srit / uicm006 prp / random processes sri ramakrishna

SRIT / UICM006 – PRP / Random Processes

SRIT / M & H / M. Vijaya Kumar 1

SRI RAMAKRISHNA INSTITUTE OF TECHNOLOGY

(AN AUTONOMOUS INSTITUTION)

COIMBATORE- 641010

UICM006 & PROBABILITY AND RANDOM PROCESSES

Random Processes

Definition: Random Process

A random process is a collection (ensemble) of random variables * ( )+ that are

functions of a real variable where , is the sample space and , is the index

set.

Remark:

If is fixed, then * ( )+ is a random variable.

If and are fixed, then * ( )+ is a number.

If is fixed, then * ( )+ is a single time function.

Classification of Random process:

Continuous random process:

If is continuous and takes any value, then ( ) is a Continuous Random Process.

Eg:

Let ( ) maximum temperature of a particular place in ( ). Here is a

continuous set and , ( ) is a continuous random process.

Discrete random process:

If assumes only discrete values and is continuous, then we call such random

process * ( )+ as Discrete Random Process.

Eg:

Let ( ) be the number of telephone calls received in the interval ( ).

Here * + and * +

( ) is a discrete random process



Continuous random sequence:

If is continuous but takes only discrete values, then ( ) is a discrete random

sequence.

Eg:

Let = temperature at the end of the hour of a day. The is a continuous set

of all possible values of temperature.

* +

* ( )+

( ) is a continuous random sequence.

Let ( ) denotes the outcome of the toss of a coin, then ( ) for

is a random sequence since the sample space * + is a discrete and hence

( ) is also discrete.

Discrete random sequence:

A random process ( ) in which both and are discrete is called discrete

random sequence.

Eg:

Let denote the outcome of the toss of a fair die. Here * +

* +

* + is a discrete random sequence.

Definition:

Strongly stationary (OR) Strict sense stationary process

A random process is called a strongly stationary process or strict sense stationary

process (SSS), if all its finite dimensional distributions are invariant under translation of

time .

( ) ( )

In general

( ) ( )

for any and any real number .



Auto correlation of a random process

Let ( ) and ( ) be the two given numbers of the random process * ( )+. The

auto correlation is

( ) , ( ) ( )-

Prove that auto correlation function is an even function of .

Proof:

By definition of auto correlation function

( ) , ( ) ( )-

Replace by – , we get

( ) , ( ) ( )-

Put

( ) , ( ) ( )-

, ( ) ( )-

( ) ( )

Which proves auto correlation function is an even function.

Mean square value:

Putting , we get

( ) , ( ) ( )- , ( )-

which is the mean square value of ( ).

Auto covariance of a random process:

( ) {[ ( ) ( ( ))][ ( ) ( ( ))]}

( ) , ( )- , ( )-

Correlation coefficient

The correlation coefficient of the random process * ( )+ is defined as

( ) ( )

√ ( )√ ( )

where ( ) denotes the auto covariance.



Cross correlation

The cross correlation of the two random process * ( )+ and * ( )+ is defined as

( ) , ( ) ( )-

Wide sense stationary (WSS)

A random process * ( )+ is called a weekly stationary process or wide – sense

stationary process if

, ( )- constant

, ( ) ( )- ( ) depends only on , where .

Evolutionary process

A random process that is not stationary in any sense is called as evolutionary

process.

First order stationary

A random process * ( )+ is said to be first order stationary, if its mean should be

constant.

, ( )-

Problem: 1

If ( ) ( ) where ‘Y’ is uniformly distributed in ( ). Show that ( ) is

wide-sense stationary.

Solution:

Given ( ) ( )

Since is uniformly distributed in ( ), then the probability density function is

( ) {

To prove ( ) is WSS, we have to show that

(i). , ( )- Constatnt

(ii). ( ) ( ).

, ( )- , ( )-



∫ ( ) ( )

∫ ( )

, ( )-

, ( ) ( )- , ( ) -

, ( ) ( )-

, ( )-

( ) , ( ) ( )-

, ( ) ( ( ) )-

, ( , ( ) -) ( , ( ) -)-

, ( ) ( )-

, ( ) ( )- , ( ) -

, ( )-

, ( )-

( )

∫ ( ) ( )

( )

6 ( )

7

( )

( ) ( )

the given ( ) ( ) is wide-sense stationary.



Problem: 2

Show that the random process ( ) ( ) where A and are constants

and is uniformly distributed in ( ), is wide-sense stationary.

Solution:

Given ( ) ( )


( ) >

( )



(ii). ( ) ( ).

, ( )- , ( )-

∫ ( ) ( )

∫ ( )

, ( )-

, ( )-

( ) , ( ) ( )-

, ( ) ( ( ) )-

, ( , ( ) -) ( , ( ) -)-

, ( ) ( )-

, ( ) ( )- , ( ) -

, ( )-

, ( )-



∫ ( ) ( )

( )

6 ( )

7

( )

( )

( ) ( )


Problem: 3

Show that the random process ( ) ( ) where A and are constants

and is random variable uniformly distributed in ( ), is first order stationary.

Solution:

Given ( ) ( )


( ) {

To prove ( ) is a first order stationary, we have to prove mean , ( )- Constatnt

, ( )- , ( )-

∫ ( ) ( )

∫ ( )

, ( )-

, ( ) ( )- , ( ) -

, ( ) ( )-

, ( )-

the given ( ) ( ) is first order stationary.



