Intensity valence

Abstract

We study a continuous one-dimensional spatial model of electoral competition with two office-motivated candidates differentiated by their “intensity valence”, the degree to which they will implement their announced policy. The model generates results that differ significantly from those obtained in models with additive valence. First, the low intensity valence candidate is supported by voters with ideal points on both extremes of the policy space. Second, there exist pure strategy Nash equilibria (PSNE) in which the high intensity valence candidate wins if the distribution of voters in the policy space is sufficiently homogeneous. If, instead, this distribution is sufficiently heterogeneous, there are PSNE in which the low intensity valence candidate wins. For moderate heterogeneity, only mixed strategy equilibria exist.

A Proofs

1.1 A.1 Proof of Proposition 1

We prove first that $\Omega _1(x_1,x_2)=(b,c)$, i.e., the set of voters who strictly prefer Candidate 1, is a bounded open interval, and find the bounds b and c. Let $u_{1}=u(a,x_{1},\lambda _{1},K)=\lambda _{1}(K-|x_{1}-a|)$ and $u_{2}=u(a,x_{2},\lambda _{2},K)=K-|x_{2}-a|$. We thus have:

$$\begin{aligned} a\in \Omega _{1}(x_{1},x_{2})\Leftrightarrow & {} u_1>u_2\nonumber \\\Leftrightarrow & {} \lambda _1(K-|x_1-a|)>K-|x_2-a|\nonumber \\\Leftrightarrow & {} (\lambda _1-1)K+|x_2-a|>\lambda _1|x_1-a| \end{aligned}$$

(A1)

Note also the three preliminary results in (A2), (A3) and (A4):

$$\begin{aligned} \text{ If } a=x_2, \text{ then } u_1>u_2\Leftrightarrow & {} \frac{(\lambda _1-1)K}{\lambda _1}>|x_1-x_2| \end{aligned}$$

(A2)

$$\begin{aligned} \text{ If } a\le \min (x_{1},x_{2}), \text{ then } u_1>u_2\Leftrightarrow & {} (\lambda _1-1)K+(x_2-a)>\lambda _1(x_1-a)\nonumber \\\Leftrightarrow & {} a>\frac{\lambda _1x_1-x_2}{\lambda _1-1}-K \end{aligned}$$

(A3)

$$\begin{aligned} \text{ If } a\ge \max (x_{1},x_{2}), \text{ then } u_1>u_2\Leftrightarrow & {} (\lambda _1-1)K+(a-x_2)>\lambda _1(a-x_1)\nonumber \\\Leftrightarrow & {} a<\frac{\lambda _1x_1-x_2}{\lambda _1-1}+K \end{aligned}$$

(A4)

Consider the three following cases: Case (A) $\left| x_{1}-x_{2}\right| \le \frac{(\lambda _1-1)K}{\lambda _1}$, Case (B) $x_2>x_1+\frac{(\lambda _1-1)K}{\lambda _1}$, and Case (B’) $x_2<x_1-\frac{(\lambda _1-1)K}{\lambda _1}$.

Case (A) :

– If $\left| x_{1}-x_{2}\right| \le \frac{(\lambda _1-1)K}{\lambda _1}$ and $a\in (\min \{x_1,x_2\},\max \{x_1,x_2\})$, then $\lambda _1|x_{1}-a|< \lambda _1|x_{1}-x_{2}| \le (\lambda _1-1)K< (\lambda _1-1)K+|x_2-a|$. Inequality (A1) is thus satisfied, so $u_1>u_2$.

– If $\left| x_{1}-x_{2}\right| \le \frac{(\lambda _1-1)K}{\lambda _1}$ and $a\le \min \{x_{1},x_{2}\}$, then $u_1>u_2\Leftrightarrow a>\frac{\lambda _1x_1-x_2}{\lambda _1-1}-K$ according to Inequality (A3).

– If $\left| x_{1}-x_{2}\right| \le \frac{(\lambda _1-1)K}{\lambda _1}$ and $a\ge \max \{x_{1},x_{2}\}$, then $u_1>u_2\Leftrightarrow a<\frac{\lambda _1x_1-x_2}{\lambda _1-1}+K$ according to Inequality (A4).

Conclusion :

 If $\left| x_{1}-x_{2}\right| \le \frac{(\lambda _1-1)K}{\lambda _1}$, then $\Omega _{1}(x_{1},x_{2})=\left( \frac{\lambda _1x_1-x_2}{\lambda _1-1} -K,\frac{\lambda _1x_1-x_2}{\lambda _1-1}+K\right)$.

Case (B) :

– If $x_2>x_1+\frac{(\lambda _1-1)K}{\lambda _1}$ and $x_1<a<x_2$, then, according to Inequality (A1), $u_1>u_2\Leftrightarrow (\lambda _1-1)K+(x_2-a)>\lambda _1(a-x_1)$, i.e., $u_1>u_2\Leftrightarrow a<\frac{(\lambda _1-1)K+\lambda _1x_1+x_2}{\lambda _1+1}$.

– If $x_2>x_1+\frac{(\lambda _1-1)K}{\lambda _1}$ and $a\le x_1$, then $u_1>u_2\Leftrightarrow a>\frac{\lambda _1x_1-x_2}{\lambda _1-1}-K$ according to Inequality (A3).

– If $x_2>x_1+\frac{(\lambda _1-1)K}{\lambda _1}$ and $a\ge x_2$, then $u_1>u_2\Leftrightarrow a<\frac{\lambda _1x_1-x_2}{\lambda _1-1}+K$ according to Inequality (A4). However, since $x_2>x_1 +\frac{(\lambda _1-1)K}{\lambda _1}$ and $a\ge x_2$, note that $\frac{\lambda _1x_1-x_2}{\lambda _1-1}+K=\frac{\lambda _1 \left[ x_1+\frac{(\lambda _1-1)K}{\lambda _1}\right] -x_2}{\lambda _1-1} <\frac{(\lambda _1-1)x_2}{\lambda _1-1}=x_2\le a$. Hence, $u_1\le u_2$ if $a\ge x_2$ and $x_2>x_1+\frac{(\lambda _1-1)K}{\lambda _1}$.

Conclusion :

 If $x_2>x_1+\frac{(\lambda _1-1)K}{\lambda _1}$, then $\Omega _{1}(x_{1},x_{2})=\left( \frac{\lambda _1x_1-x_2}{\lambda _1-1} -K,\frac{(\lambda _1-1)K+\lambda _1x_1+x_2}{\lambda _1+1}\right)$.

Case (B’) :

– If $x_2<x_1-\frac{(\lambda _1-1)K}{\lambda _1}$ and $x_2<a<x_1$, then, according to Inequality (A1), $u_1>u_2\Leftrightarrow (\lambda _1-1)K+(a-x_2)>\lambda _1(x_1-a)$, i.e., $u_1>u_2\Leftrightarrow a>\frac{(1-\lambda _1)K+\lambda _1x_1+x_2}{\lambda _1+1}$.

– If $x_2<x_1-\frac{(\lambda _1-1)K}{\lambda _1}$ and $a\le x_2$, then $u_1>u_2\Leftrightarrow a>\frac{\lambda _1x_1-x_2}{\lambda _1-1}-K$ according to Inequality (A3). However, since $x_2<x_1-\frac{(\lambda _1-1)K}{\lambda _1}$ and $a\le x_2$, note that $\frac{\lambda _1x_1-x_2}{\lambda _1-1}-K=\frac{\lambda _1 \left[ x_1-\frac{(\lambda _1-1)K}{\lambda _1}\right] -x_2}{\lambda _1-1} >\frac{(\lambda _1-1)x_2}{\lambda _1-1}=x_2\ge a$. Hence $u_1\le u_2$ if $a\le x_2$ and $x_2<x_1-\frac{(\lambda _1-1)K}{\lambda _1}$.

– If $x_2<x_1-\frac{(\lambda _1-1)K}{\lambda _1}$ and $a\ge x_1$, then $u_1>u_2\Leftrightarrow a<\frac{\lambda _1x_1-x_2}{\lambda _1-1} +K$ according to Inequality (A4).

Conclusion :

 If $x_2<x_1 -\frac{(\lambda _1-1)K}{\lambda _1}$, then $\Omega _{1}(x_{1},x_{2})=\left( \frac{(1-\lambda _1)K +\lambda _1x_1+x_2}{\lambda _1+1},\frac{\lambda _1x_1-x_2}{\lambda _1-1}+K\right)$.

Now we prove that $\Omega _2(x_1,x_2)=(-\infty ,b)\cup (c,+\infty )$, i.e., the set of voters who strictly prefer Candidate 2, is a non-convex set. In the three cases (A), (B) and (B’), it is easily shown (replacing $u_1>u_2$ by $u_1=u_2$) that $I(x_1,x_2)=\{b,c\}$. Given that $\Omega _2(x_1,x_2)=\mathbb {R}\backslash \left[ \Omega _1(x_1,x_2)\cup I(x_1,x_2)\right]$, we obtain that $\Omega _2(x_1,x_2)=(-\infty ,b)\cup (c,+\infty )$. $\square$

1.2 A.2 Proof of Lemma 1

We first show that (i) $\int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1+1)K}{\lambda _1}}f_\sigma (a)da=\int _{-\frac{(\lambda _1+1)K}{\lambda _1}}^{\frac{(\lambda _1-1)K}{\lambda _1}}f_\sigma (a)da$, and then that (ii) $0<\int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1+1)K}{\lambda _1}}f_\sigma (a)da<\int _{-K}^Kf_\sigma (a)da$, i.e., that $\int _{-K}^{K}f_\sigma (a)da-\int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1+1)K}{\lambda _1}}f_\sigma (a)da>0$.

• (i) Making the substitution $u=-a$, $du=-da$, we get:

  $$\begin{aligned} \int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1+1)K}{\lambda _1}}f_\sigma (a)da=-\int _{\frac{(\lambda _{1}-1)K}{\lambda _{1}}}^{\frac{ -(\lambda _{1}+1)K}{\lambda _{1}}}f_{\sigma }(-u)du= \int _{\frac{-(\lambda _{1}+1)K}{\lambda _{1}}}^{\frac{(\lambda _{1}-1)K}{\lambda _{1}}}f_{\sigma }(-u)du. \end{aligned}$$

  Given that $f_{\sigma }$ is an even function, we have $\int _{\frac{-(\lambda _{1}+1)K}{\lambda _{1}}}^{\frac{(\lambda _{1}-1)K}{\lambda _{1}}}f_{\sigma }(-u)du=\int _{\frac{-(\lambda _{1}+1)K }{\lambda _{1}}}^{\frac{(\lambda _{1}-1)K}{\lambda _{1}}}f_{\sigma }(u)du$.

• (ii) Remark that

  $$\begin{aligned} \int _{-K}^{K}f_\sigma (a)da =\int _{-K}^{\frac{(1-\lambda _1)K}{\lambda _1}} f_\sigma (a)da+\int _{\frac{(1-\lambda _1)K}{\lambda _1}}^Kf_\sigma (a)da, \end{aligned}$$

  (A5)

  and

  $$\begin{aligned} \int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1+1)K}{\lambda _1}} f_\sigma (a)da=\int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{K}f_\sigma (a)da +\int _{K}^{\frac{(\lambda _1+1)K}{\lambda _1}}f_\sigma (a)da. \end{aligned}$$

  (A6)

  Subtracting Eqs. (A5) and (A6), we get

  $$\begin{aligned} \int _{-K}^{K}f_\sigma (a)da-\int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1+1) K}{\lambda _1}}f_\sigma (a)da=\int _{-K}^{\frac{(1-\lambda _1)K}{\lambda _1}}f_\sigma (a)da -\int _{K}^{\frac{(\lambda _1+1)K}{\lambda _1}}f_\sigma (a)da, \end{aligned}$$

  (A7)

  $f_\sigma$ is an even function, so $\int _{-K}^{\frac{(1-\lambda _1)K}{\lambda _1}} f_\sigma (a)da=\int _{\frac{(\lambda _1-1)K}{\lambda _1}}^{K}f_\sigma (a)da$. Eq. (A7) is thus equivalent to:

  $$\begin{aligned} \int _{-K}^{K}f_\sigma (a)da-\int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1+1) K}{\lambda _1}}f_\sigma (a)da= & {} \int _{\frac{(\lambda _1-1)K}{\lambda _1}}^{K}f_\sigma (a)da -\int _{K}^{\frac{(\lambda _1+1)K}{\lambda _1}}f_\sigma (a)da\nonumber \\= & {} \int _{\frac{(\lambda _1-1)K}{\lambda _1}}^{K}\left[ f_\sigma (a)-f_\sigma \left( a+\frac{K}{\lambda _1}\right) \right] da. \end{aligned}$$

  (A8)

  The limits of integration in the right hand side of (A8) are positive, and $f_\sigma (a)$ is strictly decreasing on $\mathbb {R}_+$, so $\left[ f_\sigma (a)-f_\sigma \left( a+\frac{K}{\lambda _1}\right) \right] >0$, and (A8) is strictly positive. The result in Lemma 1 follows. $\square$

1.3 A.3 Proof of Lemma 2

Let $g_1(\sigma )=\int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1+1)K}{\lambda _1}}f_\sigma (a)da$ and $g_2(\sigma )=\int _{-K}^{K}f_\sigma (a)da$. According to Lemma 1, we know that $0<g_1(\sigma )<g_2(\sigma )$, $\forall \sigma >0$.

Recall the definition of $f_\sigma$ and note that $g_1(\sigma )=\int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1+1)K}{\lambda _1}}\frac{1}{\sigma }f_1(\frac{a}{\sigma })da$. Let $z=\frac{a}{\sigma }$. Then $g_1(\sigma )=\int _{\frac{(1-\lambda _1)K}{\lambda _1\sigma }}^{\frac{(\lambda _1+1)K}{\lambda _1\sigma }}f_1(z)dz$: one can see that when $\sigma$ increases, $g_1(\sigma )$ is an integral of $f_1$ on a smaller interval. Thus, $\sigma \mapsto g_1(\sigma )$ is a continuous and strictly decreasing function on $\mathbb {R}_+^*$, with

$$\begin{aligned} \lim _{\sigma \rightarrow 0^+}g_1(\sigma )=\int _{-\infty }^{+\infty }f_1(z)dz=1 \text{ and } \lim _{\sigma \rightarrow +\infty }g_1(\sigma )=0. \end{aligned}$$

Hence, there is a unique $\sigma ^*$ such that $g_1(\sigma ^*)=\frac{1}{2}$; moreover, $g_1(\sigma )>\frac{1}{2}\Leftrightarrow \sigma <\sigma ^*$.

Now consider $g_2(\sigma )=\int _{-K}^{K}\frac{1}{\sigma }f_1(\frac{a}{\sigma })da$. Let $z=\frac{a}{\sigma }$. Then $g_2(\sigma )=\int _{-\frac{K}{\sigma }}^{\frac{K}{\sigma }}f_1(z)dz$. When $\sigma$ increases, $g_2(\sigma )$ is an integral of $f_1$ on a smaller interval. Thus, $\sigma \mapsto g_2(\sigma )$ is a continuous and strictly decreasing function on $\mathbb {R}_+^*$, with

$$\begin{aligned} \lim _{\sigma \rightarrow 0^+}g_2(\sigma )=\int _{-\infty }^{+\infty }f_1(z)dz=1 \text{ and } \lim _{\sigma \rightarrow +\infty }g_2(\sigma )=0. \end{aligned}$$

Hence, there is a unique $\sigma ^{**}$ such that $g_2(\sigma ^{**})=\frac{1}{2}$. Moreover, $g_2(\sigma )>\frac{1}{2}\Leftrightarrow \sigma <\sigma ^{**}$.

According to Lemma 1, $g_1(\sigma )<g_2(\sigma )$, $\forall \sigma >0$, which implies $g_1(\sigma ^*)=\frac{1}{2}<g_2(\sigma ^*)$. Given that $g_2(\sigma ^{**})=\frac{1}{2}$, then $g_2(\sigma ^{**})<g_2(\sigma ^*)$. Thus, $\sigma ^{**}>\sigma ^{*}$ since $g_2$ is a strictly decreasing function. Results (i), (ii) and (iii) in Lemma 2 follow. $\square$

1.4 A.4 Proof of Proposition 2

We prove (i) first, i.e., if $\sigma <\sigma ^{*}$, Candidate 1 wins with certainty if he chooses $x_{1}^{*}=0$, $\forall x_{2}\in \mathbb {R}$. We thus have to show that if $\sigma <\sigma ^{*}$, then $S_1(0,x_{2})=\int _b^cf_\sigma (a)da >\frac{1}{2}$, $\forall x_2\in \mathbb {R}$. Since $\sigma <\sigma ^{*}$, we know from Lemma 2 that $\frac{1}{2}<\int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1+1)K}{\lambda _1}}f_\sigma (a)da<\int _{-K}^Kf_\sigma (a)da$. Let’s consider the three possible cases: Case (A) $x_2\in \left[ -\frac{(\lambda _1-1)K}{\lambda _1}, \frac{(\lambda _1-1)K}{\lambda _1}\right]$, Case (B) $x_2>\frac{(\lambda _1-1)K}{\lambda _1}$, and Case (B’) $x_2<-\frac{(\lambda _1-1)K}{\lambda _1}$ which correspond to Panels (A), (B) and (B’) in Fig. 2 when $x_1=0$.

Case (A) :

  If $x_2\in \left[ -\frac{(\lambda _1-1)K}{\lambda _1},\frac{(\lambda _1-1)K}{\lambda _1}\right]$ and $x_1=x_{1}^{*}=0$, then we know from Proposition 1 that $b=\frac{-x_2}{\lambda _1-1}-K$ and $c=\frac{-x_2}{\lambda _1-1} +K$. If so, $S_{1}(0,x_{2})=\int _{\frac{-x_2}{\lambda _1-1} -K}^{\frac{-x_2}{\lambda _1-1}+K}f_{\sigma }(a)da$ and $\frac{\partial S_1}{\partial x_2}(0,x_2)=\frac{-1}{\lambda _1 -1}f_\sigma \left( \frac{-x_2}{\lambda _1-1}+K\right) +\frac{1}{\lambda _1 -1}f_\sigma \left( \frac{-x_2}{\lambda _1-1}-K\right)$. Given that $f_{\sigma }$ is an even function, $f_\sigma \left( \frac{-x_2}{\lambda _1 -1}-K\right) =f_\sigma \left( \frac{x_2}{\lambda _1-1}+K\right)$, so $\frac{\partial S_1}{\partial x_2}=\frac{-1}{\lambda _1-1} f_\sigma \left( \frac{-x_2}{\lambda _1-1}+K\right) +\frac{1}{\lambda _1 -1}f_\sigma \left( \frac{x_2}{\lambda _1-1}+K\right)$. Furthermore, given that this even function is strictly increasing on $\mathbb {R}_{-}$, and strictly decreasing on $\mathbb {R}_{+}$, then $f_\sigma \left( \frac{x_2}{\lambda _1-1}+K\right) \ge f_\sigma \left( \frac{-x_2}{\lambda _1-1}+K\right)$ if $x_2\le 0$, while $f_\sigma \left( \frac{x_2}{\lambda _1-1}+K\right) \le f_\sigma \left( \frac{-x_2}{\lambda _1-1}+K\right)$ if $x_2\ge 0$. That is $\frac{\partial S_1}{\partial x_2}\ge 0\Leftrightarrow x_2\le 0$. Thus, $x_{2}\mapsto S_{1}(0,x_{2})$ is increasing on $\left[ -\frac{(\lambda _1-1)K}{\lambda _1},0\right]$, and decreasing on $\left[ 0,\frac{(\lambda _1-1)K}{\lambda _1}\right]$, and it has a minimum at $x_{2}=\pm \frac{(\lambda _1-1)K}{\lambda _1}$. Note that $S_{1}\left( 0, \frac{(\lambda _1-1)K}{\lambda _1}\right) =\int _{-\frac{(\lambda _1 +1)K}{\lambda _1}}^{\frac{(\lambda _1-1)K}{\lambda _1}}f_\sigma (a)da$ and $S_{1}\left( 0, -\frac{(\lambda _1-1)K}{\lambda _1}\right) =\int _{\frac{(1 -\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1+1)K}{\lambda _1}}f_\sigma (a)da$, and according to Lemma 1, we have $S_{1}\left( 0, \frac{(\lambda _1-1)K}{\lambda _1}\right) =S_{1}\left( 0, -\frac{(\lambda _1-1)K}{\lambda _1}\right)$. Now since $\sigma <\sigma ^*$ and given Lemma 2, we have $S_{1}\left( 0, -\frac{(\lambda _1-1)K}{\lambda _1}\right) =\int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1 +1)K}{\lambda _1}}f_\sigma (a)da>\frac{1}{2}$; we can thus conclude that $S_{1}(0,x_{2})\ge S_{1}\left( 0,-\frac{(\lambda _1-1)K}{\lambda _1}\right) >\frac{1}{2}$, $\forall x_{2}\in \left[ -\frac{(\lambda _1-1)K}{\lambda _1},\frac{(\lambda _1 -1)K}{\lambda _1}\right]$.

