ Y^ ¼ a0 þ a1 x þ þ an xn ¼
n
X
ai xi
(18:2:8)
i¼0
This form is known as the multiple linear regression, and the coefficients {ai} are called
linear regression coefficients. It is linear because no powers of ai are involved.
If E[Y j x] is of the form
E½Y j x ¼ g(x) ¼ Y^ ¼ ax þ b
(18:2:9)
then it is called simple linear regression. The equation for the regression line is
Y^ ¼ aX þ b
(18:2:10)
The estimator Y^ is unbiased because
^ ¼ E½aX þ b ¼ amX þ mY amX ¼ E½Y
E½Y
It is also minimum variance.
Evaluation of Regression Coefficients
We will now evaluate the linear regression coefficients a and b for the case where Y is
regressed on X, using the minimum mean-square error
J ¼ E½Y aX b2
(18:2:11)
as the criterion function. We will first solve for b by differentiating J with respect to b:
@J
¼ 2E½Y aX b ¼ 0
@b
344
Estimation Theory
Hence
b ¼ mY amX
(18:2:12)
We can now solve for a by substituting Eq. (18.2.12) into Eq.(18.2.11) and differentiating
J with respect to a:
J ¼ E½Y mY a(X mX )2
@J
¼ 2E{½Y mY a(X mX )(X mX )} ¼ 0
@a
¼ E½(X mX )(Y mY ) a(X mX )2 ¼ 0
¼ sXY as2X ¼ 0
Hence
a¼
sXY
s2X
and substituting
sXY ¼ rsX sY ,
a¼r
sY
sX
(18:2:13)
The equations obtained by setting @[email protected] ¼ 0 and @[email protected] ¼ 0 are called normal equations.
With these values of a and b the mean-square error can be found as follows:
J ¼ E½Y mY a(X mX )2
¼ s2Y 2asXY þ a2 s2X ¼ s2Y 2
¼ s2Y s2XY
¼ s2Y (1 r2 )
s2X
s2XY
þ s2XY
s2X
(18:2:14)
The simple linear regression plot shown in Fig.18.2.1 indicates the slope a and the offset b.
The line of regression divides the cluster of points into two halves such that the mean
square between the points and the line of regression is minimum.
In an analogous manner, we can also regress X on Y and find the corresponding
regression coefficients a and b in the expression
X^ ¼ aY þ b
FIGURE 18.2.1
(18:2:15)
18.2
Estimation of Random Variables
345
All the parameters such as the means, variances, and correlation coefficients will be the
same. In much the same way as in Eqs. (18.2.12) and (18.2.13), the regression coefficients
a and b are given by
a¼
sXY
s2Y
and
b ¼ mX amY
(18:2:16)
Substituting sXY ¼ rsX sY in Eq. (18.2.16), a can also be given as
a¼r
sX
sY
(18:2:17)
Multiplying the two slopes a and a, we obtain
aa ¼ r
sY sX
r
¼ r2
sX sY
(18:2:18)
Since the true means and variances are not known, Eq. (18.2.18) is a more accurate method
for computing the correlation coefficient r in a real-life situation rather than using either
Eq. (18.2.13) or (18.2.17). The sign of r can be obtained only from the physics of the
problem.
The regression coefficients for a multiple linear regression problem
Y^ ¼ a0 þ a1 X1 þ þ an Xn
can be obtained in principle by forming
"
J¼E Y
n
X
(18:2:19)
#2
ai Xi
i¼0
and solving the (n þ 1) normal equations obtained from @[email protected] ¼ 0, @[email protected] ¼
0, . . . , @[email protected] ¼ 0: A simpler method will be discussed in the next section.
Example 18.2.1 The relationship between random variables X and Y is shown in
Table 18.2.1. The parameters of X and Y are given by
mX ¼ 0:905, s2X ¼ 0:198, mY ¼ 1:939, s2Y ¼ 0:308, sXY ¼ 0:080:
TABLE 18.2.1
No.
X
Y
No.
X
Y
1
2
3
4
5
6
7
8
9
10
0.05
0.29
0.74
0.55
1.07
0.47
1.06
0.7
0.54
0.65
1.82
2.05
1.28
1.19
2.06
1.96
2.46
1.76
1.13
2.35
11
12
13
14
15
16
17
18
19
20
1.54
0.72
0.66
1.23
1.35
0.97
1.3
0.96
1.73
1.52
1.49
2.15
0.88
1.49
2.18
2.74
2.1
2.69
2.18
2.82
346
Estimation Theory
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
The correlation coefficient rXY ¼ sXY =sX sY ¼ 0:08= (0:198 0:308) ¼ 0:324: We will
find the regression coefficients a, b, a, b:
a¼
sXY
0:08
¼ 0:404, b ¼ my amX ¼ 1:939 0:404 0:905 ¼ 1:574
¼
2
0:198
sX
a¼
sXY
0:08
¼ 0:26, b ¼ mX amY ¼ 0:905 0:26 1:939 ¼ 0:40
¼
0:308
s2Y
The regression lines y^ ¼ ax þ b and x^ ¼ ay þ b are shown in Figs. 18.2.2 and 18.2.3
respectively. The minimum mean-square error (MMSE) for Y on X regression ¼ 0.308 (1 2 0.3242) ¼ 0.276, and the MMSE for X on Y regression ¼ 0.194 (1 2 0.3242) ¼
0.174.
pffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
The square root of the product aa ¼ 0:404 0:26 ¼ 0:324 ¼ rXY is the correlation
coefficient obtained earlier. In general, the means and variances of random variables are
not known a priori and we can only estimate them. We will revisit this example after discussing parameter estimation.
Orthogonality Principle. We will motivate the orthogonality principle by taking the
following expectations
E½Y aX bb
(18:2:20a)
E½(Y aX b)aX
(18:2:20b)
with a ¼ sXY =s2X and b ¼ mY amX . Equation (18.2.20a) becomes E½Y aX bb ¼ ½mY amX ðmY amX )b ¼ 0: For Eq. (18.2.20b) we have
E½(Y aX b)aX ¼ {E½XY aE½X 2 bmX }a
¼ {E½XY aE½X 2 (mY amX )mX }a
¼ {sXY as2X }a ¼ {sXY sXY }a ¼ 0
Combining both equations, we can write
Y aX b aX þ b
E
¼0
error 1
estimate Y^
FIGURE 18.2.2
18.2
Estimation of Random Variables
347
FIGURE 18.2.3
Equation (18.2.21) is the orthogonality principle, which states that under MMSE
criterion the estimate Y^ is orthogonal to the error 1. In other words
^ ¼0
E½1Y
(18:2:21)
This is a powerful result that is used extensively in estimation problems.
We can give a geometric interpretation to the orthogonality principle by
means of Fig. 18.2.4. The two observations are b and a, represented by vectors
x 0 and x 1 in a two-dimensional subspace. The unknown quantity y is not in the subspace. However, the estimate y^ , which is a linear combination of x 0, and x 1 is in the
subspace. The error vector 1 is given by 1 ¼ y y^ is also shown in Fig. 18.2.4.
From the geometry of the diagram, it can be seen that 1 is a minimum only
when it is orthogonal to the subspace spanned by x 02x 1. Hence 1 is orthogonal
to the estimate y^ .
We will use the orthogonality principle to find the regression coefficients for a
multiple linear regression problem of Eq. (18.2.19) expressed in matrix form
Y^ ¼ a0 þ a1 X1 þ þ an Xn ¼ aT X
T
(18:2:19)
where a is the (n þ 1) column vector [a0, a1, . . . , an] and X is the (n þ 1) random
variable vector [1, X1, X2, . . . , Xn]T. Using the orthogonality principle that the error
FIGURE 18.2.4
348
Estimation Theory
1 ¼ Y aT X is orthogonal to the subspace spanned by the vector X, we have
E½(Y aT X)XT ¼ 0
(18:2:22)
From Eq. (17.2.5) we can define E½XXT ¼ RX and E½XY ¼ RXY , and
Eq. (18.2.22) can be rewritten as
RX a ¼ E½XY
(18:2:23)
a ¼ R1
X E½XY
We can also find the minimum mean-square error using the orthogonality
principle:
J ¼ E½(Y aT X)(Y aT X) ¼ E½1(Y aT X)
(18:2:24)
Using orthogonality, we obtain
Jmin ¼ E½1Y ¼ E½(Y aT X)Y
¼ E½(Y 2 aT E½XY
(18:2:25)
a result arrived at with comparative ease.
Example 18.2.2 We will now solve the simple linear regression problem by using
Eq. (18.2.23). We are to estimate Y in Y ¼ b þ aX þ 1. The estimator Y^ ¼ b þ aX.
The vector a ¼ [b,a]T, and the vector X ¼ [1,X ]T. The correlation matrix RX is
given by
1 mX
1
½1 X ¼
RX ¼ E
X
mX m2
1
mY
E½XY ¼ E
Y ¼
X
E(XY)
Hence the regression coefficients from Eq. (18.2.23) are
1 mX 1 mY
b
m2
1
¼
a¼
¼ 2
sX mX
a
mX m2
E(XY)
1 m2 mY mX E(XY)
¼ 2
sX
sXY
1 mY (m2 m2X ) mX (E½XY mX mY )
¼ 2
sX
sXY
mX
1
mY
E(XY)
or
mY mX a
b
sXY
¼
a
s2X
which is the same result as obtained before but without differentiation.
We can also find the minimum mean-square error using the orthogonality principle:
12 ¼ E½(Y aX b)(Y aX b) ¼ E½1(Y aX b)
18.2
Estimation of Random Variables
349
Using orthogonality, we obtain
12min ¼ E½1Y ¼ E½(Y aX b)Y
¼ E½Y 2 aE½XY mY ½mY amX ¼ E½Y 2 m2y a{E½XY mx mY }
¼ s2Y asXY ¼ s2Y (1 r2 )
a result arrived at with comparative ease.
Example 18.2.3 A random variable Y ¼ aX 2 þ bX þ c þ 1 is to be estimated by a
linear regression line Y^ ¼ aX 2 þ bX þ c. We want to find the coefficients a,b,c. The
column vector a ¼ [a,b,c]T, and the column vector X ¼ [X 2,X,1]T. The correlation
matrix RX is
82 3
9 2
3
m4 m3 m2
< X2
=
(18:2:26)
RX ¼ E 4 X 5½ X 2 X 1 ¼ 4 m3 m2 m 5
:
;
m2 m
1
1
and
82 3 9 2
3
E½X 2 Y
< X2
=
E½XY ¼ E 4 X 5Y ¼ 4 E½XY 5
:
;
E½Y
1
(18:2:27)
Hence the regression coefficients are
2 3 2
a
m4
4 b 5 ¼ 4 m3
c
m2
31 2
3
m2
E½X 2 Y
m 5 4 E½XY 5
1
E½Y
m3
m2
m
(18:2:28)
Using Eq. (18.2.25), we can derive the minimum mean-square error as follows:
2
E½X 2 Y
3
6
7
12min ¼ E½Y 2 ½a,b,c4 E½XY 5
E½Y
¼ E½Y 2 {aE½X 2 Y þ bE½XY þ cE½Y}
(18:2:29)
Example 18.2.4 We will continue the previous example with observed data for random
variables X and Y shown in Table 18.2.2. It is desired to estimate Y from X.
The correlation matrix RX is given as
2
E½X 4 E½X 3 RX ¼ 4 E½X 3 E½X 2 E½X 2 E½X
3 2
4:53
E½X 2 5
4
¼
2:78
E½X
1:79
1
2:78
1:79
1:23
3
1:79
1:23 5
1
In addition, the cross-moments of X with Y are E[X 2Y ] ¼ 3.79, E[X Y ] ¼ 2.67, with
E[Y ] ¼ 2.21 and E[Y 2] ¼ 4.92.
350
Estimation Theory
TABLE 18.2.2
No.
X
Y
No.
X
Y
1
2
3
4
5
6
7
8
9
10
11
0.91
0.22
0.23
1.92
1.87
0.55
0.94
1.4
1.06
1.38
1.91
2.6
2.21
2.2
2.03
2.07
2.38
2.48
2.37
2.44
2.34
1.9
12
13
14
15
16
17
18
19
20
21
1.47
1.37
0.63
1.39
0.79
1.53
1.51
1.35
1.86
1.64
2.25
2.28
2.26
2.23
2.27
2.11
2.1
2.17
1.77
1.95
FIGURE 18.2.5
The regression coefficients are obtained from Eq. (18.2.28) as follows:
3 2
3
2 3 2
31 2
3:79
0:46
a
4:53 2:78 1:79
4 b 5 ¼ 4 2:78 1:79 1:23 5 4 2:67 5 ¼ 4 0:79 5
2:21
2:06
c
1:79 1:23
1
The line of regression is y ¼ 0:46x2 þ 0:79x þ 2:06. The original data y and the line of
regression are shown in Fig. 18.2.5.
The minimum mean-square error is obtained from Eq. (18.2.29) and is given by
12min ¼ E½Y 2 aE½X 2 Y þ bE½XY þ cE½Y ¼ 0:0109
B 18.3 ESTIMATION OF PARAMETERS (POINT
ESTIMATION)
A population is represented by the random variable X characterized by a parameter u,
which is an unknown constant, and whose distribution function FX(x,u) is of a known
form. We estimate this parameter by observing n realizations {x1, . . . , xn} of the
18.3
Estimation of Parameters (Point Estimation)
351
random variable X. Associated with each of these realizations is a set of random variables
{X1, . . . , Xn} having the same characteristics as the population. We form a function
^ If X is the n-vector equal to [X1, . . . , Xn]T, then u^ =
g(X1, . . . , Xn) for the estimator u.
g(X) is known as the point estimator for u. As mentioned earlier, any function g(X) is
^ we cannot conclude that
also called a statistic for u. Having determined the estimator u,
it is exactly equal to the parameter u. However, we may be able to conclude that u is in
the proximity of u^ within the interval u1 ¼ g1(X) and u2 ¼ g2(X). The interval estimator
for u is the random interval, fQ1, Q2g, such that
P{Q1 , u , Q2 } ¼ g
(18:3:1)
where g is called the confidence coefficient and the interval (u1,u2) is the g –confidence
interval. The coefficient a ¼ (1 2 g) is called the significance level. We will discuss interval estimators in the next section.
Estimation of Mean
The point estimator m^ X , called the sample mean for the mean mX of the random
variable X, can be given as the weighted sum of the independent observation random
variables fX1, . . . , Xng. Thus
m^ X ¼
n
X
a i Xi ¼ aT X
(18:3:2)
i¼1
where the weight vector a has to be determined according to the criteria of unbiasedness
and minimum variance. The unbiasedness criterion applied to Eq. (18.3.2) yields
"
#
n
n
n
X
X
X
E½m^ X ¼ E
ai Xi ¼
ai E½Xi ¼ mX
ai ¼ m X
(18:3:3)
i¼1
i¼1
i¼1
Hence we have the first condition:
n
X
ai ¼ 1
(18:3:4)
i¼1
The variance of the estimator m^ X of Eq. (18.3.2) under the assumption that the Xi values
are independent is given by
var(m^ X ) ¼ s2m^ X ¼
n
X
a2i u2X
(18:3:5)
i¼1
P
The problem is to minimize var(m^ X ) subject to the constraint ni¼1 ai ¼ 1: We make use of
Lagrange multipliers to find a solution. We treat this problem as an unconstrained minimization by adjoining the constraint equation as shown below:
!
n
n
X
X
2 2
J¼
a i sX þ l
ai 1
(18:3:6)
i¼1
i¼1
P
where l is the Lagrange multiplier that has to be determined. Since l ni¼1 ai 1 ¼ 0,
we have not changed the minimization criterion. We treat Eq. (18.3.6) as unconstrained
352
Estimation Theory
and differentiate with respect to ai:
"
!#
n
n
X
@J
@ X
¼
a 2 s2 þ l
ai 1
@ai @ai i¼i i X
i¼1
¼ 2ai s2X þ l ¼ 0
and
ai ¼
l
,
2s2X
i ¼ 1, . . . , n
(18:3:7)
We have determined faig in terms of the unknown Lagrange multiplier l. Substituting for
ai from Eq. (18.3.7) into the constraint Eq. (18.3.4), we obtain
n
X
l
¼1
2
2s
X
i¼1
or l ¼
2s2X
n
(18:3:8)
Substituting Eq. (18.3.8) back into Eq. (18.3.7), we obtain the weighting coefficients
ai as
1
ai ¼ , i ¼ 1, . . . , n
n
(18:3:9)
and the minimum variance unbiased estimator for mX is given by
m^ X ¼
n
1X
Xi
n i¼1
(18:3:10)
We can compute the variance of m^X from Eq. (18.3.5):
var(m^ X ) ¼ s2m^ X ¼
n
X
s2
i¼1
X
n2
¼
s2X
n
(18:3:11)
In Eq. (18.3.11), as n ! 1, variance of m^ X tends to 0. We have already shown in
Eq. (14.1.2) that if the variance is zero, then m^ X ! mX with probability 1. Hence we
can conclude that m^ X is a consistent estimator since it tends to mX as n ! 1. The
density of m^ X , which is centered at mX , is shown in Fig. 18.3.1 for different values of n.
FIGURE 18.3.1
18.3
Estimation of Parameters (Point Estimation)
353
Minimum Mean-Square Estimator for the Mean
Following the lines of Eq. (18.3.3), we will find a minimum mean-square estimator
(MMSE) m X such that
mX ¼
n
X
a i Xi
and
E½mX ¼
n
X
i¼1
ai E½Xi ¼ kmX
(18:3:12)
i¼1
and the corresponding bias BmX in the estimator mP
X is BmX ¼ (k 1)mX . We have to find
n
2
2 2
the coefficients
fa
g
such
that
the
variance
s
¼
i
i¼1 ai sX is minimized subject to the
mX
Pn
constraint i¼1 ai ¼ k. Using Lagrange multipliers as in the previous case, the coefficients
are found to be
k
ai ¼ ,
n
i ¼ 1, . . . , n
(18:3:13)
and the corresponding minimum variance is equal to
s2mX ¼
k2 s2X
n
(18:3:14a)
and the biased estimator in terms of the unbiased estimator is given by
k
mX ¼ m^ X
n
(18:3:14b)
We will find k that minimizes the mean-square error criterion given by
MSE ¼ 12 ¼ E½mX mX 2 ¼ E½(mX E½mX ) þ (E½mX mX )2
¼ s2mX þ B2mX ,
E½(mX E½mX ) ¼ 0
(18:3:15)
Substituting Eqs. (18.3.3) and (18.3.4) in Eq. (18.3.15), we have
12 ¼ s2mX þ B2mX ¼
k2 s2X
þ (k 1)2 m2X
n
(18:3:16)
Differentiating Eq. (18.3.16) with respect to k, we obtain
@ 2
s2
½1 ¼ 2k X þ 2(k 1)m2X ¼ 0
@k
n
or
kmin ¼
m2X
s2X
þ m2X
n
and
mX ¼
m2X
s2X
þ m2X
n
m^ X
(18:3:17)
The variance of the estimator mX is obtained by substituting for k in Eq. (18.3.14a) and
2
s2mX ¼ kmin
s2X
m4X
s2X
¼
2
n
n
s2X
þ m2X
n
(18:3:18)
354
Estimation Theory
which is less than (m^ X ) but biased with bias equal to
B mX
s2X
mX
¼ (kmin 1)mX ¼ 2n
sX
þ m2X
n
(18:3:19)
However, mX is consistent and asymptotically unbiased as n ! 1. This may not be a
useful estimator because it is dependent on s2X and mX , which are both unknown quantities.
At the very least, the ratio sX =mX must be known. This example illustrates that biased estimators in some cases may be better estimators than unbiased ones.
Estimation of Variance
Analogous to formulation of the sample mean, we can express the sample variance to
be of the form
s2X ¼
n
1X
(Xi mX )2
n i¼1
(18:3:20)
and indeed s2X will be an unbiased minimum variance estimator; but mX is unknown, and
hence s2X may not be a useful estimator. However, if we substitute the sample mean m^ X
instead of mX in Eq. (18.3.20), then we can write the new estimator s~ 2X as follows:
s~ 2X ¼
n
1X
(Xi m^ X )2
n i¼1
(18:3:21)
To determine whether s2X is unbiased, we take expectations on both sides of Eq. (18.3.21)
and write
(
"
#)
n
n
n
n
X
1X
2 X
1X
2
2
X Xi
Xj þ 2
Xj
Xk
E(s~ X ) ¼ E
n i¼1 i n j¼1
n j¼1 k¼1
8
2
39
>
>
=
n
n
n
n X
n
<1 X
X
X
X
1
1
6 2 2 2 2
7
2
¼E
(18:3:22)
X i Xj þ 2
Xj þ 2
Xj Xk 5
4 Xi Xi >
>
n
n j¼1
n j¼1
n j¼1 k¼1
;
:n i¼1
j=i
n 1X
j=k
2 n
2(n 1) 2 n(n 1) 2
2
2
¼
mX þ
1 þ 2 (sX þ mX ) mX
n i¼1
n n
n
n2
n 1X
n1 2
n 1 2 2(n 1) 2 (n 1) 2
n1 2
mX þ
mX ¼
sX
¼
sX þ
mX n i¼1
n
n
n
n
n
As we can see from Eq. (18.3.22), the estimator s~ 2X is biased. To obtain an unbiased
estimator, we have to divide Eq. (18.3.21) by (n 2 1) instead of by n and write the
unbiased minimum variance estimator s^ 2X , also called the sample variance, as
s^ 2X ¼
n
1 X
(Xi m^ X )2
n 1 i¼1
(18:3:23)
In forming an unbiased estimator for the variance, the degrees of freedom in the estimator
determines the divisor. In Eq. (18.3.20) there are n degrees of freedom since there are n
18.3
Estimation of Parameters (Point Estimation)
355
independent values of Xi, and hence the divisor
P is n. However, in Eq. (18.3.23) there are
only (n 2 1) degrees of freedom since m^ X ¼ ni¼1 (Xi =n), which takes away one degree of
freedom. Thus, the divisor is (n 2 1). The square root of the sample variance s^ X is called
the standard error.
We will now compute the variance of the sample variance
"
var(s^ 2X )
#2
n
1 X
2
¼E
(Xi m^ X ) (s2X )2
n 1 i¼1
(
)
n
n
X
X
1
2
2
¼E
(Xi m^ X )
(Xj m^ X ) s4X
(n 1)2 i¼1
j¼1
8
2
39
n
n X
n
< 1
=
X
X
4
2
25
4
s4X
¼E
(X
m
^
)
þ
(X
m
^
)
(X
m
^
)
i
i
j
X
X
X
:(n 1)2 i¼1
;
i¼1 j¼1
i=j
n
n(n 1) 4
n
s4
v þ
s s4X ¼
v þ X
2 4
2 X
2 4
n1
(n 1)
(n 1)
(n 1)
(18:3:24)
where v4 ¼ E(Xi mX )4 from Eq. (10.4.2). We can also confirm that s^ 2X is a consistent
estimator since var(s^ 2X ) tends to 0 as n tends to 1. In fact, all three estimators, s2X , s~ 2X ,
and s^ 2X , are consistent while only s2X and s^ 2X are unbiased. However, for large n all the
estimators are asymptotically unbiased.
Estimation of Covariance
The unbiased estimator for the covariance sXY between two random variables X and Y
can be derived with the following assumptions. Corresponding to each of the random variables, we have two sets of independent observations fX1, . . . , Xng and fY1, . . . , Yng with
E[Xi] ¼ mX, E[Yi] ¼ mY, E[XiYi] ¼ sXY þ mX mY, and E[XiYj] ¼ mX mY for i = j. The
estimator s^ XY for the covariance can be expressed as
s^ XY ¼
n
1X
(Xi m^ X )(Yi m^ Y )
K i¼1
(18:3:25)
where the divisor K is found from the following steps to satisfy the unbiasedness condition
E(s^ XY ) ¼ sXY :
"
#
n
X
X
E
(Xi m^ X )(Yi m^ Y ) ¼
E½Xi Yi Xi m^ Y Yi m^ X þ m^ X m^ Y i¼1
i
X 1X
1X
1 XX
Y j Yi
Xj þ 2
Xi Yj
¼
E Xi Yi Xi
n
n
n
i
"
X
Xi Yi X Xi Yj Xi Yi X Yi Xj 1 X
þ 2
Xi Yi
E Xi Yi ¼
n
n
n
n
n
i
j=i
j=1
#
1 XX
þ 2
Xi Yj
n j=i i
356
Estimation Theory
X
1
2(n 1)
n(n 1)
mX mY þ
¼
(sXY þ mX mY ) 1 mX mY
n
n
n2
i
Xn 1
n1
2(n 1)
n1
sXY þ
mX mY mX mY þ
mX mY
¼
n
n
n
n
i
(18:3:26)
¼ (n 1)sXY
From Eq. (18.3.26) K is found to be (n 2 1), and the unbiased minimum variance estimator
for the covariance is
s^ XY ¼
n
1 X
(Xi m^ X )(Yi m^ Y )
n 1 i¼1
(18:3:27)
a result very similar to that of the estimator s^ 2X of the variance.
We will now compute the variance of the sample covariance
"
#2
n
1 X
(Xi m^ X )(Yi m^ Y ) (sXY )2
var(s^ XY ) ¼ E
n 1 i¼1
(
)
n
n
X
X
1
¼E
(Xi m^ X )(Yi m^ Y )
(Xj m^ X )(Yj m^ Y ) s2XY
(n 1)2 i¼1
j¼1
8
2 n
39
P
2
>
>
>
>
½(X m^ X )(Yi m^ Y )
>
=
< 1
6 i¼1 i
7>
6
7
s2XY
¼E
n
n
6
7
P
P
24
>
5>
(n
1)
>
>
þ
½(X
m
^
)(Y
m
^
)½(X
m
^
)(Y
m
^
)
i
i
j
j
>
>
X
Y
X
Y
;
:
i¼1 j¼1
j=i
n
n(n 1) 2
n
s2
v XY þ
s s2XY ¼
vXY þ XY
2 2
2 XY
2 2
n1
(n 1)
(n 1)
(n 1)
(18:3:28)
2
where we have defined vXY
as the second cross-central
2 ¼ E ½(Xi mX )(Yi mY )
moment between X and Y. Since var(s^ XY ) tends to 0 as n tends to 1, we conclude that
(s^ XY ) is also a consistent unbiased minimum variance estimator.
In Eqs. (18.2.13) and (18.2.17) the means and variances were explicitly stated. In
general, these quantities can only be estimated. We will revisit the linear regression
problem later.
Example 18.3.1 We will find the estimated means and variances of the random variables
X and Y from the data given in Example 18.2.1. From Eq. (18.3.10) the unbiased,
minimum variance estimators for the mean of X and Y are
m^ X ¼
1
18:1
½0:05 þ 0:29 þ þ 1:52 ¼
¼ 0:905
20
20
m^ Y ¼
1
38:78
½1:82 þ 2:05 þ þ 2:82 ¼
¼ 1:939
20
20
18.3
Estimation of Parameters (Point Estimation)
357
Similarly, from Eq. (18.3.9) the estimated variances of X and Y are
1
½(0:05 0:905)2 þ (0:29 0:905)2 þ þ (1:52 0:905)2 19
3:767
¼ 0:198
¼
19
1
s^ 2Y ¼ ½(1:82 1:939)2 þ (2:05 1:939)2 þ þ (2:82 1:939)2 19
5:857
¼ 0:308
¼
19
s^ 2X ¼
The estimated covariance between X and Y is
1
½(0:05 0:905)(1:82 1:939) þ þ (1:52 0:905)(2:82 1:939)
19
1:521
¼
¼ 0:080
19
s^ X Y ¼
The estimated correlation coefficient is
r^ XY ¼
s^ XY
0:08
¼ 0:324
¼
s^ X s^ Y 0:247
These parameters are the same numbers that we used in Example 18.2.1. However, the
regression coefficients a and b will be estimates and will be governed by a probability
distribution.
Example 18.3.2 We will now find the estimated means and variances of the random
variables X and Y from the data given in Example 18.2.3. From Eq. (18.3.10) the unbiased,
minimum variance estimators for the mean of X and Y are
m^ X ¼
1
25:93
½0:91 þ 0:22 þ þ 1:64 ¼
¼ 1:235
21
21
m^ Y ¼
1
46:41
½2:6 þ 2:21 þ þ 1:95 ¼
¼ 2:21
21
21
Similarly, from Eq. (18.3.9) the estimated variances of X and Y are
1
½(0:91 1:235)2 þ (0:22 1:235)2 þ þ (1:64 1:235)2 20
5:50
¼ 0:275
¼
20
s^ 2X ¼
1
½(2:6 2:21)2 þ (2:21 2:21)2 þ þ (1:95 2:21)2 20
0:795
¼ 0:0398
¼
20
s^ 2Y ¼
358
Estimation Theory
We will now find estimated values for the higher-order moments of X. They are given by
^ 4 ¼
^ 4X ¼ E½X
m
1
95:1
½0:914 þ 0:224 þ þ 1:644 ¼
¼ 4:53
21
21
^ 3 ¼
^ 3X ¼ E½X
m
1
58:43
½0:913 þ 0:223 þ þ 1:643 ¼
¼ 2:78
21
21
^ 2 ¼
^ 2X ¼ E½X
m
1
37:52
½0:912 þ 0:222 þ þ 1:642 ¼
¼ 1:79
21
21
The estimated second moment of Y is given by
^ 2 ¼
^ 2X ¼ E½Y
m
1
103:36
½2:62 þ 2:212 þ þ 1:952 ¼
¼ 4:92
21
21
The estimated cross-moments of X with Y are
1
56:1
^
E½XY
¼ ½0:91 2:6 þ 0:22 2:21 þ þ 1:64 1:95 ¼
¼ 2:67
21
21
^ 2 Y ¼ 1 ½0:912 2:6 þ 0:222 2:21 þ þ 1:642 1:95 ¼ 79:6 ¼ 3:79
E½X
21
21
The estimated correlation matrix RX is,
2
^ 4
E½X
^ X ¼ 4 E½X
^ 3
R
^ 2
E½X
3 2
^ 3 E½X
^ 2
4:53
E½X
5 ¼ 4 2:78
^E½X 2 E½X
^
^
1:79
E½X
1
2:78
1:79
1:23
3
1:79
1:23 5
1
These values are the same values given in Example 18.2.3. Again, the regression coefficients a, b, and c will be only estimates.
Estimated Regression Coefficients from Data
As stated in previous examples, the evaluated regression coefficients will be
estimates of the true values since the means and variances are only estimates. We will
now discuss the methodology of finding the estimated regression coefficients and determining their variances from the given data. The linear regression problem for each yi can
be stated as
yi ¼ a1 xi1 þ þ am xim þ wi ¼ xTi a þ wi ,
i ¼ 1, . . . , n
with n . m. Or more succinctly
y ¼ XT a þ w
(18:3:29)
where the n-vector y is the realization of the random variable Y with E(y) ¼ X Ta where a
is an unknown m-vector parameter to be estimated, and xi is the observation m-vector
given by xi ¼ [xi1, . . . , xim]T. In linear regression, the coefficient a1 represents the offset
and hence xi1¼1 for all i. The n-vector w is a realization of the measurement
noise random variable W. The elements wi of w are assumed to be uncorrelated with
zero mean and variance E[w2i ] ¼ s2W for all i. X is an m n (n . m) matrix of
18.3
Estimation of Parameters (Point Estimation)
359
observations given by
x2
½x
2 1
x11 x21
6x
6 12 x22
6 .
..
6 .
.
.
X¼ 6
6
6 x1j x2j
6
6 .
..
6 .
4 .
.
xi
xi1
xi2
..
.
xij
..
.
xn 3 2
1
xn1
7
6
xn2 7 6 x12
6
.. 7
7 6 .
. 7 6 ..
7¼6
6
xnj 7
7 6 x1j
7
.. 7 6
6 .
. 5 4 ..
x1m x2m xim xmn
1
x22
..
.
x2j
..
.
3
1 1
xi2 xn2 7
7
.. 7
..
7
. 7
.
7 (18:3:30)
xij xnj 7
7
.. 7
..
7
. 5
.
x1m x2m xim xnm
Given this information, we have to estimate the regression coefficient vector a using
the least-squares error. Thus, the criterion function J(a) is given by
J(a) ¼ (y XTa )T (y XTa ) ¼ yT y aT Xy yT XT a þ aT XXT a
(18:3:31)
From Eq. (16.3.4) the derivative with respect to the vector a of Eq. (18.3.31) is
@
@
J(a) ¼ ½yT y aT Xy yT XT a þ aT XXT a
@a
@a
¼ 2Xy þ 2XXTa ¼ 0
(18:3:32)
From Eq. (18.3.32) the estimated regression coefficients a^ are given by
a^ ¼ (XXT )1 Xy
T
where XX expanded in terms of the
P
2
n
xi2
n
6P
P
6 xi2
x2i2
6 n
n
6
6 .
..
6 .
6 .
T
P.
XX ¼ 6 P
6 xij
xij xi2
6 n
n
6
6 .
..
6 ..
4P
P .
xim
xim xi2
n
n
(18:3:33)
observation points fxijg is given by
P
P
3
xij
xim
n
n
7
P
P
xi2 xij xi2 xim 7
7
n
n
7
7
..
..
7
7
.
.
P 2
P
7
xij
xij xim 7
7
n
n
7
7
..
..
7
.
.
P
P 2 5
xim xij xim
n
(18:3:34)
n
The equation for the regression line is given by
y^ ¼ XT a^
(18:3:35)
The estimator a^ is an unbiased estimator of a since
E(^a) ¼ (XXT )1 XE(y) ¼ (XXT )1 XXT a ¼ a
Covariance of the Estimated Regression Coefficients a^
Before finding the covariance of a, we will express E½^a a in a more convenient
form given by
E½^a a ¼ E½(XXT )1 Xy a ¼ E½(XXT )1 X(XT a þ w) a
¼ E½(XXT )1 Xw
(18:3:36)
360
Estimation Theory
The covariance matrix Ca^ of the estimated regression coefficients can now be given by
Ca^ ¼ cov(^a) ¼ cov½(^a a)(^a a)T ¼ cov{(XXT )1 XwwT XT (XXT )T }
¼ (XXT )1 X cov(wwT )XT (XXT )T
¼ (XXT )1 Xs2w IXT (XXT )T ¼ (XXT )1 s2w
(18:3:37)
where cov(wwT ) ¼ s2w Im . The variances of the regression coefficients are given by the
diagonal terms of the covariance matrix Ca^ . Since we do not know the variance s2w of
the measurement noise, we use its estimate s^ 2w , given by
s^ 2w ¼
(y y^ )T (y y^ ) (y XT a^ )(y XT a^ )
¼
nm
nm
(18:3:38)
In this equation, the divisor (n 2 m) reflects the reduction in the number of degrees of
freedom n by the number of regression coefficients m. This yields an unbiased estimator
for s2w.
Variance of the Estimated Regression line y^ i
We will now find the variance of the estimated mean y^ i , given by
y^ i ¼ a^ 1 xi1 þ þ a^ m xim ¼ a^ T xi ,
i ¼ 1, . . . , n
(18:3:39)
where y^ i is an unbiased estimator of yi since E½^yi ¼ y. The variance of y^ i can be given by
E½ y^ i yi 2 ¼ E½(^yi yi )T (^yi yi )
¼ xTi E ½(^a a)(^a a)T xi ¼ xTi cov(^a)xi
(18:3:40)
Substituting for cov(^a) from Eq. (18.3.37), we have
var( y^ i ) ¼ s2y^ i ¼ xTi (XXT )1 xi s^ 2w , i ¼ 1, . . . , n
(18:3:41)
and since we do not know the variance, we substitute the estimated variance s^ 2w and obtain
the estimated variance of y^ i as
s^ 2y^ i ¼ xTi (XXT )1 xi s^ 2w ,
i ¼ 1, . . . , n
(18:3:42)
which is a function of i, and is chi-square-distributed with (n 2 m) degrees of freedom.
Simple Linear Regression
The governing equation for a simple linear regression is
yi ¼ a1 þ a2 xi þ wi ,
i ¼ 1, . . . , n
(18:3:43)
with the usual definitions. From the data, we will derive expressions for the estimated
regression coefficients for a1 and a2. The normal equations can be obtained from Eqs.
(18.3.32) and (18.3.34) as follows:
X 3
2
2 X 3
xi yi
n
6X
6X
7 a^ 1
7
n
n
X
(18:3:44)
¼4
4
5
5
2
^
a
2
xi
xi
xi yi
n
n
n
18.3
Estimation of Parameters (Point Estimation)
361
The determinant of the rectangular matrix of Eq. (18.3.44) is
X
!2
xi X
X
n
2
X ¼n
xi xi
x2i n
n
n
X
X
¼n
x2i m2 m^ 2x ¼ n
(xi m^ x )2
n
X
xi
n
n
(18:3:45)
n
P
where m^ x ¼ (1=n) n xi . Solving Eq. (18.3.44) and substituting in Eq. (18.3.45), the estimated coefficients a^ 1 and a^ 2 are
a^ 1
a^ 2
2 X
x2i
1
6 nX
¼ X
4
(xi m^ x )2 n
xi
n
2X
n
X
xi
n
n
X
32 X
76 X
n
54
yi
xi yi
3
7
5
n
X
X
3
yi xi
xi yi
1
6 n Xn
7
n
n
X
X
¼ X
4
5
n
(xi m^ x )2
n
xi yi xi
yi
x2i
n
n
n
(18:3:46)
n
In Eq. (18.3.46)
n
X
xi yi X
n
xi
X
n
yi ¼ n
X
n
xi yi n2 m^ x m^ y ¼ n
n
X
(xi m^ x )(yi m^ y )
n
Hence
X
n
a^ 2 ¼
(xi m^ x )(yi m^ y )
X
(xi m^ x )2
(18:3:47)
n
Also, in Eq. (18.3.46)
X
x2i
n
X
n
yi X
n
xi
X
xi yi ¼ nm^ y
X
n
x2i
n
"
¼ nm^ y
X
"
xi
n
X
X
#
(xi m^ x )(yi m^ y ) þ nm^ x m^ y
n
#
x2i m^ 2X
nm^ x
n
X
(xi m^ x )(yi m^ y )
n
Hence
X
a^ 1 ¼ m^ y n
(xi m^ x )(yi m^ y )
X
m^ x ¼ m^ y a^ 2 m^ x
(x2i m^ 2X )
(18:3:48)
n
The estimated regression line equation is
y^ i ¼ a^ 1 þ a^ 2 xi , i ¼ 1, . . . , n
(18:3:49)
362
Estimation Theory
We can now compute the variances of a^ 1 and a^ 2 from Eqs. (18.3.33)
X 3
2 X 2
xi
xi
1
a^ 1
6
7 2
n
n
var
¼ X
4 X
5s^ w
a^ 2
(xi m^ x )2 n
xi
n
n
(18:3:50)
n
where we have substituted the estimated variance s^ 2w from Eq. (18.3.38) for the true
variance s2w .
Simplifying Eq. (18.3.50), we obtain the variances of a^ 1 , and a^ 2 as
var(^a1 ) ¼ s^ 2a^ 1 ¼ X
n
s^ 2w
(xi m^ x )2
2X
6
var(^a2 ) ¼ s^ 2a^ 2 ¼ s^ 2w 4
n
2
3
3
(xi m^ x )2 þ n(m^ 2x )
2
m^ x
61
7
7
X
5 ¼ s^ 2w 4 þ X
2
25
n
(xi m^ x )
(xi m^ x )
n
n
(18:3:51)
n
Finally, we can compute the estimated variance of the estimated regression line y^ i . From
Eq. (18.3.42), we have
X 3
2 X 2
xj
xj 1
6 nX
7 1 2
n
2
X
s^ y^ i ¼
1 xi 4
s^ , i ¼ 1, . . . , n (18:3:52)
5
2
xi w
(xj m^ x )
n
xj
n
n
n
Equation (18.3.52) can be simplified as follows
P 3
2 P 2
xj
xj X
X
1
n
5
x2j 2xi
xj þ nx2i
¼
½ 1 xi 4 nP
xj
n
xi
n
n
n
¼
X
2
xj m^ x þn(xi m^ x )2
(18:3:53)
n
where we have subtracted and added m^ 2x . Substituting Eq. (18.3.53) in Eq. (18.3.52), we
obtain
"
#
X
1
2
2
2
s^ y^ i ¼ P
(xj m^ x ) þ n(xi m^ x ) s^ 2w
n n (xj m^ x )2 n
1
(xi m^ x )2
þP
¼
(18:3:54)
s^ 2 , i ¼ 1, . . . , n
n
^ x )2 w
n (xj m
Example 18.3.3 From the data in Example 18.2.1 we will determine the estimated
regression coefficients and their variances using the techniques developed above. The
linear equation used to solve for the regression coefficients is
y i ¼ a1 þ a2 x i þ w i ,
i ¼ 1, . . . , 20
In Eq. (18.3.29) the 20-vector y is given by
1:82, 2:05, 1:28, 1:19, 2:06, 1:96, 2:46, 1:76, 1:13, 2:35, T
y¼
1:49, 2:15, 0:88, 1:49, 2:18, 2:74, 2:10, 2:69, 2:18, 2:82
18.3
Estimation of Parameters (Point Estimation)
363
and the 20-vector x is given by,
0:05, 0:29, 0:74, 0:55, 1:07, 0:47, 1:06, 0:7, 0:54, 0:65, T
x¼
1:54, 0:72, 0:66, 1:23, 1:35, 0:97, 1:3, 0:96, 1:73, 1:52
a is a 2-vector of regression coefficients given by a ¼ ½a1 ,a2 T and w is an unknown
measurement 20-vector noise whose variance can only be estimated. We can now form
the 2 20 matrix X, where the first row is always 1 since a1 multiplies 1 for all i.
Hence the matrix X is
82
39
1
1
1
1
1
1
1
1
1
1
>
>
>
=
<6 0:05 0:29 0:74 0:55 1:07 0:47 1:06 0:7 0:54 0:65 7>
7
X¼ 6
4 1
>
1
1
1
1
1
1
1
1
1 5>
>
>
;
:
1:54 0:72 0:66 1:23 1:35 0:97 1:3 0:96 1:73 1:52
The quantity XXT and (XXT )1 are given by
20
18:1
0:267
(XXT )1 ¼
XXT ¼
18:1 20:147
0:24
0:24
0:265
The estimated regression coefficients a^ are obtained from Eq. (18.3.33):
a^
1:574
0:267
0:24
38:78
â ¼ 1 ¼ (XXT )1 (Xy) ¼
¼
0:404
0:24
0:265 36:617
a^ 2
with
Xy ¼
38:78
36:617
where a^ 1 and a^ 2 exactly match the quantities b and a in Example 18.2.1. The equation of
the regression line
y^ i ¼ a^ 1 þ a^ 2 xi ¼ 1:574 þ 0:404xi ,
i ¼ 1, . . . , 20
has already been shown in Fig. 18.2.2. We will now compute the estimated variance of the
noise term from Eq. (18.3.38):
s^ 2W ¼
(y XT â)T (y XT â)
4:74
¼
¼ 0:263
nm
20 2
This compares within the limits of data analysis with MMSE ¼ 0.276, obtained in
Example 18.2.1. From Eq. (18.3.37), the covariance matrix of a^ is
0:0704 0:0633
0:267
0:24
Câ ¼ cov(â) ¼ (XXT )1 s^ 2W ¼
0:263 ¼
0:0633
0:0699
0:24
0:266
Hence var(^a1 ) ¼ 0:0704 and var(^a2 ) ¼ 0:0699.
We can now obtain the variance ofPy^ i using Eq. (18.3.54), where n ¼ 20, the average
value m^ x ¼ 0:905, and the summation n (xj m^ x )2 ¼ 3:7665. Substituting these values,
we obtain
2
3
2
(xi m^ x ) 7
1 (xi 0:905)2
2
26 1
þ
s^ y^ i ¼ s^ w 4 þ P
, i ¼ 1, . . . , 20
5 ¼ 0:263
20
20
3:7665
(xj m^ x )2
20
364
Estimation Theory
B 18.4 INTERVAL ESTIMATION (CONFIDENCE
INTERVALS)
In the previous section we discussed point estimators u^ for a parameter u. However, a point
estimator may not be meaningful without knowledge of the probability of how far away it
is from the true value. Hence, we would also like to know an interval about the estimator u^
and the probability that the true value u lies in that interval. Such an interval for u is the
random interval, fQ1, Q2g, such that
P{Q1 , u , Q2 } ¼ g
(18:4:1)
where g is the confidence coefficient. The realized interval (u1, u2) is the g confidence
^ In essence,
interval. The terms u1 and u2 could possibly be functions of the estimator u.
we can replace the empirical point estimator with an interval estimator that gives more
information about the population parameter. The coefficient a ¼ (1 2 g) is called the
significance level. In a two-sided confidence interval as in Eq. (18.4.1), it is usual to
split the significance level a on both sides of the density. We will apply the confidence
intervals to various parameters such as the mean and variance. We have already come
across the basic ideas of confidence intervals in Section 7.9 and Examples 14.8.1 and 14.8.4.
Example 18.4.1 In a presidential election, a news agency polled 1000 probable voters,
and 470 of them said that they would vote Democratic while 460 of them said that they
would vote Republican. The agency reported that the Democrat winning was 47% and
the Republican winning was 46%, with a margin of error of +4%. What this means is
that the true probability of the Democrat winning lies in the interval (51%, 43%) and
that of the Republican lies in the interval (50%, 42%):
P{0:43 , P(Democrat winning) , 0:51} ¼ x
P{0:42 , P(Republican winning) , 0:50} ¼ x
The conclusion arrived at is that the results are within the margin of error and hence the
election is a dead heat. What is never mentioned is x, the probability of these results
occurring!
Confidence Interval for the Unknown Mean of Population (Known
Variance). We
P
have already established in Eq. (18.3.10) that m^ X ¼ (1=n) ni¼1 Xi is an unbiased
minimum variance point estimator for the mean mX. If n is large enough, we
can assume that m^ X is Gaussian-distributed with variance s2X =n. We will find
the confidence interval under the assumption that s2X is known. The random variable Z given by
Z¼
m^ X mX
pffiffiffi
sX = n
(18:4:2)
is a standard Gaussian, and for a confidence coefficient g ¼ 1 2 a we can write
P{Za=2 , Z z1a=2 } ¼ 1 a ¼ g or
m^ m
P za=2 , X pffiffiffiX z1a=2 ¼ g
sX = n
(18:4:3)
where we have split the significance level a equally on both sides of the density
function. The values za=2 ¼ P(Z a=2) and z1a=2 ¼ P(Z 1 a=2) for a confidence level of g are extracted from the Gaussian table (Table 18.4.1) for
18.4
Interval Estimation (Confidence Intervals)
365
TABLE 18.4.1
Gaussian Tables
g%
P(Z z)
z
P(Z z)
z
99
98
95
90
85
80
75
70
65
60
55
50
0.005
0.01
0.025
0.05
0.075
0.1
0.125
0.15
0.175
0.2
0.225
0.25
22.57583
22.32635
21.95996
21.64485
21.43953
21.28155
21.15035
21.03643
20.93459
20.84162
20.75542
20.67449
0.995
0.99
0.975
0.95
0.925
0.9
0.875
0.85
0.825
0.8
0.775
0.75
2.57583
2.32635
1.95996
1.64485
1.43953
1.28155
1.15035
1.03643
0.93459
0.84162
0.75542
0.67449
probabilities of a=2 and (1 a=2) respectively. Since the Gaussian curve is symmetric, za=2 ¼ z1a=2 .
The z values for 95% confidence are z0.025 ¼ – 1.96 and z0.975 ¼ 1.96. The
corresponding z values for 99% confidence are z0.005 ¼ – 2.576 and
z0.995 ¼ 2.576 and for 90% confidence z0.05 ¼ – 1.645 and z0.95 ¼ 1.645.
From Eq. (18.4.3) we have
m^ X mX
pffiffiffi z1a=2 ¼ 1 a
P za=2 ,
sX = n
from which it follows that
za=2 sX
z1a=2 sX
P pffiffiffi , m^ X mX pffiffiffi
¼1a
n
n
and finally
z1a=2 sX
za=2 sX
¼1a
P m^ X pffiffiffi , mX m^ X pffiffiffi
n
n
(18:4:4)
Equation (18.4.4)pis
ffiffiffi to be interpreted
pffiffiffias the probability that the random interval
(m^ X z1a=2 sX = n, m^ X za=2 sX = n) will contain the true population mean
mX is 100(1 2 a)%. The probability density of m^ X centered about the true mean
mX and the confidence intervals are shown in Fig. 18.4.1.
The 99% confidence interval is given by
2:576sX
2:576sX
P m^ X pffiffiffi , mX m^ X þ pffiffiffi
¼ 0:99
(18:4:5)
n
n
In an analogous manner, the 95% confidence interval is given by
1:96sX
1:96sX
¼ 0:95
P m^ X pffiffiffi , mX m^ X þ pffiffiffi
n
n
(18:4:6)
366
Estimation Theory
FIGURE 18.4.1
and the 90% confidence interval is given by
1:645sX
1:645sX
P m^ X pffiffiffi , mX m^ X þ pffiffiffi
¼ 0:90
n
n
(18:4:7)
It can be seen from Eqs. (18.4.5)– (18.4.7) that as the confidence level g decreases
and conversely as the significance level a increases, the bounds become tighter. If
we want to have 100% confidence or 0% significance, then the confidence interval
is 1! The equations also show that as the number of samples n increases, the
bounds also become tighter. The confidence level is usually fixed a priori at
95% or 90% before sampling.
Confidence Interval for the Unknown Mean of Population (Unknown Variance). In
the previous section we assumed that the variance is known. Since the true variance
is known in very few cases, we use the estimated variance s^ 2X . In that case we have
to find the distribution of the random variable T, given by
T¼
m^ X mX
pffiffiffi
s^ X = n
(18:4:8)
which is a Student-t distribution given by Eq. (7.9.2) with n degrees of freedom.
Equation (18.4.8) can rewritten as
T¼
pffiffiffi
(m^ X mX )=sX = n
{½(n 1)(s^ X =sX )2 =(n 1)}1=2
(18:4:9)
pffiffiffi
where the term (m^ X mX )=sX = n is a standard Gaussian, and we will show that
the term
n (n 1)s^ 2X X
Xi m^ X 2
¼
sX
s2X
i¼1
is chi-square-distributed with (n – 1) degrees of freedom. Adding and subtracting mX to (Xi m^ X ) in the numerator of the righthand side of the equation
18.4
Interval Estimation (Confidence Intervals)
367
above, we have
n n X
Xi m^ X 2 X
Xi mX (m^ X mX ) 2
¼
sX
sX
i¼1
i¼1
(
)
n
X
Xi mX 2 m^ X mX 2 2(Xi mX )(m^ X mX )
¼
þ
sX
sX
s2X
i¼1
( )
n
X
Xi mX 2
m^ mX 2
¼
since
n X
sX
sX
i¼1
n
X
(Xi mX ) ¼ n(m^ X mX )
(18:4:10)
i¼1
P
The first term on the righthand side of Eq. (18.4.10), ni¼1 ½(Xi mX )=spXffiffiffi2 is
chi-square-distributed with n degrees ofpfreedom.
Since (m^ X mX )=(sX = n) is
ffiffiffi
Gaussian with mean 0 and variance sX = n, the second term
m^ mX 2
m^ X mX 2
pffiffiffi
n X
¼
sX
sX = n
P
is chi-square-distributed with one degree of freedom. Hence ni¼1 ½(Xi m^ X )=sX 2
is chi-square-distributed with (n 2 1) degrees of freedom. Further, it can be shown
that s^ 2X and m^ X are independent random variables. Hence, from Example 13.4.5, T
in Eq. (18.4.8) is indeed Student-t-distributed with (n 2 1) degrees of freedom.
We can now obtain the 100(1 – a)% confidence intervals for the population
mean mX using the Student-t as follows. Analogous to Eq. (18.4.4), we can write
(m^ X mX )
p
ffiffi
ffi
P{ta=2 , T t1a=2 } ¼ P ta=2 ,
t1a=2 ¼ 1 a ¼ g
s^ X = n
from which it follows that
pffiffiffi
pffiffiffi
P{ta=2 s^ X = n , (m^ X mX ) t1a=2 s^ X = n} ¼ 1 a ¼ g
and finally
pffiffiffi
pffiffiffi
P{m^ X t1a=2 s^ X = n , mX m^ X ta=2 s^ X = n} ¼ 1 a ¼ g
(18:4:11)
where ta=2 ¼ P(T a=2) and t1a=2 ¼ P(T 1 a=2) and ta=2 ¼ t1a=2
because of the symmetry of the Student-t density. From the Student-t table
(Table 18.4.2), t1a=2 is extracted for n – 1 degrees of freedom and probability
(1 2 a/2). Equation p
(18.4.11)
is to be read
ffiffiffi
pffiffiffi as the probability that the random interval (m^ X t1a=2 s^ X = n, m^ X ta=2 s^ X = n) includes the true population mean mX
is equal to (1 2 a).
These confidence limits are shown in Fig. 18.4.2 on a Student-t density with
(n – 1) degrees of freedom.
Example 18.4.2 The tensile strength of six samples of a material is given by 226, 223,
225, 227, 224, and 222 lb. The estimated mean m^ X ¼ 224:5 lb. The standard error,
s^ X ¼ 1:871 lb. Assuming a Gaussian distribution for the samples, the t0.95 value for
(6 – 1) ¼ 5 degrees of freedom from Table 18.4.2 is given by t0.95 ¼ 2.01505. Hence the
368
Estimation Theory
TABLE 18.4.2
Student-t Tables with t-Values Shown
v
FT (t) ¼ 0:95
FT (t) ¼ 0:975
FT (t) ¼ 0:995
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
6.31375
2.91999
2.35336
2.13185
2.01505
1.94318
1.89458
1.85955
1.83311
1.81246
1.79588
1.78229
1.77093
1.76131
1.75305
1.74588
1.73961
1.73406
1.72913
1.72472
12.70620
4.30265
3.18245
2.77645
2.57058
2.44691
2.36462
2.30600
2.26216
2.22814
2.20099
2.17881
2.16037
2.14479
2.13145
2.11991
2.10982
2.10092
2.09302
2.08596
63.65674
9.92484
5.84091
4.60409
4.03214
3.70743
3.49948
3.35539
3.24984
3.16927
3.10581
3.05454
3.01228
2.97684
2.94671
2.92078
2.89823
2.87844
2.86093
2.84534
lower confidence limit tL is
pffiffiffi
1:871
¼ 223:872
tL ¼ m^ X t0:95 s^ X = n ¼ 224:5 2:01505 6
and the upper confidence limit tU with t0.05 ¼ – t0.95 is
pffiffiffi
1:871
¼ 225:128
tU ¼ m^ X t0:05 s^ X = n ¼ 224:5 þ 2:01505 6
Hence we are 90% confident that the true mean lies in the interval (223.872, 225.128) lb.
On the other hand, if we are given that the true standard deviation sX ¼ 1.871 lb, then we
FIGURE 18.4.2
18.4
Interval Estimation (Confidence Intervals)
369
use the Gaussian distribution for confidence intervals given by Eq. (18.4.6), and the lower
and upper confidence intervals are
pffiffiffi
1:871
¼ 223:987
zL ¼ m^ X z0:95 sX = n ¼ 224:5 1:645 6
pffiffiffi
1:871
zU ¼ m^ X z0:05 sX = n ¼ 224:5 þ 1:645 ¼ 225:013
6
The confidence interval (223.987, 225.013) for the Gaussian is a little tighter than
(223.872, 225.128) as obtained for the Student-t, showing that the Student-t density is
flatter than the Gaussian as shown in Fig. 7.9.1.
Example 18.4.3 In a probability examination for a class of 16 students, the following
scores were recorded:
X ¼ ½ 94
70 57
77
45
43
70 67
Estimated mean m^ X ¼
96
89
62
98 56
68
75
71 16
1 X
Xi ¼ 71:125
16 i¼1
16
1 X
(Xi 71:125)2 ¼ 283:183
15 i¼1
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Standard error s^ X ¼ 283:183 ¼ 16:828
Estimated variance s^ 2X ¼
We have to find the 90% confidence limits for estimated mean m^ X ¼ 71:125. From the
Student-t tables (Table 18.4.2) the t value for 15 degrees of freedom and 95% probability
is 1.753. Hence the lower confidence limit is given by
pffiffiffi
16:828
¼ 63:75
tL ¼ m^ X t0:95 s^ X = n ¼ 71:125 1:753 16
and the upper confidence limit is
pffiffiffi
16:828
¼ 78:5
tL ¼ m^ X t0:05 s^ X = n ¼ 71:125 þ 1:753 16
Thus, we are 90% confident that the true mean will be in the interval (63.75, 78.5) or
within +7.375 of the estimated mean 71.125.
Confidence Interval for the Unknown Variance (Unknown Mean). We have already
seen in Eq. (18.4.10) that the random variable (n 1)s^ 2X =s2X is chi-squaredistributed with (n 2 1) degrees of freedom. This density is not symmetric, and
the interval estimate will not be centered at s2, unlike the Gaussian and Studentt densities. To determine the upper and lower confidence limits, we can write
(n 1)s^ 2X
2
2
P xa=2,(n1) ,
x1a=2,(n1) ¼ 1 a ¼ g
s2X
from which it follows that
(
)
x2a=2,(n1)
x21a=2;(n1)
1
P
,
¼g
(n 1)s^ 2X s2X
(n 1)s^ 2X
370
Estimation Theory
TABLE 18.4.3
Chi-Square Tables with u ¼ x2 Values Shown
g ¼ 0.9
v
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
g ¼ 0.95
g ¼ 0.99
FU (u) ¼ 0:05 FU (u) ¼ 0:95 FU (u) ¼ 0:025 FU (u) ¼ 0:975 FU (u) ¼ 0:005 FU (u) ¼ 0:995
0.00393
0.10259
0.35185
0.71072
1.14548
1.63538
2.16735
2.73264
3.32511
3.94030
4.57481
5.22603
5.89186
6.57063
7.26094
7.96165
8.67176
9.39046
10.11701
10.85081
3.84146
5.99146
7.81473
9.48773
11.0705
12.59159
14.06714
15.50731
16.91898
18.30704
19.67506
21.02601
22.36199
23.68475
24.99576
26.29620
27.58709
28.86928
30.14351
31.41042
and finally
0.00098
0.05064
0.21580
0.48442
0.83121
1.23734
1.68987
2.17973
2.70039
3.24697
3.81575
4.40379
5.00875
5.62873
6.26214
6.90766
7.56419
8.23075
8.90652
9.59078
(
5.02389
7.37776
9.34840
11.14329
12.83250
14.44938
16.01276
17.53455
19.02277
20.48306
21.91996
23.33660
24.73555
26.11891
27.48836
28.84532
30.19099
31.52636
32.85231
34.16959
(n 1)s^ 2X
(n 1)s^ 2X
, s2X 2
P 2
x1a=2,(n1)
xa=2,(n1)
0.00004
0.01003
0.07172
0.20699
0.41174
0.67573
0.98926
1.34441
1.73493
2.15586
2.60322
3.07382
3.56503
4.07467
4.60092
5.14221
5.69722
6.26480
6.84397
7.43384
7.87944
10.59663
12.83816
14.86026
16.74960
18.54758
20.27774
21.95495
23.58935
25.18818
26.75685
28.29952
29.81947
31.31935
32.80132
34.26719
35.71847
37.15645
38.58226
39.99685
)
¼g
(18:4:12)
where x21a=2,(n1) and x2a=2,(n1) are constants to be determined from the chi-square
table (Table 18.4.3) for (n – 1) degrees of freedom at probability levels of 1 a=2
and a=2 respectively. Equation (18.4.12) is to be read as the probability that the
random interval
!
