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CHAPTER 11
Social Choice and Political Science
This chapter deals with models of collective choice in which individual
agents’ preferences are aggregated into collective behavior. The ﬁrst class
of models use some ﬁxed method to aggregate preferences. We assume that
collective choices can be observed, but that individual agents’ preferences are
unobserved. The second class of models are more structured models of voting
in political economy and political science. A common idea in political science
is that voters’ preferences are “Euclidean”; we present the testable implications
of this notion. Finally, we consider models of individual voter behavior and
work out the corresponding observable implications.
11.1
TESTABLE IMPLICATIONS OF PREFERENCE
AGGREGATION FUNCTIONS
The main questions in this section take the following form. Suppose that a
group preference (or choice) is observable. Is this group preference consistent
with a collection of rational agents whose preferences are aggregated
according to some rule? We may, for example, wonder when a group’s
collective behavior is consistent with majority rule.
There are three ways to interpret the material that we are about to present.
First, if we know the aggregation rule that the agents use, we may want to
test the hypothesis that a society of agents behave rationally as individuals,
when the only observable data come in the form of aggregate preference.
Second, when the aggregation rule is unknown, we may want to test the joint
hypotheses that a group of agents use a certain aggregation rule, and that they
each behave rationally as individuals. Finally, a different interpretation of these
results is that we might want to characterize all possible “paradoxes” that we
might expect from using a given aggregation rule. Condorcet’s paradox (a cycle
on three alternatives) illustrates the problems that can arise from using majority
rule. The results in this section describe all possible paradoxes of this type.
The model is as follows. Let X be a set of possible alternatives. We shall
assume that we observe all possible binary comparisons of elements in X; that
is, we observe a complete binary relation ⪰on X. This assumption is similar in


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spirit to our assumption in Chapter 2 that all choice behavior is observable. The
set of all complete binary relations will be denoted C(X). Individual agents’
preferences will be strict preferences (linear orders) over X. Denote the set of
linear orders over X by L(X).
We now describe two classical aggregation methods. We ﬁx a ﬁnite set N,
which we interpret as a set of agents. An aggregation rule is a function f :
L(X)N →C(X) mapping proﬁles of strict preferences (⪰i)i∈N into a complete
binary relation f((⪰i)i∈N).
The majority rule is the aggregation rule fm deﬁned by
(x,y) ∈fm((⪰i)i∈N) iff |{i ∈N : (x,y) ∈⪰i}| ≥|N|
2 .
Note that, if we let ⪰= fm((⪰i)i∈N), then x ≻y iff there is a strict majority of
agents who strictly prefer x over y (deﬁning the strict part of ⪰in the usual
way).
The unanimity rule is the function fu deﬁned by (x,y) ∈fu((⪰i)i∈N) if there
is i ∈N for which x ⪰i y. Note that, if we let ⪰= fu((⪰i)i∈N), then x ≻y iff all
agents strictly prefer x over y (deﬁning the strict part of ⪰in the usual way).
We can now turn the above deﬁnitions into a revealed-preference exercise.
Assuming that a complete binary relation ⪰is observed, and given a rule f,
we want to know when there is a set of agents N and a preference proﬁle
for the agents in N such that ⪰is the image of the proﬁle under f. Say that
⪰0 ∈C(X) is majority rationalizable if there exist a ﬁnite N and (⪰i)i∈N for
which fm((⪰i)i∈N) =⪰0. Similarly, we will say that ⪰0 ∈C(X) is unanimity
rationalizable (or Pareto rationalizable) if there exist a ﬁnite N and (⪰i)i∈N for
which fu((⪰i)i∈N) =⪰0.
Remark 11.1
Note that we allow the freedom of choosing the cardinality of
N when we construct a rationalization. Below we turn to problems in which
the cardinality of N is ﬁxed.
The ﬁrst and most important result here is a negative result stating that any
⪰0 ∈C(X) is majority rationalizable. Therefore, the hypothesis that a social
preference arises from application of majority rule to some unknown society
is untestable.
McGarvey’s Theorem
Any ⪰0 ∈C(X) is majority rationalizable.
The negative message in McGarvey’s Theorem is not affected by our
assumption that all binary comparisons are observable. If one’s observations
were less complete, then, of course, majority rule would remain non-testable.
The conclusion in McGarvey’s Theorem carries over to cases in which our
observations are poorer than a complete binary relation.
Proof. The proof is constructive. Let ⪰0 ∈C(X). If ⪰0 is complete indifference,
simply let N consist of two agents with exactly opposed preferences. Suppose
then that there is at least one pair x,y ∈X with x ≻0 y.