Case (B) :

If $x_2>\frac{(\lambda _1-1)K}{\lambda _1}$ and $x_1=x_{1}^{*}=0$, then we know from Proposition 1 that $b=\frac{-x_2}{\lambda _1-1}-K$ and $c=\frac{(\lambda _1-1)K+x_2}{\lambda _1+1}$. We thus have $S_{1}(0,x_{2})=\int _{\frac{-x_2}{\lambda _1-1}-K}^{\frac{(\lambda _1 -1)K+x_2}{\lambda _1+1}}f_{\sigma }(a)da$ and $\frac{\partial S_1}{\partial x_2}(0,x_2)=\frac{1}{\lambda _1+1}f_\sigma \left( \frac{(\lambda _1-1)K +x_2}{\lambda _1+1}\right) +\frac{1}{\lambda _1-1}f_\sigma \left( \frac{-x_2}{\lambda _1 -1}-K\right) >0$. Then, $x_{2}\mapsto S_{1}(0,x_{2})$ is strictly increasing on $\left[ \frac{(\lambda _1-1)K}{\lambda _1},+\infty \right)$, and has a minimum at $x_{2}=\frac{(\lambda _1-1)K}{\lambda _1}$. Consequently, $S_{1}(0,x_{2})> S_{1}\left( 0,\frac{(\lambda _1-1)K}{\lambda _1}\right) >\frac{1}{2}$, $\forall x_{2}> \frac{(\lambda _1-1)K}{\lambda _1}$.

Case (B’) :

If $x_{2}< -\frac{(\lambda _1-1)K}{\lambda _1}$ and $x_1=x_{1}^{*}=0$, we know from Proposition 1 that $b=\frac{(1-\lambda _1)K+x_2}{\lambda _1+1}$ and $c=\frac{-x_2}{\lambda _1-1}+K$. We thus have $S_{1}(0,x_{2})=\int _{\frac{(1-\lambda _1)K+x_2}{\lambda _1 +1}}^{\frac{-x_2}{\lambda _1-1}+K}f_{\sigma }(a)da$ and $\frac{\partial S_1}{\partial x_2}(0,x_2) =\frac{-1}{\lambda _1-1}f_\sigma \left( \frac{-x_2}{\lambda _1-1}+K\right) -\frac{1}{\lambda _1+1}f_\sigma \left( \frac{(1-\lambda _1)K +x_2}{\lambda _1+1}\right) <0$. Then, $x_{2}\mapsto S_{1}(0,x_{2})$ is strictly decreasing on $\left( -\infty ,-\frac{(\lambda _1-1)K}{\lambda _1}\right]$ and has a minimum at $x_{2}=-\frac{(\lambda _1-1)K}{\lambda _1}$. Consequently, $S_{1}(0,x_{2})> S_{1}\left( 0,-\frac{(\lambda _1-1)K}{\lambda _1}\right) >\frac{1}{2}$, $\forall x_{2}< -\frac{(\lambda _1-1)K}{\lambda _1}$.

Conclusion :

 we have shown that if $\sigma <\sigma ^{*}$, then $S_{1}(0,x_{2})>\frac{1}{2}$, $\forall x_{2}\in \mathbb {R}$.

Now we prove (ii), i.e., if $\sigma >\sigma ^{**}$, Candidate 2 wins with certainty if he chooses $x_{2}^{*}=0$, $\forall x_{1}\in \mathbb {R}$. We thus have to show that if $\sigma >\sigma ^{**}$, then $S_1(x_1,0)=\int _b^cf_\sigma (a)da<\frac{1}{2}$, $\forall x_1\in \mathbb {R}$. Since $\sigma >\sigma ^{**}$, we know from Lemma 2 that $\int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1 +1)K}{\lambda _1}}f_\sigma (a)da<\int _{-K}^Kf_\sigma (a)da<\frac{1}{2}$. Let’s consider the three possible cases: Case (A) $x_1\in \left[ -\frac{(\lambda _1-1)K}{\lambda _1},\frac{(\lambda _1-1)K}{\lambda _1}\right]$, Case (B) $x_1<-\frac{(\lambda _1-1)K}{\lambda _1}$, and Case (B’) $x_1>\frac{(\lambda _1-1)K}{\lambda _1}$ which correspond to Panels (A), (B) and (B’) in Fig. 2 when $x_2=0$.

Case (A) :

If $x_1\in \left[ -\frac{(\lambda _1-1)K}{\lambda _1},\frac{(\lambda _1-1)K}{\lambda _1}\right]$ and $x_2=x_{2}^{*}=0$, then we know from Proposition 1 that $b=\frac{\lambda _1x_1}{\lambda _1-1}-K$ and $c=\frac{\lambda _1x_1}{\lambda _1-1}+K$. If so, $S_{1}(x_1,0)=\int _{\frac{\lambda _1x_1}{\lambda _1-1} -K}^{\frac{\lambda _1x_1}{\lambda _1-1}+K}f_{\sigma }(a)da$ and $\frac{\partial S_1}{\partial x_1}(x_1,0)=\frac{\lambda _1}{\lambda _1-1}f_\sigma \left( \frac{\lambda _1x_1}{\lambda _1-1} +K\right) -\frac{\lambda _1}{\lambda _1-1}f_\sigma \left( \frac{\lambda _1x_1}{\lambda _1-1}-K\right)$. Given that $f_{\sigma }$ is an even function, strictly increasing on $\mathbb {R}_{-}$, and strictly decreasing on $\mathbb {R}_{+}$, $f_\sigma \left( \frac{\lambda _1x_1}{\lambda _1-1}+K\right) \ge f_\sigma \left( \frac{\lambda _1x_1}{\lambda _1-1}-K\right)$ if $x_1\le 0$, while $f_\sigma \left( \frac{\lambda _1x_1}{\lambda _1-1}+K\right) \le f_\sigma \left( \frac{\lambda _1x_1}{\lambda _1-1}-K\right)$ if $x_1\ge 0$. That is $\frac{\partial S_1}{\partial x_1}\ge 0\Leftrightarrow x_1\le 0$. Thus, $x_{1}\mapsto S_{1}(x_1,0)$ is strictly increasing on $\left[ -\frac{(\lambda _1-1)K}{\lambda _1},0\right]$, and strictly decreasing on $\left[ 0,\frac{(\lambda _1-1)K}{\lambda _1}\right]$, and it has a maximum at $x_{1}=0$. $S_1(0,0)=\int _{-K}^Kf_\sigma (a)da$, and since $\sigma >\sigma ^{**}$, we know from Lemma 2 that $S_1(0,0)=\int _{-K}^Kf_\sigma (a)da<\frac{1}{2}$. We can thus conclude that $S_{1}(x_{1},0)\le S_{1}(0,0)<\frac{1}{2}$, $\forall x_{1}\in \left[ -\frac{(\lambda _1-1)K}{\lambda _1},\frac{(\lambda _1-1)K}{\lambda _1}\right]$.

Case (B) :

If $x_1<-\frac{(\lambda _1-1)K}{\lambda _1}$ and $x_2=x_{2}^{*}=0$, then we know from Proposition 1 that $b=\frac{\lambda _1x_1}{\lambda _1-1}-K$ and $c=\frac{(\lambda _1-1)K+\lambda _1x_1}{\lambda _1+1}$. If so, $S_{1}(x_{1},0)=\int _{\frac{\lambda _1x_1}{\lambda _1-1}-K}^{\frac{(\lambda _1-1) K+\lambda _1x_1}{\lambda _1+1}}f_{\sigma }(a)da$. Now remark that if $x_1<-\frac{(\lambda _1-1)K}{\lambda _1}$, then $\frac{(\lambda _1-1)K+\lambda _1x_1}{\lambda _1+1}<0$, so $S_{1}(x_{1},0)=\int _{\frac{\lambda _{1}x_{1}}{\lambda _{1}-1}-K}^{\frac{ (\lambda _{1}-1)K+\lambda _{1}x_{1}}{\lambda _{1}+1}}f_{\sigma }(a)da<\int _{ \frac{\lambda _{1}x_{1}}{\lambda _{1}-1}-K}^{0}f_{\sigma }(a)da<\int _{-\infty }^{0}f_{\sigma }(a)da=\frac{1}{2}$, i.e., $S_{1}(x_{1},0)<\frac{1}{2}$, $\forall x_{1}<-\frac{(\lambda _{1}-1)K}{ \lambda _{1}}$.

Case (B’) :

If $x_1>\frac{(\lambda _1-1)K}{\lambda _1}$ and $x_2=x_{2}^{*}=0$, then we know from Proposition 1 that $b=\frac{(1-\lambda _1)K+\lambda _1x_1}{\lambda _1+1}$ and $c=\frac{\lambda _1x_1}{\lambda _1-1}+K$. If so, $S_{1}(x_{1},0)=\int _{\frac{(1-\lambda _1)K+\lambda _1x_1}{\lambda _1+1}}^{\frac{\lambda _1x_1}{\lambda _1-1}+K}f_{\sigma }(a)da$. Now remark that if $x_1>\frac{(\lambda _1-1)K}{\lambda _1}$, then $\frac{(1-\lambda _1)K+\lambda _1x_1}{\lambda _1+1}>0$, so $S_{1}(x_{1},0)=\int _{\frac{(1-\lambda _1)K+\lambda _1x_1}{\lambda _1 +1}}^{\frac{\lambda _1x_1}{\lambda _1-1}+K}f_{\sigma }(a)da<\int _{0}^{\frac{\lambda _1x_1}{\lambda _1-1}+K}f_{\sigma }(a)da\le \int _{0}^{+\infty }f_\sigma (a)=\frac{1}{2}$, $\forall x_1>\frac{(\lambda _1-1)K}{\lambda _1}$.

Conclusion :

 we have shown that if $\sigma >\sigma ^{**}$, then $S_{1}(x_1,0)<\frac{1}{2}$, $\forall x_{1}\in \mathbb {R}$.

Finally, we prove (iii), i.e., if $\sigma ^{* }\le \sigma \le \sigma ^{**}$, there is no political equilibrium. Since $\sigma ^{*}\le \sigma \le \sigma ^{**}$, we know from Lemma 2 that $\int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1 +1)K}{\lambda _1}}f_\sigma (a)da\le \frac{1}{2}\le \int _{-K}^Kf_\sigma (a)da$. Because of Lemma 1, one of these inequalities must be strict, so without loss of generality, we consider that $\int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1+1)K}{\lambda _1}} f_\sigma (a)da<\frac{1}{2}\le \int _{-K}^Kf_\sigma (a)da$. Now let $(\overline{x}_{1},\overline{x}_{2})\in \mathbb {R}^{2}$. We want to show that $(\overline{x}_{1},\overline{x}_{2})$ is not a political equilibrium. It is sufficient to prove that there exist $x_{1}$ and $x_{2}$ such that $S_{1}(x_{1},\overline{x}_{2})\ge \frac{1 }{2}> S_{1}(\overline{x}_{1},x_{2})$. We proceed in two steps.

First step  We first show that $S_{1}(x_{1},\overline{x}_{2})\ge \frac{1}{2}$ if $x_{1}=\frac{\overline{x}_{2}}{\lambda _1}$, $\forall \overline{x}_{2}$.

• If $\overline{x}_{2}\in \left[ -K,K\right]$, then $\left| \overline{x}_{2}-x_1\right| = \left| \frac{(\lambda _1-1)\overline{x}_{2}}{\lambda _1}\right| \le \frac{(\lambda _1-1)K}{\lambda _1}$. We are thus in Case (A), where $b=\frac{\lambda _1x_1-\overline{x}_2}{\lambda _1-1}-K$ and $c=\frac{\lambda _1x_1-\overline{x}_2}{\lambda _1-1}+K$ (see Proposition 1). Given that $x_{1}=\frac{\overline{x}_{2}}{\lambda _1}$, then $b=-K$ and $c=K$, so $S_{1}\left( \frac{\overline{x}_{2}}{\lambda _1},\overline{x}_{2}\right) =\int _{-K}^{+K}f_{\sigma }(a)da\ge \frac{1}{2}$.

• If $\overline{x}_{2}>K$, then $\overline{x}_{2}-x_1=\frac{(\lambda _1-1)\overline{x}_{2}}{\lambda _1} >\frac{(\lambda _1-1)K}{\lambda _1}$. We are thus in Case (B), where $b=\frac{\lambda _1x_1-\overline{x}_2}{\lambda _1-1}-K$ and $c=\frac{(\lambda _1-1)K+\lambda _1x_1+\overline{x}_2}{\lambda _1+1}$ (see Proposition 1). Given that $x_{1}=\frac{\overline{x}_{2}}{\lambda _1}$, then $b=-K$ and $c=\frac{(\lambda _1-1)K+2\overline{x}_2}{\lambda _1+1}>K$ (since $\overline{x}_{2}>K$). Hence, $S_{1}\left( \frac{\overline{x}_{2}}{\lambda _1}, \overline{x}_{2}\right) =\int _{-K}^{\frac{(\lambda _1-1)K +2\overline{x}_2}{\lambda _1+1}}f_{\sigma }(a)da>\int _{-K}^{K}f_{\sigma }(a)da\ge \frac{1}{2}$.

• If $\overline{x}_{2}<-K$, then $\overline{x}_2-x_1 =\frac{(\lambda _1-1)\overline{x}_{2}}{\lambda _1}<-\frac{(\lambda _1-1)K}{\lambda _1}$. We are thus in Case (B’), where $b=\frac{(1-\lambda _1)K+\lambda _1x_1+\overline{x}_2}{\lambda _1+1}$ and $c=\frac{\lambda _1x_1-\overline{x}_2}{\lambda _1-1}+K$ (see Proposition 1). Given that $x_{1}=\frac{\overline{x}_{2}}{\lambda _1}$, then $b=\frac{(1-\lambda _1)K+2\overline{x}_2}{\lambda _1+1}<-K$ (since $\overline{x}_{2}<-K$) and $c=K$. Hence, $S_{1}\left( \frac{\overline{x}_{2}}{\lambda _1}, \overline{x}_{2}\right) =\int _{\frac{(1-\lambda _1)K +2\overline{x}_2}{\lambda _1+1}}^{K}f_{\sigma } (a)da> \int _{-K}^{K}f_{\sigma }(a)da\ge \frac{1}{2}$.

Second step  We now show that $S_{1}(\overline{x}_{1},x_{2})<\frac{1}{2}$ if $x_{2}$ is such that: $x_{2}=\overline{x}_{1}-\frac{(\lambda _1-1)K}{\lambda _1}$, $\forall \overline{x}_{1}\ge 0$, and $x_{2}= \overline{x}_{1}+\frac{(\lambda _1-1)K}{\lambda _1}$, $\forall \overline{x}_{1}<0$. For any $\overline{x}_{1}$, we are thus in Case (A), where $b=\frac{\lambda _1\overline{x}_1-x_2}{\lambda _1-1}-K$ and $c=\frac{\lambda _1\overline{x}_1-x_2}{\lambda _1-1}+K$ (see Proposition 1).

• If $\overline{x}_{1}\ge 0$, and given that $x_{2}=\overline{x}_{1}-\frac{(\lambda _1-1)K}{\lambda _1}$, then $b=\overline{x}_1+\frac{(1-\lambda _1)K}{\lambda _1}$ and $c=\overline{x}_{1}+\frac{(\lambda _1+1)K}{\lambda _1}$, so $S_{1}\left( \overline{x}_{1},\overline{x}_{1} -\frac{(\lambda _1-1)K}{\lambda _1}\right) =\int _{\overline{x}_1 +\frac{(1-\lambda _1)K}{\lambda _1}}^{\overline{x}_{1} +\frac{(\lambda _1+1)K}{\lambda _1}}f_{\sigma }(a)da\le \int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1+1)K}{\lambda _1}}f_{\sigma }(a)da<\frac{1}{2}$.

• If $\overline{x}_{1}<0$, and given that $x_{2}=\overline{x}_{1}+\frac{(\lambda _1-1)K}{\lambda _1}$, then $b=\overline{x}_1-\frac{(\lambda _1+1)K}{\lambda _1}$ and $c=\overline{x}_{1}+\frac{(\lambda _1-1)K}{\lambda _1}$, so $S_{1}\left( \overline{x}_{1},\overline{x}_{1}+\frac{(\lambda _1 -1)K}{\lambda _1}\right) =\int _{\overline{x}_1-\frac{(\lambda _1 +1)K}{\lambda _1}}^{\overline{x}_{1}+\frac{(\lambda _1 -1)K}{\lambda _1}}f_{\sigma }(a)da< \int _{-\frac{(\lambda _1 +1)K}{\lambda _1}}^{\frac{(\lambda _1-1)K}{\lambda _1}}f_{\sigma }(a)da$. Because of Lemma 1, we know that $\int _{-\frac{(\lambda _1+1)K}{\lambda _1}}^{\frac{(\lambda _1 -1)K}{\lambda _1}}f_{\sigma }(a)da=\int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1+1)K}{\lambda _1}}f_{\sigma }(a)da$, so $S_{1}\left( \overline{x}_{1},\overline{x}_{1}+\frac{(\lambda _1 -1)K}{\lambda _1}\right)<\int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1 +1)K}{\lambda _1}}f_{\sigma }(a)da<\frac{1}{2}$.

Conclusion: we have shown that if $\sigma ^{*}\le \sigma \le \sigma ^{**}$, then there is no political equilibrium. $\square$

1.5 A.5 Proof of Proposition 3

We prove (i) first, i.e., if $\sigma <\sigma ^{*}$, then $X_{1}^{*}=(-\alpha ,\alpha )$ where $\alpha$ is the unique positive real which satisfies $\int _{\alpha +\frac{(1-\lambda _1)K}{\lambda _1}}^{\alpha +\frac{(\lambda _1+1)K}{\lambda _1}}f_\sigma (a)da=\frac{1}{2}$. Moreover, we prove that $\alpha \in \left( 0,\frac{(\lambda _1-1)K}{\lambda _1}\right)$, $\sigma \mapsto \alpha (\sigma )$ is a decreasing function on $(0,\sigma ^*)$, with $\lim _{\sigma \rightarrow {\sigma ^*}^{-}} \alpha (\sigma )=0$ and $\lim _{\sigma \rightarrow 0^+} \alpha (\sigma )=\frac{(\lambda _1-1)K}{\lambda _1}$. We proceed in three steps.