(n 1)s^ 2X
(n 1)s^ 2X
,
x21a=2,(n1)
x2a=2,(n1)
includes the population variance s2X is equal to g ¼ (1 –a).
These confidence limits are shown in Fig. 18.4.3 on a chi-square density with (n – 1)
degrees of freedom.
Example 18.4.4 We will find the 90% confidence interval for the sample variance s^ 2X ¼
283:183 calculated in Example 18.4.3. From the chi-square table (Table 18.4.3), the x2
value for probability 0.05 and 15 degrees of freedom is 7.261, and the corresponding
value for probability of 0.95 and 15 degrees of freedom is 24.996. Hence, from Eq.
(18.4.12), the upper confidence limit is given by
x2U ¼
(n 1)s^ 2X 15 283:183
¼ 585:008
¼
7:261
x2a=2,(n1)
18.4
Interval Estimation (Confidence Intervals)
371
FIGURE 18.4.3
and the lower confidence limit is given by
x2L ¼
(n 1)s^ 2X
15 283:183
¼ 169:937
¼
2
24:996
x1a=2,(n1)
The
corresponding 90%
confidence limits for the standard error s^ X ¼ 16:828 are
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
( 169:937 ¼ 13:036, 585:008 ¼ 24:187). This result is to be interpreted as the probability of the standard deviation sX of the students’ scores will be in the interval
(13.036, 24.187) is 90%.
Confidence Interval for the Coefficients of Simple Linear Regression. Since the
residual variance s^ 2w is chi-square-distributed with (n 2 2) degrees of freedom
the estimated variances s^ 2a^ 1 and s^ 2a^ 2 for the regression coefficients a^ 1 and a^ 2 of
the regression line y^ i ¼ a^ 1 þ a^ 2 xi , i ¼ 1, . . . ; n are also chi-square-distributed
with (n 2 2) degrees of freedom as given in Eqs. (18.3.51). The terms a^ 1 and a^ 1
are unbiased
a1 2, and hence the quantities (^
pffiffiffiffiffiffiffiffiffiffiffiestimators of a1 and paffiffiffiffiffiffiffiffiffiffiffi
a1 )=(s^ a^ 1 = n 2) and (^a2 a2 )=(s^ a^ 2 = n 2) are Student-t-distributed with
(n 2 2) degrees of freedom. From Eqs. (18.4.11), we can form equations for
100(1 2 a)% confidence intervals
(^a1 a1 )
P ta=2,(n2) ,
t1a=2,(n2) ¼ 1 a ¼ g
s^ a^ 1
(18:4:13)
(^a2 a2 )
t1a=2,(n2) ¼ 1 a ¼ g
P ta=2,(n2) ,
s^ a^ 2
where t1a=2,(n2) is obtained from the t table (Table 18.4.2) for (n 2 2) degrees
of freedom and probability (1 2 a/2). Equation (18.4.13) can be stated more
formally as
P{^a1 t1a=2,(n2) s^ a^ 1 , a1 a^ 1 ta=2,(n2) s^ a^ 1 } ¼ 1 a
P{^a2 t1a=2,(n2) s^ a^ 2 , a1 a^ 2 ta=2,(n2) s^ a^ 2 } ¼ 1 a
(18:4:14)
and hence
Confidence interval for a1 ¼ (^a1 t1a=2,(n2) s^ a^ 1 , a^ 1 ta=2,(n2) s^ a^ 1 )
(18:4:15)
Confidence interval for a2 ¼ (^a2 t1a=2,(n2) s^ a^ 2 , a^ 2 ta=2,(n2) s^ a^ 2 )
372
Estimation Theory
Example 18.4.5 We will now find 90% confidence intervals for the estimated regression
coefficients of Example 18.3.3. The estimated regression coefficients are given by a^ 1 ¼
1:574 and a^ 2 ¼ 0:404, with estimated variances s^ 2a^ 1 ¼ 0:0704 s^ 2a^ 2 ¼ 0:0699. The standard
errors for a^ 1 and a^ 2 are s^ a^ 1 ¼ 0:265 and s^ a^ 2 ¼ 0:264. From the t table (Table 18.4.2), for
18 degrees of freedom and probability of 0.95, t0.95,18 ¼ 1.7341. From Eq. (18.4.15) we
can compute the 90% confidence interval for a^ 1 as
La^ 1
1:115
¼ 1:574 1:734 0:265 ¼
Ua^ 1
2:034
and the 90% confidence interval for a^ 2 is
La^ 2
1:115
¼ 0:404 1:734 0:264 ¼
Ua^ 2
0:054
Confidence Region for Simple Linear Regression Line.
line is given as follows:
y^ i ¼ a^ 1 þ a^ 2 xi ,
The estimated regression
i ¼ 1, . . . , n
(18:3:49)
Since y^ i is a Gaussian-distributed unbiased estimator of
yi with estimated variance
pffiffiffiffiffiffiffiffiffiffiffi
s^ 2y^ i given by Eq. (18.3.54), the quantity (^yi yi )=(s^ y^ i = n 2) is Student-t-distributed with (n – 2) degrees of freedom. Hence, we can form equations for 100(1 –
a)% confidence intervals as
y^ i yi
P ta=2,(n2) ,
t1a=2,(n2) ¼ 1 a ¼ g, i ¼ 1, . . . , n
(18:4:16)
s^ y^ i
where t1a=2,(n2) is obtained from the t table (Table 18.4.2) for (n – 2) degrees of
freedom and probability (1 – a/2). Equation (18.4.16) can be rewritten as
P{^yi t1a=2,(n2) s^ y^ i , yi y^ i ta=2,(n2) s^ y^ i } ¼ 1 a,
i ¼ 1, . . . , n (18:4:17)
and the lower bound of the 100(1 – a)% confidence region for yi is given by
Ly^ i ¼ y^ i t1a=2,(n2) s^ y^ i
(18:4:18)
with the upper bound of the region given by
Uy^ i ¼ y^ i ta=2,(n2) s^ y^ i
(18:4:19)
Example 18.4.6 In Example 18.4.5 we will find the 90% confidence region for the
regression line given by
y^ i ¼ 1:574 þ 0:404xi ,
i ¼ 1, . . . , 20
From Example 18.3.3 the variance of the regression line is given by
1 (xi 0:905)2
þ
s^ 2y^ i ¼ 0:263
, i ¼ 1, . . . , 20
20
3:7665
This value of s^ 2y^ i can be substituted in Eq. (18.4.18) to obtain the lower bound of the 90%
confidence region
1=2
1 (xi 0:905)2
þ
, i ¼ 1, . . . , 2
Ly^ i (xi ) ¼ 1:574 þ 0:404xi 1:7341 0:263
20
3:7665
18.5
Hypothesis Testing (Binary)
373
FIGURE 18.4.4
and the upper bound of the region can be obtained from Eq. (18.4.19):
1=2
1 (xi 0:905)2
þ
, i ¼ 1, . . . , 20
Uy^ i (xi ) ¼ 1:574 þ 0:404xi 1:7341 0:263
20
3:7665
The confidence region is shown in Fig. 18.4.4 along with the regression line y^ i and the
data points.
B 18.5
HYPOTHESIS TESTING (BINARY)
We have discussed some estimation problems in the previous sections. However, many
probability models require us to make decisions about populations on the basis of
limited knowledge. For example, a physician has to make a decision based on knowledge
of her patient whether to give a particular drug. Usually this decision is binary in nature,
that is, to give or not to give. Thus, we have a decision problem as opposed to an estimation problem. Such decision problems come under the category of hypothesis testing.
We postulate a carefully formed assertion based on available information. This assertion
is called the null hypothesis, H0. If we are interested only in accepting it or rejecting it at a
significance level a, then it is called significance testing. On the other hand, we may be
interested in postulating an alternate hypothesis H1 and finding methods of accepting
H0 or H1. This is called hypothesis testing. For example, we may be interested in
testing the null hypothesis H0 that the mean time between failures u (MTBF) of a
memory chip is u0 hours. The alternate hypothesis may be u ¼ u1 or u, u0. We are not
particularly interested in the alternate hypothesis u . u0 because it does not matter
whether the chip overperforms. On the other hand, in the communication receiver we
can hypothesize that the noise variance is u0. We are now interested in the alternate
hypothesis u . u0 because we do not want the variance to be too high, while we are
not interested in u , u0 because it does not affect the performance for low variances.
The experiment on which we test a null hypothesis is based on an n random vector X,
fXi, i ¼ 1, . . . , ng from a population as in the estimation problem. The n random vector will
span an n-dimensional Euclidean space R n. We divide this n-space of observations into
two mutually exclusive regions, R(H0) and R(H1). If the observed vector x, fxi,
i ¼ 1, . . . , ng, called the test statistic, lies in the space R(H1), we reject the null
374
Estimation Theory
TABLE 18.5.1
True Situation
H0
H0
H0
H0
true, H1 false
true, H1 false
false, H1 true
false, H1 true
Accept or Reject
Decision Type
Accept H0
Reject H0
Accept H0
Reject H0
Correct decision—no error
Type I error—a
Type II error—b
Correct decision—no error
hypothesis H0. On the other hand, if it falls in the region R(H0), we may not reject H0. The
region R(H0) is called the acceptance region, and R(H1) is called the critical region or
rejection region. Whether we accept or reject the null hypothesis, there is always the possibility of errors. The error created when H0 is rejected when it is true is called type I error
and is denoted by a, which is also known as the significance level. Other names for this
error are false alarm and producer’s risk. The error created when H0 is accepted when
it is false is called type II error and is denoted by b. Other terms for this error are
missed alarm and consumer’s risk. The four possibilities, two of which are correct
decisions and the other two, erroneous decisions, are shown in Table 18.5.1.
A simple hypothesis is one in which the parameters are specified exactly. If they
are not, it is called a composite hypothesis. A null hypothesis is almost always
simple. For example, H0 ¼ u0 is simple whereas H1 . u0 is composite. In testing
hypotheses we usually control the significance level a a priori and try to minimize
the type II error b.
Example 18.5.1 (Significance Testing) A pharmaceutical company wants to test the
effectiveness of the new antihypertensive drug that they are developing. A group of
n ¼ 25 volunteers are selected with a mean systolic pressure of m25 ¼ 150 mmHg with
a standard deviation of s25 ¼ 25 mmHg. The company wants to see whether the new
drug can bring the systolic pressure below 140 mmHg. We have to design a test to determine whether the drug is effective at a significance level of a ¼ 0.01.
We will assume that the systolic pressure is a Gaussian-distributed random variable X
with mean mX ¼ 150 and standard deviation sX ¼ 25. Hence the sample mean m^ 25 (X) of
systolic pressures afterptaking
the drug is Gaussian-distributed with mean 150 and standard
ffiffiffiffiffi
deviation sm^ X ¼ 25= 25 ¼ 5 as shown in Fig. 18.5.1. We have to form the null
FIGURE 18.5.1
18.5
Hypothesis Testing (Binary)
375
hypothesis H0 carefully. If we choose the null hypothesis that the sample mean m^ 25 (X)
after taking the drug is the same as the mean value of 150, then the drug has minimal
effect on the systolic pressure. If we now choose a region R such that
P{m^ 25 (X) [ R} ¼ P{m^ 25 (X) ra } ¼ 0:01
and the null hypothesis is rejected at a significance level of 1%, then we can conclude that the
drug is effective in controlling systolic blood pressure. From Eq. (18.4.2) the test statistic is
Z¼
m^ X (X) mX m^ X (X) mX
pffiffiffi
¼
sm^ X
sX = n
and we can write a one-sided equation:
m^ X mX
m^ X 150
p
ffiffi
ffi
za ¼ a ¼ 0:01
P
za ¼ P
5
sX = n
Referring to Table 18.4.1 the value of za corresponding to a probability of 0.01 is
22.326. Hence
m^ X 150
¼ 2:326
5
or m^ X ¼ 150 5 2:326 ¼ 138:37
so that ra ¼ 138.37 as shown in Fig. 18.5.1. Thus, we can reject the hypothesis H0 that the
drug is ineffective in controlling the systolic blood pressure if the sample mean of the pressure
measurements drops below 139.
On the other hand, if we increase the significance level to 2.5%, then z0.025 ¼ 21.96
and
m^ X 150
¼ 1:96
5
or
m^ X ¼ 150 5 1:96 ¼ 140:2
and we cannot reject the hypothesis at the higher significance level of 2.5% and conclude
that the drug is ineffective at that level. Thus, the significance level has to be fixed before
the test begins.
We shall now consider several cases of testing hypotheses. The form of the density
function of a random variable X is a known function fX (x,u) that depends on a
parameter u.
Case 1: H0 ¼ u0 and H1 ¼ u1. Assuming a Gaussian random variable, consider a
simple hypothesis H0 ¼ u0 against a simple alternate H1 ¼ u1. The decision to
accept or reject H0 is dependent on the test statistic x obtained from the population.
The probability densities fX(x,u0) and fX(x,u1) and the error regions a and b are
shown in Fig. 18.5.2. If x , c, the null hypothesis H0 will be accepted, and if
x . c, H0 will be rejected. The critical region R(H1) is x . c. The type I error a
depends on the hypothesis under test H0, and the type II error b depends on both
H0 and the alternate hypothesis H1. We can also see from Fig. 18.5.2 that we
cannot minimize both types of error simultaneously since decreasing a increases
b and vice versa.
Case 2: H0 ¼ u0 and H1 ¼ u . u0 (One-Sided Test). In this case we have a simple
hypothesis H0 ¼ u0 and a composite alternate H1 ¼ u . u0. The acceptance
region for H0 is x , c. Here u ranges from u0 to 1 and the error probability b
376
Estimation Theory
FIGURE 18.5.2
will be a function of u. The Gaussian probability densities along with the errors a
and b(u) are shown in Fig. 18.5.3. The acceptance and rejection regions are also
shown in figure. The critical region will still be x . c, but the type II error function
b(u) will depend on the closeness of u to u0. The function b(u) is called the operating characteristic (OC) (Fig.18.5.4) of the test. The OC shows the probability of a
wrong decision. It can be visualized by sliding the density function fX(x,u) under H1
to the left starting from u0 and integrating it from 1 to c, or
ðc
ðc
2
1
pffiffiffiffiffiffi e(1=2)½(xu)=s dx
b(u) ¼
fX (x,u)dx ¼
(18:5:1)
2p:s
1
1
The complementary function P(u) ¼ 1 b(u) is called the power of the test. It
shows the probability of a correct decision. The operating characteristic (OC) is
shown in Fig. 18.5.4.
Case 3: H0 ¼ u0 and H1 = u0 (Two-Sided Test). Here, we have a simple hypothesis
H0 ¼ u0 and a composite alternate H1 ¼ {u = u0 g. The acceptance region
is c1 , x , c2. The significance level a is split between the two tails of the
FIGURE 18.5.3
18.5
Hypothesis Testing (Binary)
377
FIGURE 18.5.4
FIGURE 18.5.5
FIGURE 18.5.6
density under H0 so that P{m(x)
^
, za=2 } ¼ c1 and P{m(x)
^
, z1a=2 } ¼ c2 . Here
u ranges from c1 to c2, and the error probability b will be a function of u. The
Gaussian probability densities along with the errors a and b(u) are shown in
Fig. 18.5.5.
378
Estimation Theory
The critical region is x , c1 and x . c2, but the type II error function b(u) (OC)
extends from 1 to 1 (Fig. 18.5.6) and the value will depend on the closeness of u to
u0. It can again be visualized by sliding the density function fX(x, u) under H1 to the
left, starting from c1 and integrating it from c1 to c2:
ð c2
ð c2
2
1
pffiffiffiffiffiffi e(1=2)½(xu)=s dx
b(u) ¼
fX (x,u)dx ¼
(18:5:2)
2p:s
c1
c1
The acceptance region is also shown in Fig. 18.5.5 and the two-sided OC curve, in
Fig. 18.5.6.
Procedure for Hypothesis Testing
The following procedures are established for credible testing of hypotheses:
1. Null hypothesis H0 must be carefully formulated.
2. Define the experiment to be performed.
3. Choose a proper test statistic.
4. Establish the significance level a.
5. Choose a value for type II error b.
6. With a and b specified, determine a suitable sample size n.
We will now apply hypothesis testing to the mean and variance of a population. We
will have n realizations {x1 , . . . , xn ) of a Gaussian random variable
P X whose mean m we
want to hypothesize. The sample mean m^ is, as usual, m^ ¼ (1=n) n xi , and the variance
of the sample mean s^ 2 ¼ s2 =n. The density of m^ will be given by
pffiffi
1
(1=2)½(xm)=(s= n)2
fm^ (x) ¼ pffiffiffiffiffiffi
(18:5:3)
pffiffiffi e
2p s= n
These quantities will be used to form the test statistic.
Mean m—known Variance s
We will now test the hypothesis that the mean m is H0 ¼ m0 against the alternates: (1)
H1 ¼ m , m0 , (2) H1 ¼ m . m0 , and (3) H1 ¼ m = m0 .
The test statistic we will use is the random variable
Z¼
m^ m0
pffiffiffi
s= n
(18:5:4)
with a significance level of a. Under the hypothesis H0, the random variable Z is Gaussian
with zero mean and unit variance. Note that za for which P(Z za ) ¼ a will be negative
for a , 0:5 and z1a for which P(Z z1a ) ¼ 1 a will be positive, and if the pdf is
symmetric, then za ¼ z1a .
1. H1 ¼ m , m0 : Accept H0 if Z . za. Or, from Eq. (18.5.4)
Z¼
m^ m0
pffiffiffi . za
s= n
or
pffiffiffi
m^ . m0 þ za s= n
(18:5:5)
pffiffiffi
The acceptance region R(H
pffiffiffi 0 ) ¼ {m^ . m0 þ za s= n} and thepffiffiffirejection region
R(H1 ) ¼ {m^ , m0 þ za s= n}. The critical point c ¼ m0 þ za s= n} The OC curve
18.5
Hypothesis Testing (Binary)
is obtained from Eq. (18.5.1):
ð1
ð1
pffiffi
1
(1=2)½(xm)=(s= n)2
pffiffiffiffiffiffi
b(m) ¼
fm^ (x, m)dx ¼
dx
pffiffiffi e
2p s= n
c
c
1
1 m0 m
p
ffiffi
ffi
p
ffiffi
ffi
1 erf
¼
þ za
2
2 s= n
379
(18:5:6)
where
ðx
erf(x) ¼
2
2
pffiffiffiffi ej dj
p
0
(18:5:7)
2. H1 ¼ m . m0 : In this case we accept H0 if Z . z12a. Or
Z¼
m^ m0
pffiffiffi , z1a
s= n
pffiffiffi
or m^ . m0 þ z1a s= n
(18:5:8)
pffiffiffi
The acceptance region R(Hp
^ , m0 þ z1a s= n} and the rejection
region
0 )ffiffiffi¼ {m
pffiffiffi
R(H1 ) ¼ {m^ . m0 þ z1a s= n}. The critical point c ¼ m0 þ z1a s= n
The OC curve is obtained from Eq. (18.5.1):
ðc
ðc
pffiffi
1
(1=2)½(xm)=(s= n)2
pffiffiffiffiffiffi
b(m) ¼
fm^ (x, m)dx ¼
dx
pffiffiffi e
2p s= n
1
1
¼
1
1 m0 m
pffiffiffi þ z1a
1 þ erf pffiffiffi
2
2 s= n
(18:5:9)
where erf(x) is given by Eq. (18.5.7)
3. H1 ¼ m = m0 : In the previous two cases we had one-sided hypothesis testing,
whereas here we will have two-sided hypothesis testing. As seen previously, the significance level a is split evenly between the two tails of the density under H0 so that the probability is a/2 on each tail. We now accept H0 if za=2 , Z , z1a=2 , or
za=2 ,
m^ m0
pffiffiffi , z1a=2
s= n
or
pffiffiffi
pffiffiffi
m0 þ za=2 s= n , m^ , m0 þ z1a=2 s= n (18:5:10)
pffiffiffi
pffiffiffi
The acceptance region R(H0 ) ¼ {m0 þ za=2 s=
n} and the
pffiffiffi n , m^ , m0 þ z1a=2pffiffis=
ffi
rejection region R(H1 ) ¼ {m^ p
,ffiffiffim0 þ za=2 s= n m^ . m0 p
þffiffizffi1a=2 s= n}. The critical
points are c1 ¼ m0 þ za=2 s= n and c2 ¼ m0 þ z1a=2 s= n. The OC curve is obtained
from Eq. (18.5.2):
ð c2
ð c2
pffiffi
1
(1=2)½(xm)=(s= n)2
pffiffiffiffiffiffi
fm^ (x,m)dx ¼
dx
b(m) ¼
pffiffiffi e
2p s= n
c1
c1
1
1 m m
1 m m
¼ erf pffiffiffi 0 pffiffiffi þ z1a=2 erf pffiffiffi 0 pffiffiffi þ za=2
(18:5:11)
2
2 s= n
2 s= n
where erf(x) is as given by Eq. (18.5.7).
Mean m—Unknown Variance s2
The procedure with an unknown variance is similar to the one with the known variance discussed earlier. We replace the unknown variance with the estimated variance
380
Estimation Theory
s^ 2X and use the Student-t distribution for analysis. The test statistic is
m^ m0
pffiffiffi
T¼
s=
^ n
(18:5:12)
and this is Student-t-distributed with (n 2 1) degrees of freedom. The critical regions for
the three cases discussed earlier are
1. H1 ¼ m , m0
Acceptance region:
Rejection region:
2. H1 ¼ m . m0
t(n1),a s^
pffiffiffi
R(H0 ) ¼ m^ . m0 þ
n
t(n1),a s^
pffiffiffi
R(H1 ) ¼ m^ , m0 þ
n
(18:5:13)
Acceptance region:
Rejection region:
t(n1),ð1aÞ s^
pffiffiffi
R(H0 ) ¼ m^ , m0 þ
n
t(n1),ð1aÞ s^
pffiffiffi
R(H1 ) ¼ m^ . m0 þ
n
(18:5:14)
3. H1 ¼ m = m0
t(n1),a=2 s^
t(n1),(1a=2) s^
pffiffiffi
pffiffiffi
Acceptance region: R(H0 ) ¼ m0 þ
, m^ , m0 þ
n
n
(18:5:15)
)
t(n1),(a=2) s^
t(n1),(1a=2) s^
pffiffiffi
pffiffiffi
Rejection region: R(H1 ) ¼ m^ , m0 þ
; m^ . m0 þ
n
n
(
(18:5:16)
To determine the OC curves in these three cases we have to integrate a noncentral t
density, and this is beyond the scope of our discussion. However, for large n, the t distribution becomes Gaussian and the OC curves can be approximated by Eqs. (18.5.6),
(18.5.9), and (18.5.11).
Example 18.5.2 The following diastolic blood pressure measurements were recorded
for a patient on 10 different occasions. The pressure measurements are assumed to be
Gaussian-distributed:
BP measurements
78
85
88
85
78
86
79
82
80
81
We want to find whether these measurements support the hypothesis that the pressure is
less than 80 at the significance levels of (1) 1%, (2) 2.5%, and (3) 5%.
This problem is a little different from what we discussed
earlier because both hypotheses
P
1
are composite. The estimated mean value m^ ¼ 10
(78 þ þ 81) ¼ 82:2 and the estimated standard deviation s^ ¼ 3:584. We choose the statistic given by Eq. (18.5.12)
m^ m0
pffiffiffi
T¼
s=
^ n
which is Student-t-distributed. The t values for 1%, 2.5%, and 5% significance levels for
(10 2 1) ¼ 9 degrees of freedom from the Student-t table (Table 18.4.2) are 2.8214,
18.5
Hypothesis Testing (Binary)
381
2.2622, and 1.8331 respectively. We can now formulate the composite hypotheses as (1) null
hypothesis H0 : m , m0 ¼ 80; and (2) alternate hypothesis H1 : m m0 ¼ 80:
The acceptance regions from Eq. (18.5.14) are
pffiffiffi
1% significance: R(H0 ) ¼ {m^ , m0 þ t(n1), (1a) s=
^ n}
3:5839
¼ 80 þ 2:8214 pffiffiffiffiffi ¼ 83:2
10
pffiffiffiffi
2.5% significance: R(H0 ) ¼ m^ , 80 þ 2:2622 3:5839
¼ 82:564
10
pffiffiffiffi
5% significance: R(H0 ) ¼ m^ , 80 þ 1:8331 3:5839
¼ 82:078
10
Since the estimated mean is m^ ¼ 82:2 , 82:564 , 83:2, we conclude that the hypothesis with the diastolic pressure is 80 at significance levels of 1% and 2.5% can be accepted.
It cannot be accepted at the 5% level because m^ ¼ 82:2 . 82:078. We will not calculate
the OC curve b(m) since this involves noncentral Student-t distribution.
Example 18.5.3 An energy specialist recommends that a homeowner could save at least
$15 a month in utilities if energy appliances were upgraded. Before the upgrade was
installed, the mean monthly utility expense m0 was $255 with standard deviation
s ¼ $17. After the upgrade, the mean monthly expenses (in U.S. dollars) for a 10-year
period is shown below:
243
252
246 262 244
219
228 212
264
260
We have to test whether the upgrade has saved $15 a month at significance levels of (1)
1%, (2) 2.5%, (3) 5%.
We will assume that the utility expenses are Gaussian-distributed and formulate
the null hypothesis H0 as mean monthly expense m0 ¼ $255 and the alternate hypothesis H1
as mean monthly expense m1 ¼ $240. If H0 is rejected and H1 is accepted at significance
levels of a ¼ 0.01,0.025,0.05, then the homeowner would have saved $15 a month. The
1
estimated mean value m^ of the average monthly expense m^ ¼ 10
½243 þ þ 260 ¼
243 with s given as 17. The density functions of m^ for H0 and H1 are given by
pffiffiffiffi 2
1
pffiffiffiffiffi e(1=2)½(x255)=(17= 10)
fm^ (x j H0 ) ¼ pffiffiffiffiffiffi
2p 17= 10
pffiffiffiffi 2
1
pffiffiffiffiffi e(1=2)½(x240)=(17= 10)
fm^ (x j H1 ) ¼ pffiffiffiffiffiffi
2p 17= 10
and are shown in Fig. 18.5.7.
The test statistic is chosen as the normalized Gaussian
m^ m0 m^ 255
pffiffiffiffiffi
pffiffiffi ¼
Z¼
s= n
17= 10
and the simple hypotheses H0: m ¼ m0 ¼ 255 and H1: m ¼ m1 ¼ 240 at significance levels
of 1%, 2.5%, and 5% can be tested.
The corresponding z values from the Gaussian table (Table 18.4.1) are z0.01 ¼ 22.326,
z0.025 ¼ 21.96, and z0.05 ¼ 21.645 respectively. The acceptance regions from Eq. (18.5.6) are
pffiffiffi
1% significance: R(H0 ) ¼ {m^ . m0 þ za s= n}
pffiffiffiffiffi
¼ 255 2:326 (17= 10) ¼ 242:49
382
Estimation Theory
FIGURE 18.5.7
pffiffiffi
2:5% significance: R(H0 ) ¼ {m^ . m0 þ za s= n}
pffiffiffiffiffi
¼ 255 1:96 (17= 10) ¼ 244:47
pffiffiffi
5% significance: R(H0 ) ¼ {m^ . m0 þ za s= n}
pffiffiffiffiffi
¼ 255 1:645 (17= 10) ¼ 246:16
pffiffiffi
The critical point c ¼ m0 þ za s= n.
The 5% significance region and the estimated mean m^ ¼ 243 are shown in Fig. 18.5.7.
Since the estimated mean is m^ ¼ 243 . 242:5, we conclude that the null hypothesis
m0 ¼ $255 can be accepted at the 1% level with consequent no average monthly
savings of $15. On the other hand, m^ ¼ 243 , 244:47 , 246:16, and the null hypothesis
is rejected at the 2.5% and 5% levels and the alternate hypothesis m1 ¼ 240 is accepted,
resulting in an average monthly savings of $15.
The OC curve b(m,za) for significance level a is obtained from Eq. (18.5.6) as
1
1 m0 m
1
1 255 m
pffiffiffi þ za
1 erf pffiffiffi
1 erf pffiffiffi
þ za
b(m, za ) ¼
¼
2
2
2 s= n
2 5:4259
The type II errors for a ¼ 0.01, 0.025, 0.05 are obtained by substituting m ¼ 240 and
z0.01 ¼ 22.326, z0.025 ¼ 21.96, z0.05 ¼ 21.645 in the equation above. Thus, b(240,
22.326) ¼ 0.3214, b(240, 21.96) ¼ 0.2032, b(240, 21.645) ¼ 0.1261. The operating
characteristic curves shown in Fig. 18.5.8 are obtained from plotting b(m,za) for 1%,
2.5%, and 5% values of za.
Number of Samples n for Given a and b. In testing hypotheses the usual practice is
to fix the type I error a at a value of 2.5% or 5% and minimize the type II error b.
The only way to minimize b is to increase the sample size n. For the case of H0:
m ¼ m0 and H1 ¼ m , m0, the critical point c for Gaussian distributions is found
to be
pffiffiffi
c ¼ m0 þ za s= n
(18:5:17)
If the allowable b is also specified, we can then obtain the required number of
samples n.
18.5
Hypothesis Testing (Binary)
383
FIGURE 18.5.8
Referring to Fig. 18.5.9, we require n to be such that
ð1
pffiffi
1
(1=2)½(xm1 )=(s= n)2
pffiffiffiffiffiffi
dx b
pffiffiffi e
2p s= n
c
pffiffiffi
Substituting j ¼ (x m1 )=(s= n) in this equation, we obtain
ð1
2
1
pffiffiffiffiffiffi e(1=2)j dj b
cm1
2p
pffi
s= n
(18:5:18)
(18:5:19)
If the integral in Eq. (18.5.19) is to be b, then the lower limit
c m1
pffiffiffi . z1b
s= n
(18:5:20)
where z12b is obtained from the Gaussian tables for a probability of b. Substituting
for c from Eq. (18.5.17) in Eq. (18.5.20) and simplifying, we have
m0 m1
pffiffiffi . z1b za
s= n
FIGURE 18.5.9
384
Estimation Theory
FIGURE 18.5.10
Thus the sample size n is given by
n.
s2 (z1b za )2
(m0 m1 )2
(18:5:21)
Example 18.5.4 We will use the data from Example 18.5.2 and find the required sample
size for a ¼ 0.05, and we are also given s ¼ 17, m0 ¼ 255, and m1 ¼ 240. As the number
of samples n is increased, the critical point c increases and b decreases and we stop at the
value of n that gives the desired value of b. In this example, for n ¼ 5,10,15 we obtain
c ¼ 242.49, 246.16, 247.78 and b ¼ 0.3715, 0.1261, 0.0382 respectively. These are
shown in Fig. 18.5.10. If we choose the value of b ¼ 0.1261 from the previous
example, we should get n ¼ 10. From Gaussian tables z0.05 ¼ 21.645 and
z12b ¼ z0.8739 ¼ 1.145. Substituting these values in Eq. (18.5.21), we obtain n .
½172 (1:145 þ 1:645)2 =(255 240)2 ¼ 9:9982 or n ¼ 10.
This corresponds to the 10 samples of Example 18.5.2.
B 18.6.
BAYESIAN ESTIMATION
The previous approaches of estimating a parameter u in the probability density f (x,u)
were based solely on the random samples fX1, . . . , Xng that are incorporated in the test
statistic. This methodology is called the classical or the objective approach. However,
in some problems we may have prior information about u that can be used to refine its
estimate. For example, the classical estimate of the probability u in a binomial distribution b(k;n,u), where k is the number of successes in n trials, is u^ ¼ k=n. We can
then view u as the realization of a random variable Q whose density function f (u) is
the prior information. The density of X will be a conditional density f (x j u). The joint
density of the sample fX1, . . . , Xng and the parameter Q can be given in terms of the
conditional density as
f (x1 , . . . , xn ,u) ¼ f (x1 , . . . , xn j u)f (u)
(18:6:1)
18.6. Bayesian Estimation
385
Using Bayes’ theorem, we can obtain the conditional density f (u j x1 , . . . , xn ) as
f (u j x1 , . . . ; xn ) ¼
Posterior
f (x1 , . . . , xn , u) f (x1 , . . . , xn , ju)
¼
f (u)
f (x1 , . . . , xn )
f (x1 , . . . , xn ) Prior
where the marginal density f (x1 , . . . , xn ) is given by
ð1
f (x1 , . . . , xn ) ¼
f (x1 , . . . , xn ,u) du
(18:6:2)
(18:6:3)
1
The density f (u j x1 , . . . , xn ) is called the posterior density. What we have accomplished is
modifying the prior density f(u) by knowledge of the joint density of the random samples
fX1, . . . , Xng to the posterior density. This method of estimation is called the subjective
approach to estimation or Bayesian estimation. It is used in medical imaging problems,
which are discussed in Chapter 23.
We will now present Bayes’ estimation of the probability p of an event where p is a
realization of a random variable X with probability density function fX( p) whose range is
[0,1]. Prior estimates of p can be obtained from
ð1
(18:6:4)
p^ ¼ pfX ( p)dp
0
To improve on the estimate of p, we conduct an experiment of tossing a die n times and
observing the number of aces to be k. We can apply Bayes’ theorem from Eq. (8.3.4) and
write the posterior density as
fX (p j B) ¼ Ð 1
0
P(B j X ¼ p)fX ( p)
P(B j X ¼ p)fX ( p)dp
where B ¼ fk aces in n trialsg. From binomial probability, we obtain
n k
p (1 p)nk
P(B j X ¼ p) ¼
k
Substituting Eq. (18.6.6) into Eq. (18.6.5), we obtain
n k
p (1 p)nk fX (p)
k
fX (p j B) ¼ Ð1 n
nk
k
fX (p)dp
0 k p (1 p)
(18:6:5)
(18:6:6)
(18:6:7)
The updated estimate of p can be obtained by substituting fX ( p j B) from Eq. (18.6.7) for fX
( p) in Eq. (18.6.4), and the result is
ð1
p^ post ¼ pfX (p j B)dp
(18:6:8)
0
If we now assume that fX( p) is uniformly distributed in f0,1] instead of a general
distribution in the range [0,1], Eq. (18.6.7) can be simplified. The integral
ð1
m!(n k)!
(18:6:9)
pm (1 p)nk dp ¼
(n þ m k þ 1)!
0
can be shown to be true using mathematical induction.
386
Estimation Theory
Substituting fX( p) ¼ 1 and from Eq. (18.6.9), we can evaluate fX(B) as
ð1
n!
k!(n k)!
1
n
¼
fX (B) ¼
pk (1 p)nk 1dp ¼
k
k!(n
k)!
(n
þ
1)!
n
þ
1
0
(18:6:10)
we can express the conditional density fX( p j B) as follows:
n!
pk (1 p)nk 1
(n þ 1)! k
k!(n k)!
p (1 p)nk
¼
fX (p j B) ¼
1
k!(n k)!
(n þ 1)
(18:6:11)
The posterior estimate for p is obtained from Eq. (18.6.8) as
ð
(n þ 1)! 1 kþ1
p (1 p)nk dp
p^ post ¼
k!(n k)! 0
and from Eq. (18.6.9) p^ post is given by
p^ post ¼
Example 18.6.1
(n þ 1)! (k þ 1)!(n k)! k þ 1
¼
k!(n k)!
(n þ 2)!
nþ2
(18:6:12)
We have to estimate the parameter Q in the following problem
Yi ¼ Q þ Wi ,
i ¼ 1, . . . , n
where fWig are independent identically distributed zero mean Gaussian random variables
with variance s2W and Q is also a Gaussian-distributed random variable with mean mQ and
variance s2Q . We want to find the Bayesian estimate m^ Q of Q.
From Eq. (18.6.2), the conditional density of Q conditioned on the n-vector random
variable Y can be given as
f (y, u) f (y j u)f (u)
f (y j u)f (u)
¼
¼ Ð1
f (y)
f (y)
1 f (y j u)f (u)du
Ð1
Ð1
Since 1 f (u j y)du ¼ ½1=f (y)
Ð 1 1 f (y j u)f (u)du ¼ 1, the factor f(y) is independent of u
and normalizes the integral 1 f (y j u)f (u)du to 1. The joint density function f (y j Q ¼
u) is governed by the joint density of W, and hence
f (u j y) ¼
f (y j u) ¼
n
Y
n
Y
2
1
pffiffiffiffiffiffi e(1=2)½(yi u)=sW 2psW
i¼1
Pn
1
(1=2)
½(yi u)=sW 2
i¼1
¼
e
n=2 n
(2p) sW
f (yi j u) ¼
i¼1
with
2
1
f (u) ¼ pffiffiffiffiffiffi e(1=2)(u=sQ )
2psQ
Multiplying f (y j u) and f (u), we have
" 2 #)
n
1 X
yi u 2
u
f (y j u)f (u) ¼ f (y, u) ¼
exp þ
2 i¼1 sW
sQ
(2p)(nþ1)=2 snW sQ
1
(
18.6. Bayesian Estimation
The factor
n
P
387
(yi u)2 can be written as,
i¼1
n
X
n
X
i¼1
i¼1
(yi u)2 ¼
n
X
( yi m^ Y þ m^ Y u)2 ¼
(yi m^ Y )2 þ n(m^ Y u)
i¼1
since the cross product vanishes. Substitution of this result in f (y,u) yields
(
" #)
n
1
1 X
yi m^ Y 2
f (y,u) ¼
exp 2 i¼1
sW
(2p)(nþ1)=2 snW sQ
1 n
1 2
2
exp (m^ u) þ 2 u
2 s2W Y
sQ
1 n
1 2
2
¼ C exp (m^ u) þ 2 u
2 s2W Y
sQ
2
2 1 n
2
2 nsQ þ sW
¼ C exp (m^ 2m^ Y u) þ u
2 s2W Y
s2W s2Q
ns2Q m^ 2Y
1 ns2Q þ s2W 2 2ns2Q m^ Y u
¼ C exp u 2
þ
2 s2W s2Q
nsQ þ s2W ns2Q þ s2W
where C is a function of only the sample values fyig and the parameters. Completing the
squares in u in the exponent and after some algebra, there results
8
32 9
2
2
>
>
ns
m
^
>
>
Y
>
>
>
6u 2 Q 2 7 >
=
<
7
ns
þ
s
16
W7
Q
f (y,u) ¼ C exp 6
sW s Q
7 >
> 26
>
4 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi 5 >
>
>
>
>
;
:
ns2 þ s2
W
Q
exp 1
nm^ 2Y
2
2 nsQ þ s2W
¼ K e(1=2)½(umu )=su 2
where the quantities, mu and su are defined by
mu ¼
ns2Q m^ Y
,
ns2Q þ s2W
sW sQ
su ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ns2Q þ s2W
and K is another function independent of u. The marginal density f(y) can be given as
ð1
pffiffiffiffiffiffi
pffiffiffiffiffiffi
2
1
pffiffiffiffiffiffi
f (y) ¼ K 2p su
e(1=2)½(umu )=su du ¼ K 2p su
2p su
1
Substituting for f(y,u) and f(y) in the posterior density f(ujy), we obtain
2
f (u j y) ¼
2
f (y,u) K e(1=2)½(umu )=su 1
pffiffiffiffiffiffi
¼
¼ pffiffiffiffiffiffi
e(1=2)½(umu )=su f (y)
2p su
K 2p su
388
Estimation Theory
Hence, the posterior density is also Gaussian-distributed with mean mu and variance s2u ,
and the conditional expectation of Q given the observed values of y using Bayesian
estimation techniques is
E½Q j y ¼ mu ¼
ns2Q m^ Y
ns2Q þ s2W
Note that using classical estimation techniques where there is no knowledge of the parameters of Q, the conditional expectation of Q given the observed values of y is
^ ¼ m^
E½Q j y ¼ Q
Y
obtained by substituting s2Q ¼ 1 in the Bayesian estimate mu.
Example 18.6.2 A machine of uncertain quality manufactures microchips. In
the absence of any information, we shall assume a priori that the bad microchips are
uniformly distributed in the interval [0,1]. We take a sample of 10 microchips and
find that 3 are bad. We want to estimate the probability of the bad microchips
manufactured.
3
The prior probability of bad microchips from classical method ¼ 10
¼ 0:3: We find
the posterior probability from Eq. (18.6.2). Here n ¼ 10 and k ¼ 3; hence the posterior
probability is (3 þ 1)/(10 þ 2) ¼ 0.3333, indicating that the proportion of bad chips is
worse than what we might obtain by the classical method. On the other hand, if the
2
number of bad chips out of 10 is only 2, then the prior will be 10
¼ 0:2. The Bayesian probability will yield (2 þ 1)/(10 þ 2) ¼ 0.25. In this case also, the proportion of bad chips is
worse than that obtained from the classical method.
Maximum A Posteriori Probability (MAP) Estimation
We will now discuss the important problem of maximizing the posterior probability to
arrive at a decision. A communication receiver with additive noise is given by
xi ¼ si þ wi ,
i ¼ 1, . . . , n, or
x¼sþw
(18:6:13)
where fxig is the observation process defined by the n-vector x, fsig is the signal process
defined by the n-vector s, and fwig is a zero mean random noise defined by the n-vector
w. Without loss in generality, we will assume that the signal vector is either all 0 or
1. We observe x and decide on one of the following two hypotheses:
H0 : xi ¼ wi , i ¼ 1, . . . , n,
or x ¼ w:
H1 : xi ¼ si þ wi , i ¼ 1, . . . , n or
0 transmitted
x ¼ s þ w:
1 transmited
where H0 is the null hypothesis and H1 is the alternate hypothesis. The prior probabilities
of these hypotheses are P(H0) and P(H1). The probability density functions of observing x
conditioned on H0 and H1 are fX(x j H0) and fX (x j H1). The posterior probabilities of H0
and H1 after observing x are P(H0 j x) and P(H1 j x). The decision will be based on maximizing the posterior probabilities. We choose the following decision criterion:
P(H0 j x) . P(H1 j x):
choose H0
P(H0 j x) , P(H1 j x):
choose H1
18.6. Bayesian Estimation
389
FIGURE 18.6.1
Or more succinctly, the decision criterion is
H
P(H1 jx) 1
_
P(H0 j x)
H0
1
(18:6:14)
If fHi, i ¼ 1,2g are the two hypotheses, then, from Bayes’ rule, we obtain
P(Hi j x) ¼
P(Hi , x) fX (x j Hi )P(Hi )
¼
,
fx (x)
fX (x)
i ¼ 0,1
(18:6:15)
Substituting Eq. (18.6.15) in Eq. (18.6.14), the decision rule becomes
fX (x j H1 )P(H1 )
H1
fX (x)
_
fX (x j H0 )P(H0 )
H0
fX (x)
1
or
H
fX (x j H1 )P(H1 ) 1
_
fX (x j H0 )P(H0 )
H0
1
(18:6:16)
Transposing the prior probabilities to the right of the inequality, we have
L(x) ¼
H
fX (x j H1 ) 1 P(H0 )
¼g
_
fX (x j H0 )
P(H1 )
H0
(18:6:17)
where L(x) is defined as the likelihood function or likelihood ratio and g is the threshold
value or the critical point. Since L(x) is the ratio of n-fold probability densities, we prefer
to take logarithm of L(x). The inequality of Eq. (18.6.17) will be preserved since the
logarithm is a monotone function. Thus the log likelihood ratio ln½L(x) is given by
H1 P(H0 )
ln½L(x) _ ln
¼ ln g
P(H1 )
H0
(18:6:18)
The inequality given by this equation maximizes the posterior probability.
We can now represent the decision receiver as shown in Fig. 18.6.1.
Minimization of Average Probability of Error
In the previous section the posterior probability was maximized to arrive at a decision
rule. We can approach this problem through minimizing the average probability of error.
The probabilities of missed alarm a and false alarm b are given by
a ¼ P(accept H1 j H0 true) ¼ P(H1 j H0 )
b ¼ P(accept H0 j H1 true) ¼ P(H0 j H1 )
a, b, and the acceptance and rejection regions are as shown in Fig. 18.6.2.
(18:6:19)
390
Estimation Theory
FIGURE 18.6.2
The probability of average error Pe is given by
Pe ¼ P(average error) ¼ P(H1 j H0 )P(H0 ) þ P(H0 j H1 )P(H1 )
ð
ð
¼ P(H0 )
fX (x j H0 )dx þ P(H1 )
fX (x j H1 )dx
R1
(18:6:20)
R0
If R is the entire region of integration, then R ¼ R0 þ R1 and
ð
ð
fX (x j H0 )dx ¼ fX (x j H1 )dx ¼ 1
R
(18:6:21)
R
Using Eq. (18.6.21), we can convert Eq. (18.6.20) into integration over the region R0
ð
ð
Pe ¼ P(H0 )
fX (x j H0 )dx fX (x j H0 )dx
R
R0
ð
þ P(H1 )
fX (x j H1 )dx
R0
ð
¼ P(H0 ) þ
½P(H1 )fX (x j H1 ) P(H0 )fX (x j H0 )dx
(18:6:22)
R0
where P(H0) is always positive, and minimizing Pe depends on the function g(x) ¼
P(H1 )fX (x j H1 ) P(H0 )fX (x j H0 ) under the integral sign. A plot of g(x) is shown in
Fig. 18.6.3. From the figure, we obtain
P(H1 )fX (x j H1 ) . P(H0 )fX (x j H0 ):
g(x) is positive ) x [ R1
P(H1 )fX (x j H1 ) , P(H0 )fX (x j H0 ):
g(x) is negative ) x [ R0
(18:6:23)
Hence we have the decision criterion
H
fX (x j H1 )P(H1 ) 1
_
fX (x j H0 )P(H0 )
H0
1
or
H
fX (x j H1 ) 1 P(H0 )
_
fX (x j H0 )
P(H1 )
H0
which is exactly the same as Eq. (18.6.17) derived for the MAP criterion.
(18:6:24)
18.6. Bayesian Estimation
391
FIGURE 18.6.3
Maximum-Likelihood Estimation
We have already formulated the basic ML criterion in Eq. (18.1.7) as the maximization of the loglikelihood function ln L(x,u). We will discus this important estimation
criterion more formally. X is a continuous random variable with density function
fX (x;u), where u is a k-vector given by u ¼ [u1, u2, . . . , uk]T are parameters to be estimated
by observing n (n . k) independent observations x1, x2, . . . , xn of the random variable X.
The likelihood function L(x,u) is defined by
L(x,u) ¼
n
Y
fX (xi ,u)
i¼1
and the loglikelihood function, by
ln½L(x,u) ¼
n
X
ln ½fX (xi ,u)
(18:6:25)
i¼1
Maximizing the loglikelihood function ln[L(x,u)] with respect to fui, i ¼ 1, . . . , kg and
setting them equal to zero yields k equations given by
@
{ln½L(x,u1 , . . . , uk )} ¼ 0,
@ui
i ¼ 1, . . . , k
(18:6:26)
The estimates of the k parameters (u1,u2, . . . , uk) are obtained by solving the k equations in
Eq. (18.6.26). We will briefly discuss the properties of ML estimators:
1. They provide a consistent approach to parameter estimation problems.
2. They can be used to generate hypothesis tests for parameters.
3. The likelihood equations need to be specifically worked out for any given distribution and estimation problems.
4. The numerical estimation problem is usually nontrivial. Simple solutions are possible in only a few cases.
5. Maximum-likelihood estimates can be heavily biased for small samples.
Example 18.6.3 In an election we want to estimate the proportion p of voters who
will vote Democratic. We will assign 1 if a person votes Democratic and 0 if he
392
Estimation Theory
does not vote Democratic. The probability f (x;p) for each person is a Bernoulli distribution with
f (x;p) ¼ px (1 p)1x ,
x ¼ 0,1;
0p1
The value of p has to be estimated using the ML rule based on an opinion poll of n
voters chosen at random. Thus, there will be n sample values x1, x2, . . . , xn, each with
probability
f (xi ;p) ¼ pxi (1 p)1xi ,
xi ¼ 0,1; 0 p 1; i ¼ 1, . . . , n
The likelihood function and the loglikelihood functions are given by
n
Pn
Pn
Y
pxi (1 p)1xi ¼ p i¼1 xi (1 p)n i¼1 xi
L(x;p) ¼
i¼1
ln½L(x;p) ¼
n
X
xi ln(p) þ n i¼1
n
X
!
xi ln(1 p)
i¼1
Differentiating ln[L(x;p)] with respect to p, we have
P
Pn
n ni¼1 xi
@
i¼1 xi
ln½L(x;p) ¼
¼0
@p
1p
p
We obtain the estimate p^ by solving for p in this equation:
p^ ¼
n
1X
xi ¼ m^ X
n i¼1
where p^ is an unbiased estimator because E½^p ¼ E½m^ X ¼ p 1 þ (1 p) 0 ¼ p.
Example 18.6.4 Telephone calls coming into an exchange are Poisson-distributed with
parameter l and are given by
fX (x;l) ¼ el
lx
,
x!
x ¼ 0,1, . . .
where l is to be estimated using the ML rule by observing a random sample x1, x2, . . . , xn,
of n telephone calls, each with probability
fX (xi ;l) ¼ el
lxi
,
xi !
i ¼ 1,2, . . . , n
The likelihood function and the loglikelihood functions are given by
Pn
n
xi
Y
enl l i¼1 xi
l l
L(x;l) ¼
¼ Qn
e
xi !
i¼1 xi !
i¼1
ln½L(x;l) ¼ nl þ
n
X
i¼1
xi ln l n
X
ln (xi !)
i¼1
Differentiating ln[L(x; l)] with respect to l, we have
Pn
@
xi
ln½L(x; l) ¼ n þ i¼1 ¼ 0
@l
l
18.6. Bayesian Estimation
393
We obtain the estimate l^ by solving for l in the preceding equation:
1
l^ ¼
n
n
X
xi ¼ m^ X
i¼1
^ ¼ E½m^ X ¼ l.
where l^ is an unbiased estimator because E½l
Example 18.6.5 A sample of observations fxi, i ¼ 1, . . . , ng drawn from a popu2
lation governed by a form of Weibull density is given by fX (x) ¼ 2lxelx u(x) with
parameter l. We want to estimate lusing the ML criterion. The likelihood function is
given by
½L(x;l) ¼
n
Y
fX (xi ) ¼ (2l)n
i¼1
n
Y
xi el
Pn
x2
i¼1 i
,
xi . 0
i¼1
and the loglikelihood function by
ln½L(x;l) ¼ n ln(2) þ n ln(l) þ
n
X
ln(xi ) l
i¼1
n
X
x2i ,
xi . 0
i¼1
Differentiating ln[L(x;l)] with respect to l, we obtain
n
@
n X
ln½L(x; l) ¼ x2 ¼ 0
@l
l i¼1 i
Solving for l in this equation, we obtain the estimate l^ as
n
l^ ¼ Pn
2
i¼1 xi
¼
1
b
m2
where
m2 ¼ E½X 2 ^ 2 ¼ m2 , but E½1=b
We have E½X 2 ¼ m2 ¼ 1=l and E½m
m2 = 1=E½b
m2 ¼ 1=m2 ¼ l, and
hence l^ is a biased estimator.
Example 18.6.6 A random sample of n is drawn from a Gaussian-distributed population of mean mX and variance s2X. We want to estimate the mean and the variance
using the ML rule. The likelihood function and the loglikelihood function are shown
below:
n
Y
2
1
2
pffiffiffiffiffiffi e(1=2sX )(xi mX )
2p
s
X
i¼1
Pn
2
2
1
¼
e(1=2sX ) i¼1 (xi mX )
n=2 n
(2p) sX
L(x;mX , s2X ) ¼
n
n
n
1 X
ln½L(x;mX ,s2X ) ¼ ln(2p) ln(s2X ) 2
(xi mx )2
2
2
2sX i¼1
394
Estimation Theory
Differentiating with respect to mX and s2X , and setting them equal to zero, we have
n
@
1 X
ln½L(x;mX , s2X ) ¼ 2
(xi mX ) ¼ 0
@mX
sX i¼1
n
@
n
1 X
2
ln½L(x;m
,
s
)
¼
þ
(xi mX )2 ¼ 0
X
X
@s2X
2s2X 2s4X i¼1
Solving these equations, we find the estimators as follows:
m^ X ¼
n
1X
xi
n i¼1
and
s^ 2X ¼
n
1X
(xi mX )2
n i¼1
In these estimators, while the estimator for the mean m^ X is unbiased, the estimator for
the variance s^ 2X is biased because E½s^ 2X ¼ ½(n 1)=ns2X .
ML Rule in Hypothesis Testing
In the previous section we saw that the MAP rule minimizes the average probability of
wrong decisions using the prior probabilities PfHi,i ¼ 0,1g of the two hypotheses. In many
cases the priors may not be exactly known. In such a case, we choose whichever conditional probability P(x j H0) or P(x j H1) is greater, thereby making the assumption that
both hypotheses H0 and H1 are equiprobable. We will show later in an example that
when the priors are not known precisely, the decision errors will be more than that when
the priors are equiprobable. This decision rule where the priors are assumed equiprobable
is called the maximum-likelihood (ML) rule for testing hypotheses and is stated as follows:
P(x j H0 ) . P(x j H1 ):
P(x j H0 ) , P(x j H1 ):
accept H0
accept H1
(18:6:27)
In terms of density functions Eq. (18.6.27) can be given more succinctly as
H
fx (x j H1 ) 1
_
fx (xjH0 )
H0
1
(18:6:28)
In many applications the ML rule is widely used, especially when the average probability
of error Pe is low. Further, the missed-alarm probability is much more important than the
false-alarm probability. For example, missing the diagnosis of cancer has more serious
consequences than does diagnosing cancer when there is none. The MAP rule is prone to
yielding higher miss probabilities than is the ML rule as we will see in the following
example.
Example 18.6.7 The output x of a system consists of an input s ¼ 0 or 1 and a zero mean
additive noise w with variance s2W ¼ 14. The prior probability of sending s ¼ 0 is P(s ¼ 0) ¼
3
1
4 and that of s ¼ 1 is P(s ¼ 1) ¼ 4. We have to find the following using both the MAP and
the ML rules: (1) the decision threshold g, (2) the average probability of error Pe (x), and (3)
the probabilities of missed alarm and false alarm. We can formulate the hypotheses as
395
18.6. Bayesian Estimation
FIGURE 18.6.4
follows:
3
4
1
H1 : {1 transmitted} with P(H1 ) ¼
4
H0 : {0 transmitted} with P(H0 ) ¼
The conditional probabilities of x under the two hypotheses are
2
2
fX (x j H0 ) ¼ pffiffiffiffiffiffi e(1=2)½4x 2p
2
2
fX (x j H1 ) ¼ pffiffiffiffiffiffi e(1=2)½4(x1) 2p
and
These probability density functions are shown in Fig. 18.6.4.
The likelihood ratio L(x) is given by
2
fX (x j H1 ) e(1=2)½4(x1) 2
2
¼ (1=2)½4x2 ¼ e2½x 2xþ1x ¼ e2(2x1)
L(x) ¼
fX (x j H0 )
e
MAP Criterion. The decision rule from the loglikelihood function of Eq. (18.6.18) is
ln½e
2(2x1)
H1 3=4
_ ln
1=4
H0
or
H1
1 ln 3
x_ þ
2
4
H0
or
H1
x _ 0:7747
H0
Thus, the critical point for MAP is g ¼ 0.7747 as shown in Fig. 18.6.4. The probability of average error Pe(x) for any value of x from Eq. (18.6.22) is given by
ðx
PeMAP (x) ¼ P(H0 ) þ
½P(H1 )fX (j j H1 ) P(H0 )fX (j j H0 )dj
1
ð 3
2 x 1 2(j1)2 3 2j2
þ pffiffiffiffiffiffi
e
e
dj
4
4
2p 1 4
pffiffiffi
pffiffiffiffiffiffi
1 1
¼ þ
erf½ 2(x 1) 3 erf½ 2p ;
2 8
¼
2
erf(x) ¼ pffiffiffiffi
p
ðx
0
2
ej dj
396
Estimation Theory
The minimum error is obtained when x ¼ g ¼ 0.7747 and Pe(min) ¼ 0.127. The
probability of missed alarm is obtained from the following integral
ðg
PMAMAP ¼
2
2
pffiffiffiffiffiffi e2(x1) dx ¼ 0:3261
2p
1
and the probability of false alarm from
ð1
PFAMAP ¼
g
2
2
pffiffiffiffiffiffi e2x dx ¼ 0:0607
2p
These probabilities are also shown in Fig. 18.6.4. The function gMAP (x) under the
integral, that governs the extremum for PeMAP (x) is given by
2
2 1
3
2
gMAP (x) ¼ pffiffiffiffiffiffi e2(x1) e2x
4
2p 4
ML Criterion. We will now obtain the same quantities using the ML criterion. Here
the decision criterion becomes
H1
1
x+
2
H0
and the critical point for ML is 0.5. Hence the probability of error PeML (x) is given by
ð
1 1 x
½ fX (j j H1 ) fX (j j H0 ) dj
PeML (x) ¼ þ
2 2 1
ð
2
2
1
1 x
¼ þ pffiffiffiffiffiffi
½e2(j1) e2j dj
2
2p 1
h
hpffiffiffiffiffiii
pffiffiffi
1 1
¼ þ
erf
2(x 1) erf 2x
2 4
The minimum error is obtained when x ¼ g ¼ 0.5 and Pe(min) ¼ 0.159. The
probability of missed alarm is obtained from the following integral
ð 0:5
2
2
pffiffiffiffiffiffi e2(x1) dx ¼ 0:1587
PMAML ¼
2p
1
and the probability of false alarm from,
ð1
2
2
pffiffiffiffiffiffi e2x dx ¼ 0:1587
PFAML ¼
2p
0:5
Note that both probabilities are the same for the maximum-likelihood criterion. The
function gML(x) under the integral that governs the extremum for PeML (x) is given by
i
2
2 h
2
gML (x) ¼ pffiffiffiffiffiffi e2(x1) e2x
2p
The average probabilities of error PeMAP (x) and PeML (x) are shown in Fig. 18.6.5.
Table 18.6.1 compares the two decision criteria.
18.6. Bayesian Estimation
397
FIGURE 18.6.5
TABLE 18.6.1
Category
P(H0)—probability of null hypothesis
P(H1)—probability of alternate hypothesis
g—critical point
Pe(min)—minimum probability of average error
PMA—probability of missed alarm
PFA—probability of false alarm
MAP Rule
ML Rule
0.75
0.25
0.7747
0.127
0.3261
0.0607
0.5
0.5
0.5
0.159
0.1587
0.1587
Comparing the two methods, we see that the minimum probability of average error is
smaller for the MAP than for the ML. However, the probability of the more important
missed alarm is higher for MAP than for ML. Thus, in most communication problems
the ML criterion is the preferred method of estimation. The functions gMAP (x) and gML
(x) for both MAP and ML criteria are shown in Fig. 18.6.6.
FIGURE 18.6.6
398
Estimation Theory
In Fig. 18.6.6 the function gML(x) corresponding to the ML rule has an odd symmetry,
and hence it integrates to zero. The critical point is 12, as seen before.
Example 18.6.8 (Discrete Probability) An extrusion company has contracted to
supply plastic shopping bags to supermarkets. Ordinarily, the probability of tearing in
this group A0 of bags is 0.01. Because of a production malfunction, a second group A1
of bags have a higher probability of 0.05 of tearing. The prior probabilities of choosing
A0 or A1 are 0.7 and 0.3 respectively. To find out which group of bags came from A1, the
company tests a sample of n bags sequentially until it comes across a first failure. From
this information it must decide whether the bags came from group A0 or A1. Using both
the MAP and the ML criteria, we have to find (1) the number n of bags required to be
tested, (2) the probability of average error Pe(n), and (3) probabilities of missed and
false alarms PMA and PFA.
