18MAB204T-U1-Common-Lecture Notes
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18MAB204T-Probability and Queueing Theory
Prepared by
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Dr. S. ATHITHAN
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Assistant Professor
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Department of of Mathematics
Faculty of Engineering and Technology
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SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
Kattankulathur-603203, Chengalpattu District.
SRM INSTITUTE OF SCIENCE AND TECHNOLOGY
Kattankulathur-603203, Chengalpattu District.
18MAB204T-Probability and Queueing Theory
S. ATHITHAN
Contents
1
Probability (Review)
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Discrete and Continuous Distributions
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3
Mathematical Expectation
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4
Moments and Moment Generating Function
8
5
Examples
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Function of a Random Variable
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Exercise/Practice/Assignment Problems
33
36
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18MAB204T-Probability and Queueing Theory
1
S. ATHITHAN
Probability (Review)
1.1
Definition of Probability and Basic Theorems
1.1.1
Random Experiment
An Experiment whose outcome or result can be predicted with certainty is called a deterministic
experiment.
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An Experiment whose all possible outcomes may be known in advance, the outcome of a particular performance of the experiment cannot be predicted owing to a number of unknown cases
is called a random experiment.
The Approaches to Define the Probability
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1.1.2
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Definition 1.1.1 (Sample Space). Set of all possible outcomes which are assumed to be equally
likely is called the sample space and is usually denoted by S. Any subset A of S containing
favorable outcomes is called an event.
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Definition 1.1.2 (Mathematical or Apriori Definition of Probability). Let S be a sample space
and A be an event associated with a random experiment. Let n(S) and n(A) be the number of
elements of S and A. Then the probability of event A occurring, denoted by P(A) and defined as
n(S)
Number of cases favorable to A
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=
Exhaustive number of cases in S(Set of all possible cases)
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n(A)
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P (A) =
P (A) = lim
n→∞
n1
n
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Definition 1.1.3 (Statistical or Aposteriori Definition of Probability). Let a random experiment
n1
be repeated n times and let an event A occur n1 times out of n trials. The ratio
is called
n
n1
the relative frequency of the event A. As n increases,
shows a tendency to stabilize and to
n
approach a constant value and is denoted by P(A) is called the probability of the event A. i.e.,
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Definition 1.1.4 (Axiomatic Approach/Definition of Probability). Let S be a sample space and
A be an event associated with a random experiment. Then the probability of the event A is
denoted by P(A) is defined as a real number satisfying the following conditions/axioms:
(i) 0 ≤ P (A) ≤ 1
(ii) P(S)=1
(iii) If A and B are mutually exclusive events, then P (A ∪ B) = P (A) + P (B)
Axiom (iii) can be extended to arbitrary number of events. i.e., If A1 , A2 , . . . , An , . . . be
mutually exclusive, then P (A1 ∪ A2 ∪ A3 · · · ∪ An . . . ) = P (A1 ) + P (A2 ) + P (A2 ) +
P (A3 ) + · · · + P (An ) + . . .
Theorem 1.1. The probability of the impossible event is zero. i.e., if ϕ is a subset of S containing
no event, then P (ϕ) = 0.
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
Theorem 1.2. If A be the complimentary event of A, then P (A) = 1 − P (A) ≤ 1.
Theorem 1.3 (Addition Theorem of Probability). If A and B be any two events, then P (A ∪
B) = P (A) + P (B) − P (A ∩ B) ≤ P (A) + P (B).
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The above Theorem can be extended for 3 events which is given by P (A ∪ B ∪ C) =
P (A) + P (B) + P (C) − P (A ∩ B) − P (B ∩ C) − P (C ∩ A) + P (A ∩ B ∩ C)
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Theorem 1.4. If B ⊂ A, then P (B) ≤ P (A).
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Conditional Probability and Independents Events
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1.2
P (A ∩ B)
P (A)
, provided P (A) ̸= 0
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P (B/A) =
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The conditional probability of an event B, assuming that the event A has happened, is denoted
by P (B/A) and is defined as
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Theorem 1.5 (Product Theorem of Probability). P (A ∩ B) = P (A) · P (B/A)
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The product theorem can be extended to three events such as
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P (A ∩ B ∩ C) = P (A) · P (B/A) · P (C/A and B)
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There are few properties for the conditional distribution. Please go through it in the textbook.
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When two events A and B are independent, we have P (B/A) = P (B) and the Product
Theorem takes the form P (A ∩ B) = P (A) · P (B). The converse of this statement is also.
i.e., If P (A ∩ B) = P (A) · P (B), then the events A and B are said to be independent.
This result can be extended to any number of events. i.e., If A1 , A2 , . . . , An be n no. of
events, then
P (A1 ∩ A2 ∩ A3 ∩ · · · ∩ An ) = P (A1 ) · P (A2 ) · P (A3 ) · · · · P (An )
Theorem 1.6. If the events A and B are independent, then the events A and B (and similarly
A and B, A and B) are also independent.
1.3
Total Probability
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
Theorem 1.7 (Theorem of Total Probability). If B1 , B2 , . . . , Bn be the set of exhaustive and
mutually exclusive events and A is another event associated with(or caused by)Bi , then
P (A) =
n
X
P (Bi )P (A/Bi )
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i=1
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18MAB204T-Probability and Queueing Theory
2
S. ATHITHAN
Discrete and Continuous Distributions
Definition 2.0.1. A Random Variable (abbreviated as RV) is a function that assigns a real
number X(s) to every element s ∈ S, where S is the sample space corresponding to a
random experiment E.
Discrete Random Variable
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If X is a random variable (RV) which can take a finite number or countably infinite number
of values, X is called a discrete RV. When the RV is discrete, the possible values of X may
be assumed as x1 , x2 , . . . , xn , . . . . In the finite case, the list of values terminates and in the
countably infinite case, the list goes upto infinity.
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For example, the number shown when a die is thrown and the number of alpha particles emitted
by a radioactive source are discrete RVs.
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Probability Mass Function
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If X is a discrete RV which can take the values x1 , x2 , x3 , . . . such that P (X = xi ) = pi ,
then pi is called the PROBABILITY FUNCTION or PROBABILITY MASS FUNCTION or POINT
PROBABILITY FUNCTION , provided pi (i = 1, 2, 3, . . . ) satisfy the following conditions:
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(a) pi ≥ 0, for all i, and
X
pi = 1
(b)
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The collection of pairs {xi , pi }, i = 1, 2, 3, . . . is called the probability distribution of the
RV X, which is sometimes displayed in the form of a table as given below:
x1
p1
x2
p2
x3
p3
...
...
xn
pn
...
...
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X = xi
P (X = xi )
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Continuous Random Variable
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If X is a RV which can take all values (i.e., infinite number of values) in an interval, then X is
called a continuous RV.
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For example, the length of time during which a vacuum tube installed in a circuit functions is a
continuous RV.
Probability Density Function
If X is a continuous RV such that
1
1
P x − dx ≤ X ≤ x − dx = f (x)dx
2
2
then f (x) is called the PROBABILITY
fies the following conditions:
DENSITY FUNCTION ( PDF )
of X, provided f (x) satis-
(a) f (x) ≥ 0, for all x ∈ RX (Range space of ) X, and
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
Z
f (x) dx = 1
(b)
RX
Moreover, P (a ≤ X ≤ b) or P (a < X < b) is defined as P (a ≤ X ≤ b) =
Z b
f (x) dx.
a
The curve y = f (x) is called the probability curve of the RV X.
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a
Note : When X is a continuous RV d P (X = a) = P (a < X < a) =
f (x) dx = 0
a
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This means that it is almost impossible that a continuous RV assumes a specific value. Hence
P (a ≤ X ≤ b) = P (a ≤ X < b) = P (a < X ≤ b) = P (a < X < b).
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Cumulative Distribution Function (CDF)
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If X is an RV, discrete or continuous, then P (X ≤ x) is called the CUMULATIVE DISTRI BUTION FUNCTION of X or DISTRIBUTION FUNCTION of X and denoted as F (x). If X is
discrete,
X
pj
F (x) =
x
f (x) dx
−∞
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If X is continuous, F (x)P (−∞ < X ≤ x) =
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Properties of the cdf F(x)
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(i) F (x) is a non-decreasing function of x, i.e., if x1 < x2 , then F (x1 ) ≤ F (x2 )
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(ii) F (−∞) = 0 and F (∞) = 1.
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(iii) If X is a discrete RV taking values x1 , x2 , x3 , . . . where x1 < x2 < x3 < · · · <
xi−1 < xi < . . . then P (X = xi ) = F (xi ) − F (xi−1 ).
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(iv) If X is a continuous RV, then
dx
F (x) = f (x) at all points where F (x) is differen-
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tiable.
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Note : Although we may talk of probability distribution of a continuous RV, it cannot be represented by a table as in the case of a discrete RV. The probability distribution of a continuous
RV is said to be known, if either its pdf or cdf is given.
For the problems on Discrete and Continuous Random Variables (RV’s) we have to note down
the following points from the table for understanding.
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18MAB204T-Probability and Queueing Theory
Discrete ♯ RV
(Probability Distribution Function)
S. ATHITHAN
Continuous ♯ RV
(Probability Density Function)
Z
Σ
Operator
1, 2, 3, . . .
0<x<∞
Standard Notation
P (X = x)
f (x)
F (X = x) = P (X ≤ x)
F (X = x) = P (X ≤ x)
Total Probability
Z∞
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∞
X
P (X = x) = 1
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CDF∗
f (x)dx = 1
−∞
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RV-Random Variable, *CDF-Cumulative Distribution Function or Distribution Function
Mathematical Expectation
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−∞
♯
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Takes Values of the form (Eg.)
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Definition 3.0.1 (Expectation). An average of a probability distribution of a random variable
X is called the expectation or the expected value or mathematical expectation of X and is
denoted by E(X).
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Definition 3.0.2 (Expectation for Discrete Random Variable). Let X be a discrete random variable taking values x1 , x2 , x3 , . . . with probabilities P (X = x1 ) = p(x1 ), P (X = x2 ) =
∞
X
xi p(xi ),
p(x2 ), P (X = x3 ) = p(x3 ), . . . , the expected value of X is defined as E(X) =
i=−∞
E
if the right hand side sum exists.
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Definition 3.0.3 (Expectation for Continuous Random Variable). Let X be a continuous random variable with pdf f (x) defined in (−∞, ∞), then the expected value of X is defined as
Z∞
E(X) =
xf (x)dx.
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−∞
Note: E(X) is called the mean of the distribution or mean of X and is denoted by X or µ.
3.1
Properties of Expectation
1. If c is any constant, then E(c) = c.
2. If a, b are constants, then E(aX + b) = aE(X) + b.
3. If (X, Y ) is two dimensional random variable, then E(X + Y ) = E(X) + E(Y ).
4. If (X, Y ) is a two dimensional random variable, then E(XY ) = E(X)E(Y ) only if
X and Y are independent random variables.
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18MAB204T-Probability and Queueing Theory
3.2
S. ATHITHAN
Variance and its properties
Definition 3.2.1. Let X be a random variable with mean E(X), then the variance of X is
2
defined as σ 2 = E{(X − µ)2 } and it is denoted by V ar(X) or σX
.
Note: V ar(X) = E(X 2 ) − [E(X)]2 .
Properties of Variance
1. V ar(a) = 0, where a is any constant.
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2. V ar(aX) = a2 V ar(X)
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3. V ar(aX ± b) = a2 V ar(X)
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4. V ar(aX + bY ) = a2 V ar(X) + b2 V ar(Y ) only if X and Y are independent
random variables.
Moments and Moment Generating Function
4.1
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Moments
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Definition 4.1.1 (Moments about origin). The r th moment of a random variable X about the
origin is defined as E(X r ) and is denoted by µ′r . Moments about origin are known as raw
moments.
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 n
X



