Second-best Efficiency of Allocation Rules:Strategy-proofness and Single-peaked

Preferences with Multiple Commodities∗

Hidekazu Anno†

Graduate School of Fundamental Science and EngineeringWaseda University

andHiroo Sasaki‡

School of Commerceand

Graduate School of Fundamental Science and EngineeringWaseda University

March 3, 2009 (This version: December 1, 2009)

Abstract

This paper studies the design of a strategy-proof resource allocation rule ineconomies with perfectly divisible multiple commodities and single-peaked pref-erences. It is known thatthe uniform ruleis the unique allocation rule satisfyingstrategy-proofness, Pareto efficiency, andanonymityif the number of commodityis only oneand preferences are single-peaked (Sprumont (1991)). However, if thenumber of commodities is greater than one, the situation drastically changes and atrade-off betweenstrategy-proofnessandPareto efficiencyarises. The generalizeduniform rule in multiple-commodity settings is still strategy-proof, butnot Paretoefficient. In this paper, we first investigate the existence problem of second-bestefficient rules, where a strategy-proof rule is second-best efficient if in the class ofall strategy-proof rules, there isno other strategy-proof rule that gives a “better”outcome than the considered rule in terms of Pareto domination for all preferenceprofiles. We show that in ann person andm good setup for any strategy-proofrule, there exists a second-best efficient rule that Pareto dominates the former. In

∗This paper is based on Anno (2008a,b) and Sasaki (2003), but some additional results are included in thepresent version. We would like to thank Takashi Akahoshi, Matthew Jackson, Yoichi Kasajima, Kohei Kawa-mura, Manipushpak Mitra, Shuhei Morimoto, Herve Moulin, Clemens Puppe, Toyotaka Sakai, Ken Sawada,Tadashi Sekiguchi, Shigehiro Serizawa, Koichi Tadenuma, William Thomson, Takuma Wakayama, NobuyaWatanabe, and Naoki Yoshihara for the helpful comments and discussions. Sasaki’s research is supported bythe Japan Society for the Promotion of Science Grants-in-Aid for Scientific Research (No.19530162).†e-mail: anno@toki.waseda.jp‡e-mail: hsasaki@waseda.jp

1

the proof of the theorem, Zorn’s Lemma plays an important role. Furthermore,by relaxing the requirement of Pareto efficiency, we show a possibility theorem,there is an egalitarian rational (consequently, non-dictatorial) strategy-proof rulethat is second-best efficient although there is no egalitarian rational strategy-proofrule that is Pareto efficient. Second, we show that the generalized uniform rule issecond-best efficient and give a new characterization of the generalized uniformrule with second-best efficiency in a two person andm good setup. That is, itis the unique rule satisfying the three axioms of the second-best efficiency, weakpeak-onliness, and egalitarian rationality. We also show that the three axioms aremutually independent.

Journal of Economic LiteratureClassification Numbers : D63, D71, D78.

Keywords: Strategy-proofness; Second-best efficiency; Generalized uniform rule;Single-peaked preference.

1 Introduction

Ever since Sprumont (1991), resource allocation in economies with single-peaked pref-erences has been studied by many authors. If the number of commodity is only one,Sprumont (1991) presented a characterization of a resource allocation rule that satisfiesthree axioms:strategy-proofness, Pareto efficiency, andanonymity. Strategy-proofnessmeans that announcing their true preferences is a dominant strategy for all agents inthe game of stating their preferences. Pareto efficiency requires that any allocation ob-tained by the rule is Pareto efficient with respect to the reported preference relations.Anonymity says that the rule is independent of the “names” of the agents.

Sprumont proposes a resource allocation rule calledthe uniform rule. 1 Under theuniform rule, the same amount of a single divisible good is alloted to everyone exceptpeople whose peaks are small enough if excess demand exists or large enough if excesssupply exists.2

Sprumont’s theorem essentially depends on the assumption that there is only onecommodity. If the number of goods is greater than one, we may naturally extend theuniform rule. The extended rule is referred to asthe generalized uniform rule. 3 Thegeneralized uniform rule is defined by applying the single good uniform rule commod-ity by commodity.

It is easy to show that the generalized uniform rule is strategy-proof.4 However,as shown in Example 1.1 below, the ruleviolatesPareto efficiency.

1The rule was originally introduced by Benassy (1982).2A single-peaked preference of one good may have several interpretations. One possible interpretation is

the “fixed price economy” interpretation. In this interpretation, the peak of a preference is interpreted as a“Walrasian demand” under a fixed price. Given a preference profile, if the total number of all peak points isgreater (or smaller) than the amount of a good to be alloted, it is said that there is an excess demand (or anexcess supply). The fixed price interpretation is just for expository convenience. The result obtained by thispaper may be applicable to many different interpretations.

3See Amoros (2002).4It also satisfies anonymity.

2

O1

O2

p(R2)

p(R1)

Ω1

Ω2Z

A

Figure 1

EXAMPLE 1.1. There are two agents and two goods. LetΩ1 andΩ2 be the amountsof goods 1 and 2. Figure 1 is an Edgeworth Box. In Figure 1, p(R1) and p(R2) des-ignate the peaks of Mr. 1 and Mr. 2’s preferences, respectively. The middle pointZ = (Ω1/2,Ω2/2) is the allocation where equal amounts of goods are alloted to eachagent. Since for each good j= 1,2, both agents have peaks greater thanΩ j/2, thegeneralized uniform rule assigns equal amounts of goods to both. This allocation isgiven by Z. However, if their indifference curves through Z can be drawn as in Figure1, there is room for Pareto improvement. (For example, A is better than Z for bothagents.)

The literature on strategy-proofness in economic environments has uncovered a ten-sion between strategy-proofness and Pareto efficiency. For example, Hurwicz (1972)shows that in pure exchange economies with two agents and two goods, no rule isstrategy-proof, Pareto-efficient, and individually rational. Zhou (1991) proves that inpure exchange economies with two agents and the number of goods greater than orequal to two, no rule is strategy-proof, Pareto efficient, and non-dictatorial.5 Serizawa(2002) shows that in pure exchange economies withn agents andm goods, no rule isstrategy-proof, Pareto-efficient, and individually rational. Finally, Serizawa and Wey-mark (2003) show that inn personm goods pure exchange economies, strategy-proofand Pareto efficient rules cannot guarantee a minimum consumption: for arbitrarilysmallε > 0, there exists a preference profile such that under the rule and the preferenceprofile, there is an agent whose consumption has an Euclidean norm smaller thanε.

5See also Kato and Ohseto (2002).

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The same kind of trade-off between strategy-proofness and Pareto efficiency existsin economies with single-peaked preferences and multiple commodities.6

One way of escaping this trade-off is to drop or replace an axiom. In fact, Amoros(2002) resolved this difficulty along these lines. That is, he replaced Pareto efficiencywith the axiom ofsame-sidedness. This axiom requires that for each good, the amountof the good received by everyone is located on the same side of the agent’s own peak.7

His main theorem says that if the number of agents istwo and the number of goods isgreater than or equal to two,the generalized uniform rule is the unique rule satisfyingenvy-freeness, strategy-proofness, and same-sidedness.8

Since same-sidedness is a straightforward extension of Sprumont’s efficiency con-cept,9 Amoros’ characterization theorems may be understood as a multiple-good ver-sion of Sprumont’s characterization.

In the present paper, however, we study some problems concerning efficiency andstrategy-proofness in a multiple-good setup from a much different point of view, andpropose a new direction of research. Since the trade-off between strategy-proofnessand Pareto efficiency (except when dictatorship is allowed) is inevitable in this setup,we mainly consider strategy-proofness. More precisely, we considerthe setΓSP ofall strategy-proof rules. We ask the following question: are there any strategy-proofrules that are not Pareto-dominated by any other strategy-proof rules? To answer thisquestion, we propose two concepts of second-best efficiency. The first, which is calledweak second-best efficiency among strategy-proof rules(WSESP), is the weaker ver-sion. That is, a strategy-proof rulef0 is WSESP,if for any strategy-proof rulef1 thatPareto-dominatesf0, then f0 Pareto-dominatesf1. The second concept, which is calledstrong second-best efficiency among strategy-proof rules(SSESP), is the stronger ver-sion. That is, a strategy-proof rulef0 is SSESP,if for any strategy-proof rulef1 whichPareto-dominatesf0, then f0 is equal tof1.

