Proof of Proposition 1

Let $\hat{k}_{1}$ and $\hat{k}_{2}$ be the parameters associated with simple strategies and let $\hat{k}_{1}^{\ast }$ and $\hat{k}_{2}^{\ast }$ be the parameters associated with the simple strategies of a SSMPE (as described in Definition 3). When talking about both parameters jointly, we use $\hat{k}=(\hat{k}_{1},\hat{k}_{2})$ and $\hat{k}^{\ast }=(\hat{k}_{1}^{\ast },\hat{k}_{2}^{\ast })$ . Because a simple strategy of $i\in \{1,2\}$ from Definition 3 is fully determined by $\hat{k}_{i}$ , with a slight abuse of notation, we call $\hat{k}_{i}$ the strategy of $i$ and $\hat{k}=(\hat{k}_{1},\hat{k}_{2})$ the profile of strategies.

We start by noting that any SSMPE strategy of $v$ satisfies, for an incumbent $j\in \{1,2\}$ , its policy commitment $\mathbf{q}\in X$ and a platform of the opposition $\mathbf{p}\in X$ , $\unicode[STIX]{x1D70E}_{v}^{\ast }(\,j,\mathbf{q},\mathbf{p})=\mathit{Yes}$ if and only if $d(\mathbf{p})\leqslant d(\mathbf{q})$ . From Definition 2, $\unicode[STIX]{x1D70E}_{v}^{\ast }(\,j,\mathbf{q},\mathbf{p})=\mathit{Yes}$ if and only if $u_{v}(\mathbf{p})\geqslant u_{v}(\mathbf{q})$ , which, from $u_{v}(\mathbf{x})=-d^{2}(\mathbf{x})$ for any $\mathbf{x}\in X$ , is equivalent to $d(\mathbf{p})\leqslant d(\mathbf{q})$ . Because of the simplicity of the representative voter’s behavior in any SSMPE, we suppress $\unicode[STIX]{x1D70E}_{v}$ from the notation below.

We now claim that, for any SSMPE strategy of $v$ and any simple strategy profile of the two parties $\hat{k}$ , the opposition party contests elections and wins. The former follows from $\mathbf{p}_{i}(\mathbf{q})\neq Out$ for any $i\in \{1,2\}$ , $\hat{k}_{i}\in \mathbb{R}$ and $\mathbf{q}\in X$ . To prove the latter claim, we need to show that $d(\mathbf{p}_{i}(\mathbf{q}))\leqslant d(\mathbf{q})$ for any $i\in \{1,2\}$ , $\hat{k}_{i}\in \mathbb{R}$ and $\mathbf{q}\in X$ . From Definition 3, if $k_{i}(\mathbf{q})\leqslant \hat{k}_{i}$ , then $d(\mathbf{p}_{i}(\mathbf{q}))=d(k_{i}(\mathbf{q})\mathbf{b}_{i})=k_{i}(\mathbf{q})b_{i}=\frac{d(\mathbf{q})}{b_{i}}b_{i}$ and if $k_{i}(\mathbf{q})\geqslant \hat{k}_{i}$ , then $d(\mathbf{p}_{i}(\mathbf{q}))=d(\hat{k}_{i}\mathbf{b}_{i})=\hat{k}_{i}b_{i}\leqslant k_{i}(\mathbf{q})b_{i}=d(\mathbf{q})$ . Because the opposition party contests elections and wins for any SSMPE strategy of $v$ and any simple strategy profile $\hat{k}$ , elections are always contested and incumbents always defeated in any SSMPE. This proves part 3 of the proposition.

We now prove parts 1 (characterization) and 2 (uniqueness or duplicity) of the proposition. By the one-stage-deviation principle, a profile of strategies $\hat{k}^{\ast }$ constitutes a SSMPE if, for any $i\in \{1,2\}$ and $\mathbf{q}\in X$ , a deviation by the opposition party $i$ to contest elections with $\mathbf{p}\neq \mathbf{p}_{i}(\mathbf{q})$ or to stay out is not profitable. We momentarily assume and later verify that both parties, when at the opposition, want to contest elections with platforms that guarantee their victory.

For any $\hat{k}$ , $i\in \{1,2\}$ and $\mathbf{q}\in X$ , the dynamic utility $i$ receives from running with $\mathbf{p}$ such that $d(\mathbf{p})\leqslant d(\mathbf{q})$ is $u_{i}(\mathbf{p})+r+\unicode[STIX]{x1D6FF}V_{i}^{i}(\mathbf{p}|\hat{k})$ . An opposition party $i$ thus runs with $\mathbf{p}^{\ast }\in \operatorname{argmax}_{d(\mathbf{x})\leqslant d(\mathbf{q})}u_{i}(\mathbf{x})+r+\unicode[STIX]{x1D6FF}V_{i}^{i}(\mathbf{x}|\hat{k})$ . For any $\hat{k}$ and any two policies $\mathbf{p}\in X$ and $\mathbf{p}^{\prime }\in X$ with $d(\mathbf{p})=d(\mathbf{p}^{\prime })$ , we have $V_{i}^{i}(\mathbf{p}|\hat{k})=V_{i}^{i}(\mathbf{p}^{\prime }|\hat{k})$ . This means that the optimal $\mathbf{p}^{\ast }$ has to lie on the ray starting at $\mathbf{b}_{v}=(0,0)$ and passing through $\mathbf{b}_{i}$ . It hence can be written as $\mathbf{p}^{\ast }=k\mathbf{b}_{i}$ for some $k\geqslant 0$ . Denote by $U_{i}(k|\hat{k})=u_{i}(k\mathbf{b}_{i})+r+\unicode[STIX]{x1D6FF}V_{i}^{i}(k\mathbf{b}_{i}|\hat{k})$ the dynamic utility of party $i\in \{1,2\}$ from running with policy $k\mathbf{b}_{i}$ when the parties use simple strategies characterized by $\hat{k}$ . The key properties of $U_{i}(k|\hat{k})$ are summarized in the lemma below. To facilitate its proof, we first state several identities for the policy utility of the two parties.

The policy utility $i\in \{1,2\}$ derives from $k\mathbf{b}_{j}$ where $k\geqslant 0$ and $j\in \{1,2\}$ is $u_{i}(k\mathbf{b}_{j})=-d^{2}(k\mathbf{b}_{j},\mathbf{b}_{i})$ . This is a continuous and differentiable function of $k$ . Using $-i=\{1,2\}\setminus \{i\}$ , we have

(A 1) $$\begin{eqnarray}\begin{array}{@{}ll@{}}{\displaystyle \frac{\unicode[STIX]{x2202}(-d^{2}(k\mathbf{b}_{-i},\mathbf{b}_{i}))}{\unicode[STIX]{x2202}k}} & \displaystyle =-2(kd^{2}(\mathbf{b}_{-i})-\mathbf{b}_{-i}\cdot \mathbf{b}_{i})=-2b_{i}b_{-i}\left(k{\textstyle \frac{b_{-i}}{b_{i}}}-\cos \unicode[STIX]{x1D6FC}\right)\\ {\displaystyle \frac{\unicode[STIX]{x2202}(-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i}))}{\unicode[STIX]{x2202}k}} & \displaystyle =-2(kd^{2}(\mathbf{b}_{i})-\mathbf{b}_{i}\cdot \mathbf{b}_{i})=-2b_{i}^{2}(k-1)\\ {\displaystyle \frac{\unicode[STIX]{x2202}^{2}(-d^{2}(k\mathbf{b}_{j},\mathbf{b}_{i}))}{\unicode[STIX]{x2202}^{2}k}} & \displaystyle =-2d^{2}(\mathbf{b}_{j}).\end{array}\end{eqnarray}$$

This proves that $u_{i}(k\mathbf{b}_{j})$ is a concave function of $k$ .

Lemma A1. For any $i\in \{1,2\}$ , fix $\hat{k}_{i}\in \mathbb{R}$ and $\hat{k}_{-i}\in \mathbb{R}$ . Let $\mathbf{q}\in X$ be the incumbent’s policy commitment. $U_{i}(k|\hat{k})$ , as a function of $k\in [0,k_{i}(\mathbf{q})]$ , is continuous, differentiable except when $k=\hat{k}_{i}$ or $k=\hat{k}_{-i}\frac{b_{-i}}{b_{i}}$ , strictly concave on each interval on which it is differentiable and $\frac{\unicode[STIX]{x2202}U_{i}(k|\hat{k})}{\unicode[STIX]{x2202}k}>0$ for $k<c$ , $\frac{\unicode[STIX]{x2202}U_{i}(k|\hat{k})}{\unicode[STIX]{x2202}k}=0$ for $k=c$ and $\frac{\unicode[STIX]{x2202}U_{i}(k|\hat{k})}{\unicode[STIX]{x2202}k}<0$ for $k>c$ where

(A 2) $$\begin{eqnarray}\begin{array}{@{}llll@{}}c= & \,{\textstyle \frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}} & & \text{ if }k<\min \{\hat{k}_{i},\hat{k}_{-i}{\textstyle \frac{b_{-i}}{b_{i}}}\}\\ c= & \,{\textstyle \frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}} & & \text{ if }\hat{k}_{i}<k<\hat{k}_{-i}{\textstyle \frac{b_{-i}}{b_{i}}}\\ c= & \,1 & & \text{ if }\hat{k}_{-i}{\textstyle \frac{b_{-i}}{b_{i}}}<k<\hat{k}_{i}\\ c= & \,1 & & \text{ if }\max \{\hat{k}_{i},\hat{k}_{-i}{\textstyle \frac{b_{-i}}{b_{i}}}\}<k.\\ \end{array}\end{eqnarray}$$

Proof. Throughout the proof fix $i\in \{1,2\}$ , $-i=\{1,2\}\setminus \{i\}$ , $\hat{k}_{i}\in \mathbb{R}$ , $\hat{k}_{-i}\in \mathbb{R}$ and incumbent’s policy commitment $\mathbf{q}\in X$ . The continuity is easy to see as the simple strategies characterized by $\hat{k}$ give rise to a value function, $V_{i}^{i}(\mathbf{x}|\hat{k})$ , that is continuous in $\mathbf{x}\in X$ for any $d(\mathbf{x})\leqslant d(\mathbf{q})$ . For the remaining properties, we have to derive $V_{i}^{i}(\mathbf{x}|\hat{k})$ explicitly. Because the opposition party always contests and wins elections for any $\hat{k}$ , we have, for any $\mathbf{x}\in X$ with $d(\mathbf{x})\leqslant d(\mathbf{q})$ ,

(A 3) $$\begin{eqnarray}\begin{array}{@{}rl@{}}V_{i}^{-i}(\mathbf{x}|\hat{k}) & \displaystyle =u_{i}(\mathbf{p}_{i}(\mathbf{x}))+r+\unicode[STIX]{x1D6FF}V_{i}^{i}(\mathbf{p}_{i}(\mathbf{x})|\hat{k})\\ V_{i}^{i}(\mathbf{x}|\hat{k}) & \displaystyle =u_{i}(\mathbf{p}_{-i}(\mathbf{x}))+\unicode[STIX]{x1D6FF}V_{i}^{-i}(\mathbf{p}_{-i}(\mathbf{x})|\hat{k}).\\ \end{array}\end{eqnarray}$$

Combining the two equations

(A 4) $$\begin{eqnarray}V_{i}^{i}(\mathbf{x}|\hat{k})=u_{i}(\mathbf{p}_{-i}(\mathbf{x}))+\unicode[STIX]{x1D6FF}\left[u_{i}(\mathbf{p}_{i}(\mathbf{p}_{-i}(\mathbf{x})))+r+\unicode[STIX]{x1D6FF}V_{i}^{i}(\mathbf{p}_{i}(\mathbf{p}_{-i}(\mathbf{x}))|\hat{k})\right]\end{eqnarray}$$

where

(A 5) $$\begin{eqnarray}\begin{array}{@{}llllll@{}}\mathbf{p}_{-i}(\mathbf{x}) & \displaystyle =k_{i}(\mathbf{x}){\textstyle \frac{b_{i}}{b_{-i}}}\mathbf{b}_{-i} & \quad \mathbf{p}_{i}(\mathbf{p}_{-i}(\mathbf{x})) & \displaystyle =k_{i}(\mathbf{x})\mathbf{b}_{i} & & \text{if }k_{i}(x)\leqslant \min \{\hat{k}_{i},\hat{k}_{-i}{\textstyle \frac{b_{-i}}{b_{i}}}\}\\ \mathbf{p}_{-i}(\mathbf{x}) & \displaystyle =k_{i}(\mathbf{x}){\textstyle \frac{b_{i}}{b_{-i}}}\mathbf{b}_{-i} & \quad \mathbf{p}_{i}(\mathbf{p}_{-i}(\mathbf{x})) & \displaystyle =\hat{k}_{i}\mathbf{b}_{i} & & \text{if }\hat{k}_{i}<k_{i}(x)<\hat{k}_{-i}{\textstyle \frac{b_{-i}}{b_{i}}}\\ \mathbf{p}_{-i}(\mathbf{x}) & \displaystyle =\hat{k}_{-i}\mathbf{b}_{-i} & \quad \mathbf{p}_{i}(\mathbf{p}_{-i}(\mathbf{x})) & \displaystyle =\hat{k}_{-i}{\textstyle \frac{b_{-i}}{b_{i}}}\mathbf{b}_{i} & & \text{if }\hat{k}_{-i}{\textstyle \frac{b_{-i}}{b_{i}}}<k_{i}(x)<\hat{k}_{i}\\ \mathbf{p}_{-i}(\mathbf{x}) & \displaystyle =\hat{k}_{-i}\mathbf{b}_{-i} & \quad \mathbf{p}_{i}(\mathbf{p}_{-i}(\mathbf{x})) & \displaystyle =\min \{\hat{k}_{-i}{\textstyle \frac{b_{-i}}{b_{i}}},\hat{k}_{i}\}\mathbf{b}_{i} & & \text{if }\max \{\hat{k}_{i},\hat{k}_{-i}{\textstyle \frac{b_{-i}}{b_{i}}}\}\leqslant k_{i}(x)\end{array}\end{eqnarray}$$

is easy to confirm using properties of the simple strategies along with $k_{i}(\mathbf{x})b_{i}=k_{-i}(\mathbf{x})b_{-i}$ for any $\mathbf{x}\in X$ .