First step :

First we show that there is a unique positive real $\alpha$ satisfying $\int _{\alpha +\frac{(1-\lambda _1)K}{\lambda _1}}^{\alpha +\frac{(\lambda _1+1)K}{\lambda _1}}f_\sigma (a)da=\frac{1}{2}$, and $\alpha \in \left( 0,\frac{(\lambda _1-1)K}{\lambda _1}\right)$. Since $\sigma <\sigma ^{*}$, we know from Lemma 2 that $\frac{1}{2}<\int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1+1)K}{\lambda _1}}f_\sigma (a)da<\int _{-K}^{K}f_\sigma (a)da$. Now let $g(v)=\int _{v-K}^{v+K}f_{\sigma }(a)da$, for $v\in \mathbb {R}$, we have $g^{\prime }(v)=f_{\sigma }(v+K)-f_{\sigma }(v-K)$. Given that $f_{\sigma }$ is an even function strictly increasing on $\mathbb {R}_{-}$, and strictly decreasing on $\mathbb {R}_{+}$, $f_{\sigma }\left( v+K\right) < f_{\sigma }\left( v-K\right)$ if $v> 0$, and $f_{\sigma }\left( v+K\right) > f_{\sigma }\left( v-K\right)$ if $v< 0$. It implies that g is an even continuous function, strictly increasing on $\mathbb {R}_{-}$, and strictly decreasing on $\mathbb {R}_{+}$. Since $g\left( \frac{K}{\lambda _1}\right) = \int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1+1)K}{\lambda _1}}f_\sigma (a)da>\frac{1}{2}$ and $g(K)=\int _{0}^{2K}f_{\sigma }(a)da<\int _{0}^{+\infty }f_{\sigma }(a)da=\frac{1}{2}$, there exists a unique $v^{*}>0$ such that $g(v^{*})=\frac{1}{2}$. And we must have $\frac{K}{\lambda _1}<v^{*}<K$. Now let $\alpha =v^{*}-\frac{K}{\lambda _1}$, then $0<\alpha <\frac{(\lambda _1-1)K}{\lambda _1}$. Given that $g(v^{*})=\int _{v^{*}-K}^{v^{*}+K}f_{\sigma }(a)da=\frac{1}{2}$,

$$\begin{aligned} g\left( \alpha +\frac{K}{\lambda _1}\right) =\int _{\alpha +\frac{(1-\lambda _1)K}{\lambda _1}}^{\alpha +\frac{(\lambda _1+1)K}{\lambda _1}}f_\sigma (a)da=\frac{1}{2}. \end{aligned}$$

(A9)

Second step :

  Now we show that $X_{1}^*=(-\alpha ,\alpha )$.

• First, we show that if $x_{1}\in (-\alpha ,\alpha )$, then $x_{1}\in X_{1}^{*}$, i.e., $S_{1}(x_{1},x_{2})>\frac{1}{2}$, $\forall x_{2}$. Consider the three possible cases: Case (A) $-\frac{(\lambda _1-1)K}{\lambda _1}\le x_{1}-x_{2}\le \frac{(\lambda _1-1)K}{\lambda _1}$, Case (B) $x_{2}>x_{1}+\frac{(\lambda _1-1)K}{\lambda _1}$, and Case (B’) $x_{2}<x_{1}-\frac{(\lambda _1-1)K}{\lambda _1}$.

  Case (A) :

  If $-\frac{(\lambda _1-1)K}{\lambda _1}\le x_{1}-x_{2}\le \frac{(\lambda _1-1)K}{\lambda _1}$, then $b=\frac{\lambda _1x_1-x_2}{\lambda _1-1}-K$ and $c=\frac{\lambda _1x_1-x_2}{\lambda _1-1}+K$ (see Proposition 1). So $S_{1}(x_{1},x_{2})= \int _{\frac{\lambda _1x_1-x_2}{\lambda _1-1}-K}^{\frac{\lambda _1x_1-x_2}{\lambda _1-1}+K}f_{\sigma }(a)da=g\left( \frac{\lambda _1x_1-x_2}{\lambda _1-1}\right)$.

  If $-\frac{(\lambda _1-1)K}{\lambda _1}\le x_{1}-x_{2}\le \frac{(\lambda _1-1)K}{\lambda _1}$ and $-\alpha<x_{1}<\alpha$ (so $-(\lambda _1-1)\alpha<(\lambda _1-1)x_{1}<(\lambda _1-1)\alpha$), then adding these two inequalities we obtain $-(\lambda _1-1)\left( \alpha +\frac{K}{\lambda _1}\right)<\lambda _1x_{1}-x_2<(\lambda _1-1)\left( \alpha +\frac{K}{\lambda _1}\right)$. Dividing by $(\lambda _1-1)$, then $-\left( \alpha +\frac{K}{\lambda _1}\right)<\frac{\lambda _1x_1-x_2}{\lambda _1-1}<\alpha +\frac{K}{\lambda _1}$. Since g is an even function, strictly increasing on $\mathbb {R}_{-}$, and strictly decreasing on $\mathbb {R}_{+}$, then $g\left( \frac{\lambda _1x_1-x_2}{\lambda _1-1}\right) >g\left( \alpha +\frac{K}{\lambda _1}\right) =\frac{1}{2}$ (see Eq. (A9)). Hence, if $x_{1}\in (-\alpha ,\alpha )$, $S_{1}(x_{1},x_{2})>\frac{1}{2}$, $\forall x_{2}\in \left[ x_1-\frac{(\lambda _1-1)K}{\lambda _1},x_1 +\frac{(\lambda _1-1)K}{\lambda _1}\right]$.

  Case (B) :

  If $x_2>x_1+\frac{(\lambda _1-1)K}{\lambda _1}$, then $b=\frac{\lambda _1x_1-x_2}{\lambda _1-1}-K$ and $c=\frac{(\lambda _1-1)K+\lambda _1x_1+x_2}{\lambda _1+1}$ (see Proposition 1), and $S_1(x_1,x_2)=\int _{\frac{\lambda _1x_1 -x_2}{\lambda _1-1}-K}^{\frac{(\lambda _1-1)K+\lambda _1x_1+x_2}{\lambda _1+1}}f_{\sigma } (a)da$. Thus, it is easy to see that if $x_2>x_1+\frac{(\lambda _1-1)K}{\lambda _1}$, then $S_1(x_1,x_2)$ is an increasing function of $x_2$, and $S_{1}(x_{1},x_{2})>S_{1} \left( x_{1},x_1+\frac{(\lambda _1-1)K}{\lambda _1}\right)$, with $S_{1}\left( x_{1},x_1+\frac{(\lambda _1-1)K}{\lambda _1}\right) >\frac{1}{2}$ according to Case (A). Hence, if $x_{1}\in (-\alpha ,\alpha )$, $S_{1}(x_{1},x_{2})>\frac{1}{2}$, $\forall x_{2}>x_1+\frac{(\lambda _1-1)K}{\lambda _1}$.

  Case (B’) :

  If $x_2<x_1-\frac{(\lambda _1-1)K}{\lambda _1}$, then $b=\frac{(1-\lambda _1)K+\lambda _1x_1+x_2}{\lambda _1+1}$ and $c=\frac{\lambda _1x_1-x_2}{\lambda _1-1}+K$ (see Proposition 1), and $S_1(x_1,x_2)=\int _{\frac{(1-\lambda _1)K +\lambda _1x_1+x_2}{\lambda _1+1}}^{\frac{\lambda _1x_1-x_2}{\lambda _1-1}+K}f_{\sigma }(a)da$. Thus, it is easy to see that if $x_2<x_1-\frac{(\lambda _1-1)K}{\lambda _1}$, $S_1(x_1,x_2)$ is a decreasing function of $x_2$, and $S_{1}(x_{1},x_{2})>S_{1} \left( x_{1},x_1-\frac{(\lambda _1-1)K}{\lambda _1}\right)$, with $S_{1}\left( x_{1},x_1-\frac{(\lambda _1-1)K}{\lambda _1}\right) >\frac{1}{2}$ according to Case (A). Hence, if $x_{1}\in (-\alpha ,\alpha )$, $S_{1}(x_{1},x_{2})>\frac{1}{2}$, $\forall x_{2}<x_1-\frac{(\lambda _1-1)K}{\lambda _1}$.

  Conclusion :

   we have shown that if $x_{1}\in (-\alpha ,\alpha )$, then $x_{1}\in X_{1}^{*}$.

• Second, we show that if $x_{1}\notin (-\alpha ,\alpha )$, then $x_{1}\notin X_{1}^{*}$. Assume that $x_{1}\ge \alpha$, and consider $x_{2}=x_1-\frac{(\lambda _1-1)K}{\lambda _1}$. We are then in Case (A), and $S_{1}(x_{1},x_{2})= \int _{\frac{\lambda _1x_1-x_2}{\lambda _1-1} -K}^{\frac{\lambda _1x_1-x_2}{\lambda _1-1}+K}f_{\sigma }(a)da=\int _{\frac{\lambda _1x_1-x_1+\frac{(\lambda _1-1) K}{\lambda _1}}{\lambda _1-1}-K}^{\frac{\lambda _1x_1-x_1 +\frac{(\lambda _1-1)K}{\lambda _1}}{\lambda _1-1}+K}f_{\sigma }(a)da=\int _{x_1+\frac{(1-\lambda _1)K}{\lambda _1}}^{x_1 +\frac{(\lambda _1+1)K}{\lambda _1}}f_{\sigma }(a)da=g\left( x_1+\frac{K}{\lambda _1}\right)$. Given that $x_{1}\ge \alpha$, then $x_1+\frac{K}{\lambda _1}\ge \alpha +\frac{K}{\lambda _1}$. Since g is strictly decreasing on $\mathbb {R}_+$, then $g\left( x_1+\frac{K}{\lambda _1}\right) \le g\left( \alpha +\frac{K}{\lambda _1}\right) =\frac{1}{2}$ (see Eq.  (A9)). Thus, for each $x_{1}\ge \alpha$, there exists $x_{2}$ such that $S_{1}(x_{1},x_{2})\le \frac{1}{2}$, so $x_{1}\notin X_{1}^{*}$. Similarly, if $x_{1}\le -\alpha$, consider $x_{2}=x_1+\frac{(\lambda _1-1)K}{\lambda _1}$. We are again in Case (A), and $S_{1}(x_{1},x_{2})= \int _{\frac{\lambda _1x_1-x_2}{\lambda _1-1}-K}^{\frac{\lambda _1x_1-x_2}{\lambda _1-1}+K}f_{\sigma }(a)da=\int _{\frac{\lambda _1x_1-x_1-\frac{(\lambda _1-1)K}{\lambda _1}}{\lambda _1-1}-K}^{\frac{\lambda _1x_1-x_1-\frac{(\lambda _1-1)K}{\lambda _1}}{\lambda _1-1}+K}f_{\sigma }(a)da=\int _{x_1-\frac{(\lambda _1+1)K}{\lambda _1}}^{x_1 +\frac{(\lambda _1-1)K}{\lambda _1}}f_{\sigma }(a)da=g\left( x_1-\frac{K}{\lambda _1} \right)$. We have $x_{1}\le -\alpha$, so $x_{1}-\frac{K}{\lambda _1}\le -\alpha -\frac{K}{\lambda _1}$. Now recall that g is an even function so $g\left( -\alpha -\frac{K}{\lambda _1}\right) =g\left( \alpha +\frac{K}{\lambda _1}\right) =\frac{1}{2}$ (see Eq.  (A9)). And given that g is strictly increasing on $\mathbb {R}_-$, then $g\left( x_1-\frac{K}{\lambda _1}\right) \le g\left( \alpha +\frac{K}{\lambda _1}\right) =\frac{1}{2}$. Thus, for each $x_{1}\le -\alpha$, there exists $x_{2}$ such that $S_{1}(x_{1},x_{2})\le \frac{1}{2}$, so $x_{1}\notin X_{1}^{*}$.

Conclusion :

  we have shown that if $x_{1}\notin (-\alpha ,\alpha )$, then $x_{1}\notin X_{1}^{*}$. Given that we have also shown that if $x_{1}\in (-\alpha ,\alpha )$, then $x_{1}\in X_{1}^{*}$, it implies that $X_{1}^*=(-\alpha ,\alpha )$.

Third step :

  It remains to show that $\sigma \mapsto \alpha (\sigma )$ is a decreasing function on $(0,\sigma ^*)$, with $\lim _{\sigma \rightarrow {\sigma ^*}^{-}} \alpha (\sigma )=0$ and $\lim _{\sigma \rightarrow 0^+} \alpha (\sigma ) =\frac{(\lambda _1-1)K}{\lambda _1}$ .

– First, we show that $\sigma \mapsto \alpha (\sigma )$ is a decreasing function on $(0,\sigma ^*)$. We know from Equation (A9) that $G(\sigma ,\alpha )=g\left( \alpha +\frac{K}{\lambda _1} \right) -\frac{1}{2}= \int _{\alpha +\frac{(1-\lambda _1)K}{\lambda _1}}^{\alpha +\frac{(\lambda _1+1)K}{\lambda _1}}f_\sigma (a)da-\frac{1}{2}=0$. The implicit function theorem gives $\frac{d\alpha }{d\sigma }=-\frac{\frac{\partial G}{\partial \sigma }}{\frac{\partial G}{\partial \alpha }}$. Let $z=\frac{a}{\sigma }$, so $G(\sigma ,\alpha )=\int _{\alpha +\frac{(1-\lambda _1)K}{\lambda _1}}^{\alpha +\frac{(\lambda _1 +1)K}{\lambda _1}}\frac{1}{\sigma }f_1(\frac{a}{\sigma })da-\frac{1}{2} =\int _{\frac{\alpha +\frac{(1-\lambda _1)K}{\lambda _1}}{\sigma }}^{\frac{\alpha +\frac{(\lambda _1+1)K}{\lambda _1}}{\sigma }}f_1(z)dz-\frac{1}{2}=0$ and $\frac{\partial G}{\partial \sigma }=-\frac{\alpha +\frac{(\lambda _1 +1)K}{\lambda _1}}{\sigma ^2}f_1\left( \frac{\alpha +\frac{(\lambda _1+1)K}{\lambda _1}}{\sigma }\right) +\frac{\alpha -\frac{(\lambda _1-1)K}{\lambda _1}}{\sigma ^2}f_1\left( \frac{\alpha -\frac{(\lambda _1-1)K}{\lambda _1}}{\sigma }\right)$. Now recall that we have shown that $\alpha <\frac{(\lambda _1-1)K}{\lambda _1}$ (first step of this proof), so $\frac{\partial G}{\partial \sigma }<0$. Concerning $\frac{\partial G}{\partial \alpha }=f_\sigma \left( \alpha +\frac{(\lambda _1 +1)K}{\lambda _1}\right) -f_\sigma \left( \alpha +\frac{(1-\lambda _1)K}{\lambda _1}\right)$, given that $f_\sigma$ is an even function, strictly increasing on $\mathbb {R}_-$ and strictly decreasing on $\mathbb {R}_+$, $f_\sigma \left( \alpha +\frac{(\lambda _1+1)K}{\lambda _1}\right) <f_\sigma \left( \alpha +\frac{(1-\lambda _1)K}{\lambda _1}\right)$ and $\frac{\partial G}{\partial \alpha }<0$. Since $\frac{\partial G}{\partial \sigma }<0$ and $\frac{\partial G}{\partial \alpha }<0$, then $\frac{d\alpha }{d\sigma }<0$ according to the implicit function theorem. Hence $\sigma \mapsto \alpha (\sigma )$ is a decreasing function on $(0,\sigma ^*)$.

– Second, we show that $\lim _{\sigma \rightarrow {\sigma ^*}^{-}} \alpha (\sigma )=0$. According to Lemma 2, we know that $\lim _{\sigma \rightarrow {\sigma ^*}^{-}} \int _{\frac{(1-\lambda _1)K}{\lambda _1}}^{\frac{(\lambda _1+1)K}{\lambda _1}} f_{\sigma }(a)da=\frac{1}{2}$. We also know from Eq. (A9) that $\int _{\alpha +\frac{(1-\lambda _1)K}{\lambda _1}}^{\alpha +\frac{(\lambda _1+1)K}{\lambda _1}}f_\sigma (a)da=\frac{1}{2}$. It is thus obvious that if $\sigma \rightarrow {\sigma ^*}^{-}$, then we must have $\alpha =0$.

– Third, we show that $\lim _{\sigma \rightarrow 0^{+}}\alpha (\sigma )=\frac{(\lambda _{1}-1)K}{\lambda _{1}}$. According to Eq. (A9), $\alpha$ is defined by $\int _{\alpha +\frac{(1-\lambda _1)K}{\lambda _1}}^{\alpha +\frac{(\lambda _1+1)K}{\lambda _1}}f_\sigma (a)da=\frac{1}{2}$. Let $z=\frac{a}{\sigma }$. We then obtain that $\alpha =\alpha (\sigma )$ is defined by $\int _{\frac{1}{\sigma } \left[ \alpha +\frac{(1 -\lambda _{1})K}{\lambda _{1}}\right] }^{\frac{1}{\sigma }\left[ \alpha +\frac{(\lambda _{1}+1)K}{\lambda _{1}}\right] }f_{1}(z)dz=\frac{1}{2}$. $\sigma \mapsto \alpha (\sigma )$ is a decreasing function on $\sigma \in (0,\sigma ^*)$, with $0<\alpha <\frac{ (\lambda _{1}-1)K}{\lambda _{1}}$, i.e., $\sigma \mapsto \alpha (\sigma )$ is bounded. Thus there is a limit $\alpha _{0}=\lim _{\sigma \rightarrow 0^{+}}\alpha (\sigma )$, with $\alpha _{0}\in \left[ 0,\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right]$.

We must have $\lim _{\sigma \rightarrow 0^{+}}\int _{\frac{1}{ \sigma }\left[ \alpha +\frac{(1-\lambda _{1})K}{\lambda _{1}} \right] }^{\frac{1}{\sigma } \left[ \alpha +\frac{(\lambda _{1}+1)K}{\lambda _{1}}\right] }f_{1}(z)dz=\int _{\lim _{\sigma \rightarrow 0^{+}} \frac{1}{\sigma }\left[ \alpha +\frac{(1-\lambda _{1})K}{\lambda _{1}} \right] }^{\lim _{\sigma \rightarrow 0^{+}}\frac{1}{\sigma }\left[ \alpha +\frac{(\lambda _{1}+1)K}{\lambda _{1}}\right] }f_{1}(z)dz=\int _{\lim _{\sigma \rightarrow 0^{+}}\frac{1}{\sigma }\left[ \alpha +\frac{(1-\lambda _{1})K}{\lambda _{1}}\right] }^{+\infty }f_{1}(z)dz=\frac{ 1}{2}$. If $\alpha _{0} < \frac{(\lambda _{1}-1)K}{\lambda _{1}}$, then $\int _{\lim _{\sigma \rightarrow 0^{+}}\frac{1}{\sigma }\left[ \alpha +\frac{ (1-\lambda _{1})K}{\lambda _{1}}\right] }^{+\infty }f_{1}(z)dz=\int _{-\infty }^{+\infty }f_{1}(z)dz =1\ne \frac{1}{2}$. It is impossible, thus $\alpha _{0}=\frac{(\lambda _{1}-1)K}{\lambda _{1}}$, and we have $\int _{\lim _{\sigma \rightarrow 0^{+}}\frac{1}{\sigma }\left[ \alpha +\frac{(1-\lambda _{1})K}{\lambda _{1}}\right] }^{+\infty }f_{1}(z)dz=\int _{0}^{+\infty }f_{1}(z)dz=\frac{1}{2}$.

Conclusion :

  We have shown that $\sigma \mapsto \alpha (\sigma )$ is a decreasing function on $(0,\sigma ^*)$, with $\lim _{\sigma \rightarrow {\sigma ^*}^{-}} \alpha (\sigma )=0$ and $\lim _{\sigma \rightarrow 0^{+}}\alpha (\sigma )=\frac{(\lambda _{1}-1)K}{\lambda _{1}}$.

Now we prove (ii), i.e., if $\sigma >\sigma ^{**}$, then $X_{2}^{* }=(-\beta ,\beta )$ where $\beta$ is the unique positive real which satisfies $\sup _{x_{1}\le \beta -\frac{(\lambda _1-1)K}{\lambda _1}}S_{1}(x_{1},\beta )=\frac{1}{2}$. Then we will prove that $\sigma \mapsto \beta (\sigma )$ is an increasing function on $(\sigma ^{**},+\infty )$.

