STAT 6310, Stochastic Processes
Jaimie Kwon
STAT 6310, Introduction to Stochastic Processes
Lecture Notes
Prof. Jaimie Kwon
Statistics Dept
Cal State East Bay
Disclaimer
These lecture notes are for internal use of Prof. Jaimie Kwon, but are
provided as a potentially helpful material for students taking the course.
A few things to note:
 The lecture in class always supersedes what’s in the notes
 These notes are provided “as-is” i.e. the accuracy and relevance of
the contents are not guaranteed
 The contents are fluid due to constant update during the lecture
 The contents may contain announcements etc. that are not relevant
to the current quarter
 Students are free to report typos or make suggestions on the notes
via emailing or in person to improve the material, but they need to
understand the above nature of the notes
 Do not distribute these notes outside the class
Best Practice for note-taking in class
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STAT 6310, Stochastic Processes
Jaimie Kwon
 I do not recommend students relying on this lecture notes in place of
actual notes he/she writes down
 Bring a notepad and write down materials that I go over in the class,
using this lecture notes as the independent reference; you don’t
miss a thing by not having a printout of this lecture note in (and
outside) the class
 If you still want to print these notes, it’d be better to print them 4
pages on a single page (using “pages per sheet” feature in MS
Word), preferably double sided (to save trees)
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1 Basic probability
1.1 Sample spaces and events
 A “sample space,” is the set of all possible outcomes of an
experiment. The sample points are the elements in a sample space.
 An “event” in a subset of the sample space
 Two events are mutually exclusive if …
1.2 Assignment of probabilities
 Probability axioms
 P(AB)=___
 P(Ac)=___
1.3 Simulation of events on the computer
 How do you simulate a coin tossing on a computer?
1.4 Counting techniques
 Multiplication rule
 Permutation of n distinct items taken r at a time
 Combination: an unordered arrangement of r items selected from n
distinct items
1.5 Conditional probability
 P(B|A)=___
 Multiplication rule P(AB)=___
 Law of total probability: if E1,…,Ek are a partition of the sample
space, then P(A)=___
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1.6 Independent event
 Events A and B are independent if ___
 Events E1,E2,… are mutually independent if ___
2 Discrete random variables
2.1 Random variables
 The probability mass function p(x) satisfies ___
 CDF F(x)=P(Xx)
2.2 Joint distributions and independent random variables
 The joint probability mass function p(x,y)=P(X=x,Y=y)
 The marginal probability mass function pX(x)=___
 The conditional probability mass function P(X=x|Y=y)
 Discrete random variables X and Y are independent if ___
 Discrete random variables X1,X2,…,Xn are mutually independent if
___
2.3 Expected values
 E(X)=___
 If Y is a function of a random variable X, then E(Y)=E(Q(X))=___
 E(a+bX)=___
 If Y=g1(X)+…+g2(X), then E(Y)=____
2.4 Variance and standard deviation
 VAR(X)=___ = ____
 STD(X)=___
 Chebychev’s inequality
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 VAR(a+bX) = ___
 STD(a+bX)=___
2.5 Sampling and simulation
 Random variables X1,…,Xn are a random sample if they are i.i.d.
 Empirical probability distribution; empirical probability mass function
2.6 Sample statistics
 Sample mean X =___
 Sample variance S 2 
1 n
( X i  X )2

n i 1
 Sample standard deviation S=___
2.7 Expected values of jointly distributed random variables and
the law of large numbers
 E(Q(X,Y))=___
 E(aX+bY)=___
 E(a1X1+a2X2+…+anXn)=___
 If X and Y are independent, E(XY)=___
 If X and Y are independent, VAR(aX+bY)=___
 If X1,…,Xn are independent, VAR(a1X1+…+anXn)=___
 If X1,…,Xn iid with E(Xi)= and VAR(Xi)=2, then
E ( X ),VAR( X ), STD( X )  ___
 Weak law of large number: under the i.i.d. setup as above,
lim n P(| X   |  )  0 for any >0.
2.8 Covariance and correlation
 COV(X,Y)=___
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 CORR(X,Y)=___
 VAR(aX+bY)=___
 Bivariate random sample
 The sample covariance cov(X,Y)
 The sample correlation corr(X,Y)
2.9 Conditional expected values
 E(Y|X=x)=___
 E(Y)=E(E(X|Y))
 VAR(Y|X=x)=___
 VAR(Y)=E[___]+VAR[___]
3 Special discrete random variables
3.1 Binomial random variable
 A binomial random variable is ___
 For Y~bin(n,p), p(y)=___. Also, E(Y)=___ and VAR(Y)=___
3.2 Geometric and negative binomial random variables
 The geometric random variable is ____. We write it as X~Geo(p).
In that case, p(x)=___, E(X)=___, VAR(X)=___
 Given an integer k>1, the negative binomial random variable is ___.
We write it as Y~Nbinom(k,p).
In that case, p(x)=___, E(X)=___, VAR(X)=___
3.3 Hypergeometric random variables
 Consider a jar containing the total of N balls, r of which red, the
remaining white. Randomly selecting n balls from the jar WOR, the
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Jaimie Kwon
number of red balls in the sample X~Hypergeo(N,r,n).
In that case, p(x)=___, E(X)=___, VAR(X)=___
 Approximately binomial with p=___
3.4 Multinomial random variables
 Let Y1,…,Yk denote the number of times the mutually exclusive
outcomes C1,…,Ck occur in n iid trials. Let pi=P(Ci) for i=1,…,k. Then
we say (Y1,Y2,…,Yk)~multinom(n, p1,…,pk) and
P(Y1=y1,…,Yk=yk)=______ where ____
 E(Yi)=__ and VAR(Yi)=___
 COV(Ys,Yt)=____ if st.
3.5 Poisson random variables
 X~Poisson() if p(x)=___
 E(X)=___ and VAR(X)=___
 If X~bin(n,p), P(X=x) ~ p(x) of Poisson() if ____ and =__
3.6 Moments and moment-generating functions (MGF)
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4 Markov chains
4.1 Introduction: modeling a simple queuing system
 A queue is a waiting line
 Consider the following queuing system:
 A company has an assigned technician to handle service for 5
computers.
 Each computer independently fails with probability.2 during any
day
 The technician can fix one computer per day.
 If a machine breaks down, it will be fixed the day it fails provided
there is no backlog. If there is a backlog, it will join a service
queue and wait until the technician fixes those ahead of it.
 Simulate the behavior of the system over 5 days of operation.
The system begins on Monday with no backlog of service and
ends on Friday evening.
 We observe two characteristics of the system:
 The Number of computers waiting for service at the end of
each day
 The number of days in the week the technician is idle.
 Simulation is “replicated” 1,000 times.
 See the text for the output
 The idea: many questions can be answered without any
computation from theory
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4.2 The Markov property
 Processes that fluctuate with time as a result of random events
acting upon a system are called “stochastic processes”. Formally, it
is a collection of random variables {X(n),nN}. In a typical context,
the “index set” N refers to time and the process is in “state” X(n) at
time n.
 Time can be measured on a discrete scale or on a continuous scale
 The states can be discrete or continuous
 Four combinations: {discrete, continuous}-time, {discrete,
continuous}-state process
 Discrete-time, discrete-state process
 Discrete-time, continuous-state process
 Continuous-time, discrete-state process
 Continuous-time, continuous-state process
 For now, we consider only discrete-time, discrete-state processes
 X(0) : initial state
 Two extremes:
 Independent X(n)
 X(n) depends on all the past: example?
 Middle ground?
 In a stochastic process having the “Markov property”, each outcome
depends only on the one immediately preceding it.
 Definition 4.2-1. The process {X(n), n=0,1,2,…} is said to have
Markov property, or a Markov chain, if P[X(n+1)=s(n+1)| X(n)=s(n),
X(n-1)=s(n-1),…. ,X(1)=s(1),X(0)=s(0)]
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=P[X(n+1)=s(n+1)|X(n)=s(n)]
for n=0,1,2,… and all possible states {s(n)}
 Examples: insect on a circular box with 6 room
 Position of a token on the board of many board games
 Gambler’s net worth in a simple game of chance
 One-dimensional position of a particle suspended in a fluid
(Random walk)
 Number of customers in a “queue” over time
 Total number of phone calls in each minute, when the number of
calls in each minute is independent 0, 1, 2 with probabilities 0.80,
0.15, 0.05, respectively.
 State diagram
1
2
3
4
5
6
 Markov property is more an assumption than a verifiable fact
 The “one-step transition probability” is the conditional probability
defined as
Pn(ij)=P[X(n+1)=j | X(n)=i]
 A process whose transition probabilities don’t depend on time n is
called a “time-homogeneous process”. Otherwise, the process is
called “nonhomogeneous”
 For a time-homogeneous process, we simply write P(ij)
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Jaimie Kwon
 A useful way to represent the transition probabilities of a timehomogeneous Markov chain is with a “one-step transition matrix”
 Ex.4.2-1 (insect in a circular box)
4.3 Computing probabilities for Markov chains
 We consider time homogeneous Markov chains for now
 The probability of Markov chain visits states s(1), s(2),…,s(n) at
times 1,2,…,n given that the chain begins in state s(0) can be
computed by
P[X(n)=s(n), X(n-1)=s(n-1),…. ,X(1)=s(1) | X(0)=s(0)]
= P[s(0)s(1)] P[s(1)s(2)] …P[s(n-1)s(n)]
 Proof:
 A “path” of a process is a sequence of states through which a
process may move through.
 The initial state X(0) can also be a random variable. In such cases,
the “initial distribution” of X(0) also need to be figured into the
computation of the probabilities of various paths
 Example: consider a transition matrix
1 2 3 4
1  0 1/ 3 1/ 3

