Conditional expectation of a product

Question

It is quite elementary that $E(X|X)=X$ or more generally $E(X|\mathcal{F})=X$, if $X$ is $\mathcal{F}$-measureable. What happens with a product though? Is it true that for all random variables $Y$ $$E(XY|X)=XE(Y|X)?$$ Schervish seems to be using this result in his book Theory of Statistics, in the proof of Theorem 2.86.

Answer 1

Yeah, it is (under some mild conditions); it is Theorem 5.1.7 in Durrett's Probability Theory and Examples book. (It also holds more generally; yours is a special case where $F = \sigma(X)$)

In case you cannot access the book (should be free online), the proof goes as follows; you know that the right hand side is measurable by $\sigma(X)$, so we just need to make sure that both terms agree on their integrals on any $X$ measurable set.

Consider $X$ to be first a simple function (let's say an indicator of a set.) Now, if you look at any $A \in \sigma(X)$, we have $\int_A 1_B E[Y | X] = \int_{A \cap B} E[Y | X] = \int_{A \cap B} Y = \int_{A} 1_B Y $. (To go from second to third equality use the fact that $A \cap B$ is in $\sigma(X)$ and the definition of $E[X | Y]$).

Now use linearity to go to simple $X$ and then monotone convergence from below to go to non-negative $X$, and then you can divide it to positive and negative to go to integrable $X$.