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First, let us write X = {x1,...,xm} (this is possible as X is ﬁnite). Given
any pair x,y ∈X for which x ≻0 y, we deﬁne two linear-order relations. The
ﬁrst, ⪰0
(x,y) ranks x ≻0
(x,y) y, and ranks x and y (strictly) above every element of
X\{x,y}; otherwise, for xi,xk ∈X\{x,y}, xi ≻0
(x,y) xk iff i > k. The second, ⪰1
(x,y),
ranks x ≻1
(x,y) y, and ranks x and y (strictly) below every element of X\{x,y};
otherwise, for xi,xk ∈X\{x,y}, xk ≻1
(x,y) xi iff i > k. By deﬁning N to be a set of
cardinality 2|{(x,y) ∈X2 : x ≻0 y}| and assigning, for each (x,y) ∈≻0, one agent
with preference ⪰0
(x,y) and one with preference ⪰1
(x,y), we arrive at (⪰i)i∈N for
which fm((⪰i)i∈N) =⪰0.
We now turn to a revealed preference question for the unanimity rule. Our
second result states that any quasitransitive relation is unanimity rationalizable.
Theorem 11.2
A binary relation ⪰0 ∈C(X) is unanimity rationalizable iff it
is quasitransitive.
Proof. It is easy to see that any unanimity rationalizable relation is quasitran-
sitive. Conversely, let ⪰0 ∈C(X) be quasitransitive.
We shall show that, for every x,y which are unranked in ≻0, there is a linear
order extending (≻0∪{(x,y)}). Then let N have a cardinality of
2 × |{(x,y) ∈X2 : x,yare unranked}|.
Assign, for each such pair, one agent with a linear order extending ≻0∪{(x,y)}
and one with a linear order extending ≻0∪{(y,x)}. Then it is easy to see that
fu((⪰i)i∈N) =⪰0.
Let, then, x,y ∈X be unranked according to (≻0∪=). That is, x ∼0 y
and x ̸= y. The relation (≻0∪=) is a partial order. Let ⪰′ be the transitive
closure of (≻0∪=∪{(x,y)}). We claim that ⪰′ is also a partial order. It is
clearly antisymmetric and reﬂexive; it remains to show that it is transitive.
But this kind of argument is familiar from Theorem 1.5, so we omit it here. By
Theorem 1.4, we know that there is a linear order which extends ⪰′.
The previous results deal with groups of unknown size: the size of the group
is as unknown as the agents’ preferences (see Remark 11.1). Therefore, in
constructing a rationalization, one has the freedom of using a group of any
size. The construction in each of the proofs illustrates that the groups can be
quite large.
It so happens, though, that we frequently know the size of the group, or at
least an upper bound. For example one may want to rationalize the behavior of
a given committee (such as a faculty meeting, Congress, or the United Nations)
when the number of members is known, but not their individual preferences.
It turns out that when we restrict the cardinality of the set of agents, the
problem of determining whether a binary relation is majority (or unanimity)
rationalizable is much more difﬁcult, and there are far fewer known results.
For an integer n, we say that ⪰0 is n-unanimity rationalizable if there exists
a set N of cardinality n for which there are (⪰i)i∈N such that fu((⪰i)i∈N) =⪰0.


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We could similarly deﬁne a related concept for majority rule, but as far as we
know there are no results in this direction.
Almost all results for n-unanimity rationalization concern n = 2. The next
result is due to Dushnik and Miller.
Theorem 11.3
A binary relation ⪰0 ∈C(X) is 2-unanimity rationalizable iff
⪰0 is quasitransitive, and there exists a partial order ⪰∗such that x ∼0 y iff
either x ⪰∗y or y ⪰∗x.
Proof. Suppose that ⪰0 is 2-unanimity rationalizable, say by (⪰1,⪰2). Clearly
≻0 is quasitransitive. Now, deﬁne x ⪰∗y if x ⪰1 y and y ⪰2 x. Note that ⪰∗is a
partial order which satisﬁes the property in the statement of the theorem.
Conversely, suppose that ⪰∗satisfying the property in the statement of the
theorem exists. Deﬁne a binary relation ⪰1 as follows. Let x ⪰1 y if either
x ≻0 y, or if x ∼0 y and x ⪰∗y. We claim that ⪰1 is a linear order. Completeness
is a consequence of the completeness of ⪰0 and the property of ⪰∗. To see that
it is antisymmetric, suppose that x ⪰1 y and y ⪰1 x. It cannot be that x ≻0 y, as
that would rule out y ⪰1 x. Thus we can assume that x ∼0 y; then x ⪰1 y implies
x ⪰∗y, and y ⪰1 x implies y ⪰∗x. It follows that x = y, as ⪰∗is a partial order.
Finally, ⪰1 is transitive. Let x ⪰1 y and y ⪰1 z. There is only something to
show when one of these comparisons is due to ⪰0 and the other to ⪰∗. These
cases are x ≻0 y ⪰∗z and x ⪰∗y ≻0 z. In the ﬁrst case, by completeness, if we
do not have x ⪰1 z, then we must have z ⪰1 x: (a) If z ⪰1 x is due to z ≻0 x, we
have a contradiction because the transitivity of ≻0 implies that z ≻0 y, which
contradicts y ⪰∗z; (b) If z ⪰1 x is due to z ⪰∗x, we have a contradiction because
the transitivity of ⪰∗implies that y ⪰∗x, a contradiction of x ≻0 y. We can
derive a similar contradiction in the case x ⪰∗y ≻0 z.
Similarly, we can deﬁne the binary relation ⪰2 by x ⪰2 y if x ≻0 y or y ⪰∗x.
Then ⪰2 is a linear order as well.
The linear orders ⪰1 and ⪰2 thus deﬁned provide a rationalization of ≻0. If
x ̸= y, then x ≻0 y iff x ≻1 y and x ≻2 y; and x ∼0 y iff ⪰1 and ⪰2 disagree in
how they compare x and y.
In light of Theorem 11.2, Theorem 11.3 gives the property of ⪰0 that, in
addition to being unanimity rationalizable, it is rationalizable by a group of
two agents. This condition, that ∼0 can be “oriented” to form a transitive ⪰∗,
is difﬁcult to falsify. One would need to check all possible orientations of ∼0. A
falsiﬁable characterization of when such an orientation is possible is provided
in the next result.
We ﬁrst need a simple deﬁnition. We will say x1,...,xk is an odd ∼0 cycle if
k is odd, and for all i, xi ∼0 xi+1 (as usual, addition is modulo k, and we allow
repetitions of vertices). We say the cycle is triangulated if there is i ∈{1,...,k}
for which xi ∼0 xi+2.
Theorem 11.4
A binary relation ⪰0∈C(X) is 2-unanimity rationalizable iff
it is quasitransitive and every odd ∼0 cycle is triangulated.


