Pareto E¢ ciency of the Issue-wise Majority Rule
Tugce Cuhadaroglu, Jean Lainé
January 2008
Abstract. We consider voting in multiple binary issues setting where the outcome is determined by issue-wise majority rule. We characterize the set of extension rules from vote
ballots to preferences over outcomes that yield Pareto-e¢ cient issue-wise majority winners.
We show that a separable and top-consistent extension rule is Pareto-e¢ cient for the issuewise majority rule if and only if it is weakly Hamming-consistent.
1. Introduction
Given a set of voters and a set of binary issues, the issue wise majority rule will result in the
outcome which consists of the favorite decisions for each issue, favorite in the sense that it
is favored by a majority of the voters. In this kind of multiple binary issues setting such as
referendum voting or committee elections the representative power of the issue-wise majority
winner has been questioned frequently. To put it, Lacy and Niou’s (2000) words, does the
majority winner reveal “what the people want”?
In order to be able to answer this question most of the theorists have searched for the
existence of a stable outcome which will be chosen by the majority rule, in particular a
Condorcet winner. Kadane (1972), Schwartz (1977) and Kramer (1972) proved that in this
multiple binary issues setting, if the voters have separable preferences and if a Condorcet
winner exists, then this will exactly be the outcome of the majority voting.
An alternative way to comment on the representative power of the majority winner is to
search whether it ensures Pareto-e¢ cient (Pareto-optimal) outcomes. Özkal-Sanver and
Sanver (2006) shows that, under the assumption of separable preferences, it is impossible
to guarantee Pareto optimal outcomes through any kind of anonymous referendum voting,
in particular issue-wise majority rule. This result triggers a natural question: Under which
conditions the issue-wise majority rule will yield Pareto-optimal outcomes?
Obviously, in addition to separability, ensuring Pareto-optimal majority outcomes requires
more restricted preference domains. Benoit and Kornhauser (1994) shows that in a numbered
post election, that is the issues are indexed similarly, if the order of the importance of the
issues is the same for all voters and if the preferences are separable and top-lexicographic
then a Pareto-optimal assembly is always selected. This top-lexicographicity property is
designed for a very special type of voting procedure, which is the numbered post election
with a common order of importance of the issues, and does not guarantee Pareto-optimality
in more general settings.
Therefore, the question to be asked remains the same: In multiple binary issues setting,
what is the largest domain of separable preferences that give out Pareto-optimal issue-wise
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majority winners? In this paper, we propose an answer to this question by using a symmetric distance criterion. In addition to separability we make use of top-consistency to
ensure sincere voting and characterize the set of top-consistent and separable extension rules
from vote-ballots to preferences over outcomes that yield Pareto-optimal issue-wise majority
winners. To be concrete, we stick to the committee election setting.
2. Notations and de…nitions
We consider a set of alternatives = [Q 1 f0; 1gQ , where Q stands for a (variable) number of
dichotomous issues, in this paper approval versus disapproval of a candidate to a committee.
Q-committee is an element x = (x1 ; :::; xQ ) of f0; 1gQ .
Let Q 1, x 2 f0; 1gQ and B f1; :::; Qg be a subset of Q0 candidates. The sub-committee
0
x=B is the element of f0; 1gQ de…ned by:8q 2 B, (x=B)q = xq ; e.g.; x=B is the sub-committee
of x that indicates the approval or disapproval of all candidates in B.
The set of voters is the set I of all …nite subsets of non-zero-integers. We denote by In =
f1; :::; n; :::; N g a set of N voters. Given Q and n, each n 2 In is associated with an ideal
Q
committee xn = (x1n ; :::; xQ
n ) 2 f0; 1g .
NQ
A (N; Q)-ballot is a vector X
= (x1 ; :::; xN ) 2 1 n N f0; 1gQ . Let X = [N;Q 1 X N Q .
Let N be odd, and let X N Q = (x1 ; :::; xN ) be a (N; Q)-ballot. The majority committee is
the element m(X N Q ) = (m1 ; :::; mQ ) of f0; 1gQ de…ned by: 8q = 1; :::; Q, j fn = 1; :::; N :
xqn = mq g j> N2 .
