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<blockquote data-quote="Nate River" data-source="post: 740136">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 <a href="https://math.stackexchange.com/a/2637771">here</a> 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 <a href="https://en.wikipedia.org/wiki/Projections_onto_convex_sets">algorithm</a>, popularised by von Neumann.</blockquote>

[QUOTE="Nate River, post: 740136"] [I]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 $\mathcal F$.[/I] 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 [I]not[/I] hold in general, see for example [URL='https://math.stackexchange.com/a/2637771']here[/URL] for an elementary construction where all three of the values above are different. However, I believe the following holds. [B]Question:[/B] 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$? [B]Comments:[/B] 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 [URL='https://en.wikipedia.org/wiki/Projections_onto_convex_sets']algorithm[/URL], popularised by von Neumann. [/QUOTE]

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