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Proof. We will establish necessity of the condition only. Sufﬁciency, while not
conceptually difﬁcult, is tedious. Therefore, let us suppose that ⪰1 and ⪰2 are
linear orders which 2-unanimity rationalize ⪰0.
To see that any odd ∼0 cycle is triangulated, suppose, by way of contradic-
tion, that there exists an odd ∼0 cycle x1,...,xk which is not triangulated. First,
it is the case that for all i, xi ̸= xi+1; otherwise since xi−1 ∼0 xi = xi+1, we have
xi−1 ∼0 xi+1, contradicting the fact that the cycle is not triangulated.
Suppose, without loss of generality, that x1 ≻1 x2 and x2 ≻2 x1. Now, because
x1,...,xk is not triangulated, it follows that x2 ≻2 x3 and x3 ≻1 x2; because,
if instead x2 ≻1 x3 and x3 ≻2 x2, we would have x1 ≻1 x3 and x3 ≻2 x1,
contradicting the fact that x1,...,xk is not triangulated. By continuing this
argument and using the fact that k is odd, we establish that x2 ≻1 x1 and
x1 ≻2 x2, a contradiction.
These results more or less exhaust the known characterizations of unanimity
rationalizable relations for ﬁxed and ﬁnite numbers of agents in abstract
environments. As far as we know, there are no such results for majority
rationalizable relations (or for other simple voting rules). Characterizing such
relations should be an important goal for future research. There is reason to
believe that such characterizations will be, in general, difﬁcult to come by. A
classical result in computer science states that, given a quasitransitive relation,
determining whether or not n agents sufﬁce to rationalize that relation by
unanimity rule is NP-complete for any n ≥3.
11.1.1
Utilitarian rationalizability
Suppose we have given a family of preference relations ⪰1,...,⪰n on a ﬁnite
set X. Further, let ⪰0 be an arbitrary transitive relation on X. We will say that
⪰0 is utilitarian rationalizable by ⪰1,...,⪰n if there exists, for each i, a utility
representation ui : X →R for which x ⪰i y ↔ui(x) ≥ui(y) such that
• x ≻0 y implies n
i=1 ui(x) > n
i=1 ui(y)
• x ∼0 y implies n
i=1 ui(x) = n
i=1 ui(y).1
If ⪰0 is complete, these conditions are equivalent to the function u0 =
n
i=1 ui being a utility representation for ⪰0.
Our goal is to characterize utilitarian rationalizable relations ⪰0. That is,
we want to test the hypothesis that society ranks alternatives according to
a utilitarian criterion, where the ordinal content of individual preference is
known, but not the cardinal content. The next result is due to Peter Fishburn.
Theorem 11.5
A transitive relation ⪰0 is utilitarian rationalizable by ⪰1
,...,⪰n iff for all ﬁnite disjoint sequences x1,...,xK and y1,...,yK (i.e. there is no
k,l for which xk = yl, but possibly allowing repetitions), and for all collections
1 See our discussion in 10.3 of utilitarianism in the context of bargaining.


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of permutations σi : K →K (one for each i = 1,...,n), such that for all
j = 1,...,K, xj ⪰0 yj and for all i = 1,...,n, yj ⪰i xσi(j), it follows that for all
j = 1,...,K, xj ∼0 yj and for all i = 1,...,n, yj ∼i xσi(j).
Proof. The relation ⪰0 is utilitarian rationalizable iff there exists, for each i and
each x ∈X a number ux
i ∈R, for which the following inequalities are satisﬁed:
I) If x ⪰0 y, then 
i ux
i ≥
i uy
i .
II) If x ≻0 y, then 
i ux
i > 
i uy
i .
III) If x ⪰i y, then ux
i ≥uy
i .
IV) If x ≻i y, then ux
i > uy
i .
We can easily write this in matrix form. Let B be a matrix with |X × N|
columns, and with one row for each constraint listed above. Rows of type (I)
and (II) will be speciﬁed by the vector 1N×{x} −1N×{y}, while rows of type (III)
and (IV) will be of the form 1(i,x) −1(i,y). We search for the existence of a
real-valued vector u ∈RX×N with the property that for rows m of type (I) or
(III), Bm ·u ≥0, and for rows m of type (II) or (IV), Bm ·u > 0. By Lemma 1.13,
the non-existence of such a u is equivalent to the existence of an integer-valued
vector η ≥0, such that for some row m of either type (II) or (IV), ηm > 0, and
η · B = 0. Rows are indexed by their associated relation; so a row of type (I)
will be written Bx⪰0y, for example.
It cannot be that only constraints of type (III) or (IV) are associated
with ηm > 0, as we know there always exists a utility representation for a
weak order on a ﬁnite set. So, let us consider two sequences of length K =

{m:m is of type (I) or (II)}
ηm, say x1,...,xK and y1,...,yK, where for ﬁxed (x,y) ∈X ×X,
|{k : xk = x,yk = y}| = ηx⪰0y + ηx≻0y.
Since ⪰0 is transitive, we may assume these two sequences are disjoint (that
is, there is no j,k for which xj = yk), as the rows can be canceled to remove
overlapping elements. Similarly, since each ⪰i is transitive, we can assume
that there is no triple x,y,z and i ∈N for which the rows corresponding to x Ri y
and y Qi z each have positive weight, where R and Q can be either ⪰or ≻.
Similarly, we may also assume that for any i = 1,...,n, and any x, it cannot
be the case that there are y,z for which both a row corresponding to y ⪰i x or
y ≻i x and a row corresponding to x ⪰i z or x ≻i z each have positive weight.
This we may assure by simply summing such rows to get a row involving only
y and z. We will call this assumption (*).
Now, it follows that since η · B = 0, and since the x1,...,xK and y1,...,yK
sequences are disjoint, it must be that rows of type (I) or (II) can only be
eliminated by a collection of rows of type (III) or (IV). That is, every instance
of xj contributes a positive term 1(i,xj), so in order for η · B = 0, this positive
term must be canceled by a negative term. This negative term must come from
a constraint of type (III) or type (IV), so there must exist, for agent i, a row of
type y ⪰i x or a row of type y ≻i x with positive weight. And by our assumption
(*), this introduces a positive term on 1(y,i), which must be matched by
some yk.