We substitute $\mathbf{x}=k\mathbf{b}_{i}$ into (A 4), using (A 5), $k_{i}(k\mathbf{b}_{i})=k$ , $u_{i}(\mathbf{x})=-d^{2}(\mathbf{x},\mathbf{b}_{i})$ and $V_{i}^{i}(\mathbf{x}|\hat{k})=V_{i}^{i}(k_{i}(\mathbf{x})\mathbf{b}_{i}|\hat{k})$ , which follows from $d(k_{i}(\mathbf{x})\mathbf{b}_{i})=\frac{d(\mathbf{x})}{b_{i}}d(\mathbf{b}_{i})=d(\mathbf{x})$ . After some straightforward algebra, summarizing with $\unicode[STIX]{x1D712}_{t}$ all the terms constant in $k$ , we have:

(A 6) $$\begin{eqnarray}\begin{array}{@{}llll@{}}U_{i}(k|\hat{k}) & \displaystyle ={\displaystyle \frac{-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})-d^{2}(k{\textstyle \frac{b_{i}}{b_{-i}}}\mathbf{b}_{-i},\mathbf{b}_{i})\unicode[STIX]{x1D6FF}+r}{1-\unicode[STIX]{x1D6FF}^{2}}} & & \text{if }k\leqslant \min \{\hat{k}_{i},\hat{k}_{-i}{\textstyle \frac{b_{-i}}{b_{i}}}\}\\ U_{i}(k|\hat{k}) & \displaystyle =-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})-d^{2}(k{\textstyle \frac{b_{i}}{b_{-i}}}\mathbf{b}_{-i},\mathbf{b}_{i})\unicode[STIX]{x1D6FF}+\unicode[STIX]{x1D712}_{1} & & \text{if }\hat{k}_{i}<k<\hat{k}_{-i}{\textstyle \frac{b_{-i}}{b_{i}}}\\ U_{i}(k|\hat{k}) & \displaystyle =-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})+\unicode[STIX]{x1D712}_{2} & & \text{if }\hat{k}_{-i}{\textstyle \frac{b_{-i}}{b_{i}}}<k<\hat{k}_{i}\\ U_{i}(k|\hat{k}) & \displaystyle =-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})+\unicode[STIX]{x1D712}_{3} & & \text{if }\max \{\hat{k}_{i},\hat{k}_{-i}{\textstyle \frac{b_{-i}}{b_{i}}}\}\leqslant k\\ \end{array}\end{eqnarray}$$

for any $k\in [0,k_{i}(\mathbf{q})]$ . Direct verification then shows all the remaining properties of $U_{i}(k|\hat{k})$ .Footnote ²² ☐

Using $U_{i}(k|\hat{k})$ we can rewrite the optimization problem of opposition party $i\in \{1,2\}$ regarding which policy to contest elections with, for any incumbent’s policy commitment $\mathbf{q}\in X$ , as $\max _{0\leqslant k\leqslant k_{i}(\mathbf{q})}U_{i}(k|\hat{k})$ . To find a SSMPE, we need to find a $\hat{k}^{\ast }$ such that the solution to this optimization problem under $\hat{k}^{\ast }$ , for any $\mathbf{q}\in X$ , can be described by $\hat{k}^{\ast }$ .

We first claim that in any SSMPE, $\hat{k}_{i}^{\ast }\in \{\!\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}},1\!\}$ for $i\in \{1,2\}$ . Notice $\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}\leqslant 1$ for any $\unicode[STIX]{x1D6FF}\in [0,1)$ and $\unicode[STIX]{x1D6FC}\in [0,\unicode[STIX]{x1D70B}]$ . To show the claim, suppose, toward a first contradiction, that $\hat{k}_{i}^{\ast }<\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}$ . Suppose $\hat{k}_{-i}^{\ast }\frac{b_{-i}}{b_{i}}\leqslant \hat{k}_{i}^{\ast }$ . Then from Lemma A1, there exists $\bar{\unicode[STIX]{x1D716}}>0$ such that for all $\unicode[STIX]{x1D716}\in (0,\bar{\unicode[STIX]{x1D716}})$ , $U_{i}^{\prime }(\hat{k}_{i}^{\ast }+\unicode[STIX]{x1D716}|\hat{k}^{\ast })>0$ , which implies that $U_{i}(\hat{k}_{i}^{\ast }|\hat{k}^{\ast })<U_{i}(\hat{k}_{i}^{\ast }+\unicode[STIX]{x1D716}|\hat{k}^{\ast })$ , $\mathbf{p}_{i}((\hat{k}_{i}^{\ast }+\unicode[STIX]{x1D716})\mathbf{b}_{i})=\hat{k}_{i}^{\ast }\mathbf{b}_{i}$ and $d((\hat{k}_{i}^{\ast }+\unicode[STIX]{x1D716})\mathbf{b}_{i})<d((\hat{k}_{i}^{\ast }+\bar{\unicode[STIX]{x1D716}})\mathbf{b}_{i})$ . In words, for incumbent’s policy commitment $(\hat{k}_{i}^{\ast }+\bar{\unicode[STIX]{x1D716}})\mathbf{b}_{i}$ , $i$ contests elections with $\hat{k}_{i}^{\ast }\mathbf{b}_{i}$ , despite the fact that running with $(\hat{k}_{i}^{\ast }+\unicode[STIX]{x1D716})\mathbf{b}_{i}$ would ensure its victory and higher dynamic utility, a contradiction. An identical argument leads to a contradiction when $\hat{k}_{-i}^{\ast }\frac{b_{-i}}{b_{i}}>\hat{k}_{i}^{\ast }$ . Now suppose, toward a second contradiction, that $\hat{k}_{i}^{\ast }\in (\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}},1)$ . If $\hat{k}_{-i}^{\ast }\frac{b_{-i}}{b_{i}}\leqslant \hat{k}_{i}^{\ast }$ , an argument identical to the one above leads to a contradiction. Suppose $\hat{k}_{-i}^{\ast }\frac{b_{-i}}{b_{i}}>\hat{k}_{i}^{\ast }$ . Then from Lemma A1, there exists $\bar{\unicode[STIX]{x1D716}}>0$ such that for all $\unicode[STIX]{x1D716}\in (0,\bar{\unicode[STIX]{x1D716}})$ , $U_{i}^{\prime }(\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}+\unicode[STIX]{x1D716}|\hat{k}^{\ast })<0$ , which implies that $U_{i}(\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}|\hat{k}^{\ast })>U_{i}(\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}+\unicode[STIX]{x1D716}|\hat{k}^{\ast })$ , $\mathbf{p}_{i}((\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}+\unicode[STIX]{x1D716})\mathbf{b}_{i})=(\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}+\unicode[STIX]{x1D716})\mathbf{b}_{i}$ and $d(\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}\mathbf{b}_{i})<d((\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}+\bar{\unicode[STIX]{x1D716}})\mathbf{b}_{i})$ , a contradiction. Finally suppose, toward a third contradiction, that $\hat{k}_{i}^{\ast }>1$ . Then irrespective of $\hat{k}_{-i}^{\ast }$ , Lemma A1 implies that there exists $\bar{\unicode[STIX]{x1D716}}>0$ such that, for all $\unicode[STIX]{x1D716}\in (0,\bar{\unicode[STIX]{x1D716}})$ , $U_{i}^{\prime }(1+\unicode[STIX]{x1D716}|\hat{k}^{\ast })<0$ , which implies that $U_{i}(1|\hat{k}^{\ast })>U_{i}(1+\unicode[STIX]{x1D716}|\hat{k}^{\ast })$ , $\mathbf{p}_{i}((1+\unicode[STIX]{x1D716})\mathbf{b}_{i})=(1+\unicode[STIX]{x1D716})\mathbf{b}_{i}$ and $d(\mathbf{b}_{i})<d((1+\bar{\unicode[STIX]{x1D716}})\mathbf{b}_{i})$ , a contradiction.

Having shown that $\hat{k}_{i}^{\ast }\in \{\!\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}},1\!\}$ for $i\in \{1,2\}$ , we now argue that $\hat{k}_{i}^{\ast }=\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}$ for all $i\in \{1,2\}$ cannot constitute an SSMPE unless $\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}=1$ , that is unless $\unicode[STIX]{x1D6FF}=0$ or $\unicode[STIX]{x1D6FC}=0$ . Suppose, toward a contradiction, that $\unicode[STIX]{x1D6FF}>0$ , $\unicode[STIX]{x1D6FC}>0$ and $\hat{k}_{i}^{\ast }=\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}$ . This implies that $\hat{k}_{i}^{\ast }<1$ for all $i\in \{1,2\}$ . Suppose, without loss of generality, that $b_{1}\geqslant b_{2}$ . Then Lemma A1 implies that there exists $\bar{\unicode[STIX]{x1D716}}>0$ such that for all $\unicode[STIX]{x1D716}\in (0,\bar{\unicode[STIX]{x1D716}})$ , $U_{1}^{\prime }(\hat{k}_{1}^{\ast }+\unicode[STIX]{x1D716}|\hat{k}^{\ast })>0$ , which implies that $U_{1}(\hat{k}_{1}^{\ast }|\hat{k}^{\ast })<U_{1}(\hat{k}_{1}^{\ast }+\unicode[STIX]{x1D716}|\hat{k}^{\ast })$ , $\mathbf{p}_{1}((\hat{k}_{1}^{\ast }+\unicode[STIX]{x1D716})\mathbf{b}_{1})=\hat{k}_{1}^{\ast }\mathbf{b}_{1}$ and $d((\hat{k}_{1}^{\ast }+\unicode[STIX]{x1D716})\mathbf{b}_{1})<d((\hat{k}_{1}^{\ast }+\bar{\unicode[STIX]{x1D716}})\mathbf{b}_{1})$ , which means party $1$ is not maximizing its dynamic utility when at the opposition. Furthermore, an argument similar to the one used in the second contradiction above implies that $\hat{k}_{1}^{\ast }=\hat{k}_{2}^{\ast }=1$ cannot constitute an SSMPE unless $\unicode[STIX]{x1D6FF}=0$ or $\unicode[STIX]{x1D6FC}=0$ .

This leaves three possible cases. Case 1: $\unicode[STIX]{x1D6FF}=0$ or $\unicode[STIX]{x1D6FC}=0$ and $\hat{k}_{1}^{\ast }=\hat{k}_{2}^{\ast }=1$ . When $\unicode[STIX]{x1D6FF}=0$ or $\unicode[STIX]{x1D6FC}=0$ , clearly $\hat{k}^{\ast }=(1,1)$ constitutes a SSMPE and, because $\hat{k}_{i}^{\ast }\in \{\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}},1\}$ for $i\in \{1,2\}$ , this SSMPE is unique. Case 2: $\unicode[STIX]{x1D6FF}>0$ and $\unicode[STIX]{x1D6FC}>0$ and $\hat{k}_{1}^{\ast }=1$ with $\hat{k}_{2}^{\ast }=\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}$ . Case 3: $\unicode[STIX]{x1D6FF}>0$ and $\unicode[STIX]{x1D6FC}>0$ and $\hat{k}_{1}^{\ast }=\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}$ with $\hat{k}_{2}^{\ast }=1$ .

