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De Finetti's theorem
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== Example == As a concrete example, we construct a sequence :<math>X_1, X_2, X_3, \dots </math> 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''<sub>i</sub> are independent and identically distributed and ''X''<sub>1</sub> = 1 with probability 2/3 and ''X''<sub>1</sub> = 0 with probability 1 − 2/3. Given the event ''p'' = 9/10, the conditional distribution of the sequence is that the ''X''<sub>i</sub> are independent and identically distributed and ''X''<sub>1</sub> = 1 with probability 9/10 and ''X''<sub>1</sub> = 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<sub>i</sub>=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 [[correlation|correlated]]. In view of the [[law of large numbers|strong law of large numbers]], we can say that :<math>\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} </math> 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 :<math>X_1,\dots, X_n,</math> 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 ''X''<sub>1</sub> = 1 − ''X''<sub>2</sub> and ''X''<sub>1</sub> 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.
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