Since $\sigma >\sigma ^{**}$, we know from Lemma 2 that $\int _{-K}^{+K}f_{\sigma }(a)da<\frac{1}{2}$. For a given $x_{2}$, let us set:

$$\begin{aligned} M_{A}(x_{2})= & {} \sup _{x_{1};\ \left| x_{1}-x_{2}\right| \le \frac{(\lambda _{1}-1)K}{\lambda _{1}}}S_{1}(x_{1},x_{2})=\sup _{x_{1}; \text{ Case } \text{ A }}S_{1}(x_{1},x_{2})\\ M_{B}(x_{2})= & {} \sup _{x_{1};\ x_{1}\le x_{2}-\frac{(\lambda _{1}-1)K}{ \lambda _{1}}}S_{1}(x_{1},x_{2})=\sup _{x_{1}; \text{ Case } \text{ B } }S_{1}(x_{1},x_{2})\\ M_{B^{'}}(x_{2})= & {} \sup _{x_{1};\ x_{1}\ge x_{2}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}}S_{1}(x_{1},x_{2})=\sup _{x_{1};\text {Case B'} }S_{1}(x_{1},x_{2})\\ M(x_{2})= & {} \sup _{x_{1}}S_{1}(x_{1},x_{2})=\max (M_{A}(x_{2}),M_{B}(x_{2}),M_{B^{'}}(x_{2})) \end{aligned}$$

Remark that each $\sup$ is in fact a $\max$ since $S_{1}$ is continuous and $S_{1}(x_{1},x_{2})\rightarrow 0$ if $x_{1}\rightarrow \pm \infty$. Now, consider the three cases, Case (A), Case (B), and Case (B’).

Case (A) :

If $x_{1}\in \left[ x_{2}-\frac{(\lambda _{1}-1)K}{\lambda _{1}},x_{2}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right]$, then $S_{1}(x_{1},x_{2})=\int _{\frac{\lambda _{1}x_{1}-x_{2}}{\lambda _{1}-1}-K}^{ \frac{\lambda _{1}x_{1}-x_{2}}{\lambda _{1}-1}+K}f_{\sigma }(a)da$ (see Proposition 1). Remark that the length of the interval $\Omega _1(x_1,x_2)$ is 2K for all $x_{1}\in \left[ x_{2}-\frac{(\lambda _{1}-1)K}{\lambda _{1}},x_{2}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right]$. We then have $\int _{\frac{\lambda _{1}x_{1}-x_{2}}{\lambda _{1}-1}-K}^{\frac{\lambda _{1}x_{1}-x_{2}}{\lambda _{1}-1}+K}f_{\sigma }(a)da\le \int _{-K}^{+K}f_{\sigma }(a)da$ since $[-K,+K]$ is the interval of length 2K on which the integral of $f_{\sigma }$ has the highest value. Thus $S_{1}(x_{1},x_{2})\le \int _{-K}^{+K}f_{\sigma }(a)da<\frac{1}{2}$. It implies that $M_{A}(x_{2})=\sup _{x_{1}; \text{ Case } \text{ A } }S_{1}(x_{1},x_{2})<\frac{1}{2}$.

Case (B) :

If $x_{1}\le x_{2}-\frac{(\lambda _{1}-1)K}{\lambda _{1}}$, then $S_{1}(x_{1},x_{2})=\int _{\frac{\lambda _{1}x_{1}-x_{2}}{\lambda _{1}-1}-K}^{\frac{(\lambda _{1}-1)K+\lambda _{1}x_{1}+x_{2}}{\lambda _{1}+1} }f_{\sigma }(a)da$ (see Proposition 1). Here $\frac{\partial S_{1}}{\partial x_{2}}>0$, thus $x_{2}\mapsto M_{B}(x_{2})$ is an increasing function. Moreover $M(0)<\frac{1}{2}$ according to Proposition 2 (ii), thus $M_{B}(0)\le M(0)<\frac{1}{2}$. Since $\lim _{x_{2}\rightarrow +\infty }M_{B}(x_{2})=1$ and $M_{B}$ is a continuous strictly increasing function, thus: there exists a unique $\beta >0$ such that $M_{B}(\beta )=\frac{1}{2}$.

– If $x_{2}\ge \beta$, then $M_{B}(x_{2})\ge \frac{1}{2}$, thus $M(x_{2})\ge \frac{1 }{2}$, which implies that $x_{2}\notin X_{2}^{*}$.

– If $0\le x_{2}<\beta$, then $M_{B}(x_{2})<\frac{1}{2}$.

Case (B’) :

If $x_{1}\ge x_{2}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}$ , then $S_{1}(x_{1},x_{2})=\int _{\frac{(1 -\lambda _{1})K+\lambda _{1}x_{1}+x_{2}}{\lambda _{1} +1}}^{\frac{\lambda _{1}x_{1}-x_{2}}{\lambda _{1}-1}+K}f_{\sigma }(a)da$ (see Proposition 1). Here $\frac{\partial S_{1}}{\partial x_{2}}<0$, thus $x_{2}\mapsto M_{B'}(x_{2})$ is a decreasing function. Moreover $M(0)<\frac{1}{2}$ according to Proposition 2 (ii), thus $M_{B^{\prime }}(0)\le M(0)<\frac{1}{2 }$, and $\lim _{x_{2}\rightarrow -\infty }M_{B^{\prime }}(x_{2})=1$. By symmetry (since $f_{\sigma }$ is an even function), for the same $\beta >0$, we have $M_{B^{\prime }}(\beta )=\frac{1}{2}$.

– If $x_{2}\le -\beta$, then $M_{B^{\prime }}(x_{2})\ge \frac{1}{2}$, thus $M(x_{2})\ge \frac{1}{2}$, which implies that $x_{2}\notin X_{2}^{*}$. – If $-\beta <x_{2}\le 0$, then $M_{B^{\prime }}(x_{2})<\frac{1}{2}$.

Conclusion :

 To sum up, we have:

–If $x_{2}\ge \beta$ or $x_{2}\le -\beta$, then $x_{2}\notin X_{2}^{*}$.

– If $0\le x_{2}<\beta$, then $M_{B}(x_{2})<\frac{1}{2}$. Since $M_{A}(x_{2})<\frac{1}{2}$ and $M_{B^{\prime }}(x_{2})\le M_{B}(x_{2})$ for $x_{2}\ge 0$, we can conclude that $M(x_{2})<\frac{1}{2}$, i.e., $x_{2}\in X_{2}^{*}$.

– If $-\beta <x_{2}\le 0$, then $M_{B^{\prime }}(x_{2})<\frac{1}{2}$. Since $M_{A}(x_{2})<\frac{1}{2}$ and $M_{B}(x_{2})\le M_{B^{\prime }}(x_{2})$ for $x_{2}\le 0$, we can conclude that $M(x_{2})<\frac{1}{2}$, i.e., $x_{2}\in X_{2}^{*}$.

Now we prove that $\sigma \mapsto \beta (\sigma )$ is an increasing function on $(\sigma ^{**},+\infty )$.

$\beta$ is defined by $M_{B}(\beta )=\max _{x_{1}\le \beta -\left( \frac{ \lambda _{1}-1}{\lambda _{1}}\right) K}S_{1}(x_{1},\beta )=\max _{x_{1}; \text{ Case } \text{ B }}S_{1}(x_{1},\beta )=\frac{1}{2}$. $(x_{1},x_{2})$ in Case (B) means that $x_{2}\ge x_{1}+\left( \frac{\lambda _{1}-1}{\lambda _{1}}\right) K$. Now, note that:

• $\forall x_{1}\in \mathbb {R}$, if $x_{2}=x_{1}+\left( \frac{\lambda _{1}-1}{ \lambda _{1}}\right) K$ (i.e., at the boundary between Case (A) and Case (B)), then $S_{1}(x_{1},x_{2})<\frac{1}{2}$ according to Part (ii) of Lemma 2.

• $\forall x_{1}\in \mathbb {R}$, $x_{2}\mapsto S_{1}(x_{1},x_{2})$ is an increasing continuous function on $x_{2}>x_{1}+\left( \frac{\lambda _{1}-1}{ \lambda _{1}}\right) K$, with $\lim _{x_{2}\rightarrow +\infty }S_{1}(x_{1},x_{2})=\lim _{x_{2}\rightarrow +\infty }\int _{\frac{\lambda _{1}x_{1}-x_{2}}{\lambda _{1}-1}-K}^{\frac{\lambda _{1}x_{1}+x_{2}+(\lambda _{1}-1)K}{\lambda _{1}+1}}f_{\sigma }(a)da=\int _{-\infty }^{+\infty } f_{\sigma }(a)da=1$.

Thus there exists a continuous function $\phi$ such that in Case (B):

$x_{2}>\phi (x_{1})\Leftrightarrow S_{1}(x_{1},x_{2})>\frac{1}{2}$ and $x_{2}=\phi (x_{1})\Leftrightarrow S_{1}(x_{1},x_{2})=\frac{1}{2}$

$\phi$ is a convex function according to Part (i) of Lemma 3 (in “Appendix A.8”).¹⁷

According to Part (iv) and (v) of Lemma 3, the function $x_{1}\mapsto \phi (x_{1})$ is first decreasing, then increasing. Then, $\beta$ is such that $\beta =\inf _{x_{1}\in \mathbb {R}}\phi (x_{1})=\min _{x_{1}\in \mathbb {R}}\phi (x_{1})$.

We now write $\beta (\sigma )=\min _{x_{1}\in \mathbb {R}}\phi _{\sigma }(x_{1})$ and $S_{1}^{\sigma }(x_{1},x_{2})$ to stress the impact of $\sigma$. Hence, we have $S_{1}^{\sigma }(x_{1},x_{2})=\int _{b(x_{1},x_{2})}^{c(x_{1},x_{2})}f_{\sigma }(a)da$ where $b(x_{1},x_{2})=\frac{\lambda _{1}x_{1}-x_{2}}{\lambda _{1}-1}-K\le c(x_{1},x_{2})=\frac{\lambda _{1}x_{1}+x_{2}+(\lambda _{1}-1)K}{\lambda _{1}+1}$ and we are in Case (B).

$b(x_{1},x_{2})\ge 0$ or $c(x_{1},x_{2})\le 0$ would imply $S_{1}(x_{1},x_{2})<\frac{1}{2}$, thus: $S_{1}^{\sigma }(x_{1},x_{2})\ge \frac{1}{2}\Longrightarrow b(x_{1},x_{2})<0<c(x_{1},x_{2})$.

If $x_{2}=\phi (x_{1})$, then $S_{1}^{\sigma }(x_{1},x_{2})=\int _{b(x_{1},x_{2})}^{c(x_{1},x_{2})}f_{\sigma }(a)da=\int _{ \frac{1}{\sigma }b(x_{1},x_{2})}^{\frac{1}{\sigma }c(x_{1},x_{2})}f_{1}(a)da= \frac{1}{2}$

If $\widetilde{\sigma }>\sigma$, then $\int _{\frac{1}{\widetilde{\sigma }} b(x_{1},x_{2})}^{\frac{1}{\widetilde{\sigma }}c(x_{1},x_{2})}f_{1}(a)da< \int _{\frac{1}{\sigma }b(x_{1},x_{2})}^{\frac{1}{\sigma } c(x_{1},x_{2})}f_{1}(a)da=\frac{1}{2}$ because $b(x_{1},x_{2})<0<c(x_{1},x_{2})$,

i.e., $S_{1}^{\widetilde{\sigma }}(x_{1},x_{2})<S_{1}^{\sigma }(x_{1},x_{2})= \frac{1}{2}$.

This implies that $\phi _{\widetilde{\sigma }}(x_{1})>\phi _{\sigma }(x_{1})$ , $\forall x_{1}\in \mathbb {R}$, thus $\beta (\widetilde{\sigma })>\beta (\sigma )$, $\forall \widetilde{\sigma }>\sigma$.

Conclusion: $\sigma \mapsto \beta (\sigma )$ is an increasing function on $(\sigma ^{**},+\infty )$. $\square$

1.6 A.6 Proof of Proposition 4

We proceed in three steps.

First step :

We first show that $\lambda _1\mapsto \sigma ^*(\lambda _1)$ is an increasing function on $(1,+\infty )$. Recall that according to the proof of Lemma 2 in “Appendix A.3”, $\sigma ^*$ is unique and defined by:

$$\begin{aligned} g_1(\sigma ^*)= \int _{\frac{(1-\lambda _1)K}{\sigma ^*\lambda _1}} ^{\frac{(\lambda _1+1)K}{\sigma ^*\lambda _1}}f_1(z)dz=\frac{1}{2} \end{aligned}$$

(A10)

Using the implicit function theorem, we know that $\frac{d\sigma ^*}{d\lambda _1}=-\frac{\frac{\partial g_1}{\partial \lambda _1}}{\frac{\partial g_1}{\partial \sigma ^*}}$. We have $\frac{\partial g_1}{\partial \sigma ^*}=-\frac{\frac{(\lambda _1+1) K}{\lambda _1}}{\sigma ^{*^2}}f_1\left( \frac{\frac{(\lambda _1+1)K}{\lambda _1}}{\sigma ^{*}}\right) +\frac{\frac{(1-\lambda _1)K}{\lambda _1}}{\sigma ^{*^2}}f_1\left( \frac{\frac{(1-\lambda _1)K}{\lambda _1}}{\sigma ^*}\right) <0$. Concerning $\frac{\partial g_{1}}{\partial \lambda _{1}}=-\frac{K}{\sigma ^{*}\lambda _{1}^{2}}f_{1}\left( \frac{ (\lambda _{1}+1)K}{\sigma ^{*}\lambda _{1}}\right) +\frac{K}{\sigma ^{*}\lambda _{1}^{2}}f_{1}\left( \frac{(1-\lambda _{1})K}{\sigma ^{*}\lambda _{1}}\right)$, given that $f_{1}$ is an even function, strictly increasing on $\mathbb {R}_{-}$ and strictly decreasing on $\mathbb {R}_{+}$, $f_{1}\left( \frac{(\lambda _{1}+1)K}{\sigma ^{*}\lambda _{1}}\right) <f_{1}\left( \frac{(1-\lambda _{1})K}{\sigma ^{*}\lambda _{1}}\right)$ thus $\frac{\partial g_{1}}{\partial \lambda _{1}}>0$. Since $\frac{\partial g_{1}}{\partial \sigma ^{*}}<0$ and $\frac{\partial g_{1}}{ \partial \lambda _{1}}>0$, then $\frac{d\sigma ^{*}}{d\lambda _{1}}>0$ according to the implicit function theorem.

Second step :

Now we show that $\lim _{\lambda _1\rightarrow 1^+}\sigma ^*(\lambda _1)=0$. According to Eq. (A10), $\sigma ^{*}=\sigma ^{*}(\lambda _{1})$ is defined by $\int _{\frac{(1-\lambda _{1})K}{\lambda _{1}\sigma ^{* }}}^{\frac{(\lambda _{1}+1)K}{\lambda _{1}\sigma ^{* }} }f_{1}(z)dz=\frac{1}{2}$, for all $\lambda _{1}>1$. $\lambda _{1}\mapsto \sigma ^{*}(\lambda _{1})$ is an increasing function on $\lambda _{1}>1$, with $\sigma ^{*}(\lambda _{1})>0$, i.e., $\lambda _{1}\mapsto \sigma ^{*}(\lambda _{1})$ has a lower bound on $\lambda _{1}>1$. Thus there is a limit $\sigma _{0}^{* }=\lim _{\lambda _{1}\rightarrow 1^{+}}\sigma ^{*}(\lambda _{1})$, with $\sigma _{0}^{*}\ge 0$. We must have $\lim _{\lambda _{1}\rightarrow 1^{+}}\int _{\frac{(1-\lambda _{1})K}{\lambda _{1}\sigma ^{* }}}^{\frac{(\lambda _{1}+1)K}{\lambda _{1}\sigma ^{* }}}f_{1}(z)dz=\int _{\lim _{\lambda _{1} \rightarrow 1^{+}}\frac{(1-\lambda _{1})K}{\lambda _{1} \sigma ^{* }}}^{\lim _{\lambda _{1}\rightarrow 1^{+}} \frac{(\lambda _{1}+1)K}{\lambda _{1}\sigma ^{* }}}f_{1}(z)dz=\frac{1}{2}$. If $\sigma _{0}^{*}>0$, then $\int _{\lim _{\lambda _{1}\rightarrow 1^{+}}\frac{(1-\lambda _{1})K}{\lambda _{1}\sigma ^{*}} }^{\lim _{\lambda _{1}\rightarrow 1^{+}}\frac{(\lambda _{1}+1)K}{\lambda _{1}\sigma ^{*}}}f_{1}(z)dz=\int _{0}^{\frac{2K}{\sigma _{0}^{*}}}f_{1}(z)dz<\frac{1}{2}$. It is impossible, thus $\sigma _{0}^{*}=0$, and we find $\int _{\lim _{\lambda _{1} \rightarrow 1^{+}}\frac{(1-\lambda _{1})K}{\lambda _{1}\sigma ^{*}}}^{\lim _{\lambda _{1}\rightarrow 1^{+}}\frac{(\lambda _{1}+1)K}{\lambda _{1}\sigma ^{*}}}f_{1}(z)dz =\int _{0}^{+\infty } f_{1}(z)dz=\frac{1}{2}$.

Third step :

We finally show that $\lim _{\lambda _1\rightarrow +\infty }\sigma ^*(\lambda _1)=\sigma ^{**}$, and that $\sigma ^{**}$ does not depend on $\lambda _{1}$. First, recall that according to the proof of Lemma 2 in “Appendix A.3”, $\sigma ^{**}$ is defined by $g_{2}(\sigma ^{**})=\int _{-K}^{K}\frac{1}{ \sigma ^{**} }f_{1}\left( \frac{a}{\sigma ^{**} }\right) da=\frac{1}{2}$; thus $\sigma ^{**}$ does not depend on $\lambda _{1}$. Second, $\lambda _{1}\mapsto \sigma ^{*}(\lambda _{1})$ is an increasing function on $\lambda _{1}>1$, with $0<\sigma ^{* }(\lambda _{1})\le \sigma ^{**}$, i.e., $\lambda _{1}\mapsto \sigma ^{* }(\lambda _{1})$ has a upper bound on $\lambda _{1}>1$. Thus there is a limit $\sigma _{\infty }^{*}=\lim _{\lambda _{1}\rightarrow +\infty }\sigma ^{*}(\lambda _{1})$, with $0<\sigma _{\infty }^{*}\le \sigma ^{** }$. We must have $\lim _{\lambda _{1}\rightarrow +\infty }\int _{\frac{(1-\lambda _{1})K}{\lambda _{1}\sigma ^{*}}}^{\frac{(\lambda _{1}+1)K}{\lambda _{1}\sigma ^{*}}}f_{1}(z)dz=\int _{\lim _{\lambda _{1}\rightarrow +\infty }\frac{(1-\lambda _{1})K}{\lambda _{1}\sigma ^{* } }}^{\lim _{\lambda _{1}\rightarrow +\infty }\frac{(\lambda _{1}+1)K}{\lambda _{1}\sigma ^{* }}}f_{1}(z)dz=\int _{\frac{-K}{ \sigma _{\infty }^{*}}}^{\frac{K}{\sigma _{\infty }^{* }} }f_{1}(z)dz=\frac{1}{2}$. Since $\int _{\frac{-K}{ \sigma ^{** }}}^{\frac{K}{ \sigma ^{** }}}f_{1}(z)dz=\frac{1}{2}$, then $\sigma _{\infty }^{* }=\sigma ^{**}$, i.e., $\lim _{\lambda _{1}\rightarrow +\infty }\sigma ^{*}(\lambda _{1})=\sigma ^{** }$.