We first claim that when $b_{1}>b_{2}$ , Case 3 cannot constitute a SSMPE. When $\unicode[STIX]{x1D6FF}>0$ and $\unicode[STIX]{x1D6FC}>0$ , $\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}<1$ . From $\hat{k}_{1}^{\ast }=\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}$ , $\hat{k}_{2}^{\ast }=1$ and $b_{1}>b_{2}$ , it follows $\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}b_{2}<b_{2}=\hat{k}_{2}^{\ast }b_{2}$ and $\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}b_{2}<\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}b_{1}=\hat{k}_{1}^{\ast }b_{1}$ , or $\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}<\min \{\hat{k}_{1}^{\ast }\frac{b_{1}}{b_{2}},\hat{k}_{2}^{\ast }\}$ . We can now use argument similar to the one used in the second contradiction above to establish contradiction with party $2$ maximizing its dynamic utility when in opposition.

It remains to be shown that, when $b_{1}>b_{2}$ , Case 2 constitutes a unique SSMPE and that, when $b_{1}=b_{2}$ , both Cases 2 and 3 constitute an SSMPE. We cover Case 2 irrespective of whether $b_{1}>b_{2}$ or $b_{1}=b_{2}$ . When $b_{1}=b_{2}$ , Case 3 is similar to Case 2 and is omitted.

Suppose $\unicode[STIX]{x1D6FF}>0$ , $\unicode[STIX]{x1D6FC}>0$ and $b_{1}\geqslant b_{2}$ . We need to show that $\hat{k}_{1}=1$ and $\hat{k}_{2}=\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}$ constitute an SSMPE. Take any incumbent’s policy commitment $\mathbf{q}\in X$ . The optimization problem of the opposition party regarding the policy to contest elections with is $\max _{k\in [0,k_{i}(\mathbf{q})]}U_{i}(k|\hat{k})$ . For any $\mathbf{q}\in X$ such that $k_{2}(\mathbf{q})>\hat{k}_{2}$ , from Lemma A1 we have $\lim _{k\rightarrow \hat{k}_{2}^{-}}U_{2}^{\prime }(k|\hat{k})\geqslant 0$ and $\lim _{k\rightarrow \hat{k}_{2}^{+}}U_{2}^{\prime }(k|\hat{k})\leqslant 0$ . By piece-wise strict concavity of $U_{2}$ established in the same lemma, $U_{2}(k|\hat{k})$ is, for any $\mathbf{q}\in X$ , increasing in $k$ on $[0,\min \{\hat{k}_{2},k_{2}(\mathbf{q})\}]$ and decreasing in $k$ on $[\hat{k}_{2},\max \{\hat{k}_{2},k_{2}(\mathbf{q})\}]$ . When $k_{2}(\mathbf{q})\leqslant \hat{k}_{2}$ , it is optimal for party $2$ to contest elections with $k_{2}(\mathbf{q})\mathbf{b}_{2}$ and when $k_{2}(\mathbf{q})>\hat{k}_{2}$ , it is optimal to run with $\hat{k}_{2}\mathbf{b}_{2}$ . The simple strategy with $\hat{k}_{2}$ is thus optimal for party 2. A similar argument can be used to show optimality of the simple strategy with $\hat{k}_{1}$ for party 1. The key to this claim is Lemma A1 along with $\hat{k}_{2}b_{2}<\hat{k}_{1}b_{1}$ and $\hat{k}_{2}b_{2}=\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}b_{2}<\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}b_{1}$ .

To finish the proof of parts 1 and 2 of the proposition, we have to show that for none of the parties, when at the opposition, staying out of the election is a profitable deviation. Take any $\hat{k}^{\ast }$ characterized in the three cases above and any incumbent’s policy commitment $\mathbf{q}\in X$ . The dynamic utility of opposition party $i\in \{1,2\}$ on the equilibrium path is $u_{i}(\mathbf{p}_{i}(\mathbf{q}))+r+\unicode[STIX]{x1D6FF}V_{i}^{i}(\mathbf{p}_{i}(\mathbf{q})|\hat{k}^{\ast })$ whereas the dynamic utility from staying out is $u_{i}(\mathbf{q})+\unicode[STIX]{x1D6FF}V_{i}^{-i}(\mathbf{q}|\hat{k}^{\ast })$ . Because on the equilibrium path $i$ contests elections, $V_{i}^{-i}(\mathbf{q}|\hat{k}^{\ast })=u_{i}(\mathbf{p}_{i}(\mathbf{q}))+r+\unicode[STIX]{x1D6FF}V_{i}^{i}(\mathbf{p}_{i}(\mathbf{q})|\hat{k}^{\ast })$ . We need to ensure that the on-path dynamic utility is larger than the off-path one, or $u_{i}(\mathbf{p}_{i}(\mathbf{q}))+r+\unicode[STIX]{x1D6FF}V_{i}^{i}(\mathbf{p}_{i}(\mathbf{q})|\hat{k}^{\ast })\geqslant \frac{u_{i}(\mathbf{q})}{1-\unicode[STIX]{x1D6FF}}$ . We derive an upper bound on the right hand side of the inequality and a lower bound on the left hand side of the inequality and show that the upper bound is smaller than the lower bound.

The upper bound is clearly $0\geqslant \frac{u_{i}(\mathbf{q})}{1-\unicode[STIX]{x1D6FF}}$ . We construct the lower bound as follows. $u_{i}(\mathbf{p}_{i}(\mathbf{q}))+r+\unicode[STIX]{x1D6FF}V_{i}^{i}(\mathbf{p}_{i}(\mathbf{q})|\hat{k}^{\ast })$ is the dynamic utility of opposition party $i$ from running with the optimal policy, given incumbent’s policy commitment $\mathbf{q}\in X$ . Because $\mathbf{p}_{i}$ is a simple strategy, $U_{i}(\min \{k_{i}(\mathbf{q}),\hat{k}_{i}^{\ast }\}|\hat{k}^{\ast })=u_{i}(\mathbf{p}_{i}(\mathbf{q}))+r+\unicode[STIX]{x1D6FF}V_{i}^{i}(\mathbf{p}_{i}(\mathbf{q})|\hat{k}^{\ast })$ . From Lemma A1 and the discussion that followed, $U_{i}(k|\hat{k}^{\ast })$ is increasing in $k$ on $[0,\hat{k}_{i}^{\ast }]$ in any SSMPE. Hence $U_{i}(0|\hat{k}^{\ast })\leqslant u_{i}(\mathbf{p}_{i}(\mathbf{q}))+r+\unicode[STIX]{x1D6FF}V_{i}^{i}(\mathbf{p}_{i}(\mathbf{q})|\hat{k}^{\ast })$ for any $\mathbf{q}\in X$ . From the proof of Lemma A1, $U_{i}(0|\hat{k}^{\ast })=\frac{-d^{2}(0,\mathbf{b}_{i})(1+\unicode[STIX]{x1D6FF})+r}{1-\unicode[STIX]{x1D6FF}^{2}}$ . $\frac{-d^{2}(0,\mathbf{b}_{i})(1+\unicode[STIX]{x1D6FF})+r}{1-\unicode[STIX]{x1D6FF}^{2}}\geqslant 0$ or equivalently $r\geqslant d^{2}(0,\mathbf{b}_{i})(1+\unicode[STIX]{x1D6FF})$ then ensures that none of the parties, when at the opposition, wants to stay out of the elections. This concludes the proof of parts 1 and 2.

To prove part 4 of the proposition, take the SSMPE $\hat{k}_{1}^{\ast }=1$ and $\hat{k}_{2}^{\ast }=\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}$ and $b_{1}\geqslant b_{2}$ , so that $\hat{k}_{1}^{\ast }b_{1}\geqslant \hat{k}_{2}^{\ast }b_{2}$ . Starting from $\mathbf{q}\in X$ such that $k_{2}(\mathbf{q})\leqslant \hat{k}_{2}^{\ast }$ , $k_{2}(\mathbf{q})b_{2}=k_{1}(\mathbf{q})b_{1}\leqslant \hat{k}_{2}^{\ast }b_{2}\leqslant \hat{k}_{1}^{\ast }b_{1}$ , both parties run with $k_{i}(\mathbf{q})\mathbf{b}_{i}$ when at the opposition in the first period. Since $d(k_{i}(\mathbf{q})\mathbf{b}_{i})=d(\mathbf{q})$ , the incumbent’s policy commitment at the beginning of the second period is a policy at the same distance from the origin as $\mathbf{q}$ . Thus both parties run with $k_{i}(\mathbf{q})\mathbf{b}_{i}$ when at the opposition in the second period. The same holds in any future period. Starting from $\mathbf{q}\in X$ such that $k_{2}(\mathbf{q})\geqslant \hat{k}_{2}^{\ast }$ , if party $2$ is at the opposition in the first period it runs with $\hat{k}_{2}^{\ast }\mathbf{b}_{2}$ and if party $1$ is at the opposition it runs with $\min \{k_{1}(\mathbf{q}),\hat{k}_{1}^{\ast }\}\mathbf{b}_{1}$ . In the former case, the incumbent’s policy commitment at the beginning of the second period satisfies $k_{2}(\hat{k}_{2}^{\ast }\mathbf{b}_{2})=\hat{k}_{2}^{\ast }$ , so that the policies alternate on $k_{2}(\hat{k}_{2}^{\ast }\mathbf{b}_{2})\mathbf{b}_{2}=\hat{k}_{2}^{\ast }\mathbf{b}_{2}$ and $k_{1}(\hat{k}_{2}^{\ast }\mathbf{b}_{2})\mathbf{b}_{1}$ starting from period 2. In the latter case, because $k_{2}(\min \{k_{1}(\mathbf{q}),\hat{k}_{1}^{\ast }\}\mathbf{b}_{1})=\min \{k_{1}(\mathbf{q}),\hat{k}_{1}^{\ast }\}\frac{b_{1}}{b_{2}}=\min \{k_{2}(\mathbf{q}),\hat{k}_{1}^{\ast }\frac{b_{1}}{b_{2}}\}\geqslant \hat{k}_{2}^{\ast }$ , party $2$ in the second period runs with $\hat{k}_{2}^{\ast }\mathbf{b}_{2}$ and the same alternation obtains from period 3 onwards.☐

Proof of Lemma 1

In the SSMPE from Proposition 1, the opposition party $i\in \{1,2\}$ contests elections and wins with policy $\mathbf{p}_{i}(\mathbf{q})$ , for any incumbent’s policy commitment $\mathbf{q}\in X$ . Because $u_{v}(\mathbf{x})=-d^{2}(\mathbf{x})$ is decreasing in $d(\mathbf{x})$ for any $\mathbf{x}\in X$ and $d(\mathbf{p}_{i}(\mathbf{q}))=\min \{k_{i}(\mathbf{q}),\hat{k}_{i}\}b_{i}$ for any $\mathbf{q}\in X$ , all we have to show is that $\min \{k_{i}(\mathbf{q}),\hat{k}_{i}\}b_{i}$ is nondecreasing in $\hat{k}_{2}$ , $b_{1}$ and $b_{2}$ for all $i\in \{1,2\}$ and $\mathbf{q}\in X$ and is increasing in $\hat{k}_{2}$ , $b_{1}$ and $b_{2}$ for some $i\in \{1,2\}$ and $\mathbf{q}\in X$ . The nondecreasing part is immediate. For the increasing part, we have $\min \{k_{1}(\mathbf{q}),\hat{k}_{1}\}b_{1}=\hat{k}_{1}b_{1}$ when $k_{1}(\mathbf{q})>\hat{k}_{1}$ , which is increasing in $b_{1}$ because $\hat{k}_{1}=1$ , and $\min \{k_{2}(\mathbf{q}),\hat{k}_{2}\}b_{2}=\hat{k}_{2}b_{2}$ when $k_{2}(\mathbf{q})>\hat{k}_{2}$ , which is increasing in $\hat{k}_{2}$ and $b_{2}$ because $b_{2}>0$ and $\hat{k}_{2}=\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}>0$ for any $\unicode[STIX]{x1D6FC}\in [0,\unicode[STIX]{x1D70B}]$ and $\unicode[STIX]{x1D6FF}\in [0,1)$ .☐

