Expected Value Question

• Nov 27th 2011, 04:24 PM
empyreandance
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?
• Nov 27th 2011, 11:34 PM
Moo
Re: Expected Value Question
Hello,

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

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

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

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

And hence the equality.