Equilibrium Existence in Discontinuous
Zero-sum Games with Applications to the
Downsian Model of Elections∗
PRELIMINARY
John Duggan
November 21, 2002
Abstract
A theorem on existence of mixed strategy equilibria in discontinuous zero-sum games is proved. This is applied to the multidimensional
Downsian model of elections with three voters. A nine-voter example
shows that a key condition of the existence theorem is violated for general finite numbers of voters. The example poses problems for other
approaches to existence in this class of games as well. As another
application, an alternative proof of a Kramer’s (1978) existence result
for the Downsian model with a continuum of voters is also given.
This paper contributes to a larger literature on existence of equilibrium in
games with discontinuous payoff functions, including Dasgupta and Maskin
(1986), Simon (1987), and Reny (1999), but it focuses on a special class of
games, namely, two-player zero-sum games. In that sense, it is related to
papers by Sion (1958) and Mertens (1986), who weaken continuity of payoffs
∗
This paper benefited from useful discussions and/or correspondence with Jeff Banks,
Matt Jackson, Michel Le Breton, and Phil Reny.
1
to upper semicontinuity in own strategies. The main theorem of this paper
is as follows. Assume compact strategy sets. Assume that, when one player
uses an absolutely continuous mixed strategy, the other player’s payoffs are
continuous in his/her own strategy. And make the key assumption that,
given any mixed strategy for one player, the other player can approximate
the payoff of any strategy with an absolutely continuous mixed strategy.
Then there is a mixed strategy equilibrium.
The main application motivating the theorem is to the classical Downsian
model of elections. Here, two candidates take positions in a convex policy
space before an election. Voters, who are not modelled as players, have
continuous, strictly convex preferences and are assumed to vote for their
preferred candidates. The candidates are assumed to be “office-motivated,”
seeking only to win the election rather than, say, to implement preferred policies. It is known that, when the policy space is one-dimensional (and voter
preferences are therefore “single-peaked”), the electoral game has pure strategy equilibria. Moreover, a pair of positions is a pure strategy equilibrium
if and only if each candidate locates at a median with respect to the voters’
ideal points. This characterization carries over to mixed strategies as well: a
pair of mixed strategies is an equilibrium if and only if both candidates play
medians with probability one. And if the number of voters is odd, then there
is a unique median, and the game actually has a unique equilibrium.
When the policy space is multidimensional, a pair of positions is a pure
strategy equilibrium if and only if each candidate locates at a position that
is undominated according to majority rule, i.e., if neither candidate takes a
position to which something is majority-preferred. The set of such points
is known as the “core” of the spatial voting game. A long line of work in
spatial modelling, including Plott (1967), Schofield (1983), Banks (1995),
and Saari (1997), has shown that, for multidimensional policy spaces, the
conditions on voter preferences that must be met if the core is non-empty
are very restrictive — so restrictive that they are generically (in variously
defined senses) violated, with the conclusion that pure strategy equilibria of
the electoral game almost never exist. The candidates’ payoffs are highly
discontinuous, and standard results, such as Glicksberg (1952), on existence
of mixed strategy equilibria do not apply. Furthermore, the upper semicontinuity condition of Sion (1958) and Mertens (1986) are violated, and
the conditions of the other above-cited papers on existence in discontinuous
games have proved difficult to verify.
2
It is true, of course, that mixed strategy equilibria exist if the policy space
is finite. Indeed, Laffond, Laslier, and Le Breton (1993) show that, in the absence of majority-ties, there is exactly one mixed strategy equilibrium in this
setting. When the policy space consists of divisions of a “dollar” among the
set of voters, and each voter simply prefers more of the dollar, the electoral
game reduces to the Colonel Blotto game, in which mixed strategy equilibria
are known to exist. (See Laslier and Picard (2002).) Banks, Duggan, and
Le Breton (2002) have shown that, if there is a mixed strategy equilibrium
of the electoral game, then the equilibrium strategies of the candidates must
have support contained in the uncovered set, a centrally located region of the
policy space — but existence in the Downsian model has remained elusive.
This paper takes a step toward the solution of this problem: it is shown
that, in the three-voter version of the multidimensional Downsian model, the
conditions of the existence theorem are satisfied, with the conclusion that a
mixed strategy equilibrium exists in this special case. The obvious question
is whether the analysis can be extended to an arbitrary number of voters, and
the last section of the paper demonstrates that the answer to this question
is, unfortunately, negative. A nine-voter example shows that the verification
of the key condition of the existence theorem — that a player can approximate any payoff with an absolutely continuous mixed strategy — becomes
highly problematic. The key condition is violated despite the presence of
a mixed strategy equilibrium in one version of the example, demonstrating
that the condition is not necessary for existence. In that version of the example, Reny’s (1999) diagonal better reply security is satisfied, but mixed
strategies are needed to secure deviating payoffs as called for in Reny’s condition, and those mixed strategies may be “far” from the initial deviation
being “secured.” In another version of the example, diagonal better reply
security is also violated. This suggests that, if mixed strategy equilibria of
the Downsian model do exist generally, then an analytical proof of existence
will require an improvement of current proof techniques.
Kramer (1978) proposes an alternative to the classical spatial model, assuming a continuum of voters and “plurality-motivated candidates,” meaning
that that each candidate seeks to maximize his/her votes net of the other candidate’s votes. This combination of assumptions rules out certain especially
problematic discontinuities in the candidates’ payoffs and allows Kramer to
prove existence of a mixed strategy equilibrium, but the proof is quite involved. It is shown, however, that the assumptions of the above existence
3
result for two-player zero-sum games are easily verified in Kramer’s model,
yielding a straightforward path to existence in the model with a continuum
of voters.