We will first approach the problem in generality and later substitute numbers to obtain
results. Let q0 be the tearing probability in group A0 and q1 be the tearing probability in
group A1. The two hypotheses can be defined by
H0 :
H1 :
bag came from group A0
bag came from group A1
(18:6:29)
The conditional probabilities of the waiting times upto the first tearing are geometrically
distributed and are given by
P(n j H0 ) ¼ q0 (1 q0 )n1 ;
P(n j H1 ) ¼ q1 (1 q1 )n1
(18:6:30)
MAP Criterion. According to the MAP criterion of Eq. (18.6.14), the decision rule is
P(H1 j n) H 1
_1
P(H0 j n) H 0
or
P(n j H1 )P(H1 ) H 1
_1
P(n j H0 )P(H0 ) H 0
or
P(n j H1 )P(H1 )
H1
P(n)
_1
P(n j H0 )P(H0 ) H 0
P(n)
or
(18:6:31)
P(n j H1 ) H 1 P(H0 )
_
P(n j H0 ) H 0 P(H1 )
Substituting the conditional probabilities from Eq. (18.6.30) into Eq. (18.6.31),
we have
q1 1 q1 n1 H 1 P(H0 )
_
q0 1 q0
H 0 P(H1 )
(18:6:32)
Taking logarithms on both sides of Eq. (18.6.32) and simplifying, we obtain
q0 P(H0 )
H 1 ln q1 P(H1 )
(n 1) _ 1 q1
H0
ln
1 q0
or
H1
n _ nMAP
H0
q0 P(H0 )
ln
q1 P(H1 )
¼1þ 1 q1
ln
1 q0
(18:6:33)
18.6. Bayesian Estimation
399
where nMAP is the critical value of n. Thus we have the decision criterion for n:
If n . nMAP : accept hypothesis H1
If n , nMAP : accept hypothesis H0
(18:6:34)
This MAP decision criterion gives us the minimum number n that are needed to be
tested sequentially.
Probabilities of Missed and False Alarms and Average Error. We will now calculate
the probabilities of missed and false alarms as functions of n. The probability of
missed alarm PMA(n) is given by
PMA (n) ¼ P(N . n j H1 ) ¼
1
X
q1 (1 q1 )k1
k¼nþ1
¼1
n
X
q1 (1 q1 )k1 ¼ 1 q1
k¼1
1 (1 q1 )n
¼ (1 q1 )n
1 (1 q1 )
(18:6:35)
where we have used the result for the finite summation of a geometric series. Similarly, the probability of false alarm PFA (n) is given by
PFA (n) ¼ P(N n j H0 ) ¼
n
X
q0 (1 q0 )k1
k¼1
n
¼ q0
1 (1 q0 )
¼ 1 (1 q0 )n
1 (1 q0 )
(18:6:36)
Finally, the probability of average error PeMAP(n) is given by
PeMAP (n) ¼ (1 q1 )n P(H1 ) þ ½1 (1 q0 )n P(H0 )
(18:6:37)
We obtain the various probabilities by substituting n ¼ nMAP in Eqs. (18.6.35) –
(18.6.37).
ML Criterion. For the ML criterion, the hypotheses H0 and H1 are equiprobable, and
substituting P(H0) ¼ P(H1) in Eq. (18.6.36), we obtain
q0
ln
H1
q
n _ nML ¼ 1 þ 1 (18:6:38)
1 q1
H0
ln
1 q0
with the decision criterion similar to that of Eq. (18.6.37).
Equations for the probabilities of missed and false alarms for the ML criterion
as functions of n are exactly the same as those of the MAP criterion except that we
substitute n ¼ nML. The probability of average error for the ML criterion can be
obtained from Eq. (18.6.37) by substituting P(H0) ¼ P(H1) ¼ 12, and the resulting
equation is
PeML (n) ¼
1 1
þ (1 q1 )n (1 q0 )n
2 2
(18:6:39)
Numerical Solutions. The conditional probability functions P(n j H0) ¼ 0.01 (1 2 0.01)n and P(n j H1) ¼ 0.05 (1 2 0.05)n are shown in Fig. 18.6.7.
400
Estimation Theory
FIGURE 18.6.7
Substituting for the MAP criterion, q0 ¼ 0.01, q1 ¼ 0.05, P(H0) ¼ 0.7, and
P(H1) ¼ 0.3 in Eq. (18.6.39), we obtain the critical value nMAP:
0:01 0:7
ln
0:05 0:3
¼ 19:4793
nMAP ¼ 1 þ 1 0:05
ln
1 0:01
Or, using the MAP criterion, the minimum number of bags to test sequentially
is n ¼ 20. The missed-alarm probability is obtained from Eq. (18.6.38): PMA (n) ¼
(1 q1 )n ¼ 0:95n , and the a error is 0.9520 ¼ 0.3585. The false-alarm probability
is obtained from Eq. (18.6.39): PFA (n) ¼ 1 (1 q0 )n ¼ 1 0:99n , and the b
error is 1 2 0.9920 ¼ 0.1821. The probability of average error PeMAP(n) is obtained
from Eq. (18.6.37): PeMAP (n) ¼ 0:95n 0:3 þ (1 0:99n ) 0:7 and PeMAP(min) ¼ PeMAP(20) ¼ 0.235.
For the ML criterion, we substitute q0 ¼ 0.01, q1 ¼ 0.05, P(H0) ¼ 0.5, and
P(H1) ¼ 0.5 in Eq. (18.6.38) and obtain the critical value nML:
0:01
ln
0:05
¼ 40:0233
nML ¼ 1 þ 1 0:05
ln
1 0:01
Or, using the ML criterion, the minimum number of bags to test sequentially is
n ¼ 41.
The missed-alarm probability is obtained from Eq. (18.6.38): PMA (n) ¼
1 (1 q1 )n ¼ 0:95n , and the a error is 0.9541 ¼ 0.1221. The false-alarm probability is obtained from Eq. (18.6.39): PFA (n) ¼ 1 (1 q0 )n ¼ 1 0:99n , and
the b error is 1 2 0.9941 ¼ 0.3377. The probability of average error PeMAP(n) is
obtained from Eq. (18.6.39): PeML (n) ¼ 12 þ 12 ½0:95n þ 0:99n and PeML(min) ¼
PeML(41) ¼ 0.2299.
Probabilities PeMAP (n) and PeML (n) are shown in Fig. 18.6.8 along with the
minimum number of bags needed for both the MAP and ML criteria. The results
indicate that the number of bags needed using the MAP is half the number
needed for ML. The more important missed-alarm probability for the ML criterion
18.6. Bayesian Estimation
401
FIGURE 18.6.8
is considerably lower than that of the MAP criterion. The ML criterion is more often
used in communication problems. The MAP criterion does find applications in
medical imaging problems such as single-photon emission computed tomography
(SPECT).
Hypothesis Testing with Added Cost for Errors
In the previous sections we never considered the costs associated with type I and
type II errors. We have already mentioned that the missed-alarm or type I error is more
serious than the false-alarm or type II error. We will now form a decision rule that
takes into account the costs associated with the errors. We shall postulate the true hypotheses by H0 and H1 and the define the accepted hypotheses by A0 and A1. There will be four
pairs of decisions, two of which, fA0,H0g and fA1,H1g, are correct decisions and the other
two of which, fA0,H1g (missed alarm) and fA1,H0g (false alarm) are incorrect decisions.
We will associate a cost Cij for each pair fAi,Hj,i,j ¼ 1,2g of the four decisions. The
average cost of the decisions C for the MAP criterion is
C ¼ C00 P(A0 , H0 ) þ C11 P(A1 ,H1 ) þ C01 P(A0 ,H1 ) þ C10 P(A1 ,H0 )
(18:6:40)
where P(Ai,Hj) i,j ¼ 1,2 represents the joint probability of accepting Ai, given that Hj is
true. We have to form a decision rule that minimizes the average cost. Expressing
Eq. (18.6.40) in terms of conditional probabilities, we have
C ¼ C00 P(A0 j H0 )P(H0 ) þ C01 P(A0 j H1 )P(H1 )
þ C11 P(A1 j H1 )P(H1 ) þ C10 P(A1 j H0 )P(H0 )
(18:6:41)
where C01 is the cost of the missed alarm and C10 is the cost of the false alarm. Substituting
conditional density functions for the conditional probabilities, there results
ð
ð
C MAP ¼ C00 P(H0 )
fX (x j H0 )dx þ C01 P(H1 )
fX (x j H1 )dx
R0
R0
ð
ð
þ C11 P(H1 )
fX (x j H1 )dx þ C10 P(H0 )
R1
fX (x j H0 )dx
R1
(18:6:42)
402
Estimation Theory
We can transform the integrals in Eq. (18.6.42) to be over the same region R0:
ð
ð
C MAP ¼ C00 P(H0 )
fX (x j H0 )dx þ C01 P(H1 )
fX (x j H1 )dx
R0
R0
ð
ð
fX (x j H1 )dx þ C10 P(H0 ) 1 fX (x j H0 )dx (18:6:43)
þ C11 P(H1 ) 1 R0
R0
Combining like terms in Eq. (18.6.43), we obtain
ð
CMAP ¼ C11 P(H1 ) þ C10 P(H0 ) þ ½P(H1 )(C01 C11 )fX (x j H1 )
R0
P(H0 )(C10 C00 )fX (x j H0 )dx
(18:6:44)
At this point it is natural to assume that the cost for an incorrect decision exceeds the cost
for a correct decision, or C10 . C00 and C01 . C11. With this assumption, the terms
P(H1 )ðC01 C11 ÞfX fX (x j H1 ) and P(H0 )ðC10 C00 ÞfX (x j H0 ) are both positive. Hence, if
the integrand I ¼ P(H1 )(C01 C11 ÞfX (x j H1 ) PðH0 Þ(C10 C00 )fX (x j H0 ) is negative,
then we can conclude that x [ R0 and choose H0. Thus the decision rule for the MAP
criterion is
fX (x j H1 )P(H1 )(C01 C11 ) H 1
_1
fX (x j H0 )P(H0 )(C10 C00 ) H 0
or
L(x) ¼
fX (x j H1 ) H 1 P(H0 )(C10 C11 )
_
fX (x j H0 ) H 0 P(H1 )(C01 C11 )
(18:6:45)
and the decision rule for the ML criterion can be obtained by substituting P(H0) ¼ P(H1) in
Eq. (18.6.45):
L(x) ¼
fX (x j H1 ) H 1 C10 C00
_
fX (x j H0 ) H 0 C01 C11
(18:6:46)
In Eqs. (18.6.45) and (18.6.46), C10 is the cost associated with a false alarm and C01 is for a
missed alarm. We weigh the missed alarm more heavily than the false alarm with the result
C01 . C10. Hence the ratio (C10 C00 =(C01 C11 ) will always be lesser than 1 if the costs
associated with right decisions are the same.
Example 18.6.9 We will reexamine Example 18.6.7, assigning cost functions to wrong
decisions. We will assume that the costs for the right decisions are the same,
C00 ¼ C11 ¼ 1, and the cost for a missed alarm to be C01 ¼ 4C00 and the cost for a
false alarm, C10 ¼ 2C00. The prior probabilities are P(H0) ¼ 34 and P(H1) ¼ 14. The conditional probabilities of x from Example 18.6.7 under the two hypotheses are
2
2
fX (x j H0 ) ¼ pffiffiffiffiffiffi e(1=2)½4x 2p
2
2
and fX (x j H1 ) ¼ pffiffiffiffiffiffi e(1=2)½4(x1) 2p
The likelihood ratio L(x) is
L(x) ¼
fX (x j H1 )
¼ e2(2x1)
fX (x j H0 )
18.6. Bayesian Estimation
403
From Eq. (18.6.45), the MAP criterion for minimization of cost functions is
H 1 P(H0 )(C10 C00 ) 3C00 (2 1)
¼
¼1
L(x) ¼ e2(2x1) _
C00 (4 1)
H 0 P(H1 )(C01 C11 Þ
The critical point from the loglikelihood function can be obtained as follows:
H1
2(2x 1) _ 0
H0
or
H1 1
x_
H0 2
or gMAP ¼
1
2
The average cost function C MAP (x) obtained from Eq. (18.6.44) is given by
ð
1
3
2 x
CMAP (x) ¼ C00 þ 2
C00
þ pffiffiffiffiffiffi
4
4
2p 1
2
2
1
3
(4 1)e2(j1) (2 1)e2j dx
4
4
n
h
i
pffiffiffi
pffiffiffi o
7 3
¼ C00 þ erf 2(x 1) erf 2x
4 8
Assuming that C00 ¼ 1, the minimum cost function C MAP(min) ¼ CMAP (0:5) ¼ 1:238.
The probability of missed alarm is obtained from the following integral
ð gMAP
2 2(x1)2
1
1
pffiffiffiffiffiffi e
dx ¼ 1 erf pffiffiffi ¼ 0:1587
PMAMAP ¼
2
2p
2
1
and the probability of false alarm, from
ð1
2 2x2
1
1
pffiffiffiffiffiffi e
dx ¼ 1 erf pffiffiffi ¼ 0:1587
PFAMAP ¼
2
2p
2
gMAP
Similarly, the ML criterion for minimization of cost functions can be obtained by substituting P(H0) ¼ P(H1) ¼ 12 in Eq. (18.6.44). The result is
H 1 C10 C00 C00 (2 1) 1
¼
¼
L(x) ¼ e2(2x1) _
H 0 C01 C11 C00 (4 1) 3
The critical point from the loglikelihood function can be obtained as follows:
H 1 1
H 1 1 ln (3)
2(2x 1) _ ln
or gML ¼ 0:2253
or _ 3
4
H0
H0 2
The average cost function C ML (x) is
ðx
2
2
C00
2
(1 þ 2) þ pffiffiffiffiffiffi
C00 (4 1)e2(j1) (2 1)e2j dx
2
2 2p 1
h
i 1 h pffiffiffi i
pffiffiffi
3
¼ C00 2 þ erf 2(x 1) erf 2x
4
4
CML (x) ¼
Assuming that C00 ¼ 1, the minimum cost function C ML(min) ¼ CML (0:2253) ¼ 1:254.
The probability of missed alarm is obtained from the following integral
ð gML
2
2
pffiffiffiffiffiffi e2(x1) dx ¼ 0:0607
PMAML ¼
2p
1
404
Estimation Theory
FIGURE 18.6.9
and the probability of false alarm, from
ð1
PFAML ¼
gML
2
2
pffiffiffiffiffiffi e2x dx ¼ 0:3261
2p
The probability density functions, the critical point gML, and the error regions for the ML
criterion are shown in Fig. 18.6.9. The cost functions C MAP (x) and C ML (x) along with their
critical points gMAP and gML are shown in Fig. 18.6.10.
The function gMAP(x) under the integral that governs the extremum for PeMAP(x) is
2
2
1 gMAP (x) ¼ pffiffiffiffiffiffi e2(j1) 3e2j
2 2p
FIGURE 18.6.10
18.6. Bayesian Estimation
FIGURE 18.6.11
TABLE 18.6.2
Category
P(H0)—probability of null hypothesis
P(H1)—probability of alternate hypothesis
C00 ¼ C11—cost of correct decisions
C01—cost of missed alarm
C10—cost of false alarm
g—critical point
C min —minimum cost of average error
PMA—probability of missed alarm
PFA—probability of false alarm
MAP Rule
ML Rule
0.75
0.25
1
4
2
0.5
1.238
0.1587
0.1587
0.5
0.5
1
4
2
0.2253
1.254
0.0607
0.3261
The function gML(x) under the integral, which governs the extremum for PeML(x), is
2
2
1 gML (x) ¼ pffiffiffiffiffiffi 3e2(j1) e2j
2 2p
These two functions gMAP(x) and gML(x) are shown in Fig. 18.6.11.
Table 18.6.2 compares the two decision criteria.
405
CHAPTER
19
Random Processes
B 19.1
BASIC DEFINITIONS
In the earlier sections a random variable X was defined as a function that maps every
outcome ji of points in the sample space S to a number X(ji) on the real line R. A
random process X(t) is a mapping that assigns a time function X(t,ji) to every outcome
ji of points in the sample space S. Alternate names for random processes are stochastic
processes and time series. More formally, a random process is a time function assigned
for every outcome j [ S according to some rule X(t,j), t [ T, j [ S, where T is an
index set of time. As in the case of a random variable, we suppress j and define a
random process by X(t). If the index set T is countably infinite, the random process is
called a discrete-time process and is denoted by Xn.
Referring to Fig. 19.1.1, a random process has the following interpretations:
1. X(j,t1) is random variable for a fixed time t1.
2. X(ji,t) is a sample realization for any point ji in the sample space S.
3. X(ji,t2) is a number.
4. X(j,t) is a collection or ensemble of realizations and is called a random process.
An important point to emphasize is that a random process is a finite or an infinite ensemble
of time functions and is not a single time function.
Example 19.1.1 A fair coin is tossed. If heads come up, a sine wave x1(t) ¼ sin(5pt) is
sent. If tails come up, then a ramp x2(t) ¼ t is sent. The resulting random process X(t) is an
ensemble of two realizations, a sine wave and a ramp, and is shown in Fig. 19.1.2. The
sample space S is discrete.
Example 19.1.2 In this example a sine wave is in the form X(t) ¼ A sin(vt þ F), where
F is a random variable uniformly distributed in the interval (0, 2p). Here the sample space
is continuous, and the sequence of sine functions is shown in Fig. 19.1.3.
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
406
19.1
FIGURE 19.1.1
FIGURE 19.1.2
FIGURE 19.1.3
Basic Definitions
407
408
Random Processes
Distribution and Density Functions. Since a random process is a random variable for
any fixed time t, we can define a probability distribution and density functions as
FX (x; t) ¼ P(j, t: X(j; t) x) for a fixed t
(19:1:1)
and
fX (x; t) ¼
@
FX (x þ Dx; t) FX (x; t)
FX (x; t) ¼ lim
Dx!0
@x
Dx
¼ lim P(x , X(t) x þ Dx)
Dx!0
(19:1:2)
These are also called first-order distribution and density functions, and in general,
they are functions of time.
Means and Variances. Analogous to random variables, we can define the mean of a
random process as
ð1
mX (t) ¼ E ½X(t) ¼
xfX (x; t)dx
(19:1:3)
1
and the variance as
s2X (t) ¼ E ½X(t) mX (t)2 ¼ E ½X 2 (t) m2X (t)
ð1
½x mX (t)2 fX (x; t)dx
¼
(19:1:4)
1
where
2
ð1
x2 fX (x; t)dx
E½X (t) ¼
(19:1:5)
1
Since the density is a function of time, the means and variances of random processes are also functions of time.
Example 19.1.3 We shall now find the distribution and density functions along with the
7
mean and variance for the random process of Example 19.1.1 for times t ¼ 0, 12 , 10
:
t ¼ 0,
x1 (0) ¼ 0,
x2 (0) ¼ 0
At t ¼ 0 the mapping diagram from the sample space to the real line is shown in
Fig. 19.1.4a along with the corresponding distribution and density functions.
The mean value is given by mX (0) ¼ 0; 12 þ 0 12 ¼ 0. The variance is given by
2
sX (0) ¼ (0 0)2 12 þ (0 0)2 12 ¼ 0:
1
1
1
1
t ¼ , x1
¼ 1, x2
¼
2
2
2
2
At t ¼ 12 the mapping diagram from the sample space to the real line is shown in
Fig. 19.1.4b along with the corresponding distribution and density functions.
The mean value is given by mX 12 ¼ 12 12 þ 1 12 ¼ 34 ¼ 0:75:
19.1
Basic Definitions
409
FIGURE 19.1.4
The variance is given by s2X
t¼
1
2
7
,
10
2
2
1
34 12 þ 1 34 12 ¼ 16
¼ 0:0625:
7
7
7
x1
¼ 1, x2
¼
10
10
10
¼
1
2
7
At t ¼ 10
the mapping diagram from the sample space to the real line is shown in
Fig. 19.1.4c along with the corresponding distribution and density functions.
7
3
The mean value is given by mX 10
¼ 0:15
¼ 7 1 1 1 ¼ 20
7 10 322 1 2
2 7
3 21
The variance is given by sX 10 ¼ 10 þ 20 2 þ 1 þ 20 2 ¼ 289
400 ¼ 0:7225:
Example 19.1.4 A random process, given by X(t) ¼ A sin(vt), is shown in Fig. 19.1.5,
where A is a random variable uniformly distributed in the interval (0,1].
410
Random Processes
FIGURE 19.1.5
The density and distribution functions of A are
fA (a) ¼
2
1, 0 , a 1
0, otherwise
0,
FA (a) ¼ 4 a,
1,
a0
0,a1
a.1
We have to find the distribution function FX (x; t). For any given t, x ¼ asin(vt) is an
equation to a straight line with slope sin(vt), and hence we can use the results of Examples
12.2.1 and 12.2.2 to solve for FX (x; t). The cases of sin(vt) . 0 and sin(vt) , 0 are shown
in Fig. 19.1.6.
Case 1: sin(vt) . 0. There are no points of intersection on the a axis for x 0, and
hence FX(x; t) ¼ 0. For 0 , x sin(vt) we solve x ¼ a sin(vt) and obtain
a ¼ x=½sin(vt). The region Ia for which a sin(vt) x is given by Ia ¼ {0 , a FIGURE 19.1.6
19.1
Basic Definitions
411
x=½sin(vt)} (Fig. 19.1.6). Thus
x
x
x
FX (x; t) ¼ FA
FA (0) ¼ FA
¼
sin(vt)
sin(vt)
sin(vt)
Finally for x . sin(vt), the region Ia for which asin(vt) x, is given by
Ia ¼ f0 , a 1g and FX (x; t) ¼ 1. Thus, for sin(vt) . 0, we have
2
0;
x0
6 x , 0 , x sin(vt)
FX (x; t) ¼ 4
sin(vt)
1,
x . sin(vt)
Case 2: sin(vt) , 0. The region Ia for which x . 0 is given by Ia ¼ f0 , a 1g, and
hence FX (x; t) ¼ 1. For 2jsin(vt)j , x 0, we solve x ¼ –ajsin(vt)j and obtain
a ¼ ½x=( j sin(vt)j). The region Ia for which –a jsin(vt)j x is given by
x
,a1
Ia ¼
j sin(vt)j
(Fig. 19.1.6). Thus,
FX (x; t) ¼ FA (1) FA
x
x
¼1
j sin(vt)j
j sin(vt)j
Finally, for x – jsin(vt)j, the region Ia for which –a jsin(vt)j x is given by
Ia ¼ f1 , a 1g and FX (x; t) ¼ 0. Thus, for sin(vt) , 0, we have
2
0,
x j sin(vt)j
x
61 , j sin(vt)j , x 0
FX (x; t) ¼ 4
j sin(vt)j
1,
x.0
Case 3: sin(vt) ¼ 0. The region Ia for which x . 0 is given by Ia ¼ f0 , a 1g and
FX (x; t) ¼ 1. For x 0, Ia ¼ 1 and FX (x; t) ¼ 0. Thus, for sin(vt) ¼ 0, we have
0, x 0
FX (x; t) ¼
1, x . 0
Example 19.1.5 We shall now find the distribution and density functions along with the
mean and variance for the random process of Example 19.1.2 and see how this process
differs from the previous ones.
We are given that X(t) ¼ A sin(vt þ F), where A is a constant and fF(f) ¼ 1/2p in
the interval (0, 2p) and we have to find fX (x; t). We will solve this problem by (1) finding
the distribution FX (x; t) and differentiating it, and (2) by direct determination of the
density function.
1. Determination of Distribution Function FX (x; t). The distribution function for F
is given by FF (f) ¼ (f=2p), 0 , f 2p. The two solutions for x ¼ A
sin(vt þ f) are obtained from the two equations:
sin(vt þ f1 ) ¼
x
A
and
sin(p vt f2 ) ¼
x
A
412
Random Processes
Hence the solutions are given by
x
f1 ¼ sin1
vt
A
and f2 ¼ p sin1
x
A
and are shown in Fig. 19.1.7.
For x –A, there are no points of intersection and hence FX (x; t) ¼ 0. For
2A , x A, the set of points along the f axis such that A sin(vt þ f) x is
(0, f1] < (f2, 2p]. Hence FX (x; t) is given by
FX (x; t) ¼ FF (f1 ) FF (0) þ FF (2p) FF (f2 )
h
x
io
1 n 1 x
sin
¼
vt 0 þ 2p p sin1
vt
2p
A
A
n
o
1
x
1
x
1
2 sin1
¼
þ p ¼ sin1
þ , A , x A
2p
A
p
A
2
Finally, for x . A, the entire curve A sin(vt þ f) is below x, and FX(x;t) ¼ 1.
2. Determination of Density Function fX (x; t)
(a) The two solutions to x ¼ A sin(vt þ f) have been found earlier.
(b) The absolute derivatives [email protected][email protected] and [email protected][email protected] are given by
@x
@f
f1
h
x
i
¼ A cos(vt þ f1 ) ¼ A cos vt þ sin1
vt
A
p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
h
x i
A2 x2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ A cos sin1
¼ A2 x 2
¼A
A
A
and
@x
@f
f2
h
x
i
¼ A cos(vt þ f2 ) ¼ A cos vt þ p sin1
vt
A
p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
h
x i
A2 x2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ A cos p sin1
¼A
¼ A2 x 2
A
A
FIGURE 19.1.7
19.1
Basic Definitions
413
(c) With the two solutions for x, the density function fX (x; t) is given by
Eq. (12.3.6):
1
1
1
1
þ
fX (x; t) ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi , A , x A
2
2
A x 2p 2p
p A2 x 2
Integration of fX (x; t) gives the distribution function FX (x; t)
2
3
0,
x A
x
1
61
7
FX (x; t) ¼ 4 sin1
þ , A , x A 5
p
A
2
1,
x.A
and these two solutions are exactly the same as before. For this random
process, we find that the density and the distribution functions are both
independent of time. For A ¼ 1, they become
1
fX (x; t) ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ,
1 , x 1
p 1 x2
2
3
0,
x 1
6 sin1 (x)
7
1
FX (x; t) ¼ 6
þ , 1 , x 1 7
4
5
p
2
1,
x.1
and these functions are shown in Fig. 19.1.8. Since fX (x) has even symmetry,
the mean value is 0 and the variance is obtained from
ð1
x2
1
pffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼
s2X ¼
2
2
1 p 1 x
pffiffiffi
The value of s ¼ +1= 2 is also shown in Fig. 19.1.8.
Later we will discuss random processes whose density and distribution functions are
independent of time.
FIGURE 19.1.8
414
Random Processes
Higher-Order Distribution Functions
If t1 and t2 are different times, then X1 ¼ X(t1) and X2 ¼ X(t2) are two different
random variables as shown in Fig. 19.1.9.
A second-order distribution function FX (x1 , x2 ; t1 , t2 ) for X1 and X2 can be defined as
FX (x1 , x2 ; t1 , t2 ) ¼ P½X(t1 ) x1 , X(t2 ) x2 (19:1:6)
and the second-order density function fX (x1 , x2 ; t1 , t2 ) as
fX (x1 , x2 ; t1 , t2 ) ¼
@2
FX (x1 , x2 ; t1 , t2 )
@x1 @x2
(19:1:7)
Similarly, if t1, . . . , tn are different times, then an nth-order distribution function is
defined as
FX (x1 , . . . , xn ; t1 , . . . , tn ) ¼ P½X(t1 ) x1 , . . . , X(tn ) xn (19:1:8)
and the nth-order density function as
fX (x1 , . . . , xn ; t1 , . . . , tn ) ¼
@n
FX (x1 , . . . , xn ; t1 , . . . , tn )
@x1 @xn
(19:1:9)
A random variable is completely defined if its distribution function is known.
Similarly, a random process X(t) is completely defined if its nth-order distribution is
known for all n. Since this is not feasible for all n, a random process in general cannot
be completely defined. Hence, we usually restrict the definition to second-order distribution function, in which case it is called a second-order process.
In a similar manner, we can define a joint distribution between two different random
processes X(t) and Y(t) (Fig. 19.1.10) as given below, where the joint second-order distribution function FXY (x1 , y2 ; t1 , t2 ) for X(t1) and Y(t2) is defined by
FXY (x1 , y2 ; t1 , t2 ) ¼ P½X(t1 ) x1 , Y(t2 ) y2 (19:1:10)
and the joint density function fXY (x1 , y2 ; t1 , t2 ) as
fXY (x1 , y2 ; t1 , t2 ) ¼
@2
FXY (x1 , y2 ; t1 , t2 )
@x1 @y2
FIGURE 19.1.9
(19:1:11)
19.1
Basic Definitions
415
FIGURE 19.1.10
The joint nth-order distribution function FXY (x1 , y2 ; t1 , t2 ) for X(t) and Y(t) is defined by
FXY (x1 , . . . , xn : y1 , . . . , yn ; t1 , . . . , tn )
¼ P½X(t1 ) x1 , . . . , X(tn ) xn : Y(t1 ) y1 , . . . , Y(tn ) yn (19:1:12)
and the corresponding nth-order density function, by
fXY (x1 , . . . , xn : y1 , . . . , yn ; t1 , . . . , tn )
¼
@2n
FXY (x1 , . . . , xn : y1 , . . . , yn ; t1 , . . . , tn )
@x1 @xn @y1 @yn
(19:1:13)
Second Order Moments
In Eqs. (19.1.3) and (19.1.4) defined mean and variance for a random process. The
second moment of a random process has also been defined in Eq. (19.1.5). Since X(t1)
and X(t2) are random variables, various types of joint moments can be defined.
Autocorrelation. The autocorrelation function (AC) RX (t1, t2) is defined as the
expected value of the product X(t1) and X(t2):
ð1 ð1
x1 x2 fX (x1 , x2 ; t1 , t2 )dx1 dx2
(19:1:14)
RX (t1 , t2 ) ¼ E½X(t1 )X(t2 ) ¼
1 1
By substituting t1 ¼ t2 ¼ t in Eq. (19.1.14), we can obtain the second moment or
the average power of the random process:
ð1
2
RX (t) ¼ E ½X (t) ¼
x2 fX (x; t)dx
(19:1:15)
1
Autocovariance. The autocovariance function (ACF) CX (t1, t2) is defined as the
covariance between X(t1) and X(t2):
CX (t1 , t2 ) ¼ E ½(X(t1 ) mX (t1 ))(X(t2 ) mX (t2 ))
ð1 ð1
¼
(x1 mX (t1 ))(x2 mX (t2 )) fX (x1 , x2 ; t1 , t2 )dx2 dx1
1 1
¼ E½X(t1 )X(t2 ) mX (t1 )mX (t2 )
(19:1:16)
416
Random Processes
Thus, from Eqs. (19.1.15) and (19.1.16) the interrelationships between AC and
ACF are
CX (t1 , t2 ) ¼ RX (t1 , t2 ) mX (t1 )mX (t2 )
RX (t1 , t2 ) ¼ CX (t1 , t2 ) þ mX (t1 )mX (t2 )
(19:1:17)
Normalized Autocovariance. The normalized autocovariance function (NACF)
rX (t1, t2) is the ACF normalized by the variance and is defined by
CX (t1 , t2 )
CX (t1 , t2 )
rX (t1 , t2 ) ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼
s2X (t1 )s2X (t2 ) sX (t1 )sX (t2 )
(19:1:18)
The NACF finds wide applicability in many problems in random processes, particularly in time-series analysis.
The following definitions pertain to two different random processes X(t) and Y(t):
Cross-Correlation. The cross-correlation function (CC) RXY (t1, t2) is defined as the
expected value of the product X(t1) and Y (t2):
ð1 ð1
RXY (t1 , t2 ) ¼ E½X(t1 )Y(t2 ) ¼
x1 y2 fXY (x1 , y2 ; t1 , t2 )dy2 dx1 (19:1:19)
1
1
By substituting t1 ¼ t2 ¼ t in Eq. (19.1.19), we can obtain the joint moment
between the random processes X(t) and Y(t) as
ð1 ð1
RXY (t) ¼ E½X(t)Y(t) ¼
xy fXY (x, y; t)dydx
(19:1:20)
1
1
Cross-Covariance. The cross-covariance function (CCF) CXY (t1, t2) is defined as the
covariance between X(t1) and Y(t2):
CXY (t1 , t2 ) ¼ E½(X(t1 ) mX (t1 ))(Y(t2 ) mY (t2 ))
ð1 ð1
¼
(x1 mX (t1 ))(y2 mY (t2 )) fXY (x1 ,y2 ; t1 , t2 )dy2 dx1
1 1
¼ E½X(t1 )Y(t2 ) mX (t1 )mY (t2 )
(19:1:21)
Thus, from Eqs. (19.1.19) and (19.1.21) the interrelationships between CC and
CCF are
CXY (t1 , t2 ) ¼ RXY (t1 , t2 ) mX (t1 )mY (t2 )
RXY (t1 , t2 ) ¼ CXY (t1 , t2 ) þ mX (t1 )mY (t2 )
(19:1:22)
Normalized Cross-Covariance. The normalized cross-covariance function (NCCF)
rXY(t1, t2) is the CCF normalized by the variances and is defined by
CXY (t1 , t2 )
CXY (t1 , t2 )
rXY (t1 , t2 ) ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼
2
2
s
X (t1 )sY (t2 )
sX (t1 )sY (t2 )
(19:1:23)
Some Properties of X(t) and Y(t)
Two random processes X(t) and Y(t) are independent if for all t1 and t2
FXY (x, y; t1 , t2 ) ¼ FX (x; t1 )FY (y; t2 )
(19:1:24)
19.1
Basic Definitions
417
They are uncorrelated if
CXY (t1 , t2 ) ¼ RXY (t1 , t2 ) mX (t1 )mY (t2 ) ¼ 0
or
RXY (t1 , t2 ) ¼ mX (t1 )mY (t2 )
for all t1 and t2
(19:1:25)
They are orthogonal if for all t1 and t2
RXY (t1 , t2 ) ¼ 0
(19:1:26)
Example 19.1.6 This example is slightly different from Example 19.1.4. A random
process X(t) is given by X(t) ¼ A sin(vt þ f) as shown in Fig. 19.1.11, where A is a uniformly distributed random variable with mean mA and variance s2A. We will find the mean,
variance, autocorrelation, autocovariance, and normalized autocovariance of X(t).
Mean:
E½X(t) ¼ mX (t) ¼ E ½A sin(vt þ f) ¼ mA sin(vt þ f)
Variance:
var½X(t) ¼ s2X (t) ¼ E½A2 sin2 (vt þ f) m2A sin2 (vt þ f)
¼ {E ½A2 m2A } sin2 (vt þ f) ¼ s2A sin2 (vt þ f)
Autocorrelation. From Eq. (19.1.14) we have
RX (t1 , t2 ) ¼ E½X(t1 )X(t2 ) ¼ E ½A2 sin(vt1 þ f) sin(vt2 þ f)
1
¼ E ½A2 {cos½v(t1 t2 ) cos½v(t1 þ t2 ) þ 2f}
2
FIGURE 19.1.11
418
Random Processes
Autocovariance. From Eq. (19.1.16) we have
CX (t1 , t2 ) ¼ RX (t1 , t2 ) mX (t1 )mX (t2 )
¼ E½A2 sin(vt1 þ f) sin(vt2 þ f) m2A sin(vt1 þ f) sin(vt2 þ f)
¼ s2A sin(vt1 þ f) sin(vt2 þ f)
1
¼ s2A {cos½v(t1 t2 ) cos½v(t1 þ t2 ) þ 2f}
2
Normalized Autocovariance. From Eq. (19.1.18) we have
rX (t1 , t2 ) ¼
CX (t1 , t2 )
s2 sin(vt1 þ f) sin(vt2 þ f)
¼ A2
¼1
sX (t1 )sX (t2 ) sA sin(vt1 þ f) sin(vt2 þ f)
Example 19.1.7 A random process X(t) with k changes in a time interval t, and its probability mass function is given by p(k; l) ¼ elt ½(lt)k =k!. It is also known that the joint
probability Pfk1 changes in t1, k2 changes in t2g is given by
P{k1 changes in t1 , k2 changes in t2 }
¼ P{k1 changes in t1 , (k2 k1 ) changes in (t2 t1 )}
¼ P{k1 changes in t1 }P{(k2 k1 ) changes in (t2 t1 )}
¼ elt1
(lt1 )k1 l(t2 t1 ) ½l(t2 t1 )(k2 k1 )
e
k1 !
(k2 k1 )!
We have to find the mean, variance, autocorrelation, autocovariance, and normalized autocovariance of X(t):
Mean:
E ½X(t) ¼ mX (t) ¼
1
X
kelt
k¼0
(lt)k
¼ lt
k!
Variance:
var½X(t) ¼ s2X (t) ¼
1
X
k2 elt
k¼0
(lt)k
(lt)2 ¼ lt
k!
Autocorrelation. From Eq. (19.1.18) we have
RX (t1 , t2 ) ¼ E½X(t1 )X(t2 ) ¼ E{X(t1 )½X(t2 ) X(t1 ) þ X(t1 )}
¼ E ½X 2 (t1 ) þ E ½X(t1 )E½X(t2 ) X(t1 ) (from condition given)
¼ (lt1 )2 þ lt1 þ lt1 l(t2 t1 )
¼ l2 t1 t2 þ lt1 if t2 . t1
and we have a similar result if t1 . t2:
RX (t1 , t2 ) ¼ l2 t1 t2 þ lt2
if
t1 . t2
19.1
Basic Definitions
419
Combining these two results, we have
RX (t1 , t2 ) ¼ l2 t1 t2 þ l min (t1 , t2 )
Autocovariance. From Eq. (19.1.19) we have
CX (t1 , t2 ) ¼ RX (t1 , t2 ) mX (t1 )mX (t2 )
¼ l2 t1 t2 þ l min (t1 , t2 ) l2 t1 t2 ¼ l min (t1 , t2 )
Normalized Autocovariance. From Eq. (19.1.18) we have
CX (t1 , t2 )
min (t1 , t2 ) 1
pffiffiffiffiffiffiffi ¼ min
rX (t1 , t2 ) ¼
¼
sX (t1 )sX (t2 )
l t1 t2
l
Example 19.1.8
rffiffiffiffi rffiffiffiffi
t1 t2
,
t2 t1
Two random processes X(t) and Y(t) are given by
X(t) ¼ A sin(vt þ f1 );
Y(t) ¼ B sin(vt þ f2 )
where A and B are two random variables with parameters E[A] ¼ mA, E[B] ¼ mB,
var[A] ¼ s2A, var[B] ¼ s2B, cov[AB] ¼ sAB, and correlation coefficient rAB ¼ sAB =
(sA sB ). We have to find the means and variances of X(t) and Y(t) and their crosscorrelation, cross-covariance, and normalized cross-covariance. The means and variances
can be obtained directly from Example 19.1.5.
Means:
mX (t) ¼ E½A sin(vt þ f1 ) ¼ mA sin(vt þ f1 )
mY (t) ¼ E ½B cos(vt þ f2 ) ¼ mB cos(vt þ f2 )
Variances:
s2X (t) ¼ s2A sin2 (vt þ f1 )
s2Y (t) ¼ s2B cos2 (vt þ f2 )
Cross-Correlation:
RXY (t1 , t2 ) ¼ E½X(t1 )Y(t2 ) ¼ E½AB sin(vt1 þ f1 ) cos(vt2 þ f2 )
1
¼ E ½AB{sin½v(t1 þ t2 ) þ f1 þ f2 þ sin½v(t1 t2 ) þ f1 f2 }
2
Cross-Covariance:
CXY (t1 ,t2 ) ¼ RXY (t1 ,t2 ) mX (t1 )mY (t2 )
¼ E ½ABsin(vt1 þ f1 )cos(vt2 þ f2 ) mA mB sin(vt1 þ f1 )cos(vt2 þ f2 )
¼ sAB sin(vt1 þ f1 )cos(vt2 þ f2 )
1
¼ sAB {sin½v(t1 þ t2 ) þ f1 þ f2 þ sin½v(t1 t2 ) þ f1 f2 }
2
Normalized Cross-Covariance:
rXY (t1 , t2 ) ¼
CXY (t1 , t2 )
sAB sin(vt1 þ f1 ) cos(vt2 þ f2 )
sAB
¼
¼
¼ rAB
sX (t1 )sY (t2 ) sA sB sin(vt1 þ f1 ) cos(vt2 þ f2 ) sA sB
420
Random Processes
B 19.2
STATIONARY RANDOM PROCESSES
The distribution functions of all the random processes except that in Example 19.1.4 were
dependent on time. However, many of the random processes such as Example 19.1.4 have
the important property that their statistics do not change with time, which is an important
step toward obtaining the statistics from a single sample function. The statistics in the time
interval (t1, t2) is the same as in the time interval (t1 þ t, t2 þ t). In other words, the probabilities of the samples of a random process X(t) at times t1, . . . , tn will not differ from
those at times t1 þ t, . . . , tn þ t. This means that the joint distribution function of
X(t1), . . . , X(tn) is the same as X(t1 þ t), . . . , X(tn þ t), or
FX (x1 , . . . , xn ; t1 , . . . , tn ) ¼ FX (x1 , . . . , xn ; t1 þ t, . . . , tn þ t)
(19:2:1)
and the corresponding density function may be written as
fX (x1 , . . . , xn ; t1 , . . . , tn ) ¼ fX (x1 , . . . , xn ; t1 þ t, . . . , tn þ t)
(19:2:2)
Random processes with the property of Eq. (19.2.1) or (19.2.2) are called nth-order
stationary processes. A strict-sense or strongly stationary process is a random process
that satisfies Eqs. (19.2.1) and (19.2.2) for all n. Analogously, we can also define lower
orders of stationarity.
A random process is first-order stationary if
FX (x; t) ¼ FX (x; t þ t) ¼ FX (x)
fX (x; t) ¼ fX (x; t þ t) ¼ fX (x)
(19:2:3)
and the distribution and density functions are independent of time. The random process in
Examples 19.1.2 and 19.1.5 is an example of a first-order stationary process.
A random process is second-order stationary if
FX (x1 , x2 ; t1 , t2 ) ¼ FX (x1 , x2 ; t1 þ t, t2 þ t) ¼ FX (x1 , x2 ; t)
fX (x1 , x2 ; t1 , t2 ) ¼ fX (x1 , x2 ; t1 þ t, t2 þ t) ¼ fX (x1 , x2 ; t)
(19:2:4)
The distribution and density functions are dependent not on two time instants t1 and t2 but
on the time difference t ¼ t1 – t2 only. Second-order stationary processes are also called
wide-sense stationary or weakly stationary. Hereafter, stationary means wide-sense
stationary, and strict-sense stationary will be specifically mentioned.
In a similar manner, two processes X(t) and Y(t) are jointly stationary if for all n
FXY (x1 , . . . , xn ; y1 , . . . , yn ; t1 , . . . , tn )
¼ FXY (x1 , . . . , xn ; y1 , . . . , yn ; t1 þ t, . . . , tn þ t)
(19:2:5)
fXY (x1 , . . . , xn ; y1 , . . . , yn ; t1 , . . . , tn )
¼ fXY (x1 , . . . , xn ; y1 , . . . , yn ; t1 þ t, . . . , tn þ t)
(19:2:6)
or
and they are jointly wide-sense stationary if
FXY (x1 , y2 ; t1 , t2 )
¼ FXY (x1 , y2 ; t1 þ t, t2 þ t) ¼ FXY (x1 , y2 ; t)
(19:2:7)
19.2
Stationary Random Processes
421
and the joint density function
fXY (x1 , y2 ; t1 , t2 ) ¼ fXY (x1 , y2 ; t1 þ t, t2 þ t) ¼ fXY (x1 , y2 ; t)
(19:2:8)
nth-order stationarity implies lower-order stationarities. Strict-sense stationarity implies
wide-sense stationarity.
Similar to Eqs. (19.1.24) –(19.1.26), we can enumerate the following properties
for stationary random processes X(t) and Y(t). Two random processes X(t) and Y(t) are
independent if for all x and y
FXY (x, y) ¼ FX (x)FY (y)
(19:2:9)
They are uncorrelated if for all t
CXY (t) ¼ RXY (t) mX mY ¼ 0
or
RXY (t) ¼ mX mY
(19:2:10)
They are orthogonal if for all t
RXY (t) ¼ 0
(19:2:11)
Moments of Continuous-Time Stationary Processes
We can now define the various moments for a stationary random process X(t).
Mean:
ð1
E ½X(t) ¼ mX ¼
xf (x)dx
(19:2:12)
1
Variance:
2
E ½X(t) mX ¼
s2X
ð1
¼
(x mX )2 f (x)dx ¼ E½X 2 (t) m2X
(19:2:13)
1
Autocorrelation:
ð1 ð1
RX (t) ¼ E ½X(t)X(t þ t) ¼
x1 x2 f (x1 , x2 ; t)dx1 dx2
(19:2:14)
1 1
Autocovariance:
CX (t) ¼ E{½X(t) mX ½X(t þ t) mX }
ð1 ð1
(x1 mX )(x2 mX ) f (x1 , x2 ; t)dx1 dx2
¼
1
(19:2:15)
1
¼ RX (t) m2X
Normalized Autocovariance (NACF):
rX (t) ¼
CX (t)
s2X
(19:2:16)
422
Random Processes
White-Noise Process
A zero mean stationary random process X(t) whose autocovariance or autocorrelation
is given by
CX (t) ¼ RX (t) ¼ s2X d(t)
(19:2:17)
where d(t) is the Dirac delta function, is called a white-noise process. The energy of a
white-noise process is infinite since CX (0) ¼RX (0) ¼ E[X 2(t)] ¼ 1. Hence it is an idealization. White-noise processes find extensive use in modeling communication systems.
The cross-moments of two jointly stationary processes X(t) and Y(t) are defined below:
Cross-Correlation:
ð1 ð1
RXY (t) ¼ E ½X(t)Y(t þ t) ¼
x1 y2 f (x1 , y2 ; t)dy2 dx1
1
(19:2:18)
1
Cross-Covariance:
CXY (t) ¼ E{½X(t) mX ½Y(t þ t) mY }
ð1 ð1
(x1 mX )(y2 mY ) f (x1 , y2 ; t)dy2 dx1
¼
1 1
¼ RXY (t) mX mY
(19:2:19)
Normalized Cross-Covariance:
rXY (t) ¼
CXY (t)
sX sY
(19:2:20)
A stationary random process X(t) is passed through a linear system with impulse
response h(t). The input – output relationship will be given by the convolution integral:
ð1
ð1
Y(t) ¼
X(t a)h(a)da ¼
X(t)h(t a)da
(19:2:21)
1
1
The cross-correlation function RXY (t) between the input and the output can be found as
follows:
ð1
E½X(t)Y(t þ t) ¼ E
X(t)X(t þ t a)h(a)da
(19:2:22)
1
or
ð1
RX (t a)h(a)da
RXY (t) ¼
(19:2:23)
1
Since the cross-correlation function depends only on t, the output Y(t) will also be a
stationary random process.
Properties of Correlation Functions of Stationary Processes
Autocorrelation Functions
1. RX (0) ¼ E[X 2(t)] ¼ average power 0
2. RX (t) ¼ RX (t). Or, RX (t) is an even function.
RX (t) ¼ E½X(t)X(t þ t) ¼ E ½X(t þ t)X(t) ¼ RX (t)
(19:2:24)
19.2
Stationary Random Processes
423
3. jRX (t)j RX (0). Or, the maximum value of jRX (t)j occurs at t ¼ 0 and jRX (t)j
for any t cannot exceed the value RX (0). From
E½X(t) + X(t þ t)2 0
we have
E ½X 2 (t) þ E ½X 2 (t þ t) + 2E ½X(t)X(t þ t) 0
or
RX (0) jRX (t)j
(19:2:25)
4. If a constant T . 0 exists such that RX (T ) ¼ RX (0), then the RX (t) is periodic and
X(t) is called a periodic stationary process. From Schwartz’ inequality
[Eq. (14.5.13)] we have
{E ½g(X)h(X)}2 E ½g2 (X)E½h2 (X)
Substituting g(X ) ¼ X(t) and h(X) ¼ ½X(t þ t þ T ) X(t þ t), we can write
{E½X(t)½X(t þ t þ T ) X(t þ t)}2 E½X 2 (t)E½X(t þ t þ T ) X(t þ t)2
or
{RX (t þ T ) RX (t)}2 2RX (0)½RX (0) RX (T )
Hence, if RX (T ) ¼ RX (0), then, since {RX (t þ T ) RX (t)}2 0, the result
RX (t þ T ) ¼ RX (t) follows.
5. E ½X(t þ f)X(t þ t þ f) ¼ RX (t) ¼ E½X(t)X(t þ t)
E ½X(t þ f)Y(t þ t þ f) ¼ RXY (t) ¼ E½X(t)Y(t þ t)
In the formulation of correlation functions the phase information is lost.
6. If E[X(t)] ¼ mX and Y(t) ¼ a þ X(t), where a is constant, then E[Y(t)] ¼ a þ mX
and,
RY (t) ¼ E{½a þ X(t)½a þ X(t þ t)}
¼ a2 þ aE ½X(t þ t) þ aE ½X(t) þ E ½X(t)X(t þ t)
¼ a2 þ 2amX þ RX (t) ¼ a2 þ 2amX þ m2X þ CX (t)
¼ (a þ mX )2 þ CX (t)
(19:2:26)
If E [X(t)] ¼ 0, then E[Y(t)] ¼ a, and we can obtain the mean value of Y(t) from a
knowledge of its autocorrelation function RY(t).
Cross-Correlation Functions
7. RXY (t) ¼ RYX (t): This is not an even function:
ð19:2:27Þ
424
Random Processes
The result follows from the definition of cross-correlation:
RXY (t) ¼ E½X(t)Y(t þ t) ¼ E½Y(t þ t)X(t) ¼ RYX (t)
but
RXY (0) ¼ RYX (0)
ð19:2:28Þ
8. R2XY (t) RX (0)RY (0): This result follows from Schwartz’ inequality:
{E ½X(t)Y(t þ t)}2 E ½X 2 (t) E ½Y 2 (t þ t)
9. 2jRXY (t)j RX (0) þ RY (0): The result follows from
E ½X(t) + Y(t þ t)2 ¼ RX (0) þ RY (0) + 2RXY (t) 0
(19:2:29)
10. If Z(t) ¼ X(t) þ Y(t), then
RZ (t) ¼ E ½Z(t)Z(t þ t) ¼ E{½X(t) þ Y(t)½X(t þ t) þ Y(t þ t)}
¼ RX (t) þ RY (t) þ RXY (t) þ RYX (t)
(19:2:30)
and if X(t) and Y(t) are orthogonal, then RZ (t) ¼ RX (t) þ RY (t).
_
11. If X(t)
is the derivative of X(t), then the cross-correlation between X(t) and
_X(t) is
RX X_ (t) ¼
dRX (t)
dt
(19:2:31a)
This result can be shown from the formal definition of the derivative of X(t).
Substituting for X_ ¼ lim1!0 {½X(t þ 1) X(t)=1}, we obtain
RX X_ (t) ¼ E½X(t)X(t þ t)
X(t þ t þ 1) X(t þ t)
¼ lim E X(t)
1!0
1
X(t)X(t þ t þ 1) X(t)X(t þ t)
¼ lim E
1!0
1
RX (t þ 1) RX (t) dRX (t)
¼
1!0
dt
1
¼ lim
_ is
12. The autocorrelation of X(t)
RX_ (t) ¼ d2 RX (t)
dt2
(19:2:31b)
19.2
Stationary Random Processes
425
This result can be shown following a procedure similar to that described as in (11)
above:
_ X(t
_ þ t)
RX_ (t) ¼ E½X(t)
X(t þ t þ 1) X(t þ t)
¼ lim E X(t)
1!0
1
X(t þ d) X(t) X(t þ t þ 1) X(t þ t)
¼ lim lim E
d!0 1!0
d
1
2
3
X(t þ t þ 1) X(t þ t)
X(t
þ
d)
1
6
7
1
¼ lim lim E4
X(t þ t þ 1) X(t þ t) 5
d!0 d 1!0
X(t)
1
1
RX (t d þ 1) RX (t d) RX (t þ 1) RX (t)
¼ lim lim
d!0 d 1!0
1
1
1 dRX (t d) dRX (t)
d2 RX (t)
¼ lim
¼
d!0 d
dt
dt
dt2
It can be shown that if a random process is wide-sense stationary, then it is necessary
and sufficient that the following two conditions be satisfied:
1. The expected value is a constant, E [X(t)] ¼ mX.
2. The autocorrelation function RX is a function of the time difference t2 2 t1 ¼ t and
not individual times, RX (t1 , t2 ) ¼ RX (t2 t1 ) ¼ RX (t).
We will now give several examples to establish conditions for stationarity. The first
few examples will be running examples that become progressively more difficult.
Example 19.2.1 We will revisit Example 19.1.5 and find the conditions necessary for the
random process X(t) ¼ A sin(vt þ F) to be stationary where A,v are constants and F is a
random variable:
1. Mean:
ðb
E½X(t) ¼ AE½sin(vt þ F) ¼ A
sin(vt þ f) fF (f)df
a
One of the ways the integral will be independent of t is for F to be uniformly distributed in (0,2p), in which case we have
1
2p
and E [X(t)] ¼ 0.
ð 2p
sin(vt þ f)df ¼
0
1
cos (vt þ f)
2p
2p
0
¼0
426
Random Processes
2. Autocorrelation:
E ½X(t1 )X(t2 ) ¼ A2 E ½sin(vt1 þ F) sin(vt2 þ F)
¼ A2 E{cos½v(t2 t1 ) cos½v(t2 þ t1 ) þ 2F}
¼ A2 cos½v(t2 t1 ) A2 E{cos½v(t2 þ t1 ) þ 2F}
ð
1 2p
cos½v(t2 þ t1 ) þ 2f df
E{cos½v(t2 þ t1 ) þ 2F} ¼
2p 0
¼
1 1
sin½v(t2 þ t1 ) þ 2f
2p 2
2p
0
¼0
Hence, RX (t1 , t2 ) ¼ A2 cos½v(t2 t1 ) ¼ A2 cos½vt, and X(t) is stationary if F is
uniformly distributed in (0,2p). Since RX(t) is periodic, X(t) is a periodic stationary process.
Example 19.2.2 We will modify Example 19.2.1 with both A and F as random variables
with density functions fA(a) and fF(f). We will now find the conditions under which
X(t) ¼ A sin(vt þ F) will be stationary.
1. Mean:
ðð
E ½X(t) ¼ E ½A sin(vt þ F) ¼
a sin(vt þ f) fAF (a, f)df da
The first condition for the double integral to be independent of t is for A and F to be
statistically independent, in which case we have
ðð
E ½A sin(vt þ F) ¼
a sin(vt þ f) fA (a) fF df da
and the second condition
Ð 2p is for F to be uniformly distributed in (0,2p), in which
case we have (1=2p) 0 sin(vt þ f) df ¼ 0 and E ½X(t) ¼ 0.
2. Autocorrelation:
E ½X(t1 )X(t2 ) ¼ E ½A2 sin(vt1 þ F) sin(vt2 þ F)
Since A and F are independent, we have
E ½X(t1 )X(t2 ) ¼ E ½A2 E{cos½v(t2 t1 ) cos½v(t2 þ t1 ) þ 2F}
¼ E ½A2 cos½v(t2 t1 ) E ½A2 E{cos½v(t2 þ t1 ) þ 2F}
and from the previous example E{cos½v(t2 þ t1 ) þ 2F} ¼ 0. Hence, RX (t1 , t2 ) ¼
E ½A2 cos½v(t2 t1 ) ¼ E ½A2 cos½vt and X(t) is stationary if A and F are independent and if F is uniformly distributed in (0,2p). Since RX(t) is periodic, X(t)
a periodic stationary process.
Example 19.2.3 We will now examine the conditions for stationarity for X(t) ¼ A
sin(Vt þ F) when A, V and F are all random variables with density functions fA(a),
fV(v), and fF(f) respectively.
19.2
Stationary Random Processes
427
1. Mean:
ððð
E ½X(t) ¼ E ½A sin(Vt þ F) ¼
a sin(vt þ f) fAVF (a,v,f)df dv da
This triple integral is a difficult one to evaluate. Hence we resort to
conditionalÐ expectations by fixing the variable V ¼ v. Under this condition,
E ½X(t) ¼ E ½X(t)jV ¼ v fV (v)dv and from the previous example, if A and F
are independent and F is uniformly distributed in (0, 2p), we have
E ½X(t)jV ¼ v ¼ 0, and
ð
ð
E ½X(t)jV ¼ v fV (v)dv ¼ E ½A sin(vt þ F) fV (v)dv ¼ 0
2. Autocorrelation:
RX (t1 , t2 jV ¼ v) ¼ E ½X(t1 jV ¼ v)X(t2 jV ¼ v)
¼ E ½A2 sin(vt1 þ F) sin(vt2 þ F)
and from the previous example
1
E ½A2 sin(vt1 þ F) sin(vt2 þ F) ¼ E ½A2 cos½v t
2
where t ¼ t2 t1 . Hence
ð
ð
1
E ½A2 cos½vt fV (v)dv
RX (t) ¼ RX (tjV ¼ v) fV (v)dv ¼
2
If V is uniformly distributed in (0,p), then
RX (t) ¼
1
p
ðp
0
1
E ½A2 sin(p) E ½A2 E ½A2 cos½vtdv ¼
¼
Sa(p)
2
2p
t
2
Example 19.2.4 In this example X(t) ¼ A cos(vt) þ B sin(vt), where A and B are
random variables with density functions fA(a) and fB(b). We have to find the conditions
under which X(t) will be stationary:
1. Mean:
E ½X(t) ¼ E ½A cos(vt) þ B sin(vt) ¼ cos(vt)E ½A þ sin(vt)E ½B
If E [X(t)] is to be independent of t, then E[A] ¼ E[B] ¼ 0, in which case
E [X(t)] ¼ 0.
428
Random Processes
2. Autocorrelation:
RX (t1 , t2 ) ¼ E ½X(t1 )X(t2 ) ¼ E{½A cos(vt1 )
þ B sin(vt1 )½A cos(vt2 ) þ B sin(vt2 )}
¼ E{A2 cos(vt1 ) cos(vt2 ) þ B2 sin(vt1 ) sin(vt2 )
þ AB½sin(vt1 ) cos(vt2 ) þ cos(vt1 ) sin(vt2 )}
1
¼ E ½A2 ½cos(v(t2 t1 )) þ cos(v(t2 þ t1 ))
2
1
þ E ½B2 ½cos(v(t2 t1 )) cos(v(t2 þ t1 ))
2
þ E ½AB sin(v(t2 þ t1 ))
The conditions under which this equation will be dependent only on (t2 t1 ) are
E[A 2] ¼ E[B 2] and E[AB] ¼ 0. In this case
RX (t1 ,t2 ) ¼ E ½A2 ½cos(v(t2 t1 )) ¼ E ½A2 ½cos(vt)
We can now summarize the conditions for stationarity:
(a) E[A] ¼ E[B] ¼ 0
(b) E[A 2] ¼ E[B 2]
(c) E[AB] ¼ 0
Example 19.2.5 A die is tossed, and corresponding to the dots S ¼ f1,2,3,4,5,6g, a
random process X(t) is formed with the following time functions as shown in Fig. 19.2.1:
X(2;t) ¼ 3, X(4; t) ¼ (2 t), X(6; t) ¼ (1 þ t)
X(1; t) ¼ 3, X(3;t) ¼ (2 t), X(5;t) ¼ (1 þ t)
FIGURE 19.2.1
19.2
Stationary Random Processes
429
We have to find mX (t), s2X (t), RX (t1 ,t2 ), CX (t1 , t2 ), and rX (t1 , t2 ) and check whether X(t) is
stationary:
Mean:
mX (t) ¼
6
1X
Xi (t) ¼ 3 3 þ (2 t) (2 t) þ (1 þ t) (1 þ t) ¼ 0
6 i¼1
The mean value is a constant.
