Friday, August 26, 2022

Full conditional probabilities and the Axiom of Choice

Here’s a claim that turns out to be equivalent to the Axiom of Choice:

  1. Given any non-empty set Ω and a collection M of [0,∞]-valued finitely additive measures on the powerset of Ω such that for any non-empty E ⊆ Ω there is a μ ∈ M with 0 < μ(E) < ∞, there is a full conditional probability P on the powerset of Ω definable in terms of the measures in M in the sense that for each non-empty E there is a μ ∈ M such that 0 < μ(E) < ∞ and for all A we have P(A|E) = μ(AE)/μ(E).

The easy direction of proof is from (1) to AC. Let M be the collection of all the finitely additive probability measures on Ω that assign probability one to some singleton. Clearly M has the desired properties. Now, for any non-empty E ⊆ Ω, there will be a μ ∈ M such that P(A|E) = μ(AE)/μ(E) and 0 < μ(E) < ∞. Thus, the point at which μ is concentrated must be in E. Moreover, it is clear that for each E, the measure μ must be unique. Let f(E) be the point at which μ is concentrated. This is a choice function for all subsets of Ω. Since Ω is an arbitrary non-empty set, we have AC.

The other direction follows by the method of proof of Lemma 3 here.

1 comment:

Unknown said...

Considering that conditional probabilities are necessary for the formulation of Bayes's theorem, how does this statement impact Bayesian epistemology? Does this imply that the axiom of choise is neccessary for Bayesian Epistemology?