Monday, February 13, 2023

Strictly proper scoring rules in an infinite context

This is mostly just a note to self.

In a recent paper, I prove that there is no strictly proper score s on the countably additive probabilities on the product space 2κ where we require s(p) to be measurable for every probability p and where κ is a cardinal such that 2κ > κω (e.g., the continuum). But here I am taking strictly proper scores to be real- or extended-real valued. What if we relax this requirement? The answer is negative.

Theorem: Assume Choice. If 2κ > κω and V is a topological space with the T1 property and ≤ is a total pre-order on V. Let s be a function s from the extreme probabilities on Ω = 2κ, with the product σ-algebra, concentrated on points to the set of measurable functions from Ω to V. Then there exist extreme probabilities r and q concentrated at distinct points α and β respectively such that s(r)(α) ≤ s(q)(α).

A space is T1 iff singletons are closed.
Extreme probabilities only have 0 and 1 values. An extreme probability p is concentrated at α provided that p(A) = 1 if α ∈ A and p(A) = 0 otherwise. (We can identify extreme probabilities with the points at which they are concentrated, as long as the σ-algebra separates points.) Note that in the product σ-lagebra on 2κ singletons are not measurable.

Corollary: In the setting of the Theorem, if Ep is a V-valued prevision from V-valued measurable functions to V such that (a) Ep(c⋅1A) = cp(A) always, (b) if f and g are equal outside of a set with p-measure zero, then Epf = Epg, and (c) Eps(q) is always defined for extreme singleton-concentrated p and q, then the strict propriety inequality Ers(r) > Ers(q) fails for some distinct extreme singleton-concentrated p and q.

Proof of Corollary: Suppose p is concentrated at α and Epf is defined. Let Z = {ω : f(ω) = f(α)}. Then p(Z) = 1 and p(Zc) = 0. Hence f is p-almost surely equal to f(α) ⋅ 1Z. Thus, Epf = f(α). Thus if r and q are as in the Theorem, Ers(r) = s(r)(α) and Ers(q) = s(q)(α), and our result follows from the Theorem.

Note: Towards the end of my paper, I suggested that the unavailability of proper scores in certain contexts is due to the limited size of the set of values—normally assumed to be real. But the above shows that sometimes it’s due to measurability constraints instead, as we still won’t have proper scores even if we let the values be some gigantic collection of surreals.

Proof of Theorem: Given an extreme probability p and z ∈ V, the inverse image of z under s(p), namely Lp, z = {ω : s(p) = z}, is measurable. A measurable set A depends only on a countable number of factors of 2κ: i.e., there is a countable set C ⊆ κ such that if ω, ω′ ∈ κ agree on C, and one is in A, so is the other. Let Cp ⊂ κ be a countable set such that Lp, s(p)(ω) depends only on Cp, where ω is the point p is concentrated on (to get all the Cp we need the Axiom of Choice). The set of all the countable subsets of κ has cardinality at most κω, and hence so does the set of all the Cp. Let Zp = {q : Cq = Cp}. The union of the Zp is all the extreme point-concentrated probabilities and hence has cardinality 2κ. There are at most κω different Zp. Thus for some p, the set Zp has cardinality bigger than κω (here we used the assmption that 2ω > κω).

There are q and r in Zp concentrated at α and β, respectively, such that αCp = βCp (i.e., α and β agree on Cp). For the function (⋅)∣Cp on the concentration points of the probabilities in Zp has an image of cardinality at most 2ω ≤ κω but Zp has cardinality bigger than κω. Since αCp = βCp, and Lq, s(q)(α) and Lr, s(r)(β) both depend only on the factors in Cp, it follows that s(q)(β) = s(q)(α) and s(r)(α) = s(r)(β). Now either s(q)(β) ≥ s(r)(α) or s(q)(β) ≤ s(r)(α) by totality. Swapping (q,α) with (r,β) if necessary, assume s(q)(α) = s(q)(β) ≤ s(r)(α).

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