# Rule with independent random variables and conditional expectations

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).

• The faulty step is when you assert that $E(X)E(Y1_A)=E(XY1_A)$. Here you must not only assume that $X$ is independent of $Y$ (which is not enough to conclude) but that $X$ is independent of $(Y,Z)$. (+1 for showing the steps you tried.)
– Did
Aug 4, 2011 at 8:56
• The usual counterexample works: take X and Y i.i.d. centered Bernoulli random variables and Z=XY. Then (X,Y) is independent (by definition), as are (Y,Z) and (Z,X) (easy to check), but (X,Y,Z) is not (for example P(X=Y=1,Z=-1)=0 instead of $\frac18$). And E(XY|Z)=Z although E(X)E(Y|Z)=0.
– Did
Aug 4, 2011 at 12:05
• @did I think your first comment serves as a good answer for me already. By (Y,Z) you mean $\sigma(\sigma(Y), \sigma(Z))$? And independence of Y, Z might not be enough to have independence from (Y,Z) (see my new question ) Aug 4, 2011 at 12:34
• Yes, [X is independent of (Y,Z)] means that the sigma-algebras sigma(X) and sigma(Y,Z) are independent. And sigma(Y,Z) coincides with sigma(sigma(Y),sigma(Z)).
– Did
Aug 4, 2011 at 13:37
• So I would have to have (i) X independent from (Y,Z).. I wonder if that follows if I assume Y and Z independent (which would be more natural to my application than to have to introduce the assumption (i)) - i make a new question out of this. Aug 5, 2011 at 11:47

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 $P(X=1)P(Y=1)P(Z=-1)=1/8$.