In the proof of the main results (Theorems 2.1, 2.2, 3.1, and 3.2), theoption setofeach agent plays an important role.10 For thei-th agent, given a preference profileR−i = (R1, · · · ,Ri−1,Ri+1, · · · ,Rn) of the other agents, the option set of thei-th agentunder a rulef is the set of his consumption bundles that may be assigned to him by rulef if the other agents announce the preference profleR−i . In the setΓSP of all strategy-proof rules, we prove that for any rulesf andg ∈ ΓSP, f Pareto-dominatesg if and onlyif the option set ofg is included in that off for all agents (Lemma 4.4). Thus, in the

6We see an analogy of the special case of a theorem in Serizawa (2002) in a single-peaked preferenceenvironment in Section 3. Amoros (2002) presented that there is no strategy-proof, Pareto efficient, andnon-dictatorial rule in a two person setup.

7That is, for each good, if the quantity of the good received by an agent is greater than or equal to hisown peak amount, then the quantities of the good received by the remaining people should be greater thanor equal to their own peak amounts, andvice versa. If the number of commodity is one, same-sidedness isequivalent to Pareto efficiency. However, if the number of commodity is greater than one, same-sidednessdoes not necessarily imply Pareto efficiency. Amoros (2002) called the axiomCondition E(CE).

8Amoros (2002) Theorem 2. In his Theorem 3, he replaces envy-freeness withweak anonymity. Re-cently, Morimoto, Serizawa, and Ching (2008) extended Amoros’ result considerably.

9As noted earlier, same-sidedness is equivalent to Pareto efficiency if the number of commodity is one.Sprumont (1991) assumed Pareto efficiency because some properties of same-sidedness are required in hisproof. Hence, in multiple-good economies, only same-sidedness is required for extending Sprumont’s char-acterization.

10The notion of the option set was originally introduced by Barbera and Peleg (1990).

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domain ofΓSP, the relationship of Pareto-domination is equivalent to the relationshipof set theoretic inclusion. In the proofs of the main theorems, this observation playsan essential role. For example, in Theorem 2.1, we prove the existence of a Pareto-undominated rule (a WSESP rule). In the proof of the theorem, it is enough to showthat there exists amaximaloption set with respect to the inclusion relationship in thecollection of option sets. This is because by observation, a Pareto-undominated ruleis a rule that gives the maximal element of the set of option sets under the inclusionrelationship.

This paper consists of five sections. In Section 2, we present the model and de-scribe our axioms. Moreover, we examine the existence of strategy-proof, efficient,and equitable allocation rule in ann person andm good setup. In Section 3, a newcharacterization of the generalized uniform rule is given in a two person andm goodsetup. In Section 4, we give the proofs of the main results. Section 5 concludes.

2 Second-Best Efficiency of Resource Allocation Rules

2.1 Single-peaked Preferences with Multiple Commodities

Let N = 1, . . . , n be the set of individuals. LetM = 1, · · · ,m be the set of commodi-ties. All commodities are perfectly divisible. The bundleΩ = (Ω1, · · · ,Ωm) ∈ Rm

++ de-notes a social endowment of the commodities.11 Let B = x = (x1, · · · , xn) ∈ (Rm

+ )n |∑ni=1 xi = Ω denote the set of feasible allocations. We do not allow free disposal.

The preferences of each individual are given by a complete, transitive, continuous andstrictly convex binary relation on

∏mj=1[0,Ω j ]. 12

DEFINITION 2.1. A preference R is single-peaked if there exists p(R) ∈ ∏mj=1[0,Ω j ]

such that for all x, x′ ∈∏mj=1[0,Ω j ](x , x′),[

∀ j ∈ M, x′j ≤ x j ≤ p j(R) ∨ p j(R) ≤ x j ≤ x′j]⇒ x P(R) x′.

Let R be the set of single-peaked preferences. We call each element ofRN a pref-erence profile, or simply a profile. For each profileR = (R1, · · · ,Rn) ∈ RN, eachi ∈ N,the subprofile obtained by removingi’s preference is denoted byR−i . That is,R−i =

(R1, · · · ,Ri−1,Ri+1, · · · ,Rn). It is convenient to write the profile (R1, · · · ,Ri−1, Ri ,Ri+1, · · · ,Rn)as (Ri ; R−i). A function fromRN to B is calleda rule. LetΓ denotes the set of all rules.

2.2 Pareto Efficiency and Second-Best Efficiency : Impossibility vs.Possibility Results in a Multiple Commodity Setting

In this subsection, we introduce our axioms. First, one cannot be better off by misre-porting one’s preference. Letf be our generic notation of rule.

11The symbolsN andR denote the set of natural and real numbers, respectively. LetR+ be the set ofnon-negative numbers and letR++ be the set of positive real numbers.

12For each preferenceR, P(R) and I (R) denote the asymmetric part ofR and the symmetric part ofR,respectively.

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Strategy-proofness (SP): for allR = (R1, · · · ,Rn) ∈ RN, all i ∈ N, and all Ri ∈ R,f i(R) Ri f i(Ri ; R−i).

Let ΓSP denote the set of allstrategy-proofrules. The following axiom reflects theidea that everyone should not be worse off than under equal division.

Egalitarian rationality (ER)13 : for all R = (R1, · · · ,Rn) ∈ RN and all i ∈ N,f i(R) Ri Ω

n .

Next, at the selected allocation, no one can be better off without making someoneworse off.

Pareto efficiency : for all R = (R1, · · · ,Rn) ∈ RN, there is nox = (x1, · · · , xn) ∈ Bsuch that i) xiRi f i(R) for all i ∈ N; and ii) xi0Pi0 f i0(R) for some i0 ∈ N.

The following proposition is an impossibility result about rules satisfying the ax-ioms above. It follows immediately from a powerful impossibility theorem by Serizawa(2002) or Serizawa and Weymark (2003). The proof of Proposition 2.1 is given in theAppendix.

PROPOSITION 2.1. Suppose that m≥ 2. Then, no rule satisfies strategy-proofness,Pareto efficiency and egalitarian rationality.

Note that with only one commodity, the impossibility described in Proposition 2.1does not hold because the uniform rule satisfies the stated axioms14(See Sprumont(1991) and Ching (1994)). Proposition 2.1 describes a difference between the environ-ment with only one commodity and environments with more than one commodities.

DEFINITION 2.2. The binary relation dom onΓ is defined as follows; for all f,g ∈ Γ

f dom g⇔ ∀R = (R1, · · · ,Rn) ∈ RN,∀i ∈ N, f i(R) Ri gi(R).15

That is, f domg means that for any preference profiles, the allocation selected byf is better than (or indifferent to) the one selected byg for all agents. By using the domrelation, we obtain the following equivalent expression ofPareto efficiency.

Pareto efficiency : for all g∈ Γ, g dom f⇒ f dom g.

In words,Pareto efficiency requires a rule to be a maximal element ofΓ preorderedby dom. The following two axioms are the second-best efficiency concepts we adoptin this paper. The first one requires that a rule is a maximal element ofΓSP preorderedby dom. The second one is more demanding. It requires that a rule is a maximal ele-ment ofΓSPpreordered by dom and there is no other rule that is welfare-equivalent to it.

13This axiom is sometimes referred to asthe equal division lower bound.14The rule also satisfies anonymity that is stronger thanERunder strategy-proofness.15Note that dom is reflexive and transitive. Hence it is a preorder onΓ. But it is not an order onΓ in

general. If f domg andg dom f , then we sayf andg are equivalent with respect to the welfare.

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Weak second-best efficiency among strategy-proof rules (WSESP) : f∈ ΓSPand for allg ∈ ΓSP, g dom f⇒ f dom g.