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What we have shown is that there exists, for each agent, a bijection σi :
K →K such that yj ⪰i xσi(j). And since one of the rows corresponds to type (II)
or (IV), ηm > 0, it follows that either there exists xj ≻0 yj, or there exists i ∈
1,...,n such that yj ≻i xσi(j). This is exactly what is precluded by the statement
of the theorem.
11.2
MODELS IN FORMAL POLITICAL SCIENCE
It is interesting and fruitful to analyze formal models in political science
from the perspective of revealed preference theory. There is a wealth of data
on which to test theories of political competition and voters’ behavior. Here
we shall discuss two different models, and investigate their empirical content
under different assumptions about what can be observed.
We ﬁrst discuss the standard model of “spatial” preference and investigate
circumstances where it lacks testable implications, even when we have rich
data (we observe a full preference relation). We then turn to more general
models of voter behavior, and consider more limited data.
11.2.1
Refuting Euclidean preferences
Many models in political science are based on voters having a special kind of
“spatial” preference. We consider here the testable implications of such models
of voter behavior. The idea is that policy positions can be represented as points
in some Euclidean space, Rd. We can view a vector x ∈Rd as representing
a position in each of d issues. We can then model a voter’s behavior using a
preference relation on Rd.
The standard benchmark model in political science is that of Euclidean
preferences, where each agent is endowed with an ideal point, and a vector
of policy positions is preferred to another iff it is closer to the voter’s ideal
point. Formally, for each agent i, let yi ∈Rd be i’s ideal point. Then i’s
preference relation, ⪰i, is deﬁned by x ⪰i z iff ∥x −yi∥≤∥z −yi∥, where
∥x∥= (d
i=1 x2
i )1/2 is the Euclidean norm on Rd (hence the term Euclidean
preferences).
The theory is simple, but when we observe data on voter behavior, we do not
observe choices among vectors in Rd. One basic problem is then to identify
alternatives with vectors in some Euclidean space, in a way that is consistent
with the theory.
Given are a ﬁnite set X of alternatives and a ﬁnite set of agents N. Agents are
endowed with preference relations over X, ⪰i, one for each i ∈N. A preference
proﬁle (⪰i)i∈N is Euclidean rationalizable if there exist a mapping ρ : X →Rd,
and ideal points yi ∈Rd, so that for all x,z,∈X and all i ∈N, x ⪰i z iff ∥ρ(x) −
yi∥≤∥ρ(z) −yi∥. Note that we make no requirement that ρ be one-to-one.
In revealed preference theory, we often try to understand the properties of
data that refute a given theory. Here we shall instead ask if the theory is at all


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165
refutable, without trying to understand the morphology of refutations. So we
ask, when is it the case that all preference proﬁles are Euclidean rationalizable?
The answer to this question will clearly depend on at least three things:
|X|, |N|, and d. If, for some triple (|X|,|N|,d), all preference proﬁles are
Euclidean rationalizable, then the model has, in some sense, no empirical
content. If instead there are proﬁles that are not Euclidean rationalizable, then
the theory may have some empirical bite. The following two results are due to
Bogomolnaia and Laslier.
Theorem 11.6
All proﬁles (⪰i)i∈N of preference relations are Euclidean
rationalizable iff d ≥min{|X| −1,|N|}.
Proof. We ﬁrst demonstrate that if d ≥min{|X| −1,|N|}, then the Euclidean
model has no empirical content. To this end, ﬁrst suppose that d ≥|X| −1. In
particular, let us consider the case d = |X| −1; the other cases follow trivially
from this. We shall carry out a construction for which we do not actually need
to know the proﬁle of preferences ⪰i.
We let the set of policy alternatives be the set in Rd+1 given by {x ∈
Rd+1 : d+1
j=1 xi = 1}. This set is isomorphic to Rd. Now, we can write X =
{1,...,d + 1}. Consider the mapping ρ which carries each m to the vector 1m
(as usual, 1m denotes the unit vector with a 1 in entry m, and zeros in all other
entries). The simplex deﬁned by (d) = {x ∈Rd+1 : x ≥0 and d+1
j=1 = x}
can be used to choose the ideal points, yi. In particular, for m,l ∈X, we can
consider the set H(m,l) = {x ∈(d) : xm ≥xl}, and the set H+(m,l) = {x ∈
(d) : xm > xl}. If y ∈H(m,l), then ∥1m −y∥≤∥1l −y∥, and if y ∈H+(m,l),
then ∥1m −y∥< ∥1l −y∥. So, given a preference ⪰i, we need yi ∈H(m,l)
for all m ⪰i l and yi ∈H+(m,l) for all m ≻i l. It is easy to see that this can
always be done. For example, take any utility representation ui : X →R of ⪰i
and consider −ui. Consider the function vi(x) = −ui(x) + λ, where λ > 0 is
chosen large enough so that vi(x) > 0 for all x. Then renormalize by α > 0
so that 
x∈X α(−ui(x) + λ) = 1. Then the vector yi ∈Rd+1 deﬁned by letting
yil = α(−ui(l) + λ) for each l satisﬁes the desired property.
On the other hand, suppose that d ≥|N|. We can suppose that d = |N|. The
case where d > |N| will easily follow from the argument given here. For each
i ∈N, let ui : X →R be a utility function representing ⪰i. Consider the point
ρ∗(x) ∈Rd whose ith coordinate is given by ρ∗(x)i = ui(x). Now, consider the
function F : Rd+1 →Rd given by
Fi(α,z) = −α(z · z) + zi;
with α ∈R and z ∈Rd.
Note that ∇zF(0,z) = I for all z (where ∇zF here stands for the Jacobian
matrix), and that Fi(0,ρ∗(x)) = ui(x) for x ∈X. By the implicit function
theorem (for example, see Rudin, 1976), for each x ∈X, there is a neighborhood
Ux of 0 such that for all α ∈Ux, there is a point ρα(x) for which Fi(α,ρα(x)) =
ui(x). By choosing α such that α ∈%
x∈X Ux and α > 0, we have for each x a


