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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
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