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--- world:definetti [2026/02/03 12:13]
rdouc [Proof]
+++ world:definetti [2026/02/03 19:01] (current)
rdouc
@@ Line 1: / Line 1: @@
+on{{page>:defs}}
 ====== De Finetti's Representation Theorem for Exchangeable Random Elements ======
@@ Line 5: / Line 8: @@
 <WRAP center round tip 80%>
 **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 $\mathcal{F}_\infty$ such that, **conditionally on $\mathcal{F}_\infty$**, the random variables $(X_i)_{i\in\mathbb{N}}$ are **independent and identically distributed (i.i.d.)**.
+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.)**.
 </WRAP>
+The proof is based on the paper "Uses of exchangeability" by J. F. Kingman [[https://projecteuclid.org/journalArticle/Download?urlId=10.1214%2Faop%2F1176995566|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:
 . **Reverse filtration construction:**
-    * For each $n\in\mathbb{N}$, let $\mathcal{F}_n$ 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 $x\in\mathsf{X}^{\mathbb{N}}$,
+    * 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 $x\in\mathsf{X}^{\mathbb{N}}$,
 $$
 f(x_1,\ldots,x_n,x_{n+1},\ldots)=f(x_{\pi(1)},\ldots,x_{\pi(n)},x_{n+1},\ldots).
 $$
-    * $(\mathcal{F}_n)_{n\in\mathbb{N}}$ is a reverse filtration with $\mathcal{F}_n\supseteq\mathcal{F}_{n+1}$ and
+    * $(\mathcal{G}_n)_{n\in\mathbb{N}}$ is a reverse filtration with $\mathcal{G}_n\supseteq\mathcal{G}_{n+1}$ and
 $$
-\mathcal{F}_\infty=\bigcap_{n\in\mathbb{N}}\mathcal{F}_n.
+\mathcal{G}_\infty=\bigcap_{n\in\mathbb{N}}\mathcal{G}_n.
 $$
-    * $\mathcal{F}_\infty$ is generated by all functions invariant under any finite permutation of coordinates.
+    * $\mathcal{G}_\infty$ is generated by all functions invariant under any finite permutation of coordinates.
 . **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 $\mathcal{F}_n$-measurable.
+    * 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 $A\in\mathcal{F}_n$, exchangeability implies
+    * For any $A\in\mathcal{G}_n$, exchangeability implies
 $$
 \mathbb{E}\!\left[\left(\frac1n\sum_{i=1}^n h(X_i)\right)\mathbf1_A\right]
+=\frac1n\sum_{i=1}^n \mathbb{E}\!\left[\left( h(X_i)\right)\mathbf1_A\right]
 =\mathbb{E}[h(X_1)\mathbf1_A].
 $$
-    * Hence
+    * The two previous item show the amazing formula:
 $$
-\frac1n\sum_{i=1}^n h(X_i)=\mathbb{E}[h(X_1)\mid\mathcal{F}_n], \quad a.s.
+\frac1n\sum_{i=1}^n h(X_i)=\mathbb{E}[h(X_1)\mid\mathcal{G}_n], \quad a.s.
 $$
-    * By the reverse martingale convergence theorem,
+    * By the **reverse martingale convergence** theorem (see for example [[world:forward-downward-martingale|Upcrossing Inequality and Martingale Convergence]]),
 $$
-\frac1n\sum_{i=1}^n h(X_i)\xrightarrow{\mathrm{a.s.}}\mathbb{E}[h(X_1)\mid\mathcal{F}_\infty].
+\frac1n\sum_{i=1}^n h(X_i)\xrightarrow{\mathrm{a.s.}}\mathbb{E}[h(X_1)\mid\mathcal{G}_\infty].
 $$
@@ Line 42: / Line 47: @@
     * For bounded measurable $f:\mathsf{X}^k\to\mathbb{R}$,
 $$
-\mathbb{E}[f(X_1,\ldots,X_k)\mid\mathcal{F}_\infty]
+\mathbb{E}[f(X_1,\ldots,X_k)\mid\mathcal{G}_\infty]
 =\lim_{n\to\infty}\frac1{n(n-1)\cdots(n-k+1)}
-\sum_{\substack{1\le i_1,\ldots,i_k\le n\\ i_j\neq i_\ell}}
+\sum_{\substack{i_{1:k} \in [1:n]^k \le n,\ldots,1 \le i_k\le n\\ i_j\neq i_\ell}}
 f(X_{i_1},\ldots,X_{i_k}).
 $$
     * But we have
 $$
-\lim_{n\to\infty}\frac1{n(n-1)\cdots(n-k+1)}
+\frac1{n(n-1)\cdots(n-k+1)}
 \sum_{\substack{1\le i_1,\ldots,i_k\le n\\ i_j\neq i_\ell}}
-f(X_{i_1},\ldots,X_{i_k}) = \lim_{n\to\infty}\frac1{n^k}\sum_{i_1=1}^n\cdots\sum_{i_k=1}^n f(X_{i_1},\ldots,X_{i_k}).
+f(X_{i_1},\ldots,X_{i_k}) + O\lr{\frac1n} = \frac1{n^k}\sum_{i_1=1}^n\cdots\sum_{i_k=1}^n f(X_{i_1},\ldots,X_{i_k}).
 $$
     * Hence, for product functions $f(x_1,\ldots,x_k)=f_1(x_1)\cdots f_k(x_k)$,
 $$
-\mathbb{E}[f_1(X_1)\cdots f_k(X_k)\mid\mathcal{F}_\infty]
+\mathbb{E}[f_1(X_1)\cdots f_k(X_k)\mid\mathcal{G}_\infty]
-=\prod_{\ell=1}^k\mathbb{E}[f_\ell(X_1)\mid\mathcal{F}_\infty].
+=\prod_{\ell=1}^k\mathbb{E}[f_\ell(X_1)\mid\mathcal{G}_\infty].
 $$
-    * Thus, conditionally on $\mathcal{F}_\infty$, $(X_i)$ are independent.
+    * Thus, conditionally on $\mathcal{G}_\infty$, $(X_i)$ are independent.

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