Because of an article that I'm trying to understand, I've come up with the folowing question:
Suppose we have $f:(\mathbb{R}_{\geq 0})^2 \rightarrow \mathbb{R}_{\geq 0} $ , $\ X,Y\geq 0 \ \ $ R.V.'s ; $X$ with density $g$
Is it ok this computation?
$$ \mathbb{E}(f(X,Y)) = \int_0^{+\infty}\mathbb{E}(f(x,Y)) \ \ g(x) \ dx $$
If $X,Y$ are independent, then $dP_{XY}=dP_XdP_Y$, and by Tonelli's theorem (Fubini's theorem for non-negative functions),
$E(f(X,Y))=\int_{\mathbb{R}^2} f(x,y)dP_{XY}=\int_{\mathbb{R}}\int_{\mathbb{R}} f(x,y)dP_YdP_X=\int_{\mathbb{R}} E(f(x,Y))dP_X=\int_0^\infty E(f(x,Y))g(x)dx$
The conclusion doesn't hold for general $X,Y$.For example $X=Y$, $f(x,y)=xy$, then $LHS=(EX)^2$, but $RHS=(EX)^2$, which are not equal in general.