24 — Backwards Martingales, Exchangeability, de Finetti
Backwards Martingale (BMG)
We have a filtered probability space
$(\Omega, \mathcal{F}_n, P)$ with
$\mathcal{F}_0 \supset \mathcal{F}_1 \supset \mathcal{F}_2 \supset \cdots$
(i.e., decreasing filtration).
A sequence ${X_n, \mathcal{F}n}{n\ge 0}$ is a backwards martingale if
\[E(X_n \mid \mathcal{F}_{n+1}) = X_{n+1}, \qquad n\ge 0.\]| Assume $E | X_n | <\infty$ for all $n$. |
Backwards Martingale Convergence Theorem (BMCT)
If $\mathcal{F}n \downarrow \mathcal{F}\infty := \bigcap_{n\ge 0}\mathcal{F}_n$, then
\[X_n \xrightarrow[n\to\infty]{\text{a.s.},\,L^1} X_\infty, \quad\text{where}\quad X_\infty = E(X_0 \mid \mathcal{F}_\infty).\]Also,
\[E(X_n) \to E(X_\infty).\]Application 1: Dominated Convergence for Conditional Expectations
Suppose:
- $Y_n \to Y$ a.s.
-
$ Y_n \le Z$ with $E(Z)<\infty$. - Filtration increases: $\mathcal{F}n \uparrow \mathcal{F}\infty$.
Then
\[E(Y_n \mid \mathcal{F}_n) \xrightarrow[n\to\infty]{\text{a.s.}} E(Y \mid \mathcal{F}_\infty).\]To prove, add the step:
\[E_{\mathcal{F}_n}(Y_n - Y) \to 0 \quad \text{a.s.}\]Let $W_n = Y_n - Y$.
Then $|W_n| \le 2Z$ and $W_n \to 0$.
Use:
\[|E(W_n \mid \mathcal{F}_n)| \le E(|W_n|\mid \mathcal{F}_n) \le E\left( \sup_{k\ge N} |W_k| \mid \mathcal{F}_n \right) \xrightarrow[N\to\infty]{} 0.\]Application 2: SLLN via Backwards Martingales
Let ${\xi_k}$ be i.i.d. with $E|\xi_1|<\infty$.
Let $S_n=\sum_{k=1}^n \xi_k$.
Define the tail σ-field:
\[\mathcal{F}_n = \sigma(S_n, S_{n+1}, \ldots).\]This is a decreasing filtration.
By the Hewitt–Savage 0–1 law, $\mathcal{F}_\infty$ is trivial.
Compute:
\[E(\xi_1 \mid \mathcal{F}_n) = \frac{S_n}{n}.\]Thus:
\[\frac{S_n}{n} = E(\xi_1 \mid \mathcal{F}_n) \quad\text{is a backwards martingale}.\]By BMCT:
\[\frac{S_n}{n} \xrightarrow[n\to\infty]{\text{a.s.}} E(\xi_1 \mid \mathcal{F}_\infty) = E(\xi_1).\]This proves the Strong Law of Large Numbers.
From i.i.d. to Exchangeable Sequences
A sequence ${X_n}$ is exchangeable if
\[(X_{\pi_1}, X_{\pi_2},\ldots) \overset{d}{=} (X_1, X_2, \ldots)\]for every finite permutation $\pi$ of ${1,\ldots,n}$.
Define the exchangeable σ-field:
\[\mathcal{E}_n = \{A : \pi(A) = A \text{ for all permutations of } \{1,\ldots,n\}\}.\]These satisfy:
\[\mathcal{E}_n \downarrow \mathcal{E}_\infty.\]For any bounded measurable $\varphi : \mathbb{R}^k \to \mathbb{R}$, define the average over permutations:
\[A_n(\varphi) = \frac{1}{n!}\sum_{\pi\in S_n} \varphi(X_{\pi_1},\ldots,X_{\pi_k}).\]Then $A_n(\varphi)$ is a backwards martingale in $n$, hence:
\[A_n(\varphi) \xrightarrow[n\to\infty]{\text{a.s.}} E\left( \varphi(X_1,\ldots,X_k)\mid \mathcal{E}_\infty \right).\]Given $\mathcal{E}_\infty$, the coordinates become independent:
\[E_{\mathcal{E}_\infty} \left[ \sum_{k=1}^n f_k(X_k) \right] = \sum_{k=1}^n E_{\mathcal{E}_\infty}(f_k(X_k)).\]This leads to de Finetti’s theorem:
de Finetti’s Theorem
If ${X_n}$ is exchangeable, then conditional on
$\mathcal{E}_\infty$, the sequence is i.i.d.
Example (Page 3 of the PDF):
For binary exchangeable sequences $X_n\in{0,1}$,
\[P(X_1=\dots=X_k=1,\ X_{k+1}=\dots=X_n=0) = \int_0^1 \theta^k (1-\theta)^{n-k} \, dF(\theta),\]where $F$ is some mixing distribution on $[0,1]$.
Summary
- Backwards martingales converge a.s. and in $L^1$ to their limit in the tail σ-field.
- Using BMG gives a clean proof of the SLLN.
- Exchangeable sequences yield a backwards-martingale argument for symmetric expectations.
- The limit conditional σ-field $\mathcal{E}_\infty$ turns exchangeable sequences into conditionally i.i.d., giving de Finetti’s theorem.
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