MATHEMATICAL EXPECTATION

CHAPTER 4

Week 10

2

4.13 Expected Value for Linear Function of Random

Variables

Definition 4.9:

Suppose and is the joint

probability function. If u is a linear function, then Y can be

written as

),,,( 21 nXXXuY ),,( 1 nxxf

nn

n

i

ii XaXaXaXaY

2211

1

3

Theorem 4.13:

If

where are random variables and

are constants, then

and

Note: If are independent, then

nn

n

i

ii XaXaXaXaY

2211

1

nXXX ,,, 21 naa ,,1

n

i

ii XEaYE1

)()(

ji

jijiii XXCovaaXVaraYVar ),(2)()(2

nXXX ,,, 21 0),( ji XXCov

4

Example 4.12:

Suppose are independent random variables,

each with mean p and variance p(1-p). If ,

determine the mean and variance for Y.

nXXX ,,, 21

n

X

Y

n

i

i 1

5

Solution:

1 1i

ii i

i i

X

E Y E E X E Xn n n

1 2

1 1( ) ( ) ( )nE X E X E X p p p

n n

npp

n

6

2

1( ) ( )

i

ii

i

X

Var Y Var Var Xn n

1 22

2

2

1( ) ( ) ( )

1(1 ) (1 ) (1 )

(1 ) (1 )

nVar X Var X Var Xn

p p p p p pn

np p p p

n n

7

Exercise:

Suppose are independent random variables each

with mean and variance . If Y is defined as

, find the mean and variance for Y.

nXXX ,,, 21

2n

i

i

Y X

1

8

4.14 Conditional Expectation

Definition 4.10

Let X and Y be two random variables with joint probability

function f(x,y). Let f(x|y) be the conditional probability function

of X, given the variable Y has a value y. Then the conditional

mean of X, given Y = y, is

Rx

yxfx

dxyxfx

yXE

variablesrandom discretefor )(

variablesrandom continuousfor ;)(

)(

9

The conditional mean of Y, given X = x, is

Ry

xyfy

dyxyfy

xYE

variablesrandom discretefor )(

variablesrandom continuousfor )(

)(

10

Definition 4.11

If X and Y are two random variables, and u(X) is a function of

X, then the conditional expectation of u(X) given Y= y is

If X and Y are two random variables, and u(Y) is a function of

Y, then the conditional expectation of u(Y) given X= x is

x

yxfxu

dxyxfxu

yXuE

variablesrandom discretefor ,)()(

variablesrandom continuousfor ,)()(

)(

y

xyfyu

dyxyfyu

xYuE

variablesrandom discretefor ,)()(

variablesrandom continuousfor ,)()(

)(

11

Example 4.13:

If the joint probability function is given as

Find the

10for ;2

21),( 22 yxyxyxf

)(),( yXExYE

12

Solution:

First of all we need to find the conditional probability functions:

We know that,

)( and )( xyfyxf

)(

),()(

yh

yxfyxf

)(

),()(

xg

yxfxyf

/ /( ) ,

y y

h y x y dx y x dx y y y

2 2 5 2 5 2

0 0

21 21 21 70 1

2 2 6 2

and

10),1(4

21

2

21

2

21)( 42

12

12

22

xxxdyyxdyyxxg

xx

13

Therefore,

10,0;3

2

72

21

)(2/3

2

2/5

2

yyxy

x

y

yx

yxf

10,1;1

2

)1(4

212

21

)( 2

442

2

xyxx

y

xx

yx

xyf

10;)1(3

)1(2

1

2

1

2)(

4

612

4

1

422

x

x

xdyy

xdy

x

yyxYE

xx

10;4

333)(

0

3

2/30

2/3

2

y

ydxx

ydx

y

xxyXE

yy

14

Example 4.14

The joint probability density function for random variables X

and Y is

Compute the conditional expectation and evaluate it

at

10,10);(5

6),( 2 yxyxyxf

),( yXE

2

1y

15

Solution:

11 2

2 2

0 0

2 2

6 6( )

5 5 2

6 1 6 1 0 ; 0 1

5 2 5 2

xh y x y dx xy

y y y

2

2

2

2

6

( , ) 56 1( )

5 2

2

1 2

x yf x y

f x yh y

y

x y

y

16

21 1

2

0 0

1

2 2

2

0

13 2

2

2

0

2

2

2( )

1 2

1 2 2

1 2

1 2 2

1 2 3 2

1 2 ; 0 1

1 2 3

x yE X y xf x y dx x dx

y

x xy dxy

x xy

y

y yy

2

2

1 1 2 1 11

2 3 2 1811 2

2

E X y

Thus,

17

Or

2

2

1221 4 1

2 311 22

xx

f x y

1 1

0 0

1

2

0

13 2

0

4 11 12 2 3

1 4

3

1 4

3 3 2

1 4 1 11

3 3 2 18

xE x y xf x y dx x dx

x x dx

x x

18

4.15 Conditional Variance

In addition to the conditional expectation, one may define the

conditional variance of X, given Y and the conditional variance

of Y, given X as

random variables

.

2

2

( ) ; for continuous

( )

( ) ; for discrete random variablesx R

x E X y f x y dx

Var X y

x E X y f x y

random variables

.

2

2

( ) ; for continuous

( )

( ) ; for discrete random variablesy R

y E Y x f y x dy

Var Y x

y E Y x f y x

19

It can be easily shown that the conditional variances can also be

written as;

22

22

)()()(

)()()(

xYExYExYVar

yXEyXEyXVar

20

Example 4.15:

From Example 4.13

Find

a)

b)

( )Var Y x

( )Var X y

10for ;2

21),( 22 yxyxyxf

21

Solution:

10,0;3

2

72

21

)(2/3

2

2/5

2

yyxy

x

y

yx

yxf

10,1;1

2

)1(4

212

21

)( 2

442

2

xyxx

y

xx

yx

xyf

10;)1(3

)1(2

1

2

1

2)(

4

612

4

1

422

x

x

xdyy

xdy

x

yyxYE

xx

22

10;4

333)(

0

3

2/30

2/3

2

y

ydxx

ydx

y

xxyXE

yy

Then and can be calculated as the

followings:

)( 2 yXE )( 2 xYE

22 2 4 5

3 / 2 3 / 2 3 / 2 00 0

3 3 3( )

5

y yyx

E X y x dx x dx xy y y

5

3 / 2

3 3; 0 1

5 5y y y

y

23

14

4

13

4

1

4

22

222 41

2

1

2

1

2)(

xxx

y

xdyy

xdy

x

yyxYE

10;2

)1(

)1(4

)1)(1(2

44

1

1

2 4

4

448

4

xx

x

xxx

x

Therefore the conditional variances are:

2

2 )()()( yXEyXEyXVar

10;80

3

80

4548

4

3

5

32

y

yy

yy

24

2

2 )()()( xYExYExYVar

10;)1(18

)1(8)1)(1(9

)1(3

)1(2

2

)1(24

262442

4

64

x

x

xxx

x

xx

25

Exercise:

Consider the joint probability function f(x,y) defined by:

a) conditional expectation of X, given Y = y

b) conditional expectation of Y, given X = x.

c) conditional variance of X, given Y = y

10;8),( yxxyyxf