Strong second-best efficiency among strategy-proof rules (SSESP) : f∈ ΓSP and forall g ∈ ΓSP, g dom f⇒ f = g.16

The following theorem guarantees that for anySPrule f , we have aWSESPrulethat dominatesf .

THEOREM 2.1. For any f ∈ ΓSP, there exists a rule f0 ∈ ΓSP such that f0 satisfiesweak second-best efficiency among strategy-proof rules and f0 dom f .

As a consequence of relaxing the efficiency condition, we obtain the followingpossibility result in contrast to Proposition 2.1.

THEOREM 2.2. There exists a strategy-proof rule satisfying weak second-best effi-ciency among strategy-proof rules and egalitarian rationality.

3 A New Characterization of the Generalized UniformRule

3.1 The Generalized Uniform Rule

One of the purposes of this paper is to characterize the following rule which is knownasthe generalized uniform rule.

DEFINITION 3.1. The generalized uniform rule U: RN → B is the rule defined bythe following; for allR = (R1, · · · ,Rn) ∈ RN, all j ∈ M, all i ∈ N,

U ij(R) =

minp j(Ri), λ j(R) if∑n

i=1 p j(Ri) ≥ Ω j ,

maxp j(Ri), µ j(R) if∑n

i=1 p j(Ri) ≤ Ω j ,

where λ j(R) solves the equationΩ j =∑n

i=1 minp j(Ri), λ j(R) and µ j(R) solves theequationΩ j =

∑ni=1 maxp j(Ri), µ j(R).

3.2 A Characterization of the Generalized Uniform Rule

The following theorem shows an interesting property of the generalized uniform rule.

THEOREM 3.1. Suppose that n= 2. The generalized uniform rule satisfies strongsecond-best efficiency among strategy-proof rules (SSESP).17

16Sasaki (2003) first introducedSSESP. He called this conditionΛSP-efficiency.17Theorem 3.1 is first proved in Sasaki (2003). In the next section, we provide an alternative proof of the

theorem.

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In the previous subsection, we point out that for anyn ≥ 2 there exists a rulesatisfyingWSESPandER. Since the generalized uniform rule satisfiesER, by Theorem3.1 it is one of the rules satisfyingWSESPandER.

Now we ask whether there exists a rule satisfyingWSESPandER other than thegeneralized uniform rule. The answer is yes. An example of a rule satisfyingWSESPandERother than the generalized uniform rule is provided in Example 3.3. However,Theorem 3.2 shows that if we impose an auxiliary condition calledweak peak-onlinessin addition toWSESPandER, then the generalized uniform rule is the unique rule thatsatisfies these three axioms.

Weak peak-onliness (WP)18 : for all R = (R1, · · · ,Rn) ∈ RN, all i ∈ N, all Ri ∈ R,

p(Ri) = p(Ri)⇒ f i(R) = f i(Ri ; R−i).

THEOREM 3.2. Suppose that n= 2. The generalized uniform rule is the only rulethat satisfies weak second-best efficiency among strategy-proof rules (WSESP), egali-tarian rationality (ER) and weak peak-onliness (WP).

3.3 Independence of Axioms in Theorem 3.2

In this subsection we show that Theorem 3.2 is a tight characterization. That is, drop-ping any one of three axioms leads to other rules.

EXAMPLE 3.1. An example of a rule that satisfies both ER and WP, but not WSESPis the equal division rule, E, defined as follows: for allR ∈ RN, E(R) = (Ω2 ,

Ω2 ).

Obviously E satisfies ER and WP but E not WSESP because U dom E but E does notdominate U.

EXAMPLE 3.2. Examples of rules that satisfy both WP and WSESP, but not ER arethe priority rules. Let D(i) be the priority rule in which individual i has the prioritydefined as follows: for allR = (R1,R2) ∈ RN, D(i)i(R) = p(Ri). Since D(i) is SPand Pareto efficient, it satisfies WSESP. It is also clear that D(i) satisfies WP. However,clearly D(i) does not satisfy ER.

EXAMPLE 3.3. An example of a rule that satisfies both ER and WSESP, but not WPis f0 below. Let f be the rule defined as follows. For allR = (R1,R2) ∈ RN,

f (R) =

(Ω,0) if Ω R1Ω2 and 0 R2Ω

2 ,

(Ω2 ,Ω2 ) otherwise.

Obviously f satisfies SP and ER.First we show that U does not dominate f . By Lemma 4.6 of section 4.2, there exist

R1, R2 ∈ R such that p(R1) = p(R2) = (Ω1, · · · ,Ωm−1,0) andΩ P(R1)Ω2 and0 P(R2)Ω2 .Then f(R1, R2) = (Ω, 0). Since U(R1, R2) = (Ω2 ,

Ω2 ), U does not dominate f .

18This axiom is sometimes referred to asown peak only.

8

By Theorem 2.1, there exists f0 ∈ ΓSP such that f0 is WSESP and f0 dom f .Since f satisfies ER and dom is transitive, f0 satisfies ER. f10 (R1, R2)R1Ω P(R1)Ω2 andf 20 (R1, R2)R2 0 P(R2)Ω2 because f0 dom f . Since U(R1, R2) = (Ω2 ,

Ω2 ), f0 , U. This

means that f0 does not satisfy WP, because if f0 satisfied WP, then by Theorem 3.2f0 = U, a contradiction.

By Example 3.1, 3.2 and 3.3, we have shown thatthe three axioms in Theorem 3.2are mutually independent.

4 Proofs

4.1 Proofs of Theorems 2.1 and 2.2

To proveTheorems 2.1 and 2.2, we need to use some facts of metric spaces. Let (X,d)be a metric space.

DEFINITION 4.1. We define the functiond : X × (2X \ ∅) → R by d(x,A) =inf d(x,a) | a ∈ A for all x ∈ X and all A∈ 2X \ ∅. 19

DEFINITION 4.2. We defineK(X) to be the set of all nonempty compact subsets in(X,d). 20

DEFINITION 4.3. Let dH be the function fromK(X) × K(X) toR defined by

dH(A, B) = maxmaxa∈A

d(a, B),maxb∈B

d(b,A).

It is well-known that (K(X),dH) is also a metric space. The metricdH is referredto as Hausdorff metric.

REMARK 4.1. Let A, B ∈ K(X). If A ⊆ B, d(a, B) = 0 for all a ∈ A. Hence,dH(A, B) = maxb∈B d(b,A) for all A, B ∈ K(X) with A⊆ B.

LEMMA 4.1. Suppose thatS ⊆ K(X) is given, whereS is totally ordered by⊆. 21

Let C= cld(∪S∈SS)22. Then

∀S1,S2 ∈ S, [∃x0 ∈ C s.t.d(x0,S1) > d(x0,S2) ⇒ S1 ⊆ S2].

Proof. Suppose the contrary. That is,S1 * S2. Then, sinceS is totally ordered by⊆,S2 ( S1. And

d(x0,S1) = mind(x0, y) | y ∈ S1≤ mind(x0, y) | y ∈ S2 (∵ S2 ( S1)= d(x0,S2).

This is a contradiction. 19Note thatd(·,A) is a continuous function fromX toR when we fixA ∈ 2X \ ∅ arbitrarily.20Note thatd(x,A) = mind(x,a) | a ∈ A when we restrict the domain ofd within X × K(X).21In general, (Z,≥) is anordered setif ≥ is a reflexive, transitive and anti-symmetric binary relation onZ,

where≥ is reflexive if for allx ∈ Z, x ≥ x, and≥ is anti-symmetric if for allx, y ∈ Z, if x ≥ y andy ≥ x, thenx = y. An ordered set (Z,≥) is total if for all x, y ∈ Z, x ≥ y or y ≥ x.

22Note that cld(∪S∈SS) denotes the closure of∪S∈SS in (X,d)

9

LEMMA 4.2. Suppose that(X,d) is a compact metric space. Suppose also thatS ⊆K(X) is given, whereS is non empty and totally ordered by⊆. Let C= cld(∪S∈SS) ∈K(X). Then there exists a sequence of compact sets inS that converges to C withrespect to dH.

