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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.

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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.

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  • $\begingroup$ 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. $\Pr[A\cap B|\mathscr{C}]=\Pr[A|\mathscr{C}]\Pr[B|\mathscr{C}]$? $\endgroup$ – Filippo Giovagnini Aug 28 '20 at 12:29
  • $\begingroup$ 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. $\endgroup$ – Filippo Giovagnini Aug 28 '20 at 14:13
  • $\begingroup$ 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 $\sigma(\mathcal{H},\sigma(X))$". $\endgroup$ – Filippo Giovagnini Aug 28 '20 at 14:31
  • $\begingroup$ @FilippoGiovagnini: set $\mathscr{A}=\sigma(X)$, $\mathscr{B}=\mathcal{E}$, and $\mathscr{C}=\mathcal{H}$ $\endgroup$ – Oliver Diaz Aug 28 '20 at 14:40

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