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Balancedness of social choice correspondences
Mathematical Social Sciences 102 (2019) 59–67
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Balancedness of social choice correspondences Jerry S. Kelly a , Shaofang Qi b , a b
∗
Department of Economics, Syracuse University, Syracuse, NY 13244-1020, USA School of Business and Economics, Humboldt University Berlin, Spandauer Str. 1, 10178 Berlin, Germany
article
info
Article history: Received 21 November 2018 Received in revised form 5 May 2019 Accepted 14 August 2019 Available online 16 September 2019 Keywords: Balancedness Tops-only Unanimity Monotonicity Scoring rules Borda
a b s t r a c t A social choice correspondence satisfies balancedness if, for each pair of alternatives, x and y, and each pair of individuals, i and j , whenever a profile has x adjacent to but just above y for individual i while individual j has y adjacent to but just above x, then only switching x and y in the orderings for both of those two individuals leaves the choice set unchanged. We show how the balancedness condition (satisfied by the Borda, Pareto, and Copeland rules) interacts with other social choice properties, especially tops-only. We also use balancedness to characterize the Borda rule (for a fixed number of voters) within the class of scoring rules. © 2019 Elsevier B.V. All rights reserved.
1. Introduction We consider a new condition, balancedness. A social choice correspondence satisfies balancedness if, for each pair of alternatives, x and y, and each pair of individuals, i and j, whenever a profile has x adjacent to but just above y for individual i while individual j has y adjacent to but just above x, then only switching x and y in the orderings for both of those two individuals leaves the choice set unchanged. Social choice theory often treats responsiveness conditions, like monotonicity, but balancedness is a non-responsiveness property. It is a natural equity condition that simultaneously incorporates some equal treatment for individuals short of anonymity, some equal treatment of alternatives short of neutrality, and some equal treatment of position of alternatives in orderings (for example, raising x just above y in the bottom two ranks for individual j exactly offsets lowering x just below y in the top two ranks for i). With only two alternatives, balancedness is equivalent to anonymity. With more than two alternatives, balancedness is a condition independent of anonymity. Any dictatorial correspondence, however, is unbalanced. For a dictatorial correspondence, one and only one individual is “effective”: changing preferences for every individual other than the dictator leaves the choice set unaffected. In addition to excluding dictatorship, we show that balancedness implies that any non-constant social ∗ Corresponding author. E-mail addresses: [email protected] (J.S. Kelly), [email protected] (S. Qi). https://doi.org/10.1016/j.mathsocsci.2019.08.003 0165-4896/© 2019 Elsevier B.V. All rights reserved.
choice correspondence must have every individual to be effective (Theorem 1). For the equal treatment of positions incorporated by balancedness, we present results from different perspectives. First, we consider the interaction of balancedness with a specific property, tops-only. By tops-only, any change that affects no one’s topranked option leaves the choice set unaffected. Plurality voting is an obvious tops-only rule. Unlike balancedness, tops-only clearly treats the top-rank differently from the remaining ones. The expected conflict between balancedness and tops-only is characterized through a kind of impossibility result (Theorem 2): whenever a social choice correspondence satisfies both balancedness and tops-only, it remains constant on all non-unanimous profiles (i.e., profiles without common option top-ranked by all individuals). We next examine the implication of balancedness within the class of scoring rules. A scoring rule assigns weights to ranks, lower weights to higher ranks, and the score for each option is the sum of weights corresponding to all ranks that option occupies in individual orderings. The Borda rule adds a constant value to weights from one rank to the next, and picks the alternatives with the lowest score. Balancedness holds for the Borda rule: the changes of scores from raising x just above y in the bottom two ranks for individual j is exactly offset by the changes of scores from lowering x just below y in the top two ranks for i, leaving the choice set unchanged. We show that except for a counterexample with small numbers of individuals and alternatives, balancedness actually uniquely characterizes the Borda rule within the class of scoring rules (Theorems 4–6).
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To be clear, we are not examining balancedness because we advocate it as a property that should hold for social choice correspondences. Rather we study balancedness because it is a common property of several correspondences that have been central to social choice theory. As noted, Borda satisfies balancedness. The Pareto correspondence, which at profile u selects those alternatives x such that no other alternative y that everyone prefers to x, also satisfies balancedness. The majority preference relation is unaffected by an interchange of any transposition pair. Accordingly, a social choice correspondence constructed from majority voting relations will also satisfy balancedness. This category includes examples like Copeland’s rule (which selects the alternatives that defeat the most other alternatives by simple majority vote) and the top-cycle correspondence (which selects the alternatives that are highest in the transitive closure of the majority voting relation). In fact, with more than two alternatives, anonymity, neutrality, the Pareto condition and balancedness simultaneously hold for the Pareto correspondence,1 the Borda rule, the Copeland rule, and top-cycle, while all those rules fail tops-only. On the other hand, anonymity, neutrality, the Pareto condition and tops-only simultaneously hold for plurality and union-of-the-tops, while balancedness fails for both of them. Dictatorship fails balancedness but not tops-only, and maximin fails both. Thus besides anonymity, balancedness is a condition also independent of unanimity, neutrality, the Pareto condition, and tops-only.2 Several properties related to balancedness appear in some recent work. Derya (2014) considers a variable-population setting that allows for individual indifference and introduces a “cancellation property”. Cancellation bears some similarity to balancedness. First, the definition of cancellation (p. 152–153) hinges critically on allowing for individual indifference (and is for a variable-population setting): it becomes meaningless if one only considers the domain with strict preferences (as we assume). To illustrate, consider a profile (starting with strict preferences) where x ranked above y by individual #1 and y above x by #2. Cancellation says that making x and y indifferent for #1 and y and x indifferent for #2 will not change the outcome. If we assume further that z is (originally) ranked above x for #1 and x above z for #2, then repeatedly applying cancellation to make x, y, z all indifferent for both #1 and #2 will not change the outcome. In other words, every time cancellation is applied, one actually enlarges an indifference class by bringing together two previously neighboring ones, yet can never separate an indifference class into two different ones. This implies, in particular, cancellation cannot be applied to generate a transposition pair through the process considered by the balancedness property.3 Moreover, although Derya provides a result that uses cancellation to characterize Borda rule within the class of scoring rule, the Borda rule there is a different version on the domain of weak preferences and for a variable-population setting, which is not equal to or stronger than our result in terms of either the result itself or the proof technique.4 1 For more on the definitions and discussions of these social choice rules and the properties mentioned in this paragraph, see Arrow et al. (2002) and Heckelman and Miller (2015). 2 The above examples imply that none of these properties imply balancedness. On the other hand, the following example satisfies balancedness but fails for all these properties. Let n = 2 and X = {x, y, z }. Let G pick (fixed alternative) y constantly except at profile u where individual 1 ranks x above y above z and individual 2 ranks y above z above x and G(u) = {z }. 3 Applying cancellation to each profile of a transposition pair to obtain the same profile with indifference, one can generate the conclusion of balancedness, and thus cancellation is (weakly) stronger than balancedness even adopting a fixed population. But again without allowing for indifference, cancellation becomes meaningless. 4 Derya also introduces a property, degree equality (for a variablepopulation), that is strictly stronger than cancellation, and thus is strictly stronger than balancedness (even adopting a fixed population).
Mihara (2017) introduces a property called “Positional Cancellation”, which requires that changes in the relative positions of two alternatives that cancel each other (e.g., the changes in the relative position of (x, y) if changing from xay to axy for #1 and changing from bxy to xby for #2) do not alter the social preference between the two. Mihara works with social welfare functions not social choice correspondences and allows for individual indifference. If adapted to the context of a social choice correspondence, (the modified) Positional Cancellation is a strictly stronger property than balancedness.5 Mihara uses the cancellation property, together with “reversal” (a neutrality property) and “positive responsiveness” (a monotonicity property), to characterize the Borda ranking among all ranking rules. Balancedness is also strictly weaker than “Invariance for Average-Position Preserving Reversals” (IAPPR) introduced by Sato (2017), which states that “if some voters reverse their preferences for two common alternatives, then the set of winners should be unchanged as long as this reversal preserves the average position of each alternative”.6 Sato characterizes the Borda correspondence through IAPPR, neutrality and positive responsiveness among all social choice correspondences, while we use balancedness alone to characterize the Borda correspondence but within the class of scoring rules.7 2. Definitions Let X with cardinality |X | = m ≥ 2 be the set of alternatives and let N = {1, 2, . . . , n} with n ≥ 2 be the set of individuals. A (strong) ordering on X is a complete, asymmetric, transitive relation on X (non-trivial individual indifference is disallowed). The highest ranked element of an ordering r is denoted r [1], the second highest is denoted r [2], etc. And r [1 : k] is the unordered set of alternatives in the top k ranks of r. The set of all orderings on X is L(X ). A profile u is an element (u(1), u(2), . . . , u(n)) of the Cartesian product L(X )N . A social choice correspondence G is a map from the domain L(X )N to non-empty subsets of X . The range of social choice correspondence G is the collection of all sets S such that there exists a profile u with G(u) = S. G satisfies tops-only if for each pair of profiles u, v , whenever u(i)[1] = v (i)[1] for each i, then G(u) = G(v ). We call a profile u unanimous if u(i)[1] = u(j)[1] for all individuals i, j. Then social choice correspondence G satisfies unanimity if G(u) = u(i)[1] at each unanimous profile u. At profile u, we say x is dominated by y if everyone prefers y to x. An alternative x is Pareto optimal at u if it is not dominated by any alternative y at u. Correspondence G satisfies the Pareto condition if, at every u, every alternative in G(u) is Pareto optimal at u. We say profile v is constructed from profile u by transposition pair (x, y) via individuals i and j if at u, x is immediately above y for i, and y is immediately above x for j, and profile v is just the same as u except that for i and j, alternatives x and y are transposed. A social choice correspondence G is called balanced if, for each x, y, u, v , i, and j, whenever profile v is constructed from u by transposition pair (x, y) via individuals i and j, then G(v ) = G(u). (Otherwise, G is called unbalanced.) 5 It is easy to exemplify that balancedness does not imply Positional Cancellation. For the other direction, that Positional Cancellation implies balancedness, indifference needs to be employed in the proof. 6 It is straightforward to show IAPPR implies balancedness. On the other hand, there are profiles that satisfy the condition of IAPPR and thus by IAPPR they should have the same outcome chosen, but there are no pair transpositions from the profiles and balancedness imposes no restriction on them. 7 Also, Mihara’s and Sato’s characterization results do not have an exceptional case for small numbers of m and n.