Proof. If C ∈ S, then the conclusion is trivial. We consider the caseC < S below.Step 1∀S′ ∈ S,∃S′′ ∈ S s.t.dH(S′′,C) < dH (S′,C)

2 .Let S′ ∈ S. SincedH is a metric andS′ , C, dH(S′,C) > 0. LetdH(S′,C) = η. Let

H = x ∈ C | d(x,S′) ≥ η2. The setH is not empty becausedH(S′,C) = η implies thatd(x∗,S′) = η for somex∗ ∈ C. Also H is a closed set because it is the inverse imageof continuous functiond(·,S′) with respect to [η2 ,+∞). Hence,H is compact becauseH ⊆ C ∈ K(X).

For the open coverB(x, η4) | x ∈ H, we can find finite pointsx1, · · · , xh ∈ H suchthatH ⊆ ∪h

i=1B(xi ,η4). 23 Sincex1, · · · , xh ∈ H ⊆ cld(∪S∈SS),

∀i ∈ 1, · · · ,h,∃Si ∈ S s.t. Si ∩ B(xi ,η

4) , ∅.

SinceS1, · · · ,Sh are totally ordered by⊆, there is a the maximal one. We call itS′′.In the following, we showdH(S′′,C) < η2. To this end, because of Remark 4.1, it is

sufficient to show thatd(x,S′′) < η2 for all x ∈ C.Let x ∈ C. First, suppose thatx ∈ H. Then

∃ix ∈ 1, · · · ,h s.t.x ∈ B(xix,η

4).

SinceS′′ ∩ B(xix,η4) , ∅, we can picky ∈ S′′ ∩ B(xix,

η4). By the triangle inequality

with respect tod,d(x, y) ≤ d(x, xix) + d(xix, y)

< η4 +η4

=η2 .

By the definition ofd, we haved(x,S′′) < η2.Note that we have shown that forx∗ ∈ H,

d(x∗,S′) = η >η

2> d(x∗,S′′).

Hence,S′ ⊆ S′′ by Lemma 4.1.Next suppose thatx < H. Then

η2 > d(x,S′) (∵ x < H)= mind(x, y) | y ∈ S′≥ mind(x, y) | y ∈ S′′ (∵ S′ ⊆ S′′)= d(x,S′′).

This completes the proof of step 1.Step 2

23Note thatB(x, ε) denotes the open ball centered atx with a radiusε.

10

By Step 1 and the axiom of choice, we have a functionΦ : S → S such that

∀S ∈ S,dH(Φ(S),C) <dH(S,C)

2.

Let S ∈ S. LetS1 ≡ S. For eachk ≥ 2, letSk ≡ Φ(Sk−1). SincedH(Sk,C) < dH (S,C)2k−1

for eachk ∈ N, Skk∈N satisfiesdH(Sk,C)→ 0 ask→ +∞.

REMARK 4.2. Note that the sequence obtained in Lemma 4.2 satisfies

S1 ⊆ S2 ⊆ S3 ⊆ · · · .

DEFINITION 4.4. Let f ∈ Γ and i ∈ N. For eachR−i ∈ RN\i,

Bf iR−i ≡ x ∈

m∏j=1

[0,Ω j ] | ∃Ri ∈ R s.t. fi(Ri ; R−i) = x.

We call Bf iR−i the option set of individual i under f andR−i .

Let τ(R,Y) = x ∈ Y | ∀y ∈ Y, xRy for all Y ⊆ ∏mj=1[0,Ω j ], all R ∈ R. That is,

τ(R,Y) denotes the best consumptions onY ⊆∏mj=1[0,Ω j ] with respect toR ∈ R. 24

LEMMA 4.3. Let f ∈ Γ.

f ∈ ΓSP⇔ ∀R ∈ RN,∀i ∈ N, f i(R) ∈ τ(Ri , Bf iR−i ).

Proof. Obvious.

LEMMA 4.4. Let f,g ∈ ΓSP.

f dom g⇔ ∀i ∈ N,∀R−i ∈ RN\i, BgiR−i ⊆ Bf i

R−i .

Proof. (⇒) Let i ∈ N,R−i ∈ RN\i. Let x ∈ BgiR−i . LetRi ∈ R be such thatτ(Ri ,

∏mj=1[0,Ω j ]) =

x. By Lemma 4.3,gi(Ri ; R−i) = x. Since f domg, f i(Ri ; R−i) Ri gi(Ri ; R−i). Hence,f i(Ri ; R−i) = x. This meansx ∈ Bf i

R−i .(⇐) Let i ∈ N,R ∈ RN. By Lemma 4.3,f i(R) ∈ τ(Ri , Bf i

R−i ) andgi(R) ∈ τ(Ri , BgiR−i ).

SinceBgiR−i ⊆ Bf i

R−i , f i(R) Ri gi(R).

For all f ,g ∈ ΓSP, f ∼ g if and only if f dom g andg dom f . Note that∼ is anequivalence relation onΓSP. For eachf ∈ ΓSP, [ f ] denotes the equivalence class offin the quotient setΓSP/ ∼.

24If f ∈ ΓSP, thenBf iR−i is closed set in

∏mj=1[0,Ω j ] for all i ∈ N, R−i ∈ RN\i. To see this, fixi ∈ N and

R−i ∈ RN\i, then we have a one agent social choice function fromR to∏m

j=1[0,Ω j ] induced by f . Note

that Bf iR−i is the range of the one agent social choice function. Applying Proposition 5 in Le Breton and

Weymark (1999), we have the conclusion. See also Barbera and Peleg (1990) .

11

DEFINITION 4.5. The binary relation Dom onΓSP/ ∼ is defined as follows : for all[ f ], [g] ∈ ΓSP/ ∼

[ f ] Dom[g] ⇔ f dom g.25

DEFINITION 4.6. Let (Z,≥) be an ordered set. We say that(Z,≥) is inductive ifevery non empty totally ordered subset ofZ has an upper bound.

The following theorem shows an important property of an inductive ordered set.

Zorn’s Lemma Let (Z,≥) be an ordered set which is inductive and let x0 be an elementof Z. Then

∃a ∈ Z s.t. a≥ x0 ∧ ∀x ∈ Z,¬(x > a).

LEMMA 4.5. The ordered set(ΓSP/ ∼,Dom

)is inductive.

Proof. Let S ⊆ ΓSP/ ∼. Suppose that(S,Dom |S×S

)is totally ordered. We construct

a ruleF : RN → B and show that [F] is an upper bound ofS.For anyi ∈ N, R−i ∈ RN\i, CR−i

= cld(∪[ f ]∈S Bf i

R−i).

Step 1In this step, we show that

∃[ fk]k∈N in S s.t.∀i ∈ N,∀R−i ∈ RN\i, Bf ikR−i → CR−i

.26 (1)

By Lemma 4.2, for eachi ∈ N, R−i ∈ RN\i,

∃[ fik]k∈N in S s.t.Bf iikR−i → CR−i

.

Note that by Remark 4.2, for anyR = (R1, · · · ,Rn) ∈ RN,

Bf 111R−1 ⊆ Bf 1

12R−1 ⊆ Bf 113R−1 ⊆ · · · ,

Bf 221R−2 ⊆ Bf 2

22R−2 ⊆ Bf 223R−2 ⊆ · · · ,

...

Bf nn1R−n ⊆ Bf n

n2R−n ⊆ Bf nn3R−n ⊆ · · · .

For eachk ∈ N, we have the greatest among [f1k], [ f2k], · · · , [ fnk] with respect to Dombecause[ f1k], [ f2k], · · · , [ fnk] is totally ordered by Dom. Define [fk] by it. We showthat[ fk]k∈N satisfies (1).

Let i ∈ N, R−i ∈ RN\i, k ∈ N. Becausefk dom fik, Bf iikR−i ⊆ Bf i

kR−i , we ob-tain dH(CR−i

, Bf ikR−i ) ≤ dH(CR−i

, Bf iikR−i ). SincedH(CR−i

, Bf iikR−i ) → 0 ask → +∞,

dH(CR−i, Bf i

kR−i )→ 0 ask→ +∞.

