A trade-off result for preference revelation

Journal of Mathematical Economics 34 Ž2000. 129–141 www.elsevier.comrlocaterjmateco

A trade-off result for preference revelation Donald E. Campbell a

a,)

, Jerry S. Kelly

b,1

Department of Economics and The Program in Public Policy, The College of William and Mary, Williamsburg, VA 23187-8795 USA b Department of Economics, Syracuse UniÕersity, Syracuse, NY 13245-1090 USA

Received 5 February 1997; received in revised form 31 August 1999; accepted 23 September 1999

Abstract If the social choice rule g selects from one up to k alternatives Žbut not more., then there exists a coalition H of k individuals such that for each profile r, the choice set g Ž r . is the collection of the top-most alternatives in the orderings of the individuals in H. Consequently, g is independent of the preferences of individuals not in H, forcing a disagreeable trade-off: Either some choice sets are very large, or most individuals never have any say in the social choice. q 2000 Elsevier Science S.A. All rights reserved. Keywords: Trade-off; Social choice; Preference

1. Introduction Gibbard Ž1973. and Satterthwaite Ž1975. have shown that, subject to a range condition, strategy-proofness implies dictatorship for resolute social choice procedures, i.e., for procedures where the choice set always contains just a single alternative. This paper characterizes social choice procedures that allow more than one alternative to be selected. There are two quite different reasons for relaxing resoluteness.

) 1

Corresponding author. Tel.: q1-757-2212383; fax: q1-757-2212390; e-mail: [email protected] E-mail: [email protected].

0304-4068r00r$ - see front matter q 2000 Elsevier Science S.A. All rights reserved. PII: S 0 3 0 4 - 4 0 6 8 Ž 9 9 . 0 0 0 3 7 - 3

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First, although resoluteness might be quite desirable, non-dictatorship is even more desirable; and then we ask a trade-off question: If we relax resoluteness slightly, allowing small non-singleton choice sets, is there then a way to construct strategy-proof social choice rules that are far from dictatorial? Or is it the case that getting far from dictatorship forces some choice sets to be quite large? Beyond this trade-off approach, which assumes that resoluteness is desirable, sometimes we actually want to chose more than one alternative. First, we might be interested in a social choice procedure that represents a preliminary stage in a process, say a piano competition, that ultimately chooses a singleton outcome, but it is actually desirable to have several alternatives selected at the preliminary stage. Secondly, even where we are settling final outcomes, we may desire that a non-singleton set of alternatives be selected in some situations 2 : The International Mathematical Society will select up to four Fields Medalists to be announced at their next Congress. Imagine a mathematician asked to take part in the process of determining the recipients of the four 1998 Fields Medals. One may feel comfortable writing down a ranking of the say 25 candidates proposed but feel baffled by the request to rank the 12,650 quadruples of candidates. We will assume that individuals have a complete ordering of X itself. An individual is assumed to be able to compare any two subsets of X, and the comparison will have to be consistent with the ordering of X in the sense specified in the Section 2. Section 2 provides the basic notation and definitions. In Section 3 we prove that the only strategy-proof rules that select from 1 to k alternatives identify a set of k individuals and then select the set of their top-most alternatives. Ching and Zhou Ž1997. derive dictatorship from strategy-proofness, and without any condition on the size of the choice sets. Because the rules characterized by our theorem are not dictatorial Žexcept when k max Ž g . s 1., the Ching–Zhou notion of strategy-proofness is very demanding when g is not resolute. Using a less demanding definition of strategy-proofness, Duggan and Schwartz Ž1997. prove that there is an individual whose most-preferred alternative always belongs to the set of alternatives selected by the social choice function. Our theorem provides more information, but our strategy-proofness requirement is more exacting than that of Duggan and Schwartz. Baigent Ž1998. examines strategy-proofness for non-resolute social choice rules, and in his case the domain is restricted to the set of dichotomous preferences — an alternative is either acceptable or not: Y is preferred to Z if every member of Y is acceptable and some members of Z are not acceptable, of if some members of Y are acceptable and every member of Z is unacceptable.

