De Finetti's theorem

In

of random variables is said to be exchangeable if for any natural number n and any finite sequence i₁, ..., i_n and any permutation of the sequence π:{i₁, ..., i_n } → {i₁, ..., i_n },

${\displaystyle (X_{i_{1}},\dots ,X_{i_{n}}){\text{ and }}(X_{\pi (i_{1})},\dots ,X_{\pi (i_{n})})}$

both have the same joint probability distribution.

If an identically distributed sequence is

A random variable X has a Bernoulli distribution if Pr(X = 1) = p and Pr(X = 0) = 1 − p for some p ∈ (0, 1).

De Finetti's theorem states that the probability distribution of any infinite exchangeable sequence of Bernoulli random variables is a "mixture" of the probability distributions of independent and identically distributed sequences of Bernoulli random variables. "Mixture", in this sense, means a weighted average, but this need not mean a finite or countably infinite (i.e., discrete) weighted average: it can be an integral over a measure rather than a sum.

More precisely, suppose X₁, X₂, X₃, ... is an infinite exchangeable sequence of Bernoulli-distributed random variables. Then there is some probability measure m on the interval [0, 1] and some random variable Y such that

• The probability measure of Y is m, and
• The conditional probability distribution of the whole sequence X₁, X₂, X₃, ... given the value of Y is described by saying that
  • X₁, X₂, X₃, ... are conditionally independent given Y, and
  • For any i ∈ {1, 2, 3, ...}, the conditional probability that X_i = 1, given the value of Y, is Y.

Another way of stating the theorem

Suppose ${\displaystyle X_{1},X_{2},X_{3},\ldots }$ is an infinite exchangeable sequence of Bernoulli random variables. Then ${\displaystyle X_{1},X_{2},X_{3},\ldots }$ are conditionally independent and identically distributed given the exchangeable sigma-algebra (i.e., the sigma-algebra consisting of events that are measurable with respect to ${\displaystyle X_{1},X_{2},\ldots }$ and invariant under finite permutations of the indices).

Example

Here is a concrete example. We construct a sequence

${\displaystyle X_{1},X_{2},X_{3},\dots }$

of random variables, by "mixing" two i.i.d. sequences as follows.

We assume p = 2/3 with probability 1/2 and p = 9/10 with probability 1/2. Given the event p = 2/3, the conditional distribution of the sequence is that the X_i are independent and identically distributed and X₁ = 1 with probability 2/3 and X₁ = 0 with probability 1 − 2/3. Given the event p = 9/10, the conditional distribution of the sequence is that the X_i are independent and identically distributed and X₁ = 1 with probability 9/10 and X₁ = 0 with probability 1 − 9/10.

This can be interpreted as follows: Make two biased coins, one showing "heads" with 2/3 probability and one showing "heads" with 9/10 probability. Flip a fair coin once to decide which biased coin to use for all flips that are recorded. Here "heads" at flip i means X_i=1.

The independence asserted here is conditional independence, i.e. the Bernoulli random variables in the sequence are conditionally independent given the event that p = 2/3, and are conditionally independent given the event that p = 9/10. But they are not unconditionally independent; they are positively correlated.

In view of the strong law of large numbers, we can say that

${\displaystyle \lim _{n\rightarrow \infty }{\frac {X_{1}+\cdots +X_{n}}{n}}={\begin{cases}2/3&{\text{with probability }}1/2,\\9/10&{\text{with probability }}1/2.\end{cases}}}$

Rather than concentrating probability 1/2 at each of two points between 0 and 1, the "mixing distribution" can be any probability distribution supported on the interval from 0 to 1; which one it is depends on the joint distribution of the infinite sequence of Bernoulli random variables.

The definition of exchangeability, and the statement of the theorem, also makes sense for finite length sequences

${\displaystyle X_{1},\dots ,X_{n},}$

but the theorem is not generally true in that case. It is true if the sequence can be extended to an exchangeable sequence that is infinitely long. The simplest example of an exchangeable sequence of Bernoulli random variables that cannot be so extended is the one in which X₁ = 1 − X₂ and X₁ is either 0 or 1, each with probability 1/2. This sequence is exchangeable, but cannot be extended to an exchangeable sequence of length 3, let alone an infinitely long one.

Extensions

Versions of de Finetti's theorem for finite exchangeable sequences,^[2][3] and for Markov exchangeable sequences^[4] have been proved by Diaconis and Freedman and by Kerns and Szekely. Two notions of partial exchangeability of arrays, known as separate and joint exchangeability lead to extensions of de Finetti's theorem for arrays by Aldous and Hoover.^[5]

The computable de Finetti theorem shows that if an exchangeable sequence of real random variables is given by a computer program, then a program which samples from the mixing measure can be automatically recovered.^[6]

In the setting of free probability, there is a noncommutative extension of de Finetti's theorem which characterizes noncommutative sequences invariant under quantum permutations.^[7]

Extensions of de Finetti's theorem to quantum states have been found to be useful in quantum information,^[8][9][10] in topics like quantum key distribution^[11] and entanglement detection.^[12] A multivariate extension of de Finetti’s theorem can be used to derive Bose–Einstein statistics from the statistics of classical (i.e. independent) particles.^[13]

See also

• Choquet theory
• Hewitt–Savage zero–one law
• Krein–Milman theorem
• Invariant sigma-algebra

References