J.S. Kelly and S. Qi / Mathematical Social Sciences 102 (2019) 59–67
3. Results 3.1. Balancedness and (non-)dictatorship To see one way in which balancedness reflects equal treatment of individuals, first observe again that any dictatorial correspondence (where for some i and all u, G(u) = u(i)[1]) is unbalanced. This can be extended to cover all non-constant correspondences with ineffective individuals. An individual i is ineffective for a correspondence G if for each pair of profiles u, u∗ , we have G(u) = G(u∗ ) whenever u(j) = u∗ (j) for all j ̸ = i. (Otherwise i is called effective for G.) For example, if G is dictatorial with dictator j, then every other individual is ineffective. We show that for any non-constant correspondence, balancedness implies every individual is effective. Suppose individual #2 is ineffective for G but #1 is effective. If v is obtained from u by transposition pair (x, y) via individuals #1 and #2, the outcome of G is affected just as if y rose for #1 and nothing else happened. For a non-constant correspondence, it cannot generally be true that such moves leave outcomes unchanged as required by balancedness. Theorem 1. Any non-constant social choice correspondence with an ineffective individual is unbalanced. Proof. Let i be an ineffective individual for non-constant correspondence G. Let u, u∗ be two profiles with G(u) ̸ = G(u∗ ). Since i is ineffective, we may assume that u(i) = u∗ (i). Now construct a sequence of profiles u = u1 , u2 , . . . , uT − 1 , uT = u∗ from u to u∗ such that for any two successive profiles ut −1 , ut in the sequence, ut differs from ut −1 only by a transposition of two adjacent alternatives in the ordering of a single individual other than i. For some t, it must be that G(ut ) ̸ = G(ut −1 ). Suppose that ut differs from ut −1 because x is transposed with y, ranked just below x in ut −1 (j). Construct profile u′t −1 from ut −1 by moving y adjacent to and just above x in ut −1 (i), and construct profile u′t from ut by moving x adjacent to and just above y in ut −1 (i). Then, by the ineffectiveness of i, G(u′t ) = G(ut ) and G(u′t −1 ) = G(ut −1 ), so G(u′t ) ̸ = G(u′t −1 ). But u′t differs from u′t −1 by transposition pair (x, y) via individuals i and j and so this constitutes a violation of balancedness. □ In the next subsection, we show how the balancedness condition interacts with other social choice properties, especially tops-only. In the last subsection, we show how balancedness can be used to characterize the Borda rule (for a fixed number of voters) within the class of scoring rules. 3.2. Balancedness and tops-only Balancedness incorporates some equal treatment of positions, so we might expect conflict with the tops-only property. By topsonly, all profiles u with an alternative x at everyone’s top must give the same value for G(u), though it need not be the case that G(u) = {x}. Do there exist any social choice correspondences satisfying tops-only and unanimity that are balanced? A positive answer to that question is given by: Example 1. Let G(u) be the common top at all unanimous profiles and set G(u) = X (or any other fixed set) on all non-unanimous profiles.
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Intuitively, tops-only implies invariance to any moves that affect no one’s top; balancedness implies invariance to a transposition pair via corresponding individuals. The two properties combined together generate further invariances. For instance, suppose individual #1 has ranking xyz (i.e., x above y above z); #2 has ranking zyx. Tops-only and balancedness together imply that social choice outcomes remain constant if #1’s ranking becomes yxz while #2’s ranking stays the same: first through a transposition pair (x, y) via #1 and #2 (by balancedness), then switching x and y again for #2 (by tops-only). Theorem 2 repeatedly applies the same technique to prove that the conclusion holds for all profiles that are non-unanimous. Theorem 2. For m ≥ 3, n ≥ 3, if social choice correspondence G satisfies tops-only and balancedness, then it is constant on all non-unanimous profiles. Thus such a G has at most m + 1 sets in its range. Proof. Let a and b be two fixed elements of X and let u∗ be a fixed profile with tops abb...bb. Given any profile u without a unanimous top, we will show there is a sequence of profiles u1 , u2 , . . . , uT all without a unanimous top, such that: 1. u1 = u; 2. uT = u∗ ; 3. For each t with 1 ≤ t < T , ut +1 can be constructed from ut by either a reordering of alternatives below someone’s top or by a paired transposition. Then we would have G(u) = G(u∗ ) for all u without a unanimous top. In the following, let T (u) be the set of all alternatives x such that u(i)[1] = x for some i. Case 1. T (u) = {a, b}. If u(1)[1] = a, go to the next paragraph. Suppose that u(1)[1] = b. For some i > 1 with a in the top rank, construct u2 by raising a to 1’s second rank and b to i’s second rank. Then construct u3 by transposition pair (a, b) via individuals 1 and i. Now u3 (1)[1] = a and u3 (i)[1] = b. If all other tops are now b, we are done. So suppose that some other top, say for individual j, is a. Construct u4 by raising a third alternative c to the second rank for j and raising c just above a for i. Then construct u5 by transposition pair (a, c) via i and j. Now c is at j’s top. Recall that at u5 , we have u5 (1)[1] = a, so we can move b to the second rank for individual 1, c to the third rank for individual 1, and for individual j, we move b to the second rank. Construct the next profile in the sequence by transposition pair (b, c) via 1 and j. Then b now becomes j’s top; the a has been changed to a b. Repeat this until all “a”s (except for #1) have been changed first to “c”s and then to “b”s. Case 2. Suppose T (u) contains a, b, and other alternatives. If c is a top for someone, say i, construct u2 by raising b to the second rank for i and raise b just above c for an individual j who has a on top. Then a transposition pair (b, c) via i and j reduces by one the number of individuals without a or b on top. Continue in this fashion until you reach Case 1. Case 3. T (u) contains one of a and b but not the other. Suppose that u(i)[1] = a and u(j)[1] = c. Construct u2 by raising b to second rank for j and raise b just above c for i. Then a transposition of b and c for i and j creates a profile in either Case 2 or Case 1. A similar analysis holds if it is b instead of a at someone’s top. Case 4. T (u) contains neither of a or b but does contain say c and d. Construct profile u2 by raising a to second rank for some i with c on top and a just above c for some j with d on top. A transposition pair (a, c) via i and j yields a profile in Case 3. □
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A modified version of this argument can also be constructed for n = 2. But there is a limit to this style of argument; there are not sequences to u∗ from profiles with unanimous tops as balancedness cannot be applied there. This limit on the argument cannot be overcome, as seen by the example at the beginning of this section; there, at profiles with a common top, different outcomes may occur. When there are at least three alternatives, for every option, there exist some non-unanimous profiles at which the option is Pareto dominated. Thus, if a social choice correspondence remains constant on all non-unanimous profiles, it must violate the Pareto condition. This leads to the next Corollary. Corollary 1. For m ≥ 3, if social choice correspondence G satisfies tops-only and Pareto, then G is unbalanced. Proof. The corollary is a straightforward application of Theorem 2 given that if G satisfies Pareto then G cannot be constant on all non-unanimous profiles. □ Both tops-only and Pareto are needed in the Corollary. If Pareto is not assumed, a constant correspondence satisfies both tops-only and balancedness. If tops-only is not assumed, the correspondence that selects the set of Pareto optimal alternatives satisfies Pareto and balancedness. Theorem 2 can be extended to similar results if we relax tops-only to similar properties where some higher ranks (not necessarily the top-rank) are treated differently from lower ones. The following social choice correspondence violates tops-only, but outcomes do depend only on which alternatives are ranked first or second by individuals (but not on how the top alternatives are ordered within those two ranks). Example 2. (Rigid8 ) 2-approval voting is a social choice correspondence G that is like plurality rule except that instead of selecting the alternatives with the most frequently occurring tops, it selects the alternatives with the most frequent occurrences in the top two ranks for everyone. Like plurality rule, 2-approval voting is unbalanced. We now extend Theorem 2 to cover social choice correspondences that, like 2-approval voting, depend only on the top two ranks for every individual. We first need to extend tops-only. Let u(i)[1 : 2] denote the (unordered) set of alternatives in the top two ranks for individual i at profile u. We say social choice correspondence G satisfies top-2-only if for each pair of profiles u, u∗ , G(u) = G(u∗ ) if u(i)[1 : 2] = u∗ (i)[1 : 2] for all individuals i. Let D be the subdomain of L(X )N consisting of all profiles for which it is not true that u(i)[1 : 2] = u(j)[1 : 2] for each pair of individuals i and j, that is, at least three alternatives occur in the top two ranks over all individuals. D is the analog here of the subdomain of non-unanimous profiles in Section 3.2. Theorem 3. Let m ≥ 4 and n ≥ 3. Then any social choice correspondence G satisfying balancedness and top-2-only must be constant on D. Proof. Since the argument will be divided into different cases, we mention the main idea here and leave some details to Appendix. Consider a specific profile u∗ with c in everyone’s top rank, a in #1’s second rank, and b in everyone else’s second rank. It will 8 “Rigid” because the number of ranks (here 2) is fixed to be the same for all individuals. For approval voting without rigidity, see Brams and Fishburn (1983). Rigid k-approval voting appeared in Alemente et al. (2016) where it was called approval voting (type-k).
suffice to show that for each profile u in D, there is a sequence of profiles in D from u to u∗ such that each is obtained from the previous profile by a transposition pair or an application of top-2-only, i.e., by a re-ordering below the top two ranks. We argue by induction on the number p of individuals who, at u, do not have c in their top two ranks. Basis: Suppose that we have p = 0 at u, i.e., every individual has c in their top two ranks. Construct u′ from u by (only) raising c to everyone’s top rank. Now consider the subdomain D′ of D consisting of all profiles in D where everyone has c top-ranked. Note that for each such profile at least two distinct alternatives must be second-ranked. Let G′ be the restriction of G to D′ . Social choice correspondence G′ on D′ induces a correspondence on those profiles on X \{c } which have non-unanimous tops. Then the analysis in the tops-only section shows that there is a sequence of profiles in D′ from u′ to u∗ . Induction step: Assume now that there is a non-empty set S of alternatives such that for all profiles v in D with the number of individuals who do not have c in their top two ranks being less than p we have the same outcome: G(v ) = S. Suppose that u in D is a profile where the number of individuals who at u do not have c in their top two ranks is p > 0. We will show there is a sequence of profiles (such that each is obtained from the previous profile by a transposition pair or an application of top-2-only) from u to a profile u′ in D with p − 1 individuals who do not have c in the top two ranks (so G(u′ ) = S). The analysis must consider different cases for the profile u, and details are in Appendix. □ There are straightforward generalizations to top-3-only, etc.9 3.3. Balancedness and Borda characterization Balancedness is an equity condition that incorporates some equal treatment for differences of position of alternatives in orderings. As observed at the beginning of this paper, raising x just above y in the bottom two ranks for individual j exactly offsets lowering x just below y in the top two ranks for i. This equal treatment of differences of position suggests trying to characterize the Borda rule within the class of scoring rules. Let a scoring system be given by weights s1 ≤ s2 ≤ s3 ≤ · · · ≤ sm (higher ranked alternatives will be assigned lower weights). At profile u, the score for an alternative x is the sum S(x, u) =
n ∑
s(u, i, x)
i=1
where s(u, i, x) = sk if u(i)[k] = x, i.e., the score for x is the sum of the weights corresponding to the ranks that x occupies in the individual orderings at u. The related scoring social choice correspondence G selects at u the alternatives x with lowest S(x, u) values. If s1 = s2 = s3 = · · · = sm then G is the constant social choice correspondence with G(u) = X at all u, which is not very helpful. Accordingly, we henceforth only consider systems of weights such that at least two weights are distinct. 9 For instance, for top-3-only, the analog of the subdomain of non-unanimous profiles consists of all profiles with at least four alternatives occurring in the top three ranks over all individuals. The specific profile u∗ has c in everyone’s top rank, d in everyone’s second rank, a in #1’s third rank and b in everyone else’s third rank. The proof proceeds by induction on the number p of individuals who, at u, do not have c in their top three ranks. The basis step (p = 0) follows from top-2, and the induction step also follows the same case division as those for top-2.
