Valence influence in electoral competition with rank objectives

Abstract

In this paper we examine the effects of valence in a continuous spatial voting model with two incumbent candidates and a potential entrant. All candidates are rank-motivated. We first consider the case where the low valence incumbent (LVC) and the entrant have zero valence, whereas the valence of the high valence incumbent (HVC) is positive. We show that a sufficiently large valence of HVC guarantees a unique equilibrium, where the two incumbents prevent the entry of the third candidate. We also show that an increase in valence allows HVC to adopt a more centrist policy position, while LVC selects a more extreme position. We also examine the existence of equilibrium for the cases where the LVC has higher or lower valence than the entrant.

Appendices

Appendix A: Supplementary result

The following result will be used in the proof of Lemmas 1, 2, and 3.

Lemma 7

Let Assumptions 1 and 2 be satisfied. Take $(x_H,x_L)$ such that $x_H+\delta<x_L<1$. If $1-F(x_L)\ge F(x_L)-F(\frac{x_H+x_L+\delta }{2})$, then there does not exist $x_N\in (x_H+\delta ,x_L)$ such that $V_N>V_L$. If $F(\frac{x_H+x_L+\delta }{2})\le 2F(x_H+\delta )$, then there does not exist $x_N\in (x_H+\delta ,x_L)$ such that $V_N>V_H$.

Proof of Lemma 7

Under Assumption 2, we have $m_l'(x)<1$ and $m_r'(x)<1$. Let $1-F(x_L)\ge F(x_L)-F(\frac{x_H+x_L+\delta }{2})$ and $x_N\in (x_H+\delta , x_L)$. We have $m_r(\frac{x_H+x_L+\delta }{2})\ge x_L$. That gives us

$$\begin{aligned} m_r\left( \frac{x_H+x_N+\delta }{2}\right)= & {} m_r\left( \frac{x_H +x_L+\delta }{2}\right) -\int _{\frac{x_H+x_N+\delta }{2}}^{\frac{x_H+x_L +\delta }{2}}m_r'(x)dx\\> & {} m_r\left( \frac{x_H+x_L+\delta }{2}\right) -\frac{x_H+x_L+\delta }{2} +\frac{x_H+x_N+\delta }{2}\\\ge & {} x_L-\frac{x_H+x_L+\delta }{2}+\frac{x_H+x_N+\delta }{2}=\frac{x_N+x_L}{2}. \end{aligned}$$

By definition,

$$\begin{aligned} 1-F\left( m_r\left( \frac{x_H+x_N+\delta }{2}\right) \right) = F\left( m_r\left( \frac{x_H+x_N+\delta }{2}\right) \right) -F\left( \frac{x_H+x_N +\delta }{2}\right) , \end{aligned}$$

so

$$\begin{aligned} 1-F\left( \frac{x_N+x_L}{2}\right) >F\left( \frac{x_N+x_L}{2}\right) -F\left( \frac{x_H+x_N+\delta }{2}\right) \end{aligned}$$

and $V_L>V_N$. Let $F(\frac{x_H+x_L+\delta }{2})\le 2F(x_H+\delta )$ and $x_N\in (x_H+\delta , x_L)$. We have $m_l(\frac{x_H+x_L+\delta }{2})\le x_H+\delta$. That gives us

$$\begin{aligned} m_l\left( \frac{x_N+x_L}{2}\right)= & {} m_l\left( \frac{x_H+x_L+\delta }{2}\right) +\int _{\frac{x_H+x_L+\delta }{2}}^{\frac{x_N+x_L}{2}}m_l'(x)dx\\< & {} m_l\left( \frac{x_H+x_L+\delta }{2}\right) +\frac{x_N+x_L}{2}-\frac{x_H+x_L+\delta }{2}\\\le & {} x_H+\delta +\frac{x_N+x_L}{2}-\frac{x_H+x_L+\delta }{2}=\frac{x_H+x_N+\delta }{2}. \end{aligned}$$

By definition,

$$\begin{aligned}&F\left( \frac{x_N+x_L}{2}\right) -F\left( m_l\left( \frac{x_N+x_L}{2}\right) \right) =F\left( m_l\left( \frac{x_N+x_L}{2}\right) \right) , \text{ so } \\&F\left( \frac{x_N+x_L}{2}\right) -F\left( \frac{x_H+x_L+\delta }{2}\right) <F \left( \frac{x_H+x_L+\delta }{2}\right) , \end{aligned}$$

and $V_H>V_N$. $\square$

Appendix B: Proofs of statements

Proof of Lemma 1

In this and subsequent proofs, we will drop the (J, g) notation from ${\tilde{V}}_i^{(J,g)}$, as all our statements apply to all $J=H,L$ and $g\in {{\mathcal {G}}}$. Denote ${\tilde{V}}_N=1-{\tilde{V}}_H-{\tilde{V}}_L$ to be the expected vote share of candidate N.

Suppose that $(x_H,x_L)$ is an equilibrium that does not prevent the entry of candidate N. Then we must have ${\tilde{V}}_N>{\tilde{V}}_L$ or ${\tilde{V}}_N>{\tilde{V}}_H$. HVC cannot rank below sole second place in equilibrium, because deviation $x_H'=x_L$ will guarantee him at least second place, with candidate L ranking last with zero vote share. Therefore we have ${\tilde{V}}_H>{\tilde{V}}_L$ and ${\tilde{V}}_N>{\tilde{V}}_L$. We will show that candidate H or L always has a payoff-improving deviation.

Case 1 :

$\theta <\min \{\alpha ,\alpha '\}$, where $\alpha =F(x_H-\delta )$ and $\alpha '=1-\alpha -\theta$.

Case 1A :

$\alpha \le \alpha '$. Take $x_L'=x_H-\delta -e$, where $e>0$. Denote by ${\tilde{V}}'_i={\tilde{V}}_i(x_H,x_L')$, for $i=H,L,N$. If e is small enough, the limit best response (attained from the right) will be $x_N=x_H+\delta$, with ${\tilde{V}}_L'>{\tilde{V}}_H'$, which is a contradiction.

Case 1B :

$\alpha >\alpha '$. This case is symmetric to Case 1A.

Case 2 :

$\theta \ge \min \{\alpha ,\alpha '\}$ and $\alpha <\alpha '$. Let ${\hat{x}}_L$ be such that $1-F({\hat{x}}_L)=F({\hat{x}}_L)-F(\frac{x_H+{\hat{x}}_L+\delta }{2})$.

Case 2A :

$1-F({\hat{x}}_L)\ge \frac{\alpha }{2}$.

Case 2A1 :

${\hat{x}}_L\ne x_L$. Take $x_L'={\hat{x}}_L$, and denote $V_i'=V_i(x_H,x_L',x_N)$. If $x_H>\delta$ and $x_N\in [0,x_H-\delta )$, we will have $V_N'<\alpha \le V_L'$. If $x_H\ge \delta$ and $x_N=x_H-\delta$, we have $V_N'=\frac{\alpha }{2}<V_L'$. If $x_N\in (x_H-\delta ,x_H+\delta )$, then $V_N'=0$. If $x_N\in (x_H+\delta , {\hat{x}}_L)$, we have $V_L'>V_N'$ by Lemma 7.

If $x_N=x_H+\delta$, then $V_N'=\frac{1}{2}(F({\hat{x}}_L)-F(\frac{x_H+{\hat{x}}_L+\delta }{2}))$. If $x_N={\hat{x}}_L$, then $V_N'=V_L'$ because $1-F({\hat{x}}_L)=F({\hat{x}}_L)-F(\frac{x_H+{\hat{x}}_L+\delta }{2})$. If $x_N\in ({\hat{x}}_L, 1]$, then $V_N'\le V_L'$. Thus, for all $x_N\in [0,1]$, we have $V_L\ge V_N$, so ${\tilde{V}}_N\le {\tilde{V}}_L$. Candidate L improves his rank with $x'_L=x_L$, and $(x_H,x_L)$ is not an equilibrium.

Case 2A2 :

${\hat{x}}_L=x_L$. By an argument identical to Case 2A1, we have $b(x_H,x_L)=O$, which contradicts our assumption.

Case 2B :

$1-F({\hat{x}}_L)<\alpha /2$. Consider the following cases.

Case 2B1 :

$x_L<x_H-\delta$. Then the limit best response will be $x_N= x_H+\delta$. It follows that for any $x_L'\in (x_L,x_H-\delta )$, the limit best response will also be $x_N'= x_H+\delta$, and we will have ${\tilde{V}}_L<{\tilde{V}}_L'<{\tilde{V}}_N$, with rank of candidate L remaining the same. So, $(x_H,x_L)$ is not an equilibrium.

Case 2B2 :

$x_L\in ({\bar{x}}_L,1]$, where ${\bar{x}}_L$ is the solution to $F(\frac{x_H+x_L+\delta }{2})=1-\alpha$. Then ${\tilde{V}}_L<\alpha$, with candidate L ranking last by assumption that $(x_H,x_L)$ is an equilibrium. If $x_L'=x_H-\delta -e$, then the limit best response will be $x_N=x_H+\delta$, and we will have ${\tilde{V}}_L'>{\tilde{V}}_L$ if e is small enough. So, $(x_H,x_L)$ is not an equilibrium.