This is the standard setting for committee elections with issue-wise majority rule. Given
X N Q , the issue-wise majority rule will result in the majority committee m(X N Q ) by computing the majority will candidate-wise.
In this kind of committee elections setting what is known about voters is only the ballots
they cast. In order to be able to investigate the Pareto characteristics of the voting rule,
we need to derive whole set of preorders over committees. With this aim we make use of
extension rules from a single committee to preferences over committees under certain helpful
assumptions.
Q
.
We denote by Q the set of all complete preorders on f0; 1gQ . Let R = [Q
Q 1
An extension rule is an application from to R which associates with each committee
x 2 f0; 1gQ an element (x) of Q :
The a-symmetric (resp. symmetric) part of is denoted by + (resp. s ).
An extension rule associates with each (N; Q)-ballot X N Q a preference pro…le (X N Q ) =
( (x1 ); :::; (xN )). The set N Q of all (N; Q; )-pro…les is the set of all those pro…les that
=
can be obtained from a ballot. Formally, N Q = [X N Q 2 1 n N f0;1gQ (X N Q ), and
NQ
[N;Q 1
.
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An extension rule (x) is said to satisfy top-consistency property (TC) if the following
condition holds; 8Q, 8x 2 f0; 1gQ , x + (x) y for all y 2 f0; 1gQ nx:
Top-consistency requires the single committee from which the preferences are extended to
be the …rst-ranked committee in the extended preorder. As we derive preorders from voteballots, top-consistency ensures sincere voting, in the sense that voters vote for their ideal
committees.
An extension rule (x) is said to be separable (S) if the following condition holds; 8Q,
8x; y; z 2 f0; 1gQ , (y=Qy6=z ) (x=Qy6=z ) (z=Qy6=z ) ) y (x) z; where Qy6=z = fq = 1; :::; Q :
y q 6= z q g
Under a separable extension rule, if a sub-committe is preferred to another one which has no
common decision with the …rst (e.g.; the opposite committe) in an extended preorder, then
the addition of the same elements to these two sub-committees will not alter the relative
ordering in the new extended pro…le. To put in other words, separability property ensures
the independence of the decision regarding a single candidate from the other ones.
2.1 Pareto E¢ ciency
Given X N Q , the outcome of the issue-wise majority rule will be the majority committee
m(X N Q ): The issue-wise majority rule is said to be Pareto-e¢ cient for the extension rule
if, for any pair fN; Qg of non-zero integers, and any (N; Q)-ballot X N Q , the majority
committee is not Pareto-dominated by another committee. To put it formally, for any N
and Q, any (N; Q)-ballot X N Q = (x1 ; :::; xN ), there is no x 2 f0; 1gQ such that x (xn )
m(X N Q ) for all n and x + (xn ) m(X N Q ) for at least one n .
2.2 Weak Hamming Consistency
The Hamming distance de…ned on two committes is simply the number of the candidates
they di¤er about. Formally, 8x = (x1 ; :::; xQ ), y = (y 1 ; :::; y Q ) 2 f0; 1gQ , d(x; y) =j fq =
1; :::; Q : xq 6= y q g j. Brams, Kilgour and Sanver (2001) show that the issue-wise majority
winner will be the committee that minimizes the total Hamming-distance from the ideals.
Obviously, Hamming distance criterion induces a very natural ordering of preferences over
committees such that the closer to the ideal will be ranked higher. Formally;
Hamming extension rule Ham is de…ned by: 8Q, 8x; y; z 2 f0; 1gQ ; d(x; y) < d(x; z) , y
+Ham
z and d(x; y) = d(x; z) , y sHam z: Note that Ham is a seperable and top-consistent
extension rule that yields out preference pro…les for which the issue-wise majority rule is
Pareto-e¢ cient. One can easily proove this.