Problem: 4

Show that the process ( ) is wide-sense stationary, A and B

are random variables if (i) , - , - (ii) ( ) ( ) (iii) ( ) .

Solution:

Given ( )



(ii). ( ) ( ).

, ( )- , -

, - , -

, ( )- , ( ) ( ) -

( ) , ( ) ( )-

,( )( ( ) ( ))-

, ( )- , ( )-

, ( )- , ( )-

, - ( ) , - ( )

, - ( ) , - ( ) , ( ) -

, - ( ) , - ( )

, ( ) ( )- , ( ) ( ) -

, ( )- , ( ) -

, -

( ) , ( ) -

( ) ( )

the given ( ) is wide-sense stationary.



Problem: 5

Let the random process * ( )+ and * ( )+ be defined as ( )

( ) , where A and B are random variables, is constant. If

( ) ( ) ( ) and ( ) ( ), prove that * ( )+ and * ( )+ are

jointly WSS.

Solution:

To show that * ( )+ and * ( )+ are jointly WSS, we have to prove that

(i) * ( )+ is WSS

(ii) * ( )+ is WSS

(iii) Cross correlation ( ) ( )

To prove * ( )+ as WSS:

( )

, ( )- , -

, - , -

, ( )- , ( ) ( ) -

( ) , ( ) ( )-

,( )( ( ) ( ))-

, ( )- , ( )-

, ( )- , ( )-

, - ( ) , - ( ) , ( ) -

, - ( ) , - ( )

, - ( ) , - ( )

, ( ) ( )- , ( ) ( ) -

, ( ( )) - , ( ) -

, ( ) -

( ) , ( ) -



( ) ( )


Similarly we can prove that ( ) is wide-sense stationary.

To prove * ( )+ and * ( )+ are jointly WSS:

( ) , ( ) ( )-

( ) ,( )( ( ) ( ))-

, ( )- , ( )-

, ( )- , ( )-

, - ( ) , - ( ) , ( ) -

, - ( ) , - ( )

( ) ( )

, ( ) ( )-

[ ( ( ))] , ( ) -

( )

( ) , ( ) -

( )

( ) ( )

Therefore * ( )+ and * ( )+ are jointly WSS.

Problem: 6

Consider the random process ( ) ( ) ( ) where * ( )+ is WSS process,

is a random process independent of ( ) and is uniformly distributed in ( ) and

is constant. Prove that * ( )+ is wide-sense stationary.

Solution:

Given ( ) ( ) ( )




( ) {



(ii). ( ) ( ).

, ( )- , ( ) ( )-

, ( )- , ( )-

, ( )- ∫ ( ) ( )

, ( )-

, ( )-

, ( )-

( ) , ( ) ( )-

, ( ) ( ) ( ) ( ( ) )-

, ( ) ( )- , ( ) ( )-

( )

, ( ) ( )-

( )

, ( ) ( )-

( )

[ , ( )- , ( )-]

( )

[ ( ) ∫ ( ) ( )

]

( )

6 ( )

4 ( )

5

7

( )

, ( ) -

( ) ( )

( ) ( )




Problem: 7

The process * ( )+ whose probability distribution under certain conditions is given by

, ( ) -

{

( )

( )

Solution: Given that

( ) 0 1 2 3 …

( ( ))

( )

( )

( )

( ) …

, ( )- ∑ ( )

( )

( )

( )

( )

( ) 6

( )

( ) 7

( ) 6 (

* (

*

7

( ) [ (

*]

, ( ) -

( ) [

]

( ) [

]

( )

( )

, ( )-

, ( )- ( ) , ( )-

, ( )- ∑ ( )

∑, - ( )



∑, ( ) - ( )

∑ ( ) ( ) ∑ ( )

6

( )

( )

( )

( ) 7

( ) 6

( )

( ) 7

( ) 6 (

* (

*

7

( ) [ (

*]

, ( ) -

( ) [

]

( ) [

]

[

( )

( ) ] , -

[

( ) ]

( ) [

]

, ( )-

, ( )- ( ) , ( )-

, ( )-

To prove ( ) is stationary; we have to show that the mean, variance and moments

are constants. But the variance is not a constant; hence the given function is not a

stationary.



Problem: 8

* ( )+

* ( )+

Solution: Given that

( )

, ( ) -

To prove * ( )+ is satationary, we need to show mean and variance should be

constant.

To find Mean:

, ( )- ∑

(

* (

*

To find Variance:

, ( )- ∑

( ) (

* (

*

, ( )- * , ( )-+

* ( )+ is stationary process.



Problem: 9

A random processes * ( )+ is defined by * ( )+

where and are independent random variables each of which has the value with

* ( )+

stationary process.

Solution:

First we need to find the mean values of the random variable and .

To find Mean value of ( ) , -:

( )

, ( )-

, - ∑

(

* (

*

, -

Simillarly , - .

To find Variance:

, - ∑

( ) (

* (

*

, -

Simillarly , - .

Given and are independent random variables.

, - , - , -



Also given ( ) ;



(ii). ( ) ( ).

