Public Information and Electoral Bias∗
Curtis R. Taylor † and Huseyin Yildirim‡
March 2005
Abstract
We present a theory of strategic voting that predicts elections are more likely to be
close and voter turnout is more likely to be high when citizens possess better public
information about the composition of the electorate. These findings are disturbing because
they suggest that providing more information to potential voters about aggregate political
preferences (e.g., through polls, political stock markets, or expert forecasts) may actually
undermine the democratic process. We show that if the distribution of preferences is
common knowledge, then strategic voting leads to a stark neutrality result in which the
probability that either alternative wins the election is 12 . This occurs because members
of the minority compensate exactly for their smaller group size by voting with higher
frequency. By contrast, when citizens are symmetrically ignorant about the distribution
of types, the majority is more likely to win the election and expected voter turnout is lower.
Indeed, when the population is large and voting costs are small, the majority wins with
probability arbitrarily close to one in equilibrium. Welfare is, therefore, unambiguously
higher when citizens possess less information about the distribution of political preferences.
∗
We thank Han Hong and Oksana Loginova for helpful suggestions. Taylor’s research was supported in
part by NSF grant SES-0417737.
†
Department of Economics, Duke University, Box 90097, Durham, NC 27708-0097, email: [email protected]
‡
Department of Economics, Duke University, Box 90097, Durham, NC 27708-0097, email:
[email protected]
1
The time has come, I think, to advocate the unspeakable in a forthright and
unapologetic manner, and not in the facetious or peripheral way that tradition
and circumstance have heretofore demanded. By our rules of procedure, and by
any scientific method of counting or reckoning under these rules, the race for the
presidency has ended in a flat tie and should be decided by the toss of the coin . . .
—Stephen Jay Gould, Boston Globe, November 30, 2000.
1
Introduction
The U.S. presidential election of 2000 between George W. Bush and Al Gore resulted in a
virtual tie. Gore won the national popular election by 532,994 votes or just 0.35%. Bush
won the popular election in the decisive state of Florida by 537 votes or just 0.008%. Bush’s
ultimate margin of victory in the Electoral College was a mere 5 of 537 votes, the thinnest
margin since Hayes beat Tilden by a single vote in 1876. While the U.S. presidential contest
of 2000 is the most notable election in recent history to result in a virtual tie, it is not the only
one. The Washington State gubernatorial race of 2004, for example, was decided by 129 votes
in a hand recount of over 2.9 million ballots.1 Similarly, the 1997 race for the U.K. House
of Commons seat for Winchester resulted in a deadlock, the official count separating the top
two candidates by only 2 votes out of 52,198 cast. The two vote margin was disputed and the
election was ultimately decided by a court-ordered special runoff.2
In this paper we present a theory of strategic voting that predicts that elections are more
likely to be close and voter turnout is more likely to be high when citizens possess better public
information about the composition of the electorate. These findings are disturbing because
they indicate that providing more information to potential voters about aggregate political
preferences (e.g., through polls, political stock markets, or expert forecasts) may actually
undermine the democratic process.3 Specifically, two types of inefficiency may obtain. First,
candidates and/or issues may win elections even though they were preferred by only a minority
of the citizens (democratic inefficiency). Second, aggregate voter turnout may be excessive in
the sense that too many citizens expend resources in casting votes (economic inefficiency).
Our theory is built on a model in which citizens possess private valuations over electoral
1
See http://www.cnn.com/2005/allpolitics/01/12/washington.governor.ap/#contentarea.
See http://news.bbc.co.uk/vote2001/hi/english/voting system/newsid 1171000/1171887.stm.
3
For empirical evidence that polls do influence voter turnout, see Sudman (1986) and West (1991).
2
2
outcomes and in which voting is costly.4 Within this framework we explore two informational regimes, one in which the distribution of political preferences is common knowledge
(the informed-voter setting) and one in which citizens are symmetrically ignorant about this
distribution (the uninformed-voter setting). In each regime we are able to characterize a
unique symmetric Bayesian Nash Equilibrium (BNE) in which all citizens randomize between
voting for their preferred alternative and abstaining. We further show that this is the only
behavior consistent with a symmetric BNE of the voting game when the population of citizens
is sufficiently large.
We compare expected equilibrium outcomes across the two informational environments
and find stark differences. First, in the informed-voter setting, the probability that either
alternative wins the election under the mixed-strategy BNE equals
1
2
regardless of the distri-
bution of political preferences or the cost of voting. This neutrality result stems from the fact
that individuals who expect to be in the minority vote with higher probability than those who
expect to be in the majority in equilibrium. In other words, individuals who expect to be in
the minority suffer less from the free-rider problem. Indeed, their lower expected numbers are
offset exactly by their greater expected participation, so that the expected equilibrium number
of votes for each alternative is the same regardless of the actual distribution of preferences.
In the uninformed-voter setting, however, citizens are not able to base their voting decisions on the distribution of political preferences, since they know only their own types.
In fact, given a symmetric common prior over the parameter governing the distribution of
tastes, it follows that all citizens vote with the same probability regardless of type. This
equilibrium probability of voting is strictly positive, and the majority group, therefore, wins
the election with probability strictly greater than 12 . Importantly, it is also possible to show
that expected equilibrium voter turnout is lower in the uninformed-voter setting. Hence, the
uninformed-voter setting gives rise to elections involving higher expected social benefits and
lower expected social costs. In other words, when citizens possess less information about the
political landscape, elections are both more democratically efficient and more economically
efficient.
Since many (if not most) important elections involve a large number of potential voters,
it is crucial to know whether the uninformed-voter setting continues to yield higher welfare
in the limit as the number of citizens tends to infinity. In this context, it is possible to show
that the equilibrium number of votes for each alternative correspond to independent random
4
Citizens in our private-values model are differentiated by their intrinsic preferences over political alternatives. Hence, we do not study the information aggregation problem that is the focus of common-value models
such as Feddersen and Pesendorfer (1997, 1998).
3
variables following Poisson distributions with endogenously determined means.5 Armed with
this fact, it is straightforward to verify the asymptotic superiority of the uninformed voter
setting. Indeed, our strongest result holds in the limit as the number of citizens tends to
infinity and the relative cost of voting approaches zero. In this key situation, the alternative
favored by the majority wins the election with probability arbitrarily close to 1 when citizens
are uninformed but only with probability
1
2
when they are informed.
Our private-values costly-voting model follows in the tradition of the pioneering works by
Ledyard (1984) and Palfrey and Rosenthal (1983, 1985). Ledyard (1984) is primarily concerned
with formalizing political competition in a setting with rational voting and endogenously
determined political alternatives.6 Palfrey and Rosenthal (1983, 1984) focus on the issue of
equilibrium voter turnout. In their 1983 paper, Palfrey and Rosenthal consider a costly-voting
model with complete information and characterize a large number of equilibria, some of which
involve significant voter turnout. In their 1985 paper, however, they consider a modified model
in which citizens are privately informed about their cost of voting. This gives rise to a unique
equilibrium in which only citizens with costs below an endogenous threshold vote.7
More recent papers in the private-values costly-voting paradigm include Campbell (1999)
and Borgers (2004).8 Campbell (1999) studies a model in which members of the minority
group possess stronger political preferences (higher values or lower costs) than members of the
majority. He presents a limit result that is similar in spirit to our neutrality finding in the
informed-voter setting. Specifically, Campbell finds that the minority group wins the election
with probability no less than
1
2
when the number of citizens tends to infinity. While Campbell’s
findings are both intriguing and enlightening, there are some key differences between his
analysis and ours. First, our neutrality result holds even in small populations for a range of
parameters. Second, Campbell is primarily concerned with explaining the disproportionate
influence of zealous minority groups on democratic outcomes rather than on the informational
issues at the core of our analysis. Hence, Campbell does not consider the impact of public
information on electoral outcomes or voter turnout.
Borgers (2004) investigates a version of our informed-voter setting in which it is common knowledge that the distribution of political preferences is symmetric (i.e., each citizen
5
This finding is reminiscent of Myerson (1998, 2000). There are, however, two important differences. First,
the means of the Poisson distributions in our model are determined endogenously. Second, they are finite when
voting costs are positive.
6
For recent work on strategic candidacy, see Dutta, Jackson and Le Breton (2001).
7
Aldrich (1993) surveys the three main theories of voter turnout, including strategic models with costly
voting.
8
See also Osborne, Rosenthal and Turner (2000) who investigate strategic costly participation in meetings
rather than elections.
4
is equally likely to prefer either alternative). In this context, Borgers shows that equilibrium
voter turnout maybe excessive, and he argues cogently that compulsory-voting laws, therefore,
make little sense. While Borgers’ model is elegant and his findings regarding equilibrium expected voter turnout are significant, his analysis differs markedly from ours. First, because it
focuses on symmetrically distributed types, Borgers’ model cannot address the electoral bias
at the heart of our analysis. Second, Borgers (2004) does not consider the impact of public
information on expected turnout or electoral outcomes. Finally, Borgers restricts attention to
small populations of potential voters while our results are most powerful in large populations.
In the next section we set out the model. Sections 3 and 4 contain the analysis of the
informed-voter setting and the uninformed-voter setting respectively. In Section 5 we analyze
the uninformed-voter setting in the important case as the number of citizens tends to infinity.
Some brief concluding comments appear in Section 6. All proofs not in the text have been
relegated to the Appendix.
2
The Model
Suppose that there are n ≥ 2 risk-neutral agents who may cast a vote in an election between
two alternatives, A and B. Each agent is one of two types, one who prefers A or one who
prefers B. A type t agent receives a gross payoff normalized to 1, if alternative t is implemented
and 0 otherwise for t = A, B. The cost of voting is c ∈ (0, 1] for all agents. Hence, each agent
possesses two (relevant) actions, to abstain or to vote for his preferred alternative.9 The ex
post payoff of a type t agent is given in Table 1.
Abstain
Vote
t wins
1
1−c
t loses
0
0−c
Table 1: Ex Post Payoff to a Type t Agent
Agents simultaneously choose whether to vote. The election is decided by majority rule
and ties are broken by a fair coin toss. Each agent privately knows his type but believes that
the preferences of the other agents are determined by realizations of i.i.d. random variables
where the probability that an agent is type A is λ ∈ (0, 1).
We compare symmetric Bayesian-Nash Equilibrium (BNE) outcomes of the class of voting
games just described across two informational settings. In the informed-voter setting, the
value of λ is common knowledge among all agents. In the uninformed-voter setting, all agents
possess a non-degenerate common prior over possible values of λ.
9
Abstaining strictly dominates voting for one’s less preferred alternative.
5
3
Informed Voters
Let φt ∈ [0, 1] be the probability that a type-t agent votes (i.e., he abstains with probability
1 − φt ). A symmetric BNE in this setting is a pair of probabilities (φA , φB ) such that it is
optimal for a type t agent to vote with probability φt when all other agents adhere to this
strategy. To find such an equilibrium, note that a type A agent compares the expected payoff
from abstaining, UA0 , to the expected payoff from voting, UA1 . Let k denote the number of
other agents who prefer alternative A (i.e., n − 1 − k agents prefer B), and let kA be the
number of other type A agents who vote and kB be the number of type B agents who vote.
The expected payoffs to a type A agent from abstaining and voting can then be written as
UA0
=
n−1
X
k=0
k X
n−1 k
k
n−1−k
λ (1 − λ)
φkA (1 − φA )k−kA
k
kA A