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point ρα(x) for which Fi(α,ρα(x)) = ui(x). Thus, x →Fi(α,ρα(x)) represents
⪰i on X. Finally, we claim that for each i ∈N, z →Fi(α,z) represents a
Euclidean preference. We have
Fi(α,z) = −(α(z · z) −zi),
which is ordinally equivalent to
−
 
z −1i
2α
!
·
 
z −1i
2α
!
,
which represents a Euclidean preference with ideal point 1i
2α. Let ρ(x) = ρα(x),
and we are done.
To show the converse, for every d, we will consider an environment with
d + 1 individuals and d + 2 alternatives which is not consistent with the
Euclidean model. We will demonstrate a list of preferences which cannot be
represented. Let us write out the alternatives as {x0,x1,...,xd+1}. Individual i ∈
N will have a preference which ranks xi strictly above all other alternatives, and
ranks the remaining alternatives as indifferent. We will argue by contradiction.
The ﬁrst point to mention is that, by virtue of the preference proﬁle under
consideration, it follows that for all j ̸= k, ρ(xj) ̸= ρ(xk). That is, ρ is
one-to-one.
Though the following proof will hold for arbitrary d, it helps to establish it
ﬁrst in the special case of d = 1.
First, for d = 1, it is simple to show that there is no representation of this
environment. We have two individuals, and ρ(x0),ρ(x1), and ρ(x2) all lie on a
straight line. Since i = 1 is indifferent between ρ(x0) and ρ(x2), ρ(x1) must be
in between ρ(x0) and ρ(x2). And since i = 2 is indifferent between ρ(x0) and
ρ(x1), it follows that ρ(x1) must be in between ρ(x0) and ρ(x1). This is clearly
impossible.
The proof for d ≥2 relies on an interesting geometric fact. Consider the
function ψ : Rd\{0} →Rd given by
ψ(x) =
x
∥x∥2 .
This function, known as an inversion function in geometry, has two very
interesting properties. First, it satisﬁes ψ(ψ(x)) = x for all x ∈Rd\{0}. Second,
it maps every sphere containing the origin (on its boundary, and of course
excluding the origin) to a hyperplane which does not intersect the origin,
and of course, therefore maps every hyperplane not containing the origin to a
sphere containing the origin (on its boundary). Every hyperplane which passes
through the origin is mapped to itself. Lastly, for any point in the interior of
any sphere containing the origin, this point is mapped to the opposite side of
the origin from the hyperplane which is the image of this sphere. For more
on the interesting geometry behind this function, one should consult Coxeter
(1969), Chapter 6.


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11.2 Models in formal political science
167
So, now suppose that d ≥2. It is clear that without loss of generality, we
may assume that ρ(x0) = 0.
We know that any (d + 1)-tuple of elements of {0,ρ(x1),ρ(x2),...,ρ(xd+1)}
containing 0 must lie on some sphere (as there is some agent who is
indifferent between them all and 0). As a consequence, any set of
{ψ(ρ(x1)),...,ψ(ρ(xd+1))}\{ψ(ρ(xi))}
lies
on
some
hyperplane
which
does not intersect the origin (the image of i’s indifference surface), and having
the property that ψ(ρ(xi)) lies on the opposite side of the origin from this
hyperplane.
We
will
study
the
intersection
of
all
hyperplanes
containing
{ψ(ρ(x1)),...,ψ(ρ(xd+1))}, namely H = {d+1
i=1 λiψ(ρ(xi)) : 
i λi = 1}.2
We
will
also
make
use
of
the
afﬁne
set
generated
by
the
set
{ψ(ρ(x1)),...,ψ(ρ(xd+1))}\{ψ(ρ(xi))}, call this Hi. Hi lies in a hyperplane
which does not pass through the origin, on the other side of which is ψ(ρ(xi)).
We have two possibilities: either H = Hi for some i, or not.
Suppose H = Hi for some i. Without loss of generality, assume that H =
Hd+1. This implies that on any sphere on which 0,ρ(x1),...,ρ(xd) all lie,
ρ(xd+1) must also lie, so that the preference ⪰d+1 cannot be represented. This
is a contradiction.
For the second case, we note that we may conclude that 0 ∈H.3 In particular,
there are λi for which d+1
i=1 λi = 1 such that 0 = d+1
i=1 λiψ(ρ(xi)). We claim
that each λi ≤0, which will be a contradiction.
We illustrate the case of i = d + 1. To see this, remember we said that each
Hd+1 lies on a hyperplane (the image of the indifference surface for agent
d+1), the other side of the origin from which is ψ(ρ(xd+1)). Let the normal of
this hyperplane be pd+1. Formally, we have that there is αd+1 > 0 such that for
all j = 1,...,d, pd+1 · ψ(ρ(xj)) = αd+1 > 0, and pd+1 · ψ(ρ(xd+1)) > αd+1. We
know that d+1
j=1 λjψ(ρ(xj)) = 0; consequently, d+1
j=1 λjpd+1 · ψ(ρ(xj)) = 0.
However, d+1
j=1 λjpd+1 · ψ(ρ(xj)) = αd+1(d
j=1 λj) + λd+1pd+1 · ψ(ρ(xd+1)).
If λd+1 ≥0, then we have αd+1(d
j=1 λj) + λd+1pd+1 · ψ(ρ(xd+1)) ≥
αd+1
d+1
j=1 λj = αd+1 > 0, a contradiction. Consequently, λd+1 ≤0.
2 That all hyperplanes containing {ψ(ρ(xi)),...,ψ(ρ(xd+1))} also contain H is obvious. That
each w ∈H is contained in all such hyperplanes can be proved via Lemma 1.12. That is,
for vectors y1,...,ym, p · y = 0 is a consequence of p · yi = 0 for all i iff there is λ ∈Rm
for which y = m
i=1 yi. Now w is in every hyperplane containing each ψ(ρ(xi)) if for all
p ∈Rd+1 and α ∈R, p · ψ(ρ(xi)) −α(1) = 0 implies p · w −α(1) = 0. So apply this result to
the vectors (ψ(ρ(xi)),−1) and (w,−1), so that there is λ ∈Rd+1 for which d+1
i=1 λi = 1 and
w = d+1
i=1 λiψ(ρ(xi)).
3 To see this, note that by dimensionality of the space, there is μ ∈Rd+1\{0} for which
d+1
i=1 μiψ(ρ(xi)) = 0 (the vectors ψ(ρ(xi)) cannot be linearly independent). We must show
that d+1
i=1 μi ̸= 0. But, since μ ̸= 0, the converse would imply that there exists some i for
which μi ̸= 0 and −μi = 
j̸=i μj. Then −μiψ(ρ(xi)) = 
j̸=i μjψ(ρ(xj)). Dividing each
side by −μi now results in ψ(ρ(xi)) ∈Hi, a contradiction.


