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.
Conditional Expectation
3
$\begingroup$
probability-theory
probability
-
0@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. – 2011-03-26
2 Answers
3
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.
-
0@Eric: Very good! Glad to have helped. – 2011-03-26
2
Hint: Both are equal to $E(E(Y|G)E(Z|G))$.