×
UA1
=
kA =0
kX
A −1 
n − 1 − k kB
1 n − 1 − k kA
φB (1 − φB )n−1−k−kB +
φB (1 − φB )n−1−k−kA  ,
kB
2
kA
kB =0
n−1
X
k=0

k X
n−1 k
k
n−1−k
λ (1 − λ)
φkA (1 − φA )k−kA
k
kA A
kA =0
n − 1 − k kB
n − 1 − k kA
n−1−k−kB
φB (1 − φB )
φB (1 − φB )n−1−k−kA
+
kB
kA
kB =0
1 n − 1 − k kA +1
n−2−k−kA
− c.
+
φB (1 − φB )
2 kA + 1
×
kX
A −1 Apart from the cost of voting, these expressions differ in two ways. First, when kB = kA ,
the agent in question is pivotal because his vote would break a tie in favor of alternative A
(raising the probability that A wins from
1
2
to 1). Second, when kB = kA + 1, the agent is
also pivotal because his vote would create a tie when alternative B would have otherwise won
(raising the probability that A wins from 0 to 12 ). The net expected benefit to a type A agent
from voting is, therefore, the probability that he is pivotal times one half minus his voting
cost:
k X
n−1 k
k
n−1−k
∆A =
λ (1 − λ)
φkA (1 − φA )k−kA
−
=
k
kA A
k=0
kA =0
n − 1 − k kA
1
×
φB (1 − φB )n−1−k−kA
kA
2
n − 1 − k kA +1
1
+
φB (1 − φB )n−2−k−kA
− c.
kA + 1
2
UA1
UA0
n−1
X
6
(1)
In order to write this expression in a more useful form, let the ex ante probability that an
agent votes for alternative A and B be denoted respectively by αA ≡ λφA and αB ≡ (1−λ)φB .
Hence, the ex ante probability of abstaining is 1 − αA − αB . Also, recall that the number of
ways kA agents can vote for A, kB can vote for B, and n − 1 − kA − kB can abstain is given
by the trinomial coefficient
n−1
kA , kB , n − 1 − kA − kB
≡
(n − 1)!
.
kA !kB !(n − 1 − kA − kB )!
Lemma 1 The net expected utility to a type t agent from voting can be written as
1
Pt (αt , αt0 ) − c,
2
(2)
where
Pt (αt , α ) ≡
t0
n−1
bX
2 c
k=0
n−1
αk αk0 (1 − αt − αt0 )n−1−2k
k, k, n − 1 − 2k t t
+
n−2
bX
2 c
k=0
for t = A, B, t 6=
t0 ,
(3)
n−1
αk αk+1
(1 − αt − αt0 )n−2−2k ,
0
k, k + 1, n − 2 − 2k t t
and b·c is the usual operator that rounds a number to the lower integer
when necessary.
This lemma has a very intuitive interpretation. It says that the net expected utility to
a type t agent from voting is composed of three parts. Specifically, Pt (αt , αt0 ) is the ex ante
probability that his vote is pivotal. The first summation in (3) is the probability that his vote
breaks a tie (i.e., the event that k of the other agents vote for each alternative and n − 1 − 2k
of them abstain). The second summation is the probability that his vote creates a tie (i.e.,
the event that k of the other agents vote for alternative t, k + 1 vote for t0 , and n − 2 − 2k
abstain). The third term in equation (2) is simply the agent’s cost of voting.
A noteworthy feature of Lemma 1 is that the net expected utility from voting can be
expressed entirely in terms of the ex ante probabilities, αA and αB . This fact gives rise to the
following result.
Proposition 1 (Neutrality) Suppose (φA , φB ) is a symmetric BNE in totally mixed strategies; i.e., 0 < φt < 1 for t = A, B.
(i) The ex ante probability that an agent votes for alternative A equals the ex ante probability
that he votes for B,
λφA = (1 − λ)φB .
7
(ii) Both outcomes are equally likely in equilibrium,
1
Pr{A wins|λ} = Pr{B wins|λ} = .
2
Proof of Proposition 1: In a symmetric BNE in totally mixed strategies, we must have
∆t = 0, or equivalently
1
Pt (αt , αt0 ) − c = 0
2
(4)
for t, t0 = A, B and t 6= t0 . Solving (4) for αt , we obtain
2c −
αt0 =
b n−1
2 c
P
k=0
b n−2
2 c
P
k=0
k k
n−1
k,k,n−1−2k αt αt0 (1
− αt − αt0 )n−1−2k
.
k k
n−1
k,k+1,n−2−2k αt αt0 (1
− αt − αt0 )n−2−2k
Hence, αA = αB , or equivalently λφA = (1 − λ)φB .
To prove part (ii), let αt = α in equilibrium. Note that the probability that alternative A
wins given that an odd number k of citizens vote is
k−1
P 2
n
n−k
k
(2α)
j=0
k (1 − 2α)
P
k
n
n−k (2α)k
j=0
k (1 − 2α)
k
j
αk−j αj
k k−j j
α
j α
.
Factoring αk out of the summations and cancelling like terms in the numerator and denominator yields
k
j=0 j
Pk
k
j=0 j
P k−1
2
1
= .
2
If an even number k > 0 of citizens vote, then a tie occurs in the event that j =
k
2
of them vote
for alternative B, in which case the election is decided by a coin toss. Hence, the probability
that alternative A wins is
k
k 1
+
k
j
2
2
Pk
k
j=0 j
P k2 −1
j=0
=
1
.
2
Finally, if k = 0 citizens vote, then A also wins with probability 12 . This result is disturbing. It says that in a mixed-strategy symmetric BNE, the probability
that either alternative is implemented does not depend on λ. Hence, the alternative that is
preferred by the expected majority wins the election with the same probability as the one that
is preferred by the expected minority.
8
The intuition underlying this result rests on a delicate tradeoff between an agent’s incentive
to win the election and his incentive to free ride on other voters who likely share his preferences.
For instance, if λ >
1
2,
then a type A agent knows that he is most likely a member of the
majority and he has a relatively high incentive to free ride (i.e., he votes with lower probability).
On the other hand, a type B agent knows that he is most likely a member of the minority
and he has a relatively low incentive to free ride (i.e., he votes with higher probability).
Remarkably, these effects exactly balance in equilibrium leading to a situation in which the
inherently stronger position of a type A agent is completely neutralized.10
Since the neutrality result in Proposition 1 presumes that the agents play a symmetric
totally mixed-strategy BNE, it is important to determine the conditions under which such an
equilibrium exists. Let α denote the ex ante probability that an agent votes for one of the
two alternatives in a symmetric totally mixed-strategy BNE (when one exists). Since the ex
ante probability that he votes at all is 2α, it must be that 0 < α ≤
1
2.
In order to find a
mixed-strategy equilibrium, it is, therefore, necessary to find a solution in this range to the
polynomial indifference equation
1
P (α, n) − c = 0,
2
(5)
where
P (α, n) ≡ Pt (α, α) =
n−1
bX
2 c
k=0
+
n−1
α2k (1 − 2α)n−1−2k
k, k, n − 1 − 2k
(6)
n−2
bX
2 c
k=0
n−1
α2k+1 (1 − 2α)n−2−2k .
k, k + 1, n − 2 − 2k
The function P (α, n) gives the probability that a given agent’s vote is pivotal when the other
n − 1 agents vote for each alternative with ex ante probability α and abstain with probability
1 − 2α.
Lemma 2 P (α, n) is strictly decreasing in α ∈ [0, 12 ].
10
Partial evidence in favor of our neutrality result comes from the frequent failure of pre-election polls in
predicting the election outcomes. For instance, Jowell, Hedges, Lynn, Farrant and Heath (1993) report that
in the 1992 British election, while all polls predicted a very close finish, with the Labor Party victory, the
Conservatives in fact won with a significant lead of 7.6 percent. Similarly, Durand, Blais and Vachon (2001)
record that in the 1998 general elections in Quebec, public polls overwhelmingly predicted an easy victory for
the ruling Parti Quebecois, a party dedicated to Quebec sovereignty. Yet, the Quebec Liberal Party ended
up winning. Finally, Durand, Blais and Larochelle (2004) note that in the 2002 French presidential elections,
although polls consistently predicted a matchup between the incumbent president, Jacques Chirac, and the
incumbent prime minister, Lionel Jospin, in the second round, Jean-Marie Le Pen from an extremist right-wing
party instead finished second. Interestingly, a common defense by the pollsters has been the weak turnout by
the expected majority on the election day.
9
Hence, the probability that a given agent’s vote is pivotal is lower when all other agents
vote with higher ex ante probability. An important implication of this result is that there is
at most one solution to the polynomial indifference equation (5). Furthermore, there exists a
solution if and only if 12 P (0, n) > c and 12 P ( 12 , n) ≤ c. Let α∗ (c, n) denote the solution when it
exists. Note that existence of α∗ (c, n) is necessary but not sufficient for existence of a totally
mixed-strategy BNE. In particular, existence of a totally mixed-strategy BNE also requires
both α∗ (c, n)/λ and α∗ /(1 − λ) < 1.
Proposition 2 (Characterization) There exists a unique symmetric mixed-strategy BNE
if and only if
c(n) ≤ c <
1
2
(7)
and
α∗ (c, n) < λ < 1 − α∗ (c, n),
(8)
n
 1
n−1


