Show that $E[f(X_1,X_2)\mid \mathcal{F}_n]=\frac{2}{n(n-1)}\sum_{ 1 \leq pLet $(X_k)_k$ be a sequence of i.i.d random variables and $f:\mathbb{R}^2 \to \mathbb{R}$ be a measurable function such that $f(X_1,X_2) \in L^1$ and for every $(x,y) \in \mathbb{R}^2,f(x,y)=f(y,x).$ Let for $n \geq 2,Y_n=\frac{2}{n(n-1)}\sum_{ 1 \leq p<q \leq n}f(X_p,X_q)$ and $\mathcal{F}_n=\sigma(Y_n,Y_{n+1},...)$.
Show that for $n \geq 2, Y_n=E[f(X_1,X_2)\mid \mathcal{F}_n].$
We have that
\begin{align}
\sum_{1 \leq p<q \leq n}f(X_p,X_q)&=E\left[\sum_{1 \leq p<q \leq n}f(X_p,X_q)\mid Y_n \right]
\\&=\sum_{1 \leq p <q \leq n}E [f(X_p,X_q)\mid Y_n ]
\\&=\sum_{1 \leq p<q \leq n}E [f(X_1,X_2)\mid Y_n ]
\\&=\frac{n(n-1)}{2}E [f(X_1,X_2)\mid Y_n ]
\end{align}
which is true since for $1 \leq p < q \leq n, P_{(X_p,X_q,U_n)}=P_{(X_1,X_2,U_n)}.$
How to verify that $Y_n=E[f(X_1,X_2)\mid \mathcal{F}_n ]$ ?
 A: Since the $\sigma$- algebra $\mathcal{F}_{n}$ is generated by sets of the form $\{Y_{n}\in{A_{n}}, Y_{n+1}\in{A_{n+1}}, ..., Y_{n+k}\in{A_{n+k}}\}$ where $A_{n}, A_{n+1}, ..., A_{n+k}$ are Borel subsets of $\mathbb{R}$, to verify that $\mathbb{E}(f(X_{1},X_{2})|\mathcal{F}_{n})=Y_{n}$, it suffices to show that:
$$\int_{Y_{n}\in{A_{n}}, Y_{n+1}\in{A_{n+1}}, ..., Y_{n+k}\in{A_{n+k}}} f(X_{1},X_{2})d\mathbb{P}=\int_{Y_{n}\in{A_{n}}, Y_{n+1}\in{A_{n+1}}, ..., Y_{n+k}\in{A_{n+k}}} Y_{n}d\mathbb{P}$$
To this effect, notice that the values of $Y_{n}, Y_{n+1}, Y_{n+2}, ...$ are invariant under permutation of $X_{1}, X_{2}, ..., X_{n}$. Thus:
$$ \int_{Y_{n}\in{A_{n}}, Y_{n+1}\in{A_{n+1}}, ..., Y_{n+k}\in{A_{n+k}}} f(X_{1},X_{2})d\mathbb{P}=\int_{Y_{n}\in{A_{n}}, Y_{n+1}\in{A_{n+1}}, ..., Y_{n+k}\in{A_{n+k}}} f(X_{i},X_{j})d\mathbb{P}$$
for any choice of $i,j$ such that $1\leq{i}<j\leq{n}$. Summing over all such $i$ and $j$ on both sides of the equation above and then dividing by $\frac{n(n-1)}{2}$ then gives us the desired result.
A: Let $\mathcal F_{n}^m$ be the $\sigma$-algebra generated by the random variables $U_n,\dots,U_m$. Then $(\mathcal F_n^m)_{m\geqslant n}$ is a filtration. By the martingale convergence theorem, it suffices to show that for each fixed $m$,$\mathbb E[f(X_1,X_2)\mid \mathcal F_n^m]=U_n$.
Let $\mathcal C$ be the collection of the sets of the form $\bigcap_{i=n}^m \{U_i\in B_i\}$, where $B_i$ is a Borel subset of the real line. Then $\mathcal C$ is a $\pi$-systm generating $\mathcal F_n^m$. Therefore, it suffices to prove that
$$\mathbb E\left[f(X_1,X_2)\mathbb 1_{\bigcap_{i=n}^m \{U_i\in B_i\}}\right]=\mathbb E\left[U_n\mathbb 1_{\bigcap_{i=n}^m \{U_i\in B_i\}}\right]
$$
for each Borel subsets $B_n,\dots,B_m$ of the real line. This comes from the fact that $U_n=\frac 1{C^2_n}f_n(X_1,\dots,X_n)$ with $f_n(x_1,\dots,x_n)=\frac 1{\binom n2}\sum_{1\leqslant i<j\leqslant n}f(x_i,x_j)=\frac 1{2\binom n2}\sum_{1\leqslant i\neq j\leqslant n}f(x_i,x_j)$ by symmetry of $f$. As a consequence, $U_n=g_n(x_1,\dots,x_n)$, where $g_n(x_1,\dots,x_n)=\frac 1{2\binom n2}\sum_{1\leqslant i\neq j\leqslant n}f(x_i,x_j)$ and $g_n$ is symmetric in all its arguments. Consequently, $(X_1,X_2,U_n,\dots,U_m)$ has the same distribution as $(X_p,X_q,U_n,\dots,U_m)$ for $1\leqslant p<q\leqslant n$ and it follows that
$$\mathbb E\left[f(X_1,X_2)\mathbb 1_{\bigcap_{i=n}^m \{U_i\in B_i\}}\right]=\mathbb E\left[f(X_p,X_q)\mathbb 1_{\bigcap_{i=n}^m \{U_i\in B_i\}}\right].$$
