Question

Conditional expectation, can someone check my solution? I don't have the answers and I don't know if I'm doing this right.

Answer 1

Given a random vector (X,Y) with pmf below: \[P_{(X,Y)}(i,j)=1/10 | 1 \le i \le j \le 4\]find \[E(X|Y=y),E(Y|X=x)\]

Answer 2

Here's the distribution

Answer 3

\[P(X|Y=1) = 1, \space x=1; \space 0, \space otherwise\]\[P(X|Y=2)=1/2, \space 1\le x\le2; \space 0, \space otherwise\]

Answer 4

\[P(X|Y=3) = 1/3, \space 1\le x \le 3; \space 0 \space otherwise\]
\[P(X|Y=4) = 1/4, \space 1\le x \le 4; \space 0 \space otherwise\]

Answer 5

Thus,\[P(X|Y=y)=1/y, \space 1 \le x \le y; \space 0 \space otherwise\]

Answer 6

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

Answer 7

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

Answer 8

So,
\[E(X|Y=1)=1/1 *(0+1)\]\[E(X|Y=2)=1/2 * (1+2)\]\[E(X|Y=3)=1/3 *(1+2+3)\]\[E(X|Y=4)=1/4 * (1+2+3+4)\]Thus\[E(X|Y=y)=1/y\sum_{i=1}^{y}i\]

Answer 9

Now for\[E(Y|X=x)\]\[P(Y|X=1) = 1/4, \space 1 \le y \le 4; \space 0, \space otherwise\]\[P(Y|X=2) = 1/3, \space 2 \le y \le 4; \space 0, \space otherwise\]\[P(Y|X=3) = 1/2, \space 3\le y \le 4; \space 0, \space otherwise\]\[P(Y|X=4) = 1, \space y=4; \space 0, \space otherwise\]

Answer 10

From the above it follows that\[P(Y|X=x)=1/(5-x), \space x\le y \le 4; \space 0 \space otherwise\]\[E(Y|X=1)=\sum_{y}{y*P(y|X=1)}=\sum_{j=1}^{4}{y*P(y|X=1)=}\]\[=1*1/4+2*1/4+3*1/4+4*1/4=1/4*(1+2+3+4)\]

Answer 11

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

Answer 12

Thus,\[E(Y|X=x)=\frac{1}{5-x}*\sum_{j=x}^4{j}\]

Answer 13

@Zarkon @ganeshie8 can either of you check this when you get a chance?

Answer 14

Hey

Answer 15

hey @ganeshie8 can you see if my E(Y|X=x) and E(X|Y=y) are correct?

Answer 16

@Zarkon , sorry to bother you again, but can you just verify if this is correct or not?

Answer 17

@Directrix can you double check please?

Answer 18

@kropot72 will know.

Answer 19

thanks!

Answer 20

@ganeshie8 can you take a look please?

Answer 21

Still havent gotten a confirmation from anyone