Poisson voting games under proportional rule

Abstract

We analyze strategic voting under proportional rule and two parties, embedding the basic spatial model into the Poisson framework of population uncertainty. We prove that there exists a unique Nash equilibrium. We show that it is characterized by a cutpoint in the policy space that is always located between the average of the two parties’ positions and the median of the distribution of voters’ types. We also show that, as the expected number of voters goes to infinity, the equilibrium converges to that of the case with deterministic population size.

1 Introduction

Citizens often act strategically as voters. Their expected behavior influences the competing parties’ positioning choices, and their actions determine the electoral outcome and, consequently, the policy that is implemented according to the electoral rule in force. In parliamentary systems elections are usually conducted under proportional representation. In this case the policy outcome may be represented as a convex combination of the policies proposed by each party, with weights proportional to the number of votes that they obtain. Such an outcome function reflects the fact that every parliamentary member participates to the decisional process with the same power. In this paper we study voters’ strategic behavior under this proportional rule, considering a unidimensional policy space and two parties.

In a mass election a voter is typically unaware of the exact number of other voters in the population but may have only some probabilistic information about it. To capture this kind of uncertainty we use the Poisson model introduced by Myerson (1998), in which the number of agents is not common knowledge but is a Poisson distributed random variable. Being realistic to assume that the size of the electorate is random from the viewpoint of an individual, voting settings were the immediate application of Poisson games (Myerson 2000, 2002) and the Poisson framework has become the standard tool to model strategic voting in large elections (see, e.g., Feddersen and Pesendorfer 1999; Castanheira 2003; Krishna and Morgan 2011, 2012; Bouton and Castanheira 2012; Bouton 2013; Bouton and Gratton 2015; Herrera et al. 2014).¹ In particular, Myerson (2002) analyzes strategic voting in single-winner elections under different scoring rules, while Bouton (2013) and Bouton and Gratton (2015) study strategic voting in runoff elections. Herrera et al. (2014) examine turnout in proportional electoral systems with a costly rational voting model à la Ledyard (1984), where the strategic aspect regards the choice between voting and abstaining. So far, strategic voting under proportional representation focused on the choice of which party to vote for has not been explored in the context of population uncertainty. This work aims to fill such a lack in the literature.

We consider the basic spatial model in which the positions of two parties are exogenously given and voters have to choose which party to vote for. Voters can be of different types, each type being characterized by a bliss point in the policy space. We analyze the polar case in which the policy that is implemented after the voting stage is a weighted average of the two parties’ positions, where weights are equal to the corresponding shares of votes.² The further the implemented policy is from a type’s bliss point, the worse off that type is.

In the context of deterministic population size, voters’ strategic behavior in a pure proportional system with multiple parties has been analyzed in De Sinopoli and Iannantuoni (2007). They show that, as the number of voters grows to infinity, in equilibrium basically voters split in two and only the two extremist parties take votes. The policy outcome is precisely the “cutpoint outcome”, that is, the outcome that is implemented when all the voters whose bliss points are on its left vote for the leftmost party and all the voters whose bliss points are on its right vote for the rightmost one. This result has been applied to other electoral systems exhibiting positive degrees of power sharing in Meroni (2017), where the limit equilibrium outcome is characterized for the two-party case. In the complete information model, the equilibrium and the equilibrium outcome are unique at the limit, when there is a continuum of voters. Before the limit, when the number of players is finite and individuals’ incentives to act strategically can be fully investigated, there is a plethora of equilibria and the equilibrium outcome is indeterminate.

The population uncertainty model turns out to solve such an indeterminacy. In fact, we show that for every expected number of voters there exists a unique Nash equilibrium. This is precisely due to the environmental equivalence property of Poisson games, for which every player’s environment (including everyone in the game except herself) is the same and which does not hold if the population size is deterministic.³ The equilibrium is characterized by a “cutpoint type” so that voters divide exactly in two, those to the left of the cutpoint voting for the leftist party and those to its right voting for the rightist one.⁴ This is due to the fact that, in the Poisson environment, if a type prefers the rightist party to the leftist one for a given behavior of the population, then all the types located on her right will strictly do so.⁵ Moreover, if a type is indifferent between the two parties given a strategy of the others, she will strictly prefer the rightist party to the leftist one given another strategy such that the leftist party’s vote share is larger. Then, the type indifferent against this second strategy will be on her left. A consequence is that the best response correspondence is a continuous decreasing function and this ensures the existence of a unique Nash equilibrium.

We examine the characteristics of the equilibrium with respect to the parameters of the model and show that the cutpoint always lies between the average of the two parties’ positions and the median of the distribution of types. Furthermore, we study the behavior of the equilibrium cutpoint as the expected number of voters increases and show that it converges to the equilibrium cutpoint of the deterministic model with a continuum of voters.

Finally, we remove the hypothesis that parties’ platforms are exogenously given and explore in an example their strategic decision of which position to take in the policy space when they are policy-motivated. Under such an assumption, the form of the outcome function implies that parties potentially face a trade-off. The more moderate is the policy that they choose, the larger is their share of votes but the less extreme is the position to which such a weight is assigned. We show that, if the expected number of voters is sufficiently large, parties choose extreme positions in equilibrium, in line with the results of the model with a continuum of strategic voters (Meroni 2017). However, this does not hold for every population size, depending on the distribution of voters’ types. As the expected number of voters vanishes, for instance, parties’ equilibrium positions may converge if the distribution is highly concentrated at a point in the policy space, in line with the results of the model with a continuum of sincere voters (see Saporiti 2014; Matakos et al. 2016).⁶

The paper is organized as follows. We describe the model in Sect. 2. In Sect. 3 we analyze its equilibrium and in Sect. 4 we examine how this varies as the population size increases. We present an example of parties’ political competition in Sect. 5.

2 The model

The policy space is represented by the closed interval ${\mathbb {X}}=[0,1]$. Two parties L and R are respectively located at positions $x_L,x_R\in {\mathbb {X}}$, with $x_L < x_R$. Every voter chooses which party to vote for, and the resulting electoral outcome determines the policy that is implemented according to the proportional rule. Precisely, let v be the share of votes gained by party L. The policy outcome $x(v,x_L,x_R)$ is a weighted average of the two parties’ positions where weights are given by the corresponding vote shares, i.e.

$$\begin{aligned} x(v,x_L,x_R)= v x_L + (1-v) x_R. \end{aligned}$$

The number of voters is assumed to be a random variable drawn from a Poisson distribution with mean n. The probability that the actual number of voters is k, then, is equal to

$$\begin{aligned} p(k \mid n)=\frac{e^{-n}n^k}{k!}. \end{aligned}$$

Each voter has a type that determines her preferences over the possible outcomes. We identify the set of types with $\Theta =[0,1]$, where $\theta _t\in \Theta$ represents the bliss point of type t voters, and we denote with $\theta$ the typical element of $\Theta$. Each voter’s type is independently drawn from $\Theta$ according to some probability distribution F. That is, for every element $\theta \in \Theta$, the probability that a randomly sampled voter is of type $\theta _t\le \theta$ is given by $F(\theta )$ and it is independent of the number and types of all the other voters. The decomposition property of the Poisson distribution implies that the number of voters with type smaller than or equal to $\theta$ is a Poisson random variable with parameter $nF(\theta )$. We assume that F is continuous and strictly increasing.

