# Conditional expectation and martingales

I have a few questions concerning martingales. Let $Y\in \mathcal{L}^1(\Omega,\mathcal{F},\mathbb{P})$ be given, and $(\mathcal{F}_n)$ a filtration, and define $X_n:=\mathbb{E}[Y|\mathcal{F}_n]$.

We know there exists $X_{\infty}$ such that $X_n\to X_{\infty}$ a.s.

I want to show that for $Y\in \mathcal{L}^2(\Omega,\mathcal{F},\mathbb{P})$ we have $X_n \overset{\mathcal{L}^2}{\to} X_{\infty}$. I hope the following line of reasning is correct.

\begin{align*}\int_{\Omega}|X_n-X_{\infty}|^2d\mathbb{P} &= \int_{\Omega}|X_n^2+X_{\infty}^2-2X_nX_{\infty}|d\mathbb{P}\\&\leq \int_{\Omega}|X_n||X_n-X_{\infty}|d\mathbb{P}+\int_{\Omega}|X_{\infty}||X_{\infty}-X_n|d\mathbb{P}\end{align*} which converges to 0 by $\mathcal{L}^1$-convergence.

Then I want to find a condition s.t. $X_{\infty}= Y$, but isn't this simply saying that $\lim_n \mathcal{F}_n = \mathcal{F}$. ?

Also, i want to find an example where $\mathbb{P}(X_{\infty}= Y)=0$. So, this means $\lim_n \mathbb{P}(\mathbb{E}[Y|\mathcal{F}_n]= Y)=0$. Hence $\mathbb{P}\left\{\omega:\mathbb{E}[Y|\mathcal{F}_n](\omega)= Y(\omega)\right\}=0$ for all n.

How do I find such $Y$ and $\mathcal{F}_n$?

Thank you kindly in advance

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Actually $X_\infty$ is explicitly given by $E[Y\mid \mathcal{F}_\infty]$, where $\mathcal{F}_\infty =\bigvee_n\mathcal{F}_n$. So if $Y$ is $\mathcal{F}_\infty$-measurable, then $X_\infty=Y$ a.s. – Stefan Hansen Oct 29 '12 at 14:02
So does that mean if $\mathbb{P}(E[Y|\mathcal{F}_{\infty}]=Y) = 0$ then Y is not $\mathcal{F}_{\infty}$-measurable?. And does that mean that $\mathcal{F}_{\infty}$ is generated by Null-sets? – DinkyDoe Oct 29 '12 at 15:23

The sequence $\{X_n\}$ form a martingale for the filtration $\{\mathcal F_n\}$, and by Jensen's inequality, $E(X_n^2)\leq EY^2<\infty$ for each $n$. So, if $M_n:=\max_{0\leq k\leq n}|X_k|$, we get by $L^2$-maximal inequality that $$EM_n^2\leq 2E(X_n^2)\leq 2E(Y^2),$$ proving that the random variable $M:=\sup_{k\in\Bbb N}|X_k|^2$ is integrable. As $|X_{\infty}-X_n|^2\leq 4M,$ the result follows from the dominated convergence theorem.
As Stefan Hansen says, we have $X_{\infty}=E[Y\mid\mathcal F_{\infty}]$ a.s., where $\mathcal F_{\infty}$ is the $\sigma$-algebra generated by $\mathcal F_k$, $k\in\Bbb N$.