0
P  2 1 / 2 0
3 1 / 2 0
0

4 1 / 3 1 / 3 1 / 3
1 / 3
1 / 2 :
1 / 2

0 
 What’s the probability of
 Path 1242 given it begins on 1?
 Path 2134 given it begins on 2?
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 Path 1242 if the probability of ¼ of starting in any of the
compartments?
 What’s the probability of the rat ending up in compartment 1 two
moves after it starts from compartment 1?
 Definition 4.3-1. The “k-step transition probability” is defined as
P(k)(ij) = P[X(n+k)=j | X(n)=i]
 For going from state i to j in two steps, the list of all possible paths
can be written as isj where s ranges over all possible states.
Thus
P ( 2) (i  j )   P(i  s) P( s  j )
s
 The above equation shows that the “two-step transition matrix” is
given by P2. In general,
 Theorem 4.3-1. The k-step transition probabilities are obtained by
raising the one-step transition matrix to k’th power
 The matrix Pk is called the “k-step transition matrix”.
 Example. If the rat begins in compartment 1, what’s the probabilities
of it being in compartments 1,2,3 or 4 after 4 moves?
 Example. Signal transmission through consecutive noisy channels
 Example. A communication channel is sample each minute. Let
X(n)= Busy(1) or not busy(2) at time n and let the transition matrix
be
.6 .4
P

