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(⋅|A^c), 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 A^c.

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 E_ps(p) < E_ps(q) for another probability q. For t ≥ 0, let p_t be such that tq + (1−t)p_t = p, i.e., let p_t = (p−tq)/(1−t). Since p is regular, for t sufficiently small, p_t 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 p_t in place of p₂, we can set up a situation where the Bayesian update will have expected score:

• tE_qs(q) + (1−t)E_pts(p_t)

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

• tE_qs(p_t) + (1−t)E_pts(q).

Given anti-anti-Bayesianism, we must have

• tE_qs(p_t) + (1−t)E_pts(q) ≤ tE_qs(q) + (1−t)E_pts(p_t).

Letting t → 0 and using continuity, we get:

• E_p0s(q) ≤ E_p0s(p₀).

But p₀ = p. So we have propriety.