1
Existence in Zero-sum Games
A two-player zero-sum game is a tuple (X, X 0 , µ, µ0 , u), where X and X 0 are
subsets of Hausdorff topological spaces, µ and µ0 are finite Borel measures
on X and X 0 , the product X × X 0 is endowed with the product topology,
and u : X × X 0 → < is a bounded, Borel measurable payoff function. We say
the game is symmetric if X = X 0 and, for all x, x0 ∈ X, we have u(x, x0 ) =
−u(x0 , x). A pair (x, x0 ) is a pure strategy equilibrium if, for all y ∈ X and all
y 0 ∈ X 0 , we have u(y, x0 ) ≤ u(x, x0 ) ≤ u(x, y 0 ). A mixed strategy, denoted σ
or σ 0 , is a Borel probability measure on X or X 0 . Let Σ denote the set of all
mixed strategies on X and Σ0 the set of mixed strategies on X 0 . We extend
the payoff function u to mixed strategy pairs as follows:
Z Z
0
U (σ, σ ) =
u(x, x0 ) σ(dx)σ 0 (dx0 ).
If σ is degenerate on x, we write simply U (x, σ 0 ) for this payoff. Because u is
bounded, we may view this as defining a mapping U (·, σ 0 ) : X → <, which,
by Fubini’s theorem, is Borel measurable. We say σ is absolutely continuous
if it is absolutely continuous with respect to µ, and we let Σ◦ denote the set
of absolutely continuous mixed strategies on X. (We use similar terminology
for σ 0 , x0 , µ0 , and Σ◦0 .) A pair (σ, σ 0 ) is a mixed strategy equilibrium if, for
all y ∈ X and all y 0 ∈ X 0 ,
U (y, σ 0 ) ≤ U (σ, σ 0 ) ≤ U (σ, y 0 ).
That existence of a mixed strategy equilibrium does not follow from the
previous assumptions, even when strategy sets are assumed to be compact,
is illustrated in an example, covered in the next section, due to Sion and
Wolfe (1957).
Four conditions will play important roles in what follows.
(C0) The sets X and X 0 are compact.
4
(C1) For all σ 0 ∈ Σ◦0 , the function U (·, σ 0 ) : X → < is continuous, and for
all σ ∈ Σ◦ , the function U (σ, ·) : X 0 → < is continuous.
(C2) For all σ 0 ∈ Σ0 and for all x ∈ X,
U (x, σ 0 ) > 0 ⇒ sup U (σ, σ 0 ) > 0.
σ∈Σ◦
(C3) For all σ 0 ∈ Σ0 and for all x ∈ X,
sup U (σ, σ 0 ) ≥ U (x, σ 0 ),
σ∈Σ◦
and for all σ ∈ Σ and for all x0 ∈ X 0 ,
inf U (σ, σ 0 ) ≤ U (σ, x0 ).
σ 0 ∈Σ◦0
The first of these conditions is standard. The second is a minimal continuity
assumption and is self-explanatory. The third captures the idea that any
profitable deviation from a strategy pair with payoff zero can be replaced
by an absolutely continuous deviation. The fourth strengthens this, saying
that both players can approximate any deviation payoff by using absolutely
continuous strategies.
The main result of this paper is the following theorem on existence of
equilibria in symmetric games.
Theorem 1 In any two-player, symmetric, zero-sum game satisfying (C0)(C2), there exists a mixed strategy equilibrium.
To prove the theorem, let
U = {U (·, σ 0 ) | σ 0 ∈ Σ◦ }
denote the set of payoff functions for one player induced by absolutely continuous mixed strategies of the other player. This set is non-empty, because
1
0
σ 0 = µ0 (X
0 ) µ is absolutely continuous; it is convex, since convex combinations
of absolutely continuous strategies are absolutely continuous; and, by (C1),
5
it is a subset of C(X), the vector space of real-valued continuous functions
on X. Let
N = {f ∈ C(X) | for all x ∈ X, f (x) ≤ 0}
denote the non-positive orthant of C(X), which is also convex. Note that, if
U (·, σ̂) ∈ U ∩ N , then (σ̂, σ̂) is a mixed strategy equilibrium: U (σ̂, σ̂) = 0 by
symmetry, and then
U (y, σ̂) ≤ 0 ≤ U (σ̂, y)
for all y ∈ X. Suppose, then, that U ∩ N = ∅. The function that is identically negative one is an internal point of N , and the separating hyperplane
theorem (Aliprantis and Border, 1999, Theorem 5.46) yields a non-zero linear
functional p : C(X) → < that properly separates U and N , i.e., for all f ∈ N
and all f 0 ∈ U , p(f ) ≤ p(f 0 ), with strict inequality for at least one f and
one f 0 . Since the function that is identically zero is in N , we have 0 ≤ p(f 0 )
for all f 0 ∈ U, and it follows that p(f ) ≤ 0 for all f ∈ N . Therefore, p is a
positive linear function. By (C0), each f 0 ∈ C(X) has compact support, so
the Riesz representation theorem (Rudin, 1987, Theorem 2.14) yields a Borel
measure ν on X such that
Z
p(f ) =
f dν
for all f ∈ C(X). Since p properly separates U and N , it must be that
ν(X) > 0. Furthermore, ν(X) < ∞, since otherwise the function that is
1
identically one would not be integrable. Letting σ̂ = ν(X)
ν, I claim that
(σ̂, σ̂) is a mixed strategy equilibrium. By construction, for all σ 0 ∈ Σ◦ , we
have
Z
1
0
U (σ̂, σ ) =
f 0 (x)σ̂(dx) =
p(f 0 ) ≥ 0,
ν(X)
where f 0 = U (·, σ 0 ). By (C2) and symmetry, U (x, σ̂) ≥ 0 for all x0 ∈ X.