We denote each voter’s action set with $C=\{L,R\}$. An action profile $z=(z_L,z_R)\in {\mathbb {Z}}_+^2$ is a vector that specifies for each action $c\in C$ the number of voters who choose that action, $z_c$. The payoff to a voter of type t who chooses action c when the realization of the rest of the population’s behavior is z is a function of the distance between her bliss point and the induced policy.⁷ That is,

$$\begin{aligned} {\hat{u}}_t(c,z)=f(|\theta _t - {\hat{x}}(c,z)|), \end{aligned}$$

where ${\hat{x}}(c,z)$ is the outcome that the voter induces choosing c when $z=(z_L,z_R)$ is the action profile of the other voters, i.e

$$\begin{aligned} {\hat{x}}(L,z)=\frac{z_L+1}{z_L+z_R+1} x_L + \frac{z_R}{z_L+z_R+1} x_R \end{aligned}$$

and

$$\begin{aligned} {\hat{x}}(R,z)=\frac{z_L}{z_L+z_R+1} x_L + \frac{z_R+1}{z_L+z_R+1} x_R. \end{aligned}$$

Note that ${\hat{x}}(R,z) > {\hat{x}}(L,z)$ for every z. The function $f:[0,1]\rightarrow {\mathbb {R}}$ is assumed to be bounded, continuous, strictly decreasing, and strictly concave in its argument.

Given $x_L$ and $x_R$, with abuse of notation we denote with x(z) the outcome that is implemented when z is the entire population’s realized behavior, i.e.

$$\begin{aligned} x(z):= x\left( \frac{z_L}{z_L+z_R}, x_L, x_R\right) . \end{aligned}$$

Moreover, we denote with $u_t(x(z))$ the corresponding payoff to type t voters. Clearly, $u_t(x(z))= {\hat{u}}_t(L,z')={\hat{u}}_t(R,z'')$ where $z'=(z_L-1,z_R)$ and $z''=(z_L,z_R-1)$. The function $u_t$ is strictly concave because f is strictly concave, and it is assumed to be differentiable.⁸

A strategy function (or simply strategy) is a measurable function $\sigma :\Theta \rightarrow \Delta (C)$ which associates to each type a probability distribution on C, i.e. a mixed action.⁹ The mixed action employed by players of type t is denoted $\sigma _t$. Strategy $\sigma$ induces the average behavior $\tau (\sigma )\in \Delta (C)$, which specifies for each action $c\in C$ the marginal probability that a randomly sampled voter chooses c, $\tau _c(\sigma )$, when players are expected to play according to $\sigma$. It is possible to see that $\tau _c(\sigma )$ is also the expected share of votes received by party c.¹⁰ The number of voters who choose c is a Poisson random variable with mean $n\tau _c(\sigma )$ and it is independent of the number of voters who choose the other action.

Then, under the population’s average behavior $\tau \in \Delta (C)$, the probability of the action profile $z\in Z(C)$ is given by

$$\begin{aligned} P(z \mid \tau )=e^{-n}n^{z_L+z_R}\frac{{\tau _L}^{z_L}{\tau _R}^{z_R}}{z_L!z_R!}. \end{aligned}$$

From the perspective of any player of any type, the number of other players (not including herself) who choose action c is also a Poisson random variable with mean $n\tau _c$ due to the environmental equivalence property of Poisson games. Thus, the expected payoff to a type t voter who votes for c when the other voters vote according to $\tau$ is given by

$$\begin{aligned} U_t(c,\tau )=\sum _{z\in Z(C)}P(z \mid \tau ) u_t(c,z). \end{aligned}$$

Following standard terminology, we say that action c is a pure best response for type t voters against the population’s behavior $\tau$ if $U_t(c,\tau ) \ge U_t(c',\tau )$ for $c'\ne c$. We denote the set of such actions ${{\,\mathrm{PBR}\,}}_t(\tau )$. The set of type t voters’ best responses against $\tau$ is ${{\,\mathrm{BR}\,}}_t(\tau )=\Delta ({{\,\mathrm{PBR}\,}}_t(\tau ))$, and ${{\,\mathrm{BR}\,}}(\tau )$ is the collection of strategy functions such that $\sigma _t\in {{\,\mathrm{BR}\,}}_t(\tau )$ for every t.

Definition 1

The strategy function $\sigma$ is a Nash equilibrium if $\sigma \in {{\,\mathrm{BR}\,}}(\tau (\sigma ))$.

3 Equilibrium analysis

In this section we prove that the model has a unique Nash equilibrium. We show that in equilibrium voters split in two, those to the left of a “cutpoint type” voting for party L and those to its right voting for party R.

Formally, we say that strategy $\sigma$ is a cutpoint strategy if there is a value $\theta \in \Theta$ such that $\sigma _t(L)=1$ if $\theta _t<\theta$ and $\sigma _t(R)=1$ if $\theta _t>\theta$. That is, all the types to the left of type $\theta$ vote for party L and all the types to the right of $\theta$ vote for party R. We will usually refer to a cutpoint strategy with the correspondent cutpoint. Note that a given $\theta$ identifies an infinite number of strategy functions that differ only at the point $\theta _t=\theta$, which are equivalent given the continuity of the function F.

We begin proving that, for any average behavior $\tau$, the best response against $\tau$ is a cutpoint strategy and is unique, i.e. the best response correspondence is a function. In fact, the assumptions on the utility function imply that, given an average population behavior, if a type prefers to vote for party R than for party L then all the types on her right will strictly do so.¹¹ Not only this holds for every average behavior, but it holds for every realization of the population behavior. We prove first such a stronger result.

Lemma 1

For every $z\in Z(C)$ and $\theta _{t'}>\theta _{t}$, if type t prefers R to L given z then type $t'$ strictly prefers R to L given z.

Proof

Fix $z\in Z(C)$ and let $\Delta _t={\hat{u}}_t(R,z) - {\hat{u}}_t(L,z)$ for every t. Recall that ${\hat{x}}(R,z) > {\hat{x}}(L,z)$ and consider $\theta _{t'}>\theta _{t}$. There are six possible ways in which the points ${\hat{x}}(R,z)$, ${\hat{x}}(L,z)$, $\theta _{t'}$, and $\theta _{t}$ are ordered in the interval [0, 1]. For each case, the strict decreasingness and strict concavity of the function f imply the following:

• 1. if ${\hat{x}}(L,z)< {\hat{x}}(R,z)\le \theta _t < \theta _{t'}$ then $0<\Delta _t<\Delta _{t'}$;

• 2. if $\theta _t< \theta _{t'}\le {\hat{x}}(L,z) < {\hat{x}}(R,z)$ then $\Delta _t<\Delta _{t'}<0$;

• 3. if $\theta _t \le {\hat{x}}(L,z) < {\hat{x}}(R,z) \le \theta _{t'}$ then $\Delta _t< 0 < \Delta _{t'}$;

• 4. if ${\hat{x}}(L,z) \le \theta _t < \theta _{t'} \le {\hat{x}}(R,z)$ then either $\Delta _t< 0<\Delta _{t'}$ or $0\le \Delta _t<\Delta _{t'}$ or $\Delta _t<\Delta _{t'} \le 0$;

• 5. if ${\hat{x}}(L,z) \le \theta _t \le {\hat{x}}(R,z) \le \theta _{t'}$ then either $\Delta _t \le 0 < \Delta _{t'}$ or $0<\Delta _t<\Delta _{t'}$;

• 6. if $\theta _t \le {\hat{x}}(L,z) \le \theta _{t'} \le {\hat{x}}(R,z)$ then either $\Delta _t < 0 \le \Delta _{t'}$ or $\Delta _t<\Delta _{t'}<0$.