Case 2B3 :

$x_L\in (x_H+\delta ,{\bar{x}}_L)$. Because $\alpha /2>1-F({\hat{x}}_L)$, we have $x_L<{\hat{x}}_L$ and $1-F(x_L)>F(x_L)-F(\frac{x_H+x_L+\delta }{2})$. Take $x_H'=x_H+e$. If e is small enough, we will have $1-F(x_L)>F(x_L)-F(\frac{x_H'+x_L+\delta }{2})$.

Because of this and $\theta \ge \alpha$, by Lemma 7 we will have $V_L'>V_N'$ and $V_H'>V_N'$ for all $x_N\in (x_H+\delta ,x_L)$. The limit best response will be $x_N=x_L$ for both $x_H$ and $x_H'$. Hence ${\tilde{V}}_H(x_H',x_L)>{\tilde{V}}_H(x_H,x_L)$, with rank of candidate H remaining the same. So, $(x_H,x_L)$ is not an equilibrium.

Case 2B4 :

$x_L={\bar{x}}_L$. Candidate N has limit best response $x_N=x_L$ (argument is similar to one for Case 2B3). Let $x_L'={\bar{x}}_L+e<{\hat{x}}_L$. Then candidate N will have limit best response $x_N=x_H-\delta$ (attained from the left). We will have ${\tilde{V}}_L'>{\tilde{V}}_L$, with candidate L ranking last in either case.

Case 3 :

$\alpha '\le \alpha$ and $\theta \ge \min \{\alpha ,\alpha '\}$ This case is symmetric to Case 2.

$\square$

Proof of Lemma 2

Conditions (14) and (15) imply that $x_H+\delta<x_L<1$. Indeed, if $x_H+\delta >x_L$, then $\gamma <0$, and (14) is violated, as $\phi \ge 0$. If $x_H+\delta =x_L$, then $\gamma =\phi =0$, so $x_L=1$ and $\alpha >0$, as, by assumption, we have $\delta <\frac{1}{2}$. This contradicts (15). As $x_H+\delta <x_L$, we must have $\beta >0$ and $\gamma =\phi >0$, so $x_L<1$.

We are first going to show that, if the conditions (14), (15), and (16) are satisfied, then there does not exist $x_N\in [0,1]$ such that $V_N>V_H$ or $V_N>V_L$. As a consequence, there does not exist a best response $x_N\in [0,1]$ or a limit best response for candidate N. Indeed, if $x_N\in [0,x_H-\delta )$, then we have $V_N=F(\frac{x_H+x_N-\delta }{2})<\alpha \le \theta +\beta <V_H$ and $V_N<\gamma +\phi =V_L$. If $x_N=x_H-\delta$, then $V_N=\frac{\alpha }{2}<V_H=\frac{\alpha }{2}+\theta +\beta$ and $V_N<V_L=\gamma +\phi$. If $x_N\in (x_H-\delta ,x_H+\delta )$, then $V_N=0$. If $x_N=x_H+\delta$, then $V_N=\frac{\beta }{2}<V_H=\frac{\beta }{2}+\theta +\alpha$ and $V_N<V_L=\gamma +\phi$. If $x_N\in (x_H+\delta ,x_L)$, then, as $\gamma =\phi$ and $\beta \le \alpha +\theta$, by Lemma 7 we have $V_L>V_N$ and $V_H>V_N$. Suppose that $x_N=x_L$. We have $V_L=V_N=\gamma$ and $V_H=\alpha +\theta +\beta$. We have $V_H\ge V_N$ because $\alpha +\theta +\beta =\lim _{x\rightarrow x_L^-}V_H(x_H,x_L,x)$ and $\gamma =\lim _{x\rightarrow x_L^-}V_N(x_H,x_L,x)$, but $V_H(x_H,x_L,x_N)>V_N(x_H,x_L,x_N)$ for all $x_N\in (x_H+\delta ,x_L)$. Finally, if $x_N\in (x_L,1]$, then we have $V_N=1-F(\frac{x_L+x_N}{2})<\gamma \le \alpha +\theta +\beta =V_H$ and $V_N<V_L$, as $V_N+V_L=2\gamma$.

We will now show that if conditions (14), (15), and (16) are violated, there will exist $x_N\in [0,1]$ such that $V_N>V_H$ or $V_N>V_L$, that trigger the entry by candidate N. Let $x_H+\delta <x_L$. If $\gamma <\phi$, we will have $V_N>V_L$ if $x_N=x_L+e$ and $e>0$ is small enough. If $\gamma >\phi$, we will have $V_N>V_L$ if $x_N=x_L-e$ and $e>0$ is small enough. If $\alpha >\gamma +\phi$, we will have $V_N>V_L$ if $x_N=x_H-\delta -e$ and $e>0$ is small enough. If $\alpha >\theta +\beta$, we will have $V_N>V_H$ if $x_N=x_H-\delta -e$ and $e>0$ is small enough. Finally, if $\beta >\theta +\alpha$, we will have $V_N>V_H$ if $x_N=x_H+\delta +e$ and $e>0$ is small enough.

Let $x_H+\delta =x_L<1$. Then $V_N>V_L$ if $x_N=x_H+\delta +e$ and $e>0$ is small enough. If either $x_H+\delta =x_L=1$ or $x_H+\delta >x_L$, then $V_L=0$ for all $x_N\in [0,1]$, and there exists $e>0$ such that $V_N>0$ if either $x_N=x_H-\delta -e$ or $x_N=x_H+\delta +e$. $\square$

Proof of Lemma 3

Let the conditions of Lemma 2 be satisfied. Then, clearly, $b(x_H,x_L)=O$. Suppose that $\alpha <\gamma +\phi$ and $\alpha =\beta +\theta$. Consider the following cases.

Case 1 :

$\phi <\alpha$. Let $x_L'=x_L-e$. If e is small enough, candidate N will have limit best response $x_N=x_H-\delta$, with ${\tilde{V}}_L'=1-F(\frac{x_L'+x_H+\delta }{2})>1-F(\frac{x_L+x_H+\delta }{2})={\tilde{V}}_L$, ${\tilde{V}}_H'<\beta +\theta<\gamma +\phi <{\tilde{V}}_L'$, and $\alpha ={\tilde{V}}_N'<{\tilde{V}}_L'$, so candidate L improves his payoff.

Case 2 :

$\phi =\alpha$. Then we have ${\tilde{V}}_H={\tilde{V}}_L$. Let $x_H'=x_H-e$. Then candidate N has limit best response $x_N=x_L$. We have ${\tilde{V}}_H'<{\tilde{V}}_H$, but, if e is small enough, we will have ${\tilde{V}}_H'>{\tilde{V}}_N'$ and ${\tilde{V}}_H'>{\tilde{V}}_L'$, so candidate H will improve his rank.

Case 3 :

$\phi >\alpha$. Let $x_H'=x_H+e$. Then candidate N has limit best response $x_N=x_L$. We have ${\tilde{V}}_H'=F(\frac{x_L+x_H'+\delta }{2})>F(\frac{x_L+x_H+\delta }{2})={\tilde{V}}_H$. Recall from the proof of Lemma 1 that $\alpha +\beta +\theta \ge \gamma =\phi$. By that we will have ${\tilde{V}}_H'>{\tilde{V}}_N'$ and ${\tilde{V}}_H'>{\tilde{V}}_L'$.

It follows that either candidate H or candidate L will be able to improve his payoff.

Suppose that $\alpha <\gamma +\phi$ and $\alpha <\beta +\theta$. Let $x_H'=x_H+e$. Similarly to Case 3 above, candidate H will be able to improve his payoff.

It follows that, if the conditions of Lemma 2 are satisfied, condition $\alpha =\gamma +\phi$ is necessary for no payoff-improving deviations for candidates H and L to exist. Suppose that this condition is satisfied. We will show that candidate H has no payoff-improving deviations, and candidate L has a payoff-improving deviation if and only if condition (13) is violated.

Note that candidate H is ranked first and consider $x_H'\ne x_H$. If $x_H'\in [0,x_H)$, we will have ${\tilde{V}}_H'<{\tilde{V}}_H$, because ${\tilde{V}}_H'\le F(\frac{x_H'+x_L+\delta }{2})$. If $x_H'\in (x_H,1]$, then candidate N will have limit best response $x_N=\min \{x_H-\delta ,x_L\}$. This gives us ${\tilde{V}}_N'\ge \alpha$, ${\tilde{V}}_L'<\alpha$, and ${\tilde{V}}_H'\le 1-\alpha ={\tilde{V}}_H$. So, there is no $x_H'$ such that ${\tilde{V}}_H'>{\tilde{V}}_H$.

Consider $x_L'\ne x_L$:

Case 1 :

$x_L'\in [0,x_H-\delta )$. We have ${\tilde{V}}_L'<\alpha =\gamma +\phi ={\tilde{V}}_L$. If candidate N has limit best response $x_N=x_H+\delta$, we will have ${\tilde{V}}_L'<{\tilde{V}}_N$, as $\alpha <\beta +\gamma +\phi$. If $b(x_H,x_L')=O$, then ${\tilde{V}}_L<\theta +\beta +\gamma +\phi <{\tilde{V}}_H$, so candidate L will not improve his rank by deviating to $x_L'$, but will reduce his share of vote.

Case 2 :

$x_L'=x_H-\delta$. We have ${\tilde{V}}_L'=\frac{\alpha }{2}<{\tilde{V}}_L$, with candidate L not ranking higher than second for same reasons as with $x_L'\in [0,x_H-\delta )$.

Case 3 :

$x_L'\in (x_H-\delta ,x_H+\delta )$. ${\tilde{V}}_L=0$.