.1 .9
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Jaimie Kwon
 We assumed we know the initial state. If the initial state is random,
for example, we know that
P[ X (1)  i]   P[ X (0)  s]P( s  i)
s
 Let (n) be the row vector whose elements are probabilities
P[X(n)=s] for all possible states. Then the above formula can be
written as:
(1) = (0) P.
 Likewise, the probabilities for the chain after k steps is:
(k) = (0) Pk.
4.4 The simple queuing system revisited
 Model the queuing system as a Markov chain!
 Let X(n) be the backlog on day n
 What is P? i.e. what are P(ij)s?
p <- 0.2
P <- matrix(c(sum(dbinom(0:1, 5, p)), dbinom(2:5, 5, p),
dbinom(0:4, 4, p),
0, dbinom(0:3, 3, p),
0,0, dbinom(0:2, 2, p),
0,0,0, dbinom(0:1, 1, p)),
byrow=TRUE, ncol=5)
P
> P
[,1]
[,2]
[,3]
[,4]
[,5]
[1,] 0.73728 0.2048 0.0512 0.0064 0.00032
[2,] 0.40960 0.4096 0.1536 0.0256 0.00160
[3,] 0.00000 0.5120 0.3840 0.0960 0.00800
[4,] 0.00000 0.0000 0.6400 0.3200 0.04000
[5,] 0.00000 0.0000 0.0000 0.8000 0.20000
 That’s the Monday’s backlog
 Compute two, three, four, five-step transition probability matrix:
P%*%P
P%*%P%*%P
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P%*%P%*%P%*%P
P%*%P%*%P%*%P%*%P
round(P%*%P, 4)
round(P%*%P%*%P, 4)
round(P%*%P%*%P%*%P, 4)
round(P%*%P%*%P%*%P%*%P, 4)
[1,]
[2,]
[3,]
[4,]
[5,]
[,1]
0.5153
0.4828
0.4192
0.3288
0.2202
[,2]
0.2991
0.3077
0.3233
0.3415
0.3533
[,3]
0.1452
0.1613
0.1934
0.2405
0.3009
[,4]
0.0368
0.0437
0.0578
0.0799
0.1117
[,5]
0.0036
0.0045
0.0064
0.0094
0.0139
 Since we begin at X(0)=0, the top row of P5 gives the probabilities
of having 0-4 as backlog after five steps (on Friday)
4.5 Simulating the behavior of a Markov chain
 Why simulate?
 “Easy” way
 “Only” way
 As a part of a larger, complex system
 Example 4.5-2: deterioration of an automobile over states “excellent,
good, fair, and poor” over years. What’s the amount of time it will
take an auto to reach state 4?
> P
[1,]
[2,]
[3,]
[4,]
[,1] [,2] [,3] [,4]
0.7 0.3 0.0 0.0
0.0 0.6 0.4 0.0
0.0 0.0 0.5 0.5
0.0 0.0 0.0 1.0
4.6 Steady-state probabilities
 For the rat problem,
> P100
[,1] [,2] [,3] [,4]
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[1,]
[2,]
[3,]
[4,]
0.3
0.3
0.3
0.3
0.2
0.2
0.2
0.2
0.2
0.2
0.2
0.2
Jaimie Kwon
0.3
0.3
0.3
0.3
 All the rows of the 100th-step transition matrix are (almost) the same.
Regardless of the starting state, after many moves, the rat has
probabilities ____ of being in states 1,2,3, and 4, respectively.
 Let’s limit ourselves further to only finite state Markov chains
 Definition 4.6-1. A finite state Markov chain is said to be regular if
the k-step transition matrix has all nonzero entries for some value of
k>0
 Example: insect in a circular box {regular, not regular}
 Example: rat in the maze
 Example: busy/free of a communication channel
 Theorem 4.6-1. Let X(n), n=0, 1, 2, …, be a regular Markov chain
with one-step transition matrix P. Then there exists a matrix 
having identical rows with nonzero entries such that,
lim k  P k  
 Let  be the common row vector of the limiting matrix .. Then  is
the steady-state probability vector, whose elements are steady state
probabilities.
 Consider the fraction of visits a Markov chain makes to a particular
state in k transitions, i.e.
V(j,k) = (Number of visits to state j in k transitions)/k
 Example. If a chain visits 1,2,1,2,2 in the first 5 times, then
V(2,5)=___ and V(1,5)=___.
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 Q: what’s E(V(j,k))=?
 Theorem 4.6-2. Let X(n), n=0,1,2,… be a regular Markov chain. Let
j be the steady state probability for state j. Then
lim k  E (V ( j , k ))   j .
 i.e., the long-run fraction of visits to state j
= steady-state probability of j
 Q: how to find  without matrix multiplication?
 Theorem 4.6-3. Let X(n), n=0,1,2,… be a regular Markov chain with
one-step transition matrix P. Then the steady-state probability vector
=(1, 2, …,S ) may be found by solving the system of equations
P=, 1+ 2+ …+S=1..
 Informal proof: Note Pk-1P = Pk. Send k on both sides to infinity
then we have P=. The result follows.
 Example: Rat in the maze
 Example: busy/not state of a communication line
 If the chain is currently in state 1 and the path (12321) is
observed, the return time is __ transitions.
 Theorem 4.6-4. Let X(n), n=0,1,2,… be a regular Markov chain. Let
=(1, 2, …,S ) be the steady-state probability vector for the chain,
and let Tj denote the time it takes to return to state j given that the
chain is currently in state j, j=1,2,…,S. Then,
E(Tj)=1/j.
 Beginning at a random initial state X(0) with initial probability vector
(0), the k-step probability vector is (k)=(0)Pk. If (0) happens to
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be the same as , it follows that
 =(0)= (1)= (2)=…
For this reason, the steady-state distribution is also called a
stationary distribution
4.7 Absorbing states and first passage times
 Not all Markov chains are regular.
 Ex. Automobile deterioration over years
 Definition 4.7-1. A state j is said to be an absorbing state if P(jj)=1;
otherwise it is a non-absorbing state.
 We assume there is a path that leads from each non-absorbing
state to an absorbing state.
 Q: beginning at a non-absorbing state, how long does it take to
reach an absorbing state?
 Theorem 4.7-1. Let A denote the set of absorbing states in a finite-
state Markov chain. Assume that A has at least one state and that
there is a path from every non-absorbing state to at least one state
in A. Let Ti denote the number of transitions it takes to go from the
non-absorbing state
i
to A. Then
PTi  k    P ( k ) i  j 
jA
 Ti is called the time to absorption.
 The above result gives the cumulative distribution function of Ti
 How about the mean time to absorption?
 We can use the above result to obtain PTi  k  up to a reasonably
large k.
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 Or, we can use the following result:
 Let a chain consists of non-absorbing states 1,2,…,r and the set of
absorbing states A. Let 1 , 2 ,..., r be the mean time to absorption
from states 1,2,…,r respectively. It is easy to see that:
i   Pi  j    Pi  j 1   j  .
r
jA
j 1
Combining this with
r
 Pi  j    Pi  j   1 , we obtain:
jA
j 1
_____.
Thus, we have proved,
 Theorem 4.7-2. Let Q denote the matrix consisting of transition
probabilities among the non-absorbing states. The mean times to
absorption satisfy the system of equations
 1  
 1 
   1
 
 2      Q 2 
   

  
 
  r  1
r 
 1 

 
1
 Or,  2   I  Q 1   . Even more compactly, μ  I  Q 11 for column


 

1
 r 
vectors  and 1.

