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Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 35 / 2 / i / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 35 2 i
Created 2026-10-03 Updated 2026-10-06  0 By others on same topic  0 Discussions Create my own version
Two exchangeable random variables satisfy (X,Y)=d(Y,X): their joint probability distribution is unchanged by swapping the coordinates. Thus for every measurable set A,
P(X∈A)=P((X,Y)∈A×Y)=P((Y,X)∈A×Y)=P(Y∈A).
(1)
Exchangeability implies identical marginal distributions, but does not imply independence.

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