Conclusion :

 We have shown that $\lambda _{1}\mapsto \sigma ^{* }(\lambda _{1})$ is an increasing function on $(1,+\infty )$, with $\lim _{\lambda _{1}\rightarrow 1^{+}}\sigma ^{*}(\lambda _{1})=0$ and $\lim _{\lambda _{1}\rightarrow +\infty }\sigma ^{*}(\lambda _{1})=\sigma ^{** }$, and that $\sigma ^{**}$ does not depend on $\lambda _{1}$. $\square$

1.7 A.7 Proof of Proposition 5

We prove (i) first, i.e., if it exists an interval of length $2\frac{(\lambda _1-1)K}{\lambda _1}$ wherein strictly more than 50 percent of the voters are located, Candidate 1 wins with certainty if he chooses the midpoint of this interval. Consider that z is the midpoint of this interval $\Delta =\left[ z-\frac{(\lambda _1 -1)K}{\lambda _1},z+\frac{(\lambda _1-1)K}{\lambda _1}\right]$ of length $2\frac{(\lambda _1-1)K}{\lambda _1}$ wherein strictly more than 50 percent of the voters are located, so $\int _{\Delta }f(a)da=\int _{z-\frac{(\lambda _1-1)K}{\lambda _1}}^{z +\frac{(\lambda _1-1)K}{\lambda _1}}f(a)da>\frac{1}{2}$. We have to show that Candidate 1 wins with certainty if he chooses $x_1^*=z$, $\forall x_2\in \mathbb {R}$. Consider the three possible cases: Case (A) $x_2\in \left[ z-\frac{(\lambda _1-1)K}{\lambda _1},z +\frac{(\lambda _1-1)K}{\lambda _1}\right]$, Case (B) $x_2>z+\frac{(\lambda _1-1)K}{\lambda _1}$, and Case (B’) $x_2<z-\frac{(\lambda _1-1)K}{\lambda _1}$ which correspond to Panels (A), (B) and (B’) in Fig. 2 when $x_1=z$.

Case (A) :

If $x_2\in \left[ z-\frac{(\lambda _1-1)K}{\lambda _1},z +\frac{(\lambda _1-1)K}{\lambda _1}\right]$ and $x_1=x_{1}^{*}=z$, then we know from Proposition 1 that $b=\frac{\lambda _1z-x_2}{\lambda _1-1}-K$ and $c=\frac{\lambda _1z-x_2}{\lambda _1-1}+K$. If so, $\Omega _1(z,x_2) =\left( \frac{\lambda _1z-x_2}{\lambda _1-1}-K,\frac{\lambda _1z-x_2}{\lambda _1-1} +K\right)$. Note that $\frac{\lambda _1z-x_2}{\lambda _1-1}-K\le z-\frac{(\lambda _1-1)K}{\lambda _1}\Leftrightarrow x_2\ge z-\frac{(\lambda _1-1)K}{\lambda _1}$ which is always true in Case (A), i.e., the lower bound of $\Omega _1(z,x_2)$ is lower than the lower bound of $\Delta$. Note also that $\frac{\lambda _1z-x_2}{\lambda _1-1}+K\ge z+\frac{(\lambda _1-1)K}{\lambda _1}\Leftrightarrow x_2\le z+\frac{(\lambda _1-1)K}{\lambda _1}$ which is always true in Case (A), i.e., the upper bound of $\Omega _1(z,x_2)$ is higher than the upper bound of $\Delta$. If so, $\Delta \subseteq \Omega _1(z,x_2)$; hence, $S_1(z,x_2)>\frac{1}{2}$ if $\int _{z-\frac{(\lambda _1-1)K}{\lambda _1}}^{z +\frac{(\lambda _1-1)K}{\lambda _1}}f(a)da>\frac{1}{2}$, $\forall x_2\in \left[ z -\frac{(\lambda _1-1)K}{\lambda _1},z+\frac{(\lambda _1-1)K}{\lambda _1}\right]$.

Case (B) :

If $x_2>z+\frac{(\lambda _1-1)K}{\lambda _1}$ and $x_1=x_{1}^{*}=z$, then $b=\frac{\lambda _1z-x_2}{\lambda _1-1}-K$ and $c=\frac{(\lambda _1-1)K+\lambda _1z+x_2}{\lambda _1+1}$ (see Proposition 1). If so, $\Omega _1(z,x_2)=\left( \frac{\lambda _1z-x_2}{\lambda _1 -1}-K,\frac{(\lambda _1-1)K+\lambda _1z+x_2}{\lambda _1+1}\right)$. Note that $\frac{\lambda _1z-x_2}{\lambda _1-1}-K\le z-\frac{(\lambda _1-1)K}{\lambda _1}\Leftrightarrow x_2\ge z-\frac{(\lambda _1-1)K}{\lambda _1}$ which is always true in Case (B), i.e., the lower bound of $\Omega _1(z,x_2)$ is lower than the lower bound of $\Delta$. Note also that $\frac{(\lambda _1-1)K+\lambda _1z+x_2}{\lambda _1+1}> z+\frac{(\lambda _1-1)K}{\lambda _1}\Leftrightarrow x_2> z+\frac{(\lambda _1-1)K}{\lambda _1}$ which is always true in Case (B), i.e., the upper bound of $\Omega _1(z,x_2)$ is higher than the upper bound of $\Delta$. If so, $\Delta \subseteq \Omega _1(z,x_2)$; hence, $S_1(z,x_2)>\frac{1}{2}$ if $\int _{z-\frac{(\lambda _1-1)K}{\lambda _1}}^{z+\frac{(\lambda _1-1)K}{\lambda _1}}f(a)da>\frac{1}{2}$, $\forall x_2>z+\frac{(\lambda _1-1)K}{\lambda _1}$.

Case (B’) :

If $x_2<z-\frac{(\lambda _1-1)K}{\lambda _1}$ and $x_1=x_{1}^{*}=z$, then $b=\frac{(1-\lambda _1)K+\lambda _1z+x_2}{\lambda _1+1}$ and $c=\frac{\lambda _1z-x_2}{\lambda _1-1}+K$ (see Proposition 1). If so, $\Omega _1(z,x_2)=\left( \frac{(1-\lambda _1) K+\lambda _1z+x_2}{\lambda _1+1},\frac{\lambda _1z-x_2}{\lambda _1-1}+K\right)$. Note that $\frac{(1-\lambda _1)K+\lambda _1z+x_2}{\lambda _1+1}<z -\frac{(\lambda _1-1)K}{\lambda _1}\Leftrightarrow x_2< z -\frac{(\lambda _1-1)K}{\lambda _1}$ which is always true in Case (B’), i.e., the lower bound of $\Omega _1(z,x_2)$ is lower than the lower bound of $\Delta$. Note also that $\frac{\lambda _1z-x_2}{\lambda _1-1}+K> z+\frac{(\lambda _1-1)K}{\lambda _1}\Leftrightarrow x_2<z-\frac{(\lambda _1-1)K}{\lambda _1}$ which is always true in Case (B’), i.e., the upper bound of $\Omega _1(z,x_2)$ is higher than the upper bound of $\Delta$. If so, $\Delta \subseteq \Omega _1(z,x_2)$; hence, $S_1(z,x_2)>\frac{1}{2}$ if $\int _{z-\frac{(\lambda _1-1)K}{\lambda _1}}^{z+\frac{(\lambda _1 -1)K}{\lambda _1}}f(a)da>\frac{1}{2}$, $\forall x_2<z-\frac{(\lambda _1-1)K}{\lambda _1}$.

Conclusion :

 We have shown that if it exists an interval of length $2\frac{(\lambda _1-1)K}{\lambda _1}$ wherein strictly more than 50 percent of the voters are located, Candidate 1 wins with certainty if he chooses the midpoint of this interval.

Now we prove (ii), i.e., if all the possible intervals of length 2K have strictly less than 50% of the voters, Candidate 2 wins with certainty if he chooses the median location. Without loss of generality, we consider that the median location is 0, so we have to show that in such a case, Candidate 2 wins with certainty if he chooses $x_2^*=0$, $\forall x_1\in \mathbb {R}$. Consider the three possible cases: Case (A) $x_1\in \left[ -\frac{(\lambda _1-1)K}{\lambda _1},\frac{(\lambda _1-1)K}{\lambda _1}\right]$, Case (B) $x_1<-\frac{(\lambda _1-1)K}{\lambda _1}$, and Case (B’) $x_1>\frac{(\lambda _1-1)K}{\lambda _1}$ which correspond to Panels (A), (B) and (B’) in Fig. 2 when $x_2=0$.

Case (A) :

If $x_1\in \left[ -\frac{(\lambda _1-1)K}{\lambda _1},\frac{(\lambda _1-1)K}{\lambda _1}\right]$ and $x_2=x_{2}^{*}=0$, then we know from Proposition 1 that $b=\frac{\lambda _1x_1}{\lambda _1-1}-K$ and $c=\frac{\lambda _1x_1}{\lambda _1-1}+K$. If so, $S_{1}(x_1,0)=\int _{\frac{\lambda _1x_1}{\lambda _1 -1}-K}^{\frac{\lambda _1x_1}{\lambda _1-1}+K}f(a)da$. Remark that the length of the interval $\Omega _1(x_1,0)$ is 2K for all $x_{1}\in \left[ -\frac{(\lambda _{1}-1)K}{\lambda _{1}} \frac{(\lambda _{1}-1)K}{\lambda _{1}}\right]$. Hence, given that all the intervals of length 2K have strictly less than 50 percent of the voters, we can conclude that $S_{1}(x_{1},0)<\frac{1}{2}$, $\forall x_{1}\in \left[ -\frac{(\lambda _1-1)K}{\lambda _1},\frac{(\lambda _1 -1)K}{\lambda _1}\right]$.

Case (B) :

If $x_1<-\frac{(\lambda _1-1)K}{\lambda _1}$ and $x_2=x_{2}^{*}=0$, then $S_{1}(x_{1},0)=\int _{\frac{\lambda _1x_1}{\lambda _1-1}-K}^{\frac{(\lambda _1-1)K+\lambda _1x_1}{\lambda _1+1}}f(a)da$ (see Proposition 1). If $x_1<-\frac{(\lambda _1-1)K}{\lambda _1}$, then $\frac{(\lambda _1-1)K+\lambda _1x_1}{\lambda _1+1}<0$, so $S_{1}(x_{1},0)=\int _{\frac{\lambda _{1}x_{1}}{\lambda _{1}-1}-K}^{\frac{ (\lambda _{1}-1)K+\lambda _{1}x_{1}}{\lambda _{1}+1}}f(a)da<\int _{\frac{\lambda _{1} x_{1}}{\lambda _{1}-1}-K}^{0}f(a)da<\int _{-\infty }^{0}f(a)da=\frac{1}{2}$, i.e., $S_{1}(x_{1},0)<\frac{1}{2}$, $\forall x_{1} <-\frac{(\lambda _{1}-1)K}{\lambda _{1}}$.

Case (B’) :

If $x_1>\frac{(\lambda _1-1)K}{\lambda _1}$ and $x_2=x_{2}^{*}=0$, then $S_{1}(x_{1},0)=\int _{\frac{(1-\lambda _1)K +\lambda _1x_1}{\lambda _1+1}}^{\frac{\lambda _1x_1}{\lambda _1-1}+K}f(a)da$ (see Proposition 1). If $x_1>\frac{(\lambda _1-1)K}{\lambda _1}$, then $\frac{(1-\lambda _1)K+\lambda _1x_1}{\lambda _1+1}>0$, so $S_{1}(x_{1},0) =\int _{\frac{(1-\lambda _1)K+\lambda _1x_1}{\lambda _1+1}}^{\frac{\lambda _1x_1}{\lambda _1-1}+K}f(a)da<\int _{0}^{\frac{\lambda _1x_1}{\lambda _1-1}+K}f(a)da \le \int _{0}^{+\infty }f_\sigma (a)=\frac{1}{2}$, $\forall x_1>\frac{(\lambda _1 -1)K}{\lambda _1}$.

Conclusion :

 We have shown that if all the possible intervals of length 2K have strictly less than 50 percent of the voters, then $S_{1}(x_1,0)<\frac{1}{2}$, $\forall x_{1}\in \mathbb {R}$, i.e., Candidate 2 wins with certainty if he chooses the median location. $\square$

1.8 A.8 Proof of Proposition 6

We consider that $\sigma \in [\sigma ^{* },\sigma ^{** })$. If so, we know from Proposition 2 that there is no political equilibrium, i.e., no pure strategy equilibrium, and we are looking for a set of mixed strategy equilibria. We will proceed in three steps.

First step :

In the space $\mathbb {R}^{2}$, we will study the shape of:

$$\begin{aligned} H_{1}= & {} \left\{ (x_{1},x_{2})\in \mathbb {R}^{2}; \pi _{1}(x_{1},x_{2})=1\right\} \\ H_{2}= & {} \left\{ (x_{1},x_{2})\in \mathbb {R}^{2}; \pi _{2}(x_{1},x_{2})=1\right\} \\ H_{0}= & {} \left\{ (x_{1},x_{2})\in \mathbb {R}^{2}; \pi _{1}(x_{1},x_{2})=\pi _{2}(x_{1},x_{2})=\frac{1}{2}\right\} \end{aligned}$$

Second step :

We will eliminate weakly dominated strategies.

Third step :

We will determine a set of mixed strategy equilibria.

First step :

Clearly we have $H_{1}\cup H_{2}\cup H_{0}=\mathbb {R}^{2}$. If Candidate 1 plays $x_{1}$ and Candidate 2 plays $x_{2}$, then: Candidate 1 wins if $(x_{1},x_{2})\in H_{1}$; Candidate 2 wins if $(x_{1},x_{2})\in H_{2}$; there is a tie if $(x_{1},x_{2})\in H_{0}$.

We want to draw the graph in $\mathbb {R}^{2}$ delimiting $H_{1}$, $H_{2}$, $H_{0}$. Consider Cases (A), (B) and (B’) of Proposition 1. Then, Case (A) corresponds to the domain of $\mathbb {R}^{2}$ between lines $D_{1}$ and $D_{2}$, Case (B) to the domain above line $D_{2}$, and Case (B’) to the domain below line $D_{1}$, where:

– $D_{1}$ is the line of Equation $x_{1}-x_{2}=\left( \frac{\lambda _{1}-1}{ \lambda _{1}}\right) K$.

– $D_{2}$ is the line of Equation $x_{1}-x_{2}=-\left( \frac{\lambda _{1}-1}{ \lambda _{1}}\right) K$.

Case (A) :

If $\left| x_{1}-x_{2}\right| \le \left( \frac{\lambda _{1}-1}{\lambda _{1}}\right) K$, and recalling that according to the proof of Part (i) of Proposition 3 in “Appendix A.5” (first step, and Case (A) in the second step) $S_{1}(x_{1},x_{2})=g\left( \frac{\lambda _1x_1-x_2}{\lambda _1-1}\right)$, then

$$\begin{aligned} (x_{1},x_{2})\in & {} H_{1}\Leftrightarrow S_{1}(x_{1},x_{2})>\frac{1}{2}\\ (x_{1},x_{2})\in & {} H_{1}\Leftrightarrow g\left( \frac{\lambda _{1}x_{1} -x_{2}}{\lambda _{1}-1}\right) >\frac{1}{2}\\ (x_{1},x_{2})\in & {} H_{1}\Leftrightarrow \left| \frac{\lambda _{1}x_{1}-x_{2}}{\lambda _{1}-1}\right|<v^{*}\\ (x_{1},x_{2})\in & {} H_{1}\Leftrightarrow -(\lambda _{1}-1)v^{*}<\lambda _{1}x_{1}-x_{2}<(\lambda _{1}-1)v^{*} \end{aligned}$$

Let $\Delta _{1}$ be the line of Equation $\lambda _{1}x_{1}-x_{2}=(\lambda _{1}-1)v^{*}$, and $\Delta _{2}$ be the line of Equation $\lambda _{1}x_{1}-x_{2}=-(\lambda _{1}-1)v^{*}$, then

$(x_{1},x_{2})\in H_{1}\Leftrightarrow (x_{1},x_{2})$ is between $\Delta _{1}$ and $\Delta _{2}$

$(x_{1},x_{2})\in H_{0}\Leftrightarrow (x_{1},x_{2})\in \Delta _{1}\cup \Delta _{2}$ Denote by E and F the points such that: $\{E\}=D_{2}\cap \Delta _{1}$ and $\{F\}=D_{2}\cap \Delta _{2}$

If $(x_{1},x_{2})\in D_{2}$ (i.e., just at the boundary between Case (A) and Case (B)), then:

$$\begin{aligned} \pi _{1}(x_{1},x_{2})= & {} 1 \text{ if } (x_{1},x_{2})\in (F,E) \nonumber \\ \pi _{2}(x_{1},x_{2})= & {} 1 \text{ if } (x_{1},x_{2})\notin [ F,E]\nonumber \\ \pi _{1}(x_{1},x_{2})= & {} \pi _{2}(x_{1},x_{2})=\frac{1}{2} \text{ if } (x_{1},x_{2})\in \{F,E\} \end{aligned}$$

(A11)

Case (B) :

If $x_{2}\ge x_{1}+\left( \frac{\lambda _{1}-1}{\lambda _{1}}\right) K$, then $(x_{1},x_{2})\in H_{1}\Leftrightarrow S_{1}(x_{1},x_{2})>\frac{1}{2}$ where $S_{1}(x_{1},x_{2}) =\int _{\frac{\lambda _{1}x_{1}-x_{2}}{\lambda _{1}-1}-K}^{\frac{\lambda _{1}x_{1}+x_{2} +(\lambda _{1}-1)K}{\lambda _{1}+1}}f_{\sigma }(a)da$. For $x_{1}$ given, $x_{2}\mapsto S_{1}(x_{1},x_{2})$ is a continuous increasing function, with $\lim _{x_{2}\rightarrow +\infty }S_{1}(x_{1},x_{2})=\int _{-\infty }^{+\infty }f_{\sigma }(a)da=1$. It means that for any given $x_{1}$, we have $S_{1}(x_{1},x_{2})>\frac{1}{2}$ for $x_{2}$ high enough; hence, $\pi _{1}(x_{1},x_{2})=1$. From this result and from Eq. (A11), there is a function $\phi :\mathbb {R\rightarrow R}$ such that (in Case (B)):

$*$ If $x_{1}\le x_{1}(F)$ or $x_{1}\ge x_{1}(E)$, then

$\pi _{1}(x_{1},x_{2})=1\Leftrightarrow x_{2}>\phi (x_{1})$

$\pi _{1}(x_{1},x_{2})=\frac{1}{2}\Leftrightarrow x_{2}=\phi (x_{1})$

$*$ If $x_{1}\in (x_{1}(F),x_{1}(E))$, then $\pi _{1}(x_{1},x_{2})=1$.

It means that $H_{1}$ is the domain above the graph of $\phi$ (for $x_{1}\le x_{1}(F)$ or $x_{1}\ge x_{1}(E)$ ) and above $D_{2}$ (for $x_{1}\in [ x_{1}(F),x_{1}(E)]$).

Lemma 3

• (i) $H_{1}$ is a convex set inside Case (B).

• (ii) $\phi$ is a convex function on $x_{1}\le x_{1}(F)$ and on $x_{1}\ge x_{1}(E)$.

• (iii) The set $H_{1}$ does not intersect the line $x_{2}=0$ inside Case (B).

• (iv) The line of Equation $\lambda _{1}x_{1}+x_{2}+(\lambda _{1}-1)K=0$ is an asymptote of $\phi$ when $x_{1}\rightarrow -\infty$ and $x_{2}\rightarrow +\infty$.

• (v) The line of Equation $\lambda _{1}x_{1}-x_{2}-(\lambda _{1}-1)K=0$ is an asymptote of $\phi$ when $x_{1}\rightarrow +\infty$ and $x_{2}\rightarrow +\infty$.