In every case we have $\Delta _{t'}>\Delta _t$, that is, ${\hat{u}}_{t'}(R,z) - {\hat{u}}_{t'}(L,z) > {\hat{u}}_t(R,z) - {\hat{u}}_t(L,z)$. It follows that if ${\hat{u}}_t(R,z) \ge {\hat{u}}_t(L,z)$ then ${\hat{u}}_{t'}(R,z) > {\hat{u}}_{t'}(L,z)$. $\square$

Recall that the environmental equivalence property of Poisson games implies that, for every average behavior $\tau$ and action profile z, each type attaches the same probability $P(z \mid \tau )$ to z. Thus, it follows readily from Lemma 1 that, for every $\tau$ and every $\theta _{t'} > \theta _{t}$, if $U_t(R,\tau ) \ge U_t(L,\tau )$ then $U_{t'}(R,\tau ) > U_{t'}(L,\tau )$.

Lemma 2

For every $\tau \in \Delta (C)$ and $\theta _{t'}>\theta _{t}$, if type t prefers R to L given $\tau$ then type $t'$ strictly prefers R to L given $\tau$.

A direct consequence of Lemma 2 is that ${{\,\mathrm{BR}\,}}(\tau )$ is a cutpoint strategy. In fact the previous result implies that, for every given $\tau$, there must be a cutpoint such that all the voters to its right strictly prefer R over L while all the voters to its left strictly prefer L over R.¹² The type whose bliss point is exactly the cutpoint is indifferent between L and R given $\tau$. In addition, the above lemma excludes that there could be more than one indifferent type. Thus, for every $\tau$, the cutpoint that characterizes ${{\,\mathrm{BR}\,}}(\tau )$ is unique.

Proposition 1

For every $\tau \in \Delta (C)$, ${{\,\mathrm{BR}\,}}(\tau )$ is a cutpoint strategy and is unique.

It follows that every equilibrium of the model is a cutpoint strategy and is therefore identified by a point in $\Theta$. We are going to prove the existence of a unique Nash equilibrium exploiting this fact, which allows to consider the restriction of the best response correspondence to cutpoint strategies. That is, we consider the function ${{\,\mathrm{BR}\,}}(\theta ):[0,1]\rightarrow [0,1]$, and we show that it is continuous and decreasing.

Theorem 1

There exists a unique Nash equilibrium.

Proof

Given a cutpoint strategy $\theta$, note that the probability that a randomly sampled voter votes for party L (resp. R) is given by $\tau _L(\theta )=F(\theta )$ (resp. $\tau _R(\theta )=1-F(\theta )$). By Proposition 1, ${{\,\mathrm{BR}\,}}(\theta )$ is characterized by the point $\theta _t\in \Theta$ such that $U_t(R,\tau (\theta ))-U_t(L,\tau (\theta ))=0$, that is,

$$\begin{aligned} \sum _{z\in Z(C)} e^{-n}n^{z_L+z_R}\frac{F(\theta )^{z_L}[1-F(\theta )]^{z_R}}{z_L!z_R!} [{\hat{u}}_t(R,z)-{\hat{u}}_t(L,z)]=0. \end{aligned}$$

It is easy to see that ${{\,\mathrm{BR}\,}}(\theta )$ is continuous in $\theta$ given that $F(\theta )$ is continuous in $\theta$. Therefore, the Brouwer fixed point theorem readily implies that ${{\,\mathrm{BR}\,}}(\theta )$ has at least one fixed point, that is, an equilibrium exists.¹³

We can now prove that ${{\,\mathrm{BR}\,}}(\theta )$ is decreasing in $\theta$. In fact we can show that, given two cutpoint strategies $\theta$ and $\theta '$ such that $\theta '>\theta$, if $U_t(R,\tau (\theta ))=U_t(L,\tau (\theta ))$ then $U_t(R,\tau (\theta '))>U_t(L,\tau (\theta '))$. That is, the type t who is indifferent between R and L given $\theta$ strictly prefers R to L given $\theta '$. By Lemma 2, then, the indifferent type against $\theta '$ will be on the left of type t, i.e.$BR(\theta ')<BR(\theta).$

First note that, for every t and every couple of realizations z and $z'$ such that $z=(z_L,z_R)$ and $z'=(z_L+1,z_R-1)$, we have

$$\begin{aligned} {\hat{u}}_t(R,z') - {\hat{u}}_t(L,z') > {\hat{u}}_t(R,z)-{\hat{u}}_t(L,z). \end{aligned}$$

(1)

Indeed, since ${\hat{u}}_t(R,z')={\hat{u}}_t(L,z)=u_t(x(z_L+1,z_R))$, the above condition is equivalent to

$$\begin{aligned} u_t(x(z_L+1,z_R)) > \frac{1}{2} u_t(x(z_L+2,z_R-1)) + \frac{1}{2} u_t(x(z_L,z_R+1)). \end{aligned}$$

Given that

$$\begin{aligned} x(z_L+1,z_R)=\frac{1}{2} x(z_L+2,z_R-1) + \frac{1}{2} x(z_L,z_R+1), \end{aligned}$$

condition (1) follows directly from the strict concavity of the utility function.

Then, note that for every t and $\theta \in \Theta$ we can write

$$\begin{aligned}&U_t(R,\tau (\theta ))-U_t(L,\tau (\theta )) =\sum _{m=0}^{\infty } \frac{e^{-n}n^m}{m!} \sum _{z_L=0}^m \left( {\begin{array}{c}m\\ z_L\end{array}}\right) F(\theta )^{z_L} \\&\quad \times [1-F(\theta )]^{m-z_L}\left[ {\hat{u}}_t(R,(z_L,m-z_L)) - {\hat{u}}_t(L, ( z_L,m-z_L))\right] . \end{aligned}$$

Fix the realization of the population $m>0$. For every t and $z_L \le m$, let

$$\begin{aligned} \Delta _t(z_L):={\hat{u}}_t(R,(z_L,m-z_L)) - {\hat{u}}_t(L, (z_L,m-z_L)). \end{aligned}$$