Case 4 :

$x_L'\in [x_H+\delta ,x_m)$: the limit best response of candidate N will be $x_N=x_L'$, with ${\tilde{V}}_N>{\tilde{V}}_L$ and ${\tilde{V}}_H>{\tilde{V}}_L$.

Case 5 :

$x_L'\in (x_m,x_L)$. Put

$$\begin{aligned} x_m'=\frac{x_H+x_L'+\delta }{2}. \end{aligned}$$

Take $x_N\in [0,x_H-\delta ]$. Then $V_N'<\alpha$. We have $\lim _{x\rightarrow (x_H-\delta )^-}V_N(x_H,x_L',x)=\alpha =\gamma +\phi <\lim _{x\rightarrow (x_H-\delta )^-}V_L(x_H,x_L',x)$ and $\lim _{x\rightarrow (x_H-\delta )^-}V_H(x_H,x_L',x)=F(x_m')-\alpha$. If $x_N\in (x_H-\delta ,x_H+\delta )$, then $V_N'=0$. As $\alpha +\theta \ge \beta >F(x_m')-F(x_H+\delta )$, by Lemma 7 we have $V_N'<V_H'$ whenever $x_N\in (x_H+\delta ,x_L')$. We also have $V_N'<V_H'$ when $x_N=x_H+\delta$. If $x_N=x_L'$, then $V_N'=V_L'=\frac{1-F(x_m')}{2}$.

As $x_L=m_r(x_m)$, we have

$$\begin{aligned} m_r(x_m')=x_L-\int _{x_m'}^{x_m}m'_r(x)dx>\frac{x_L+x_L'}{2}>x_L', \end{aligned}$$

so $F(x_m')-F(x_L')<1-F(x_L')$. Thus, by Lemma 7, we have $V_N'<V_L'$ whenever $x_N\in (x_H+\delta ,x_L')$. We have $\lim _{x\rightarrow x_L^{'+}}V_N(x_H,x_L',x)=1-F(x_L')>F(x_m')-F(x_L')=\lim _{x\rightarrow x_L^{'+}}V_L(x_H,x_L',x)$ and

$$\begin{aligned} \lim _{x\rightarrow x_L^{'+}}V_H\left( x_H,x_L',x\right) =V_H\left( x_H,x_L',x_L'\right) . \end{aligned}$$

As $\alpha =\gamma +\phi$, we have $\lim _{x\rightarrow x_L^{'+}}V_N(x_H,x_L',x)<\lim _{x\rightarrow (x_H-\delta )^-}V_N(x_H,x_L',x)$. Limit best response of candidate N will be as follows:

• 1. If $\alpha >F(x_m')-\alpha$, candidate N will have limit best response $x_N=x_H-\delta$,

• 2. If $\alpha \le F(x_m')-\alpha$, candidate N will have limit best response $x_N=x_L'$.

Candidate L will not deviate to $x_L'\in (x_m,x_L)$ if and only if $2\alpha \le F(x_m')$. This will be true for all $x_L'\in (x_m,x_L)$ if and only if condition (13) holds.

Case 6 :

$x_L'=x_m$. Consider a limiting case $x_N\rightarrow _-x_H-\delta$. Then we have $V_N=\alpha$, $V_H=F(x_m')-\alpha$, and $V_L=1-F(x_m')$. The ranking of the candidates will be as follows. If $F(x_m')>2\alpha$, then $V_L>V_N$ and $V_H>V_N$. If $F(x_m')=2\alpha$, then $V_L>V_H=V_N$. If $F(x_m')<2\alpha$, then $V_L>V_N>V_H$. In another limiting case, $x_N\rightarrow _+x_L'$, we have $V_N=\alpha$, $V_H=F(x_m')$, and $V_L=1-F(x_m')-\alpha$, with $V_H>V_N>V_L$. It follows that candidate N will have limit best response $x_N=x_L'$ if $F(x_m')\ge 2\alpha$.

Case 7 :

$x_L'\in (x_L,1]$. We will have ${\tilde{V}}_L'<{\tilde{V}}_L$. As $\gamma =\phi$, for $x_N=x_L-e$ we will have $V_N'>V_L'$ if e is small enough. So, candidate N will enter. We will also have ${\tilde{V}}_L'<{\tilde{V}}_H$ and ${\tilde{V}}_L'<{\tilde{V}}_N'$.

It follows that, when $\alpha =\gamma +\phi$, candidate L will not be able to improve his payoff if and only if (13) holds. $\square$

Proof of Lemma 4

Denote $f_L=f(x_L)$, $f_{H-}=f(x_H-\delta )$, and $f_{m}=f(x_m)$.

Define

$$\begin{aligned} H_1=1+F\left( \frac{x_H+x_L+\delta }{2}\right) -2F(x_L) \end{aligned}$$

(28)

and

$$\begin{aligned} H_2=F(x_H-\delta )+2F(x_L)-2. \end{aligned}$$

(29)

Conditions (11) then become $H_1=0$ and $H_2=0$. We have the following derivatives:

$$\begin{aligned} \begin{array}{ll} \frac{\partial H_1}{\partial \delta }=\frac{1}{2}f_m&{}\frac{\partial H_2}{\partial \delta }=-f_{H-}\\ \frac{\partial H_1}{\partial x_H}=\frac{1}{2}f_m&{}\frac{\partial H_2}{\partial x_H}=f_{H-}\\ \frac{\partial H_1}{\partial x_L}=\frac{1}{2}f_m-2f_L&{}\frac{\partial H_2}{\partial x_L}=2f_L \end{array} \end{aligned}$$

(30)

Let $x_L^2(x_H)$ be a solution to $H_2=0$. Since $f(x)>0$ for all $x\in [0,1]$, by the Implicit Function Theorem we have

$$\begin{aligned} \frac{\partial x_L^2(x_H)}{\partial x_H}=-\frac{f_{H-}}{2f_L}<0. \end{aligned}$$

(31)

We have $x_L^2(\delta )=1$ and, since $\frac{\partial H_2}{\partial x_L}<0$, we must have $x_L^2(1-\delta )<1$. As $H_1(\delta ,1)=F(\frac{1}{2}+\delta )-1<0$ and $H_1(1-\delta ,x_L^2(1-\delta ))=1+F(\frac{1+x_L^2(1-\delta )}{2})-2F(x_L^2(1-\delta ))>0$, for some $x_H\in (\delta ,1-\delta )$ we have $H_1(x_H,x_L^2(x_H))=0$. It follows that system (11) has a solution.

Now let Assumption 2 be satisfied. Let $x_L^1(x_H)$ be a solution to $H_1=0$. Since $f(x)>0$ for all $x\in [0,1]$, by the Implicit Function Theorem we have

$$\begin{aligned} \frac{\partial x_L^1(x_H)}{\partial x_H}=-\frac{\frac{1}{2}f_m}{\frac{1}{2}f_m-2f_L}>0, \end{aligned}$$

(32)

since from Assumption 2 it follows that $2f_L>\frac{1}{2}f_m$. Therefore there exists only one $x_H$ such that $x_L^1(x_H)=x_L^2(x_H)$. It follows that system (11) has a unique solution. $\square$

Proof of Lemma 5

As $\gamma ^*=\phi ^*$, we have

$$\begin{aligned} 1-F(x_L^*)=F(x_L^*)-F\left( \frac{x_H^*+x_L^*+\delta }{2}\right) . \end{aligned}$$

We have $\lim _{x_N\rightarrow _+ x_H^*+\delta _H}V_N=\beta ^*$ and $\lim _{x_N\rightarrow _+ x_H^*+\delta _H}V_L=\alpha ^*$. Hence, by Lemma 7, we must have $\beta ^*\le \alpha ^*$. $\square$

Proof of Lemma 6

Denote $f_{H+}=f(x_H+\delta )$. Let

$$\begin{aligned} M=\left( \begin{array}{cc} \frac{\partial H_1}{\partial x_H}&{}\frac{\partial H_1}{\partial x_L}\\ \frac{\partial H_2}{\partial x_H}&{}\frac{\partial H_2}{\partial x_L} \end{array} \right) \end{aligned}$$

(33)

be the Jacobian matrix of $(H_1 H_2)$. We have

$$\begin{aligned} |M|=f_mf_L-f_{H-}\left( \frac{1}{2}f_m-2f_L\right) , \end{aligned}$$

(34)

and a sufficient condition for $|M|>0$ is $4f_L\ge f_m$, which is satisfied because of Assumption 2’. The derivatives of $x_H^*$ and $x_L^*$ with respect to $\delta$ will be given by the the Implicit Function Theorem:

$$\begin{aligned} \left( \begin{array}{c} \frac{\partial x_H^*}{\partial \delta }\\ \frac{\partial x_L^*}{\partial \delta } \end{array} \right) =-\frac{1}{|M|}\left( \begin{array}{cc} 2f_L&{}2f_L-\frac{1}{2}f_m\\ -f_{H-}&{}\frac{1}{2}f_m \end{array}\right) \left( \begin{array}{c} \frac{1}{2}f_m\\ -f_{H-} \end{array}\right) , \end{aligned}$$

with all densities evaluated at $(x_H^*,x_L^*)$. It follows that

$$\begin{aligned} \frac{\partial x_H^*}{\partial \delta }= & {} -\frac{1}{|M|}\left( f_L(f_m-2f_{H-}) +\frac{1}{2}f_mf_{H-}\right) ,\nonumber \\ \frac{\partial x_L^*}{\partial \delta }= & {} \frac{1}{|M|}f_mf_{H-}. \end{aligned}$$

