How to prove this identity for discrete random variables?

Question

How to prove that $E(ξE(η|G)) = E(ηE(ξ|G))$, where $ξ$, $G$ and $η$ are random discrete variables. Of course, if both parts exist.

Answer 1

In this answer $G$ must be looked at as a $\sigma$-algebra (eventually generated by a random variable).

Characteristic for $\mathbb E[\xi\mid G]$ is that it satisfies:$$\int_A\xi(\omega)P(d\omega)=\int_A\mathbb E[\xi\mid G](\omega)P(d\omega)\text{ whenever }A\text{ is }G\text{-measurable}$$

Or equivalently:$$\mathbb E[\xi\mathbf1_A]=\mathbb E[\mathbb E[\xi\mid G]\mathbf1_A]\text{ whenever }A\text{ is }G\text{-measurable}$$

This can be expanded to the more general statement that: $$\mathbb E[\xi\psi]=\mathbb E[\mathbb E[\xi\mid G]\psi]\text{ whenever }\psi\text{ is }G\text{-measurable}$$

Now note that $\mathbb E[\eta\mid G]$ is by definition $G$-measurable so that we are allowed to conclude that:$$\mathbb E[\xi\mathbb E[\eta\mid G]]=\mathbb E[\mathbb E[\xi\mid G]\mathbb E[\eta\mid G]]\tag1$$

Similarly we have:$$\mathbb E[\eta\mathbb E[\xi\mid G]]=\mathbb E[\mathbb E[\eta\mid G]\mathbb E[\xi\mid G]]\tag2$$

Now note that $(1)$ and $(2)$ have equal RHS.

Answer 2

From the law of total expectation

\begin{align*} E\left(\xi E\left(\eta|G\right)\right) & =E\left(E\left(\xi E\left(\eta|G\right)|G\right)\right)\\ & =E\left(E\left(\eta|G\right)E\left(\xi|G\right)\right)\\ & =E\left(E\left(\eta E\left(\xi|G\right)|G\right)\right)\\ & =E\left(\eta E\left(\xi|G\right)\right) \end{align*}