Given the result above, we have $\Delta _t(z_L) > \Delta _t(z_L - 1)$ for every $0<z_L \le m$. Let also

$$\begin{aligned} B(z_L \mid \theta ):=\left( {\begin{array}{c}m\\ z_L\end{array}}\right) F(\theta )^{z_L}[1-F(\theta )]^{m-z_L} \end{aligned}$$

denote the probability that the number of voters choosing L is equal to $z_L$ given that the population size is m and given the cutpoint strategy $\theta \in \Theta$, and consider $\theta '>\theta$. The binomial distribution with parameters m and $F(\theta ')$ first order stochastically dominates the one with parameters m and $F(\theta )$, that is,

$$\begin{aligned} \sum _{z_L=0}^{\bar{z}} B(z_L \mid \theta ') < \sum _{z_L=0}^{\bar{z}} B(z_L \mid \theta ) \end{aligned}$$

for every $\bar{z} < m$. Therefore, for every t, we have

$$\begin{aligned}&\sum _{z_L=0}^m B(z_L\mid \theta )\Delta _t(z_L) \\&\quad< B(0\mid \theta ')\Delta _t(0) + [B(1\mid \theta ) + B(0\mid \theta ) - B(0\mid \theta ')] \Delta _t(1) + \sum _{z_L=2}^m B(z_L\mid \theta )\Delta _t(z_L) \\&\quad< B(0\mid \theta ')\Delta _t(0) + B(1\mid \theta ')\Delta _t(1) + [B(2\mid \theta ) + B(0\mid \theta ) - B(0\mid \theta ') + B(1\mid \theta )\\&\qquad - B(1\mid \theta ') ] \Delta _t(2) + \sum _{z_L=3}^m B(z_L\mid \theta )\Delta _t(z_L)< \cdots \\&\quad <\sum _{z_L=0}^{m-1} B(z_L\mid \theta ')\Delta _t(z_L) + \left[ B(m\mid \theta ) + \sum _{z_L=0}^{m-1} B(z_L\mid \theta ) - \sum _{z_L=0}^{m-1} B(z_L\mid \theta ')\right] \Delta _t(m) \\&\quad = \sum _{z_L=0}^{m} B(z_L\mid \theta ')\Delta _t(z_L) + \left[ \sum _{z_L=0}^{m} B(z_L\mid \theta ) - \sum _{z_L=0}^{m} B(z_L\mid \theta ')\right] \Delta _t(m) \\&\quad = \sum _{z_L=0}^{m} B(z_L\mid \theta ')\Delta _t(z_L), \end{aligned}$$

where the first inequality derives from $B(0\mid \theta ) - B(0\mid \theta ')>0$ and $\Delta _t(1) >\Delta _t(0)$, the second inequality derives from $B(0\mid \theta ) + B(1\mid \theta ) - B(0\mid \theta ') - B(1\mid \theta ')>0$ and $\Delta _t(2) >\Delta _t(1)$, and so on until the last inequality, which derives from $\sum _{z_L=0}^{m-1} B(z_L\mid \theta ) - \sum _{z_L=0}^{m-1} B(z_L\mid \theta ')>0$ and $\Delta _t(m)>\Delta _t(m-1)$.¹⁴ As this holds for every realization $m>0$, we can conclude that

$$\begin{aligned} U_t(R,\tau (\theta ))-U_t(L,\tau (\theta )) < U_t(R,\tau (\theta '))-U_t(L,\tau (\theta ')) \end{aligned}$$

for every t.¹⁵

Thus, if $U_t(R,\tau (\theta ))=U_t(L,\tau (\theta ))$ then $U_t(R,\tau (\theta '))>U_t(L,\tau (\theta '))$, so $BR(\theta ')<BR(\theta )$. Since the best response function is (strictly) decreasing, the Nash equilibrium is unique. $\square$

We conclude this section showing some characteristics of the Nash equilibrium relatively to the parameters of the model. In particular, we can prove that the equilibrium cutpoint is always located between the average of the two parties’ positions and the median of the distribution of voters’ types. To this end, let $\theta ^*$ be the equilibrium cutpoint, let ${\bar{\theta }}=\frac{x_L+x_R}{2}$ and $F(\theta _m)=\frac{1}{2}$.

Proposition 2

If $\theta _m>{\bar{\theta }}$ then ${\bar{\theta }}<\theta ^*<\theta _m$, while if $\theta _m<{\bar{\theta }}$ then $\theta _m<\theta ^*<{\bar{\theta }}$.

Proof

Consider the cutpoint strategy $\theta _m$, which induces the average behavior $\tau$ such that $\tau _L(\theta _m)=\tau _R(\theta _m)=\frac{1}{2}$. Any two symmetric realizations, i.e. any two realizations z and $z'$ such that $z=(z_1,z_2)$ and $z'=(z_2,z_1)$, have the same probability given $\theta _m$. Let t be the type whose bliss point is ${\bar{\theta }}$. For every realization z such that $z_L=z_R$ we have ${\hat{u}}_t(R,z) = {\hat{u}}_t(L,z)$, while for every symmetric realizations z and $z'$ we have ${\hat{u}}_t(R,z) - {\hat{u}}_t(L,z) = {\hat{u}}_t(L,z') - {\hat{u}}_t(R,z')$. It follows that type t is indifferent between choosing L and choosing R given $\theta _m$, that is, $BR(\theta _m)={\bar{\theta }}$. Since the best response function is continuous and strictly decreasing, both when $\theta _m>{\bar{\theta }}$ and when $\theta _m<{\bar{\theta }}$ its fixed point lies between these two values. Of course, if $\theta _m={\bar{\theta }}$ then $\theta ^*=\theta _m={\bar{\theta }}$. $\square$

The results presented in this section hold for every value of the expected number of voters n. In the next section we study the behavior of the Nash equilibrium as the expected population size increases.

4 Large electorate

If we fix the parties’ platforms $x_L$ and $x_R$ and the distribution of voters’ types F, the Nash equilibrium is a function of the expected number of voters n, $\theta ^*(n)$. In this section we study the behavior of this function as the electorate becomes large. In particular we show that, as n goes to infinity, the equilibrium cutpoint converges to that of the deterministic model with a continuum of voters studied in De Sinopoli and Iannantuoni (2007).

To this end, let the cutpoint outcome ${\hat{\theta }}$ be the unique policy outcome implemented when all the types to its left vote for party L and all the types to its right vote for party R, i.e. the unique solution to

$$\begin{aligned} {\hat{\theta }}= F({\hat{\theta }})x_L + [1-F({\hat{\theta }})]x_R. \end{aligned}$$

De Sinopoli and Iannantuoni (2007) prove that ${\hat{\theta }}$ is the unique equilibrium of the model with deterministic population size in which there is a continuum of voters, when this is seen as the limit of a sequence of finite games.¹⁶ We are going to prove that ${\hat{\theta }}$ is the limit point of the equilibrium cutpoint of the Poisson model as the expected number of voters grows to infinity.

First, we present some results that derive from the properties of the Poisson distribution. We begin reminding that there exist exponentially decreasing bounds for the tail probabilities of a Poisson random variable.¹⁷ This directly implies that, in a Poisson game, the probability of the population realizations that are smaller than a given value becomes smaller and smaller as n increases.

Lemma 3

For every $\delta >0$ and $\bar{m}>0$ there exists a value $\bar{n} \in {\mathbb {R}}$ such that, for every $n \ge \bar{n}$, $\sum _{m=0}^{\bar{m}}p(m \mid n)<\delta$.

The second result regards the parties’ shares of votes. Given the average behavior $\tau$ and fixed the realization of the population m, the vote share of party L is a random variable that takes value $\frac{z_L}{m}$ with probability $\left( {\begin{array}{c}m\\ z_L\end{array}}\right) {\tau _L}^{z_L}(1-\tau _L)^{m-z_L}$ (similarly for the vote share of party R). Let us denote such a variable ${\tilde{s}}(m)$. Note that

$$\begin{aligned} {\tilde{s}}(m)=\sum _{i=1}^m \frac{{\tilde{\omega }}_i}{m}, \end{aligned}$$

where ${\tilde{\omega }}_i$ is a player’s vote for party L given $\tau$, i.e. the random variable that takes value 1 with probability $\tau _L$ and 0 with probability $1-\tau _L$. Let $\text {Pr}(|{\tilde{s}}(m)-\tau _L| > \varepsilon )$ denote the probability that ${\tilde{s}}(m)$ is further than $\varepsilon$ from its expected value $\tau _L$. The next result follows directly from the weak law of large numbers, which states that the sample mean converges in probability to the expected value.