 Note: the textbook notation Ti is a bit confusing since it’s used for
both
 The return time for the regular Markov chain
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 The time to absorption
 We talked about the expected return time for regular Markov chains.
For regular Markov chains, how about the first passage time from
state i to state j (where ij)?
 The trick is to define a new chain that is identical to the original
chain except that the state j is redefined to be an absorbing state.
Time to absorption to the state from state i is the first passage
time from state i to state j.
 Example: 4 speakers example. If speaker A just finished talking,
Expected time until speaker B speaks?
 Let f(ij) denote the probability that the chain is eventually
absorbed into state j given that it starts in state i. Under the
assumption that there is a path from every non-absorbing state to
the set of absorbing states, we have
lim k  P ( k ) i  j   f i  j  .
 If j is non-absorbing state, f(ij) = ___
 If j is the only abosrbing state, f(ij) = ____
 Interesting cases are when _____
 Theorem 4.7-3. Let the one-step transition matrix of a finite Markov
chain have the form
P
I 0
.
Non - absorbing states  R Q
Absorbing states
Let F be the matrix whose (i,j)th element is f(ij). Then
F  I  Q  R
1
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Jaimie Kwon
 Theorem 4.7-4. Let i and j be nonabsorbing states. Let ij denote
the expected number of visits to state j beginnning in state i, before
the chain reaches an abosrbing state. (If we are interested in returns
to state i itself, the initial state i is counted as one visit.) Let U be the
matrix whose ij’th element is ij . Then
- 20 -
U  I  Q 
1
.
STAT 6310, Stochastic Processes
Quiz 1 result:
> mean(x)
[1] 27.70732
> median(x)
[1] 28
> length(x)
[1] 41
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Jaimie Kwon
STAT 6310, Stochastic Processes
- 22 -
Jaimie Kwon
STAT 6310, Stochastic Processes
Jaimie Kwon
5 Continuous random variables
 Probability density functions
 Expected value and distribution of a function of a random variable
 Simulating continuous random variables
 Joint probability distributions
6 Special continuous random variables
 Exponential random variable
 Normal random variable
 Gamma random variable
 The Weibull random variable
 MGFs
 Method of moment estimation
- 23 -
STAT 6310, Stochastic Processes
 Midterm #1 result (out of 90)
> mean(x)
[1] 75.30952
> median(x)
[1] 81
> sd(x)
[1] 14.28711
>
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Jaimie Kwon
STAT 6310, Stochastic Processes
Jaimie Kwon
7 Markov counting and queuing processes
7.1 Bernoulli counting process
 Definition 7.1-1. A stochastic process is called a “counting process”
if
i) The possible states are the nonnegative integers.
ii) For each state i, the only possible transitions are
i i, ii+1, ii+2,…
 If a counting process that has the Markov propoerty is called
“Markov counting process”
 Assume that we have divided a continuous-time interval into
discrete disjoint subintervals of equal lengths. The subintervals are
called “frames.”
 Definition 7.1-2. A counting process is said ot be a “Bernoulli
counting process” if
i) the number of successes that can occur in each frame is either 0
or 1.
ii) The probability, p, that a success occurs during any frame is the
same for all frames.
iii) Successes in nonoverlapping frames are independent of one
another.
 Theorem 7.1-1. Let X(n) denote the total # of successes in a
Bernoulli counting process at the end of the n’th frame, n=1,2,… Let
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STAT 6310, Stochastic Processes
Jaimie Kwon
the initial state be X(0)=0. The probability distribution of X(n) is
n
P( X (n)  x)    p x (1  p) n x .
 x
 What’s the one-step transition matrix for X(n)?
 For realistic modeling,
 The frame length should be determined by practical
considerations and
 The success probability can be comptued by the formula
p

n
where
 , the rate of success, is the expected number of successes in
a unit time
 n is the number of frames in this unit of time
 In practice, we estimate  by
ˆ 
Number of successes in t unit of time
t
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STAT 6310, Stochastic Processes
Jaimie Kwon
Table for the example 7.1-2, 3, 5
 = 3 per hour

5 min or
1 min or
… =1/n
5/60 hour
1/60 hour
p = 
1/4
1/20
= /n = 3/n
n (for t=1 hour)
12
60
Send to 
E(X(n)) = np
3
3
3 (= mean of
Poisson(3  t))
Var(X(n)) = np(1-
12*1/4*3/4 60*1/20*19/
3 (= variance of
p)
= 2.25
20 = 2.85
Poisson(3  t))
Distribution of Y =
Geo(1/4)
Geo(1/20)

“# of frames
between one
success to the
next”
E(Y) = 1/p
4 (frames) 20 (frames)

V(Y) = (1-p)/p2
12
380

SD(Y)
3.46
19.5
(frames)
(frames)
T = Y  = “Time
Exp() distribution
between one
success to the
next”
E(T) = 1/
1/3
1/3
- 27 -
1/3
STAT 6310, Stochastic Processes
V(T) =
1  
2
(1-3/12) /
(1-3/60) / 32
32 = 0.083 = 0.106
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Jaimie Kwon
1/32 = 0.111…
STAT 6310, Stochastic Processes
Jaimie Kwon
Empty Table for the example 7.1-2, 3, 5
 = 3 per hour

5 min or
1 min or
5/60 hour
1/60 hour
… =1/n  0
p = 

n (for t=1 hour)
E(X(n)) = np
Var(X(n)) = np(1p)
Distribution of Y =
“# of frames
between one
success to the
next”
E(Y) = 1/p
V(Y) = (1-p)/p2
SD(Y)
T = Y  = “Time
between one
success to the
next”
E(T) = 1/
V(T) =
1  
2
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STAT 6310, Stochastic Processes
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Jaimie Kwon
STAT 6310, Stochastic Processes
Jaimie Kwon
 We denote the frame length by  (=1/n)
 p  
 Example: consider a call center that receives 3 calls per hour. If we
model the number of hits as a Bernoulli counting process,
What would be a good time unit? Hour
What would be a good frame length ? 1/60 hour
What is ? 3 calls per hour
What is p?
What’s the effect of using different  on p? on E(X(n))? On V(X(n))?
 The Bernoulli counting process is time homogeneous
 Theorem 7.1-2. Let Y denote the number of frames from one
success to the next. Then,
P(Y>y)=P(Y=y)=(1-p)y-1p, y=1,2,…
 (typo in the book! P.272)
 E(Y) = 1/p, V(Y)=(1-p)/p2
 Theorem 7.1-3. The amount of time T, that passes from one
success to the next, is given by T=Y. Thus,
 E(T)=1/
 V(T)=
1  
2
7.2 The Poisson process
 Definition 7.2-1. Let N(t) denote the # of successes in the interval
[0,t]. Assume that it is a continuous-time counting process and that
the count begins at zero. N(t) is said to be a “Poisson process” if
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STAT 6310, Stochastic Processes
Jaimie Kwon
i) Successes in nonoverlapping intervals occur independently of one
another.
ii) The probability distribution of the number of successes depends
only on the length of the interval and not on the starting point of the
interval.
iii) The probability of x successes in an interval of length t is
P N (t )  x   e t
t x , x  0,1,2,... where  is the expected number of
x
successes per unit of time.
 Consider a Bernoulli counting process with n frames in the interval
[0,t], each with length . The probability distribution of the # of
successes in [0,t] is approximately Poisson(), where  = np = t,
i.e.,
n
t x .
x
P( X (n)  x)    p x (1  p) n x  e  
 e  t
x!
x!
 x
This approximation holds for any interval of length t.
 Example: recall the telephone call example ( = 3 calls per hour)
 Let T be the time until the first success to occur in a Poisson
process. Note that:
 P(T > t) = P(N(t) = 0) = e-t, Or,
 FT(t) = P(T  t) = 1  P(N(t) = 0) = 1  e-t
 Theorem 7.2-1. Let T be the time between consecutive successes in
a Poisson process with rate . Then T ~ Exp(), with pdf
fT(t) = e-t
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STAT 6310, Stochastic Processes
Jaimie Kwon
 Note that E(T)=1/ and V(T)=1/2. They are limits of those in the
Bernoulli counting process as 0.
7.3 Exponential random variables and the Poisson process
 Let Ti, i=1,2,… denote the interarrival times in a Poisson process
with rate . Then Ti ~ iid Exp()
 Useful result for simulation
 Let Wn be the waiting time until it takes to observe n successes in a
Poisson process.
 Theorem 7.3-2. Note that
FWn t 
n 1
t x
x 0
x!
=P(Wn  t) = 1  P(Wn > t) = 1  P(N(t)  n  1)=1   e t
and
f (t ) 
n t n1
(n  1)!
e t .
The distribution is called the Erlang distribution and
is a special case of the Gamma distribution
(Gamma(,)=Gamma(n, 1/))