2

A companion paper investigates cases where we neÕer want a single alternative to be selected. ŽCampbell and Kelly, 1999. For example, exactly k organs are available for transplant.

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2. Foundations We take as given a set X of mutually exclusive alternatives with < X < ) 2 and a set N s  1, . . . ,n4 of individuals, n ) 1. A strong order % Žon X . is transitive and complete; in this paper x % y only if x / y, and when we say that % is complete we mean that either x % y or y % x holds if x / y. ŽNote that we do not have x % x.. Where r Ž i . is a strong order, we designate the set containing just its top-most element Žif it has one. by r Ž i .w1x, the set containing just its second by r Ž i .w2x, and the set consisting of the jth ranked alternative by r Ž i .w j x. A profile r s Ž r Ž1., r Ž2., . . . , r Ž n.. assigns a strong order r Ž i . to each i g N. Let ` be the collection of all possible profiles. A social choice function g maps ` into the family of non-empty subsets of X. For H : N, we say that g is independent of H if for any two admissible profiles p and r p Ž i . s r Ž i . for all i g N _ H implies g Ž p . s g Ž r . .

3

The range of g is X g s  Y : X :Y s g Ž r . for some r g ` 4 . We let k minŽ g . and k max Ž g . be the cardinalities of the smallest and largest sets in X g . The analog here of the Gibbard–Satterthwaite assumption that all alternatives are in the range of g is the basis for the following definition: A social choice function g is regular if all subsets of X of size k minŽ g . are in X g . If k max Ž g . s 1 and X g has more than two members, then g is resolute and the Gibbard–Satterthwaite result tells us there is an individual i such that g Ž r . is the alternative in X g that ranks highest in r Ž i . for arbitrary r in `. We turn our attention then to rules where 1 - k max Ž g .. There are three cases to be considered depending on the size of k minŽ g . relative to 1 and k max Ž g .: 1 s k min Ž g . - k max Ž g . ;

Ž A.

1 - k min Ž g . - k max Ž g . ;

Ž B.

1 - k min Ž g . s k max Ž g . .

Ž C.

The first of these cases is addressed in this paper; the other two cases are dealt with in Campbell and Kelly Ž1999.. An individual’s input to the social choice function is an ordering on X; but for purposes of defining manipulability, we need an individual to use rankings on X g . We assume that these rankings are derived from orderings on X by means of ‘‘extension principles’’. An extension principle D associates with each r Ž i . a 3

Equivalently, in the language of Deb et al. Ž1997., only the members of N _ H can have a ‘‘say’’ in g.

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partial strong ordering DŽ r Ž i .. on non-empty subsets of X; each DŽ r Ž i .. is irreflexive, antisymmetric, and transitive. ŽNot every extended preference is complete.. For any S : X, and any profile r we let r Ž i .< S denote the restriction of r Ž i . to S, while r < S represents the n-tuple Ž r Ž1.
4

Pattanaik Ž1978. introduced maximin extension to the study of the preference revelation problem.

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Note that if g is strategy-proof with respect to a D that supports leximin, then it is strategy-proof under the single context Ž L, L, . . . , L.. This follows from the fact that if g is not strategy-proof with respect to Ž L, L, . . . , L., then there is an individual i and profiles r and r U such that r U is an i-variant of r and g Ž r U . LŽ r Ž i .. g Ž r ., where L is leximin extension. If D is a collection of contexts that contains D s Ž D 1 , D 2 , . . . , Dn . where Di is an extension principle that coincides with L on the two sets g Ž r U . and g Ž r ., then g Ž r U . Di Ž r Ž i .. g Ž r ., so g is not strategy-proof with respect to D. Accordingly, throughout this paper, without further comments, whenever we want to prove a characterization result that assumes that g is strategy-proof with respect to a D that supports leximin, we will confine our attention to just the single context Ž L, L, . . . , L..

