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Let $(X_n)_{n \ge 1}$ be an exchangeable random sequence and let $\alpha$ be the corresponding de-Finetti random measure. I want to show that the empirical distribution of the sequence converges to $\alpha$ almost surely.

Note: For an iid-P random sequence $(X_n)_{n \ge 1}$, Glivenko Cantelli lemma states that the empirical distribution converges to $P$ almost surely.

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1 Answer 1

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I wonder why this question has gone unanswered this long.

By de Finetti's theorem any random sequence $(\xi_n)_{n\geq1}$ in a Borel space $S$ is conditionally i.i.d., say, given $\mathcal{F}$. Fix $f\geq0$ measurable on $S$. Then \begin{eqnarray*} P\left(\frac{1}{n}\sum_{k\leq n}f(\xi_k)\rightarrow\int fd\alpha \right)&=&EE\left(1\left\{\frac{1}{n}\sum_{k\leq n}f(\xi_k)\rightarrow\int fd\alpha \right\}\bigg|\mathcal{F}\right)\\ &=&E \int1\left\{\frac{1}{n}\sum_{k\leq n}f(x_k)\rightarrow\int fd\alpha \right\}d\alpha^\infty(x)=E1=1, \end{eqnarray*} where the law of large numbers was used (conditionally) in the next to last equality. An approximation (since $S$ is Borel) now implies \begin{gather}\label{empirical} \eta_n\equiv n^{-1}\sum_{k\leq n}1_B(\xi_k)\rightarrow \alpha B\:\:\: a.s., \:\:\:\:\: B\in\mathcal{S}. \end{gather} For an alternate proof one can also exploit the fact that the $(\eta_n)$ form a reverse martingale. For a proof of this fact, a good reference is $\textit{Probabilistic Symmetries and Invariance Principles}$ by Olav Kallenberg, where also the above result can be found.

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