Proof of Proposition 3

By Proposition 1, $\hat{k}_{1}=1$ and $\hat{k}_{2}=\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}$ . Notice that $\hat{k}_{2}\in (0,1]$ for any $\unicode[STIX]{x1D6FF}\in [0,1)$ and $\unicode[STIX]{x1D6FC}\in [0,\unicode[STIX]{x1D70B}]$ . To show part 1, using the transformation of polar coordinated to Cartesian ones — $\hat{k}_{2}\mathbf{b}_{2}=(\hat{k}_{2}b_{2}\cos \unicode[STIX]{x1D6FC},\hat{k}_{2}b_{2}\sin \unicode[STIX]{x1D6FC})$ — $d_{p}$ can be expressed as $d_{p}=2b_{2}\hat{k}_{2}\sin {\textstyle \frac{\unicode[STIX]{x1D6FC}}{2}}$ . $d_{p}$ is clearly nondecreasing in $b_{2}$ , increasing in $b_{2}$ when $\unicode[STIX]{x1D6FC}>0$ and constant in $b_{1}$ . To show part 2,

(A 7) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}\left[2b_{2}{\textstyle \frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}}\sin {\textstyle \frac{\unicode[STIX]{x1D6FC}}{2}}\right]}{\unicode[STIX]{x2202}\unicode[STIX]{x1D6FC}}=\frac{b_{2}\cos {\textstyle \frac{\unicode[STIX]{x1D6FC}}{2}}}{1+\unicode[STIX]{x1D6FF}}\left[1-2\unicode[STIX]{x1D6FF}+3\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}\right]\end{eqnarray}$$

where the first term is positive unless $\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D70B}$ . Solving $1-2\unicode[STIX]{x1D6FF}+3\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}=0$ gives $\unicode[STIX]{x1D6FC}=\arccos \left[\frac{2\unicode[STIX]{x1D6FF}-1}{3\unicode[STIX]{x1D6FF}}\right]$ . From $\{x|\arccos x\in \mathbb{R}\}=[-1,1]$ , $-1\leqslant \frac{2\unicode[STIX]{x1D6FF}-1}{3\unicode[STIX]{x1D6FF}}\leqslant 1$ holds when $\unicode[STIX]{x1D6FF}\geqslant \frac{1}{5}$ . Because $\arccos x\in [0,\unicode[STIX]{x1D70B})$ for $x\in (-1,1]$ , $\arccos \left[\frac{2\unicode[STIX]{x1D6FF}-1}{3\unicode[STIX]{x1D6FF}}\right]<\unicode[STIX]{x1D70B}$ when $\unicode[STIX]{x1D6FF}>\frac{1}{5}$ . Defining $\unicode[STIX]{x1D6FC}^{\prime }=\arccos \left[\frac{2\unicode[STIX]{x1D6FF}-1}{3\unicode[STIX]{x1D6FF}}\right]$ when $\unicode[STIX]{x1D6FF}\geqslant \frac{1}{5}$ and $\unicode[STIX]{x1D6FC}^{\prime }=\unicode[STIX]{x1D70B}$ when $\unicode[STIX]{x1D6FF}\in [0,\frac{1}{5})$ , $\unicode[STIX]{x1D6FC}^{\prime }<\unicode[STIX]{x1D70B}$ when $\unicode[STIX]{x1D6FF}>\frac{1}{5}$ . That $d_{p}$ is increasing in $\unicode[STIX]{x1D6FC}$ when $\unicode[STIX]{x1D6FC}\in [0,\unicode[STIX]{x1D6FC}^{\prime }]$ and decreasing in $\unicode[STIX]{x1D6FC}$ when $\unicode[STIX]{x1D6FC}\in [\unicode[STIX]{x1D6FC}^{\prime },\unicode[STIX]{x1D70B}]$ then follows from the fact that $\cos \unicode[STIX]{x1D6FC}$ , and hence $1-2\unicode[STIX]{x1D6FF}+3\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}$ , is decreasing in $\unicode[STIX]{x1D6FC}$ .☐

B.1 Model with general utility functions

The model analyzed in this section is identical to the one in the main part of the paper except for the stage utility player $i\in \{1,2,v\}$ derives from policy $\mathbf{p}$ which now is

(B 1) $$\begin{eqnarray}u_{i}(\mathbf{p})=f(d(\mathbf{p},\mathbf{b}_{i}))\end{eqnarray}$$

where $f:[0,\infty )\rightarrow \mathbb{R}$ is a continuous, decreasing and concave function in $d(\mathbf{p},\mathbf{b}_{i})$ . We also assume that $f$ is twice continuously differentiable on $(0,\infty )$ . Notice that these assumptions allow $f^{\prime }(0)=0$ and $f(x)=-x^{2}$ , so that the model in the paper is a special case of the model analyzed here.

For space considerations, we refrain from repeating the arguments leading to Proposition 1 and stress only those aspects of the analysis that differ considerably. An argument similar to the one from the proof of Proposition 1 shows that, since parties use simple strategies, it is (i) optimal for $i\in \{1,2\}$ to contest elections with policies located only on the $\mathbf{b}_{i}$ -ray and (ii) optimal for $v$ to vote for the opposition party running with $\mathbf{p}$ when the incumbent’s policy commitment is $\mathbf{q}$ if and only if $d(\mathbf{p})\leqslant d(\mathbf{q})$ .

In an SSMPE, the extent of moderation of party $i\in \{1,2\}$ is the $k$ which maximizes $\tilde{U} _{i}(k)$ where, for $k\geqslant 0$ ,

(B 2) $$\begin{eqnarray}\begin{array}{@{}ll@{}}\tilde{U} _{i}(k) & \displaystyle =u_{i}(k\mathbf{b}_{i})+u_{i}(kb_{ir}\mathbf{b}_{-i})\unicode[STIX]{x1D6FF}\\ & \displaystyle =f(d(k\mathbf{b}_{i},\mathbf{b}_{i}))+f(d(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i}))\unicode[STIX]{x1D6FF}.\\ \end{array}\end{eqnarray}$$

$\tilde{U} _{i}(k)$ is the dynamic utility party $i$ receives, devoid of any constant terms, when it runs with policy $k\mathbf{b}_{i}$ and party $-i$ , not moderating to a larger extent, runs with $kb_{ir}\mathbf{b}_{-i}$ , where we are using the shorthand $b_{ir}=\frac{b_{i}}{b_{-i}}$ . The derivation of $\tilde{U} _{i}(k)$ is similar to the derivation of (A 6) and is not repeated here.

Proof. The continuity in part 1 follows from the continuity of $f$ , $d(k\mathbf{b}_{i},\mathbf{b}_{i})$ and $d(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i})$ . The differentiability follows from the derivatives below, which can be easily checked

$$\begin{eqnarray}\begin{array}{@{}llll@{}}d(k\mathbf{b}_{i},\mathbf{b}_{i}) & \displaystyle =b_{i}|1-k| & \quad d(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i}) & \displaystyle =b_{i}\sqrt{(k-\cos \unicode[STIX]{x1D6FC})^{2}+\sin ^{2}\unicode[STIX]{x1D6FC}}\\ d^{\prime }(k\mathbf{b}_{i},\mathbf{b}_{i}) & \left\{\begin{array}{@{}ll@{}}{<}0\quad & \text{if }k\in [0,1)\\ \nexists \quad & \text{if }k=1\\ {>}0\quad & \text{if }k\in (1,\infty )\end{array}\right. & \quad d^{\prime }(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i}) & \left\{\begin{array}{@{}ll@{}}{<}0\quad & \text{if }k<\cos \unicode[STIX]{x1D6FC}\\ =0\quad & \text{if }k=\cos \unicode[STIX]{x1D6FC}\wedge \unicode[STIX]{x1D6FC}\neq 0\\ \nexists \quad & \text{if }k=\cos \unicode[STIX]{x1D6FC}\wedge \unicode[STIX]{x1D6FC}=0\\ {>}0\quad & \text{if }k>\cos \unicode[STIX]{x1D6FC}\end{array}\right.\\ d^{\prime \prime }(k\mathbf{b}_{i},\mathbf{b}_{i}) & \left\{\begin{array}{@{}ll@{}}=0\quad & \text{if }k\neq 1\\ \nexists \quad & \text{if }k=1\end{array}\right. & \quad d^{\prime \prime }(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i}) & \left\{\begin{array}{@{}ll@{}}{>}0\quad & \text{if }\unicode[STIX]{x1D6FC}\notin \{0,\unicode[STIX]{x1D70B}\}\\ 0\quad & \text{if }\begin{array}{@{}l@{}}\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D70B}\,\vee \\ (k\neq 1\wedge \unicode[STIX]{x1D6FC}=0)\end{array}\\ \nexists \quad & \text{if }k=1\wedge \unicode[STIX]{x1D6FC}=0.\end{array}\right.\\ \end{array}\end{eqnarray}$$

For parts 2 and 3, we have

(B 3) $$\begin{eqnarray}\tilde{U} _{i}^{\prime }(k)=f^{\prime }(d(k\mathbf{b}_{i},\mathbf{b}_{i}))d^{\prime }(k\mathbf{b}_{i},\mathbf{b}_{i})+f^{\prime }(d(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i}))d^{\prime }(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i})\unicode[STIX]{x1D6FF}\end{eqnarray}$$

and direct verification shows that $\tilde{U} _{i}^{\prime }(k)>0$ for $k<\cos \unicode[STIX]{x1D6FC}$ , $\lim _{k\rightarrow \cos \unicode[STIX]{x1D6FC}^{-}}\tilde{U} _{i}^{\prime }(k)\geqslant 0$ , $\tilde{U} _{i}^{\prime }(k)<0$ for $k>1$ and $\lim _{k\rightarrow 1^{+}}\tilde{U} _{i}^{\prime }(k)\leqslant 0$ .

For the strict concavity for $k\in (\max \{0,\cos \unicode[STIX]{x1D6FC}\},1)$ in part 4, we have

(B 4) $$\begin{eqnarray}\displaystyle \tilde{U} _{i}^{\prime \prime }(k) & = & \displaystyle f^{\prime \prime }(d(k\mathbf{b}_{i},\mathbf{b}_{i}))\left[d^{\prime }(k\mathbf{b}_{i},\mathbf{b}_{i})\right]^{2}+f^{\prime }(d(k\mathbf{b}_{i},\mathbf{b}_{i}))d^{\prime \prime }(k\mathbf{b}_{i},\mathbf{b}_{i})\nonumber\\ \displaystyle & & \displaystyle +\,\unicode[STIX]{x1D6FF}f^{\prime \prime }(d(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i}))\left[d^{\prime }(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i})\right]^{2}+f^{\prime }(d(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i}))d^{\prime \prime }(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i})\unicode[STIX]{x1D6FF}.\end{eqnarray}$$

$\tilde{U} _{i}^{\prime \prime }(k)\leqslant 0$ is a consequence of the fact that all the summands of the expression are nonpositive. To see that $\tilde{U} _{i}^{\prime \prime }(k)<0$ under Conditions B1 and B2, note that the last summand is either zero (when $d^{\prime \prime }(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i})=0\Rightarrow \unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D70B}$ ), which implies that the next to last summand is negative ( $\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D70B}\Rightarrow f^{\prime \prime }<0$ by Condition B2), or negative.

For part 5, which claims that $\tilde{U} _{i}(k)$ is increasing on $(\max \{0,\cos \unicode[STIX]{x1D6FC}\},1)$ when Condition B1 fails, notice that the failure of the condition implies either $\unicode[STIX]{x1D6FF}=0$ or $\unicode[STIX]{x1D6FC}=0$ . In the latter case the $(\max \{0,\cos \unicode[STIX]{x1D6FC}\},1)$ interval is empty so assume $\unicode[STIX]{x1D6FF}=0$ and $\unicode[STIX]{x1D6FC}>0$ . Substituting $\unicode[STIX]{x1D6FF}=0$ into (B 2) gives $\tilde{U} _{i}(k)=f(d(k\mathbf{b}_{i},\mathbf{b}_{i}))$ so that $\tilde{U} _{i}(k)$ is increasing on $(\max \{0,\cos \unicode[STIX]{x1D6FC}\},1)$ . Finally, when Condition B2 in part 6 fails, we have $\unicode[STIX]{x1D6FC}=\unicode[STIX]{x1D70B}$ and $f^{\prime \prime }=0$ so that $d^{\prime }(k\mathbf{b}_{i},\mathbf{b}_{i})=-d^{\prime }(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i})=-b_{i}$ for $k\in (0,1)$ . This implies that $\tilde{U} _{i}^{\prime }(k)=(1-\unicode[STIX]{x1D6FF})c$ where $c>0$ so that $\tilde{U} _{i}(k)$ is increasing on $(0,1)$ .☐

Proposition B2 (SSMPE Comparative Static with General Utility).

