Economics Letters 118 (2013) 10–12

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Economics Letters

journal homepage: www.elsevier.com/locate/ecolet

Anonymity, monotonicity, and limited neutrality: Selecting a single alternativefrom a binary agenda✩

Donald E. Campbell a,∗, Jerry S. Kelly b

a Department of Economics and The Program in Public Policy, The College of William and Mary, Williamsburg, VA 23187-8795, United Statesb Department of Economics, Syracuse University, Syracuse, NY 13244-1090, United States

a r t i c l e i n f o

Article history:Received 12 March 2012Received in revised form7 August 2012Accepted 15 August 2012Available online 18 September 2012

JEL classification:D70D71

Keywords:AnonymityMonotonicityNeutralitySocial choice function

a b s t r a c t

Anonymity and neutrality conflict if the number of individuals is even and a single alternative is selected.Limited neutrality, anonymity, and monotonicity imply majority rule when the agenda is a two-elementset.

© 2012 Elsevier B.V. All rights reserved.

It is well known that anonymity and neutrality cannot bothbe satisfied by a rule that selects a single alternative from a two-element set if the number of individuals is even: If exactly halfof the agents prefer x to y, and x is selected, then neutralityimplies that y must be selected when x and y switch positions ineach individual preference relation. But this new configuration ofindividual preferences can also be obtained from the original bypermuting the individuals, in which case anonymity requires thatx still be selected. (The same conclusion emerges if the number ofindividuals is odd and someone expresses indifference. Theorem 1,in Moulin, 1983, p. 25, characterizes the relationship betweenthe number of individuals and the number of alternatives if bothneutrality and anonymity are to be satisfied.)

We seek an axiomatic foundation for majority rule that isvalid whether the number of voters is even or odd. Theorem1 in Campbell and Kelly (2011) establishes that monotonicity,neutrality, and a weak version of anonymity, defined in thatpaper, imply majority rule. This note proposes a weak form ofneutrality and proves that it implies majority rule in the presenceof monotonicity and the standard anonymity condition. (When

✩ We are grateful for the input of our referees and associate editor.∗ Corresponding author. Tel.: +1 757 221 2383; fax: +1 757 221 1175.

E-mail addresses: decamp@wm.edu (D.E. Campbell), jskelly@syr.edu (J.S. Kelly).

0165-1765/$ – see front matter© 2012 Elsevier B.V. All rights reserved.doi:10.1016/j.econlet.2012.08.028

ties are possible there are different versions of majority rulebecause different formulas for breaking a tie can be used.We applythe term majority rule to any social choice function that selectsthe majority winner when there is no tie.) We employ a fixedpopulation model that assumes away individual indifference, andthen a variable population model that countenances individualindifference between the feasible alternatives.

The two-element set X = {1, −1} represents the set of feasibleoutcomes and N = {1, 2, . . . , n} is the set of individuals. We letP = {1, 0, − 1} denote the set of possible preferences on X . Amember p of Pn is called a profile, with p(i) = 1 (resp.,−1) denotinga strict preference for 1 over −1 (resp., −1 over 1) by individual i,and 0 denoting indifference. For two profiles p and r wewrite p ≥ rif p(i) ≥ r(i) for all i ∈ N . Let t(p) be the number of i ∈ N suchthat p(i) = 1. Given profile p, the profile −p is the one for whichindividual i’s preference is −p(i), for arbitrary i ∈ N . Then t(−p)is the number of individuals i for whom p(i) = −1. We say thatprofile p is oriented if p(i) = 0 for any i ∈ N and, for any j ∈ N suchthat p(j) = 1, we have p(i) = 1 for all i ≤ j. In other words, anoriented profile is a string of 1 s followed by a string of−1 s, unlessp(i) = 1 for all i, or p(i) = −1 for all i.

