Question

The joint pmf of ? and ? is given by ??,? (?, ?) = (? + ?)/ 27 ??? ? = 0, 1,2; ? = 1, 2, 3, and ??,? (?, ?) = 0 otherwise. a. Find ?(?|? = ?) for all ? = 0,1, 2. b. Find ?(3 + 0.2?|? = 2).

Answer 1

P?,? (?, ?) = (? + ?) / 27

So the joint distribution table looks like

X Y	1	2	3
0	1/27	2/27	3/27
1	2/27	3/27	4/27
2	3/27	4/27	5/27

a)

Distrition of X-

X	0	1	2
P(X)	1/27+2/27+3/27=6/27	2/27+3/27+4/27=9/27	3/27+4/27+5/27=12/27

conditional distribution table of Y given x is P(?|? = ?)

For x=0

Y	1	2	3
P(Y|X=0)	P(Y=1,X=0)/P(X=0) =(1/27)/(6/27)	P(Y=2,X=0)/P(X=0) (2/27)/(6/27)	P(Y=3,X=0)/P(X=0) = (3/27)/(6/27)

Y	1	2	3
P(Y|X=0)	1/6	1/3	1/2

Expected value of (Y|X=0) = 1*1/6 + 2 * 1/3 + 3* 1/2 = 7/3 [Ans]

For x=1

Y	1	2	3
P(Y|X=1)	P(Y=1,X=1)/P(X=1) =(2/27)/(9/27)	P(Y=2,X=1)/P(X=1) (3/27)/(9/27)	P(Y=3,X=1)/P(X=1) = (4/27)/(9/27)

Y	1	2	3
P(Y|X=1)	2/9	1/3	4/9

Expected value of (Y|X=1) = 1*2/9 + 2 * 1/3 + 3* 4/9 = 20/9 [Ans]

For x=2

Y	1	2	3
P(Y|X=2)	P(Y=1,X=2)/P(X=2) =(3/27)/(12/27)	P(Y=2,X=2)/P(X=2) (4/27)/(12/27)	P(Y=3,X=2)/P(X=2) = (5/27)/(12/27)

Y	1	2	3
P(Y|X=2)	1/4	1/3	5/12

Expected value of (Y|X=2) = 1*1/4 + 2 * 1/3 + 3* 5/12 = 13/6 [Ans]

b)

?(3 + 0.2?|? = 2) = 3 + .2 * E(Y|X=2) =3 + .2 * 13/6 = 3.433 [Ans]

If you find my answer useful then please support me by putting thumbs-up. Thank you.