Conditional expectation expressions

Question

I'm deriving the law of total variance and come across these two expression:

$$E[Y*E[Y|X]]$$ $$E[E[Y|X]*E[Y|X]]$$

$Y$ and $X$ are random variables.

Intuitively I think they should be equal, though I have no idea of how to prove it.

Answer 1

The definition I am familiar with is the following for $L^1$ functions, which is derived from the definition for $L^2$ functions using Hilbert Space. We necessarily need $E(Y|X)$ to be in $L^2$ for this problem.

Definition: Let $Y\in L^1(\Omega, \mathcal{F}, P)$ and $X$ a random variable. Then the conditional expectation of $Y$ given $X$ is the (unique) random variable denoted by $E(Y|X) \in L^1(\Omega, \sigma(X), P)$ where $$E(YZ) = E(E(Y|X)Z)$$ for all $Z$ bounded and $\sigma(X)$-measurable. Equivalently, $$E(Yf(X)) = E(E(Y|X)f(X))$$ for all $f$ Borel.

Hence, from simply the definition, let $Z = E(Y|X)$ and you are done if $E(Y|X)$ is bounded. If not, consider $E(Y|X) \wedge n$, which is bounded. Then $E(Y|X) \wedge n \cdot E(Y|X) \to E^2(Y|X)$ and $|E(Y|X) \wedge n \cdot E(Y|X)| \leq E^2(Y|X) \in L^1$ so by Lebesgue Dominated Convergence Theorem, $$E(E(Y|X)\cdot E(Y|X)\wedge n) \to E(E(Y|X)E(Y|X))$$ Similarly, $YE(Y|X)\wedge n \to YE(Y|X)$ and $|YE(Y|X)\wedge n| \leq |YE(Y|X)| \in L^1$, so $$E(YE(Y|X)\wedge n) \to E(YE(Y|X))$$ Since $E(Y|X) \wedge n$ bounded and $\sigma(X)$-measurable, $$ E(E(Y|X)\cdot E(Y|X)\wedge n) = E(YE(Y|X)\wedge n)$$ so take limit and use the above two results.