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?

Answer 1

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?