A quick question on Conditional Expectation

Question

Let $(\Omega,\mathcal{F},P)$ be a probability space and $\mathcal{G}\subset\mathcal{F}$ be a sub-$\sigma$-field. Also let $X$ be an integrable random variable, and $E(X|\mathcal{G})$ the conditional expectation w.r.t. $\mathcal{G}$.

Show that for any bounded $\mathcal{G}$-measurable r.v. $Z$, $E(XZ)=E(E(X|\mathcal{G})Z)$. -(*)

By "bounded" I take it they mean that for some $M>0$, we have $\{|Z|>M\}=\emptyset$ or is it "bounded almost surely", so $\{|Z|>M\}$ is contained in a null set, either way will it make a difference??

Anyway I think the need $Z$ to be bounded as to ensure the product is still integrable, is this right?

And my idea to show the proposition is to show it is 1st true for simple functions converging towards $Z$ and this will follow as by definition of conditional expectation (*) holds for indicator functions (partial averaging property), then just apply Monotone Convergence theorem. Is this right? If not please let me know what to do, or if there is a simpler way to show it please tell me.

Thanks in advance. Any help and hints greatly appreciated.

Answer 1

Using the tower property, the claim follows easily: Since

$$\mathbb{E}(X Z) = \mathbb{E}\big( \mathbb{E}(XZ \mid \mathcal{G}) \big)$$

we get

$$\mathbb{E}(X Z) = \mathbb{E} \big( Z \mathbb{E}(X \mid \mathcal{G}) \big)$$

using that $Z$ is $\mathcal{G}$-measurable.

Concerning your remaining questions:

• Yes, the boundedness ensures the integrability of $X \cdot Z$. However, the claim holds true for any ($\mathcal{G}$-measurable) random variable $Z$ such that $X \cdot Z \in L^1$.
• "Boundedness" means (at least in this context) that $Z$ is almost surely bounded, i.e. $\mathbb{P}(|Z| \geq M)=0$ for $M$ sufficiently large. However, since the set $\{|Z| \geq M\}$ has measure $0$, we can always assume without loss of generality that $Z$ is bounded (in the sense that $|Z(\omega)| \leq M$ for all $\omega$). Indeed, if we set $$\tilde{Z}(\omega) := Z(\omega) 1_{[-M,M]}(Z(\omega)),$$ then $\tilde{Z}$ is bounded and $$\int f(X,Z) \, d\mathbb{P} = \int f(X,\tilde{Z}) \, d\mathbb{P}$$ for any measurable function $f$ such that $f(X,Z)$ is integrable.