A social choice function g on domain D ⊂ Pn maps each p ∈

D into X . Note that a social choice function cannot declare a tiebetween x and y. An important special case is the domain Ln,denoting the set of p ∈ Pn such that p(i) = 0 for any i ∈ N . We saythat g is a majority rule on D if, for any p ∈ D, we have g(p) = 1 if

D.E. Campbell, J.S. Kelly / Economics Letters 118 (2013) 10–12 11

t(p) > t(−p) and g(p) = −1 if t(p) < t(−p). Amajority tie occurswhen t(p) = t(−p). Of course, this can happen if n is even or n isarbitrary and the domain allows for the possibility of individualindifference. When ties occur there are many ways of breaking atie, some of which will violate anonymity and some will violateneutrality. But if n is odd and t(p) + t(−p) = n for every profile pin the domain then there is only one majority rule.

If p ≥ r implies g(p) ≥ g(r) we say that g is monotonic. Therule g is anonymous if g(p) = g(r) whenever t(p) = t(r) andt(−p) = t(−r). (Equivalently, g is anonymous if g(r) = g(p)whenever there is a permutation π on N such that r(i) = p(π(i))for all i ∈ N .) And g is neutral if g(p) = −g(−p) for all p ∈ D.Finally, g satisfies limited neutrality if there is at most one orientedprofile p such that g(p) = g(−p). (If g(p) = g(−p), and t(p)and t(−p) are both positive, then of course anonymity implies thatthere are many profiles r at which g(r) = g(−r). But there isonly one oriented profile with that property if g satisfies limitedneutrality.)

Our first result assumes the domain Ln. We will show thatevery social choice function satisfying anonymity, monotonicity,and limited neutrality is a majority rule. If n is odd there isonly one majority rule, but if n is even there will be ties, andall ties are broken in favor of the same alternative by virtue ofanonymity. (Ifn is evennot everymajority rule satisfies anonymity,monotonicity, and limited neutrality because ties can be broken ina non-anonymous manner.)

Let g be any social choice function on Ln satisfying anonymityand monotonicity. Let c(g) be the minimum of { t(p) : p ∈

Ln and g(p) = 1}. Then anonymity and monotonicity imply thatg(p) = 1 if and only if t(p) ≥ c(g). If c(g) = 0 then g is constantand g(p) = g(−p) for every profile p. If c(g) = 1, in which case 1is selected unless everyone prefers −1, we have g(p) = −g(−p)if and only if p = (1, 1, . . . , 1) or (−1, −1, . . . ,−1). If c(g) =

2 < n/2 then it is easy to confirm that g(p) = g(−p) if and only if2 ≤ t(p) ≤ n−2. In all three cases there is more than one orientedprofile for which g(p) = g(−p). If n is even and g is a majorityrule then either c(g) = n/2 or c(g) = n/2 + 1. In either case,g(p) = g(−p) if t(p) = n/2, but g(p) = −g(−p) if t(p) = n/2,and thus majority rule satisfies limited neutrality.

Theorem 1. Assume that there are n ≥ 2 individuals. If g : Ln → Xsatisfies monotonicity, anonymity, and limited neutrality then g is amajority rule.

Proof. Let g be an arbitrary social choice function satisfyingmonotonicity and anonymity. Wewill show that g does not satisfylimited neutrality if c(g) < n/2 or c(g) > n/2 + 1.

Suppose that c(g) < n/2. Then c(g) ≤ n/2 −12 , whether n is

even or odd. If t(p) = n/2 or n/2 +12 , depending on whether n

is even or odd, then we have g(p) = 1 = g(−p). If c(g) < n/2and t(p) = c(g) we have g(p) = 1. And we also have g(−p) = 1because t(−p) = n − c(g) > n/2 > c(g). Therefore, c(g) < n/2implies that g(p) = g(−p) for at least two oriented profiles.(Anonymity allows us to rearrange any profile to make it orientedwithout changing the selected alternative.)