Applying symmetry again, we then have
U (x, σ̂) ≤ 0 ≤ U (σ̂, x0 )
for all x, x0 ∈ X, establishing the claim.
Theorem 1 extends easily to non-symmetric zero-sum games, if we strengthen
(C2) to (C3).
6
Corollary 1 In any two-player zero-sum game satisfying (C0)-(C3), there
exists a mixed strategy equilibrium.
For the proof of the corollary, we note that any non-symmetric game can
be converted into a symmetric one by first letting each player have strategy
set X̂ = X × X 0 , with one player’s strategies denoted (x, y) and the other’s
denoted (x0 , y 0 ). Here, we endow X̂ with the product topology and corresponding Borel field, and we extend the product measure µ̂ = µ × µ0 to this
field. We then define the “symmetrized” payoff function
û((x, y), (x0 , y 0 )) = u(x, y 0 ) − u(x0 , y),
which is Borel measurable. Thus, we essentially let the players play the
original game twice, once where the players assume their original roles and
once where the players switch roles. Clearly, ((x, y), (x0 , y 0 )) is a pure strategy
equilibrium of the symmetrized game if and only if (x, y 0 ) and (x0 , y) are
equilibria of the original game.
We extend û to mixed strategy pairs (σ̂, σ̂ 0 ) as above. It is straightforward
to verify that
Z Z
0
Û (σ̂, σ̂ ) =
û((x, y), (x0 , y 0 ))σ̂(d(x, y))σ̂ 0 (d(x0 , y 0 ))
= U (σ̂1 , σ̂20 ) − U (σ̂10 , σ̂2 ),
where a subscript 1 denotes the marginal probability measure on X and a
subscript 2 denotes the marginal probability measure on X 0 . Thus, (σ̂, σ̂ 0 )
is a mixed strategy equilibrium of the symmetrized game if and only if the
corresponding marginals form mixed strategy equilibria of the original game.
Note that the symmetrized game satisfies (C0)-(C3) if the original game does.
To see that this is true of (C3), take any σ̂ 0 ∈ Σ̂0 , and any (x, y) ∈ X̂. Then
Û ((x, y), σ̂ 0 ) = U (x, σ̂20 ) − U (σ̂10 , y),
and, for any ² > 0, (C3) in the original game yields σ1 ∈ Σ◦ and σ2 ∈ Σ◦0
such that
²
²
U (σ1 , σ̂20 ) ≥ U (x, σ̂20 ) −
and U (σ̂10 , σ2 ) ≤ U (σ̂10 , y) + .
2
2
Defining σ̂ = σ1 × σ2 as the product measure, we therefore have σ̂ ∈ Σ̂◦ and
Û (σ̂, σ̂ 0 ) = Û ((x, y), σ̂ 0 ) − ²,
7
which delivers (C3) for the symmetrized game. Theorem 1 then yields a
mixed strategy equilibrium of the symmetrized game, and the corollary then
follows from the preceding observations.
2
Discussion
In the case of symmetric two-player zero-sum games, (C1) and (C2) together
imply Reny’s (1999) diagonal better reply security of the mixed extension of
the game. In fact, they imply his stronger condition of better reply security,
which is defined as follows. Let
Γ = {(σ, σ 0 , u) ∈ X × X × < | U (σ, σ 0 ) = u},
and let Γ denote the closure of Γ, where X × X × < is given the product
topology. Then the mixed extension satisfies better reply security if, for all
(σ, σ 0 , u) ∈ Γ such that (σ, σ 0 ) is not a mixed strategy equilibrium, either:
1. there exists σ̂ ∈ Σ and an open set G ⊆ Σ0 such that σ 0 ∈ G and, for
all σ̂ 0 ∈ G, U (σ̂, σ̂ 0 ) > u, or
2. there exists σ̂ 0 ∈ Σ0 and an open set G ⊆ Σ such that σ ∈ G and, for
all σ̂ ∈ G, U (σ̂, σ̂ 0 ) < u.
That is, if (σ, σ 0 ) is not an equilibrium, then some player can secure a payoff
strictly above the closure payoff (either u or −u, depending on the player),
even when the other player’s strategy is allowed to vary within some open
set.
Assume (C1) and (C2) hold for a symmetric game. Let (σ, σ 0 , u) belong
to Γ, and suppose (σ, σ 0 ) is not an equilibrium. Then, by symmetry and
interchangability, either (σ, σ) is not an equilibrium or (σ 0 , σ 0 ) is not. In the
first case, there exists x ∈ X such that U (x, σ) > U (σ, σ) = 0. By (C2),
there exists σ̂ ∈ Σ◦ such that U (σ̂, σ 0 ) > 0. By (C1), U (σ̂, ·) : X → < is a
continuous function, so there is an open set G ⊆ Σ0 such that σ 0 ∈ G and,
for all σ̂ 0 ∈ G, U (σ̂, σ̂ 0 ) > 0. If u ≤ 0, then we are done. Otherwise, u > 0.
If (σ 0 , σ 0 ) is not an equilibrium, then we can argue as above that the second
player has an absolutely continuous strategy that keeps U below zero, and
8
so below u, over some open set of the first player’s strategies, as required. If
(σ 0 , σ 0 ) is an equilibrium, then σ 0 keeps U below zero for all strategies of the
first player, and we can set σ̂ 0 = σ 0 to fulfill better reply security.