Lemma 4

For every $\delta >0$ and $\varepsilon >0$ there exists a value $\bar{m} \in {\mathbb {R}}$ such that, for every $m \ge \bar{m}$, $\text {Pr}(|{\tilde{s}}(m)-\tau _L| > \varepsilon ) < \delta$.

We can use Lemmas 3 and 4 to prove that the sample mean converges in probability to the expected value also when the size of the population is a Poisson random variable. This gives a clear intuition of the fact that the policy outcome, which is a continuous function of the shares of votes, remains sufficiently close to its expected value with probability sufficiently high if n is large enough. Such a fact is at the basis of the result in Theorem 2.¹⁸

Lemma 5

For every $\delta >0$ and $\varepsilon >0$ there exists a value $\bar{n} \in {\mathbb {R}}$ such that, for every $n \ge \bar{n}$,

$$\begin{aligned} \sum _{m=0}^{\infty } p(m\mid n) \left[ \text {Pr}(|{\tilde{s}}(m)-\tau (L)| > \varepsilon ) \right] < \delta . \end{aligned}$$

Proof

Given $\delta >0$ and $\varepsilon >0$, let $\delta _1=\delta _2={\delta }/{2}$. By Lemma 4, there exists a value $\bar{m}$ such that, for every $m \ge \bar{m}$, $\text {Pr}(|{\tilde{s}}(m)-\tau (L)| > \varepsilon ) < \delta _1$. Given $\bar{m}$, then, there exists by Lemma 3 a value $\bar{n}$ such that, for every $n \ge \bar{n}$, $\sum _{m=0}^{\bar{m}}p(m\mid n)<\delta _2$. Therefore, for $n \ge \bar{n}$ we have

$$\begin{aligned}&\sum _{m=0}^{\infty } p(m\mid n) \left[ \text {Pr}(|{\tilde{s}}(m)-\tau (L)|> \varepsilon ) \right] \\&\quad =\sum _{m=0}^{\bar{m}} p(m\mid n) \left[ \text {Pr}(|{\tilde{s}}(m)-\tau (L)|> \varepsilon ) \right] + \sum _{m=\bar{m}+1}^{\infty } p(m \mid n) \left[ \text {Pr}(|{\tilde{s}}(m)-\tau (L)| > \varepsilon ) \right] \\&\quad< \sum _{m=0}^{\bar{m}} p(m\mid n) + \sum _{m=\bar{m}+1}^{\infty } p(m \mid n) \delta _1 < \delta _2 + \delta _1 = \delta . \end{aligned}$$

$\square$

Given the average behavior $\tau \in \Delta (C)$, let $X(\tau )$ denote the expected outcome, i.e.¹⁹

$$\begin{aligned} X(\tau )= \tau _L x_L + (1 - \tau _L) x_R. \end{aligned}$$

We can now prove that, for n sufficiently large, the equilibrium cutpoint lies in a neighborhood of the expected equilibrium outcome which shrinks to zero as n goes to infinity. Since the equilibrium cutpoint and the expected equilibrium outcome are always located on opposite sides relative to the cutpoint outcome, it follows that the equilibrium cutpoint lies in a neighborhood of the cutpoint outcome of the same length. Thence, the limit point of the equilibrium cutpoint $\theta ^*(n)$ as n goes to infinity is the cutpoint outcome ${\hat{\theta }}$.

Let $\sigma ^*(n)$ be the Nash equilibrium characterized by the cutpoint $\theta ^*(n)$, and let $\tau ^*(n)=\tau (\sigma ^*(n))$.

Theorem 2

For every $\delta >0$ there exists a value $\bar{n} \in {\mathbb {R}}$ such that, for every $n \ge \bar{n}$, $|\theta ^*(n) - X(\tau ^*(n))| < \delta$. Furthermore, $|{\hat{\theta }} - \theta ^*(n)| < \delta$ and $|{\hat{\theta }} - X(\tau ^*(n))| < \delta$.

Proof

See Appendix A. $\square$

Remark 1

From the results of this section it follows that the model with large electorate is completely analogous to the limit model with deterministic population size, as the number of voters goes to infinity. In particular, if we extend our analysis to the case in which there are more than two parties as in De Sinopoli and Iannantuoni (2007), we have that as n goes to infinity in equilibrium only the two most extreme parties take votes. With deterministic population size, indeed, in equilibrium all the voters except those in a neighborhood of the equilibrium outcome choose the extremist parties, in order to drag the policy outcome towards their bliss point. That neighborhood reduces as the number of voters increases. An equivalent result is true in the Poisson environment where, as the expected population size increases, the probability of low realizations goes rapidly to zero. Before the limit there is multiplicity of equilibria notwithstanding environmental equivalence because, intuitively, central voters can “coordinate” on different behaviors.²⁰

5 Parties’ political competition: an example

In this section we remove the hypothesis that parties’ platforms are exogenous and discuss an example of the political competition that happens before the voting stage; that is, parties’ strategic decision of which position to take in the policy space. In particular, we assume that parties L and R are purely policy-motivated with the same preferences as voters, and are characterized respectively by ideal policies 0 and 1 for simplicity. We let $x_L,x_R \in {\mathbb {X}}$ denote the positions that they choose, removing the assumption $x_L<x_R$.²¹ We assume that the distribution of voters’ types F is also differentiable.

In the deterministic model with a continuum of strategic voters, Meroni (2017) shows that two policy-motivated parties choose extreme positions in equilibrium, in contrast with the results of the model in which voters vote sincerely (Saporiti 2014; Matakos et al. 2016). The same result is true in the model where the number of voters is Poisson distributed when its expected value is sufficiently large. In fact we can show that, loosely speaking, by changing the chosen platform a party moves the implemented policy in the same direction of the change. Hence, in equilibrium, the centripetal force due to the incentive to obtain a larger vote share is dominated by the centrifugal force due to the incentive to assign that weight to a more extreme platform, and parties’ positions diverge.

Proposition 3

There exists a value $\bar{n} \in {\mathbb {R}}$ such that, for every $n \ge \bar{n}$, the unique Nash equilibrium is $(x_L,x_R)=(0,1)$.

Proof

See Appendix B. $\square$

Divergence of parties’ positions does not extend to every expected number of voters, depending on the distribution of voters’ types F. To see this note that if n is sufficiently small then the equilibrium of the voting game remains sufficiently close to the average of parties’ positions.²² Differently from the cutpoint outcome, that value does not depend on the distribution function F, so in this case parties’ positions can converge for some specifications of that function.