xr P (X = x) if X is discrete



x=0


Z∞




xr f (x)dx if X is continuous



x=0
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µ′r = E(X r ) =
Note:By moments we mean the moments about origin or raw moments.
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The first four moments about the origin are given by
1. µ′1 = E(X)=Mean
2. µ′2 = E(X 2 )
3. µ′3 = E(X 3 )
4. µ′4 = E(X 4 )
2
Note: V ar(X) = E(X 2 ) − [E(X)]2 =pµ′2 − µ′1 =Second moment - square of the first
moment. Standard Deviation SD = σX = V ar(X).
Definition 4.1.2 (Moments about mean or Central moments). The r th moment of a random
variable X about the mean µ is defined as E[(X − µ)r ] and is denoted by µr .
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
The first four moments about the mean are given by
1. µ1 = E(X − µ) = E(X) − E(µ) = µ − µ = 0
2. µ2 = E[(X − µ)2 ] = V ar(X)
3. µ3 = E[(X − µ)3 ]
4. µ4 = E[(X − µ)4 ]
Definition 4.1.3 (Moments about any point a). The r th moment of a random variable X about
any point a is defined as E[(X − a)r ] and we denote it by m′r .
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The first four moments about a point ‘a’ are given by
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1. m′1 = E(X − a) = E(X) − a = µ − a
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2. m′2 = E[(X − a)2 ]
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3. m′3 = E[(X − a)3 ]
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4. m′4 = E[(X − a)4 ]
Relation between moments about the mean and moments about any arbitrary point a
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Let µr be the r th moment about mean and m′r be the r th moment about any point a. Let µ be
the mean of X.
E[(X − µ)r ]
E[(X − a) − (µ − a)]r
E[(X − a) − m′1 ]r
E (X − a)r − r C1 (X − a)r−1 m′1 + r C2 (X − a)r−2 (m′1 )2 − · · · + (−1)r (m′1 )r
E(X − a)r − r C1 E(X − a)r−1 m′1 + r C2 E(X − a)r−2 (m′1 )2
−r C3 E(X − a)r−3 (m′1 )3 + r C4 E(X − a)r−4 (m′1 )4 − · · · + (−1)r (m′1 )r
= m′r − r C1 m′r−1 m′1 + r C2 m′r−2 (m′1 )2 − r C3 m′r−3 (m′1 )3 + r C4 m′r−4 (m′1 )4
− · · · + (−1)r (m′1 )r
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∴ µr =
=
=
=
=
We define(fix) m′0 = 1, then we have
µ1 = m′1 − m′0 m′1 = 0
µ2 = m′2 − 2 C1 m′1 · m′1 + (m′1 )2
= m′2 − (m′1 )2
µ3 = m′3 − 3 C1 m′2 · m′1 + 3 C2 m′1 · (m′1 )2 − (m′1 )3 ·
= m′3 − 3m′2 · m′1 + 2(m′1 )3
µ4 = m′4 − 4 C1 m′3 · m′1 + 4 C2 m′2 · (m′1 )2 − 4 C3 m′1 · (m′1 )3 + (m′1 )4 ·
= m′4 − 4m′3 · m′1 + 6m′2 · (m′1 )2 − 3(m′1 )4
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18MAB204T-Probability and Queueing Theory
4.2
S. ATHITHAN
Moment Generating Function (MGF)
Definition 4.2.1 (Moment Generating Function (MGF)). The moment generating function of a
random variable X is defined as E(etX ) for allt ∈ (−∞, ∞). It is denoted by M (t) or
MX (t).
MX (t) = E(etX ) =