, ( )- , -

, - , -

( ) , ( ) ( )-

,( )( ( ) ( ))-

, ( )- , ( )-

, ( )- , ( )-

, - 4 , - , -

5 , , - , - -

, - 4 , - , -

5

4 , - , -

5 4

, - , -

5

, - , - , - , -

, -

( ) ( )




Ergodic Random process

Time average

The time average of a random process is defined by

∫ ( )

Ensemble average

The ensemble average of a random process * ( )+ is the expected value of the

random variable at a time .

Ergodic random process

A random process * ( )+ is called ergodic if all its ensemble average equals to

appropriate time averages.

Mean Ergodic process

A random process * ( )+ is said to be mean Ergodic, if

, ( )-

:

∫ ( )

; , ( )-

Mean Ergodic theorem

Let * ( )+ be a random process with constant mean and let be its time average.

Then * ( )+ is mean ergodic if

Correlation Ergodic random process

A random process * ( )+ is said to be correlation Ergodic, if the process * ( )+ is

mean ergodic where

( ) ( ) ( )

:

∫ ( ) ( )

; , ( )-

where is the time average of ( ).



Problem: 10

Prove that the random processes ( ) ( ), A and are constants and

is uniformly distributed random variable in ( ) is mean ergodic.

Solution:

Given that ( ) ( )

Since is uniformly distributed in ( ), we have

( ) {

, ( )- , ( )-

∫ ( )

( )

, ( )-

, ( )-

∫ ( )

∫ ( )

6 ( )

7

4

6 ( ) ( )

75

, ( )-

Therefore * ( )+ is mean ergodic.



Problem: 11

Consider the random process * ( )+ with ( ) ( ), where is a

uniformly distributed random variable in ( ). Prove that * ( )+ is correlation

ergodic.

Solution:

Given that ( ) ( )


( ) {

( ) ( ) ( )

( ) ( ( ) )

, ( ( ) ) ( ( ) )-

, ( ) ( )-

( )

, ( ) ( )-

To show that * ( )+ is correlation ergodic, we have to show that * ( )+is mean ergodic.

, ( )-

To find Mean , ( )-:

, ( )- 8

, ( ) ( )-9

, ( ( )) ( ( ))-

[ ( ) ∫ ( )

( ) ]

6 ( )

4 ( )

5

7

( ( ) )

, ( )-

( )



∫ ( )

∫

, ( ) ( )-

6 ( )

7

( ) , -

6 ( ) ( )

7

( )

4

6 ( ) ( )

7

( )5

( )

, ( )-

* ( )+ is correlation Ergodic.

Problem: 12


uniformly distributed random variable with ( ). Prove that * ( )+ is correlation

ergodic.

Solution:

Given that ( ) ( )

Since is uniformly distributed in ( ), we have

( ) {

To show that * ( )+ is correlation ergodic, we have to show that * ( )+ is mean ergodic.

( ) ( ) ( )

( ) ( ( ) )

, ( ) ( )-

, ( ) ( )-

( ) , ( ) ( )- , ( ) -



To find Mean , ( )-:

, ( )- [ , ( ) ( )-]

* ( ( )) ( ( ))+

[ ( ) ∫ ( )

( ) ]

6 ( )

4 ( )

5

7

, ( ) -

, ( )- ( )

∫ ( )

∫ , ( ) ( )-

6 ( )

7

( ) , -

6 ( ) ( )

7

( )

( )

, ( )-

* ( )+ is correlation ergodic.

Problem: 13

Let * ( )+ be a WSS process with zero mean and the auto correlation function

( ) | |

* ( )+ over ( ).

Solution:

( ) | |

Also given * ( )+ be a WSS process with zero mean, hence we have



, ( )-

By definition of time average,

∫ ( )

In the interval ( ), we have

∫ ( )

, - <

∫ ( )

=

∫ , ( )-

, -

, -

∫ ( )

( )

By definition ( ) ( ) , ( )- , ( )-

( )

( ) ( )

, -

∫ ( )

( )

∫( ( ))

∫(

| |

*

∫ .

/

, -



<.

/

. /

=

<

=

[

]

, -

By mean Ergodic theorem, we have

, -

So we conclude that * ( )+ is not mean ergodic.

Markov Process and Markov chain

Markov Process

A random process * ( )+ is called a Markov process if , ( ) ( )

( ) ( ) ( ) - , ( ) ( ) - for all

.

In other words, if the future behavior of a process depends on the present value but not

on the past, then the process is called a Markov process.

Markov chain

If the above condition is satisfied for all , then the process * ( )+ is called a Markov

chain and the constants * + are called the states of the Markov chain.

In other words, a discrete parameter Markov process is called a Markov chain.

One-step Transition Probability

The conditional transition probability [ ] is called the one-step

transition probability from state to state at the step and is denoted by

( ).



Homogeneous Markov Chain

If the one-step transition probability does not depend on the step

( ) ( )

the Markov chain is called a homogeneous Markov chain.

Transition Probability Matrix (t.p.m)

When the Markov chain is homogeneous, the one-step transition probability is

denoted by . The matrix ( ) is called the transition probability matrix satisfying

the conditions (i) and (ii) ∑ for all , i.e., the sum of the elements of any

row of the t.p.m is 1.

n-Step Transition Probability

The conditional transition probability that the process is in state at step , given

that it was in step at step 0, [ ] is called the n-step transition

probability and is denoted by ( )

.