Thus, Theorem 1, which is stated for symmetric games, follows directly
from Reny’s Theorem 3.1 (or from his Theorem 4.1). The proof in the previous section is included for completeness and because the simple geometric
approach used there may be of independent interest. Corollary 1 can then
be proved as above, of course, but it is interesting that the result does not
appear to follow directly from Reny’s: in contrast to symmetric games, (C1)
and (C3) do not appear to imply better reply security of the mixed extension
for general two-player zero-sum games. The argument given in the previous
paragraph obviously relies on symmetry to conclude that either (σ, σ) is
not an equilibrium or (σ 0 , σ 0 ) is not. Without that conclusion, since some
player (say the first) can deviate profitably from (σ, σ 0 ), we may still find
an absolutely continuous deviation σ̂ such that U (σ̂, ·) exceeds U (σ, σ 0 ) over
some open set G around σ 0 . If u ≤ U (σ, σ 0 ), then we are done. Otherwise,
u > U (σ, σ 0 ), but now we have no information about possible deviations by
the second player to complete the argument.
Thus, Corollary 1 gives sufficient conditions for existence weaker in one
respect than Reny’s better reply security: those conditions imply that, given
an out-of-equilibrium strategy pair, some player has a deviation that is robust
to variations in the other’s strategy, but they do not require the deviation to
improve on all closure payoffs of the deviating player.
Since (C0)-(C3) are sufficient for existence, those conditions obviously
cannot hold in two-player zero-sum games without mixed strategy equilibria.
A classic example is due to Sion and Wolfe (1957). In this game, we let
X = X 0 = [0, 1] denote the strategy space of two players, and we let u denote
the payoffs to player 1 and −u the payoffs to player 2. Given x, y ∈ X, define

1 if y < x




 0 if x = y
−1 if x + 12 > y > x
u(x, y) =


0 if y = x + 12



1 if y > x + 12 .
9
The corresponding payoff matrix, with payoffs for player 1 (the “column”
player), is depicted in Figure 1.
[ Figure 1 here. ]
Since this game satisfies (C0) and (C1), yet has no mixed strategy equilibrium, an implication of Corollary 1 is that it violates (C2). To see this,
let player 2 use the mixed strategy σ 0 with equal probability on 1 and 1/2,
i.e., σ 0 ({1}) = σ 0 ({1/2}) = 1/2. Consider the pure strategy x = 0 for player
1, and note that u(x, σ 0 ) = 1/2. Note, however, that
sup u(σ, σ 0 ) = 0,
σ∈Σ◦
contrary to (C2).
A simpler example was suggested by Michel Le Breton (personal correspondence). Let X = X 0 = [0, 1], and define

if x > y and either x = 1 or y = 0
 1
0
if x = y
u(x, y) =

−1 if x < y and x = 0 and y = 1.
This defines a symmetric two-player zero-sum game with no mixed strategy
equilibrium. To see this, consider any symmetric equilibrium. If the equilibrium mixed strategy places positive probability on [0, 1), then either player
can obtain a positive expected payoff by moving that mass up arbitrarily
close to (but below) 1; and if the equilibrium puts probability one on 1, then
either player can obtain a positive expected payoff by choosing 0, a contradiction. Since the game has no symmetric equilibrium, it cannot have any
equilibrium. Here, (C2) is violated when x = 0 and σ 0 ({1}) = 1. In this
case U (x, σ 0 ) = 1, but if σ̂ is an absolutely continuous mixed strategy, then
U (σ̂, σ 0 ) = −1.
This example is especially relevant, because if u(x, y) = 1 is interpreted
as x being majority-preferred to y, then this is an example of the Downsian
electoral model with specific majority-preferences. It shows that, without
some restrictions on voter preferences (and thereby on majority preferences),
we are not guaranteed the existence of an equilibrium. In what follows, standard assumptions will be imposed on voter preferences, including continuity,
to preclude the latter example.
10
3
Downsian Elections with Three Voters
Now let X = X 0 ⊆ <d be a compact, convex subset of Euclidean space
representing the set of possible public policies, and let µ = µ0 be Lebesgue
measure. Assume a set N = {1, 2, 3} of three voters, where the policy preferences of each voter i are represented by a strictly pseudo-concave utility
function ui : X → <, i.e., ui is differentiable and, for all distinct x, y ∈ X
with x interior to X, if ui (y) ≥ ui (x), then ∇ui (x) · (y − x) > 0. Note
the implication that each ui is strictly quasi-concave on the interior of X.
For simplicity, assume that the boundary points of X are Pareto dominated.
These assumptions capture the classical special case in which voters have
Euclidean preferences and ideal points interior to X.
Define the strict majority preference relation, P , as xP y if and only if
ui (x) > ui (y) for at least two voters; define weak majority preference, R, as
xRy if and only if not yP x, i.e., ui (x) ≥ ui (y) for at least two voters; and
define majority indifference, I, as xIy if and only if xRy and yRx. For any
relation Q, which may range over P , R, or I, we let Q(x) = {y ∈ X | yQx}
denote the upper section of Q, and we let Q−1 (x) = {y ∈ X | xQy} denote
the lower section of Q. Note that, by continuity of voter preferences, P is an
open relation, while R and I are closed. Furthermore, if xIy, then there is
some voter i such that ui (x) = ui (y), implying that I(x) has zero Lebesgue
measure.