3. A trade-off theorem In this section, we will show that if g is a regular social choice function that is strategy-proof with respect to a D that supports leximin and if 1 s k minŽ g . k max Ž g ., then there is a coalition H Ži.e., a subset of N . of cardinality k max Ž g . such that for every admissible profile r g Ž r . s j i g H r Ž i . w1x Remark 1. A collection D of contexts having the property that for every individual i, every profile r, and every pair of sets A, B, if the leximin extension of r Ž i . has A LŽ r ŽŽ i .. B, then there exists a D s Ž D 1 , D 2 , . . . , Dn . g D with A Di Ž r Ž i .. B may not have any strategy-proof social choice functions. All we are showing is that if one exists, then it must have the form

g Ž r . s j i g H r Ž i . w1x But if such a rule is manipulable at some other context in D, then there do not exist any rules strategy-proof with respect to D. We next need to establish some notation. Given two profiles r and r U , we will often need to refer to the sequence of profiles rs

r0 s r1 s r2 s ...

rU s

rn s

Ž r Ž 1. , Ž rU Ž 1. , Ž rU Ž 1. ,

r Ž 2. ,

r Ž 3. ,

...,

r Ž n. .

r Ž 2. ,

r Ž 3. ,

...,

r Ž n. .

rU Ž 2. ,

r Ž 3. ,

..., ...

r Ž n. .

Ž rU Ž 1. ,

rU Ž 2. ,

rU Ž 3. ,

...,

rU Ž n . . .

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This is the standard sequence from r to r U . Two successive entries in this sequence are r jy1 s Ž r U Ž 1 . ,r U Ž 2 . , . . . ,r U Ž j y 1 . ,r Ž j . , . . . ,r Ž n . . , and r j s Ž r U Ž 1 . ,r U Ž 2 . , . . . ,r U Ž j y 1 . ,r U Ž j . , . . . ,r Ž n . . in which r j is created from r jy1 by replacing the jth component of r jy1 by r U Ž j . and leaving all other components of r jy1 unchanged. Then, for appropriately chosen profiles r and r U we will show that g is manipulable by j by showing that either g Ž r j . L Ž r Ž j . . g Ž r jy1 . at r jy1 ; or g Ž r jy1 . L Ž r U Ž j . . g Ž r j . at r j . The key step in our proof is to associate with g a social welfare function and then show that the associated social welfare function must have an oligarchy. Before we can attempt that, we need five preliminary results that provide the machinery we use in the proof of our theorem. In the first preliminary result we connect g values at r with the g values at r U , where r U differs from r by bringing all the elements of g Ž r ., and possibly a few more, to the top of everyone’s ordering. Given a profile r and a subset S of X, r Ž S . denotes a new profile constructed in the following way: For each r Ž i . in r, the alternatives in S are raised above all the elements in X _ S. Within S and within X _ S, alternatives are ordered as they were in r Ž i .; that is r Ž S .< S s r < S and r Ž S .< X _ S s r < X _ S. When S is small, we will drop the set bracket notation, referring to r Ž x . and r Ž x, y . instead of r Ž x 4. and r Ž x, y4.. Lemma 1 (The Completeness Lemma). If g is strategy-proof with respect to a D that supports leximin and g(r) : S : X, then g(r(S)) s g(r). Proof: Let r U s r Ž S . and construct the standard sequence from r to r U . Let j be the smallest integer such that g Ž r j . / g Ž r .. If g Ž r j . ≠ S, then g is manipulable by j at r j since g Ž r jy1 . LŽ r U Ž j .. g Ž r j .. Therefore, g Ž r j . is a subset of S. Under leximin, the set orders are complete: either g Ž r jy1 . LŽ r Ž j .. g Ž r j . or g Ž r j . LŽ r Ž j .. g Ž r jy1 .; similarly for LŽ r U Ž j ... But since the restrictions to S satisfy r U Ž j .< S s r Ž j .< S, LŽ r Ž j .. and LŽ r U Ž j .. order the sets g Ž r jy1 . and g Ž r j . in the same way. Therefore, either g Ž r jy1 . LŽ r U Ž j .. g Ž r j . or g Ž r j . LŽ r Ž j .. g Ž r jy1 .. Then I g is manipulable by j at either r j or r jy1. The next preliminary lemma tells us that if we bring the elements of a sufficiently large set Y to the top of everyone’s ordering at a profile, then the elements chosen at that profile by g will all be in Y.