(35)

Denote $f_t=f((3x_H+x_L+3\delta )/4)$. By the Implicit Function Theorem and Assumption 2, D varies continuously with $\delta$:

$$\begin{aligned} \frac{\partial D(x_H^*,x_L^*)}{\partial \delta }= & {} \frac{1}{4}\left( 3\frac{\partial x_H^*}{\partial \delta }+\frac{\partial x_L^*}{\partial \delta }+3\right) f_t-2\left( \frac{\partial x_H^*}{\partial \delta }+1\right) f_{H-}\nonumber \\= & {} \frac{1}{|M|}\left( f_{H-}f_m\left( 4f_L-\frac{f_t}{4}\right) +f_tf_L\left( 2f_{H-}-\frac{f_m}{2}\right) \right) . \end{aligned}$$

(36)

If Assumption 2’ is also satisfied, this derivative is positive. As $D<0$ for $\delta$ close to 0 and $D>0$ for $\delta$ close to $\frac{1}{2}$, we have $D<0$ if $\delta <\delta _0$ and $D>0$ if $\delta >\delta _0$ for some $\delta _0\in (0,\frac{1}{2})$ under Assumption 2’. $\square$

Proof of Theorem 3

The second of the derivatives (35) is positive. Now suppose that $f_L=af_m$ and $f_{H-}=bf_m$. We have $\frac{\partial x_H^*}{\partial \delta }=-f^2_m/|M|\cdot (a(1-2b)+\frac{b}{2})$. We must find the minimum $c<1$ for which inequality $(a(1-2b)+\frac{b}{2})\ge 0$ will hold for all a, b such that $a\in [c,1/c]$, $b\in [c,1/c]$, and $a/b\in [c,1/c]$. The solution to this problem is $c=\frac{3}{4}$. Hence, the derivative is positive as long as Assumption 2’ is true. $\square$

Proof of Theorem 4

Assuming distribution function $\mathbf{F_a}(x,a)$ and differentiating (28) and (29) at $a=0$, we get

$$\begin{aligned} \frac{\partial H_1}{\partial a}=0, \frac{\partial H_2}{\partial a}=1. \end{aligned}$$

(37)

Similarly, at $b=0$ we have

$$\begin{aligned} \frac{\partial H_1}{\partial b}=2, \frac{\partial H_2}{\partial b}=-2. \end{aligned}$$

(38)

By the Implicit Function Theorem, (30) and (33), at $a=b=0$ we have

$$\begin{aligned} \left( \begin{array}{cc} \frac{\partial x_H}{\partial a}&{}\frac{\partial x_L}{\partial a}\\ \frac{\partial x_H}{\partial b}&{}\frac{\partial x_L}{\partial b}\\ \end{array}\right) =-M^{-1}\cdot \left( \begin{array}{cc} \frac{\partial H_1}{\partial a}&{}\frac{\partial H_1}{\partial b}\\ \frac{\partial H_2}{\partial a}&{}\frac{\partial H_2}{\partial b}\\ \end{array}\right) =\left( \begin{array}{cc} f_m-4f_L&{}-2f_m\\ -f_m&{}4f_{H-}+2f_m \end{array}\right) \cdot \frac{1}{2|M|}, \end{aligned}$$

where |M| is given by (34), and all densities are evaluated at $(x_H^*,x_L^*)$. The required signs of partial derivatives follow from Assumption 2. $\square$

Proof of Theorem 5

Assuming distribution functions $\mathbf{F_a}(x,a)$ and $\mathbf{F_b}(x,b)$, we differentiate D with respect to a and b:

$$\begin{aligned} \frac{\partial D}{\partial a}=-2, \frac{\partial D}{\partial b}=0. \end{aligned}$$

(39)

Let

$$\begin{aligned} {\bar{M}}=\left( \begin{array}{ccc} \frac{\partial H_1}{\partial x_H}&{}\frac{\partial H_1}{\partial x_L}&{}\frac{\partial H_1}{\partial \delta }\\ \frac{\partial H_2}{\partial x_H}&{}\frac{\partial H_2}{\partial x_L}&{}\frac{\partial H_2}{\partial \delta }\\ \frac{\partial D}{\partial x_H}&{}\frac{\partial D}{\partial x_L}&{}\frac{\partial D}{\partial \delta } \end{array} \right) , \end{aligned}$$

(40)

where

$$\begin{aligned} \frac{\partial D}{\partial x_H}=\frac{3}{4}f_t-2f_{H-},\quad \frac{\partial D}{\partial x_L}=\frac{1}{4}f_t,\quad \text{ and } \quad \frac{\partial D}{\partial \delta }=\frac{1}{4}f_t+2f_{H-}. \end{aligned}$$

(41)

Taking $D=0$, at $a=b=0$ we have

$$\begin{aligned} \left( \begin{array}{cc} \frac{\partial x_{Ha}^*}{\partial a}&{}\frac{\partial x_{Hb}^*}{\partial b}\\ \frac{\partial x_{La}^*}{\partial a}&{}\frac{\partial x_{Lb}^*}{\partial b}\\ \frac{\partial \delta _a}{\partial a}&{}\frac{\partial \delta _b}{\partial b} \end{array}\right) = - {\bar{M}}^{-1}\cdot \left( \begin{array}{cc} \frac{\partial H_1}{\partial a}&{}\frac{\partial H_1}{\partial b}\\ \frac{\partial H_2}{\partial a}&{}\frac{\partial H_2}{\partial b}\\ \frac{\partial D}{\partial a}&{}\frac{\partial D}{\partial b}\end{array}\right) \end{aligned}$$

(42)

where $\frac{\partial D}{\partial \delta }$ is evaluated keeping $x_H$, $x_L$ constant. Evaluating this expression, we get

$$\begin{aligned} \frac{\partial \delta _a}{\partial a}= & {} \left( 2f_mf_L - \frac{1}{4}f_mf_t + \frac{3}{2}f_Lf_t\right) \frac{1}{|M|}\end{aligned}$$

(43)

$$\begin{aligned} \frac{\partial \delta _b}{\partial b}= & {} \left( -2f_mf_{H-} + \frac{1}{2}f_mf_t - \frac{1}{2}f_{H-}f_t\right) \frac{1}{|M|}, \end{aligned}$$

(44)

where

$$\begin{aligned} |{\bar{M}}|=4f_{H-}f_mf_L - \frac{1}{4}f_{H-}f_mf_t + 2f_{H-}f_Lf_t - \frac{1}{2}f_mf_Lf_t, \end{aligned}$$

and all densities are evaluated at $(x_H^*,x_L^*)$. If Assumption 2’ is satisfied, then all derivatives have the required signs. $\square$

Proof of Theorem 6

Suppose that candidates have strategies $(x_H^*,x_L^*)$ defined by (21), (22), and candidate H has no payoff-improving deviations. Candidate L has positive vote share if $x_H^*+\delta _H<x_L^*+\delta _L$, or

$$\begin{aligned} \delta _L<\frac{1-2\delta _H}{3}. \end{aligned}$$

(45)

Candidate N should not be able to outrank candidate H with any $x_N$ to the left of $x_H^*-\delta _H$. The corresponding condition, derived similarly to (9), is

$$\begin{aligned} \delta _H-\delta _L\ge \frac{1}{12}. \end{aligned}$$

(46)

We now derive conditions (24). Candidate L cannot make a deviation $x_L'\in (x_H^*+\delta _H-\delta _L,x_L)$ if after such a deviation candidate N can enter with $x_N$ to the right of $x_L'+\delta _L$. Candidate N will not follow through with such a move only if he can instead enter with $x_N$ to the left of $x_H^*-\delta _H$, pushing candidate H to third place, and, at the same time, getting a higher share of the votes than with any $x_N$ to the right of $x_L'+\delta _L$. To derive a condition similar to (10) and (13) that disallows such deviations by candidate L we need to consider two cases.

Case 1.:

If $x_H^*\ge \frac{1}{2}$ or

$$\begin{aligned} \delta _L\ge \frac{1-2\delta _H}{8}, \end{aligned}$$

(47)

then, for any $x_L'\in (x_H^*+\delta _H-\delta _L,x_L)$, candidate N can choose a position to the left of $x_H^*-\delta _H$ that will give him a higher vote share than any position to the right of $x_L'+\delta _L$. Hence, no payoff-improving deviation for Candidate L will exist if and only if candidate N cannot rank above candidate H for any $x_N$ to the left of $x_H^*-\delta _H$ and for any $x_L'\in (x_H^*+\delta _H-\delta _L,x_L^*)$. That amounts to $x_H^*-\delta _H\le 2\delta _H-\delta _L$, or

$$\begin{aligned} \delta _L\le \frac{14\delta _H-2}{9}. \end{aligned}$$

(48)

If this condition holds, any attempt by candidate L to move closer to candidate H will result in candidate N entering to the right of candidate L, pushing him to the third place.