25Note that Dom is an order onΓSP/ ∼.26Bf i

kR−i k∈N is a sequence in (K(∏m

j=1[0,Ω j ]),dH). Bf ikR−i → CR−i

means the convergence of the se-quence with thedH metric.

12

Step 2In this step, we defineF. We can pick a representativefk for each equivalence

class [fk] (∵ Axiom of choice.). Later on, we fix fkk∈N. For eachR ∈ RN, we obtainthe sequence fk(R)k∈N in B. SinceB is compact, there exists a convergent subse-quence fR(k)(R)k∈N 27. Now we defineF : RN → B as follows : for eachR ∈ RN,F(R) = limk→+∞ fR(k)(R). Obviously,F ∈ Γ.

Step 3∀i ∈ N,∀R−i ∈ RN\i, BF iR−i = CR−i

.

First we showBF iR−i ⊆ CR−i

. Let xi ∈ BF iR−i . Then

∃Ri ∈ R s.t.F i(Ri ; R−i) = xi .

Let R = (Ri ; R−i). By the definition ofF, F(R) = limk→+∞ fR(k)(R). For all k ∈ N,

by thestrategy-proofnessof fR(k), f iR(k)

(R) ∈ τ(Ri , Bf iR(k)R−i ) ⊆ B

f iR(k)R−i ⊆ CR−i

. Hence,

f iR(k)

(R)k∈N is a convergent sequence inCR−i. SinceCR−i

is closed in∏m

j=1[0,Ω j ],

xi ∈ CR−i.

Next we showCR−i ⊆ BF iR−i . Let xi ∈ CR−i

. Let Rid ∈ R be a preference represented

by the utility functionud defined byud(z) = −‖xi − z‖. Let Rd = (Rid; R−i). We show

that F i(Rd) = xi by contradiction. Suppose thatxi , F i(Rd) = limk→+∞ f iRd(k)(Rd).

Then, without loss of generality, we may assume

∃ε > 0,∃K1 ∈ N s.t.∀` ≥ K1, fiRd(`)(Rd) < B(xi , ε). (2)

SinceBf ikR−i → CR−i

, there existsK2 ∈ N such thatdH(CR−i, B

f iK2R−i ) < ε. By Remark

4.1,Bf iK2R−i ∩ B(xi , ε) , ∅. Furthermore, by Remark 4.2,

∀` ≥ K2, Bf i`R−i ∩ B(xi , ε) , ∅. (3)

Let L be the positive integer such thatL = maxK1,K2. By (2) and (3),

f iRd(L)(Rd) < B(xi , ε) and B

f iRd(L)R−i ∩ B(xi , ε) , ∅.

Now let y ∈ Bf iRd(L)R−i ∩ B(xi , ε) and letRi

d ∈ R such thatτ(Rid,∏m

j=1[0,Ω j ]) = y.Then by Lemma 4.3,f i

Rd(L)(Rid; R−i) = y. This implies thatfRd(L) is manipulable byi at

(Rid; R−i) via Ri

d, a contradiction. HenceF i(Rd) = xi . This impliesxi ∈ BF iR−i .

Step 4 The ruleF is strategy-proof.

Let R ∈ RN, i ∈ N. First we prove that

∀k ∈ N, F i(R) Ri f ik(R). (4)

27R(·) is an operator fromN to N creating a subsequence. In general, there may exist more than oneconvergent subsequences. In this case, let us choose an arbitrary convergent subsequence, and define it as fR(k)(R)k∈N.

13

Because for allk ∈ N, Bf ikR−i ⊆ Bf i

k+1R−i , f ik+1(R) Ri f i

k(R). SinceRi is continuous andF i(R) = limk→+∞ f i

R(k+1)(R), we obtain (4).Next, we proveF ∈ ΓSP by contradiction. Suppose the contrary, that is

∃xi ∈ BF iR−i s.t.xiP(Ri)F i(R).

Sincexi ∈ BF iR−i , F i(Ri ; R−i) = xi for someRi ∈ R. Let R = (Ri ; R−i). By the definition

of F,xi = lim

k→+∞f iR(k)

(R). (5)

Sincexi ∈ SUC(Ri , F i(R)) 28 andRi is continuous,B(xi , ε) ⊆ SUC(Ri , F i(R)) for someε > 0. By (5), for sufficiently largeL ∈ N, f i

R(L)(R) ∈ SUC(Ri , F i(R)). By (4),

SUC(Ri , F i(R)) ⊆ SUC(Ri , f iR(L)

(R)). This implies thatf iR(L)

(R) ∈ SUC(Ri , f iR(L)

(R)).

This means thatf iR(L)

is manipulable byi at R = (Ri ; R−i) via Ri . This contradicts the

strategy-proofnessof f iR(L)

. HenceF ∈ ΓSP.

By Step 4, [F] ∈ ΓSP/ ∼. By Step 3and the definition ofCR−i,

∀[ f ] ∈ S, [F] Dom [ f ].

Hence [F] is an upper bound ofS.

Proof of Theorem 2.1.By Lemma 4.5 and Zorn’s lemma, for anyf ∈ ΓSP, there existsf0 ∈ ΓSP such thatf0 dom f and [f0] is a maximal element of

(ΓSP/ ∼,Dom

).

Proof of Theorem 2.2.Let E be the rule defined by

∀R ∈ RN,∀i ∈ N,Ei(R) =Ω

n.

ObviouslyE satisfiesSPandER. By Theorem 2.1, there exists a ruleE0 which satisfiesWSESPandE0 domE. Since dom is transitive,E0 satisfiesER.

4.2 Proofs of Theorems 3.1 and 3.2

First, we introduce an useful lemma provided in Amoros (2002).

LEMMA 4.6. (Amoros (2002)) If x∗, x′, x′′ ∈∏mj=1[0,Ω j ] satisfy[

∃ j ∈ M s.t. |x∗j − x′′j | < |x∗j − x′j |]∨[∃ j ∈ M s.t.(x∗j − x′′j )(x∗j − x′j) < 0

],

then there exists R∈ R such that p(R) = x∗ and x′′P(R)x′.

For the purpose of reference, we prepare an equivalent expression of Lemma 4.6.

28Let UC(R, x) = y ∈ ∏mj=1[0,Ω j ] | yRx, SUC(R, x) = y ∈ ∏m

j=1[0,Ω j ] | yP(R)x and LC(R, x) = y ∈∏mj=1[0,Ω j ] | xRy for all x ∈∏m

j=1[0,Ω j ], and allR ∈ R.

14

LEMMA 4.7. If x∗, x′, x′′ ∈∏mj=1[0,Ω j ](x′ , x′′) satisfy

¬[∀ j ∈ M, x∗j ≤ x′j ≤ x′′j ∨ x′′j ≤ x′j ≤ x∗j ],

then there exists R∈ R such that p(R) = x∗ and x′′P(R)x′.

Before we prove Theorem 3.1, we show the following lemma about the shape ofthe option set. Note that Lemma 4.8 holds in the environment with more than twoindividuals.

LEMMA 4.8. Suppose that f satisfies SP and WP. Then

∀i ∈ N,∀R−i ∈ RN\i,∀ j ∈ M,∃a j ,b j ∈ [0,Ω j ] s.t. Bf iR−i =

m∏j=1

[a j ,b j ].

Proof. Fix i ∈ N andR−i ∈ RN\i arbitrarily. Proof is done by two steps.

Step 1 The option setBf iR−i is convex.

To this end, suppose the contrary. That is, we assume∃v, w ∈ Bf i

R−i ,∃λ ∈ (0, 1) s.t. λv+ (1− λ)w < Bf iR−i .

Obviously

∃v,w ∈ Bf 1R2 ,∀λ ∈ (0,1) s.t. λv+ (1− λ)w < Bf 1

R2 .Let x = 1

2v+ 12w and letR ∈ R be a preference that satisfiesp(R) = x.

