A theory of strategic voting with non-instrumental motives

Abstract

Empirical studies have documented non-instrumental motives for voting. However, the theoretical literature on strategic voting has largely ignored these motives. In this paper, we examine voter behavior in multi-candidate elections in the presence of ethical, expressive, and instrumental concerns. Voters in our model derive utility from both the election outcome and the action of voting. A fraction of voters are ethical, who follow a group-welfare maximizing voting rule. The rule may require them to misalign their votes, that is, to vote for a candidate who is not their most preferred. We characterize the optimal rule for the ethical voters and provide comparative statics with respect to various electoral parameters. In particular, we find that the degree of misaligned voting is increasing in the importance of the election but is non-monotonic in the popularity of the Condorcet loser.

Mathematical appendix

First, we include some preliminary calculations used in the proofs. Take the expected aggregate welfare of centrist voters (i.e., the objective function in Eq. (1)) is denoted by

$$\begin{aligned} F(\sigma _{c})\equiv wP(\sigma _{c})+\int _{0}^{1}q_{c}\cdot B_{E}(\sigma _{c})+(1-q_{c})\cdot B_{S}dq_{c}. \end{aligned}$$

where $B_{E}(\sigma _{c})$ is the per capita expressive benefit derived by ethical voters and $B_{S}$ is the per capita expressive benefit derived by sincere voters. First, it can be shown that:

• $B_{E}(\sigma _{c})=\theta - \theta (\sigma _{c}^{2}-\sigma _{c}+\frac{1}{2}) = \theta (\frac{1}{2}-\sigma _{c}^{2}+\sigma _{c})$, and

• $B_{S}=\frac{3}{4}\theta$.

Therefore, it follows that:

$$\begin{aligned} \int _{0}^{1}q_{c}\cdot B_{E}(\sigma _{c})+(1-q_{c})\cdot B_{S}dq_{c}=\frac{\theta }{2}\left( \frac{1}{2}-\sigma _{c}^{2}+\sigma _{c}\right) +\frac{3}{8}\theta \end{aligned}$$

Now, lets compute $P(\sigma _{c})$, the probability of a centrist victory induced by voting rule $\sigma _{c}$. Since we restrict our attention on $\sigma _{c}\in [0,\frac{1}{2}]$, $P(\sigma _{c})$ is equivalent to the probability that the center-right receives more votes than the extremist. That is:

$$\begin{aligned} P(\sigma _{c})&=\text {Pr}\left( (1-k_{e})\left[ (1-q_{c})\frac{1}{2}+q_{c}(1-\sigma _{c})\right] \ge k_{e}\right) \\ &=\text {Pr}\left( q_{c}\ge \frac{\frac{k_{e}}{1-k_{e}} -\frac{1}{2}}{\frac{1}{2}-\sigma _{c}}\right) \end{aligned}$$

Since $q_{c}$ is distributed uniformly on [0, 1], it follows that:

$$\begin{aligned} P(\sigma _{c})=\max \left\{ 0,1-\frac{\frac{k_{e}}{1-k_{e}}-\frac{1}{2}}{\frac{1}{2}-\sigma _{c}}\right\} \end{aligned}$$

Therefore, the objective function $F(\sigma _{c})$ can be written as:

$$\begin{aligned} F(\sigma _{c})=w\cdot \max \left\{ 0,1-\frac{\frac{k_{e}}{1-k_{e}}-\frac{1}{2}}{\frac{1}{2}-\sigma _{c}}\right\} +\frac{\theta }{2}\left( \frac{1}{2}-\sigma _{c}^{2}+\sigma _{c}\right) +\frac{3}{8}\theta . \end{aligned}$$

(5)

We will use Eq. 5 in the following proofs.

Proposition 1

Proof

Since

$$\begin{aligned} 1-\frac{{\tilde{k}}_{e}-\frac{1}{2}}{\frac{1}{2} -\sigma _{c}}\le 0\,\,\forall \sigma _{c}\in \left[ 1-\tilde{k_{e}},\frac{1}{2}\right] , \end{aligned}$$

the objective function becomes $F(\sigma _{c})=\frac{\theta }{2}(\frac{1}{2}-\sigma _{c}^{2}+\sigma _{c})+\frac{3}{8}\theta$ within the interval $[1-\tilde{k_{e}},\frac{1}{2}]$, and it achieves its maximum at $\frac{1}{2}.$ Therefore, the choice domain of the maximization problem can be reduced to $[0,1-\tilde{k_{e}})\cup \{\frac{1}{2}\}$.