Case 2.:

If $x_H^*<\frac{1}{2}$ or

$$\begin{aligned} \delta _L<\frac{1-2\delta _H}{8}, \end{aligned}$$

(49)

then for all $x_L'\in (x_H^*+\delta _H-\delta _L,\frac{x_H^*+\delta _H+x_L^*-3\delta _L}{2})$, candidate N can obtain a higher vote share if he enters to the right of $x_L'+\delta _L$ rather than to the left of $x_H^*-\delta _H$. It follows that in equilibrium, candidate N should not be able to outrank candidate H if $x_L'=\frac{x_H^*+\delta _H+x_L^*-3\delta _L}{2}$, or

$$\begin{aligned} \delta _H-\delta _L\ge \frac{3}{26}. \end{aligned}$$

(50)

As candidate L’s vote share will be zero whenever $x_L\in (x_H-\delta _H+\delta _L,x_H+\delta _H-\delta _L)$, there will be no payoff-improving deviation $x_L'\in [x_H^*-\delta _H+\delta _L,1]$ if either (47) and (48), or (49) and (50), are satisfied. Together with (45) and the excessive (46) this gives us conditions (24).

Suppose now that conditions (24) are satisfied. Our goal is to derive the conditions on $\delta _H$ and $\delta _L$ under which candidate L cannot improve his payoff with a deviation $x_L'\in [0,x_H-\delta _H+\delta _L)$.

Before deriving the conditions, we will introduce some notation for vote shares of the three candidates, depending on $x_L'$ and the possible response of candidate N. Let

$$\begin{aligned} {\bar{V}}_L={\tilde{V}}_L(x_H^*,x_L^*)=1-\frac{x_H^*+x_L^*+\delta _H -\delta _L}{2}=\frac{2-4\delta _H+4\delta _L}{5} \end{aligned}$$

be candidate L’s vote share with (21), (22). For $x_L'\in [0,x_H^*-\delta _H+\delta _L)$, let

$$\begin{aligned} {\hat{V}}_N= & {} \lim _{x_N\rightarrow _+x_H^*+\delta _H}V_N\left( x_L',x_H^*,x_N\right) =\frac{3-6\delta _H-4\delta _L}{5},\\ {\hat{V}}_L(x_L')= & {} \lim _{x_N\rightarrow _+x_H^*+\delta _H}V_L\left( x_L',x_H^*,x_N\right) =\frac{x_L'}{2}+\frac{2-4\delta _H+9\delta _L}{10},\\ {\hat{V}}_H(x_L')= & {} \lim _{x_N\rightarrow _+x_H^*+\delta _H}V_H\left( x_L',x_H^*,x_N\right) =\frac{2+16\delta _H-\delta _L}{10}-\frac{x_L'}{2}. \end{aligned}$$

These are the limits of the vote shares of the three candidates if candidate N chooses a position immediately to the right of $x_H^*+\delta _H$. Let

$$\begin{aligned} V_{L0}={\hat{V}}_L(0)=\frac{2-4\delta _H+9\delta _L}{10} \quad \text{ and } \quad V_{L1}=\lim _{x_L'\rightarrow x_H^*-\delta _H+\delta _L}{\hat{V}}_L(x_L')=\frac{2-4\delta _H+9\delta _L}{5} \end{aligned}$$

be the smallest and largest (in the limit) vote shares that candidate L can obtain with $x_L'\in [0,x_H^*-\delta _H+\delta _L)$, given such response from candidate N. Similarly define

$$\begin{aligned} V_{H0}={\hat{V}}_H(0)=\frac{2+16\delta _H-\delta _L}{10}\quad \text{ and } \quad V_{H1}=\lim _{x_L'\rightarrow x_H^*-\delta _H+\delta _L}{\hat{V}}_H(x_L')=2\delta _H-\delta _L. \end{aligned}$$

Now suppose that, whenever $x_L'\in [0,x_H^*-\delta _H-\delta _L]$, candidate N chooses a position immediately to the right of $x_L'+\delta _L$, and does not enter when $x_L'\ge x_H^*-\delta _H-\delta _L$. Define

$$\begin{aligned} V_L'(x_L')=\left\{ \begin{array}{l@{\quad }l} \lim _{x_N\rightarrow _+ x_L'+\delta _L}V_L(x_L',x_H^*,x_N),&{}x_L'\in [0, x_H^*-\delta _H-\delta _L]\\ V_L(x_L',x_H^*,O),&{}x_L'\in \left( x_H^*-\delta _H-\delta _L,x_H^*-\delta _H+\delta _L\right) , \end{array}\right. \end{aligned}$$

or

$$\begin{aligned} V_L'(x_L')=\left\{ \begin{array}{l@{\quad }l} \delta _L+x_L',&{}x_L'\in [0,\frac{2-4\delta _H-\delta _L}{5}]\\ \frac{2-4\delta _H+9\delta _L}{10}+\frac{x_L'}{2},&{}x_L' \in \left( \frac{2-4\delta _H-\delta _L}{5},\frac{2-4\delta _H+9\delta _L}{5}\right) . \end{array}\right. \end{aligned}$$

Similarly, let

$$\begin{aligned} V_N'(x_L')=\left\{ \begin{array}{l@{\quad }l} \frac{2-4\delta _H-\delta _L}{10}-\frac{x_L'}{2},&{}x_L'\in [0,\frac{2-4 \delta _H-\delta _L}{5}]\\ 0,&{}x_L'\in \left( \frac{2-4\delta _H-\delta _L}{5},\frac{2-4\delta _H+9\delta _L}{5}\right) . \end{array}\right. \end{aligned}$$

Note that the vote share of candidate H must be greater than one half whenever $x_L'\in [0,x_H^*-\delta _H+\delta _L)$ and $x_N<x_H^*-\delta _H$ (or $x_N=O$). This is true because

$$\begin{aligned} x_H^*-\delta _H+\delta _L=\frac{2-4\delta _H+9\delta _L}{5}<\frac{1}{2} \text{ whenever } \delta _L<\frac{1+8\delta _H}{18}, \end{aligned}$$

which must hold if (24) are satisfied. Let

$$\begin{aligned} x'=\frac{2-4\delta _H-11\delta _L}{15} \end{aligned}$$

be such that $V_L'(x')=V_N'(x')$. Note that $V_L'(x_L')<V_N'(x_L')$ if $x_L'\in [0,x')$, and $V_L'(x_L')> V_N'(x_L')$ if $x_L'\in (x',x_H^*-\delta _H-\delta _L)$.

Finally, suppose that candidate N chooses a position immediately to the left of $x_L'-\delta _L$ whenever $x_L'>\delta _L$, and $x_N=O$ otherwise. Define

$$\begin{aligned} V_{L}''(x_L')=\left\{ \begin{array}{l@{\quad }l} V_L(x_H^*,x_L',O),&{}x_L'\in [0,\delta _L]\\ \lim _{x_N\rightarrow _{-}x_L-\delta _L}V_L(x_H^*,x_L',x_N) &{}x_L'\in (\delta _L,x_H^*-\delta _H+\delta _L) \end{array}\right. \end{aligned}$$

or

$$\begin{aligned} V_L''(x_L')=\left\{ \begin{array}{l@{\quad }l} \frac{2-4\delta _H+9\delta _L}{10}+\frac{x_L'}{2},&{}x_L'\in [0,\delta _L]\\ \frac{2-4\delta _H+19\delta _L}{10}-\frac{x_L'}{2},&{}x_L'\in (\delta _L, \frac{2-4\delta _H+9\delta _L}{5}). \end{array}\right. \end{aligned}$$

Similarly, let

$$\begin{aligned} V_N''(x_L')=\left\{ \begin{array}{l@{\quad }l} 0,&{}x_L'\in [0,\delta _L]\\ x_L'-\delta _L,&{}x_L'\in (\delta _L,\frac{2-4\delta _H+9\delta _L}{5}). \end{array}\right. \end{aligned}$$

Let

$$\begin{aligned} x''=\frac{2-4\delta _H+29\delta _L}{15} \end{aligned}$$

be such that $V_L''(x'')=V_N''(x'')$. Note that $V_L''(x_L')>V_N''(x_L')$ if $x_L'\in [\delta _L,x'')$, and $V_L''(x_L')<V_N''(x_L')$ if $x_L'\in (x'',x_H-\delta _H+\delta _L)$. We also have $x''>x'$.

Note that $V_{H0}>V_{L0}$ as $\delta _L<2\delta _H$ in any equilibrium. Similarly, we always have ${\bar{V}}_L>V_{L0}$ as $\delta _L<2-4\delta _H$, and we should always have ${\bar{V}}_L<V_{L1}$. As ${\hat{V}}_H(x_L')$ is decreasing in $x_L'$ and ${\hat{V}}_L(x_L')$ is increasing in $x_L'$, we have ${\hat{V}}_N>V_{L1}\Rightarrow {\hat{V}}_N>V_{L0}$ and ${\hat{V}}_N>V_{H0}\Rightarrow {\hat{V}}_N>V_{H1}$.

Any $(\delta _H,\delta _L)$ satisfying conditions (24) falls into one of the nine cases (see Fig. 4):

Fig. 4

Equilibrium conditions (24)

Case 1 :

${\hat{V}}_N\ge V_{L1}$ and ${\hat{V}}_N>V_{H0}$, or

$$\begin{aligned} \delta _L\le \frac{1-2\delta _H}{13}\quad \text{ and } \quad \delta _L<\frac{4-28\delta _H}{7}. \end{aligned}$$

(51)

There exists$x_L'\in [0,x_H^*-\delta _H+\delta _L)$ that improves the payoff of candidate L. Take $x_L'=x_H^*-\delta _H+\delta _L-e$. Candidate N will have the limit best response $x_N'=x_H^*+\delta _H$, ranking first. If $e>0$ is sufficiently small, we will have ${\hat{V}}_{L}(x_L')>{\hat{V}}_H(x_L')$. This will be true because condition ${\hat{V}}_N>V_{H0}$ (which can be rewritten as $\delta _H<\frac{1}{7}-\frac{\delta _L}{4}$) implies $V_{L1}>V_{H1}$ (or $\delta _H<\frac{1}{7}-\delta _L$). We will also have ${\hat{V}}_L(x_L')>{\bar{V}}_L$ if e is sufficiently small, because ${\hat{V}}_{L1}>{\bar{V}}_L$. So, candidate L will improve his vote share without changing his rank.

