# conditional expectation and order statistic

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space.Let $\ X=(X_1,..,X_n)$ a random vector, with$\ n$ independents random variables whose law is $\mu$ on $\mathbb{R}$. We define $T:\mathbb{R^n}\to\mathbb{R^n}$ such that $T(X_1,..,X_n)=(X_{(1)},..,X_{(n)})$ with $X_{(k)}$ the k-th order statistic.Let $S_n$ set of all permutations $\sigma$ of set $\left\{ 1,..,n \right\}$ and for any $f:\mathbb{R^n}\to\mathbb{R}$ we define $Sf(x_1...x_n)=\sum_{\sigma \in S^n}f(x_{\sigma(1)},..,x_{\sigma(n)})$.

$\\$ Prove that: $\\$

$\mathbb{E}(f(X)|T(X))=S(f(X))$ a.s.

and if $V(x_1,..,x_n)=(\sum_{i=1}^nx_i,\sum_{i<j;i,j=1}^nx_ix_j,....,x_1x_2..x_n)$ then conclude that $\mathbb{E}(f(X)|V(X))=S(f(X))$ a.s.

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Nice exercise. / Is this homework? / The RHS of the definition of $Sf$ should probably be divided by $n!$. / For starters, you might wish to show that $S(f(X))$ is $T(X)$-measurable since $S(f(X))=S(f(T(X)))$ and that $T$ is invariant by $S_n$ in the sense that $T(X)=T(X^\sigma)$ for every $\sigma$ in $S_n$, where $(X^\sigma)_i=X_{\sigma(i)}$. – Did Oct 11 '12 at 21:35