How to show $E(XE(Y\mid F)=E(E(X\mid F)Y)$?

Question

Let $X,Y\in\mathcal{L}^2$ and let $F$ be a $\sigma$-algebra.

How to show that $E(XE(Y\mid F)=E(E(X\mid F)Y)$?

Maybe you can give me some help?

It's more conventional to use $\mathscr{F}$ to denote a general $\sigma$-algebra. "$F$" may be confused with the random variable named $F$. — Zhanxiong, Jul 02 '15 at 15:02

score 3 · Accepted Answer · edited Jul 02 '15 at 14:51

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The trick is iterated expectation. $$ E[XE(Y\mid F)]=E[E(XE(Y\mid F)\mid F)]=E[E(Y\mid F)E(X\mid F)] $$ where the second equality is because $E(Y\mid F)$ is $F$-measurable. Now can you do the same with $E[E(X\mid F)Y]$ to produce the rightmost expression above?

edited Jul 02 '15 at 14:51

Michael Hardy

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answered Jul 02 '15 at 14:13

Kim Jong Un

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$E(E(X|F)Y)=E(E(E(X|F)Y|F))=E(E(X|F)E(Y|F))$? – Rhjg Jul 02 '15 at 14:29
@Rhjg Yes, that's right. – Kim Jong Un Jul 02 '15 at 14:43

How to show $E(XE(Y\mid F)=E(E(X\mid F)Y)$?

1 Answers1