( )

[ ]

CHAMPMAN –KOLMOGOROV THEOREM

If is the transition probability matrix of a Homogeneous Markov chain and the

step tpm ( ) , then the power of the tpm.

( )

( )

( )

0 ( )

1 [ ]

Regular matrix

A stochastic matrix is said to be regular matrix, if all the entries of (for some

+ve integer ) are positive.

If the transition matrix is regular, then we say that the Homogeneous Markov Chain

is regular.



Classification of states of Markov Chain:

Irreducible:

A Markov chain is said to be irreducible if every state can be reached from every

other state, where ( )

for some and for all and .

Return state:

If ( )

, for some , then we call the state of the Markov chain as return

state.

Period:

Let ( )

for all . Let be a return state. Then we define the period as follows

2 ( )

3

State is periodic with period if and aperiodic if .

Problem: 14

The transition probability matrix of the Markov chain with three states 0, 1, 2 is

[

]

, -

Find (i). , - (ii) , - (iii) , -

Solution:

[

]



( )

[

]

From the definition of conditional probability,

( ) , - ∑ , - , -

, ⁄ - , - , ⁄ - , -

, - , -

( )

, - ( )

, - ( )

, -

(

*

(

*

(

*

(

*

, -

( ) , -

( )

( )

( )

, -

(

* (

* (

* (

*

( ) , - ( )

Problem: 15

The t.p.m of a Markov chain * + having three states 1, 2, 3 is

[

] and the initial state distributions of the chain is

( ) ( ). Find (i). , - (ii) , -

Solution:



( ) ( )

, - , - , -

[

]

( )

[

]

( ) , - ∑ , - , -

, ⁄ - , - , ⁄ - , -

, ⁄ - , -

( )

, - ( )

, - ( )

, -

( ) ( ) ( )

, -

( ) , -

( )

( )

( )

, -

( )( )( )( )

Problem: 16

A salesman territory consists of three cities A, B and C. He never sells in the same city

on successive days. If he sells in city A, then the next day, he sells in city B. However, if he

sells in either B or C, the next day he is twice as likely to sell in city A as in the other city. In

the long run, how often does he sell in each of the cities?

Solution:



The transition probability matrix of the Markov chain is

[

]

The given problem describes a Markov Chain with three states and three cities A, B, C.

Let ( ) be the steady state probability distribution.

Also ( ) and

( )

[

]

( )

(

)

( )

( )

( )

( )

Solving ( ) ( ) ( ), we get

The steady state probability distribution is

(

* ( )

Thus in the long run, he sells 40% of the time in city A, 45% of the time in the city B

and 15% of the time in city C.



Problem: 17

A man either drives a car or catches the train to go to office each day. He never goes

two days in row by train; but if he drives one day, then the next day he is just as likely to

drive again as he is to travel by train. Now suppose that on the first day of the week, the

man tossed a die and drove to work if and only if a 6 appeared. Find (i) the probability

that he takes a train on the third day (ii) the probability that he drives to work in the long

run?

Solution:

The travel problem is a Markov chain with state space ( )

where Train & Car


[

]

Probability of travelling by a car , -

The initial state probability distribution is

( )

(

*

( ) ( ) (

*(

+

(

*

( ) ( )

(

*(

+

(

*

, -



We find the steady state probability distributions of the Markov chain.


Also ( ) and

( ) (

+ ( )

(

, ( )

( )

( )

Solving ( ) ( ), we get

, -

Problem: 18

An engineer analyzing a series of digital signals generated by a testing system

observes that only 1 out of 15 highly distorted signals follow a highly distorted signal, with

no recognizable signal between, whereas 20 out of 23 recognizable signals follow

recognizable signals, with no highly distorted signal between. Given that only highly signals

are not recognizable, find the fraction of signals that are highly distorted.

Answer:


[

]




Also ( ) and

( ) [

] ( )

(

, ( )

( )

( )

Solving ( ) ( ), we get

Hence of signals generated by the testing systems are highly distorted.

Problem: 19

The probability of a dry day following a rainy day is and that the probability of a

rainy day following a dry day is

. Given that May 1st is a dry day. Find the probability that

May 3rd is a dry day and also May 5th is a dry day.

Answer:


[

] , -

Since May 1st is a dry day; ( ) and ( )

The initial state distribution is ( ) ( )



( ) ( ) ( ) [

] (

*

( ) ( ) (

* [

] (

*

( ) ( ) (

* [

] (

*

( ) ( ) (

* [

] (

*

Problem: 20

Consider the Markov chain with the transition probability matrix

[

]. Find the limiting probabilities of the system.

Answer:

Given the transition probability matrix [

].


Also ( ) and

( ) [

] ( )

(

+ ( )



( )

( )

( )

Solving ( ) ( ) ( ), we get

The steady state probability distribution is

(

* ( )

Poisson Process

Poisson Postulates

If * ( )+ denotes the number of occurrences of a certain event in ( ), then the

discrete random process * ( )+ is called the Poisson process, provided the following

postulates are satisfied.

i. , ( )-

ii. , ( )-

iii. , ( )-

iv. ( ) is independent of the number of occurrences of the event in any interval prior

or after the interval ( ).

v. The probability that the event occurs a specified number of times in ( )

depends only on t, but not on



Probability law for the Poisson process * ( )+

Let be the rate of occurrences or number of occurrences per unit time and ( )

be the probability of occurrences of the event in the interval ( ) is a Poisson process

with parameter .

