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Let $(\Omega, \mathcal{A},P)$ be a probability space, and let $(\mathcal{F}_k)_{k \geq 1}$ be a filtration which converges to $\mathcal{A}$. I suppose it is true that $$ E \left( \big(E \left( X | \mathcal{F}_k \right) \big)^2 \right) \to E \left(X^2 \right). $$ How to prove this? I guess one will need Jensen's inequality and a convergence theorem for submartingales, but cannot find the right reference (I am working on number theory, not probability).

(If you know an answer, please also provide a citable reference.)

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    $\begingroup$ When you say that the filtration $\left(\mathcal F_k\right)_{k\geqslant 1}$ converges to $\mathcal A$, you mean that $\mathcal A$ is the $\sigma$-algebra generated by $\bigcup_k\mathcal F_k$, right? In this case, the martingale convergence theorem ensures that $\mathbb E\left[X\mid \mathcal F_k\right]\to \mathbb E\left[X\mid \mathcal A\right]$ both almost surely and in $\mathbb L^2$, which gives the wanted result using the convergence of the $\mathbb L^2$ norms. $\endgroup$
    – Davide Giraudo
    Commented Dec 9, 2016 at 15:03
  • $\begingroup$ Yes, your comment on $\mathcal{A}$ is what I mean. But still I don't understand the situation. Is the result that you mean what I can find in the section "Discrete-time results" on the Wikipedia page en.wikipedia.org/wiki/…? If yes, then why does it say there that there exists $\textbf{some}$ limit, while I require the limit to be X? $\endgroup$
    – Kurisuto Asutora
    Commented Dec 9, 2016 at 20:03
  • $\begingroup$ I meant the results of the section "Convergence of conditional expectations: Lévy's zero–one law". $\endgroup$
    – Davide Giraudo
    Commented Dec 9, 2016 at 22:10
  • $\begingroup$ But the result there is stated for almost sure convergence and $L^1$ convergence - so it is not applicable in my situation. Right? $\endgroup$
    – Kurisuto Asutora
    Commented Dec 12, 2016 at 9:49
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    $\begingroup$ It also holds in $L^p$ spaces, see Corollary 2. 21 of these notes: statslab.cam.ac.uk/~ps422/mynotes.pdf. $\endgroup$
    – Davide Giraudo
    Commented Dec 12, 2016 at 10:13

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Let us state Corollary 2.1 of these notes.

Let $p>1$, $X\in\mathbb L^p$ and let $\left(\mathcal F_n\right)_{n\geqslant 1}$ be a filtration. Denote by $\mathcal F$ the $\sigma$-algebra generated by $\bigcup_{n\geqslant 1}\mathcal F_n$. Then the convergence $$\lim_{n\to +\infty} \mathbb E\left[X\mid\mathcal F_n\right]=\mathbb E\left[X\mid\mathcal F\right],$$ where the convergence holds almost everywhere and in $\mathbb L^p$.

The wanted result follows from this corollary and the elementary fact that if $\left\lVert Y_n-Y\right\rVert_2\to 0$ then $\left\lVert Y_n\right\rVert_2^2\to \left\lVert Y\right\rVert_2$.

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