J.S. Kelly and S. Qi / Mathematical Social Sciences 102 (2019) 59–67
The Borda correspondence, GB , uses weights 1 < 2 < 3 < · · · < m. But of course other weights also generate Borda. If social
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3.3.2. Borda characterization for n > 3 We begin by looking at m = 3.
choice correspondence G is generated by weights s1 ≤ s2 ≤ s3 ≤ · · · ≤ sm then G is also generated by linearly transformed weights t1 ≤ t2 ≤ t3 ≤ · · · ≤ tm where ti = α + β si for real numbers α , β , with β > 0. Lemma 1. For m ≥ 3 and n ≥ 2: if G is a scoring social choice correspondence for scoring system given by
Theorem 4. For m = 3 and n ≥ 4: if G is a scoring social choice correspondence and G satisfies balancedness, then G is the Borda correspondence. Proof. We separate the analysis into two cases, n is even and n is odd. (n is even) We examine the following profile v :
s1 ≤ s2 ≤ s3 ≤ · · · ≤ sm with at least two weights distinct and if G satisfies balancedness, then s1 ̸ = s2 . Proof. If s1 = s2 , then there exists a k, 2 ≤ k < m, such that s1 = s2 = · · · = sk < sk+1 . Construct profile u such that u(i)[1] = y and u(i)[2] = x for all i < n while u(n)[k] = x and u(n)[k + 1] = y. Then x ∈ G(u) and y ∈ / G(u). If profile u∗ is constructed from u by transposition pair (x, y) via individuals 1 and n, then x ∈ / G(u∗ ) and y ∈ G(u∗ ). So G(u∗ ) ̸= G(u) and G fails balancedness. □ As a consequence of Lemma 1, we assume from now on that the weights have been linearly transformed so that s1 = 1 and s2 = 2. For profiles of strong orderings, we want to show that generally if a scoring social choice correspondence satisfies balancedness, it must be the Borda rule. However, the next subsection shows a limitation on this objective. 3.3.1. A counterexample when m = n = 3 There is an exception to our general result that any balanced scoring rule must be the Borda rule. When m = n = 3, the profiles where there are no transposition pairs, and so where the balancedness condition has no force, are exactly the voting paradox profiles, displaying unusually great symmetry. For example, all other profiles have a majority voting winner. The profiles where balancedness does have force have just enough asymmetry to allow for scoring rules different from Borda. Example 3. Let m = n = 3 and set scoring weights to be 1, 2, and 3.1. The related correspondence, G, differs from Borda. At profile u: 1 x y z
2 x y z
1
2
3
4
n−1
n
x y z
z y x
x y z
z y x
x y z
z y x
···
Consider a scoring rule G with weights: 1 < 2 ≤ s3 (using Lemma 1). We examine the scores for x, y, z: For x, 2n (1 + s3 ); For z, 2n (1 + s3 ); For y, 2n. Under this scoring rule, either x and z are both in G(v ) or neither x nor z is in G(v ). Now we consider a sequence of transposition pairs (x, y) first via individuals 1 and 2, and then via 3 and 4, . . . , and so on until via (n-1) and n, which yields the profile u: 1
2
3
4
y x z
z x y
y x z
z x y
···
n−1
n
y x z
z x y
We examine again the scores for x, y, z: For x, 2n; For y, 2n (1 + s3 ); For z, 2n (1 + s3 ). This time, either y and z are both in G(u) or neither y nor z is in G(u). But u is obtained from v by a sequence of transposition pairs, and balancedness implies G(u) = G(v ), which means both must be {x, y, z }, which, in turn, means at each profile, all three scores must be the same. Therefore, 2n (1 + s3 ) = 2n, and thus s3 = 3, i.e., G is the Borda correspondence. (n is odd) We examine the following profile v (where the construction of the first n − 3 profiles is the same as the case of n even and the last three profiles is a voting paradox profile):
3
1
2
n−4
n−3
n−2
n−1
n
y z x
x y z
z y x
x y z
z y x
y x z
x z y
z y x
the Borda rule has a tie between x and y, and z is Paretodominated by y so GB (u) = {x, y}. But with scoring weights 1, 2, and 3.1, the scores of x and y are 5.1 and 5 respectively: G(u) = {y}. G ̸ = GB . Nevertheless, G here is balanced, as can be checked.10 So, for m = n = 3, balancedness of a scoring rule does not imply Borda. This use of small numbers of individuals and alternatives is critical in Example 3, as we will see. 10 Because all scoring rules are neutral and anonymous, to check balancedness, we may assume that the pair interchanged is {x, y} and that x is just above y for #1 and just below y for #2, and the interchange is for those two individuals. Without loss of generality, we assume #3 prefers x to y (although x and y need not to be adjacent in u(3)). There are 12 profiles satisfying those conditions. But for six of those profiles, either z is on top for both #1 and #2 or on bottom for both, and it is easy to see that balancedness is satisfied. So we need only consider the remaining six profiles and the details are omitted.
···
Consider a scoring rule G with 1 < 2 ≤ s3 (using Lemma 1). We examine the scores for x, y, z: 1 For x, n− (1 + s3 ) + 2; 2 n−1 For z, 2 (1 + s3 ) + 2; For y, 2n − 3 + s3 . So under this scoring rule, either x and z are both in G(v ) or neither x nor z are in G(v ). Now we consider a sequence of transposition pairs (x, y) first via individuals 1 and 2, and then via 3 and 4, . . . , and so on until via (n − 4) and (n − 3) (so the voting paradox of the last three individuals remains unchanged), which yields the profile u: 1
2
y x z
z x y
···
n−4
n−3
n−2
n−1
n
y x z
z x y
y x z
x z y
z y x
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We examine again the scores for x, y, z: For x, 2n − 3 + s3 ; 1 (1 + s3 ) + 2; For y, n− 2 n−1 For z, 2 (1 + s3 ) + 2. This time, either y and z are both in G(u) or both are out of G(u). But by balancedness, G(u) = G(v ), so at each profile all scores must be the same. Therefore n−1 2n − 3 + s3 = (1 + s3 ) + 2 2 which has the unique solution s3 =
3n−9 n−3
Clearly GB (v ) = X . For the correspondence G with weights 1, 2, 3, . . . , m − 1, w > m, the scores are 2n (1 + w ) for x and z, 2n (m + 1) for the rest. Because w > m, we have G(v ) = X \{x, z }. Construct u by transposing x and y first via individuals 1 and 2, and then via 3 and 4, . . . , and so on until via (n-1) and n, which yields the profile u:
= 3, (note that we have
n ≥ 4), i.e., G is the Borda correspondence.11
□
Proof. For m = 3, the proof is given by Theorem 4. For m ≥ 4, suppose that we have linearly transformed weights 1 < 2 ≤ s3 ≤ · · · ≤ sm such that for some j ≥ 3, it is the case that sj ̸ = j. Let G be the corresponding scoring social choice correspondence. We show that there is a profile v such that G(v ) ̸ = GB (v ) and that G must fail balancedness. Consider the smallest integer j such that sj ̸ = j. In what follows, we first assume that the smallest integer j with sj ̸ = j is that j = m. We then extend the analysis to the case where j < m. When j = m, the weights are: 1 < 2 < 3 < · · · < m − 1 ≤ w where w ̸ = m. Analysis here is for w > m. The same profile applies to the case m − 1 ≤ w < m. Again, we separate the analysis for even n and odd n, both of which extend the proof of Theorem 4. And since the extension follows the very same spirit, we treat even n here with details, and briefly comment on the case for odd n at the end. (n is even) For even n in Theorem 4 (where m = 3), we previously looked at profile: 3
4
x y z
z y x
x y z
z y x
···
n−1
n
x y z
z y x
Now to obtain v , we insert additional alternatives repeatedly: at each stage of the construction, we add another alternative, placing it just above z for odd numbered individuals, just below z for even numbered individuals; any two neighboring profiles are reverse orderings: 1
2
n−1
n
x y
z am−3
x y
z am−3
a2
a1 a2
a2
a1 a2
.. .
am−3 z
.. .
a1 y x
···
.. .
am−3 z
n−1
n
y x
z am−3
y x
z am−3
a2
a1 a2
a2
.. .
.. .
Theorem 5. For m ≥ 3 and n ≥ 4: if G is a scoring social choice correspondence and G satisfies balancedness, then G is the Borda correspondence.
2
2
a1 a2
We next take up larger m.
1
1
.. .
a1 y x
11 This proof provides insight into why Example 3 works. That example starts with a profile that has no transposition pairs.
am−3 z
···
.. .
.. .
a1 x y
am−3 z
a1 x y
This yields scores 2n (1 + w ) for y and z, 2n (m + 1) for the rest. Again since w > m, we now have G(u) = X \{y, z }, a failure of balancedness. (n is odd) Similar to the case for even n, we construct the profile v based on the one in Theorem 4: 1
2
n−4
n−3
n−2
n−1
n
x y
z am−3
x y
z am−3
y x
x z
z y
a2
a1 a2
a2
z a1
y a1
x a1
a1
a2
y x
.. .
.. .
.. .
am−3
am−3
am−3
a1 a2
.. .
am−3 z
.. .
···
a1
.. .
y x
am−3 z
.. .
