Conditional Expected Values

Example 1| Example 2 | Useful Web Resources| Solutions

 

 

The conditional expectation of Y given X = x, denoted E(Y | X = x) is found by computing the conditional probability mass function for Y given X = x and taking the expected value of this distribution.

Theorem 2.9-1:

For discrete random variables X and Y the conditional expectation of Y given X = x is

 

Example 1:

Let X and Y have the following joint probability mass function:

x
y
p(x,y)
1
2
3
Py(y)
1
0.25
0.15
0.10
0.50
2
0.05
0.35
0.10
0.50
 
Px(x)
0.30
0.50
0.20
1

 

The conditional proabability mass function of Y given X=1 is

P(Y =1 | X=1)= 0.25/0.30 =0.834

P(Y =2 | X=1) = 0.05/0.30=0.167

E(Y | X=1) = 1(0.834) +2 (0.167) =1.168

Compute the probability mass function of Y given X=2.

 

Round all answers to 1 decimal place.

A: What is P(Y=1|X=2)?

B: What is P(Y=2|X=2)?

 

C:What is E(Y|X=2)?

 

 

D: What is P(Y=1| X=3)?

E: What is P(Y=2|X=3)?

F: What is E(Y|X=3)?

 

 

Solution

 

Theorem 2.9-1

For random variables X and Y,

E(Y) = E ( E ( Y | X ) )

Example 2.9-2:

Refer to Example 2.9-2 on page 96. A game consists of first tossing a die. If the face value on the die is X, then a coin is tossed X times. Let Y be the number of heads.

So X takes on the values 1,2,3,4,5 or 6. Therefore Px(x)= 1/6 for x=1,2,...,6.

P(Y |X=1) is defined as the probability of getting a Head, given the face value on the dice is 1. The probability of getting a Head is 1/2 so, P(Y|X=1)= P(Y| X=2)= ... P(Y|X=6) = (1/2) because the two events are independent.

E(Y|X=1)= 1(1/2) =1/2 , E(Y|X=2)= 2(1/2)= 2/2

E(Y|X=3) =3(1/2)=3/2 , E(Y| X=4) = 4(1/2) =4/2

E(Y|X=5) =5(1/2)= 5/2, E(Y|X=6) =6(1/2) =6/2

From Theorem 2.9-1 we know that

So E(Y)=E(E(Y|X))= S E(Y | X=x)Px(x) = (1/2)(1/6) +(2/2)(1/6)+ ...+(6/2)(1/6) = S (x/2)(1/6) = 1.75 for x=1,2...,6 .

Definition 2.9-2:

For discrete random variables X and Y,

Analogously to Theorem 2.4-3:

Example 2:

Lets refer to Example 1 again.

VAR( Y |X=1)= (1-1.168)^2P(Y=1|X=1)+ (2-1.168)^2P(Y=2|X=1)

= (1-1.168)^2(0.834) + (2-1.168)^2(0.167)

=0.02354+0.11560

=0.14

 

Round all answers to 2 decimal places.

A: What is VAR(Y|X=2)?

B: What is STD(Y|X=2)?

C: What is VAR(Y|X=3)?

D: What is STD(Y|X=3)?

     

Solutions

 

Solution to Example 1

    x
    y
    p(x,y)
    1
    2
    3
    Py(y)
    1
    0.25
    0.15
    0.10
    0.50
    2
    0.05
    0.35
    0.10
    0.50
     
    Px(x)
    0.30
    0.50
    0.20

     

    A: P(Y=1 |X=2) =0.15/0.50=0.30

    B: P(Y=2 |X=2)= 0.35/0.5=0.70

    C: E(Y | X=2)= 1(0.3) +2(0.7)=1.70

    D: P(Y=1 | X=3)= 0.10/0.2= 0.5

    E: P(Y=2 | X=3)= 0.10/0.2= 0.5

    F: E(Y|X=3) = 1(0.5)+ 2(0.5)= 1.5

    Solution to Example 2

    VAR (Y|X=2) = (1-1.7)^2P(Y=1|X=2) +(2-1.7)^2P(Y=2|X=2)= (1-1.7)^2( 0.30) + (2-1.7)^2 (0.70) =0.147+0.063=0.21

    STD(Y|X=2)= sqrt(0.21)=0.458=0.46

    VAR(Y|X=3)= (1-1.5)^2P(Y=1|X=3) +(2-1.5)^2P(Y=2|X=3) = (1-1.5)^2(0.5) + (2-1.5)^2(0.5)=0.125+0.125=0.25

    STD(Y|X=3) =sqrt(0.25)=0.5

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