Case 1. f1(R,R2) = v∨ f 1(R,R2) = wFor eachd = (d1, · · · ,dm) ∈ ∏m

j=1ej ,−ej and eachy ∈ ∏mj=1[0,Ω j ], E(y, d) =

z ∈ ∏mj=1[0,Ω j ] | ∃γ1, · · · , γm ∈ R+ s.t. z = y +

∑mj=1 γ jd j, whereej denotes

the m-dimensional vector in whichjth coordinate is 1 and other coordinates are 0.Without loss of generality, we may assumef 1(R,R2) = v. Suppose that ford =(d1, · · · ,dm),d′ = (d′1, · · · , d′m) ∈ ∏m

j=1ej ,−ej, v ∈ E(p(R),d) andw ∈ E(p(R),d′).Suppose also thatd,d′ satisfy that

∀ j ∈ M,[v ∈ E(p(R), (−d j ,d− j)) ∨ w ∈ E(p(R), (−d′j ,d

′− j))⇒ d j = d′j

], (6)

where (−d j ,d− j) = (d1, · · · ,d j−1,−d j ,d j+1, · · · , dm) and (−d′j ,d′− j) is defined in the

same manner. Obviously, there existsj′ ∈ M such thatd j′ = −d′j′ becausep(R) is themidpoint ofv andw. Sincev j′ < x j′ < w j′ or w j′ < x j′ < v j′ , by Lemma 4.6 there existsR1 ∈ R such thatp(R1) = x andwP(R1)v. Hencef 1(R1,R2) , v by Lemma 4.3. Sincef satisfiesWP, this is a contradiction.

Case 2. f1(R,R2) , v∧ f 1(R,R2) , wLet c = f 1(R,R2). If

[∃ j ∈ M s.t. |x j − v j | < |x− c j | ∨ (x− v j)(x− c j) < 0]

or[∃ j ∈ M s.t. |x j − w j | < |x− c j | ∨ (x− w j)(x− c j) < 0],

15

then by Lemma 4.6,[∃Rv ∈ R s.t. p(Rv) = x∧ vP(Rv)c

]∨[∃Rw ∈ R s.t. p(Rw) = x∧ wP(Rb)c

].

This contradicts the fact thatc = f 1(R,R2) and f is weakly peak-only. Hence

[∀ j ∈ M, |x j − v j | ≥ |x− c j | ∧ (x− v j)(x− c j) ≥ 0]

and[∀ j ∈ M, |x j − w j | ≥ |x− c j | ∧ (x− w j)(x− c j) ≥ 0].

Suppose thatd, d′ ∈ ∏mj=1ej ,−ej satisfy v ∈ E(p(R),d), w ∈ E(p(R), d′) and the

condition (6) inCase 1. Fix j ∈ M arbitrarily. If d j = −d′j , thenc j = x j because(x− v j)(x− c j) ≥ 0 and (x− w j)(x− c j) ≥ 0. If d j = d′j , thenv j andw j can be repre-

sentedv j = x j + λd j andw j = x j + λ′d j for someλ, λ′. Sincex = 1

2v + 12w, λ = λ′.

Hencev j = w j = x j . Since|x j − v j | ≥ |x − c j |, |x − c j | = 0. Hencec j = x j . We have

shown thatc = x. Sincex < Bf 1R2 andc ∈ Bf 1

R2 , a contradiction.

Step 2∀ j ∈ M,∃a j ,b j ∈ [0,Ω j ] s.t. Bf 1R2 =∏m

j=1[a j ,b j ].For each j ∈ M, let Prj denote the projection with respect tojth coordinate.

Since Prj is continuous andBf 1R2 is compact, Prj(B

f 1R2 ) ⊆ [0,Ω j ] is compact. Let

a j = min Prj(Bf 1R2 ) and b j = max Prj(B

f 1R2 ). We show that

∏mj=1a j ,b j ⊆ Bf 1

R2 by

contradiction. We only show that (b1, · · · ,bm) ∈ Bf 1R2 . We can prove other cases in

the same manner. Suppose the contrary. That is, (b1, · · · ,bm) < Bf 1R2 . Let ˜R ∈ R be a

preference that satisfiesp( ˜R) = (b1, · · · ,bm). Let h ∈ Bf 1R2 satisfy f 1( ˜R,R2) = h. Since

h , (b1, · · · ,bm),∃ j′ ∈ M s.t.h j′ < b j′ .

Sinceb j′ = max Prj(Bf 1R2 ), there existsh′ ∈ Bf 1

R2 such thath′j′ = b j′ . Hence, by Lemma4.6,

∃R1 ∈ R s.t. p(R1) = (b1, · · · , bm) ∧ h′P(R1)h

because|b j′ − h′j′ | < |b j′ − h j′ |. However this implies that

f 1(R1,R2) , h.

This is a contradiction becausef satisfiesweak peak-onliness.

Proof of Theorem 3.1. Let U be the generalized uniform rule. Note thatU satisfiesweak peak-onliness and strategy-proofness. We concludes by contradiction. Suppose

that there existsg ∈ ΓSP such thatg domU but U , g. By Lemma 4.4,BU1R2 ⊆ Bg1

R2

andBU2R1 ⊆ Bg2

R1 for all (R1,R2) ∈ RN. By the hypothesis,

∃(R1,R2) ∈ RN s.t.BU1R2 ( Bg1

R2 ∨ BU2R1 ( Bg2

R1 . (7)

If not, thenBU1R2 = Bg1

R2 andBU2R1 = Bg2

R1 for all (R1,R2) ∈ RN. Note thatBU1R2 is a

direct product of closed intervals by Lemma 4.8. This implies by single-peakedness

16

that #τ(R1, BU1R2 ) = 1 for all (R1,R2) ∈ RN. Then, by Lemma 4.3 we haveU1(R1,R2) =

g1(R1,R2) for all (R1,R2) ∈ RN. By the feasibility condition,U2(R1,R2) = g2(R1,R2)for all (R1,R2) ∈ RN. Hence,U = g, a contradiction. Hence (7) holds. Without loss ofgenerality, suppose that there existsR2 ∈ R such thatBU1

R2 ( Bg1R2 .

We have ˜x ∈ ∏mj=1[0,Ω j ] such that ˜x ∈ Bg1

R2 and x < BU1R2 . Let R ∈ R satisfy

p(R) = x. By Lemma 4.8, for eachj ∈ M, there exista j ,b j ∈ [0,Ω j ] such thatBU i

R =∏m

j=1[a j ,b j ]. Then for eachj ∈ M, one of the following three holds;

(i) x j < a j (⇔ Ω j − a j < Ω j − x j),

(ii) x j ∈ [a j ,b j ] (⇔ Ω j − x j ∈ [Ω j − b j ,Ω j − a j ]),

(iii) b j < x j (⇔ Ω j − x j < Ω j − b j).

Let y ∈ BU1R2 be defined by the following; for eachj ∈ M, y j = a j if (i) holds, y j = x j

if (ii) holds andy j = b j if (iii) holds. Obviously,τ(R, BU1R2 ) = y. Hence,

U1(R,R2) = y and g1(R,R2) = x.

Now let us consider the allotment for individual 2. For allY ⊆ ∏mj=1[0,Ω j ], define

sym(Y) = y ∈∏mj=1[0,Ω j ] | ∃x ∈ Y s.t.y = Ω−x. LetR∗ ∈ R be such thatp(R∗) = Ω−

p(R2). Then, by the definition of the generalized uniform rule,Ω− p(R2) = U1(R∗,R2).

Hence,Ω − p(R2) ∈ BU1R2 . Hencep(R2) ∈ sym(BU1

R2 ) =∏m

j=1[Ω j − b j ,Ω j − a j ]. By thedefinition ofy andx, for eachj ∈ M,

(i) ⇒ Ω j − y j = Ω j − a j ,

(ii) ⇒ Ω j − y j = Ω j − x j ,

(iii) ⇒ Ω j − y j = Ω j − b j .

Sincep j(R2) ∈ [Ω j − b j ,Ω j − a j ] for all j ∈ M,

∀ j ∈ M, p j(R2) ≤ Ω j − y j ≤ Ω j − x j ∨ Ω j − x j ≤ Ω j − y j ≤ p j(R

2).