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Social Choice and Political Science
Theorem 11.6 says that rationalization by Euclidean preferences may
require many dimensions. Our next result deals with a weaker theory: voters
have convex preferences over policies, but they do not need to be Euclidean. It
turns out that two dimensions sufﬁce in this case.
Formally, we say that a proﬁle (⪰i)i∈N is convex rationalizable if there exists
a mapping ρ : X →Rd, and a convex preference relation ⪰∗
i on Rd for all i ∈N,
such that for all x,y ∈X, x ⪰i y iff ρ(x) ⪰∗
i ρ(y).
Theorem 11.7
If d > 1, |X| = 2, or |N| = 1, then any proﬁle (⪰i)i∈N of
preference relations is convex rationalizable.
We omit the proof of Theorem 11.7, but the basic idea is to let ρ : X →Rd be
any mapping carrying each X to a distinct vertex of some regular polytope.
Then, for each preference ⪰i, we let U(x) = conv{ρ(y) : y ⪰i x}, where
conv denotes the convex hull. These are nested. The preferences ⪰∗
i to be
constructed must be such that their upper contour sets at x contain U(x). It
is always possible to construct such ⪰∗.
11.2.2
Rational voting when policy positions are known
We now turn to the case when policy alternatives are already given by vectors
in Rd. A voter has preferences over Rd; we are going to consider convex and
Euclidean preferences.
Suppose that we observe the behavior of a single voter. A voting record V
is a ﬁnite collection of pairs, {(yk,nk)}K
k=1, where each yk,nk ∈Rd and yk ̸= nk.
The interpretation is that out of the pair (yk,nk), yk ∈Rd was chosen (voted
“yes”) while nk ∈Rd was not.
The environment is a special case of choice theory, as described in Chapter 2.
For each k, yk is chosen over (or “revealed preferred” to) nk. The set of
“budgets” consist of pairs; and we assume that we observe the entire choice
function over those budgets (which happens to be single-valued).
By our results in Chapters 2 and 3, we know that any voting record is weakly
rationalizable by some utility function (namely, complete indifference). And a
voting record is strongly rationalizable iff its revealed preference pair admits
no cycles. In this case, we know the revealed preference pair is given by
yk ⪰c nk for all k, and yk ≻c nk for all k; that is, ⪰c = (≻c∪=), so acyclicity
of the order pair is the same as acyclicity of ≻c.4 The ﬁrst result highlights a
reformulation of acyclicity.
A voting record is strongly pair rationalizable if there is a utility function
u : Rd →R such that, for all k, u(yk) > u(nk). The word “pair” in “strongly pair
rationalizable” is meant to remind us that data come in the form of pairwise
comparisons.
4 Of course, we also have yk ⪰c yk for all k.


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11.2 Models in formal political science
169
For any V′ ⊆V, we deﬁne Y(V′) = {yk : (yk,nk) ∈V′} and N(V′) = {nk :
(yk,nk) ∈V′}. These are the sets of elements that were chosen from V′ and
rejected from V′ respectively. Finally, X(V′) = Y(V′) ∪N(V′).
Proposition 11.8
A voting record V is strongly pair rationalizable iff for all
nonempty V′ ⊆V, Y(V′) ̸= N(V′).
Proof. Suppose that V is strongly rationalizable, and suppose, toward a
contradiction, that there is some V′ ⊆V for which Y(V′) = N(V′). We will
show how to construct a ≻c cycle. Let nk1 ∈N(V′) be arbitrary; we know that
yk1 ≻c nk1. Moreover, yk1 ∈N(V′), so there is (yk2,nk2) ∈V′ for which nk2 = yk1.
But then we have yn2 ≻c nk2 = yk1 ≻c nk1. By continuing this construction
inductively, and since V′ is ﬁnite, we must eventually construct a cycle.
Conversely, suppose that we have Y(V′) ̸= N(V′) for all nonempty V′ ⊆V,
but that V is not strongly rationalizable. Then we know there is a cycle
ykl ≻c nkl = ykl−1 ≻c nkl−1 ...nk1 = ykl.
By setting V′ = {(yki,nki)}, we have Y(V′) = N(V′), a contradiction.
An extreme point of a set X ⊆Rd is a point x ∈X which cannot be written as
a convex combination of points in X\{x}. The set of extreme points of X will
be denoted E(X). The following theorem is due to Tasos Kalandrakis (as was
Proposition 11.8).
Theorem 11.9
Let V be a voting record. The following conditions are
equivalent:
I) For all nonempty V′ ⊆V, there is x ∈E(X(V′))\Y(V′).
II) There exists a concave utility function strongly pair rationalizing V.
III) There exists a strictly concave utility function strongly pair rational-
izing V.
IV) There exists a quasiconcave utility function strongly pair rationaliz-
ing V.
The main interest is in condition (I). Notice that it is a stronger condition than
acyclicity. In fact, we included Proposition 11.8 to highlight the difference
between Condition (I) and acyclicity as formulated in Proposition 11.8. If
for all nonempty V′ ⊆V there is x ∈E(X(V′))\Y(V′), then, of course, it is
impossible that Y(V′) = N(V′). The following is a simple example of a strongly
pair rationalizable voting record which is not rationalizable by a concave, or a
quasiconcave, utility function.
Example 11.10
Suppose X = R, and that the voting record consists of V =
{(0,1),(2,1)}. Then this voting record can never be strongly rationalized by a
concave utility function. This is obvious, but to see this via Theorem 11.9, note
that E(X(V)) = {0,2}, and Y(V) = {0,2}, a contradiction to condition (I).