Z∞




etx f (x)dx if X is continuous



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x=0
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 n
X



etx P (X = x) if X is discrete




x=0
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If X is a random variable. Its MGF MX (t) is given by
tX
(tX)2
(tX)3
(tX)r
tX
MX (t) = E(e ) = E 1 +
+
+
+ ··· +
+ ...
1!
2!
3!
r!
(t)2
(t)3
(t)r
t
E(X 2 ) +
E(X 3 ) + · · · +
E(X r ) + . . .
= 1 + E(X) +
1!
2!
3!
r!
(t)2 ′
(t)3 ′
(t)r ′
t
µ2 +
µ3 + · · · +
µ + ...
= 1 + µ′1 +
1!
2!
3!
r! r
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i.e.
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µ′1 = Coefft. of t in the expansion of MX (t)
t2
µ′2 = Coefft. of
in the expansion of MX (t)
2!
..
.
tr
µ′r = Coefft. of
in the expansion of MX (t)
r!
..
.
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∴ MX (t) generates all the moments about the origin. (That is why we call it as Moment
Generating Function (MGF)).
Another phenomenon which we often use to find the moments is given below:
(r)
µ′r = MX (0)
We have
MX (t) = E(etX ) = 1 +
t
1!
µ′1 +
(t)2
2!
µ′2 +
(t)3
3!
µ′3 + · · · +
(t)r
r!
µ′r + . . .
Differentiate with respect to t, we get
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18MAB204T-Probability and Queueing Theory
′
MX
(t)
=
µ′1
+
tµ′2
+
′′
MX
(t) = µ′2 + tµ′3 +
(t)2
2!
(t)2
2!
µ′3
+ ··· +
µ′4 + · · · +
(t)r−1
(r − 1)!
(t)r−2
(r − 2)!
S. ATHITHAN
µ′r + . . .
µ′r + . . .
S
..
.
(r)
MX (t) = µ′r + terms of higher powers of t
..
.
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Putting t = 0, we get
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′
MX
(0) = µ′1
′′
MX
(0) = µ′2
..
.
(r)
MX (0) = µ′r
..
.
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′
MX
(0)
1!
+
t2
2!
′′
MX
(0) + . . .
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MX (t) = MX (0) +
t
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The Maclaurin’s series expansion given below will give all the moments.
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The MGF of X about its mean µ is MX−µ (t) = E et(X−µ) .
Similarly, the MGF of X about any point a is MX−a (t) = E et(X−a) .
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Properties of MGF
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1. McX (t) = MX (ct)
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2. MX+c (t) = ect MX (t)
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3. MaX+b (t) = ebt MX (at)
4. If X and Y are independent RV’s then, MX+Y (t) = MX (t) · MY (t).
5
Examples
Example: 5.1. From 6 positive and 8 negative numbers, 4 are chosen at random(without replacement) and multiplied. What is the probability that the product is negative?
Solution: If the product is negative(-ve), any one number is -ve (or) any 3 numbers are -ve.
No. of ways choosing 1 negative no. = 6 C3 ·8 C1 = 20 · 8 = 160.
No. of ways choosing 3 negative nos. = 6 C1 ·8 C3 = 6 · 56 = 336.
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
Total no. of ways choosing 4 nos. out of 14 nos. =14 C4 = 1001.
∴ P (the product is -ve) =
160 + 336
1001
=
496
1001
.
Example: 5.2. A box contains 6 bad and 4 good apples. Two are drawn out from the box at a
time. One of them is tested and found to be good. What is the probability that the other one is
also good.
Aliter: Probability of the first drawn apple is good=
C2 /10 C2
4
10
4
6/45
=
A
P (A)
4
=
4/10
=
1
3
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P (A ∩ B)
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∴ The required probability is P (B/A) =
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Solution: Let A be an event that the first drawn apple is good and let B be an event that the
other one is also good.
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.
10
Once the first apple is drawn, the box contains only 9 apples in total and 3 good apples.
1
AT
3
9
=
3
.
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∴ Probability of the second drawn apple is also good=
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S
Example: 5.3. A random variable X has the following distribution
x
0 1 2
3
4
5
6
7
2
2
2
P[X=x] 0 k 2k 2k 3k k
2k
7k + k
Find (i) the value of k,(ii) the Cumulative Distribution Function (CDF) (iii) P (1.5 < X <
1
4.5/X > 2) and (iv) the smallest value of α for which P (X ≤ α) > .
2
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Solution:
P [X = xi ] = 1.
i
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Here,
X
C
(i) We know that
P [X = 0] + P [X = 1] + P [X = 2] + P [X = 3]+
P [X = 4] + P [X = 5] + P [X = 6] + P [X = 7]
i.e. 0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k
i.e. 10k2 + 9k − 1
i.e. (10k − 1)(k + 1)
1
=⇒ k =
or k = −1
10
Since the value of the probability is not negative, we take the value k =
=
=
=
=
1
10
1
1
0
0
.
∴ The probability distribution is given by
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18MAB204T-Probability and Queueing Theory
x
P [X = x]
0
0
P [X = x]
0
1
k
1
2
2k
2
3
2k
2
4
3k
3
10 10 10 10
(ii) The Cumulative Distribution Function (CDF)
S. ATHITHAN
5
k2
1
6
2k2
2
7
7k + k
17
100
100
100
2
F (x) = P (X ≤ x)
F (0) = P (X ≤ 0) = P (X = 0) = 0
1
10
1
=
10
1
S
F (1) = P (X ≤ 1) = P (X = 0) + P (X = 1) = 0 +
F (2) = P (X ≤ 1) = P (X = 0) + P (X = 1) + P (X = 2) = 0 +
+
2
3
10
AT
H
IT
H
A
N
10
10
..
.
F (7) = P (X ≤ 7) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 2)
+P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7)
100
=
=1
100
=
0
1
2
3
P [X = x]
0
k
2k
S
P [X = x]
0
F [X = x]
2k
TE
10
6
7
3k
k2
2k2
7k2 + k
2
2
3
1
2
17
10
3
10
5
10
8
100
81
100
83
10
10
10
100
100
100
100
=1
100
O
10
1
5
E
0
1
4
O
x
N
F
The detailed CDF F (X = x) is given in the table.
TU
R
(iii)
P [(1.5 < X < 4.5) ∩ (X > 2)]
P (X > 2)
C
P (1.5 < X < 4.5/X > 2) =
LE
P (1.5 < X < 4.5) = P (X = 3) + P (X = 4) =
10
P (X > 2) = 1 − P (X ≤ 2)
= 1 − {P (X = 0) + P (X = 1) + P (X = 2)}
3
7
= 1−
=
10
10
P (1.5 < X < 4.5/X > 2) =
5
10
7
10
=
5
7
(iv) The smallest value of α for which P (X ≤ α) >
Page 13 of 38
5
1
2
= 0.5.
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
5
From the CDF table, we found that P (X ≤ 3) =
= 0.5. But we need the probability
10
8
= 0.8 > 0.5. ∴ α = 4 satisfies
which is more than 0.5, for this we have P (X ≤ 4) =
10
the given condition.
Solution:
P [X = x]
2
5k
5
3
7k
7
4
9k
9
5
11k
11
6
13k
13
81
81
81
81
81
81
81
81
81
81
H
3
7k
7
4
9k
9
5
11k
11
6
13k
13
7
15k
15
8
17k
17
81
81
=1
81
O
TE
2
5k
5
81
9
81
16
81
25
81
36
81
49
81
64
81 81
(iii) P (0 < X < 3/X > 2)
81
81
81
81
81
81
TU
R
E
81
4
F [X = x]
81
1
1
3k
3
N
P [X = x]
0
k
1
S
(ii) the Distribution Function (CDF)
x
P [X = x]
8
17k
17
F
1
7
15k
15
IT
1
3k
3
O
(i) The value of k =
0
k
1
AT
x
P [X = x]
H
A
N
S
Example: 5.4. A random variable X has the following distribution
x
0 1
2
3
4
5
6
7
8
P[X=x] k 3k 5k 7k 9k 11k 13k 15k 17k
Find (i) the value of k,(ii) the Distribution Function (CDF) (iii) P (0 < X < 3/X > 2) and
1
(iv) the smallest value of α for which P (X ≤ α) > .
2
LE
C
P (0 < X < 3/X > 2) =
P [(0 < X < 3) ∩ (X > 2)]
P (X > 2)
(iv) The smallest value of α for which P (X ≤ α) >
From the CDF table, we found that P (X ≤ 5) =
1
2
36
= 0(∵ there is no common elemen
= 0.5.
< 0.5. But we need the probability
81
49
> 0.5. ∴ α = 6 satisfies the
which is more than 0.5, for this we have P (X ≤ 6) =
81
given condition.
Page 14 of 38
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18MAB204T-Probability and Queueing Theory