Consider for example a distribution F that is highly concentrated in a sufficiently small neighborhood of $\theta =\frac{1}{3}$ and suppose that parties choose positions $x_L=0$ and $x_R=1$. If n is close to zero, the equilibrium cutpoint of the voting game is close to ${\bar{\theta }}=\frac{1}{2}$. Therefore, the expected vote share of party L is close to one, and with probability close to one the outcome is close to 0. It follows that party R has now the incentive to deviate to a more moderate position to increase her expected vote share and bring the outcome closer to 1.²³

Remark 2

The assumption of extreme parties’ ideal policies is needed only to simplify the proof of the above divergence result. In the proof it is shown that at the limit the behavior of voters and the implemented outcome converge to those of the model with deterministic population size (see Appendix B). That proof can be extended to general ideal policies for parties at the cost of a much heavier notation and more cumbersome steps. In particular, in line with Meroni (2017), if n is sufficiently large then in the unique equilibrium at least one party chooses an extreme position while the other party chooses a position that is always more extreme than its ideal policy. In fact, if the cutpoint ${\hat{\theta }}_{0,1}$ when parties’ chosen platforms are 0 and 1 lies between the two parties ideal policies $\theta _L$ and $\theta _R$, then in equilibrium parties choose positions 0 and 1. On the other hand, if $\theta _L<\theta _R<{\hat{\theta }}_{0,1}$ then party L chooses position 0 and party R chooses a position between $\theta _R$ and 1, while if ${\hat{\theta }}_{0,1}<\theta _L<\theta _R$ then party L chooses a position between 0 and $\theta _L$ and party R chooses position 1.

Appendices

Appendix A: Proof of Theorem 2

Theorem 2

For every $\delta >0$ there exists a value $\bar{n} \in {\mathbb {R}}$ such that, for every $n \ge \bar{n}$, $|\theta ^*(n) - X(\tau ^*(n))| < \delta$. Furthermore, $|{\hat{\theta }} - \theta ^*(n)| < \delta$ and $|{\hat{\theta }} - X(\tau ^*(n))| < \delta$.

Proof

We start showing that, for every $\delta >0$, there exists a value $\bar{n}_L \in {\mathbb {R}}$ such that, for every $n \ge \bar{n}_L$, if $\theta _t \le X(\tau ^*(n)) - \delta$ then $\sigma ^*_t(n)(L)=1$. This, together with an analogous and symmetric result for $\theta _t \ge X(\tau ^*(n)) - \delta$, implies the first statement of the theorem.

Fix $\delta$, take a type t such that $\theta _t \le X(\tau ^*(n)) - \delta$ and consider the difference of utility that she gets choosing R instead of L given the equilibrium behavior $\tau ^*(n)$. We have

$$\begin{aligned}&U_t(R,\tau ^*(n)) - U_t(L,\tau ^*(n)) \nonumber \\&\quad =\sum _{m=0}^{\infty } p(m\mid n) \sum _{z_L=0}^m \left( {\begin{array}{c}m\\ z_L\end{array}}\right) {\tau ^*_L(n)}^{z_L}(1-\tau ^*_L(n))^{m-z_L} [u_t({\hat{x}}(R,z_L,m-z_L)) - u_t({\hat{x}}(L,z_L,m-z_L))] \nonumber \\&\quad = \sum _{m=0}^{\infty } p(m\mid n) \sum _{z_L=0}^m \textit{Pr}\left( {\tilde{s}}(m)=\frac{z_L}{m} \mid \tau ^*(n)\right) \bigg [u_t({\hat{x}}(R,z_L,m-z_L)) - u_t\left( {\hat{x}}(R,z_L,m-z_L)-\frac{x_R -x_L}{m+1}\right) \bigg ] \end{aligned}$$

(2)

as ${\hat{x}}(R,z_L,m-z_L) - {\hat{x}}(L,z_L,m-z_L) = \frac{x_R - x_L}{m+1}$, where $\textit{Pr}\left( {\tilde{s}}(m)=\frac{z_L}{m} \mid \tau ^*(n) \right)$ denotes the probability that the vote share of party L is equal to $\frac{z_L}{m}$ given $\tau ^*(n)$.

Since the utility function $u_t$ is continuous on [0, 1] and differentiable on (0, 1), the mean value theorem implies that for each m and $z_L$ there exists a value

$$\begin{aligned} \bar{x} \in \left[ {\hat{x}}(R,z_L,m-z_L)-\frac{x_R -x_L}{m+1},{\hat{x}}(R,z_L,m-z_L) \right] \end{aligned}$$

such that

$$\begin{aligned} u_t({\hat{x}}(R,z_L,m-z_L)) - u_t\left( {\hat{x}}(R,z_L,m-z_L)-\frac{x_R -x_L}{m+1}\right) = \frac{x_R -x_L}{m+1} \cdot \frac{\partial u_t(x)}{\partial x} \biggr {|} _{x=\bar{x}}. \end{aligned}$$

We can exploit this fact to upperly bound the expression in (2), using bounds for $\frac{\partial u_t(x)}{\partial x} \big {|}_{x=\bar{x}}$. Indeed, as the utility function f is strictly concave, its slope is the greatest (in absolute value) when its argument is the maximum, so

$$\begin{aligned} \max _{\theta _t \in \Theta } \left( \max _{x\in [0,1]} \frac{\partial u_t(x)}{\partial x} \right) = \big {|}f'(1)\big {|} := G(1). \end{aligned}$$

For every m, $G(1)\frac{x_R-x_L}{m+1}$ is an upper bound of the gain that any player can get choosing R instead of L for any given induced outcome. Furthermore, the slope of f is the lower in absolute value the smaller is its argument. Therefore, for outcomes strictly on the right of a voter’s bliss point, we have

$$\begin{aligned} \min _{x \ge \theta _t + \eta } \Bigg {|}\frac{\partial u_t(x)}{\partial x} \Bigg {|}= \big {|}f'(\eta )\big {|} := G(\eta ) \end{aligned}$$

for every $\theta _t \in [0,1-\eta ]$ and $\eta > 0$. For every m, $G(\eta )\frac{x_R-x_L}{m+1}$ is a lower bound of the loss that a voter incurs choosing R instead of L when the induced outcome is on the right of her bliss point plus $\eta$.

Now, we have to determine a neighborhood of the expected vote share of party L and to bound the probability that the actual vote share is in that neighborhood. By Lemma 4, indeed, the distribution of the other m votes remains sufficiently close to its expected value with sufficiently high probability if m is sufficiently large. Let $\eta < \delta$ and let $\varepsilon _m = \max \left\{ 0, \delta - \eta - \frac{1}{m+1}\right\}$ for every m. It is possible to see that, when the vote share of party L induced by the other m votes is not further than $\varepsilon _m$ from its expected value, every outcome that type t can induce is on the right of $\theta _t + \eta$.²⁴ To this end, consider ${\tilde{s}}(m)=\tau ^*_L(n) + \varepsilon _m$ and, with abuse of terminology, let ${\hat{x}}\left( L,\tau ^*_L(n) + \varepsilon _m\right)$ denote the outcome that the voter induces choosing L.²⁵ We have

$$\begin{aligned}&{\hat{x}}\left( L,\tau ^*_L(n) + \varepsilon _m\right) \\&\quad = \frac{m}{m+1} \left\{ \left[ \tau ^*_L(n) + \varepsilon _m\right] x_L + \left[ 1 - \tau ^*_L(n) - \varepsilon _m\right] x_R \right\} + \frac{1}{m+1} x_L \\&\quad =\frac{m}{m+1}X(\tau ^*(n)) - \frac{m}{m+1} \varepsilon _m (x_R -x_L) + \frac{1}{m+1} x_L \\&\quad> X(\tau ^*(n)) - \frac{1}{m+1} - \frac{m}{m+1} \varepsilon _m \\&\quad > X(\tau ^*(n)) - \frac{1}{m+1} - \varepsilon _m \\&\quad =X(\tau ^*(n)) - \delta + \eta \\&\quad \ge \theta _t + \eta , \end{aligned}$$

where the first inequality derives from $x_R -x_L \le 1$, $x_L \ge 0$, and $X(\tau ^*(n)) < 1$, the second from $\frac{m}{m+1}<1$, while the last one from the assumption that $\theta _t \le X(\tau ^*(n)) - \delta$. Clearly, also the outcome that type t induces choosing R will be greater than $\theta _t + \eta$.

