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I'm having trouble establishing this identity below.
Suppose $Y$ and $Z$ are random variables on $(\Omega,F,P)$ and $Y \in L^1$. Suppose $Z$ is bounded and let $G \subset F$ be a sub-$\sigma$-field. Show that $$\mathbb{E}(\mathbb{E}(Y|G)Z)=\mathbb{E}(\mathbb{E}(Z|G)Y).$$
Is there a way I can attack it using the Cauchy-Schwarz inequality: $$\mathbb{E}(XY|G)^2 \leq \mathbb{E}(X^2|G)\mathbb{E}(Y^2|G)?$$ Any help would be appreciated.

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What is the relevance of $X \in L^1$? –  PEV Mar 26 '11 at 4:04
@PEV: I think $X = Y$ is intended. You need some condition on $Y$ in order for the conditional expectation to be defined. By Hölder, the conditions $Y \in L^1$ and $Z \in L^\infty$ ensure that both sides of the desired equality are in fact finite. –  t.b. Mar 26 '11 at 4:17

2 Answers 2

up vote 3 down vote accepted

I don't think Cauchy-Schwarz can be of any help because $L^1 \supsetneqq L^2$, usually. Since $\mathbb{E}(Y|G)$ and $\mathbb{E}(Z|G)$ are $G$-measurable we have $\mathbb{E}(\mathbb{E}(Y|G)Z|G) = \mathbb{E}(Y|G)\mathbb{E}(Z|G) = \mathbb{E}(Y\mathbb{E}(Z|G)|G)$. Now take (unconditional) expectation on both sides.

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@Theo: what about the boundedness of $Z$? –  user8280 Mar 26 '11 at 4:25
@Eric: Sorry, what exactly is your question? –  t.b. Mar 26 '11 at 4:29
@Theo: I guess I'd like to know why $Z$ was assumed to be bounded? Thanks. –  user8280 Mar 26 '11 at 4:31
@Eric: As I said in my comment to PEV above, the conditions $Y \in L^1$ and $Z \in L^\infty$ ensure via Hölder that all the functions appearing in my answer are integrable and thus their (conditional) expectations are defined and finite. If you want me to elaborate, I should know under what conditions exactly conditional expectations is defined in your course. –  t.b. Mar 26 '11 at 4:40
@Theo: Thanks very much. I get it now. –  user8280 Mar 26 '11 at 4:52

Hint: Both are equal to $E(E(Y|G)E(Z|G))$.

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