1. ## Expectation

I have attached the question on this thread.

I have managed to attain the following:

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

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

Yet I cannot obtain the fact that

$\displaystyle E[XY]=1$

I'm not sure how to derive this using my results.

2. ## 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.

3. ## Re: Expectation

$\displaystyle E[Y|X]=\int^\infty_0 yxe^{-xy} dy=[-ye^{-xy}]^\infty_0-\frac{1}{x}[e^{-xy}]^\infty_0=\frac{1}{x}$

And is follows that:

$\displaystyle xE[Y|X]=1$

and then we just take the expectation

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

Am I correct?

4. ## 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