On a representation theorem for finitely exchangeable random vectors

A random vector X=(X1,…,Xn) with the Xi taking values in an arbitrary measurable space (S,S) is exchangeable if its law is the same as that of (Xσ(1),…,Xσ(n)) for any permutation σ. We give an alternative and shorter proof of the representation result (Jaynes [6] and Kerns and Székely [9]) stating t...

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Bibliographic Details
Published inJournal of mathematical analysis and applications Vol. 442; no. 2; pp. 703 - 714
Main Authors Janson, Svante, Konstantopoulos, Takis, Yuan, Linglong
Format Journal Article
LanguageEnglish
Published Elsevier Inc 15.10.2016
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Summary:A random vector X=(X1,…,Xn) with the Xi taking values in an arbitrary measurable space (S,S) is exchangeable if its law is the same as that of (Xσ(1),…,Xσ(n)) for any permutation σ. We give an alternative and shorter proof of the representation result (Jaynes [6] and Kerns and Székely [9]) stating that the law of X is a mixture of product probability measures with respect to a signed mixing measure. The result is “finitistic” in nature meaning that it is a matter of linear algebra for finite S. The passing from finite S to an arbitrary one may pose some measure-theoretic difficulties which are avoided by our proof. The mixing signed measure is not unique (examples are given), but we pay more attention to the one constructed in the proof (“canonical mixing measure”) by pointing out some of its characteristics. The mixing measure is, in general, defined on the space of probability measures on S; but for S=R, one can choose a mixing measure on Rn.
ISSN:0022-247X
1096-0813
1096-0813
DOI:10.1016/j.jmaa.2016.04.070