The solution to the maximization problem can be found by first finding the local optimal ${\tilde{\sigma }}_{c}$ on the interval $[0,1-{\tilde{k}}_{e}]$. If ${\tilde{\sigma }}_{c}=1-\tilde{k_{e}}$ , then the global optimum is $\sigma _{c}^{*}=\frac{1}{2}$ since $F(1-{\tilde{k}}_{e})<\frac{3}{4}\theta =F(\frac{1}{2})$. Otherwise, a comparison between $F({\tilde{\sigma }}_{c})$ and $F(\frac{1}{2})$ determines the global optimum.

It is easy to verify that the second order derivative of F within the interval $[0,1-\tilde{k_{e}}]$ is negative because $\tilde{k_{e}}>\frac{1}{2}$ and $\sigma _{c}<\frac{1}{2}$. Consequently, the first order condition is sufficient and necessary for ${\tilde{\sigma }}_{c}$.

Let $F^{*}(w,\theta ,{\tilde{k}}_{e})=\max _{\sigma _{c}\in [0,1-{\tilde{k}}_{e}]}F$. By Berge’s theorem, $F^{*}$ is continuous in its arguments, in particular it is continuous in ${\tilde{k}}_{e}.$ We will show below the existence of a ${\tilde{k}}_{e}$ for which $F^{*}$ exceed $\frac{3}{4}\theta$ and another ${\tilde{k}}_{e}$ for which $F^{*}$ is less than $\frac{3}{4}\theta$. This will allow us to apply the intermediate value theorem and conclude that there exists ${\bar{k}}_{e}$ such that $F^{*}(w,\theta ,{\bar{k}}_{e})=\frac{3}{4}\theta =F(\sigma _{c}=\frac{1}{2})$. This ${\bar{k}}_{e}$ is the threshold we desire because $F^{*}$ is monotonic in ${\tilde{k}}_{e}$.

It is straightforward to see that for ${\tilde{k}}_{e}$ sufficiently low, $F^{*}(w,\theta ,{\tilde{k}}_{e})>\frac{3}{4}\theta =F(\sigma _{c}=\frac{1}{2}).$ When ${\tilde{k}}_{e}$ sufficiently low, a positive probability of a centrist victory can be obtained with an infinitesimal loss of expressive benefit. Next, we would like to show that the reverse inequality holds for sufficiently high ${\tilde{k}}_{e}$. Note that the inequality $F^{*}(w,\theta ,{\tilde{k}}_{e})<\frac{3}{4}\theta$ is necessary and a sufficient condition for $\frac{1}{2}$ being optimum. Hence, a sufficient condition for the optimality of $\frac{1}{2}$ is:

$$\begin{aligned} \frac{3}{4}\theta \ge \max _{\sigma _{c}\in [0,1-\tilde{k_{e}}]}\left\{ wP(\sigma _{c})\right\} +\max _{\sigma _{c}\in [0,1-\tilde{k_{e}}]}\left\{ \frac{\theta }{2}\left( \frac{1}{2}-\sigma _{c}^{2}+\sigma _{c}\right) \right\} +\frac{3}{8}\theta \end{aligned}$$

Because $\frac{1}{2}=\text {argmax}_{\sigma _{c}}\frac{\theta }{2}\left( \frac{1}{2}-\sigma _{c}^{2}+\sigma _{c}\right)$, $\max _{\sigma _{c}\in [0,1-\tilde{k_{e}}]}\left\{ \frac{\theta }{2}\left( \sigma _{c}^{2}-\sigma _{c}+\frac{1}{2}\right) \right\} =\frac{3}{8}\theta -\epsilon$ for some positive $\epsilon$. Additionally, $\max _{\sigma _{c}\in [0,1-\tilde{k_{e}}]}\left\{ wp(\sigma _{c})\right\} =w(2-2{\tilde{k}}_{e})$. Rewriting the sufficient condition, we have:

$$\begin{aligned} \frac{3}{4}\theta \ge w(2-2{\tilde{k}}_{2})+\frac{3}{4}\theta -\epsilon \iff {\tilde{k}}_{e}\ge 1-\frac{\epsilon }{2w} \end{aligned}$$

Thus, for ${\tilde{k}}_{e}$ sufficiently high, $\frac{1}{2}$ is the optimum and equivalently, $F^{*}(w,\theta ,{\tilde{k}}_{e})<\frac{3}{4}\theta$. $\square$