Case 2 :

$V_{H1}\ge V_{L1}$ and ${\hat{V}}_N>V_{H1}$, or

$$\begin{aligned} \delta _L\le \delta _H-\frac{1}{7} \quad \text{ and } \quad \delta _L>\max \{0,16\delta _H-3\}. \end{aligned}$$

There does not exist$x_L'\in [0,x_H^*-\delta _H+\delta _L)$ that improves the payoff of candidate L. For any $x_L'\in [0,x_H^*-\delta _H+\delta _L)$, candidate N will have the limit best response $x_N'=x_H^*+\delta _H$. Let ${\bar{x}}$ be such that ${\hat{V}}_N({\bar{x}})={\hat{V}}_H({\bar{x}})$. Then, for all $x_L'\in [0,{\bar{x}})$, we will have ${\hat{V}}_H(x_L')\ge {\hat{V}}_N>{\hat{V}}_L(x_L')$, and for any $x_L'\in ({\bar{x}},x_H^*-\delta _H+\delta _L)$, we will have ${\hat{V}}_N>{\hat{V}}_H(x_L')>{\hat{V}}_L(x_L')$. So, the rank of candidate L will decrease for any $x_L'\in [0,x_H^*-\delta _H+\delta _L)$.

Case 3 :

${\hat{V}}_N\ge V_{L1}$, $V_{H1}<V_{L1}$, and ${\hat{V}}_N\le V_{H0}$, or

$$\begin{aligned} \delta _L\le \frac{1-2\delta _H}{13}, \delta _L>\delta _H-\frac{1}{7}, \text{ and } \delta _L\ge \frac{4-28\delta _H}{7}. \end{aligned}$$

There exists$x_L'\in [0,x_H^*-\delta _H+\delta _L)$ that improves the payoff of candidate L. Take $x_L'=x_H^*-\delta _H+\delta _L-e$. If $e>0$ is sufficiently small, then candidate N will have the limit best response $x_N'=x_H^*+\delta _H$, with ${\hat{V}}_N>{\hat{V}}_L(x_L')>{\hat{V}}_H(x_L')$ and ${\hat{V}}_L(x_L')>\max \{{\bar{V}}_L,{\hat{V}}_H(x_L')\}$. So, candidate L will be better off.

Case 4 :

$V_{H1}\ge {\hat{V}}_N$ and ${\hat{V}}_N\ge V_{L1}$, or

$$\begin{aligned} \delta _L\le 16\delta _H-3 \quad \text{ and } \quad \delta _L\le \frac{1-2\delta _H}{13}. \end{aligned}$$

There does not exist$x_L'\in [0,x_H^*-\delta _H+\delta _L)$ that improves the payoff of candidate L. For any such $x_L'$, candidate N will have the limit best response $x_N'=x_H^*+\delta _H$, so ${\hat{V}}_H(x_L')>{\hat{V}}_N>{\hat{V}}_L(x_L')$, and the rank of candidate L will decrease.

Case 5 :

$V_{H1}< {\hat{V}}_N$ and ${\hat{V}}_N< V_{L1}$, or

$$\begin{aligned} \delta _L> 16\delta _H-3 \quad \text{ and } \quad \delta _L>\frac{1-2\delta _H}{13}. \end{aligned}$$

There exists$x_L'\in [0,x_H^*-\delta _H+\delta _L)$ that improves the payoff of candidate L. We have $\delta _L<\frac{1-2\delta _H}{8}$, so $x_H^*<\frac{1}{2}$. Take $x_L'=x_H^*-\delta _H+\delta _L-e$. If $e>0$ is sufficiently small, we have $V_N''(x_L')\approx x_H^*-\delta _H<{\hat{V}}_N$. It follows that candidate N has the limit best response $x_N'=x^*_H+\delta _H$ that yields ${\hat{V}}_L(x_L')>{\hat{V}}_N>{\hat{V}}_H(x_L')$, so candidate L will be better off.

Case 6 :

${\hat{V}}_{N}< V_{L1}$, $V_{H1}\ge V_{L1}$, and ${\hat{V}}_N\ge V_{L0}$, or

$$\begin{aligned} \delta _L>\frac{1-2\delta _H}{13},\quad \delta _L\le \delta _H-\frac{1}{7} \quad \text{ and } \quad \delta _L\le \frac{1-2\delta _H}{4.25}. \end{aligned}$$

There exists$x_L'\in [0,x_H^*-\delta _H+\delta _L)$ that improves the payoff of candidate L if and only if$\delta _L>\frac{1-2\delta _H}{8}$. Indeed, such $x_L'$ must satisfy the following conditions:

• 1. ${\hat{V}}_L(x_L')>{\bar{V}}_L$, or $x_L'\in (\frac{2-4\delta _H-\delta _L}{5},x_H^*-\delta _H+\delta _L)$.

• 2. ${\hat{V}}_L(x_L')\ge {\hat{V}}_N$, or $x_L'\in [\frac{4-8\delta _H-17\delta _L}{5},x_H^*-\delta _H+\delta _L)$. Otherwise, there exists $x_N'>x_H^*+\delta _H$ such that $V_H(x_L',x_H^*,x_N')>V_N(x_L',x_H^*,x_N')>V_L(x_L',x_H^*,x_N')$.

• 3. $V'_L(x_L')\ge V'_N(x_L)$, or $x_L'\in (x',x_H^*-\delta _H+\delta _L)$. Otherwise, there exists $x_N'>x_L'+\delta _L$ such that $V_H(x_L',x_H^*,x_N')>V_N(x_L',x_H^*,x_N')>V_L(x_L',x_H^*,x_N')$.

• 4. $V''_L(x_L')\ge V''_N(x_L)$, or $x_L'\in [0,x'']$. Otherwise, there exists $x_N'<x_L'-\delta _L$ such that $V_H(x_L',x_H^*,x_N')>V_N(x_L',x_H^*,x_N')>V_L(x_L',x_H^*,x_N')$.

We have $x'<x''$. It is true that $x''<\frac{2-4\delta _H-\delta _L}{5}<\frac{4-8\delta _H-17\delta _L}{5}$ if $\delta _L<\frac{1-2\delta _H}{8}$, $x''>\frac{2-4\delta _H-\delta _L}{5}>\frac{4-8\delta _H-17\delta _L}{5}$ if $\delta _L<\frac{1-2\delta _H}{8}$, and $x''=\frac{2-4\delta _H-\delta _L}{5}=\frac{4-8\delta _H-17\delta _L}{5}$ if $\delta _L=\frac{1-2\delta _H}{8}$.

Case 7. :

$V_{H1}< V_{L1}$ and $V_{H1}\ge {\hat{V}}_{N}$, or

$$\begin{aligned} \delta _L>\delta _H-\frac{1}{7} \quad \text{ and } \quad \delta _L\le 16\delta _H-3. \end{aligned}$$

There exists$x_L'\in [0,x_H^*-\delta _H+\delta _L)$ that improves the payoff of candidate L if and only if$\delta _L>\frac{1-2\delta _H}{8}$. The argument here is identical to Case 6.

Case 8. :

${\hat{V}}_{N}< V_{L0}$ and $V_{H1}< V_{L1}$, or

$$\begin{aligned} \delta _L> \frac{1-2\delta _H}{4.25} \quad \text{ and } \quad \delta _L>\delta _H-\frac{1}{7}. \end{aligned}$$

There exists$x_L'\in [0,x_H^*-\delta _H+\delta _L)$ that improves the payoff of candidate L Take $x_L'=x''$. Then, $x_N=O$ and ${\hat{V}}_{L}(x_L')>{\bar{V}}_L$, so candidate L is better off.

Case 9. :

${\hat{V}}_{N}< V_{L0}$ and $V_{H1}\ge V_{L1}$, or

$$\begin{aligned} \delta _L> \frac{1-2\delta _H}{4.25}\quad \text{ and } \quad \delta _L\le \delta _H-\frac{1}{7}. \end{aligned}$$

There exists$x_L'\in [0,x_H^*-\delta _H+\delta _L)$ that improves the payoff of candidate L Take $x_L'=x''$. The argument here is identical to Case 8.

Combining Cases 1–9, we obtain conditions (23). $\square$

Proof of Theorem 7

The proof of this theorem is analogous to the proof of Theorem 2. An additional condition (26) has to be satisfied; this condition is more strict than (13).