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Social Choice and Political Science
Hence, in the voting model, concavity imposes testable restrictions over and
above the restrictions contained in strong rationalizability by a utility function.
Contrast this with Afriat’s Theorem of Chapter 3, which says that concavity
has no testable implications above rationality alone. The difference arises from
the kind of data we have assumed here. In the demand theory environment
of Chapter 3, we could never directly compare two arbitrary consumption
bundles. In the present environment, such comparisons are not only possible,
they are the only types of comparisons allowed.
Proof. That the existence of a concave strict rationalization implies
Equation (I) is well known, as at least one minimizer of a concave function
always occurs at an extreme point. And the minimizer cannot occur at a point
yk ∈Y(V′), as the corresponding point nk ∈N(V′) is ranked strictly lower.
We will show that Equation (I) implies the existence of a concave utility
function strictly rationalizing V. To see this, consider the set X. We will show
that for each x ∈X(V), there is ux ∈R and px ∈Rd such that for all (x,y) ∈V,
ux > uy, and for all x,y ∈X(V), uy ≤ux + px · (y −x). By deﬁning u : Rd →R
by u(z) = minx∈X(V) ux + px · (z −x), we will have a rationalization. This is the
typical Afriat construction; see Chapter 3.
In fact, we shall set up a system like that in the proof of Afriat’s Theorem.
Deﬁne a matrix B as follows. The matrix has |X(V)| + d|X(V)| columns. First
are |X(V)| columns, one labeled with each element of X(V). Then a second
collection of d|X(V)| columns, these come in groups of d columns, each group
is labeled with an element of X(V).
For each pair (x,y) ∈V, we have a row with a zero in each entry, with the
exception of a 1 in the ﬁrst column labeled with x and a −1 in the ﬁrst column
devoted to y. For each distinct pair (x,y) ∈X(V) × X(V), we have a row with
a zero in each entry, with the exception of 1 in the ﬁrst column labeled with x,
−1 in the ﬁrst column devoted to y, and in the second set of d columns labeled
with x we include the vector (y −x). That is, in the ﬁrst such column we write
y1 −x1; in the second column y2 −x2, and so on.
We are looking for a utility function u : X →R and vector px for each x ∈X
that solve the system of inequalities we described above. We can view a utility
function as a vector u ∈RX(V), and stack the d-dimensional vectors px in a way
that is congruent with the columns of B. Let P be the vector (px)x∈X stacked in
a manner congruent with the columns of B. Then we can write the system of
linear inequalities as
B ·
4 u
P
5
≥0,
where the ﬁrst V inequalities corresponding to u must be strict.
Now suppose that there is no solution to the system of inequalities. By
Lemma 1.12, we conclude that for each (x,y) ∈V, there exists λ(x,y) ≥0 and
for each (x,y) ∈X(V)×X(V), there is μ(x,y) ≥0, where at least one λ is strictly


--- Page 14 ---
11.2 Models in formal political science
171
positive, and such that for each x ∈X(V),

y∈X(V):(x,y)∈V
λ(x,y) −

y∈X(V):(y,x)∈V
λ(y,x) +

y∈X(V)
μ(x,y) −

y∈X(V)
μ(y,x) = 0 (11.1)
and

y∈X(V)
μ(x,y)(y −x) = 0.
(11.2)
Consider all pairs (x,y) for which either λ(x,y) > 0 or μ(x,y) > 0. Call this set
M, and consider X(M). The set M is nonempty; it is not necessarily a subset
of V. Note that Equation (11.1) implies that, for any x ∈X(M), there is y with
λ(z,y) > 0 or μ(z,y) > 0 (or both).
Next, consider an extreme point z of the set X(M). There is at least one
extreme point because X(M) is ﬁnite. As z ∈X(M), there is y with λ(z,y) > 0 or
μ(z,y) > 0. Now, we cannot have μ(z,y) > 0 for any y because equation (11.2)
would imply that z is not an extreme point of X(M).
Therefore, z is associated with a positive λ(z,y), for some y. But this implies,
in fact, that (z,y) ∈V. Deﬁne V′ ⊆V by (x,y) ∈V′ iff λ(x,y) > 0. We know
that all extreme points of X(M) are therefore elements of Y(V′). But as a
consequence, all extreme points of X(V′) are also elements of Y(V′), which is a
contradiction.
Going back to the model discussed in Section 11.2.1, we can ask when a voting
record V is rationalizable by a preference relation that is not only convex, but
Euclidean. Data V = {(yk,nk)}K
k=1 are rationalizable by the Euclidean model if
there is y ∈Rd such that, for all k, ∥yk −y∥< ∥nk −y∥.
The following result provides an answer. It states that whenever a weighted
average of the yk vectors and the nk vectors coincide, it must be the case that
the weighted average of the squared lengths of the yk vectors is strictly less
than the weighted average of squared lengths of the nk vectors.
Theorem 11.11
Let V = {(yk,nk)}K
k=1 be a voting record. Then there is a
Euclidean preference strongly rationalizing V iff for all λ ∈Rk for which
λ ≥0 and K
k=1 λk = 1, if 
k λkyk = 
k λknk, then 
k λk

yk · yk
< 
k λk

nk · nk
.
Proof. The proof is based on an application of a lemma related to Lemma 1.14.
V is rationalizable by a Euclidean preference iff there exists b ∈Rd such that
for all (yk,nk) ∈V,
−(yk −b) · (yk −b) > −(nk −b) · (nk −b).
Rewriting this equation, we seek the existence of b ∈Rd such that for all k,
b ·

yk −nk
> (yk · yk) −(nk · nk)
2
.