Variance:
s2X (t) ¼
6
1X
1
2
Xi2 (t) ¼ ½32 þ (2 t)2 þ (1 þ t)2 ¼ ½t2 t þ 7
6 i¼1
3
3
Autocorrelation:
1
1
RX (t1 , t2 ) ¼ ½9 þ (1 þ t1 )(1 þ t2 ) þ (2 t1 )(2 t2 ) ¼ ½14 t2 t1 þ 2t1 t2 3
3
Autocovariance. Since the mean value is zero, CX (t1 , t2 ) ¼ RX (t1 , t2 ).
Normalized Autocovariance:
CX (t1 , t2 )
½14 t2 t1 þ 2t1 t2 rX (t1 , t2 ) ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
CX (t1 )CX (t2 )
(t12 t1 þ 7)(t22 t2 þ 7)
This process is not stationary.
Example 19.2.6 (Random Binary Wave) A sample function of a random binary wave
X(t) consisting of independent rectangular pulses p(t), each of which is of duration T, is
shown in Fig. 19.2.2. The height H of the pulses is a random variable with constant amplitudes, which are equally likely to be +A. The time of occurrence of X(t) after t ¼ 0 is
another random variable Z, which is uniformly distributed in (0,T ). We have to find the
mean, variance, autocorrelation, autocovariance, and the normalized autocovariance
of X(t).
The random process X(t) is given by
X(t) ¼
1
X
Hp(t kT Z)
k¼1
FIGURE 19.2.2
430
Random Processes
The probabilities of the random variable H are P(H ¼ A) ¼ 12 and P(H ¼ A) ¼ 12. Since
Z is uniformly distributed in (0,T ), the following probabilities can be formulated
for 0 , t1 , t2 , T:
P(t1 , Z , t2 ) ¼
t2 t1
,
T
P(0 , Z , t1 ) ¼
t1
,
T
P(t2 , Z , T ) ¼ 1 t2
T
Mean. Since X(t) assumes only þA or 2A with equal probability,
E ½X(t) ¼
1
1
A þ (A) ¼ 0
2
2
Variance. The variance is given by
var½X(t) ¼
s2X
1 2 1
A þ (A)2 ¼ A2
¼ E ½X (t) ¼
2
2
2
Autocorrelation. Determining RX (t1 , t2 ) is a bit more involved. The product X(t1)X(t2)
in various intervals of time (t1, t2) can be found:
t1 , Z , t2 T
Here t1 and t2 lie in adjacent pulse intervals as shown in Fig. 19.2.3. In Fig. 19.2.3a
X(t1) ¼ 2A and X(t2) ¼ þA and X(t1)X(t2) ¼ 2A 2. In Fig. 19.2.3b X(t1) ¼ þA and
X(t2) ¼ þA and X(t1)X(t2) ¼ A 2. The values 2A 2 and A 2 will occur with equal
probability:
0 , Z , t1 T: In this case, t1 and t2 lie in the same pulse interval as shown in
Fig. 19.2.3c. Here, X(t1) ¼ þA and X(t2) ¼ þA and X(t1)X(t2) ¼ A 2.
t2 , Z , T: Here also t1 and t2 lie in the same pulse interval as shown in Fig. 19.2.3d.
However, X(t1) ¼ 2A and X(t2) ¼ 2A and X(t1)X(t2) ¼ A 2.
(t1 2 t2) . T: In this case t1 and t2 lie in different pulse intervals and
X(t1)X(t2) ¼ +A 2.
We can now find the autocorrelation function RX (t1 , t2 ) ¼ E ½X(t1 )X(t2 ). For
(t1 2 t2) . T, X(t1) and X(t2) are in different pulse intervals and invoking the independence of the pulses E [X(t1)X(t2)] ¼ E [X(t1)]E [X(t2)] ¼ 0 since E [X(t)] ¼ 0.
For (t1 2 t2) T, we have the following, using conditional expectations:
E ½X(t1 )X(t2 ) ¼ E ½X(t1 )X(t2 )jt1 , Z , t2 P(t1 , Z , t2 )
þ E ½X(t1 )X(t2 )j0 , Z , t1 P(0 , Z , t1 )
þ E ½X(t1 )X(t2 )jt1 , Z , TP(t1 , Z , T )
The conditional expectations can be determined using the probabilities and the products X(t1)X(t2) found earlier for the various intervals. Since A 2 and 2A 2 occur
with equal probability in the interval t1 , Z , t2 T, the first conditional expectation term becomes
1 t2 t1
1 t2 t1
A2 ¼0
E ½X(t1 )X(t2 )jt1 , Z , t2 P(t1 , Z , t2 ) ¼ A2 2
2
T
T
19.2
FIGURE 19.2.3
Stationary Random Processes
431
432
Random Processes
FIGURE 19.2.4
Hence
"
RX (t1 , t2 ) ¼
A2
t
t2
t2 t1
1
þ1
¼ A2 1 ,
T
T
T
(t2 t1 ) T
(t2 t1 ) . T
0,
This equation was derived under the assumption t1 , t2. For arbitrary values of t1
and t2 with t ¼ (t2 2 t1), the autocorrelation function RX(t1, t2) is given by
8 jtj
< 2
A 1
, jtj , T
RX (t) ¼
T
:
0,
otherwise
The autocorrelation function for the process X(t) is shown in Fig. 19.2.4.
Autocovariance. Since the mean value is 0, CX(t) ¼ RX(t).
Normalized Autocovariance. The NACF is given by
(
jtj
CX (t)
1 , jtj , T
rX (t) ¼ 2 ¼
T
sX
0;
otherwise
Example 19.2.7 (Random Telegraph Wave) This example is a little different from the
previous one. A sample function of a random telegraph wave is shown in Fig. 19.2.5. The
wave assumes either of the values 1 or 0 at any instant of time.
FIGURE 19.2.5
19.2
Stationary Random Processes
433
The probability of k changes from 0 to 1 in a time interval t is Poisson-distributed with
probability mass function p(k; l) given by
(lt)k lt
e ,
k!
p(k; l) ¼
t0
where l is the average number of changes per unit time. We have to find the mean,
variance, autocorrelation, autocovariance, and normalized autocovariance.
Mean. Since X(t) assumes only 1 or 0 with equal probability, we obtain
E ½X(t) ¼ mX ¼ 1 P(x ¼ 1) þ 0 P(x ¼ 0) ¼
1
2
Variance. The variance is given by
var½X(t) ¼
s2X
2
¼ E ½X (t) m2X
1 2 1 2
1 1
1 þ 0 ¼
¼
2
2
4 4
Autocorrelation. The autocorrelation function can be given in terms of the joint
density functions with t2 . t1:
RX (t1 , t2 ) ¼ E ½X(t1 ) ¼ 0 X(t2 ) ¼ 0 þ E ½X(t1 ) ¼ 0 X(t2 ) ¼ 1
þ E ½X(t1 ) ¼ 1 X(t2 ) ¼ 0 þ E ½X(t1 ) ¼ 1 X(t2 ) ¼ 1
¼ (0 0)P½X(t1 ) ¼ 0 X(t2 ) ¼ 0 þ (0:1)P½X(t1 ) ¼ 0 X(t2 ) ¼ 1
þ (1 0)P½X(t1 ) ¼ 1 X(t2 ) ¼ 0 þ (1:1)P½X(t1 ) ¼ 1 X(t2 ) ¼ 1
or
RX (t1 , t2 ) ¼ P½X(t1 ) ¼ 1 X(t2 ) ¼ 1
Expressing the joint probability in terms of conditional probabilities, we have
RX (t1 , t2 ) ¼ P½X(t2 ) ¼ 1 j X(t1 ) ¼ 1P½X(t1 ) ¼ 1
The conditional probability in this equation is the probability of even number of
changes and using the Poisson distribution:
RX (t1 , t2 ) ¼
1
1 X
½l(t2 t1 )k l(t2 t1 )
e
2 k¼0
k!
k even
(
)
1
1
el(t2 t1 ) 1 X
½l(t2 t1 )k X
½l(t2 t1 )k
þ
¼
2 k¼0
2
k!
k!
k¼0
el(t2 t1 ) el(t2 t1 ) þ el(t2 t1 )
¼
2
2
1
¼ {1 þ e2l(t2 t1 ) },
4
t2 . t1
A similar equation holds good for t2 , t1. Hence, substituting jt2 t1 j ¼ jtj in the
equation above, RX(t) can be given by
1
RX (t) ¼ {1 þ e2ljtj }
4
434
Random Processes
FIGURE 19.2.6
The autocorrelation function for the process X(t) is shown in Fig. 19.2.6 for
l ¼ 0.5.
Autocovariance:
CX (t) ¼ RX (t) m2X ¼
e2ljtj
4
Normalized Autocovariance:
rX (t) ¼
CX (t)
¼ e2ljtj
s2X
Example 19.2.8 (Modulation) A random process Y(t) is given by Y(t) ¼ X(t)
cos(vt þ F), where X(t), a zero mean wide-sense stationary random process with autocorrelation function RX (t) ¼ 2e2ljtj is modulating the carrier cos(vt þ F). The random variable F is uniformly distributed in the interval (0,2p), and is independent of X(t). We have
to find the mean, variance, and autocorrelation of Y(t):
Mean. The independence of X(t) and F allows us to write
E ½Y(t) ¼ E ½X(t)E ½cos(vt þ F)
and with E ½X(t) ¼ 0 and E ½cos(vt þ F) ¼ 0 from Example 19.2.1, E[Y(t)] ¼ 0.
Variance. Since X(t) and F are independent, the variance can be given by
s2Y ¼ E ½Y 2 (t) ¼ E ½X 2 (t) cos2 (vt þ F) ¼ s2X E ½cos2 (vt þ F)
However
1
1
E ½cos2 (vt þ F) ¼ E ½1 þ cos(2vt þ 2FÞ ¼
2
2
and hence s2Y ¼ s2X =2 ¼ 1.
and s2X ¼ CX (0) ¼ RX (0) ¼ 2
19.2
Stationary Random Processes
435
FIGURE 19.2.7
Autocorrelation:
RY (t) ¼ E ½Y(t)Y(t þ t) ¼ E ½X(t) cos(vt þ F)X(t þ t) cos(vt þ vt þ F)
1
¼ RX (t) E ½cos(vt) þ cos(2vt þ vt þ 2F)
2
RX (t)
RX (t)
cos(vt) þ
E ½cos(2vt þ vt þ 2F)
¼
2
2
From Example 19.2.1 E ½cos(2vt þ vt þ 2F) ¼ 0, and hence
RY (t) ¼
RX (t)
cos(vt) ¼ e2ljtj cos(vt)
2
A graph of RY(t) is shown in Fig. 19.2.7 with l ¼ 0.5 and v ¼ 2p.
Example 19.2.9 (Cross-Correlation) Two random processes X(t) and Y(t) are given
by X(t) ¼ A cos(vt) þ B sin(vt) and Y(t) ¼ A sin(vt) þ B cos(vt), where A and B
are random variables with density functions fA(a) and fB(b). We have to find the
cross-correlation, cross-covariance, and the normalized cross-covariance between X(t)
and Y(t).
From Example 19.2.4 the processes X(t) and Y(t) are stationary if E [A] ¼ E [B] ¼ 0,
E [A 2] ¼ E [B 2], and E [AB] ¼ 0. Hence the mean values mX ¼ mY ¼ 0. The variances of
these processes are E ½A2 ¼ E ½B2 ¼ s2 .
Cross-Correlation. The cross-correlation function RXY(t1, t2) can be written as
RXY (t1 , t2 ) ¼ E ½X(t1 )Y(t2 )
¼ E{½A cos(vt1 ) þ B sin(vt1 )½A sin(vt2 ) þ B cos(vt2 )}
¼ E{ A2 cos(vt1 ) sin(vt2 ) þ B2 sin(vt1 ) cos(vt2 )
þ AB½cos(vt1 ) cos(vt2 ) sin(vt1 ) sin(vt2 )}
436
Random Processes
Substituting the values E [A 2] ¼ E [B 2] ¼ s2 and E [AB]¼0 in this equation, we
obtain
RXY (t1 ,t2 ) ¼ s2 sin½v(t1 t2 ) ¼ s2 sin½vt
where we have substituted t ¼ (t2 2 t1). The term RXY(t) is shown in Fig. 19.2.8 for
s2¼4 and v ¼ 2p.
Here RXY(t) is an odd function, unlike the autocorrelation function. Since
RXY (t)¼0 for t ¼ 0, we conclude that X(t) and Y(t) are orthogonal.
Cross-Covariance. Since mX ¼ mY ¼ 0, CXY (t) ¼ RXY(t).
Normalized Cross-Covariance:
rXY (t) ¼
CXY (t) s2 sin½vt
¼
¼ sin½vt
s2
s2
Example 19.2.10 A random telegraph wave X(t) as in Example 19.2.7 is passed through
a linear system with impulse response h(t) ¼ e – btu(t), where u(t) is a unit step function.
The output of the system is Y(t). It is desired to find the cross-correlation function RXY (t).
From Example 19.2.7 the autocorrelation function RX (t) ¼ 14 {1 þ e2ljtj }. Hence,
from Eq. (19.2.23), we have
ð1
1
(1 þ e2ljtaj )eba da
4
ð
ð
8
1
1 t 2l(ta) ba
1 1 2l(at) ba
>
>
þ
e
e da þ
(e
)e da
<
4b 4 0
4 t
ð1
¼
>
1
1
>
:
þ
e2l(at) eba da
4b 4 0
RXY (t) ¼
0
FIGURE 19.2.8
t.0
t0
19.2
Stationary Random Processes
437
Evaluating the integrals and substituting l ¼ 1 and b ¼ 0.5, we obtain
8
lebt
e2lt
>
>
; t.0
>
>
2
2
< 4l b
4(2l b)
1
þ
RXY (t) ¼
4b >
>
e2lt
>
>
:
,
t0
4(2l þ b)
or
8
4 (t=2) 1 2t
>
>
e
e ,
>
< 15
6
1
RXY (t) ¼ þ
2 >
>
>
: 1 e2t ,
10
t.0
t0
The cross-correlation function RXY (t) is shown in Fig. 19.2.9.
RXY (t) does not possess any symmetry, unlike RX (t).
Moments of Discrete-Time Stationary Processes
In actual practice observations are made on the sample function of a stationary
random process X(t) at equally spaced time intervals fti, i ¼ 0, +1, . . .gwith corresponding sequence of random variables fXi, i ¼ 0, +1, . . .g. These observation random variables will not be independent. Since fXig are samples of a stationary random process,
the means and variances of these samples are the same as in the original process:
E ½Xi ¼ mX ; var½Xi ¼ s2X ,
i ¼ 0, +1, . . .
(19:2:32)
Analogous to the continuous case, we can define the various second moments:
Autocovariance:
CX (h) ¼ E ½(Xi mX )(Xiþh mX ),
i ¼ 0,+1, . . .
(19:2:33)
Autocorrelation:
RX (h) ¼ E ½Xi Xiþh , i ¼ 0, + 1, . . .
FIGURE 19.2.9
(19:2:34)
438
Random Processes
Normalized Autocovariance (NACF):
rX (h) ¼
CX (h) CX (h)
¼
CX (0)
s2X
(19:2:35)
If fYi, i ¼ 0, +1, . . .g is the sequence obtained from a second stationary random
process Y(t) with mean mY and variance s2Y, then we can define the cross-moments as
follows:
Cross-Covariance:
CXY (h) ¼ E ½(Xi mX )(Yiþh mY i ¼ 0,+1, . . .
(19:2:36)
Cross-Correlation:
RXY (h) ¼ E ½Xi Yiþh i ¼ 0,+1, . . .
(19:2:37)
Normalized Cross-Covariance (NCCF):
CXY (h)
CXY (h)
rXY (h) ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼
sX sY
CX (0)CY (0)
Example 19.2.11
(19:2:38)
A discrete zero mean stationary random process Xi is given by
Xi ¼ fXi1 þ vi ,
i ¼ 1, . . .
(19:2:39)
where vi is a zero mean Gaussian random process with variance s2v . We want to find the
variance of Xi and the NACF rX(h).
Multiplying both sides of Eq. (19.2.39) by Xi and taking expectations, we have
E ½Xi2 ¼ fE ½Xi Xi1 þ E ½Xi vi or
s2X ¼ fCX (1) þ E ½Xi vi (19:2:40)
The cross-correlation E ½Xi vi can be computed as follows:
E ½Xi vi ¼ E ½(fXi1 þ vi )vi ¼ s2v
(19:2:41)
since vi occurs after Xi21. Hence, substituting Eq. (19.2.41) in Eq. (19.2.40) and dividing
throughout by s2X , we obtain
1 ¼ frX (1) þ
s2v
s2X
and s2X ¼
s2v
1 frX (1)
(19:2:42)
Premultiplying both sides of Eq. (19.2.39) by Xih and taking expectations, we have
E ½Xih Xi ¼ E ½fXih Xi1 þ E ½Xih vi (19:2:43)
In Eq. (19.2.43) E ½Xih vi ¼ 0 since vi occurs after Xih for h . 0; hence
CX (h) ¼ fCX (h 1),
h.0
(19:2:44)
Dividing Eq. (19.2.44) by CX (0) ¼ s2X , we obtain an equation for the NACF rX (h)
rX (h) ¼ frX (h 1),
h.0
(19:2:45)
19.3
Ergodic Processes
439
and solving for rX(h) with initial condition rX(0) ¼ 1, we obtain
rX (h) ¼ fh ,
h.0
(19:2:46)
Substituting Eq. (19.2.46) in Eq. (19.2.42), we have
s2X ¼
s2v
1 f2
(19:2:47)
The process defined by Eq. (19.2.39) is called an autoregressive process of order 1.
B 19.3
ERGODIC PROCESSES
The ensemble average of a random process X(t) is the mean value mX(t) defined by,
ð1
mX (t) ¼
xfX (x; t)dx
(19:3:1)
1
Finding the ensemble average mX(t) requires storing a multiplicity of sample functions and
finding the average. In many instances this process may be nontrivial. Given a sample
function of any random process X(t), the time average m^ X is defined by
m^ X ¼
1
2T
ðT
X(t)dt
(19:3:2)
T
A reasonable question to ask is whether the ensemble average can be obtained from a
much easier time average. We observe that the ensemble average is not a random variable
but is a function of time, whereas the time average is a random variable that is not a function of time. If these averages are to be equal, then the first condition that we have to
impose is that the random process X(t) be stationary, which removes the time factor in
the mean value. Under these conditions, we can write
E
1
2T
ðT
X(t)dt ¼ E ½m^ X ¼
T
1
2T
ðT
E ½X(t)dt ¼ mX
(19:3:3)
T
and m^ X is an unbiased estimator of mX.
If a random variable is to be equal to a constant, then the second condition from
Example 14.1.1 is that the variance of m^ X must tend to zero as T ! 1. Such random processes are said to satisfy the ergodic hypothesis. We will now derive the conditions for a
random process to be mean-ergodic and correlation-ergodic.
Mean-Ergodic
A stationary random process X(t) is called mean-ergodic if the ensemble average is
equal to the time average of the sample function x(t). We will assume that the following
conditions are satisfied by X(t):
X(t) is stationary, implying that it has a constant mean mX, and the autocorrelation
function RX (t, t þ t) is a function of t only, or RX (t, t þ t) ¼ RX (t).
RX (0) ¼ E [X 2(t)] is bounded, or RX(0) ,1. Hence CX (0) ¼ RX (0) 2 m2X ,1.
440
Random Processes
We have to find the conditions under which the variance s2m^ X of the time average m^ X
goes to 0. The variance s2m^ X is given by
s2m^ x ¼ E ½(m^ X mX )2 ¼ E
(
1
2T
ðT
2
)
½X(t) mX dt
T
(
)
ð ð
1 2 T T
¼E
½X(t) mX ½X(s) mX dt ds
2T
T T
2 ð T ð T
1
¼
CX (t,s)dt ds
2T
T T
2 ð T ð T
1
¼
CX (s,t)dt ds, X(t) is stationary
2T
T T
ð19:3:4Þ
Substitution of t ¼ s 2t and u ¼ t in Eq. (19.3.4), transforms the coordinate system from
[s,t] to [t,u]. The Jacobian kJk of the transformation is given by
2
3
@t
1
@s 7
¼
@s 5
0
@s
@t
6 @t
kJk ¼ 4
@s
@t
1
1
¼1
(19:3:5)
Thus
s2m^ X ¼
1
2T
2 ð ð
CX (t)dt du
(19:3:6)
The integration has to be carried out over the parallelogram shown in Fig. 19.3.1 in regions
I and II.
Region I: t 0: Limits for t are from – 2T to 0; limits for u are from – T to T þ t.
Region II: t . 0: Limits for t are from 0 to 2T; limits for u are from – T þ t to T.
FIGURE 19.3.1
19.3
Ergodic Processes
Substituting these limits in Eq. (19.3.6), we obtain the variance s2m^ X :
2 ð T ð 2T
ð Tþt ð 0
1
2
sm^ X ¼
CX (t)dt du þ
CX (t)dt du
2T
Tþt
0
T
2T
ð 2T
ð0
1
CX (t)½2T tdt þ
CX (t)½2T þ tdt
¼ 2
4T
0
2T
ð 2T
ð 2T
1
1
jtj
¼ 2
CX (t)½2T jtjdt ¼
CX (t) 1 dt
4T
2T 2T
2T
2T
441
(19:3:7)
Finally, if the random process is to be mean-ergodic, the following condition has to be
satisfied:
ð 2T
1
jtj
2
CX (t) 1 lim s ¼ lim
dt ! 0
(19:3:8)
T!1 m^ X
T!1 2T
2T
2T
Since
ð 2T
ð 2T
jtj
jtj
dt CX (t) 1 jCX (t)j 1 jCX (t)jdt
(19:3:9)
dt 2T
2T
2T
2T
2T
Ð1
hence, if 1 jCX (t)jdt is bounded, then limT!1 s2m^ X does indeed go to zero. A necessary
and sufficient condition for a random process to be mean-ergodic is
ð1
jCX (t)jdt , 1
(19:3:10)
ð 2T
1
Even though we have derived necessary and sufficient condition for a mean-ergodic
process, we cannot use it to test the ergodicity of any process since it involves prior knowledge of the autocovariance function CX(t). However, if a partial knowledge of the ACF
such as jCX(t)j goes to 0 as t ! 1, then we can conclude that the process is ergodic.
Although ergodic processes must be stationary, stationary processes need not be
ergodic. Since it is rather difficult to show that any arbitrary random process is ergodic,
we will assume, unless otherwise stated, that a process that is stationary is also ergodic.
Correlation-Ergodic
The time autocorrelation function of a stationary random process X(t) is defined by
ð
1 T
R^ X (l) ¼
X(t)X(t þ l)dt
(19:3:11)
2T T
A stationary random process X(t) is correlation-ergodic if
ðT
ð
1 T ^
1
RX (l)dt ¼ lim
lim
X(t)X(t þ l)dt ¼ RX (l)
T!1 2T T
T!1 2T T
(19:3:12)
for all l. Clearly R^ X (l) is a random variable, but
ð T
ð T
1
1
E
E ½R^ X (l) ¼
X(t)X(t þ l)dt ¼
RX (l)dt ¼ RX (l)
2T
2T T
T
and R^ X (l) is an unbiased estimator of RX (l). As in the case of mean-ergodic processes,
here also the variance of R^ X (l) must tend to 0 as T ! 1. Hence we can form the
442
Random Processes
variance s2R^ l of R^ X (l) for any fixed l:
X
( ð
)
h
2
i2 T
1
2
½X(t)X(t þ l) RX (l)dt
sR^ l ¼ E R^ X (l) RX (l)
¼E
X
2T T
2 ð T ð T
1
¼
E ½ X(t)X(t þ l) RX (l)½ X(s)X(s þ l) RX (l) dt ds
2T
T T
2 ð T ð T
1
¼
CRlX (t,s)dt ds
2T
T T
2 ð T ð T
1
CRlX (s t)dt ds
¼
2T
T T
(19:3:13)
Substituting (s – t) ¼ t in Eq. (19.3.13), we obtain a result similar to Eq. (19.3.8), namely
ð 2T
1
jtj
lim s2R^ l ¼ lim
CRlX (t) 1 dt ! 0
(19:3:14)
T!1
T!1 2T
X
2T
2T
and the necessary and sufficient condition for X(t) to be correlation-ergodic is
ð1
CRlX (t) dt , 1
(19:3:15)
1
where
CRlX (t) ¼ RRlX (t) R2X (l)
RRlX ¼ E ½X(t)X(t þ l)X(t þ t)X(t þ t þ l)
Thus the autocorrelation ergodicity involves the fourth-order stationary moments of X(t).
As a corollary to this result, we can define mean-square or power ergodicity as
ðT
1
lim
X 2 (t)dt ¼ RX (0)
(19:3:16)
T!1 2T T
and the necessary and sufficient conditions for mean-square ergodicity are
ð 2T
1
jtj
lim s2R^ 0 ¼ lim
CR0 X (t) 1 dt ! 0
T!1
T!1 2T
X
2T
2T
(19:3:17)
or
ð1
1
jCR0 X (t)jdt , 1
(19:3:18)
where
CR0 X (t) ¼ R0RX (t) R2X (0)
R0RX (t) ¼ E{½X(t)X(t þ t)2 }
Example 19.3.1 We will revisit Example 19.2.1, where X(t) ¼ A sin(v0t þ F), where
A and F are independent and A is uniformly distributed over (0,1) and F is uniformly
distributed over (0,2p) and find whether it is mean- and power-ergodic.
19.3
Ergodic Processes
443
We have already seen in Example 19.2.1 that
E ½X(t) ¼ E ½AE ½sin(v0 t þ F) ¼ 0
RX (t) ¼ E ½X(t)X(t þ t) ¼ E ½A1 E ½sin(v0 t þ F) sin(v0 t þ v0 t þ F)
1
¼ E ½A2 cos(v0 t)
2
and X(t) is indeed stationary. In fact, it is also strict-sense stationary. We will check first
whether it is mean-ergodic. Since the mean value is zero, CX (t) ¼ 12 E ½A2 cos(v0 t). The
variance of the time average s2m^ X is given by
s2m^ X
E ½A2 ¼
2T
ð 2T jtj cos(v0 t)
dt
1
2T
2
2T
We can use the result from Fourier transform that time multiplication is frequency convolution and evaluate the integral shown above. Substituting the Fourier transforms of the
triangular and the cosine functions
jtj
sin(vT ) 2
cos(v0 t)
p
1
()
½d(v þ v0 ) þ d (v v0 )
and
() 2T
2T
vT
2
2
in the equation for the variance, we obtain
ð
1 E ½A2 1
sin(vT ) 2 p
s2m^ X ¼
½d(v þ v0 ) þ d(v v0 )dv
2T
2p 2T 1
vT
2
E ½A2 sin(v0 T ) 2
¼
2p
v0 T
2
and as T ! 1, sm^ X ! 0, for v0 = 0. Thus X(t) is mean-ergodic.
Substituting E ½A2 ¼ 13 in the variance equation, we obtain
1 sin(v0 T ) 2
2
sm^ X ¼
6p
v0 T
To check for power ergodicity, it is simpler to use Eq. (19.3.16) rather than Eq. (19.3.17).
Thus
1
lim
T!1 2T
ðT
1
X (t) dt ¼ lim
T!1
2T
T
2
ðT
T
A2 cos2 (v0 t þ F) dt ¼
A2
2
for
v0 = 0
Since this expression does not converge to RX (0) ¼ 16, the process X(t) is neither powerergodic nor correlation-ergodic.
Example 19.3.2 A random process X(t) ¼ A, where A is random variable uniformly distributed over (0,1]. Since E ½A ¼ 12 and RA (t) ¼ E ½A2 ¼ 13, this process is stationary. To
check for ergodicity in the mean, we apply Eq. (19.3.2) for the time average and check
whether it tends to the ensemble average as T tends to 1. Thus
ð
ð
1 T
1 T
m^ X ¼
X(t)dt ¼
A dt ¼ A as T ! 1
2T T
2T T
444
Random Processes
is not equal to the ensemble average mX ¼ 12. Hence X(t) is neither mean-ergodic nor
correlation-ergodic.
Example 19.3.3 We now consider a random process X(t) ¼ B þ A sin(v0t þ F), where
B, A, and F are independent random variables. A and B are uniformly distributed over
(0,1] and (0,2] respectively. We will check whether X(t) is mean-ergodic.
We first determine whether X(t) is stationary. The mean value mX is given by
E ½B þ A sin(v0 t þ F) ¼ mB þ E ½A sin(v0 t þ F) ¼ mB
since E ½A sin(v0 t þ F) ¼ 0 from previous examples. The autocorrelation function is
RX (t) ¼ E{½B þ A sin(v0 t þ F)½B þ A sin(v0 t þ v0 t þ F)}
¼ E ½B2 þ 0 þ 0 þ E ½A2 E ½sin(v0 t þ F)A sin(v0 t þ v0 t þ F
1
¼ E ½B2 þ E ½A2 E ½cos(v0 t) cos(2v0 t þ v0 t þ 2F
2
1
¼ E ½B2 þ E ½A2 cos(v0 t)
2
Hence the process X(t) is stationary. The autocovariance function CX(t) is given by
1
CX (t) ¼ RX (t) m2X ¼ E ½B2 þ E ½A2 cos(v0 t) m2B
2
1
¼ s2B þ E ½A2 cos(v0 t)
2
and substituting this equation in Eq. (19.3.7), we obtain the variance of the time average as
ð 2T 1
1
jtj
s2B þ E ½A2 cos(v0 t) 1 dt
s2m^ X ¼
2T 2T
2
2T
"
#
2
1
E ½A2 sin(v0 T ) 2
2
2 2T sin(v0 T )
2
2TsB þ E ½A ¼ sB þ
¼
2T
2p
v0 T
2p
v0 t
and
(
lim s2
T!1 m^ X
¼
lim s2B
T!1
)
E ½A2 sin(v0 T ) 2
þ
¼ s2B
2p
v0 T
if
v0 = 0
We find that even though X(t) is stationary, it is not mean-ergodic. Hence the time average
is not a good estimator of the ensemble average. We can now substitute E ½A2 ¼ 13 and
s2B ¼ 13 in the variance equation and write
s2m^ X ¼ s2B þ
E ½A2 sin(v0 T ) 2 1
sin(v0 T )
¼ 1þ
2p
v0 T
3
2pv0 T
Example 19.3.4 The autocovariance of a zero mean Gaussian random process X(t) is
given by CX (t) ¼ s2X e2ajtj , where s2X is its variance. We will check whether this
19.4
Estimation of Parameters of Random Processes
445
process is mean-ergodic and correlation-ergodic. The variance of the time average from
Eq. (19.3.7) is
s2m^ X
ð 2T
jtj
1
jtj
2 2ajtj
CX (t) 1 s e
1
dt ¼
dt
2T
2T 2T X
2T
2T
s2X
1 e4aT
1
¼
2aT
4aT
1
¼
2T
ð 2T
and limT!1 s2m^ X ¼ 0. Hence the process is mean-ergodic. To determine whether it is
correlation-ergodic, the fourth moment RRlX (t) ¼ E ½X(t)X(t þ l)X(t þ t)X(t þ t þ l)
is calculated from the following result. If fX1,X2,X3,X4) are zero mean Gaussiandistributed random variables, then E ½X1 X2 X3 X4 ¼ E ½X1 X2 E ½X3 X4 þ E ½X1 X3 E ½X2 X4 þ
E ½X1 X4 E ½X2 X3 . Hence
E ½X(t)X(t þ l)X(t þ t)X(t þ t þ l) ¼ E ½X(t)X(t þ l)E ½X(t þ t)X(t þ t þ l)
þ E ½X(t)X(t þ t)E ½X(t þ l)X(t þ t þ l)
þ E ½X(t)X(t þ t þ l)E ½X(t þ l)X(t þ t)
¼ R2X (l) þ R2X (t) þ RX (t þ l)RX (t l)
The covariance CRlX (t) ¼ RRlX (t) R2X(l) ¼ R2X (t) þ RX (t þ l)RX (t l), and substituting for
the correlation functions, we obtain CRlX ðtÞ as
CRlX (t) ¼ s4X e4ajtj þ s4X e2ajtþlj e2ajtlj ¼ 2s4X e4ajtj , t . l
Substituting for CRlX (t) in Eq. (19.3.14), we obtain
lim s2^ l
T!1 RX
1
¼ lim
T!1 2T
¼ lim
s4X
T!1 aT
ð 2T
2s4X e4ajtj
2T
1
jtj
1
dt
2T
1 e8aT
!0
8aT
A similar result holds good for t , l. We conclude from the preceding result that the
process X(t) is also correlation-ergodic and hence power-ergodic.
B 19.4 ESTIMATION OF PARAMETERS OF
RANDOM PROCESSES
19.4.1
Continuous-Time Processes
The ergodic hypothesis detailed in the previous section enables us to define estimators for
the parameters of a random process from its sample realization. If X(t) is a sample realization of a stationary ergodic random process available only for a time duration T, then the
estimators for the parameters, mean, variance, autocorrelation, autocovariance, and normalized autocovariance can be defined.
446
Random Processes
Mean:
m^ X ¼
1
T
ðT
X(t)dt
(19:4:1)
0
We note that m^ X is an unbiased estimator of mX since
ðT
ð
1
1 T
X(t)dt ¼
E½X(t)dt ¼ mX
E ½m^ X ¼ E
T 0
T 0
The variance of m^ X is very similar to Eq. (19.3.7) and is given by
ð
1 T
jtj
¼
CX (t) 1 var½m^ X ¼
dt
T T
T
ð
h
2 T
ti
CX (t) 1 dt , even function
¼
T 0
T
s2m^ X
(19:4:2)
Variance:
s^ 2X
1
¼
T
ðT
½X(t) m^ X 2 dt
(19:4:3)
0
If the mean value mX is known, then s^ 2X will be an unbiased estimator. We will
determine the conditions under which s^ 2X given by Eq. (19.4.3) can be an unbiased
estimator:
ðT
1
½X(t) m^ X 2 dt
T 0
ðT
1
^ mX ) (m^ X mX )2 dt
¼E
½X(t)
T 0
( ð "
# )
1 T (X(t) mX )2 þ (m^ X mX )2
¼E
dt
T 0 2(X(t) mX )(m^ X mX )
( ð
)
1 T
2
¼E
½(X(t) mX ) dt þ E (m^ X mX )2
T 0
s2
s2
E ½s^ 2X ¼ E
X
2
E
T2
m^ X
ð T ð T
½X(t) mX ½X(s) mX dsdt
0
0
and from Eq. (19.4.2) we have
E ½s^ 2X ¼ s2X þ s2m^ X 2
T
ð T
jtj
CX (t) 1 dt
T
T
s2m^
¼
s2X
s2m^ X
X
(19:4:4)
Hence s^ 2X cannot be an unbiased estimator for finite T with the bias equal to s2m^ X .
However, it is asymptotically unbiased.
19.4
Estimation of Parameters of Random Processes
447
Autocovariance. Two estimators for the autocovariance function are given by
ð Tt
1
C X (t) ¼
½X(t) m^ X ½X(t þ t) m^ X dt, 0 , t T
(19:4:5)
T t 0
ð
1 Tt
½X(t) m^ X ½X(t þ t) m^ X dt, 0 , t T
(19:4:6)
C^ X (t) ¼
T 0
If the mean mX is known, then CX (t) given by Eq. (19.4.5) is an unbiased estimator
since
ð Tt
1
E ½CX (t) ¼
E{½X(t) mX ½X(t þ t) mX } dt
T t 0
¼ CX (t)
T t
¼ CX (t), 0 , t T
T t
and C^ X (t) is a biased estimator as shown below:
ð
1 Tt
^
½X(t) mX ½X(t þ t) mX dt
CX (t) ¼
T 0
h
T t
ti
¼ CX (t) 1 , 0 , t T
¼ CX (t)
T
T
with bias equal to CX (t)(t=T ).
In actual practice only the sample mean m^ X is known, and we will determine whether
C X (t) can be an unbiased estimator. Equation (19.4.5) can be rewritten as
ð Tt1
1
E ½C X (t1 ) ¼ E
½X(t) m^ X ½X(t þ t1 ) m^ X dt
T t1 0
1
¼E
T t1
ð Tt1
½(X(t) mX ) ðm^ X mX )
0
½(X(t þ t1 ) mX ) (m^ X mX )dt
,
0 , t1 T
ÐT
and substituting for m^ X ¼ (1=T ) 0 X(s)ds in the equation above, we have
ð Tt1
1
2
½(X(t) mX )(X(t þ t1 ) mX ) þ (m^ X mX ) dt
E ½CX (t1 ) ¼ E
T t1 0
ð Tt1
1
E
½(X(t) mX )(m^ X mX )
T t1 0
þ ½(X(t þ t1 ) mX )(m^ X mX )dt
ð Tt1 ð T
1
¼ CX (t1 ) þ s2m^ X CX (s t)ds dt
T(T t1 ) 0
0
ð Tt1 ð T
1
CX (s t t1 )ds dt, 0 , t1 T
T(T t1 ) 0
0
where
E ½(m^ X mX )2 ¼ s2m^ X ¼
1
T
jtj
CX (t) 1 dt [from Eq. (19:4:2)
T
T
ðT
(19:4:7)
448
Random Processes
FIGURE 19.4.1
In Eq. (19.4.7) we make the transformation t ¼ (s – t) and u ¼ t. As the Jacobian of the
transformation is 1, we can use techniques similar to those employed in Eqs. (19.3.5)–
(19.3.7) to determine E ½C X (t1 ). This transformation amounts to finding the integral in the
parallelogram shown in Fig. 19.4.1.
The integration along the t axis is from – T to (T – t1). Thus
ð ð Tt1
1
E ½CX (t1 ) ¼ CX (t1 ) þ s2m^ X du
½CX (t) þ CX (t t1 )dt
(19:4:8)
T(T t1 )
T
Ð
The limits of integration for du are determined as follows in the three regions:
Region I: 2T , t 2 t1
jtj
du ¼ T þ t ¼ T 1 T
(19:4:9a)
t1
du ¼ T t1 ¼ T 1 T
(19:4:9b)
t þ t1
du ¼ T t t1 ¼ T 1 T
(19:4:9c)
ð tþT=2
T=2
Region II: 2t1 , t 0
ð T=2t1
T=2
Region III: 0 , t (T 2 t1)
ð T=2t1
tT=2
Substituting Eqs. (19.4.9) in Eq. (19.4.8), we obtain E ½C X (t1 ) as follows:
E½CX (t1 ) ¼ CX (t1 ) þ s2m^ X
9
8 ð t1 jtj
>
>
>
>
1
(t)
þ
C
(t
t
)dt
½C
X
X
1
>
>
>
>
T
>
>
T
>
>
>
>
ð
=
<
i
0 h
1
t1
, 0 , t1 T
1 ½CX (t) þ CX (t t1 )dt
þ
>
(T t1 ) >
T
t1
>
>
>
>
ð
>
>
Tt1 h
>
>
>
>
t þ t1 i
>
;
: þ
1
½CX (t) þ CX (t t1 )dt >
T
0
(19:4:10)
19.4
Estimation of Parameters of Random Processes
449
For large values of T or for low values of the lag t1, the terms within braces can be approximated as 2s2m^ X , and the approximation for E ½CX (t1 ) is
E ½C X (t1 ) CX (t1 ) s2m^ X ,
0 , t1 T
(19:4:11)
Thus, CX (t) is a biased estimator with the bias B s2m^ X .
In an analogous manner the expected value of C^ X (t1 ) given by Eq. (19.4.6) can be
evaluated as follows:
ð Tt1
1
^
E ½CX (t1 ) ¼ E
½(X(t) mX )(X(t þ t) mX ) þ E(m^ X mX )2 dtg
T 0
ð Tt1
1
E
½(X(t) mX )(m^ X mX ) þ ½(X(t þ t1 ) mX )(m^ X mX )dt
T 0
ð
ð
T t1
1 Tt1 T
½CX (t1 ) þ s2m^ X 2
¼
CX (s t)ds dt
T 0
T
0
ð
ð
1 Tt1 T
CX (s t t1 )ds dt, 0 , t1 T
2
T 0
0
(19:4:12)
By substituting t ¼ (s – t) and u ¼ t and performing the integration in the parallelogram
shown in Fig. 19.4.1, the final result similar to Eq. (19.4.10) is
T t1
E ½C^ X (t1 ) ¼
½CX (t1 ) þ s2m^ X T
9
8 ð t1 jtj
>
>
>
>
1
(t)
þ
C
(t
t
)
dt
½C
X
X
1
>
>
>
>
T
>
>
T
>
>
>
>
ð
=
<
i
0 h
1
t1
, 0 , t1 T
1
þ
½CX (t) þ CX (t t1 ) dt
>
T>
T
t1
>
>
>
>
ð Tt1 h
>
>
>
>
>
>
t þ t1 i
>
;
: þ
1
½CX (t) þ CX (t t1 ) dt >
T
0
(19:4:13)
For large values of T or for low values of the lag t1, the terms within braces can be approximated as ½(T t1 )=T2s2m^ X , and the approximation for E ½C^ X (t1 ) is
h
t1 i
E ½C^ X (t1 ) ½CX (t1 ) s2m^ X 1 (19:4:14)
, 0 , t1 T
T
Thus, C^ X (t1 ) is also a biased estimator with a larger bias than C X (t) with the bias
h
t1
t1 i
B CX (t1 ) s2m^ X 1 , 0 , t1 T
T
T
We will show later that C^ X (t) is the preferred estimator as it gives better minimum meansquare error.
Example 19.4.1 The autocovariance function of a random process X(t) is CX (t) ¼ eajtj ,
with a ¼ 0.5. The data interval T ¼ 50. We will find the variance of the estimated mean
450
Random Processes
value m^ X as a function of T, compare the true value CX(t) to E½CX ðtÞ given by
Eq. (19.4.10) and E ½C^ X (t) given by Eq. (19.4.13), and finally compare the approximations
to E ½C^ X (t) given by Eq. (19.4.14).
Variance s2m^ X (T ): From Eq. (19.4.2), we obtain
s2m^ X (T )
2
¼
T
¼
ð T
2½e
ð
h
ti
2 T at h
ti
CX (t) 1 dt ¼
e
1 dt
T
T 0
T
0
aT
þ aT 1
a2 T 2
A graph of s2m^ X (T ) is shown in Fig. 19.4.2 for a ¼ 0.5, and the estimate of the mean
m^ X is asymptotically unbiased since limT!1 s2m^ X (T ) ! 0.
Means of Estimated Autocovariances. C X (t) and C^ X (t). From Eq. (19.4.10) the
mean value of C X (t) is given by
2 e50a þ 50a 1
E CX (t) ¼ e
þ
2500a2
ðt
ab
1
jbj
e
þ ea(bt) 1 db
50 t 50
50
ð0
ab
1
t
e
þ ea(bt) 1 db
50 t t
50
ð 50t
ab
1
bþt
e
þ ea(bt) 1 db,
50 t 0
50
at
FIGURE 19.4.2
0 , t 50
19.4
Estimation of Parameters of Random Processes
451
and from Eq. (19.4.13) the mean value of C^ X (t) is given by
h
i 50 t 2 e50a þ 50a 1
at
^
e
E CX (t) ¼
þ
2500a2
50
ðt
ab
1
jbj
a(bt)
e
þe
1
db
50 50
50
ð
1 0 ab
t
e
þ ea(bt) 1 db
50 t
50
ð
1 50t ab
bþt
e
þ ea(bt) 1 db,
50 0
50
0 , t 50
In graphing the estimated autocovariance functions, it is usual to plot the lag length
t upto 10– 20% of the length T. Expected values E ½C X (t) and E ½C^ X (t) are shown
along with CX (t) in Fig. 19.4.3 for t ¼ 0– 10 and tabulated in Table 19.4.1. The
bias as shown in the figure is approximately equal to B s^ 2mX (T ) ¼ 0:0768.
FIGURE 19.4.3
TABLE 19.4.1.
t
CX (t)
E[C X (t)]
CX (t)
E[ĈX (t)]
ĈX (t)]
0
1
2
3
4
5
6
7
8
9
10
1
0.606531
0.367879
0.22313
0.135335
0.082085
0.049787
0.030197
0.018316
0.011109
0.006738
0.9232
0.529171
0.291129
0.147612
0.061327
0.009859
20.020499
20.038005
20.047664
20.052531
20.054435
0.9232
0.529731
0.291079
0.14633
0.058535
0.005285
20.027013
20.046603
20.058484
20.065691
20.070062
0.9232
0.518588
0.279484
0.138755
0.056421
0.008873
20.018039
20.032684
20.040038
20.044075
20.043548
0.9232
0.519136
0.279436
0.13755
0.053852
0.004756
20.023771
20.040078
20.049127
20.053867
20.05605
452
Random Processes
The figure also shows that there is very little difference between E ½C X (t) and
E ½C^ X (t). Table 19.4.1 also shows the approximation E ½C X (t) given by
Eq. (19.4.11) and the approximation E ½C^ X (t) given by Eq. (19.4.14).
Covariance of ACF Estimators. Before finding the covariance of the estimator C X (t)
at two different lags t1 and t2 , we will first find the covariance cov ½X(t)X(tþt1 ),
X(t)X(s þ t2 ):
cov½X(t)X(t þ t1 ), X(s)X(s þ t2 )
¼ E{½(X(t) mX )(X(t þ t1 ) mX )½(X(s) mX )(X(s þ t2 ) mX )}
¼ E ½(X(t) mX )(X(t þ t1 ) mX )(X(s) mX )(X(s þ t2 ) mX )
E ½(X(t) mX )(X(t þ t1 ) mX )E ½(X(s) mX )(X(s þ t2 ) mX )
(19:4:15)
Expressing E ½(X(t) mX )(X(t þ t1 ) mX )(X(s) mX )(X(s þ t2 ) mX ) of Eq.
(19.4.15) in terms of the fourth cumulant k4 from Eq. (11.6.8), we obtain
E ½(X(t) mX )(X(t þ t1 ) mX )(X(s) mX )(X(s þ t2 ) mX )
¼ k4 (s t, t1 , t2 ) þ CX (t1 )CX (t2 )
þ CX (s t)CX (s t þ t2 t1 ) þ CX (s t þ t2 )CX (s t t1 )
and substituting in Eq. (19.4.15), we obtain
cov½X(t)X(t þ t1 ), X(s)X(s þ t2 )
¼ k4 (s t, t1 , t2 ) þ CX (s t)CX (s t þ t2 t2 ) þ CX (s t þ t2 )CX (s t t1 )
(19:4:16)
which is an exact expression. The presence of the fourth cumulant k4 in
Eq. (19.4.16) renders evaluation of the covariance intractable. However, if X(t)
is a Gaussian random process, then the fourth cumulant k4 is zero and the covariance of [X(t)X(t þ t1), X(s)X(s þ t2)] becomes
cov½X(t)X(t þ t1 ), X(s)X(s þ t2 )
¼ CX (s t)CX (s t þ t2 t2 ) þ CX (s t þ t2 )CX (s t t1 )
(19:4:17)
Assuming without loss of generality that 0 t1 t2, the covariance of the estimator C X (t) at two different lags t1 and t2 can be given by
cov½C X (t1 ), CX (t2 )
ð Tt1
ð Tt2
1
1
¼ cov
X(t)X(t þ t1 ) dt
X(s)X(s þ t2 ) ds
(T t1 ) 0
(T t2 ) 0
ð Tt2 ð Tt1
1
¼
cov½X(t)X(t þ t1 )X(s)X(s þ t2 )dt ds
(T t1 )(T t2 ) 0
0
(19:4:18)
19.4
Estimation of Parameters of Random Processes
453
FIGURE 19.4.4
Substituting for the covariance in Eq. (19.4.18) from Eq. (19.4.16), we obtain
cov½CX (t1 ), C X (t2 )
¼
1
(T t1 )(T t2 )
ð Tt2 ð Tt1
0
0
CX (s t)CX (s t þ t2 t1 )
þ CX (s t þ t2 )CX (s t t1 )dt ds
(19:4:19)
We now make the transformation t ¼ (s 2 t) and u ¼ t, and as the Jacobian of the
transformation is 1, we can use techniques similar to those employed in Eq. (19.4.8)
to determine the covariance. This transformation amounts to finding the integral in
the parallelogram shown in Fig. 19.4.4. The integration along the t axis is from
2(T 2 t2) to (T 2 t1). Thus, for a Gaussian process X(t), we obtain
cov½CX (t1 ), CX (t2 )
ð ð Tt1
CX (t)CX (t þ t2 t1 )
1
du
¼
(T t1 )(T t2 )
(Tt2 ) þ CX (t þ t2 )CX (t t1 )d t
(19:4:20)
Ð
The integral du in Eq. (19.4.20) is calculated in three regions as shown in Fig.
19.4.4 in a manner similar to that used in Eqs. (19.4.9):
Region I: 2(T 2 t2) , t 0
ð tþT=2t2
T=2
jtj t2
du ¼ T þ t t2 ¼ T 1 T
(19:4:21a)
Region II: 0 (t2 2 t1)
ð tþT=2t2
tT=2
t2
du ¼ T t2 ¼ T 1 T
(19:4:21b)
454
Random Processes
Region III: (t2 2 t1) , t (T 2 t1)
ð T=2t1
tT=2
t þ t1
du ¼ T t t1 ¼ T 1 T
(19:4:21c)
Substituting Eqs. (19.4.21) into Eq. (19.4.20), the covariance between adjacent values t1
and t2 of the sample autocovariance for a Gaussian process X(t) is given by
1
(T t1 )(T t2 )
8 ð0
(T jtj t2 )½CX (t)CX (t þ t2 t1 )
>
>
>
>
>
þCX (t þ t2 )CX (t t1 )dt
>
>
> Tt
ð t22 t1
<
(T t2 )½CX (t)CX (t þ t2 t1 )
þ
>
þCX (t þ t2 )CX (t t1 )dt
0
>
>
ð Tt1
>
>
(T t t1 )½CX (t)CX (t þ t2 t1 )
>
>
>
:þ
þCX (t þ t2 )CX (t t1 )dt
t2 t1
cov½CX (t1 ), C X (t2 ) ¼
9
>
>
>
>
>
>
>
>
=
>
>
>
>
>
>
>
>
;
(19:4:22)
Although Eq. (19.4.22) was derived with the condition 0 t1 t2, it can be shown that it
holds good for all t1 and t2. The covariance
Ð is generally difficult to evaluate. For large T,
(T 2 t1) T (T 2 t2), and the integral du T in Eq. (19.4.21). With these approximations the distinction between CX (t) and C^ X (t) becomes moot and Eq. (19.4.22)
becomes
cov½CX (t1 ), C X (t2 ) cov½C^ X (t1 ) C^ X (t2 )
ð
1 1
½CX (t)CX (t þ t2 t1 ) þ CX (t þ t2 )CX (t t1 )dt
T 1
(19:4:23)
The variance of the sample autocovariance function is obtained by substituting
t1 ¼ t2 ¼Ð s in Eqs. (19.4.20) and (19.4.22). With this substitution Eqs. (19.4.21)
become du ¼ T 2 s 2 jtj, and the general form for the variance is
var½C X (s) ¼
ð Ts
1
(T jsj)2
(T s jtj)½CX2 (t) þ CX (t þ s)CX (t s) dt,
s,T
(19:4:24)
(Ts)
and the corresponding variance for the estimator C^ X (t) is obtained by substituting T 2 for
(T jsj)2 in the denominator of Eq. (19.4.24). The estimator C^ X (t) is more often used
since it yields a minimum mean-square estimator as shown in Example 19.4.2.
The approximations for the variances CX (s) and C^ X (t) corresponding to Eq. (19.4.16)
for large T are
ð
1 1
^
var½CX (s) var½CX (s) ½C 2 (t) þ CX (t þ s)CX (t s)dt
(19:4:25)
T 1 X
Example 19.4.2 We will continue Example 19.4.1 assuming that X(t) is a zero mean
Gaussian random process with autocovariance function CX(t) ¼ e – ajtj. Since we are
19.4
Estimation of Parameters of Random Processes
455
given that X(t) is zero mean, C X(t) will be an unbiased estimator. We have to find the
variances of the two estimators C X(t) and ĈX(t) and calculate the minimum meansquare error for both. Substituting CX(t) ¼ e – ajtj in Eq. (19.4.24), the variance of C X(t)
is given by
var CX (s) ¼
1
ðT jsjÞ2
ð Ts
(T s jtj) e2ajtj þ eajtþsj eajtsj dt
(19:4:26)
(Ts)
Since the integrand is even, Eq. (19.4.26) can be rewritten as
var C X (s) ¼
2
ð T sÞ 2
ð Ts
(T s t) e2at þ eajtþsj eajtsj dt
(19:4:27)
0
where we have removed the absolute sign from the quantities where it is clear that t is
always positive. Equation (19.4.27) can integrated numerically to obtain the variance of
the unbiased estimator. However, it is instructive to find a closed-form solution. The absolute value signs in the product term e – ajtþsj e – ajt2sj has to be carefully removed. In
Fig. 19.4.5a, s , T/2 and the range of integration for t contains two segments (0,s)
and (s, T 2 s).
In Fig. 19.4.5b, s . T/2 and the range of integration for t contains only one segment
(0, T – s).
Hence we can write
0,s
e
T
2
ajtþsj ajtsj
e
(
ea(tþs) ea(ts) ¼ e2as ,
0,ts
ea(tþs) ea(ts) ¼ e2at ,
s,tT s
¼
T
,sT
2
eajtþsj eajtsj ¼ ea(tþs) ea(ts) ¼ e2as ,
0,tT s
(19:4:28)
Substituting Eq. (19.4.28) in Eq. (19.4.27), the variance of the unbiased estimator ĈX(s)
can written as follows:
9
8 ðs
2at
>
>
2as
>
>
(T s t) e
þe
dt
>
>
>
>
>
>
0
>
>
>
>
ð
<
Ts
T=
2
2at
2at
þ
(T
s
t)
e
þ
e
dt
,
0
,
s
(19:4:29)
var C X (s) ¼
2>
ð T sÞ 2 >
s
>
>
>
> ð Ts
>
>
>
>
T
>
>
>
;
:
,sT>
(T s t) e2at þ e2ats dt,
2
0
456
Random Processes
FIGURE 19.4.5
Integrating Eq. (19.4.29), we obtain the variance of the estimator CX(s) as
var CX (s)
9
8 2a(Ts)
e
þ 2a(T s) 1 þ e2a(Ts)
>
>
>
>
>
>
< þe2as 2a(T 2s) 1 þ 2a2 s(2T 3s),
0 , s T2 =
1
¼
2
>
2a2 ðT sÞ2 >
>
>
> e2at(Ts) þ 2a(T s) 1 þ e2as T s , T , s T >
;
:
T
2
ð19:4:30Þ
The variance of the biased estimator ĈX(s) obtained by replacing (T 2 s)2 by T 2 in the
denominator of Eq. (19.4.30). The minimum mean-square errors for both the estimators
19.4
Estimation of Parameters of Random Processes
457
FIGURE 19.4.6
are given by
2
1 2U (s) ¼ E C X (s) CX (s) ¼ var CX (s)
h
i2
h
i
1 2 (s) ¼ E C^ X (s) CX (s) ¼ var C^ X (s) þ B2
(19:4:31)
B
The mean-square error for the unbiased estimator ĈX(s) is the same as its variance.
The bias B for ĈX(s) as shown earlier for a zero mean process is B ¼ 2CX(t)fjtj/T],
and substituting for B in Eq. (19.4.31), the mean-square error for the biased estimator
ĈX(s) is
h
i t
1 2B (s) ¼ var C^ X (s) þ
T
2
e2as
(19:4:32)
The mean-square errors and the variances are shown in Fig. 19.4.6 for T ¼ 4, a ¼ 0.9.
It is seen from the figure that the mean-square error for a biased estimator ĈX(s) is less
than that of the unbiased one ĈX(s).
Autocorrelation. If m̂X is zero in Eqs. (19.4.5) and (19.4.6), we obtain two corresponding estimators for the autocorrelation function RX(t):
ð Tjtj
1
X(t)X(t þ t)dt, 0 t T
(19:4:33a)
Unbiased: RX (t) ¼
T jtj 0
ð
1 Tjtj
^
X(t)X(t þ t)dt
(19:4:34a)
Biased: RX (t) ¼
T 0
RX(t) is an unbiased estimator since
ð Tjtj
1
X(t)X(t þ t)dt
E RX (t) ¼ E
T jtj 0
ð Tjtj
1
E½ X(t)X(t þ t)dt ¼ RX (t)
¼
T jtj 0
(19:4:33b)
458
Random Processes
and E [R̂X(t)] is a biased estimator as shown below:
ð
i
T jtj Tjtj
jtj
^
E RX (t) ¼ E
X(t)X(t þ t)dt ¼ 1 RX (t)
T
T
0
h
(19:4:34b)
However, R̂X(t), although biased, gives a smaller mean-square error as shown in
Example 19.4.2 and hence is the preferred estimator.
Normalized Autocovariance:
r^ X (t) ¼
C^ X (t)
C^ X (0)
(19:4:35)
r̂X(t) will not be an unbiased estimator.
19.4.2 Discrete-Time Processes
In actual practice observations are made on the sample function of a stationary ergodic
random process X(t) at equally spaced time intervals t1, t2, . . . , tn with corresponding
random variables X1, X2, . . . , Xn. These observation random variables will not be independent and will be used in the estimation of mean and autocovariance of the random process.
Since fXig are samples of a stationary random process, the means and variances of these
samples are the same as those of the original process:
E ½Xi ¼ mX : var½Xi ¼ s2X , i ¼ 1, . . . ; n
(19:4:36)
Estimation of the Mean
In the previous chapter we estimated the mean m̂X of a random variable by minimizing
its variance from a set of independent observation random variables fX1, . . . , Xng. We can
use the same techniques to estimate the mean m̂X of the random process X(t) except that the
set fX1, . . . , Xng is not independent. We define the estimated mean m̂X as the weighted sum
of the dependent observations fX1, . . . , Xng:
m^ X ¼
n
X
ai Xi ¼ aT X ¼ XT a
(19:4:37)
i¼1
We have to determine the weight vector a if possible under the criteria of unbiasedness and
minimum variance. The unbiasedness criterion yields the condition
n
X
ai ¼ aT 1 ¼ 1
(19:4:38)
i¼1
where 1 is a unity vector defined by 1 ¼ [1, . . . ,1]T. The variance of the estimator m̂X is
given by
var(m^ X ) ¼ s2m^ X ¼ E aT (X mX )(X mX )T a ¼ aT CX a
(19:4:39)
19.4
Estimation of Parameters of Random Processes
459
where CX is the autocovariance matrix of the vector random variable X, defined by
CX ¼ E (X mX )(X mX )T
2
3
CX (1)
CX (2)
CX (n 1)
CX (0)
6 CX (1)
CX (0)
CX (1)
CX (n 2) 7
6
7
6
7
C
(2)
C
(1)
C
(0)
C
(n
3)
6
7 ð19:4:40Þ
X
X
X
X
¼6
7
..
..
..
..
6
7
4
5
.
.
.
.