It remains to be shown that, for small $\delta _L$, (26) is satisfied if and only if $\delta _H$ is large enough. Take $\delta _L=0$ and denote

$$\begin{aligned} D_2=F(x_H^*+\delta _H)-2F(x_H^*-\delta _H). \end{aligned}$$

(52)

By the Implicit Function Theorem and Assumption 2’, we have

$$\begin{aligned} \frac{\partial D_2}{\partial \delta _H}= & {} f_{H+}\left( \frac{\partial x_H^*}{\partial \delta _H}+1\right) -2f_{H-}\left( \frac{\partial x_H^*}{\partial \delta _H}-1\right) \nonumber \\= & {} \frac{f_{H-}}{|M|}(4f_{H+}f_L-f_{H+}f_m+4f_Lf_m)>0. \end{aligned}$$

(53)

We have $D_2<0$ for $\delta _H$ near 0 and $D_2>0$ for $\delta _H$ near $\frac{1}{2}$. It follows that there exists $\delta _{H0}\in (0,\frac{1}{2})$ such that $D_2<0$ if $\delta _H<\delta _H0$ and $D_2>0$ if $\delta _H>\delta _{H0}$. $\square$

Proof of Theorem 8

Consider the following cases.

Case 1.:

$F(x_H-\delta _H)<1-F(x_H+\delta _H)$.

Case 1A.:

$x_L\in [0,x_H-\delta _H+\delta _L)$. Candidate N has the limit best response $x_N=x_H+\delta _H$. If $x_H-\delta _H+\delta _L>0$, then candidate L has deviation $x_L'\in (x_L,x_H-\delta _H+\delta _L)$. Let $x_L=0$ and $x_H=\delta _H-\delta _L$. We have ${\tilde{V}}_H=F(2\delta _H-\delta _L)$, ${\tilde{V}}_L=0$, and $\frac{\partial {\tilde{V}}_H}{\partial x_H}=f(2\delta _H-\delta _L)-\frac{1}{2}f(0)$. If $\frac{\partial {\tilde{V}}_H}{\partial x_H}>0$, candidate H can increase his payoff with some $x_H'>x_H$. If $\frac{\partial {\tilde{V}}_H}{\partial x_H}\le 0$, then the peak of f must lie to the left of $2\delta _H-\delta _L$, so f is decreasing on $[2\delta _H-\delta _L,1]$. Take $x_L'=1$. Then we have $V_N=F(\frac{x_N+1-\delta _L}{2})-F(\frac{2\delta _H-\delta _L+x_N}{2})$, which is decreasing at $x_N=1+\delta _L$. So, candidate N will enter to the left of $1+\delta _L$, and candidate L will have a positive vote share.

Case 1B.:

$x_L=x_H-\delta _H+\delta _L$ if $x_H-\delta _H+\delta _L>0$. Candidate N has the limit best response $x_N=x_H+\delta _H$. Let $x_L'=x_H-\delta _H+\delta _L-e$, with $e>0$. Then, candidate N will still have the limit best response $x_N=x_H+\delta _H$, and ${\tilde{V}}_L'=x_L'>F(\frac{x_L}{2})=\frac{x_L}{2}={\tilde{V}}_L$ if e is small enough.

Case 1C.:

$x_L\in (x_H-\delta _H+\delta _L,x_H+\delta _H-\delta _L)$. Same as Case 1A. Candidate N has the limit best response $x_N=x_H+\delta _H$. We have ${\tilde{V}}_H=F(x_H+\delta _H)$, which is increasing in $x_H$.

Case 1D.:

$x_L\in (x_H+\delta _H-\delta _L,1)$. Let $V_{N1}=F(x_H-\delta _H)=\lim _{x_N\rightarrow _{-}x_H-\delta _H}V_N$, $V_{N2}=1-F(\frac{x_H+x_L+\delta _H+\delta _L}{2})=\lim _{x_N\rightarrow _{+}x_L+\delta _L}V_N$, and $V_{N3}=\sup _{x_N\in (x_H+\delta _H,x_L+\delta _L)}V_N$.

Consider the following subcases.

Case 1D1:

$V_{N2}>\frac{1}{2}$. Then candidate N will obtain the sole first place. Consider the following subcases.

Case 1D1a:

$V_{N2}>V_{N3}$. Then candidate N has limit best response at $x_{N}=x_L+\delta _L$. If we take $x_H'=x_H+e$ and $e>0$ is sufficiently small, we will have $V_{N2}'>\max \{\frac{1}{2},V_{N3}'\}$, with ${\tilde{V}}_H'>{\tilde{V}}_H$ and candidate H ranking second, as previously.

Case 1D1b:

$V_{N2}=V_{N3}$. Suppose that N has best response $x_N=b(x_H,x_L)\in (x_H+\delta _H,x_L+\delta _L)$. In that case, we must have $f(x_1)=f(x_2)$, where $x_1=\frac{x_H+x_N+\delta _H}{2}$ and $x_2=\frac{x_L+x_N-\delta _L}{2}$. Moreover, by Assumption 3, we should have $f'(x_1)>0$ and $f'(x_2)<0$. By the Implicit Function Theorem, we should have $\frac{\partial x_1}{\partial x_H}>0$. As a result, if we take $x_H'=x_H+e$, we should have ${\tilde{V}}_H'\ge F(\frac{x_H+b(x_H',x_L)+\delta _H}{2})>F(\frac{x_H+b(x_H,x_L)+\delta _H}{2})={\tilde{V}}_H$ if $e>0$ is sufficiently small.

Now let the set $b_0(x_H,x_L)$ contain two limit points $x_N=x_H+\delta _H$ and $x_N=x_L+\delta _L$. If $J=H$, then limit point $x_N=x_H+\delta _H$ is chosen by N, and the vote share ${\tilde{V}}_H$ is increasing in $x_H$. If $J=L$, then limit point $x_N=x_L+\delta _L$ is chosen by N. Take $x_L'=x_L+e$. If e is sufficiently small, we should have $V_{N3}'>V_{N3}=V_{N2}>V_{N2}'$, so candidate N will have limit best response $x_N'=x_H+\delta _H$ or best response $b(x_H,x_L')\in (x_H+\delta _H,x_L'+\delta _L)$. That should increase the vote share of L from zero to some positive value.

Case 1D1c:

$V_{N2}<V_{N3}$. Take $x_H'=x_H+e$. Candidate N has either best response $b(x_H,x_L)\in (x_H+\delta _H,x_L+\delta _L)$ or limit best response $x_N=x_H+\delta _H$. In either case, $x_H'=x_H+e$ will increase ${\tilde{V}}_H$ if e is small (by the argument above).

Case 1D2:

$V_{N2}=\frac{1}{2}$. Candidate N will have limit best response at $x_N=x_L+\delta _L$ if $V_H\ge V_{N3}$ (with $V_H$ corresponding to $V_N=V_{N3}$) and a best response $b(x_H,x_L)\in (x_H+\delta _H,x_L+\delta _L)$ or a limit best response $x_N=x_H+\delta _H$ if $V_H< V_{N3}$ (in the latter case, candidate N will rank first). In any case, by the above argument candidate H can increase his rank/vote share with $x_H'=x_L+\delta _L+e$, if e is small enough.

Case 1D3:

$V_{N2}<\frac{1}{2}$. We first show that candidate N cannot obtain a sole first place in any equilibrium. Suppose that it is not the case, and candidate N has a best response $b(x_H,x_L)\in (x_H+\delta _H,x_L+\delta _L)$ such that it gives him the sole first place. As $V_{N2}<\frac{1}{2}$, take $x_L'=x_L-e$. We have $\frac{\partial x_2}{\partial x_L}>0$ by argument similar Case 1D1b. So, ${\tilde{V}}_L$ will increase and the rank ordering will be preserved if $e>0$ is sufficiently small. If candidate N has a limit best response $x_N=x_H+\delta _H$ or $x_N=x_H-\delta _H$ such that gives him the sole first place, candidate L can also take $x_L'=x_L-e$ and increase his vote share. As a consequence, candidate N cannot rank first in any equilibrium.

Next we show that candidate N cannot share first place with candidate H in any equilibrium. Consider the contrary. As $V_{N2}<\frac{1}{2}$, candidate N cannot have limit best response at $x_N=x_L+\delta _L$. So, one of the following must be the case. First, candidate N can have limit best response at $x_N=x_H-\delta _H$. Take $x_L'=x_L-e$. Then N will rank first with $x_N=x_H-\delta _H$, and candidate L will increase his vote share. Second, candidate N can have limit best response at $x_N=x_H+\delta _H$ or best response at $x_N\in (x_H+\delta _H,x_L+\delta _L)$. We must have $V_{N1}<{\tilde{V}}_{H}=V_{N3}={\tilde{V}}_N$. Take $x_H'=x_H+e$. If e is small, then candidate N will not have a limit best response at $x_N'\ne x_H'-\delta _H$ because if $V_{N1'}<V_{N3'}$, so ${\tilde{V}}_H'>{\tilde{V}}_H$.

We have established that, in any equilibrium, we must have ${\tilde{V}}_H>{\tilde{V}}_N>{\tilde{V}}_H$. We are now ready to consider different cases of $V_{N1}$, $V_{N2}$, and $V_{N3}$.

Case 1D3a:

$V_{N1}>\max \{V_{N2},V_{N3}\}$. Then, N has limit best response at $x_N=x_H-\delta _H$. Take $x_H'=x_H-e$. If $e>0$ is small enough, N will continue to have limit best response at $x_N'=x_H'-\delta _H$, with ${\tilde{V}}_H'>{\tilde{V}}_H$.

Case 1D3b:

$V_{N2}>\max \{V_{N1},V_{N3}\}$. Then, N has limit best response at $x_L=x_L+\delta _L$. Take $x_H'=x_H+e$. If $e>0$ is small enough, N will continue to have the same limit best response, with ${\tilde{V}}_H'>{\tilde{V}}_H$.