Proof of Lemma 3

• (i) To show that $H_{1}$ is a convex set, assume that $M_{0}=\left( x_{1}(0),x_{2}(0)\right) \in H_{1}$ and $M_{1}=\left( x_{1}(1),x_{2}(1)\right) \in H_{1}$. For $\alpha \in [ 0,1]$, set $M_{\alpha }=\alpha M_{1}+(1-\alpha )M_{0}$, i.e., $M_{\alpha }=\left( x_{1}(\alpha ),x_{2}(\alpha )\right) =\left( \alpha x_{1}(1)+(1-\alpha )x_{1}(0),\alpha x_{2}(1)+(1-\alpha )x_{2}(0)\right)$. $M_{\alpha }\in [ M_{0},M_{1}]$ and we want to prove that $M_{\alpha }\in H_{1}$. Let us denote by $b_{0}$, $c_{0}$ the values of b and c at $M_{0}$; by $b_{1}$, $c_{1}$ the values of b and c at $M_{1}$ and by $b_{\alpha }$, $c_{\alpha }$ the values of b and c at $M_{\alpha }$. We have $b_{\alpha }=\frac{\lambda _{1}x_{1}(\alpha )-x_{2}(\alpha )}{\lambda _{1}-1} -K=\alpha b_{1}+(1-\alpha )b_{0}$ and $c_{\alpha }=\frac{\lambda _{1}x_{1}(\alpha )+x_{2}(\alpha )+(\lambda _{1}-1)K}{\lambda _{1}+1}=\alpha c_{1}+(1-\alpha )c_{0}$

  We set $L(\alpha )=S_{1}(x_{1}(\alpha ),x_{2}(\alpha ))=\int _{b_{\alpha }}^{c_{\alpha }}f_{\sigma }(a)da=\int _{\alpha b_{1}+(1-\alpha )b_{0}}^{\alpha c_{1}+(1-\alpha )c_{0}}f_{\sigma }(a)da$.

  We know that $L(0)>\frac{1}{2}$ and $L(1)>\frac{1}{2}$ since $M_{0}\in H_{1}$ and $M_{1}\in H_{1}$.

  $L^{\prime }(\alpha )=(c_{1}-c_{0})f_{\sigma }(c_{\alpha })-(b_{1}-b_{0})f_{\sigma }(b_{\alpha })$

  $L\text {''}(\alpha )=(c_{1}-c_{0})^{2}f_{\sigma }^{\prime }(c_{\alpha })-(b_{1}-b_{0})^{2}f_{\sigma }^{\prime }(b_{\alpha })$

  We are in Case (B) with $S_{1}(x_{1}(0),x_{2}(0))>\frac{1}{2}$ and $S_{1}(x_{1}(1),x_{2}(1))>\frac{1}{2}$, then $b_{0}<0<c_{0}$ and $b_{1}<0<c_{1}$, thus $b_{\alpha }<0<c_{\alpha }$. It implies that $f_{\sigma }^{\prime }(b_{\alpha })>0>f_{\sigma }^{\prime }(c_{\alpha })$, thus $L\text {''}(\alpha )<0$.

  $\alpha \mapsto L(\alpha )$ is a concave function on [0, 1], with $L(0)> \frac{1}{2}$ and $L(1)>\frac{1}{2}$, thus $L(\alpha )>\frac{1}{2}$, i.e., $\pi _{1}(x_{1}(\alpha ),x_{2}(\alpha ))=1$ for every $\alpha \in [0,1]$.

  Conclusion: inside Case (B), $H_{1}$ is a convex set.

• (ii) Part (ii) of Lemma 3 is implied straightfully by Part (i).

• (iii) To show Part (iii) of Lemma 3, note that for $x_{2}=0$, $S_{1}(x_{1},x_{2})=S_{1}(x_{1},0)=\int _{\frac{\lambda _{1}x_{1}}{\lambda _{1}-1}-K}^{\frac{\lambda _{1}x_{1}+(\lambda _{1}-1)K}{ \lambda _{1}+1}}f_{\sigma }(a)da$

  with (Case B): $x_{1}+\left( \frac{\lambda _{1}-1}{\lambda _{1}}\right) K\le x_{2}=0$, i.e., $\lambda _{1}x_{1}+(\lambda _{1}-1)K\le 0$,

  thus $S_{1}(x_{1},0)\le \int _{\frac{\lambda _{1}x_{1}}{\lambda _{1}-1} -K}^{0}f_{\sigma }(a)da\le \frac{1}{2}$ i.e., $\pi _{1}(x_{1},0)<1$, then $(x_{1},0)\notin H_{1}$.

• (iv) To show Part (iv) of Lemma 3, note that $S_{1}(x_{1},x_{2})=\int _{\frac{\lambda _{1}x_{1}-x_{2}}{ \lambda _{1}-1}-K}^{\frac{\lambda _{1}x_{1}+x_{2}+(\lambda _{1}-1)K}{\lambda _{1}+1}}f_{\sigma }(a)da$ tends to $\int _{-\infty }^{0}f_{\sigma }(a)da= \frac{1}{2}$ if $\lambda _{1}x_{1}+x_{2}+(\lambda _{1}-1)K\rightarrow 0$ as $x_{1}\rightarrow -\infty$ and $x_{2}\rightarrow +\infty$.

• (v) To show Part (v) of Lemma 3, note that in Case (B), $S_{1}(x_{1},x_{2})=\int _{\frac{\lambda _{1}x_{1}-x_{2}}{ \lambda _{1}-1}-K}^{\frac{\lambda _{1}x_{1}+x_{2}+(\lambda _{1}-1)K}{\lambda _{1}+1}}f_{\sigma }(a)da$ tends to $\int _{0}^{+\infty }f_{\sigma }(a)da= \frac{1}{2}$ if $\lambda _{1}x_{1}-x_{2}-(\lambda _{1}-1)K\rightarrow 0$ as $x_{1}\rightarrow +\infty$ and $x_{2}\rightarrow +\infty$. $\square$ (end of the Proof of Lemma 3)

Case (B’) :

If $x_{2}\le x_{1}-\left( \frac{\lambda _{1}-1 }{\lambda _{1}}\right) K$, we are in a case symmetric to Case (B) with respect to point $O=(0,0)$. We denote by $E^{\prime }$, $F^{\prime }$ the symmetric points of E, F with respect to O. By symmetry, all properties of Case (B) are also satisfied in Case (B’).

Conclusion :

 This first step permits to obtain Fig. 4 which shows the partition of $\mathbb {R}^{2}$ in $H_{1}$, $H_{2}$, $H_{0}$. We have seen that $\phi$ is a convex function. Moreover, it can be easily shown that $\phi$ is an increasing function on $x_{1}\ge x_{1}(E)$, as in Fig. 4. If $\phi ^{\prime }(x_{1}(F))<0$, then $\phi$ is a decreasing function on $x_{1}\le x_{1}(F)$. If $\phi ^{\prime }(x_{1}(F))>0$, then $\phi$ is first decreasing then increasing on $x_{1}\le x_{1}(F)$.

Fig. 4

Partition of $\mathbb {R}^2$ in $H_1$, $H_{2}$ and $H_{0}$

Second step :

We proceed by successive eliminations of weakly dominated strategies. We assume to simplify that $\phi ^{\prime }(x_{1}(F))<0$, i.e., $\phi$ is decreasing on $x_{1}<x_{1}(F)$ as in Fig. 4 (a very similar proof is possible for $\phi ^{\prime }(x_{1}(F))>0$). We consider that $\delta _{1}$ and $\delta _{2}$ are mixed strategies for Candidate 1 and Candidate 2, respectively. With the help of Fig. 4, we find that

– For $\varepsilon _1 >0$, $x_{1}(F)+\varepsilon _1$ dominates all $x_{1}$ with $x_{1}>x_{1}(F)+\varepsilon _1$ (and symmetrically $-x_{1}(F)-\varepsilon _1$ dominates all $x_{1}$ with $x_{1}<-x_{1}(F)-\varepsilon _1$). Thus we can restrain the study to $\delta _{1}(\varepsilon _1)$ with $\text{ supp }(\delta _{1}(\varepsilon _1))\subset [-x_{1}(F)-\varepsilon _1 ,x_{1}(F)+\varepsilon _1 ]$.

– Moreover, $x_{2}(F)$ dominates all $x_{2}$ with $x_{2}>x_{2}(F)$ (and symmetrically $-x_{2}(F)$ dominates all $x_{2}$ with $x_{2}<-x_{2}(F)$). Thus we can restrain the study to $\delta _{2}$ with $\text{ supp }(\delta _{2})\subset [-x_{2}(F),x_{2}(F)]$.

Third step :

The objective is to find a set of mixed strategy equilibria $(\delta _{1}^{* }(\varepsilon _1),\delta _{2}^{* }(\varepsilon _2))$, where $\varepsilon _1>0$ and $\varepsilon _2\ge 0$, and with $\text{ supp }(\delta _{1}^{* }(\varepsilon _1))\subset [-x_{1}(F)-\varepsilon _1 ,x_{1}(F)+\varepsilon _1 ]$ and $\text{ supp }(\delta _{2}^{* }(\varepsilon _2))\subset [-x_{2}(F),x_{2}(F)]$. Since $\{F\}=D_{2}\cap \Delta _{2}$, we have $x_{1}(F)=\frac{K}{\lambda _{1}} -v^{*}$, and $x_{2}(F)=K-v^{*}$. We set $d=2\left( \frac{\lambda _{1}-1}{\lambda _{1}}\right) v^{*}$ and $D=2(\lambda _{1}-1)v^{*}=\lambda _{1}d$. By assumption, $v^{*}\le \frac{K}{\lambda _{1}}$ thus $x_{1}(F)\ge 0$ and $x_{2}(F)>0$. Let k be the highest integer ($k\ge 0$) such that $x_{1}(F)-kd\ge 0$. Then: $x_{2}(F)-kD=K-v^{*}-kD=\lambda _{1}\left( \frac{K}{\lambda _{1}} -v^{*}\right) +(\lambda _{1}-1)v^{*}-k\lambda _{1}d$, i.e., $x_{2}(F)-kD=\lambda _{1}\left[ x_{1}(F)+\frac{d}{2}-kd\right]$. We distinguish two cases:

First case :

  $x_{1}(F)-kd\in \left[ 0,\frac{d}{2}\right)$. Then $x_{1}(F)+\frac{d}{2}-kd\in \left[ \frac{d}{2},d\right)$, thus $x_{2}(F)-kD\in \left[ \frac{D}{2},D\right)$, and for $\varepsilon _1 >0$ small enough, $x_{1}(F)+\varepsilon _1 -kd\in \left( 0,\frac{ d}{2}\right)$.

Second case :

 $x_{1}(F)-kd\in \left[ \frac{d}{2},d\right)$. Then $x_{1}(F)+\frac{d}{2}-kd\in \left[ d,\frac{3d}{2}\right)$, thus $x_{2}(F)-kD\in \left[ D, \frac{3D}{2}\right)$, and for $\varepsilon _1 >0$ small enough, $x_{1}(F)+\varepsilon _1 -kd\in \left( \frac{d }{2},d\right)$.

To show the existence of mixed strategy equilibria, it is sufficient to prove Lemma 4.

Lemma 4

• (i) (First case) If $x_{1}(F)-kd\in \left[ 0,\frac{d}{2} \right)$, let $\delta _{1}^{*}(\varepsilon _{1})$ and $\delta _{2}^{* }(\varepsilon _{2})$ be the mixed strategies such that:

  $\text{ supp }(\delta _{1}^{* }(\varepsilon _1))=\{x_{1}(F)+\varepsilon _1 -\ell d;0\le \ell \le k\} \cup \{-\left( x_{1}(F)+\varepsilon _1 -\ell d\right) ;0\le \ell \le k\}$

  $\text{ supp }(\delta _{2}^{* }(\varepsilon _{2}))=\{x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D;0\le \ell \le k\} \cup \{-\left( x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D\right) ;0\le \ell \le k\}$

  where for any j ($j=1,2$) each element of $\text{ supp }(\delta _{j}^{*} (\varepsilon _{j}))$ is played with the same probability $\frac{1}{2k+2}$. For $\varepsilon _{1}>0$, $\varepsilon _{2}\ge 0$ such that $\varepsilon _{1}\le \varepsilon _{1}^{\max }=(k+\frac{1}{2})d-x_{1}(F)$ and $\varepsilon _{2}\le \varepsilon _{2}^{\max }=\frac{x_{2}(F)}{\lambda _{1}}-(k+\frac{1}{2})d$, then $(\delta _{1}^{*}(\varepsilon _{1}),\delta _{2}^{*}(\varepsilon _{2}))$ is a mixed equilibrium for the game played by Candidates 1 and 2.

• (ii) (Second case) If $x_{1}(F)-kd\in \left[ \frac{d}{2} ,d\right)$, let $\delta _{1}^{* }(\varepsilon _{1})$ and $\delta _{2}^{*}(\varepsilon _{2})$ be the mixed strategies such that:

  $\text{ supp }(\delta _{1}^{*}(\varepsilon _{1}))=\{0\} \cup \{x_{1}(F)+\varepsilon _{1}-\ell d;0\le \ell \le k\} \cup \{-\left( x_{1}(F)+\varepsilon _{1}-\ell d\right) ;0\le \ell \le k\}$

  $\text{ supp }(\delta _{2}^{*}(\varepsilon _{2}))=\{0\} \cup \{x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D;0\le \ell \le k\} \cup \{-\left( x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D\right) ;0\le \ell \le k\}$

  where for any j ($j=1,2$) each element of $\text{ supp }(\delta _{j}^{*}(\varepsilon _{j}))$ is played with the same probability $\frac{1}{2k+3}$. For $\varepsilon _{1}>0$, $\varepsilon _{2}\ge 0$ such that $\varepsilon _{1}\le \varepsilon _{1}^{\max }=(k+1)d-x_{1}(F)$ and $\varepsilon _{2}\le \varepsilon _{2}^{\max }=\frac{x_{2}(F)}{\lambda _{1}}-(k+1)d$, then $(\delta _{1}^{* }(\varepsilon _{1}),\delta _{2}^{* }(\varepsilon _{2}))$ is a mixed equilibrium for the game played by Candidates 1 and 2.

Proof of Lemma 4

• (i) To prove the first case of Lemma 4, we just have to show that

  $$\begin{aligned} (I)\ \left\{ \begin{array}{ll} \pi _{1}(x_{1},\delta _{2}^{*}(\varepsilon _{2}))\le \frac{1}{2k+2} , &{}\quad \forall x_{1}\in [-(x_{1}(F)+\varepsilon _{1}),x_{1}(F)+\varepsilon _{1}] \\ \pi _{1}(x_{1},\delta _{2}^{*}(\varepsilon _{2}))=\frac{1}{2k+2}, &{}\quad \forall x_{1}\in \text{ supp }(\delta _{1}^{*}(\varepsilon _{1})) \end{array}\right. \end{aligned}$$

  and

  $$\begin{aligned} (II)\ \left\{ \begin{array}{ll} \pi _{2}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})\le \frac{2k+1}{2k+2} ,&{}\quad \forall x_{2}\in [-x_{2}(F),x_{2}(F)] \\ \pi _{2}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})=\frac{2k+1}{2k+2} ,&{}\quad \forall x_{2}\in \text{ supp }(\delta _{2}^{*}(\varepsilon _{2})) \end{array}\right. \end{aligned}$$

  where $\pi _{2}=1-\pi _{1}$.

  Note that for $(x_{1},x_{2})\in [-(x_{1}(F)+\varepsilon _{1}),x_{1}(F)+\varepsilon _{1}]\times [-x_{2}(F),x_{2}(F)]$, we have $\pi _{1}(x_{1},x_{2})=1\Leftrightarrow -\frac{D}{2}<\lambda _{1}x_{1}-x_{2}< \frac{D}{2}$ and $\pi _{1}(x_{1},x_{2})=\frac{1}{2}\Leftrightarrow \lambda _{1}x_{1}-x_{2}=\pm \frac{D}{2}$.

  • $*$We prove first (I).

    $\pi _{1}(x_{1},\delta _{2}^{*}(\varepsilon _{2}))=\frac{1}{2k+2} \sum _{\ell =0}^{k}[ \pi _{1}(x_{1},x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D)+\pi _{1}(x_{1},-( x_{2}(F)$ $-\lambda _{1}\varepsilon _{2}-\ell D) )]$

    where $\pi _{1}(x_{1},x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D)=1\Leftrightarrow -\frac{D}{2}<\lambda _{1}x_{1}-( x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D) <\frac{D}{2}$. Since $x_{2}(F)=\lambda _{1}x_{1}(F)+\frac{D}{2}$, we have:

    $\pi _{1}(x_{1},x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D)=1\Leftrightarrow -\frac{D}{2}<\lambda _{1}x_{1}-\lambda _{1}x_{1}(F)- \frac{D}{2}+\lambda _{1}\varepsilon _{2}+\ell D<\frac{D}{2}$

    $\pi _{1}(x_{1},x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D)=1\Leftrightarrow \lambda _{1}x_{1}(F)-\lambda _{1}\varepsilon _{2}-\ell D<\lambda _{1}x_{1}<\lambda _{1}x_{1}(F)-\lambda _{1}\varepsilon _{2}-(\ell -1)D$

    i.e., $\pi _{1}(x_{1},x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D)=1\Leftrightarrow x_{1}(F)-\varepsilon _{2}-\ell d<x_{1}<x_{1}(F)-\varepsilon _{2}-(\ell -1)d$

    Moreover, $\pi _{1}(x_{1},x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D)= \frac{1}{2}\Leftrightarrow x_{1}\in \{x_{1}(F)-\varepsilon _{2}-\ell d;x_{1}(F)-\varepsilon _{2}-(\ell -1)d\}$

    Similarly,

    $\pi _{1}(x_{1},-\left( x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D\right) )=1\Leftrightarrow -\frac{D}{2}<\lambda _{1}x_{1}+( x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D) <\frac{D}{2}$

    $\pi _{1}(x_{1},-\left( x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D\right) )=1\Leftrightarrow -\frac{D}{2}<\lambda _{1}x_{1}+\lambda _{1}x_{1}(F)+\frac{D}{2}-\lambda _{1}\varepsilon _{2}-\ell D<\frac{D}{2}$

    $\pi _{1}(x_{1},-\left( x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D\right) )=1\Leftrightarrow -\lambda _{1}x_{1}(F)+\lambda _{1}\varepsilon _{2}+(\ell -1)D<\lambda _{1}x_{1}<-\lambda _{1}x_{1}(F)+\lambda _{1}\varepsilon _{2}+\ell D$

    i.e.,

    $\pi _{1}(x_{1},-\left( x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D\right) )=1\Leftrightarrow -x_{1}(F)+\varepsilon _{2}+(\ell -1)d<x_{1}<-x_{1}(F)+\varepsilon _{2}+\ell d$

    Moreover

    $\pi _{1}(x_{1},-\left( x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D\right) )=\frac{1}{2}\Leftrightarrow x_{1}\in \{-x_{1}(F)+\varepsilon _{2}+(\ell -1)d,-x_{1}(F)+\varepsilon _{2}+\ell d\}$

    Notation: For all real a, b with $a<b$, we denote by $\chi _{[ a,b]}$ the function from $\mathbb {R}$ to $\mathbb {R}$, such that

    $\chi _{[ a,b]}(x)=1$ if $a<x<b$, and $\chi _{[ a,b]}(x)=\frac{1}{2}$ if $x=a$ or $x=b$, and $\chi _{[ a,b]}(x)=0$ if $x\notin [ a,b]$.

