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world:definetti [2026/02/03 13:44]
rdouc [Proof]
world:definetti [2026/02/03 19:01] (current)
rdouc
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 +on{{page>:​defs}}
 +
 +
 ====== De Finetti'​s Representation Theorem for Exchangeable Random Elements ====== ====== De Finetti'​s Representation Theorem for Exchangeable Random Elements ======
  
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 </​WRAP>​ </​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 ===== ===== 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: 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:
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 $$ $$
 \mathbb{E}\!\left[\left(\frac1n\sum_{i=1}^n h(X_i)\right)\mathbf1_A\right] \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]. =\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{G}_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{G}_\infty]. \frac1n\sum_{i=1}^n h(X_i)\xrightarrow{\mathrm{a.s.}}\mathbb{E}[h(X_1)\mid\mathcal{G}_\infty].
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 \mathbb{E}[f(X_1,​\ldots,​X_k)\mid\mathcal{G}_\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)} =\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}). f(X_{i_1},​\ldots,​X_{i_k}).
 $$ $$
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 \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}} \sum_{\substack{1\le i_1,​\ldots,​i_k\le n\\ i_j\neq i_\ell}}
-f(X_{i_1},​\ldots,​X_{i_k}) + 0\lr{\frac1n} = \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 $f(x_1,​\ldots,​x_k)=f_1(x_1)\cdots f_k(x_k)$,     * Hence, for product functions $f(x_1,​\ldots,​x_k)=f_1(x_1)\cdots f_k(x_k)$,
world/definetti.1770122647.txt.gz · Last modified: 2026/02/03 13:44 by rdouc