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world:definetti
Table of Contents
on
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De Finetti's Representation Theorem for Exchangeable Random Elements
Theorem
De Finetti's Theorem: Let $(X_i)_{i\in\mathbb{N}}$ be a family of exchangeable random elements defined on a measurable space $(\mathsf{X},\mathcal{X})$. Then, there exists a $\sigma$-field $\mathcal{G}_\infty$ such that, conditionally on $\mathcal{G}_\infty$, the random variables $(X_i)_{i\in\mathbb{N}}$ are independent and identically distributed (i.i.d.).
The proof is based on the paper “Uses of exchangeability” by J. F. Kingman Click here to see the paper.
Proof
Without loss of generality, we model $(X_i)_{i\in\mathbb{N}}$ as the coordinate projections on the canonical probability space $(\mathsf{X}^{\mathbb{N}},\mathcal{X}^{\otimes\mathbb{N}},\mathbb{P})$. We proceed as follows:
1. Reverse filtration construction:
- For each $n\in\mathbb{N}$, let $\mathcal{G}_n$ be the $\sigma$-field generated by all measurable functions $f:\mathsf{X}^{\mathbb{N}}\to\mathbb{R}$ invariant under any permutation of the first $n$ coordinates, i.e. for any permutation $\pi$ of $\{1,\ldots,n\}$ and any $x\in\mathsf{X}^{\mathbb{N}}$,
$$ f(x_1,\ldots,x_n,x_{n+1},\ldots)=f(x_{\pi(1)},\ldots,x_{\pi(n)},x_{n+1},\ldots). $$
- $(\mathcal{G}_n)_{n\in\mathbb{N}}$ is a reverse filtration with $\mathcal{G}_n\supseteq\mathcal{G}_{n+1}$ and
$$ \mathcal{G}_\infty=\bigcap_{n\in\mathbb{N}}\mathcal{G}_n. $$
- $\mathcal{G}_\infty$ is generated by all functions invariant under any finite permutation of coordinates.
2. Conditional expectation and empirical averages:
- For any bounded measurable $h:\mathsf{X}\to\mathbb{R}$, the empirical average $\frac1n\sum_{i=1}^n h(X_i)$ is $\mathcal{G}_n$-measurable.
- For any $A\in\mathcal{G}_n$, exchangeability implies
$$ \mathbb{E}\!\left[\left(\frac1n\sum_{i=1}^n h(X_i)\right)\mathbf1_A\right] =\frac1n\sum_{i=1}^n \mathbb{E}\!\left[\left( h(X_i)\right)\mathbf1_A\right] =\mathbb{E}[h(X_1)\mathbf1_A]. $$
- The two previous item show the amazing formula:
$$ \frac1n\sum_{i=1}^n h(X_i)=\mathbb{E}[h(X_1)\mid\mathcal{G}_n], \quad a.s. $$
- By the reverse martingale convergence theorem (see for example Upcrossing Inequality and Martingale Convergence),
$$ \frac1n\sum_{i=1}^n h(X_i)\xrightarrow{\mathrm{a.s.}}\mathbb{E}[h(X_1)\mid\mathcal{G}_\infty]. $$
3. Multivariate functions:
- For bounded measurable $f:\mathsf{X}^k\to\mathbb{R}$,
$$ \mathbb{E}[f(X_1,\ldots,X_k)\mid\mathcal{G}_\infty] =\lim_{n\to\infty}\frac1{n(n-1)\cdots(n-k+1)} \sum_{\substack{i_{1:k} \in [1:n]^k \le n,\ldots,1 \le i_k\le n\\ i_j\neq i_\ell}} f(X_{i_1},\ldots,X_{i_k}). $$
- But we have
$$ \frac1{n(n-1)\cdots(n-k+1)} \sum_{\substack{1\le i_1,\ldots,i_k\le n\\ i_j\neq i_\ell}} f(X_{i_1},\ldots,X_{i_k}) + O\lr{\frac1n} = \frac1{n^k}\sum_{i_1=1}^n\cdots\sum_{i_k=1}^n f(X_{i_1},\ldots,X_{i_k}). $$
- Hence, for product functions $f(x_1,\ldots,x_k)=f_1(x_1)\cdots f_k(x_k)$,
$$ \mathbb{E}[f_1(X_1)\cdots f_k(X_k)\mid\mathcal{G}_\infty] =\prod_{\ell=1}^k\mathbb{E}[f_\ell(X_1)\mid\mathcal{G}_\infty]. $$
- Thus, conditionally on $\mathcal{G}_\infty$, $(X_i)$ are independent.
world/definetti.txt · Last modified: 2026/02/03 19:01 by rdouc