The Downsian model of elections posits two candidates who, prior to an
election, take positions x and x0 in the policy space X. Voters, who are
not modelled as players, are assumed to vote for candidates with preferred
positions, generating a symmetric two-player zero-sum game between the
candidates with payoffs defined as

 1 if xP x0
0
0 if xIx0
u(x, x ) =

−1 if x0 P x,
consistent with the assumption that each candidate seeks only to win the
election. Note that these payoffs are rather stylized: if, for example, one
voter prefers x to x0 and the others are indifferent, then u(x, x0 ) = 0. It is
not assumed that indifferent voters abstain, in which case the first candidate
would win with x and should receive a payoff of 1; and it is not assumed that
11
indifferent voters flip coins, in which case the first candidate would win with
probability 3/4. The payoffs specified are standard,1 however, and are relatively nicely behaved — the alternatives introduce additional discontinuities
into the model, making them less attractive as starting points.
Mixed strategies, Borel probability measures on X, are denoted by σ and
σ , and u is extended to pairs of mixed strategies as in Section 1.
0
Theorem 2 In the Downsian model of elections with three voters, there exists a mixed strategy equilibrium.
To prove the theorem, it suffices to verify conditions (C1) and (C2) from
Theorem 1, and, by symmetry, it suffices to check them for the first candidate.
Because existence of a pure strategy equilibrium is well-known when the core
is non-empty, we simplify the analysis of (C2) by considering only the case of
an empty core. First, given any x ∈ X, note that I(x) has Lebesgue measure
zero. Now fix σ 0 ∈ Σ◦ , and consider a sequence {xn } converging to x. Letting
χY denote the indicator function of set Y ⊆ X, note that
U (x, σ 0 ) = σ 0 (P (x)) − σ 0 (P −1 (x))
Z
Z
0
=
χP (x) (y) σ (dy) − χP −1 (x) (y) σ 0 (dy),
and similarly for each xn . By Lebesgue’s dominated convergence theorem, it
suffices to show that χP (xn ) converges to χP (x) almost everywhere-σ 0 , and that
χP −1 (xn ) converges to χP −1 (x) almost everywhere-σ 0 . Since σ 0 (I(x)) = 0, we
may take any y ∈ P (x) ∪ P −1 (x). If y ∈ P (x), i.e., χP (x) (y) = 1, then, since
P is open, we must have y ∈ P (xn ) for high enough n, i.e., χP (xn ) (y) = 1 for
high enough n. Thus, χP (xn ) (y) → χP (x) (y). If y ∈ P −1 (x), so χP (x) (y) = 0,
then, since P is open, we must have y ∈ P −1 (xn ) for high enough n, so
χP (xn ) (y) = 0 for high enough n. Thus, χP (xn ) (y) → χP (x) (y) again. The
argument that χP −1 (xn ) (y) converges to χP −1 (x) (y) is essentially the same.
We conclude that (C1) is satisfied.
For (C2), take any σ 0 ∈ Σ and any x ∈ X such that U (x, σ 0 ) > 0. Since
we have assumed that boundary points of X are Pareto dominated, we focus
1
See McKevley (1986), for example.
12
on the case in which x is interior to X. I will construct, for each ² > 0, a
non-empty open ball G² ⊆ X near x such that, for all z ∈ G² ,
U (z, σ 0 ) ≥ U (x, σ 0 ) − ².
Once that is done, the absolutely continuous mixed strategy σ ² defined by
the uniform distribution on the ball G² satisfies
U (σ ² , σ 0 ) ≥ U (x, σ 0 ) − ²,
as required. Define
Z
P
0
u(z, y) σ 0 (dy)
U (z, σ ) =
P (x)∪P −1 (x)
and
Z
I
0
u(z, y) σ 0 (dy),
U (z, σ ) =
I(x)
and note that
U (z, σ 0 ) = U P (z, σ 0 ) + U I (z, σ 0 ).
The argument of the previous paragraph shows that U P (·, σ 0 ) is continuous
at x, so we may specify an open ball G around x such that, for all z ∈ G,
²
U P (z, σ 0 ) ≥ U P (x, σ 0 ) − .
2
If σ 0 (I(x)) = 0, then we may set G² = G, and we are done.
Suppose that σ 0 (I(x)) > 0. Note that, for all w ∈ I(x), there exists i ∈ N
such that ui (x) = ui (w). Let φ : I(x) \ {x} → N be any measurable mapping
such that, for all w ∈ I(x) \ {x}, uφ(w) (x) = uφ(w) (w), and, for each i ∈ N ,
let αi = σ 0 (φ−1 (i)). That is, (α1 , α2 , α3 ) is a distribution on voters induced
by the mixed strategy σ 0 restricted to the points majority-indifferent to x,
where αi is roughly the probability of a point toward which, compared to x,
voter i is indifferent. Here, if several voters are indifferent between x and
some w, then we allocate the probability to one of them arbitrarily.
Construct a non-zero vector p∗ ∈ <d as follows. If there exists p ∈ <d
such that p · ∇ui (x) > 0 for each voter i, then, selecting one such vector, let
13
p∗ = p. If not, then there is a vector p ∈ <d such that p · ∇ui (x) > 0 for
at least two voters, say 1 and 2. Otherwise, were there no such vector, we
would have either ∇ui (x) = 0 for two voters or ∇ui (x) = 0 for one voter
and ∇uj (x) = −∇uk (x) for the remaining two. In the former case, by strict
pseudo-concavity, x is the ideal point of two voters, while in the latter case,
Plott’s (1967) symmetry condition is satisfied at x. Either way, x is a core
point, contradicting our maintained assumption in this proof that the core
is empty. Thus, the claim follows. In case α1 + α2 ≥ α3 , set p∗ = p. In case
α3 > α1 + α2 , there is some w ∈ I(x) \ {x} such that u3 (w) = u3 (x). Since
xIw, it must be that ui (w) ≥ ui (x) for at least one other voter, say voter 2.