CX ½(n 1) CX ½(n 2) CX ½(n 3) with CX(h) defined as the discrete autocovariance given by
CX (h) ¼ E (Xi mX )(Xiþh mX
CX (0)
(19:4:41)
and CX(h) ¼ CX( –h) since CX is a real symmetric matrix. Equation (19.4.40) has to be
minimized subject to the constraint of Eq. (19.4.38). The unconstrained function to be
minimized after adjoining with the Lagrange multiplier l is given by
J ¼ aT CX a þ l(aT 1 1)
(19:4:42)
T
where 1 is an n vector of ones given by 1 ¼ [1, 1, . . . , 1] . Differentiating with respect to
the vector a, we obtain
@J
@ T
¼
a CX a þ l(aT 1 1) ¼ 2CX a þ l1 ¼ 0
@a @a
Hence
l
a ¼ C1
1
2 X
(19:4:43)
Solving for l from the constraint equation [Eq. (19.4.38)] and substituting this result in
Eq. (19.4.43), we obtain
a¼
C1
X
1
1T C1
X 1
(19:4:44)
The solution for the coefficients fai, i ¼ 1, . . . , ng will not be the same for every n. If the Xi
values are independent, then CX ¼ I and Eq. (19.4.44) reduces to
1
I
1
1
s2X
a¼
1¼ T 1¼ 1
1
n
1 1
1T 2 I:1
sX
(19:4:45)
and all coefficients have the same weight equaling 1/n, a result already obtained in
Eq. (18.3.9). As it is, each ai in Eq. (19.4.44) will be different and the solution becomes
cumbersome. To make the solution intuitively appealing we will assume that all the coefficients faig have the same weight ¼ 1/n and the estimator is given by
n
1X
m^ X ¼
Xi
(19:4:46)
n i¼1
The estimator given by Eq. (19.4.46) will be unbiased but will not be minimum variance.
The variance of the estimator m̂X can be obtained from Eq. (19.4.39)
1
(19:4:47)
var(m^ X ) ¼ s2mX ¼ 2 1T CX 1
n
460
Random Processes
or, in expanded form
var(m^ X ) ¼
s2mX
"
#
n1 X
1
h
CX (0) þ 2
¼
1 CX (h)
n
n
h¼1
(19:4:48)
The variance given by Eq. (19.4.48) is not minimum. For independent Xis, CX(h) ¼ 0,
h = 0, and the variance s2m^ X , given by
s2m^ X ¼
1 T
nCX (0) s2X
1 CX ¼
¼
2
n
n2
n
is a minimum, a result that has been derived previously [in Eq. (18.3.11)].
Example 19.4.3 The autocovariance function of a random process X(t) is
CX (h) ¼ eajhj , with a ¼ 0.5 as in Example 19.4.1. The number of data points n ¼ 50.
We will find the variance of the estimated value m^ x . From Eq. (19.4.48) we can write
the following equation for the estimated variance:
"
#
n1 X
1
h
1þ2
var(m^ X ) ¼ s2m^ X (n) ¼
1 eah
n
n
h¼1
A closed-form solution for the summation given above can be obtained as
a
s2m^ X (n) ¼
1 (n þ 2)e2a þ (n 2)e ne3a 2ea(n2) þ 2ea(n1) n
n2
(1 ea )(1 2ea þ e2a )
and substituting a ¼ 0.5 in this equation, we obtain
s2m^ X (n) ¼
pffiffiffi
1
½7:3258e(n=2) (e e) þ 4:083n 7:83541
n2
and this is graphed in Fig. 19.4.7 along with the value CX(0)/n as the variance of the estimated mean for independent observations. The variance for n ¼ 50 is s2m^ X (50) ¼ 0:0785.
We can see from the figure that the estimated mean m^ X is asymptotically unbiased since
the variance s2m^ X (n) ! 0 as n ! 1.
FIGURE 19.4.7
19.4
Estimation of Parameters of Random Processes
461
Estimation of the Autocovariance
The autocovariance of a stationary ergodic random process X(t) has already been
defined [Eq. (19.2.15)] as
CX (h) ¼ E{½X(t) mX ½X(t þ h) mX }
(19:4:49)
where t has been replaced by h for convenience. The estimator CX (h) for the ACF from the
observed random variables set {X1 , . . . , Xn } will be of the form
CX (h) ¼
nh
X
ai aiþh (Xi m^ X )(Xiþh m^ X )
(19:4:50)
i¼1
where m^ X is the unbiased estimator of the mean from Eq. (19.4.46). We have to choose, if
possible, the coefficients {ai , i ¼ 1, . . . , n} such that E ½CX (h) ¼ CX (h). Equation
(19.4.50) can be rewritten as follows:
CX (h) ¼
nh
X
ai aiþh ½(Xi mX ) (m^ X mX )½(Xiþh mX ) (m^ X mX )
i¼1
Expanding this equation and taking expectations, we obtain
"
#
nh
X
(Xi mX )(Xiþh mX ) þ (m^ X mX )2
E ½CX (h) ¼
ai aiþh E
(Xi mX )(m^ X mX ) (Xiþh mX )(m^ X mX )
i¼1
¼
nh
X
ai aiþh {CX (h) þ s2m^ X } i¼1
"
¼
ai aiþh E ½(Xi mX )(m^ X mX )
i¼1
nh
X
ai aiþh E(Xiþh mX )(m^ X mX )
i¼1
nh
X
nh
X
#
ai aiþh (CX (h) þ
s2m^ X )
(S1 þ S2 )
(19:4:51)
i¼1
where the summations S1 and S2 have been defined as follows:
S1 ¼
nh
X
ai aiþh E ½(Xi mX )(m^ X mX )
i¼1
S2 ¼
nh
X
(19:4:52)
ai aiþh E ½(Xiþh mX )(m^ X mX )
i¼1
Substituting for m^ X from Eq. (19.4.46) in the summation S1 in Eq. (19.4.52), we can write
S1 ¼
nh X
n
nh X
n
1X
1X
ai aiþh E ½(Xi mX )(X j mX ) ¼
ai aiþh CX ( j i)
n i¼1 j¼1
n i¼1 j¼1
(19:4:53)
This equation represents the sum of every term of the (n 2 h) n matrix shown in
Fig. 19.4.8.
462
Random Processes
FIGURE 19.4.8
The summation is carried out in three stages. The shaded parallelogram in region II
consists of the sum of (n 2 h) (h þ 1) terms, and the shaded triangles in regions I and
III consists of the remaining (n 2 h) (n 2 h 2 1) terms. Equation (19.4.53) can be
rewritten in terms of these summations as follows:
S1 ¼
8
h X
nh
X
>
>
<
ai aiþh CX ( j)
1
n>
>
:
j¼0 i¼1
nh1
nh
X X
þ
ai aiþh CX ( j)
j¼1 i¼jþ1
Region II
nh1
X nhj
X
þ
j¼1
9
>
>
ai aiþh CX ( j þ h)=
i¼1
Region I
Region III
>
>
;
(19:4:54)
An equation similar to Eq. (19.4.53) for S2 is
S2 ¼
¼
nh X
n
1X
ai aiþh E ½(Xiþh mX )(X j mX )
n i¼1 j¼1
nh X
n
1X
ai aiþh CX ( j i h)
n i¼1 j¼1
(19:4:55)
A summation matrix similar to that in Fig. 19.4.8 is shown in Fig. 19.4.9.
From Fig. 19.4.9 we can write the summation in much the same way as we did in
Eq. (19.4.54):
S2 ¼
9
8 h nh
nh1
nh1
nh
XX
X
X X
X nhj
>
>
>
=
ai aiþh CX ( j) þ
ai aiþh CX ( j) þ
ai aiþh CX ( j þ h)>
1<
n>
>
:
j¼0 i¼1
j¼1
Region II
j¼1 i¼jþ1
i¼1
Region III
Region I
>
>
;
(19:4:56)
19.4
Estimation of Parameters of Random Processes
463
FIGURE 19.4.9
Substituting Eqs. (19.4.55) and (19.4.56) in Eq. (19.4.51), we can write E ½CX (h) as
"
!#
nh
h
X
2X
2
E ½CX (h) ¼
ai aiþh CX (h) þ sm^ X CX ( j)
n j¼0
i¼1
"
#"
#
nhj
nh1
X
1 X
(ai aiþh þ aiþj aiþhþj )
CX ( j) þ CX (h þ j)
n i¼1
j¼1
(19:4:57)
where the variance of the sample mean m^ X is as follows:
"
#
n1 X
1
h
CX (0) þ 2
s2m^ X ¼
1 CX (h)
n
n
h¼1
(19:4:48)
The analysis becomes intractable finding the coefficients faig such that E[CX(h)] ¼ CX (h).
In the spirit of Eqs. (19.4.5) and (19.4.6), two estimators for the ACF CX (h) can be defined
by substituting ai aiþh ¼ ½1=(n h) or ai aiþh ¼ (1=n) in Eq. (19.4.50):
CX (h) ¼
nh
1 X
(Xi m^ X )(Xiþh m^ X )
n h i¼1
1
C^ X (h) ¼
n
nh
X
(Xi m^ X )(Xiþh m^ X )
(19:4:58)
(19:4:59)
i¼1
The expected value E ½C X (h) is obtained by substituting ai aiþh ¼ ½1=(n h) in
Eq. (19.4.57)
"
#
h
2X
2
E ½CX (h) ¼ CX (h) þ sm^ X CX ( j)
n j¼0
"
#
nh1
X
2
(n h j)½CX ( j) þ CX (h þ j)
(19:4:60)
n(n h) j¼1
464
Random Processes
and the expected value E ½C^ X (h) is obtained by substituting ai aiþh ¼ (1=n) in Eq. (19.4.57)
"
!#
h
X
h
2
E ½C^ X (h) ¼
1
CX ( j)
CX (h) þ s2m^ X n
n j¼0
"
#
X
2 nh1
2
(n h j)½CX ( j) þ CX (h þ j)
n
j¼1
(19:4:61)
where s2m^ X in these equations is as given by Eq. (19.4.48):
s2m^ X
"
#
n1 X
1
h
CX (0) þ 2
¼
1 CX (h)
n
n
h¼1
(19:4:48)
Example 19.4.4 We will follow Example 19.4.3 and find E ½C X (h) and E ½C^ X (h) for the
random process whose ACF CX (h) ¼ eajhj with a ¼ 0.5 and the number of data points
n ¼ 50. In Example 19.4.3, we have calculated s2m^ X for 50 data points as
s2m^ X (50) ¼ 0:0785. E ½C X (h) for n ¼ 50 is obtained from Eq. (19.4.60) as
"
#
h
X
2
eaj
E ½C X (h) ¼ eah þ s2m^ X (50) 50 j¼0
"
#
4X
9h
2
aj
a(hþj)
(50 h j)½e þ e
a ¼ 0:5
50(50 h) j¼1
Similarly, E ½C^ X (h) is obtained from Eq. (19.4.61):
"
#
h
h
2 X
ah
2
aj
^
þ sm^ X (50) e
E ½CX (h) ¼ 1 e
50
50 j¼0
"
#
9h
2 4X
aj
a(hþj)
2
(50 h j)½e þ e
a ¼ 0:5
50 j¼1
E ½C X (h), E ½C^ X (h) and the true value CX (h) are shown in Fig. 19.4.10 for lags h ¼ 10
(10% of data points). It is seen that both estimators are biased but there is very little difference between them for the 10 lags shown here.
Variance of Autocovariance Estimators
Finding the variances of C X (h) and C^ X (h) is extremely tedious and is not worth the
effort, particularly when the number of data points n is large. Hence the sample mean
m^ X can be replaced by the true mean mX in Eqs. (19.4.58) and (19.4.59). The unbiased
estimator corresponding to Eq. (19.4.58) is given by
C X (h) ¼
nh
1 X
(Xi mX )(Xiþh mX )
n h i¼1
(19:4:62)
19.4
Estimation of Parameters of Random Processes
465
FIGURE 19.4.10
and E ½C X (h) ¼ CX (h). The biased estimator corresponding to Eq. (19.4.59) is
CX (h) ¼
nh
1X
(Xi mX )(Xiþh mX )
n i¼1
(19:4:63)
As a first step toward finding the variance C X (h), we will define a random variable Zi as
Zi ¼ (Xi mX )(Xiþh mX ), and Eq. (19.4.62) becomes
CX (h) ¼
nh
1 X
Zi
n h i¼1
From Eq. (19.4.48) the variance of C X (h) can be written as follows:
"
#
nh1
X
1
g
Cz (0) þ 2
1
var½CX (h) ¼
CZ (g)
nh
nh
g¼1
(19:4:64)
(19:4:65)
The mean value of Zi is CX (h) for all i and the autocovariance of Zi can be defined by
CZ (g) ¼ E ½(Zi CX (h))(Ziþg CX (h)) ¼ E ½Zi Ziþg CX2 (h)
(19:4:66)
E ½Zi Ziþg ¼ E ½(Xi mX )(Xiþh mX )(Xiþg mX )(Xiþgþh mX )
(19:4:67)
where
Since E(Xk mX ) ¼ 0 for all k, we can write E ½Zi Ziþg in terms of the fourth cumulant k4
from Eq. (11.6.8):
E ½Zi Ziþg ¼ k4 (h,g) þ CX2 (h) þ CX2 (g) þ CX (g þ h)CX (g h)
(19:4:68)
and substituting this equation in Eq. (19.4.66), we obtain
CZ (g) ¼ k4 (h,g) þ CX2 (g) þ CX (g þ h)CX (g h)
(19:4:69)
CZ (0) ¼ k4 (h,0) þ CX2 (0) þ CX2 (h)
(19:4:70)
466
Random Processes
Substituting Eqs. (19.4.69) and (19.4.70) in Eq. (19.4.65), we obtain
var½CX (h)
9
8
k4 (h,0) þ CX2 (0) þ CX2 (h)
>
>
>
>
>
>
=
1 <
nh1
¼
X
g
n h>
>
>
1
½k4 (h,g) þ CX2 (g) þ CX (g þ h)CX (g h) >
>
>
;
: þ2
n
h
g¼1
(19:4:71)
For a Gaussian process the fourth cumulant k4 is zero, and hence Eq. (19.4.71) becomes
9
8 2
CX (0) þ CX2 (h)
>
>
>
>
>
>
=
<
1
nh1
var½CX (h) ¼
X
g
n h>
>
>
1
½CX2 (g) þ CX (g þ h)CX (g h) >
>
>
;
:þ2
n
h
g¼1
(19:4:72)
Or, more compactly
1
var½CX (h) ¼
(n h)2
(
nh1
X
)
(n h jgj)½CX2 (g)
h,n
þ CX (g þ h)CX (g h) ,
g¼(nh1)
(19:4:73)
Equation (19.4.73) is the discrete counterpart of the continuous process of Eq. (19.4.24)
and represents the variance of an unbiased estimator C X (h) of a Gaussian process with
known mean value mX.
In a similar manner the variance of the biased estimator C^ X (h) of a Gaussian process
with known mean value mX given by Eq. (19.4.63) can be obtained as follows:
9
8 2
CX (0) þ CX2 (h)
>
>
>
>
>
>
=
<
n
h
nh1
var½C^ X (h) ¼ 2
X
g
n >
>
>
1
½CX2 (g) þ CX (g þ h)CX (g h) >
>
>
;
: þ2
n
h
g¼1
(19:4:74)
Example 19.4.5 We will continue the same way as in Example as 19.4.4 and find
the variances of the unbiased estimator CX (h) and the biased estimator C^ X (h). We
are given CX (h) ¼ e 2ajhj with a ¼ 0.5 and n ¼ 50. From Eq. (19.4.73), var½CX (h) is
given by
9
8
1 þ e2ah
>
>
=
<
1
nh1
X
g
var½CX (h) ¼
2ag
a(jgþhj) a(jghj)
1
þe
e
>
½e
n h>
;
: þ2
nh
g¼1
As in Example 19.4.2, the absolute value signs in the product term eajgþhj eajghj have to
be carefully removed. Following the procedure in that example, we can write
19.4
0h
n1
:
2
eajgþhj eajghj ¼
Estimation of Parameters of Random Processes
ea(gþh) ea(gh) ¼ e2ah ,
ea(gþh) ea(gh) ¼ e2ag ,
467
0gh
h,gnh1
n1
, h n:
2
eajgþhj eajghj ¼ ea(gþh) ea(gh) ¼ e2ah ,
0gnh1
The variance equation CX (h), after substituting the equations above, is
var½C X (h)
9
8
h X
g
>
>
>
>
2ah
2ag
2ah
>
>
þ2
1
þe
1þe
½e
>
>
>
>
n
h
>
>
>
>
g¼1
>
>
>
>
>
>
nh1
=
<
X
g
n
1
1
2ag
2ag
1
þe
, 0 h þ2
½e
¼
nh
2 >
n h>
>
> g¼hþ1
>
>
>
>
>
>
>
> nh1
X
>
>
g
n
1
>
>
2ag
2ah
>
>
>
>
,
h
n
2
1
þ
e
,
½e
;
:
nh
2
g¼1
The variance of the biased estimator C^ X (h) is calculated by substituting (n h)=n2 for
1=(n h) in the preceding equation. Both variances are shown in Fig. 19.4.11.
The result is very similar to the one for the continuous case given in Fig. 19.4.6, and
the biased estimator gives a better minimum mean-square error.
Estimation of Autocorrelation Functions
The two estimators for the autocorrelation function are obtained by substituting m̂ ¼ 0
in the two estimators for the autocovariance function given by Eqs. (19.4.58) and (19.4.59):
R X (h) ¼
nh
1 X
Xi Xiþh
n h i¼1
FIGURE 19.4.11
(19:4:75a)
468
Random Processes
RX (h) is an unbiased estimator since
E ½RX (h) ¼
nh
nh
1 X
1 X
E ½Xi Xiþh ¼
RX (h) ¼ RX (h)
n h i¼1
n h i¼1
1
R^ X (h) ¼
n
nh
X
Xi Xiþh
(19:4:75b)
(19:4:76a)
i¼1
R^ X (h) is biased estimator since
nh
nh 1 X
1
^
E ½Xi Xiþh ¼ 1 RX (h)
E ½RX (h) ¼
n n h i¼1
n
(19:4:76b)
with the bias B ¼ {½RX (h)=n}.
Further analyses of these estimators are the same as those of the ACF estimators C X (h)
and C^ X (h) with the estimated mean m^ X set equal to 0.
Estimation of Normalized Autocovariance Functions (NACFs)
Estimation of Mean As mentioned earlier, the normalized autocovariance rX (h)
(NACF) plays a key role in modeling time series. The NACF for a stationary random
process for discrete time can be defined by an equation analogous to Eq. (19.2.16) as follows:
rX (h) ¼
CX (h)
CX (0)
and CX (h) ¼ rX (h)CX (0)
(19:4:77)
The unbiased estimator C X (h) for large n is defined in Eqs. (19.4.62). The estimated NACF
r X (h) for large sample n can be defined as
r X (h) ¼
C X (h)
C X (0)
(19:4:78)
resulting in a ratio of random variables, and the analysis becomes rather involved. We will
now determine the mean of r X (h). Since CX (h) is an unbiased estimator, we can define its
0
0
variations about the mean value CX (h) as C X (h) with E ½C X (h) ¼ 0 and write CX (h) ¼
0
CX (h)þ C X (h):
"
#
0
CX (h)
CX (h) þ CX (h)
E ½rX (h) ¼ E
¼E
0
CX (0)
CX (0) þ CX (0)
8
"
#1 9
0
>
>
>
C
(0)
>
0
>
>
>
>½CX (h) þ C X (h) 1 þ X
>
>
=
<
CX (0)
¼E
>
>
CX (0)
>
>
>
>
>
>
>
>
;
:
(19:4:79)
19.4
Estimation of Parameters of Random Processes
469
0
Assuming that the variations C X (h) are small, we can expand Eq. (19.4.79) in Taylor
series, and neglecting second-order terms, we have
"
#
0
1
CX (h)C X (0)
CX (h)
0
E ½rX (h) E CX (h) þ CX (h) ¼E
¼ rX (h)
CX (0)
CX (0)
CX (0)
(19:4:80)
0
Thus, we have an unbiased estimator under the assumption of small variations in C X (h).
Estimation of Covariance of NACF Finding the covariance between adjacent
values of the estimated NACF r X (h) is very involved and is detailed elsewhere in the
literature [8,24,31]. The final result for a Gaussian process X(t) is as follows:
cov½rX (h), r X (h þ g)
9
8
1
>
>
½rX (k)rX (k þ g) þ rX (k þ h þ g)rX (k h) >
>
>
>
>
>
nh
>
>
>
>
>
>
2
>
>
>
>
r
(h)r
(k)r
(k
h
g)
>
>
X
X
X
=
<
1
X
nhg
¼
>
> 2 r (h þ g)r (k)r (k h)
>
k¼1>
>
>
X
X
>
>
nh X
>
>
>
>
>
>
>
>
>
>
2
>
>
;
: þ r2 (k)r (h)r (h þ g)
X
X
X
n
(19:4:81)
The variance of r X (h) is obtained by substituting g ¼ 0 in Eq. (19.4.81). The result is
9
8
1
>
1 >
=
<
½r2X (k) þ rX (k þ h)rX (k h)
X
nh
(19:4:82)
var½rX (h) ¼
4
2
>
>
k¼1: rX (h)rX (k)rX (k h) þ r2X (k)r2X (h) ;
nh
n
Large-Lag Approximations
For most stationary random processes the NACF becomes negligibly small after some
lag q, that is, rX (h) 0, jhj . q. Hence, under the assumption rX(h) ¼ 0 for jhj . q in
Eq. (19.4.81), we can write the large-lag covariance as
cov½rX (h), r X (h þ g) q
1 X
r (k)rX (k þ g),
n h k¼q X
jhj . q
(19:4:83)
Similarly, the large-lag variance after substituting rX(h)¼0 for jhj . q in Eq. (19.4.82) is
var½rX (h) q
1 X 2
r (k),
n h k¼q X
jhj . q
(19:4:84)
If the estimator for ACF is C^ X (h), then the divisor (n 2 h) in Eqs. (19.4.83) and (19.4.84) is
replaced by the divisor n. Thus the variance equation [Eq. (19.4.84)] becomes
var½^rX (h) q
1X 2
r (k),
n k¼q X
jhj . q
(19:4:85)
470
Random Processes
It has been shown [3,25] that if the true NACF rX(h) is zero for jhj . q, then the estimated
NACF rX(h) is normally distributed with zero mean and variance given by Eq. (19.4.84).
This property is called Anderson’s theorem.
Discrete White-Noise Process
A zero mean stationary discrete white-noise process fXig has the following property:
E ½Xi Xj ¼ dji where the Krönecker delta dji ¼ 1, j ¼ i and dji ¼ 0, j = i, an equation
analogous to Eq. (19.2.17) for continuous white noise. As a consequence, the NACF rX(h)
has the property
1, h ¼ 0
(19:4:86)
rX (h) ¼
0, h = 0
Hence the variance of the estimated r X (h) for white noise is obtained by substituting
rX (k) ¼ 0 for k = 0 in Eq. (19.4.84) to yield
var½rX (h) 1
nh
(19:4:87)
According to Anderson’s theorem, the estimated NACF r X (h) of a stationary discrete
white-noise process is Gaussian-distributed with zero mean and variance 1=(n h). As
a result, we can construct a test for white noise from knowledge
of r X (h). Since the
pffiffiffiffiffiffiffiffiffiffiffi
variance is 1=(n h), the 95% confidence interval (+2s) is +2= n h.
If the estimated NACF is given by
r^ X (h) ¼
C^ X (h)
C^ X (0)
(19:4:88)
then, from Eq. (19.4.85), the variance of the estimator r^ X (h) is
1
n
2ffiffi
p
and the corresponding confidence interval is + n.
var½^rX (h) (19:4:89)
Example 19.4.6 We have generated 100 points of a zero mean unit variance Gaussian
white noise fXig by the computer. The plot is shown in Fig. 19.4.12.
The estimated NACFs r X (h) and r^ X (h) of this sequence were determined from
Eqs. (19.4.78) and (19.4.88) and plotted in Fig. 19.4.13 along with their confidence
intervals for 50 lags.
In the plot of the estimator r X (h), only two points out of 50 (4%) exceed the confidence limits, whereas in the estimator r^ X (h), no points exceed the confidence limits.
Both estimators pass the white-noise test at the 5% significance level comfortably, and
hence we conclude that the process fXig is indeed a white-noise sequence.
Example 19.4.7 We will generate 100 points of the discrete-time process for f ¼ 0.8
and 20.8 in Example 19.2.11 with s2X ¼ 1
Xi ¼ fXi1 þ ni , i ¼ 1, . . . , 100
and estimate the NACF and compare it to the true NACF.
The computer-generated random sequence is shown in Fig. 19.4.14 for f ¼ 0.8
and 20.8. The figure shows that for f ¼ 20.8, the process is more oscillatory. The
19.4
Estimation of Parameters of Random Processes
FIGURE 19.4.12
FIGURE 19.4.13
FIGURE 19.4.14
471
472
Random Processes
FIGURE 19.4.15
estimated NACFs r X (h) and r^ X (h) are determined from Eqs. (19.4.78) and (19.4.88)
respectively. They are shown in Fig. 19.4.15 for f ¼ 0.8 and 20.8 for 10 lags.
Here also the NACFs for f ¼ 20.8 show oscillatory behavior. For the 10 lags
shown,there is very little difference between the two estimators r X (h) and r^ X (h).
However, the discrepancy between the true NACF and the estimators for higher lags is
to be expected.
B 19.5
POWER SPECTRAL DENSITY
19.5.1 Continuous Time
In signal analysis power spectra are associated with Fourier transforms that transform
signals from the time domain to the frequency domain. The same concept is also
applicable to stationary random processes. The correlation functions represent stationary
processes in the time domain. We can transform them to the frequency domain by
taking their Fourier transforms. The power spectral density (psd) function SX (v) of a
real stationary random process X(t) is defined as the Fourier transform of the autocorrelation function:
ð1
RX (t)ejvt dt
(19:5:1)
SX (v) ¼ FT½RX (t) ¼
1
From the Fourier inversion theorem we can obtain the autocorrelation function from the
power spectral density:
ð
1 1
RX (t) ¼ IFT½SX (v) ¼
SX (v)e jvt dv
(19:5:2)
2p 1
Equations (19.5.1) and (19.5.2) are called the Wiener –Khinchine theorem.
Since RX(0) ¼ E [X 2(t)], the average power in the random process, we obtain the following from Eq. (19.5.2):
RX (0) ¼ E ½X 2 (t) ¼
1
2p
ð1
ð1
SX (v)dv ¼
1
SX ( f )df
1
(19:5:3)
19.5
Power Spectral Density
473
Thus, SX ( f ) represents the average power per hertz, and hence the term power spectral
density. Since RX (t) is an even function, we can rewrite Eq. (19.5.1) as
ð1
SX (v) ¼
ð1
RX (t)½cos(vt) þ j sin(vt)dt ¼
RX (t) cos(vt)dt
1
(19:5:4)
1
and SX (v) is also an even function. Hence Eq. (19.5.2) can be rewritten as
RX (t) ¼
1
2p
ð1
SX (v) cos(vt)dv
(19:5:5)
1
The cross-spectral density (csd) SXY(v) of two real stationary random processes X(t)
and Y(t) is defined as the Fourier transform of the cross-correlation function, RXY(t)
ð1
RXY (t)ejvt dt
SXY (v) ¼ FT½RXY (t) ¼
(19:5:6)
1
and the inverse Fourier transform of SXY(v) gives the cross-correlation function:
RXY (t) ¼ IFT½SXY (v) ¼
1
2p
ð1
SXY (v)e jvt dv
(19:5:7)
1
The cross-spectral density SXY(v) will be complex, in general, even when the random processes X(t) and Y(t) are real.
The Fourier transforms can also be obtained from the tables in Appendix A.
Example 19.5.1 The power spectral density (psd) of the random binary wave of
Example 19.2.6 is to be determined. The autocorrelation of the random process X(t) is
given by
8
<
jtj
A 1
,
RX (t) ¼
T
:
0,
2
jtj , T
otherwise
The psd SX (v) given by
ð0
SX (v) ¼
ðT t
t
A2 1 þ ejvt dt þ A2 1 ejvt dt
T
T
T
0
A2
½(1 e jvT þ jvT ) þ (1 ejvT jvT )
v2 T
sin(vT=2) 2
¼ A2 T
vT=2
¼
is shown in Fig. 19.5.1. As we can see from figure, the psd is an even function.
Example 19.5.2 We will find the psd from the autocorrelation RX (t) ¼ 14½1 þ e2ljtj of
the random telegraph wave given in Example 19.2.7. Taking the Fourier transform, the psd
474
Random Processes
FIGURE 19.5.1
is given by
SX (v) ¼
1
4
ð1
1
ejvt dt þ
1
4
ð0
e2lt ejvt dt þ
1
1
4
ð1
e2lt ejvt dt
0
1
4l
p
l
¼ 2pd(v) þ 2
¼ d(v) þ 2
2
4
2
4l þ v
4l þ v2
The impulse function (p=2)d(v) represents the direct-current (dc) value of RX (t) ¼ 14. The
psd is shown in Fig. 19.5.2 for l ¼ 1.
Example 19.5.3 We will find the psd from the autocorrelation RX (t) ¼ e2ljtj cos(v0 t)
of Example 19.2.8. We can take the Fourier transform of RX(t) directly, but it is easier to
FIGURE 19.5.2
19.5
Power Spectral Density
475
evaluate the FT by using the frequency convolution property of the FT as follows:
FT½e2ljtj ¼
4l2
4l2 þ v2
and
FT½cos(v0 t) ¼ p½d(v þ v0 ) þ d(v v0 )
Using the frequency convolution property x(t)y(t) , (1=2p)X(v) Y(v), we have
e2ljtj cos(v0 t) ()
or
1
4l2
p½d(v þ v0 ) þ d(v v0 )
2p 4l2 þ v2
1
1
þ
SX (v) ¼ 2l
4l2 þ (v þ v0 )2 4l2 þ (v v0 )2
and simplifying, we obtain
SX (v) ¼
4l(v2 þ v20 þ 4l2 )
v4 2(v20 4l2 )v2 þ (v20 þ 4l2 )2
With l ¼ 1 and v0 ¼ 2p, the psd becomes
SX (v) ¼
4(v2 þ 4p2 þ 4l2 )
v4 8(p2 1)v2 þ 16(p2 þ 1)2
The psd SX(v) is shown in Fig. 19.5.3.
Example 19.5.4 (Bandlimited Process) The autocorrelation function RX (t) ¼
sin(v0 t)=pt of a stationary random process X(t) is shown in Fig. 19.5.4 for v0 ¼ 2p.
We have to find the psd SX(v).
From item 2 in the FT table, the Fourier transform of RX (t) can be obtained as follows:
sin(v0 t)
() pv0 (v), hence SX (v) ¼ pv0 (v)
pt
The psd SX(v) is shown in Fig. 19.5.5 for v0 ¼ 2p.
The random process that has a psd as in Fig. 19.5.5 is called a bandlimited signal since
the frequency spectrum exists only between –2p and 2p.
FIGURE 19.5.3
476
Random Processes
FIGURE 19.5.4
FIGURE 19.5.5
Example 19.5.5 (Bandlimited Process)
RX (t) ¼
The autocorrelation function RX(t)
sin½v0 (t t0 ) sin½v0 (t þ t0 )
þ
p(t t0 )
p(t þ t0 )
of a bandlimited random process X(t) is shown in Fig. 19.5.6 for v0 ¼ 2p and t0 ¼ 3.
FIGURE 19.5.6
19.5
Power Spectral Density
477
Since sin(v0 t)=pt , pv0 (v) using the time-shifting property x(t + t0 ) , X(v)e+jvt0
of the FT, we have
sin½v0 (t t0 )
() pv0 (v)ejvt0 ,
p(t t0 )
sin½v0 (t þ t0 )
() pv0 (v)e jvt0
p(t þ t0 )
or
sin½v0 (t t0 ) sin½v0 (t þ t0 )
þ
() pv0 (v)½e jvt0 þ ejvt0 p(t t0 )
p(t þ t0 )
and
SX (v) ¼ 2pv0 (v) cos(vt0 )
The psd is shown in Fig. 19.5.7 for v0 ¼ 2p and t0 ¼ 3.
Example 19.5.6 (Bandpass Process)
RX (t) ¼
The autocorrelation function RX(t)
2 sin(v0 t)
cos(vc t)
pt
of a bandpass random process X(t) is shown in Fig. 19.5.8 for v0 ¼ 2p and vc ¼ 16p.
We will find the psd using the frequency convolution property of the FT:
xðtÞyðtÞ ()
1
XðvÞ YðvÞ
2p
Since 2 sin(v0 t)=pt , 2pv0 (v) and cos(vc t) , p½d(v þ vc ) þ d(v vc Þ, we have
2 sin(v0 t)
1
cos(vc t) ()
2pv0 (v) p½d(v þ vc ) þ d(v vc )
pt
2p
FIGURE 19.5.7
478
Random Processes
FIGURE 19.5.8
and the psd
SX (v) ¼ pv0 (v þ vc ) þ pv0 (v vc )
The function SX(v) is shown in Fig. 19.5.9 for v0 ¼ 2p and vc ¼ 16p.
Example 19.5.7 (White Noise) A white-noise process has the autocorrelation function
RX (t) ¼ s2X d(t) given in Eq. (19.2.17). The psd of white noise is given by
SX (v) ¼ s2X ,
1 , v , 1
and has a flat spectrum. Since the process has all frequencies with equal power, it is
called “white Ðnoise,” analogously to white light. It has infinite energy since
1
RX (0) ¼ (1=2p) 1 s2X dv ¼ 1, and hence it is an idealization.
Example 19.5.8 The cross-correlation function of two random processes X(t) ¼
A cos(v0 t) þ B sin(v0 t) and Y(t) ¼ A sin(v0 t) þ B cos(v0 t) in Example 19.2.9 is
RXY (t) ¼ s2 sin(v0 t), E[A 2] ¼ E[B 2] ¼ s2. We will find the cross-spectral density
SXY(v).
Taking FT of s2 sin(v0 t), from tables, we have
SXY (v) ¼ jps2 ½d(v þ v0 ) d(v v0 )
and this is shown in Fig. 19.5.10.
Note that SXY(v) is neither even nor real.
FIGURE 19.5.9
19.5
Power Spectral Density
479
FIGURE 19.5.10
Example 19.5.9 The cross-spectral density is to be found for the cross-correlation
function RXY(t) derived in Example 19.2.10:
8
> 4 (t=2) 1e2t , t . 0
1 < 15 e
6
RXY (t) ¼ þ
1
2 >
2t
: e ,
t0
10
Taking FT of each term in RXY(t), we have
4 (t=2)
4
1
e
;
()
15
15 1=2 þ jv
1
() pd(v);
2
1 2t
1 1
e
()
6
6 2 þ jv
1 2t
1
1
e ()
10
10 2 jv
The cross-spectral density SXY(v) is obtained by adding the FT terms shown above:
SXY (v) ¼ pd(v) þ
4
1
1 1
1
1
1 8 þ 2v2 þ 3jv
þ
¼
þ pdðvÞ
þ
15 12 þ jv 6 2 þ jv 10 2 jv 3 (1 þ 2jv)(4 þ v2 )
The cross-spectral density SXY(v) is complex and possesses no symmetry, unlike the
power spectral density. The constant term 12 in RXY(t) gives rise to the impulse function
in the frequency domain. The amplitude jSXY (v)j and the phase argfSXY(v)g spectra are
graphed in Fig. 19.5.11.
Properties of Power Spectral Densities of Stationary Random Processes
1. SX(v) is a real function. In general, the Fourier transform X(v) of any function X(t)
will be complex. However, the ACF RX(t) is an even function and satisfies the
relation RX(t) ¼ RX(2t). Hence, from the definition of psd, we obtain
ð1
ð1
RX (t)ejvt dt ¼
RX (t)½cos(vt) þ j sin(vt)dt
SX (v) ¼
1
1
ð1
RX (t) cos(vt)dt
¼
1
Ð1
since the imaginary part 1 RX (t) j sin(vt)dt ¼ 0 because an even function multiplying an odd function results in an odd function and the integral of an odd function over (21,1) is zero.
2. From property 1 the psd SX(v) is an even function and, hence it is a function of v2.
As a consequence SX(v) ¼ SX(2v).
(19.5.8)
3. SX(v) 0: The psd is a nonnegative function of v.
(19.5.9)
480
Random Processes
FIGURE 19.5.11
4. From the Fourier transform properties:
ð
ð1
1 1
SX (v)dv ¼
SX ( f )df ¼ E ½X 2 (t)
RX (0) ¼
2p 1
1
ð1
RX (t)dt
SX (0) ¼
(19:5:10)
1
5. If Z(t) ¼ X(t) þ Y(t), then from Eq. (19.2.30), we have
RZ (t) ¼ RX (t) þ RY (t) þ RXY (t) þ RYX (t)
SZ (v) ¼ SX (v) þ SY (v) þ SXY (v) þ SYX (v)
If X(t) and Y(t)
SZ (v) ¼ SX (v) þ SY (v)
are
orthogonal,
then
Eq.
(19.5.11)
(19:5:11)
reduces
to
Alternate Form for Power Spectral Density
The psd can also be obtained directly from the stationary random process X(t). We
will truncate X(t) in the interval (2T, T ) and define the random process XT (t) as
X(t), T t T
XT (t) ¼
(19:5:12)
0,
otherwise
ÐT
The FT XT (v) of XT (t), given by XT (v) ¼ T X(t)ejvt dt, is a random variable. The quantity ST (v), defined by
ð
2
1 T
jXT (v)j2 XT (v)XT (v)
(19:5:13)
¼
X(t)ejvt dt ¼
ST (v) ¼
2T T
2T
2T
is called the periodogram. The periodogram represents the power of the sample function
X(t) at the frequency v. Equation (19.5.13) can be expanded as follows:
ð ð
1 T T
ST (v) ¼
X(t)X(s)ejv(ts) dt ds
(19:5:14)
2T T T
We can make the transformation t ¼ t 2 s and u ¼ s in Eq. (19.5.14) and perform the integration in the (t, u) plane. The Jacobian k J k of the transformation from Eq. (19.3.5) is 1,
19.5
Power Spectral Density
481
and Eq. (19.5.14) can be transformed as follows:
ST (v) ¼
1
2T
ð ð
X(u)X(u þ jtj)du ejvt dt
(19:5:15)
We can obtain the limits of integration from Fig. 19.3.1 and write Eq. (19.5.15) as
ð 2T ð T
1
ST (v) ¼
X(u)X(u þ t)du ejvt dt
2T Tþt
0
ð 0 ð Tjtj
1
X(u)X(u þ t)du ejvt dt
þ
2T 2T T
ð0
ð 2T
R^ X (t)ejvt dt þ
R^ X (t)ejvt dt
¼
0
2T
ð 2T
R^ X (t)ejvt dt
¼
(19:5:16)
2T
where we have used the definition for R^ X (t) as in Eq. (19.4.34a). Using Eq. (19.4.34b), the
expected value of ST(v) can be written as
ð 2T h
ð 2T
i
jtj jvt
jvt
^
E½ST (v) ¼
E RX (t) e
dt ¼
RX (t) 1 dt
(19:5:17)
e
2T
2T
2T
Taking the limit of E½ST (v) in Eq. (19.5.17) as T ! 1, we have
ð 1
jtj jvt
RX (t) 1 dt
lim E½ST (v) ¼ lim
e
T!1
T!1
2T
1
jtj
¼ lim FT RX (t) 1 T!1
2T
(19:5:18)
Using the frequency convolution property of the FT, we obtain
1
1
X(v) Y(v) ¼
x(t)y(t) ()
2p
2p
ð1
X(p)Y(v p)dp
1
and with RX (t) , SX (v) and
jtj
sin2 vT
1
, 2T
2T
(vT )2
Eq. (19.5.18) can be rewritten as follows:
jtj
lim E½ST (v) ¼ lim FT RX (t) 1 T!1
T!1
2T
ð1
2
sin (v p)T
SX (p)T
dp
¼ lim
T!1 1
p½(v p)T 2
(19:5:19)
482
Random Processes
From Ref. 36,
lim T
T!1
sin2 (vT )
! d(v) and
p(vT )2
lim T
T!1
sin2 (v p)T
¼ d(v p)
p½(v p)T 2
and substituting this result in Eq. (19.5.19), we have
ð1
lim E½ST (v) ¼
T!1
SX (p)d(v p)dp ¼ SX (v)
(19:5:20)
1
Thus, the expected value of the periodogram as T ! 1 yields the psd SX(v).
Estimation of Power Spectral Density
We have defined an estimator R^ X (t) for the autocorrelation function in Eq. (19.4.34a).
It is intuitive to express the estimator S^ X (v) for the psd as the FT½R^ X (t) as follows:
S^ XT ðvÞ ¼
ðT
R^ X ðtÞejvt dt : S^ X ðvÞ ¼ lim S^ XT ðvÞ
T!1
T
(19:5:21)
This equation corresponds to Eq. (19.5.16). The periodogram is an asymptotically
unbiased estimator for SX(v) since limT!1 E½S^ X (v) ¼ SX (v). We may be tempted to conclude that since R^ X (t) is a consistent estimator of RX(t), the Fourier transform of R^ X (t) will
also be a consistent estimator of SX(v). This is not true because S^ X (v) fails to converge to
SX(v) for T ! 1. We will show heuristically that var½S^ XT (v) does not approach 0 as
T ! 1. From Eq. (19.5.14) the second moment E ½S^ XT (v)2 is given by
ð ð ð ð
h
i2
1 T T T T
E S^ XT (vÞ ¼ 2
E ½X(t)X(s)X(v)X(u)
4T T T T T
ejv(tsþvu) dt ds dv du
(19:5:22)
If the process X(t) is Gaussian, then from Eq. (11.6.8) or Example 19.3.4, we obtain
E ½X(t)X(s)X(v)X(u) ¼ RX (t s)RX (v u) þ RX (t v)RX (s u)
þ RX (t u)RX (s v)
(19:5:23)
Substituting Eq. (19.5.23) into Eq. (19.5.22), we have
1
E ½S^ XT (v)2 ¼ 2
4T
ðT ðT ðT ðT
½RX (t s)RX (v u)
T
T
T
T
þ RX (t u)RX (s u) þ RX (t u)RX (s v)
ejv(tsþvu) dt ds dv du
(19:5:24)
19.5
Power Spectral Density
483
Rearranging Eq. (19.5.24), we obtain
1
E ½S^ XT (v)2 ¼
2T
ðT ðT
T
1
2T
þ
1
2T
1
2T
þ
1
2T
1
2T
RX (t s)ejv(ts) dt ds
T
ðT ðT
T
RX (v u)ejv(vu) dv du
T
ðT ðT
T
T
T
T
T
T
T
T
ðT ðT
ðT ðT
ðT ðT
RX (t u)ejv(tu) dt du
RX (v s)ejv(vs) dv ds
RX (t v)ejv(tþv) dt dv
RX (s u)e jv(sþu) ds du
(19:5:25)
Using Eq. (19.5.21), we can simplify Eq. (19.5.25) as follows:
(
1
E ½S^ XT (v) ¼ 2{E ½S^ XT (v)}2 þ
4T 2
2
ðT ðT
T
2
RX (t v)ejv(tþv) dt dv
)
(19:5:26)
T
As T ! 1, the second term on the righthand side of Eq. (19.5.26) tends to zero. Hence
E ½S^ XT (v)2 2{E ½S^ XT (v)}2 ,
v=0
Subtracting {E ½S^ XT (v)}2 from both sides of this equation, we obtain
E ½S^ XT (v)2 {E ½S^ XT (v)}2 {E ½S^ XT (v)}2 ,
v=0
or
var½S^ XT (v) {E ½S^ XT (v)}2 , v = 0
(19:5:27)
Substitution of Eq. (19.5.20) limT!1 E ½SXT (v) ¼ SX (v), in Eq. (19.5.27) results in
lim var½S^ XT (v) S2X (v),
T!1
v=0
(19:5:28)
Thus, as T ! 1, var½S^ XT (v) does not go to zero but is approximately equal to the psd
S2X (v), and hence S^ X (v) is not a consistent estimator. To make S^ X (v) consistent, spectral
windowing [25] is employed. However, with spectral windowing the asymptotically
unbiased nature of the estimator is lost.
484
Random Processes
19.5.2 Discrete Time
As in Eq. (19.5.1), we can also define power spectral density for a discrete-time
stationary ergodic random process fXi ¼ X(ti), i ¼ 0, +1, . . .g, where the intervals are
equally spaced. The autocorrelation function RX(h) has been defined in Eq. (19.2.34).
The psd SX(v) is the Fourier transform of RX(h) and is given by
1
X
SX (v) ¼ FT½RX (h) ¼
RX (h)ejvh
(19:5:29)
h¼1
Since the discrete-time Fourier transforms are periodic, the psd for discrete-time random
process Xi as given by Eq. (19.5.29) is periodic with period 2p.
The inverse FT of SX(v) is the autocorrelation function RX(h), given by
ð
1 p
SX (v)e jvh dv
(19:5:30)
RX (h) ¼ IFT½SX (v) ¼
2p p
Example 19.5.10 (Discrete Analog of Example 19.5.1) The autocorrelation RX(h) of
the discrete-time process fXig corresponding to the continuous-time process X(t) of
Example 19.5.1 is given by
8 jhj
< 2
A 1
, jhj , n
RX (h) ¼
n
:
0,
otherwise
and is shown in Fig. 19.5.12 for A ¼ 1 and n ¼ 10.
The psd is obtained from Eq. (19.5.29) for finite n as
n1
X
jhj jvh
2
SX (v) ¼
A 1
e
n
h¼(n1)
"
#
1
n1 X
h jvh X
h jvh
2
¼A 1þ
1þ e
þ
1 e
n
n
h¼(n1)
h¼1
"
#
"
#
n1 n1 X
X
h
h
¼ A2 1 þ
1
1
e jvh þ ejvh ¼ A2 1 þ 2
cos(vh)
n
n
h¼1
h¼1
A2 1 cos(nv)
A2 sin(nv=2) 2
¼
¼
n 1 cos(v)
n sin(v=2)
and
SX (0) ¼
A 2 n2
¼ A2 n
n
The psd is graphed in Fig. 19.5.13 for A ¼ 1 and n ¼ 5 for values of v between – 2p and
2p, and the periodic nature of the psd is evident.
Figure 19.5.13 is similar to Fig. 19.5.1 except for the periodicity.
Example 19.5.11 (Discrete Analog of Example 19.5.2) The discrete AC function RX(h)
corresponding to Example 19.5.2 is given by
RX (h) ¼ s2X 1 þ e2ljhj , 1 , h , 1
19.5
Power Spectral Density
485
FIGURE 19.5.12
where 1 is a discrete unity function. The psd is determined as follows:
SX (v) ¼
1
X
s2X 1 þ e2ljhj ejvh
h¼1
"
¼
1
X
s2X
h¼1
(
¼
s2X
(
¼
s2X
1
X
h¼1
1:e
jvh
þ
1
X
e
2lh jvh
e
h¼1
þ
1
X
#
e
2lh jvh
e
h¼1
1 X
2pd(v 2ph) þ 1 þ
eh(2lþjv) þ eh(2ljv)
h¼1
1
X
1 e4l
2pd(v 2ph) þ
4l
1þe
2e2l cos(v)
k¼1
)
)
where d(v) is the Dirac delta function. The delta function train corresponds the discrete
constant s2X ¼ 14 in the frequency domain. The plot of SX(v) is shown in Fig. 19.5.14
FIGURE 19.5.13
486
Random Processes
FIGURE 19.5.14
with s2X ¼ 15 and l ¼ 15. The graph is very similar to Fig. 19.5.2 except that it is periodic
with period equal to 2p.
Alternate Form for Power Spectral Density
The discrete psd can also be obtained from the stationary discrete-time random
process fXkg. We will truncate this process in the interval (0, N – 1) and write
0k N1
Xk ,
(19:5:31)
XkN ¼
0,
otherwise
PN1
The discrete-time FT of the sequence fXkNg is given by XN (v) ¼ k¼0
Xk ejvk , which is a
random variable. We now define the quantity SN(v) as
2
1
1 NX
jXN (v)j2 XN (v)XN (v)
jvk
¼
Xk e
¼
SN (v) ¼
N k¼0
N
N
(19:5:32)
which is the periodogram for the discrete random process. Equation (19.5.32) can be
expanded as follows:
SN (v) ¼
N 1 N
1
X
1X
Xk Xm ejv(km)
N m¼0 k¼0
(19:5:33)
Substituting h ¼ (k 2 m) and j ¼ m in Eq. (19.5.33), the summation is carried out along
the diagonal with limits found by referring to Fig. 19.4.1, resulting in
"
#
(N1)=2
N1
X
X
1
Xm Xmþh ejvh
SN (v) ¼
N m¼½(N1)=2þh
h¼0
(19:5:34)
"
#
½(N1)=2jhj
0
X
X
1
jvh
þ
Xm Xmþh e
N m¼½(N1)=2
h¼(N1)
The estimated autocorrelation R^ X (h) from Eq. (19.4.76) is
X
1 Nh1
R^ X (h) ¼
Xi Xiþh
N i¼0
(19:5:35)
19.5
Power Spectral Density
487
and substituting Eq. (19.5.35) in Eq. (19.5.34), we obtain
SN (v) ¼
N1
X
h¼0
R^ X (h)ejvh
h¼(N1)
N
1
X
¼
0
X
R^ X (h)ejvh þ
R^ X (h)ejvh
(19:5:36)
h¼(N1)
Taking expectations and substituting Eq. (19.4.76b) in Eq. (19.5.36), we obtain
E ½SN (v) ¼
N1
X
E ½R^ X (h)ejvh ¼
h¼(N1)
N1 X
jhj
1
RX (h)ejvh
N
h¼(N1)
(19:5:37)
Taking the limit as of Eq. (19.5.37) as N ! 1, we have
N 1 X
jhj
RX (h)ejvh
N
h¼(N1)
jhj
¼ lim DTFT RX (h) 1 N!1
N
lim E ½SN (v) ¼ lim
N!1
N!1
1
(19:5:38)
We use the frequency convolution property of the DTFT
ð
1
1 p
X(v) Y(v) ¼
X(p)Y(v p)dp
xn yn ()
2p
2p p
with RX (h) , SX (v) and from Example 19.5.10
jhj
1 sin(vN=2)
1
()
N
N sin(v=2)
2
Eq. (19.5.38) can be written as
jhj
lim E ½SN (v) ¼ lim DTFT RX (h) 1 N!1
N!1
N
ðp
1
1 sin½(v p) N=2
¼ lim
SX (p)
dp
N!1 2p p
N sin½(v p)=2
(19:5:39)
Using the result
1
X
1 sin(vN=2)
d(v 2kp)
lim
!
N!1 N
sin(v=2)
k¼1
in Eq. (19.5.39), we obtain
1
N!1 2p
ðp
lim E ½SN (v) ¼ lim
N!1
SX (p)d(v p)dp ¼ SX (v)
p
(19:5:40)
488
Random Processes
Thus, the periodogram SN(v) is an asymptotically unbiased estimator of the psd SX(v). The
variance of SN(v) can be obtained by using techniques similar to the continuous case, and
we can show a result similar to Eq. (19.5.28):
lim var½SN (v) S2X (v), v = 0
N!1
(19:5:41)
This equation shows that the estimator SN(v) is not a consistent estimator of SX(v).
Example 19.5.12 To check the validity of Eq. (19.5.41), a computer simulation of discrete white noise of zero mean and unit variance was performed. The power spectral
density of the white noise is the variance, or SX(v) ¼ s2X ¼ 1. The number 2m of data
points with m ¼ 7,9,10,11 were chosen so that they could fit into the discrete Fourier
transform algorithm. The psd S(v) was estimated for N ¼ 128, 512, 1024, and 2048
points using Eq. (19.5.32). Because of the symmetry about N/2, the estimates S128(v),
FIGURE 19.5.15
TABLE 19.5.1.
#
PSD
N
Mean
Variance
1
2
3
4
5
SX(v)
S128(v)
S512(v)
S1024(v)
S2048(v)
—
128
512
1024
2048
1
1.0045
0.9944
0.9971
0.9987
0
0.9214
0.8908
0.9941
1.0240
19.5
Power Spectral Density
489
S512(v), S1024(v), and S2048(v) are graphed for only half the number of data points in Figs.
19.5.15a –19.5.15d.
The estimated psd’s and their variances are shown in Table 19.5.1.
The table shows that the estimator SN(v) is unbiased because the means for all N are
nearly equal to 1 as expected. However, the variances of the psd for all N are nearly the
same, equaling the derived value s4X ¼ 1, which is the expected result according to
Eq. (19.5.41). Thus, the simulation of discrete white noise confirms the desired theoretical
result.
CHAPTER
20
Classification of Random
Processes
B 20.1
SPECIFICATIONS OF RANDOM PROCESSES
In the last chapter we discussed the fact that a random process X(t) is a random variable
indexed on a time parameter and the random processes were classified as only nonstationary and stationary. We can also give other specifications of random processes
depending on time t and the state X. If the timepoints are a set of points like
{tn , n ¼ 0, +1, +2, . . .}, they are discrete, and if they are intervals of time on the
real line such as {t T} or {T1 , t , T2 }, they are continuous. The state of a
process is the set of values of a sample function Xi or X(t). If this set is finite or countably
infinite, the state space is discrete and if it is uncountably infinite, the state space is
continuous. On the basis of these definitions, we can specify four sets of random processes
as follows.
20.1.1 Discrete-State Discrete-Time (DSDT) Process
Example 20.1.1 The random variable Xn denotes the position of a particle at time n. At
each time i the particle undergoes a jump of þ1 with probability p and a jump of 21 with
probability q and stays in the same position with probability 1 2 p 2 q. The jumps are
independent. If Yi is a random variable representing the jump at the ith step, then the
sequence {Yi , i ¼ 1, 2, . . .} are independent with P{Yi ¼ 1} ¼ p, P{Yi ¼ 1} ¼ q and
P{Yi ¼ 0} ¼ 1 p q. Initially the particle is at X0 ¼ 0. The DSDT process Xi is
given by
Xi ¼ Xi1 þ Yi ,
i ¼ 1, . . . , n
This process, called random walk, is represented in Fig. 20.1.1.
Probability and Random Processes, by Venkatarama Krishnan
Copyright # 2006 John Wiley & Sons, Inc.
490
(20:1:1)
20.1 Specifications of Random Processes
491
FIGURE 20.1.1
20.1.2
Discrete-State Continuous-Time (DSCT) Process
Example 20.1.2 A random process N(t) has an increment of unity at each random time tk
where the random times {tk } are Poisson-distributed with l as the occurrence of events per
unit time.
The probability of occurrence of k events in the time interval (t1, t2) is given by
P{N(t2 ) N(t1 ) ¼ k} ¼
l(t2 t1 )k l(t2 t1 )
e
,
k!
k ¼ 0, 1, . . . ,
(20:1:2)
Such a process is known as a Poisson or counting process and is shown in Fig. 20.1.2.
20.1.3
Continuous-State Discrete-Time (CSDT) Process
Example 20.1.3 In Example 20.1.1 if the jumps are continuously distributed with
probability density function fX(x) as shown in Fig. 20.1.3, then we have a continuousstate discrete-time process. This process is also a random walk with continuous state
space. It also characterizes a sample function discretized at equal intervals of time.
20.1.4
Continuous-State Continuous-Time (CSCT) Process
Example 20.1.4 In this process both the state and the time are continuous. For example,
in the extrusion of plastic bags, the thickness of the plastic sheet will be continuously
FIGURE 20.1.2
492
Classification of Random Processes
FIGURE 20.1.3
FIGURE 20.1.4
varying with respect to time with the statistics being constant over long periods of time. In
the random-walk example (Example 20.1.1), if both the differential time Dt and the differential state Dx tend to zero, and if certain limiting conditions are satisfied, then we get a
Brownian motion whose sample function is shown in Fig. 20.1.4.
We will now discuss different types of random processes.
B 20.2
POISSON PROCESS
A random process {N(t), t 0} represents the total number of events (detection of photons
or electrons) that have occurred prior to time t. Denoting the time of occurrence of an event
as a random variable T, we make the following assumptions about the process {N(t), t 0}:
1. Pfsingle event T in the interval t, t þ Dtg ¼ P{N(t þ Dt) N(t) ¼ 1}
¼ P{N(Dt) ¼ 1} ¼ lDt þ o(Dt), where any function g(x) is of o(Dt) if
limDt!0 ½g(Dt)=Dt ! 0:
2. P{no event in the interval t,t þ Dt} ¼ P{N(t þ Dt) N(t) ¼ 0} ¼ P{N(Dt) ¼
0} ¼ 1 lDt þ o(Dt):
3. P{more than two events in the interval t,t þ Dt} ¼ P{N(Dt) 2} ¼ o(Dt):
4. Events occurring in nonoverlapping time intervals are statistically independent:
P{½N(t2 ) N(t1 )½N(t4 ) N(t3 )} ¼ P{½N(t2 t1 )½N(t4 t3 )}
¼ P{N(t2 t1 )}P{N(t4 t3 )}
for
t1 , t2 , t3 , t4
(20:2:1)
20.2
Poisson Process
493
FIGURE 20.2.1
Processes satisfying Eq. (20.2.1) are called independent increment processes. Given these
assumptions, we will find the probability of n events occurring in time interval t:
P{n events in the time interval t} ¼ P{N(t) ¼ n} ¼ PN (n, t)
We will first find the probability of no event in the interval t þ Dt ¼ PN {0, t þ Dt}.
The interval (0,t þ Dt) can be partitioned as (0,t) and (t,t þ Dt) as shown in Fig. 20.2.1.
Thus, PN {0, t þ Dt} can be given as probability of no event in (0,t) and probability of
no event in (t,t þ Dt). Since these two intervals are nonoverlapping, we can write from
assumption 4
PN {0, t þ Dt} ¼ PN {0, t}PN {0, Dt}
(20:2:2)
and from assumption 2
PN (0, t þ Dt) ¼ PN (0, t)(1 lDt) þ o(Dt)
or
PN (0, t þ Dt) PN (0, t) ¼ PN (0, t)lDt þ o(Dt)
(20:2:3)
Dividing Eq. (20.2.3) by Dt and taking the limit as Dt ! 0, we have
dPN (0, t)
¼ lPN (0, t)
dt
(20:2:4)
The initial condition for this first-order differential equation is obtained from Pfno events
in zero timeg equals 1. Hence the solution to Eq. (20.2.4) with initial condition PN(0,
0) ¼ 1 is given by
PN (0, t) ¼ elt u(t)
which is the probability that no event T occurs in the time interval (0, t). Or
P{N(t) ¼ 0} ¼ PN (0, t) ¼ P(T . t) ¼ elt u(t)
(20:2:5)
Hence, the probability of occurrence of an event T in the interval (0, t) is the complement
of Eq. (20.2.5) and is written as
P(T t) ¼ FT (t) ¼ (1 elt )u(t)
and the density function is
fT (t) ¼ lelt u(t)
Equation (20.2.5) will serve as the initial condition for the solution of PN {n, t}.
(20:2:6)
494
Classification of Random Processes
FIGURE 20.2.2
From Fig. 20.2.2 it can be observed that n events occurring in the interval (0, t þ Dt)
can be partitioned as n events in (0, t) AND no event in (t, t þ Dt) OR (n 2 1) events in (0, t)
AND 1 event in (t, t þ Dt). From assumptions 1 –4, we can write
PN (n, t þ Dt) ¼ PN (n, t)½1 l Dt þ o(Dt) þ PN (n 1, t) l Dt þ o(Dt)
(20:2:7)
Rearranging terms in Eq. (20.2.7), dividing throughout by Dt, and taking the limit as
Dt ! 0, we obtain a differential – difference equation
PN (n, t þ Dt) PN (n, t)
¼ lim ½PN (n, t)l þ PN (n 1, t)l þ o(Dt)
Dt!0
Dt!0
Dt
lim
or
dPN (n, t)
¼ lPN (n, t) þ lPN (n 1, t)
dt
(20:2:8)
Equation (20.2.8) is a differential – difference equation and the two initial conditions for
solving the are
1. PN (0, t) ¼ elt u(t)
2. PN (n, 0) ¼ 0, n . 0
Equation (20.2.8) can be solved recursively as follows:
n ¼ 1:
dPN (1, t)
¼ lPN (1, t) þ lelt
dt
n ¼ 2:
dPN (2, t)
¼ lPN (2, t) þ lPN (1, t) or
dt
or PN (1, t) ¼ ltelt ,
PN (2, t) ¼
t0
(lt)2 lt
e ,
2
t0
Proceeding in a similar manner, we obtain the result for n events occurring in interval t as
PN (n, t) ¼
ðlt)n lt
e ,
n!
t0
(20:2:9)
The discrete-state continuous-time (DSCT) random process satisfying Eq. (20.2.9) is
called a Poisson random process and has been shown in Fig. 20.1.2.