Case 1D3c:

$V_{N3}>\max \{V_{N1},V_{N2}\}$. Then, N has limit best response at $x_N=x_H+\delta _H$ or a best response $b(x_H,x_L)\in (x_H+\delta _H,x_L+\delta _L)$. As $\frac{\partial x_2}{\partial x_L}>0$, candidate L will increase his share of vote with $x_L'=x_L-e$ if $e>0$ is small enough.

Case 1D3d:

$V_{N1}=V_{N2}>V_{N3}$. If $J=H$,¹⁷ then N has limit best response at $x_N=x_H-\delta _H$, and candidate H can deviate $x_H'=x_H-e$. If $e>0$ is small enough, candidate N will have limit best response $x_N'=x_L+\delta _L$, increasing H’s vote share. If $J=L$, then N has limit best response at $x_N=x_L+\delta _L$, and candidate L can deviate $x_L'=x_L+e$. If $e>0$ is small enough, candidate N will have limit best response $x_N'=x_H-\delta _H$, increasing L’s vote share.

Case 1D3e:

$V_{N1}=V_{N2}>V_{N3}$. If $J=H$, then N has limit best response at $x_N=x_H-\delta _H$. If $J=L$, then N will have a best response $b(x_H,x_L)\in (x_H+\delta _H,x_L+\delta _L)$ or a limit best response $x_N=x_H+\delta _H$. In any case, candidate N can increase his share of vote with $x_L'=x_L-e$ if $e>0$ is small enough.

Case 1D3f:

$V_{N2}=V_{N3}>V_{N1}$. If $J=H$, then N will have a best response $b(x_H,x_L)\in (x_H+\delta _H,x_L+\delta _L)$ or a limit best response $x_N=x_H+\delta _H$. Then, the vote share of candidate H will increase if $x_H'=x_H+e$ if $e>0$ is small enough. If $J=L$, then N will have limit best response $x_N=x_L+\delta _L$. If $x_L'=x_L+e$, candidate N will have limit best response $x_N'=x_H+\delta _H$ or a best response $b(x_H,x_L')\in (x_H+\delta _H,x_L+\delta _L)$, increasing L’s share of vote from zero.

Case 1D3g:

$V_{N2}=V_{N3}=V_{N1}$. Suppose that candidate N has best response $x_N=b(x_H,x_L)\in (x_H+\delta _H,x_L+\delta _L)$. Then we have $f(\frac{x_H+\delta _H+x_N}{2})=f(\frac{x_N+x_L-\delta _L}{2})$. We have $V_{N2}=1-F(\frac{x_H+x_L+\delta _H+\delta _L}{2})$. As $\frac{x_N+x_L-\delta _L}{2}-\frac{x_H+x_L+\delta _H+\delta _L}{2}>|\delta _L|$ and the distribution f is single-peaked, we have

$$\begin{aligned} {\tilde{V}}_{L}<V_2-|\delta _L|f\left( \frac{x_N+x_L-\delta _L}{2}\right)<V_1-|\delta _L|f(x_H-\delta _H)<{\tilde{V}}_L' \end{aligned}$$

with $x_L'=x_H-\delta _H-e$, if $e>0$ is small enough.¹⁸ Now suppose that N does not have a best response in the interval $(x_H+\delta _H,x_L+\delta _L)$. If $J=L$, then N will have a limit best response $x_N=x_L+\delta _L$. If $x_L'=x_L+e$, with $e>0$, candidate N will have limit best response $x_N'=x_H+\delta _H$ or a best response $b(x_H,x_L')\in (x_H+\delta _H,x_L+\delta _L)$, increasing L’s share of vote from zero. If $J=H$, we cannot easily determine a payoff-improving deviation for either candidate H or candidate L.

Case 1E:

$x_L=1$. In this case, only $V_{N1}$ and $V_{N3}$ are well-defined. Consider the following cases:

Case 1E1:

$V_{N1}>V_{N3}$. In that case, candidate N has limit best response at $x_N=x_H-\delta _H$, and candidate H can increase his vote share with $x_H=x_H-e$, if $e>0$ is small enough.

Case 1E2:

$V_{N1}=V_{N3}$. If $J=H$, then candidate N has limit best response at $x_N=x_H-\delta _H$, and candidate L can increase his vote share with $x_L'=1-e$, if $e>0$ is small enough. Such a deviation will also increase L’s vote share if $J=L$ and $\lim _{x_N\rightarrow 1-\delta _L}V_N<V_{N3}$. If the latter is satisfied as equality, then candidate N should have limit best response at $x_N=1+\delta _L$, with ${\tilde{V}}_L=0$, and also we should have $f(x_2)\le f(1)$. As f is assumed to be single-peaked, it must be maximized within the interval $[x_2,1]$. Thus we should have $1-\frac{x_H+1+\delta _H+\delta _L}{2}>x_H-\delta _H$, or $x_H-\delta _H+\delta _L>1-2x_H>0$. It follows that candidate L can choose position $x_L'=x_H-\delta _H+\delta _L-e$ such that ${\tilde{V}}_L'<0$ if $e>0$ is small enough.

Case 1E3:

$V_{N1}<V_{N3}$. In that case, candidate N has limit best response at $x_N=x_H+\delta _H$, a limit best response $x_N=1+\delta _L$, or a best response $b(x_H,x_L)\in (x_H+\delta _H,x_L+\delta _L)$. In all cases, candidate H can increase his share of vote with $x_H'=x_H+e$.

Case 2:

$F(x_H-\delta _H)=1-F(x_H+\delta _H)$.

Case 2A:

$x_L\in [0,x_H-\delta _H+\delta _L)$. Candidate N has the limit best response $x_N=x_H+\delta _H$. We have ${\tilde{V}}_L=F(\frac{x_H-\delta _H+x_L+\delta _L}{2})$, which is increasing in $x_L$; then moving (slightly) to the right, L increases his vote share and does not decrease the rank.

Case 2B:

$x_L\in [x_H-\delta _H+\delta _L,x_H+\delta _H-\delta _L]$. Candidate N has two limit best responses: $x_N=x_H+\delta _H$ and $x_N=x_H-\delta _H$. We have ${\tilde{V}}_L=0$ if $x_L\in (x_H-\delta _H+\delta _L,x_H+\delta _H-\delta _L)$. If $x_L=x_H-\delta _H+\delta _L$, we have (assuming that $x_H>\delta _H-\delta _L$) ${\tilde{V}}_L=\frac{1}{2}F(x_H-\delta _H+\delta _L)$. If we take $x_L'=x_L-e$, we must get ${\tilde{V}}_L'=F(x_H-\delta _H+\delta _L-e)$, which will be greater than ${\tilde{V}}_L$ if $e>0$ is small enough.

Case 2C:

$x_L\in (x_H+\delta _H-\delta _L,1]$. This is identical to Case 2A. $\square$

Proof of Theorem 9

From Theorem 8 it follows that equilibrium in case $J=H$ is only possible if $V_{N1}=V_{N2}=V_{N3}$. If $V_{N1}=V_{N2}$ and the distribution of voter ideal points is symmetric, then we should have $x_H-\delta _H=1-\frac{x_H+x_L+\delta _H+\delta _L}{2}$, or $x_L=2-3x_H+\delta _H-\delta _L$.

Let $x_H\in [\frac{1+\delta _H-\delta _L}{3},\frac{1}{2}]$ and $x_L=2-3x_H+\delta _H-\delta _L$. Then both $V_{N1}=V_{N2}$ increase with $x_H$, while $V_{N3}=F(\frac{x_H+x_L+\delta _H-\delta _L}{2})-F(x_H+\delta _H)$ decreases with $x_H$. At the same time, for $x_H= \frac{1+\delta _H-\delta _L}{3}$ we have $x_H-\delta _H=\frac{x_H+x_L+\delta _H-\delta _L}{2}-x_H-\delta _H$, so $V_{N1}<V_{N3}$, as the density is single-peaked. Finally, for $x=\frac{1}{2}$ we have $V_{N3}=F(\frac{1}{2}+\delta _H-\delta _L)-F(\frac{1}{2}+\delta _H)<F(\frac{1}{2}-\delta _H)=V_{N1}$. It follows that there exists a unique $x_H$ such that $V_{N1}=V_{N2}=V_{N3}$ with $x_L>x_H$.

In that case, candidate N will have limit best response $x_N=x_H-\delta _H$. It is then straightforward to show that candidate H has no payoff-improving deviations. We have ${\tilde{V}}_H=F(\frac{x_H+\delta _H+x_L-\delta _L}{2})-F(x_H-\delta _H)$. As f is symmetric and $x_H-\delta _H=1-\frac{x_H+\delta _H+x_L+\delta _L}{2}$, ${\tilde{V}}_H'$ decreases with $x_H'$, because for $x_H'>x_H$ candidate N will have limit best response $x_N'=x_H'-\delta _H$. Any $x_H'<x_H$ will result in candidate N entering to the right of H, reducing his share of vote. Similarly for L, we have $x_H-\delta _H+\delta _L=1-\frac{x_H+\delta _H+x_L-\delta _L}{2}$, so ${\tilde{V}}_L'<{\tilde{V}}_L$ for any $x_L'\in [0,x_H-\delta _H+\delta _L)$. Any deviation $x_L'\in (x_H+\delta _H-\delta _L,x_L)$ will result in candidate N entering with limit best response $x_N'=x_L+\delta _L$ and ${\tilde{V}}_L'=0$. $\square$