kx,



k,
Example: 5.5. Find the value of k for the pdf f (x) =

3k − kx,



0,
find (a) the CDF of X (b) P (1.5 < X < 3.2/0.5 < X < 1.8)
S. ATHITHAN
when 0 ≤ x ≤ 1
when 1 ≤ x ≤ 2
. Also
when 2 ≤ x ≤ 3
otherwise
Solution:
f (x)dx = 1
Z0
Z1
0 · dx +
i.e.
1
−∞
2
Z2
kx · dx +
3
Z3
k · dx +
1
f (x)dx = 1
Z∞
(3k − kx)dx +
2
0 · dx = 1
3
2
+k
[x]21
+ 3k · x − k ·
S
1
0
x2
2
3
+0 = 1
2
i.e. 2k = 1
N
O
i.e. 0 + k
x2
TE
O
F
0
f (x)dx +
H
0
f (x)dx +
AT
−∞
f (x)dx +
Z∞
Z3
H
f (x)dx +
i.e.
Z2
IT
Z1
A
−∞
N
We know that
Z0
S
Z∞
k=
1
2
E
=⇒
LE
C
TU
R
x

,



2




1



,
1
(a) Since k = , f (x) = 2

2

1


(3 − x),



2




0,
Page 15 of 38
when 0 ≤ x ≤ 1
when 1 ≤ x ≤ 2
when 2 ≤ x ≤ 3
otherwise
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
Zx
The CDF of X is F (x) = P (X ≤ x) =
f (x)dx
−∞
When x < 0, F (x) = 0, since f (x) = 0 for x < 0
Zx
When 0 ≤ x < 1, F (x) =
f (x)dx
−∞
=
Zx
f (x)dx =
2
dx =
A
H
f (x)dx
dx +
TE
x
O
N
=
2
E
R
0
Zx
dx +
1
2
Zx
dx +
0
0
TU
(2x − 1)
S
f (x)dx
Z1
4
O
Zx
0
2
dx =
F
0
When 2 ≤ x < 3, F (x) =
1
1
AT
2
H
=
Zx
x
IT
0
Z1
4
0
Zx
When 1 ≤ x < 2, F (x) =
x2
N
0
x
S
Zx
1
2
(3 − x) dx =
1
4
(−x2 + 6x − 5)
0
When x ≥ 3, F (x) = 1
whenx < 0
LE
C

0







x2


,



2



1
F (x) =
(2x − 1),

4





1


(−x2 + 6x − 5),



4




1,
∴
when 0 ≤ x ≤ 1
when 1 ≤ x ≤ 2
when 2 ≤ x ≤ 3
when x ≥ 3
(b) P (1.5 < X < 3.2/0.5 < X < 1.8) =
Page 16 of 38
P [(1.5 < X < 3.2) ∩ (0.5 < X < 1.8)]
P (0.5 < X < 1.8)
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
P [(1.5 < X < 3.2) ∩ (0.5 < X < 1.8)] = P (1.5 < X < 1.8)
Z1.8
1
3
=
dx =
2
20
1.5
Z1
x
P (0.5 < X < 1.8) =
2
Z1.8
dx +
0.5
1
2
dx =
3
16
1
S
P [(1.5 < X < 3.2) ∩ (0.5 < X < 1.8)]
P (1.5 < X < 3.2/0.5 < X < 1.8) =
N
P (0.5 < X < 1.8)
A
16
=
AT
H
IT
H
20
TE
S
O
F
Example: 5.6. Experience has shown that while walking in a certain park, the time X (in
minutes) duration between seeing two people smoking has a density function
(
kxe−x , when x > 0
f (x) =
0,
otherwise
N
O
Find the (a) value of k (b) distribution function of X (c) What is the probability that a person,
who has just seen a person smoking will see another person smoking in 2 to 5 minutes, in at
least 7 minutes?
C
TU
R
E
Solution: Given the pdf of X is
LE
∴ f (x) ≥ 0 ∀ x
=⇒
(
kxe−x ,
f (x) =
0,
when x > 0
otherwise
k > 0.
Z∞
f (x) dx = 1
(a) We know that
−∞
Z∞
i.e.
kxe−x dx = 1
0
i.e. k x
e−x
−1
−
e−x
∞
(−1)2
0
= 1
=⇒ k = 1
Page 17 of 38
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18MAB204T-Probability and Queueing Theory
(
xe−x ,
∴ f (x) =
0,
S. ATHITHAN
when x > 0
otherwise
(b) The distribution function (CDF) is given by
Zx
F (x) = P (X ≤ x) =
f (x) dx
Here we have x > 0, F (x) =
xe−x dx = 1 − (1 + x)e−x , x > 0
N
Zx
S
−∞
A
0
H
(c)
Z5
O
2
xe−x dx
F
=
AT
2
H
f (x) dx
P (2 < X < 5) =
IT
Z5
TE
S
= −6e−5 + 3e−2
= −0.04 + 0.406 = 0.366
Z∞
P (X ≥ 7) =
TU
C
LE
Page 18 of 38
f (x) dx
7
R
E
N
O
and
Z∞
=
xe−x dx
7
= 8e−7
= 0.007
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
Example: 5.7. The sales of convenience store on a randomly selected day are X thousand
dollars, where X is a random variable with a distribution function


