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 2^κ where we require s(p) to be measurable for every probability p and where κ is a cardinal such that 2^κ > κ^ω (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 2^κ > κ^ω 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 = 2^κ, 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 2^κ singletons are not measurable.

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

Proof of Corollary: Suppose p is concentrated at α and E_pf is defined. Let Z = {ω : f(ω) = f(α)}. Then p(Z) = 1 and p(Z^c) = 0. Hence f is p-almost surely equal to f(α) ⋅ 1_Z. Thus, E_pf = f(α). Thus if r and q are as in the Theorem, E_rs(r) = s(r)(α) and E_rs(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 L_p, z = {ω : s(p) = z}, is measurable. A measurable set A depends only on a countable number of factors of 2^κ: 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 C_p ⊂ κ be a countable set such that L_{p, s(p)(ω)} depends only on C_p, where ω is the point p is concentrated on (to get all the C_p we need the Axiom of Choice). The set of all the countable subsets of κ has cardinality at most κ^ω, and hence so does the set of all the C_p. Let Z_p = {q : C_q = C_p}. The union of the Z_p is all the extreme point-concentrated probabilities and hence has cardinality 2^κ. There are at most κ^ω different Z_p. Thus for some p, the set Z_p has cardinality bigger than κ^ω (here we used the assmption that 2^ω > κ^ω).

There are q and r in Z_p concentrated at α and β, respectively, such that α∣_Cp = β∣_Cp (i.e., α and β agree on C_p). For the function (⋅)∣_Cp on the concentration points of the probabilities in Z_p has an image of cardinality at most 2^ω ≤ κ^ω but Z_p has cardinality bigger than κ^ω. Since α∣_Cp = β∣_Cp, and L_{q, s(q)(α)} and L_{r, s(r)(β)} both depend only on the factors in C_p, 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).