Rule with independent random variables and conditional expectations

Question

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).

Answer 1

The faulty step is when you assert that $E(X)E(Y1_A)=E(XY1_A)$. Here one must not only assume that $X$ is independent on $Y$ but that $X$ is independent on $(Y,Z)$.

To see that there is a difference, the usual example works here: take $X$ and $Y$ i.i.d. centered Bernoulli random variables and $Z=XY$. Then $X$ and $Y$ are independent (by definition), as are $Y$ and $Z$, as are $Z$ and $X$ (easy to check), but $X$, $Y$ and $Z$ are not*. And $E(XY|Z)=Z$ although $E(X)E(Y|Z)=0$.

*For example, $P(X=1,Y=1,Z=-1)=0$ but $P(X=1)P(Y=1)P(Z=-1)=1/8$.