# Balancedness of Social Choice Correspondences

A social choice correspondence satisfies balancedness if, for every pair of alternatives, x and y, and every 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 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.

## Authors

• 1 publication
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## 1. Introduction

We consider the interaction of several social choice properties with a new condition, balancedness. A social choice correspondence satisfies balancedness if, for every pair of alternatives, and , and every pair of individuals, and , whenever a profile has adjacent to but just above for individual while individual has adjacent to but just above , then only switching and in the orderings for both of those two individuals leaves the choice set unchanged.

Social choice theory often considers 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 for differences of position of alternatives in orderings (for example, raising just above in the bottom two ranks for individual exactly offsets lowering just below in the top two ranks for ).

Let with cardinality be the set of alternatives and let with be the set of individuals. A (strong) ordering on is a complete, asymmetric, transitive relation on (non-trivial individual indifference is disallowed). The highest ranked element of an ordering is denoted , the second highest is denoted , etc.  And  is the unordered set of alternatives in the top ranks of .  The set of all orderings on is . A profile is an element of the Cartesian product .

A social choice correspondence is a map from the domain  to non-empty subsets of . The range of social choice correspondence is the collection of all sets such that there exists a profile with .

satisfies tops-only if for all profiles , , whenever for all , then .

We say profile is constructed from profile  by transposition pair  via individuals  and if at , is immediately above for , and is immediately above for , and profile is just the same as except that for and , alternatives and are transposed. A social choice correspondence will be called balanced if, for all , , , , , and , whenever profile is constructed from by transposition pair via individuals and , then . (Otherwise, will be called unbalanced).

If , balancedness is equivalent to anonymity. For , balancedness holds for the Pareto correspondence111For definitions and discussions of these social choice rules and the properties mentioned in this paragraph, see Arrow, Sen, and Suzumura (2002) and Heckelman and Miller (2015)., the Borda rule, the Copeland rule, and top-cycle (which selects the maximal set of the transitive closure of simple majority voting). (Thus balancedness is consistent with anonymity, neutrality, and the Pareto condition.) Those rules all fail tops-only. Balancedness fails for the tops-only correspondences that are dictatorship, plurality, and union-of-the-tops. (Thus unbalancedness is also consistent with unanimity, anonymity, neutrality, and the Pareto condition.)  In fact, balancedness fails for almost all tops-only correspondences as Section 1 will show. Balancedness also fails for the maximin rule.

To see one way balancedness reflects equal treatment of individuals, first observe again that any dictatorial correspondence (where for some and all , ) is unbalanced.  This can be extended to cover all correspondences with ineffective individuals.  An individual is ineffective for a correspondence if for all profiles , , we have whenever for all .  For example, if is dictatorial with dictator , then every other individual is ineffective. We show that balancedness implies every individual is effective.

Theorem 1.  Any non-constant social choice correspondence with an ineffective individual is unbalanced.

Proof:  Let be an ineffective individual for non-constant correspondence .  Let , be two profiles with .  Since is ineffective, we may assume that .  Now construct a sequence of profiles

 u=u1,u2,...,uT−1,uT=u∗

from to such that for any two successive profiles , in the sequence, differs from only by a transposition of two adjacent alternatives in the ordering of a single individual other than .  For some , it must be that .  Suppose that differs from because is transposed with , ranked just below in .

Construct profile from by moving adjacent to and just above in , and construct profile from by moving adjacent to and just above in . Then, by the ineffectiveness of , ( and , so .  But differs from by transposition pair via individuals and and so this constitutes a violation of balancedness.

In the first half of this paper, we show how the balancedness condition interacts with other social choice properties, especially tops-only.  In the second half, we show how balancedness can be used to characterize the Borda rule (for a fixed number of voters) within the class of scoring rules.

## 2. Tops-only

Balancedness incorporates some equal treatment of positions, so we might expect conflict with the tops-only property.  By tops-only, all profiles with an alternative at everyone’s top must give the same value for , though it need not be the case that .  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 be the common top at all unanimous profiles and set (or any other fixed set) on all non-unanimous profiles.

That example illustrates the following result.

Theorem 2.  For , , if social choice correspondence satisfies tops-only and balancedness, then it is constant on all non-unanimous profiles.

Thus such has at most sets in its range.

Proof:  Let and be two fixed elements of and let be a profile with tops .  Given any profile without a unanimous top, we will show there is a sequence of profiles all without a unanimous top, such that:

1.  ;

2.  ;

3.  For all with , can be constructed from by either a reordering of alternatives below someone’s top or by a paired transposition.

Then for all without a unanimous top.

In the following, let be the set of all alternatives such that for some .