Solution of Eq. (20.2.7) Using Generating Functions
The differential –difference equation [Eq. (20.2.8)] can also be solved using generating functions. From Eq. (11.3.1) we can define the generating function for PN (n, t) as
G(z, t) ¼
1
X
n¼0
PN (n, t)zn
(20:2:10)
20.2
Poisson Process
495
Multiplying Eq. (20.2.8) by z n and summing over all n from 0 to 1, we have
1
1
1
X
X
dX
PN (n, t)zn ¼ l
PN (n, t)zn þ l
PN (n 1, t)zn
dt n¼0
n¼0
n¼0
or
d
G(z, t) ¼ l(z 1)G(z, t)
dt
(20:2:11)
For a fixed z, Eq. (20.2.11) is an ordinary differential equation in t. The initial condition
G(z, 0) is obtained as follows:
G(z, 0) ¼
1
X
PN (n, 0)zn ¼ PN (0, 0) ¼ 1;
since PN (n, 0) ¼ 0 for n 1
n¼0
With this initial condition, the solution to Eq. (20.2.11) is
G(z, t) ¼ elt(z1) ,
t0
(20:2:12)
The power-series expansion of Eq. (20.2.12) yields
lt
(lt)2
(ltÞ n
þ þ zn
þ G(z, t) ¼ elt 1 þ z þ z2
1!
2!
n!
Hence, PN(n, t) is the coefficient of zn in the preceding equation and is given by
PN (n, t) ¼ P{N(t) ¼ n} ¼ elt
(lt)n
,
n!
t0
(20:2:13)
The statistics of the Poisson process as derived in Example 19.1.7 are
Mean:
E½N(t) ¼ mN (t) ¼ lt
Variance:
var½N(t) ¼ s2N (t) ¼ lt
Autocorrelation:
E½N(t1 )N(t2 ) ¼ RN (t1 , t2 ) ¼ l2 t1 t2 þ lmin (t1 , t2 )
Autocovariance:
E{½N(t1 ) mN (t1 )½N(t2 ) mN (t2 )} ¼ CN (t1 , t2 ) ¼ lmin (t1 , t2 )
Normalized Autocovariance:
CN (t1 , t2 Þ
1
¼ min
rN (t1 , t2 ) ¼
sN (t1 )sN ðt2 Þ l
rffiffiffiffi rffiffiffiffi
t1 t2
,
t2 t1
Since the mean is a function of time t and the autocorrelation is a function of t1 and t2, the
Poisson process is a nonstationary process.
Distribution of Interarrival Times
We will now find the distribution of interarrival times for the Poisson process N(t). Let
T1 be the random time of occurrence of the first event after time t ¼ 0, T2 the random time
496
Classification of Random Processes
FIGURE 20.2.3
between the occurrence of the first and second events, and Tn the random time between the
occurrence of the (n 2 1)st and nth events as shown in Fig. 20.2.3. The sequence of times
fTi, i ¼ 1, . . . , ng is called interarrival times. Let {ti , i ¼ 1, . . . , n} be the realizations of
these random interarrival times. Let Sn be the random variable representing the arrival
time of the nth event or the waiting time until the nth event:
S1 ¼ T1 , S2 ¼ T1 þ T2 , . . . , Sn ¼ T1 þ T2 þ þ Tn ¼
n
X
Ti
(20:2:14)
i¼1
If {ti , i ¼ 1, . . . , n} are the realizations of the random arrival times {Si , i ¼ 1, . . . , n}, then
t1 ¼ t1 , t2 ¼ t1 þ t2 , . . . , tn ¼ t1 þ t2 þ þ tn ¼
n
X
ti
(20:2:15)
i¼1
We will find the distributions of these interarrival times {Ti } and the waiting time for the
nth event Sn.
The Poisson process can also be specified in terms of the arrival times {Si } as
N(t) ¼
1
X
u(t Si )
(20:2:16a)
i¼1
where u(t) is a unit step function. Referring to Fig. 20.2.3, N(t) is the total number of
events that have occurred in the time interval t, and each event causes an increment
by 1. Hence Eq. (20.2.16a) can be simplified as
N(t) ¼
1
X
i¼1
u(t Si ) ¼
N(t)
X
i¼1
uðt Si Þ ¼
N(t)
X
1,
t0
(20:2:16b)
i¼1
By definition, the first event {T1 . t} occurs after time t, and hence there are no
events occurring in the interval (0, t). Hence from Eq. (20.2.5), we obtain
P{T1 . t} ¼ P{N(t) ¼ 0} ¼ PN (0, t) ¼ elt u(t)
(20:2:17)
We will show that the interarrival times T1,T2, . . . , Tn are independent identically
distributed (i.i.d.) random variables with density function elt u(t). The second event
20.2
Poisson Process
497
{T2 . t} occurs after the first event T1 ¼ t1:
P{T2 . t j T1 ¼ t1 } ¼ P{T2 . t j S1 ¼ t1 } ¼ P ½no events in (t1 , t1 þ t) = S1 ¼ t1
(20:2:18)
From the independence of nonoverlapping intervals (assumption 4) of the Poisson process,
Eq. (20.2.18) can be written as
P{T2 . t j S1 ¼ t1 } ¼ P{T2 . t} ¼ P{no events in (t1 , t1 þ t)} ¼ elt u(t)
(20:2:19)
Hence T1 and T2, are independent.
In a similar manner, for the nth arrival time Tn, we can write
P{Tn . t j Tn1 ¼ tn1 , . . . , T1 ¼ t1 } ¼ P{Tn . t j Sn1 ¼ tn1 , . . . , S1 ¼ t1 } (20:2:20)
Again, from the independent increment property, Eq. (20.2.20) can simplified to
P{Tn . t j Sn1 ¼ tn1 , . . . , S1 ¼ t1 } ¼ P{Tn . t} ¼ P{no events in (tn , t}
¼ elt u(t)
(20:2:21)
Thus the interarrival times T1, T2, . . . , Tn are i.i.d. random variables with FT (t) ¼ P(Tn t) ¼ (1 elt )u(t) and density function fT (t) ¼ lelt u(t). As a consequence,
the probP
ability density function for the waiting time for the nth event Sn ¼ ni¼1 Ti , n 1 is an
n-fold convolution of lelt u(t), yielding an Erlang density with n-degrees of freedom
similar to Eq. (7.4.1):
fSn (t) ¼
(lt)n1 lelt
,
(n 1)!
l . 0, t . 0, n: integer . 0
(20:2:22)
Example 20.2.1 A message sending station sends two independent bitstreams X and Y
with Poisson rates l and m. The receiving station receives the composite stream
W ¼ X þ Y. We have to find (1) the rate of the composite stream and (2) the probability
that stream Y arrives first. The density functions of the bitstreams X and Y are
fX (x) ¼ elx
(lx)k
(my)m
u(x) and fY (y) ¼ emy
u(y)
k!
m!
1. We will find the rate of W ¼ X þ Y using generating functions. The generating
functions for X and Y are
PX (z) ¼ elðz1Þ
and
PY (z) ¼ em(z1)
Since X and Y are independent, the pdf fW (w) is the convolution of fX(x) and fY (x),
and the generating function PW (z) for fW (w) is given by the product PX(z)PY (z), or
PW (z) ¼ e(lþm)(z1) . Hence
fW (w) ¼ eðlþm)w
½(l þ m)wk
u(w)
k!
2. Let the times of arrival of streams X and Y be the random variables S and T. From
Eq. (20.2.6) the interarrival times are exponentially distributed and hence
fS (s) ¼ lels u(s) and
fT (t) ¼ memt u(t)
498
Classification of Random Processes
Since the streams are independent, the arrival times are also independent, and the
joint density of the arrival times S and T is
fST (s, t) ¼ lme(ltþms) ,
s 0, t 0
The event that stream Y arrives before stream X is the same as the arrival time of Y
is before the arrival time of X. Hence P{Y X} ¼ P{T S}. We will find
P(T S). The region {T S} is shown in Fig. 20.2.4:
1
ð1 ð1
ð1
lt e
P(T S) ¼
lme(ltþms) dt ds ¼
lmems l
0 s
0
s
ð1
m
¼m
e(lþm)s ds ¼
lþm
0
Hence
P(Y X) ¼
m
lþm
Example 20.2.2 A Poisson process N(t) with rate l consists of a subprocess N1(t) with
probability p and rate l1 and another subprocess N2(t) with probability (1 2 p) and rate
l2. We have to find the probabilities P{N1 (t) ¼ n} and P{N2 (t) ¼ m} and their rates l1
and l2 .
We are given the following:
N(t) elt
(lt)k
,
k!
N1 (t) el1 t
(l1 t)k
(l2 t)k
, N2 (t) el2 t
k!
k!
From the statement of the problem N(t) ¼ N1 (t) þ N2 (t), and if N1 (t) ¼ n and N2 (t) ¼ m,
then N(t) ¼ n þ m. The joint probability P{N1 (t) ¼ n, N2 (t) ¼ m} can be given in terms of
conditional probabilities as
P{N1 (t) ¼ n, N2 (t) ¼ m} ¼ P{½N1 (t) ¼ n, N2 (t) ¼ m j N(t) ¼ m þ n}elt
(lt)mþn
(m þ n)!
where P{N(t) ¼ m þ n} ¼ elt (lt)mþn =½(m þ n)! has been substituted in the equation
above. The event ½N1 (t) ¼ n, N2 (t) ¼ m j N(t) ¼ m þ n} can be considered as (m þ n)
FIGURE 20.2.4
20.2
Poisson Process
499
Bernoulli trials with success defined as N1 (t) ¼ n with probability p and failure as
N2 (t) ¼ m with probability (1 2 p). Thus, using the binomial distribution [Eq. (4.2.2)],
we have
mþn n
p (1 p)m
P{½N1 (t) ¼ n, N2 (t) ¼ m j N(t) ¼ m þ n} ¼
n
and P{N1 (t) ¼ n, N2 (t) ¼ mg can be given by
P{N1 (t) ¼ n, N2 (t) ¼ m} ¼
(m þ n)! n
(lt)mþn
p (1 p)m elt
n!m!
(m þ n)!
¼ eltp elt(1p)
(ltp)n ½lt(1 p)m
n!
m!
The marginal distribution P{N1 (t) ¼ n} can be given as follows:
P{N1 (t) ¼ n} ¼ eltp
1
(ltp)n X
½lt(1 p)m
(ltp)n
¼ eltp
elt(1p)
n! m¼0
m!
n!
Similarly
P{N2 (t) ¼ m} ¼ elt(1p)
½lt(1 p)m
m!
Hence, the Poisson rates for N1(t) and N2(t) are l1 ¼ lp and l2 ¼ l(1 2 p).
Let S1n and S2m be the waiting times for the nth and mth events of the Poisson
subprocesses N1(t) and N2(t) respectively. We will find the probabilities of (1) the
waiting time for the first event in N1(t) is greater than the waiting time for the first
event in N2(t), or P{S11 . S21 }, and (2) the waiting time for the second event in N1(t)
is less than the waiting time for the first event in N2(t), or P{S12 , S21 }.
From Eq. (20.2.14) the waiting times can be given in terms of the interarrival times, or
S11 ¼ T11, S21 ¼ T21 and S12 ¼ T11 þ T12.
1. From Example 20.2.1
P{S11 . S21 } ¼ P{T11 . T21 } ¼
l1
lp
¼p
¼
l1 þ l2 lp þ l(1 pÞ
2. Since S11 , S12, we can write P{S12 , S21 } ¼ P{S12 , S21 ; S11 , S21 } or
P{S12 , S21 } ¼ P{S12 , S21 jS11 , S21 }P{S11 , S21 }
From Example 20.2.1 PfS11 , S21 g ¼ l 2 =ðl1 þ l2 Þ and
PfS12 , S21 j S11 , S21 g ¼ PfT11 þ T12 , T21 j T11 , T21 g ¼ PfT12 , T21 g
¼
l2
l1 þ l2
Hence
PfS12
l2
, S12 g ¼
¼ ð1 pÞ2
l1 þ l2
Example 20.2.3 Advance booking for a World Series baseball game is open on weekdays only from 10:00 a.m. to 3:00 p.m. During the lunch hour from 12:00 to 1:00 p.m.,
500
Classification of Random Processes
customers arrive at the rate of l ¼ 8 h21 (8 per hour) and for the rest of the time at the rate
of l ¼ 4h21. The arrival time N(t) of customers to the booking counter is a random
process given by
P{N(t) ¼ k} ¼ eLt
(Lt)k
k!
where the rate L is a binary random variable assuming values of 4 and 8 with probabilities 0.2 and 0.8 respectively. We have to find E[N(t)], var[N(t)] and determine
the variance to mean ratio s2 =m. If this ratio equals 1, we can conclude that N(t) is
Poisson.
We note that the distribution of N(t) conditioned on either L¼8 or 4 is Poisson.
Hence E½N(t) j L ¼ 8 ¼ 8 and E½N(t) j L ¼ 4 ¼ 4. This result is used in the following
equation:
E½N(t)j ¼ E½NðtÞ j L ¼ 8P(L ¼ 8) þ E½N(t) j L ¼ 4P(L ¼ 4)
¼ 8 0:2 þ 4 0:8 ¼ 1:6 þ 3:2 ¼ 4:8
In a similar manner, we can find the second moment E[N 2(t)] with conditional second
moments E½N 2 (t) j L ¼ 8 ¼ l þ m2 ¼ 8 þ 64 ¼ 72 and E½N 2 (t) j L ¼ 4 ¼ 4 þ 16 ¼ 20.
From the conditional second moments, E[N 2(t)] can be given as follows:
E½N 2 (t) ¼ E½N 2 (t) j L ¼ 8P(L ¼ 8) þ E½N 2 (t) j L ¼ 4P(L ¼ 4)
¼ 72 0:2 þ 20 0:8 ¼ 30:4
Hence the variance of N(t) is given by
var(N(t)) ¼ E½N 2 (t) {E½N(t)}2 ¼ 30:4 4:82 ¼ 7:36
The variance : mean ratio s2 =m ¼ 7:36=4:8 ¼ 1:53. Since this ratio is not equal
to 1, N(t) is not a Poisson process.
Example 20.2.4 Customers arrive at a checkout counter at a Poisson rate of l per
minute. The service time is a random variable Y in minutes. We consider two cases: (1)
Y is a constant ¼ k minutes and (2) Y is exponentially distributed with density function
fY (y) ¼ memy u(y). For both cases we will find the probabilities that (1) a customer does
not have to wait and (2) the mean value of a customer’s waiting time.
From Eq. (20.2.6) the interarrival time T of customers is distributed as
fT (t) ¼ lelt u(t), and from Eq. (20.2.19) the interarrival times are independently distributed. The service time Y is distributed as fY (y) ¼ memy u(y). If the interarrival time T is
greater than the service time Y, then there will be no waiting time for the customer.
1. Service time is a constant: Y ¼ k minutes
(a) Pfno waiting time for the customerg ¼ PfT . kg. Hence
1
ð1
lt le
P{T . k} ¼
lelt dt ¼
¼ elk
l
k
k
(b) If T k, then the customer has to wait (k 2 T ) minutes and the average
waiting time is given by
ðk
1
E½k T ¼ (k t)lelt dt ¼ k (1 ekl )
l
0
20.2
Poisson Process
501
2. Service time Y exponentially distributed: fY (y) ¼ memy u(y)
(a) Again, P fno waiting time for the customerg ¼ PfT . Yg and
ð1
P{T . Y j Y ¼ t}fY (t)dt
P{T . Y} ¼
0
ð1
¼
elt memt dt ¼
0
m
lþm
(b) Here also, if T Y, then the customer has to wait (Y 2 T ) minutes. We have to
find E[Y 2 T ]. In terms of conditional expectation, we can write
ð1
E½Y T ¼
E½Y T j Y ¼ y fY (y)dy
0
and from the result 1b (above) the conditional expectation E½Y T j Y ¼ y is
ðy
1
E½Y T j Y ¼ y ¼ (y t)lelt dt ¼ y (1 ely )
l
0
Substituting this equation into the previous equation, we obtain
ð1
1
l
ly
E½Y T ¼
y (1 e ) m emy dy ¼
l
m(l þ m)
0
Poisson White Noise
The Poisson process given by Eq. (20.2.16) can be further generalized by associating
with each event T a random weight A. The weights are i.i.d. random variables with density
function fA(a) and are also independent of the arrival times S. The weighted Poisson
process X(t) is defined by
X(t) ¼
1
X
Ai u(t Si ),
t0
(20:2:23a)
i¼1
which can be simplified from Eq. (20.2.16.b) as follows:
X(t) ¼
1
X
Ai u(t Si ) ¼
i¼1
N(t)
X
Ai u(t Si ) ¼
i¼1
N(t)
X
Ai ,
t0
(20:2:23b)
i¼1
The pdf fX (x; t) can be expressed in terms of the conditional expectation as
X
fX ½x; tjN(t)P{N(t)}
fX (x; t) ¼
(20:2:24)
N(t)
The conditional probability fX (x; t j N(t) ¼ i can be evaluated from Eqs. (20.2.23):
i ¼ 1:
fX ½x; t j N(t) ¼ 1 ¼ fA (a)
(20:2:25)
i ¼ 2:
fX ½x; t j N(t) ¼ 2 ¼ fA (a1 þ a2 )
(20:2:26)
Since the Ai terms are i.i.d. random variables, fA (a1 þ a2 ) in Eq. (20.2.26) is a convolution
of fA (a) with itself; or
fX ½x; t j N(t) ¼ 2 ¼ fA (a) fA (a)
Proceeding in this fashion, fX ½x; t j N(t) ¼ i is an i-fold convolution of fA (a) and
fX ½x; t j N(t) ¼ i ¼ fA (a) fA (a) fA (a) ¼ fA(i) (a)
(20:2:27)
502
Classification of Random Processes
In Eq. (20.2.4), P{N(t) ¼ i} is a Poisson distribution given by Eq. (20.10.13):
P{N(t) ¼ i} ¼ PN (i, t) ¼ elt
(lt)i
i!
(20:2:28)
Substituting Eqs. (20.2.27) and (20.2.28) in Eq. (20.2.24) we obtain the pdf of X(t) as
fX (x; t) ¼ elt
1
X
fA(i) (a)
i¼1
(lt)i
i!
(20:2:29)
The mean and autocorrelation of the generalized Poisson process X(t) can be obtained
from the properties of expectations as follows:
Mean:
E½X(t) ¼ E{E½X(t) j N(t) ¼ n}
( "
#)
N(t)¼n
1 N(t)¼n
X
X
X
¼E E
Ai
E½Ai PN (n, t)
¼
n¼1
i¼1
¼
1
X
i¼1
1
X
(lt) lt
(lt)n
e ¼ mA elt
n!
(n 1)!
n¼1
n
mA n
n¼1
¼ mA ltelt ¼
1
X
(lt)n
m¼0
m!
¼ mA lt ¼ mX (t),
t0
(20:2:30)
Autocorrelation:
E½X(t1 )X(t2 ) ¼ E{X(t1 )½X(t2 ) X(t1 ) þ X(t1 )}
¼ E½X 2 (t1 ) þ E½X(t1 )E½X(t2 ) X(t1 )
¼ E{E½X 2 (t1 ) j N(t1 )}
þ E½X(t1 ) j N(t1 )E½(X(t2 ) X(t1 )) j N(t2 t1 )
(20:2:31)
We will evaluate E{E½X 2 (t1 ) j N(t1 )}:
(
)
n X
n
X
2
E½Ai A j E{E½X (t1 ) j N(t1 ) ¼ n} ¼ E
i¼1 j¼1
¼E
8
>
>
n
<X
>
>
: i¼1
E½A2i þ
n X
n
X
E½Ai E½A j i¼1 j¼1
i=j
9
>
>
=
>
>
;
¼ E{nE½A2 þ n(n 1)m2A }
¼
1
X
E½A2 nelt1
n¼1
¼ E½A2 lt1 elt1
1
(lt1 )n X
(lt1 )n
þ
m2A n(n 1)elt1
n!
n!
n¼1
1
X
(lt1 )m
m¼0
m!
þ m2A l2 t12 elt1
1
X
(lt1 )m
m¼0
m!
¼ E½A2 lt1 þ m2A l2 t12
(20:2:32)
20.2
Poisson Process
503
Substituting Eqs. (20.2.30) and (20.3.32) in Eq. (20.2.31) and E½A2 ¼ s2A þ m2A , we obtain
E½X(t1 )X(t2 ) ¼ E½A2 l(t1 ) þ m2A l2 t12 þ mA lt1 ½mA lt2 mA lt1 ¼ ½s2A þ m2A lt1 þ m2A l2 t1 t2 ¼ RX (t1 , t2 )
(20:2:33a)
Equation (20.2.33a) was obtained on the premise that t1 , t2. We can derive a similar
equation for t2 , t1. Hence the autocorrelation function RX(t1, t2) is given by
RX (t1 , t2 ) ¼ ½s2A þ m2A l min(t1 , t2 ) þ m2A l2 t1 t2 , t1 , t2 , 0
(20:2:33b)
The autocovariance function CX (t1, t2) is given by
CX (t1 , t2 ) ¼ (s2A þ m2A )l min(t1 , t2 ), t1 , t2 0
(20:2:34)
Example 20.2.5 The events of a Poisson process with l ¼ 12 are weighted with i.i.d.
random variables {Ai }, resulting in a weighted Poisson process X(t). The weights are
uniformly distributed in the interval (0,2). We will find the mean, autocorrelation, and
autocovariance of X(t). The mean value of the weights mA ¼ 1. The variance of the
weights s2A ¼ 13. X(t) given by Eq. (20.2.23) is shown in Fig. 20.2.5.
Mean. From Eq. (20.2.30) the mean value of X(t) is mX (t) ¼ mA lt ¼ 1 12 t ¼ t=2.
Autocorrelation. The second moment of A is E½A2 ¼ s2A þ m2A ¼ 13 þ 12 ¼ 43. Hence,
from Eq. (20.2.33), the autocorrelation function is given by
2
41
2
1
2 1
min(t1 , t2 ) þ 1
RX (t1 , t2 ) ¼
t1 t2 ¼ min(t1 , t2 ) þ t1 t2
32
2
3
4
Autocovariance. From Eq. (20.2.34) the ACF is CX (t1 , t2 ) ¼ 23 min(t1 , t2 ):
Formal Derivative of Poisson Process
The generalized Poisson process given by Eq. (20.2.23) shown in Fig. 20.2.5 consists
of a series of step functions occurring at Poisson-distributed random times with random
FIGURE 20.2.5
504
Classification of Random Processes
weights. The formal derivative of X(t) yields a Poisson impulse process V(t) consisting of
a series of Dirac delta functions:
1
1
X
d
dX
X(t) ¼
Ai u(t Si ) ¼
Ai d(t Si ),
dt
dt i¼1
i¼1
t0
(20:2:35)
The mean of V(t) can be calculated as
mV (t) ¼
dmX (t) dmA lt
¼
¼ mA l,
dx
dt
t0
(20:2:36)
The autocorrelation function of V(t) is
RV (t1 , t2 ) ¼
@2 RX (t1 , t2 )
@2
¼
{½s2A þ m2A l min(t1 , t2 ) þ m2A l2 t1 t2 }
@t1 @t2
@t1 @t2
¼ ½s2A þ m2A ld(t2 t1 ) þ m2A l2
¼ RV (t2 t1 ) ¼ RV (t)
(20:2:37)
Since mV is independent of time and RV (t1 , t2 ) is a function of the time difference t2 2 t1,
we conclude that V(t) is a stationary process.
The autocovariance function is
CV (t1 , t2 ) ¼ ½s2A þ m2A ld(t2 t1 ) ¼ CV (t2 t1 ) ¼ CV (t)
(20:2:38)
The FT SV 0 (v) of the acf CV (t) of Eq. (20.2.38) is given by
SV 0 (v) ¼ FT{½s2A þ m2A ld(t2 t1 )} ¼ l½s2A þ m2A (20:2:39)
This is the psd of the process V 0 (t) ¼ V(t) mA l. Since the spectrum is constant over the
entire frequency range, the process V 0 (t) is called Poisson white noise.
Example 20.2.6 We will determine the derivative process V(t) of Example 20.2.5, where
l ¼ 12 and the weight process A . 0 is uniformly distributed in the interval (0,2). The
process V(t) is shown in Fig. 20.2.6.
The mean mV is obtained from Eq. (20.2.36) as mV ¼ mA l ¼ 1 12 ¼ 12.
The autocorrelation function from Eq. (20.2.37) is
4
1
RV (t) ¼ ½s2A þ m2A d(t2 t1 ) þ m2A l2 ¼ d(t) þ
3
4
FIGURE 20.2.6
20.3
Binomial Process
505
and the autocovariance from Eq. (20.2.38) is CV (t) ¼ 43 d(t). The psd equals FT½RV ðtÞ
given by SV (v) ¼ 43 þ ðp=2Þd(v).
B 20.3
BINOMIAL PROCESS
Bernoulli Process
In Section 4.1 we discussed Bernoulli trials as repeated functionally and statistically
i.i.d. trials, each of which consists of only two events s and f with probabilities p and
q ¼ (1 2 p) respectively. Consider a countably infinite sequence {Xn , n ¼ 0, 1, . . .} of
random variables defined by
Xn ¼
1
0
for success s in the nth trial
for failure f in the nth trial
with probabilities
P(Xn ¼ 1) ¼ p
P(Xn ¼ 0) ¼ 1 p
(20:3:1)
Thus Xn represents a Bernoulli process, shown in Fig. 20.3.1.
We will now find the statistics of this process:
Mean:
E½Xn ¼ mX ¼ 1 P(Xn ¼ 1) þ 0 P(XN ¼ 0) ¼ p
(20:3:2)
Variance:
var½Xn ¼ s2X ¼ E½X 2 m2X ¼ 12 p þ 02 (1 p) p2 ¼ p(1 p)
(20:3:3)
Autocorrelation:
(
E½Xn1 Xn2 ¼ RX (n1 , n2 ) ¼
12 p þ 02 (1 p) ¼ p,
2
E½Xn1 E½Xn2 ¼ p ,
n1 n2 ¼ 0
(20:3:4)
n1 n2 = 0
Autocovariance:
CX (n1 ; n2 ) ¼ RX (n1 ; n2 ) m2X
(
12 p þ 02 (1 p) p2 ¼ p(1 p),
¼
E½Xn1 E½Xn2 ¼ p2 ¼ 0, n1 n2 = 0
FIGURE 20.3.1
n1 n2 ¼ 0
(20:3:5)
506
Classification of Random Processes
Normalized Autocovariance:
rX (n1 , n2 ) ¼
1,
0,
n1 n2 ¼ 0
n1 n2 = 0
(20:3:6)
Since the mean is independent of time n and the autocorrelation depends only on the time
difference (n1 – n2), the process Xn is stationary.
Binomial Process
We now define a new process Yn of the sum of the random variables {Xi } in a
Bernoulli process as discussed earlier. This is called the binomial process, given by
Yn ¼
n
X
Xi ,
n.0
(20:3:7)
i¼1
This process, corresponding to the Bernoulli process of Fig. 20.3.1, is shown in Fig. 20.3.2
and is a discrete-state discrete-time (DSDT) process.
The probability of k successes in n trials has already been derived in Eq. (4.2.2) and is
given by a binomial distribution:
n k
p (1 p)nk
Probability{k successes in any sequence in n trials} ¼ P(Yn ¼ k) ¼
k
The statistics of the binomial process are:
Mean. Since the probability of each success is p, the mean value of Yn can be obtained
using the linearity of the expectation operator:
"
#
n
n
X
X
E ½Yn ¼ E
Xi ¼
E ½Xi ¼ np, n . 0
(20:3:8)
i¼1
i¼1
Variance. From the independence of Xi, the second moment of Yn can be given by
"
#2
n
n
n X
n
X
X
X
2
Xi ¼
E½Xi2 þ
E ½Xi X j ¼ np þ n(n 1)p2 , n . 0
E½Yn ¼ E
i¼1
s2Y
¼
E ½Yn2 i¼1
i¼1 j¼1
2
{E ½Yn } ¼ np þ n(n 1)p2 n2 p2 ¼ np(1 p),
n.0
(20:3:9)
FIGURE 20.3.2
20.4
Independent Increment Process
507
Autocorrelation: For n1 n2, we have
"
#
n2 X
n1
X
E ½Yn1 Yn2 ¼ RY (n1 , n2 ) ¼ E
Xi X j
j¼1 i¼1
"
¼E
n1
X
i¼1
#
Xi2
þ
n1 X
n1
X
E ½Xi X j þ
j¼1 i¼1
i=j
nX
n1
2 n1 X
E ½Xi X j j¼1 i¼1
i=j
¼ n1 p þ n1 (n1 1)p2 þ n1 (n2 n1 )p2
or
RY (n1 , n2 ) ¼ n1 p(1 p) þ n1 n2 p2 ;
for
n1 n2
(20:3:10)
Similarly for n2 n1 we have
RY (n1 , n2 ) ¼ n2 p(1 p) þ n1 n2 p2 ;
for
n2 n1
(20:3:11)
n1 , n2 . 0
(20:3:12)
Combining Eqs. (20.3.10) and (20.3.11), we obtain
RY (n1 , n2 ) ¼ p(1 p) min(n1 , n2 ) þ n1 n2 p2 ,
Autocovariance:
CY (n1 , n2 ) ¼ p(1 p) min (n1 , n2 ), n1 , n2 . 0
(20:3:13)
Unlike the Bernoulli process, the mean of the binomial process is dependent on
time n and the autocorrelation function is dependent on both n1 and n2 and
hence is a nonstationary process.
B 20.4
INDEPENDENT INCREMENT PROCESS
A random process X(t) is called an independent increment process if for a sequence of
times t1 , t2 , . . . , tn the increments X(t1 ), {X(t2 ) X(t1 )}, . . . , {X(tn ) X(tn1 )} of
the process X(t) are a sequence of independent random variables:
E{½X(ti ) X(ti1 )½X(tiþ1 ) X(ti )}
¼ E ½X(ti ) X(ti1 ) E ½X(tiþ1 ) X(ti )
(20:4:1)
Further, if all order distributions of the increments fX(t) 2 X(s)g, t , s are dependent only
on the time difference t ¼ t s, then X(t) is called a stationary independent increment
process.
We will show that a stationary independent increment process is not a stationary
process, and its mean value and variance are given by
E ½X(t) ¼ m0 þ m1 t
(20:4:2a)
where m0 ¼ E ½X(0) and m1 ¼ E ½X(1) E ½X(0) and
var½X(t) ¼ s2X(t) ¼ s20 þ s21 t
where s20 ¼ E ½X(0) m0 2 and s21 ¼ E ½X(1) m1 2 s20 .
(20:4:2b)
508
Classification of Random Processes
We will verify Eq. (20.4.2a). Defining g(t) ¼ E ½X(t) X(0), for any t and t, we have
g(t þ t) ¼ E ½X(t þ t) X(0) ¼ E{½X(t þ t) X(t) þ ½X(t) X(0)}
¼ E ½X(t þ t) X(t) þ E ½X(t) X(0)
(20:4:3)
In Eq. (20.4.3)
E ½X(t) X(0) ¼ g(t) by definition
(20:4:4)
and from the stationary independent increment property we can write
E ½X(t þ t) X(t) ¼ E ½X(t) X(0) ¼ g(t)
(20:4:5)
Substituting Eqs. (20.4.5) and (20.4.4) in Eq. (20.4.3), we obtain
g(t þ t) ¼ g(t) þ g(t)
(20:4:6)
Differentiating both sides of Eq. (20.4.6) independently with respect to t and t, we have
d
d
g(t þ t) ¼ g(t);
dt
dt
d
d
g(t þ t) ¼ g(t)
dt
dt
(20:4:7)
Hence g0 (t) ¼ g0 (t) and in particular at t ¼ 1, g(t)jt¼1 ¼ E ½X(1) X(0) ¼ k1 . Substituting k1 for g0 (t) and solving the differential equation g0 (t) ¼ k1 , we obtain the solution
g(t) ¼ k1t þ k0. Substitution of the initial condition g(0) ¼ E ½X(0) X(0) ¼ 0 in the
solution results in g(t) ¼ k1t, or
E ½X(t) X(0) ¼ E ½X(1) X(0)t
E ½X(t) ¼ m0 þ m1 t
(20:4:8)
(20:4:2a)
thus verifying Eq. (20.4.2a).
In a similar manner we can verify Eq. (20.4.2b). Hence, a random process with
stationary independent increments is nonstationary since the mean and variance are
functions of time.
Uncorrelated and Orthogonal Increment Processes
A random process X(t) is called an uncorrelated increments process if for a sequence
of times t1 , t2 , t3 , t4 the increments fX(t2) 2 X(t1)g and fX(t4) 2 X(t3)g of the process
X(t) are uncorrelated random variables, or
E{½X(t2 ) X(t1 )½X(t4 ) X(t3 )} ¼ E ½X(t2 ) X(t1 )E ½X(t4 ) X(t3 )
(20:4:9)
Similarly, the random process X(t) is called an orthogonal increments process if
E{½X(t2 ) X(t1 )½X(t4 ) X(t3 )} ¼ 0
(20:4:10)
The term independent increment process implies an uncorrelated increment process, but
the converse is not true. If a process has uncorrelated increments, it need not necessarily
have independent increments. Also, if a process X(t) has uncorrelated increments, then the
process Y(t) ¼ X(t) E ½X(t) has orthogonal increments.
Example 20.4.1 (Poisson Process) A Poisson process N(t) with random arrival
times fSk , k ¼ 1, . . . , n, . . .} and random interarrival times {Tk , k ¼ 1, . . . , n, . . .} is an
20.4
Independent Increment Process
509
independent increment process since
P{Tk1 s, Tk t} ¼ P{½N(Sk1 ) N(Sk2 ) ¼ 1, ½N(Sk ) N(Sk1 ) ¼ 1}
¼ P ½N(Sk1 ) N(Sk2 ) ¼ 1 P ½N(Sk ) N(Sk1 ) ¼ 1
¼ (1 elt )ð1 elt Þu(s)u(t)
The independent increments are stationary since
P ½N(t) N(s) ¼ 1 ¼ l(t s)el(ts)
depends only on the time difference (t 2 s) and not the individual times t and s. The mean
of the Poisson process can be obtained by substituting in Eq. (20.4.2a)
m0 ¼ E ½N(0) ¼
1
X
k el:0
k¼0
(l:0)k
¼0
k!
and with
E½N(1) ¼
1
X
k el
k¼0
lk
¼l
k!
m1 ¼ l 0 ¼ l and hence E½N(t) ¼ lt, a result that has been obtained previously. The
variance of the process can be obtained by substituting in Eq. (20.4.2b) s20 ¼ 0 and
s21 ¼
1
X
k2 el
k¼0
lk
(lt)2 ¼ l
k!
Example 20.4.2 (Binomial Process)
Eq. (20.3.7) as
Yn ¼
and hence var ½N(t) ¼ lt
A binomial process has been defined in
n
X
Xi ,
n.0
i¼1
where n is the number of trials and Xi represents either 1 for success or 0 for failure at count
i. Since each Xi is independent of Xj, j = i, we conclude that the process Yn is an independent increment process. The independent increments are stationary because
k
nm k
P ½(Yn Ym ¼ k) ¼
p (1 p)(nm) , n . m
k
is dependent only on the count difference (n 2 m) and not on individual counts n and m.
In the binomial process m0 ¼ E½Y0 ¼ 0 by definition and E ½Y1 ¼ E ½X1 ¼ p. Substitution of E ½Y1 in m1 results in m1 ¼ p 0 ¼ p, and hence, substituting m0 ¼ 0, m1 ¼ p,
and t ¼ n in Eq. (20.4.2a), we obtain E ½Yn ¼ np, a result obtained previously.
In a similar manner the variance of the process Yn can be obtained by substituting in
Eq. (20.4.2b) s20 and s21 ¼ p(1 p), resulting in var ½Yn ¼ npð1 pÞ, another result
obtained previously.
Elementary Random-Walk Process
The Bernoulli process considered earlier is now modified as shown in Fig. 20.4.1 so
that the failure is now represented by 21 instead of 0. Under this modification we define a
random variable Zn
Zn ¼
1
1
for success in the nth trial
for failure in the nth trial
(20:4:11a)
510
Classification of Random Processes
FIGURE 20.4.1
with probabilities
P(Zn ¼ 1) ¼ p
P(Zn ¼ 1) ¼ 1 p ¼ q
(20:4:11:b)
If p ¼ 12, this random process is called a symmetric random walk.
The mean and variance of Zn are
Mean:
mZ ¼ E ½Zn ¼ 1 P(Xn ¼ 1) 1 P(XN ¼ 1) ¼ p (1 p) ¼ 2p 1
(20:4:12)
Second Moment:
E ½Zn2 ¼ 1 p þ 1 (1 p) ¼ 1
(20:4:13)
Variance:
s2Z ¼ var½Zn ¼ 12 p þ (1)2 (1 p) (2p 1)2 ¼ 4p(1 p)
(20:4:14)
In much the same way as we did in Eq. (20.3.7), we define a new process Wn as the sum of
the random variables fZig in the modified Bernoulli process. This is called an elementary
random-walk process, given by
Wn ¼
n
X
Zi ,
n.0
(20:4:15)
i¼1
This process, shown in Fig. 20.4.2, is also a discrete-state discrete-time (DSDT) process.
The probability that there are k successes and (n 2 k) failures in n trials is again a binomial
FIGURE 20.4.2
20.4
Independent Increment Process
511
distribution given by
n k
Probability{k successes in any sequence in n trials} ¼ P(Wn ¼ k) ¼
p (1 p)nk
k
The statistics of this elementary random walk are
Mean:
"
mW ¼ E ½Wn ¼ E
n
X
#
Zi ¼
i¼1
n
X
E ½Zi ¼ n(2p 1),
n.0
(20:4:16)
i¼1
Variance. From the independence of Zi, the second moment of Wn can be given by
"
#2
n
n
n X
n
X
X
X
2
E ½Wn ¼ E
Zi ¼
E ½Zi2 þ
E ½Zi Z j i¼1
i¼1
i¼1 j¼1
2
¼ n þ n(n 1)(2p 1)
n.0
Hence
s2W ¼ E ½Wn2 {E ½Wn }2 ¼ n þ n(n 1)(2p 1)2
n2 (2p 1)2 ¼ 4np(1 p),
n.0
(20:4:17)
Autocorrelation. For n1 n2, we obtain
"
#
n2 X
n1
X
Zi Z j
E ½Wn1 Wn2 ¼ RW (n1 , n2 ) ¼ E
j¼1 i¼1
"
¼E
n1
X
i¼1
#
Zi2 þ
n1 X
n1
X
j¼1 j¼1
i=j
E ½Zi Z j þ
nX
n1
2 n1 X
E ½Zi Z j (20:4:18)
j¼1 i¼1
i=j
Substituting for the expectation operators from Eqs. (20.4.12) and (20.4.13), we
obtain E ½Wn1 Wn2 ¼ 4p(1 p)n1 þ n1 n2 (2p 1)2 with a similar expression for
n2 n1 given by E ½Wn1 Wn2 ¼ 4p(1 p)n2 þ n1 n2 (2p 1)2 . Combining these
two equations, we have
E ½Wn1 Wn2 ¼ RW (n1 , n2 Þ ¼ 4p(1 p) min(n1 , n2 ) þ n1 n2 (2p 1)2
(20:4:19)
Autocovariance:
CW (n1 , n2 ) ¼ 4p(1 p) min(n1 , n2 )
(20:4:20)
Since the mean is a function of time n and the autocorrelation is a function of both
n1 and n2, we conclude that the random-walk process is also a nonstationary
process.
Example 20.4.3 (Random-Walk Process) Since the random-walk process is a
takeoff from the Bernoulli process and with distribution function similar to the
binomial process, we can conclude that this process is also a stationary independent increment process. We will show that the random-walk process satisfies
Eqs. (20.4.2a) and (20.4.2b). In this process m0 ¼ E ½W0 ¼ 0 by definition and
512
Classification of Random Processes
m1 ¼ E ½W1 ¼ E ½Z1 ¼ 2p 1. Substituting for m0 ¼ 0, m1 ¼ (2p 2 1) and t ¼ n in
Eq. (20.4.2a), we obtain E ½Wn ¼ mW ¼ n(2p 1), which is the same as Eq. (20.4.16).
In a similar manner, s20 ¼ E ½W0 m0 2 ¼ 0 and s21 ¼ E ½W12 m21 ¼ 1 (2p 1)2 ¼
4p(1 p). Hence var½Wn ¼ 4np(1 p), which is the same as Eq. (20.4.17).
B 20.5
RANDOM-WALK PROCESS
An elementary random walk has been discussed in the previous section. We will generalize this concept a little further.
A particle is initially at rest at i ¼ 0 with W0 ¼ 0. At times fi ¼ 1,2, . . . , n, . . .g it
undergoes jumps fZi, i ¼ 1, 2, . . . , n, . . .g, where fZig is a sequence of independent identically distributed random variables with probabilities
9
P(Zi ¼ þ1Þ ¼ p
=
P(Zi ¼ 1Þ ¼ q
(20:5:1)
i ¼ 1, . . . , n, . . . , P(Zi Z j ) ¼ P(Zi )P(Z j ), i = j
;
P(Zi ¼ 0Þ ¼ 1 p q
If Wn is the position of the particle after n jumps, then it can be written in terms of Zi as
Wn ¼
n
X
Zi , n . 0, and W0 ¼ 0
(20:5:2)
i¼1
The random processes Zi and Wn are shown in Fig. 20.5.1.
If the particle tends to move indefinitely in either the positive or negative direction,
then this random walk is called unrestricted. On the other hand, the particle may be
restricted to þa in the positive direction and 2b in the negative direction where a and
b are positive, in which case a and b are called absorbing barriers. Let the position of
the particle at time n be equal to k where k ¼ 0, +1, +2, . . . , +n. We have to find
the probability P(Wn ¼ kg. We observe that to reach point k, the particle undergoes r1
positive jumps, r2 negative jumps, and r3 zero jumps with the following constraining
conditions:
r1 þ r2 þ r3 ¼ n,
k ¼ r1 r2 ,
FIGURE 20.5.1
r3 ¼ n r1 r2
(20:5:3)
20.5 Random-Walk Process
513
With r1 as an independent variable P(Wnr1 ¼ k} is a multinomial distribution given by
P(Wnr1 ¼ k) ¼
n!
pr1 qr2 (1 p q)r3
r1 !r2 !r3 !
(20:5:4)
Substituting the constraints of Eq. (20.5.3) in Eq. (20.5.4), we obtain
P(Wnr1 ¼ k) ¼
n!
pr1 qr1 k (1 p q)n2r1 þk
r1 !(r1 k)!(n 2r1 þ k)!
(20:5:5)
From Eq. (20.5.5), the constraints for r1 are r1 k and r1 ðn þ kÞ=2. In other words,
kr
nþk
2
(20:5:6)
where n þ2 k represents the integer closest to but not exceeding ðn þ kÞ=2.
Since the events {Wnr1 ¼ k} for each r1 are mutually exclusive, we can sum over all
possible values of r1 in Eq. (20.5.6) and write P(Wn ¼ kg as follows:
X
P(Wnr1 ¼ k)
P(Wn ¼ k) ¼
r1
¼
bðnþkÞ=2c
X
r1 ¼k
n!
r1 !(r1 k)!(n 2r1 þ k)!
pr1 qr1 k (1 p q)n2r1 þk
(20:5:7)
The probability P(Wn ¼ k) can be more easily determined by using the generating function
(GF) defined in Eq. (11.3.1). The generating function for the jumps Zi is
X
zk P(Zi ¼ k) ¼ pz þ qz1 þ (1 p q)z0
PZ (z) ¼ E ½zZi ¼
k
¼
pz2 þ (1 p q)z þ q
z
(20:5:8)
Since Wn is the sum of n independent jumps, the GF for Wn is the n-fold multiplication of
the GF PZ (z), or
2
n
pz þ (1 p q)z þ q
n
Wn
, n.0
(20:5:9)
PWn (z) ¼ E ½z ¼ ½PZ (z) ¼
z
and since W0 ¼ 0, the GF for W0 can be defined as follows:
PW0 (z) ¼ E ½zW0 ¼ ½PZ (z)0 ¼ 1
(20:5:10)
With PW0 (z) ¼ 1 from Eq. (20.5.10), we can take the s transform of PWn (z) ¼ ½PZ (z)n
and write
G(z, s) ¼
1
X
sn ½PZ (z)n ¼
n¼0
¼
z
s( pz2
1
1 sPZ (z)
z
þ (1 p q)z þ q)
for
jsPZ (z)j , 1
The probability P(Wn ¼ k) is the coefficient of s nz k in Eq. (20.5.11).
(20:5:11)
514
Classification of Random Processes
3
Example 20.5.1 In a random-walk process p ¼ 12 and q ¼ 10
. We will determine the
probability P(Wn ¼ k) of the particle being at k ¼ 3 in n ¼ 10 trials. We will solve this
problem using the direct approach of Eq. (20.5.5) and the generating function approach
of Eq. (20.5.11).
Direct Approach. In Eq. (20.5.5) the known quantities are n ¼ 10, k ¼ 3, p ¼ 12, and
3
q ¼ 10
. We have to find P(W10 ¼ 3) for all possible values of r1 the positive jumps.
The values of r1 from Eq. (20.5.6) are 3,4,5,6. From Eq. (20.5.7) P(W10 ¼ 3) is
given by
P(W10 ¼ 3) ¼
6
X
10!
1r1 3 r1 3 1132r1
r !(r 3)!(13 2r1 )! 2 10
5
r ¼3 1 1
1
or
P(W10 ¼ 3) ¼ 0:000192 þ 0:00756 þ 0:0567 þ 0:070875 ¼ 0:135327:
3
Generating Function Approach. Substituting the values of n ¼ 10, p ¼ 12, and q ¼ 10
in Eq. (20.5.11), we obtain
z
1
¼
G(z, s) ¼
1 2 1
3
s(5z2 þ 2z þ 3)
zs z þ zþ
1
2
5
10
10z
The required probability P(W10 ¼ 3) is the coefficient of s 10z 3 in the power-series
expansion of G(z, s). The power-series expansion for G(z,s) in terms of s is given by
k
1 X
s(5z2 þ 2z þ 3)
G(z, s) ¼
10z
k¼0
and the coefficient of s 10 is
Coefficient of s10 ¼
2
10
5z þ 2z þ 3
10z
This coefficient of s 10 will be expanded as a power series in z. Expanding the preceding equation, the coefficient of z 3 is
135,327
Coefficient of z3 ¼
1,000,000
135, 327
Hence, P(W10 ¼ 3) ¼ 1,000,000
, agreeing with the previous result obtained from the
direct method.
Random Walk with Two Absorbing Barriers a and 2b
We will now consider a random walk with two absorbing barriers. The particle starts
at an initial value of j and terminates at either a or 2b with a,b . 0. The initial value of the
random walk is
W0 ¼ j, b , j , a
It can be shown from heuristic arguments that as n ! 1, the ultimate probability of
absorption at either a or 2b is certain. We will define the probability of absorption as
Pa (n, j) ¼ P{absorption at a at time n starting from j},
n ¼ 1, 2; . . .
(20:5:12)
20.5 Random-Walk Process
515
The probability Pa(n, j ) also represents the first passage of the particle into a at time n
starting from j, without having entered either 2b or a at any previous times:
Pa (n, j) ¼ P{b , W1 , a, . . . , b , Wn1 , a,
Wn ¼ a j W0 ¼ j},
n ¼ 1, 2, . . .
(20:5:13)
At n ¼ 0, we have the initial probabilities
Pa (0, j) ¼
1,
0,
if j ¼ a (absorption occurs at a at 0 when it starts at a
if j = a (absorption cannot occurs at a at 0 not starting at a
(20:5:14)
and at j ¼ a and j ¼ 2b the end probabilities for n ¼ 1, 2, . . . , are
Pa (n, a) ¼ 0,
Pa (n, b) ¼ 0,
n ¼ 1, 2, . . .
n ¼ 0, 1, 2, . . .
(20:5:15)
We define for convenience an event An as
An ¼ {absorption at a at time n} and P{An jW0 ¼ j} ¼ Pa ðn, j)
(20:5:16)
Starting from j at n ¼ 0, after the first step at n ¼ 1, absorption at a at time (n 2 1) will
occur for the cases when the state can be at either ( j þ 1) with probability p or ( j 2 1)
with probability q, or remain at j with probability (1 2 p 2 q). Since these are mutually
exclusive events, we have
P{An j W0 ¼ j} ¼ pP{An1 j W0 ¼ j þ 1} þ qP{An1 j W0 ¼ j 1}
þ (1 p q)P{An1 j W0 ¼ j}
(20:5:17)
Substituting the definition of Pa(n, j ) from Eq. (20.5.16) in Eq. (20.5.17), we obtain
Pa (n, j) ¼ pPa (n 1, j þ 1) þ qPa (n 1), j 1)
þ (1 p q)Pa (n 1, j), b , j , a,
n ¼ 1, 2, . . .
(20:5:18)
Equation (20.5.15) is a difference equation in the two variables n and j. We can eliminate
the time variable n by taking the z transform of Pa(n, j ), or
Ga (z, j) ¼
1
X
zn Pa (n, j)
(20:5:19)
n¼0
with the initial probability Pa(0, j ) given by Eq. (20.5.14). Multiplying Eq. (20.5.18) by z n
and summing over all n, we obtain
Ga (z, j) ¼ z{pGa (z, j þ 1) þ qGa (z, j 1) þ (1 p q)Ga (z, j)}
(20:5:20)
Equation (20.5.20) is a second-order difference equation in j. The end conditions from
Eqs. (20.5.14) and (20.5.15) are
Ga (z, a) ¼ 1,
Gb (z, a) ¼ 0
(20:5:21)
We can take another z transform for the variable j. Since j is bounded by a, we can
solve the difference equation [Eq. (20.5.20)] more easily by assuming a solution of the
form Ga(z, j ) ¼ lj(z). Thus
l j (z) þ z pl jþ1 (z) þ ql j1 (z) þ (1 p q)l j (z) ¼ 0
516
Classification of Random Processes
or
z pl2 (z) ½1 z(1 p q)l(z) þ zq ¼ 0
The solutions to the quadratic equation [Eq. (20.5.22)] are
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 z(1 p q) þ ½1 z(1 p q)2 4z2 pq
l1 (z) ¼
2zp
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 z(1 p q) ½1 z(1 p q)2 4z2 pq
l2 (z) ¼
2zp
(20:5:22)
(20:5:23)
with the relationship between the roots l1 (z)l2 (z) ¼ (zq=zp) ¼ (q=p) and l1 (z) . l2 (z).
We will assume that z is positive and real. If l1(z) and l2(z) are to be real, then the
discriminant in Eq. (20.5.23) must be positive. Hence we have the following inequality
½1 z(1 p q)2 4z2 pq . 0
or
z2 ½(1 p q)2 4pq 2z(1 p q) þ 1 . 0
(20:5:24)
The factors of the preceding quadratic equation in z are
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2(1 p q) + 4(1 p q)2 4½(1 p q)2 4pq
z1 ; z2 ¼
2½(1 p q)2 4pq
pffiffiffiffiffi
1 p q + 2 pq
1
1
¼
¼
pffiffiffiffiffiffi ¼
pffiffiffiffi pffiffiffi
(1 p q)2 4pq 1 p q + 2 pq 1 ( p + q)2
or
"
1
z1 pffiffiffiffi pffiffiffi 2
1 ( p q)
#"
#
1
z2 .0
pffiffiffiffi pffiffiffi
1 ( p þ q) 2
(20:5:25)
If z is positive, the inequality of Eq. (20.5.25) will be satisfied only when both
1
1
pffiffiffiffi pffiffiffi 2 and z2 ,
pffiffiffiffi pffiffiffi 2
1 ( p q)
1 ( p þ q)
pffiffiffi pffiffiffi 2
pffiffiffi pffiffiffi 2
Since ( p þ q) . ( p q) , Eq. (20.2.25) is equivalent to
z1 ,
0,z,
1
pffiffiffiffi pffiffiffi
1 ( p þ q)2
The general solution of the difference Eq. (20.5.2) can be given by
Ga (z, j) ¼ c1 (z)lj1 (z) þ c2 (z)lj2 (z)
(20:5:26)
where c1(z) and c2(z) are arbitrary functions of z. Substituting the initial conditions of
Eq. (20.5.21) in Eq. (20.5.26), we can solve for c1(z) and c2(z) as follows:
c1 (z) ¼
lb1 (z)lb
2 (z)
a
b
l2 (z) l2 (z)laþb
1 (z)
and
c2 ðzÞ ¼
1
la2 (z)
aþb
lb
2 (z)l1 (z)
(20:5:27)
20.5 Random-Walk Process
517
Substituting Eq. (20.5.27) into Eq. (20.5.36), we obtain the solution Ga(z, j ) as
Ga (z, j) ¼
j
j
lb1 (z)lb
l1jþb (z) l2jþb (z)
2 (z)l1 (z) þ l2 (z)
¼ aþb
a
b
aþb
l2 (z) l2 (z)l1 (z)
l1 (z) laþb
2 (z)
(20:5:28)
To obtain the probability of absorption at a at time n starting from j, we have to express
Eq. (20.5.8) as a power series in z, and the coefficient of z n will yield Pa(n,j ). These
involved calculations can be found in Refs. 13 and 17. However, the events of eventual
absorption at a or – b are mutually exclusive, and their probabilities can be calculated
easily. We will find the probability of eventual absorption at a starting from j.
Equation (20.5.28) is rewritten as
Ga (z, j) ¼
l1jþb (z) l2jþb (z)
aþb
laþb
1 (z) l2 (z)
(20:5:29)
The probability of eventual absorption at a starting at j will be given by the sum of the
probabilities Pa(n, j ) over all n and can be obtained by substituting z ¼ 1 in Eq. (20.5.29):
Pa ( j) ¼ P{absorption at a starting at j} ¼
1
X
Pa (n, j)
n¼0
¼ Ga (1, j) ¼
l1jþb (1) l2jþb (1)
aþb
laþb
1 (1) l2 (1)
(20:5:30)
Substituting z ¼ 1 in the roots l1(z) and l2(z), since l1(z) . l2(z), we have
p , q, l1 (1) ¼
( p þ q) ( p q) q
( p þ q) þ ( p q)
¼ . l2 (1) ¼
¼1
2p
p
2p
p . q, l1 (1) ¼
( p þ q) þ ( p q)
( p þ q) ( p q) q
¼ 1 . l2 (1) ¼
¼
2p
2p
p
(20:5:31)
p ¼ q, l1 (1) ¼ l2 (1) ¼ 1
Substituting Eq. (20.5.31) in Eq. (20.5.30), we obtain
Pa ( j) ¼ G(1, j) ¼ paj
p jþb q jþb
,
paþb qaþb
p=q
(20:5:32a)
Using l’Hôpital’s rule when p ¼ q, we have
Pa (j) ¼
jþb
,
aþb
p¼q
(20:5:32b)
In a similar manner we can obtain the solution G2b(z, j ) for the absorption at 2b starting
from j. The result is
Gb (z, j) ¼
l1ja (z)
(aþb)
l1
(z)
l2ja (z)
l(aþb)
(z)
2
The probability of absorption at 2b starting at j is given by
Gb (1, j) ¼ Pb ( j) ¼ 1 Pa ( j) ¼ 1 Ga (1, j)
(20:5:33)
518
Classification of Random Processes
or
8
paj qaj
>
>
< qbþj aþb
,
p
qaþb
Pb (0) ¼ Gb (1, 0)
>
aj
>
:
, p¼q
aþb
p=q
(20:5:34)
Example 20.5.2 (Gambler’s Ruin) This is a classic problem that has wide applications
in testing the efficacy of drugs, insurance payoffs, and so on. Two gamblers A and B wager
on the flip of a biased coin. If heads show up, A wins 1 dollar from B, and if tails show up,
A loses one dollar to B. The probability of heads is p and tails q ¼ 1 2 p. Initially A starts
with k dollars and B with N 2 k dollars. The game stops when either A loses all the money
or is ruined or A wins all the money and B is ruined. We want to calculate the probabilities
of A either winning all the money or losing all the money.
This problem can be cast in the random-walk framework with absorbing barriers. We
will solve this problem from fundamentals and show that we get the same result from using
generating functions. Each toss is an i.i.d. random variable Zi with probabilities
P(Zi ¼ þ1) ¼ p
i ¼ 1, . . . , n, . . . , P(Zi Zj ) ¼ P(Zi )P(Zj ),
P(Zi ¼ 1) ¼ q ¼ (1 p)
i=j
and the cumulative sum Wn is given by Eq. (20.5.2) with initial value W0 ¼ k. Or,
Wn ¼
n
X
Zi , n . 0
and
W0 ¼ k
(20:5:35)
i¼1
The game terminates when Wn ¼ N or Wn ¼ 0. The absorbing barriers are either N or 0.
We define an event An as absorption at N at time n and the probability of absorption at N at
time n on condition that the starting state is k, as follows:
P{An jW0 ¼ k} ¼ PN (n, k)
Starting from k at n ¼ 0, after the first step at n ¼ 1, absorption at N at time (n 2 1) will
occur for the cases when the state can be at either (k þ 1) with probability p or (k 2 1) with
probability q. Since these are mutually exclusive events, we have
P{An jW0 ¼ k} ¼ pP{An1 jW0 ¼ k þ 1} þ qP{An1 jW0 ¼ k 1}
and in terms of the defined probability PN(n, k)
PN (n, k) ¼ pPN (n 1, k þ 1) þ qPN (n 1, k 1),
k ¼ 1, 2, . . . , N 1
with end conditions PN(n, N ) ¼ 1 and PN(n, 0) ¼ 0, for n ¼ 1, 2, . . . , N 2 1.
We are interested in the eventual probability of absorption in N and hence we can
eliminate n from the equation above and write
PN (k) ¼ pPN (k þ 1) þ qPN (k 1),
k ¼ 1, 2, . . . , N 1
with end conditions PN(N ) ¼ 1 and PN(0) ¼ 0,
Since p þ q ¼ 1, the preceding equation can be rewritten as
PN (k)( p þ q) ¼ pPN (k þ 1) þ qPN (k 1)
20.5 Random-Walk Process
519
and rearranging terms, we have
q
PN (k þ 1) PN (k) ¼ ½PN (k) PN (k 1),
p
k ¼ 1, 2, . . . , N 1
We solve this equation recursively as follows from the initial condition PN(0) ¼ 0
k ¼ 1:
k ¼ 2:
q
q
PN (2) PN (1) ¼ ½PN (1) PN (0) ¼ PN (1)
p
p
2
q
q
PN (1)
PN (3) PN (2) ¼ ½PN (2) PN (1) ¼
p
p
k ¼ k 1:
q
PN (k) PN (k 1) ¼ ½PN (k 1) PN (k 2)
p
k1
q
¼
PN (1)
p
k ¼ N 1:
q
PN (N) PN (N 1) ¼ ½PN (N 1) PN (N 2)
p
N1
q
PN (1)
¼
p
Adding the first k21 equations, we obtain
"
2
k1 #
q
q
q
þ þ
PN (k) PN (1) ¼ PN (1) þ
p
p
p
or
"
2
k1 #
q
q
q
PN (k) ¼ PN (1) 1 þ þ
þ þ
p
p
p
k
q
1
p
¼ PN (1)
q ,
1
p
PN (k) ¼ PN (1)k,
p=q
p¼q
Substituting k ¼ N in the equation above, we can obtain PN(1) by applying the boundary
condition PN(N ) ¼ 1 as shown below:
q
1
p
PN (1) ¼
N , p = q
q
1
p
and
PN (1) ¼
1
, p¼q
N
520
Classification of Random Processes
Substituting for PN(1) in the equation for PN(k), we obtain
k
k
q
q
q
1
1
1
p
p
p
PN (k) ¼
N ¼
N ,
q
q
q
1
1
p 1
p
p
¼
k
,
N
p¼q
p=q
(20:5:36)
PN(k) is the probability that A starting from k dollars will eventually reach N dollars.