, if n is odd

n−1
2
2
c(n) ≡
n−1
1 n


, if n is even.

n
2
2
(9)
where
Proof of Proposition 2: Since, from Lemma 2, P (α, n) is strictly decreasing in α ∈
[0, 12 ], there is at most one solution to
only if
1
2 P (0, n)
1
1
2 P ( 2 , n)
− c > 0 and
1
1
2 P ( 2 , n)
1
2 P (α, n)
− c = 0. Moreover, a solution exists if and
− c ≤ 0. It is easy to verify that P (0, n) = 1 and
= c(n) as given in (9). Hence, there is a unique solution α∗ (c, n) to equation (5) if
and only if c(n) ≤ c < 12 . For α∗ (c, n) to be part of an equilibrium, it also needs to satisfy: (1)
0 < φA =
α∗ (c,n)
λ
< 1, and (2) 0 < φB =
α∗ (c,n)
1−λ
< 1, or equivalently α∗ (c, n) < λ < 1−α∗ (c, n).
Condition (7) specifies the range of voting costs for which a solution α∗ (c, n) ∈ (0, 12 ] to
(6) exists. Specifically, if c ≥ 12 , then the unique BNE is for all agents to abstain. A deviating
agent would surely be pivotal (P (0, n) = 1), but deviating is still not profitable ( 12 P (0, n) ≤ c).
On the other hand, if c ≤ c(n), then at least one type of agent must vote with probability
1 in equilibrium since α∗ (c, n) ≥
1
2.
When (7) is satisfied, existence of a symmetric totally
mix-strategy BNE also requires that λ not be too extreme. In particular, if (8) fails, then the
agents in the expected minority will be unable to neutralize the expected majority even when
voting with probability 1.
10
As an illustration of Proposition 2, suppose n = 2. In this case α∗ = 1 − 2c. Hence, a
symmetric mixed-strategy equilibrium exists if and only if c ∈ ( 14 , 12 ) and λ ∈ (1 − 2c, 2c). In
fact, this example corresponds to the most stringent setting possible. In particular, as the
number of agents increases, the region of parameter values for c and λ satisfying (7) and (8)
grows. This is formalized in the following lemma.
Lemma 3
(i) c(n) is decreasing, and converges to zero as n → ∞.
(ii) α∗ (c, n) is decreasing in n, and converges to zero as n → ∞.
This lemma indicates that the region of the parameter space where the totally mixedstrategy BNE prevails grows as the population of agents increases. This is important in
view of the fact that many key elections involve a relatively large number of potential voters.
Combining Lemma 3 with Propositions 1 and 2 yields the following key result.
Theorem 1 (Neutrality in Large Elections) For any voting cost c ∈ (0, 12 ) and any distribution of preferences λ ∈ (0, 1), there exists a critical population size n such that n ≥ n
implies the existence of a unique symmetric BNE (φA , φB ). Moreover, in this equilibrium
λφA = (1 − λ)φB and
1
Pr{A wins|λ} = Pr{B wins|λ} = .
2
Proof of Theorem 1: This directly follows from Proposition 2 and Lemma 3. This theorem shows that the neutrality result identified in Proposition 1 is endemic to
elections with large populations. In other words, when the distribution of voter preferences
is common knowledge, large elections are likely to be close even when the expected fraction
of the population preferring one of the alternatives is arbitrarily close to one and the cost of
voting is arbitrarily close to zero.
The disturbing neutrality results given in Proposition 1 and Theorem 1 are a consequence
of the strategic voting behavior that arises when agents have good information about voter
preferences. When agents do not know the actual value of λ, they are unable to tailor their
voting behavior precisely to match the environment, and the alternative favored by the majority is more likely to win. This is demonstrated in the following two sections, which consider
a setting with uninformed voters in the respective cases where n is finite and infinite.
11
4
Uninformed Voters
Consider a setting in which λ is not common knowledge. Specifically, suppose that before
learning their types, the agents’ beliefs about λ correspond to a non-degenerate common prior
distribution. For ease of exposition, assume that the prior is symmetric and defined over
a finite set of values, λ1 < λ2 < · · · < λr , where r ≥ 2. Symmetry of the prior requires
λi = 1 − λr+1−i and Pr{λ = λi } = Pr{λ = 1 − λi } = θi ∈ (0, 1), for i = 1, . . . , r. This implies
that E[λ] = 12 . Also, after learning their types, agents’ updated beliefs are
Pr{λ = λi |A} = 2θi λi
and
Pr{λ = λi |B} = 2θi (1 − λi ).
As in the informed-voter setting, a symmetric BNE in this context corresponds to a pair
of voting probabilities (φA , φB ). Moreover, (φA , φB ) is a symmetric BNE in totally mixed
strategies if and only if it satisfies the indifference conditions
1
E[PA (λφA , (1 − λ)φB )|A] − c = 0
2
(10)
1
E[PB ((1 − λ)φB , λφA )|B] − c = 0,
2
(11)
and
where Pt (·, ·) is defined in (3).
Proposition 3 (Non-Neutrality) Suppose (φA , φB ) is a symmetric BNE in totally mixed
strategies.
(i) Both types of agents vote with the same probability, φA = φB = φ.
(ii) For n < ∞, the alternative favored by the expected majority is more likely to win the
election,
Pr{A wins |λ} >
1
1
⇔ λ> .
2
2
The first part of this result indicates that any symmetric BNE in totally mixed strategies
must be strongly symmetric in the sense that all agents vote with the same probability regardless of type. The intuition is straightforward. When the prior is symmetric, an agent who
discovers that he is type A learns as much about the environment as an agent who discovers
12
that he is type B. Hence, both types vote with the same probability in equilibrium. This
contrasts sharply with part (i) of Proposition 1 where it is shown that agents in the expected
minority vote with higher probability than those in the expected majority. This effect is absent
in the uninformed-voter setting because no agent expects to be in the minority conditional on
learning only his own type. In particular, E[λ|A] = 1 − E[λ|B] = 2E[λ2 ] > 12 .
Next, consider part (ii) of Proposition 3. Because both types of agents vote with the same
probability, it follows that the alternative with the expected majority is more likely to win the
election. Again, this contrasts with Proposition 1 where it was shown that each alternative
was equally likely to win when λ was common knowledge.
In order to characterize a symmetric BNE in totally mixed strategies in the uninformedvoter setting, recall from Proposition 3 that all agents vote with the same probability φ. Thus,
given λ, the probability that an agent’s vote is pivotal is
P̃A (φ|λ, n) ≡ PA (λφ, (1 − λ)φ)
=
bXc +
n−2
bX
2 c
k=0
k=0
(12)
n−1
λk (1 − λ)k φ2k (1 − φ)n−1−2k
k, k, n − 1 − 2k
n−1
2
n−1
λk (1 − λ)k+1 φ2k+1 (1 − φ)n−2−2k
k, k + 1, n − 2 − 2k
if he is type A, and
P̃B (φ|λ, n) ≡ PB ((1 − λ)φ, λφ)
=
bXc +
b Xc k=0
n−2
2
k=0
(13)
n−1
λk (1 − λ)k φ2k (1 − φ)n−1−2k
k, k, n − 1 − 2k
n−1
2
n−1
λk+1 (1 − λ)k φ2k+1 (1 − φ)n−2−2k
k, k + 1, n − 2 − 2k
if he is type B. Under the symmetric prior, each type of agent expects to be pivotal with the
same probability, namely
Q(φ, n) ≡ E[P̃t (φ|λ, n)|t] =
r+1
bX
2 c
i=1
13
2θi T (φ, n, λi ),
(14)
t = A, B, where for future use we define
T (φ, n, λi ) ≡ λi P̃A (φ|λi , n) + (1 − λi )P̃A (φ|1 − λi , n)
=
(15)
bXc n−1
2
k=0
+2
n−1
λk (1 − λi )k φ2k (1 − φ)n−1−2k
k, k, n − 1 − 2k i
n−2
bX
2 c
k=0
n−1
λk+1 (1 − λi )k+1 φ2k+1 (1 − φ)n−2−2k
k, k + 1, n − 2 − 2k i
for λi 6= 12 , and let T (φ, n, 12 ) = 12 P̃A (φ| 12 , n) for consistency. Hence, the indifference conditions
for a BNE in totally mixed strategies given in (10) and (11) reduce to the single polynomial
indifference equation
1
Q(φ, n) − c = 0.
2
(16)
In other words, a symmetric BNE in totally mixed strategies corresponds to a solution φ∗ (n) ∈
(0, 1) to (16). The following result gives a necessary and sufficient condition for the existence
of such a solution and shows that if one exists, then it is unique.
Proposition 4 (Characterization) There exists a unique symmetric BNE in totally mixed
strategies if and only if c(n) < c < 12 , where

