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!

  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ Thanks a lot for your help! $\endgroup$ – Higgs88 Jun 26 '13 at 21:28

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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

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


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

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 $X_{\infty}=E[X|\mathcal{F}_{\infty}]$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.