Case 1.  .  If , go to the next paragraph.  Suppose that . For some with in the top rank, construct by raising to 1’s second rank and to ’s second rank.  Then construct by transposition pair via individuals 1 and .  Now and .

If all other tops are now , we are done.  So suppose that some other top, say for individual , is .  Construct by raising a third alternative to the second rank for and raising just above for .  Then construct by transposition pair via and . Now is at ’s top. Recall that at , we have , so we can move to the second rank for individual 1, to the third rank for individual 1, and for individual , we move to the second rank. Construct the next profile in the sequence by transposition pair via 1 and . Then now becomes ’s top; the has been changed to a . Repeat this until all ””s have been changed first to ””s and then to ””s.

Case 2.  Suppose contains , , and other alternatives.  If is a top for someone, say , construct by raising to the second rank for and raise just above for an individual who has on top.  Then a transposition pair via and reduces by one the number of individuals without or on top.  Continue in this fashion until you reach Case 1.

Case 3.   contains one of and but not the other.  Suppose that and .  Construct by raising to second rank for and raise just above for .  Then a transposition of and for and creates a profile in either Case 2 or Case 1.  A similar analysis holds if it is instead of at someone’s top.

Case 4.   contains neither of or but does contain say and .  Construct profile by raising to second rank for some with on top and just above for some with on top.  A transposition pair via and yields a profile in Case 3.

A modified version of this argument works also for .

But there is a limit to this style of argument; there are not sequences to from profiles with unanimous tops as balancedness cannot be applied there. This limit on the argument can not be overcome, as seen by the example at the beginning of this section; there, at profiles with a common top, different outcomes may occur.

Corollary.  For , if social choice correspondence satisfies tops-only and Pareto, then is unbalanced.

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.

## 3. Top-2

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.  (Rigid222”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 Alemante, Campbell, and Kelly (2015) where it was called approval voting (type-k).) -approval voting is a social choice correspondence 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, it is unbalanced.

We now extend Theorem 1 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.   is the (unordered) set of alternatives in the top two ranks for individual at profile .  Then we say social choice correspondence satisfies top-2-only if for all profiles , , if for all individuals .

Let D be the subdomain of consisting of all profiles for which it is not true that for all individuals and , 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 1.

Theorem 3.  Let and .  Then any social choice correspondence satisfying balancedness and top-2-only must be constant on D.

Proof:  Consider a specific profile with in everyone’s top rank, in #1’s second rank, and in everyone else’s second rank.  It will suffice to show that for every profile in D, there is a sequence of profiles in D from to such that each is obtained from the previous profile by a transposition pair or an application of top-2-only.

We argue by induction on the number of individuals who, at , do not have in their top two ranks.

Basis:  Suppose that we have at , i.e., every individual has in their top two ranks.  Construct from by (only) raising to everyone’s top rank.  Now consider the subdomain D of D consisting of all profiles in D where everyone has top-ranked.  Note that for each such profile at least two distinct alternatives must be second-ranked.  Let be the restriction of to D.  Social choice correspondence on D induces a correspondence on those profiles on which have non-unanimous tops.  Then the analysis in the tops-only section shows that there is a sequence of profiles in D from to .

Induction step:  Assume now that there is a non-empty set S of alternatives such that for all profiles in D with the number of individuals who do not have in their top two ranks being less than we have the same outcome: . Suppose that in D is a profile where the number of individuals who at do not have in their top two ranks is .  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 to a profile in D with individuals who do not have in the top two ranks (so ).  Without loss of generality suppose that it is the first individuals with ranked in the top two ranks.  We may assume that has been raised to the top for each of those individuals so that is:

Here , , and are distinct from (though these four need not all be distinct from one another).

Case 1.  At least one of the alternatives, say , in the top two ranks for is ranked second by one of .  Without loss of generality, let be an individual with in the top two and let 1 be the individual with on top and in the second rank:

Then if necessary, lower to the second rank for and raise to the third rank for that individual.

Now a transposition pair via individuals 1 and yields a profile in D with individuals with in the top two ranks and so only individuals with not in the top two ranks.

Case 2.  None of the alternatives in the top two ranks for ,…, is ranked second by one of 1,…,.

Subcase 2A.  There are two individuals among with the same alternative ranked second:

For one of those individuals, say #2, move , an alternative in (moved for to second rank if necessary) up to the third rank and then raise to the third rank for :

Now a transposition pair via individuals 2 and yields a profile in D still with individuals with not in the top two ranks but now back in Case 1 (since is in the top two ranks for individual and is ranked second for individual #1).

Subcase 2B.  No element in the top two ranks for is second for any of , and no two of have the same second element. (Recall .)