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Lemma 2. Suppose that g is a regular social choice function that is strategy-proof with respect to a D that supports leximin. Let Y be a non-empty subset of X that contains a set Z in the range of g and let r be a profile such that for all i, all y g Y and x g X _ Y, yr(i)x. Then g(r) : Y. Proof: Suppose that to the contrary we have g Ž r . ≠ Y. By the range assumption, there is a profile r U such that g Ž r U . s Z : Y. Construct the standard sequence from r to r U and let j be the largest integer such that g Ž r j . ≠ Y. Then j - n and g Ž r jq1 . : Y. But r j Ž j q 1. s r Ž j q 1. and g Ž r jq1 . LŽ r Ž j q 1.. g Ž r j ., so g is I manipulable by j q 1 at r j . We will use Lemma 2 for sets Y of one, two, or three elements without explicitly citing it. In particular, we use Lemma 2 to associate with g a social welfare function f g that maps each profile r to a social relation, K . We start with a profile r and then construct r Ž x, y .. Lemma 2 tells us that g Ž r Ž x,y.. :  x, y4 and so we can map each r to the following binary relation K on X: for all x, we have x K x; and if x and y are distinct, x K y if and only if x g g Ž r Ž x, y ... We associate with g the function f g that maps r to K . The asymmetric part of K is denoted by % . Our third preliminary lemma establishes an important independence property of g. Lemma 3. Let Y be a non-empty subset of X and consider two profiles, r and p such that for all y g Y and x g X _ Y, yr(i)x and yp(i)x for all i. If r < Y s p < Y, then g(r) s g(p). Proof: If g Ž r . / g Ž p . construct the standard sequence from r to p. There must be some j such that g Ž r j . / g Ž r jq1 .. By Lemma 2, g Ž r j . : Y and g Ž r jq1 . : Y. But since r Ž j q 1. and pŽ j q 1. agree on Y, under the leximin extension principle we must have LŽ r Ž j q 1.. and LŽ pŽ j q 1.. ordering g Ž r j . and g Ž r jq1 . in the I same way. But then g is manipulable either at r j or at r jq1. Lemma 4. For any nonempty subset I of N and any x g X, if there exists a profile r such that r(i)[1] s {x} for all i g I, x ranks at the bottom of r(i) for all i g N _ I, and g(r) s {x}, then g(p) s {x} for any profile p such that p(i)[1] s {x} for all i g I. Proof: Let  rt 4 be the standard sequence from r to p. Suppose g Ž rt . s  x 4 . If g Ž rtq1 . /  x 4 and t q 1 g I then person t q 1 would manipulate at rtq1. If g Ž rtq1 . /  x 4 and t q 1 g N _ I then t q 1 would manipulate at rt . Then g Ž p . s I g Ž rn . s  x 4 by induction. By definition, profile r Ž x, y . has alternative x ranked either first or second for every i g N and y in first or second place for every i g N. If g Ž r Ž x, y .. s  x 4 , we