Theorem 7.3-3. Wn  T1  T2  ...  Tn and thus E(Wn) = n/ and V(Wn) =
n/2
 Example: 2 thunderstorms
7.4 The Bernoulli single-server queuing process
 A queueing process: a stochastic process of the number of persons
or objects in the system
 Arrivals and services
 A finite or limited-capacity queue vs. an unlimited-capacity queue
 We move from discrete-time to continuous-time as before
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STAT 6310, Stochastic Processes
Jaimie Kwon
 We move from singler-server process to multiple-server process
 Definition 7.4-1. A queuing process is said to be a single-server
Bernoulli queuing process, with unlimited capacity, if
i) The arrivals occur as a Bernoulli counting process, with arrival
probability PA in each frame.
ii) When the server is busy, departures or completions of service
occur as a Bernoulli counting process, with service completion
probability PS in each busy frame.
ii) Arrivals occur independently of services
 Convention: The earlist a customber can complete service is one
frame after arriving
 The # in the queuing system: those being served and those in line
 States: the # of customers in the system
 Note that:
 P(0  0) = 1- PA, P(0  1) = PA
 For i  0,
P(i  i  1) = PS(1  PA),
P(i  i) = (1 –PA) (1  PS) + PA PS,
P(i  i+1) = PA (1  PS)
 Example: a single server queue with PA = .10 and probability service
PS = .15. What’s the one-step transition matrix?
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STAT 6310, Stochastic Processes
Jaimie Kwon
 The number of frames from the arrival of one customer to the next ~
Geo(PA)
 The number of frames it takes to service a customer ~ Geo(PS)
 PA 
 PS 
A
n
S
n
  A  and
 S 
 The expected service time S=1/S.

 Suppose the system has a capacity of C customers. P(ij) is same
as before for i < C. If the system is full, we assume that potential
customers continue to arrive but do not join the system unless a
service has occurred.
 P(C  C) = P(0 arrivals and 0 services) +
P(1 arrival and 1 service) +
P(1 arrival and 0 service)
= (1 – PA)(1– PA) + _____ + ______
= ________
 P(C  C – 1) = P(0 arrivals and 1 service) = ________
 For i < C, identical as above
 Example: a telephone has the capability of keeping one caller on
hold while another is talking. While a caller is on hold, a new attempt
to call is lost (busy signal). Thus, at most two calls can be in the
system at any time. Suppose the probability that a call arrives during
a frame on 1 minute is PA = .10 and probability that a call is
completed during a frame is PS = .15. For this limited-capacity
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STAT 6310, Stochastic Processes
Jaimie Kwon
Bernoulli queuing process, what is the one-step transition matrix P?
 What is the steady-state probability vector?
0.453 0.336 0.211
7.5 The M/M/1 queuing process
 Aim: to derive a continuous-time queuing process by letting   0 in
the single-server Bernoulli queuing process.
 Such process is called M/M/1 queuing process
 In general, we consider {M, G, D}/{M, G, D}/k queuing process
 M: Markov
 G: general
 D: deterministic
 k: # of servers
 Let’s recall that:
 Note that:
 P(0  0) = 1- PA, P(0  1) = PA
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STAT 6310, Stochastic Processes
Jaimie Kwon
 For i  0,
P(i  i  1) = PS(1  PA),
P(i  i) = (1-PA) (1  PS) + PA PS,
P(i  i+1) = PA (1  PS)

PA 

PS 
A
n
S
n
  A  and
 S 
 Convert them to:
 P(0  0) = 1 – A, P(0  1) = A
 For i > 0,
P(i  i  1) = S (1  A)  _________,
P(i  i) = (1 – A) (1  S) + AS  __________,
P(i  i+1) = A (1  S)  _________
 Approximate transition probabilities for the M/M/1 queuing
process in a small time interval of length 
 Transitions other than the above have negligible probability when
the frame size is small
 Definition 7.5-1. A continuous-time, single-server queuing process is
said to be M/M/1 queuing process if
i) for sufficiently small time intervals of length , transition
probabilities can be approximated by the above equaitons.
ii) Transitions occurring in nonoverlapping intervals are
independent of one another
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STAT 6310, Stochastic Processes
Jaimie Kwon
 Theorem 7.5-1. The time between the arrivals of customers in
M/M/1 ~ Exp(A), and given that a customer is being served, the
time to complete service is ~ Exp(S)
 Recall that Exp(A) is an exponential distribution with mean A =
1/A, and Exp(S) an exponential distribution with mean S = 1/S.
 Aim: derive the steady-state distribution for the number of
customers in an M/M/1 queuing system.
 Idea: if there are i>0 customers at time t + , there could have been
only i – 1, i + 1, or i customers at time t, if  is made small enough.
 Let i denote the steady-state probability that there are i > 0
customers in the system. If the system is in the steady-state
condition, then for all state i,
P(i customers at time t + ) = P(i customers at time t) = i
which leads to:
i = i-1 A + i+1 S + i (1 – A – S)
and
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STAT 6310, Stochastic Processes
Jaimie Kwon
0 = 1 S + 0 (1 – S)
 Simplifying, we have:
0A = 1 S
1A = 2 S
…
iA = i+1 S, i = 0, 1, ….
This makes sense and is called the “balance equation”
 Why is it called the banance equation:
i A = i+1 S 
 i P(i  i + 1) = i + 1 P(i + 1  i),
i.e., In steady state, mass leaving from i to i + 1  mass leaving from
i + 1 to i (otherwise, it wouldn’t be steady)
 
