On a representation theorem for finitely exchangeable random vectors

TL;DR

This paper provides a shorter, more straightforward proof of a representation theorem for finitely exchangeable random vectors, showing their law as a mixture of product measures with a signed mixing measure, avoiding measure-theoretic complexities.

Contribution

It offers an alternative, simplified proof of the exchangeability representation theorem, emphasizing a finitistic, linear algebra approach and analyzing the properties of the canonical mixing measure.

Findings

Law of exchangeable vectors is a mixture of product measures.

The mixing signed measure is not unique.

For real-valued spaces, the mixing measure can be on b.

Abstract

A random vector $X=(X_1,\ldots,X_n)$ with the $X_i$ taking values in an arbitrary measurable space $(S, \mathscr{S})$ is exchangeable if its law is the same as that of $(X_{\sigma(1)}, \ldots, X_{\sigma(n)})$ for any permutation $\sigma$. We give an alternative and shorter proof of the representation result (Jaynes \cite{Jay86} and Kerns and Sz\'ekely \cite{KS06}) 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 $S=\mathbb{R}$, one can choose a mixing measure on $\mathbb{R}^n$.