Assume Conditions B1, B2 and B3 hold. Then $\unicode[STIX]{x1D705}_{i}\in (\max \{0,\cos \unicode[STIX]{x1D6FC}\},1)$ for $i\in \{1,2\}$ in Proposition B1 and (assuming $\unicode[STIX]{x1D6FC}\neq \unicode[STIX]{x1D70B}$ for the first relation)

(B 5) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D705}_{i}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D6FC}}<0\quad \frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D705}_{i}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D6FF}}<0.\end{eqnarray}$$

Proof. From Proposition B1, if Conditions B1 and B2 hold, $\unicode[STIX]{x1D705}_{i}=\operatorname{argmax}_{k\geqslant 0}\tilde{U} _{i}(k)\in [\max \{0,\cos \unicode[STIX]{x1D6FC}\},1]$ for $i\in \{1,2\}$ . We need to show that Condition B3 implies $\unicode[STIX]{x1D705}_{i}\in (\max \{0,\cos \unicode[STIX]{x1D6FC}\},1)$ for $i\in \{1,2\}$ . We show that Condition B3 is required only for $\unicode[STIX]{x1D705}_{i}<1$ and that $\unicode[STIX]{x1D705}_{i}>\max \{0,\cos \unicode[STIX]{x1D6FC}\}$ holds in general. Fix $i\in \{1,2\}$ . By Lemma B1 part 4, $\tilde{U} _{i}(k)$ is strictly concave for $k\in (\max \{0,\cos \unicode[STIX]{x1D6FC}\},1)$ . To show that $\unicode[STIX]{x1D705}_{i}<1$ , it thus suffices to show that $\lim _{k\rightarrow 1^{-}}\tilde{U} _{i}^{\prime }(k)<0$ . Substituting

(B 6) $$\begin{eqnarray}\begin{array}{@{}rclrcl@{}}\displaystyle \lim _{k\rightarrow 1^{-}}(d(k\mathbf{b}_{i},\mathbf{b}_{i})) & = & 0 & \quad \displaystyle \lim _{k\rightarrow 1^{-}}(d^{\prime }(k\mathbf{b}_{i},\mathbf{b}_{i})) & = & -b_{i}\\ \displaystyle \lim _{k\rightarrow 1^{-}}(d(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i})) & = & b_{i}\sqrt{2(1-\cos \unicode[STIX]{x1D6FC})} & \quad \displaystyle \lim _{k\rightarrow 1^{-}}(d^{\prime }(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i})) & = & b_{i}\sin {\textstyle \frac{\unicode[STIX]{x1D6FC}}{2}}\end{array}\end{eqnarray}$$

into (B 3) gives $\lim _{k\rightarrow 1^{-}}\tilde{U} _{i}^{\prime }(k)=-b_{i}[f^{\prime }(0)-f^{\prime }(b_{i}\sqrt{2(1-\cos \unicode[STIX]{x1D6FC})})\unicode[STIX]{x1D6FF}\sin \frac{\unicode[STIX]{x1D6FC}}{2}]<0$ , where the inequality follows from Condition B3. To show that $\unicode[STIX]{x1D705}_{i}>\max \{0,\cos \unicode[STIX]{x1D6FC}\}$ , it suffices to show that $\lim _{k\rightarrow 0^{+}}\tilde{U} _{i}^{\prime }(k)>0$ when $\unicode[STIX]{x1D6FC}\geqslant \frac{\unicode[STIX]{x1D70B}}{2}$ and $\lim _{k\rightarrow \cos \unicode[STIX]{x1D6FC}^{+}}\tilde{U} _{i}^{\prime }(k)>0$ when $\unicode[STIX]{x1D6FC}<\frac{\unicode[STIX]{x1D70B}}{2}$ . When $\unicode[STIX]{x1D6FC}\geqslant \frac{\unicode[STIX]{x1D70B}}{2}$ , substituting

(B 7) $$\begin{eqnarray}\begin{array}{@{}rclrcl@{}}\displaystyle \lim _{k\rightarrow 0^{+}}(d(k\mathbf{b}_{i},\mathbf{b}_{i})) & = & b_{i} & \quad \displaystyle \lim _{k\rightarrow 0^{+}}(d^{\prime }(k\mathbf{b}_{i},\mathbf{b}_{i})) & = & -b_{i}\\ \displaystyle \lim _{k\rightarrow 0^{+}}(d(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i})) & = & b_{i} & \quad \displaystyle \lim _{k\rightarrow 0^{+}}(d^{\prime }(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i})) & = & -b_{i}\cos \unicode[STIX]{x1D6FC}\end{array}\end{eqnarray}$$

into (B 3) gives $\lim _{k\rightarrow 0^{+}}\tilde{U} _{i}^{\prime }(k)=-b_{i}f^{\prime }(b_{i})(1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC})>0$ . When $\unicode[STIX]{x1D6FC}<\frac{\unicode[STIX]{x1D70B}}{2}$ , substituting

(B 8) $$\begin{eqnarray}\begin{array}{@{}rclrcl@{}}\displaystyle \lim _{k\rightarrow \cos \unicode[STIX]{x1D6FC}^{+}}(d(k\mathbf{b}_{i},\mathbf{b}_{i})) & = & b_{i}(1-\cos \unicode[STIX]{x1D6FC}) & \quad \displaystyle \lim _{k\rightarrow \cos \unicode[STIX]{x1D6FC}^{+}}(d^{\prime }(k\mathbf{b}_{i},\mathbf{b}_{i})) & = & -b_{i}\\ \displaystyle \lim _{k\rightarrow \cos \unicode[STIX]{x1D6FC}^{+}}(d(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i})) & = & b_{i}\sin \unicode[STIX]{x1D6FC} & \quad \displaystyle \lim _{k\rightarrow \cos \unicode[STIX]{x1D6FC}^{+}}(d^{\prime }(kb_{ir}\mathbf{b}_{-i},\mathbf{b}_{i})) & = & 0\end{array}\end{eqnarray}$$

into (B 3) gives $\lim _{k\rightarrow \cos \unicode[STIX]{x1D6FC}^{+}}\tilde{U} _{i}^{\prime }(k)=-b_{i}f^{\prime }(b_{i}(1-\cos \unicode[STIX]{x1D6FC}))>0$ .

Since $\unicode[STIX]{x1D705}_{i}\in (\max \{0,\cos \unicode[STIX]{x1D6FC}\},1)$ under the conditions of the proposition, it is implicitly defined by $\tilde{U} _{i}^{\prime }(\unicode[STIX]{x1D705}_{i})=0$ . From the implicit function theorem $\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D705}_{i}}{\unicode[STIX]{x2202}x}=\frac{\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}x}\tilde{U} _{i}^{\prime }(\unicode[STIX]{x1D705}_{i})}{-\tilde{U} _{i}^{\prime \prime }(\unicode[STIX]{x1D705}_{i})}$ . The denominator of this expression is positive by strict concavity of $\tilde{U} _{i}(k)$ . The numerator of this expression for $x\in \{\unicode[STIX]{x1D6FC},\unicode[STIX]{x1D6FF}\}$ is

(B 9) $$\begin{eqnarray}\begin{array}{@{}rcl@{}}{\textstyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D6FC}}}\tilde{U} _{i}^{\prime }(\unicode[STIX]{x1D705}_{i}) & = & f^{\prime \prime }(d(\unicode[STIX]{x1D705}_{i}b_{ir}\mathbf{b}_{-i},\mathbf{b}_{i}))d^{\prime }(\unicode[STIX]{x1D705}_{i}b_{ir}\mathbf{b}_{-i},\mathbf{b}_{i}){\displaystyle \frac{\unicode[STIX]{x2202}d(\unicode[STIX]{x1D705}_{i}b_{ir}\mathbf{b}_{-i},\mathbf{b}_{i})}{\unicode[STIX]{x2202}\unicode[STIX]{x1D6FC}}}\unicode[STIX]{x1D6FF}\\ & & +\,f^{\prime }(d(\unicode[STIX]{x1D705}_{i}b_{ir}\mathbf{b}_{-i},\mathbf{b}_{i})){\displaystyle \frac{\unicode[STIX]{x2202}d^{\prime }(\unicode[STIX]{x1D705}_{i}b_{ir}\mathbf{b}_{-i},\mathbf{b}_{i})}{\unicode[STIX]{x2202}\unicode[STIX]{x1D6FC}}}\unicode[STIX]{x1D6FF}\\ {\textstyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D6FF}}}\tilde{U} _{i}^{\prime }(\unicode[STIX]{x1D705}_{i}) & = & f^{\prime }(d(\unicode[STIX]{x1D705}_{i}b_{ir}\mathbf{b}_{-i},\mathbf{b}_{i}))d^{\prime }(\unicode[STIX]{x1D705}_{i}b_{ir}\mathbf{b}_{-i},\mathbf{b}_{i})\end{array}\end{eqnarray}$$

where both expressions are negative since $\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D6FC}}d(\unicode[STIX]{x1D705}_{i}b_{ir}\mathbf{b}_{-i},\mathbf{b}_{i})=\frac{b_{i}\unicode[STIX]{x1D705}_{i}\sin \unicode[STIX]{x1D6FC}}{\sqrt{(\unicode[STIX]{x1D705}_{i}-\cos \unicode[STIX]{x1D6FC})^{2}+\sin ^{2}\unicode[STIX]{x1D6FC}}}>0$ and $\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D6FC}}d^{\prime }(\unicode[STIX]{x1D705}_{i}b_{ir}\mathbf{b}_{-i},\mathbf{b}_{i})=\frac{b_{i}\sin \unicode[STIX]{x1D6FC}(1-\unicode[STIX]{x1D705}_{i}\cos \unicode[STIX]{x1D6FC})}{\left(\sqrt{(\unicode[STIX]{x1D705}_{i}-\cos \unicode[STIX]{x1D6FC})^{2}+\sin ^{2}\unicode[STIX]{x1D6FC}}\right)^{3}}>0$ .Footnote ²⁵ ☐

B.2 Forward-looking representative voter

In this section, we study a version of the model from the main part of the paper in which the representative voter is forward-looking. We will show that for any SSMPE identified in Proposition 1, there exists an equilibrium with forward-looking $v$ and that this equilibrium generates a comparative static on the representative voter’s welfare with respect to antagonism and extremism which is identical to the one stated in Corollary 1.

Throughout this section assume $\unicode[STIX]{x1D6FF}_{v}\in [0,1)$ . The model from the main part requires the following changes. The utility $v$ derives from a sequence of policies $\mathbf{P}=\{\mathbf{p}_{0},\mathbf{p}_{1},\ldots \}$ is the discounted sum of payoffs from each period

(B 10) $$\begin{eqnarray}U_{v}(\mathbf{P})=\mathop{\sum }_{t=0}^{\infty }\unicode[STIX]{x1D6FF}_{v}^{t}u_{v}(\mathbf{p}_{t}).\end{eqnarray}$$

Given a path of policies generated by play according to $\unicode[STIX]{x1D70E}=(\unicode[STIX]{x1D70E}_{1},\unicode[STIX]{x1D70E}_{2},\unicode[STIX]{x1D70E}_{v})$ starting from an incumbent $j\in \{1,2\}$ committed to $\mathbf{q}$ , $\mathbf{P}(\,j,\unicode[STIX]{x1D70E},\mathbf{q})=\{\mathbf{p}_{0},\mathbf{p}_{1},\ldots \}$ , the dynamic utility of $v$ is

(B 11) $$\begin{eqnarray}V_{v}^{j}(\mathbf{q}|\unicode[STIX]{x1D70E})=U_{v}(\mathbf{P}(\,j,\unicode[STIX]{x1D70E},\mathbf{q}))=\mathop{\sum }_{t=0}^{\infty }\unicode[STIX]{x1D6FF}_{v}^{t}u_{v}(\mathbf{p}_{t}).\end{eqnarray}$$

In Definition 2 of SMPE, we now require $\unicode[STIX]{x1D70E}_{v}^{\ast }(\,j,\mathbf{q},\mathbf{p})=\mathit{Yes}$ if and only if

(B 12) $$\begin{eqnarray}u_{v}(\mathbf{p})+\unicode[STIX]{x1D6FF}_{v}V_{v}^{i}(\mathbf{p}|\unicode[STIX]{x1D70E}^{\ast })\geqslant u_{v}(\mathbf{q})+\unicode[STIX]{x1D6FF}_{v}V_{v}^{j}(\mathbf{q}|\unicode[STIX]{x1D70E}^{\ast })\end{eqnarray}$$

for any incumbent $j\in \{1,2\}$ committed to $\mathbf{q}\in X$ and the opposition $i=\{1,2\}\setminus \{j\}$ contesting elections with $\mathbf{p}\in X$ . Given a profile of strategies $\unicode[STIX]{x1D70E}$ , assuming the initial incumbent’s policy commitment $\mathbf{q}^{0}$ is distributed according to $F(\mathbf{q}^{0})$ with strictly positive density on $X$ , and denoting with $\unicode[STIX]{x1D70C}\in (0,1)$ the probability that party $1$ is the first-period incumbent, the representative voter’s ex ante welfare from the dynamic electoral competition game with parties’ extremism $b_{1}$ and $b_{2}$ and antagonism $\unicode[STIX]{x1D6FC}$ is