a2
a2
It is easy to check that GB (v ) = {x, y, z } by noting that for the first n-3 individuals any two neighboring profiles are reverse orderings. For the correspondence G with weights 1, 2, 3, . . . , m− 3 3 (1 +w ) + 6 for x and z, n− (1 + m) + 6 1, w > m, the scores are n− 2 2 for y, and the rest a1 , a2 , . . . have higher scores. Because w > m, G(v ) = {y}. Construct u by transposing x and y for 1 and 2, and then via 3 and 4, . . . , and so on until via (n − 4) and (n − 3), which yields the profile u. And it is easy to check that at u, the scores are n−3 3 (1 +w ) + 6 for y and z, n− (1 + m) + 6 for x, with the remaining 2 2 scores unchanged (and higher). Because w > m, G(u) = {x}, a failure of balancedness. Finally, we extend the above analysis to the case where the smallest integer j with sj ̸ = j is that j < m. The weights are: 1 < 2 < · · · < j − 1 ≤ sj ≤ · · · ≤ sm where sj ̸ = j. We look at the restriction of G to the domain D where (m − j) alternatives are set at everyone’s bottom in a fixed order. G restricted to this domain D induces a scoring social choice correspondence G′ on j alternatives with weights 1, 2, . . . , (j − 1), sj and sj ̸ = j (while clearly G′ also satisfies balancedness). And this violates the above analysis (for the case where the smallest integer j with sj ̸ = j is that j = m). □ 3.3.3. Borda characterization for n = 3, m > 3 Let us return to the case n = 3, where we learned in Example 1 that for m = 3 not every scoring social choice correspondence satisfying balancedness is the Borda rule. For the case m = 4 and n = 3, a natural analog of Example 3 has scoring weights 1, 2, 3,
J.S. Kelly and S. Qi / Mathematical Social Sciences 102 (2019) 59–67
and 4.1. But that rule is unbalanced, as can be seen at profile u: 1
2
3
a y x b
b a x y
x y b a
Here the scores are S(a) = 7.1, S(b) = 8.1, S(x) = 7, S(y) = 8.1. The smallest of these scores is 7 and so G(u) = {x}. If v is constructed from profile u by transposition pair (x, y) via individuals 1 and 2, then the scores become S(a) = 7.1, S(b) = 8.1, S(x) = 7.1, S(y) = 8. So G(v ) = {x, a}, a failure of balancedness. Theorem 6. For m > 3 and n = 3: if G is a scoring social choice correspondence and G satisfies balancedness, then G is the Borda correspondence. Proof. The proof follows a similar pattern of Theorem 5’s proof. Starting with m = 4, new alternatives are inserted into a profile to show unbalancedness of a scoring rule with weights different from those of the Borda rule. Assume first m = 4. Let us consider a scoring rule G but not the Borda correspondence with linearly transformed weights: 1 < 2 ≤ s3 ≤ s4 . We consider the following profile u: 1
2
3
a y x b
b a x y
x y b a
2
3
a x y b
b a y x
x y b a
We examine the scores for a, b, x, y: For a, 3 + s4 ; For x, 3 + s4 ; For b, 1 + s3 + s4 ; For y, 2 + 2s3 . By balancedness, G(u) = G(v ). Since at profile v , alternatives a and x have the same score, and at u, at least a ∈ G(u) or x ∈ G(u), we have a and x are both in G(u) and G(v ). So (from u), S(a) = 3 + s4 = S(x) = 1 + 2s3 , i.e., 2s3 − s4 = 2
(1)
Again, consider a transposition pair (x, y) via individual 2 and 3 that yields the profile v ′ : 1
2
3
a x y b
b a x y
y x b a
We examine the scores for a, b, x, y: For a, 3 + s4 ; For x, 4 + s3 ; For b, 1 + s3 + s4 ; For y, 1 + s3 + s4 By balancedness, a, x ∈ G(v ′ ), so 4 + s3 = 3 + s4 , or s3 − s4 = −1
(2)
Eqs. (1) and (2) have the unique solution s3 = 3, s4 = 4, i.e., G is the Borda correspondence. Now we extend the analysis to m > 4. Suppose we have weights 1 < 2 ≤ s3 ≤ · · · ≤ sm , and consider the smallest integer j such that sj ̸ = j. We need to consider the case where j = m and where j < m. But similar to the proof of Theorem 5, the case where j < m can be implied by the analysis for j = m by looking at the restriction of G to the domain D where (m − j) alternatives are set at everyone’s bottom in a fixed order, which induces a scoring social choice correspondence on j alternatives. And for the case where j = m, we extend the above profile by inserting new alternatives to obtain a profile vˆ such that G(vˆ ) ̸ = GB (vˆ ) and show that G must fail balancedness. Thus, consider the linearly transformed weights: 1 < 2 < 3 < · · · < m − 1 ≤ w where w ̸ = m. Analysis here is for w > m. The same profile works for w < m. We first show one stage. Look at profile v above. Now insert alternative c just above b for #1 and just below b for #2. The third individual has c at the bottom, and we obtain vˆ :
We examine the scores for a, b, x, y: For a, 3 + s4 ; For x, 1 + 2s3 ; For b, 1 + s3 + s4 ; For y, 4 + s4 . Since S(a) < S(y) and S(x) ≤ S(b) under this scoring rule, at least a ∈ G(u) or x ∈ G(u) (or both). Now we consider a transposition pair (x, y) via individual 1 and 2 that yields the profile v : 1
65
1
2
3
a x y c b
b c a y x
x y b a c
The Borda scores for a and x are the lowest, so GB (vˆ ) = {a, x}. With weights 1, 2, 3, 4, w5 , with w5 > 5, the scores at vˆ are: S(a) = 8, S(b) = 4+w5 , S(c) = 6+w5 , S(x) = 3+w5 , and S(y) = 9, so G(vˆ ) = {a}. After transposing x and y for individuals 1 and 2 to create profile vˆ ′ , the scores become: S(a) = 8, S(b) = 4 + w5 , S(c) = 6 + w5 , S(x) = 8, and S(y) = 4 + w5 , so G(vˆ ′ ) = {a, x} and balancedness is violated. Eventually the profile looks like vˆ ∗ : 1
2
3
a x
b cm−4
x y
c2 c1
b a c1
y c1 c2
.. .
cm−4 b
.. .
a y x
c2
.. .
cm−4
The Borda scores for a and x are the lowest, so GB (vˆ ∗ ) = {a, x}. With weights 1, 2, . . . , m − 1, w , where w > m, the scores at vˆ ∗ are: S(a) = m + 3, S(b) = w + 4, S(x) = w + 3, S(y) = m + 4, and the rest have higher scores. So G(vˆ ∗ ) = {a} given w > m. After transposing x and y for individuals 1 and 2 to create profile vˆ ∗′ ,
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J.S. Kelly and S. Qi / Mathematical Social Sciences 102 (2019) 59–67
the scores become: S(a) = m + 3, S(b) = w + 4, S(x) = m + 3, S(y) = w+4, and the remaining scores are unchanged, so G(vˆ ∗′ ) = {a, x} and balancedness is violated. □
1
4. Conclusion
.. .
c x
We have introduced a new condition, balancedness. Balancedness holds for several correspondences that have been central to social choice theory, such as Borda, Pareto, and Copeland rules. We show that the interaction of balancedness with tops-only together imply invariance of outcomes over most of the domain. We also obtain a new characterization of the Borda correspondence within the class of scoring rules. Hence we can get a characterization of Borda’s rule with a fixed set of individuals by appending balancedness to a characterization of scoring rules for fixed populations (see Fishburn, 1973a,b). While Young (1974), Hansson and Sahlquist (1976), Coughlin (1979–1980), Nitzan and Rubinstein (1981), and Debord (1992) have characterizations of Borda’s rule, they work in a different context than ours; for these authors, a rule has to work for a variable number of individuals and they use variable population properties like that of separability introduced by Smith (1973).
We complete the proof for Theorem 3 by including the details omitted in Section 2. We will not repeat the basis step already in Section 2 and start from the induction step below. Induction step: Assume now that there is a non-empty set S of alternatives such that for all profiles v in D with the number of individuals who do not have c in their top two ranks being less than p we have the same outcome: G(v ) = S. Suppose that u in D is a profile where the number of individuals who at u do not have c in their top two ranks is p > 0. We will show there is a sequence of profiles (such that each is obtained from the previous profile by a transposition pair or an application of top-2-only) from u to a profile u′ in D with p − 1 individuals who do not have c in the top two ranks (so G(u′ ) = S). Without loss of generality suppose that it is the first q = n − p individuals with c ranked in the top two ranks. We may assume that c has been raised to the top for each of those q = n − p individuals so that u is: 2
c
c
.. .
···
.. .
c x
.. .
c
.. .
···
1
2
c
c
t
.. .
1
2
c
c
t
t
.. .
x
w .. .
q
q+1
c
x
z
y
w .. .
···
.. .
n
w .. .
.. .
s
y
w .. .
t
n
···
q
q+1
c
x
z
y
w .. .
.. .
t
.. .
···
.. .
n
Construct: 1
2
c
c
s
n
t
x
z
y
y
w .. .
Then if necessary, lower x to the second rank for q + 1 and raise c to the third rank for that individual.
z
.. .
···
Subsubcase 2Bi. p < n − 1 (and so q > 1).
y
···
q+1 x
Subcase 2B. No element in the top two ranks for q + 1, . . . , n is second for any of 1, . . . , q, and no two of 1, . . . , q have the same second element. (Recall p = n − q.)
c
.. .
q c
Now a transposition pair (t , y) via individuals 2 and q + 1 yields a profile in D still with p individuals with c not in the top two ranks but now back in Case 1 (since t is in the top two ranks for individual q + 1 and is ranked second for individual #1).
2
···
···
y
c
.. .
z
.. .
t
1
.. .
···
.. .
z
c
y
.. .
n
Subcase 2A. There are two individuals among 1, . . . , q with the same alternative t ranked second:
n
q+1
c
···
Case 2. None of the alternatives in the top two ranks for q + 1, . . . , n is ranked second by one of 1, . . . , q.
q+1
q
q+1
Now a transposition pair (c , x) via individuals 1 and q + 1 yields a profile in D with q + 1 individuals with c in the top two ranks and so only p − 1 individuals with c not in the top two ranks.
x
Case 1. At least one of the alternatives, say x, in the top two ranks for q+1, . . . , n is ranked second by one of 1, . . . , q. Without loss of generality, let q + 1 be an individual with x in the top two and let 1 be the individual with c on top and x in the second rank: 2
q
c
q
Here x, y, z and w are distinct from c (though these four need not all be distinct from one another).
1
c
.. .
c
.. .
···
For one of those individuals, say #2, move y, an alternative in u(q + 1)[1 : 2] (moved for q + 1 to second rank if necessary) up to the third rank and then raise t to the third rank for q + 1:
Appendix. Omitted proof for Theorem 3
1
2
t
.. .
···
q
q+1
c
x
z
y
w .. .
.. .
t
.. .
···
n
where t is 2’s second and y is in q + 1’s top two (moved to second rank if necessary). Now a transposition pair (t , y) via 1 and q + 1 takes us back to Case 1 (since t will be in individual q + 1’s top two and in #2’s second rank).
J.S. Kelly and S. Qi / Mathematical Social Sciences 102 (2019) 59–67
Subsubcase 2Bii. p = n − 1 (so q = 1). 1
2
c s
x y
.. .
···
say #2. Move z to second rank for #2, n
1
z
x y
w .. .
.. .
If not all of 2, . . . , n have the same top two alternatives, say n has z in the top two but 2 does not, move z to second place for individual n and raise c to n’s third rank while raising c and z to third and fourth rank for #2: 1
2
c s
x y
.. .
···
n
c
Now a transposition pair (c , z) via 2 and n takes us back to p = n − 2 in D. On the other hand if 2, . . . , n all have the same two alternatives in the top two ranks: 1
2
c s
x y
···
n x y
.. .
.. .
Raise y to third rank for #1 and s to third rank for #2 1
2
c s
x y
y
s
.. .
.. .
···
n x y
.. .
Now a transposition pair (s, y) via individuals 1 and 2 takes us back (in D) to Case 1. Case 3. p = n (so q = 0). Because we are in D, there must be an alternative z, distinct from x and y (#1’s top two), in the top two ranks for someone,
···
n
z
.. .
1
2
x y c
z c
z
.. .
2
Then raise c and z to ranks three and four for #1 and c to third rank for #2:
z
z
.. .
.. .
w
c
67
···
n
.. .
Now transposition pair (c , z) via individual 1 and 2 takes us back (in D) to Case 2Bii. □
References Alemente, F., Campbell, D.E., Kelly, J.S., 2016. Characterizing the resolute part of monotonic social choice correspondences. Econom. Theory 62, 765–783. Arrow, K.J., Sen, A.K., Suzumura, K. (Eds.), 2002. Handbook of Social Choice and Welfare. Elsevier, Amsterdam. Brams, S.J., Fishburn, P.C., 1983. Approval Voting. Birkhauser, Boston. Coughlin, P.J., 1979–1980. A direct characterization of Black’s first Borda count. Econom. Lett. 4, 131–133. Debord, B., 1992. An axiomatic characterization of Borda’s k-choice function. Soc. Choice Welf. 9, 337–343. Derya, A.M., 2014. Essays in Collective Decision Making (Ph.D. dissertation). Bilkent University. Fishburn, P.C., 1973a. Summation social welfare functions. Econometrica 41, 1183–1196. Fishburn, P.C., 1973b. The Theory of Social Choice. Princeton University Press, Princeton. Hansson, B., Sahlquist, H., 1976. A proof technique for social choice with variable electorate. J. Econom. Theory 13, 193–200. Heckelman, J.C., Miller, N.R. (Eds.), 2015. Handbook of Social Choice and Voting. Elgar, Cheltenham. Mihara, H.R., 2017. Characterizing the Borda ranking rule for a fixed population, Working paper, https://mpra.ub.uni-muenchen.de/78093/. Nitzan, S.I., Rubinstein, A., 1981. A further characterization of Borda ranking method. Publ. Choice 36, 153–158. Sato, N., 2017. A simple characterization of Borda rule for fixed electorate, Working paper, https://ssrn.com/abstract=2917013. Smith, J.H., 1973. Aggregation of preferences with variable electorate. Econometrica 41, 1027–1041. Young, H.P., 1974. An axiomatization of Borda’s rule. J. Econom. Theory 9, 43–52.