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will see that strategy-proofness gives the coalition of individuals who rank x first at r Ž x, y . a great deal of power in many other situations as well. The first step is to show that g Ž r X . s  x 4 if each individual who ranks x first at r Ž x, y . also ranks x first at r X , and r X Ž x, y, z . s r X for some z. Lemma 5. For arbitrary x, y g X, suppose that g(r) s {x} for some profile r such that r(x,y) s r. Then for arbitrary z g X _ {x,y} we haÕe g(r Y ) s {x} for any profile r Y such that r Y s r Y (x,y,z) and x r(i) y implies r Y (i)[1] s {x} for arbitrary i g N. Proof: Let I s  i g N: xr Ž i . y4 . Let p be the profile for which xpŽ i . ypŽ i . z for all i g I, and ypŽ j . zpŽ j . x for all j g N _ I, with pŽ j . otherwise the same as r Ž j .. Then p s pŽ x, y, z . and profile p is obtained from r by moving z just below y for all i g I and moving z between y and x for every member of N _ I, as displayed below. I x y z ...

N_I y z x ...

Let  rt 4 be the standard sequence from r to p. We have g Ž r 0 . s g Ž r . s  x 4 . Suppose that g Ž rty1 . s  x 4 . We will show that g Ž rt . s  x 4 , but first we have to establish that z f g Ž rt .. Let s be the same as rt except that we move y to the top of sŽ h. for all h g N. Let  s j 4 be the standard sequence from s to rt . We have g Ž s0 . s g Ž s . s  y4 . Suppose that z f g Ž s jy1 .. If z g g Ž s j . then we have g Ž s j . :  x, y, z 4 Žby Lemma 2. and g Ž s j . / g Ž s jy1 .. This implies that j g I and z ranks below x and y in s j Ž j .. Therefore, person j would manipulate at s j via s jy1Ž t .. Then z f g Ž s j . and thus z f g Ž sn . s g Ž rt . by induction. Suppose that g Ž rt . s  y4 or  x, y4 . If t g I then person t would manipulate at rt via rty1Ž t .. And if t g N _ I then person t would manipulate at rty1 via rt Ž t . because g Ž rty1 . s  x 4 and yr Ž t . x. Then we must have g Ž rt . s  x 4 by Lemma 2, and thus g Ž rn . s g Ž p . s  x 4 by induction. But x ranks last in pŽ j .< x, y, z 4 for any j g N _ I. Therefore, Lemmas 2 and 4 imply that g Ž r Y . s  x 4 for any r Y such that r Y s r Y Ž x, y, z . and x is at the top of r Y Ž i . for all i g I. I Following the lead of Gibbard Ž1973., we will derive properties of the associated social welfare function f g from the strategy-proofness of g and then use these properties to determine the extent to which g is sensitive to individuals’ preferences. Four more lemmas are required. Lemma 6. f g satisfies Independence of IrreleÕant AlternatiÕes. Proof: This is immediate from Lemma 3.

I

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Lemma 7. f g satisfies the Pareto condition. Proof: If at r, everyone prefers x to y, then at r Ž x, y ., everyone will have x at the top of their ordering. By Lemma 2, g Ž r Ž x, y .. s  x 4 , so x % y and Pareto is satisfied. I Binary relation K satisfies quasitransitiÕity if for every triple  x, y, z 4 in X x % y and y % z imply x % z. Lemma 8. f g is defined at all possible profiles of strong orders and yields at each profile an ordering K that is reflexiÕe, complete, and quasitransitiÕe. Proof: It is clear that f g Ž r . can be constructed at every profile of strong orders. That it is reflexive is immediate from the definition of K , and completeness follows from Lemma 2. It remains to prove quasitransitivity. We need to prove that for any profile r and any three alternatives x, y, and z, g Ž r Ž x, y .. s  x 4 and g Ž r Ž y, z .. s  y4 implies g Ž r Ž x, z .. s  x 4 . Lemma 3 implies that g Ž r Ž x, y, z .Ž x, z .. s g Ž r Ž x, z .., and therefore we assume that r s r Ž x, y, z .. Because g Ž r Ž x, y .. s  x 4 , Lemma 5 implies that g Ž p . s  x 4 for any p such that p s pŽ x, y, z . and x is at the top of pŽ j . for all i g I '  i g N: xr Ž j . y4 . Similarly, g Ž r Ž y, z .. s  y4 implies that g Ž pX . s  y4 for any pX such that pX s pX Ž x, y, z . and y is at the top of pX Ž j . for all j g J '  j g N: yr Ž j . z 4 . Now, investigate the power of the coalition I l J. Consider the profile r X in which r X Ž x, y, z . s r X , every i g I _ J has zr X Ž i . xr X Ž i . y, every h g I l J has xr X Ž h. yr X Ž h. z, and every j g J _ I has yr X Ž j . zr X Ž j . x. I_J z x y ...