limt   P(X(t) = i) P(X(t + ) = i + 1 | X(t + ) = i ) =
limt   P(X(t + 1) = i) P(X(t + ) = i | X(t + ) = i +1 )
 The solution of the balance equation can be found as follows:
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STAT 6310, Stochastic Processes
Jaimie Kwon
 Theorem 7.5-2 Let r = A/S denote the arrival/service ratio of an
M/M/1 queuing process. The steady-state distribution exists iff r < 1
and the steady-state probability mass function for the number in the
system is
i = (1 – r)ri, i=0, 1, 2, …
 As a consequence:
P(X > x) = _____
P(X  x) = _____
 Example: 30 orders per day; service time of 12 mintes per order;
proportion of idle time period; P(X  2); P(X > 9); expected # of
orders in the system and the SD?
 Let Y have the steady state distribution of the # of customers in the
M/M/1 queuing system. Then X = Y + 1 ~ Geo(1 – r)
 E(Y) = ____
 SD(Y) = _____
 Theorem 7.5-3. The steady-state mean and SD of the number of
customers in an M/M/1 queuing system with arrival/service ratio r <
1 are
=
r
1 r
,=
r
1 r
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STAT 6310, Stochastic Processes
Jaimie Kwon
 What happens if r < 1 but near 1?
 What happens if r  1?
 Theorem 7.5-4. Assume the M/M/1 queuing system is in steady
state. Let T denote the amount of time a customer spends in the
system and r = A/S. Then
E(T) = ______________
1
 r
 1
 1


 1  r  S (1  r )S
 Example

7.6 K-server queuing process
 Definition 7.6-1. A process with k-servers and unlimited capacity is
called a “k-server Bernoulli queuing process” if:
i) The arrivals occur as a Bernoulli counting process with PA.
ii) For each busy server, service completions occur as a Bernoulli
counting process with PS, which is same for all servers.
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STAT 6310, Stochastic Processes
Jaimie Kwon
iii) Arrivals are independent of services; each busy servers function
independently of one another
n
 P(j service completions | n busy servers) =   PSj 1  PS n j , j  n .
j
 
 Recall that for a singler server:
 P(0  0) = 1- PA, P(0  1) = PA
 For i  0,
P(i  i  1) = PS(1  PA),
P(i  i) = (1-PA) (1  PS) + PA PS,
P(i  i+1) = PA (1  PS)

PA 

PS 
A
n
S
n
  A  and
 S 
 For two-servers: (if i = 0 or 1, behave just like a singler server)
 P(0  0) = 1- PA, P(0  1) = PA
 P(1  0) = PS(1  PA),
P(1  1) = (1-PA) (1  PS) + PA PS,
P(1  2) = PA (1  PS)
 For i  2,
P(i  i + 1)
= PA (1  PS)2
 P(i  i  1)
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STAT 6310, Stochastic Processes
Jaimie Kwon
= P(one arrival and two services OR no arrival and one service)
= (1  PA)[2(1 – PS) PS] + PA PS2
(Bernoulli probability)
P(i  i  2) =
P(i  i) =
 Now convert using PA 
A
n
  A  and PS 
S
n
 S 
 A continuous-time process is an M/M/k queuing process with
unlimited capacity if it satisfies the following conditions:
i) when there are n busy servers and i customers in the system, the
following approximations apply to the possible transitions ina
sufficiently small frame of length :
P(i  i + 1)  A;
P(i  i - 1)  n S;
P(i  i)  1 – A – nS
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STAT 6310, Stochastic Processes
Jaimie Kwon
ii) transitions in nonoverlapping frames are independent of one
another
 Rules of thumb: Estimating A and S for practical problems:
- To estimate A, use the method of moment, or
̂ A = (# of arrivals observed over a time period) /
(amount of the time period)
- To estimate S, solve
1
̂S
= sample mean of service completion times
This is the method of moment, too. (why?)
7.7 Balance equations and steady-state probabilities
 Idea: want to let the arrival rate A and the service rate S depend
on the current number of customers in the system
Examples?
 Definition 7.7-1 Let ai and si denote the arrival rate and system
service rate, respectively, of a queuing process with i customers in
the system. The process is called a “general Markov queuing
process” if the following conditions hold: In a sufficiently small time
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STAT 6310, Stochastic Processes
Jaimie Kwon
frame of length , we have :
P(i  i + 1)  ai 
P(i  i - 1)  si 
P(i  i)  1 – ai  – si .
The error of approximation is negligible compared with the terms
involving , when  is small. The transitions in nonoverlapping
intervals are independent of one another.
 Example: For M/M/k queuing process,
aj = ___ and
sj = ____
sj = j S, j = 0, 1, 2, …, k – 1,
s j = k S, j  k
 Theorem 7.7-1. Let aj, j = 0, 1, 2, … be the arrival rates and let sj, j =
1, 2, … be the service rates of a general Markov queuing process.
Let j, j = 0, 1, 2, …, denote the steady-state probability distribution
of the process. Then, the j’s satisfy the system of balance
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STAT 6310, Stochastic Processes
Jaimie Kwon
equations:
 j a j   j 1 s j 1 , j = 0, 1, 2, …
 Verification is similar to that of M/M/1 queuing process (Ex. 7.7-9)
 Theorem 7.7-2. The steady-state probability distribution for the
general Markov queuing process exists if and only if
1
a 0 a 0 a1 a 0 a1 a 2


 ...  
s1 s1 s 2
s1 s 2 s3
and the steady-state probability distribution is:
 j  0
a0 a1 ...a j 1
s1 s 2 ...s j
, j  1,2,...
and
 a

a a a aa
 0  1  0  0 1  0 1 2  ...
s1 s1 s 2 s1 s 2 s3


1
 Proof:
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STAT 6310, Stochastic Processes
Jaimie Kwon
 When does M/M/k queuing process satisfy the the theorem?
 Answer: When
r A

 1.
k kS
 In M/M/k, what is the general form of the steady state probability
distribution?

 k 1 r j

rk
Answer:  0  1   

 j 1 j! k!1  r / k 

rj

,
jk
0

j!

