# Why $\mathbb{E}[X\mid \sigma(\mathcal{H},\mathcal{E})]=\mathbb{E}[X\mid \mathcal{H}]$?

My problem:

Suppose $$\mathcal{E}$$ and $$\mathcal{H}$$ are sub-$$\sigma$$-algebras of the $$\sigma$$-algebra $$\mathcal{F}$$. Let $$X \in L^1(\Omega,\mathcal{F},\mathbb{P})$$ and $$\sigma(X)=\{X^{-1}(A): A \in \mathcal{B}(\mathbb{R}) \}$$. Suppose that $$\mathcal{E}$$ is independent from $$\sigma(\mathcal{H},\sigma(X))$$.

Then $$\mathbb{E}[X\mid \sigma(\mathcal{H},\mathcal{E})]=\mathbb{E}[X\mid \mathcal{H}]$$

My attempt:

I tried using the characterisation $$\mathbb{E}[XZ]=\mathbb{E}[\mathbb{E}[X\mid \mathcal{H}]Z]$$ for all $$\mathcal{H}$$-measurable and bounded random variable or $$\mathbb{E}[XZ]=\mathbb{E}[\mathbb{E}[X\mid \sigma(\mathcal{H},\sigma(X))]Z]$$ for all $$\sigma(\mathcal{H},\sigma(X))$$-measurable and bounded random variable.

This is a ell known result by Doob.

Theorem: Let $$\mathscr{A}$$, $$\mathscr{B}$$ and $$\mathscr{C}$$ be sub--$$\sigma$$--algebras of $$\mathscr{F}$$. $$\mathscr{A}\perp_\mathscr{C} \mathscr{B}$$ iff \begin{align} \Pr[A|\sigma(\mathscr{C},\mathscr{B})]=\Pr[A|\mathscr{C}]\tag{1}\label{doob-independence} \end{align} for all $$A\in \mathscr{A}$$.

Here is a shot proof:

Suppose that $$\mathscr{A}$$ and $$\mathscr{B}$$ are conditional independent given $$\mathscr{C}$$, that is $$\Pr[A\cap B|\mathscr{C}]=\Pr[A|\mathscr{C}] \Pr[B|\mathscr{C}]$$ for all $$A\in \mathscr{A}$$ and $$B\in \mathscr{B}$$. Then, for any $$A\in\mathscr{A}$$, $$\mathscr{B}$$ and $$C\in\mathscr{C}$$ we have \begin{align} \Pr\big[A\cap\big(C\cap B)\big]&=\Pr\big[ \mathbb{1}_C\Pr[A\cap B|\mathscr{C}]\big]= \Pr\big[\mathbb{1}_C\Pr[A|\mathscr{C}]\Pr[B|\mathscr{C}]\big]\\ &= \Pr\big[\Pr[A|\mathscr{C}]\Pr[B\cap C|\mathscr{C}]\big]= \Pr\Big[\Pr\big[\Pr[A|\mathscr{C}]\mathbb{1}_{B\cap C}\big|\mathscr{C}\big]\Big]\\ &= \Pr\big[\Pr[A|\mathscr{C}]\mathbb{1}_{B\cap C}\big] \end{align} Since $$\sigma(\mathscr{B},\mathscr{C})=\sigma\Big(\{B\cap C: B\in\mathscr{B}, C\in\mathscr{C}\}\Big)$$, a monotone class argument shows that \begin{align} \Pr[A\cap H]=\Pr\big[\Pr[A|\mathscr{C}]\mathbb{1}_H \big] \end{align} for all $$H\in\sigma(\mathscr{B},\mathscr{C})$$. This means that $$\Pr[A|\sigma(\mathscr{B},\mathscr{C})]=\Pr[A|\mathscr{C}]$$

Conversely, suppose that $$\eqref{doob-independence}$$ holds. For any $$A\in\mathscr{A}$$ and $$B\in\mathscr{B}$$ we have \begin{align*} \Pr[A\cap B|\mathscr{C}]=\Pr\Big[\mathbb{1}_{B}\Pr[A|\sigma(\mathscr{B},\mathscr{C})]\Big| \mathscr{C}\Big]= \Pr\Big[\mathbb{1}_B\Pr[A|\mathscr{C}]\Big|\mathscr{C}\Big] =\Pr[A|\mathscr{C}]\Pr[B|\mathscr{C}] \end{align*} This shows that $$\mathscr{A}$$ and $$\mathscr{B}$$ are independent given $$\mathscr{C}$$.

The extension to random variables is done by expanding first to simple functions and then by the usual monotone approximation by simple functions.

• But why can you conclude $\Pr\big[ \mathbb{1}_C\Pr[A\cap B|\mathscr{C}]\big]= \Pr\big[\mathbb{1}_C\Pr[A|\mathscr{C}]\Pr[B|\mathscr{C}]\big]$ i.e. $\Pr[A\cap B|\mathscr{C}]=\Pr[A|\mathscr{C}]\Pr[B|\mathscr{C}]$? – Filippo Giovagnini Aug 28 '20 at 12:29
• I understand, but can we get it from the assumption "$\mathscr{E}$ is indipendent from $\sigma(\mathcal{H},\sigma(X))$"? Intuitively I understand it but I am not able to write it formally. – Filippo Giovagnini Aug 28 '20 at 14:13
• What I am asking you is if we can obtain this property: $\Pr[A\cap B| \mathscr{C}] =\Pr[A| \mathscr{C}]\Pr[B| \mathscr{C}]$ from this one: "$\mathscr{E}$ is indipendent from $\sigma(\mathcal{H},\sigma(X))$". – Filippo Giovagnini Aug 28 '20 at 14:31
• @FilippoGiovagnini: set $\mathscr{A}=\sigma(X)$, $\mathscr{B}=\mathcal{E}$, and $\mathscr{C}=\mathcal{H}$ – Oliver Diaz Aug 28 '20 at 14:40