Subsubcase 2Bi.   (and so ).

Construct:

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 via 1 and takes us back to Case 1 (since will be in individual ’s top two and in #2’s second rank).

Subsubcase 2Bii.   (so ).

If not all of have the same top two alternatives, say has in the top two but 2 does not, move to second place for individual and raise to ’s third rank while raising and to third and fourth rank for #2:

Now a transposition pair via 2 and takes us back to in D.

On the other hand if all have the same two alternatives in the top two ranks:

Raise y to third rank for #1 and s to third rank for #2

Now a transposition pair via individuals 1 and takes us back (in D) to Case 1.

Case 3.   (so ).

Because we are in D, there must be an alternative , distinct from and (#1’s top two), in the top two ranks for someone, say #2.  Move to second rank for #2,

Then raise and to ranks three and four for #1 and to third rank for #2:

Now transposition pair via 1 and takes us back (in D) to Case 2Bii.

There are straightforward generalizations to top-3-only, etc.

## 4. Balancedness and Borda

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 just above in the bottom two ranks for individual exactly offsets lowering just below in the top two ranks for .  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

At profile , the score for an alternative is the sum

 S(x,u)=n∑i=1s(u,i,x)

where if , i.e., the score for is the sum of the weights corresponding to the ranks that occupies in the individual orderings at .  The related scoring social choice correspondence selects at the alternatives with lowest values.  If

 s1=s2=s3=⋯=sm

then is the constant social choice correspondence with at all , which is not very helpful.  Accordingly, we henceforth only consider systems of weights such that at least two weights are distinct.

The Borda correspondence, , uses weights .  But of course other weights also generate Borda.  If social choice correspondence is generated by weights

 s1≤s2≤s3≤⋯≤sm

then

is also generated by linearly transformed weights

 t1≤t2≤t3≤⋯≤tm

where for real numbers , , with .

Lemma 0.  For and : if is a scoring social choice correspondence for scoring system given by

 s1≤s2≤s3≤⋯≤sm

with at least two weights distinct and if satisfies balancedness, then .

Proof:  If , then there exists a , , such that .  Construct profile such that and for all while and .  Then and .  If profile is constructed from by transposition pair via individuals and , then and .  So and fails balancedness.

As a consequence of Lemma 0, we assume from now on that the weights have been transformed so that and .

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 section shows a limitation on this objective.

## 5. Borda: m = n = 3

Example 3:  Let and set scoring weights to be , and .

The related correspondence, , differs from Borda.  At profile :

the Borda rule has a tie between and and is Pareto-dominated by so .  But with scoring weights , and , the scores of and are 5.1 and 5 respectively: .  .

Nevertheless, is balanced, as can easily be checked.  So, for , 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.

## 6. Borda: n > 3

Our analysis proceeds by induction on m.  We begin by looking at and a few small values of .

Lemma 1.  For and , 5, or 6: if is a scoring social choice correspondence and satisfies balancedness, then is the Borda correspondence.

Proof: We show the following: if is a scoring rule but not the Borda correspondence, then violates balancedness.

Basis case, , for Lemma 1

()  We first treat the case .  Consider the following profile :

Consider a scoring rule with (using Lemma 0).  We examine the scores for :

For , ;

For , ;

For , .

So under this scoring rule, either and are both in or neither nor are in .

Now we consider a transposition pair via individuals 1 and 2 that yields the profile :

We examine again the scores for :

For , ;

For , ;

For , .

This time, either and are both in or both are out of .  But by balancedness, , so at each profile all scores must be the same.  Therefore , which has the unique solution , i.e., is the Borda correspondence.

()  Next, we consider . We examine the following profile .

Consider a scoring rule with weights: (using Lemma 0). We examine the scores for :

For , ;

For , ;

For , .

Under this scoring rule, either and are both in or neither nor is in .

Now we consider a sequence of transposition pairs first via individuals 1 and 2, and then via 3 and 4, which yields the profile :

We examine again the scores for :

For , ;

For , ;

For , .

This time, either and are both in or neither nor is in .  But is obtained from by a sequence of transposition pairs, and balancedness implies , which means both must be , which, in turn, means at each profile, all three scores must be the same.  Therefore, , i.e., is the Borda correspondence.

()  Finally, the analysis for proceeds just as for , but starting from the profile :

and this time doing three transposition pairs (for 1 and 2, then 3 and 4, and then 5 and 6).

For , merely observe that any such is equal to one of 4, 5, or 6 plus some multiple of 3.  Using profiles above that are expanded by that many multiples of a voting paradox profile (on which all alternatives have the same score regardless of the weighting scheme) provides profiles showing .