The probability of ruin P0(k) of A is
Nk
p
1
q
P0 (k) ¼ 1 PN (k) ¼
N , p = q
p
1
q
¼
Nk
,
N
p¼q
(20:5:37)
The probability of ruin P0(k) can be easily obtained from Eq. (20.5.34) by substituting for
the absorbing barriers b ¼ 0, a ¼ N and j ¼ k, which results in Eq. (20.5.37).
Discussion of Results. We will investigate what happens when A plays against an infinitely rich adversary B. If p . q, then ðq=pÞ , 1, and as N ! 1 in Eq. (20.5.36),
we have
k
q
k
1
q
p
p . q: lim PN (k) ¼ lim
¼
1
.0
(20:5:38)
N
N!1
N!1
p
q
1
p
This results in a finite probability that A will get infinitely rich. On the other hand, if
p q, then (q=p) . 1, and as N ! 1 in Eq. (20.5.36), we have
k
q
Nk
1
p
p
p , q: lim PN (k) ¼ lim
¼
lim
! 0
N
N!1
N!1
N!1 q
q
(20:5:39)
1
p
p ¼ q:
lim PN (k) ¼ lim
N!1
k
N!1 N
! 0
This results in A getting ruined with probability 1 if p q. The interesting part is
that even if it is a fair game with equal probability of winning, a rich player will
always ruin a poorer player. On the other hand, if p . q, A can get infinitely rich
even against an infinitely rich adversary. The gambling halls all over the world
make sure that the probability of any player winning is always less than 0.5!
Example 20.5.3 An insurance company earns $1 per day from premiums with probability p. However, on each day it pays out a claim of $2 each day independent of the
past with probability q ¼ 1 2 p. When a claim is paid, $2 is removed from the reserve
20.6
Gaussian Process
521
of money of $k. Hence we can assume that on the ith day, the income for that day is a
random variable Zi equal to 1 with probability p and 21 with probability q as in
Example 20.5.2. The cumulative income for the company is Wn as in Eq. (20.5.35). If it
starts initially with a reserve capital of $k, then we have to find the probability that it
will eventually run out of money. The solution is a reformulation of Eqs. (20.5.38) and
(20.5.39). If p . q, then the probability that it will not run out of money is given by
1 (q=p)k . 0. If k is large, then it has very high probability of staying in business. On
the other hand, if p q, it will always run out of money as indicated in Eq. (20.5.39).
B 20.6
GAUSSIAN PROCESS
Many of the physically occurring random phenomena can be modeled as a Gaussian
process. Further, according to the central limit theorem, random variables characterized
individually by arbitrary distributions can be modeled as Gaussian if a sufficient
number of them can be added. The distribution of a more commonly used white noise
is Gaussian.
We have already stated (in Section 6.4) that a random variable X is Gaussiandistributed if its density function is given by
fX (x) ¼
2
1
pffiffiffiffiffiffi eð1=2Þ½(xmX )=sX sX 2p
(20:6:1)
where mX is the mean value and s2X is the variance.
A second-order random process X(t) is a Gaussian random process if for any collection of times ft1, t2, . . . , tng, the random variables, X1 ¼ X(t1), X2 ¼ X(t2), . . . , Xn ¼ X(tn)
are jointly Gaussian distributed random variables for all n with the joint density given by
fX1 X2 ...Xn (x1 , x2 , . . . , xn ) ¼
T 1
1
e(1=2)(xmX ) CX (xmX )
(2p)n=2 jCX j
where the mean vector mX and the covariance
2
2
3
mX (t1 )
CX (t1 )
6 CX (t2 , t1 )
6 mX (t2 ) 7
6
6
7
mX ¼ 6 . 7; CX ¼ 6
..
4
4 .. 5
.
CX (tn , t1 )
mX (tn )
(20:6:2)
matrix CX are given by
CX (t1 , t2 )
CX (t2 )
..
.
CX (tn , t2 )
3
CX (t1 , tn )
CX (t2 , tn ) 7
7
7
..
5
.
CX (tn )
(20:6:3)
Properties
1. A Gaussian process is completely determined if the mean vector mX and the
covariance matrix CX are known.
2. Sum of Gaussian processes are also Gaussian.
3. If a Gaussian process is weakly stationary, it is also strictly stationary.
White Gaussian Noise
A white-noise process has been defined in Eq. (19.2.17) as a stationary process whose
autocorrelation is a Dirac delta function. The white-noise process that is widely used for
522
Classification of Random Processes
FIGURE 20.6.1
modeling communication problems is the Gaussian process. A standard Gaussian whitenoise process V(t) has zero mean and unit variance with probability density
1
2
fV (v) ¼ pffiffiffiffiffiffi e(v =2) , 1 , v , 1
2p
(20:6:4)
The autocorrelation function and power spectral densities are given by
RV (t) ¼ d(t);
SV (v) ¼ 1, 1 , v , 1
(20:6:5)
Similarly, a standard white Gaussian noise sequence fVng has zero mean and unit variance
with the sequences coming from a density function given by Eq. (20.6.4). However, its
autocorrelation function and power spectral densities are given by
RV (h) ¼ dh ;
SV (v) ¼ 1, 1 , v , 1
(20:6:6)
In this study, 10,000 points of a Gaussian white noise fVng was generated from a computer
using the Box – Mueller method. The noise sequence was standardized to zero mean and
unit variance. This sequence is shown in Fig. 20.6.1a for only 200 points for clarity,
and its histogram illustrating the Gaussian nature of the density is shown in Fig. 20.6.1b.
Example 20.6.1 [AR(1) Process]
random process Xi was given by
In Example 19.2.11 a discrete zero mean stationary
Xn ¼ fXn1 þ Vn , n ¼ 1, . . . , X0 ¼ V0
where fVn, n ¼ 0, 1. . .g is a Gaussian process. This process is called an autoregressive (1)
process. We will investigate this process assuming that Vn is standard Gaussian white
noise and the coefficient f ¼ 0.8.
We will show that Xn is also Gaussian-distributed. By defining B k as a backward
difference operator defined by Bk Xn ¼ Xnk , k 0, we can write the given equation as
(1 fB)Xn ¼ Vn
or
Xn ¼ (1 fB)1 Vn ¼ (1 þ fB þ f2 B2 þ þ fk1 Bk1 )Vn
¼ Vn þ fVn1 þ f2 Vn2 þ þ fk1 Vnkþ1 þ Since Xn is a sum of a weighted Gaussian-distributed random sequence, it is also a
Gaussian-distributed continuous-state discrete-time (CSDT) random process. The
20.7
Wiener Process (Brownian Motion)
523
FIGURE 20.6.2
variance of this process as given by Eq. (19.2.47) is s2X ¼ s2X =(1 f2 ) ¼ 1=ð1 0:64Þ ¼
2:778 with mean mX ¼ 0.
This process Xn simulated on a computer with 10,000 points of the Gaussian noise
sequence generated previously is shown in Fig. 20.6.2a for only 200 points for clarity.
The variance of the simulated process is 2.905, which is very close to the true value
of 2.778. The histogram of the data of 10,000 points is shown in Fig. 20.6.2b. On
the same histogram plot is shown the true pdf of Xn and the pdf of true white-noise
sequence Vn.
B 20.7
WIENER PROCESS (BROWNIAN MOTION)
The random motion of a particle in a fluid subject to collisions and influence of other particles is called Brownian motion. One of the mathematical models of this motion is the
Wiener process.
If the following conditions are satisfied, W(t) is a Wiener process with parameter s2:
1. W(0) ¼ 0.
2. W(t) is a Gaussian process with sample continuous functions.
3. E[W(t)] ¼ 0.
4. The autocovariance function CW(t, s) ¼ RW(t, s) ¼ E[W(t), W(s)] ¼ s2 min(t, s).
5. The variance of W(t) from (4) is s2W (t) ¼ s2 t.
The pdf of the Wiener process is given by
1
2
2
fW (w, t) ¼ pffiffiffiffiffiffiffiffiffiffi e(1=2)(w =s t)
2pst
(20:7:1)
The covariance matrix of the random variable vector WT ¼ ½W(t1 ), W(t2 ), . . . , W(tn )T
for times 0 , t1 , t2 , . . . , tn can be given from condition 4 as
2
3
t1 t1 t1 t1
6 t1 t2 t2 t2 7
6
7
T
2 6 t1 t2 t3
t3 7
(20:7:2)
CW ¼ E ½WW ¼ s 6
7
6. . .
.. 7
4 .. .. ..
5
.
t1 t2 t3 tn
524
Classification of Random Processes
and this matrix is positive definite. Since W(t) is Gaussian, the finite-dimensional density
function using Eq. (20.7.2) is given by
fW (w1 , t1 ; w2 , t2 ; . . . ; wn , tn ) ¼
T 1
1
e(1=2)W CW W
1=2
(2p) jCW j
n=2
(20:7:3)
The determinant jCW j ¼ t1 (t2 t1 ) (tn1 tn ) and Eq. (20.7.3) can be rewritten as
n
Y
1
1 (wk wk1 )2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi exp fW (w1 , t1 ; w2 , t2 ; . . . ; wn , tn ) ¼
2
2 s2 (tk tk1 )
k¼1 2p(tk tk1 )s
(20:7:4)
with t0 ¼ 0 ¼ w0. The density function of Eq. (20.7.4) shows that the sequence
{Wt1 , (Wt2 Wt1 ), . . . , (Wtn Wtn1 )}
is a collection of independent random variables for increasing f0 , t1 , t2 , . . . , tng
and the density depends only on the time difference, showing that the Wiener process
is a process of stationary independent increments. Hence, it is nonstationary and is a
continuous-state continuous-time (CSCT) process.
Derivation of Wiener Process
We will derive the Wiener process as the limiting form of the elementary randomwalk process discussed in Section 20.4, which will be restated in a form where limits
can be taken. A particle starts from origin at t ¼ 0. In a small interval Dt it undergoes a
small step +Dw with equal probabilities, thus constituting a symmetric random walk.
The number n of Bernoulli trials in the time interval (0, t) is equal to the greatest
integer less than or equal to t/Dt, or
jtk
n¼
Dt
where bxc represents the greatest integer less than or equal to x, called the “floor function.”
Similar to the definition in Eq. (20.4.11), the Bernoulli random variable Zi is
defined by
Zi ¼
þ1
1
if the ith step of Dw is upward
if the ith step of Dw is downward
with probabilities
1
2
1
P(Zi ¼ 1) ¼
2
P(Zi ¼ þ1) ¼
(20:7:5)
The mean and variance of Zi are given by E ½Zi ¼ 0 and var ½Zi ¼ 1.
If the position of the particle at time nDt is W(nDt), then
W(nDt) ¼ Dw
n
X
i¼1
Zi u(t iDt)
(20:7:6)
20.7
Wiener Process (Brownian Motion)
525
The mean and variance of the process W(nDt) are
E ½W(nDt) ¼ 0 and
var ½W(nDt) ¼ E ½W 2 (nDt) ¼ Dw2 n ¼ Dw2
jtk
Dt
(20:7:7)
We now let Dw and Dt tend to zero in such a way that Eq. (20.7.7) is nontrivial. Hence we
have to set
pffiffiffiffiffi
Dw ¼ s Dt
(20:7:8)
where s is a positive constant so that as Dt ! 1,
W(t) ¼ lim s
Dt!0
1
X
pffiffiffiffiffi
Zi u(t iDt) Dt
i¼1
and
lim Dw2
Dt!0
jtk
! s2 t
Dt
(20:7:9)
W(t) is the desired Wiener process, and a sample function is shown in Fig. 20.7.1.
The statistics of the Wiener process are as follows:
Mean. Since E[Zi] ¼ 0, we have
E ½W(t) ¼ mW (t) ¼ 0
(20:7:10)
Variance. From Eq. (20.7.7), we obtain
var½W(t) ¼ E ½W 2 (t) ¼ s2W (t) ¼ s2 t
(20:7:11)
Autocorrelation. Since E[W(t)] ¼ 0, RW (t1, t2) ¼ CW (t1, t2).
For t1 , t2:
RW (t1 , t2 ) ¼ CW (t1 , t2 ) ¼ E½W(t1 )W(t2 )
¼ E{½W(t2 ) W(t1 ) þ W(t1 )W(t1 )}:
From linearity of expectations: RW (t1 , t2 ) ¼ E{½W(t2 ) W(t1 )W(t1 )} þ E ½W 2 (t1 ).
From independent increments: RW (t1 , t2 ) ¼ E ½W 2 (t1 ) ¼ s2 t1 .
Similarly, for t2 , t1, we have RW (t1 , t2 ) ¼ E ½W 2 (t2 ) ¼ s2 t2 .
FIGURE 20.7.1
526
Classification of Random Processes
Hence
RW (t1 , t2 ) ¼ CW (t1 , t2 ) ¼ s2 min(t1 , t2 )
(20:7:12)
The Wiener process has the following properties:
1. From the properties of the random walk from which W(t) was constructed, we
conclude that for (0 , t , s) the increments W(t) 2 W(0) and W(s) 2 W(t) are
independent.
2. From Eq. (20.7.1), for t2 , t1 the density function of the increment W(t2) 2 W(t1) is
2
2
1
2
fW (w1 , t1 ; w2 , t2 ) ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi eð1=2Þ½(w2 w1 ) =s (t2 t1 )
2
2p(t2 t1 )s
(20:7:13)
and since it depends only on the time difference (t2 2 t1), W(t) is a stationary independent increment process.
3. Since W(t) is the limit of a symmetric random walk, it can be conjectured that it is a
continuous process for all t . 0. Indeed it can be shown [35,37] that it is continuous everywhere with probability 1.
4. It can be shown [35,37] that W(t) is of unbounded variation, and as a result it is
differentiable nowhere in the conventional sense.
Property 4 has far-reaching consequences. If we have any function g[W(t)] of another
nonrandom function W(t), then g[W(t þ dt)] can be expanded in a Taylor series as
g½W(t þ dt) ¼ g½W(t) þ
dg½W(t)
1 d2 g½W(t)
dW(t) þ
dW 2 (t) þ dw
2 dw2
(20:7:14)
Neglecting higher-order terms Eq. (20.7.14), we can write dg[W(t)] as
g½W(t þ dt) g½W(t) ¼ dg½W(t) ¼
dg½W(t)
dW(t)
dw
(20:7:15)
However, if W(t) is a Wiener process, the term dW 2(t) cannot be neglected because
from Eq. (20.7.8) it is of the order of dt. Hence the differential dg[W(t)] can be
written only as
g½W(t þ dt) g½W(t) dg½W(t)
1 d2 g½W(t)
dW(t) þ
dW 2 (t)
dw
2 dw2
dg½W(t)
1 d 2 g½W(t)
dW(t) þ s2
dt
dw
2
dw2
(20:7:16)
neglecting only third and higher-order terms. Thus the differential dg[W(t)] does not obey
the ordinary laws of calculus! It is also interesting to note that if W(t) is a deterministic
process, then s2 is equal to zero and Eq. (20.7.16) reverts to the usual calculus. For more
details on this interesting aspect, see Refs. 35 and 37.
Formal Derivative of Wiener Process
We have stated in properties 3 and 4 that the Wiener process is continuous everywhere
but differentiable nowhere in the conventional sense. However, we will formally
20.8
Markov Process
527
differentiate it and obtain its autocorrelation function. If Z(t) ¼ W(t) is the formal derivative of the Wiener process, it can be shown from Eq. (19.2.31b) that the autocorrelation
function of Z(t) ¼ W(t) is given by
RZ (t, s) ¼ RW (t, s) ¼ E ½W(t)W(s) ¼
@2 RW (t, s)
@t @s
(20:7:17)
and if s ¼ t þ t, we obtain Eq. (19.2.31b). We substitute RW (t, s) ¼ s2 min(t, s) from
Eq. (20.7.12) in Eq. (20.7.17) and obtain
RW (t, s) ¼ RZ (t, s) ¼ E ½Z(t)Z(s) ¼
@2 2
s min(t, s) ¼ s2 d(t s)
@t @s
(20:7:18)
From Eq. (20.7.18) we conclude that since the autocorrelation of the derivative of the
Wiener process yields a Dirac delta function, it exists only in the generalized sense.
However, the autocorrelation of Z(t) ¼ W(t) is a function of the time difference (t 2 s).
It is also a zero mean Gaussian process. Hence the formal derivative Z(t) of a Wiener
process is a zero mean stationary Gaussian process with power spectral density s2 that
is constant in the entire frequency range. This is the Gaussian white noise that has been
used extensively to model communication and time-series problems. The similarity
between the Gaussian and Poisson white noises [Eq. (20.2.39)] is to be noted.
B 20.8
MARKOV PROCESS
It was stated in Section 19.1 that a random process X(t) is completely specified if the
nth-order joint density given by Eq. (19.1.9) for an ordered set of time functions
ft1 , t2 , t3 , . . . , tng
fX (xn , . . . , x1 ; tn , . . . , t1 ) ¼
@n
FX (xn , . . . , x1 ; tn , . . . , t1 )
@xn @x1
(20:8:1)
is specified for all positive integers n. The joint density given by Eq. (20.8.1) can be
decomposed into n first-order conditional densities for the same set of ordered time functions as shown below:
fX (x1 ; t1 )
fX (x2 ; t2 j x1 ; t1 )
fX (x3 ; t3 j x2 , x1 ; t2 , t1 )
fX (xn1 ; tn1 j xn2 , . . . , x2 , x1 ; tn2 , . . . , t2 , t1 )
(20:8:2)
fX (xn ; tn j xn1 , . . . , x2 , x1 ; tn1 , . . . , t2 , t1 )
By repeated application of the definition of conditional probability, the n-dimensional
joint density of Eq. (20.8.1) can be obtained from Eq. (20.8.2). Thus, for the random
process to be completely determined, we should have n ! 1 of these first-order
conditional densities. However, the specifications for a class of random processes can
be considerably simplified if we can express their conditional densities as
fX (xn ; tn jxn1 , . . . , x2 , x1 ; tn1 , . . . , t2 , t1 ) ¼ fX (xn ; tn jxn1 ; tn1 )
(20:8:3)
What Eq. (20.8.3) conveys is that the probabilistic behavior of the present given all past
history is dependent only on the immediate past. The property exhibited by Eq. (20.8.3) is
called the Markov property, and the class of processes having this property is known as
528
Classification of Random Processes
Markov processes. Using the Markov property the n-dimensional density can be given
as follows:
fX (xn , . . . , x1 ; tn , . . . , t1 ) ¼ f (xn ; tn j xn1 ; tn1 )fX (xn1 , . . . , x1 ; tn1 , . . . , t1 )
(20:8:4)
By applying the Markov property repeatedly to lower-order density functions, we can
write the n-dimensional density as
fX (xn , . . . , x1 ; tn , . . . , t1 ) ¼ fX (x1 ; t1 )
n
Y
f (xi ; ti j xi1 ; ti1 )
(20:8:5)
i¼2
Hence all finite-dimensional densities of a Markov process are completely determined by
the initial marginal density fX (x1 ; t1 ) and the set of first-order conditional densities,
{f (xi ; ti j xi1 ; ti1 ), i ¼ 2, 3, . . .}. The conditional probability density f (xi ; ti j xi1 ; ti1 )
is called the transition probability density. For a discrete-time Markov process, the
equation corresponding to Eq. (20.8.3) can be written as follows:
P{X(tn ) xn j X(tn1 ) xn1 , . . . , X(t1 ) x1 } ¼ P{X(tn ) xn j X(tn1 ) xn1 }
(20:8:6)
What we have shown is that given a Markov process, its n-dimensional density can be
expressed by the initial marginal density and a set of first-order conditional densities.
However, to construct a Markov process from the given first-order conditional densities,
the following consistency conditions have to be satisfied:
ð1
1.
fX (x2 ; t2 ) ¼
fX (x2 ; t2 j x1 ; t1 ) fX (x1 ; t1 ) dx2 , t1 , t2
(20:8:7)
1
This equation is the usual equation that relates a joint density to its marginal
density.
ð1
(x
;
t
j
x
;
t
)
¼
fX (x3 ; t3 j x2 ; t2 ) fX (x2 ; t2 j x1 ; t1 ) dx2 , t1 , t2 , t3
(20:8:8)
f
2. X 3 3 1 1
1
This equation is called the Chapman– Kolmogorov equation, which all Markov
processes must satisfy. We will show that Markov processes satisfy this equation.
The joint pdf fX (x3 , x1 ; t3 , t1 ) with t1 , t2 , t3 can be written as
ð1
fX (x3 , x1 ; t3 , t1 ) ¼
fX (x3 , x2 , x1 ; t3 , t2 , t1 ) dx2
(20:8:9)
1
Using the property of conditional expectation twice, we can rewrite Eq. (20.8.9) as,
ð1
fX (x3 , x1 ; t3 , t1 Þ ¼
fX (x3 ; t3 j x2 , x1 ; t2 , t1 )fX (x2 , x1 ; t2 , t1 ) dx2
1
ð1
fX (x3 ; t3 j x2 , x1 ; t2 , t1 )fX (x2 ; t2 j x1 ; t1 )fX (x1 ; t1 ) dx2
¼
1
(20:8:10)
Using the Markov property fX (x3 ; t3 j x2 , x1 ; t2 , t1 ) ¼ fX (x3 ; t3 j x2 ; t2 ) in
Eq. (20.8.10), we obtain
ð1
fX (x3 , x1 ; t3 , t1 ) ¼ fX (x1 ; t1 )
fX (x3 ; t3 j x2 ; t2 )fX (x2 ; t2 j x1 ; t1 ) dx2
(20:8:11)
1
20.8
Markov Process
529
Dividing Eq. (20.8.11) throughout by fX (x1 ; t1 ), for t1 , t2 , t3 , we have
fX (x3 , x1 ; t3 , t1 )
¼ fX (x3 , t3 j x1 ; t1 )
fX (x1 ; t1 )
ð1
¼
fX (x3 ; t3 j x2 ; t2 ) fX (x2 ; t2 j x1 ; t1 ) dx2
(20:8:8)
1
which is the desired Chapman–Kolmogorov (C–K) equation.
This derivation also holds good for discrete-time processes, and the result is
X
P{X(t3 ) x3 j X(t1 ) x1 } ¼
P{X(t3 ) x3 j X(t2 ) x2 }
x2
P{X(t2 ) x2 j X(t1 ) x1}
(20:8:12)
The state fx2, t2) represents the intermediate state, and operation of the C – K equation is
shown in Fig. 20.8.1.
The C – K equation is a sufficient condition for a Markov process in the sense that nonMarkovian processes may also satisfy this equation.
We can also define a Markov process in terms of conditional expectations. If g() is
any function with maxx jg(x)j , 1, then X(t) is a Markov process for an ordered sequence
ft1 , t2 , . . . , tng if and only if
E{g½X(tn ) j X(tn1 ), X(tn2 ), . . . , X(t1 )} ¼ E{g½X(tn ) j X(tn1 )}
(20:8:13)
If the conditional probability P{X(tn ) xn j X(tn1 ) xn1 } of a Markov process is
dependent only on the time difference (tn tn1 ), it is called a homogeneous process.
In other words, if (t2 t1 ) ¼ (tn tn1 ), then homogeneity implies
P{X(tn ) xn j X(tn1 ) xn1 } ¼ P{X(t2 ) x2 j X(t1 ) x1 }
Generally, a homogenous Markov process may not be stationary since PfX(t1) x1g may
depend on t1.
We have shown earlier that a stationary independent increment process is always nonstationary in light of Eqs. (20.4.2a) and (20.4.2b). We will now show that this process is
always a Markov process. Let X(t) be a stationary independent increment process with
FIGURE 20.8.1
530
Classification of Random Processes
t0 ¼ 0 and X(t0) ¼ x0 ¼ 0. For all ti , tj, the events
{X(ti ) ¼ xi , X(tj ) ¼ xj
and
{X(ti t0 ) ¼ xi x0 , X(tj ti ) ¼ xj xi }
are equal, and hence their joint densities are also equal. As a consequence
fX (xj ; tj , xi ; ti ) ¼ fX (xj xi ; tj ti , xi x0 ; ti t0 ),
i,j
(20:8:14)
With x0 ¼ t0 ¼ 0 and using the independent increment property, Eq. (20.8.14) can be
rewritten after dropping the time difference (tj 2 ti), due to stationarity:
fX (xj ; tj , xi ; ti ) ¼ fX (xj xi ) fX (xi ; ti ),
i,j
(20:8:15)
Analogous to Eq. (20.8.15), for the sequence ftn , tn21 , . . . , t1 , t0 ¼ 0g, we
can write
fX (xn , . . . ; x1 , x0 ; tn , . . . , t1 , t0 ) ¼ fX (xn xn1 , . . . , x2 x1 , x1 x0 ;
tn tn1 , . . . , t2 t1 , t1 t0 )
¼ fX (xn xn1 ) fX (x2 x1 ) fX (x1 ; t1 )
(20:8:16)
From Eq. (20.8.15) we can substitute
fX (xk xk1 ) ¼
fX (xk , xk1 ; tk , tk1 )
,
fX (xk1 ; tk1 )
k ¼ n, n 1, . . . , 2
in Eq. (20.8.16) and write
fX (xn , . . . x1 ; tn , . . . , t1 ) ¼
fX (xn , xn1 ; tn , tn1 ) fX (x2 , x1 ; t2 , t1 )
fX (xn1 ; tn1 ) fX (x2 ; t2 )
(20:8:17)
Equation (20.8.17) expresses the nth-order joint density of a stationary independent increment process X(t) in terms of the second- and first-order densities. From the definition of
conditional density, we can write
fX (xn ; tn j xn1 , . . . , x1 ; tn1 , . . . , t1 ) ¼
fX (xn , xn1 x1 ; tn , tn1 , . . . , t1 )
fX (xn1 x1 ; tn1 , . . . , t1 )
(20:8:18)
Applying Eq. (20.8.17) to both numerator and denominator of Eq. (20.8.17), we obtain
fX (xn ; tn j xn1 , . . . , x1 ; tn1 , . . . , t1 )
fX (xn , xn1 ; tn , tn1 ) fX (x2 , x1 ; t2 , t1 )
fX (xn1 ; tn1 ) fX (x2 ; t2 )
¼
fX (xn1 ; tn , tn1 ) fX (x2 , x1 ; t2 , t1 )
fX (xn2 ; tn2 ) fX (x2 ; t2 )
¼
(20:8:19)
fX (xn , xn1 ; tn , tn1 )
¼ fX (xn ; tn j xn1 ; tn1 )
fX (xn1 ; tn1 )
which is the defining equation of a Markov process. However, the converse is not true, or,
not every Markov process is a stationary independent increment process.
Summary of Properties of Markov Processes
1.
fX (xn ; tn j xn1 ; tn1 , . . . , x1 ; t1 ) ¼ fX (xn ; tn j xn1 ; tn1 )
(20:8:20)
20.8
Markov Process
531
Applying the chain rule for conditional expectations from Eq. (20.8.5), we
can write
fX (xn ; tn j xn1 ; tn1 , . . . , x1 ; t1 ) ¼
fX (xn ; tn , xn1 ; tn1 , . . . , x1 ; t1 )
fX (xn1 ; tn1 , . . . , x1 ; t1 )
¼ fX (xn ; tn j xn1 ; tn1 )
(20:8:21)
2. This rule can be generalized to any sequence of times ftnk . tnk21 . . . . . tn1g:
fX (xnk ; tnk j xnk1 ; tnk1 , . . . , xn1 ; tn1 ) ¼ fX (xnk ; tnk j xnk1 ; tnk1 )
3. A time-reversed Markov process is also a Markov process:
fX (xn ; tn j xnþ1 ; tnþ1 , . . . , xnþk ; tnþk ) ¼ fX (xn ; tn j xnþ1 ; tnþ1 )
(20:8:22)
4. Given the present information, the past and the future are conditionally
independent. Or, for k , m , n, we obtain
fX (xn ; tn , xk ; tk j xm ; tm ) ¼ fX (xn ; tn j xm ; tm ) f (xk ; tk j xm ; tm )
(20:8:23)
We will show that this result is true:
fX (xn ; tn , xk ; tk j xm ; tm ) ¼
¼
fX (xn ; tn , xm ; tm , xk ; tk )
fX (xm ; tm )
fX (xn ; tn j xm ; tm ) fX (xm ; tm j xk ; tk ) fX (xk ; tk )
fX (xm ; tm )
(chain rule)
Substituting for
fX (xm ; tm j xk ; tk ) ¼
fX (xk ; tk j xm ; tm ) fX (xm ; tm )
fX (xk ; tk )
in the preceding equation, we obtain Eq. (20.8.23).
5. Taking conditional expectations in property 3, we have
E ½X(tn )X(tk ) j X(tm ) ¼ xm ¼ E ½X(tn ) j X(tm ) ¼ xm ¼ E ½X(tk ) j X(tm ) ¼ xm (20:8:24)
6. The future, conditioned on the past history upto the present, is the future conditioned on the present:
E{X(tnþ1 ) j X(tn ), X(tn1 ), . . . , X(t1 )} ¼ E{X(tnþ1 ) j X(tn )}
This equation is a restatement of Eq. (20.8.20).
(20:8:25)
532
Classification of Random Processes
7. If t1 , t2 , t3, then
E ½X(t3 ) j X(t1 ) ¼ E{E ½X(t3 ) j X(t2 )jX(t1 )}
(20:8:26)
This equation is a restatement of the C –K property in terms of conditional
expectations.
Example 20.8.1 (Poisson Process) The Poisson process N(t) is an independent
increment process from Eq. (20.2.10.). It is also a stationary independent increment
process, because the probability of the increment, N(t2) 2 N(t1), given by
P{N(t2 ) N(t1 ) ¼ k2 k1 } ¼
½l(t2 t1 )k2 k1 l(t2 t1 )
e
(k2 k1 )!
is dependent only on time difference (t2 2 t1). From the result of Eq. (20.8.19), we can
conclude that the Poisson process is a homogeneous Markov process. However, it will
be instructive to show the Markov property from the definition of Eq. (20.8.20). The
second-order probability function is given by
P{N(t2 ) ¼ k2 , N(t1 ) ¼ k1 } ¼ P{N(t2 ) ¼ k2 jN(t1 ) ¼ k1 }P{N(t1 ) ¼ k1 }
The conditional probability N(t2 ) ¼ k2 given that N(t1 ) ¼ k1 is the probability that there
are k2 2 k1 events occurring in the time interval t2 2 t1:
P{N(t2 ) ¼ k2 jN(t1 ) ¼ k1 } ¼ P{N(t2 ) N(t1 ) ¼ k2 k1 }
¼
½l(t2 t1 )k2 k1 l(t2 t1 )
e
,
(k2 k1 )!
k2 k1
Hence the second-order probability function P{N(t2 ) ¼ k2 , N(t1 ) ¼ k1 } is given by
P{N(t2 ) ¼ k2 , N(t1 ) ¼ k1 } ¼
½lt1 )k1 ½l(t2 t1 )k2 k1 lt2
e ,
k1 !(k2 k1 )!
k2 k1
Proceeding in a similar manner, we can obtain the third-order probability function as
P{N(t3 ) ¼ k3 , N(t2 ) ¼ k2 , N(t1 ) ¼ k1 }
¼
½lt1 )k1 ½l(t2 t1 )k2 k1 ½l(t3 t2 )k3 k2 lt3
e ,
k1 !(k2 k1 )!(k3 k2 )!
k3 k2 k1
The second-order probability function P{N(t3 ) ¼ k3 , N(t2 ) ¼ k2 } can be evaluated as
P{N(t3 ) ¼ k3 , N(t2 ) ¼ k2 } ¼ P{N(t3 ) N(t2 ) ¼ k3 k2 }P{N(t2 ) ¼ k2 }
¼
½lt2 )k2 ½l(t3 t2 )k3 k2 lt3
e ,
k2 !(k3 k2 )!
k3 k2
20.8
Markov Process
533
Combining all these equations, we have
P{N(t3 ) ¼ k3 j N(t2 ) ¼ k2 } ¼
P{N(t3 ) ¼ k3 , N(t2 ) ¼ k2 }
P{N(t2 ) ¼ k2 }
½lt2 k2 ½l(t3 t2 )k3 k2 lt3
e
½l(t3 t2 )k3 k2 l(t3 t2 )
k2 !(k3 k2 )!
e
¼
¼
k2
(k3 k2 )!
½lt2 lt2
e
k2 !
P{N(t3 ) ¼ k3 j N(t2 ) ¼ k2 , N(t1 ) ¼ k1 }
¼
P{N(t3 ) ¼ k3 , N(t2 ) ¼ k2 , N(t1 ) ¼ k1 }
P{N(t2 ) ¼ k2 , N(t1 ) ¼ k1 }
½lt1 )k1 ½l(t2 t1 )k2 k1 ½l(t3 t2 )k3 k2 lt3
e
k1 !(k2 k1 )!(k3 k2 )!
¼
½lt1 )k1 ½l(t2 t1 )k2 k1 lt2
e
k1 !(k2 k1 )!
¼
½l(t3 t2 )k3 k2 l(t3 t2 )
e
(k3 k2 )!
Hence P{N(t3 ) ¼ k3 j N(t2 ) ¼ k2 , N(t1 ) ¼ k1 } ¼ P{N(t3 ) ¼ k3 j N(t2 ) ¼ k2 }, showing that
N(t) is indeed a Markov process. Since the conditional density
P{N(t2 ) ¼ k2 jN(t1 ) ¼ k1 } ¼
½l(t2 t1 )k2 k1 l(t2 t1 )
e
,
(k2 k1 )!
k2 k1
is a function of the time difference (t2 2 t1), we conclude further that the Poisson process is
also homogeneous Markov process.
Example 20.8.2 (Wiener Process)
process as given by Eq. (20.7.4) is
The n-dimensional density function of a Wiener
(
)
1
1 (wk wk1 )2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi exp fW (w1 , w2 , . . . , wn ; t1 , t2 , . . . , tn ) ¼
2
2 s2 (tk tk1 )
k¼1 2p(tk tk1 )s
n
Y
with t0 ¼ 0 ¼ w0 and the sequence
{Wt1 , (Wt2 Wt1 ), . . . , (Wtn Wtn1 )}
534
Classification of Random Processes
is a collection of independent random variables for increasing f0 , t1 , t2 , . . . , tng. We
have to show that
fW (wn j wn1 , . . . , w1 ) ¼ fW (wn j wn1 )
or
fW (wn , wn1 , . . . , w1 ) fW (wn , wn1 )
¼
fW (wn1 , . . . , w1 )
fW (wn1 )
1
1 (wk wk1 )2
ffi exp k¼1 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2 s2 (tk tk1 )
fW (wn , wn1 , . . . , w1 )
2p(tk tk1 )s2
¼
fW (wn1 , . . . , w1 )
Qn1
1
1 (wk wk1 )2
p
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
exp
k¼1
2 s2 (tk tk1 )
2p(tk tk1 )s2
1
1 (wn wn1 )2
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi exp 2 s2 (tn tn1 )
2p(tn tn1 )s2
Qn
1
1 (wn wn1 )2
exp
2p(tn tn1 )(tn1 tn2 )s4
2 s2 (tn tn1 )
1 (wn1 wn2 )2
exp fW (wn , wn1 )
2 s2 (tn1 tn2 )
¼
fW (wn1 )
1
1 (wn1 wn2 )2
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi exp 2 s2 (tn1 tn2 )
2p(tn1 tn2 )s2
1
1 (wn wn1 )2
¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi exp 2 s2 (tn tn1 )
2p(tn tn1 )s2
which shows that W(t) is a homogeneous Markov process with stationary independent
increments.
Example 20.8.3 (AR Process)
given by
In Example 20.6.1 an autoregressive(1) process was
Xn ¼ fXn1 þ Vn ,
n ¼ 1, . . . , X0 ¼ V0
where fVn, n ¼ 0, 1. . .g is a Gaussian process with zero mean and variance s2V. The conditional density fX (xn j Xn21 ¼ xn21) is also Gaussian-distributed with mean fxn21 and
variance s2V. This is a Markov process since the conditional density function depends
only on the Xn21 and not any of the previous values. Thus
fX (xn j Xn1 ¼ xn1 , . . . , X0 ¼ x0 ) ¼ fX (xn j Xn1 ¼ xn1 )
Example 20.8.4 (Polya Urn Model) Initially at n ¼ 0, an urn contains b blue balls and r
red balls. In each trial, a ball is drawn at random and its color recorded. It is then replaced
with c balls of the same color. Let Xn be the random variable representing the color of the
ball at the nth draw. We will also assume that Xk ¼ 1 if the ball selected at the kth trial is
20.8
Markov Process
535
blue and Xk ¼ 0 if it is red. The probability of X1 ¼ 1 is
P{X1 ¼ 1} ¼
b
bþr
The probability of X2 ¼ 1 conditioned on X1 ¼ 1 is
P{X2 ¼ 1 j X1 ¼ 1} ¼
bþc
bþcþr
Hence, the joint probability PfX2 ¼ 1, X1 ¼ 1g
P{X2 ¼ 1, X1 ¼ 1} ¼
b
bþc
bþr bþcþr
The probability of X3 ¼ 1 conditioned on X2 ¼ 1 and X1 ¼ 1 is
b þ 2c
b þ 2c þ r
P{X3 ¼ 1 j X2 ¼ 1, X1 ¼ 1} ¼
and the joint probability PfX3 ¼ 1, X2 ¼ 1, X1 ¼ 1g is
P{X3 ¼ 1, X2 ¼ 1, X1 ¼ 1} ¼
b þ 2c
bþc
b
b þ 2c þ r b þ c þ r b þ r
The realization fX6 ¼ 0, X5 ¼ 1, X4 ¼ 0, X3 ¼ 0, X2 ¼ 1, X1 ¼ 1g can be interpreted as the
event of drawing 3 blue balls in 6 trials, and the probability of this event is p(3, 6). Thus
P{X6 ¼ 0, X5 ¼ 1, X4 ¼ 0, X3 ¼ 0, X2 ¼ 1, X1 ¼ 1} ¼ p(3, 6)
¼
r þ 2c
b þ 2c
rþc
r
bþc
b
b þ 5c þ r b þ 4c þ r b þ 3c þ r b þ 2c þ r b þ c þ r b þ r
Extending these results, the probability of a sequence of k blue balls and (n 2 k) red balls
defined by p(k, n) can be determined as follows:
p(k, n) ¼ P{Xn ¼ xn , Xn1 ¼ xn1 , . . . , X1 ¼ x1 }
¼
b(b þ c)(b þ 2c) ½b þ (k 1)cr(r þ c)(r þ 2c) ½r þ (n k 1)c
(b þ r)(b þ c þ r)(b þ 2c þ r) ½b þ r þ (n 1)c
This random sequence is an example of a discrete-state discrete-time (DSDT) random
process. It is nonstationary because the probability is dependent on time, n.
From the discussion above, we have
P{X3 ¼ 1jX2 ¼ 1} ¼
bþc
bþcþr
but
P{X3 ¼ 1jX2 ¼ 1, X1 ¼ 1} ¼
b þ 2c
b þ 2c þ r
Since P{X3 ¼ 1jX2 ¼ 1, X1 ¼ 1} = P{X3 ¼ 1jX2 ¼ 1}, the sequence fXng is not a
Markov process.
536
Classification of Random Processes
B 20.9
MARKOV CHAINS
When the state of a Markov process is discrete, we call it a Markov chain (MC). If the time
is also discrete, it is called a discrete-time Markov chain. The binomial process (Example
20.4.2) and the random-walk process (Example 20.4.3) are illustrations of a discrete
Markov chain. If the time is continuous, it is called a continuous-time Markov chain.
The Poisson process discussed in detail in Section 20.2 is an example of a continuoustime Markov chain. These processes have been extensively studied, and many physical
phenomena are modeled by them. We will now discuss discrete Markov chains. For
continuous Markov chains, see Refs. 13, 30, and 54.
Discrete Markov Chain
A random process fXn, n ¼ 0, 1, . . .g with discrete state space S is a Markov chain if
the following probability relation holds for each f j, in, . . . , i0g [ S:
P{Xnþm ¼ j j Xn ¼ in , . . . , X0 ¼ i0 } ¼ P{Xnþm ¼ j j Xn ¼ in } ¼ p(m)
in j (n)
(20:9:1)
The quantity P{Xnþm ¼ j j Xn ¼ in } ¼ p(m)
in j (n) is the conditional probability of transition
from state in at time n to state j at time n þ m. It is referred to as the m-step transition
probability, which in general will be a function of time n. However, if the transition
from state i to state j is dependent only on the time difference m ¼ n þ m 2 n and not
on the current time n, then the transition probability is stationary and the Markov chain
is homogeneous, or
P{Xnþm ¼ j j Xn ¼ i} ¼ P{Xm ¼ j j X0 ¼ i} ¼ p(m)
ij
The one-step transition probability of a homogeneous Markov chain
defined as
pij ¼ P{Xnþ1 ¼ j j Xn ¼ i} ¼ P{X1 ¼ j j X0 ¼ i}
(20:9:2)
p(1)
ij
¼ pij can be
(20:9:3)
If the number of states N is finite, the Markov chain is finite; otherwise it is infinite. The
one-step probabilities fpijg can be expressed as either a transition matrix P or a state
transition flow diagram.
To make these ideas clear, we will consider the modified Bernoulli process given in
Eq. (20.4.11), where the random variable Xn at time n is in state 1 with probability p(n)
0 or in
state 21 with probability p(n)
1 as given below:
P(Xn ¼ 1) ¼ p(n)
0
(20:9:4)
P(Xn ¼ 1) ¼ p(n)
1
Further, if the particle is in state 1 at time n, then the probability of transition to 21 is a at
time (n þ 1), and if it is in state 21 at n, then the probability of transition to 1 is b at the
time (n þ 1). We are considering a two-state homogeneous Markov chain. The transition
matrix P is shown in Eq. (20.9.5):
1
P¼
1
1 1a
a
1
b
1b
The state transition flow diagram is shown in Fig. 20.9.1.
(20:9:5)
20.9
Markov Chains
537
FIGURE 20.9.1
If there are Nþ1 states and the transition probability from state i to state j is pij, then
the one-step transition matrix P is given by
2
0
p00
8
0
>
>
> 6
p10
>
>
16
>
6 .
>
>
6
< 6 ..
6
P¼
6p
>
i
>
6 i0
>
> 6 .
>
>
6 .
>
>
: 4 .
N
pN0
1
p01
j
p0j
p11
..
.
pi1
..
.
p1j
..
.
pij
..
.
pN1
pNj
N
3
p0N
p1N 7
7
.. 7
7
. 7
7
piN 7
7
.. 7
7
. 5
(20:9:6)
pNN
The matrix P is called a stochastic matrix and has the property that the row sum equals 1:
N
X
pij ¼ 1, i ¼ 0, 1, . . . , N
(20:9:7)
j¼0
The m-step transition probability for all i, j defined in Eq. (20.9.2) can also be given in
matrix form as follows:
P
(m)
0
8 2 (m)
0 p00
>
>
>
>
>
16
>
6 p(m)
>
10
>
> 6
>
6 .
>
< 6 ..
6
¼
6 (m)
>
6 pi0
>
>
6
i
>
>
6 .
>
>
6 .
>
>
4 .
>
>
:
N p(m)
N0
1
p(m)
01
p(m)
11
j
p(m)
0j
p(m)
1j
..
.
..
.
p(m)
i1
p(m)
ij
..
.
..
.
p(m)
N1
p(m)
Nj
N
p(m)
0N
p(m)
1N
3
7
7
7
.. 7
. 7
7
(m) 7
piN 7
7
.. 7
7
. 5
p(m)
NN
(20:9:8)
Let us consider the two-step transition from state i at time 0 through an intermediate
state k at time m 2 1 to state j at time m. From the property of conditional expectations
we can write
X
P{Xm ¼ j, X0 ¼ i} ¼
P{Xm ¼ j, Xm1 ¼ k, X0 ¼ i}
k
¼
X
P{Xm ¼ j j Xm1 ¼ k, X0 ¼ i}
k
P{Xm1 ¼ k j X0 ¼ i}Pf X0 ¼ i}
(20:9:9)
Dividing Eq. (20.9.9) throughout by PfX0 ¼ ig and using the Markov property,
P{Xm ¼ j j Xm1 ¼ k, X0 ¼ i} ¼ P{Xm ¼ j j Xm1 ¼ k}, we obtain the following equation
538
Classification of Random Processes
for all i, j:
P{Xm ¼ j j X0 ¼ i} ¼ p(m)
ij ¼
X
P{Xm ¼ j j Xm1 ¼ k}P{Xm1 ¼ k j X0 ¼ i}
k
¼
X
pkj p(m1)
ik
(20:9:10)
k
This equation is a restatement of the Chapman –Kolmogorov equation [Eq. (20.8.12)].
Similar equations can be written for p(m1)
in terms of pkj and p(m2)
. Thus, in matrix
ij
ik
form
P(m) ¼ PP(m1)
or P(m) ¼ Pm
Eq. (20.9.11) is an important result that shows the m-step transition matrix P
mth power of the one-step transition matrix P.
(20:9:11)
(m)
as the
State Occupancy Probabilities
(n)
(n)
Let the row vector p(n) ¼ ½p(n)
represent the state occupancy
0 , p0 , . . . , pN
(0)
(0)
(0)
probabilities at time n and the row vector p ¼ ½p(0)
0 , p1 , . . . , pN represent the initial
probability of the states at time 0. Since p is the state transition probability matrix, we
can obtain the state occupancy probabilities p (n) of any state at time n from a knowledge
of the probability of the state at time n21 from the following compact matrix relation:
p(n) ¼ p(n1) P
(20:9:12)
This equation can be solved recursively, and the probabilities of the states at time n can be
related to the initial probabilities of the states at time 0 as given below:
p(n) ¼ p(n1) P ¼ p(n2) P2 ¼ ¼ p(0) Pn
(20:9:13)
The next logical question that may arise is whether after a sufficiently long time a statistical equilibrium can be established in the sense that the occupancy probabilities of the
states are independent of the initial probabilities. To answer that question, we set
limn!0 p(n) ¼ p in Eq. (20.9.12) and obtain
lim p(n) ¼ lim p(n1) P
n!0
n!0
or
p ¼ pP
or p(I P) ¼ 0
(20:9:14)
The matrix equation p(I P) is a homogeneous system of linear equations and has a solution only if the determinant of (I 2 P) is equal to 0. This is very similar to the eigenvector
[Eq. (16.2.2)]. Solution of Eq. (20.9.14) will give the row vector p only P
upto a mulitiplicative constant, and it can be made unique by the normalization equation Ni¼0 pi ¼ 1. By
utilizing the techniques of Chapter 16, we can obtain the solution for a finite Markov chain
by diagonalizing the matrix P.
The eigenvalues of the matrix P are obtained by forming the characteristic equation
jlI 2 Pj ¼ lN þ a1lN21 þ . . . þ aN ¼ 0 and solving for the N eigenvalues. If we substitute l ¼ 1 in the eigenequation lI 2 P ¼ 0, we obtain I 2 P ¼ 0. Since the row sum of
the stochastic matrix is 1 and all the terms are positive, we conclude that l ¼ 1 is
always an eigenvalue of the stochastic matrix P and all the other eigenvalues will be
less than 1. We can diagonalize the matrix P assuming that the eigenvalues are distinct
and write
M1 PM ¼ L
or
P ¼ MLM1
(20:9:15)
20.9
Markov Chains
539
where M is the modal matrix of eigenvectors and L is a diagonal matrix of distinct eigenvalues of P. From Eq. (20.9.15) we can write the limn!1 Pn as follows:
P ¼ lim Pn ¼ lim ½MLM1 n ¼ M lim Ln M1
n!1
n!1
(20:9:16)
n!1
As n ! 1, since all except one of the eigenvalues in L are less than 1, they all tend to 0
as shown below:
3
2
3 2
1 0 0
1 0 0
6 1 l2 0 7 6 0 0 0 7
7
6
7 6
(20:9:17)
¼6. .
lim Ln ¼ lim 6 . .
.. 7
.. 7
n!1
n!14 .
.5
. 5 4 .. ..
. ..
0
0
0
lN
0
0
It can be shown that for a stochastic matrix with distinct
thepeigenvector
corpffiffiffiffi eigenvalues,
pffiffiffiffi
ffiffiffiffi
responding to the eigenvalue 1 is given by m ¼ ½1= N , 1= N , . . . , 1= N T .
Substituting Eq. (20.9.17) in Eq. (20.9.16) and expanding, we obtain
P ¼ lim Pn ¼ P1
n!1
2 1
pffiffiffiffi m01
6 N
6
6 1
6 pffiffiffiffi m11
6
¼6 N
6 .
..
6 .
.
6 .
4 1
pffiffiffiffi mN1
N
m0N
3
72
7 1 0
7
6
m1N 7
76 0 0
76
4
.. 7
7
. 7 0 0
5
mNN
0
32 m0
00
0
6 m0
7
0 76 10
76
.
56
4 ..
0
m0N0
m001
m011
..
.
m0N1
3
m00N
m01N 7
7
.. 7
7
. 5
m0NN
(20:9:18)
21
0
where the primes on m denote the elements of the inverse matrix M ¼ M of the modal
matrix M. Equation (20.9.18) can be simplified as
3
3
2
2 0
1
m00 m001
m00N
pffiffiffiffi
p
ffiffiffiffi
p
ffiffiffiffi
p
ffiffiffiffi
6 N7
6 N
N
N7
7
7
6
6
6 1 7
6 m0
0
0 7
m01
m0N 7
6 pffiffiffiffi 7
6 p00
6 N 7 0
6 ffiffiffiffi pffiffiffiffi pffiffiffiffi 7
0
0
7
6
N
N7
(20:9:19a)
P¼6
7
6 . 7 m00 m01 m0N 6 N
6 . 7
6 ..
.. 7
..
6 . 7
6 .
. 7
.
7
7
6
6 0
4 1 5
4 m00 m001
m00N 5
pffiffiffiffi pffiffiffiffi pffiffiffiffi
pffiffiffiffi
N
N
N
N
or
2
p0
6 p0
6
P¼6 .
4 ..
p0
p1
p1
..
.
p1
3
pN
pN 7
7
.. 7
. 5
pN
(20:9:19b)
is the steady-state probability transition matrix. Note that in the P matrix each column has
the same elements. Substitution of Eq. (20.9.19b) in Eq. (20.9.14) results in
lim p(n) ¼ p(0) P1 ¼ p(0) P ¼ ½ p0
n!1
p1
pN ¼ p
(20:9:20)
This equation shows that the limiting state occupancy probabilities are independent of the
initial state probability row vector p(0).
540
Classification of Random Processes
Example 20.9.1 A two-state Markov chain with probability transition matrix P as in
Eq. (20.9.5) is shown below along with the initial probability row vector p(0):
1a
a
P¼
; p(0) ¼ p(0)
¼ ½ p 1p
p(0)
0
1
b
1b
We want to find the equilibrium probability distributions of the occupancy of states 1, 21
(0)
and show that they are independent of the initial occupancy p(0)
0 and p1 .
The eigenvalues of P are obtained by solving the characteristic equation given by
l 1 þ a
a
¼ (l 1)(l 1 a þ b) ¼ 0
jlI Pj ¼ b
l 1 þ b
and the eigenvalues l1 ¼ 1 and l2 ¼ 1 2 a 2 b. The unnormalized eigenvector c1
corresponding to l1 ¼ 1 is obtained from the eigenequation
a a c21
c11
1
¼ 0 or
¼
b b
c22
c12
1
and the unnormalized eigenvector c2 corresponding to l2 ¼ 1 2 a 2 b is obtained from
b
b
a
a
c21
c22
¼0
or
c21
c22
"
¼
1
b
a
#
The matrices of unnormalized eigenvectors MU and its inverse M21
U are
2
3
b
a
"
#
6bþa bþa7
1 1
6
7
b
MU ¼
¼6
M1
7
U
1
4 a
a 5
a
bþa bþa
The diagonal eigenvalue matrix is L ¼
n
(n)
lim P(n)
0
0
1ab
1 0
0
0 .
, and since 1 2 a 2 b , 1, the limit as
¼
Hence, as n ! 1, P n is given by
2
3 2
3
b
a
b
a
"
#
6
7 6
7
1
1
1 0 6 bþa bþa 7 6 bþa bþa 7
b
¼
6
7¼6
7
0 0 4 a
1 a 5 4 a
a 5
a
bþa bþa
bþa bþa
n ! 1 of L is given by limn!1 L
n!1
1
and
2
(1)
p¼p
(0) 1
¼p P ¼½p
b
6b þ a
6
1 p 6
4 b
bþa
a 3
b þ a7 b
7
7¼
bþa
a 5
bþa
which is independent of the initial probability vector p (0) ¼ [ p
a
bþa
1 2 p].
Example 20.9.2 A book representative visits three universities a, b, and c each week.
However, there is a finite probability that she may stay over for another week in any of
the three places. The transition probability matrix P and the initial probability row
20.9
Markov Chains
541
FIGURE 20.9.2
vector p (0) of being in states a, b, and c at time 0 are shown below:
2 a
a 0:2
6
P ¼ b 4 0:1
b
0:15
0:4
c 3
0:65
7
0:5 5
c
0:25
0:05
0:7
2
p(0)
3T
3T
0:3
6
7
6
7
¼ 4 p(0)
b 5 ¼ 4 0:45 5
0:25
p(0)
c
p(0)
a
2
We have to find (1) the probabilities of the representative being in the universities a, b, and
c after 3 weeks and (2) the steady-state probability of being in any of the universities. The
state transition flow diagram for this three-state Markov chain is shown in Fig. 20.9.2.
1. We can apply Eq. (20.9.14) to obtain the probabilities that she will be in the
various universities after 3 weeks. Thus
2
p(3)
a
3T
2
0:3
3T 2
0:2
0:15
0:65
33
6
7
6
7 6
7
(0) 3
p(3) ¼ 4 p(3)
0:4
0:5 5
b 5 ¼ p P ¼ 4 0:45 5 4 0:1
0:25
0:7 0:25 0:05
p(3)
c
2
3T 2
3 2
3
0:3
0:293 0:237 0:47
0:347
6
7 6
7 6
7
¼ 4 0:45 5 4 0:315 0:254 0:431 5 ¼ 4 0:251 5
0:25
2. The eigenvalues of the matrix P are
The modal matrix M is:
2
0:57735
M ¼ 4 0:57735
0:57735
0:47
0:262
0:268
0:402
l1 ¼ 1, l2 ¼ 0.23197, l3 ¼ 20.58197.
0:37197
0:89737
0:22574
3
0:56601
0:32757 5
0:75652
and
P1 ¼ML1 M1
2
32
32
3
0:57735 0:37197 0:56601
100
0:63437 0:43479 0:66288
6
76
76
7
¼ 4 0:57735 0:89737 0:32757 54 0 0 0 54 0:65635 0:80072 0:14437 5
0:57735 0:22574 0:75652
000
0:67997 0:0929
0:77287
2
3
0:36626 0:25103 0:38272
6
7
¼ 4 0:36626 0:25103 0:38272 5
0:36626 0:25103 0:38272
542
Classification of Random Processes
FIGURE 20.9.3
and the steady-state probability p ¼ p(1) ¼ p(0)P ¼ [0.36626 0.25103 0.38272].
The evolution of state occupancy probabilities p(n) for all three states are shown in
Fig. 20.9.3 for n ¼ 1, 2, ... ,12. From the figure it is seen that steady-state equilibrium
has been obtained in about 9 weeks.
Example 20.9.3 A new automobile can be classified into five states: 1 ¼ excellent,
2 ¼ good, 3 ¼ fair, 4 ¼ poor, 5 ¼ scrap. The transition probabilities of going from one
state to another in each year is given by the transition probability matrix P. Clearly, the
initial probability row vector p will be given by p ¼ ½ 1 0 0 0 0 . At the end of
the first year the probabilities of the automobile being in excellent condition (state
1) ¼ 0.6, in good condition (state 2) ¼ 0.25, in fair condition (state 3) ¼ 0.1 and in poor
condition (state 4) ¼ 0.05. Similarly, if the automobile is in scrap condition (state 5), the
probabilities of it continuing to be in scrap condition ¼ 0.85, in poor condition ¼ 0.1,
and in fair condition ¼ 0.05. We will find the probabilities of the condition of the car in
(1) 5 years and (2) steady-state equilibrium for the state occupancy probabilities:
2 1
1 0:6
26
6 0:2
P ¼ 36
6 0:05
44 0
5
0
2
0:25
0:45
0:15
0:05
0
3
0:1
0:2
0:4
0:15
0:05
4
0:05
0:1
0:25
0:35
0:1
The state transition flow diagram is shown in Fig. 20.9.4.
FIGURE 20.9.4
5 3
0
0:05 7
7
0:15 7
7
0:45 5
0:85
20.9
Markov Chains
543
1. From Eq. (20.9.13) the state probabilities after n ¼ 5 years is given by
p(5) ¼ p(0) P5 , where p(0) is the row vector p(0) ¼ ½ 1 0 0 0 0 . Hence
p(5)
2 3T 2
0:2034
1
6 0 7 6 0:1506
6 7 6
6 7 6
7 6
¼ p(0) P5 ¼ 6
6 0 7 6 0:0863
6 7 6
4 0 5 4 0:0419
0:0204
0
2
3T
0:2034
7
6
6 0:1984 7
7
6
7
¼6
6 0:1736 7
7
6
4 0:1502 5
0:2744
0:1984
0:1618
0:1736 0:1502
0:1641 0:1571
0:1090
0:1447 0:1615
0:0656
0:0415
0:1215 0:1572
0:1061 0:1518
3
0:2744
0:3664 7
7
7
0:4985 7
7
7
0:6138 5
0:6802
Thus, after 5 years there is a probability of 0.2034 that the car is still in excellent condition
and a probability of 0.2744 that the car is in scrap condition. It is also interesting that with
the given transition probabilities, there is a better chance that the car is in excellent condition than in either good, fair, or poor condition. We will now check what happens to
these probabilities after a long period of time has elapsed.
