Continuity and completeness under risk

Suppose somenon-degenerate preferencesR, with strict part P , over risky outcomes satisfy Independence. Then, when they satisfy any two of the following axioms, they satisfy the third. Herstein–Milnor: for all lotteries p, q, r , the set of a’s for which ap + (1 − a)qRr is closed. Archimedean: for all p, q, r there exists a > 0 such that if pPq, then ap + (1 − a)rPq. Complete: for all p, q, either pRq or qRp. © 2010 Elsevier B.V. All rights reserved. Let X be a finite set, and for m = |X |, let P = {p ∈ Rm + : ∑ i pi = 1} be the set of lotteries over X . Let ≽ be a transitive and reflexive binary relation on P . As usual define s ≻ t if s ≽ t and not t ≽ s, and s ∼ t if s ≽ t and t ≽ s. We say that ≽ is non trivial if there exist s and t in P such that s ≻ t . The relation ≽ satisfies: Independence, if for all p, q, r ∈ P and λ ∈ (0, 1), p ≽ q if and only if λp + (1 − λ)r ≽ λq + (1 − λ)r; Herstein–Milnor, if for all p, q, r ∈ P the set {α ∈ [0, 1] : αp + (1 − α)q ≽ r} is closed;1 Archimedean, if for all p, q, r ∈ P , p ≻ q implies λp+ (1−λ)r ≻ q for some λ ∈ (0, 1); Completeness, if for all p and q, either p ≽ q or q ≽ p. In this note I prove the following theorem. Theorem 1. Suppose ≽ is a transitive, reflexive, non-trivial binary relation onP , that satisfies Independence. If ≽ satisfies any two of the following axioms, it satisfies the third: Herstein–Milnor, Archimedean and Completeness. Schmeidler (1971) proved an analogous theorem for the case in which≽ is a preference relation on a set, not necessarily involving lotteries. He proved that if ≽ on a connected topological set Z is such that for some x and y, x ≻ y, then closed weak upper and lower contour sets and open strict upper and lower contour sets imply completeness. That Completeness and Independence imply that HM Continuity and Archimedean are equivalent is trivial and was first claimed ✩ I thank Edi Karni, Efe Ok, and a referee in this journal for their comments. E-mail address: [email protected]. 1 The proof below shows that under Independence, this version of HM implies the stronger version which also requires that {α : r ≽ αp + (1 − α)q} is closed. A similar argument applies to the definition of the Archimedean axiom. by Aumann (1962, p. 453). Karni (2007) proved that under a property weaker than Independence (Local Mixture Dominance), Completeness and Archimedean imply HM Continuity. Hence, we only need to prove that under the assumptions of the theorem, HM Continuity and Archimedean imply Completeness; to do so I will prove the following lemma, which together with Schmeidler’s theorem will establish the desired result. Lemma 1. Suppose X is finite, that ≽ is a transitive, reflexive binary relation on the space P of lotteries over X, and that ≽ satisfies Independence. (a) If ≽ satisfies HM Continuity then for all p, {q : q ≽ p} and {q : p ≽ q} are closed. (b) If ≽ satisfies the Archimedean Axiom, then for all p, {q : q ≻ p} and {q : p ≻ q} are open in the relative topology in P . Aversion of part (a) of the lemmawas established in Proposition 1 in Dubra et al. (2004), but with slightly different axioms: a weaker Independence, and a stronger continuity: Double Mixture Continuity: for any p, q, r, s in P the following set is closed T = {λ ∈ [0, 1] : λp + (1 − λ)r ≽ λq + (1 − λ)s}. Part (a) of the lemma is relevant, despite Proposition 1 in Dubra et al., because Double Mixture Continuity is not a standard axiom, and the Independence axiom in this paper is standard. Also, the proof is similar, but simpler. To the best of my knowledge, part (b) is new. Both (a) and (b) could be proved in a more cumbersome manner by appealing to the well known equivalence between algebraic closedness (HM Continuity) and topological closedness (and similarly for openness). Proof. Proof of (a). In order to show that for all v the set S = {r : r ≽ v} is closed take any q in its boundary. If S is a singleton, there is nothing to prove, and if it is not, by the Independence axiom it 0165-4896/$ – see front matter© 2010 Elsevier B.V. All rights reserved. doi:10.1016/j.mathsocsci.2010.11.001 Author's personal copy J. Dubra / Mathematical Social Sciences 61 (2011) 80–81 81 is a convex set and therefore has a nonempty relative interior. Pick any p in the relative interior of S. Let B be the open unit ball in the linear space generated by S − v, endowed with the relative topology. Fix any λ ∈ (0, 1) and any ε > 0. For any b ∈ B, pick δ > 0 small enough that εb + δB ⊂ εB. Since q is in the boundary of S, there exists w ∈ S such that ‖w − q‖ < δ, which implies εb + (1 − λ)(q − w) ∈ εB and therefore λp + (1 − λ) q + εb = λp + (1 − λ) w + (1 − λ) (q − w) + εb ∈ λp + (1 − λ) S + εB. (1) For a fixedλ, since p is in the relative interior of S, there exists ε > 0 small enough such that p+ ε λ B ⊆ S. Since Eq. (1) was true for all ε, we obtain λp + (1 − λ)q + εB ⊆ λ  p + ε