j 
j k
j
r
 r
 k  , j  k
 0 k!k j k
k
1

 In M/M/2 with r = A / S = .25, what is the steady state probability
distribution?
 Formula for the expected # of customers in an M/M/k system in
steady state:
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STAT 6310, Stochastic Processes
E(N )  r 
Jaimie Kwon
r k / k
1  r / k 2
 Example: a single fast server vs. two slow servers. Consider
M/M/1 : S = 80 customers per minute
M/M/2 : S = 40 customers per minute
The arrival rate A = 10 customers per minute is same for both
cases.
Compute the steady state probability for both cases.
Find and compare the mean number of customers in each system.
Lesson?
 Example: effect of adding a server. Consider
M/M/1 system with r = .9.
What’s the mean number of customers in the system?
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How does it change if we add two servers with the same service
rate each? (or M/M/3 system)
The expected time for a customer to spend in each system? [Need
to know A . Let’s assume A = 6 customers/minute]
 Theorem 7.7-3 (Little’s formula) Assume that an M/M/k system has
reached steady-state condition. Let T denote the amount of time a
customer spends in the system, and let N denote the number of
customers in the system. Then
A E(T) = E(N)
 Why does this make sense?
7.8 More Markov queuing processes
 For the steady-state probabilities for M/M/k process as k becomes
large,
1


rj
lim  0  1     e r
k 
 j 1 j! 
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Also,
rj
lim  j  e
, j  1,2,...
k 
j!
r
Doesn’t this look familiar?
 Mega-mart with a large # of checkout stations.
4 customers arrive per minute.
Each customer spends 1.5 minute being served.
Arrival/service ratio = _____
Average # of busy check-out stations in the steady state = _____
P(more than 10 busy lines) = ______
 Other cases:
 M/M/k process with finite capacity
 M/M/k process with ai 
A
i 1
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7.9 Simulating an M/M/k queuing process
8 The distribution of sums of random variables
8.1 Sums of random variables
8.2 Sums of random variables and the CLT
 Theorem 8.2-1. CLT for iid random sample
 Proof
8.3 Confidence intervals for means
8.4 A random sum of random variables
 Definition 8.4-1. Let N(t) be a Poisson process with rate  and Y1, Y2,
… Yn be iid with mean  and SD , independent of N(t). A process
X(t) is a “compound Poisson process” if
N (t )
X (t )   Yi
i 1
 Examples?
 Theorem 8.4-1. For a compound Poisson process X(t),
E[X(t)] = ____ and Var[X(t)] = ________
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STAT 6310, Stochastic Processes
 x = 75.9, medina = 78.5, sd = 5.14


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STAT 6310, Stochastic Processes
Jaimie Kwon
9 Selected systems models
9.1 Distribution of extremes
9.2 Scheduling problems
9.3 Sojourns and transitions for continuous-time Markov
processes
 Definition 9.3-1: a process is a continuous-time Markov process if
the following conditions are satisfied:
i) There exist nonnegative rates ij, i  j, such that for frames of
sufficiently small length ,
P(i  j)  ij ,
P(i  i)  1 –
 
i j
ij
where the approximation errors are negligible for small 
ii) Transitions that occur in nonoverlapping frames are independent
of one another
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 Theorem 9.3-1. Let Pij(t) = P[Markov process in state j at time t
given that it starts in state i]. If the chain can visit every state, then
there exists  = (1, …, S) such that
lim t  Pij (t )   j .
The  satisfies
the matrix equation   = 0 and j j = 1, where  = {ij} with ii = j  i ij.
 Informal Proof:
 A “sojourn time”: the time spent in a state on one visit
 A “conditional transition probability”: the probability of a transition
given that a transition to a new state is made
 Theorem 9.3-2. The sojourn time for a state i in a continuous-time
Markov process ~ Exp(j  i ij). Thus its mean i = 1/(j  i ij).
Given that a transition is made from state i to j, j  i, the transition
occurs according to a Markov chain with one-step transition
probabilities:
pij 
ij
.
 ij'
j 'i
 Theorem 9.3-3. Suppose that the conditional probabilities pij above
define a regular Markov chain and pi the steady-state probability for
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this regular Markov chain. Then
i 
pi i
S
p
i 1
i
i
 Informal proof
 Sojourn times and conditional transition probabilities make the
simulation possible
9.4 Sojourns and transitions for continuous-time semi-Markov
processes
 “Semi-Markov process”: continuous-time process in which the
sojourn time distributions are not exponential but the conditional
transition probabilities follow the Markov chain

9.5 Sojourns and transitions for queuing processes
 The M/M/k queuing process and its variations are continuous-time
Markov processes with
i ,i1  ai
i ,i1  si
ii  ai  si
 Theorem 9.5-1. In M/M/k queuing process, the sojourn time for state
i ~ Exp(ai + si) with mean i = 1/(ai + si). When a transition is made to
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Jaimie Kwon
a new state, it occurs as a Markov chain with
P (i  i  1) 
ai
ai  si
P (i  i  1) 
si
ai  si

10 Reliability models
10.1 The reliability function
 If T denote the time to failure of a system, then the “reliability at time
t” or “reliability function” R(t) is defined to be:
R(t) = P(T > t)
 It’s easy to see that
R(t) = 1 – F(t)
 For a system whose time to failure ~ Exp(), the reliability function
R(t) = _________
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 Theorem 10.1-1. Let a system consist of n components with
reliability functions R1(t), …,Rn(t) and assume they fail independently
of one another. The reliability function of the series system is
RS(t) = ______
The reliability function of the parallel system is
RP(t) = ______
n
 RS (t )   Ri (t )
i 1
 Many systems consist of series and parallel subsystems. The
reliability can be comptued via repeated application of the above
theorem.
10.2 Hazard Rate
 Definition 10.2-1. Let f(t) be the probability density function of T, the
time to failure of a system. Then the “hazard rate” (“failure-rate” or
“intensity function”) is
h(t ) 
f (t )
R(t )
 Motivation
 The hazard rate h(t) can be modeled as:
 Increasing function of t
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STAT 6310, Stochastic Processes
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 Decreasing fucntion of t
 Constant
 Definition 6.4-1. A random variable X ~ Weibull(, ) for ,  > 0, if
its cdf is given by

F ( x)  1  exp  x /  

 for x  0.
The pdf of Weibull is
f(x) = ______

  1 

 
1

 
 2   2 1 
 

2
1 
   2 1  

  
 Hazard rate of Weibull(, ) distribution?
 Consider a parallel system consisting of two parts that fail
independently of each other with Exp(2) lifetime distribution. Find
RP(t) =
F(t) =
f(t) =
h(t) =
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in turn. Draw h(t) and discuss the result.
 Theorem 10.2-1 R(t )  exp   0 h( s)ds 
t