2. We follow the procedure of finding the eigenvalues and eigenvectors of the transition matrix P. The eigenvalue matrix L, the corresponding modal matrix of P, and its
inverse are as follows:
2
1
60
6
6
L¼6
60
6
40
0
0
0:4529
0
0
0
0
0
0
0
0:2417
0
0
0
0
0
0
2
0:4472 0:7139
6
6 0:4472 0:1188
6
M¼6
6 0:4472 0:6081
6
4 0:4472 0:2899
0:4472
2
M1
0:1114
6
6 0:4614
6
¼6
6 0:3828
6
4 0:5697
0:0048
0:1496
0:1555
0:7999
0
0
0:1555
3
7
7
7
7
7
7
5
0:4491
0:8065
0:7891
0:5471
0:2789
0:2566
0:2238
0:0293
0:0651
0:1647
0:2695
0:1919 0:5895
0:7069 0:0846
0:3446
0:5495
0:7773
0:5102
0:2363
0:0127
0:1395
0:6005
0:9562
2
1
60
6
6
1
L ¼6
60
6
40
0
3
0:1265
7
0:4190 7
7
0:7403 7
7
7
0:5099 5
0:0201
1:3551
3
7
7
7
7
7
7
1:3288 5
0:4999
0:8695
0:5378
0 0
0 0
0
0
0 0
0
0 0
0 0
0
0
3
0
07
7
7
07
7
7
05
0
544
Classification of Random Processes
Hence from Eq. (20.9.18) we have
2
P ¼ P1 ¼ ML1 M1
0:0499
6 0:0499
6
¼6
6 0:0499
4 0:0499
0:0499
0:0695
0:0695
0:0695
0:0695
0:0695
0:1205
0:1205
0:1205
0:1205
0:1205
0:1541
0:1541
0:1541
0:1541
0:1541
3
0:6060
0:6060 7
7
0:6060 7
7
0:6060 5
0:6060
and the steady-state probabilities row vector p is
p ¼ p(0) P ¼ ½ 0:0499 0:0695
0:1205 0:1541
0:6060 These are more reasonable numbers since the long-term probabilities of being in various
states are excellent 4.99%, good 6.95%, fair 12.05%, poor 15.41%, and finally scrap
60.6%. The conclusion is that there is 60.6% chance that the automobile will be a scrap
in the long run with the given probability transition matrix.
First-Passage Probabilities and Classification of States
A Markov chain is called irreducible if every state can be reached from every other
state in a finite number of steps. In this case, the probability of going from every state i to
every other state j in the state space S of the chain will be given by the probability,
p(m)
ij . 0, where m is a finite integer. In this case, the states are called intercommunicating.
On the other hand, if E is a proper subset of S, E , S, and if i and j are states in E, then the
subset E is closed if
X
pij ¼ 1
for all
i[E
j[E
In other words, no transition is possible from a state i [ E to a state k [ E c. In this case the
Markov chain is called reducible. If there are no other proper subsets of E that are closed,
then it is called an irreducible sub-Markov chain. If i [ E is the only state, then it is called
an absorbing state, in which case pii ¼ 1.
Let us consider that the Markov chain is initially in state i at time n. We define the
probability of first return to i at time n þ m as fii(m), or
fii(m) ¼ P{Xnþ1 = i, . . . , Xnþm1 = i, Xnþm ¼ i j Xn ¼ i}
(20:9:21)
with one-step transition fii(1) ¼ P{Xnþ1 ¼ i j Xn ¼ i} ¼ p(1)
ii ¼ pii similar to Eq. (20.9.2).
For a homogeneous Markov chain, Eq. (29.9.21) can also be written as
fii(m) ¼ P{X1 = i, . . . , Xm1 = i, Xm ¼ i j X0 ¼ i}
(20:9:22)
with one-step transition fii(1) ¼ P{X1 ¼ i j X0 ¼ i} ¼ p(1)
ii ¼ pii .
In a similar manner we can also define the probability of first passage from state i at
time n to state j at time n þ m as fij(m) , or
fij(m) ¼ P{Xnþ1 = j, . . . , Xnþm1 = j, Xnþm ¼ j j Xn ¼ i}
(29:9:23)
20.9
Markov Chains
545
with one-step transition fij(1) ¼ P{Xnþ1 ¼ j j Xn ¼ i} ¼ p(1)
ij ¼ pij in conformity with
Eq. (20.9.2). For a homogeneous Markov chain, Eq. (20.9.23) is equivalent to
fij(m) ¼ P{X1 = j, . . . , Xm1 = j, Xm ¼ j j X0 ¼ i}
(20:9:24)
with the corresponding one-step probability fij(1) ¼ P{X1 ¼ j j X0 ¼ i} ¼ p(1)
ij ¼ pij .
The probabilities of eventual first return to state i or the first passage to state j starting
from state i are given by
fi ¼
1
X
fii(m)
fij ¼
m¼1
1
X
fij(m)
(20:9:25)
m¼1
It is now possible to have more classifications of the Markov chains depending on
the nature of fi. If fi ¼ 1, the state i is called recurrent; if fi , 1, the state i is called
transient. If the chain starts at time 0 in state i and returns to state i only at times a,
2a, . . . , with period a . 1, then the state i is called periodic with period a. If a ¼ 1,
then i is aperiodic.
If fi ¼ 1 and fij ¼ 1, we can define a mean recurrence time and a mean first-passage
time from state i to state j as follows:
mi ¼
1
X
mfi
mij ¼
m¼1
1
X
mfij
(20:9:26)
m¼1
With this knowledge we can classify the states even further. If the mean recurrence time mi
is finite, then the state is called positive recurrent; if it is infinite, it is called null recurrent.
Finally, a state i is ergodic if it is aperiodic and positive recurrent, or a ¼ 1, fi ¼ 1 and
mi , 1. A finite-state aperiodic irreducible Markov chain is always ergodic, and the occupancy probability of the states as n ! 1 will tend to an equilibrium distribution independent of the initial probabilities as in Examples 20.9.1 – 20.9.3. In transient states the
equilibrium distribution will be 0.
Example 20.9.4
The transition matrix P of a four-state Markov chain is given below:
1
8 2
1 1
>
>
>
<26 0
6
P¼
6
>
3
> 4 0
>
:
4 0:3
2
0
0:4
3
0
0:6
0:8
0
0:2
0:2
4
3
0
0 7
7
7
0 5
0:5
and the state transition flow diagram is shown in Fig. 20.9.5.
We can obtain the classification of the states by inspection of the transition flow
diagram of Fig. 20.9.5. From the flow diagram state 1 is absorbing because p11 ¼ 1,
and once the transition occurs to state 1, it never gets out of it. It is also a recurrent
state. States 2 and 3 are recurring states because the transition occurs only between
these two states. If the transition occurs out of state 4, it can never get back in. Hence,
state 4 is a transient state. We will determine the equilibrium probabilities P from
546
Classification of Random Processes
FIGURE 20.9.5
Eq. (20.9.18). The diagonal eigenvalue matrix L is given by
2
0:5
60
L¼6
40
0
0
1
0
0
0
0
0:4
0
3
0
07
7
05
1
2
0
60
n
lim L ¼ 6
40
n!1
0
0
1
0
0
3
0
07
7
05
1
0
0
0
0
The presence of two 1s for eigenvalues indicates the presence of sub-Markov chains.
The modal matrix and its inverse are
2
0
0
0
0:8575
3
7
7
7
5
60
6
M¼6
40
0:6804 0:5907
0:6804
0:7877
0
0
1
0:2722 0:1750
0:5145
2
0:6
6 0
6
M1 ¼ 6
4 0
1:1622
0:3556
0:0444
0:8398
0:6299
0:7255
0
0:7255
0
1
3
07
7
7
05
0
and the equilibrium probability P is given by
P ¼ M lim Ln M1
n!1
1
2
8 2
1 1
0
>
>
> 6
<
2 6 0 0:5714
¼
6
>
> 3 4 0 0:5714
>
:
4 0:6 0:2286
3
4
0
0
3
0:4286 0 7
7
7
0:4286 0 5
0:1714 0
State 4 is transient since p4j ¼ 0, j ¼ 1, . . . , 4. State 1 is absorbing since p11 ¼ 1. States 2
and 3 are recurrent with p23 ¼ 0.4286 and p32 ¼ 0.5714.
20.9
Markov Chains
547
Example 20.9.5 We will slightly change the transition probabilities by introducing a
path between states 2 and 1 by inserting p21 ¼ 0.02 and modifying p23 ¼ 0.58:
1
2
8 2
1
1
0
>
>
> 6
<
2 6 0:02 0:4
P¼
6
>
340
0:8
>
>
:
4 0:3
0
3
0
4
3
0
0:58 0 7
7
7
0:2
0 5
0:2
0:5
The state transition flow diagram is shown in Fig. 20.9.6.
This small change makes a big difference in the properties of the Markov chain. From
the flow diagram, the only state that is recurrent is 1. The other states (2 –4) are transient
because everything flows into state 1 directly or indirectly. This will be confirmed from the
equilibrium probabilities P of the states. The diagonal eigenvalue matrix L and the
limn!1 are given by
2
0:5
60
L¼6
40
0
0
0:9885
0
0
0
0
0:3885
0
2
3
0
07
7
05
1
0
60
n
lim L ¼ 6
40
n!1
0
0
0
0
0
0
0
0
0
3
0
07
7
05
1
Since there is only one eigenvalue of 1, there are no sub-Markov chains.
The modal matrix and its inverse are
2
0
60
M¼6
40
1
3
0
0
0:5
0:6739 0:5830 0:5 7
7
0:6837
0:7926 0:5 5
0:28
0:1784 0:5
2
3
0:5853 0:3687 0:0461 1
6 1:4748
0:8497
0:6251 0 7
7
M1 ¼ 6
4 0:0106 0:7330
0:7225 0 5
2
0
0
0
FIGURE 20.9.6
548
Classification of Random Processes
and the equilibrium probability P is given by
1
2
1 1
26
61
P ¼ M lim Ln M1 ¼ 6
n!1
341
2 3 4
3
0 0 0
0 0 07
7
7
0 0 05
4 1 0 0 0
The equilibrium probability matrix shows that states 2 –4 are transient because the probabilities are zero. State 1 is the absorbing recurrent state.
Example 20.9.6
A state transition matrix P for a seven-state Markov chain is given by
1
8 2
1
>
>
> 6 0:6
>
>
2 6 0:7
>
>
>
6
>
>
6
>
<360
6
P ¼ 460
>
6
>
>
6
>
>560
>
>
6
>
>
6 4 0:2
>
>
:
7 0
2
3
4
5
6
7
0:4
0
0
0
0
0
0:3
0
0
0:7
0
0
0:3 0
0
0
0
0
0
0
0
0
0:4
0:5
0:1
0:3
0:3
0:4
0:1
0:2
0:3
0:1
0:1
0:4
3
7
7
7
7
7
7
0
0 7
7
0
0 7
7
0:2 0:3 7
5
0:1
The corresponding state transition flow diagram is shown in Fig. 20.9.7.
FIGURE 20.9.7
20.9
Markov Chains
549
We will classify the states by inspection from Fig. 20.9.7. The flow diagram shows
that states 1 and 2 form a sub-Markov chain since there are no paths out of that pair.
Hence they form recurring states. States 3– 5 form another sub-Markov chain, and there
are no paths leading out of these three states. Hence they form another set of recurring
states. All paths lead out of states 6 and 7 and hence are transient states. We can
confirm these findings by computing the equilibrium probability P.
The diagonal eigenvalue matrix L and the limn!1 Ln are given by
2
0:2
60
6
6
60
6
6
L ¼ 60
6
60
6
6
40
0
0
0
0
0
0
0:1
0
0
1
0
0
0
0
0
0
0
0
0
0
0:2303
0
0
0:1303
0
0
0
0
0
0
1
0
0
0
2
0
0
0
60
6
6
60
6
6
n
lim L ¼ 6 0
n!1
6
60
6
6
40
0
3
7
7
7
7
7
7
0 7
7
0 7
7
7
0 5
0:1
3
0
07
7
7
07
7
7
07
7
07
7
7
05
0
0
0
0
0
0 0
0
0
0
1
0
0
0 0
0 0
0
0
0
0 0
0
0
0
0
0
0
0 0
0 1
0 0
0
0
0 0
0
Since there are two eigenvalues of 1, we conclude that there is a sub-Markov chain. The
modal matrix M and its inverse M 21 are given by
2
0
0
0
0
0
0:6963 0:4710
0
0
0
0
0:6963
60
6
6
60
6
6
M ¼ 60
6
60
6
6
41
0
0
0:4682 0:028
0:4682
0:0438
0:1665
0:4607
0
0
0
0:9487
0:4682
0:3511
0:0271
0:9946
0:4388
0:4821
0
0:1741
0
0:3162
0:4682
0:0861
0:5789
2
M
1
0:0833
6 0
6
6
6 0
6
6
¼6 0
6
6 0
6
6
4 0:9139
0:7720
0
18:25
2:1082
8:75
4:2164
13:25
1:0541
0
1:2521
0:6629
0:2210
0
0
16:6314
0:3084
5:0356
1:0184
11:5958
1:3268
0:3333
0
0:5222
0:7720
0
0
0
0
0
0
3
0:8243 7
7
7
7
0
7
7
0
7
7
7
0
7
7
0:3140 5
0
3
1
3
0 3:1623 7
7
7
7
0
0
7
7
0
0
7
7
7
0
0
7
7
0
0
5
0
0
550
Classification of Random Processes
The equilibrium probability P is given by
P ¼ M lim Ln M1
n!1
1
8 2
1 0:6364
>
>
>
>
>
26
>
6 0:6364
>
>
> 6
>
6
>
<360
6
¼ 460
>
6
>
>560
>
>
6
>
>
6
>
>
6
4 0:1591
>
>
:
7 0
2
3
4
5
6
7
0:3636
0
0
0
0
0
0:3636
0
0
0:5862
0
0:3103
0
0
0:1035 0
0
0:5862
0:3103
0:1035 0
0
0:0909
0:5862
0:4396
0:3103
0:2328
0:1035 0
0:0776 0
0
0:5862
0:3103
0:1035 0
3
07
7
7
07
7
7
07
7
07
7
7
05
0
The matrix P shows that there are two closed sets of recurring states f1,2g and f3,4,5g with
equilibrium probabilities. States 6 and 7 have zero probabilities, thus confirming that they
are transient states. An interesting point to note is that there is no direct transition from
state 6 to state 2 in the state transition matrix but there is a direct transition from state 6
to state 2 in the equilibrium probability distribution.
Properties of Markov Chains
1. Let X ¼ fXn: n ¼ 0, 1, . . . , Ng be a Markov chain with state space S ¼ fi, j, k, . . .g.
The m-step transition probability p(m)
ij (n) is the probability of Xnþ1 ¼ j given that
Xn¼i, or
P{Xnþm ¼ jjXn ¼ i} ¼ P{Xnþm ¼ jjXn ¼ i} ¼ p(m)
ij (n)
The Markov chain is homogeneous when
P{Xnþm ¼ jjXn ¼ i} ¼ P{Xm ¼ jjX0 ¼ i} ¼ p(m)
ij
The transition probabilities are independent of time n.
2. If P is the matrix of one-step transition probabilities, then pij is the (i, j)th element
in P as shown below:
0
1
...
8 2
p00 p01 >
>
>06
>
>
p
p11 10
>
>
16
>
6 .
>
..
>
< 6
6 ..
.
6
P¼
6
>
> i 6 pi0 pi1 >
>
6
>
..
> 6 ..
>
>
4 .
.
>
>
:N
pN0 pN1 j
... N
p0j
p1j
..
.
pij
..
.
pNj
p0N
p1N
..
.
3
7
7
7
7
7
7
piN 7
7
.. 7
7
. 5
pNN
P is called the stochastic matrix or Markov matrix with row sum equal to 1, or
N
X
pij ¼ 1, i ¼ 0, 1, . . . , N
j¼0
3. The Markov chain satisfies the Chapman –Kolmogorov equation
X
X
P{Xm ¼ jjXm1 ¼ k}P{Xm1 ¼ kjX0 ¼ i} ¼
pkj p(m1)
p(m)
ij ¼
ik
k
k
20.10 Martingale Process
551
4. Let the row
of the initial
probability of the states of a Markov chain at time 0
vector
(0)
(0) be p(0) ¼ p(0)
,
p
,
.
.
.
,
p
.
Then
the row vector of state occupancy probabilities
N
0
1
(n)
(n) ,
p
,
.
.
.
,
p
at time n, p(n) ¼ p(n)
N , is given by
0
1
p(n) ¼ p(n1) P ¼ p(0) Pn
5. A Markov chain with Markov matrix P is called irreducible if every state can be
reached from every other state in a finite number of steps. The long-term statistical
equilibrium occupancy probabilities p of the states can be obtained from solving
p ¼ pP
or p(I P) ¼ 0
6. Only one eigenvalue of an irreducible Markov chain is 1, and the absolute values
of all other eigenvalues are less than 1.
7. Starting from a state i at time n, the probability of first return to i at time n þ m as
fii(m) is defined by
fii(m) ¼ P{Xnþ1 = i, . . . , Xnþm1 = i, Xnþm ¼ ijXn ¼ i}
and starting from a state i at time n the probability of first passage to j at time n þ m
as fij(m) is defined by
fij(m) ¼ P{X1 = j, . . . , Xm1 = j, Xm ¼ jjX0 ¼ i}
The probabilities of eventual first return to state i or first passage to state j are
fi ¼
1
X
fii(m)
m¼1
fij ¼
1
X
fij(m)
m¼1
If fi or fij is equal to 1, then the state i or j is certain, and both are called recurrent.
If they are less than 1, then the states are called transient.
8. The mean recurrence time and the mean first passage time can be defined as
mi ¼
1
X
mfi
m¼1
mij ¼
1
X
mfij
m¼1
If mi , 1, then the state i is positive recurrent, and if mi ¼ 1, then i is null recurrent and similar definitions apply to state j if mij , 1, or mij ¼ 1
9. In a Markov chain with transitionP
matrix P, if E is a proper subset of the state space
S, or if E , S, then E is closed if j[E pij ¼ 1 for all i [ E. The states in E form a
sub-Markov chain in which case there will be two eigenvalues of 1.
10. A closed set of states E , S containing no proper subsets that are closed is called
irreducible. If the number of states within an irreducible set is finite, then each
state in E is recurrent. For any irreducible set, every state communicates with
every other state.
11. If all the states of a Markov chain are irreducible, then a steady-state row vector
defined by p can be determined from p ¼ pP, or p(I P) ¼ 0.
B 20.10
MARTINGALE PROCESS
The meaning of a martingale in both British and American dictionaries refers to a strap
preventing a horse from rearing. This has no relevance in the mathematical sense,
552
Classification of Random Processes
where it conveys the sense of a fair game. It is a stochastic process where the best estimate
of the future value conditioned on the past, including the present, is the present value. It is
a process without trend and hence unpredictable. Martingale concepts are widely used in
engineering and stochastic finance. Many filtering problems in engineering such as the
Kalman filter [27] can be cast in a martingale framework. The option pricing model developed by Black and Scholes [7] in 1973 for pricing stock options can be formulated within
the framework of martingale theory. Just like a Markov process, conditional expectations
play a key role in martingale processes. For detailed study of martingales, see Refs. 16, 26,
35, 37, and 40.
A random sequence fYn, n 0g is a discrete martingale with respect to the sequence
fXm, m ng if the following conditions are satisfied:
(1) EjYn j , 1
(2) E ½Yn jXm , Xm1 , . . . , X0 ¼ Ym ; m n
(20:10:1)
If the second condition in Eq. (20.10.1) is modified as
(2a) E ½Yn jXm , Xm1 , . . . , X0 Ym , m n
(20:10:2)
then the random sequence fYn, n 0g is a submartingale with respect to the sequence
fXm, m 0g and if modified as
(2b) E ½Yn jXm , Xm1 , . . . , X0 Ym , m n
(20:10:3)
then the random sequence fYn, n 0g is a supermartingale with respect to the sequence
fXm, m 0g.
A random process fY(t), t 0g is a continuous martingale with respect to the process
fX(s), 0 s , tg if the following conditions are satisfied:
(3) EjY(t)j , 1
(4) E½Y(t)jX(s), 0 s , t ¼ Y(s)
(20:10:4)
If the second condition in Eq. (20.10.1) is modified as
(4a) E ½Y(t)jX(s), 0 s , t Y(s)
(20:10:5)
then the random process fY(t), t 0g is a continuous submartingale with respect to the
process fX(s), 0 s , tg and if modified as
(4b) E ½Y(t)jX(s), 0 s , t Y(s)
(20:10:6)
then the random process fY(t), t 0g is a continuous supermartingale with respect to the
underlying process fX(s), 0 s , tg.
We can explain the concept of a martingale by a simple example. A player tosses a fair
coin p ¼ q ¼ 12 and wins a dollar if heads come up and loses a dollar if tails come up. This
is the modified Bernoulli process described by Eq. (20.4.11). Let Wn ¼ Z1 þ Z2 þ . . . þ Zn
be his fortune at the end of n tosses. The average fortune of the player at the (n þ 1)st toss
is the same as his current fortune Wn, and this is not affected by how he arrived at his
current fortune, that is, E ½wnþ1 jZn , Zn1 , . . . , Z1 ¼ Wn , thus capturing the notion that
the game is fair. In the same spirit we can think of the game being biased in favor of
the player (submartingale) or biased against the player (supermartingale).
Unless otherwise stated, we will assume that the condition EjYn j , 1 or EjY(t)j , 1
is always satisfied in the following examples.
20.10 Martingale Process
553
Example 20.10.1 Random-Walk Process Let fZig be the modified Bernoulli process
described by Eq. (20.4.11) with mZ ¼ p 2 q and s2Z ¼ 4pq. The corresponding random-walk process is a modified form of Eq. (20.4.15):
Wn ¼
n
X
Zi ¼
n1
X
i¼1
Zi þ Zn ¼ Wn1 þ Zn
(20:10:7)
i¼1
The mean value of Wn from Eq. (20.4.16) is mW ¼ n( p 2 q), and the variance of Wn from
Eq. (20.4.17) is s2W ¼ 4npq. The conditional expectation E ½Wnþ1 jZn , Zn1 , . . . , Z1 is
given by
E ½Wnþ1 jZn , Zn1 , . . . , Z1 ¼ E ½Znþ1 þ Wn jZn , Zn1 , . . . , Z ¼ ( p q) þ Wn
(20:10:8)
If we now define Yn ¼ Wn 2 n( p 2 q) in Eq. (20.10.5), we can write
E ½Ynþ1 jZn , Zn1 , . . . , Z1 ¼ E ½Znþ1 ( p q) þ Yn jZn , Zn1 , . . . , Z ¼ Yn
(20:10:9)
Hence Yn is a discrete martingale with respect to the sequence fZk, 1 k ng.
Example 20.10.2 The characteristic function of Wn in Eq. (20.10.4) is FWn (v) ¼
E ½ejvWn , and we define
Yn ¼
e jvWn
FWn (v)
and
E ½Yn ¼ 1
For m n, we can write for the following conditional expectation:
jv(Wn Wm ) jvWm e
e
Wm , . . . ,W1
E ½Yn jWm , . . . ,W1 ¼ E
FWn (v)
Because of the independent increment property, (Wn 2 Wm) is independent of Wm and
hence E ½e jv (Wn Wm ) jWm , . . . , W1 ¼ E ½e jv (Wn Wm ) and the preceding equation can be
written as
E ½Yn jWm , . . . ,W1 ¼
E ½e jv(Wn Wm ) e jvWm
e jvWm
¼
¼ Ym ,
FWnm (v) FWm (v) FWm (v)
mn
Hence Yn is a martingale with respect to the sequence fWm, . . . , W1g.
Example 20.10.3 The sequence fZig of Example 20.10.2 is now a set of i.i.d. random
variables with p ¼ q ¼ 12 with variance s2Z. Here the mean value mZ ¼ 0. We form the
sequence
!2
n
X
Yn ¼
Zi ns2Z
i¼1
Since E(
Pn
i¼1
2
ns2Z ,
we have E[Yn] ¼ 0. For m n, we can write
82
3
!2
nm
< X
Z
s2Z (n m)5
E ½Yn jZm , . . . , Z1 ¼ E 4
: i¼m i
Zi ) ¼
2
þ4
m
X
i¼1
!2
Zi
9
3
=
ms2Z 5Zm , . . . , Z1
;
554
Classification of Random Processes
Pnm
2
s2Z (n m) is independent of the sequence fZm, . . . , Z1g we have
2
3
2
3
!2
!2
nm
nm
X
X
E4
Zi s2Z (n m)Zm , . . . , Z1 5 ¼ E4
Zi s2Z (n m)5 ¼ 0
i¼m
i¼m
Since E
i¼m
Zi
and hence
2
E ½Yn jZm , . . . , Z1 ¼ E4
m
X
i¼1
!2
Zi
3
!2
m
X
2
msZ Zm , . . . , Z1 5 ¼
Zi ms2Z ¼ Ym
i¼1
and Yn is a discrete martingale with respect to the sequence fZk, 1 k ng.
Example 20.10.4 (Poisson Martingale) The Poisson process fN(t), t 0g with mean
mN(t) ¼ lt is given by the probability function
P{N(t) ¼ k} ¼ elt
(lt)k
k!
Let Y(t) ¼ N(t) 2 lt. We will show that Y(t) is a martingale. For s t we can write
E ½Y(t) j Y(s), s t ¼ E ½Y(t) Y(s) þ yðsÞ j Y(s), s t
Since Y(t) is an independent increment process, Y(t) 2 Y(s) is independent of Y(s). Hence
E ½Y(t) Y(s) j Y(s), s t ¼ E ½Y(t) Y(s), s t ¼ 0
Hence we have
E ½Y(t)jY(s) ¼ Y(s),
st
showing that Y(t) is a martingale with respect to the process fY(s), s tg.
Example 20.10.5 (Wiener Martingale) The Wiener process fW(t), t 0g has been
defined in Section 20.7 as a zero mean independent increment process with density
function given by Eq. (20.7.1):
1
2
2
fW (w, t) ¼ pffiffiffiffiffiffiffiffiffiffi eð1=2Þ(w =s t)
2pst
Using the same procedure as in previous examples, we can write
E ½W(t)jW(s), s t ¼ E ½W(t) W(s) þ W(s)jW(s), s t
Since W(t) is an independent increment process, W(t) 2 W(s) is independent of W(s).
Hence
E ½W(t) W(s)jW(s), s t ¼ E ½W(t) W(s), s t ¼ 0
Hence we have
E ½W(t)jW(s) ¼ W(s),
st
showing that W(t) is a martingale with respect to the process fW(s), s tg.
In all the five examples given above the underlying random process was an independent increment process. The question arises as to whether all independent increment
processes with zero mean are martingales. We will show that this is indeed the case. If
20.10 Martingale Process
555
X(t) is an independent increment process, then from Eq. (20.4.1), we have
E{½X(ti ) X(ti1 )½X(tiþ1 ) X(ti )}
¼ E ½X(ti ) X(ti1 )E ½X(tiþ1 Þ X(ti )
(20:4:1)
If the mean of X(t) is mX(t), then we can define a zero mean independent increment process
Y(t) ¼ X(t) 2 mX(t) and write the conditional expectation E ½Y(t)jY(s) for 0 s t as
follows:
E ½Y(t) j Y(s) ¼ E ½Y(t) Y(s) j Y(s) þ E ½Y(s) j Y(s),
0st
(20:10:10)
Using the independent increment property, we obtain
E ½Y(t) Y(s) j Y(s) ¼ E ½Y(t) Y(s) ¼ 0,
0st
(20:10:11)
0st
(20:10:12)
Substituting Eq. (20.10.11) in Eq. (20.10.10), we obtain
E ½Y(t) Y(s) j Y(s) ¼ E ½Y(t) Y(s) ¼ 0,
thus showing that a zero mean independent increment process is a martingale with respect
to the process fY(s), 0 s tg. However, not all martingales are independent increment
processes, as shown in the following example.
Example 20.10.6 (Likelihood Ratio Martingale) The likelihood ratio under the two
hypotheses H0 and H1 in Eq. (18.6.17) is modified as
LX (n) ¼
fX (Xn , . . . , X1 j H1 )
fX (Xn , . . . , X1 j H0 )
We will show that under the hypothesis H0, the likelihood ratio LX(n) is a martingale with
respect to the sequence fXn, . . . , X1g. The conditional expectation E[LX(n þ 1)jXn, . . . ,
X1jH0] can be written as
E ½LX (n þ 1) j Xn , . . . , X1 j H0 ð1
fX (xnþ1 , xn , . . . , x1 j H1 )
fX (xnþ1 , xn , . . . , x1 j H0 )dxnþ1
¼
f
1 X (xnþ1 , xn , . . . , x1 j H0 )
From the definition of conditional probabilities, we can substitute the following in the
equation above
fX (xnþ1 , xn , . . . , x1 j H0 ) ¼ fX (xnþ1 j xn , . . . , x1 j H0 ) fX (xn , . . . , x1 j H0 )
and write
E ½LX (n þ 1) j Xn , . . . , X1 j H0 ð1
fX (xnþ1 , xn , . . . , x1 j H1 ) fX (xnþ1 j xn , . . . , x1 j H0 )
dxnþ1
¼
fX (xn , . . . , x1 j H0 ) fX (xnþ1 j xn , . . . , x1 j H0 )
1
ð1
1
¼
fX (xnþ1 , xn , . . . , x1 j H1 )dxnþ1 ¼ LX (n)
fX (xn , . . . , x1 j H0 ) 1
556
Classification of Random Processes
thus showing that LX (n) is a martingale with respect to the sequence fXn, . . . , X1g under the
hypothesis H0.
Example 20.10.7 The Wiener process fW(t), t 0g has mean 0 and variance s2t. We
form the function Y(t) with W(t) as
Y(t) ¼ ekW(t)k
2
(s2 t=2)
t.0
,
where k is an arbitrary constant. We will show that Y(t) is a martingale with respect to the
Wiener process fW(t), t 0g. We form the conditional expectation with s t:
E ½Y(t) j W(s), 0 s , t ¼ ek
2
(s2 t=2)
E½ekW(t) j W(s)
¼ ek
2
(s2 t=2)
E ½ek½W(t)W(s) eW(s) j W(s)
Using the independent increment property E ½ek½W(t)W(s) jW(s) ¼ E ½ek½W(t)W(s) in the
preceding equation, we find that
E ½Y(t) j W(s), 0 s , t ¼ ekW(s) ek
2
(s2 t=2)
E½ek½W(t)W(s) But E ½ek½W(t)W(s) is the moment generating function (MGF) of the process W(t) 2 W(s),
which has zero mean and variance s2(t 2 s). From Example 11.5.2 the MGF is given by
2 2
E ½ek½W(t)W(s) ¼ ek ½s (ts)=2 , and substituting this equation, we obtain
E ½Y(t) j W(s), 0 s , t ¼ ekW(s) ek
2
ðs2 t=2) k2 ½s2 ðtsÞ=2
e
¼ e kWðsÞ ek
2
ðs2 s=2Þ
¼ Y(s)
thus showing the required result.
Example 20.10.8 The random process is given by Y(t) ¼ W 2(t), where W(t) is a Wiener
process with variance s2t. We will show that Y(t) is a submartingale.
We form the conditional expectation with s , t:
E ½Y(t) j W(s) ¼ E ½W 2 (t) W 2 (s) W 2 (s) j W(s),
0s,t
With 0 s , t and, using the independent increment property, this equation can be
written as
E ½Y(t) j W(s) ¼ E ½W 2 (t) W 2 (s) W 2 (s) ¼ s2 (t s) W 2 (s) . Y(s)
showing that Y(t) is indeed a submartingale. However, we can show that V(t) ¼
W 2(t) 2 s2t is a martingale as shown below:
E ½V(t) j W(s) ¼ E ½W 2 (t) s2 t W 2 (s) s2 s þ V(s) ¼ V(s),
0 s , t
Thus, V(t) is a martingale with respect to the Wiener process fW(t), t 0g.
Example 20.10.9 If fYn, n 0g is a martingale with respect to the sequence fXm, m 0g
and f (.) is some convex function, then f(Yn) is submartingale with respect to fXm, m 0g
as shown below:
From Jensen’s inequality [Eq. (14.6.3), E ½h(X) h(E ½X), we can write
E ½f(Yn )jXm f{E ½Yn jXm } ¼ f(Ym )
Showing that f(Yn) is a submartingale.
20.11 Periodic Random Process
557
As particular cases the following convex functions of a martingale Yn are all submartingales: (1) jYnj, (2) Y2n, (3) jYnja. Similarly, convex functions of a continuous martingale
Y(t) are submartingales as shown below:
1. Poisson martingale N(t) 2 lt. [N(t) 2 lt]2 is a submartingale.
2. Wiener martingale W(t). We have already seen in Example 20.10.8 that W 2(t) is a
submartingale.
3. W 2(t) 2 s2t is a martingale. [W 2(t) 2 s2t]2 is a submartingale.
A martingale process is based on expected values and a Markov process on conditional
probabilities, or
E ½X(t) j X(k), 0 k s ¼ X(s)
P ½X(t) j X(k), 0 k s ¼ P ½X(t) j X(s)
martingale
Markov
Hence a martingale process need not be a Markov process or vice versa. However, all independent increment processes are both martingales and Markov processes.
Decomposition of Submartingales
Doob [16] and Meyer [41] have shown that under certain regularity conditions any submartingale can be uniquely decomposed into a martingale and an increasing predictable
process. Heuristically, a process A(t) or An is predictable if it can be completely determined
from a knowledge of either fX(s), 0 s , t)g or fXk, 0 , k ng. All deterministic functions
are predictable. The Doob–Meyer decomposition states that if Yt is a submartingale with
respect to an underlying random process fXt, t 0g, then the unique decomposition is
given by
Yt ¼ Mt þ At
(20:10:13)
where Mt is a martingale with respect to the same underlying random process fXt, t 0g and
At is an increasing predictable process. It is the decomposition aspect of the submartingales
that gave rise to an elegant proof [19] for the nonlinear filtering problem of which the Kalman
linear filtering problem is a special case. Complete details can be found in Refs. 26, 35, and 37.
B 20.11
PERIODIC RANDOM PROCESS
The continuity of the sample paths of a periodic random processes is of concern to us for
expanding them in Fourier series. Among the different types of continuity the mean-square
continuity is of importance. A random process X(t) is continuous in the mean-square sense
if the following condition is satisfied:
lim EjX(t) X(s)j2 ! 0
s!t
(20:11:1)
The mean-square continuity of the random process is tied to the autocorrelation RX(t,s) of
the process. We will show that a random process X(t) is mean-square continuous if and
only if RX(t,s) is continuous at the diagonal point s ¼ t ¼ t:
EjX(t) X(s)j2 ¼ RX (t, t) 2RX (t, s) þ RX (s, s)
(20:11:2)
558
Classification of Random Processes
Since RX(t,s) is continuous at s ¼ t, by assumption the righthand side is zero at s ¼ t, thus
showing the necessity. Conversely
RX (t, t) RX (s, s) ¼ E ½X(t)X(t) E ½Xs2 ¼ E{½X(t) X(s)X(t)} þ E{½X(t) X(s)X(s)}
From the preceding equation, using Schwarz’s inequality of Eq. (14.5.13), we can form
RX (t, t) RX (s,s)
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
E ½jX(t) X(s)j2 E ½X 2 (t) þ E ½jX(t) X(s)j2 E ½X 2 (s)
(20:11:3)
From the assumption of mean-square continuity of X(t) the righthand side goes to zero as
t, s ! t and hence RX(t,s) is continuous at s ¼ t.
A random process X(t) is periodic in the wide sense with period Tc if
mX (t þ kTc ) ¼ mX (t) for all t and integer k
RX (t, s) ¼ RX (t þ kTc , s) ¼ RX (t, s þ kTc ) for all t, s, and integer k
(20:11:4)
The autocorrelation function RX(t,s) is periodic in both arguments t and s and is shown in
Fig. 20.11.1 in terms of constant-value contours.
If X(t) is a stationary process, then with t ¼ t 2 s, Eq. (20.11.4) simplifies to
RX (t) ¼ RX (t þ kTc )
for all t and integer k
(20:11:5)
and the mean-square continuity of X(t) guarantees that RX(t) is uniformly continuous in t.
Such wide-sense periodic stationary processes are shown in Examples 19.2.1 –19.2.3.
As a consequence of the continuity, the Fourier series of RX(t) given by
RX (t) ¼
1
X
Rk e jkvc (st) ,
t¼st
(20:11:6)
n¼1
converges uniformly to RX(t).
As a consequence of the uniform continuity of RX(t), the zero mean stationary periodic random process X(t) with fundamental frequency vc ¼ 2p/Tc can be represented in
FIGURE 20.11.1
20.11 Periodic Random Process
559
the mean-square sense by a Fourier series
1
X
X(t) ¼
Xn e jnvc t , X0 ¼ 0 and
Xn ¼
n¼1
1
Tc
ð Tc
X(t)ejnvc t dt
(20:11:7)
0
The Fourier coefficients fXng are in general complex random variables, and we will
investigate their properties. Since X(t) is a zero mean process, E[Xn] ¼ 0. The crosscorrelation between two of the coefficients can be calculated using Eq. (20.11.7):
E ½Xm Xn 1
¼ 2
Tc
1
¼ 2
Tc
ð Tc ð Tc
0
E ½X(s)X(t)ejvc (msnt) ds dt
0
ð Tc ð Tc
0
RX (s t)ejvc (msmt) dt ds
(20:11:8)
0
Substituting for RX(s 2 t) from Eq. (20.11.6) in Eq. (20.11.8), we obtain
E ½Xm Xn 1
¼ 2
Tc
¼
ð Tc ð Tc X
1
0
Rk e jkvc (st) ejvc (msnt) dt ds
0 n¼1
ð Tc
ð Tc
1
1 X
jvs(km)
R
e
ds
e jvt(nk) dt
k
T 2 n¼1
0
0
(20:11:9)
The complex exponentials {ejnvc s } and {ejnvc t } are orthogonal sets and hence
1
Tc
ð Tc
e
jvs(km)
ds ¼ dkm ¼
0
1,
0,
¼ dnk ¼
k¼m
1
:
T
k=m
c
1, n ¼ k
0,
ð Tc
e jvt(nk) ds
0
(20:11:10)
n=k
Substituting Eq. (20.11.10) into Eq. (20.11.9), we obtain the result
E ½Xm Xn 1
1 X
Rm ,
¼ 2
Rk dkm dnk ¼ Rm dmn ¼
0,
Tc n¼1
n¼m
n=k
or
E ½Xm2 ¼ Rm
(20:11:11)
Thus, the Fourier coefficients of a zero mean periodic stationary random process X(t) are
orthogonal and uncorrelated. If it is a Gaussian process, then these coefficients are also
independent.
Example 20.11.1
given by
The autocorrelation of a zero mean white Gaussian noise W(t) is
E ½W(s)W(t) ¼ RW (t s) ¼ s2 d(t s)
560
Classification of Random Processes
where s2 is the variance parameter. The process may be considered periodic with period
Tc ! 1. Substituting for RW(t 2 s) in Eq. (20.11.8), we obtain
s2
Tc !1 Tc2
E ½Wm Wn ¼ lim
ð Tc ð Tc
0
s2
dmn
Tc !1 Tc2
d(s t)ejvc (msnt) dt ds ¼ lim
0
or
s2
¼ s2W
Tc !1 Tc2
E ½Wm2 ¼ lim
and the Fourier coefficients Wm and Wn are independent for any m and n provided
m = n.
Cyclostationary Process
A random process X(t) is called strictly cyclostationary [21] if the joint distribution
function for any set of samples at times t1,t2, . . . ,tn is the same for any integer multiples
of some period T, that is, if the cdf and pdf satisfy the equations
FX (x1 , . . . , xn ; t1 , . . . , tn ) ¼ FX (x1 , . . . , xn ; t1 þ kT, . . . , tn þ kT)
fX (x1 , . . . , xn ; t1 , . . . , tn ) ¼ fX (x1 , . . . , xn ; t1 þ kT, . . . , tn þ kT)
(20:11:12)
for any n and k. This is somewhat restrictive, and hence we define a wide-sense cyclostationary process if the mean mX(t) and autocorrelation RX(t1,t2) are periodic:
mX (t þ kT) ¼ mX (t)
RX (t1 þ kT, t2 þ kT) ¼ RX (t1 , t2 )
(20:11:13)
The autocorrelation function of a wide-sense cyclostationary process is shown in
Fig. 10.11.2. The difference between Figs. 20.11.2 and 20.11.1 should be noted.
Example 20.11.2 A random process X(t) is defined by X(t) ¼ A cos(vc t) þ B sin(vc t),
where A and B are two zero mean independent Gaussian random variables with variances
FIGURE 20.11.2
20.11 Periodic Random Process
561
s2A and s2B respectively. The mean value of this process is
E ½X(t) ¼ mX (t) ¼ E ½A cos(vc t) þ E ½B sin(vc t) ¼ 0,
E ½A ¼ E ½B ¼ 0
The autocorrelation function RX(t,s) is given by
RX (t, s) ¼ E ½X(t)X(s) ¼ E{½A cos(vc t) þ B sin(vc t)½A cos(vc s) þ B sin(vc s)}
¼ E ½A2 cos(vc t) cos(vc s) þ E ½B2 sin(vc t) sin(vc s)
þ E ½AB½cos(vc t) sin(vc s) þ sin(vc tÞ cos(vc s)
Since A and B are zero mean and independent, E ½A2 ¼ s2A , E ½B2 ¼ s2B , and E ½AB ¼
E ½AE ½B ¼ 0 and hence
1
(s t, s þ t) ¼ s2A ½cos (vc (s t)) þ cos(vc (s þ t))
2
1
þ s2B ½cos(vc (s t)) cos(vc (s þ t))
2
¼
s2A þ s2B
s2 s2B
cos(vc (s t)) þ A
cos(vc (s þ t))
2
2
Since the autocorrelation function RX(s2t, s þ t) is dependent on both t and s, the process
is not stationary. However, with mX(t) ¼ 0 and Tc ¼ 2p/vc , we have RX (s t, s þ t) ¼
RX (s t þ kTc , s þ t þ kTc ) in the preceding equation. Hence the process is wide-sense
cyclostationary.
Example 20.11.3
A pulse-amplitude-modulated random process is of the form
1
X
X(t) ¼
Ak p(t kTc )
k¼1
where fAkg is a stationary digital sequence of random variables with mean mA and autocorrelation function RA (h) ¼ E ½Ak Akþh . p(t) is a deterministic pulse of duration Tc. We want
to find the mean and the autocorrelation function of X(t):
Mean:
mX (t):
1
X
E ½X(t) ¼ mX (t) ¼
1
X
E ½Ak p(t kTc ) ¼ mA
k¼1
p(t kTc )
k¼1
and the mean is a periodic function with period Tc.
Autocorrelation:
E ½X(t)X(s) ¼ RX (t, s) ¼
1
1
X
X
E ½Ak Aj p(t kTc )p(s jTc )
k¼1 j¼1
¼
1
1
X
X
RA (k, j)p(t kTc )p(s jTc )
k¼1 j¼1
¼ RX (t kTc , s kTc ), k ¼ +1, +2, . . .
562
Classification of Random Processes
FIGURE 20.11.3
Since both conditions of Eq. (20.11.13) are satisfied, we conclude that X(t) is a
cyclostationary process.
We will make the result more concrete by assuming that p(t) is a rectangular pulse of
unit height and width Tc. The random variable sequence fAkg is an independent set of equiprobable þ1 or 21 representing the symbols 1 or 0 respectively. Since Ak is equiprobable,
mA ¼ 0, and we obtain mX(t) ¼ 0. A sample waveform of X(t) corresponding to sending
symbols f0100110010100g is shown in Fig. 20.11.3.
Since the Ak values are independent with zero mean, we have
E ½X(t)X(s) ¼ Rx (t, s) ¼
1
X
E ½A2k p(t kTc )p(s kTc )
k¼1
1
1
With E ½A2k ¼ 1:1 þ (1) (1) ¼ 1, we obtain
2
2
2
3
kTc t , (k þ 1)Tc
1,
kTc s , (k þ 1)Tc 5
RX (t, s) ¼ 4
0, otherwise
The function RX (t, s) is shown in Fig. 20.11.4, which corresponds to Fig. 20.11.2, showing
that X(t) is a cyclostationary process.
FIGURE 20.11.4
20.12 Aperiodic Random Process (Karhunen – Loeve Expansion)
563
B 20.12 APERIODIC RANDOM PROCESS
(KARHUNEN – LOEVE EXPANSION)
In the last section the coefficients of the Fourier expansion of a mean-square continuous
periodic random process were found to be orthogonal random variables. We can
attempt a similar Fourier expansion in the interval [a,b] for a mean-square continuous
aperiodic random process X(t) as
X(t) ¼
1
X
Xn fn (t),
atb
(20:12:1)
n¼1
where the sequence ffn(t)g is an orthonormal set in the interval [a,b] defined by
ðb
a
fn (t)fm (t)dt
¼
1,
0,
m¼n
m=n
(20:12:2)
and the generalized Fourier coefficients (FCs) Xn are given by
ðb
Xn ¼
a
X(t)fn (t)dt,
n ¼ 0, 1, . . .
(20:12:3)
If the sequence ffn(t)g is any orthonormal set, then the coefficients fXng will not be orthogonal, unlike the periodic random processes as shown in the following example.
Example 20.12.1 Let X(t), 0 t T, be a zero mean stationary Gaussian random
process with continuous ACF RX(t). A Fourier series expansion of X(t) is given by
X(t) ¼ lim
N!1
N
X
Xn e jnv0 t :
v0 ¼
n¼N
2p
T
The Fourier coefficient Xn is given by
1
Xn ¼
T
ð T=2
X(t)ejnv0 t dt
with
E ½Xn ¼ 0,
E ½X(t) ¼ 0
T=2
We will show that the random variables fXng are not orthogonal. E½Xn Xm can be written as
E ½Xn Xm 1
¼ 2
T
ð T=2 ð T=2
E½X(t)X(s)e jv0 (ntms) dt ds
T=2 T=2
Substituting m ¼ n 2 k in this equation, we obtain
E ½Xn Xm 1
¼ 2
T
ð T=2 ð T=2
RX (t s)e jv0 n(ts) ejv0 ks dt ds
T=2 T=2
Substituting t ¼ t 2 s in this equation, yields
E ½Xn Xm 1
¼ 2
T
ð T=2 ð T=2s
RX (t)e
T=2
T=2s
jv0 nt
dt ejv0 ks ds
(A)
564
Classification of Random Processes
We will define the quantity in Eq. (A):
ð T=2s
RX (t)ejv0 nt dt ¼ SX (n, s) and
lim SX (n, s) ¼ SX (nv0 )
T!1
T=2s
Substituting the preceding equation in Eq. (A), we find that
ð
1 T=2
E½Xn Xm ¼ 2
SX (n, s)e jv0 ks ds = 0
T T=2
Hence we have shown that if X(t) is expanded in any orthonormal set, the resulting Fourier
coefficients fXng are not orthogonal. However as T ! 1, we see that SX(n, s) tends to
SX(nv0) and hence
ð
1 T=2
SX (nv0 )e jv0 ks ds ! 0
lim 2
T!1 T
T=2
Thus, the Fourier coefficients are orthogonal as T ! 1.
We are interested in finding the necessary and sufficient conditions for that orthonormal set that will yield an orthogonal Fourier coefficient set. We first multiply Eq. (20.12.1)
by Xm and take expectations
E½X(t)Xm ¼
1
X
E½Xn Xm fn (t),
atb
(20:12:4)
n¼1
Under the orthogonality condition of the FCs we can substitute E½Xn Xm ¼ 0, n = m in
Eq. (20.12.4) and obtain
E½X(t)Xm ¼ EjXm j2 fm (t),
atb
(20:12:5)
We can obtain a second equation by multiplying the following equation for Xm
ðb
Xm ¼ X (s)fm (s)ds
a
by X(t) and taking expectations. Or, for a t b
ðb
ðb
E½X(t)Xm ¼ E½X(t)X (s)fm (s)ds ¼ RX (t, s)fm (s)ds
(20:12:6)
Comparing Eqs. (20.12.5) and (20.12.6), we can write
ðb
E½X(t)Xm ¼ RX (t, s)fm (s)ds ¼ EjXm2 jfm (t),
(20:12:7)
a
a
atb
a
Defining the signal energy EjXn2 j ¼ ln , Eq. (20.12.7) can be rewritten as
ðb
RX (t, s)fn (s)ds ¼ ln fn (t), a t b
a
or
ðb
RX (t, s)f(s)ds ¼ lf(t),
a
atb
(20:12:8)
20.12 Aperiodic Random Process (Karhunen – Loeve Expansion)
565
This equation which is called the Karhunen – Loeve (K-L) integral equation, which
gives the necessary and sufficient conditions for the random process X(t) to be
expanded as a series in terms of the orthonormal set ffm(t)g such that the generalized
Fourier coefficients Xm, called Karhunen –Loeve (K-L) coefficients, are orthogonal. The
expansion of Eq. (20.12.1) is called the Karhunen –Loeve (K-L) expansion. The orthonormal functions ffm(t)g are called the eigenfunctions and flmg are called the eigenvalues of the K-L integral equation. It is to be particularly noted that the process
X(t) can be nonstationary.
Summary:
K-L equation satisfied ) Fourier coefficients{Xn } are orthogonal
Fourier coefficients{Xn } are orthogonal ) K-L equation is satisfied
Example 20.12.2 We are given a white-noise process with autocorrelation function
RV(t, s) ¼ s2d(t 2 s). Even though this process violates the second-order condition in
the sense that the autocorrelation function is not finite, we can formally find the orthonormal set such that the Fourier coefficients are orthogonal in the interval [2a, a]. Substituting for RX(t, s) in Eq. (20.12.8), we obtain
ðb
s2 d(t s)fn (s)ds ¼ ln fn (t)
a
or
s2 fn (t) ¼ ln fn (t) and
s2 ¼ ln ,
at a
This result conveys that any orthonormal set ffn(t)g with eigenvalues s2 will generate
K-L coefficients Vn for the white-noise process in the interval [2a, a] with
E[V2n] ¼ ln ¼ s2.
Example 20.12.3
A random process Y(t) not necessarily stationary is given by
Y(t) ¼ X(t) þ V(t)
where V(t) is a zero mean white Gaussian noise uncorrelated with X(t). We want to find the
eigenfunctions ffn(t)g in the interval [2a, a] that will give K-L coefficients. The autocorrelation function of Y(t) is
RY (t, s) ¼ RX (t, s) þ s2 d(t s)
The eigenfunctions ffX(t)g and ffY (t)g for representing X(t) and Y(t) can be found
by applying the K-L integral equation Eq. (20.12.8) to RX(t,s) and RY(t,s) as given
below:
ða
RX (t, s)fX (s)ds ¼ lX fX (t),
a,ta
RY (t, s)fY (s)ds ¼ lY fY (t),
a,t a
a
ða
a
566
Classification of Random Processes
Since X(t) and V(t) are uncorrelated, we can substitute RY (t, s) ¼ RX (t, s) þ s2 d(t s)
in the preceding integral equation for RY(t, s) and obtain
ða
½RX (t, s) þ s2 d(t s)fY (s)ds ¼ lY fY (t), a , t a
a
or
ða
RX (t, s)fY (s)ds ¼ ½lY s2 fY (t),
a,t a
a
From the uniqueness of the K-L equation, fX (t) ¼ fY (t) and lY ¼ lX þ s2 . Hence, we can
use the same eigenfunction set ffX (t)g for the series expansion of both X(t) and Y(t), and
from Example 20.12.1 the same can be used for expanding V(t). Thus, all three signals
X(t), Y(t), and V(t) can be represented for 2a , t a in the same function space as
X(t) ¼
1
X
Xn fn (t),
Y(t) ¼
n¼1
1
X
Yn fn (t),
V(t) ¼
n¼1
1
X
Vn fn (t)
n¼1
The previous two examples are simple ones. However, solving the general K-L integral
equation can be difficult. In particular cases, closed-form solutions can be obtained as
shown in the following examples.
Example 20.12.4 (Nonstationary Process)
Wiener process W(t) is
The autocorrelation function RW(t, s) of the
RW (t, s) ¼ s2 min(t, s)
(20:12:9)
The series expansion for W(t) for the time of observation [0, T ] can be obtained from the
K-L integral equation [Eq. (20.12.8)] as follows:
ðT
s2 min(t, s)f(s)ds ¼ lf(t)
0
or
s2
ð t
ðT
sf(s)ds þ t
0
f(s)ds ¼ lf(t),
0 , t, s T
(20:12:10)
t
The usual procedure to solve integral equations of this type is to differentiate them as many
times as possible so that they are converted into differential equations. The resulting differential equations are then solved from the boundary conditions obtained from their form.
Differentiating Eq. (20.12.10) with respect to t, there results
ðT
2
s tf(t) þ f(s)ds tf(t) ¼ lf(t)
t
or
s2
ðT
f(s)ds ¼ lf(t),
0 , t, s T
(20:12:11)
t
Differentiating again with respect to t [Eq. (20.12.11)], we obtain a second-order
homogeneous differential equation given by
f(t) þ
s2
¼ 0,
l
0 , t T, l = 0
(20:12:12)
20.12 Aperiodic Random Process (Karhunen – Loeve Expansion)
567
pffiffiffi
The characteristic equation is s 2 þ (s2/l) ¼ 0, and the roots are s1 , s2 ¼ +(s= l). The
solution to the differential equation is
st
st
f(t) ¼ A cos pffiffiffi þ B sin pffiffiffi
(20:12:13)
l
l
There are three unknowns—A, B, and l—to be resolved. The boundary conditions are
obtained
as follows. Substituting t ¼ 0 þ in Eq. (20.12.10), we can write
Ð 0þ
s2 0 min (t, s)f(s)ds ¼ lf(0 þ), or lf(0 þ) ¼ 0 and f(0 þ) ¼ 0. Substituting this
initial condition in Eq. (20.12.13), we have A ¼ 0 and the solution becomes
st
f(t) ¼ B sin pffiffiffi
(20:12:14)
l
ÐT
Substituting t ¼ T2 in Eq. (20.12.11), we obtain s2 T f(s)ds ¼ lf(T ), or
pffiffiffi pffiffiffi
f(T ) ¼ 0. Differentiating Eq. (20.12.14), f(t) ¼ B cos (st= l)s= l and we have
2
sT
sT
B cos pffiffiffi ¼ 0 from which ln ¼
,n 1
(20:12:15)
(2n 1)p=2
l
The constant B is obtained from the normalization integral given by
ðT
B2 sin2 (kt)dt ¼ 1
(20:12:16)
0
where, for convenience, we have defined
s
s(2n 1)p (2n 1)p
¼
k ¼ pffiffiffiffiffi ¼
2sT
2T
ln
The integral in Eq. (20.12.16) yields
ðT
ðT 2
B
B2 T sin(2kT)
2
2
¼1
½1 cos(2kt)dt ¼
B sin (kt)dt ¼
2
2k
0
0 2
(20:12:17)
Withpkffiffiffiffiffiffiffiffi
¼ (2n 1)p=2T, the term sin (2kT)=2k in Eq. (20.12.17) goes to zero and
B ¼ 2=T .
Hence the eigenfunctions fn(t) are
rffiffiffiffi 2
(2n 1)pt
n1
fn (t) ¼
sin
,
(20:12:18)
0,tT
T
2T
The K-L expansion of the Wiener process W(t) is given by
rffiffiffiffi 1
X
2
(2n 1)pt
W(t) ¼
Wn
sin
, 0,tT
T
2T
n¼1
where the coefficients Wn are given by an equation similar to Eq. (20.12.3)
rffiffiffiffi ðT
2
(2n 1)pt
n1
Wn ¼ W(t)
sin
dt,
0,tT
T
2T
0
(20:12:19)
(20:12:20)
with the property E ½Wn2 ¼ ln , n 1.
Example 20.12.5 A sinusoid with random phase is given by X(t) ¼ A cos(vt þ F),
where A and v are constants and F is uniformly distributed in (2p, p). The
568
Classification of Random Processes
autocorrelation function as derived in Example 19.2.1 is RX(t, s) ¼ (A 2/2) cos [v(t 2 s)].
We have to find the K-L expansion for X(t) in the period [2T/2, T/2] and show that the
K-L coefficients Xn are orthogonal.
Substituting for the autocorrelation function in the K-L equation [Eq. (20.12.8)], we
obtain
ð
A2 T=2
T
T
cos½v(t s)f(s)ds ¼ lf(t), t, s 2
2
2 T=2
Taking derivative with respect to t once, we get
ð T=2
A2
v
sin½v(t s)f(s)ds ¼ lf(t),
2
T=2
T
T
t, s 2
2
Taking the derivative with respect to t again, we have
ð
A2 2 T=2
v
cos½v(t s)f(s)ds ¼ lf(t)
2
T=2
or
v2 lf(t) ¼ lf(t) or
f(t) þ v2 f(t) ¼ 0,
T
T
t, s 2
2
The solution to the second-order homogeneous differential equation above is given by
f(t) ¼ C cos(vt) þ D sin(vt)
From the form of f(t) we conclude that the eigenfunctions are
rffiffiffiffi
rffiffiffiffi
2
2
f1 (t) ¼
cos(vt)
f2 (t) ¼
sin(vt)
T
T
There are only two terms in the finite K-L expansion given by
rffiffiffiffi
rffiffiffiffi
2
2
X(t) ¼ X1
cos(vt) þ X2
cos(vt)
T
T
We can find these two K-L coefficients by equating like coefficients in the expansion of
X(t) given by
X(t) ¼ A cos(vt) cos(F) A sin(vt) sin(F)
Thus
rffiffiffiffi
T
cos(F)
X1 ¼ A
2
rffiffiffiffi
T
and X2 ¼ A
sin(F)
2
With F uniformly distributed in (2p, p), the joint moment E[X1X2] is
E ½X1 X2 ¼
A2 T
A2 T
E ½cos (F) sin (F) ¼
E ½sin (2F) ¼ 0
2
2
showing that the K-L coefficients are orthogonal.
We will now find the eigenvalue l1 by substituting f1(t) in the K-L equation:
rffiffiffiffi
rffiffiffiffi
ð
A2 T=2
2
2
cos½v(t s)
cos(vs)ds ¼
cos(vt)
T
T
2 T=2
20.12 Aperiodic Random Process (Karhunen – Loeve Expansion)
569
or
A2
2
ð T=2
1
{cos(vt) þ cos½v(t 2s)}ds ¼ l1 cos(vt)
T=2 2
or
A2 T
cos(vt) ¼ l cos(vt) and
4
l1 ¼
A2 T
4
In a similar manner, substituting f2(t) in the K-L equation, l2 ¼ A 2T/4. The energy
values in the K-L coefficients are
E ½X12 ¼
A2 T
A2 T
E ½cos2 (F) ¼
¼ l1
2
4
E ½X12 ¼
A2 T
A2 T
E ½sin2 (F) ¼
¼ l2
2
4
Example 20.12.6 The autocorrelation function of a random process X(t) is
RX (t, s) ¼ e2jtsj . The signal is observed in the interval [22, 2]. We want to find the
eigenvalues and eigenfunctions.
The function RX (t, s) ¼ e2jtsj is shown in Fig. 20.12.1.
This example a little more involved than the previous ones, and since X(t) is not specified, we cannot obtain the K-L coefficients directly.
The K-L equation is obtained by substituting RX(t,s) in Eq. (20.12.8). The result is
ðt
e2(ts) f (s)ds þ
s,t
2
ð2
t
e2(ts) f (s)ds ¼ lf(t), 2 t 2
st
Differentiating this equation with respect to t, we obtain
ðt
e
2
2
2(ts)
ð2
f(s)ds þ 2
e2(ts) f(s)ds ¼ lf(t)
t
FIGURE 20.12.1
570
Classification of Random Processes
Differentiating again yields
ð t
ð2
4f(t) þ 4
e2(ts) f(s)ds þ e2(ts) f(s)ds ¼ lf(t)
t
2
The quantity within braces is the K-L equation and equals lf(t). Hence we obtain
lf(t) þ 4(1 l)f(t) ¼ 0,
2t 2
a second-order homogeneous differential equation. The characteristic equation is
s2 þ 4(1 l)=l ¼ 0, and the oscillatory roots are given by v2 ¼ 4(1 l)=l. We have
an implicit relationship between v and l with l ¼ 4=(4 þ v2 ).
The two eigenfunctions are of the form f1 (t) &frac1