finding expectation

Printable View

• Jan 2nd 2013, 10:09 PM
alphabeta89
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!(Happy)
• Jan 2nd 2013, 10:29 PM
chiro
Re: finding expectation
Hey alphabeta89.

To start off with: what did you get for your region of integration?
• Jan 2nd 2013, 11:15 PM
alphabeta89
Re: finding expectation
Quote:

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
• Jan 3rd 2013, 12:12 AM
chiro
Re: finding expectation
Hint: Integrate out the x variable first and use the properties of the Gamma Function:

Gamma function - Wikipedia, the free encyclopedia
• Jan 3rd 2013, 07:33 AM
alphabeta89
Re: finding expectation
Do we need to find out what is $f_{X|Y}{(x|y)}$? Then use $E[X|Y] = \int_{-\infty}^{\infty} {x f_{X|Y}{(x|y)}}\mathrm{d}x$?
• Jan 3rd 2013, 02:45 PM
chiro
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.
• Jan 4th 2013, 07:41 AM
alphabeta89
Re: finding expectation
Hi, just to check, is this correct?

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

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

$E{[X|Y]}&=\int_{-\infty}^{\infty} {x f_{X|Y}{(x|y)}}\mathrm{d}x=\int_{-y}^{y}{x\cdot{\frac{1}{2y}}}\mathrm{d}x=0$
• Jan 4th 2013, 05:36 PM
chiro
Re: finding expectation
Can you show the step by step integration techniques to get f_y(Y)?

The steps and approach look good.
• Jan 5th 2013, 11:19 PM
alphabeta89
Re: finding expectation
Quote:

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?
• Jan 5th 2013, 11:37 PM
chiro
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.