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Let $X$ a random variable, $\mathcal{A}$, $\mathcal{B}$, two $\sigma$-algebras such that $\sigma(X,\mathcal{A})$ is independent of $\mathcal{B}$.

Is it always true that $\mathbb{E}[X|\sigma(\mathcal{A}, \mathcal{B})]=\mathbb{E}[X|\mathcal{A}]$?

I can show it for $\mathcal{A}$, $\mathcal{B}$ finite algebras, but I do not know how to extend it to general sigma-algebras.

user88163
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1 Answers1

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What we have to show is that the equation $$\int_CX\mathrm d\mu=\int_C\mathbb E[X\mid\mathcal A]\mathrm d\mu$$ holds true for any $C\in\sigma(\mathcal A,\mathcal B)$. One can see that by the assumption of independence, it's true when $C=A\cap B$ for some $A\in\mathcal A$ and $B\in\mathcal B $. The class $\mathcal C=\{A\cap B,A\in\mathcal A,B\in\mathcal B\}$ is stable by finite intersections, and we can conclude by Dynkin's theorem.

Davide Giraudo
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  • how would be the first step of the proof? I thought at: $\int_A XdP \int_BXdP = \int_A E[X|\mathscr A] \int_B E[X|\mathscr A]$. and then? What happen when I integrate for some B$\in \mathscr B$ an $\mathscr A$ measurable r.v.? – Buddy_ Oct 12 '19 at 17:19