Then pseudo-concavity implies (w −x)·∇u2 (x) > 0 and (w −x)·∇u3 (x) > 0,
and we set p∗ = w − x. This construction yields, along with p∗ , two voters,
j and k (with the third voter denoted by l), such that αj + αk ≥ αl , that
p∗ · ∇uj (x) > 0, and that p∗ · ∇uk (x) > 0.
For each δ > 0, now define xδ = x + δp∗ . By strict pseudo-concavity, for
δ small enough, we have uj (xδ ) > uj (x) and uk (xδ ) > uk (x), and so xδ P x.
Furthermore, for all w ∈ φ−1 (j), we may take δ small enough that xδ P w. The
argument is as follows. We have xIw and uj (x) = uj (w) by construction. If
uk (x) ≥ uk (w), then uj (xδ ) > uj (x) = uj (w) and uk (xδ ) > uk (x) ≥ uk (w) for
small enough δ implies xδ P w. Otherwise, if uk (x) < uk (w), then xIw implies
either that ul (x) > ul (w), in which case continuity yields ul (xδ ) > ul (w) for
δ small enough, or that ul (x) = ul (w). In the latter case, by strict pseudoconcavity, we have (w − x) · ∇ui (x) > 0 for each voter i, and then, by
construction, we have p∗ · ∇ui (x) > 0 for each voter i. This implies that, for
small enough δ and for each voter i, we have ui (xδ ) > ui (x), which yields
uj (xδ ) > uj (x) = uj (w) and ul (xδ ) > ul (x) = ul (w). We then have xδ P w for
small enough δ, as claimed. A similar claim holds for all w ∈ φ−1 (k). Thus,
χφ−1 (j)\P −1 (xδ )
and
χφ−1 (k)\P −1 (xδ )
converge to zero pointwise as δ goes to zero. Therefore, by Lebesgue’s dominated convergence theorem, we can choose δ small enough that xδ P x, that
σ 0 (φ−1 (j) \ P −1 (xδ )) <
²
8
and
and that xδ ∈ G.
14
σ 0 (φ−1 (k) \ P −1 (xδ )) <
²
8
Assuming zP x, we may write
U I (z, σ 0 )
Z
0
= σ ({x}) +
u(z, y) σ 0 (dy)
−1
φ (j)
Z
Z
0
+
u(z, y) σ (dy) +
u(z, y) σ 0 (dy)
−1
−1
φ (k)
φ (l)
Z
= σ 0 ({x}) +
u(z, y) σ 0 (dy)
φ−1 (j)∩P −1 (xδ )
Z
Z
0
+
u(z, y) σ (dy) +
u(z, y) σ 0 (dy)
−1
−1
δ
−1
−1
δ
Zφ (j)\P (x )
Zφ (k)∩P (x )
+
u(z, y) σ 0 (dy) +
u(z, y) σ 0 (dy).
φ−1 (k)\P −1 (xδ )
φ−1 (l)
Let {xn } be any sequence converging to xδ , and take any y ∈ φ−1 (j)∩P −1 (xδ ),
so that u(xδ , y) = 1. Recalling that xδ P x and P is open, we have xn P x and
u(xn , y) = 1 for high enough n. Similar remarks hold for y ∈ φ−1 (k) ∩
P −1 (xδ ). Then, by Lebesgue’s dominated convergence theorem,
lim inf U I (xn , σ 0 )
n→∞
0
≥ σ (φ−1 (j) ∩ P −1 (xδ )) + σ 0 (φ−1 (k) ∩ P −1 (xδ ))
−σ 0 (φ−1 (j) \ P −1 (xδ )) − σ 0 (φ−1 (k) \ P −1 (xδ )) − σ 0 (φ−1 (l))
²
= αj + αk − − αl
4
²
≥ − ,
4
where the last inequality uses αj + αk ≥ αl . Therefore, there is an open ball
G0 around xδ such that
²
U I (z, σ 0 ) ≥ −
2
0
for all z ∈ G .
Finally, let G² be any open ball around xδ contained in G ∩ G0 . Using
U (x, σ 0 ) = 0, we then have
I
U (z, σ 0 ) = U P (z, σ 0 ) + U I (z, σ 0 )
≥ U P (x, σ 0 ) + U I (x, σ 0 ) − ²
= U (x, σ 0 ) − ²
15
for all z ∈ G² , as desired.
4
More than Three Voters
Unfortunately, Theorem 1 does not readily yield the existence of mixed strategy equilibria in the Downsian model more generally, because the key condition (C2) becomes extremely difficult to verify. The way it was verified in the
proof of Theorem 2 was, for a given point x and mixed strategy σ 0 , to move
away from x an arbitrarily small amount to a point xδ that (i) is strictly
majority-preferred to x and (ii) is strictly majority-preferred to close to half
of the points in I(x) in the support of σ 0 . This approach is depicted in Figure
2, where the points majority-preferred to x form three petals with tips y, z,
and w, and the points majority-indifferent to x form the boundary of this set.
Suppose σ 0 puts at least as much probability mass on the indifference curves
of voters 1 and 2 in I(x) (indicated by dark curves) as on voter 3’s indifference curve in I(x). Any point in the y-petal will be preferred by voters 1 and
2 to x, and such points close enough to x will be strictly majority-preferred
to a set that contains, in the limit, the indifference curves of voters 1 and
2 in I(x), as required. Once the point xδ is found, it is possible to find a
sufficiently small open ball around it consisting of points also satisfying (i)
and (ii), and then the uniform distribution on this ball fulfills (C2).