(B 13) $$\begin{eqnarray}W(b_{1},b_{2},\unicode[STIX]{x1D6FC}|\unicode[STIX]{x1D70E})=\int _{X}\unicode[STIX]{x1D70C}V_{v}^{1}(\mathbf{z}|\unicode[STIX]{x1D70E})+(1-\unicode[STIX]{x1D70C})V_{v}^{2}(\mathbf{z}|\unicode[STIX]{x1D70E})d(F(\mathbf{z})).\end{eqnarray}$$

Proof. Fix the SSMPE $\unicode[STIX]{x1D70E}^{\prime }=(\unicode[STIX]{x1D70E}_{1}^{\prime },\unicode[STIX]{x1D70E}_{2}^{\prime },\unicode[STIX]{x1D70E}_{v}^{\prime })$ from Proposition 1. We construct an SMPE $\unicode[STIX]{x1D70E}^{\ast }=(\unicode[STIX]{x1D70E}_{1}^{\ast },\unicode[STIX]{x1D70E}_{2}^{\ast },\unicode[STIX]{x1D70E}_{v}^{\ast })$ for the model with $\unicode[STIX]{x1D6FF}_{v}\in [0,1)$ and then show that it possesses the properties from parts 1 through 3. Assume that $\unicode[STIX]{x1D70E}^{\prime }$ is such that party $2$ moderates, that is $\unicode[STIX]{x1D70E}_{1}^{\prime }(\mathbf{q})=\mathbf{p}_{1}(\mathbf{q})$ with $\hat{k}_{1}=1$ and $\unicode[STIX]{x1D70E}_{2}^{\prime }(\mathbf{q})=\mathbf{p}_{2}(\mathbf{q})$ with $\hat{k}_{2}=\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}$ for any $\mathbf{q}\in X$ . For the ‘mirror’ SSMPE identified in Proposition 1 part 2, the proof is similar and omitted.Footnote ²⁶

Denote $\tilde{k}_{2}=\sqrt{\left(\frac{b_{1}}{b_{2}}\right)^{2}(1-\unicode[STIX]{x1D6FF}_{v})+\unicode[STIX]{x1D6FF}_{v}\hat{k}_{2}^{2}}$ and note that, because $\hat{k}_{2}\leqslant 1$ and $\frac{b_{1}}{b_{2}}\geqslant 1$ , $\hat{k}_{2}\leqslant \tilde{k}_{2}\leqslant \frac{b_{1}}{b_{2}}$ . For any $\mathbf{q}\in X$ define

(B 14) $$\begin{eqnarray}\tilde{p}_{1}(\mathbf{q})=\left\{\begin{array}{@{}ll@{}}k_{1}(\mathbf{q})\quad & \text{for }k_{2}(\mathbf{q})\leqslant \hat{k}_{2}\\ k_{1}(\mathbf{q})\sqrt{{\displaystyle \frac{1-\unicode[STIX]{x1D6FF}_{v}\left({\displaystyle \frac{\hat{k}_{2}b_{2}}{d(\mathbf{q})}}\right)^{2}}{1-\unicode[STIX]{x1D6FF}_{v}}}}\quad & \text{for }k_{2}(\mathbf{q})\in (\hat{k}_{2},\tilde{k}_{2})\\ 1\quad & \text{for }k_{2}(\mathbf{q})\geqslant \tilde{k}_{2}.\end{array}\right.\end{eqnarray}$$

Standard arguments show that $\tilde{p}_{1}(\mathbf{x})=\tilde{p}_{1}(\mathbf{y})$ for any $\mathbf{x}\in X$ and $\mathbf{y}\in X$ if $d(\mathbf{x})=d(\mathbf{y})$ and that $\tilde{p}_{1}(\mathbf{x})$ is continuous in $d(\mathbf{x})$ and increasing in $d(\mathbf{x})$ if $k_{2}(\mathbf{x})\in [0,\tilde{k}_{2}]$ .

We now construct $\unicode[STIX]{x1D70E}^{\ast }$ . For party $2$ set $\unicode[STIX]{x1D70E}_{2}^{\ast }=\unicode[STIX]{x1D70E}_{2}^{\prime }$ , that is $\unicode[STIX]{x1D70E}_{2}^{\ast }$ is a simple strategy $\mathbf{p}_{2}$ from Definition 3 with $\hat{k}_{2}=\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}$ . For party $1$ and any $\mathbf{q}\in X$ , set $\unicode[STIX]{x1D70E}_{1}^{\ast }(\mathbf{q})=\tilde{p}_{1}(\mathbf{q})\mathbf{b}_{1}$ and note that $\unicode[STIX]{x1D70E}_{1}^{\ast }=\unicode[STIX]{x1D70E}_{1}^{\prime }$ when $\unicode[STIX]{x1D6FF}_{v}=0$ . For $v$ set $\unicode[STIX]{x1D70E}_{v}^{\ast }$ such that $\unicode[STIX]{x1D70E}_{v}^{\ast }(\,j,\mathbf{q},\mathbf{p})=\mathit{Yes}$ if and only if, when $j=1$

(B 15) $$\begin{eqnarray}\begin{array}{@{}llll@{}}d(\mathbf{p}) & {\leqslant}d(\mathbf{q}) & \quad \text{for }k_{2}(\mathbf{q}) & {\leqslant}\hat{k}_{2}\\ d(\mathbf{p}) & {\leqslant}\sqrt{d^{2}(\mathbf{q})(1-\unicode[STIX]{x1D6FF}_{v})+\unicode[STIX]{x1D6FF}_{v}(\hat{k}_{2}b_{2})^{2}} & \quad \text{for }k_{2}(\mathbf{q}) & \in (\hat{k}_{2},{\textstyle \frac{b_{1}}{b_{2}}})\\ d(\mathbf{p}) & {\leqslant}\sqrt{d^{2}(\mathbf{q})-\unicode[STIX]{x1D6FF}_{v}b_{1}^{2}+\unicode[STIX]{x1D6FF}_{v}(\hat{k}_{2}b_{2})^{2}} & \quad \text{for }k_{2}(\mathbf{q}) & {\geqslant}{\textstyle \frac{b_{1}}{b_{2}}}\end{array}\end{eqnarray}$$

and when $j=2$

(B 16) $$\begin{eqnarray}\begin{array}{@{}rclll@{}}d(\mathbf{p}) & {\leqslant} & d(\mathbf{q}) & & \quad \text{for }k_{2}(\mathbf{q})\leqslant \hat{k}_{2}\\ d(\mathbf{p}) & {\leqslant} & \tilde{p}_{1}(\mathbf{q})b_{1} & & \quad \text{for }k_{2}(\mathbf{q})\in (\hat{k}_{2},\tilde{k}_{2})\\ d(\mathbf{p}) & {\leqslant} & \sqrt{d^{2}(\mathbf{q})+\unicode[STIX]{x1D6FF}_{v}b_{1}^{2}-\unicode[STIX]{x1D6FF}_{v}(\hat{k}_{2}b_{2})^{2}} & & \quad \text{for }k_{2}(\mathbf{q})\geqslant \tilde{k}_{2}\end{array}\end{eqnarray}$$

and note that $\unicode[STIX]{x1D70E}_{v}^{\ast }=\unicode[STIX]{x1D70E}_{v}^{\prime }$ when $\unicode[STIX]{x1D6FF}_{v}=0$ .

We now argue that $\unicode[STIX]{x1D70E}^{\ast }=(\unicode[STIX]{x1D70E}_{1}^{\ast },\unicode[STIX]{x1D70E}_{2}^{\ast },\unicode[STIX]{x1D70E}_{v}^{\ast })$ constitutes an SMPE. From (B 15) and $\unicode[STIX]{x1D70E}_{2}^{\ast }(\mathbf{q})=\min \{k_{2}(\mathbf{q}),\hat{k}_{2}\}\mathbf{b}_{2}$ , party $2$ contests and wins elections for any policy commitment $\mathbf{q}\in X$ of party $1$ . From (B 16) and $\unicode[STIX]{x1D70E}_{1}^{\ast }(\mathbf{q})=\tilde{p}_{1}(\mathbf{q})\mathbf{b}_{1}$ , party $1$ contests and wins elections for any policy commitment $\mathbf{q}\in X$ of party $2$ . Because elections are always contested and incumbents always defeated in $\unicode[STIX]{x1D70E}^{\prime }$ as well as in $\unicode[STIX]{x1D70E}^{\ast }$ , because $\unicode[STIX]{x1D70E}_{2}^{\ast }(\mathbf{q})=\unicode[STIX]{x1D70E}_{2}^{\prime }(\mathbf{q})$ and $k_{2}(\unicode[STIX]{x1D70E}_{2}^{\ast }(\mathbf{q}))\leqslant \hat{k}_{2}$ for any $\mathbf{q}\in X$ and because $\unicode[STIX]{x1D70E}_{1}^{\ast }(\mathbf{q})=\unicode[STIX]{x1D70E}_{1}^{\prime }(\mathbf{q})$ for any $\mathbf{q}\in X$ such that $k_{2}(\mathbf{q})\leqslant \hat{k}_{2}$ , it follows that $V_{1}^{1}(\mathbf{q}|\unicode[STIX]{x1D70E}^{\ast })=V_{1}^{1}(\mathbf{q}|\unicode[STIX]{x1D70E}^{\prime })$ for any $\mathbf{q}\in X$ and $V_{2}^{2}(\mathbf{q}|\unicode[STIX]{x1D70E}^{\ast })=V_{2}^{2}(\mathbf{q}|\unicode[STIX]{x1D70E}^{\prime })$ for any $\mathbf{q}\in X$ such that $k_{2}(\mathbf{q})\leqslant \hat{k}_{2}$ . In addition, because $\tilde{p}_{1}(\mathbf{q})\geqslant k_{1}(\mathbf{q})$ for any $\mathbf{q}\in X$ such that $k_{2}(\mathbf{q})\leqslant \frac{b_{1}}{b_{2}}$ and because $\tilde{p}_{1}(\mathbf{q})=\hat{k}_{1}=1$ for any $\mathbf{q}\in X$ such that $k_{2}(\mathbf{q})\geqslant \frac{b_{1}}{b_{2}}$ , it follows that $V_{2}^{2}(\mathbf{q}|\unicode[STIX]{x1D70E}^{\ast })\leqslant V_{2}^{2}(\mathbf{q}|\unicode[STIX]{x1D70E}^{\prime })$ for any $\mathbf{q}\in X$ such that $k_{2}(\mathbf{q})>\hat{k}_{2}$ .

From the proof of Proposition 1 we know that $u_{1}(k\mathbf{b}_{1})+\unicode[STIX]{x1D6FF}V_{1}^{1}(k\mathbf{b}_{1}|\unicode[STIX]{x1D70E}^{\prime })$ is increasing in $k$ for $k\in [0,1]$ and decreasing in $k$ for $k\geqslant 1$ . Since $V_{1}^{1}(k\mathbf{b}_{1}|\unicode[STIX]{x1D70E}^{\ast })=V_{1}^{1}(k\mathbf{b}_{1}|\unicode[STIX]{x1D70E}^{\prime })$ for any $k\in \mathbb{R}_{{\geqslant}0}$ , $\unicode[STIX]{x1D70E}_{1}^{\ast }$ is optimal for party $1$ . For a policy commitment $\mathbf{q}\in X$ of party $2$ such that $k_{2}(\mathbf{q})\leqslant \tilde{k}_{2}$ , the largest policy on the $\mathbf{b}_{1}$ -ray that generates a victory of party $1$ is $\tilde{p}_{1}(\mathbf{q})\mathbf{b}_{1}$ , and the same policy maximizes the dynamic utility of party $1$ . For a policy commitment $\mathbf{q}\in X$ of party $2$ such that $k_{2}(\mathbf{q})>\tilde{k}_{2}$ , running with policy $\mathbf{b}_{1}$ guarantees the electoral victory for party $1$ , and the same policy maximizes the dynamic utility of party $1$ .