0,
when x < 0



2

x

,
when 0 ≤ x < 1
F (x) =
2


k(4x − x2 ) − 1, when 1 ≤ x < 2



1,
when x ≥ 2
A
N
S
Suppose that this convenience store’s total sales on any given day is less than 2000 units(Dollars
or Pounds or Rupees). Find the (a) value of k (b) Let A and B be the events that tomorrow the
store’s total sales between 500 and 1500 units respectively. Find P (A) and P (B). (c) Are A
and B are independent events?
H
IT
when x < 0
when 0 ≤ x < 1
when 1 ≤ x < 2
when x ≥ 2
F
AT

0,




d
x,
F (x) =
f (x) = F ′ (x) =

dx
k(4 − 2x),



0,
H
Solution: The pdf of X is given by
TE
S
O
Since f (x) is a pdf,
Z∞
O
Z1
f (x)dx +
R
TU
−∞
Z0
LE
C
i.e.
−∞
0
2
Z∞
k(4 − 2x) dx +
0
1
x2
2
1
0 · dx = 1
2
2
+ k 4x − x2 1 + 0 = 1
0
=⇒ k =

0,



x,
f (x) =

(2 − x),



0,
Page 19 of 38
f (x)dx = 1
Z2
x dx +
i.e. 0 +
f (x)dx +
1
Z1
0 · dx +
Z∞
Z2
f (x)dx +
E
i.e.
−∞
N
Z0
f (x)dx = 1
we have
1
2
when x < 0
when 0 ≤ x < 1
when 1 ≤ x < 2
when x ≥ 2
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
Since the total sales X is in thousands of units, the sales between 500 and 1500 units is the
1
3
event A which stands for
= 0.5 < X <
= 1.5 and the sales over 1000 units is the
2
2
event B which stands for X > 1. =⇒ A ∩ B = 1 < X < 1.5
Now
Z1.5
P (A) = P (0.5 < X < 1.5) =
f (x)dx
(2 − x) dx =
x dx +
4
1
H
A
0.5
3
N
=
Z1.5
S
0.5
Z1
f (x)dx
H
P (B) = P (X > 1) =
IT
Z2
AT
1
Z2
(2 − x) dx =
=
2
O
F
1
1
1
O
TE
S
Z1.5
P (A ∩ B) = P (1 < X < 1.5) =
f (x)dx
1
R
E
N
Z1.5
3
=
(2 − x) dx =
8
LE
C
TU
The condition for independent events: P (A) · P (B) = P (A ∩ B)
3
3 1
Here, P (A) · P (B) = · = = P (A ∩ B)
4 2
8
∴ A and B are independent events.
Example: 5.8. If a random variable X has the following probability distribution, find
x
−1 0
1
2
E(X), E(X 2 ), V ar(X), E(2X + 1), V ar(2X + 1).
p(x) 0.3 0.1 0.4 0.2
Solution: Here X is a discrete RV. ∴
∞
X
E(X) =
xi p(xi )
i=−∞
= (−1) × 0.3 + 0 × 0.1 + 1 × 0.4 + 2 × 0.2
= −0.3 + 0 + 0.4 + 0.4 = 0.5
Page 20 of 38
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18MAB204T-Probability and Queueing Theory
∞
X
2
E(X ) =
S. ATHITHAN
x2i p(xi )
i=−∞
2
= (−1) × 0.3 + 02 × 0.1 + 12 × 0.4 + 22 × 0.2
= 0.3 + 0 + 0.4 + 0.8 = 1.5
S
V ar(X) = E(X 2 ) − [E(X)]2
= (1.5) − (0.5)2
H
IT
E(2X + 1) = 2E(X) + 1
= 2 × (0.5) + 1
H
A
N
= 1.5 − 0.25 = 1.25
AT
= 1+1= 2
F
V ar(2X + 1) = 22 V ar(X)
N
random variable X
E(X 2 ), V ar(X),
3
4
7
9
has the following probability disE(3X − 4), V ar(3X − 4).
TU
R
E
Example:
5.9. If a
tribution, find E(X),
x
0
1
2
1
3
5
p(x)
25 25 25
O
TE
S
O
= 4 × (1.25) = 5
25
25
LE
C
Solution: Here X is a discrete RV. ∴
E(X) =
X
xi p(xi )
i
=
E(X 2 ) =
14
5
X
x2i p(xi )
i=
=
Page 21 of 38
46
5
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
V ar(X) = E(X 2 ) − [E(X)]2
2
46
14
=
−
5
5
34
=
25
E(3X − 4) = 3E(X) − 4
S
22
5
N
= =
H
IT
306
25
AT
H
=
A
V ar(3X − 4) = 32 V ar(X)
O
TE
S
O
F
Example: 5.10. A test engineer found that the distribution function of the lifetime X (in years)
of an equipment follows a distribution function which is given by
(
0,
when x < 0
F (x) =
−x
1 − e 5 , when x ≥ 0
N
Find the pdf, mean and variance of X.
E
Solution: The pdf of X is given by
when x < 0
when x ≥ 0
LE
C
TU
R

0,
d
f (x) = F ′ (x) =
F = 1 −x
 e 5,
dx
5
Z∞
The Mean E(X) =
xf (x)dx
−∞
Z0
=
1 x
x e− 5 dx
5
−∞
= −
1
5
x
(5x + 25)e− 5
∞
0
= 5
V ar(X) = E(X 2 ) − [(X)]2
Page 22 of 38
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18MAB204T-Probability and Queueing Theory
Z∞
Now E(X 2 ) =
S. ATHITHAN
x2 f (x)dx
−∞
Z0
=
1 x
x2 e− 5 dx
5
−∞
V ar(X) = 50 − [5]2 = 25
H
A
N
∴
1
x ∞
(5x2 + 50x + 250)e− 5 0 = 50
5
S
= −
AT
H
IT
Example: 5.11. Find the mean and standard deviation of the distribution
(
kx(2 − x), when 0 ≤ x ≤ 2
f (x) =
0,
otherwise
TE
S
O
F
Solution: Given that the continuous RV X whose pdf is given by
(
kx(2 − x), when 0 ≤ x ≤ 2
f (x) =
0,
otherwise
N
O
Since f (x) is a pdf,
Z∞
f (x)dx = 1
E
we have
TU
R
−∞
Z0
f (x)dx +
LE
C
i.e.
−∞
−∞
f (x)dx +
0
Z0
i.e.
Z∞
Z2
2
Z∞
Z2
0 · dx +
kx(2 − x) dx +
0
0 · dx = 1
2
2
i.e. 0 + k x −
x3
3
2
+0 = 1
0
=⇒ k =
Page 23 of 38
f (x)dx = 1
3
4
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
Z∞
The Mean E(X) =
xf (x)dx
−∞
Z0
3
x x(2 − x)dx
4
=
−∞
3
2
x
x4
= 1
−
= − 2
4
3
4 0
N
H
3
x2 x(2 − x)dx
4
F
=
AT
−∞
Z0
O
−∞
3
N
V ar(X) =
6
5
4
− [1]2 =
2
x4
1
5
4
−
x5
5
2
=
0
6
5
r
=⇒
S.D. =
1
5
TU
R
E
∴
O
TE
S
= −
IT
x2 f (x)dx
H
Now E(X 2 ) =
A
V ar(X) = E(X 2 ) − [(X)]2
Z∞
S
3
LE
C
Example: 5.12. If X has the distribution function

