I want to use a rule for conditional expectation I found in (German) wikipedia, not in my script/textbook of probability theory, I guess it should be simple and follow more or less straight from the general definition (I want have a proof to be sure that I don't build up on a wikipedia mistake)
Let X be independent of Z and of Y (XY integrable and X,Y,Z random variables) $$ E(XY|Z) = E(X) E(Y|Z) $$
My idea: Showing that the rhs meets the conditions of the general definition of E(XY|Z), that is (i) it should be $\sigma(Z)$-measureable, check. (ii) $E\left( E(X) E(Y|Z) 1_A \right) \stackrel{!}{=} E(XY 1_A) \forall A \in \sigma(Z)$ Now the lhs$=E(X) E(E(Y|Z)1_A) = E(X)E(Y 1_A)$ (according to (ii) of the definition of $E(Y|Z)$, for all $A\in \sigma(Z)$)
$= E(XY1_A)$ as wished (X, Y are independent).
But, in this proof I did not use that X,Z are independent, so it would follow as well $E(XY|Z)=E(Y)E(X|Z)=E(Y)E(X)=E(XY)$ which shouldn't be this way.
Maybe it's all much simpler and I have just the wrong point of view on it. Q: Does anybody see the flaw in my proof? Can anybody hint me to a proof or a reference to a proof?
(@Didier, i try to get Williams book, I have to see if my library can get it for me).