Hamming-consistent extension rule HamC is de…ned by; 8Q, 8x; y; z 2 f0; 1gQ , d(x; y) <
d(x; z) ) y +HamC (x) z: While Hamming extension rule partitions the preorder into indi¤erent classes consistent with the distance from the best committee, Hamming-consistent
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extension rules allows ordering within these classes, hence increasing the number of attainable preorders. For instance consider n 2 N with xn = (1; 1; 1); a committee of three
candidates. The extended pro…le that Hamming-extension rule yields will be;
(1; 1; 1) +Ham (1; 1; 0) sHam (1; 0; 1) sHam (0; 1; 1) +Ham (0; 0; 1) sHam (1; 0; 0) sHam (0; 1; 0)
+Ham
(0; 0; 0); while Hamming-consistency allows 169 extended pro…les, 36 of them being
strict orders.
Propositon 2.2.1: Any Hamming-consistent extension rule HamC is Pareto-e¢ cient for
the issue-wise majority rule.
Proof: Take any X N Q 2 X and any Hamming-consistent extension rule HamC : Suppose
for a contradiction that m(X N Q ) is Pareto dominated by a distinct y 2 f0; 1gQ :
Formally, y HamC (xn ) m(X N Q ) for all n and y +HamC (xn ) m(X N Q ) for at least one n : Due
to Hamming-consistency, for all n; d(xn ; y)
d(xn ; m(X N Q )); resulting in n2N d(xn ; y)
NQ
)): As proved by Brams, Kilgour and Sanver (2004), the committee that
n2N d(xn ; m(X
minimizes the total distance to the ideals is the issue-wise majority winner.
Hence, n2N d(xn ; y) = n2N d(xn ; m(X N Q )) and together with Hamming-consistency for all
n; d(xn ; y) = d(xn ; m(X N Q )):
Let B = fq = 1; :::; Q : mq (X N Q ) 6= y q g with j B j= Q0 < Q: Q0 must be an even integer.
Suppose w.l.g. mq (X N Q ) = 0 for all q. Then, for all q 2 B; (m(X N Q )=B)q = 0; (y=B)q = 1
and for all n; j q 2 B : (xn =B)q = 0 j= j q 2 B : (xn =B)q = 1 j= Q0 =2; implying for all n;
q
0
q
0
q2B (xn =B) = Q =2 =)
n2N q2B (xn =B) = N Q =2 (1)
By de…nition of m(X N Q ) and by construction for all q 2 B; j n 2 N : (xn =B)q = 1 j< N=2:
Hence, for all q 2 B; n2N (xn =B)q < N=2 =) q2B n2N (xn =B)q < Q0 N=2 creating a direct
contradiction with (1).•
What Propositon 2.2.1 shows is simply that the issue-wise majority rule yields Pareto-e¢ cient
outcomes if the preferences of the voters are can be justi…ed by a symmetric distance criterion.
However, Hamming-consistency property is not su¢ cient to characterize the set of all topconsistent and separable extension rules for which the issue-wise majority rule is Paretoe¢ cient. This set is much larger than all Hamming-consistent extension rules. Therefore,
we introduce a weakening of the Hamming consistency property. Its de…nition requires the
following notation: for any committee x = (x1 ; :::; xN ) 2 f0; 1gQ , the opposite committee
( x) is de…ned by: 8q 2 B, ( x)q 6= xq
The weak Hamming-consistency property is de…ned by:
(WHC) 8Q, 8 < Q2 , 8x; y 2 f0; 1gQ , d(x; y)
) y + (x) ( y)
The weak Hamming-consistency property only determines the ranking of a committee relative to its opposite committee by comparing the distances to the ideal. It should be
clear that Hamming-consistency implies weak Hamming consistency. Unlike Hammingconsistent extension rules, weak Hamming-consistency does not guarantee separability or
top-consistency of the extended pro…les. But, if we take seperability and top-consistency as
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assumptions, then the set of weakly Hamming-consistent extension axioms will exactly be
the same as Hamming-consistent extension axioms when Q=3. But, even if we increase the
number of candidates by 1, then, the number of attainable individual pro…les under weak
Hamming-consistency will be 224 more than the number of attainable individual pro…les
under Hamming-consistency, for sure without harming seperability and top-consistency.
