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Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, $\mathcal{G},\mathcal{H} \subseteq \mathcal{F}$ be sub-$\sigma$-algebras of $\mathcal{F}$ and let $X$ be a real-valued random variable that is independent of $\mathcal{H}$. Does $$ \mathbb{E}[X\mid\sigma(\mathcal{G},\mathcal{H})]=\mathbb{E}[X\mid\sigma(\mathcal{G})] $$ hold?

Intuitively, I think this should hold, by I can not think of a proof for this. I would be grateful for help.

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3 Answers 3

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This is not true. Let $U,V$ be i.i.d. $N(0,1)$, $X=U+V, Y=U$ and $Z=U-V$. It is well known (and easy to prove) that $X$ and $Z$ are independent. But $E[X|Y,Z]=U+V$ and $E[X|Y]=E[U+V|U]=U$. So $\mathcal G=\sigma (Y), \mathcal H=\sigma (Z)$ provides a counter-example.

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Here is another counterexample: Let $Y,Z$ be independent variables taking the values $\pm 1$ with probability $1/2$ each. Then $X=YZ$ is independent of $Z$, yet it satisfies $E[X|Y,Z]=X$ and $E[X|Y]=0$.

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Yuval Peres and geetha290krm already provide counter-examples. It is worth mentioning that the additional assumption that $\mathcal H$ is independent of $\sigma(X)\vee\mathcal G$ makes equality $$ \mathbb{E}[X\mid\sigma(\mathcal{G},\mathcal{H})]=\mathbb{E}[X\mid\sigma(\mathcal{G})] $$ true, as already shown here.

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  • $\begingroup$ That is really helpful, thanks a lot! $\endgroup$
    – Learner
    Sep 21, 2022 at 14:15

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