1. ## finding expectation

The joint density function of X and Y is given by
$f(x, y) = \frac{1}{2\sqrt{2\pi}}\frac{e^{-y/2}}{y^{3/2}}$, where $|x| < y < \infty$.

Find $E[X|Y]$.

How do I go about doing this question? Any help is appreciated. Thanks in advance!

2. ## Re: finding expectation

Hey alphabeta89.

To start off with: what did you get for your region of integration?

3. ## Re: finding expectation

Originally Posted by chiro
Hey alphabeta89.

To start off with: what did you get for your region of integration?
-y<x<y, 0<y<\infty

4. ## Re: finding expectation

Hint: Integrate out the x variable first and use the properties of the Gamma Function:

Gamma function - Wikipedia, the free encyclopedia

5. ## Re: finding expectation

Do we need to find out what is $f_{X|Y}{(x|y)}$? Then use $E[X|Y] = \int_{-\infty}^{\infty} {x f_{X|Y}{(x|y)}}\mathrm{d}x$?

6. ## Re: finding expectation

You could find the conditional distribution or you could just condition Y on the appropriate values and get the expectation.

They will both give you the same result and the rigorous way to prove it is to use Fubini's Theorem.

7. ## Re: finding expectation

Hi, just to check, is this correct?

$f_Y{(y)}=\int_{-y}^{y}K\frac{e^{-y/2}}{y^{3/2}}\mathrm{d}x= K\frac{2e^{-y/2}}{\sqrt{y}}=\frac{e^{-y/2}}{\sqrt{2\pi{y}}}$

$f_{X|Y}{(x|y)}=\frac{f_{X,Y}{(x,y)}}{f_Y{(y)}}= \frac{\left (\frac{1}{2\sqrt{2\pi}}\frac{e^{-y/2}}{y^{3/2}}\right )}{\left (\frac{e^{-y/2}}{\sqrt{2\pi{y}}} \right )}=\frac{1}{2y}$

$E{[X|Y]}&=\int_{-\infty}^{\infty} {x f_{X|Y}{(x|y)}}\mathrm{d}x=\int_{-y}^{y}{x\cdot{\frac{1}{2y}}}\mathrm{d}x=0$

8. ## Re: finding expectation

Can you show the step by step integration techniques to get f_y(Y)?

The steps and approach look good.

9. ## Re: finding expectation

Originally Posted by chiro
Can you show the step by step integration techniques to get f_y(Y)?

The steps and approach look good.
$f_Y{(y)}=\int_{-\infty}^{\infty}f_{X,Y}{(x,y)}\mathrm{d}x=\int_{-y}^{y}\frac{1}{2\sqrt{2\pi}}\frac{e^{-y/2}}{y^{3/2}}\mathrm{d}x= \frac{1}{2\sqrt{2\pi}}\frac{2e^{-y/2}}{\sqrt{y}}=\frac{e^{-y/2}}{\sqrt{2\pi{y}}}$.

Is this correct?

10. ## Re: finding expectation

That looks right algebraically.

One way to confirm the theoretical results is to simulate the random variables in a statistical package if you so choose to do.