    Taking into account the last equivalences, we have

    $$\begin{aligned} \pi _{1}(x_{1},\delta _{2}^{*}(\varepsilon _{2}))= & {} \frac{1}{2k+2} \sum _{\ell =0}^{k}\left[ \chi _{[x_{1}(F)-\varepsilon _{2}-\ell d,x_{1}(F)-\varepsilon _{2}-(\ell -1)d]}(x_{1})\right. \\&+\left. \chi _{[-x_{1}(F)+\varepsilon _{2}+(\ell -1)d,-x_{1}(F) +\varepsilon _{2}+\ell d]}(x_{1})\right] \end{aligned}$$

    So, $\pi _{1}(x_{1},\delta _{2}^{*}(\varepsilon _{2}))=\frac{1}{2k+2}[ \chi _{[x_{1}(F)-\varepsilon _{2}-kd, x_{1}(F)-\varepsilon _{2}+d]}(x_{1})$$+\chi _{[-x_{1}(F)+\varepsilon _{2}-d, -x_{1}(F)+\varepsilon _{2}+kd]}(x_{1})]$

    where $x_{1}(F)-kd\in [0,\frac{d}{2})$ as we are in the first case. Since $0\le \varepsilon _{2}\le \frac{x_{2}(F)}{\lambda _{1}}-(k+\frac{1}{2})d$ (and $x_{2}(F)=\lambda _{1}x_{1}(F)+\frac{D}{2}$), then $x_{1}(F)-kd\ge \varepsilon _{2}\ge 0$. This implies that $\pi _{1}(x_{1},\delta _{2}^{*}(\varepsilon _{2}))\le \frac{1}{2k+2}$. This proves the first point of (I) in the first case.

    To prove the second point of (I) in the first case, i.e., $\pi _{1}(x_{1},\delta _{2}^{*}(\varepsilon _{2}))=\frac{1}{2k+2},\ \forall x_{1}\in \text{ supp }(\delta _{1}^{*}(\varepsilon _{1}))$, assume that $x_{1}\in \text{ supp }(\delta _{1}^{*}(\varepsilon _{1}))$.

    • $\bullet$ If $x_{1}=x_{1}(F)+\varepsilon _{1}-\ell d$, with $0\le \ell \le k$,

      $\pi _{1}(x_{1},\delta _{2}^{*}(\varepsilon _{2}))=\frac{1}{ 2k+2}\chi _{[x_{1}(F)-\varepsilon _{2}-kd,x_{1}(F)-\varepsilon _{2}+d]}(x_{1})=\frac{1}{2k+2}$ (for $0<\varepsilon _{1}<d$ and $0\le \varepsilon _{2}<d-\varepsilon _{2}$).

    • $\bullet$ If $x_{1}=-\left( x_{1}(F)+\varepsilon _{1}-\ell d \text{ and } \ 0\le \varepsilon _{2}<d-\varepsilon _{1}\right)$, with $0\le \ell \le k$,

      $\pi _{1}(x_{1},\delta _{2}^{*}(\varepsilon _{2}))=\frac{1}{2k+2}\chi _{[-x_{1}(F)+\varepsilon _{2}-d,-x_{1}(F)+\varepsilon _{2}+kd]}(x_{1})=\frac{1}{2k+2}$ (for $\varepsilon _{1}>0$ and $0\le \varepsilon _{2}\le d-\varepsilon _{1}$ ). This proves the second point of (I) in the first case.

  • $*$Now we prove (II) in the first case. Given that $\pi _{2}=1-\pi _{1}$, (II) in the first case can be rewritten as:

    $$\begin{aligned} (II)\ \left\{ \begin{array}{ll} \pi _{1}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})\ge \frac{1}{2k+2} ,&{}\quad \forall x_{2}\in [-x_{2}(F),x_{2}(F)] \\ \pi _{1}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})=\frac{1}{2k+2}, &{}\quad \forall x_{2}\in \text{ supp }(\delta _{2}^{*}(\varepsilon _{2})) \end{array}\right. \end{aligned}$$

    $\pi _{1}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})=\frac{1}{2k+2} \sum _{\ell =0}^{k}[ \pi _{1}(x_{1}(F)+\varepsilon _{1}-\ell d,x_{2})+\pi _{1}(-( x_{1}(F)$$+\varepsilon _{1}-\ell d) ,x_{2}) ]$ where

    $\pi _{1}(x_{1}(F)+\varepsilon _{1}-\ell d,x_{2})=1\Leftrightarrow -\frac{D}{2}<\lambda _{1}\left( x_{1}(F)+\varepsilon _{1}-\ell d\right) -x_{2}<\frac{D}{2}$

    Since $\lambda _{1}x_{1}(F)=x_{2}(F)-\frac{D}{2}$, we have:

    $\pi _{1}(x_{1}(F)+\varepsilon _{1}-\ell d,x_{2})=1\Leftrightarrow -\frac{D}{2}<x_{2}(F)-\frac{D}{2}+\lambda _{1}\varepsilon _{1}-\ell D-x_{2}<\frac{D}{2}$

    i.e.,

    $\pi _{1}(x_{1}(F)+\varepsilon _{1}-\ell d,x_{2})=1\Leftrightarrow x_{2}(F)+\lambda _{1}\varepsilon _{1}-(\ell +1)D<x_{2}<x_{2}(F)+\lambda _{1}\varepsilon _{1}-\ell D$

    Moreover $\pi _{1}(x_{1}(F)+\varepsilon _{1}-\ell d,x_{2})=\frac{1}{2} \Leftrightarrow x_{2}\in \{x_{2}(F)+\lambda _{1}\varepsilon _{1}-(\ell +1)D,x_{2}(F)+\lambda _{1}\varepsilon _{1}-\ell D\}$. Similarly,

    $$\begin{aligned} \pi _{1}(-\left( x_{1}(F)+\varepsilon _{1}-\ell d\right) ,x_{2})=1\Leftrightarrow & {} -\frac{D}{2}<-\lambda _{1}\left( x_{1}(F)+\varepsilon _{1}-\ell d\right) -x_{2}<\frac{D}{2} \\ \pi _{1}(-\left( x_{1}(F)+\varepsilon _{1}-\ell d\right) ,x_{2})=1\Leftrightarrow & {} -\frac{D}{2}<-x_{2}(F)+\frac{D}{2}-\lambda _{1}\varepsilon _{1}+\ell D-x_{2}<\frac{D}{2} \end{aligned}$$

    i.e.,

    $\pi _{1}(-\left( x_{1}(F)+\varepsilon _{1}-\ell d\right) ,x_{2})=1\Leftrightarrow -x_{2}(F)-\lambda _{1}\varepsilon _{1}+\ell D<x_{2}<-x_{2}(F)-\lambda _{1}\varepsilon _{1}+(\ell +1)D$

    Moreover

    $\pi _{1}(-\left( x_{1}(F)+\varepsilon _{1}-\ell d\right) ,x_{2})=\frac{1}{2} \Leftrightarrow x_{2}\in \{-x_{2}(F)-\lambda _{1}\varepsilon _{1}+\ell D,-x_{2}(F)-\lambda _{1}\varepsilon _{1}+(\ell +1)D\}$

    Taking into account the last equivalences, we have

    $$\begin{aligned} \pi _{1}\left( \delta _{1}^{*}(\varepsilon _{1}),x_{2}\right)= & {} \frac{1}{2k+2} \sum _{\ell =0}^{k}\left[ \right. \chi _{[x_{2}(F)+\lambda _{1}\varepsilon _{1}-(\ell +1)D,x_{2}(F)+\lambda _{1}\varepsilon _{1}-\ell D]}(x_{2})\\&+\chi _{[-x_{2}(F)-\lambda _{1}\varepsilon _{1}+\ell D,-x_{2}(F)-\lambda _{1}\varepsilon _{1}+(\ell +1)D]}(x_{2})\left. \right] \\= & {} \frac{1}{2k+2}\left[ \chi _{[x_{2}(F)+\lambda _{1}\varepsilon _{1}-(k+1)D,x_{2}(F)+\lambda _{1}\varepsilon _{1}]}(x_{2})\right. \\&+\left. \chi _{[-x_{2}(F)-\lambda _{1}\varepsilon _{1},-x_{2}(F)-\lambda _{1}\varepsilon _{1}+(k+1)D]}(x_{2}) \right] \end{aligned}$$

    As we are in the first case, we know that $x_{2}(F)-kD\in \left[ \frac{D}{2} ,D\right)$, i.e., $x_{2}(F)-(k+1)D\in \left[ -\frac{D}{2},0\right)$. Since $0<\varepsilon _{1}\le (k+\frac{1}{2})d-x_{1}(F)$ (and $x_{1}(F)= \frac{x_{2}(F)}{\lambda _{1}}-\frac{d}{2}$), then $0<\lambda _{1}\varepsilon _{1}\le (k+1)D-x_{2}(F)$. This implies that:

    $\pi _{1}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})\ge \frac{1}{2k+2} \left( \chi _{[0,x_{2}(F)+\lambda _{1}\varepsilon _{1}]}(x_{2})+\chi _{[-x_{2}(F)-\lambda _{1}\varepsilon _{1},0]}(x_{2})\right)$, so $\pi _{1}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})\ge \frac{1}{ 2k+2}\chi _{[-x_{2}(F)-\lambda _{1}\varepsilon _{1},x_{2}(F)+\lambda _{1}\varepsilon _{1}]}(x_{2})$.

    Then $\pi _{1}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})\ge \frac{1}{2k+2 }$, $\forall x_{2}\in [-x_{2}(F),x_{2}(F)]$. This proves the first point of (II).

    To prove the second point of (II), assume that $x_{2}\in \text{ supp } (\delta _{2}^{*}(\varepsilon _{2}))$.

    • $\bullet$ If $x_{2}=x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D$, with $0\le \ell \le k$, then

      $\pi _{1}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})=\frac{1}{2k+2}\chi _{[x_{2}(F)+\lambda _{1}\varepsilon _{1}-(k+1)D,x_{2}(F)+\lambda _{1}\varepsilon _{1}]}(x_{2})=\frac{1}{2k+2}$ (for $0<\varepsilon _{1}<d$ and $0\le \varepsilon _{2}<d-\varepsilon _{1}$).

    • $\bullet$ If $x_{2}=-\left( x_{2}(F)-\ell D\right)$, with $0\le \ell \le k$, then

      $\pi _{1}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})=\frac{1}{2k+2}\chi _{[-x_{2}(F)-\lambda _{1}\varepsilon _{1},-x_{2}(F)-\lambda _{1}\varepsilon _{1}+(k+1)D]}(x_{2})=\frac{1}{2k+2}$ (for $0<\varepsilon _{1}<d$ and $0\le \varepsilon _{2}<d-\varepsilon _{1}$).

    This proves the second point of (II).

• (ii) To prove the second case of Lemma 4, we just have to show that

  $$\begin{aligned} (I)\ \left\{ \begin{array}{ll} \pi _{1}\left( x_{1},\delta _{2}^{*}(\varepsilon _{2})\right) \le \frac{1}{2k+3} ,&{}\quad \forall x_{1}\in [-(x_{1}(F)+\varepsilon _{1}),x_{1}(F)+\varepsilon _{1}] \\ \pi _{1}(x_{1},\delta _{2}^{*}(\varepsilon _{2}))=\frac{1}{2k+3}, &{}\quad \forall x_{1}\in \text{ supp }\left( \delta _{1}^{*}(\varepsilon _{1})\right) \end{array}\right. \end{aligned}$$

  and

  $$\begin{aligned} (II)\ \left\{ \begin{array}{ll} \pi _{2}\left( \delta _{1}^{*}(\varepsilon _{1}),x_{2}\right) \le \frac{2k+2}{2k+3} , &{}\quad \forall x_{2}\in [-x_{2}(F),x_{2}(F)] \\ \pi _{2}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})=\frac{2k+2}{2k+3}, &{}\quad \forall x_{2}\in \text{ supp }\left( \delta _{2}^{*}(\varepsilon _{2})\right) \end{array}\right. \end{aligned}$$

  where $\pi _{2}=1-\pi _{1}$.

  • $*$We prove first (I) in the second case.

    $$\begin{aligned} \pi _{1}\left( x_{1},\delta _{2}^{*}(\varepsilon _{2})\right)= & {} \frac{1}{2k+3}\pi _{1}(x_{1},0)+\frac{1}{2k+3}\sum _{\ell =0}^{k}\left[ \pi _{1}(x_{1},x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D)\right. \\&\left. +\pi _{1}(x_{1},-\left( x_{2}(F)-\lambda _{1} \varepsilon _{2}-\ell D\right) ) \right] \end{aligned}$$

    where

    $\sum _{\ell =0}^{k}\left[ \pi _{1}(x_{1},x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D)+\pi _{1}(x_{1},-\left( x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D\right) )\right]$

    $=\chi _{[x_{1}(F)-kd-\varepsilon _{2},x_{1}(F)+d-\varepsilon _{2}]}(x_{1})+\chi _{[-x_{1}(F)-d+\varepsilon _{2},-x_{1}(F)+kd+\varepsilon _{2}]}(x_{1})$ and $\pi _{1}(x_{1},0)=\chi _{[-\frac{d}{2},\frac{d}{2}]}(x_{1})$

    Then $\pi _{1}(x_{1},\delta _{2}^{*}(\varepsilon _{2}))=\frac{1}{2k+3} \left[ \chi _{[x_{1}(F)-\varepsilon _{2}-kd,x_{1}(F)-\varepsilon _{2}+d]}(x_{1})+\chi _{[-\frac{d}{2},\frac{d}{2}]}(x_{1})\right.$

    $\left. +\chi _{[-x_{1}(F)+\varepsilon _{2}-d,-x_{1}(F)+\varepsilon _{2}+kd]}(x_{1}) \right]$

    where $x_{1}(F)-kd\in \left[ \frac{d}{2},d\right)$ as we are in the second case. Since $0\le \varepsilon _{2}\le \frac{x_{2}(F)}{\lambda _{1}} -(k+1)d$ (and $x_{2}(F)=\lambda _{1}x_{1}(F)+\frac{D}{2}$), then $0\le \varepsilon _{2}\le x_{1}(F)-(k+\frac{1}{2})d$. This implies that $\pi _{1}(x_{1},\delta _{2}^{*}(\varepsilon _{2}))\le \frac{1}{2k+3}$. This proves the first point of (I).

    To prove the second point of (I), assume that $x_{1}\in \text{ supp }(\delta _{1}^{*}(\varepsilon _{1}))$.

    • $\bullet$ If $x_{1}=x_{1}(F)+\varepsilon _{1}-\ell d$, with $0\le \ell \le k$, then $\pi _{1}(x_{1},\delta _{2}^{*}(\varepsilon _{2}))= \frac{1}{2k+3}\chi _{[x_{1}(F)-\varepsilon _{2}-kd,x_{1}(F) -\varepsilon _{2}+d]}(x_{1})=\frac{1}{2k+3}$ (for $\varepsilon _{1}>0$ and $0\le \varepsilon _{2}\le d-\varepsilon _{1}$).

    • $\bullet$ If $x_{1}=-\left( x_{1}(F)+\varepsilon _{1}-\ell d\right)$, with $0\le \ell \le k$, then

      $\pi _{1}(x_{1},\delta _{2}^{*}(\varepsilon _{2}))=\frac{1}{2k+3}\chi _{[-x_{1}(F)+\varepsilon _{2}-d,-x_{1}(F)+\varepsilon _{2}+kd]}(x_{1})=\frac{1}{2k+3}$ (for $\varepsilon _{1}>0$ and $0\le \varepsilon _{2}\le d-\varepsilon _{1}$).

    • $\bullet$ If $x_{1}=0$, then $\Pi _{1}(x_{1},\delta _{2}^{*}(\varepsilon _{2}))=\frac{1}{2k+3}\chi _{[-\frac{d}{2},\frac{d}{2} ]}(x_{1})=\frac{1}{2k+3}$

    This proves the second point of (I).

  • $*$Now we prove (II) in the second case. Given that $\pi _{2}=1-\pi _{1}$, (II) in the second case can be rewritten as:

    $$\begin{aligned}&(II)\ \left\{ \begin{array}{ll} \pi _{1}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})\ge \frac{1}{2k+3} ,&{}\quad \forall x_{2}\in [-x_{2}(F),x_{2}(F)] \\ \pi _{1}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})=\frac{1}{2k+3}, &{}\quad \forall x_{2}\in \text{ supp }(\delta _{2}^{*}(\varepsilon _{2})) \end{array}\right. \\&\pi _{1}(\delta _{1}^{*}(\varepsilon _{1}),x_{2}) =\frac{1}{2k+3}\pi _{1}(0,x_{2})\\&\qquad \qquad \qquad \qquad \quad +\frac{1}{2k+3}\sum _{\ell =0}^{k}[ \pi _{1}(x_{1}(F)+\varepsilon _{1}-\ell d,x_{2})\\&\qquad \qquad \qquad \qquad \qquad +\pi _{1}(-\left( x_{1}(F)+\varepsilon _{1}-\ell d\right) ,x_{2})] \end{aligned}$$

    where $\pi _{1}(x_{1}(F)+\varepsilon _{1}-\ell d,x_{2})+\pi _{1}(-\left( x_{1}(F)+\varepsilon _{1}-\ell d\right) ,x_{2})$

    $=\chi _{[x_{2}(F)+\lambda _{1}\varepsilon _{1}-(k+1)D,x_{2}(F)+\lambda _{1}\varepsilon _{1}]}(x_{2})+\chi _{[-x_{2}(F)-\lambda _{1}\varepsilon _{1},-x_{2}(F)-\lambda _{1}\varepsilon _{1}+(k+1)D]}(x_{2})$

    and $\pi _{1}(0,x_{2})=\chi _{[-\frac{D}{2},\frac{D}{2}]}(x_{2})$. Then

    $$\begin{aligned} \pi _{1}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})= & {} \frac{1}{2k+3}\Big [ {} \chi _{[x_{2}(F)+\lambda _{1}\varepsilon _{1}-(k+1)D,x_{2}(F)+\lambda _{1}\varepsilon _{1}]}(x_{2})\\&+\chi _{[-\frac{D}{2},\frac{D}{2}]}(x_{2})+\chi _{[-x_{2}(F)-\lambda _{1}\varepsilon _{1},-x_{2}(F)-\lambda _{1}\varepsilon _{1}+(k+1)D]}(x_{2})\Big ] \end{aligned}$$

    As we are in the second case, we know that $x_{2}(F)-kD\in [D,\frac{3D }{2})$. Since $0<\varepsilon _{1}\le (k+1)d-x_{1}(F)$ (and $x_{1}(F)= \frac{x_{2}(F)}{\lambda _{1}}-\frac{d}{2}$), then $0<\lambda _{1}\varepsilon _{1}\le (k+\frac{3}{2})D-x_{2}(F)$. This implies that $\pi _{1}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})\ge \frac{1}{2k+3} \chi _{[-x_{2}(F)-\lambda _{1}\varepsilon _{1},x_{2}(F)+\lambda _{1}\varepsilon _{1}]}(x_{2})$.

    Then $\pi _{1}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})\ge \frac{1}{2k+3 }$, $\forall x_{2}\in [-x_{2}(F),x_{2}(F)]$. This proves the first point of (II).

    To prove the second point of (II), assume that $x_{2}\in \text{ supp } (\delta _{2}^{* }(\varepsilon _{2}))$.

    • $\bullet$ If $x_{2}=x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D$, with $0\le \ell \le k$, then

      $\pi _{1}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})=\frac{1}{2k+3}\chi _{[x_{2}(F)+\lambda _{1}\varepsilon _{1}-(k+1)D,x_{2}(F)+\lambda _{1}\varepsilon _{1}]}(x_{2})=\frac{1}{2k+3}$ (for $\varepsilon _{1}>0$ and $0\le \varepsilon _{2}<d-\varepsilon _{1}$).

    • $\bullet$ If $x_{2}=-\left( x_{2}(F)-\lambda _{1}\varepsilon _{2}-\ell D\right)$, with $0\le \ell \le k$, then

      $\pi _{1}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})=\frac{1}{2k+3}\chi _{[-x_{2}(F)-\lambda _{1}\varepsilon _{1},-x_{2}(F)-\lambda _{1}\varepsilon _{1}+(k+1)D]}(x_{2})=\frac{1}{2k+3}$ (for $\varepsilon _{1}>0$ and $0\le \varepsilon _{2}<d-\varepsilon _{1}$ ).

