Wednesday, February 1, 2023

From anti-anti-Bayesianism to propriety

Let’s work in the setting of my previous post, including technical assumption (3), and also assume Ω is finite and that our scoring rules are all continuous.

Say that an anti-Bayesian update is when you take a probability p, receive evidence A, and make your new credence be p(⋅|Ac), i.e., you conditionalize on the complement of the evidence. Anti-Bayesian update is really stupid, and you shouldn’t get rewarded for it, even if all you care about are events other than A and Ac.

Say that an H-scoring rule s is anti-anti-Bayesian providing that the expected score of a policy of anti-Bayesian update on an event A whose prior probability is neither zero nor one is never better than the expected score of a policy of Bayesian update.

I claim that given continuity, anti-anti-Bayesianism implies that the scoring rule is proper.

First, note that by continuity, if it’s proper at all the regular probabilities (ones that do not assign 0 or 1 to any non-empty set) on H, then it’s proper (I am assuming we handle infinities like in this paper, and use Lemma 1 there).

So all we need to do is show that it’s proper at all the regular probabilities on H. Let p be a regular probability, and contrary to propriety suppose that Eps(p) < Eps(q) for another probability q. For t ≥ 0, let pt be such that tq + (1−t)pt = p, i.e., let pt = (ptq)/(1−t). Since p is regular, for t sufficiently small, pt will be a probability (all we need is that it be non-negative). Using the trick from the Appendix of the previous post with q in place of p1 and pt in place of p2, we can set up a situation where the Bayesian update will have expected score:

  • tEqs(q) + (1−t)Epts(pt)

and the anti-Bayesian update will have the expected score:

  • tEqs(pt) + (1−t)Epts(q).

Given anti-anti-Bayesianism, we must have

  • tEqs(pt) + (1−t)Epts(q) ≤ tEqs(q) + (1−t)Epts(pt).

Letting t → 0 and using continuity, we get:

  • Ep0s(q) ≤ Ep0s(p0).

But p0 = p. So we have propriety.

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