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Expectation
I have attached the question on this thread.
I have managed to attain the following:
$\displaystyle P_x(x)=\alpha^2xe^{\alpha x}$
$\displaystyle P_{yx}(yx)=xe^{xy}$
Yet I cannot obtain the fact that
$\displaystyle E[XY]=1$
I'm not sure how to derive this using my results.

Re: Expectation
Hello,
Consider that E[XY]=E[E[XYX]]
But E[XYX]=X*E[YX].
Since you know the pdf of YX, 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.

Re: Expectation
$\displaystyle E[YX]=\int^\infty_0 yxe^{xy} dy=[ye^{xy}]^\infty_0\frac{1}{x}[e^{xy}]^\infty_0=\frac{1}{x}$
And is follows that:
$\displaystyle xE[YX]=1$
and then we just take the expectation
$\displaystyle E[XE[YX]]=E[1]=1=E[XY]$
Am I correct?

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 YX, you should respectively add that it's for x>0 and y>0