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Let $(\Omega, \mathcal{F},\mathbb P)$ be a probability space and let $X$ be a random variable in $L^1$. Let $(\mathcal{F}_k)_k$ be any filtration, and define $\mathcal{F}_{\infty}$ to be the minimal $σ$-algebra generated by $(\mathcal{F}_k)_k$. Then

$$E[X|\mathcal{F}_k]\rightarrow E[X|\mathcal{F}_{\infty}],\ \ k\rightarrow\infty$$

both $\mathbb P$-almost surely and in $L^1$.

Could someone give me a proof of this proposition or a reference? Many thanks!

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    $\begingroup$ Here are some suggestive questions. Are you familiar with the Levy Martingale $X_k = E[X \mid \mathcal{F}_k]$ given an $L^1$-random variable $X$? Do you know the Martingale Convergence Theorem? This will give you almost-sure convergence of the Levy martingale. Now how can you conclude $L^1$ convergence? (Hint: use Uniform Integrability). $\endgroup$ – A Blumenthal Jun 25 '13 at 21:05
  • $\begingroup$ Thank you for your reply. By Martingale Convergence Theorem we can conclude that $X_k$ converge a.s and $L^1$ to $X_{\infty}$, but how could we show $X_{\infty}=E[X|\mathcal{F}_{\infty}]$? For example, I do not know how to prove that $X_{\infty}$ is measurable to $\mathcal{F}_{\infty}$, could you give some more hints about that? $\endgroup$ – Higgs88 Jun 25 '13 at 21:34
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I decided this part was too long for a comment. We know that with $X_k = E[X \mid \mathcal{F}_k]$, we have $$ X_k \rightarrow X_{\infty} $$ in $L^1$ and almost surely for some $L^1$ random variable $X_{\infty}$.

Recall that $\mathcal{F}_{\infty} = \sigma \left(\cup_k \mathcal{F}_k \right)$. To show that $X_{\infty} = E[X \mid \mathcal{F}_{\infty}]$, we'll appeal to the defining property of conditional expectation: we must show that $$ (*) ~~~~~E[X I_A ] = E[X_{\infty} I_A] $$ for all $A \in \mathcal{F}_{\infty}$. Convince yourself that $(*)$ holds for all $A \in \mathcal{F}_k$ for any $k$ (Hint: use Dominated Convergence), and thus that $(*)$ holds for all $A \in\cup_k \mathcal{F}_k$, as this is an increasing union.

There's a technical problem here, that $\cup_k \mathcal{F}_k$ is merely an algebra, and not generally a sigma-algebra. Resolving this is a good exercise in applying the Dynkin $\pi-\lambda$ theorem. Try to work this out using the fact that $\cup_k \mathcal{F}_k$ is a $\pi$-system, and show that the set of $A \in \mathcal{F}_{\infty}$ for which $(*)$ holds is a $\lambda$-system.

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  • $\begingroup$ Thanks a lot for your help! $\endgroup$ – Higgs88 Jun 26 '13 at 21:28
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Let $Y_n = E(X| \mathcal{F}_n)$. Then it $Y_n$ is a martingale, and $$\sup_n E(|Y_n|) = \sup_n E(|E(X| \mathcal{F}_n)|) \leq \sup_n E(E(|X||\mathcal{F}_n)) = E(|X|) $$ where the bound in the middle is due the conditional Jensen inequality.

Now the heavy artillery, by Doob's Convergence theorem $Y_\infty := \lim_{n \to \infty} Y_n$ exists almost surely. And since the sequence is dominated by $X$ (again conditional Jensen) we conclude the $L^1$ convergence and thus the convergence in probability.

You can find the Doob's Convergence theorem in Williams' "Probability with Martingales" Thm. 11.5. Is a rather important result based on a "band argument" and it can be extended to continuous time martingales.

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  • $\begingroup$ Thanks for your help, I will also be 2013 Phd student at Polytechnique, who is your advisor and which is your speciality? $\endgroup$ – Higgs88 Jun 26 '13 at 21:30
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    $\begingroup$ Cool! I work with Prof. Graham. My work can be described roughly as applications of point processes to telecomunication network. À bientôt! $\endgroup$ – Bunder Jun 27 '13 at 6:54
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Thanks so much for A Blumenthal and Bunder, I could accomplish the proof.

Firstly we show that $X_{\infty}$ is measurable with respect to $\mathcal{F}_{\infty}$. For every $X_n$ is measurable in $\sigma$-algebra $\mathcal{F}_n$, so in $\mathcal{F}_{\infty}$, then the limit a.s. $X_{\infty}$ is measurable in $\mathcal{F}_{\infty}$.

Then we have to show

$$(1)\ \ E[1_{A}X]=E[1_{A}X_{\infty}],\ \ \forall A\in\cup_n\mathcal{F}_n$$

implies

$$(2)\ \ E[1_{A}X]=E[1_{A}X_{\infty}],\ \ \forall A\in\mathcal{F}_{\infty}=\sigma(\cup_n\mathcal{F}_n)$$

Let $\mathcal{F}:=\{A\in\mathcal{F}_{\infty}: (1) \text{ holds}\}$, then we see that $\mathcal{F}$ is a $\lambda$-system:

  1. $\Omega\in\mathcal{F}$;

  2. if $A\subset B\in\mathcal{F}$, then $B\A\in\mathcal{F}$;

  3. if $A_n$ is a sequence of sets in $\mathcal{F}$ s.t $A_n\subset A_{n+1}$, then by dominated convergence theorem we have $\cup_nA_n\in\mathcal{F}$

and $\cup_n\mathcal{F}_n$ is a $\pi$-system:

  1. $\cup_n\mathcal{F}_n$ is not empty;

  2. if $A$, $B\in\mathcal{F}$, then $A\cap B\in\mathcal{F}$

By Dynkin System Theorem we have $\mathcal{F}=\mathcal{F}_{\infty}$, so by definition we prove that $X_{\infty}=E[X|\mathcal{F}_{\infty}]$.

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