From the proof of Proposition 1 we know that $u_{2}(k\mathbf{b}_{2})+\unicode[STIX]{x1D6FF}V_{2}^{2}(k\mathbf{b}_{2}|\unicode[STIX]{x1D70E}^{\prime })$ is increasing in $k$ for $k\in [0,\hat{k}_{2}]$ and decreasing in $k$ for $k\geqslant \hat{k}_{2}$ . Since $V_{2}^{2}(k\mathbf{b}_{2}|\unicode[STIX]{x1D70E}^{\ast })=V_{2}^{2}(k\mathbf{b}_{2}|\unicode[STIX]{x1D70E}^{\prime })$ for $k\in [0,\hat{k}_{2}]$ and $V_{2}^{2}(k\mathbf{b}_{2}|\unicode[STIX]{x1D70E}^{\ast })\leqslant V_{2}^{2}(k\mathbf{b}_{2}|\unicode[STIX]{x1D70E}^{\prime })$ for $k>\hat{k}_{2}$ , $\unicode[STIX]{x1D70E}_{2}^{\ast }$ is optimal for party $2$ . For a policy commitment $\mathbf{q}\in X$ of party $1$ such that $k_{2}(\mathbf{q})\leqslant \hat{k}_{2}$ , the largest policy on the $\mathbf{b}_{2}$ -ray that generates the victory of party $2$ is $k_{2}(\mathbf{q})\mathbf{b}_{2}$ , and the same policy maximizes the dynamic utility of party $2$ . For a policy commitment $\mathbf{q}\in X$ of party $1$ such that $k_{2}(\mathbf{q})>\hat{k}_{2}$ , running with policy $\hat{k}_{2}\mathbf{b}_{2}$ guarantees the electoral victory for party $2$ , and the same policy maximizes the dynamic utility of party $2$ .

It remains is to be shown that $\unicode[STIX]{x1D70E}_{v}^{\ast }$ satisfies the definition of SMPE. From the definition, for any incumbent $j\in \{1,2\}$ committed to $\mathbf{q}\in X$ and the opposition $i=\{1,2\}\setminus \{j\}$ contesting elections with $\mathbf{p}\in X$ , $\unicode[STIX]{x1D70E}_{v}^{\ast }(\,j,\mathbf{q},\mathbf{p})=\mathit{Yes}$ if and only if

(B 17) $$\begin{eqnarray}u_{v}(\mathbf{p})+\unicode[STIX]{x1D6FF}_{v}V_{v}^{i}(\mathbf{p}|\unicode[STIX]{x1D70E}^{\ast })\geqslant u_{v}(\mathbf{q})+\unicode[STIX]{x1D6FF}_{v}V_{v}^{j}(\mathbf{q}|\unicode[STIX]{x1D70E}^{\ast }).\end{eqnarray}$$

Using $\unicode[STIX]{x1D70E}_{1}^{\ast }$ , $\unicode[STIX]{x1D70E}_{2}^{\ast }$ and $u_{v}(\mathbf{q})=-d^{2}(\mathbf{q})$ for any $\mathbf{q}\in X$ , straightforward algebra gives

(B 18) $$\begin{eqnarray}\begin{array}{@{}ll@{}}V_{v}^{1}(\mathbf{q}|\unicode[STIX]{x1D70E}^{\ast }) & \displaystyle =\left\{\begin{array}{@{}ll@{}}-{\textstyle \frac{d^{2}(\mathbf{q})}{1-\unicode[STIX]{x1D6FF}_{v}}}\quad & \text{ for }k_{2}(\mathbf{q})\leqslant \hat{k}_{2}\\ -{\textstyle \frac{(\hat{k}_{2}b_{2})^{2}}{1-\unicode[STIX]{x1D6FF}_{v}}}\quad & \text{ for }k_{2}(\mathbf{q})>\hat{k}_{2}\end{array}\right.\\ V_{v}^{2}(\mathbf{q}|\unicode[STIX]{x1D70E}^{\ast }) & \displaystyle =\left\{\begin{array}{@{}ll@{}}-{\textstyle \frac{d^{2}(\mathbf{q})}{1-\unicode[STIX]{x1D6FF}_{v}}}\quad & \text{ for }k_{2}(\mathbf{q})\,\leqslant \tilde{k}_{2}\\ -b_{1}^{2}-\unicode[STIX]{x1D6FF}_{v}{\textstyle \frac{(\hat{k}_{2}b_{2})^{2}}{1-\unicode[STIX]{x1D6FF}_{v}}}\quad & \text{ for }k_{2}(\mathbf{q})>\tilde{k}_{2}\end{array}\right.\end{array}\end{eqnarray}$$

which can be used to derive $\unicode[STIX]{x1D70E}_{v}^{\ast }$ as stated in (B 15) and (B 16). This concludes the proof that $\unicode[STIX]{x1D70E}^{\ast }$ constitutes an SMPE.

Part 1 of the proposition, that elections are always contested and incumbents always defeated under $\unicode[STIX]{x1D70E}^{\ast }$ , has already been noted. Part 2 claims that policies converge to identical alternation under $\unicode[STIX]{x1D70E}^{\prime }$ and $\unicode[STIX]{x1D70E}^{\ast }$ . From Proposition 1, this means alternation between $k_{1}(\mathbf{q})\mathbf{b}_{1}$ and $k_{2}(\mathbf{q})\mathbf{b}_{2}$ when $k_{2}(\mathbf{q})\leqslant \hat{k}_{2}$ and alternation between $k_{1}(\hat{k}_{2}\mathbf{b}_{2})\mathbf{b}_{1}$ and $\hat{k}_{2}\mathbf{b}_{2}$ when $k_{2}(\mathbf{q})>\hat{k}_{2}$ . When $k_{2}(\mathbf{q})\leqslant \hat{k}_{2}$ , $\unicode[STIX]{x1D70E}^{\ast }$ clearly generates the identical alternation since $\unicode[STIX]{x1D70E}_{2}^{\ast }=\unicode[STIX]{x1D70E}_{2}^{\prime }$ and $\unicode[STIX]{x1D70E}_{1}^{\ast }(\mathbf{q})=k_{1}(\mathbf{q})\mathbf{b}_{1}$ . When $k_{2}(\mathbf{q})>\hat{k}_{2}$ , the first time party $2$ contests elections it does so with $\unicode[STIX]{x1D70E}_{2}^{\ast }(\mathbf{q})=\hat{k}_{2}\mathbf{b}_{2}$ and from then on the equilibrium policies alternate between $k_{1}(\hat{k}_{2}\mathbf{b}_{2})\mathbf{b}_{1}$ and $\hat{k}_{2}\mathbf{b}_{2}$ . To show part 3, taking derivatives with respect to $\unicode[STIX]{x1D6FC}$ and $b_{i}$ for $i\in \{1,2\}$ of the expressions in (B 18) shows that $\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D6FC}}V_{v}^{j}(\mathbf{q}|\unicode[STIX]{x1D70E}^{\ast })\geqslant 0$ and $\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}b_{i}}V_{v}^{j}(\mathbf{q}|\unicode[STIX]{x1D70E}^{\ast })\leqslant 0$ for any $i\in \{1,2\}$ , any $j\in \{1,2\}$ and any $\mathbf{q}\in X$ . At the same time $\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}b_{1}}V_{v}^{2}(\mathbf{q}|\unicode[STIX]{x1D70E}^{\ast })<0$ when $k_{2}(\mathbf{q})>\tilde{k}_{2}$ , $\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}b_{2}}V_{v}^{1}(\mathbf{q}|\unicode[STIX]{x1D70E}^{\ast })<0$ when $k_{2}(\mathbf{q})>\hat{k}_{2}$ and $\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}\unicode[STIX]{x1D6FC}}V_{v}^{1}(\mathbf{q}|\unicode[STIX]{x1D70E}^{\ast })>0$ when $k_{2}(\mathbf{q})>\hat{k}_{2}$ and $\unicode[STIX]{x1D6FF}>0$ .☐

B.3 Unique SPE with finite horizon

In this section, we analyze a finite horizon version of the model studied so far. We show that there exists a unique SPE and describe the policy dynamics generated by this equilibrium strategy profile.

Assume the parties compete in $T$ consecutive elections, where $T<\infty$ . For simplicity, we assume that $b_{1}=b_{2}$ and that the two parties have lexicographic preferences over office rents and policy outcomes. This implies that the opposition party, for any incumbent’s policy commitment $\mathbf{q}\in X$ , contests the election with a winning policy, if such a policy exists. Denote $\unicode[STIX]{x1D705}=\frac{1+\unicode[STIX]{x1D6FF}\cos \unicode[STIX]{x1D6FC}}{1+\unicode[STIX]{x1D6FF}}$ and note that $\unicode[STIX]{x1D705}\in (0,1]$ for any $\unicode[STIX]{x1D6FF}\in [0,1)$ and $\unicode[STIX]{x1D6FC}\in [0,\unicode[STIX]{x1D70B}]$ . We use the notation $\mathbf{p}_{i}(\mathbf{q}|\hat{k}_{i})$ for the simple strategies from Definition 3, in order to make explicit their dependence on $\hat{k}_{i}$ .

In any SPE, the voting strategy of the representative voter in any period has to satisfy, for any incumbent party $j\in \{1,2\}$ , any incumbent’s policy commitment $\mathbf{q}\in X$ and any electoral platform of the opposition party $\mathbf{p}\in X$ , $\unicode[STIX]{x1D70E}_{v}(\,j,\mathbf{q},\mathbf{p})=\mathit{Yes}$ if and only if $d(\mathbf{p})\leqslant d(\mathbf{q})$ .

We proceed by backward induction. Consider the last period $T$ . For any incumbent’s policy commitment $\mathbf{q}\in X$ , the opposition party $i$ contests elections with policy $\mathbf{p}^{\ast }\in \operatorname{argmax}_{\mathbf{p}\in \{\mathbf{x}\in X|d(\mathbf{x})\leqslant d(\mathbf{q})\}}-d^{2}(\mathbf{p},\mathbf{b}_{i})$ . Clearly, $\mathbf{p}^{\ast }$ has to lie on the $\mathbf{b}_{i}$ -ray and, hence, can be written as $k\mathbf{b}_{i}$ for some $k\geqslant 0$ . Denote $\tilde{U} _{i,T}(k)=-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})$ . Then the optimization problem of party $i$ can be written as $\max _{k\in [0,k_{i}(\mathbf{q})]}\tilde{U} _{i,T}(k)$ . This problem has a unique solution $k=1$ if $k_{i}(\mathbf{q})\geqslant 1$ and $k=k_{i}(\mathbf{q})$ if $k_{i}(\mathbf{q})\leqslant 1$ . In other words, $\mathbf{p}_{i}(\mathbf{q}|1)$ is the uniquely optimal strategy of the opposition party $i$ for all $i\in \{1,2\}$ and $\mathbf{q}\in X$ . Denote $\mathbf{p}_{i,T}(\mathbf{q})=\mathbf{p}_{i}(\mathbf{q}|1)$ for $i\in \{1,2\}$ .

Consider the period $T-1$ . For any incumbent’s policy commitment $\mathbf{q}\in X$ , the opposition party $i$ contests elections with a policy on the $\mathbf{b}_{i}$ -ray because both parties use simple strategies in period $T$ that depend on $d(\mathbf{q})$ but not on the exact location of $\mathbf{q}$ . The optimization problem of the opposition party $i$ is $\max _{k\in [0,k_{i}(\mathbf{q})]}\tilde{U} _{i,T-1}(k)$ where $\tilde{U} _{i,T-1}(k)=-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})-d^{2}(\mathbf{p}_{-i,T}(k\mathbf{b}_{i}),\mathbf{b}_{i})\unicode[STIX]{x1D6FF}$ . Because $\mathbf{p}_{-i,T}(k\mathbf{b}_{i})=k\mathbf{b}_{-i}$ for $k\in [0,1]$ and $\mathbf{p}_{-i,T}(k\mathbf{b}_{i})=\mathbf{b}_{-i}$ for $k\geqslant 1$ ,

(B 19) $$\begin{eqnarray}\begin{array}{@{}ll@{}}\tilde{U} _{i,T-1}(k) & \displaystyle =-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})-d^{2}(k\mathbf{b}_{-i},\mathbf{b}_{i})\unicode[STIX]{x1D6FF}\\ \tilde{U} _{i,T-1}(k) & \displaystyle =-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})-d^{2}(\mathbf{b}_{-i},\mathbf{b}_{i})\unicode[STIX]{x1D6FF}\end{array}\end{eqnarray}$$

when $k\in [0,1]$ and $k\geqslant 1$ , respectively. An identical argument to the one used to prove Lemma A1 then shows that $\tilde{U} _{i,T-1}(k)$ has a unique maximum at $k=\unicode[STIX]{x1D705}$ , is increasing on $[0,\unicode[STIX]{x1D705}]$ and decreasing on $[\unicode[STIX]{x1D705},\infty )$ . Thus, the optimization problem of the opposition party $i$ has a unique solution $k=\unicode[STIX]{x1D705}$ if $k_{i}(\mathbf{q})\geqslant \unicode[STIX]{x1D705}$ and $k=k_{i}(\mathbf{q})$ if $k_{i}(\mathbf{q})\leqslant \unicode[STIX]{x1D705}$ . In other words, $\mathbf{p}_{i}(\mathbf{q}|\unicode[STIX]{x1D705})$ is the uniquely optimal strategy of the opposition party $i$ for all $i\in \{1,2\}$ and $\mathbf{q}\in X$ . Denote $\mathbf{p}_{i,T-1}(\mathbf{q})=\mathbf{p}_{i}(\mathbf{q}|\unicode[STIX]{x1D705})$ for $i\in \{1,2\}$ .