Contents lists available at ScienceDirect
Mathematical Social Sciences journal homepage: www.elsevier.com/locate/mss
Balancedness of social choice correspondences Jerry S. Kelly a , Shaofang Qi b , a b
∗
Department of Economics, Syracuse University, Syracuse, NY 13244-1020, USA School of Business and Economics, Humboldt University Berlin, Spandauer Str. 1, 10178 Berlin, Germany
article
info
Article history: Received 21 November 2018 Received in revised form 5 May 2019 Accepted 14 August 2019 Available online 16 September 2019 Keywords: Balancedness Tops-only Unanimity Monotonicity Scoring rules Borda
a b s t r a c t A social choice correspondence satisfies balancedness if, for each pair of alternatives, x and y, and each pair of individuals, i and j , whenever a profile has x adjacent to but just above y for individual i while individual j has y adjacent to but just above x, then only switching x and y in the orderings for both of those two individuals leaves the choice set unchanged. We show how the balancedness condition (satisfied by the Borda, Pareto, and Copeland rules) interacts with other social choice properties, especially tops-only. We also use balancedness to characterize the Borda rule (for a fixed number of voters) within the class of scoring rules. © 2019 Elsevier B.V. All rights reserved.
1. Introduction We consider a new condition, balancedness. A social choice correspondence satisfies balancedness if, for each pair of alternatives, x and y, and each pair of individuals, i and j, whenever a profile has x adjacent to but just above y for individual i while individual j has y adjacent to but just above x, then only switching x and y in the orderings for both of those two individuals leaves the choice set unchanged. Social choice theory often treats responsiveness conditions, like monotonicity, but balancedness is a non-responsiveness property. It is a natural equity condition that simultaneously incorporates some equal treatment for individuals short of anonymity, some equal treatment of alternatives short of neutrality, and some equal treatment of position of alternatives in orderings (for example, raising x just above y in the bottom two ranks for individual j exactly offsets lowering x just below y in the top two ranks for i). With only two alternatives, balancedness is equivalent to anonymity. With more than two alternatives, balancedness is a condition independent of anonymity. Any dictatorial correspondence, however, is unbalanced. For a dictatorial correspondence, one and only one individual is “effective”: changing preferences for every individual other than the dictator leaves the choice set unaffected. In addition to excluding dictatorship, we show that balancedness implies that any non-constant social ∗ Corresponding author. E-mail addresses: [email protected] (J.S. Kelly), [email protected] (S. Qi). https://doi.org/10.1016/j.mathsocsci.2019.08.003 0165-4896/© 2019 Elsevier B.V. All rights reserved.
choice correspondence must have every individual to be effective (Theorem 1). For the equal treatment of positions incorporated by balancedness, we present results from different perspectives. First, we consider the interaction of balancedness with a specific property, tops-only. By tops-only, any change that affects no one’s topranked option leaves the choice set unaffected. Plurality voting is an obvious tops-only rule. Unlike balancedness, tops-only clearly treats the top-rank differently from the remaining ones. The expected conflict between balancedness and tops-only is characterized through a kind of impossibility result (Theorem 2): whenever a social choice correspondence satisfies both balancedness and tops-only, it remains constant on all non-unanimous profiles (i.e., profiles without common option top-ranked by all individuals). We next examine the implication of balancedness within the class of scoring rules. A scoring rule assigns weights to ranks, lower weights to higher ranks, and the score for each option is the sum of weights corresponding to all ranks that option occupies in individual orderings. The Borda rule adds a constant value to weights from one rank to the next, and picks the alternatives with the lowest score. Balancedness holds for the Borda rule: the changes of scores from raising x just above y in the bottom two ranks for individual j is exactly offset by the changes of scores from lowering x just below y in the top two ranks for i, leaving the choice set unchanged. We show that except for a counterexample with small numbers of individuals and alternatives, balancedness actually uniquely characterizes the Borda rule within the class of scoring rules (Theorems 4–6).
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To be clear, we are not examining balancedness because we advocate it as a property that should hold for social choice correspondences. Rather we study balancedness because it is a common property of several correspondences that have been central to social choice theory. As noted, Borda satisfies balancedness. The Pareto correspondence, which at profile u selects those alternatives x such that no other alternative y that everyone prefers to x, also satisfies balancedness. The majority preference relation is unaffected by an interchange of any transposition pair. Accordingly, a social choice correspondence constructed from majority voting relations will also satisfy balancedness. This category includes examples like Copeland’s rule (which selects the alternatives that defeat the most other alternatives by simple majority vote) and the top-cycle correspondence (which selects the alternatives that are highest in the transitive closure of the majority voting relation). In fact, with more than two alternatives, anonymity, neutrality, the Pareto condition and balancedness simultaneously hold for the Pareto correspondence,1 the Borda rule, the Copeland rule, and top-cycle, while all those rules fail tops-only. On the other hand, anonymity, neutrality, the Pareto condition and tops-only simultaneously hold for plurality and union-of-the-tops, while balancedness fails for both of them. Dictatorship fails balancedness but not tops-only, and maximin fails both. Thus besides anonymity, balancedness is a condition also independent of unanimity, neutrality, the Pareto condition, and tops-only.2 Several properties related to balancedness appear in some recent work. Derya (2014) considers a variable-population setting that allows for individual indifference and introduces a “cancellation property”. Cancellation bears some similarity to balancedness. First, the definition of cancellation (p. 152–153) hinges critically on allowing for individual indifference (and is for a variable-population setting): it becomes meaningless if one only considers the domain with strict preferences (as we assume). To illustrate, consider a profile (starting with strict preferences) where x ranked above y by individual #1 and y above x by #2. Cancellation says that making x and y indifferent for #1 and y and x indifferent for #2 will not change the outcome. If we assume further that z is (originally) ranked above x for #1 and x above z for #2, then repeatedly applying cancellation to make x, y, z all indifferent for both #1 and #2 will not change the outcome. In other words, every time cancellation is applied, one actually enlarges an indifference class by bringing together two previously neighboring ones, yet can never separate an indifference class into two different ones. This implies, in particular, cancellation cannot be applied to generate a transposition pair through the process considered by the balancedness property.3 Moreover, although Derya provides a result that uses cancellation to characterize Borda rule within the class of scoring rule, the Borda rule there is a different version on the domain of weak preferences and for a variable-population setting, which is not equal to or stronger than our result in terms of either the result itself or the proof technique.4 1 For more on the definitions and discussions of these social choice rules and the properties mentioned in this paragraph, see Arrow et al. (2002) and Heckelman and Miller (2015). 2 The above examples imply that none of these properties imply balancedness. On the other hand, the following example satisfies balancedness but fails for all these properties. Let n = 2 and X = {x, y, z }. Let G pick (fixed alternative) y constantly except at profile u where individual 1 ranks x above y above z and individual 2 ranks y above z above x and G(u) = {z }. 3 Applying cancellation to each profile of a transposition pair to obtain the same profile with indifference, one can generate the conclusion of balancedness, and thus cancellation is (weakly) stronger than balancedness even adopting a fixed population. But again without allowing for indifference, cancellation becomes meaningless. 4 Derya also introduces a property, degree equality (for a variablepopulation), that is strictly stronger than cancellation, and thus is strictly stronger than balancedness (even adopting a fixed population).
Mihara (2017) introduces a property called “Positional Cancellation”, which requires that changes in the relative positions of two alternatives that cancel each other (e.g., the changes in the relative position of (x, y) if changing from xay to axy for #1 and changing from bxy to xby for #2) do not alter the social preference between the two. Mihara works with social welfare functions not social choice correspondences and allows for individual indifference. If adapted to the context of a social choice correspondence, (the modified) Positional Cancellation is a strictly stronger property than balancedness.5 Mihara uses the cancellation property, together with “reversal” (a neutrality property) and “positive responsiveness” (a monotonicity property), to characterize the Borda ranking among all ranking rules. Balancedness is also strictly weaker than “Invariance for Average-Position Preserving Reversals” (IAPPR) introduced by Sato (2017), which states that “if some voters reverse their preferences for two common alternatives, then the set of winners should be unchanged as long as this reversal preserves the average position of each alternative”.6 Sato characterizes the Borda correspondence through IAPPR, neutrality and positive responsiveness among all social choice correspondences, while we use balancedness alone to characterize the Borda correspondence but within the class of scoring rules.7 2. Definitions Let X with cardinality |X | = m ≥ 2 be the set of alternatives and let N = {1, 2, . . . , n} with n ≥ 2 be the set of individuals. A (strong) ordering on X is a complete, asymmetric, transitive relation on X (non-trivial individual indifference is disallowed). The highest ranked element of an ordering r is denoted r [1], the second highest is denoted r [2], etc. And r [1 : k] is the unordered set of alternatives in the top k ranks of r. The set of all orderings on X is L(X ). A profile u is an element (u(1), u(2), . . . , u(n)) of the Cartesian product L(X )N . A social choice correspondence G is a map from the domain L(X )N to non-empty subsets of X . The range of social choice correspondence G is the collection of all sets S such that there exists a profile u with G(u) = S. G satisfies tops-only if for each pair of profiles u, v , whenever u(i)[1] = v (i)[1] for each i, then G(u) = G(v ). We call a profile u unanimous if u(i)[1] = u(j)[1] for all individuals i, j. Then social choice correspondence G satisfies unanimity if G(u) = u(i)[1] at each unanimous profile u. At profile u, we say x is dominated by y if everyone prefers y to x. An alternative x is Pareto optimal at u if it is not dominated by any alternative y at u. Correspondence G satisfies the Pareto condition if, at every u, every alternative in G(u) is Pareto optimal at u. We say profile v is constructed from profile u by transposition pair (x, y) via individuals i and j if at u, x is immediately above y for i, and y is immediately above x for j, and profile v is just the same as u except that for i and j, alternatives x and y are transposed. A social choice correspondence G is called balanced if, for each x, y, u, v , i, and j, whenever profile v is constructed from u by transposition pair (x, y) via individuals i and j, then G(v ) = G(u). (Otherwise, G is called unbalanced.) 5 It is easy to exemplify that balancedness does not imply Positional Cancellation. For the other direction, that Positional Cancellation implies balancedness, indifference needs to be employed in the proof. 6 It is straightforward to show IAPPR implies balancedness. On the other hand, there are profiles that satisfy the condition of IAPPR and thus by IAPPR they should have the same outcome chosen, but there are no pair transpositions from the profiles and balancedness imposes no restriction on them. 7 Also, Mihara’s and Sato’s characterization results do not have an exceptional case for small numbers of m and n.