i
h
n−1
 c(n)2n E λ n+1
2 (1 − λ) 2
, if n is odd
i
h
c(n) =
n
 c(n)2n E (λ(1 − λ)) 2 ,
if n is even.
Similar to Proposition 2 of the preceding section, this result characterizes the region of
the parameter space where the unique symmetric BNE in totally mixed strategies obtains.
Specifically, if c > 12 , then no agents vote in equilibrium, and if c < c(n), then they all vote
with certainty.11
There are two important questions regarding the value of c(n). First, in order to facilitate
comparison with the informed-voter setting, it is necessary to know how c(n) compares with
c(n). Second, since a major focus of this investigation is elections with a large number of
agents, it is important to know the limiting value of c(n). Both questions are answered in the
following result.
Corollary 1 For any n, c(n) < c(n),and hence lim c(n) = 0.
n→∞
11
Recall, however, that in the informed-voter setting, if c < c(n), then agents in the minority group vote with
certainty, whereas agents in the majority might still mix.
14
Proof of Corollary 1: When n is odd, note that
c(n) = c(n)2
n
r
X
n+1
2
θi λi
(1 − λi )
i=1
n−1
2
< c(n)2
n
r
X
i=1
n−1
1 2
n1
= c(n)2
= c(n).
2 4
n−1
1 2
θi λi
4
The same argument proves the case for n even. Finally, Lemma 3 established that lim c(n) =
n→∞
0, and hence, lim c(n) = 0. n→∞
This result is helpful because it indicates that if a symmetric BNE in totally mixed strategies obtains in the informed-voter setting, then an analogous BNE obtains in the uninformedvoter setting. Moreover, for n sufficiently large, existence and uniqueness of these equilibria
are assured. With this result in hand, it is possible to compare the equilibrium expected voter
turnout across the two informational regimes.
Proposition 5 (Expected Turnout) Suppose (7) and (8) hold so that a symmetric totallymixed strategy BNE obtains in either informational setting. Then, the expected equilibrium
voter turnout is higher in the informed-voter setting than in the uninformed-voter setting.
Proof of Proposition 5: The expected equilibrium voter turnout is 2nα∗ and nφ∗ in the
∗
informed and uninformed cases, respectively. Hence, we must show α∗ > φ . Define φb = φ
2
b i , n) ≡ P̃A (2φ|λ
b i , n). Then,
and PbA (φ|λ
b i , n) =
PbA (φ|λ
+
n−1
bX
2 c
n−1
b n−1−2k
22k λki (1 − λi )k φb2k (1 − 2φ)
k, k, n − 1 − 2k
k=0
n−2
bX
2 c
n−1
b n−2−2k .
22k+1 λki (1 − λi )k+1 φb2k+1 (1 − 2φ)
k, k + 1, n − 2 − 2k
k=0
Since 2λi (2(1 − λi )) < 1, we have
b i , n) <
PbA (φ|λ
n−1
bX
2 c
k=0
n−1
b n−1−2k
φb2k (1 − 2φ)
k, k, n − 1 − 2k
+ 2(1 − λi )
n−2
bX
2 c
k=0
Moreover, since
r
P
i=1
θi λi =
1
2
and
r
X
i=1
r
P
i=1
n−1
b n−2−2k .
φb2k+1 (1 − 2φ)
k, k + 1, n − 2 − 2k
θi λi 2(1 − λi ) < 12 , it follows that
b i , n) =
θi λi PbA (φ|λ
1
b n) < 1 P (φ).
b
Q(2φ,
2
2
15
2
b n)
Finally, Lemma 2 and the proof of Proposition 4 established that both P (φ̂, n) and Q(2φ,
∗
∗
b implying that α∗ > φ where 1 P (α∗ , n)−c = 0 and 1 Q(2( φ ), n)−c = 0. are decreasing in φ,
2
2
2
2
The intuition behind this result is easily grasped. For any λ, the expected equilibrium
voter turnout is 2α∗ n in the informed-voter setting and φ∗ n in the uninformed-voter setting.
To see that 2α∗ > φ∗ , consider a situation in which agents vote with the same ex ante
probability, 2α = φ, in each setting. In this case, a voter in the informed situation has a
higher probability of being pivotal than one in the uninformed situation, P (α, n) > Q(φ, n).
The reason is that when agents have better information, they are better positioned to vote
strategically. Specifically, with better information, agents who expect to be in the minority
vote with higher probability and agents who expect to be in the majority vote with lower
probability. This voter-composition effect leads to a closer election and higher probability of
being pivotal in the informed-voter setting when the ex ante probability of voting is the same
across regimes. Of course, in equilibrium the probability of being pivotal must equal 2c in
both settings. This, however, requires agents to vote with higher ex ante probability in the
informed-voter setting, 2α∗ > φ∗ .
This discussion indicates that agents are on average more likely to vote when they have
better information. It turns out that this intuition holds not only across informational regimes,
but within the uninformed-voter setting itself as is now demonstrated.
Definition 1 (Mean-Preserving Spread) Let θ 0 = (θ10 , . . . , θ 0 r+1 , . . . , θr0 ) and θ = (θ1 , . . . ,
b 2 c
θb r+1 c , . . . , θr ) be two symmetric distributions over {λ1 , . . . , λb r+1 c , . . . , λr }. Distribution θ 0
2
2
is said to be a mean-preserving spread of θ if there is some i0 ∈ {1, . . . , r+1
2 } such that
θi0 ≥ θi ,
θi0 ≤ θi ,
if i = 1, . . . , i0 .
if i = i0 + 1, . . . , r+1
2
Proposition 6 (Uncertainty) Let θ 0 and θ be two symmetric priors that induce respective
equilibrium voting probabilities φ0∗ and φ∗ . If θ 0 is a mean-preserving spread of θ, then φ0∗ ≤ φ∗ .
Proof of Proposition 6: Let θ 0 and θ be two symmetric distributions such that θ 0 is a
mean-preserving spread over θ as defined above. ¿From (14) and (16), it is evident that to
show φ0∗ ≤ φ∗ , it suffices to show that for a fixed φ ∈ (0, 1),
r+1
bX
2 c
i=1
θi0 T (φ, n, λi ) −
r+1
bX
2 c
θi T (φ, n, λi ) ≤ 0.
(17)
i=1
It is easy to verify from (15) that T (φ, n, λi ) is increasing in λi for λi < 12 , stationary at
16
λi = 12 , and decreasing in λi for λi > 12 . Hence,
r+1
bX
2 c
θi0 T (φ, n, λi ) −
i=1
r+1
bX
2 c
θi T (φ, n, λi )
i=1
=
i0
X
(θi0 − θi )T (φ, n, λi ) +
i=1
≤
i0
X
i=1
r+1
bX
2 c
(θi0 − θi )T (φ, n, λi )
i=i0 +1
(θi0 − θi )T (φ, n, λi0 ) +
r+1
bX
2 c
(θi0 − θi )T (φ, n, λi0 )
i=i0 +1