( ) ( )

Proof:

( ) , ( ) -

( ) , ( ) -

{( ) ( ) ( )

( ) ( ) ( )}

( ) ( ) ( )( )

( ) ( ) ( ) ( )

( ) ( ) , ( ) ( )-

( ) ( )

, ( ) ( )-

Taking limit as , we get

4 ( ) ( )

5

, ( ) ( )-

( ) , ( ) ( )- ( )

Assume that the solution of equation (1) is

( ) ( )

( ) ( )

Differentiating with respect to ‘t’ on both sides, we get

( )

, ( ) ( ) - ( )

Using (2) and (3) in equation (1), we get

( )

, ( ) ( ) - 6

( )

( ) ( )

( )

( )7

( )

( )

( ) ( )

( )



( )

( ) ( )

( ) ( )

( )

( )

( ) ( )

( ) ( )

( )

( )

( )

( ) ( )

( ) ( )

( )

( )

Integrating both sides, we get

( )

Taking exponential on both sides, we get

( )

( ) ( )

Using (4) in equation (2), we get

( ) ( )

( )

Theorem:

Prove that the sum of two independent Poisson process is also a Poisson process.

Proof:

Let ( ) ( ) ( )

, ( ) - [

( ) ]

∑ , ( ) - , ( ) -

∑ ( )

( )

( )



∑

( )

( )

( )

( )

∑

( )

, -

( )

∑

( )

( )

( )

( )

( )

,( ) -

which is also a Poisson process with parameter ( ) .

Theorem:

The difference between two independent Poisson processes is not a Poisson process.

Proof:

Let ( ) ( ) ( )

, ( )- , ( ) ( )-

, ( )- , ( )-

( )

, ( )- (, ( ) ( )- )

, ( )

( ) ( ) ( )-

, ( )- ,

( )- , ( )- , ( )-

( ) (

) (( )( )) , ( ) -

(

) ( )

, ( )- ,( ) - ( )

Hence the difference between two independent Poisson processes is not a Poisson process.



Property:

Is the Poisson process first order stationary?

Proof:

Mean of the Poisson Process

, ( )- ∑ ( )

∑ ( )

∑ ( )

( )

, ( ) -

∑( )

( )

6( )

( )

( )

7

6 ( )

( )

7

6

7

, ( )-

Since mean is not constant, therefore Poisson process is not first order stationary.

Result:

, ( )-

Property:

The inter arrival time of a Poisson process with parameter has an exponential

distribution with parameter .

Proof:

Let us consider the two consecutive calls of the event in and .



Let T be the interval time between the two calls in and , then T follows continuous

random variable.

( ) , ( )-

, ( ) -

( )

( )

By definition of cumulative distribution, we have

( ) , -

, -

( )

The probability density function is

( )

( )

( )

( )

which is an exponential distribution with parameter .

Auto Correlation function of a Poisson Process

Let ( ) and ( ) be any two Poisson process.

( ) , ( ) ( )-

, ( )* ( ) ( ) ( )+- , ( )-

, ( )* ( ) ( )+ ( )-

( ) , ( )* ( ) ( )+- , ( )-

Since and are independent, then

( ) ( ) ( )

( ) , ( )- , ( ) ( )- , ( )-

( , ( )- , ( )-) ( )

( ) ( )



Theorem:

Prove that Poisson process is a Markov Process.

Answer:

Let us take the conditional probability distribution of ( ) given that past values of

( ) and ( ). Assume that and .

Consider , ( ) ( ) ⁄ -

, ( ) ( ) ( ) -

, ( ) ( ) -

( ) ( )

( )

, ( ) ( ) ⁄ -

by using second and third order probability function

hence the conditional probability distribution of ( ) given the values of the process ( )

and ( ) depends only on the most recent value ( ) of the process. Hence the Poisson

process is a Markov Process.

Problem: 21

Suppose that customer arrives at a bank counter according to Poisson process with

mean rate 3 per minute. Find the probability that during a time interval of two minutes

(i) Exactly four customers arrive

(ii) Greater than four customers arrive

(iii) Fewer than four customers arrive

Solution:

Given that mean rate per minute. Inter arrival time minutes.

By definition, we have

( ) , ( ) - ( )

, ( ) - ( )

( )

, - , ( ) -



( )

, - , ( ) -

, ( ) -

{ , ( ) - , ( ) - , ( ) -

, ( ) - , ( ) -}

8 ( )

( )

( )

( )

( )

9

8

9

* +

, - , ( ) -

* , ( ) - , ( ) - , ( ) - , ( ) -+

8 ( )

( )

( )

( )

9

8

9

* +

Problem: 22

Customers arrive at a complaint department at the rate of 5 per hour for male

customers and 10 per hour for female customers. If the arrivals in each case follow Poisson

process, calculate the probabilities that (1) atmost 4 male customers (2) atmost 4 female

customers will arrive in a 30 minute period.

Solution:


( ) , ( ) - ( )



Given that mean arrival rate for male customers is per hour.

, ( ) -

.

/ .