Corollary 1

Proof

Recall from the proof of Proposition 1 that the threshold ${\bar{k}}_{e}$ is the solution to the equation

$$\begin{aligned} F^{*}(w,\theta ,{\bar{k}}_{e})=\frac{3}{4}\theta = F\left( \sigma _{c}=\frac{1}{2}\right) . \end{aligned}$$

(6)

where $F^{*}(w,\theta ,{\tilde{k}}_{e})=\max _{\sigma _{c}\in [0,1-{\tilde{k}}_{e}]}F$ is the maximum aggregate social welfare conditioned on positive levels of misaligned voting. Moreover, by the Berge’s maximum theorem, $F^{*}(\cdot ,\cdot ,\cdot )$ is continuous in its arguments. Moreover, if we rewrite equality 6 as $F^{*}(w,\theta ,{\bar{k}}_{e})-\frac{3}{4}\theta =0$, then by the implicit function theorem, we have that ${\bar{k}}_{e}$ is a continuous function of w and $\theta$. Observe that by the envelop theorem, $F^{*}(w,\theta ,{\bar{k}}_{e})-\frac{3}{4}\theta$ is increasing in w , decreasing in $\theta$ and decreasing in ${\tilde{k}}_{e}$. it follows that ${\bar{k}}_{e}$ must be increasing in w and decrease in $\theta$. Finally, note that the expression $\left( F^{*}(w,\theta ,{\bar{k}}_{e})-\frac{3}{4}\theta \right) \frac{1}{w}$, seen as a function of $\frac{\theta }{w}$, is decreasing in $\frac{\theta }{w}$. And therefore it follows that ${\bar{k}}_{e}$ must be decreasing in $\frac{\theta }{w}$. $\square$

Proposition 2

Proof

Note first that, $F(0)\ge F(\frac{1}{2})\iff w\left( 1-\frac{{\tilde{k}}_{e}-\frac{1}{2}}{\frac{1}{2}}\right) +\frac{\theta }{4}\ge \frac{3}{8}\theta \iff 1-\frac{\theta }{16w}\ge {\tilde{k}}_{e}$. This implies that $1-\frac{\theta }{16w}\ge {\tilde{k}}_{e}$ is sufficient for misaligned voting. Furthermore, $\frac{\partial F(0)}{\partial \sigma _{c}}=-\frac{w({\tilde{k}}_{e}-\frac{1}{2})}{\frac{1}{4}}+\frac{\theta }{2}\le 0 \iff \frac{1}{2}+\frac{\theta }{8w}\le {\tilde{k}}_{e}$. This means that for $\frac{1}{2}+\frac{\theta }{8w} \le {\tilde{k}}_{e} \le \frac{\theta }{16w}\ge {\tilde{k}}_{e}$, $\sigma ^{*}=0$. $\square$

Proposition 3

Proof

We shall use the monotone comparative statics results from Milgrom and Shannon (1994): If the cross derivative of F with respect to the choice variable (i.e. $\sigma _{c}$) and the parameter of interest, and if the cross partial is positive (negative), then the solution is increasing (decreasing) in that parameter.

Observe that for voting rules $\sigma _{c}\in [0,1-{\tilde{k}}_{e}),\text { the probability of centrist victory is }P(\sigma _{c})=1-\frac{{\tilde{k}}_{e}-\frac{1}{2}}{\frac{1}{2}-\sigma _{c}}$. Therefore the aggregate welfare (see expression (5)) is

$$\begin{aligned} F(\sigma _{c})=w\cdot \left( 1-\frac{{\tilde{k}}_{e}-\frac{1}{2}}{\frac{1}{2}-\sigma _{c}}\right) +\frac{\theta }{2} \left( \frac{1}{2}-\sigma _{c}^{2}+\sigma _{c}\right) +\frac{3}{8}\theta . \end{aligned}$$

From this, one can easily verify that $\frac{\partial ^{2}F}{\partial w\partial \sigma _{c}}<0,$$\frac{\partial ^{2}F}{\partial \theta \partial \sigma _{c}}>0$ and $\frac{\partial ^{2}F}{\partial {\tilde{k}}_{e}\partial \sigma _{c}}<0$. $\square$