3. A Characterization Result
Theorem 3.1: A top-consistent and separable extension rule will be Pareto-consistent for
the issue-wise majority rule if and only if it is weakly Hamming-consistent.
Proof of the su¢ ciency part:
Let be a weakly Hamming-consistent extension rule. Let Q; N be any non-zero integers and
let X N Q be a (N; Q)-ballot such that m(X N Q ) is Pareto-dominated by x = (x 1 ; :::; x Q ) in
the pro…le (X N Q ) = ( (x1 ); :::; (xN )).
Step 1: 8Q; N , 9n 2 f1; :::; N g such that d(xn ; m(X N Q )) < Q2
Indeed, it follows from the de…nition of m(X N Q ) that j fxqi : xqi = mq (X N Q )g j> N2Q (*).
Suppose that for any n, one has j fq = 1; :::; Q : xqi = mq (X N Q )g j Q2 . Then one get
an immediate contradiction with (*). Thus, there exists n such that j fq = 1; :::; Q : xqi =
mq (X N Q )g j> Q2 , which means that d(xn ; m(X N Q )) < Q2 . Furthermore, it follows from the
de…nition of (WHC) that x 6= m(X N Q ).
Step 2: m(X N Q ) is not Pareto-dominated by ( m(X N Q ))
This is an immediate consequence of Step 1 together with weak Hamming-consistency.
Step 3: 8Q; N , m(X N Q ) is not Pareto-dominated
Suppose that y Pareto-dominates m(X N Q ). Let B = fq = 1; :::Q : y q 6= mq (X N Q )g, with
j B j= Q0 < Q. It follows from construction that y=B = ( m(X N Q )=B). Moreover, it
Q0
. Thus, weak
follows from step 1 that 9n 2 f1; :::; N g such that d(xn =B; m(X N Q )=B)
2
+Ham
NQ
Hamming-consistency ensures that m(X )=B
(x=B) y=B. The conclusion follows
from the separability property of .
Proof of the necessary part:
Q
Let be an extension rule that is not weakly Hamming-consistent. Thus, 9Q, 9
two
2
Q
integers such that 9x; y 2 f0; 1g which verify d(x; y) = and ( y) (x) y. One may assume
without loss of generality that y q = 1 for all q (through a relevant relabelling of issues). We
claim that there exists a (N; Q)-ballot X N Q such that m(X N Q ) is Pareto-dominated.
Q
1):::(Q
+1)
Let N = + 1, where
= (
) = Q(Q
. The set of ideal committees for
2:3:4:::: (
1)
individuals in f1; :::; g, that is [1
n
fxn g, is fx 2 f0; 1gQ :j fq = 1; :::; Q : xq = 0g j= g,
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while xqN = 0 for all q = 1; :::; Q. Consider any candidate q, and let us compute the number
Q 1
1):::(Q
)
n(q) of approvals given to q. One get n(q) = (
) = (Q
= (QQ ) : .
2:3:4:::: (
1)
Q
Suppose that Q is even. Thus,
1 ) (QQ ) : > ( 12 + Q1 ): . Since n(q) is an integer,
2
then n(q) > +1
: indeed, if is even, then n(q) > ( 12 + Q1 ): ) n(q) > 2 + 1 > +1
; if is
2
2
+1
+1
1
1
odd, then n(q) > ( 2 + Q ): ) n(q) > 2 + Q > 2 .
Q 1
1
Suppose that Q is odd. Then < Q2 )
) (QQ ) : > ( 21 + 2Q
): . Since n(q) is an
2
1
integer, then n(q) > +1
: indeed, if is even, then n(q) > ( 12 + 2Q
): ) n(q) > 2 + 1 > +1
2
2
+1
+
>
.
; if is odd, then n(q) > ( 21 + Q1 ): ) n(q) > +1
2
2Q
2
This proves that y = m(X N Q ). It follows from the de…nition of , together with the lack of
weak Hamming-consistency, that ( y) (xn ) y for all n = 1; :::; . Moreover, it follows from
top-consistency that xN = ( y) + (xN ) y. Hence, ( m(X N Q )) Pareto-dominates m(X N Q ),
which completes the proof •
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