---
tag: W19129PVKQ
kind: statement
role: theorem
name: De Finetti's theorem
msc: [60G09, 62F15]
uses: [4MR4YJS1ZD, 00MRN0FARV, 059TJN0NZ1, 0WQXY32Q8D, 0386Y0Z7Y7]
about: [4MR4YJS1ZD]
sources:
  - url: https://en.wikipedia.org/wiki/De_Finetti%27s_theorem
same_as:
  wikidata: Q4408070
  wikipedia: De Finetti's theorem
---

## Statement

Let $S$ be a [[Polish space|584GXXY73M]] with its [[Borel σ-algebra|0WQXY32Q8D]], and let $\mathcal{P}(S)$ be the set of [[probability measures|059TJN0NZ1]] on $S$.
Give $\mathcal{P}(S)$ the smallest σ-algebra that makes the maps $m \mapsto m(A)$ measurable for all Borel sets $A \subseteq S$.
Let $X_1, X_2, \dots$ be an infinite [[exchangeable|4MR4YJS1ZD]] sequence of random variables with values in $S$.
Then there is a unique probability measure $\nu$ on $\mathcal{P}(S)$ such that for every $n \ge 1$ and all Borel sets $A_1, \dots, A_n \subseteq S$,
$$P(X_1 \in A_1, \dots, X_n \in A_n) = \int_{\mathcal{P}(S)} m(A_1) \cdots m(A_n) \, \nu(dm).$$

## Remarks

In words, the sequence is a mixture of [[i.i.d.|00MRN0FARV]] sequences: first draw a law $m$ from $\nu$, then draw $X_1, X_2, \dots$ i.i.d. with law $m$.
For $S = \set{0, 1}$ the theorem says that there is a unique probability measure $\mu$ on $[0, 1]$ with
$$P(X_1 = x_1, \dots, X_n = x_n) = \int_0^1 p^k (1 - p)^{n - k} \, \mu(dp), \qquad k = x_1 + \dots + x_n,$$
so the sequence is a mixture of [[Bernoulli|0386Y0Z7Y7]] sequences.
In this case the random parameter is the almost sure limit of the frequencies $(X_1 + \dots + X_n)/n$, and its law is $\mu$.
One proof shows that, given the σ-algebra of events invariant under finite permutations, the variables are conditionally i.i.d.; a reverse martingale argument gives the limit of the empirical measures.
The theorem fails for finite sequences: drawing both balls without replacement from an urn with one red and one blue ball gives an exchangeable pair $(X_1, X_2)$ with $P(X_1 = X_2 = 1) = 0$ and $P(X_1 = X_2 = 0) = 0$.
A mixture would need $\int p^2 \, \mu(dp) = 0$ and $\int (1 - p)^2 \, \mu(dp) = 0$, which is impossible.
In the $\set{0,1}$ case, $\operatorname{Cov}(X_1, X_2) = \int p^2 \, d\mu - \left(\int p \, d\mu\right)^2 \ge 0$, so infinite exchangeable sequences of this kind are never negatively correlated.

## Explanations

### L1 human

Suppose you toss a coin of unknown bias many times, and you believe that the order of the tosses does not matter.
De Finetti's theorem says that this belief alone forces a simple picture.
There is a hidden bias $p$, drawn once from some law, and given $p$ the tosses are independent with success probability $p$.
So symmetry under reordering is the same as independence given an unknown parameter.

### L2 human

The theorem is the main justification for the Bayesian model "parameter with a prior, data i.i.d. given the parameter": the measure $\nu$ plays the role of the prior.
The [[Pólya urn model|DHPXDFEEMF]] gives an example: its colour sequence is exchangeable, and the mixing measure is a [[beta distribution|0EW824Z7G2]].
For general exchangeable sequences, the [[Dirichlet process|XT4SWG6J7N]] is a standard choice of $\nu$ in nonparametric Bayesian statistics.
The [[Hewitt–Savage zero–one law|JTXNFQNQ4J]] is a related result: for an i.i.d. sequence, the exchangeable σ-algebra is trivial, which matches $\nu$ being a point mass.