Suppose that c(g) > n/2 + 1 and n is even. If t(p) = n/2 theng(p) = −1 = g(−p). If t(p) = n/2 + 1 we have g(p) = −1. Butg(−p) = −1 also because t(−p) = n/2 − 1 < c(g). Supposethat c(g) > n/2 + 1 and n is odd. If t(p) = n/2 ±

12 then

g(p) = −1 = g(−p) because n/2−12 < c(g) and n/2+

12 < c(g).

Whether n is even or odd we have g(p) = g(−p) for at least twooriented profiles.

If n/2 ≤ c(g) ≤ n/2 + 1 then c(g) =12 (n + 1) if n is odd,

and c(g) = n/2 or n/2 + 1 if n is even. Therefore, a social choicefunction on Ln can satisfy monotonicity, anonymity, and limitedneutrality only if it is a majority rule. �

On the domain Ln, as c(g) moves away from n/2 the set ofprofiles for which g(p) = g(−p) expands, as we now show.

(i) Suppose that c(g) < n/2. Then g(p) = g(−p) if and only ifc(g) ≤ t(p) ≤ n−c(g), as we now show. Suppose t(p) < c(g).Then g(p) = −1. And t(−p) = n − t(p) > n − c(g) > c(g)and hence g(−p) = 1. If t(p) > n − c(g) then we havet(p) > n − c(g) > n − n/2 = n/2 > c(g) and thus g(p) = 1.And t(−p) = n − t(p) < c(g) which implies g(−p) = −1.However, if c(g) ≤ t(p) ≤ n − c(g) then g(p) = 1. Andt(−p) = n − t(p) ≥ c(g) and thus g(−p) = 1.

(ii) Suppose that c(g) > n/2 + 1. We have g(p) = g(−p) ifand only if n − c(g) < t(p) < c(g), as we now show. Ift(p) ≥ c(g) then g(p) = 1. And t(−p) = n−t(p) ≤ n−c(g) ≤

n/2 + 1 < c(g). Then g(−p) = −1. If t(p) ≤ n − c(g) thent(−p) = n − t(p) ≥ c(g) and thus g(−p) = 1. On the otherhand,

t(p) ≤ n − c(g) < n/2 − 1 < n/2 + 1 < c(g)

and thus g(p) = −1. Finally, if c(g) > n/2+ 1 and n− c(g) <t(p) < c(g) we have g(p) = −1 and

t(−p) = n − t(p) < n − [n − c(g)] = c(g)

and hence g(−p) = −1.

In order to derive majority rule when individuals may expressindifference we extend the model to allow for a variablepopulation. There is a difficulty with n = 2. Consider the socialchoice function g : P2

\ {(0, 0)} → X defined by setting g(p) = 1if and only if p(1) = −1 or p(2) = −1. Then g(1, −1) = g(−1, 1)and (1, −1) is the only oriented profile p for which g(p) = g(−p).This social choice function satisfies anonymity, monotonicity,and limited neutrality. But g is not a majority rule becauseg(−1, 0) = 1. It is eliminated when we add an independenceof indifference condition: If the population is augmented by anindividual who is indifferent between the two alternatives thenthe selected alternative does not change. (This is defined formallybelow.) Consider the restriction f of our two-person rule g to thesubdomain L1 by setting f (x) = g(x, 0) for x ∈ {−1, 1}. Then fis constant. Because n = 1 there is only one equation for whichf (p) = f (−p) holds. But f fails to satisfy limited neutrality becausethere are two oriented profiles, (1) and (−1), and f (p) = f (−p) forp = (1) and p = (−1). We could circumvent the difficulty posedby our two-person example by restricting attention to profilesfor which there are at least three non-zero entries but we wouldstill have to impose independence of indifference to obtain ourmajority rule theorem for the variable population model.

Return to the general case. The domain for the variable popula-tion model is V, the union of the sets Pn

\ {(0, 0, . . . , 0)} over alln ≥ 2. In other words, each profile in V is a non-zero n-vector in{−1, 0, 1}n for some n ≥ 2.