IlJ x y z ...

J_I y z x ...

ŽThe preferences of the members of N _ I j J are arbitrary, and hence not displayed.. We establish g Ž r X . s  x 4 . To rule out y g g Ž r X . construct r Y from r X by moving x up to the top for each i g I _ J. We have g Ž r Y . s  x 4 because everyone in I ranks x at the top. The standard sequence from r Y to r X will show that y f g Ž r X .. To rule out z g g Ž r X . construct qY from r X by moving y up to the top for each i g I l J. We have g Ž qY . s  y4 because everyone in J ranks y at the top. The standard sequence from qY to r X will show that z f g Ž r X .. Therefore, g Ž r X . s  x 4 by Lemma 2. Then g Ž r X Ž x, z .. s  x 4 , by Lemma 1. Lemma 5 and the fact that  i g N: xr Ž i . z 4 is a superset of I l J s  i g N: xr X Ž i . yr X Ž i . z 4 establish I g Ž r Ž x, z .. s  x 4 , confirming the quasitransitivity of f g Ž r ..

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We know that f g satisfies Independence of Irrelevant Alternatives and the Pareto criterion, and that f g Ž r . is quasitransitive for each profile r. Therefore, Žsee Mas-Colell and Sonnenschein Ž1972.., f g is oligarchical. This means that there is a subset H of N such that f g Ž r . ranks x above y if everyone in H ranks x above y. In addition, every member of H has veto power. Ž H is the oligarchy, and an individual i is said to have veto power if xr Ž i . y implies that y cannot rank above x in f g Ž r ... Now that we know there exists an oligarchy H for f g , we need to come back to g and see what power H has there. Lemma 9. If indiÕidual i is a member of the oligarchy H of f g then r (i)[1] : g(r). Proof: Without loss of generality, we assume that i s 1 and r Ž1.w1x s  x 4 . Suppose that x f g Ž r .. Let r U differ from r by reordering r Ž j . for each j ) 1 so the restrictions to X _  x 4 satisfy r U Ž j .<Ž X _  x 4. s r Ž1.<Ž X _  x 4. and x is at the bottom of each r U Ž j .. Then x f g Ž r U .. For if x g g Ž r U ., construct the standard sequence from r to r U and let t be the smallest value such that x g g Ž rt .. Of course, t ) 1. Since x f g Ž rty1 . and x is at the bottom of r U Ž t ., g is manipulable by t at rt . Now let y be the second element in r Ž1. so that r U is the profile 1 2 3 ... n x y y y y ... ... ... ... x x x Then g Ž r U . s  y4 because person 1 can guarantee that  y4 is the outcome by reporting an ordering with y at the top ŽLemma 2., and with preference r U Ž1., person 1 prefers  y4 to any subset of X not containing x. From g Ž r U . s  y4 , Lemma 1 gives g Ž r U Ž x, y .. s  y4 , contrary to the fact that person 1 has veto power. I Theorem. Suppose that g is a regular social choice function with 1 s k m i n(g) k m a x (g) and g is strategy-proof with respect to a D that supports leximin. Let H be the oligarchy of f g . Then for eÕery admissible profile r g Ž r . s j i g H r Ž i . w1x . Proof: Lemma 9 establishes that j i g H r Ž i .w1x : g Ž r .. It remains to how that g Ž r . is a subset of j i g H r Ž i .w1x. We do this in two steps. First we show that the claim is true if everyone in H has the same top-ranking alternative. This preliminary result is then used to show how we are led to a contradiction if g Ž r . is not a subset of j i g H r Ž i .w1x for all r. We begin by showing that g Ž r . s  x 4 if r Ž i .w1x s  x 4 for all if i g H. If X s  x, y, z 4 then g Ž r . s  x 4 , by Lemma 5 because H is the oligarchy for f g and

