Probability Theory – Conditional Expectation of Symmetric Statistic with iid Random Variables

conditional probabilityconditional-expectationprobability theory

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 $\mathbb{E}(S(X_1,…,X_n)|X_i) \overset{d}{=} \mathbb{E}(S(X_1,…,X_n)|X_j)?$}$, i.e. the conditional expectations have the same distribution?

Related link: Conditional expectation of iid random variables

$\textbf{motivation:}$ Efron and Stein (1981) used this "identity" https://pdodds.w3.uvm.edu/research/papers/others/1981/efron1981a.pdf

Best Answer

Let $X,Y$ be independent identically distributed real valued random variables. Then for any admissible $g:\mathbb{R}\to \mathbb{R}$, $E[g(X)]=E[g(Y)]$ by identical distribution, and for any admissible $h:\mathbb{R}^2\to \mathbb{R}$ also have this. So (we talk about a.s. equalities when dealing with conditional expectations) $$\begin{aligned}E[f(X,Y)|X]&=E[f(x,Y)]|_{x=X}\sim\\ &\sim E[f(y,Y)]|_{y=Y}=\\ &=E[f(y,X)]|_{y=Y}=\\ &=E[f(Y,X)|Y]\end{aligned}$$ We conclude by using the assumption $f(x,y)=f(y,x)$.

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