It is straightforward to extend our conditions to the variablepopulation model. Variable population anonymity (resp., variablepopulation monotonicity) on the domain V applies anonymity(resp., monotonicity) to the restriction of g to Pn for each n ≥

2. Variable population limited neutrality also applies the originalcondition to the restriction of g to Pn for each n ≥ 2. It has forceonly for oriented profiles, and hence for profiles which do not haveany zero entries. Therewill be nodanger of confusion ifwedrop thepreface ‘‘variable population’’. We say that g : V → X is majorityrule if the restriction of g to Pn is a majority rule for each n ≥ 2.

We employ one new condition. For any n, any profile p =

(p(1), p(2), . . . , p(n)) ∈ Pn, and any x ∈ P we let (p, x) denotethe profile (p(1), p(2), . . . , p(n), x) ∈ Pn+1 for which individual iprefers p(i) for all i ≤ n and x is the preference of individual n+ 1.

We say that g : V → X satisfies independence of indifference ifg(p, 0) = g(p) for all p ∈ V.

12 D.E. Campbell, J.S. Kelly / Economics Letters 118 (2013) 10–12

Theorem 2. If g : V → X satisfies monotonicity, anonymity, limitedneutrality, and independence of indifference then g is a majority rule.

Proof. Theorem 1 implies that for any n ≥ 2 the restriction ofg to the subdomain Ln is a majority rule. Let p be any profilein V. Anonymity allows us to assume that p(j) = 0 for some jimplies p(i) = 0 for all i ≥ j. Let m = t(p) + t(−p). Thenp′

= (p(1), p(2), . . . , p(m)) belongs to Lm. If t(p) = t(−p) theng(p′) is the majority winner, and thus by repeated application ofindependence of indifference, g(p) is the majority winner at p.

To complete the proof we show that the restriction of g toL1 is majority rule. A constant social choice function f on L1satisfies anonymity and monotonicity, but it does not satisfylimited neutrality: f (−1) = f (1) and f (1) = f (−1), and both(−1) and (1) are oriented profiles. If f is non-constant and satisfiesmonotonicity then f (1) = 1 and f (−1) = −1, which is majorityrule. �

Theorem1 in Campbell and Kelly (2011) implies that g : V → Xis majority rule if it satisfies monotonicity, partial anonymity, andneutrality. Partial anonymity requires the selected alternative tobe unchanged by a permutation of individuals at any profile p suchthat t(p) + t(−p) is odd. (In fact, invariance is only required underpermutations that map an indifferent individual i into i itself.)Therefore, we can now claim the following:

Theorem 3. If g : V → X satisfies monotonicity, independenceof indifference, and either anonymity plus limited neutrality orneutrality plus partial anonymity then g is a majority rule.

A majority rule on V can violate both ‘‘limited neutrality plusanonymity’’ and ‘‘partial anonymity plus neutrality’’. For instance,let g(p) be the majority winner if there is one, and in the case ofa tie set g(p) = p(j) for the smallest j such that p(j) = 0 if t(p) iseven, and set g(p) = 1 if t(p) is odd.We could impose a consistency

condition to disqualify such examples, and thereby get a necessaryand sufficient condition for majority rule on V. But that would berather contrived.

We conclude with some remarks on related results. May (1952)characterized themajority rule social welfare function, which ranksthe alternatives as a function of individual preferences, bymeans ofanonymity, neutrality, and a rather strong version ofmonotonicity.Tied alternatives have the same rank in the social ordering. In otherwords, May did not incorporate tie breaking into his formalizationof the social choice process. (Fishburn, 1973, Aşan and Sanver,2002, Maskin, 1995, Campbell and Kelly, 2000, Woeginger, 2003and Yi, 2005, consider the case of three or more alternatives.)Dasgupta and Maskin (2008) require the selection of a singlealternative but they assume a continuum of voters, and ruleout the possibility that an individual is indifferent between twoalternatives.

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