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thus g Ž r X . s  x 4 if r X s r X Ž x, y . and r X Ž i .w1x s x for all i g H. Suppose, then, that X has at least four members. Choose any w, x, y, and z in X. Let p be a profile such that pŽ w, x, y, z . s p, every one in H ranks x first, y second, and z third, and everyone in N _ H ranks z first, y second, and x third. Then g Ž p . s  x 4 by Lemma 5 and the fact that H is the oligarchy for f g . Let q be a profile such that every one in H ranks y first, z second, w third, and x fourth, and everyone in N _ H ranks z first, y second, w third, and x last. Lemma 5 implies that g Ž q . s  y4 . We will compare profiles p and q to the profile s defined by sŽ i . s pŽ i . for all i g H and sŽ i . s q Ž i . for all i g N _ H, to establish that g Ž s . s  x 4 , and then use that fact to show that g Ž r . s  x 4 . Profiles p, q, and s are displayed in the next table. profile p H x y z w ...

N_H z y x w ...

profile q H y z w x .. .

N_H z y w ... x

profile s H x y z w ...

N_H z y w ... x

Let  qt 4 be the standard sequence from q to s. We have g Ž q . s  y4 and sŽ i . s q Ž i . for all i g N _ H, and y ranks second in sŽ i . for all i g H. Therefore, g Ž qt . cannot contain an alternative that ranks below y in sŽ i . for any i g H and all t ) 0. Therefore, g Žs. s g Ž qn . :  x, y4 by induction on t. We next prove that g Ž s . s  x 4 . Let  st 4 be the standard sequence from s to p. If g Ž s . s  y4 then an induction argument will show that g Ž st . does not contain an alternative that ranks below y in pŽ i . for any i g N _ H or t G 0. That is, g Ž st . :  z, y4 for all t G 0. But g Ž sn . s g Ž p . s  x 4 , a contradiction. Assume that g Ž s . s  x, y4 . Suppose that g Ž st . s  x, y4 . We will show that g Ž stq1 . s  x, y4 . If t q 1 g H then g Ž stq1 . s g Ž st .. Suppose, then, that t q 1 g N _ H. We must have x g g Ž stq1 . or else t q 1 would manipulate at st . Individual t q 1 would manipulate at stq1 if g Ž stq1 . s  x 4 or g Ž stq1 . contains x and another alternative that ranks below y in pŽ t q 1.. If g Ž stq1 . s  x, y, z 4 or  x, z 4 then t q 1 would manipulate at st . Suppose that g Ž stq1 . is a proper superset of  x, y4 , but g Ž stq1 . /  x, y, z 4 . Then g Ž stq1 . contains an alternative Žother than x . that ranks below y in pŽ t q 1., and hence individual t q 1 would manipulate at stq1. We are forced to conclude that g Ž stq1 . s  x, y4 if g Ž st . s  x, y4 , and thus g Ž sn . s  x, y4 by induction. But this contradicts g Ž p . s  x 4 . Therefore, g Ž s . s  x 4 . By Lemma 4, for any profile r such that r Ž i .w1x s r Žj.w1x for all i, j g H we have g Ž r . s r Ž i .w1x for i g H. Finally, suppose that j i g H r Ž i .w1x contains more than one alternative but is a proper subset of g Ž r . s S. Lemma 1 allows us to assume that r s r Ž S .. Now derive r U from r so that for each i in H, the alternatives in S _ j i g H r Ž i .w1x are