Thus, given expression (2) and the above discussion, we have

$$\begin{aligned}&U_t(R,\tau ^*(n)) - U_t(L,\tau ^*(n)) <\sum _{m=0}^{\infty } p(m\mid n) \nonumber \\&\quad \times \Biggl \{ \textit{Pr}\big {(}|{\tilde{s}}(m)-\tau ^*_L(n)| > \varepsilon _m \big {)} \frac{x_R-x_L}{m+1} G(1) - \textit{Pr}\big {(}|{\tilde{s}}(m)-\tau ^*_L(n)| \le \varepsilon _m \big {)} \frac{x_R-x_L}{m+1} G(\eta ) \Biggr \}. \end{aligned}$$

(3)

Now, let

$$\begin{aligned} {\bar{\delta }}= \frac{1}{2} \cdot \frac{G(\eta )}{G(1)+G(\eta )} > 0. \end{aligned}$$

It follows from Lemma 4 that, given ${\bar{\delta }}$ and $\varepsilon _m>0$, there exists a value $\bar{m}$ such that, for $m \ge \bar{m}$, $\textit{Pr}\left( |{\tilde{s}}(m)-\tau ^*_L(n)|>\varepsilon _m\right) < {\bar{\delta }}$.²⁶ Hence, the right hand side of (3) is smaller than

$$\begin{aligned} \sum _{m=0}^{\bar{m}-1}p(m\mid n) \frac{x_R-x_L}{m+1} G(1) + \sum _{m=\bar{m}}^{\infty } p(m\mid n) \frac{x_R-x_L}{m+1} \left[ {\bar{\delta }} G(1) - (1- {\bar{\delta }}) G(\eta )\right] \end{aligned}$$

(4)

where, for the values $m<\bar{m}$, we have used the upper bound 1 for $\textit{Pr}(|{\tilde{s}}(m)-\tau ^*_L(n)| > \varepsilon _{m})$ and the lower bound 0 for $\textit{Pr}(|{\tilde{s}}(m)-\tau ^*_L(n)| \le \varepsilon _{m})$. Note that, given the formula of ${\bar{\delta }}$, we have

$$\begin{aligned} {\bar{\delta }} G(1) - (1- {\bar{\delta }}) G(\eta ) = -\frac{G(\eta )}{2} < 0, \end{aligned}$$

so expression (4) is equivalent to

$$\begin{aligned}&\sum _{m=0}^{\bar{m}-1}p(m \mid n) \frac{x_R-x_L}{m+1} G(1) - \sum _{m=\bar{m}}^{\infty } p(m\mid n) \frac{x_R-x_L}{m+1} \frac{G(\eta )}{2} \nonumber \\&\quad =\sum _{m=0}^{\bar{m}-1} \left[ p(m\mid n) \frac{x_R-x_L}{m+1} G(1) - p(m+\bar{m} \mid n)\frac{x_R-x_L}{m+\bar{m}+1} \frac{G(\eta )}{2} \right] \nonumber \\&\qquad - \sum _{m=2\bar{m}}^{\infty } p(m\mid n) \frac{x_R-x_L}{m+1} \frac{G(\eta )}{2}. \end{aligned}$$

(5)

We can finally prove that, for n sufficiently large, the above expression is smaller than zero. Precisely, let

$$\begin{aligned} \bar{n}_L = \left[ 2\cdot \frac{(2 \bar{m})!}{\bar{m}!} \cdot \frac{G(1)}{G(\eta )} \right] ^{\frac{1}{\bar{m}}}. \end{aligned}$$

We can show that, for $n \ge \bar{n}_L$, expression (5) is negative since every term of its first summation is negative. That is, for every $m\in [0,\bar{m}-1]$,

$$\begin{aligned} \frac{p(m \mid n)}{p(m+\bar{m} \mid n)} \cdot \frac{m+\bar{m}+ 1}{m+1}\le \frac{1}{2} \cdot \frac{G(\eta )}{G(1)}. \end{aligned}$$

(6)

This is due to the fact that, as n increases, every population realization m becomes relatively less likely than the higher realization $m+\bar{m}$. In fact, note that the left hand side of (6) attains its maximum for $m=\bar{m}-1$, as

$$\begin{aligned} {{\,\mathrm{argmax}\,}}_{m\in [0,\bar{m}-1]} \frac{p(m \mid n)}{p(m+\bar{m} \mid n)} \cdot \frac{m+\bar{m} + 1}{m+1}&= {} {{\,\mathrm{argmax}\,}}_{m\in [0,\bar{m}-1]} \frac{(m+\bar{m} + 1)!}{(m+1)!} n^{-\bar{m}} \nonumber \\&= {} \bar{m} - 1, \end{aligned}$$

and that

$$\begin{aligned} \frac{(2 \bar{m})!}{\bar{m}!} {n}^{-\bar{m}} \le \frac{1}{2} \cdot \frac{G(\eta )}{G(1)} \end{aligned}$$

for $n \ge \bar{n}_L$.

We can conclude that, for $n \ge \bar{n}_L$, $U_t(L,\tau ^*(n)) > U_t(R,\tau ^*(n))$ for every $\theta _t \le X(\tau ^*(n)) - \delta$. An analogous argument implies that there exists a value $\bar{n}_R$ such that, for $n \ge \bar{n}_R$, $U_t(R,\tau ^*(n)) > U_t(L,\tau ^*(n))$ for every $\theta _t \ge X(\tau ^*(n)) + \delta$. Setting $\bar{n}=\max \{\bar{n}_L, \bar{n}_R\}$, we obtain the desired result.

For the second statement of the theorem note that, for any n, if $\theta ^*(n) < {\hat{\theta }}$ then $X(\tau ^*(n)) > {\hat{\theta }}$, since $\tau _L^*(n) = F(\theta ^*(n)) < F({\hat{\theta }})$. Hence, we have $|\theta ^*(n) - X(\tau ^*(n))| \ge \max \{ |{\hat{\theta }} - \theta ^*(n)|, |{\hat{\theta }} - X(\tau ^*(n))|\}$.²⁷ The result directly follows. $\square$

Appendix B: Proof of Proposition 3

Proposition 3

There exists a value $\bar{n} \in {\mathbb {R}}$ such that, for every $n \ge \bar{n}$, the unique Nash equilibrium is $(x_L,x_R)=(0,1)$.