Proposition 4

Proof

The result with respect to w and $\theta$ follows immediate from Proposition 1 and 3 . We will focus instead on the comparative statics with respect to ${\tilde{k}}_{e}$. When $0<\sigma _{c}^{*}<\frac{1}{2}$, the first order conditions characterizes the optimum and the following expression is obtained: $\sigma _{c}^{*}=\frac{1}{2}-\root 3 \of {\frac{w}{\theta }\left( {\tilde{k}}_{e}-\frac{1}{2}\right) }.$ Thus, we see that $\frac{\partial \sigma _{c}^{*}}{\partial {\tilde{k}}_{e}}=-\frac{w}{3\theta }\left( \frac{w}{\theta }({\tilde{k}}_{e}-\frac{1}{2})\right) ^{-\frac{2}{3}}$. From the expression for the probability of the center winning the election, we obtain:

$$\begin{aligned} \frac{\partial P(\sigma _{c}^{*})}{\partial {\tilde{k}}_{e}} =-\frac{\frac{1}{2}-\sigma _{c}^{*}+({\tilde{k}}_{e} -\frac{1}{2})\frac{\partial \sigma _{c}^{*}}{\partial {\tilde{k}}_{e}}}{(\frac{1}{2}-\sigma _{c}^{*})^{2}} \end{aligned}$$

Replacing our previous values for $\sigma _{c}^{*}$ and $\frac{\partial \sigma _{c}^{*}}{\partial {\tilde{k}}_{e}}$, we obtain that the numerator is equal to $\frac{2}{3}\root 3 \of {\frac{w}{\theta }\left( {\tilde{k}}_{e}-\frac{1}{2}\right) }>0$. Since the denominator is also positive and the fraction is multiplied by -1, we get the probability of winning is decreasing when $\sigma ^{*}$ is interior. The results extends easily to the case where $\sigma ^{*}=0$ (refer to Proposition 1 and 2) $\square$

Proposition 5

Proof

Let $F(\sigma _{c})$ denote the expected social welfare (i.e., expression (2)). It can be shown that

$$\begin{aligned} F(\sigma _{c})={\left\{ \begin{array}{ll} p(\sigma _{c}){\mathbb {E}}(w_{r})-\frac{\theta }{2}\left( \sigma _{c}^{2} -\sigma _{c}+\frac{1}{2}\right) -\frac{\theta }{8} &{} \text { if }\sigma _{c}\le \frac{1}{2}\\ p(\sigma _{c}){\mathbb {E}}(w_{l})-\frac{\theta }{2} \left( \sigma _{c}^{2}-\sigma _{c}+\frac{1}{2}\right) -\frac{\theta }{8} &{} \text { if }\sigma _{c}>\frac{1}{2} \end{array}\right. } \end{aligned}$$

where ${\mathbb {E}}(w_{r})$ is the expectation of $w_{r}(\cdot )$ taken with respect to x (similarly for ${\mathbb {E}}(w_{l})$). First, note that $F(\sigma _{c})$ is not symmetric around $\frac{1}{2}$ unless ${\mathbb {E}}(w_{r})={\mathbb {E}}(w_{l}).$ It is the generically the case ${\mathbb {E}}(w_{r})\ne {\mathbb {E}}(w_{l})$. Now, clearly if ${\mathbb {E}}(w_{r})>{\mathbb {E}}(w_{l})$, then for any $\sigma _{c}>\frac{1}{2}$, $F(\sigma _{c})<F(1-\sigma _{c})$ and therefore the ethical voting rule must be$\sigma _{c}^{*}\in [0,\frac{1}{2}]$. Similarly, if ${\mathbb {E}}(w_{r})<{\mathbb {E}}(w_{l})$, then it must be the case that $\sigma _{c}^{*}\in [\frac{1}{2},1]$. $\square$

Proposition 6

Proof

The ethical voting rule must minimize the aggregate opportunity cost of misaligned voting conditional on a certain probability of center-right victory. Hence, we first need to identify the opportunity cost of voting for the certain-right. If the voter would have abstained otherwise (i.e. $\max {1-x,x}-t\le 0$), the opportunity cost of voting for the center right is $t-x$. If the voter would have turnout and vote for the center-left (i.e. $1-x-t>0, x \le \frac{1}{2}$), the opportunity cost of switching his vote is $1-2x$. The aggregate opportunity cost will be minimized when $\max _{(x,t) \in R}t-x = \max _{(x,t) \in R} 1 - 2x$. Otherwise, the same probability of victory can be achieved at a lower cost by making abstainers head to the poll instead of center-left voters switching their votes to the center-right or viceversa.

Say that ethical voting rule $\sigma _c^* \in [0,\frac{1}{2}]$ requires all voters with $x\ge \sigma _c^*$ who are already casting a ballot for the center-left to switch their votes to the center-right. This means that $\max _{(x,t) \in R}1-2x = 1-2\sigma _c^*$. As $\max _{(x,t) \in R}t-x = \max _{(x,t) \in R} 1 - 2x$, $\max _{(x,t) \in R} t-x = 1 -2\sigma _c^*$. Hence, ethical voting rule $\sigma _c^*$ requires abstainers with $t \le 1 - 2 \sigma _c^* + x$ to vote for the center-right.

