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Notation: We write $\mathbb E_{\mathcal F} X$ for the conditional expectation $\mathbb E[X|\mathcal F]$ of a random variable $X$ with respect to a $\sigma$-algebra $\mathcal F$.

Let $X$ be an integrable random variable with associated $\sigma$-algebra $\mathcal X$, and let $\mathcal G, \mathcal H$ be sub $\sigma$-algebras of $\mathcal X$. It is known that the seemingly intuitive identity

$$\mathbb E_{\mathcal G} \mathbb E_{\mathcal H} X = \mathbb E_{\mathcal H} \mathbb E_{\mathcal G} X = \mathbb E_{\mathcal G \cap \mathcal H} X$$

does not hold in general, see for example here for an elementary construction where all three of the values above are different.

However, I believe the following holds.

Question: Is it true that the sequence

$$\mathbb E_{\mathcal H} X, \mathbb E_{\mathcal G} \mathbb E_{\mathcal H} X, \mathbb E_{\mathcal H} \mathbb E_{\mathcal G} \mathbb E_{\mathcal H} X, \mathbb E_{\mathcal G} \mathbb E_{\mathcal H} \mathbb E_{\mathcal G} \mathbb E_{\mathcal H} X, \dots$$

converges almost surely to $ \mathbb E_{\mathcal G \cap \mathcal H} X$?

Comments: The desired result seems to be true if $X$ is a random variable taking finitely many values. Indeed, viewing the conditional expectation operator as a projection, the convergence is guaranteed by the following algorithm, popularised by von Neumann.

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    $\begingroup$ Von Neumann's theorem about alternating orthogonal projections in Hilbert spaces yields $L^2$-convergence for $X\in L^2$. $\endgroup$ Commented Aug 2 at 7:38
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    $\begingroup$ One certainly has strong convergence in $L^2$. (This is true for general orthogonal projections on Hilbert space, not only for conditional expectations, and is, I think, also due to von Neumann. It should be quite straightforward to prove by using the spectral theorem for self-adjoint operators.) I don't know about the almost everywhere convergence, though. $\endgroup$ Commented Aug 2 at 7:39
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    $\begingroup$ Ok, the other Jochen was a minute earlier. ;-) $\endgroup$ Commented Aug 2 at 7:40

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Here is a partial answer: It is at least true under the slightly stronger integrability assumption that $X \in L^p$ for some $p > 1$ (I don't know about the case $X \in L^1$, though).

First, as mentioned in the comments, strong convergence on $L^2$ (on hence on $L^1$ by density) follows from a Theorem a von Neumann for alternating orthogonal projections on Hilbert spaces.

The almost everywhere convergence for $X \in L^p$ with $p > 1$ follows from the following result of Stein (see Corollary 1 on page 1895 in this classical article):

Theorem. Let $(\Omega,\mu)$ be a measure space and let $P$ be a self-adjoint and positive semi-definite bounded linear operator on $L^2(\Omega,\mu)$. Assume that $P$ extrapolates to a linear operator of norm $\le 1$ on $L^1$. Then $P^nf$ converges almost everywhere as $n \to \infty$ if $f \in L^p(\Omega,\mu)$ for some $p \in (1,\infty)$.

Now, to obtain the claim for $X \in L^p$ with $p > 1$, apply the theorem to each of the two operators $P_1 := \mathbb{E}_{\mathcal{H}} \mathbb{E}_{\mathcal{G}} \mathbb{E}_{\mathcal{H}}$ and $P_2 := \mathbb{E}_{\mathcal{G}} \mathbb{E}_{\mathcal{H}} \mathbb{E}_{\mathcal{G}}$ and use von Neumann's theorem to see that the almost everywhere limits of $P_1^n X$ and $P_2^n \mathbb{E}_{\mathcal{H}} X$ coincide.

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  • $\begingroup$ Thanks for the answer! I may be missing something obvious, but how does one obtain the final conclusion from the fact that the two limits coincide? $\endgroup$
    – Nate River
    Commented Aug 2 at 11:23
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    $\begingroup$ @NateRiver: The functions $P_1^n X$ are the elements of your sequence with the odd indices and the functions $P_2^n \mathbb{E}_{\mathcal{H}} X$ are the elements of your sequence with the even indices. If the "odd" and the "even" subsequence both converge almost everywhere to the same function, the entire sequence converges almost everywhere. $\endgroup$ Commented Aug 2 at 13:42
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    $\begingroup$ ... I missed that $\mathbb E_{\mathcal G} \mathbb E_{\mathcal G} = \mathbb E_{\mathcal G}$. Nice! $\endgroup$
    – Nate River
    Commented Aug 2 at 14:10

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