J.S. Kelly and S. Qi / Mathematical Social Sciences 102 (2019) 59–67
3. Results 3.1. Balancedness and (non-)dictatorship To see one way in which balancedness reflects equal treatment of individuals, first observe again that any dictatorial correspondence (where for some i and all u, G(u) = u(i)[1]) is unbalanced. This can be extended to cover all non-constant correspondences with ineffective individuals. An individual i is ineffective for a correspondence G if for each pair of profiles u, u∗ , we have G(u) = G(u∗ ) whenever u(j) = u∗ (j) for all j ̸ = i. (Otherwise i is called effective for G.) For example, if G is dictatorial with dictator j, then every other individual is ineffective. We show that for any non-constant correspondence, balancedness implies every individual is effective. Suppose individual #2 is ineffective for G but #1 is effective. If v is obtained from u by transposition pair (x, y) via individuals #1 and #2, the outcome of G is affected just as if y rose for #1 and nothing else happened. For a non-constant correspondence, it cannot generally be true that such moves leave outcomes unchanged as required by balancedness. Theorem 1. Any non-constant social choice correspondence with an ineffective individual is unbalanced. Proof. Let i be an ineffective individual for non-constant correspondence G. Let u, u∗ be two profiles with G(u) ̸ = G(u∗ ). Since i is ineffective, we may assume that u(i) = u∗ (i). Now construct a sequence of profiles u = u1 , u2 , . . . , uT − 1 , uT = u∗ from u to u∗ such that for any two successive profiles ut −1 , ut in the sequence, ut differs from ut −1 only by a transposition of two adjacent alternatives in the ordering of a single individual other than i. For some t, it must be that G(ut ) ̸ = G(ut −1 ). Suppose that ut differs from ut −1 because x is transposed with y, ranked just below x in ut −1 (j). Construct profile u′t −1 from ut −1 by moving y adjacent to and just above x in ut −1 (i), and construct profile u′t from ut by moving x adjacent to and just above y in ut −1 (i). Then, by the ineffectiveness of i, G(u′t ) = G(ut ) and G(u′t −1 ) = G(ut −1 ), so G(u′t ) ̸ = G(u′t −1 ). But u′t differs from u′t −1 by transposition pair (x, y) via individuals i and j and so this constitutes a violation of balancedness. □ In the next subsection, we show how the balancedness condition interacts with other social choice properties, especially tops-only. In the last subsection, we show how balancedness can be used to characterize the Borda rule (for a fixed number of voters) within the class of scoring rules. 3.2. Balancedness and tops-only Balancedness incorporates some equal treatment of positions, so we might expect conflict with the tops-only property. By topsonly, all profiles u with an alternative x at everyone’s top must give the same value for G(u), though it need not be the case that G(u) = {x}. Do there exist any social choice correspondences satisfying tops-only and unanimity that are balanced? A positive answer to that question is given by: Example 1. Let G(u) be the common top at all unanimous profiles and set G(u) = X (or any other fixed set) on all non-unanimous profiles.
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Intuitively, tops-only implies invariance to any moves that affect no one’s top; balancedness implies invariance to a transposition pair via corresponding individuals. The two properties combined together generate further invariances. For instance, suppose individual #1 has ranking xyz (i.e., x above y above z); #2 has ranking zyx. Tops-only and balancedness together imply that social choice outcomes remain constant if #1’s ranking becomes yxz while #2’s ranking stays the same: first through a transposition pair (x, y) via #1 and #2 (by balancedness), then switching x and y again for #2 (by tops-only). Theorem 2 repeatedly applies the same technique to prove that the conclusion holds for all profiles that are non-unanimous. Theorem 2. For m ≥ 3, n ≥ 3, if social choice correspondence G satisfies tops-only and balancedness, then it is constant on all non-unanimous profiles. Thus such a G has at most m + 1 sets in its range. Proof. Let a and b be two fixed elements of X and let u∗ be a fixed profile with tops abb...bb. Given any profile u without a unanimous top, we will show there is a sequence of profiles u1 , u2 , . . . , uT all without a unanimous top, such that: 1. u1 = u; 2. uT = u∗ ; 3. For each t with 1 ≤ t < T , ut +1 can be constructed from ut by either a reordering of alternatives below someone’s top or by a paired transposition. Then we would have G(u) = G(u∗ ) for all u without a unanimous top. In the following, let T (u) be the set of all alternatives x such that u(i)[1] = x for some i. Case 1. T (u) = {a, b}. If u(1)[1] = a, go to the next paragraph. Suppose that u(1)[1] = b. For some i > 1 with a in the top rank, construct u2 by raising a to 1’s second rank and b to i’s second rank. Then construct u3 by transposition pair (a, b) via individuals 1 and i. Now u3 (1)[1] = a and u3 (i)[1] = b. If all other tops are now b, we are done. So suppose that some other top, say for individual j, is a. Construct u4 by raising a third alternative c to the second rank for j and raising c just above a for i. Then construct u5 by transposition pair (a, c) via i and j. Now c is at j’s top. Recall that at u5 , we have u5 (1)[1] = a, so we can move b to the second rank for individual 1, c to the third rank for individual 1, and for individual j, we move b to the second rank. Construct the next profile in the sequence by transposition pair (b, c) via 1 and j. Then b now becomes j’s top; the a has been changed to a b. Repeat this until all “a”s (except for #1) have been changed first to “c”s and then to “b”s. Case 2. Suppose T (u) contains a, b, and other alternatives. If c is a top for someone, say i, construct u2 by raising b to the second rank for i and raise b just above c for an individual j who has a on top. Then a transposition pair (b, c) via i and j reduces by one the number of individuals without a or b on top. Continue in this fashion until you reach Case 1. Case 3. T (u) contains one of a and b but not the other. Suppose that u(i)[1] = a and u(j)[1] = c. Construct u2 by raising b to second rank for j and raise b just above c for i. Then a transposition of b and c for i and j creates a profile in either Case 2 or Case 1. A similar analysis holds if it is b instead of a at someone’s top. Case 4. T (u) contains neither of a or b but does contain say c and d. Construct profile u2 by raising a to second rank for some i with c on top and a just above c for some j with d on top. A transposition pair (a, c) via i and j yields a profile in Case 3. □
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A modified version of this argument can also be constructed for n = 2. But there is a limit to this style of argument; there are not sequences to u∗ from profiles with unanimous tops as balancedness cannot be applied there. This limit on the argument cannot be overcome, as seen by the example at the beginning of this section; there, at profiles with a common top, different outcomes may occur. When there are at least three alternatives, for every option, there exist some non-unanimous profiles at which the option is Pareto dominated. Thus, if a social choice correspondence remains constant on all non-unanimous profiles, it must violate the Pareto condition. This leads to the next Corollary. Corollary 1. For m ≥ 3, if social choice correspondence G satisfies tops-only and Pareto, then G is unbalanced. Proof. The corollary is a straightforward application of Theorem 2 given that if G satisfies Pareto then G cannot be constant on all non-unanimous profiles. □ Both tops-only and Pareto are needed in the Corollary. If Pareto is not assumed, a constant correspondence satisfies both tops-only and balancedness. If tops-only is not assumed, the correspondence that selects the set of Pareto optimal alternatives satisfies Pareto and balancedness. Theorem 2 can be extended to similar results if we relax tops-only to similar properties where some higher ranks (not necessarily the top-rank) are treated differently from lower ones. The following social choice correspondence violates tops-only, but outcomes do depend only on which alternatives are ranked first or second by individuals (but not on how the top alternatives are ordered within those two ranks). Example 2. (Rigid8 ) 2-approval voting is a social choice correspondence G that is like plurality rule except that instead of selecting the alternatives with the most frequently occurring tops, it selects the alternatives with the most frequent occurrences in the top two ranks for everyone. Like plurality rule, 2-approval voting is unbalanced. We now extend Theorem 2 to cover social choice correspondences that, like 2-approval voting, depend only on the top two ranks for every individual. We first need to extend tops-only. Let u(i)[1 : 2] denote the (unordered) set of alternatives in the top two ranks for individual i at profile u. We say social choice correspondence G satisfies top-2-only if for each pair of profiles u, u∗ , G(u) = G(u∗ ) if u(i)[1 : 2] = u∗ (i)[1 : 2] for all individuals i. Let D be the subdomain of L(X )N consisting of all profiles for which it is not true that u(i)[1 : 2] = u(j)[1 : 2] for each pair of individuals i and j, that is, at least three alternatives occur in the top two ranks over all individuals. D is the analog here of the subdomain of non-unanimous profiles in Section 3.2. Theorem 3. Let m ≥ 4 and n ≥ 3. Then any social choice correspondence G satisfying balancedness and top-2-only must be constant on D. Proof. Since the argument will be divided into different cases, we mention the main idea here and leave some details to Appendix. Consider a specific profile u∗ with c in everyone’s top rank, a in #1’s second rank, and b in everyone else’s second rank. It will 8 “Rigid” because the number of ranks (here 2) is fixed to be the same for all individuals. For approval voting without rigidity, see Brams and Fishburn (1983). Rigid k-approval voting appeared in Alemente et al. (2016) where it was called approval voting (type-k).
suffice to show that for each profile u in D, there is a sequence of profiles in D from u to u∗ such that each is obtained from the previous profile by a transposition pair or an application of top-2-only, i.e., by a re-ordering below the top two ranks. We argue by induction on the number p of individuals who, at u, do not have c in their top two ranks. Basis: Suppose that we have p = 0 at u, i.e., every individual has c in their top two ranks. Construct u′ from u by (only) raising c to everyone’s top rank. Now consider the subdomain D′ of D consisting of all profiles in D where everyone has c top-ranked. Note that for each such profile at least two distinct alternatives must be second-ranked. Let G′ be the restriction of G to D′ . Social choice correspondence G′ on D′ induces a correspondence on those profiles on X \{c } which have non-unanimous tops. Then the analysis in the tops-only section shows that there is a sequence of profiles in D′ from u′ to u∗ . Induction step: Assume now that there is a non-empty set S of alternatives such that for all profiles v in D with the number of individuals who do not have c in their top two ranks being less than p we have the same outcome: G(v ) = S. Suppose that u in D is a profile where the number of individuals who at u do not have c in their top two ranks is p > 0. We will show there is a sequence of profiles (such that each is obtained from the previous profile by a transposition pair or an application of top-2-only) from u to a profile u′ in D with p − 1 individuals who do not have c in the top two ranks (so G(u′ ) = S). The analysis must consider different cases for the profile u, and details are in Appendix. □ There are straightforward generalizations to top-3-only, etc.9 3.3. Balancedness and Borda characterization Balancedness is an equity condition that incorporates some equal treatment for differences of position of alternatives in orderings. As observed at the beginning of this paper, raising x just above y in the bottom two ranks for individual j exactly offsets lowering x just below y in the top two ranks for i. This equal treatment of differences of position suggests trying to characterize the Borda rule within the class of scoring rules. Let a scoring system be given by weights s1 ≤ s2 ≤ s3 ≤ · · · ≤ sm (higher ranked alternatives will be assigned lower weights). At profile u, the score for an alternative x is the sum S(x, u) =
n ∑
s(u, i, x)
i=1
where s(u, i, x) = sk if u(i)[k] = x, i.e., the score for x is the sum of the weights corresponding to the ranks that x occupies in the individual orderings at u. The related scoring social choice correspondence G selects at u the alternatives x with lowest S(x, u) values. If s1 = s2 = s3 = · · · = sm then G is the constant social choice correspondence with G(u) = X at all u, which is not very helpful. Accordingly, we henceforth only consider systems of weights such that at least two weights are distinct. 9 For instance, for top-3-only, the analog of the subdomain of non-unanimous profiles consists of all profiles with at least four alternatives occurring in the top three ranks over all individuals. The specific profile u∗ has c in everyone’s top rank, d in everyone’s second rank, a in #1’s third rank and b in everyone else’s third rank. The proof proceeds by induction on the number p of individuals who, at u, do not have c in their top three ranks. The basis step (p = 0) follows from top-2, and the induction step also follows the same case division as those for top-2.