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all brought down below the alternatives in j i g H r Ž i .w1x but above the alternatives in X _ S and in such a way that orderings within j i g H r Ž i .w1x, within S _ j i g H r Ž i .w1x, and within X _ S are all preserved. Construct the standard sequence  rt 4 from r to r U . Lemma 2 implies that g Ž rt . : S for all t. If g Ž r U . / S, let t be the smallest value such that g Ž rt . / S. Then t g H, and Lemma 9 implies that j i g H rt Ž i .w1x : g Ž rt .. Therefore, g Ž rt . LŽ rty1Ž t .. g Ž rty1 ., contradicting the fact that g is strategy-proof. Hence, g Ž r U . s S. Choose some h g H and construct pU from r U so that for each i in H, r Ž h.w1x is brought to the top of i’s ordering, which is otherwise left unchanged. We have g Ž pU . s r Ž h.w1x because j i g H pU Ž i .w1x is a singleton. Let  rtU 4 be the standard sequence from r U to pU . We have g Ž rnU . s r Ž h.w1x. Suppose g Ž rtU . : U . : j i g H r Ž i .w1x, otherwise person t would manipuj i g H r Ž i .w1x. Then g Ž rty1 U late at rty1. Therefore, g Ž r U . s g Ž r . : j i g H r Ž i .w1x by induction, contradicting g Ž r U . s S. Hence, we have to drop the supposition that j i g H r Ž i .w1x is a proper subset of g Ž r .. I In characterizing all regular social choice rules that are strategy-proof with respect to a collection of contexts that supports leximin, we have uncovered a dismal trade-off: Either some choice sets are very large, or most individuals have no say in the social choice. Allowing choice sets to be pairs as well as singletons permits only two individuals’ preferences to be consulted; if triples can be chosen, that allows no more than three individuals to have a say in the social choice. If you want the preferences of thousands to be consulted, then the social choice set will have to contain thousands of alternatives at some profiles. Remark 2. We can define leximax extension analogously to leximin. The rule g Ž r . s jr Ž i .w1x is strategy-proof with respect to leximax. But strategy-proofness with respect to leximax does not force g Ž r . to be the union of the tops of the member of a Žsmall. coalition. In fact, g can be regular and strategy-proof but not independent of anyone’s preferences. To see this, let X s  x,y,z 4 . Set g Ž r . s  z 4 if everyone has z at the top of her ordering. Otherwise g Ž r . :  x,y 4 . If fact, g Ž r . s  x,y 4 unless either Ž i . everyone prefers x to y, in which case g Ž r . s  x 4 , or Ž ii . everyone prefers y to x, in which case g Ž r . s  y4 . It is straightforward to show that g is strategy-proof with respect to leximax. But if even one person, say individual 1, employs the leximin extension principle there will be an opportunity for manipulation. Suppose 1 x z y

2 z y x

3 z y x

...

n z y x

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Then g Ž r . s  x, y4 . But if person 1 reports z as her top-ranked alternative then g will select  z 4 . Because leximin gives  z 4 LŽ r Ž1..  x, y4 , g is manipulable by 1 at r. However, when there are more than three alternatives, we have not been able to find a regular rule with k minŽ g . s 1 - k max Ž g . that is strategy-proof with respect to leximax.

Acknowledgements We thank The Rockefeller Foundation for supporting us at the Bellagio Study Center where this work was initiated in August of 1995. We are indebted to Matt Jackson, John Duggan, and Salvador Barbera` for helpful observations on a 1995 draft of this paper. Thanks also to Jana Ballin for a helpful suggestion while we were proving our theorem.

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