/

, - , ( ) -

, ( ) - , ( ) - , ( ) - , ( ) - , ( ) -

.

/. /

.

/. /

.

/. /

.

/. /

.

/. /

.

/* +

.

/, -

.

Given that mean arrival rate for female customers is per hour.

, ( ) -

. /

.

/

, - , ( ) -

, ( ) - , ( ) - , ( ) - , ( ) - , ( ) -

.

/. /

.

/. /

.

/. /

.

/. /

.

/. /

* +

, -



Problem: 23

Assume that the no. of messages input to a communication channel in an interval of

duration ‘t’ seconds is a Poisson process with mean rate of , compute

The probability that exactly three messages will arrive during a ten second interval.

The probability that the no. of messages arrival in an interval of duration five seconds is

between three and seven.

Solution:

Given that mean rate per minute.


( ) , ( ) - ( )

( )

, -

, ( ) -

( )( )

, -

, ( ) -

, ( ) - , ( ) - , ( ) - , ( ) - , ( ) -

( )( ( ))

( )( ( ))

( )( ( ))

( )( ( ))

( )( ( ))

, -



Problem: 24

If particles are emitted from a radioactive source at the rate of 20 per hour. Find

i. The probability that exactly five particles are emitting during a 15 minute period.

ii. The probability that fewer than three particles are emitting during a 12 minute period.

Solution:

Given that mean rate per hour.


( ) , ( ) - ( )

( )

, -

, ( ) -

.

/. /

, -

, ( ) -

, ( ) - , ( ) - , ( ) -

.

/. /

.

/. /

.

/. /

6

7



Problem: 25

A fisher man ctaches a fish at a poisson rate 2 per hour from a large lake with lots of

fish. If he starts fishing at 10.00 AM. What is the probability that he catches one fish by

10.30 AM and three fishes by noon?

Answer:

Given that mean rate per hour.


( ) , ( ) - ( )

, ( ) - ( )

[

] [ (

* ]

. /.

/

[

( ) ] , ( ) -

( )( )

Problem: 26

If customers arrive at a counter accordance of Poisson process with mean rate of 2

per minute, find the probability that the interval between two consecutive arrivals is

i. more than 1 minute

ii. between 1 minute and 2 minute

iii. four minute or less



Solution:

Given that mean rate per minute.

By property, the inter arrival time of a Poisson process with parameter has an

exponential distribution with parameter .

The probability density function of exponential distribution is

( )

( )

, -

, -

∫

6

7

[ ∫

]

( )

, -

, -

, -

∫

6

7

( )

( )

, -

, -

∫



6

7

( )

( )

Gaussian process or normal process

Definition:

A real valued random process * ( )+ is said to be Gaussian process or Normal

process, if the random variables ( ) ( ),… ( ) are jointly normal for every

and for any set of .

The order density of a Gaussian process is given by

( )

( ) | |

>

| |∑∑| | ( )( )

?

( ( )) ( )

{ ( ) ( )} | | | |

Problem: 27

Let * ( )+ is a Gaussian random process with * ( ) + and

( ) | |. Find (i) , ( ) - (ii) ,| ( ) ( )| -

Solution:

Since * ( )+ is a Gaussian random process, any number of * ( )+ is a random variable.

( ) ( ) , ( ) ( )-

( ) ( ) , ( ) ( )-

( ) , ( )-

* ( )+ ( )

( ) | | ( )

Let in ( ), we get



( ) | |

( ) ( )

Using (3) in (1), we get

, ( )- ( ) ( )

* ( )+ is a random variable with mean 10 and variance 16.

, ( ) -:

√

, ( ) - , -

( )

(| ( ) ( )| ):

( ) ( )

Then is also a random variable.

( ) , ( ) ( )-

, ( )- , ( )-

, ( ) -

( )

( ) 0( ( ) ( )) 1

, ( ) ( ) ( ) ( )-

, ( )- , ( )- , ( )- , ( )-

( ) , ( )- , ( )- ( )

( ) , - , ( )-

0.5 0.5

0

0

𝟎 𝟓

𝐏(𝒁 𝟎 𝟓)



( ) , ( )- , ( )- ( )

( ) ( ) ( )

| | | | | |

( )

√

(| ( ) ( )| ) (| | )

( )

, -:

, - , -

( )

( )

Problem: 28

Suppose * ( )+ is a normal random process with mean ( ) and

( ) | |. Find the probability that (i) ( ) (ii) | ( ) ( )|

Solution:

Since * ( )+ is a Normal or Gaussian random process.

Given that ( ) | |

* ( )+ ( )

| |

0.5 0.5

0

0

𝟎 𝟕𝟏

𝐏( 𝟎 𝟕𝟏 𝒛 𝟎 𝟕𝟏)

𝟎 𝟕𝟏



, ( ) -:

, ( ) - , -

( )

(| ( ) ( )| ):

( ) ( ), Then is also a random variable.

( ) , ( ) ( )-

, ( )- , ( )-

, ( ) -

( )

( ) 0( ( ) ( )) 1

, ( ) ( ) ( ) ( )-

, ( )- , ( )- , ( )- , ( )-

( ) , ( )- , ( )- ( )

( ) , - , ( )-

( ) , ( )- , ( )- ( )

( ) ( ) ( )

| | | | ( | |)

( )

0.5 0.5

0

0

𝟎 𝟓

𝐏(𝒁 𝟎 𝟓)



√

(| ( ) ( )| ) (| | )

( )

, -:

, - , -

( )

Sine wave random Processes

A random process * ( )+ of the form ( ) ( ) where any non empty

combination are random variables is called sine wave process.