Consider now the period $T-2$ . Without repeating the obvious details, for any incumbent’s policy commitment $\mathbf{q}\in X$ , the optimization problem of the opposition party $i$ writes $\max _{k\in [0,k_{i}(\mathbf{q})]}\tilde{U} _{i,T-2}(k)$ where

(B 20) $$\begin{eqnarray}\tilde{U} _{i,T-2}(k)=-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})-d^{2}(\mathbf{p}_{-i,T-1}(k\mathbf{b}_{i}),\mathbf{b}_{i})\unicode[STIX]{x1D6FF}-d^{2}(\mathbf{p}_{i,T}(\mathbf{p}_{-i,T-1}(k\mathbf{b}_{i})),\mathbf{b}_{i})\unicode[STIX]{x1D6FF}^{2}.\end{eqnarray}$$

Because $\mathbf{p}_{-i,T-1}(k\mathbf{b}_{i})=k\mathbf{b}_{-i}$ for $k\in [0,\unicode[STIX]{x1D705}]$ and $\mathbf{p}_{-i,T-1}(k\mathbf{b}_{i})=\unicode[STIX]{x1D705}\mathbf{b}_{-i}$ for $k\geqslant \unicode[STIX]{x1D705}$ , we have $\mathbf{p}_{i,T}(\mathbf{p}_{-i,T-1}(k\mathbf{b}_{i}))=k\mathbf{b}_{i}$ if $k\in [0,\unicode[STIX]{x1D705}]$ and $\mathbf{p}_{i,T}(\mathbf{p}_{-i,T-1}(k\mathbf{b}_{i}))=\unicode[STIX]{x1D705}\mathbf{b}_{i}$ if $k\geqslant \unicode[STIX]{x1D705}$ . Hence

(B 21) $$\begin{eqnarray}\begin{array}{@{}ll@{}}\tilde{U} _{i,T-2}(k) & \displaystyle =-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})-d^{2}(k\mathbf{b}_{-i},\mathbf{b}_{i})\unicode[STIX]{x1D6FF}-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})\unicode[STIX]{x1D6FF}^{2}\\ \tilde{U} _{i,T-2}(k) & \displaystyle =-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})-d^{2}(\unicode[STIX]{x1D705}\mathbf{b}_{-i},\mathbf{b}_{i})\unicode[STIX]{x1D6FF}-d^{2}(\unicode[STIX]{x1D705}\mathbf{b}_{i},\mathbf{b}_{i})\unicode[STIX]{x1D6FF}^{2}\end{array}\end{eqnarray}$$

when $k\in [0,\unicode[STIX]{x1D705}]$ and $k\geqslant \unicode[STIX]{x1D705}$ , respectively. We first note that $\tilde{U} _{i,T-2}(k)$ is increasing on $[0,\unicode[STIX]{x1D705}]$ , which follows from $\tilde{U} _{i,T-2}^{\prime }(k)=\tilde{U} _{i,T-1}^{\prime }(k)-2b_{i}^{2}\unicode[STIX]{x1D6FF}^{2}(k-1)>\tilde{U} _{i,T-1}^{\prime }(k)>0$ for any $k\in (0,\unicode[STIX]{x1D705})$ . Furthermore, $\tilde{U} _{i,T-2}(k)$ is clearly increasing on $[\unicode[STIX]{x1D705},1]$ and decreasing on $[1,\infty )$ . Thus the optimization problem of the opposition party $i$ has a unique solution $k=1$ if $k_{i}(\mathbf{q})\geqslant 1$ and $k=k_{i}(\mathbf{q})$ if $k_{i}(\mathbf{q})\leqslant 1$ . In other words, $\mathbf{p}_{i}(\mathbf{q}|1)$ is the uniquely optimal strategy of the opposition party $i$ for all $i\in \{1,2\}$ and $\mathbf{q}\in X$ . Denote $\mathbf{p}_{i,T-2}(\mathbf{q})=\mathbf{p}_{i}(\mathbf{q}|1)$ for $i\in \{1,2\}$ .

Consider the period $T-3$ . For any $\mathbf{q}\in X$ , the optimization problem of the opposition party $i$ writes $\max _{k\in [0,k_{i}(\mathbf{q})]}\tilde{U} _{i,T-3}(k)$ , where similar arguments as in the previous period show

(B 22) $$\begin{eqnarray}\begin{array}{@{}ll@{}}\tilde{U} _{i,T-3}(k) & \displaystyle =-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})-d^{2}(k\mathbf{b}_{-i},\mathbf{b}_{i})\unicode[STIX]{x1D6FF}-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})\unicode[STIX]{x1D6FF}^{2}-d^{2}(k\mathbf{b}_{-i},\mathbf{b}_{i})\unicode[STIX]{x1D6FF}^{3}\\ \tilde{U} _{i,T-3}(k) & \displaystyle =-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})-d^{2}(k\mathbf{b}_{-i},\mathbf{b}_{i})\unicode[STIX]{x1D6FF}-d^{2}(\unicode[STIX]{x1D705}\mathbf{b}_{i},\mathbf{b}_{i})\unicode[STIX]{x1D6FF}^{2}-d^{2}(\unicode[STIX]{x1D705}\mathbf{b}_{-i},\mathbf{b}_{i})\unicode[STIX]{x1D6FF}^{3}\end{array}\end{eqnarray}$$

when $k\in [0,\unicode[STIX]{x1D705}]$ and $k\geqslant \unicode[STIX]{x1D705}$ , respectively. Denoting by $\unicode[STIX]{x1D710}_{i}(k)=-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})-d^{2}(k\mathbf{b}_{-i},\mathbf{b}_{i})\unicode[STIX]{x1D6FF}$ , $\tilde{U} _{i,T-3}(k)=(1+\unicode[STIX]{x1D6FF}^{2})\unicode[STIX]{x1D710}_{i}(k)$ if $k\in [0,\unicode[STIX]{x1D705}]$ and $\tilde{U} _{i,T-3}(k)=\unicode[STIX]{x1D710}_{i}(k)+c$ if $k\geqslant \unicode[STIX]{x1D705}$ , where $c$ is constant in $k$ . Because $\unicode[STIX]{x1D710}_{i}(k)$ is increasing in $k$ on $[0,\unicode[STIX]{x1D705}]$ and decreasing in $k$ on $[\unicode[STIX]{x1D705},\infty )$ , $\tilde{U} _{i,T-3}(k)$ has a unique maximum at $k=\unicode[STIX]{x1D705}$ . By the now familiar arguments, $\mathbf{p}_{i}(\mathbf{q}|\unicode[STIX]{x1D705})$ is the uniquely optimal strategy of the opposition party $i$ for all $i\in \{1,2\}$ and $\mathbf{q}\in X$ . Denote $\mathbf{p}_{i,T-3}(\mathbf{q})=\mathbf{p}_{i}(\mathbf{q}|\unicode[STIX]{x1D705})$ for $i\in \{1,2\}$ .

So far, we have shown that it is uniquely optimal for the opposition party $i$ to contest elections with $\mathbf{p}_{i}(\mathbf{q}|\hat{k}_{i})$ , where $\hat{k}_{i}=1$ for $T$ and $T-2$ and $\hat{k}_{i}=\unicode[STIX]{x1D705}$ for $T-1$ and $T-3$ . Suppose that this patterns repeats for periods up to $T-s+2$ and $T-s+1$ where $s$ is even.

Consider period $T-s$ . We need to show that it is uniquely optimal for the opposition party $i$ to contest elections with $\mathbf{p}_{i}(\mathbf{q}|1)$ for all $i\in \{1,2\}$ and $\mathbf{q}\in X$ . Because in $T-s+1$ the opposition party contests elections with $\mathbf{p}_{i}(\mathbf{q}|\unicode[STIX]{x1D705})$ ,

(B 23) $$\begin{eqnarray}\begin{array}{@{}ll@{}}\tilde{U} _{i,T-s}(k) & \displaystyle =\unicode[STIX]{x1D710}_{i}(k)\mathop{\sum }_{t=0}^{(s-2)/2}\unicode[STIX]{x1D6FF}^{2t}-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})\unicode[STIX]{x1D6FF}^{s}\\ \tilde{U} _{i,T-s}(k) & \displaystyle =-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})+c\end{array}\end{eqnarray}$$

when $k\in [0,\unicode[STIX]{x1D705}]$ and $k\geqslant \unicode[STIX]{x1D705}$ , respectively. Because $\unicode[STIX]{x1D710}_{i}(k)$ is increasing in $k$ on $[0,\unicode[STIX]{x1D705}]$ and decreasing in $k$ on $[\unicode[STIX]{x1D705},\infty )$ , and because $-d^{2}(k\mathbf{b}_{i},\mathbf{b}_{i})$ is increasing in $k$ on $[0,1]$ and decreasing in $k$ on $[1,\infty )$ , $\tilde{U} _{T-s}(k)$ has a unique maximum $k=1$ . Hence, it is uniquely optimal for the opposition party $i$ to contest elections with $\mathbf{p}_{i}(\mathbf{q}|1)$ for all $i\in \{1,2\}$ and $\mathbf{q}\in X$ .

Consider $T-s-1$ . We need to show that it is uniquely optimal for the opposition party $i$ to contest elections with $\mathbf{p}_{i}(\mathbf{q}|\unicode[STIX]{x1D705})$ for all $i\in \{1,2\}$ and $\mathbf{q}\in X$ . Because in $T-s$ the opposition party contests elections with $\mathbf{p}_{i}(\mathbf{q}|1)$ and in $T-s+1$ with $\mathbf{p}_{i}(\mathbf{q}|\unicode[STIX]{x1D705})$ ,

(B 24) $$\begin{eqnarray}\begin{array}{@{}ll@{}}\tilde{U} _{i,T-s-1}(k) & \displaystyle =\unicode[STIX]{x1D710}_{i}(k)\mathop{\sum }_{t=0}^{s/2}\unicode[STIX]{x1D6FF}^{2t}\\ \tilde{U} _{i,T-s-1}(k) & \displaystyle =\unicode[STIX]{x1D710}_{i}(k)+c\end{array}\end{eqnarray}$$

when $k\in [0,\unicode[STIX]{x1D705}]$ and $k\geqslant \unicode[STIX]{x1D705}$ , respectively. Because $\unicode[STIX]{x1D710}_{i}(k)$ is increasing in $k$ on $[0,\unicode[STIX]{x1D705}]$ and decreasing in $k$ on $[\unicode[STIX]{x1D705},\infty )$ , $\tilde{U} _{T-s-1}(k)$ has a unique maximum $k=\unicode[STIX]{x1D705}$ . Hence, it is uniquely optimal for the opposition party $i$ to contest elections with $\mathbf{p}_{i}(\mathbf{q}|\unicode[STIX]{x1D705})$ for all $i\in \{1,2\}$ and $\mathbf{q}\in X$ .

To summarize, there exists a unique SPE in which the opposition party $i$ contests elections with $\mathbf{p}_{i}(\mathbf{q}|\hat{k}_{i})$ for all $i\in \{1,2\}$ and $\mathbf{q}\in X$ . $\hat{k}_{i}=1$ for all periods $T-s$ with $s\in \mathbb{N}_{{\geqslant}0}$ and $s$ even and $\hat{k}_{i}=\unicode[STIX]{x1D705}$ for all periods $T-s$ with $s\in \mathbb{N}_{{\geqslant}1}$ and $s$ odd. With an appropriately chosen initial-period incumbent party, the SPE just described generates identical policy dynamics to the one generated by the SSMPE from Proposition 1, for any incumbent’s policy commitment $\mathbf{q}\in X$ . When party $2$ is at the opposition in the initial period of the infinite horizon model, both parties moderate, if at the opposition, in the initial period of the finite horizon model with $T$ even. When party $1$ is at the opposition in the initial period of the infinite horizon model, none of the parties moderates, if at the opposition, in the initial period of the finite horizon model with $T$ odd.