bXc
bXc


= T (φ, n, λi0 ) 
θi0 −
θi  ≤ 0. 
r+1
2
r+1
2
i=1
i=1
This result indicates that agents are less likely to vote when there is more uncertainty
about the distribution of voter preferences. This makes sense. If the prior is more diffuse,
then agents believe that the election is less likely to be close and that their votes are, therefore,
less likely to be pivotal. Hence, they vote with lower probability in equilibrium.
At this point, it is possible to combine several findings to prove the following important
welfare result.
Theorem 2 (Welfare) Suppose (7) and (8) hold so that a symmetric totally-mixed strategy
BNE obtains in either informational setting. Then, for any given λ, expected equilibrium
welfare is higher in the uninformed-voter setting than in the informed-voter setting.
Proof of Theorem 2: Without loss of generality, suppose λ ≥ 12 . By part (ii) of Proposition
1, equilibrium expected welfare in the informed-voter setting is
1
n − c2α∗ n.
2
Similarly, expected equilibrium welfare in the uninformed-voter setting is
Pr{A wins |λ}λn + Pr{B wins |λ}(1 − λ)n − cφ∗ n.
By part (ii) of Proposition 3
1
Pr{A wins |λ}λ + Pr{B wins |λ}(1 − λ) ≥ .
2
Moreover, by Proposition 5 φ∗ < 2α∗ , from which the result follows. This is a central finding of the paper. For a fixed n, expected welfare is unambiguously
higher when agents are uninformed about the composition of the electorate. In particular, in
17
the uninformed-voter setting, the alternative preferred by the majority is more likely to win
the election and expected voting costs are strictly less.
An implication of this finding is that the public release of information such as publication
of pre-election polls that resolves all uncertainty about the distribution of voter preferences is
welfare-reducing. The reason for this inefficiency is twofold. First, when the distribution of
voter preferences becomes common knowledge, each agent knows exactly whether he belongs to
the expected minority or the expected majority group. In this instance, as explained following
Proposition 1, each agent tailors his voting strategy to his expected group size so that in
equilibrium each group is equally likely to win the election. When no information about
the distribution of voter preferences is provided, however, each agent votes with the same
equilibrium probability, making the expected majority more likely to win, which is obviously
better for expected welfare. Second, as Proposition 5 reveals, the expected total cost of voting
is smaller when voters are uninformed. The reason is that when uninformed, all agents put
more weight on being in the majority and hence possess less incentive to vote.
A key question is whether this welfare result continues to hold in the limit as n tends to
infinity. In particular, it is important to know the limiting expected turnout and the limiting
probability that the majority wins the election. These questions are addressed in the next
section.
5
Large Elections
In this section the uninformed-voter setting is studied in the important case as the number
of agents tends to infinity. The particular object of interest is the asymptotic equilibrium
probability that the alternative favored by the majority wins the election. In the informedvoter setting, Theorem 1 demonstrated that this probability is
1
2.
In the uninformed-agent
setting, individuals all vote with the same equilibrium probability φ∗ (n) implying that the
probability that the expected majority wins is greater than
infinity, however, Lemma 4 below shows that
φ∗ (n)
1
2
for finite n. As n tends to
goes to zero, so that more analysis is
required to determine whether the majority group continues to win with probability greater
than
1
2
in the limit.
In order to address this question, recall from (14) and (16) that the equilibrium condition
is given by
r+1
bX
2 c
θi T (φ, n, λi ) − c = 0,
i=1
where T (φ, λi , n) is defined in (15).
18
(18)
As was discussed in the previous section, when c ∈ c(n), 12 , condition (18) implicitly
defines the unique symmetric BNE voting probability φ∗ (n) employed by both types of agent.
Some important properties of this probability are summarized in the following lemma.
Lemma 4
(i) φ∗ (n) is decreasing.
(ii) limn→∞ φ∗ (n) = 0.
(iii) limn→∞ nφ∗ (n) < ∞.
Next, for a given n and λ, let XA and XB be the random variables representing the
equilibrium number of votes for alternatives A and B. Furthermore, let X0 = n − XA − XB be
the number of abstentions. Note that (XA , XB , X0 |n) ∼ Multinomial(λφ∗ (n), (1−λ)φ∗ (n), 1−
φ∗ (n)|n). The following lemma gives the asymptotic marginal distributions for XA and XB
and reveals that these random variables are statistically independent in the limit.
Lemma 5 Let limn→∞ nφ∗ (n) = m. Then, the limiting marginal distributions of XA and XB
are independent Poisson distributions with respective means λm and (1 − λ)m.
∗
λφ (n)
Proof of Lemma 5: Note that (XA |XB ) ∼ Binomial(n − XB , 1−(1−λ)φ
∗ (n) ). Since
lim
n→∞
λφ∗ (n)
(n − XB )
= λm,
1 − (1 − λ)φ∗ (n)
it follows (see Billingsley 1995, Theorem 23.2) that
D
(XA |XB ) −→ Poisson(λm),
which is independent of XB . The same argument shows
D
(XB |XA ) −→ Poisson((1 − λ)m).
Hence, the limiting distributions of XA and XB are independent. For notational ease, denote the Poisson p.d.f. with mean µ by f (x|µ). Now, in order to determine the limit of T (φ∗ (n), n, λi ), suppose that (XA , XB , X0 |n−1) ∼ Multinomial(λi φ∗ (n), (1−
λi )φ∗ (n), 1 − φ∗ (n)|n − 1). Then, inspection of (15) reveals that
T (φ∗ (n), n, λi ) = Pr{XB = XA } + 2λi Pr{XB = XA + 1}.
19
(19)
Lemma 5 indicates that in the limit XA and XB are independent random variables that follow
Poisson distributions with respective means λi m and (1 − λi )m. Applying these facts yields
lim T (φ∗ (n), n, λi ) = lim [Pr{XB = XA } + 2λi Pr{XB = XA + 1}]
n→∞
n→∞
=
∞
X
f (k|λi m)f (k|(1 − λi )m) + 2λi
k=0
= e−m
"
∞
X
f (k|λi m)f (k + 1|(1 − λi )m)
k=0
∞
X
k=0
#
∞
2k
2k+1
X
m
m
(λi (1 − λi ))k
+2
(λi (1 − λi ))k+1
≡ J(m, λi ).
(k!)2
k!(k + 1)!
k=0
Thus, in the limit, expected equilibrium voter turnout, m∗ , is determined by
r+1
bX
2 c
θi J(m, λi ) − c = 0,
(20)
i=1
which is the analogue of (18) when n = ∞. Before proving existence and uniqueness of m∗ , it
is necessary to establish the following important technical result.
Lemma 6 Let XA and XB be two independent random variables that follow Poisson distributions with respective means λm and (1 − λ)m. Then:
(i) limm→∞ Pr{XA = XB + k} = 0 for any finite integer k;
(ii) limm→∞ Pr{XA > XB } = 1 if and only if λ > 12 .
To interpret this result, suppose expected voter turnout, m, is exogenous. Part (i) of
Lemma 6 implies that limm→∞ Pr{XA = XB } = limm→∞ Pr{XA = XB + 1} = 0. Hence,
as m tends to infinity, the probability that a given agent’s vote is pivotal tends to 0. In
addition, as expected turnout becomes infinite, the probability that the alternative favored by
the majority wins the election goes to 1.
Lemma 7
(i) For c ∈ (0, 12 ), there exists a unique and finite number m∗ satisfying (20).
(ii) m∗ is strictly decreasing in c.
(iii) For any M < ∞ there exists c(M ) > 0 such that c < c(M ) implies m∗ > M .
Proof of Lemma 7: Each part is proven in turn.
20
(i) First observe that for c ∈ (0, 12 ), m∗ must be finite. Otherwise, the equilibrium condition
(20) would be violated as limm→∞ J(m, λi ) = 0 by part (i) of Lemma 6.
Next, we show that m∗ is unique. To do so, we establish that J(m, λi ) is strictly
decreasing in m. Differentiation yields
"∞
∞
2k−1
2k
X
X
∂
−m
k 2km
k+1 (2k + 1)m
(λi (1 − λi ))
+
2
(λ
(1
−
λ
))
J(m, λi ) = e
i
i
∂m
(k!)2
k!(k + 1)!
k=1
k=0
#
∞
∞
2k+1
X
X
m2k
k+1 m
−
(λi (1 − λi ))k
−
2
(λ
(1
−
λ
))
i
i
(k!)2
k!(k + 1)!
k=0
k=0
"∞
∞
X
X
2(k + 1)m2(k+1)−1
(2k + 1)m2k
= e−m
(λi (1 − λi ))k+1
+
2
(λi (1 − λi ))k+1
2
((k + 1)!)
k!(k + 1)!
k=0
k=0
#
∞
∞
2k
2k+1
X
X
k m
k+1 m
−
(λi (1 − λi ))
−
2
(λ
(1
−
λ
))
.
i
i
(k!)2
k!(k + 1)!
k=0
k=0
Since the first and last summations cancel out, we have
" ∞
#
∞
2k
2k
X
X
∂
−m
k+1 (2k + 1)m
k m
2
(λi (1 − λi ))
(λi (1 − λi ))
J(m, λi ) = e
−
∂m
k!(k + 1)!
(k!)2
k=0
k=0
∞ X
2λi (1 − λi )(2k + 1)
m2k
= e−m
− 1 (λi (1 − λi ))k
k+1
(k!)2
k=0
"
#
∞
X
2( 14 )(2k + 1)
m2k
−m
≤e
< 0.
− 1 (λi (1 − λi ))k
k+1
(k!)2
k=0
It is easy to verify that J(0, λi ) = 1. Moreover, since
r+1
bX
2 c
θi J(0, λi ) − c =
i=1
r+1
bX
2 c
θi (1) − c ≥
i=1
1
− c > 0,
2
and (by Lemma 6)
lim
m→∞
r+1
bX
2 c
θi J(m, λi ) − c = −c < 0,
i=1
there is a unique solution m∗ ∈ (0, ∞) to (20).
(ii) The fact that
∂
∂m J(m, λi )
< 0 implies that m∗ is strictly decreasing in c.
(iii ) Fix M < ∞ and define
c(M ) ≡
r+1
bX
2 c
i=1
21
θi J(M, λi ).
Since M is finite and
lim
m→∞
r+1
bX
2 c
θi J(m, λi ) = 0,
i=1
c(M ) > 0. Furthermore, since m∗ is strictly decreasing in c ∈ (0, 12 ), the result follows. Lemma 7 is of independent interest. First, it demonstrates that there exists a unique
limiting expected voter turnout m∗ . Moreover, m∗ is higher the lower is the cost of voting c,
and as c approaches zero, m∗ goes to infinity. In other words, for sufficiently small c, expected
turnout is arbitrarily high.
It is now posible to combine Lemmas 6 and 7 to prove the central result of this section.
Theorem 3 (Non-Neutrality in Large Elections)
(i) If n → ∞, then the alternative favored by the majority is more likely to win the election,
Pr{A wins|λ} >
1
1
⇔ λ> .
2
2
(ii) If n → ∞, then the probability that the alternative favored by the majority wins is arbitrarily close to one when the cost of voting is sufficiently small:
1
lim Pr{A wins|λ} = 1 ⇔ λ > .
c→0
2
Proof of Theorem 3: Each part is proven in turn.
(i) Observe that XA ∼ f (·|λm∗ ) and XB ∼ f (·|(1 − λ)m∗ ). Now,
Pr{A wins|λ} >
1
2
1
1
Pr{XA = XB } > Pr{XB > XA } + Pr{XB = XA }
2
2
⇔ Pr{XA > XB } > Pr{XB > XA }
∞ X
∞ X
∞
∞
X
X
∗
∗
⇔
f (j|(1 − λ)m )f (k|λm ) >
f (j|λm∗ )f (k|(1 − λ)m∗ ).
⇔ Pr{XA > XB } +
j=0 k=j+1
Now, suppose λ >
λ>
1
2
j=0 k=j+1
and consider the following sequence of implications where k 6= j:
1
⇔ λ > (1 − λ) ⇔ λk−j > (1 − λ)k−j
2
⇔ (λm∗ )k ((1 − λ)m∗ )j > (λm∗ )j ((1 − λ)m∗ )k
⇔ f (j|(1 − λ)m∗ )f (k|λm∗ ) > f (j|λm∗ )f (k|(1 − λ)m∗ )
∞
∞
∞ X
∞ X
X
X
∗
∗
⇔
f (j|(1 − λ)m )f (k|λm ) >
f (j|λm∗ )f (k|(1 − λ)m∗ ).
j=0 k=j+1
j=0 k=j+1
22
The same arguement shows that Pr{A wins|λ} <
=
1
2
1
2
if and only if λ < 12 , and Pr{A wins|λ}
if and only if λ = 12 .
(ii) This follows from part (ii) of Lemma 6 and part (iii) of Lemma 7. Together, theorems 1, 2 and 3 tell a compelling story. Specifically, whether elections are
large or small, a setting where agents are uninformed about the distribution of voter tastes
results in electoral outcomes possessing higher expected social benefits an lower expected
social costs. The difference between the informational regimes is most stark in the case of
large elections in which the relative cost of voting is small. In this case, Theorem 3 indicates
that the alternative favored by the majority will win the election with probability close to
one in equilibrium. When λ is common knowledge, however, the probability of an efficient
electoral outcome is only 12 .12
As an illustration of Theorem 3, suppose that the prior assigns probability
1
2
to each of
two possible values for λ ∈ {λL , λH } and that the relative cost of voting is c = 0.001. Table 2
presents the limiting equilibrium expected voter turnout and the asymptotic probability that
the majority wins the election for various values of λH :
λH
0.6
0.7
0.8
0.9
m∗
169.375
48.4721
22.1975
11.9319
Pr{A wins|λ}
0.995528
0.997723
0.998461
0.998824
Table 2: Large Efficient Elections
Consistent with Proposition 6, as λH increases and λL symmetrically decreases, the limiting
expected equilibrium turnout, m∗ falls. Nevertheless, the probability that the majority wins
the election remains very high so long as the relative cost of voting is small.
6
Conclusion
In this paper we have explored the impact of public information about the composition of
the electorate on equilibrium voting behavior. Our theoretical findings demonstrate that
providing more information of this kind to potential voters harms the democratic process.
It biases electoral outcomes toward the alternative preferred by the minority and it leads to
higher expected aggregate voting costs.
12
Analysis along the same lines as presented above confirms that as n tends to infinity, the number of votes
for each alternative in the informed-voter setting follow independent Poisson distributions with equal means.
23
Our results pertain to a setting where voters possess intrinsic preferences (i.e., private
values) over electoral alternatives.13 Clearly, the release of public information about voter intentions in a common-values environment would have different effects. Nevertheless, it seems
plain that in many elections voters are motivated primarily by their fundamental ideologies.
Party affiliations and political labels such as Liberal and Conservative obviously connote intrinsic differences in ideological principles. Hence, in elections where ideologies are especially
important, the findings presented in this paper sound a precautionary note.
To the extent that opinion polls, political stock markets, and expert forecasts contain real
information about voter sentiment, their publication may actually impede efficiency, giving rise
to closely contested elections with extraordinary voter turnout. Before firm policy conclusions
can be drawn, however, future research must investigate the endogeneity of the source of
information about political preferences. In particular, given the feedback we find between
information and equilibrium voting behavior, it is important to understand the incentives for
individuals to report their true preferences to pollsters and the incentives for pollsters and
pundits to disclose publically and fully any information they obtain.14
13
See Hansen, Palfrey and Rosenthal (1987) for empirical evidence in favor of private-values costly-voting
models.
14
Ottaviani and Sorensen (2005) consider competition among forecasters whose payoffs depend on the ex
post accuracy of their forecasts. The difference in an election context is that accuracy depends on the voters’
endogenous behavior.
24
Appendix
Proof of Lemma 1: Recall that αA = λφA and αB = (1 − λ)φB . Define βA = λ(1 − φA )
and βB = (1 − λ)(1 − φB ). By substituting these and noting the following facts,
λk = λkA λk−kA
(1 − λ)n−1−k = (1 − λ)kA (1 − λ)n−1−k−kA
(1 − λ)n−1−k = (1 − λ)kA +1 (1 − λ)n−2−k−kA ,
equation (1) reduces to

