Chapter 2

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Probability and Stochastic Processes

A Friendly Introduction for Electrical and Computer Engineers SECOND EDITION

Roy D. Yates David J. Goodman

08_1 Yates Chapter 2 1

Probability and Stochastic Processes

A Friendly Introduction for Electrical and Computer Engineers SECOND EDITION

Roy D. Yates David J. Goodman

Definitions, Theorems, Proofs, Examples, Quizzes, Problems, Solutions

Chapter 2

(2)

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08_1 Yates Chapter 2 3 s

.

x X(s)=x

S

real line

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Random Variable and Event Space

A = for every set , A = event = the set of s in S such that the values assumed by the random variable function , for those as its argument, X(×) = , are less than or equal to the given number on the real line.

¿ Note : A should be identified for all .

{ ^{s X s} ^: ^{( )} ^£ ^x }

^A ^Ì ^S

( ) X × z

xÎR1

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Valid and Invalid Random Variable

Example (Valid and Invalid Random Variable)

Consider throwing a dice once and reading the face value.

(1)Define

(Note that a sample has to be mapped to a unique value the same as a function.) Then

x(×) is a valid random variable.

{

1. ,2 3, 4, 5, 6

}

S = f f f f f f ^E ⁼

{

^f^{, ,{}^{S even}^},{^odd^}

}

1 3 5

2 4 6

1 for

x( ) 2 for , s f , f , f

s s f f , f

ì =

= íî =

Example (Valid and Invalid Random Variable)

Consider throwing a dice once and reading the face value.

(1)Define

(Note that a sample has to be mapped to a unique value the same as a function.) Then

x(×) is a valid random variable.

1 3 5

2 4 6

1 for

x( ) 2 for , s f , f , f

s s f f , f

ì =

= íî =

{ }

{ } { } { }

{ } { }

1 3 5

1 2 3 4 5 6

(i) for 1, : x( )

(ii) for 1 2, : x( ) , ,

(iii) for 2, : x( ) , , , , ,

x A s s x

x A s s x f f f odd

x A s s x f f f f f f S

f

< = £ =

£ < = £ = =

³ = £ = =

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Example (continued)

(2) Define y such that, for

y(×) is an invalid random variable.

y( )s =i

s = fi

{ }

{ } { }

1

(i) for 1, y( ) (ii) for 1 2, y( )

y A s y

y A s y f event

f

< = £ =

£ < = £ = ¹

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Remember P(AB) = P(A)P(B) if A and B are independent!

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Poisson Random Variable as a Limit of Binomial RV

We place “at random” n points in the (0,T) interval. What is the probability that k of these points will lie in the interval ?

Define

A(event) = { When we place a single point, it is placed in the interval }

Then,

P{ k points in the interval } = P { A occurs k times } = . If

(Poisson Theorem)

1 2

( , )t t

1 2

( , )t t

2 1

( ) t t , 1 ( ^C)

P p q p P

T

A = - º = - = A

We place “at random” n points in the (0,T) interval. What is the probability that k of these points will lie in the interval ?

Define

A(event) = { When we place a single point, it is placed in the interval }

Then,

P{ k points in the interval } = P { A occurs k times } = . If

(Poisson Theorem)

k n k

n p q k

æ ö -

ç ÷è ø

1, 1

n>> p<<

( )

!

k

k n k np

n np

p q e

k k

- -

æ ö »

ç ÷è ø

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Poisson Theorem

a

®

¥

® p np

n , 0,

If , then

(Proof)

, hence and

) ! 1

(

lim k

p e k p

n

_k _n _k ^k

n

a

- -

¥

®

÷÷ - =

ø ö çç è æ

k n

k p

k p k n

x

P ÷÷ - ^-

ø ö ççè

=æ

= ) (1 )

( p^k p ⁿ ^k

k

k n n

n _-

+ - -

= - (1 )

!

) 1 (

) 1

( L

If , then

(Proof)

, hence and

k n

k p

k p k n

x

P ÷÷ - ^-

ø ö ççè

=æ

= ) (1 )

( p^k p ⁿ ^k

k

k n n

n _-

+ - -

= - (1 )

!

) 1 (

) 1

( L

a

=

np

^p ⁼ ^a_n

n p n-a

= - 1

k k n

k n k

n n

n k n

k

n n

k n

k

n n n k

k n n

k n x P

- -

-

÷ø ç ö

è æ -

÷ø ç ö

è æ - úû ù êë

é -

- -

=

úû ù êë

é - úû ù êë

é -

- -

=

÷ø ç ö

è

÷ æ - ø ç ö è + æ -

= -

=

a a

a

a a

1 1

1) 1

( 1) 1 )(

1

! (

1 1) 1

( 1) 1 )(

1

! (

!

) 1 (

) 1 ) (

(

L L L

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Poisson Theorem (continued)

as

Therefore, when

.

a -a

¥

® - =

¥

® e

n , lim(1 n)ⁿ

n

¥

® n

) !

(x k e k

P

ak a

= -

=

) !

(x k e k

P

ak a

= -

=

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Exercise (Poisson Theorem)

Problem: In a large hotel it is known that 99% of all guests return room keys when

checking out. If 250 engineers check out after a large conference, what is the probability that not more than three will fail to return their keys?

Solution:

Let N = total number of engineers = 250;

p = probability to fail to return keys = 0.01;

k = number of engineer to fail to return keys.

Conditions are met to apply the Poisson Theorem, i.e.,

Problem: In a large hotel it is known that 99% of all guests return room keys when

checking out. If 250 engineers check out after a large conference, what is the probability that not more than three will fail to return their keys?

Solution:

Let N = total number of engineers = 250;

p = probability to fail to return keys = 0.01;

k = number of engineer to fail to return keys.

Conditions are met to apply the Poisson Theorem, i.e., 250 1

N = ?

0.01 1

p = =

2.5.

Np =a =

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Exercise (continued)

0 2.5

(0 fail to return key) (1 ) 2.5 0.0821

! 0!

k

k N k

N e e

p p p

k k

a ^-a ^-

æ ö -

=ç ÷ - » = =

è ø

1 2.5

(1 fail to return key) 2.5 0.2052 1!

p e

-

= =

2 2.5

(2 fail to return key) 2.5 0.2565 2!

p e

-

= =

3 2.5

(3 fail to return keys) 2.5 0.2138 3!

p e

-

= =

3 2.5

(3 fail to return keys) 2.5 0.2138 3!

p e

-

= =

(no more than 3 fail to return keys) 0.0821 0.2052 0.2565 0.2138 0.7576.

p = + + + =

Chapter 2

Probability and Stochastic Processes

Probability and Stochastic Processes

Definitions, Theorems, Proofs, Examples, Quizzes, Problems, Solutions

Chapter 2

.

Random Variable and Event Space

{ s X s : ( ) £ x }

Valid and Invalid Random Variable

{

}

{

}

{ }

{ } { } { }

{ } { }

Example (continued)

{ }

{ } { }

Poisson Random Variable as a Limit of Binomial RV

Poisson Theorem

) ! 1

(

lim k

p e k p

n

a

÷÷ - =

ø ö çç è æ

a

=

np

Poisson Theorem (continued)

Exercise (Poisson Theorem)

Exercise (continued)

{ ^{s X s} ^: ^{( )} ^£ ^x }