Question

Problem 4 (Conditional Expectation and Variance). Suppose the joint distri- bution of (X, Y) is given by the following contingency (row represents x) table 20 points (x,y) 2 4 6 1 0.3 0 0.1 2 0 0.2 0 3 0.1 0 0.3 A) Compute the marginal distributions of X and Y B) Are X and Y independent? Explain. C) Find the conditional distribution of Y given X -1 D) Compute E[Y|X 1] E) Compute EY|X= 2] F Compute E[exp(X)Y|x 2

Answer 1

x,y	2	4	6	total
1	0.3	0	0.1	0.4
2	0	0.2	0	0.2
3	0.1	0	0.3	0.4
total	0.4	0.2	0.4	1
A) marginal distribution of Y
Y	2	4	6	total
P(Y=y)	0.4	0.2	0.4	1
marginal distribution of X
X	1	2	3	total
P(X=x)	0.4	0.2	0.4	1

B)

X and Y are independent if for each possible values of X and Y following is true:

$P(X=x,Y=x)=P(X=x)P(Y=y)$

Here P(Y=2) = 0.4

P(X=1) = 0.4

and P(X=1, Y=2) = 0.3

P(X=1)P(Y=2) = 0.4*0.4 =0.16

Since P(X=1)P(Y=2) is not equal to P(X=1, Y=2) so X and Y are not independent.

c)

P(Y|X=1)=P(X=1,Y)/P(X=1)

Y	2	4	6	total
P(Y|X=1)	0.3/0.4=0.75	0	0.1/0.4=0.25	1

d)

Y	2	4	6	total
P(Y|X=1)	0.75	0	0.25	1
Y*P(Y|X=1)=	1.5	0	1.5	3

E[Y|X=1] = ΣY*P(Y|X=1) = 3

e)

P(X=2)=0.2

Y	2	4	6	total
P(Y|X=2)	0/0.2=0	0.2/0.2=1	0/0.2=0	1
Y*P(Y|X=2)=	0	4	0	4

E[Y|X=2] = ΣY*P(Y|X=2) = 4

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