0, when x < 1






1


, when 1 ≤ x < 4



3


1
, when 4 ≤ x < 6
F (x) =
2




5



, when 6 ≤ x < 10


6




1, when x ≥ 10
Find (a) the probability distribution of X (b) Find the mean and variance of X.
Page 24 of 38
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18MAB204T-Probability and Queueing Theory
Solution: Given that the CDF of X is

0,






1


,



3


1
,
F (x) =
2




5



,


6




1,
S. ATHITHAN
when x < 1
when 1 ≤ x < 4
when 4 ≤ x < 6
when 6 ≤ x < 10
N
S
when x ≥ 10
H
A
We know that P (X = xi ) = F (xi ) − F (xi−1 ), i = 1, 2, 3, . . . , where F is constant in
xi−1 ≤ x ≤ xi .
AT
H
IT
The CDF changes its values at x = 1, 4, 6, 10. ∴ The probability distribution takes its values
as follows:
O
F
P (X = 1) = F (1) − F (0) =
TE
S
P (X = 4) = F (4) − F (1) =
3
1
2
5
6
N
O
P (X = 6) = F (6) − F (4) =
1
−0=
−
−
1
3
1
2
5
6
3
=
=
=
1
6
1
3
1
6
TU
R
E
P (X = 10) = F (10) − F (6) = 1 −
1
x
p(x)
4
1
6
1
10
1
3
6
3
6
LE
C
∴ The probability distribution is given by
1
1
14
Mean=E(X) = Σxp(x) =
.
3
95
E(X 2 ) = Σx2 p(x) =
.
3
285
196
89
V ar(X) = E(X 2 ) − [E(X)]2 =
−
=
.
9
9
9
Page 25 of 38
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
N
S
Example: 5.13. If X has the distribution function

0, when x < 0





1


, when 0 ≤ x < 2



6


1
, when 2 ≤ x < 4
F (x) =
2




5



, when 4 ≤ x < 6


8




1, when x ≥ 6
A
Find (a) the probability distribution of X (b) Find the mean and variance of X.
H
IT
when x < 0
AT
when 0 ≤ x < 2
F
when 2 ≤ x < 4
O
when 4 ≤ x < 6
when x ≥ 6
O
TE
S

0,






1


,



6


1
,
F (x) =
2




5



,


8




1,
H
Solution: Given that the CDF of X is
N
We know that P (X = xi ) = F (xi ) − F (xi−1 ), i = 1, 2, 3, . . . , where F is constant in
xi−1 ≤ x ≤ xi .
LE
C
TU
R
E
The CDF changes its values at x = 1, 4, 6, 10. ∴ The probability distribution takes its values
as follows:
P (X = 0) =
1
6
P (X = 2) = F (2) − F (0) =
P (X = 4) = F (4) − F (2) =
1
2
5
8
−
−
P (X = 6) = F (6) − F (4) = 1 −
Page 26 of 38
1
6
1
2
5
8
=
=
=
1
3
1
8
3
8
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18MAB204T-Probability and Queueing Theory
x
∴ The probability distribution is given by
p(x)
37
.
Mean=E(X) = Σxp(x) =
12
101
E(X 2 ) = Σx2 p(x) =
.
6
2
2
6
−
37
2
1
4
1
6
3
6
3
8
8
2
12
= 16.83 − 9.51 = 7.32.
A
N
S
V ar(X) = E(X ) − [E(X)] =
101
0
1
S. ATHITHAN
IT
H
Example: 5.14. The first four moments about X = 5 are 2,5,12 and 48. Find the first four
central moments.
AT
H
Solution: Let m′1 , m′2 , m′3 , m′4 be the first four moments about X = 5. Given m′1 =
1, m′2 = 5, m′3 = 12, m′4 = 48 be the first four moments about X = 5 m′1 = E(X −
5) = 1 =⇒ E(X) − 5 = 1 =⇒ E(X) = µ1 = 6
E
N
O
TE
S
O
F
µ2 = m′2 − (m′1 )2 = 5 − 1 = 4
µ3 = m′3 − 3m′2 · m′1 + 2(m′1 )3
= 12 − 3 × 5 × 1 + 2(1) = −1
µ4 = m′4 − 4m′3 · m′1 + 6m′2 · (m′1 )2 − 3(m′1 )4
= 48 − 4(12)(1) + 6(5)(1) − 3(1) = 27
TU
R
Example: 5.15. The first four moments about x = 4 are 1,4,10 and 45. Find the first four
moments about the mean.
Solution:
LE
C
Let m′1 , m′2 , m′3 , m′4 be the first four moments about X = 4. Given m′1 = 1, m′2 =
4, m′3 = 10, m′4 = 45 be the first four moments about X = 4 m′1 = E(X − 4) =
1 =⇒ E(X) − 4 = 1 =⇒ E(X) = µ1 = 5
µ2 = m′2 − (m′1 )2 = 4 − 1 = 3
µ3 = m′3 − 3m′2 · m′1 + 2(m′1 )3
= 10 − 3 × 4 × 1 + 2(1) = 0
µ4 = m′4 − 4m′3 · m′1 + 6m′2 · (m′1 )2 − 3(m′1 )4
= 45 − 4(10)(1) + 6(4)(1) − 3(1) = 26
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
Example: 5.16. A random variable X has the pdf f (x) = kx2 e−x , x ≥ 0. Find the r th
moment and hence find the first four moments.
Solution: We know that
Z∞
f (x) dx = 1
−∞
Z∞
i.e.,
kx2 e−x dx = 1
e−x x3−1 dx = 1
N
i.e., k
S
0
Z∞
∵ Γ(n) =
H
i.e., kΓ(3) = 1

