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Let $X$ be a random variable and $\mathcal H$ be a $\sigma$-algebra.

The Doob martingale property says $E(X \mid E(X \mid \mathcal{H}))=E(X \mid \mathcal{H})$ (proved by using the tower property).

Now suppose that we have two $\sigma$-algebras $\mathcal H_1$, $\mathcal H_2$.

Can we say something about

$$E(X \mid E(X \mid \mathcal{H_1}), E(X \mid \mathcal{H_2}))$$

My intuition: the Doob martingale property says that the best approximation of $X$ knowing the approximation $E(X \mid \mathcal{H})$ is $E(X \mid \mathcal{H})$. That would mean that $E(X \mid E(X \mid \mathcal{H_1}), E(X \mid \mathcal{H_2})) = E(X \mid \mathcal{H_2})$ if $\mathcal H_1 \subseteq \mathcal H_2$ since $E(X \mid \mathcal{H_2})$ contains more information about $X$. What do you think?

Jose Avilez
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W. Volante
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  • as far as I know you cannot prove that $E(X | E(X | \mathcal{H}))=E(X|\mathcal{H})$ just using the power property, because $\sigma (E(X|\mathcal{H}))\subset \mathcal{H}$ – Masacroso Feb 05 '22 at 16:44
  • Sorry, what exactly is the question? Are you asking whether it is true that if $\mathcal H_1 \subseteq \mathcal H_2$ then $E[X \mid E[X \mid \mathcal{H_1}], E[X \mid \mathcal{H_2}]) = E[X \mid \mathcal{H_2}]$? That is true and follows pretty directly from the tower property. – Nate Eldredge Feb 05 '22 at 17:10
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    @Masacroso: Yes, I think you can. Let's let $\mathcal{G} = \sigma(E[X \mid \mathcal{H}]) \subseteq \mathcal{H}$. Then $E[X \mid \mathcal{G}] = E[E[X \mid \mathcal{H}] \mid \mathcal{G}]$ by tower, but this equals $E[X \mid \mathcal{H}]$ because $E[X \mid \mathcal{H}]$ is already $\mathcal{G}$-measurable. And the claim at the end of the question has exactly the same proof, letting instead $\mathcal{G} = \sigma(E[X \mid \mathcal{H}_1], E[X \mid \mathcal{H}_2]) \subseteq \mathcal{H}_2$. – Nate Eldredge Feb 05 '22 at 17:14
  • @Nate ah! you are right. I didnt see that $E[X|\mathcal{H}]$ is $\mathcal{G}$ measurable! – Masacroso Feb 05 '22 at 17:47
  • @NateEldredge Thank you, I just needed to see that $\sigma\left(E\left[X \mid \mathcal{H}{1}\right], E\left[X \mid \mathcal{H}{2}\right]\right) \subseteq \mathcal{H}_{2}$! – W. Volante Feb 05 '22 at 18:37
  • @W.Volante: Yeah, it look me a while to realize as well. I thought about several possible counterexamples, then finally decided to just try to prove the statement directly from the definition of conditional expectation, and then it all became immediately clear. – Nate Eldredge Feb 05 '22 at 18:48
  • @NateEldredge Actually, an alternative way would simply be: $H_1 \subseteq H_2$ implies that $\sigma (E[X| H_1]) \subseteq \sigma (E[X|H_2])$ ? Is this true ? – W. Volante Feb 05 '22 at 20:11
  • @W.Volante: That is not true, same counterexample as my previous answer. – Nate Eldredge Feb 05 '22 at 20:12
  • @NateEldredge Correct! Thank you. Here is another one that is kinda related: https://math.stackexchange.com/questions/4376124/on-conditional-independence-under-transformation – W. Volante Feb 07 '22 at 19:26

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