Problem: 29

For the sine wave process ( ) , the amplitude is a

random variable with uniform distribution in the interval to . Determine whether the

process is stationary or not.

Answer:

For a stationary process the averages are time invariant. The mean is given by

, ( )- ∫

( )


( ) {

, ( )- ∫

0.5 0.5

0

0

𝟎 𝟓𝟑

𝐏( 𝟎 𝟕𝟏 𝒛 𝟎 𝟕𝟏)

𝟎 𝟓𝟑



4

5

Hence the mean is not constant. Therefore the processes is not stationary.

Problem: 30

Consider the processes ( ) ( ) where is uniformly distributed in

the interval ( ). etermine whether the process is stationary or not.

Answer:

For a stationary process the averages are time invariant. The mean is given by

, ( )- ∫ ( )

( )


( ) >

( )

, ( )- ∫ ( )

, ( )-

, ( ) ( )-

, ( ) ( )-

Hence the mean is not constant.

, ( )- ∫ , ( )-

∫ ( )

∫ 4

( )

5



∫ ( ( ))

, ( )-

, ( ) -

Hence the first and second moments are constant. Therefore the process * ( )+ is

satationary.

Two Marks

1. Define random process. (or) Define stochastic processes

Answer:

A random process is a collection (ensemble) of random variables * ( )+ that are

functions of a real variable where , is the sample space and , is the index

set.

2. Define a wide sense stationary process

Answer:

A random process * ( )+ is called a weekly stationary process or wide – sense

stationary process if

, ( )- constant

, ( ) ( )- ( ) depends only on , where .

3. Define a strictly stationary random process (SSS).

Answer:

A random process is called a strongly stationary process or strict sense stationary

process (SSS), if all its finite dimensional distributions are invariant under translation of time

.

( ) ( )

In general ( ) ( )

for any and any real number .



4. Prove that a first order stationary process has a constant mean.

Proof:

Let us consider the random process * ( )+ at two different times and .

, ( )- ∫ ( )

( ) is the density function of the random processes * ( )+

, ( )- ∫ ( )

( ) is the density function of the random processes * ( )+

Let

, ( )- ∫ ( )

∫ ( )

Since ( ) is first order stationary and hence ( ) ( )

, ( )-

, ( )- , ( )-

Thus mean of the random process * ( )+ Mean of the random process * ( )+.

5. When is a random process said to be mean ergodic?

Answer:

A random process * ( )+ is said to be mean Ergodic, if

, ( )-

:

∫ ( )

; , ( )-

6. When is a random process said to be correlation ergodic?

Answer:

A random process * ( )+ is said to be correlation ergodic if the process * ( )+ is

mean ergodic where ( ) ( ) ( )



:

∫ ( ) ( )

; , ( )-

where is the time average of ( ).

7. Consider the random process ( ) ( ), where is a random variable with

( )

sense stationary.

Solution:

To prove ( ) is WSS, we need to prove , ( )- Constatnt & ( ) ( ).

, ( )- , ( )-

∫ ( )

, ( )-

0 .

/ .

/1

0 .

( )/ .

/1

, ( ) - , ( ) -

, ( )-

the given processes is not WSS process.

8. Define a Markov chain and give an Problem.

Answer:

A random process * ( )+ is called a Markov process if , ( ) ( )

( ) ( ) ( ) - , ( ) ( ) - for all

.



If the above condition is satisfied for all , then the process * ( )+ is called a Markov

chain and the constants * + are called the states of the Markov chain.


[

]

9. State the postulates of a Poisson process.

Answer:

If * ( )+ denotes the number of occurrences of a certain event in ( ), then the

discrete random process * ( )+ is called the Poisson process, provided the following

postulates are satisfied.

, ( )-

, ( )-

, ( )-

( ) is independent of the number of occurrences of the event in any interval prior

or after the interval ( ).

The probability that the event occurs a specified number of times in ( )

depends only on t, but not on

10. Write down any two properties of Poisson process.

Answer:

The inter arrival time of a Poisson process with parameter has an exponential

distribution with parameter .

The difference between two independent Poisson processes is not a Poisson process.

The that Poisson process is a Markov Process.

11. Prove that sum of two independent Poisson processes is again a Poisson process.

Proof:

Let ( ) ( ) ( )

, ( ) - [

( ) ]



∑ , ( ) - , ( ) -

∑ ( )

( )

( )

∑

( )

( )

( )

( )

∑

( )

, -

( )

∑

( )

( )

( )

( )

( )

,( ) -

which is also a Poisson process with parameter ( ) .

Problem: 1

If ( ) ( ) where ‘Y’ is uniformly distributed in ( ). Show that ( ) is

wide-sense stationary.

Problem: 2

Show that the random process ( ) ( ) where A and are constants and

is uniformly distributed in ( ), is wide-sense stationary.

Problem: 3

Show that the random process ( ) ( ) where A and are constants and

is random variable uniformly distributed in ( ), is first order stationary.

