1. ## Expected Value Question

Hello everyone,

I have the following question. Suppose that X and Y are independent and f(x,y) is nonnegative. Put g(x)=E[f(x,Y)] and show E[g(X)]=E[f(X,Y)]. Show more generally that Integral over X in A of g(X) dP = Integral over X in A of f(X,Y) dP. Extend to f that may be negative. I've had no issues, except with the extension to negative f part. Any suggestions?

2. ## Re: Expected Value Question

Hello,

I suggest you write the expectations with respect to which rv you consider it : $E_X[m(X)]=\int m(x) ~dP_X$ for example.

So we actually have $g(x)=E_Y[f(x,Y)]$, we want to prove that $E_X[g(X)]=E_{(X,Y)}[f(X,Y)]$

But $E_X[g(X)]=E_X[E_Y[f(X,Y]]=E_X\left[\int_\Omega f(X,y) ~dP_Y\right]=\int_\Omega \int_\Omega f(x,y)~dP_Y dP_X$

And we also have $E_{(X,Y)}[f(X,Y)]=\int_{\Omega^2} f(x,y)~dP_{(X,Y)}$
But since X and Y are independent, $dP_{(X,Y)}=dP_X dP_Y$
So $E_{(X,Y)}[f(X,Y)]=\int_\Omega \int_\Omega f(x,y) ~dP_Y dP_X$
(I think this is where the positiveness of f intervenes, in order to apply Fubini's theorem)

And hence the equality.