Show that $E[X \mid \sigma (\mathscr{G},\mathscr{H})] = E[X \mid \mathscr{G}]$ if $\mathscr{H}$ independent of $\sigma (\sigma (X),\mathscr{G})$

Question

The problem I'm working on is:

If $X \in L^1(\Omega,\mathscr{F},P)$, and $\mathscr{G},\mathscr{H}$ are $\sigma$ sub-algebras of $\mathscr{F}$, with $\mathscr{H}$ independent of $\sigma (\sigma (X),\mathscr{G})$, then show that $E[X \mid \sigma (\mathscr{G},\mathscr{H})] = E[X \mid \mathscr{G}]$

Here is where I'm at:

Since $E[X \mid \sigma (\mathscr{G},\mathscr{H})$ is the random variable in $L^2$ that is $\sigma(\mathscr{G},\mathscr{H})$ measurable that minimizes $E[(X-Y)^2]$, and $E[X \mid \mathscr{G}]$ is the random variable in $L^2$ that is $\mathscr{G}$ measurable and minimizes $E[(X-Y)^2]$, I believe what I need to do here is show that for any random variable $Y$ that is $\sigma (\mathscr{G},\mathscr{H})$ measurable that minimizes $E[(X-Y)^2]$, it is also $\mathscr{G}$ measurable. This would show $E[X \mid \sigma (\mathscr{G},\mathscr{H})] \geq E[X \mid \mathscr{G}]$, and $\leq$ I think follows easily since any thing that is $\mathscr{G}$ measurable is certainly $\sigma (\mathscr{G},\mathscr{H})$ measurable.

I'm not sure how to show this, or if this is the right approach.

Answer 1

Recall that a $\mathcal{E}$-measurable $L^2$-integrable random variable $Y$ equals $\mathbb{E}(X \mid \mathcal{E})$ if, and only if,

$$\int_{D} X \, d\mathbb{P} = \int_D Y \, d\mathbb{P} \tag{1}$$

for all $D \in \mathcal{D}$ where $\mathcal{D}$ is a $\cap$-stable generator of the sub-algebra $\mathcal{E}$.

---

Here, we set

$$\mathcal{E} := \sigma(\mathcal{G},\mathcal{H}) \qquad \mathcal{D} := \{G \cap H; G \in \mathcal{G}, H \in \mathcal{H}\}.$$

First of all, we note that $\mathcal{D}$ is in fact a $\cap$-stable generator of $\mathcal{E}=\sigma(\mathcal{G},\mathcal{H})$. Moreover, we have due to the independence of $\mathcal{H}$ and $\sigma(\sigma(X),\mathcal{G})$:

$$\begin{align*} \int_{G \cap H} \mathbb{E}(X \mid \mathcal{G}) \, d\mathbb{P} &= \int 1_H \underbrace{\mathbb{E}(1_G X \mid \mathcal{G})}_{\sigma(\sigma(X),\mathcal{G})-\text{measurable}} \, d\mathbb{P} \\ &= \int 1_H \, d\mathbb{P} \cdot \int 1_G \mathbb{E}(X \mid \mathcal{G}) \, d\mathbb{P} \\ &= \int 1_H \, d\mathbb{P} \int_G X \, d\mathbb{P} \\ &= \int 1_H \, d\mathbb{P} \int_G X \, d\mathbb{P} \\ &= \int 1_G 1_H X \, d\mathbb{P}.\end{align*}$$

This shows that $(1)$ holds for any $D=G \cap H \in \mathcal{D}$. Consequently, the claim follows.