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.
Best Answer
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.