 Example: If h(t) = et, t > 0, then
R(t) = ______
F(t) = ______
f(t) = _______
10.3 Renewal Processes
 Definition 10.3-1. A “renewal process” is a counting process in
which the times between the counted outcomes are iid nonnegative
random variables.
 N(t): the number of “renewals” in an interval of length t
 “renewal times”: T1, T2, …, with E(Ti) =  and SD(Ti) = .
Ti : time between the (i-1)th and ith renewal.
 Define the time of n’th renewal: Sn = T1 + …+ Tn
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 Theorem 10.3-1.
P(Sn  t) = P(N(t)  n)
 The lifetime of a printer ribbon T1 ~ Uniform(1, 3) in months. Note
that N(24) = # of ribbon replacement over 24 months. Find:
P(N(24)  15) =
95% confidence interval of N(24) P(What is the number of ribbon
 CLT on Sn for large n => CLT on N(t) for large t

 Theorem 10.3-2. For large t,
 t t 2 
N (t ) ~ N  N (t ) ,  N2 (t )  N  , 3 
  


approximately.
 Informal proof
10.4 Maintained Systems
 Consider a system that is either up or down, going through up-down
cycle.
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 TU(i) and TD(i) : “uptime” and “downtime” for i’th cycle. Assume both
are iid, continuous random variables.
 Theorem 10.4-1. lim P(system is up at time t) =
U
U   D
where E(TU(i)) = U and E(TD(i)) = D.
 “maintained system”: a system consisting of k components that can
fail independently of one another. Broken components are repaired
on a first-come, first-served basis. Interested in the number of
components that are up at any given time. For this system:
i,i – 1 = ___
i,i + 1 = ___
where : “failure rate” of an individial component
: “service rate”
i,i – 1 = i , i=1, 2,…, k;
i,i + 1 = , i=0, 1,…, k - 1 ;
Behaves like finite-capacity (with at most k customers in the system)
M/M/k system
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Solve balance equations to find the steady state distribution of the
number of “up” components
 The system can be extended to the case when there are c ( k)
technicians working independently of each other.
10.5 Nonhomogeneous Poisson Process and Reliability Growth
 In the Bernoulli counting process (Poisson process), what if the
frame success probability (rate) changes over time?
 Definition 10.5-1. The counting process N(t) is said to be a
nonhomogeneous Poisson process if
i) N(0) = 0
ii) The counts in nonoverlapping intervals are independent.
iii) There exists a differentiable, increasing “mean function” m(t)
such that m(0) = 0 and
N (t )  N ( s) ~ Pois (m(t )  m( s)) for s < t
 The “intensity function”  (t ) 
d
m(t )
dt
 The Poisson process is a special case with m(t) = _______
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STAT 6310, Stochastic Processes
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 The number of failures of a system can be modeled by the
nonhomogeneous Poisson process.
 “reliability growth” = “(t) is decreasing function of t”

 A nonhomogenous Poisson process is called a “Weibull process” if
 (t )  t  1 and
m(t )  t 
 The intensity function (t) increases with t if _____
The intensity function (t) decreases with t if _____ (reliability
growth)
 The time to the first failure T1 ~ Weibull distribution (what
parameters?)
 Unlike the renewal process, the time between the (n – 1)th and n’th
failure depends on when the (n – 1)th failure occurred.
____
 Example: the daily failures of a new equipment is modeled as a
Weibull process with  = 2 and  = .5. Find:
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P(N(10) > 5) = ________
P(N(20) – N(10) > 5) =_______
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STAT 6310, Stochastic Processes
Jaimie Kwon
STAT/MATH 6310, Stochastic Processes, Spring 2006 Course Note
Dr. Jaimie Kwon
March 8, 2016
Table of Contents
1
2
Basic probability .............................................................................3
1.1
Sample spaces and events ......................................................3
1.2
Assignment of probabilities .......................................................3
1.3
Simulation of events on the computer.......................................3
1.4
Counting techniques .................................................................3
1.5
Conditional probability ..............................................................3
1.6
Independent event....................................................................4
Discrete random variables ..............................................................4
2.1
Random variables ....................................................................4
2.2
Joint distributions and independent random variables ..............4
2.3
Expected values .......................................................................4
2.4
Variance and standard deviation ..............................................4
2.5
Sampling and simulation ..........................................................5
2.6
Sample statistics ......................................................................5
2.7
Expected values of jointly distributed random variables and the
law of large numbers .........................................................................5
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4
Jaimie Kwon
2.8
Covariance and correlation .......................................................5
2.9
Conditional expected values .....................................................6
Special discrete random variables ..................................................6
3.1
Binomial random variable .........................................................6
3.2
Geometric and negative binomial random variables .................6
3.3
Hypergeometric random variables ............................................6
3.4
Multinomial random variables ...................................................7
3.5
Poisson random variables ........................................................7
3.6
Moments and moment-generating functions (MGF)..................7
Markov chains ................................................................................8
4.1
Introduction: modeling a simple queuing system ......................8
4.2
The Markov property ................................................................9
4.3
Computing probabilities for Markov chains .............................11
4.4
The simple queuing system revisited ......................................13
4.5
Simulating the behavior of a Markov chain .............................14
4.6
Steady-state probabilities .......................................................14
4.7
Absorbing states and first passage times ...............................17
5
Continuous random variables .......................................................23
6
Special continuous random variables ...........................................23
7
Markov counting and queuing processes......................................25
7.1
Bernoulli counting process .....................................................25
7.2
The Poisson process ..............................................................31
7.3
Exponential random variables and the Poisson process .........33
7.4
The Bernoulli single-server queuing process ..........................33
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7.5
The M/M/1 queuing process ...................................................36
7.6
K-server queuing process .......................................................41
7.7
Balance equations and steady-state probabilities ...................44
7.8
More Markov queuing processes ............................................49
7.9
Simulating an M/M/k queuing process ....................................51
8
The distribution of sums of random variables ...............................51
8.1
Sums of random variables ......................................................51
8.2
Sums of random variables and the CLT .................................51
8.3
Confidence intervals for means ..............................................51
8.4
A random sum of random variables ........................................51
9
Selected systems models .............................................................53
9.1
Distribution of extremes ..........................................................53
9.2
Scheduling problems ..............................................................53
9.3
Sojourns and transitions for continuous-time Markov processes
53
9.4
Sojourns and transitions for continuous-time semi-Markov
processes ........................................................................................55
9.5
10
Sojourns and transitions for queuing processes .....................55
Reliability models ......................................................................56
10.1
The reliability function ..........................................................56
10.2
Hazard Rate ........................................................................57
10.3
Renewal Processes .............................................................59
10.4
Maintained Systems ............................................................60
10.5
Nonhomogeneous Poisson Process and Reliability Growth 62
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