Question

Similar question to the one before, conditional expectation

Answer 1

Let v be a vector (X,Y), with prob mass fn \[P_{(X,Y)}(i,j)=1/10 | 1 \le i \le j \le 4\]Find E(X|Y=y)

Answer 2

that was such that \[1 \le i \le j \le 4\]Here's what I did so far:
\[P(X|Y=y) = 1/10, X=y\]\[P(X|Y=y) = 0, X > y\]Stumped on the case for X<y

Answer 3

Isn't it 1/10 again?
So\[P(X|Y=y) = 1/10, X \le y; 0 \space X > y\]

Answer 4

Okay so here's what i have:\[E(X|Y=1)=\sum_{x=1}^{4}x*P(x|Y=1)=1*1/10+2*0+3*0+4*0=1/10\]\[E(X|Y=2)=\sum_{x=1}^{4}x*P(x|Y=2)=1*1/10+2*1/10+3*0+4*0=(1+2)/10\]

Answer 5

\[E(X|Y=3)=\sum_{x=1}^{4}x*P(x|Y=3)=1*1/10+2*1/10+3*1/10+4*0=(1+2+3)/10\]

Answer 6

\[E(X|Y=4)=\sum_{x=1}^{4}x*P(x|Y=4)=(1+2+3+4)/10\]

Answer 7

Thus,
\[E(X|Y=y) = (1/10)*\sum_{i=1}^{y}i\]

Answer 8

@Zarkon @ganeshie8 does this look correct?

Answer 9

Now, find \[E(Y|X=x)\]\[P(Y|X=x)=1/10, \space Y \ge x; \space \space 0, \space Y < x\]

Answer 10

Joint probability distribution table that i drew up:

Answer 11

\[E(Y|X=1) = \sum_{y}{y*P(y|X=1)}=\sum_{y=1}^{4}y*P(y|X=1)=\]\[=1*1/10+2*1/10+3*1/10+4*1/10\]
\[E(Y|X=2) = \sum_{y}{y*P(y|X=2)}=\sum_{y=1}^{4}y*P(y|X=2)=\]\[=1*0+2*1/10+3*1/10+4*1/10\]
\[E(Y|X=3) = \sum_{y}{y*P(y|X=3)}=\sum_{y=1}^{4}y*P(y|X=3)=\]\[=1*0+2*0+3*1/10+4*1/10\]
\[E(Y|X=4) = \sum_{y}{y*P(y|X=4)}=\sum_{y=1}^{4}y*P(y|X=4)=\]\[=1*0+2*0+3*0+4*1/10\]

Answer 12

So we have:
\[E(Y|X=x) = 1/10 * \sum_{i=x}^{4}i\]

Answer 13

Do these look right? I'd love it if I could just check the answer somewhere, but I won't know the answer until this gets graded, and I haven't found anything online.

Answer 14

Hmm something's not right..