J.S. Kelly and S. Qi / Mathematical Social Sciences 102 (2019) 59–67
The Borda correspondence, GB , uses weights 1 < 2 < 3 < · · · < m. But of course other weights also generate Borda. If social
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3.3.2. Borda characterization for n > 3 We begin by looking at m = 3.
choice correspondence G is generated by weights s1 ≤ s2 ≤ s3 ≤ · · · ≤ sm then G is also generated by linearly transformed weights t1 ≤ t2 ≤ t3 ≤ · · · ≤ tm where ti = α + β si for real numbers α , β , with β > 0. Lemma 1. For m ≥ 3 and n ≥ 2: if G is a scoring social choice correspondence for scoring system given by
Theorem 4. For m = 3 and n ≥ 4: if G is a scoring social choice correspondence and G satisfies balancedness, then G is the Borda correspondence. Proof. We separate the analysis into two cases, n is even and n is odd. (n is even) We examine the following profile v :
s1 ≤ s2 ≤ s3 ≤ · · · ≤ sm with at least two weights distinct and if G satisfies balancedness, then s1 ̸ = s2 . Proof. If s1 = s2 , then there exists a k, 2 ≤ k < m, such that s1 = s2 = · · · = sk < sk+1 . Construct profile u such that u(i)[1] = y and u(i)[2] = x for all i < n while u(n)[k] = x and u(n)[k + 1] = y. Then x ∈ G(u) and y ∈ / G(u). If profile u∗ is constructed from u by transposition pair (x, y) via individuals 1 and n, then x ∈ / G(u∗ ) and y ∈ G(u∗ ). So G(u∗ ) ̸= G(u) and G fails balancedness. □ As a consequence of Lemma 1, we assume from now on that the weights have been linearly transformed so that s1 = 1 and s2 = 2. For profiles of strong orderings, we want to show that generally if a scoring social choice correspondence satisfies balancedness, it must be the Borda rule. However, the next subsection shows a limitation on this objective. 3.3.1. A counterexample when m = n = 3 There is an exception to our general result that any balanced scoring rule must be the Borda rule. When m = n = 3, the profiles where there are no transposition pairs, and so where the balancedness condition has no force, are exactly the voting paradox profiles, displaying unusually great symmetry. For example, all other profiles have a majority voting winner. The profiles where balancedness does have force have just enough asymmetry to allow for scoring rules different from Borda. Example 3. Let m = n = 3 and set scoring weights to be 1, 2, and 3.1. The related correspondence, G, differs from Borda. At profile u: 1 x y z
2 x y z
1
2
3
4
n−1
n
x y z
z y x
x y z
z y x
x y z
z y x
···
Consider a scoring rule G with weights: 1 < 2 ≤ s3 (using Lemma 1). We examine the scores for x, y, z: For x, 2n (1 + s3 ); For z, 2n (1 + s3 ); For y, 2n. Under this scoring rule, either x and z are both in G(v ) or neither x nor z is in G(v ). Now we consider a sequence of transposition pairs (x, y) first via individuals 1 and 2, and then via 3 and 4, . . . , and so on until via (n-1) and n, which yields the profile u: 1
2
3
4
y x z
z x y
y x z
z x y
···
n−1
n
y x z
z x y
We examine again the scores for x, y, z: For x, 2n; For y, 2n (1 + s3 ); For z, 2n (1 + s3 ). This time, either y and z are both in G(u) or neither y nor z is in G(u). But u is obtained from v by a sequence of transposition pairs, and balancedness implies G(u) = G(v ), which means both must be {x, y, z }, which, in turn, means at each profile, all three scores must be the same. Therefore, 2n (1 + s3 ) = 2n, and thus s3 = 3, i.e., G is the Borda correspondence. (n is odd) We examine the following profile v (where the construction of the first n − 3 profiles is the same as the case of n even and the last three profiles is a voting paradox profile):
3
1
2
n−4
n−3
n−2
n−1
n
y z x
x y z
z y x
x y z
z y x
y x z
x z y
z y x
the Borda rule has a tie between x and y, and z is Paretodominated by y so GB (u) = {x, y}. But with scoring weights 1, 2, and 3.1, the scores of x and y are 5.1 and 5 respectively: G(u) = {y}. G ̸ = GB . Nevertheless, G here is balanced, as can be checked.10 So, for m = n = 3, balancedness of a scoring rule does not imply Borda. This use of small numbers of individuals and alternatives is critical in Example 3, as we will see. 10 Because all scoring rules are neutral and anonymous, to check balancedness, we may assume that the pair interchanged is {x, y} and that x is just above y for #1 and just below y for #2, and the interchange is for those two individuals. Without loss of generality, we assume #3 prefers x to y (although x and y need not to be adjacent in u(3)). There are 12 profiles satisfying those conditions. But for six of those profiles, either z is on top for both #1 and #2 or on bottom for both, and it is easy to see that balancedness is satisfied. So we need only consider the remaining six profiles and the details are omitted.
···
Consider a scoring rule G with 1 < 2 ≤ s3 (using Lemma 1). We examine the scores for x, y, z: 1 For x, n− (1 + s3 ) + 2; 2 n−1 For z, 2 (1 + s3 ) + 2; For y, 2n − 3 + s3 . So under this scoring rule, either x and z are both in G(v ) or neither x nor z are in G(v ). Now we consider a sequence of transposition pairs (x, y) first via individuals 1 and 2, and then via 3 and 4, . . . , and so on until via (n − 4) and (n − 3) (so the voting paradox of the last three individuals remains unchanged), which yields the profile u: 1
2
y x z
z x y
···
n−4
n−3
n−2
n−1
n
y x z
z x y
y x z
x z y
z y x
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J.S. Kelly and S. Qi / Mathematical Social Sciences 102 (2019) 59–67
We examine again the scores for x, y, z: For x, 2n − 3 + s3 ; 1 (1 + s3 ) + 2; For y, n− 2 n−1 For z, 2 (1 + s3 ) + 2. This time, either y and z are both in G(u) or both are out of G(u). But by balancedness, G(u) = G(v ), so at each profile all scores must be the same. Therefore n−1 2n − 3 + s3 = (1 + s3 ) + 2 2 which has the unique solution s3 =
3n−9 n−3
Clearly GB (v ) = X . For the correspondence G with weights 1, 2, 3, . . . , m − 1, w > m, the scores are 2n (1 + w ) for x and z, 2n (m + 1) for the rest. Because w > m, we have G(v ) = X \{x, z }. Construct u by transposing x and y first via individuals 1 and 2, and then via 3 and 4, . . . , and so on until via (n-1) and n, which yields the profile u:
= 3, (note that we have
n ≥ 4), i.e., G is the Borda correspondence.11
□
Proof. For m = 3, the proof is given by Theorem 4. For m ≥ 4, suppose that we have linearly transformed weights 1 < 2 ≤ s3 ≤ · · · ≤ sm such that for some j ≥ 3, it is the case that sj ̸ = j. Let G be the corresponding scoring social choice correspondence. We show that there is a profile v such that G(v ) ̸ = GB (v ) and that G must fail balancedness. Consider the smallest integer j such that sj ̸ = j. In what follows, we first assume that the smallest integer j with sj ̸ = j is that j = m. We then extend the analysis to the case where j < m. When j = m, the weights are: 1 < 2 < 3 < · · · < m − 1 ≤ w where w ̸ = m. Analysis here is for w > m. The same profile applies to the case m − 1 ≤ w < m. Again, we separate the analysis for even n and odd n, both of which extend the proof of Theorem 4. And since the extension follows the very same spirit, we treat even n here with details, and briefly comment on the case for odd n at the end. (n is even) For even n in Theorem 4 (where m = 3), we previously looked at profile: 3
4
x y z
z y x
x y z
z y x
···
n−1
n
x y z
z y x
Now to obtain v , we insert additional alternatives repeatedly: at each stage of the construction, we add another alternative, placing it just above z for odd numbered individuals, just below z for even numbered individuals; any two neighboring profiles are reverse orderings: 1
2
n−1
n
x y
z am−3
x y
z am−3
a2
a1 a2
a2
a1 a2
.. .
am−3 z
.. .
a1 y x
···
.. .
am−3 z
n−1
n
y x
z am−3
y x
z am−3
a2
a1 a2
a2
.. .
.. .
Theorem 5. For m ≥ 3 and n ≥ 4: if G is a scoring social choice correspondence and G satisfies balancedness, then G is the Borda correspondence.
2
2
a1 a2
We next take up larger m.
1
1
.. .
a1 y x
11 This proof provides insight into why Example 3 works. That example starts with a profile that has no transposition pairs.
am−3 z
···
.. .
.. .
a1 x y
am−3 z
a1 x y
This yields scores 2n (1 + w ) for y and z, 2n (m + 1) for the rest. Again since w > m, we now have G(u) = X \{y, z }, a failure of balancedness. (n is odd) Similar to the case for even n, we construct the profile v based on the one in Theorem 4: 1
2
n−4
n−3
n−2
n−1
n
x y
z am−3
x y
z am−3
y x
x z
z y
a2
a1 a2
a2
z a1
y a1
x a1
a1
a2
y x
.. .
.. .
.. .
am−3
am−3
am−3
a1 a2
.. .
am−3 z
.. .
···
a1
.. .
y x
am−3 z
.. .
