Conditional expectation of symmetric statistic when conditioning on multiple iid random variables have equal distribution This is one step further than the related link: Conditional expectation of symmetric statistic has the same distribution when conditioning on iid random variables
Suppose I have iid random variables $\{X_1,...,X_n\}$ and a statistic $S(X_1,...,X_n)$ that is symmetric (i.e. if $ (\sigma_{i})_{i=1}^n $ is a permutation of $\{1,...,n \} $, then $S(x_1,...,x_n) = S(x_{\sigma_1},...,x_{\sigma_n} )$).
$\textbf{How do I show that $\mathbb{E}(S(X_1,...,X_n)|X_i,X_j) \overset{d}{=} \mathbb{E}(S(X_1,...,X_n)|X_i,X_\ell)$}$ for $j \neq \ell$ i.e. the conditional expectations have the same distribution?
 A: If $U$ and $V$ are independent vectors say of size $k$ and $\ell$ and $f\colon\mathbb R^k\times\mathbb R^\ell\to\mathbb R$ a measurable function, then
$$
\mathbb E\left[f(U,V)\mid V\right]=g(V),\mbox{ where }g(v)=\mathbb E\left[f(U,v)\right].
$$
This is a consequence of Fubini's theorem.
Then in the context of the question, denote $V=(X_i,X_j)$ and $U=(X_k)_{k\in\{1,\dots,n\}\setminus \{i,j\}}$ and $f=S$. Then
$$
\mathbb E\left[S(X_1,\dots,X_n)\mid X_i,X_j\right]=g(X_i,X_j)
$$
and using symmetry of $S$, \begin{align}g(v_1,v_2)&=\mathbb E\left[S(X_1,\dots,X_{i-1},v_1,X_{i+1},\dots,X_{j-1},v_2,X_{j+1},\dots,X_n)\right]\\
&= \mathbb E\left[S(v_1,v_2,X_1,\dots,X_{i-1},X_{i+1},\dots,X_{j-1},X_{j+1},\dots,X_n)\right].
\end{align}
Using the fact that the vectors $(X_1,\dots,X_{i-1},X_{i+1},\dots,X_{j-1},X_{j+1},\dots,X_n)$ and $(X_1,\dots,X_{n-2})$ have the same distribution, we finally get that
$$
g(v_1,v_2)=\mathbb E\left[S(v_1,v_2,X_1,\dots,X_{n-2})\right].
$$
Therefore, if $i\neq j$ and $i'\neq j'$,  $\mathbb E\left[S(X_1,\dots,X_n)\mid X_i,X_j\right]\overset{\mathcal D}=\mathbb E\left[S(X_1,\dots,X_n)\mid X_{i'},X_{j'}\right]$ because $(X_i,X_j)$ have the same distribution as $(X_{i'},X_{j'})$.