[ Figure 2 here. ]
The problem with more than three voters is that the sets of voters indifferent between x and y, between x and z, and between x and w may be
pairwise disjoint. Term these voters “xy-indifferent,” and so on. Moreover,
it may be necessary, in order to deviate from x and win against any of these
alternatives, to move to a point preferred to x by all of the corresponding
indifferent voters. This contrasts with the three-voter example above, where
a point in the y-petal is preferred by voter 1, not voter 3, but is still strictly
majority-preferred to z. Using nine voters, I specify preferences so that, in
order to perturb x in a way preferred by all xy-indifferent voters, at least
one xz-indifferent voter and at least one xw-indifferent voter must be made
worse off. A similar problem is encountered in moving to points preferred to
x by all xz-indifferent voters or all xw-indifferent voters.
16
Assume X is two-dimensional and nine voters have preferences as depicted
in Figure 3. Though preferences are not specified completely here, it is clear
that the indifference curves depicted can be generated by continuous, strictly
quasi-concave utility functions. Letting Pi denote the strict preference of
voter i, majority preferences are defined as before, namely, xP y if and only if
#{i ∈ N | xPi y} ≥ 5, and xIy if neither xP y nor yP x. The payoff function
of the Downsian model is

if xP x0
 1
0
if xIx0
u(x, x0 ) =

−1 if x0 P x,
which again defines a symmetric two-player zero-sum game.
[ Figure 3 here. ]
The key in this example is the voters’ preferences over four points: x, y,
z, and w. Preferences restricted to that set are as follows, where parentheses
indicate that no restriction is imposed over a pair.
1
z
xw
y
2
3
z w
yx zx
w y
4
y
zx
w
5
(zw)
x
y
6
7
(yz) w
x
yx
w
z
8
y
xw
z
9
(yw)
x
z
Note that three voters, {1, 3, 5}, prefer x to y and four voters, {4, 6, 8, 9},
prefer y to x, with two, {2, 7}, indifferent. In order to find points arbitrarily
close to x that are strictly majority-preferred to y, it is necessary to make
both xy-indifferent voters, 2 and 7, better off, meaning that such points must
be in the shaded y-petal in Figure 3. Similarly, three voters, {7, 8, 9}, prefer
x to z and four voters, {1, 2, 5, 6}, prefer z to x, with two, {3, 4}, indifferent.
To find points arbitrarily close to x that are strictly majority-preferred to
z, it is necessary to make both xz-indifferent voters, 3 and 4, better off,
meaning that such points must be in the shaded z-petal. And finally, three
voters, {2, 4, 6}, prefer x to w and four voters, {3, 5, 7, 9}, prefer w to x,
with two, {1, 8}, indifferent. To find points arbitrarily close to x that are
strictly majority-preferred to w, it is necessary to make both xw-indifferent
voters, 1 and 8, better off, meaning that such points must be in the shaded
w-petal. Because the sets of xy-, xz-, and xw-indifferent voters are pairwise
17
disjoint, it will not be possible to find a point close to x that is strictly
majority-preferred to two of y, z, and w.
Now suppose that the first candidate takes position x and the second uses
the mixed strategy σ 0 that puts small probability ² > 0 on a and puts equal
weight on y, z, and w, i.e., σ 0 ({y}) = σ 0 ({z}) = σ 0 ({w}) = (1 − ²)/3. As
suggested by Figure 3, assume that voters 1-6 strictly prefer x to a, so in
particular xP a and U (x, σ 0 ) = ² > 0. The approach to verifying (C2) used
in the previous section is to find a small open ball around xδ , arbitrarily
close to x, consisting of points that defeat the second candidate’s positions
most of the time. In order to do that in this example, it is necessary to find
points arbitrarily close to x that are strictly majority-preferred to at least
two of y, z, and w. Such points must be preferred to x by at least two sets
of indifferent voters, which, as argued above, is impossible, and we conclude
that (C2) cannot be verified in this way.
Can (C2) be verified in other ways? If more structure is imposed on the
example, the answer is negative. Let X consist exactly of the shaded petals
in Figure 3 together with a,2 and further specify voter preferences as follows:
assume zI5 w, zI6 y, and yI9 w; assume 4 and 6 strictly prefer y to the interior
of the z-petal; assume 1 and 5 strictly prefer z to the interior of the w-petal;
and assume 7 and 9 prefer w to the interior of the y-petal. Figure 4 depicts
additional “indifference curves” to clarify these restrictions. This means that
y is majority-preferred to every alternative in the shaded z- and w-petals,
excluding x and z and w themselves: for example, voters 4, 6, 7, 8, and 9
strictly prefer y to the entire z-petal, except for x and z. Now there is no way
to verify (C2), because it is actually violated: every absolutely continuous
mixed strategy of the first candidate puts probability one on the interiors of
these petals, and the expected payoff of every alternative in the interior of a
petal is no greater than (4² − 1)/3 < 0.
[ Figure 4 here. ]
Note that voter preferences between y, z, and w, respectively, and a
have not been specified yet. Suppose that the three alternatives are strictly
majority-preferred to a, as would be the case in Figure 3 if a were moved
a sufficient distance from x to the left. Then there is a mixed strategy
2
This set is non-convex, of course, but that is not usually an issue for existence of mixed
strategy equilibria.
18
equilibrium in this model: let each candidate use the mixed strategy σ ∗ that
puts probability 1/3 each on y, z, and w. Clearly, (C2) is not necessary
for existence. Furthermore, we can verify that Reny’s (1999) diagonal better
reply security is satisfied. Note that the expected payoff of the first candidate
from using σ ∗ against any alternative in X \ {x, y, z, w} is positive. Now take
any pair (σ, σ 0 ) such that σ = σ 0 , and suppose that there is some alternative
v such that U (v, σ 0 ) > U (σ, σ 0 ) = 0. If σ 0 puts positive probability on X \
{x, y, z, w}, then σ ∗ fulfills the requirement of diagonal better reply security.
