1
$\begingroup$

I'm trying to solve this problem in my homework assignment and I get different result from the answer. I know the answer is right, but at the same time I also don't see where I did wrong in my solution. So here's the problem:

Let X and Y have a joint uniform distribution on the triangle with vertices (0,0), (3,0), (0,3). Find: (i) E(X|Y) and E(Y|X) (ii) Var(X|Y) and Var(Y|X) (iii) EX and Var(X)

I've correctly completed the first and second sub-problem, but for the third one, I struggle to get it right. I'm trying to use the fact that EX = E(E(X|Y)) to derive EX. And the way I approach it is: EX = $$ \int {E(X|Y)*f(Y)} dY$$ where f(Y) is the pdf of Y being a certain value From the sub-problem 1, I have: E(X|Y) = (3-Y)/2 and f(Y) = 3-Y Hence, I tried to solve integration: $$\int{\frac{(3-Y)^2}{2}} dY $$ where 0<=Y<=3 and I get result 4.5 whereas the correct answer is 1.

Can someone please show where I did wrong? I'd much appreciate your help!

Thanks in advance!

$\endgroup$
2
$\begingroup$

The joint density is $\frac{1}{4.5}$ over the triangle. That has not been used in calculating the pdf of $Y$, which is $\frac{3-y}{4.5}$ on $[0,3]$ and $0$ elsewhere.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ how come I didn't see this?! I kept using 1 as the joint distribution. Thanks a lot for pointing it out! $\endgroup$ – Vol_Smile Oct 6 '13 at 7:11
  • $\begingroup$ Because quite properly you were concentrating on the ideas. $\endgroup$ – André Nicolas Oct 6 '13 at 7:15
  • $\begingroup$ I'm stuck again, on the part Var(X). I get result 1/2 where as the answer id 1/6. I used VAR(X) = E(X^2) - [E(X)]^2 in this case. Since I already know E(X)=1, I only need to solve E(X^2), which is equivalent to E(E(X^2|Y)). From previous step, I get E(X^2|Y)=(3-Y)^2 / 3 and since pdf of Y is (3-Y)/4.5, I can use both of them and solve the integral to get 3/2 as E(X^2). But that leads to Var(X) = 3/2 - 1 = 1/2 $\endgroup$ – Vol_Smile Oct 6 '13 at 8:00
  • $\begingroup$ The (unconditional) expectation of $X^2$ is indeed $\frac{3}{2}$. Since I am error prone, I checked it two ways. The variance is not $\frac{1}{6}$. $\endgroup$ – André Nicolas Oct 6 '13 at 17:07
  • $\begingroup$ Thanks a lot for confirming my result:) I'll let my professor and class know. $\endgroup$ – Vol_Smile Oct 7 '13 at 4:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.