e−x xn−1 dx
IT
Z∞

A
0
.
[∵ Γ(n) = (n − 1)!]
O
F
Now, The r th moment is given by
2
AT
i.e., k · 2! = 1 =⇒ k =
H
0
1
−∞
Z∞
=
E
N
O
TE
S
µ′r = E(X r )
Z∞
=
xr f (x) dx
LE
C
TU
R
0
Z∞
= k
e−x xr+3−1 dx
0
=
Now, First Moment µ′1 =
Second Moment µ′2 =
Third Moment µ′3 =
Fourth Moment µ′4 =
Page 28 of 38
xr kx2 e−x dx
1
2
1
2
1
2
1
2
1
2
Γ(r + 3) =
1
2
(r + 2)!
(3)! = 3
(4)! = 12
(5)! = 60
(6)! = 360
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18MAB204T-Probability and Queueing Theory
Example: 5.17. A random variable X has the pdf f (x) =
1
S. ATHITHAN
x
e− 2 , x ≥ 0. Find the
2
MGF(Moment Generating Function) and hence find its mean and variance.
Solution: The MGF of X is given by
MX (t) = E(etX ) =
Z∞
etx f (x) dx
−∞
Z∞
2
x
etx e− 2 dx
S
=
1
1
2
1
e− 2 (1−2t)x dx
A
=
N
0
Z∞
∴ MX (t) =
0
1
1 − 2t
1
IT
2 − 12 (1 − 2t)
H
#∞
e− 2 (1−2t)x
AT
=
1
1
H
0
"
if t <
2
.
O
F
= (1 − 2t)−1 = 1 + 2t + 4t2 + 8t3 + . . .
TE
S
Now, Differentiating w.r.to t, we get
=
=
=
=
=
2 + 8t + 24t2 + . . .
8 + 48t + . . .
′
(0) = 2
MX
′′
MX (0) = 8
E(X 2 ) − [E(X)]2 = 8 − 4 = 4.
LE
C
TU
R
E
N
O
′
MX
(t)
′′
MX (t)
Now, First Moment=Mean= E(X) = µ′1
Second Moment= E(X 2 ) = µ′2
∴ V ar(X)
1
, x = 1, 2, 3, . . . . Find the
2x
MGF(Moment Generating Function) and hence find its mean and variance.
Example: 5.18. A random variable X has the pmf p(x) =
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
Solution: The MGF of X is given by
tX
MX (t) = E(e
∞
X
) =
=
etx p(x)
x=−∞
∞
X
tx
e
∞ t x
X
e
1
∞ t x
X
e
=
=
2x
2
x=1
x=1
"
#
t t 2
e
e
+
1+
+ ...
2
2
t −1
e
1−
2
2
N
2
et
2−
et
if et ̸= 2.
Now, Differentiating w.r.to t, we get
2et
O
F
′
MX
(t) =
AT
H
IT
∴ MX (t) =
et
A
=
2
H
=
et
S
x=1
TE
S
′′
MX
(t) =
(2 − et )2
2[(2 − et )et + 2e2t ]
(2 − et )3
TU
R
E
N
O
′
(0) = 2
Now, First Moment=Mean= E(X) = µ′1 = MX
2
′
′′
Second Moment= E(X ) = µ2 = MX (0) = 6
∴ V ar(X) = E(X 2 ) − [E(X)]2 = 6 − 4 = 2.
LE
C
Example: 5.19. A random variable X has the r th moment of the form µ′r = (r + 1)!2r . Find
the MGF(Moment Generating Function) and hence find its mean and variance.
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
Solution: The MGF of X is given by
t
1!
+
2+
t2
2!
t2
2!
µ′2
2
+
3!2 +
t3
3!
µ′3
t3
3!
+ ··· +
tr
r!
µ′r + . . .
4!23 + . . .
IT
= 1+
1!
µ′1
S
) = 1+
t
N
∴ MX (t) = E(e
(r + 1)!2r
2!2
3!22
4!23
A
tX
=
=
=
=
..
.
H
Given µ′r
∴ µ′1
µ′2
µ′3
AT
H
= 1 + 2(2t) + 3(2t)2 + 4(2t)3 + . . .
F
∴ MX (t) = (1 − 2t)−2 .
=
=
=
=
=
−2(1 − 2t)−3 (−2)
6(1 − 2t)−4 (−2)2
′
MX
(0) = 4
′′
MX (0) = 24
E(X 2 ) − [E(X)]2 = 24 − 16 = 8.
TU
R
E
N
O
TE
S
′
MX
(t)
′′
MX (t)
Now, First Moment=Mean= E(X) = µ′1
Second Moment= E(X 2 ) = µ′2
∴ V ar(X)
O
Now, Differentiating w.r.to t, we get
1 3 1
, , re8 4 8
spectively. Evaluate P {|X − µ| ≥ 2σ} and compare it with the upper bound given by
Tchebycheff’s(Chebychev’s) inequality.
LE
C
Example: 5.20. A random variable X takes the values -1,0,1 with probabilities
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
Solution:
E(X) = 0
1
E(X 2 ) =
4
1
V ar(X) =
4
∴ P {|X − µ| ≥ 2σ} = P {|X| ≥ 1}
= P {X = −1 or X = 1} =
N
σ2
P {|X − µ| ≥ c} ≤
c2
A
Choosing c = 2σ
H
1
P {|X − µ| ≥ 2σ} ≤
4
S
By Chebychev’s inequality
1
IT
4
AT
H
In this problem both the values coincide.
O
TE
S
O
F
Note: Alternative form of Chebychev’s inequality: If we put c = kσ, where k > 0, then
it takes the form
1
|X − µ|
≥σ ≤ 2
P
k
k
N
P
|X − µ|
k
<σ
≥1−
1
k2
TU
R
E
Example: 5.21. A fair dice is tossed 720 times. Use Tchebycheff’s(Chebychev’s) inequality to
find a lower bound for the probability of getting 100 to 140 sixes.
LE
C
Solution:
p = P {getting ‘6’ in a single toss} =
q = 1−
1
6
=
5
6
1
6
and
n = 720
X follows binomial distribution with mean np = 120 and variance npq = 100. i.e. µ =
120 and σ = 100.
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
By Alternate form of Chebychev’s inequality
P {|X − 120| ≤ 10k} ≥ 1 −
P {120 − 10k ≤ X ≤ 120 + 10k} ≥ 1 −
Choosing k = 2,
k2
1
k2
N
A
IT
H
4
.
Function of a Random Variable
O
6
F
AT
4
1
H
3
k2
we get,
3
P {100 ≤ X ≤ 140} ≥
∴ Required lower bound is
1
S
P {|X − µ| ≤ kσ} ≥ 1 −
TU
R
E
N
O
TE
S
In the analysis of electrical systems, we will be often interested in finding the properties of
a signal after it has been subjected to certain processing operations by the system, such as
integration, weighted averaging, etc. These signal processing operations may be viewed as
transformations of a set of input variables to a set of output variables. If the input is a set of
random variables (RVs), then the output will also be a set of RVs. In this chapter, we deal with
techniques for obtaining the probability law (distribution) for the set of output RVs when the
probability law for the set of input RVs and the nature of transformation are known.
Finding fY (y), when fX (x) is given or known
LE
C
Let us now derive a procedure to find fY (y), the pdf of Y , when Y = g(X), where X is a
continuous RV with pdf fX (x) and g(x) is a strictly monotonic function of x.
Case (i): g(x) is a strictly increasing function of x.
fY (y) =
=
=
=
P [Y ≤ y], where fY (y) is the cdf of Y
P [g(X) ≤ y]
P [X ≤ g −1 (y)]
fX [g −1 (y)]
Differentiating both sides with respect to y,
fY (y) = fX (x)
Page 33 of 38
dx
dy
, where x = g −1 (y)
(6.1)
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
Case (ii): g(x) is a strictly decreasing function of x.
P [Y ≤ y], where fY (y) is the cdf of Y
P [g(X) ≤ y]
P [X ≥ g −1 (y)]
1 − P [X ≤ g −1 (y)]
1 − fX [g −1 (y)]
fY (y) =
=
=
=
=
dy
, where x = g −1 (y)
(6.2)
N
dx
A
fY (y) = −fX (x)
S
Differentiating both sides with respect to y,
dy
dx
H
dx
AT
fY (y) = fX (x)
IT
H
Combining (6.1) and (6.2), we get
|g ′ (x)|
O
F
fY (y) = fX (x)
S
Note : The above formula for fY (y) can be used only when x = g −1 (y) is single valued.
N
O
TE
When x = g −1 (y) takes finitely many values x1 , x2 , x3 , . . . , xn , i.e., when the roots of the
equation y = g(x) are x1 , x2 , x3 , . . . , xn , the following extended formula should be used to
find fY (y):
dx1
E
TU
R
fY (y) = fX (x1 )
LE
C
fY (y) = fX (x1 )
dy
dx1
+ fX (x1 )
|g ′ (x1 )|
dxn
+ fX (x2 )
+ · · · + fX (xn )
dy
dx2
|g ′ (x2 )|
dy
+ · · · + fX (xn )
Example: 6.1. Let X be a random variable with pdf f (x) =
2
dxn
2
9
or
dxn
|g ′ (xn )|
(x + 1), −1 < x < 2. Find
the pdf of the random variable Y = X .
Solution:
Divide the interval into two parts (-1,1) and (1,2).
dx
fY (y) =
fX (x)
dy
1
2
2
2
√
√
When −1 < x < 1, fY (y) = √
(1 + y) + (1 − y) = √ , 0 < y <
2 y 9
9
9 y
1.
1
2
√
When 1 < x < 2, fY (y) = √
(1 + y) , 1 < y < 4.
2 y 9
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18MAB204T-Probability and Queueing Theory
Example: 6.2. Let X be a random variable with pdf f (x) =
of the random variable Y = 2X − 3.
Solution:
fY (y) =
dx
dy
fX (x) =
1
2
·
When 1 < x < 5, fY (y) =
x
12
S. ATHITHAN
, 1 < x < 5. Find the pdf
x
12
y+3
48
, −1 < y < 7.
dx
fX (x) =
1
1
·
1
N
A
π/2 − (−π/2)
=
π
IT
Now, fY (y) =
1
H
Solution:
Given that fX (x) is uniformly distributed. ∴ fX (x) =
S
Example: 6.3.
If X is uniformly distributed in (−π/2, π/2), find the pdf of Y = tan X.
(1 + y 2 )
1
When x ∈ (−π/2, π/2), fY (y) =
, −∞ < y < ∞.
π(1 + y 2 )
π
LE
C
TU
R
E
N
O
TE
S
O
F
AT
H
dy
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18MAB204T-Probability and Queueing Theory
7
S. ATHITHAN
Exercise/Practice/Assignment Problems
1. Find the probability of drawing two red balls in succession from a bag containing 3 red
and 6 black balls when (i) the ball that is drawn first is replaced, (ii) it is not replaced.
2. A bag contains 3 red and 4 white balls. Two draws are made without replacement; what
is the probability that both the balls are red.
O
TE
S
O
F
AT
H
IT
H
A
N
S
3. A random variable X has the following distribution
X
-2 -1
0
1
2
3
P[X=x] 0.1 k 0.2 2k 0.3 3k
Find (i) the value of k,(ii) the Distribution Function (CDF) (iii) P (0 < X < 3/X <
1
2) and (iv) the smallest value of α for which P (X ≤ α) > .
2
1
Ans: k =
15
4. A random variable X has the following distribution
X
0 1
2
3
4
P[X=x] k 2k 5k 7k 9k
Find (i) the value of k,(ii) the Distribution Function (CDF) (iii) P (0 < X < 3/X <
1
2) and (iv) the smallest value of α for which P (X ≤ α) > .
3
1
Ans: k =
24
5. A random variable X has a pdf
(
3x2 , when 0 ≤ x ≤ 1
f (x) =
0,
otherwise
LE
C
TU
R
E
N
Find the (i) Find a, b such that P (X ≤ a) = P (X > a) and P (X > b) = 0.05,(ii)
r
3 1
the Distribution Function (CDF) (iii) P (0 < X < 0.5/X < 0.8). Ans: a =
,
2
r
3 19
b=
20
6. The amount of bread (in hundred kgs) that a certain bakery is to sell in a day is a random
variable X with a pdf


