Chapter 4

(1)

Probability and Stochastic Processes

A Friendly Introduction for Electrical and Computer Engineers SECOND EDITION

Roy D. Yates David J. Goodman

08_1 Yates Chapter 4 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 4

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1 2 1 2 1 2 2 1 2 1

[ x , y ] [ x , y ] [ x , y ]

P x < £ x y < £ y = P x < £ x £ y - P x < £ x £ y

x ,y

( ,

2 2

)

x ,y

( ,

1 2

)

x ,y

( ,

2 1

)

x ,y

( ,

1 1

)

F x y F x y F x y F x y

= - - +

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Let the random variable be defined as a function of and : ( , ).

The cdf of is found similar to the functions of single random variable. If there exists such that:

{ } { ( , )

z

Z X Y

Z g X Y

Z D

Z z g X Y z

=

£ = £

_z

z ,

} {( , ) }, then, the cdf of is found by,

( ) { } {( , ) } ( , ) .

The pdf of is found by taking the derivative of ( ) with respect to z such that,

( ) ( ).

z

Z X Y

D

Z

Z Z

x y D

Z

F z P Z z P X Y D f x y dxdy

Z F z

f z d F z

dz

= Î

= £ = Î =

=

òò

Functions of Two Random Variables

Let the random variable be defined as a function of and : ( , ).

The cdf of is found similar to the functions of single random variable. If there exists such that:

{ } { ( , )

z

Z X Y

Z g X Y

Z D

Z z g X Y z

=

£ = £

_z

z ,

} {( , ) }, then, the cdf of is found by,

( ) { } {( , ) } ( , ) .

The pdf of is found by taking the derivative of ( ) with respect to z such that,

( ) ( ).

z

Z X Y

D

Z

Z Z

x y D

Z

F z P Z z P X Y D f x y dxdy

Z F z

f z d F z

dz

= Î

= £ = Î =

=

òò

(44)

,

( ) ( , ) ( , )

( ) ( ) ( , )

( , ) (*) I

z

Z X Y

D z y

X Y Z z y

Z X Y

X Y

F z f x y dxdy

f x y dxdy

f z F z f x y dxdy

z z

f z y y dy

¥ -

-¥ -¥

¥ -

-¥ -¥

¥ -¥

=

¶ ¶

= =

¶ ¶

= -

òò ò ò

ò ò ò

,

f x and y are independent, the following hold ( , ) ( ) ( ).

In this case, Eq. (*) reads

( ) ( ) ( ) . This is a convolution integral.

X Y X Y

Z X Y

f z y y f z y f y

f z

^¥

f z y f y dy

-¥

- = -

= ò -

1. Z = X + Y

,

( ) ( , ) ( , )

( ) ( ) ( , )

( , ) (*) I

z

Z X Y

D z y

X Y Z z y

Z X Y

X Y

F z f x y dxdy

f x y dxdy

f z F z f x y dxdy

z z

f z y y dy

¥ -

-¥ -¥

¥ -

-¥ -¥

¥ -¥

=

¶ ¶

= =

¶ ¶

= -

òò ò ò

ò ò ò

,

f x and y are independent, the following hold ( , ) ( ) ( ).

In this case, Eq. (*) reads

( ) ( ) ( ) . This is a convolution integral.

X Y X Y

Z X Y

f z y y f z y f y

f z

^¥

f z y f y dy

-¥

- = -

= ò -

(45)

Example (Z = X + Y)

z

0

( ) 0

and are independent r.v. with

( ) ( ); ( ) ( ).

Find ( ) when = + . ( ) ( ) ( )

(

x y

X Y

z

Z X Y

z z y y

X Y

f x e u x f y e u y

f z Z X Y

f z f z y f y dy

e e dy

a b

ab ab b a

- -

- - -

= =

= -

=

= -

ò ò

2

) , .

( ( ) 0 0 ( ) 0 ( ) 0 0)

If ,

( ) .

In this case, note that x and y are exponential random variables and z

z z

X X Y

z Z

e e

f x for x f z y for y z and f y for y

f z ze

a b

a

b a

a b

a

- -

-

- ¹

= < => - = > = <

=

is an Erlang random variable with parameter m=2. Let's plot f

_X

( ) and ( ) for x f

_Z

z a = 1.

z

0

( ) 0

and are independent r.v. with

( ) ( ); ( ) ( ).

Find ( ) when = + . ( ) ( ) ( )

(

x y

X Y

z

Z X Y

z z y y

X Y

f x e u x f y e u y

f z Z X Y

f z f z y f y dy

e e dy

a b

ab ab b a

- -

- - -

= =

= -

=

= -

ò ò

2

) , .

( ( ) 0 0 ( ) 0 ( ) 0 0)

If ,

( ) .

In this case, note that x and y are exponential random variables and z

z z

X X Y

z Z

e e

f x for x f z y for y z and f y for y

f z ze

a b

a

b a

a b

a

- -

-

- ¹

= < => - = > = <

=

is an Erlang random variable

with parameter m=2. Let's plot f

_X

( ) and ( ) for x f

_Z

z a = 1.

(46)

0 1 2 3 4 5 6 7 8 9 10 0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

f_X(x) fZ(z)

0 1 2 3 4 5 6 7 8 9 10

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

f_X(x) fZ(z)

(47)

2. Z = X/Y

,

0

, ,

0

, ,

0

0 ,

( ) ( , )

( , ) ( , )

( , 0 ; , 0 )

( ) ( , ) ( , )

( , )

Z

Z X Y

D yz

X Y yz X Y

yz

Z X Y yz X Y

X Y

F z f x y dxdy

f x y dxdy f x y dxdy

x x

z y x yz z y x yz

y y

f z d f x y dxdy f x y dxdy

dz

y f zy y dy

¥ ¥

-¥ -¥

¥ ¥

-¥ -¥

¥

=

= +

£ > => £ £ < => ³

= +

òò

ò ò ò ò

ò

⁰ ^,

,

( ) ( , )

| | ( , )

X Y

y f zy y dy y f zy y dy

-¥

¥ -¥

-

=

ò ò

,

0

, ,

0

, ,

0

0 ,

( ) ( , )

( , ) ( , )

( , 0 ; , 0 )

( ) ( , ) ( , )

( , )

Z

Z X Y

D yz

X Y yz X Y

yz

Z X Y yz X Y

X Y

F z f x y dxdy

f x y dxdy f x y dxdy

x x

z y x yz z y x yz

y y

f z d f x y dxdy f x y dxdy

dz

y f zy y dy

¥ ¥

-¥ -¥

¥ ¥

-¥ -¥

¥

=

= +

£ > => £ £ < => ³

= +

òò

ò ò ò ò

ò

⁰ ^,

,

( ) ( , )

| | ( , )

X Y

y f zy y dy y f zy y dy

-¥

¥ -¥

-

=

ò

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three values of r .

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because the integral in the final expression is Var X [ ] = s

1²

.

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