# Expectation

• May 12th 2012, 05:59 AM
iPod
Expectation
I have attached the question on this thread.

I have managed to attain the following:

$P_x(x)=\alpha^2xe^{-\alpha x}$

$P_{y|x}(y|x)=xe^{-xy}$

Yet I cannot obtain the fact that

$E[XY]=1$

I'm not sure how to derive this using my results.
• May 12th 2012, 09:44 AM
Moo
Re: Expectation
Hello,

Consider that E[XY]=E[E[XY|X]]

But E[XY|X]=X*E[Y|X].

Since you know the pdf of Y|X, you'll easily get the expectation. Then you'll get E[something with respect to X], which is again easy because you have the pdf of X.
• May 12th 2012, 11:23 AM
iPod
Re: Expectation
$E[Y|X]=\int^\infty_0 yxe^{-xy} dy=[-ye^{-xy}]^\infty_0-\frac{1}{x}[e^{-xy}]^\infty_0=\frac{1}{x}$

And is follows that:

$xE[Y|X]=1$

and then we just take the expectation

$E[XE[Y|X]]=E[1]=1=E[XY]$

Am I correct?
• May 13th 2012, 11:35 PM
Moo
Re: Expectation
Yes, it's correct.

A few remarks :
- are you aware of the properties of the conditional expectation that have been used ?
- in the pdf of X and of Y|X, you should respectively add that it's for x>0 and y>0