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Social Choice and Political Science
By introducing a variable α ∈R, we see that there exists such a b ∈Rd iff there
exists (b,α) ∈Rd+1 for which
b ·

yk −nk
−α
4(yk · yk) −(nk · nk)
2
5
> 0
and α > 0. By Lemma 1.12, it can be shown that there is a solution to
these inequalities iff for all λ ∈Rk where λ ≥0 and K
k=1 λk > 0, we have
K
k=1 λk(yk −nk) = 0 implies K
k=1 λk

(yk · yk) −(nk · nk)

< 0. Note that we
may simply renormalize so that 
k λk = 1.
Remark 11.12
It can be similarly shown that a voting record V is strongly
rationalizable by a linear preference iff for all λ ∈Rk for which λ ≥0
and K
k=1 λk = 1, we have K
k=1 λkyk ̸= K
k=1 λknk. Thus, the condition
of Euclidean rationalizability is strictly weaker than the condition of linear
rationalizability. This should not be surprising, as a linear preference is like a
Euclidean preference with an ideal point “at inﬁnity.” The situation would, of
course, change if we allowed indifference into a voting record.
Section 11.2.1 also addressed the question of when the Euclidean model
has no testable implications whatsoever. Instead of trying to describe the
rationalizable datasets, we found properties on the parameters of the problems
such that no data could refute the Euclidean model.
In the present context, we can ask a similar question. The difference
compared to the environment in 11.2.1 is that now the policy positions are
observed as vectors in Rd. Let us deﬁne an election to be a binary set E =
{x,z} ⊆Rd, where x ̸= z. For any ﬁnite collection of elections E1,...,Ek, a voting
record V is consistent with this collection if for all (x,z) ∈V, {x,z} ∈Ei for some
i, and for all i, if {x,z} = Ei, then either (x,z) ∈V or (z,x) ∈V. That is, a voting
record is consistent with a sequence of elections iff it describes a sequence of
possible votes over those elections. The next result is due to Degan and Merlo.
Theorem 11.13
Suppose k ≤d. Then there is an open and dense set
of election sequences of length k such that any consistent voting record
is rationalizable by Euclidean preferences. If d < k, then for all election
sequences there are voting records that are not rationalizable by Euclidean
preferences.
Proof. Let E be an election {x,z}. Let λ ∈Rd be deﬁned by λ = x −z and let
c ∈R be deﬁned by c = (x·x−z·z)
2
. A simple calculation reveals that a Euclidean
preference with ideal point y ranks x ≻z if and only if λ·y > c and ranks z ≻x
iff λ · y < c.
Each election Ei = {xi,zi} is identiﬁed with a pair (λi,ci), deﬁned as above.
Given a voting record V, there is y such that the Euclidean preferences with
ideal point y rationalize V iff there is y ∈Rd such that λi · y > ci if (xi,zi) ∈V,
and λi · y < ci if (zi,xi) ∈V. We may write in matrix form  ∈Rk×d, where 
collects the vectors λi, and c ∈Rk.


--- Page 16 ---
11.3 Chapter references
173
Now, with k ≤d, for an open and dense set of , the range of the function
y → · y −c is k-dimensional. And by the preceding discussion, for any such
 and c, and any V, there is y rationalizing V.
Suppose instead that d < k. We know then that λ1,...,λk cannot be linearly
independent; without loss of generality, let us suppose that λk = k−1
i=1 αiλi, and
let us suppose again without loss that αi ≥0 for all i (we can always relabel xi
as zi and zi as xi). Now consider the value c = k−1
i=1 αici. There are two cases.
Either c ≤ck or c > ck. In the ﬁrst case, whenever λi · y ≤ci for all i ≤k −1, it
follows that λk ·y ≤ck. In the second case, whenever λi ·y ≥ci for all i ≤k−1,
it follows that λk · y > ck. Either way, there are voting records that cannot be
rationalized.
The message of Theorem 11.13 is similar to that of Theorem 11.6; it
characterizes the environments where the Euclidean model is empirically
vacuous.
11.3
CHAPTER REFERENCES
Theorem 11.1 is due to McGarvey (1953). Theorems 11.2 and 11.3 are
due to Dushnik and Miller (1941). Theorem 11.4 is due to Ghouila-Houri
(1962) and Gilmore and Hoffman (1964). The NP-completeness result is due
to Yannakakis (1982). Results generalizing McGarvey’s Theorem for other
choice rules include Deb (1976) and Kalai (2004). Sprumont (2001) describes
necessary and sufﬁcient conditions for 2-unanimity rationalization by weak
orders in certain economic environments. Theorem 11.5 appears in Fishburn
(1969) without proof; a much more general statement can be found in Fishburn
(1973a). Related to these results is the paper of Knoblauch (2005), which
implicitly uses model-theoretic ideas (see Chapter 13) to describe the length
of potential axiomatizations of Pareto rationalizability.
The results and proofs in Section 11.2.1 are due to Bogomolnaia and Laslier
(2007). Bogomolnaia and Laslier (2007) actually establish a slightly different
result than Theorem 11.6, that even when one admits linear preferences
(preferences represented by a linear function), the result does not change.
Note that the function ρ described in the proof of Theorem 11.6 is allowed
to depend on the proﬁle ⪰1,...,⪰n. In general, we may not wish this to be the
case. Bogomolnaia and Laslier (2007) discuss this issue.
As for a proof of Theorem 11.7, the construction of preferences ⪰∗we
suggest can be established as in Richter and Wong (2004).
Some related papers are useful to mention. Knoblauch (2010) provides
a polynomial-time algorithm (and construction) for understanding when a
proﬁle is Euclidean rationalizable for d = 1. Azrieli (2011) studies Euclidean
preferences with a “valence” dimension. Ballester and Haeringer (2011)
describes conditions ensuring that a preference proﬁle is rationalizable as a
single-peaked proﬁle for some ordering of the alternatives.


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The discussion in Section 11.2.2 draws on Kalandrakis (2010) and Degan
and Merlo (2009). Most of the technical ideas in this section can be ascribed
to Richter and Wong (2004); we have followed Richter and Wong in the proof
of Theorem 11.9, which appears in Kalandrakis (2010) with a different proof.
Theorem 11.13 is due to Degan and Merlo (2009).
Richter and Wong (2004) deal with the following problem. Suppose we have
a ﬁnite subset K ⊆Rd, and a complete and transitive binary relation ⪰on K.
When is it the case that there exists a complete, transitive, extension of ⪰
which can be represented by a concave utility function? (One difference with
Kalandrakis (2010) is that he discards the completeness condition.)
Finally, Gomberg (2011, 2014) presents the testable implications of group
behavior when the group can vary, and is assumed to be composed of rational
individuals choosing according to a scoring rule.