Proof

First, note that the assumptions on parties’ preferences and on the outcome function imply that in equilibrium $x_L < x_R$. In fact, if $x_L=x_R$ then the induced outcome coincides always with the two policies, while if $x_L\ne x_R$ then the induced outcome lies between them for every voters’ action profile, and every possible profile realizes with positive probability given the equilibrium of the voting game.²⁸

Given parties’ choices $x_L$ and $x_R$, let ${\hat{\theta }}_{x_L,x_R}$ be the associated cutpoint outcome, ${\theta }^*_{x_L,x_R}(n)$ the equilibrium cutpoint of the voting game when the expected number of voters is n, and $X(\theta ^*_{x_L,x_R}(n))$ the corresponding expected outcome, i.e. $X(\theta ^*_{x_L,x_R}(n))= F(\theta ^*_{x_L,x_R}(n)) x_L + [1 - F(\theta ^*_{x_L,x_R}(n))] x_R$. Note that, for every $x_L < x_R$, the cutpoint outcome ${\hat{\theta }}_{x_L,x_R}$ is strictly increasing in $x_L$ and in $x_R$. In fact, we have

$$\begin{aligned} \frac{\partial {\hat{\theta }}_{x_L,x_R}}{\partial x_L}=\frac{F({\hat{\theta }}_{x_L,x_R})}{1+ F'({\hat{\theta }}_{x_L,x_R}) (x_R-x_L)} \quad \text { and } \quad \frac{\partial {\hat{\theta }}_{x_L,x_R}}{\partial x_R}=\frac{1-F({\hat{\theta }}_{x_L,x_R})}{1+ F'({\hat{\theta }}_{x_L,x_R}) (x_R-x_L)}. \end{aligned}$$

Both derivatives are strictly positive when ${\hat{\theta }}_{x_L,x_R} \notin \{0,1\}$ since F is strictly increasing. By Theorem 2, we know that for every ${\delta }>0$ there exists a value $\bar{n} \in {\mathbb {R}}$ such that, for every $n\ge \bar{n}$,

$$\begin{aligned} \left| {\hat{\theta }}_{x_L,x_R} - X(\theta ^*_{x_L,x_R}(n)) \right| < {\delta }. \end{aligned}$$

Furthermore, let ${\tilde{X}}(\theta ^*_{x_L,x_R}(n))$ denote the random policy outcome given the equilibrium cutpoint $\theta ^*_{x_L,x_R}(n)$. Given the continuity of the outcome function in the shares of votes, Lemma 5 implies that ${\tilde{X}}(\theta ^*_{x_L,x_R}(n))$ converges in probability to its expected value. That is, for every $\delta >0$ and $\varepsilon >0$ there exists a value $\bar{n} \in {\mathbb {R}}$ such that, for every $n \ge \bar{n}$,

$$\begin{aligned} \text {Pr}\left( \left| {\tilde{X}}(\theta ^*_{x_L,x_R}(n)) - X(\theta ^*_{x_L,x_R}(n))\right| > \varepsilon \right) < \delta . \end{aligned}$$

We can now show that, for n sufficiently large and for every choice $x_R$ of party R, choosing a position $x_L>0$ is never a best response for party L, as $x_L=0$ gives her a higher payoff. To this end, fix party R’s choice $x_R$ and let $\bar{x}_L > 0$. Since the cutpoint outcome is strictly increasing in $x_L$, we have ${\hat{\theta }}_{\bar{x}_L,x_R} > {\hat{\theta }}_{0,x_R}$. Let $\eta = {\hat{\theta }}_{\bar{x}_L,x_R} - {\hat{\theta }}_{0,x_R}$. We are going to split $\eta$ into five intervals (all equal for simplicity) and let n increase. Two intervals will bound the distances between the two expected outcomes, given $\bar{x}_L$ and given 0, and the corresponding cutpoint outcomes. Two other intervals will bound the distances between the two actual outcomes and the corresponding expected outcomes. So the last interval will be the minimum distance between any two realized outcomes given the two different choices of party L. Formally, given the results above, there exists a value $n_1 \in {\mathbb {R}}$ such that, for every $n\ge n_1$,

$$\begin{aligned} \left| {\hat{\theta }}_{\bar{x}_L,x_R} - X(\theta ^*_{\bar{x}_L,x_R}(n)) \right|< \frac{\eta }{5} \quad \text { and } \quad \left| {\hat{\theta }}_{0,x_R} - X(\theta ^*_{0,x_R}(n)) \right| < \frac{\eta }{5}. \end{aligned}$$

Moreover, for every $\delta >0$ there exists a value $n_2 \in {\mathbb {R}}$ such that, for every $n \ge n_2$,

$$\begin{aligned} \text {Pr}\left( \left| {\tilde{X}}(\theta ^*_{\bar{x}_L,x_R}(n)) - X(\theta ^*_{\bar{x}_L,x_R}(n)) \right|< \frac{\eta }{5} \cup \left| {\tilde{X}}(\theta ^*_{0,x_R}(n)) - X(\theta ^*_{0,x_R}(n)) \right| < \frac{\eta }{5} \right) > 1 - \delta . \end{aligned}$$

Let $u_L(x)$ be the payoff that party L attains when the realized outcome is x, and $U_L(x_L,x_R)$ her expected payoff of choosing $x_L$ given party R’s choice $x_R$. Let $n \ge {n}_{\bar{x}_L,x_R} = \max \{n_1,n_2\}$. Note that $u_L \left( {\hat{\theta }}_{0,x_R} + \frac{2}{5}\eta \right) - u_L \left( {\hat{\theta }}_{\bar{x}_L,x_R} -\frac{2}{5}\eta \right) = k > 0$ is a lower bound of the gain that party L gets choosing 0 instead of $\bar{x}_L$ when the outcomes realized in the two cases are no further than $\frac{2}{5}\eta$ from the corresponding cutpoints. Given the above discussion, this happens with probability greater than $1-\delta$. On the other hand, $u_L(1) - u_L(0) = h < 0$ is (in absolute value) an upper bound of the loss that the party can incur in every other case, which happens with probability smaller than $\delta$. Thus, let ${\delta } = \frac{k}{2(k-h)}>0$. We have

$$\begin{aligned} U_L(0,x_R) - U_L(\bar{x}_L,x_R)> & {} (1-{\delta }) \left[ u_L \left( {\hat{\theta }}_{0,x_R} +\frac{2}{5}\eta \right) - u_L \left( {\hat{\theta }}_{\bar{x}_L,x_R} -\frac{2}{5}\eta \right) \right] \\&+\, {\delta } [u_L(1) - u_L(0)] = (1-{\delta })k + {\delta } h = \frac{k}{2} > 0, \end{aligned}$$

so party L prefers to choose 0 rather than $\bar{x}_L$ given party R’s choice $x_R$. Since this holds for every $x_R$ and every $\bar{x}_L>0$, it follows that there exists a value $\bar{n}_L\in {\mathbb {R}}$ such that, for $n\ge \bar{n}_L$, party L has a unique best response against any choice of party R, that is choosing $x_L=0$. An analogous argument proves that there exists a value $\bar{n}_R\in {\mathbb {R}}$ such that, for $n\ge \bar{n}_R$, party R has a unique best response against any choice of party L, that is choosing $x_R=1$. It follows that, for $n \ge \bar{n} = \max \{ \bar{n}_L, \bar{n}_R \}$, (0, 1) is the unique Nash equilibrium. $\square$