$\lim_{n \to \infty}E[f(X_1,X_2)|\mathcal{F}_n]=E[f(X_1,X_2)]$ a.s.

Question

The exercise is related to a question that was asked before: Show that $E[f(X_1,X_2)\mid \mathcal{F}_n]=\frac{2}{n(n-1)}\sum_{ 1 \leq p<q \leq n}f(X_p,X_q)$, where we proved that for every $n \geq 2, Y_n=E[f(X_1,X_2)\mid \mathcal{F}_n].$

From the martingale convergence theorem, we deduce that $Y_n$ converges a.s. and in $L^1$ to $Y_{\infty}=E[f(X_1,X_2)|\bigcap_{n}\mathcal{F}_n].$

How can we prove that $Y_{\infty}$ is a.s. equal to an $\bigcap_{n}\sigma(X_n,X_{n+1},...)$-measurable random variable? And therefore to deduce from this that $Y_{\infty}=E[f(X_1,X_2)]$ a.s.

Attempt: tried to consider $\limsup_{n}Y_n$ but I do not know if this will work.

Answer 1

The equality $Y_{\infty}=E[f(X_1,X_2)]$ a.s. can be deduced from the following reasoning. First, $$ \mathbb E\left\lvert Y_n-\frac1{\binom n2}\sum_{k_0\leqslant i<j\leqslant n}f(X_i,X_j)\right\rvert\leqslant \frac{k_0n}{\binom n2}\mathbb E\left[\left\lvert f(X_1,X_2)\right\rvert\right] $$ hence for each $k_0$, $Y_\infty$ is in the limit in $L^1$ of the sequence $\left(\frac1{\binom n2}\sum_{k_0\leqslant i<j\leqslant n}f(X_i,X_j)\right)_{n \geqslant 1}$, which is $\sigma(X_j,j\geqslant k_0)$-measurable. As a consequence $Y_\infty$ should be almost surely equal to a $\bigcap_{k_0\geqslant 1}\sigma(X_j,j\geqslant k_0)$-measurable random variable hence constant.

Answer 2

You already know that $Y_n$ converges almost surely, so you already know that $Y_{\infty} = \lim_{n\to\infty} Y_n$, therefore no need to consider the limes superior. It is clear that $Y_n$ is $\sigma(X_1,\dots,X_n)$-measurable by definition. A priori, $Y_{\infty}$ is therefore almost surely equal to a $\sigma(X_1,X_2,\dots)$-measurable random variable (I'll drop the disclaimer about almost surely equal sometimes in the following for the sake of brevity). However, if we leave away all the terms involving $X_1$ in the definition of $Y_n$, the limit is unchanged (I'd advise to make sure you understand that and check it), so $Y_{\infty}$ is even $ \sigma(X_2,X_3,\dots)$-measurable. Proceeding inductively, we see that $Y_{\infty}$ is actually $\sigma(X_n,X_{n+1},\dots)$-measurable for every $n \in \mathbb{N}$, i.e., $Y_{\infty}$ is $\cap_{n} \sigma(X_n,X_{n+1},\dots)$-measurable. By Kolmogorov's 0-1 law, this implies that $Y_{\infty}$ is almost surely equal to a constant. Since $Y_{n}$ converges in $L^1$, we have $$ |\mathbb{E}[Y_n] - \mathbb{E}[Y_{\infty}]| \leq \mathbb{E}[|Y_n - Y_{\infty}|] \to 0 $$ Since $\mathbb{E}[Y_n] = \mathbb{E}[f(X_1,X_2)]$ and $Y_{\infty}$ is almost surely constant, it follows that $Y_{\infty} = \mathbb{E}[f(X_1,X_2)]$ almost surely.