finding expectation

**alphabeta89** · January 2nd 2013, 09:09 PM

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!

**chiro** · January 2nd 2013, 09:29 PM

Hey alphabeta89.

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

**alphabeta89** · January 2nd 2013, 10:15 PM

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

**chiro** · January 2nd 2013, 11:12 PM

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

Gamma function - Wikipedia, the free encyclopedia

**alphabeta89** · January 3rd 2013, 06:33 AM

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

**chiro** · January 3rd 2013, 01:45 PM

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.

**alphabeta89** · January 4th 2013, 06:41 AM

Hi, just to check, is this correct?

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

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

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

**chiro** · January 4th 2013, 04:36 PM

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

The steps and approach look good.

**alphabeta89** · January 5th 2013, 10:19 PM

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?

**chiro** · January 5th 2013, 10:37 PM

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.

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