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

Table of contents
- Solution i

Solution

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 \in A) = P ((X, Y) \in A \times Y) = P ((Y, X) \in A \times Y) = P (Y \in A) .

Exchangeability implies identical marginal distributions, but does not imply independence.

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