a2
a2
It is easy to check that GB (v ) = {x, y, z } by noting that for the first n-3 individuals any two neighboring profiles are reverse orderings. For the correspondence G with weights 1, 2, 3, . . . , m− 3 3 (1 +w ) + 6 for x and z, n− (1 + m) + 6 1, w > m, the scores are n− 2 2 for y, and the rest a1 , a2 , . . . have higher scores. Because w > m, G(v ) = {y}. Construct u by transposing x and y for 1 and 2, and then via 3 and 4, . . . , and so on until via (n − 4) and (n − 3), which yields the profile u. And it is easy to check that at u, the scores are n−3 3 (1 +w ) + 6 for y and z, n− (1 + m) + 6 for x, with the remaining 2 2 scores unchanged (and higher). Because w > m, G(u) = {x}, a failure of balancedness. Finally, we extend the above analysis to the case where the smallest integer j with sj ̸ = j is that j < m. The weights are: 1 < 2 < · · · < j − 1 ≤ sj ≤ · · · ≤ sm where sj ̸ = j. We look at the restriction of G to the domain D where (m − j) alternatives are set at everyone’s bottom in a fixed order. G restricted to this domain D induces a scoring social choice correspondence G′ on j alternatives with weights 1, 2, . . . , (j − 1), sj and sj ̸ = j (while clearly G′ also satisfies balancedness). And this violates the above analysis (for the case where the smallest integer j with sj ̸ = j is that j = m). □ 3.3.3. Borda characterization for n = 3, m > 3 Let us return to the case n = 3, where we learned in Example 1 that for m = 3 not every scoring social choice correspondence satisfying balancedness is the Borda rule. For the case m = 4 and n = 3, a natural analog of Example 3 has scoring weights 1, 2, 3,
J.S. Kelly and S. Qi / Mathematical Social Sciences 102 (2019) 59–67
and 4.1. But that rule is unbalanced, as can be seen at profile u: 1
2
3
a y x b
b a x y
x y b a
Here the scores are S(a) = 7.1, S(b) = 8.1, S(x) = 7, S(y) = 8.1. The smallest of these scores is 7 and so G(u) = {x}. If v is constructed from profile u by transposition pair (x, y) via individuals 1 and 2, then the scores become S(a) = 7.1, S(b) = 8.1, S(x) = 7.1, S(y) = 8. So G(v ) = {x, a}, a failure of balancedness. Theorem 6. For m > 3 and n = 3: if G is a scoring social choice correspondence and G satisfies balancedness, then G is the Borda correspondence. Proof. The proof follows a similar pattern of Theorem 5’s proof. Starting with m = 4, new alternatives are inserted into a profile to show unbalancedness of a scoring rule with weights different from those of the Borda rule. Assume first m = 4. Let us consider a scoring rule G but not the Borda correspondence with linearly transformed weights: 1 < 2 ≤ s3 ≤ s4 . We consider the following profile u: 1
2
3
a y x b
b a x y
x y b a
2
3
a x y b
b a y x
x y b a
We examine the scores for a, b, x, y: For a, 3 + s4 ; For x, 3 + s4 ; For b, 1 + s3 + s4 ; For y, 2 + 2s3 . By balancedness, G(u) = G(v ). Since at profile v , alternatives a and x have the same score, and at u, at least a ∈ G(u) or x ∈ G(u), we have a and x are both in G(u) and G(v ). So (from u), S(a) = 3 + s4 = S(x) = 1 + 2s3 , i.e., 2s3 − s4 = 2
(1)
Again, consider a transposition pair (x, y) via individual 2 and 3 that yields the profile v ′ : 1
2
3
a x y b
b a x y
y x b a
We examine the scores for a, b, x, y: For a, 3 + s4 ; For x, 4 + s3 ; For b, 1 + s3 + s4 ; For y, 1 + s3 + s4 By balancedness, a, x ∈ G(v ′ ), so 4 + s3 = 3 + s4 , or s3 − s4 = −1
(2)
Eqs. (1) and (2) have the unique solution s3 = 3, s4 = 4, i.e., G is the Borda correspondence. Now we extend the analysis to m > 4. Suppose we have weights 1 < 2 ≤ s3 ≤ · · · ≤ sm , and consider the smallest integer j such that sj ̸ = j. We need to consider the case where j = m and where j < m. But similar to the proof of Theorem 5, the case where j < m can be implied by the analysis for j = m by looking at the restriction of G to the domain D where (m − j) alternatives are set at everyone’s bottom in a fixed order, which induces a scoring social choice correspondence on j alternatives. And for the case where j = m, we extend the above profile by inserting new alternatives to obtain a profile vˆ such that G(vˆ ) ̸ = GB (vˆ ) and show that G must fail balancedness. Thus, consider the linearly transformed weights: 1 < 2 < 3 < · · · < m − 1 ≤ w where w ̸ = m. Analysis here is for w > m. The same profile works for w < m. We first show one stage. Look at profile v above. Now insert alternative c just above b for #1 and just below b for #2. The third individual has c at the bottom, and we obtain vˆ :
We examine the scores for a, b, x, y: For a, 3 + s4 ; For x, 1 + 2s3 ; For b, 1 + s3 + s4 ; For y, 4 + s4 . Since S(a) < S(y) and S(x) ≤ S(b) under this scoring rule, at least a ∈ G(u) or x ∈ G(u) (or both). Now we consider a transposition pair (x, y) via individual 1 and 2 that yields the profile v : 1
65
1
2
3
a x y c b
b c a y x
x y b a c
The Borda scores for a and x are the lowest, so GB (vˆ ) = {a, x}. With weights 1, 2, 3, 4, w5 , with w5 > 5, the scores at vˆ are: S(a) = 8, S(b) = 4+w5 , S(c) = 6+w5 , S(x) = 3+w5 , and S(y) = 9, so G(vˆ ) = {a}. After transposing x and y for individuals 1 and 2 to create profile vˆ ′ , the scores become: S(a) = 8, S(b) = 4 + w5 , S(c) = 6 + w5 , S(x) = 8, and S(y) = 4 + w5 , so G(vˆ ′ ) = {a, x} and balancedness is violated. Eventually the profile looks like vˆ ∗ : 1
2
3
a x
b cm−4
x y
c2 c1
b a c1
y c1 c2
.. .
cm−4 b
.. .
a y x
c2
.. .
cm−4
The Borda scores for a and x are the lowest, so GB (vˆ ∗ ) = {a, x}. With weights 1, 2, . . . , m − 1, w , where w > m, the scores at vˆ ∗ are: S(a) = m + 3, S(b) = w + 4, S(x) = w + 3, S(y) = m + 4, and the rest have higher scores. So G(vˆ ∗ ) = {a} given w > m. After transposing x and y for individuals 1 and 2 to create profile vˆ ∗′ ,
66
J.S. Kelly and S. Qi / Mathematical Social Sciences 102 (2019) 59–67
the scores become: S(a) = m + 3, S(b) = w + 4, S(x) = m + 3, S(y) = w+4, and the remaining scores are unchanged, so G(vˆ ∗′ ) = {a, x} and balancedness is violated. □
1
4. Conclusion
.. .
c x
We have introduced a new condition, balancedness. Balancedness holds for several correspondences that have been central to social choice theory, such as Borda, Pareto, and Copeland rules. We show that the interaction of balancedness with tops-only together imply invariance of outcomes over most of the domain. We also obtain a new characterization of the Borda correspondence within the class of scoring rules. Hence we can get a characterization of Borda’s rule with a fixed set of individuals by appending balancedness to a characterization of scoring rules for fixed populations (see Fishburn, 1973a,b). While Young (1974), Hansson and Sahlquist (1976), Coughlin (1979–1980), Nitzan and Rubinstein (1981), and Debord (1992) have characterizations of Borda’s rule, they work in a different context than ours; for these authors, a rule has to work for a variable number of individuals and they use variable population properties like that of separability introduced by Smith (1973).
We complete the proof for Theorem 3 by including the details omitted in Section 2. We will not repeat the basis step already in Section 2 and start from the induction step below. Induction step: Assume now that there is a non-empty set S of alternatives such that for all profiles v in D with the number of individuals who do not have c in their top two ranks being less than p we have the same outcome: G(v ) = S. Suppose that u in D is a profile where the number of individuals who at u do not have c in their top two ranks is p > 0. We will show there is a sequence of profiles (such that each is obtained from the previous profile by a transposition pair or an application of top-2-only) from u to a profile u′ in D with p − 1 individuals who do not have c in the top two ranks (so G(u′ ) = S). Without loss of generality suppose that it is the first q = n − p individuals with c ranked in the top two ranks. We may assume that c has been raised to the top for each of those q = n − p individuals so that u is: 2
c
c
.. .
···
.. .
c x
.. .
c
.. .
···
1
2
c
c
t
.. .
1
2
c
c
t
t
.. .
x
w .. .
q
q+1
c
x
z
y
w .. .
···
.. .
n
w .. .
.. .
s
y
w .. .
t
n
···
q
q+1
c
x
z
y
w .. .
.. .
t
.. .
···
.. .
n
Construct: 1
2
c
c
s
n
t
x
z
y
y
w .. .
Then if necessary, lower x to the second rank for q + 1 and raise c to the third rank for that individual.
z
.. .
···
Subsubcase 2Bi. p < n − 1 (and so q > 1).
y
···
q+1 x
Subcase 2B. No element in the top two ranks for q + 1, . . . , n is second for any of 1, . . . , q, and no two of 1, . . . , q have the same second element. (Recall p = n − q.)
c
.. .
q c
Now a transposition pair (t , y) via individuals 2 and q + 1 yields a profile in D still with p individuals with c not in the top two ranks but now back in Case 1 (since t is in the top two ranks for individual q + 1 and is ranked second for individual #1).
2
···
···
y
c
.. .
z
.. .
t
1
.. .
···
.. .
z
c
y
.. .
n
Subcase 2A. There are two individuals among 1, . . . , q with the same alternative t ranked second:
n
q+1
c
···
Case 2. None of the alternatives in the top two ranks for q + 1, . . . , n is ranked second by one of 1, . . . , q.
q+1
q
q+1
Now a transposition pair (c , x) via individuals 1 and q + 1 yields a profile in D with q + 1 individuals with c in the top two ranks and so only p − 1 individuals with c not in the top two ranks.
x
Case 1. At least one of the alternatives, say x, in the top two ranks for q+1, . . . , n is ranked second by one of 1, . . . , q. Without loss of generality, let q + 1 be an individual with x in the top two and let 1 be the individual with c on top and x in the second rank: 2
q
c
q
Here x, y, z and w are distinct from c (though these four need not all be distinct from one another).
1
c
.. .
c
.. .
···
For one of those individuals, say #2, move y, an alternative in u(q + 1)[1 : 2] (moved for q + 1 to second rank if necessary) up to the third rank and then raise t to the third rank for q + 1:
Appendix. Omitted proof for Theorem 3
1
2
t
.. .
···
q
q+1
c
x
z
y
w .. .
.. .
t
.. .
···
n
where t is 2’s second and y is in q + 1’s top two (moved to second rank if necessary). Now a transposition pair (t , y) via 1 and q + 1 takes us back to Case 1 (since t will be in individual q + 1’s top two and in #2’s second rank).
J.S. Kelly and S. Qi / Mathematical Social Sciences 102 (2019) 59–67
Subsubcase 2Bii. p = n − 1 (so q = 1). 1
2
c s
x y
.. .
···
say #2. Move z to second rank for #2, n
1
z
x y
w .. .
.. .
If not all of 2, . . . , n have the same top two alternatives, say n has z in the top two but 2 does not, move z to second place for individual n and raise c to n’s third rank while raising c and z to third and fourth rank for #2: 1
2
c s
x y
.. .
···
n
c
Now a transposition pair (c , z) via 2 and n takes us back to p = n − 2 in D. On the other hand if 2, . . . , n all have the same two alternatives in the top two ranks: 1
2
c s
x y
···
n x y
.. .
.. .
Raise y to third rank for #1 and s to third rank for #2 1
2
c s
x y
y
s
.. .
.. .
···
n x y
.. .
Now a transposition pair (s, y) via individuals 1 and 2 takes us back (in D) to Case 1. Case 3. p = n (so q = 0). Because we are in D, there must be an alternative z, distinct from x and y (#1’s top two), in the top two ranks for someone,
···
n
z
.. .
1
2
x y c
z c
z
.. .
2
Then raise c and z to ranks three and four for #1 and c to third rank for #2:
z
z
.. .
.. .
w
c
67
···
n
.. .
Now transposition pair (c , z) via individual 1 and 2 takes us back (in D) to Case 2Bii. □
References Alemente, F., Campbell, D.E., Kelly, J.S., 2016. Characterizing the resolute part of monotonic social choice correspondences. Econom. Theory 62, 765–783. Arrow, K.J., Sen, A.K., Suzumura, K. (Eds.), 2002. Handbook of Social Choice and Welfare. Elsevier, Amsterdam. Brams, S.J., Fishburn, P.C., 1983. Approval Voting. Birkhauser, Boston. Coughlin, P.J., 1979–1980. A direct characterization of Black’s first Borda count. Econom. Lett. 4, 131–133. Debord, B., 1992. An axiomatic characterization of Borda’s k-choice function. Soc. Choice Welf. 9, 337–343. Derya, A.M., 2014. Essays in Collective Decision Making (Ph.D. dissertation). Bilkent University. Fishburn, P.C., 1973a. Summation social welfare functions. Econometrica 41, 1183–1196. Fishburn, P.C., 1973b. The Theory of Social Choice. Princeton University Press, Princeton. Hansson, B., Sahlquist, H., 1976. A proof technique for social choice with variable electorate. J. Econom. Theory 13, 193–200. Heckelman, J.C., Miller, N.R. (Eds.), 2015. Handbook of Social Choice and Voting. Elgar, Cheltenham. Mihara, H.R., 2017. Characterizing the Borda ranking rule for a fixed population, Working paper, https://mpra.ub.uni-muenchen.de/78093/. Nitzan, S.I., Rubinstein, A., 1981. A further characterization of Borda ranking method. Publ. Choice 36, 153–158. Sato, N., 2017. A simple characterization of Borda rule for fixed electorate, Working paper, https://ssrn.com/abstract=2917013. Smith, J.H., 1973. Aggregation of preferences with variable electorate. Econometrica 41, 1027–1041. Young, H.P., 1974. An axiomatization of Borda’s rule. J. Econom. Theory 9, 43–52.