Problem: 4

Show that the process ( ) is wide-sense stationary, A and B are

random variables if (i) , - , - (ii) ( ) ( ) (iii) ( ) .



Problem: 5

Let the random process * ( )+ and * ( )+ be defined as ( )

( ) , where A and B are random variables, is constant. If

( ) ( ) ( ) and ( ) ( ), prove that * ( )+ and * ( )+ are jointly

WSS.

Problem: 6

Consider the random process ( ) ( ) ( ) where * ( )+ is WSS process,

is a random process independent of ( ) and is uniformly distributed in ( ) and is

constant. Prove that * ( )+ is wide-sense stationary.

Problem: 7

The process * ( )+ whose probability distribution under certain conditions is given by

, ( ) -

{

( )

( )

Problem: 8

* ( )+

* ( )+

Problem: 9

A random processes * ( )+ is defined by * ( )+ where

and are independent random variables each of which has the value with

* ( )+

stationary process.

Problem: 10

Prove that the random processes ( ) ( ), A and are constants and is

uniformly distributed random variable in ( ) is mean ergodic.

Problem: 11


uniformly distributed random variable in ( ). Prove that * ( )+ is correlation ergodic.



Problem: 12


uniformly distributed random variable with ( ). Prove that * ( )+ is correlation ergodic.

Problem: 13

Let * ( )+ be a WSS process with zero mean and the auto correlation function

( ) | |

* ( )+ over ( ).

Problem: 14


[

]

, -

Find (i). , - (ii) , - (iii) , -

Problem: 15

The t.p.m of a Markov chain * + having three states 1, 2, 3 is

[

] and the initial state distributions of the chain is

( ) ( ). Find (i). , - (ii) , -

Problem: 16

A salesman territory consists of three cities A, B and C. He never sells in the same city on

successive days. If he sells in city A, then the next day, he sells in city B. However, if he sells in

either B or C, the next day he is twice as likely to sell in city A as in the other city. In the long

run, how often does he sell in each of the cities?

Problem: 17

A man either drives a car or catches the train to go to office each day. He never goes two

days in row by train; but if he drives one day, then the next day he is just as likely to drive

again as he is to travel by train. Now suppose that on the first day of the week, the man tossed

a die and drove to work if and only if a 6 appeared. Find (i) the probability that he takes a

train on the third day (ii) the probability that he drives to work in the long run?



Problem: 18

An engineer analyzing a series of digital signals generated by a testing system observes

that only 1 out of 15 highly distorted signals follow a highly distorted signal, with no

recognizable signal between, whereas 20 out of 23 recognizable signals follow recognizable

signals, with no highly distorted signal between. Given that only highly signals are not

recognizable, find the fraction of signals that are highly distorted.

Problem: 19

The probability of a dry day following a rainy day is

and that the probability of a rainy

day following a dry day is

. Given that May 1st is a dry day. Find the probability that May 3rd is

a dry day and also May 5th is a dry day.

Problem: 20

Consider the Markov chain with the transition probability matrix

[

]. Find the limiting probabilities of the system.

Problem: 21

Suppose that customer arrives at a bank counter according to Poisson process with

mean rate 3 per minute. Find the probability that during a time interval of two minutes

(iv) Exactly four customers arrive

(v) Greater than four customers arrive

(vi) Fewer than four customers arrive

Problem: 22

Customers arrive at a complaint department at the rate of 5 per hour for male

customers and 10 per hour for female customers. If the arrivals in each case follow Poisson

process, calculate the probabilities that (1) atmost 4 male customers (2) atmost 4 female

customers will arrive in a 30 minute period.

Problem: 23

Assume that the no. of messages input to a communication channel in an interval of

duration ‘t’ seconds is a Poisson process with mean rate of , compute

The probability that exactly three messages will arrive during a ten second interval.

The probability that the no. of messages arrival in an interval of duration five seconds is

between three and seven.



Problem: 24

If particles are emitted from a radioactive source at the rate of 20 per hour. Find

iii. The probability that exactly five particles are emitting during a 15 minute period.

iv. The probability that fewer than three particles are emitting during a 12 minute period.

Problem: 25

A fisher man ctaches a fish at a poisson rate 2 per hour from a large lake with lots of fish.

If he starts fishing at 10.00 AM. What is the probability that he catches one fish by 10.30 AM

and three fishes by noon?

Problem: 26

If customers arrive at a counter accordance of Poisson process with mean rate of 2 per

minute, find the probability that the interval between two consecutive arrivals is

iv. more than 1 minute

v. between 1 minute and 2 minute

vi. four minute or less

Problem: 27

Let * ( )+ is a Gaussian random process with * ( ) + and

( ) | |. Find (i) , ( ) - (ii) ,| ( ) ( )| -

Problem: 28

Suppose * ( )+ is a normal random process with mean ( ) and

( ) | |. Find the probability that (i) ( ) (ii) | ( ) ( )|

Problem: 29

For the sine wave process ( ) , the amplitude is a random

variable with uniform distribution in the interval to . Determine whether the process is

stationary or not.

Problem: 30

Consider the processes ( ) ( ) where is uniformly distributed in the

interval ( ). etermine whether the process is stationary or not.

“Education is not only a ladder of opportunity, but it is also an

investment in our future”

-Ed Markey.

srit / uicm006 prp / random processes sri ramakrishna

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