    • $\bullet$ If $x_{2}=0$, then $\pi _{1}(\delta _{1}^{*}(\varepsilon _{1}),x_{2})=\frac{1}{2k+3}\chi _{[-\frac{D}{2},\frac{D }{2}]}(x_{2})=\frac{1}{2k+3}$

    This proves the second point of (II). $\square$ (end of the Proof of Lemma 4)

To finish the proof of Proposition 6, it remains to show that r is a non decreasing function of $\sigma$ on $\sigma \in [ \sigma ^{* },\sigma ^{** })$, with $r=2$ for $\sigma =\sigma ^{* }$, and $\lim _{ \sigma \rightarrow {\sigma ^{** }}^{-}}r=+\infty$.

According to the definition of r in Proposition 6, and Lemma 4 which provides the probabilities by which each element of $\text{ supp }(\delta _{j}^{* }(\varepsilon _j))$ is played in the first and second cases, we have $r=2k+2$ in the first case, and $r=2k+3$ in the second case. We know that k is the integer part of $\frac{x_{1}(F)}{d}$, thus in both cases r is the integer part of $\frac{2x_{1}(F)}{d}+2$. We have $\frac{x_{1}(F)}{d}=\frac{1}{2(\lambda _{1}-1)}\left[ \frac{K}{v^{*}} -\lambda _{1}\right]$ where $v^{*}$ is a decreasing function of $\sigma$, thus $\frac{x_{1}(F)}{d}$ is an increasing function of $\sigma$ on $\sigma \in [\sigma ^{*},\sigma ^{** })$. This implies that k and r are non decreasing functions of $\sigma$ on $\sigma \in [ \sigma ^{*},\sigma ^{** })$.

Moreover, for $\sigma =\sigma ^{*}$, we hav e $v^{*}=\frac{K}{\lambda _{1}}$ thus $\frac{x_{1}(F)}{d}=0$ and $k=0$ which gives $r=2$.

$\lim _{\sigma \rightarrow \sigma ^{**-}}v^{*}=0$ thus $\lim _{\sigma \rightarrow \sigma ^{**-}}\frac{x_{1}(F)}{d}=+\infty$, which gives $\lim _{\sigma \rightarrow \sigma ^{**-}}k=+\infty$ and $\lim _{\sigma \rightarrow \sigma ^{**-}}r=+\infty$. $\square$

1.9 A.9 Proof of Proposition 7

The objective function of Candidate 1 is $\max _{x_1}S_1(x_1,x_2)$, while the one of Candidate 2 is $\min _{x_2}S_1(x_1,x_2)$. To show that there is no pure strategy Nash equlibrium when candidates maximize vote share, we proceed in three steps. First, we find Candidate 2’s best response to an arbitrary strategy by Candidate 1. Second, we find Candidate 1’s best response to an arbitrary strategy by Candidate 2. Third, we show that the two best responses do not intersect in a $(x_{1},x_{2})$ space.

• First step: Candidate 2’s best response  Consider Cases (A), (B) and (B’) of Proposition 1.

Case (A) :

If $x_{2}\in \left[ x_{1}-\frac{(\lambda _{1}-1)K}{\lambda _{1}},x_{1}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right]$, then $S_{1}(x_{1},x_{2})=\int _{\frac{\lambda _{1}x_{1}-x_{2}}{\lambda _{1}-1}-K}^{ \frac{\lambda _{1}x_{1}-x_{2}}{\lambda _{1}-1}+K}f_{\sigma }(a)da$. We have $\frac{\partial S_1}{\partial x_2}=-\frac{1}{\lambda _1-1}f_\sigma \left( \frac{\lambda _1x_1-x_2}{\lambda _1-1}+K\right) +\frac{1}{\lambda _1-1}f_\sigma \left( \frac{\lambda _1x_1-x_2}{\lambda _1-1}-K\right)$. Given that $f_\sigma$ is an even function strictly increasing on $\mathbb {R}_{-}$, and strictly decreasing on $\mathbb {R}_{+}$, then $\frac{1}{\lambda _1-1}f_\sigma \left( \frac{\lambda _1x_1-x_2}{\lambda _1-1} -K\right) \ge \frac{1}{\lambda _1-1}f_\sigma \left( \frac{\lambda _1x_1-x_2}{\lambda _1 -1}+K\right)$ if $\frac{\lambda _1x_1-x_2}{\lambda _1-1}\ge 0$. That is $\frac{\partial S_1}{\partial x_2}\le 0\Leftrightarrow \lambda _1x_1\le x_2$.

Now, remark:

$$\begin{aligned} \lambda _1x_1>x_{1}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\Leftrightarrow & {} x_1>\frac{K}{\lambda _1}\\ \lambda _1x_1<x_{1}-\frac{(\lambda _{1}-1)K}{\lambda _{1}}\Leftrightarrow & {} x_1<-\frac{K}{\lambda _1} \end{aligned}$$

Hence, there are three subcases in Case (A).

–If $x_1<-\frac{K}{\lambda _1}$, then $\lambda _1x_1<x_{1}-\frac{(\lambda _{1}-1)K}{\lambda _{1}}$. Given that $x_{2}\in \left[ x_{1}-\frac{(\lambda _{1}-1)K}{\lambda _{1}},x_{1}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right]$ in Case (A), if $x_1<-\frac{K}{\lambda _1}$, then $x_2\ge \lambda _1x_1$, and $x_2\mapsto S_1(x_1,x_2)$ is decreasing on $\left[ x_{1}-\frac{(\lambda _{1}-1)K}{\lambda _{1}},x_{1}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right]$. Hence, if $x_1<-\frac{K}{\lambda _1}$, Candidate 2 plays $x_2=x_{1}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}$.

– If $x_1\in \left[ -\frac{K}{\lambda _1},\frac{K}{\lambda _1}\right]$, then $\lambda _1x_1\in \left[ x_{1}-\frac{(\lambda _{1}-1)K}{\lambda _{1}},x_{1}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right]$. Given that $\frac{\partial S_1}{\partial x_2}\le 0\Leftrightarrow \lambda _1x_1\le x_2$, then $x_2\mapsto S_1(x_1,x_2)$ is increasing on $\left[ x_{1}-\frac{(\lambda _{1} -1)K}{\lambda _{1}},\lambda _1x_1\right]$ and decreasing on $\left[ \lambda _1x_1, x_{1}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right]$. Thus, Candidate 2 plays $x_2=x_{1}-\frac{(\lambda _{1}-1)K}{\lambda _{1}}$ if $S_1\left( x_1,x_{1}-\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right) <S_1\left( x_1,x_{1}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right)$; and Candidate 2 plays $x_2=x_{1}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}$ if $S_1\left( x_1,x_{1}-\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right) >S_1\left( x_1,x_{1}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right)$. Recalling that $S_1\left( x_1,x_{1}-\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right) =\int _{\frac{\lambda _1x_1-x_1+\frac{(\lambda _1-1)K}{\lambda _1}}{\lambda _1 -1}-K}^{\frac{\lambda _1x_1-x_1+\frac{(\lambda _1-1)K}{\lambda _1}}{\lambda _1-1}+K} f_\sigma = \int _{x_1+\frac{K}{\lambda _1}-K}^{x_1+\frac{K}{\lambda _1}+K}f_\sigma =g\left( x_1+\frac{K}{\lambda _1}\right)$ (see the Proof of Part (i) of Proposition 3 in “Appendix A.5”), and $S_1\left( x_1,x_{1}+\frac{(\lambda _{1} -1)K}{\lambda _{1}}\right) =\int _{\frac{\lambda _1x_1-x_1-\frac{(\lambda _1-1)K}{\lambda _1}}{\lambda _1-1}-K}^{\frac{\lambda _1x_1-x_1-\frac{(\lambda _1-1)K}{\lambda _1}}{\lambda _1-1}+K}f_\sigma = \int _{x_1-\frac{K}{\lambda _1}-K}^{x_1-\frac{K}{\lambda _1}+K} f_\sigma =g\left( x_1-\frac{K}{\lambda _1}\right)$, and given that g is an even function, strictly increasing on $\mathbb {R}_{-}$, and strictly decreasing on $\mathbb {R}_{+}$, then $g\left( x_1+\frac{K}{\lambda _1}\right) \ge g\left( x_1-\frac{K}{\lambda _1}\right) \Leftrightarrow x_1\le 0$. If so, then for $x_1\le 0$, Candidate 2 plays $x_2=x_{1}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}$, and for $x_1\ge 0$, Candidate 2 plays $x_2=x_{1}-\frac{(\lambda _{1}-1)K}{\lambda _{1}}$.

– If $x_1>\frac{K}{\lambda _1}$, then $\lambda _1x_1>x_{1} +\frac{(\lambda _{1}-1)K}{\lambda _{1}}$. Given that $x_{2}\in \left[ x_{1}-\frac{(\lambda _{1}-1)K}{\lambda _{1}},x_{1}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right]$ in Case (A), if $x_1>\frac{K}{\lambda _1}$, then $x_2<\lambda _1x_1$, and $x_2\mapsto S_1(x_1,x_2)$ is increasing on $\left[ x_{1}-\frac{(\lambda _{1}-1)K}{\lambda _{1}},x_{1}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right]$. Hence, if $x_1>\frac{K}{\lambda _1}$, Candidate 2 plays $x_2=x_{1}-\frac{(\lambda _{1}-1)K}{\lambda _{1}}$. To sum up, we have shown that in Case (A), for a given $x_1$, the best response of Candidate 2 is:

$$\begin{aligned} x_2=\left\{ \begin{array}{ll} x_{1}+\frac{(\lambda _{1}-1)K}{\lambda _{1}} &{} \quad \text{ if } x_1\le 0 \\ x_1-\frac{(\lambda _{1}-1)K}{\lambda _{1}} &{} \quad \text{ if } x_1\ge 0 \end{array} \right. \end{aligned}$$

Case (B) :

If $x_{2}\ge x_{1}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}$, then $S_{1}(x_{1},x_{2})=\int _{\frac{\lambda _{1}x_{1}-x_{2}}{\lambda _{1}-1}-K}^{\frac{(\lambda _{1}-1)K+\lambda _{1}x_{1}+x_{2}}{\lambda _{1}+1} }f_{\sigma }(a)da$. If so, $\frac{\partial S_1}{\partial x_2}=\frac{1}{\lambda _1+1}f_\sigma \left( \frac{(\lambda _{1}-1)K+\lambda _{1}x_{1}+x_{2}}{\lambda _{1}+1}\right) +\frac{1}{\lambda _1-1}f_\sigma \left( \frac{\lambda _{1}x_{1}-x_{2}}{\lambda _{1}-1}-K\right) >0$. Thus, if $x_{2}>x_{1}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}$, Candidate 2 plays $x_{2}=x_{1}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}$.

Case (B’) :

If $x_{2}\le x_{1}-\left( \frac{\lambda _{1}-1 }{\lambda _{1}}\right) K$, then $S_{1}(x_{1},x_{2}) =\int ^{\frac{\lambda _{1}x_{1}-x_{2}}{\lambda _{1}-1}+K}_{\frac{-(\lambda _{1}-1)K+\lambda _{1}x_{1}+x_{2}}{\lambda _{1}+1} }f_{\sigma }(a)da$. If so, $\frac{\partial S_1}{\partial x_2}<0$. Thus, if $x_{2}<x_{1}-\frac{(\lambda _{1}-1)K}{\lambda _{1}}$, Candidate 2 plays $x_{2}=x_{1}-\frac{(\lambda _{1}-1)K}{\lambda _{1}}$.

Conclusion :

 To sum up, we have shown that in Cases (A), (B) and (B’, for a given $x_1$, Candidate 2’s best response is:

$$\begin{aligned} x_2=\left\{ \begin{array}{ll} x_{1}+\frac{(\lambda _{1}-1)K}{\lambda _{1}} &{}\quad \text{ if } x_1\le 0 \\ x_1-\frac{(\lambda _{1}-1)K}{\lambda _{1}} &{}\quad \text{ if } x_1\ge 0 \end{array} \right. \end{aligned}$$

(A12)

Figure 5 depicts Candidate 2’s best response in the $(x_1,x_2)$ space (the two blue solid lines). Remark that Candidate 2’s best response is inside Case (A).

Fig. 5

Candidate 2’s best response in Cases (A), (B) and (B’), and Candidate 1’s best response in Case (A)

• Second step: Candidate 1’s best response Consider Case (A) of Proposition 1. If $x_{1}\in \left[ x_{2}-\frac{(\lambda _{1}-1)K}{\lambda _{1}},x_{2}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right]$, then $S_{1}(x_{1},x_{2})=\int _{\frac{\lambda _{1}x_{1}-x_{2}}{\lambda _{1}-1}-K}^{ \frac{\lambda _{1}x_{1}-x_{2}}{\lambda _{1}-1}+K}f_{\sigma }(a)da$. We have $\frac{\partial S_1}{\partial x_1}=\frac{1}{\lambda _1-1}f_\sigma \left( \frac{\lambda _1x_1-x_2}{\lambda _1-1}+K\right) -\frac{1}{\lambda _1-1}f_\sigma \left( \frac{\lambda _1x_1-x_2}{\lambda _1-1}-K\right)$. Given that $f_\sigma$ is an even function strictly increasing on $\mathbb {R}_{-}$, and strictly decreasing on $\mathbb {R}_{+}$, then $\frac{1}{\lambda _1-1}f_\sigma \left( \frac{\lambda _1x_1-x_2}{\lambda _1-1}+K\right) \ge \frac{1}{\lambda _1-1}f_\sigma \left( \frac{\lambda _1x_1-x_2}{\lambda _1-1}-K\right)$ if $\frac{\lambda _1x_1-x_2}{\lambda _1-1}\le 0$. That is $\frac{\partial S_1}{\partial x_1}\ge 0\Leftrightarrow x_1\le \frac{x_2}{\lambda _1}$. Now, remark:

  $$\begin{aligned} \frac{x_2}{\lambda _1}>x_{2}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\Leftrightarrow & {} x_2<-K\\ \frac{x_2}{\lambda _1}<x_{2}-\frac{(\lambda _{1}-1)K}{\lambda _{1}}\Leftrightarrow & {} x_2>K \end{aligned}$$

  Hence, there are three subcases in Case (A). – If $x_2<-K$, then $\frac{x_2}{\lambda _1}>x_{2}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}$. Given that $x_{1}\in \left[ x_{2}-\frac{(\lambda _{1}-1)K}{\lambda _{1}},x_{2}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right]$ in Case (A), if $x_2<-K$, then $\frac{x_2}{\lambda _1}\ge x_1$, and $x_1 \mapsto S_1(x_1,x_2)$ is increasing on $\left[ x_{2}-\frac{(\lambda _{1}-1)K}{\lambda _{1}},x_{2}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right]$. Hence, if $x_2<-K$, Candidate 1 plays $x_1=x_{2}+\frac{(\lambda _{1} -1)K}{\lambda _{1}}$. – If $x_2\in \left[ -K,K\right]$, then $\frac{x_2}{\lambda _1}\in \left[ x_{2}-\frac{(\lambda _{1}-1)K}{\lambda _{1}},x_{2}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right]$. Given that $\frac{\partial S_1}{\partial x_1}\ge 0\Leftrightarrow x_1\le \frac{x_2}{\lambda _1}$, then $x_1\mapsto S_1(x_1,x_2)$ is increasing on $\left[ x_{2}-\frac{(\lambda _{1}-1)K}{\lambda _{1}},\frac{x_2}{\lambda _1}\right]$ and decreasing on $\left[ \frac{x_2}{\lambda _1},x_{2}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right]$. Thus, if $x_2\in \left[ -K,K\right]$, Candidate 1 plays $x_1 =\frac{x_{2}}{\lambda _1}$. – If $x_2>K$, then $\frac{x_2}{\lambda _1}<x_{2}-\frac{(\lambda _{1}-1)K}{\lambda _{1}}$. Given that $x_{1}\in \left[ x_{2}-\frac{(\lambda _{1}-1)K}{\lambda _{1}},x_{2}+\frac{(\lambda _{1}-1)K}{\lambda _{1}}\right]$ in Case (A), if $x_2>K$, then $x_1>\frac{x_2}{\lambda _1}$, and $x_1\mapsto S_1(x_1,x_2)$ is decreasing on $\left[ x_{2} -\frac{(\lambda _{1}-1)K}{\lambda _{1}},x_{2}+\frac{(\lambda _{1} -1)K}{\lambda _{1}}\right]$. Hence, if $x_2>K$, Candidate 1 plays $x_1=x_{2}-\frac{(\lambda _{1}-1)K}{\lambda _{1}}$.

To sum up, we have shown that in Case (A), for a given $x_2$, the best response function of Candidate 1 is:

$$\begin{aligned} x_1=\left\{ \begin{array}{ll} x_{2}+\frac{(\lambda _{1}-1)K}{\lambda _{1}} &{}\quad \text{ if } x_2<-K \\ \frac{x_2}{\lambda _1} &{}\quad \text{ if } x_2\in [-K,K]\\ x_2-\frac{(\lambda _{1}-1)K}{\lambda _{1}} &{} \quad \text{ if } x_2>K \end{array} \right. \end{aligned}$$

(A13)

Now, for a given $x_{2}$, if $x_{1}$ is a best response of Candidate 1, then:

• Either $\left| x_{2}-x_{1}\right| >\frac{(\lambda _{1}-1)K}{ \lambda _{1}}$, i.e., $(x_{1},x_{2})$ is in the interior of Case (B) or Case (B’).

• Either $\left| x_{2}-x_{1}\right| <\frac{(\lambda _{1}-1)K}{ \lambda _{1}}$, i.e., $(x_{1},x_{2})$ is in the interior of Case (A). It satisfies then Eq. (A13).

• Either $\left| x_{2}-x_{1}\right| =\frac{(\lambda _{1}-1)K}{ \lambda _{1}}$, at the boundary of Cases (A) and (B) (or (A) and (B’)). But $S_{1}(x_{1},x_{2})=\max _{\overline{x}_{1}\in \mathbb {R}}S_{1}( \overline{x}_{1},x_{2})\ge \max _{\overline{x}_{1};(\overline{x} _{1},x_{2})\in \text{ Case } (\text{ A })}S_{1}(\overline{x}_{1},x_{2})$, i.e., $x_{1}$ is also a best response to $x_{2}$ restrained to Case (A).Thus $(x_{1},x_{2})$ satisfies (A13).

Conclusion :

 To sum up, we have shown that in Case (A), for a given $x_2$, the best response of Candidate 1 is:

$$\begin{aligned} x_1=\left\{ \begin{array}{ll} x_{2}+\frac{(\lambda _{1}-1)K}{\lambda _{1}} &{}\quad \text{ if } x_2<-K \\ \frac{x_2}{\lambda _1} &{}\quad \text{ if } x_2\in [-K,K]\\ x_2-\frac{(\lambda _{1}-1)K}{\lambda _{1}} &{}\quad \text{ if } x_2>K, \end{array}\right. \end{aligned}$$

(A14)

while in Cases (B) and (B’), $(x_1,x_2)$ is in the interior of Case (B) or (B’). Figure 5 depicts Candidate 1’s best response in the $(x_1,x_2)$ space (the red dashed line) in Case (A).

• Third step: Conclusion To have a pure strategy Nash equilibrium, it is necessary that the two best responses intersect. However, the intersection is empty: Candidate 2’s best response is always inside Case (A) (the blue solid lines), and Candidate 1’s best response in Case (A) (the red dashed line) does not intersect it; when Candidate 1’s best response is in the interior of Case (B) or (B’), it cannot intersect Candidate 2’s best response which is always inside Case (A). Hence, there is no pure strategy Nash equilibrium. $\square$