Lemma 2.  For and , 5, or 6: if is a scoring social choice correspondence and satisfies balancedness, then is the Borda correspondence.

Proof:  We prove by induction on .  The basis step is given by Lemma 1.  For , suppose that the lemma holds for all and we have linearly transformed weights

such that for some , it is the case that .  Let be the corresponding scoring social choice correspondence.  We show that there is a profile such that and that must fail balancedness.  Consider the smallest integer such that .  If , just take a profile that works for alternatives (using the induction hypothesis) and append additional alternatives to everyone’s bottom.

So we need only consider the case where the first such that is .  The weights are:

where .

Analysis here is for . The same profile applies to the case . We treat , but the same construction works for 4 or 6 individuals. (See the footnote below, for example, for .)  For we previously looked at profile:

Now at each stage of the construction, we add another alternative, placing it just above and for #2, just below and for #1, and at the bottom for everyone else. At the first stage we insert :

(after which #1 and #2 have opposite rankings).333So for , we also look at the profile for in the basis step in Lemma 1, then insert just above and for #2, just below and for #1, and at the bottom for #3 and #4.  The Borda scores are 11 for , , and and 17 for : .  For the correspondence with weights 1, 2, 3, and , the scores are for and , 11 for and for .  Because , the smallest of these is and .  Construct by transposing and for #1 and #2; this yields score 11 for , and for both and and for .  Again the smallest of these is , so , a failure of balancedness.

Eventually the profile looks like :

(after which #1 and #2 have opposite rankings).  It is easy to check that: .  For the correspondence with weights 1, 2, 3, …, , the scores are for and , for , and the rest have higher scores.  Because , the smallest of these is and .

Construct by transposing and for #1 and #2.  This yields scores for , and for both and , with the remaining scores unchanged.  Since , we now have , a failure of balancedness.

## 7. Borda: n = 3, m > 3

Let’s return to the case , where we learned in Example 1 that for not every scoring social choice correspondence satisfying balancedness is the Borda rule.  For the case and , a natural analog of Example 1 has scoring weights 1, 2, 3, and .  But that rule is unbalanced, as can be seen at profile :

Here the scores are , , , .  The smallest of these scores is 7 and so .  If is constructed from profile  by transposition pair  via individuals  and , then the scores become , , , .  So , a failure of balancedness.

Theorem 5.  For and : if is a scoring social choice correspondence and satisfies balancedness, then is the Borda correspondence.

Proof:  As in the proof of Lemma 2, we argue by induction on .  Starting with a basis case, new alternatives are inserted into a profile to unbalancedness of a scoring rule with weights different from those of the Borda rule.  For the basis case here, with , the scoring rule has weights 1, 2, , .

Basis case, , for Theorem 5

Proof. Let’s consider a scoring rule but not the Borda correspondence: . We consider the following profile :

We examine the scores for :

For , ;

For , ;

For , ;

For , .

Since and under this scoring rule, at least or (or both).

Now we consider a transposition pair via individual 1 and 2 that yields the profile :

We examine the scores for :

For , ;

For , ;

For , ;

For , .

By balancedness, . Since at profile , alternatives and have the same score, and at , at least or , we have and are both in and . So (from ), , i.e.,

(1)

Again, consider a transposition pair via individual 2 and 3 that yields the profile :

We examine the scores for :

For , ;

For , ;

For , ;

For ,

By balancedness, , so , or

(2)

Equations (1) and (2) have the unique solution , , i.e., is the Borda correspondence.

Now suppose the Theorem holds for all and we have weights

such that for some , it is the case that .  Let be the corresponding scoring social choice correspondence.  We show that there is a profile such that and that must fail balancedness.  Consider the first such that .  If , just take a profile that works for alternatives (using the induction hypothesis) and append additional alternatives to everyone’s bottom.

So we need only consider the case where the first such that is .  The weights are:

where .

Analysis here is for . The same profile works for .

We first show one stage.  Look at the profile just above. Now insert alternative just above for #1 and just below for #2.  The third individual has at the bottom:

The Borda scores for and are the lowest, so . With weights 1, 2, 3, 4, , with , the scores at are: , , , , and , so .  After transposing and for individuals 1 and 2 to create profile , the scores become: , , , , and , so and balancedness is violated.

Eventually the profile looks like :

With weights 1, 2, …, , , with , the scores at are: , , , , and the rest have higher scores. So given .  After transposing and for individuals 1 and 2 to create profile , the scores become: , , , , and the remaining scores are unchanged, so and balancedness is violated.

We obtain from this result a new characterization of the Borda correspondence.  While Young (1974), Hansson and Sahlquist (1976), Coughlin (1979/80), 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). Here 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, 1973b)].

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