Monday, May 2, 2022

An argument for probabilism without assuming strict propriety

Suppose that s is a proper scoring rule on a finite space Ω continuous on probabilities and suppose that for no probability p is the expectation Eps(p) infinitely bad (i.e., no probability is infinitely bad by its own lights). Suppose that s is probability distinguishing: there isn’t a non-probability c and probability p such that s(c) = s(p) everywhere. Then any non-probability credence c is weakly s-dominated by some probability p: i.e., s(p)(ω) is at least as good as s(c)(ω) for all ω, and strictly better for at least one ω. (This follows from the fact that Lemma 1 of this short piece holds with the same proof when q is a non-probability.)

If one thinks that one should always switch to a weakly dominating option, then this conclusion provides an argument for probabilism.

One might, however, reasonably think that it is only required to switch to a weakly dominating option when one assigns non-zero probability of the weakly dominating option being better. If so, then we get a weaker conclusion: your credences should either be irregular (i.e., assign zero to some non-empty set) or probabilistic. But a view that permits violations of the axioms of probability but only when one has irregular credences seems really implausible. So your credences should be probabilistic.

The big question is whether probability distinguishing is any more plausible as a condition on a scoring rule than strictness of propriety. I think it has some plausibility, but I am not quite sure how to argue for it.

1 comment:

Alexander R Pruss said...

I had to add the not-infinitely-bad-by-own-lights condition.