aa r X i v : . [ m a t h . P R ] F e b PROPER SCORING RULES AND DOMINATION
ALEXANDER R. PRUSS
Abstract.
I generalize a theorem of Predd, et al. (2009) on dominationand strictly proper scoring rules to the case of non-additive scoring rules.
Let Ω be a finite space. Let C be the set of all functions from the powerset P Ω to the reals: these we call credence functions. Let P be the subsetof C which consists of the functions satisfying the axioms of probability. Ascoring or inaccuracy rule is a function s from C to [0 , ∞ ] Ω . Thus, s ( c ) for a c ∈ C is a function from Ω to [0 , ∞ ], and its value at ω ∈ Ω represents howinaccurate having credence c is if the true state is ω .Given a probability p ∈ P , let E p be the expected value with respect to p . Then a scoring rule s is said to be proper provided that for all p ∈ P and c ∈ C , E p s ( p ) ≤ E p s ( c ), where E p f = X ω ∈ Ω ,p ( { ω } ) =0 p ( { ω } ) f ( ω )is the mathematical expectation with a tweak to ensure well-definition incase f is infinite at some points with probability zero. If the inequality E p s ( p ) ≤ E p s ( c ) is always strict, the rule is said to be strictly proper.Enumerate the elements of Ω as ω , . . . , ω n . Let T be the set of vectors v = ( v , . . . , v n ) in R n with v i ≥ P i v i = 1. For v ∈ T , let p v be theprobability function such that p v ( { ω i } ) = v i for all i . Say that a scoring ruleis continuous on probabilities if the function ˆ s ( v ) = s ( p v ) is a continuousfunction from T to [0 , ∞ ] Ω , where the latter space is equipped with theproduct topology.Say that a member of [0 , ∞ ] Ω is finite if it never takes on the value ∞ ;otherwise, say it’s infinite. Let S s = { s ( p ) : p ∈ P} be the set of all scoresof probabilities, and let F s = S s ∩ [0 , ∞ ) Ω be the set of finite scores. Theorem 1.
Suppose s is strictly proper and that S s is the closure of F s .Then for any c ∈ P\C , there is a probability p such that s ( p ) < s ( c ) every-where on Ω . In other words, given the stated conditions, every score of a non-probability is dominated by the score of a probability (lower scores arebetter).
Corollary 1.
Suppose s is strictly proper and continuous on the probabil-ities. Then for any c ∈ P\C , there is a probability p such that s ( p ) < s ( c ) everywhere on Ω . In the special case of additive scoring rules, this was proved by Predd,et al. (2009). This result in the case of strict propriety was announcedby Richard Pettigrew, but the proof appears deficient. Michael Nielsen hasdiscovered a proof of this domination result using different methods duringapproximately the same time period as I did.
Proof of Corollary 1.
The set S s is the image of the compact set T underthe continuous mapping v s ( p v ) so it is compact.It remains to show that it is equal to F s . Note that [0 , ∞ ] Ω is metrizable(e.g., one can use the metric d ( a, b ) = P i | arctan a i − arctan b i | ), so we canwork in terms of sequences. Suppose ( s n ) is a convergent sequence in F s .Then s n = ˆ s ( v n ) for a sequence ( v n ) in T . Passing to a subsequence, bycompactness of T , we may assume v n converges to v ∈ T is compact. Thus, s n converges to ˆ s ( v ) by continuity of ˆ s , and hence the limit of s n is in S s .Thus, the closure of F s is a subset of S s .Next, observe that if v ∈ T ◦ , so that all the coordinates v i of v are non-zero and c is any member of C\{ p v } , then by strict propriety E p v (ˆ s ( v )) Fix a point z ∈ [0 , ∞ ] n . Let D be a subset of R n . Suppose thatfor all v ∈ C , the support function σ D ( v ) is finite. Let H v = { w : h z, v i ≤ σ D ( v ) } . Suppose further that z ∈ ( H ◦ v ) ∗ for every non-zero v ∈ C . Thenthere is a z ′ in the convex hull Conv( D ) such that z ′ i < z i for all i . Lemma 2. Let D be a subset closed in R n such that for every non-zero v ∈ C , there is a unique closed half-space H v with outward normal v suchthat D ⊆ H v and the boundary of H v intersects D at exactly one point.Then for any z ∈ Conv( D ) , there is a z ′ ∈ D such that z ′ i ≤ z i for all i .Proof of Lemma 1. Let Q = { w ∈ R n : ∀ i ( w i < z i ) } . Suppose first that Q does not intersect the convex hull Conv( D ). Then let H be a half-spacewith outward normal v that separates Conv( D ) from Q . It follows thatthere is an a ∈ R such that for every w ∈ Q , we have h w, v i ≥ a . Suppose v j > j . Choose any member w ′ of Q . Let b = h w ′ , v i . Let w i = w ′ i for i = j and let w i = w ′ i − ( b − a + 1) /v j . Then w ∈ Q and h w, v i = b − ( b − a + 1) = a − < a , a contradiction. Thus, v j ≤ j ,and hence v ∈ C .Then H v and H have the same outward normal, and since H v is thesmallest closed half-space with that outward normal to contain D , we musthave H v ⊆ H . It follows that H v separates Conv( D ) from Q . Let a = σ D ( v ).Then H v = { z : h z, v i ≤ a } , and for all w ∈ Q , we have h w, v i ≥ a . Taking asequence of members of Q converging to z and using the continuity of h· , v i ,we conclude that h z, v i ≥ a . But this contradicts the claim that z ∈ ( H ◦ v ) ∗ .Thus, Q intersects Conv( D ). Let z ′ be any element of their intersection. (cid:3) Proof of Lemma 2. Observe that Conv( D ) ∩ ( z + C ) is bounded. To see this,let v = ( − /n, ..., − /n ). Then Conv( D ) ⊆ H v . But it is easy to see thatthe H v ∩ ( w + C ) must be bounded.Thus, Conv( D ) ∩ ( w + C ) is a compact set. Let z ′ be the point of thatset furthest from z . Any point in z ′ + C other than z ′ will be even furtherfrom z , so Conv( D ) ∩ ( z ′ + C ) is empty.Let H ′ be a closed half-space that separates Conv( D ) from z ′ + C . Since C is self-dual, H ′ has outward normal v ′ in C . Then D is contained in H v ′ .Then only way z ′ can be a convex combination of members of D , then, is ifit is a convex combination of members of D on the boundary of H v ′ . Butthere is only one member of D on the boundary of H v ′ for v ′ ∈ C , andhence z ′ must be equal to that one member. Thus, z ′ ∈ D , and the proof iscomplete. (cid:3) Proof of Theorem 1. Let φ : ( −∞ , ∞ ] Ω → ( −∞ , ∞ ] n be the homeomor-phism such that φ ( f ) = u where u i = f ( ω i ). Observe that E p v ( f ) = h φ ( f ) , v i for v ∈ T . Let s ∗ ( v ) = φ (ˆ s ( v )).Let D = φ [ F s ]. Since all scores are non-negative, if u ∈ C and v ∈ D , we have h v, u i ≤ 0, and so σ D ( u ) ≤ < ∞ for all u ∈ C . Fix any ALEXANDER R. PRUSS c ∈ C\P . Note that For each s ∗ ( v ) ∈ D for v ∈ T , we can let H v = { w ∈ ( −∞ , ∞ ] n : h w, − v i ≤ h s ∗ ( v ) , − v i} . Strict propriety together with theclosure assumption then ensures that if z = φ ( s ( c )), then the conditions ofLemma 1 are satisfied. Let z ′ be as in the conclusion of Lemma 1.Then the conditions of Lemma 2 yield a z ′′ ∈ D such that z ′′ i ≤ z ′ i for all i . Letting p ∈ P be such that z ′′ = φ ( s ( p )), our proof is complete. (cid:3) References [1] Joel B. Predd, Robert Seiringer, Elliott H. Lieb, Daniel N. Osherson, H. VincentPoor, and Sanjeev R. Kulkarni. 2009. “Probabilistic Coherence and Proper ScoringRules”,