k n−1
X
n−1 X k
1

∆A =
αkA β k−kA
k
kA A A
2
k=0
kA =0
n − 1 − k kA n−1−k−kA
n − 1 − k kA +1 n−2−k−kA
α B βB
−c
×
+
α B βB
kA
kA + 1
b n−1
2 c n−1 1 X X n−1
k
n − 1 − k kA kA k−kA n−1−k−kA
=
α A αB βA
βB
k
kA
kA
2
kA =0 k=0
b n−2
2 c n−1 k
n − 1 − k kA kA +1 k−kA n−2−k−kA
1 X X n−1
+
βB
− c.
α A α B βA
k
kA
kA + 1
2
kA =0 k=0
Moreover, given the following two combinatorial identities
n−1
k
n−1−k
n−1
n − 1 − 2kA
=
k
kA
kA
kA , kA , n − 1 − 2kA
k − kA
and
n−1
k
n−1−k
n − 2 − 2kA
n−1
,
=
k
kA
kA + 1
kA , kA + 1, n − 2 − 2kA
k − kA
∆A simplifies to
b n−1
n−1
2 c
X n − 1 − 2kA k−k n−1−k−k
n−1
1 X
kA kA
A
∆A =
α α
β A A βB
kA , kA , n − 1 − 2kA A B
2
k − kA
kA =0
k=0
c
b
n−1
X n − 2 − 2kA k−k n−2−k−k
n−1
1 X
A
αkAA αkBA +1
β A A βB
−c
kA , kA + 1, n − 2 − 2kA
2
k − kA
n−2
2
+
kA =0
k=0
c
b
1 X
n−1
αk αk (βA + βB )n−1−2k
k, k, n − 1 − 2k A B
2
n−1
2
=
k=0
1
+
2
n−2
bX
2 c
k=0
n−1
αk αk+1 (βA + βB )n−2−2k − c,
k, k + 1, n − 2 − 2k A B
25
where, w.l.o.g., we change index of summations to k.
The expressions in (2) and (3) then follows by simply observing that βA +βB = 1−αA −αB .
Relabelling A and B, we find a symmetric expression for ∆B . Proof of Lemma 2: We simply differentiate P (α, n) with respect to α

n−1
bX
2 c
∂
(n − 1)!

P (α, n) = 
2kα2k−1 (1 − 2α)n−1−2k
2
∂α
(k!) (n − 1 − 2k)!
k=1
−
n−1
bX
2 c
k=0

(n − 1)!

α2k 2(n − 1 − 2k)(1 − 2α)n−2−2k 
(k!)2 (n − 1 − 2k)!

b n−2
2 c
 X
+
k=0
−
n−2
bX
2 c
k=0
(n − 1)!
(2k + 1)α2k (1 − 2α)n−2−2k
k!(k + 1)!(n − 2 − 2k)!

(n − 1)!

α2k 2(n − 2 − 2k)(1 − 2α)n−3−2k  .
k!(k + 1)!(n − 2 − 2k)!
Next, we combine the first summation with the last one and the second summation with
the third one,

b n−1
2 c
∂
2k
 X
P (α, n) ≤ (n − 1)! 
α2k−1 (1 − 2α)n−1−2k
2
∂α
(k!) (n − 1 − 2k)!
k=1
−
n−2
bX
2 c
k=0

2(n − 2 − 2k)

α2k+1 (1 − 2α)n−3−2k 
k!(k + 1)!(n − 2 − 2k)!
+ (n − 1)!
n−2
bX
2 c
k=0

2k + 1
2(n − 1 − 2k)
−
k!(k + 1)!(n − 2 − 2k)! (k!)2 (n − 1 − 2k)!
−1
b n−1
2 c
 X
= (n − 1)! 
k=0
−
n−2
bX
2 c
k=0
α2k (1 − 2α)n−2−2k
2(k + 1)
α2k+1 (1 − 2α)n−3−2k
((k + 1)!)2 (n − 1 − 2(k + 1))!

2(n − 2 − 2k)

α2k+1 (1 − 2α)n−3−2k 
k!(k + 1)!(n − 2 − 2k)!
+ (n − 1)!
n−2
bX
2 c
k=0
2k + 1
2
−
2
k!(k + 1)!(n − 2 − 2k)! (k!) (n − 2 − 2k)!
26
α2k (1 − 2α)n−2−2k
≤ (n − 1)!
−1 b n−1
2 c
X
+ (n − 1)!
b Xc k=0
n−2
2
k=0
2(n − 2 − 2k)
2(k + 1)
−
2
k!(k + 1)!(n − 2 − 2k)! ((k + 1)!) (n − 1 − 2(k + 1))!
2
2k + 1
−
2
k!(k + 1)!(n − 2 − 2k)! (k!) (n − 2 − 2k)!
α2k+1 (1 − 2α)n−3−2k
α2k (1 − 2α)n−2−2k .
Because
((k +
2(n − 2 − 2k)
2(k + 1)
−
=0
− 1 − 2(k + 1))! k!(k + 1)!(n − 2 − 2k)!
1)!)2 (n
and
2k + 1
2
1
−
,
=−
2
k!(k + 1)!(n − 2 − 2k)! (k!) (n − 2 − 2k)!
k!(k + 1)!(n − 2 − 2k)!
it follows that
n−2
bX
2 c
∂
1
P (α, n) ≤ −(n − 1)!
α2k (1 − 2α)n−2−2k < 0.
∂α
k!(k + 1)!(n − 2 − 2k)!
k=0
Proof of Lemma 3: To prove part (i), note that
n
c(n + 2)
n+1 , if n is odd
=
n+1
c(n)
n+2 , if n is even,
which, given c(2) = c(3) = 14 , reveals





