Thursday, February 2, 2023

From strict anti-anti-Bayesianism to strict propriety

In my previous post, I showed that a continuous anti-anti-Bayesian accuracy scoring rule on probabilities defined on a sub-algebra of events satisfying the technical assumption that the full algebra contains an event logically independent of the sub-algebra is proper. However, I couldn’t figure out how to prove strict propriety given strict anti-anti-Bayesianism. I still don’t, but I can get closer.

First, a definition. A scoring rule on probabilities on the sub-algebra H is strictly anti-anti-Bayesian provided that one expects it to penalize non-trivial binary anti-Bayesian updates. I.e., if A is an event with prior probability p neither zero nor one, and Bayesian conditionalization on A (or, equivalently, on Ac) modifies the probability of some member of H, then the p-expected score of finding out whether A or Ac holds and conditionalizing on that is strictly better than if one adopted the procedure of conditionalizing on the complement of the actually obtaining event.

Suppose we have continuity, the technical assumption and anti-anti-Bayesianism. My previous post shows that the scoring rule is proper. I can now show that it is strictly proper if we strengthen anti-anti-Bayesianism to strict anti-anti-Bayesianism and add the technical assumption that the scoring rule satisfies the finiteness condition that Eps(p) is finite for any probability p on H. Since we’re working with accuracy scoring rules and these take values in [−∞,M] for finite M, the only way to violate the finiteness condition is to have Eps(p) =  − ∞, which would mean that s is very pessimistic about p: by p’s own lights, the expected score of p is infinitely bad. The finiteness condition thus rules out such maximal pessimism.

Here is a sketch of the proof. Suppose we do not have strict propriety. Then there will be two distinct probabilities p and q such that Eps(p) ≤ Eps(q). By propriety, the inequality must be an equality. By Proposition 9 of a recent paper of mine, it follows that s(p) = s(q) everywhere (this is where the finiteness condition is used). Now let r = (p+q)/2. Using the trick from the Appendix here, we can find a probability p′ on the full algebra and an event Z such that r is the restriction of p to H, p is the restriction of the Bayesian conditionalization of p′ on Z to H, and q is the restriction of the Bayesian conditionalization of q on Zc to H. Then the scores of p and q will be the same, and hence the scores of Bayesian and anti-Bayesian conditionalization on finding out whether Z or Zc is actual are guaranteed to be the same, and this violates strict anti-anti-Bayesianism.

One might hope that this will help those who are trying to construct accuracy arguments for probabilism—the doctrine that credences should be probabilities. The hitch in those arguments is establishing strict propriety. However, I doubt that what I have helps. First, I am working in a sub-algebra setting. Second, and more importantly, I am working in a context where scoring rules are defined only for probabilities, and so the strict propriety inequality I get is only for scores of pairs of probabilities, while the accuracy arguments require strict propriety for pairs of credences exactly one of which is not a probability.

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