when 0 ≤ x < 5
Ax,
f (x) = A(10 − x), when 5 ≤ x < 10


0,
otherwise
Find the (i) the value of A,(ii) the Distribution Function (CDF) (iii) P (X > 5/X <
5), P (X > 5/2.5 < X < 7.5). (iv) the probability that in a day the sales is (a) more
1
than 500 kgs (b) less than 500 kgs (c) between 250 and 750 kgs. Ans: A =
25
7. The Cumulative Distribution Function (CDF) of a random variable X is given by

1 − 4 , when x > 2
F (x) =
x2
0,
otherwise
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
Find (i) the pdf of X,(ii) P (X > 5/X < 5), P (X > 5/2.5 < X < 7.5) (iii)
P (X < 3), P (3 < X < 5).
8. A coin is tossed until a head appears. What is the expected value of the number of tosses?.
Also find its variance.
9. The pdf of a random variable X is given by
(
a + bx,
f (x) =
0,
when 0 ≤ x ≤ 1
otherwise
N
S
Find the (i) the value of a, b if the mean is 1/2,(ii) the variance of X (iii) P (X >
0.5/X < 0.5)
H
A
10. The first three moments about the origin are 5,26,78. Find the first three moments about
the value x=3. Ans: 2,5,-48
H
AT
when 0 ≤ x ≤ 1
otherwise
F
12. The pdf of a random variable X is given by
(
k(1 − x),
f (x) =
0,
IT
11. The first two moments about x=3 are 1 and 8. Find the mean and variance. Ans: 4,7
S
O
Find the (i) the the value of k,(ii) the r th moment about origin (iii) mean and variance.
Ans: k = 2
O
TE
13. An unbiased coin is tossed three times. If X denotes the number of heads appear, find
the MGF of X and hence find the mean and variance.
N
14. Find the MGF of the distribution whose pdf is f (x) = ke−x , x > 0 and hence find its
mean and variance.
LE
C
TU
R
E
15. The pdf of a random variable X is given by


when 0 ≤ x ≤ 1
x,
f (x) = 2 − x, when 1 < x ≤ 2


0,
otherwise
For this find the MGF and prove that mean and variance cannot be find using this MGF
and then find its mean and variance using expectation.
16. The pdf of a random variable X is given by f (x) = ke−|x| , −∞ < x < ∞ Find the
(i) the the value of k,(ii) the r th moment about origin (iii) the MGF and hence mean and
1
variance. Ans: k =
2
17. Find the MGF of the RV whose moments are given by µ′r = (2r)!. Find also its mean
and variance.
18. Use Tchebycheff’s(Chebychev’s) inequality to find how many times a fair coin must be
tossed in order to the probability that the ratio of the number of heads to the number of
tosses will lie between 0.45 and 0.55.
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18MAB204T-Probability and Queueing Theory
S. ATHITHAN
19. If X denotes the sum of the numbers obtained when 2 dice are thrown, obtain an upper
bound for P {|X − 7| ≥ 4} (Use Chebychev’s inequality). Compare with the exact
probability.
20. A RV X has mean µ = 12 and variance σ 2 = 9 and unknown probability distribution.
Find P [6 < X < 8].
√ √
3σ
.
21. If a RV X is uniformly distributed over (− 3, 3), compute P |X − µ| ≥
2
N
A
H
when 0 ≤ x ≤ 1
otherwise
IT
23. Given a RV X with density function
(
2x,
f (x) =
0,
S
22. A RV X is exponentially distributed with parameter 1. Use Chebychev’s inequality to
show that P (−1 < X < 3) ≥ 3/4. Find the actual probability to compare.
H
find the pdf of Y = 8X 3 .
AT
24. If X has an exponential distribution with parameter a, find the pdf of Y = log X.
√
25. If X has an exponential distribution with parameter 1, find the pdf of Y = X.
O
F
26. If X is uniformly distributed in (1, 2), find the pdf of Y = 1/X.
TE
S
27. If X is uniformly distributed in (−2π, 2π), find the pdf of (i) Y = X 3 and (ii) Y =
X 4.
LE
C
TU
R
E
N
O
S OLVE /P RACTICE M ORE P ROBLEMS FROM THE T EXTBOOK AND R EFERENCE B OOKS
Figure 7.1: Values of e−λ .
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