Because ˜x , y, Ω − x , Ω − y. By the single-peakedness ofR2, (Ω − y)P(R2)(Ω − x).Feasibility implies that

U2(R,R2) = Ω − y and g2(R,R2) = Ω − x.

However, this contradicts thatg domU.

LEMMA 4.9. Suppose that f satisfies SP, ER and WP. Then U dom f .

Proof. We show thatBf iR ⊆ BU i

R for all i ∈ N and allR ∈ R. Then we obtain theconclusion by Lemma 4.4. Without loss of generality, suppose thati = 1. LetR2 ∈ R.

ThenERand the feasibility condition imply thatBf 1R2 ⊆ sym(UC(R2, Ω2 )).

Step 1∀R2 ∈ R,[p(R2) = p(R2) and UC(R2, Ω2 ) ⊆ UC(R2, Ω2 )⇒ Bf 1

R2 ⊆ sym(UC(R2, Ω2 ))].

17

Suppose not. We haveR2 ∈ R such thatp(R2) = p(R2), UC(R2, Ω2 ) ⊆ UC(R2, Ω2 )

and Bf 1R2 * sym(UC(R2, Ω2 )). Then there exists a consumption bundlex such that

x ∈ Bf 1R2 andx < sym(UC(R2, Ω2 )). Obviously we can takeRx ∈ R such thatp(Rx) = x.

By Lemma 4.3,f (Rx,R2) = (x,Ω − x). SinceBf 1R2 ⊆ sym(UC(R2, Ω2 )), thenx < Bf 1

R2 .Hence,f (Rx, R2) , (x,Ω − x). Then, f satisfiesWP, a contradiction.

Step 2∀x ∈ sym(UC(R2, Ω2 ))\BU1R2 ,∃R2 ∈ R s.t.p(R2) = p(R2) andΩ− x < UC(R2, Ω2 ).

For eachj ∈ M, define

a j =

Ω j − p j(R2) if Ω j

2 ≤ p j(R2),Ω j

2 otherwise,b j =

Ω j

2 if Ω j

2 ≤ p j(R2),

Ω j − p j(R2) otherwise,

thenBU1R2 =∏m

j=1[a j ,b j ]. Hence,sym(BU1R2 ) =

∏mj=1[Ω j − b j ,Ω j − a j ].

Let x ∈ sym(UC(R2, Ω2 ))\BU1R2 . Note thatx , Ω2 becauseΩ2 ∈ BU1

R2 . HenceΩ − x ,Ω2 . We show the following by contradiction.

¬[∀ j ∈ M,

Ω j

2≤ Ω j − x j ≤ p j(R

2) ∨ p j(R2) ≤ Ω j − x j ≤

Ω j

2

]. (8)

Suppose not. Then for allj ∈ M, if Ω j

2 ≤ Ω j − x j ≤ p j(R2), thenΩ j − p j(R2) ≤ x j and

x j ≤ Ω j

2 . This is equivalent tox j ∈ [a j ,b j ]. Similarly we can show that for allj ∈ M,

if p j(R2) ≤ Ω j − x j ≤ Ω j

2 , thenx j ∈ [a j , b j ]. Hence we have shown thatx ∈ BU1R2 , a

contradiction. We have (8).By Lemma 4.7, there existsR2 ∈ R such that

p(R2) = p(R2) andΩ

2P(R2)(Ω − x).

Now we showBf 1R2 ⊆ BU1

R2 by contradiction. Suppose that we have a consumption

bundle x in Bf 1R2 but not in BU1

R2 . ThenER and the feasibility condition imply thatx ∈ sym(UC(R2, Ω2 )). By Step 2, we have a preferenceR2 ∈ R such that p(R2) =p(R2) andΩ−x < UC(R2, Ω2 ). LetRx be a preference whose peak isx. Then, by Lemma4.3 and feasibility conditionf (Rx,R2) = (x,Ω − x). By WP, f (Rx, R2) = (x,Ω − x).f 2(Rx, R2) = Ω − x < UC(R2, Ω2 ) but f satisfiesER, a contradiction.

Proof of Theorem 3.2. ObviouslyU satisfiesERandWP. By Theorem 3.1,U satisfiesSSESPwhich is stronger requirement thanWSESP. Next, we show the converse. Sup-pose thatf satisfiesWSESP, WPandER. By Lemma 4.9,U dom f . Since f satisfiesWSESP, f domU. SinceU satisfiesSSESP, f = U.

5 Concluding Remarks

In an environment with multiple commodities and single-peaked preferences, it isknown that there is no allocation rule satisfying strategy-proofness, Pareto efficiency,

18

and egalitarian rationality (Proposition 2.1 and Amoros (2002)). In contrast to thiswell-known fact, in this paper, we proposed second-best efficiency concepts and showedthat WSESP is compatible with strategy-proofness and egalitarian rationality in annpersonmgood economy (Theorem 2.2).

In addition, we showed that in a two personm good economy, the generalizeduniform rule is the only rule that satisfiesER, WP, andWSESP(Theorem 3.2). As aconclusion, we present some open questions concerning second-best efficiency.

Whether the generalized uniform rule in economies withn person andm goodsatisfiesWSESPis still open.

Furthermore, in pure exchange economies with non-satiated preferences, we mayraise a similar question. Barbera and Jackson (1995) investigate some incentive is-sues of the exchange economies in which the prices are rigid. To study whether theirstrategy-proof allocation rules are second-best efficient seems to be a very importantissue. In fact, Sasaki (2006) shows that fixed-price trading29 satisfies SSESP in a twoperson two good pure exchange economy. Extending Sasaki’s result to then personmgood setting would be interesting. In addition, Barbera and Jackson (1995) investigatea class of rules called fixed-proportion trading, a generalization of fixed-price trading,and characterize the rules with strategy-proofness and individual rationality. However,the class of fixed-proportional trading rules includes some apparently inefficient rules.(For example, the trivial rule, which is the rule that assigns the initial endowment, isincluded in the class of fixed-proportion trading rules.) With the second-best efficiencyconcepts proposed in this paper, we may characterize a subclass of fixed-proportiontrading rules.

More generally, as discussed in theIntroduction, it is known that there are severalimpossibility theorems suggesting the existence of a trade-off between Pareto efficiencyand strategy-proofness in various kinds of resource allocation problems. If the require-ment of Pareto efficiency is weakened as was done in the present paper, the same kindof positive results as in this paper may be obtained in different settings.

Appendix

In the appendix, we prove Proposition 2.1. Lemma A.1 says that if the peak of apreferenceR is Ω, then we can easily find an indifference curve ofR that intersectswith all rays onRm

+ from the origin.

LEMMA A.1. Suppose that R∈ R satisfies p(R) = Ω. Then

∃x0 ∈m∏j=1

[0,Ω j ]\0 s.t.∀h ∈ x ∈ Rm+ | ‖x‖ = 1, (UC(x0,R) ∩ LC(x0,R)

) ∩ `(0,h) , ∅,

where`(0,h) = µh ∈ Rm | µ ∈ R+.

Proof. For eachj ∈ M, o j denotes the vector defined by

29See Barbera and Jackson (1995).

19

o jk =

Ω j if k = j,

0 if k , j.

Let u denote a continuous utility representation ofR. Since coo1, · · · ,om30 iscompact,

∃x0 ∈ coo1, · · · ,oms.t.∀x ∈ coo1, · · · ,om, u(x) ≥ u(x0).

SinceR is strictly monotonic on∏m

j=1[0,Ω j ], x0 , 0.Let h be in x ∈ Rm

+ | ‖x‖ = 1. Obviously,`(0,h) ∩ co(o1, · · · ,om) , ∅. Picky ∈ `(0,h) ∩ co(o1, · · · ,om) arbitrarily. Then we have

u(y) ≥ u(x0) > u(0).

Since co(0, y) is connected, by the intermediate value theorem,

∃z ∈ co(0, y) s.t.u(z) = u(x0).

Obviously,z is an element of(UC(x0,R) ∩ LC(x0,R)

) ∩ `(0,h).