Otherwise, σ 0 has support on {x, y, z, w}, and v must be belong to X \
{x, y, z, w, a}. Without loss of generality, suppose v lies in the y-petal, so that
σ 0 ({x, y}) > 1/2, and then v itself fulfills the requirement of the condition.
Note that, for the mixed strategy σ 0 that puts probability ² on a and
(1−²)/3) each on y, z, and w, no pure strategy v is such that U (v, σ 0 ) > 0 and
there exists an open set G ⊆ Σ0 with σ 0 ∈ G and, for all σ̂ 0 ∈ G, U (v, σ̂ 0 ) > 0.
The reason is that U (v, σ 0 ) > 0 implies v ∈ {x, z}. In either case, the
mixed strategies σ̂ 0 that moves mass from z toward x satisfies U (v, σ̂ 0 ) < 0,
and such a strategy can be found arbitrarily close to σ 0 . Thus, the use of
mixed strategies in securing deviation payoffs is crucial. Moreover, these
mixed strategies may not be close to the original deviation. As explained
above, U (x, σ 0 ) > 0, but, because the alternatives near x generate a negative
expected payoff, no mixed strategy close to the point mass on x will generate
a positive payoff.
Now suppose that a is strictly majority-preferred to y and w. This would
be the case in Figure 3 if voters 3, 5, 7, 8, and 9 strictly preferred a to y and
if voters 4, 6, 7, 8, and 9 strictly preferred a to w. Then there is no mixed
strategy equilibrium in this model. To see this, suppose σ ∗ is an equilibrium
strategy. If σ ∗ puts greater probability on Ly ∪ Lw ∪ {a} than on Lz , then
U (z, σ ∗ ) ≥ σ ∗ (Ly ∪ Lw ∪ {a}) − σ ∗ (Lz ) > 0,
a contradiction. Therefore, σ ∗ (Lz ) ≥ σ ∗ (Ly ∪ Lw ∪ {a}). If σ ∗ puts greater
probability on Lz than on a, then
U (y, σ ∗ ) = σ ∗ (Lz ∪ Lw ) − σ ∗ (Ly ∪ {a}) > 0,
a contradiction. Therefore, σ ∗ (Lz ) = σ ∗ ({a}) and σ ∗ (Ly ∪ Lw ) = 0. Letting
v be an element of the z-petal sufficiently close to x that vP a, it follows
that U (v, σ ∗ ) = σ ∗ (Lz ) + σ ∗ ({a}), so that Lz and a must be chosen with
19
probability zero. Therefore, σ ∗ has support on {x, y, z, w}. Since aP y and
aP w, it must be that σ ∗ ({x, z}) ≥ 1/2.
5
Kramer’s Model
References
[1] C. Aliprantis and K. Border (1999) Infinite Dimensional Analysis: A
Hitchhiker’s Guide, 2nd ed., Springer-Verlag: New York.
[2] J. Banks (1995) “Singularity Theory and Core Existence in the Spatial
Model,” Journal of Mathematical Economics, 24: 523-536.
[3] J. Banks, J. Duggan, and M. Le Breton (2002) “Bounds for Mixed Strategy Equilibria and the Spatial Model of Elections,” Journal of Economic
Theory, 103: 88-105.
[4] P. Dasgupta and E. Maskin (1986) “The Existence of Equilibrium in
Discontinuous Economic Games, I: Theory,” Review of Economic Studies, 53: 1-26.
[5] I. Glicksberg (1952) “A Further Generalization of the Kakutani Fixed
Point Theorem, with Application to Nash Equilibrium,” Proceedings of
the American Mathematical Society, 3: 170-174.
[6] G. Kramer (1978) “Existence of Electoral Equilibrium,” in P. Ordeshook, ed., Game Theory and Political Science, NYU Press: New
York.
[7] G. Laffond, J.-F. Laslier, and M. Le Breton (1993) “The Bipartisan Set
of a Tournament Game,” Games and Economic Behavior, 5: 182-201.
[8] J.-F. Laslier and N. Picard (2002) “Distributive Politics and Electoral
Competition,” Journal of Economic Theory, 103: 106-130.
[9] J.-F. Mertens (1986) “The Minmax Theorem for U.S.C.-L.S.C. Payoff
Functions,” International Journal of Game Theory, 15: 237-250.
20
[10] R. McKelvey (1986) “Covering, Dominance, and the Institution-free
Properties of Social Choice,” American Journal of Political Science, 30:
283-314.
[11] C. Plott (1967) “A Notion of Equilibrium and its Possibility under Majority Rule,” American Economic Review, 57: 787-806.
[12] P. Reny (1999) “On the Existence of Pure and Mixed Strategy Nash
Equilibria in Discontinuous Games,” Econometrica, 67: 1029-1056.
[13] W. Rudin (1987) Real and Complex Analysis, 3rd ed., WCB/McGraw
Hill: New York.
[14] D. Saari (1997) “The Generic Existence of a Core for q-Rules,” Economic
Theory, 9: 219-260.
[15] N. Schofield (1983) “Generic Instability of Majority Rule,” Review of
Economic Studies, 50: 695-705.
[16] L. Simon (1987) “Games with Discontinuous Payoffs,” Review of Economic Studies, 54: 569-597.
[17] M. Sion (1958) “On General Minimax Theorems,” Pacific Journal of
Mathematics, 8: 171-176.
[18] M. Sion and P. Wolfe (1957) “On a Game without a Value,” Contributions to the Theory of Games, III, Princeton: Annals of Mathematical
Studies No 39, 299-306.
21