c(n) =




1
2
n−1
−1
2
Q
2k+1
2k+2 ,
k=0
n
−1
2
1 Q 2k+1
2
2k+2 ,
k=0
if n is odd
(A-1)
if n is even.
Since 2k+1
2k+2 < 1, c(n) is decreasing. Moreover, c(n) ≤ b(n) for some sequence b(n) ≡
n q
Q
k+1
√1
k+2 . Simple algebra shows b(n) = n+2 , which implies lim b(n) = 0. Hence, lim c(n) =
n→∞
k=0
n→∞
0.
To prove that α∗ (c, n) is decreasing in n, it suffices to prove P (α, n) is decreasing in n. By
definition,
P (α, n) − P (α, n + 1) =
n−1
bX
2 c
k=0
+
n−2
bX
2 c
−
n−1
bX
2 c
k=0
k=0
(n − 1)!
α2k (1 − 2α)n−1−2k
− 1 − 2k)!
(k!)2 (n
(A-2)
n
bX
2c
(n − 1)!
n!
2k+1
n−2−2k
(1 − 2α)
−
α
α2k (1 − 2α)n−2k
k!(k + 1)!(n − 2 − 2k)!
(k!)2 (n − 2k)!
k=0
n!
α2k+1 (1 − 2α)n−1−2k .
k!(k + 1)!(n − 1 − 2k)!
27
Before signing this expression, we suppose that n is odd, and re-write the third summation:
n
bX
2c
k=0
n!
α2k (1 − 2α)n−2k
(k!)2 (n − 2k)!
2 X
n−1
=
k=0
2k
(n − 1)!
1+
α2k (1 − 2α)n−2k
2
n − 2k (k!) (n − 1 − 2k)!
n−1
= (1 − 2α)
2
X
k=0
n−1
2
X
+ 2α
k=1
(n − 1)!
α2k (1 − 2α)n−1−2k
− 1 − 2k)!
(k!)2 (n
(n − 1)!
α2k−1 (1 − 2α)n−2k
(k − 1)!k!(n − 2k)!
n−1
= (1 − 2α)
2
X
k=0
n−1
−1
2
X
+ 2α
k=0
(n − 1)!
α2k (1 − 2α)n−1−2k
− 1 − 2k)!
(k!)2 (n
(n − 1)!
α2k+1 (1 − 2α)n−2−2k .
k!(k + 1)!(n − 2 − 2k)!
Inserting this into (A-2) and cancelling terms yield
n−1
P (α, n) − P (α, n + 1) = 2α
2
X
k=0
n−1
−1
2
+ (1 − 2α)
X
k=0
n−1
2
−
X
k=0
X k=0
n−1
2
−
X
k=0
(n − 1)!
α2k+1 (1 − 2α)n−2−2k
k!(k + 1)!(n − 2 − 2k)!
n!
α2k+1 (1 − 2α)n−1−2k
k!(k + 1)!(n − 1 − 2k)!
n−1
−1
2
=
2
1
+
n − 1 − 2k k + 1
(n − 1)!
α2k+1 (1 − 2α)n−1−2k
− 2 − 2k)!
(k!)2 (n
n!
2(n − 1)! n
α2k+1 (1 − 2α)n−1−2k +
α
n−1 2
k!(k + 1)!(n − 1 − 2k)!
2 !
n−1
−1
2
=
(n − 1)!
α2k (1 − 2α)n−1−2k
(k!)2 (n − 1 − 2k)!
X
((n + 1) − n)
k=0
2(n − 1)! n
+
α −
n−1 2
!
2
(n − 1)!
α2k+1 (1 − 2α)n−1−2k
(k!)2 (n − 1 − 2k)!
n!
n−1 n−1
2 !( 2
+ 1)!
αn > 0.
Given that α∗ (c, n) is decreasing in n and bounded below by 0, it converges to some α` ≥ 0.
1
P (α` , n)
n→∞ 2
Suppose α` > 0. Since P (α, n) is continuous in α, it follows that lim
28
− c = 0, or
lim P (α` , n) = 2c > 0.
n→∞
We now argue that for α` > 0, lim P (α` , n) = 0. To see this, note that since P (α` , n)
n→∞
is decreasing in n, it converges to some h(α` ) ≥ 0, and so does its subsequence P (α` , n + 1),
where, by definition
n
P (α` , n + 1) =
2
X
k=0
n!
α` 2k (1 − 2α` )n−2k
(k!)2 (n − 2k)!
n−1
2
X
n!
α` 2k+1 (1 − 2α` )n−1−2k
k!(k + 1)!(n − 1 − 2k)!
k=0
 n
2
X
n
(n − 1)!
= (1 − 2α` ) 
α` 2k (1 − 2α` )n−1−2k
n − 2k (k!)2 (n − 1 − 2k)!
+
k=0
n−1
2
+
X
k=0