REMARK A.1. Note that(UC(x0,R) ∩ LC(x0,R)

) ∩ `(0,h) is singleton because R isstrictly monotonic on

∏mj=1[0,Ω j ].

Lemma A.2 states that if the peak of a preferenceR is Ω, then we have an utilityrepresentation ofR whose values on the diagonal line of the consumption set dependsonly on the distance from the origin.

LEMMA A.2. Suppose that R∈ R satisfies p(R) = Ω. Then there exists a continuousutility representation u of R that satisfies

∀x ∈ co(0,Ω), u(x) = ‖x‖

Proof. Obvious.

Lemma A.3 guarantees that if a single-peaked preferenceR with p(R) = Ω is “ho-mothetic” on

∏mj=1[0,Ω j ] can be extended to a strictly monotonic homothetic prefer-

ence onRm+ . Note that for each strictly monotonic homothetic preference onRm

+ , itsrestriction on

∏mj=1[0,Ω j ] is a “homothetic” single-peaked preference withp(R) = Ω.

LEMMA A.3. Let R∈ R be a preference such that(1) p(R) = Ω,(2)∀x, x′ ∈∏m

j=1[0,Ω j ],∀λ ∈ R+, [xI(R)x′ andλx, λx′ ∈∏mj=1[0,Ω j ] ⇒ (λx)I (R)(λx′)].

Then, there existsR⊆ Rm+ × Rm

+ such that(1’) R is complete31, transitive32, continuous33, strictly monotonic34 and homoth-

etic35,30co(o1, · · · ,om) denotes the convex hull ofo1, · · · ,om31For all x, y ∈ Rm

+ , xRyor yRx.32For all x, y, z ∈ Rm

+ , xRyandyRzimpliesxRz.33For all x ∈ Rm

+ , UC(x,R) and LC(x,R) are closed sets inRm+ .

34For all x, y ∈ Rm+ (x , y), if x ≥ y, thenxP(R)y.

35For all x, y ∈ Rm+ , all λ ∈ R+, xI(R)y implies (λx)I (R)(λy).

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(2’) R |∏mj=1[0,Ω j ]×

∏mj=1[0,Ω j ]= R.

MoreoverR is unique.

Proof. Let u :∏m

j=1[0,Ω j ] → R be a continuous utility representation ofR whosevalues on co(0,Ω) are distance from the origin. Suppose that the indifference curvethat includesx0 ∈

∏mj=1[0,Ω j ]\0 intersects with all rays from the origin (∵ Lemma

A.1). Without loss of generality, we can suppose thatx0 ∈ `(0,Ω). For eachh ∈ x ∈Rm+ | ‖x‖ = 1, let xh be the point in

(UC(x0,R) ∩ LC(x0,R)

) ∩ `(0,h).36

Defineu : Rm+ → R by the following.u(0) = 0 and for eachx ∈ Rm

+\0,

u(x) =‖x‖‖xh(x)‖

‖x0‖,

whereh(x) = x‖x‖ . Let R⊆ Rm

+ × Rm+ be the binary relation defined by for allx, y ∈ Rm

+ ,

xRy⇔ u(x) ≥ u(y).

R |∏mj=1[0,Ω j ]×

∏mj=1[0,Ω j ]= R is trivial. R is trivially homothetic, too.R is complete and

transitive because ˜u is a real-valued function.

Claim 1. R is continuous.To this end, we show that ˜u is continuous function. Letu : Rm

+ → R be the functionsuch thatu(0) = 0 and˜u(x) = ‖x‖

‖xh(x)‖ for all x ∈ Rm+\0. Obviously, if ˜u is continuous,

then u is continuous because‖x0‖ is constant. Hence, we show that˜u is continuous.Note thath is continuous onRm

+\0 since norm is a continuous function. By RemarkA.1, the restriction ofhon UC(x0,R)∩LC(x0,R) is a bijection. Hence,h |UC(x0,R)∩LC(x0,R)

is a homeomorphism between UC(x0,R) ∩ LC(x0,R) andx ∈ Rm+ | ‖x‖ = 1. 37 Let

f = h |UC(x0,R)∩LC(x0,R). Let g be the Euclidean norm. Thenu = gg f −1h on Rm

+\0.Hence,˜u |Rm

+\0 is continuous. Note that we have shown that˜u : Rm+ → R is continuous

at eachx ∈ Rm+\0. 38

Next we show thatu is continuous at 0. Letxn be a convergent sequence inRm+

whose limit is 0. First, we supposexn is a sequence inRm+\0. Note that

∃α ∈ R+ s.t.∀x ∈ UC(x0,R) ∩ LC(x0,R), α ≥ ‖x‖,

since UC(x0,R) ∩ LC(x0,R) is compact. Hence,u(xn) → 0 asn → +∞. 39 Now weconsider the general case. First, we show that ˜u(xn) is bounded. Since the sequenceis defined onR+, we show that the sequence is bounded above. Suppose not. Thenwe have a increasing subsequence ˜u(xk(n)) of ˜u(xn) that diverges. Sinceu(x) = 0if and only if x = 0, we can suppose thatxk(n) is in Rm

+\0. We have shown that

36See Remark A.1.37If a bijection from a compact space to a Hausdorff space is continuous, then it is a homeomorphism.38Let xn be a convergent sequence inRm

+ whose limit isx′ ∈ Rm+\0. Since every subsequence of a

convergent sequence converges to the limit of original sequence,xn is equal to 0 at most finitely manyn ∈ N.Hence, without loss of generality, we can suppose thatxn is a convergent sequence inRm

+\0. Since wehave shown thatu |Rm

+ \0 is continuous,u(xn)→ ˜u(x′). This proves thatu is continuous at everyx′ ∈ Rm+\0.

39Note thatα‖xn‖ → 0 asn→ +∞. Note also thatα‖xn‖ ≥ ˜u(xn) for all n ∈ N.

21

for any convergent sequence onRm+\0 with xk(n) → 0, ˜u(xk(n)) → 0, a contradiction.

Hence, ˜u(xn) is bounded. Suppose that ˜u(xn) is in [0, δ], whereδ > 0 is a sufficientlylarge real number. We show that ˜u(xn) converges to a number in [0, δ]. Suppose thecontrary. Then ˜u(xn) has at least two limit points in [0, δ]. 40 Suppose thatβ, γ ∈ [0, δ]with β , γ are limit points of ˜u(xn). Then at least one ofβ andγ is not 0. Withoutloss of generality, suppose thatβ is not 0. Then, we have a subsequence ˜u(xk′(n)) thatconverges toβ. We can easily construct a subsequence ˜u(xt(k′(n))) of ˜u(xk′(n)) thatsatisfies

∀n ∈ N, xt(k′(n)) ∈ Rm+\0,

a contradiction. Hence, ˜u(xn) converges to a number in [0, δ]. If the limit of ˜u(xn)is not equal to 0, we have a contradiction same as above. This completes the proof ofclaim 1.

Claim 2. R is strictly monotonic.To this end, we show that ˜u is strictly monotonic. Note that by the definition of ˜u,

u is strictly monotonic on all rays from the origin. Letx, x′ ∈ Rm+ satisfyx ≥ x′ and

x , x′. If x′ = 0, thenu(x) > u(x′). We supposex′ , 0 below. For sufficiently smallλ > 0, λx, λx′ ∈ ∏m

j=1[0,Ω j ], λx ≥ λx′ andx , x′. By (2’), u(λx) > u(λx′) > u(0). Bythe intermediate value theorem, we havey ∈ co(0, λx) such that ˜u(y) = u(λx′). Sinceu is strictly monotonic on(0, x), λx ≥ y andλx , y. Hence, ˜u(x) > u( y

λ) becausex ≥ y

λ

andx , yλ. On the other hand, ˜u(x′) = u( y

λ) sinceR is homothetic. This completes the

proof of claim 2.

The uniqueness ofR is obvious by (2’). We complete the proof of Lemma A.3.

Proof of Proposition 2.1.Obvious from Lemma A.3 and Theorem in Serizawa (2002).

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