## Wednesday, April 12, 2023

### Generating qualitative probabilities

A (partial) qualitative probability is a relation defined on an algebra of sets and satisfying the following axioms:

1. Preorder: ≾ is reflexive and transitive

2. Zero: ⌀ ≾ A

3. Additivity: if A ∪ B and C are disjoint, then A ≾ B if and only A ∪ C ≾ B ∪ C.

It’s just occurred to me that there is a nice way to construct qualitative probabilities out of a family of finitely additive probabilities. Fix an algebra F of subsets of Ω. Let Q be a non-empty set of finitely additive probability functions on F taking values in totally ordered fields. Then say that AQB if and only if p(A) ≤ p(B) for all p in Q. This clearly satisfies the three axioms above.

This may seem like a major extension of the concept of a total qualitative probability p generated by a single probability function p, but it’s not as much an extension as it may seem.

Remark 1: Not every qualitative probability, not even every total qualitative probability, can be constructed as Q for some Q.

At least one 1959 Kraft, Pratt and Seidenberg counterexample to the thesis that every total qualitative probability is generated by a single real-valued probability function trivially extends to show this.

Now let ≺ be the strict order generated by ≾ (and similarly with subscripts): A ≺ B iff A ≾ B but not B ≾ A.

Theorem 1: Assume Choice. Given a non-empty set Q of finitely additive probability functions on F there is a probability function p taking values in some ordered field such that ≾p extends ≾Q and ≺p extends Q.

Sketch of proof: All the ordered fields that the members of Q take values in can be embedded in the surreals. There will be a set-sized field that contains the ranges of all the embeddings. So we can assume that all the members of Q take values in a single field. The rest follows by an ultrafilter construction. Let H be the set of non-empty finite subsets of Q ordered by inclusion. Given K in H, let pK be the sum of the members of K. Then let p be the ultraproduct of the pK with respect to some ultrafilter on H. Verifying finite additivity and the fact that ≾p extends ≾Q is trivial. Verifying that p extends Q is only slightly harder. Suppose AQB. Then AQB. For some q in Q we must have q(A) < q(B). Then for any K in H containing {q}, we have pK(A) < pK(B), and so p(A) < p(B).

Theorem 2: Assume Choice. Given a non-empty set Q of finitely additive real-valued probability functions on F there is a real-valued probability function p such that p extends Q.

Sketch of proof: Let H be the set of non-empty finite subsets of Q ordered by inclusion. For K in H, let pK be the sum of the members of K. By compactness, pK has a limit point p.

Remark 2: One cannot require that p extend Q in Theorem 2. For let Ω have cardinality greater than the continuum. Then there is no regular real-valued finitely additive probability on the powerset of Ω (a probability is regular if P(A) > 0 for every non-empty A), since if there were, then each Ωz = {x ∈ Ω : x < z} would have different probabilities, and so the probability would have more values than the continuum. Let Q be all finitely additive real-valued probabilities on the powerset of Ω. Then ⌀≺QA for any non-empty A (since 0 < q(A) for q concentrated on some point of A). But if we had ⌀≺pA, then p would be regular. I am not sure what to say if Ω has continuum cardinality.