n
(n − 1)!
α` 2k+1 (1 − 2α` )n−2−2k  .
n − 1 − 2k k!(k + 1)!(n − 2 − 2k)!
n
n→∞ n−2k
Since, for a fixed k, lim
n
n→∞ n−1−2k
= lim
= 1, we have lim P (α` , n + 1) = (1 −
n→∞
2α` ) lim P (α` , n), or, equivalently 2α` h(α` ) = 0, yielding h(α` ) = lim P (α` , n) = 0. This
n→∞
n→∞
contradicts lim P (α` , n) = 2c 6= 0. Hence, α` = 0. The same exact arguments hold for an
n→∞
even n. Proof of Proposition 3: To prove part (i), suppose (φA , φB ) is a symmetric BNE in
totally mixed strategies. Then, (10) and (11) imply
E[PA (λφA , (1 − λ)φB )|A] = E[PB ((1 − λ)φB , λφA )|B].
(A-3)
E[PB (1 − λ)φB , λφA )|B] = E[PA (λφB , (1 − λ)φA )|A].
(A-4)
Next, we establish
To see this, note from (3) that
E[PB ((1 − λ)φB , λφA )|B] =
n−1
bX
2 c k=0
×
r
X
+
n−2
bX
2 c ×
k=0
r
X
n−1
k, k, n − 1 − 2k
k
k
2θi (1 − λi )(λi φA ) ((1 − λi )φB ) (1 − λi φA − (1 − λi )φB )
n−1−2k
#
i=1
n−1
k, k + 1, n − 2 − 2k
2θi (1 − λi )(λi φA )
k+1
k
((1 − λi )φB ) (1 − λi φA − (1 − λi )φB )
i=1
29
n−2−2k
#
.
Now, by symmetry of the prior
r
X
2θi (1 − λi )(λi φA )k ((1 − λi )φB )k (1 − λi φA − (1 − λi )φB )n−1−2k
i=1
=
r h
X
2θi λr+1−i (λr+1−i φB )k ((1 − λr+1−i )φA )k (1 − λr+1−i φB − (1 − λr+1−i )φA )n−1−2k
i
i=1
and
r
X
2θi (1 − λi )(λi φA )k+1 ((1 − λi )φB )k (1 − λi φA − (1 − λi )φB )n−2−2k
i=1
=
r h
X
i
2θi λr+1−i (λr+1−i φB )k ((1 − λr+1−i )φA )k+1 (1 − λr+1−i φB − (1 − λr+1−i )φA )n−2−2k .
i=1
Substituting these equalities into the above expression for E[PB ((1 − λ)φB , λφA )|B] yields
(A-4). Next, combine (A-3) and (A-4) to get
E[PA (λφA , (1 − λ)φB )|A] = E[PA (λφB , (1 − λ)φA )|A].
To derive a contradiction to this, suppose w.l.o.g. that φA > φB . We show that this implies
E[PA (λφA , (1 − λ)φB )|A] − E[PA (λφB , (1 − λ)φA )|A] < 0.
¿From (3) we have
E[PA (λφA , (1 − λ)φB )|A] − E[PA (λφB , (1 − λ)φA )|A]
n−1 "
bX
X
r
2 c n−1
=
2θi λi (λi (1 − λi )φA φB )k
k, k, n − 1 − 2k
i=1
k=0
i
× (1 − λi φA − (1 − λi )φB )n−1−2k − (1 − λi φB − (1 − λi )φA )n−1−2k
+
n−2 "
bX
2 c X
r
n−1
2θi λi (λi (1 − λi )φA φB )k
k, k + 1, n − 2 − 2k
i=1
k=0
i
× (1 − λi )φB (1 − λi φA − (1 − λi )φB )n−2−2k − (1 − λi )φA (1 − λi φB − (1 − λi )φA )n−2−2k .
Both of the double summations in this expression are negative. To see that the first one is
negative, consider the inner summation
r
X
i=1
2θi λi (λi (1 − λi )φA φB )k
× (1 − λi φA − (1 − λi )φB )n−1−2k − (1 − λi φB − (1 − λi )φA )n−1−2k .
30
Use symmetry of the prior to combine terms 1 and r, 2 and r − 1, and so on, to get
r+1
bX
2 c
i=1
2θi (2λi − 1)(λi (1 − λi )φA φB )k
× (1 − λi φA − (1 − λi )φB )n−1−2k − (1 − λi φB − (1 − λi )φA )n−1−2k .
(If r is odd, the expression is still valid in this case since the term involving λb r+1 c = 12
2
is automatically zero.) If k = n−1
,
then
each
term
in
this
summation
is
evidently
zero. If
2
k <
n−1
2 ,
then each term is negative. To see this, note that 2λi − 1 < 0 because we are
summing over the left half of the distribution. Moreover,
φA > φB ⇒ (1 − 2λi )φA > (1 − 2λi )φB
⇒ 1 − λi φA − (1 − λi )φB > 1 − λi φB − (1 − λi )φA
⇒ (1 − λi φA − (1 − λi )φB )n−1−2k > (1 − λi φB − (1 − λi )φA )n−1−2k .
A similar argument shows that the second double summation in the above expression is also
negative.
The proof of part (ii) closely follows the proof of its counterpart in Proposition 1. Given
φA = φB = φ from part (i), the probability that alternative A wins when an odd number k of
citizens vote is
n
k (1
−
n
k (1
− φ)n−k φk
φ)n−k φk
k−1
P 2
j=0
P
k
k−j ((1
j (λφ)
k
k
k−j ((1
j=0 j (λφ)
−
λ)φ)j
− λ)φ)j
.
Factoring φk out of the summations and cancelling like terms in the numerator and denominator yields
P k−1
2
k k−j
(1 − λ)j
j=0 j λ
.
Pk
k k−j
(1 − λ)j
j=0 j λ
Since
k k−j
(1
j=0 j λ
Pk
− λ)j = 1, the probability that A wins is equal to
2 X
k
k−1
j=0
which is greater than
1
2
j
λk−j (1 − λ)j ,
if and only if λ > 12 .
If an even number k > 0 of citizens vote, then a tie occurs in the event that j =
k
2
of
them vote for alternative B, in which case the election is decided by a coin toss. Hence, the
31
probability that alternative A wins is
2
X
k
k
−1
λ
j
j=0
k−j
j
(1 − λ) +
k
k
2
1
λ (1 − λ)
.
2
k
2
k
2
To evaluate this probability, note that
k X
k
j=0
j
−1 2
X
k
k
λ
k−j
j
(1 − λ) =
j=0
j
λ
k−j
j
(1 − λ) +
k
k
2
1
λ (1 − λ)
2
k
2
k
2
k
X
k
k
k k−j
1
k
j
+
λ (1 − λ) + k λ 2 (1 − λ) 2
= 1.
j
2
2
k
j= 2 +1
Since
k
j
=
k k−j ,
the probability that A wins is strictly greater than
1
2
if and only if λ > 12 .
Finally, if k = 0 citizens vote, then A wins with probability 12 . Proof of Proposition 4: Recall from (14)
Q(φ, n) =
r+1
bX
2 c
2θi T (φ, n, λi ),
i=1
where T (φ, n, λi ) is defined in (15). First, we show that Q(φ, n) is decreasing in φ for φ ∈ (0, 1).
To do so, it suffices to show each term T (φ, n, λi ) is decreasing in φ. Differentiating yields
n−1
bX
2 c
∂
(n − 1)!
T (φ, n, λi ) = 2
λk (1 − λi )k φ2k−1 (1 − φ)n−1−2k
∂φ
(k − 1)!k!(n − 1 − 2k)! i
k=1
n−2
2
−
b Xc
k=0
(n − 1)!
λk (1 − λi )k φ2k (1 − φ)n−2−2k
(k!)2 (n − 2 − 2k)! i
+ 2λi (1 − λi )
n−2
bX
2 c
− 2λi (1 − λi )
n−3
bX
2 c
k=0
k=0
(2k + 1)(n − 1)!
λk (1 − λi )k φ2k (1 − φ)n−2−2k
k!(k + 1)!(n − 2 − 2k)! i
(n − 1)!
λk (1 − λi )k φ2k+1 (1 − φ)n−3−2k .
k!(k + 1)!(n − 3 − 2k)! i
Since
2
n−1
bX
2 c
k=1
(n − 1)!
λk (1 − λi )k φ2k−1 (1 − φ)n−1−2k
(k − 1)!k!(n − 1 − 2k)! i
= 2λi (1 − λi )
−1
b n−1
2 c
X
k=0
(n − 1)!
λk (1 − λi )k φ2k+1 (1 − φ)n−3−2k ,
(k)!(k + 1)!(n − 3 − 2k)! i
32
we have
n−2
bX
2 c
∂
(n − 1)!
T (φ, n, λi ) = −
λk (1 − λi )k φ2k (1 − φ)n−2−2k
2
∂φ
(k!) (n − 2 − 2k)! i
k=0
n−2
2
+ 2λi (1 − λi )
b Xc
k=0
≤−
n−2
bX
2 c
(n − 1)!
λk (1 − λi )k φ2k (1 − φ)n−2−2k
− 2 − 2k)! i
(k!)2 (n
k=0
+
1
2
n−2
bX
2 c
k=0
(2k + 1)(n − 1)!
λk (1 − λi )k φ2k (1 − φ)n−2−2k
k!(k + 1)!(n − 2 − 2k)! i
(2k + 1)(n − 1)!
λk (1 − λi )k φ2k (1 − φ)n−2−2k
k!(k + 1)!(n − 2 − 2k)! i
b n−2
2 c
(n − 1)!
1 X
=−
λk (1 − λi )k φ2k (1 − φ)n−2−2k < 0.
2
k!(k + 1)!(n − 2 − 2k)! i
k=0
where we use the fact that 2λi (1 − λi ) ≤ 12 .
Given that Q(φ, n) is decreasing in φ ∈ (0, 1), (16) has a unique solution φ∗ (n) ∈ (0, 1)
if and only if
c(n) =
1
2 Q(0, n)
1
2 Q(1, n),
− c > 0 and
1
2 Q(1, n)
− c ≤ 0. Noting Q(0, n) = 1 and defining
the result follows. Proof of Lemma 4: Each part is proven in turn.
(i) Since Q(φ, n) is decreasing in φ, it suffices to show Q(φ, n) is decreasing in n for a fixed
φ ∈ (0, 1). To do so, we prove that each term T (φ, n, λi ) is decreasing in n. But this just
mimics the proof of the similar result for α∗ (n) in part (ii) of Lemma 3. In particular,
T (φ, n, λi ) − T (φ, n + 1, λi ) =
n−1
bX
2 c
k=0
n−2
2
+2
bXc
k=0
n
2
−
bXc
k=0
−2
(n − 1)!
λk (1 − λi )k φ2k (1 − φ)n−1−2k
− 1 − 2k)! i
(k!)2 (n
(n − 1)!
λk+1 (1 − λi )k+1 φ2k+1 (1 − φ)n−2−2k
k!(k + 1)!(n − 2 − 2k)! i
n!
λk (1 − λi )k φ2k (1 − φ)n−2k
(k!)2 (n − 2k)! i
n−1
bX
2 c
k=0
n!
λk+1 (1 − λi )k+1 φ2k+1 (1 − φ)n−1−2k .
k!(k + 1)!(n − 1 − 2k)! i
33
For an odd n, note that
n
bX
2c
k=0
n!
λk (1 − λi )k φ2k (1 − φ)n−2k
− 2k)! i
(k!)2 (n
2 X
n−1
=
1+
k=0
2k
n − 2k
(n − 1)!
λk (1 − λi )k φ2k (1 − φ)n−2k
(k!)2 (n − 1 − 2k)! i
n−1
= (1 − φ)
2
X
k=0
n−1
−1
2
+ 2φ
X
k=0
(n − 1)!
λk (1 − λi )k φ2k (1 − φ)n−1−2k
− 1 − 2k)! i
(k!)2 (n
(n − 1)!
λk+1 (1 − λi )k+1 φ2k+1 (1 − φ)n−2−2k .
k!(k + 1)!(n − 2 − 2k)! i
Inserting this fact and simplifying terms yield
n−1
T (φ, n, λi ) − T (φ, n + 1, λi ) =
2
X
k=0
n−1
−1
2
+ 2λi (1 − λi )
X
k=0
n−1
2
− 2λi (1 − λi )
X
k=0
(n − 1)!
λk (1 − λi )k φ2k+1 (1 − φ)n−1−2k
− 1 − 2k)! i
(k!)2 (n
(n − 1)!
λk (1 − λi )k φ2k+1 (1 − φ)n−1−2k
k!(k + 1)!(n − 2 − 2k)! i
n!
λk (1 − λi )k φ2k+1 (1 − φ)n−1−2k > 0.
k!(k + 1)!(n − 1 − 2k)! i
The same exact argument follows for n even.
(ii) Similar arguments for lim α∗ (n) = 0 show lim φ∗ (n) = 0.
n→∞
n→∞
(iii) Suppose, by way of contradiction, that limn→∞ nφ∗ (n) = ∞. Then, φ∗ (n) =
a(n)
n
for some sequence a(n) such that limn→∞ a(n) = ∞.Now, we argue that in this case
limn→∞ T (φ∗ (n), n, λi ) = 0.
Consider each term in the first summation. For a fixed
k ≥ 0 and large n, the Stirling’s approximation implies
(n − 1)!
≈ (n − 1)2k .
(n − 1 − 2k)!
Moreover, since lim φ∗ (n) = 0 and φ∗ (n) =
a(n)
n ,
(1 − φ∗ (n))n−1−2k ≈ e−a(n) .
Hence,
(n − 1)!
∗
2k
∗
n−1−2k
lim
φ (n) (1 − φ (n))
n→∞ (n − 1 − 2k)!
2k
a(n)2k
2k a(n)
= lim (n − 1)
e−a(n) = lim a(n) = 0.
n→∞
n→∞ e
n
34
where the second line uses the hypothesis that limn→∞ a(n) = ∞. This means the limit
of the first summation in T (φ∗ (n), n, λi ) is 0. The same line of argument reveals that
the limit for the second summation is also 0. Hence, limn→∞ T (φ∗ (n), n, λi ) = 0. But
this violates the equilibrium condition given in (18). Thus, limn→∞ nφ∗ (n) < ∞. Proof of Lemma 6: Define W = XA − XB . Applying the Central Limit Theorem,
W − (2λ − 1)m D
√
−→ N (0, 1)
m
as m → ∞. Thus,
k+ε−(2λ−1)m
√
m
Z
lim Pr{XA = XB + k} = lim Pr{W = k} ≈ lim
m→∞
m→∞
m→∞
z2
1
√ e− 2 dz
2π
k−ε−(2λ−1)m
√
m
for some ε ∈ (0, 1). Therefore, limm→∞ Pr{XA = XB + k} = 0.
Similarly,
lim Pr{XA > XB } = lim Pr{W > 0} ≈ lim
m→∞
m→∞
Z∞
m→∞
z2
1
√ e− 2 dz
2π
ε−(2λ−1)m
√
m
for some ε ∈ (0, 1). Therefore, limm→∞ Pr{XA > XB } = 1 if and only if λ > 12 . 35
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