Finally, notice that $1-2\sigma _c^* + x = 1 - x$ when $x=\sigma _c^*$. This means that the restriction that the ethical voting rule imposes on center-left voter and that on abstainers intersect at the boundary that separates center-left voters from abstainers (i.e. $1-x-t=0$). Hence, we obtain that, under ethical voting rule $\sigma _c^* \in [0,\frac{1}{2}]$ , voters of type (x,t)

• 1. vote for the center-right candidate if $\sigma _{c}^{*}\le x$ and $t\le 1-2\sigma _{c}^{*}+x$.

• 2. vote for the center-left if $x<\sigma _{c}^{*}$ and $t<1-x$.

• 3. abstain otherwise.

$\square$

Proposition 7

Proof

First, we obtain the expression for the probability of victory for the centrist as a function of the ethical voting rule i.e., $P(\sigma _{c})$. Note that for $\tau \le \frac{1-k_{e}}{2}$, the probability of centrist victory is 1. On the other hand, if $\sigma _{c}>1-\frac{k_{e}}{1-k_{e}}$ and $\tau >(1-\sigma _{c})(1-k_{e})$, then we have the probability of victory being 0. Thus, we have that:

If $\sigma _{c}\le 1-\frac{k_{e}}{1-k_{e}}=1-{\tilde{k}}_{e}$,

$$\begin{aligned} P(\sigma _{c})&=\frac{1-k_{e}}{2k_{e}}+ \int _{\frac{1-k_{e}}{2}}^{k_{e}} \left( 1-\frac{\frac{2x}{1-k_{e}}-1}{1-2\sigma _{c}}\right) \frac{1}{k_{e}}dx\\&=\frac{1-k_{e}}{2k_{e}}+\left. \frac{2(1-\sigma _{c}) \tau -\frac{\tau ^{2}}{1-k_{e}}}{(1-2\sigma _{c}) k_{e}}\right| _{\frac{1-k_{e}}{2}}^{k_{e}}\\&=1+\frac{1-{\tilde{k}}_{e}}{1-2\sigma _{c}}-\frac{1}{4(1-2\sigma _{c}){\tilde{k}}_{e}} \end{aligned}$$

If $\sigma _{c}>1-{\tilde{k}}_{e},$

$$\begin{aligned} P(\sigma _{c})&=\frac{1-k_{e}}{2k_{e}} +\int _{\frac{1-k_{e}}{2}}^{(1-\sigma _{c})(1-k_{e})} \left( 1-\frac{\frac{2x}{1-k_{e}}-1}{1-2\sigma _{c}}\right) \frac{1}{k_{e}}dx\\&=\frac{1-k_{e}}{2k_{e}}+\frac{(1-\sigma _{c})^{2} (1-k_{e})}{1-2\sigma _{c}}-\frac{\left( \frac{3}{4}-\sigma _{c}\right) (1-k_{e})}{(1-2\sigma _{c})k_{e}}\\&=\frac{(1-\sigma _{c})^{2}}{\left( 1-2\sigma _{c}\right) {\tilde{k}}_{e}} -\frac{1}{4(1-2\sigma _{c}){\tilde{k}}_{e}} \end{aligned}$$

Recall $F(\sigma _{c})$ is the ethical voter’s objective function as defined in Eq. 1. With some algebra, one can show that

$$\begin{aligned} F'(\sigma _{c})={\left\{ \begin{array}{ll} -\frac{(1-2{\tilde{k}}_{e})^{2}}{2(1-2\sigma _{c})^{2}{\tilde{k}}_{e}} +\frac{\theta }{w}(1-2\sigma _{c}) &{} \quad \forall \sigma _{c}\le 1-{\tilde{k}}_{e}\\ -\frac{1}{2{\tilde{k}}_{e}}+\frac{\theta }{w}(1-2\sigma _{c}) &{} \quad \forall \sigma _{c}>1-{\tilde{k}}_{e} \end{array}\right. } \end{aligned}$$

It can be verified that the second order condition is satisfied and so the FOC is sufficient and necessary the optimal ethical voting rule. In addition, the cross partial with respect to $\sigma _{c}$ and ${\tilde{k}}_{e}$ satisfies

$$\begin{aligned} \frac{\partial ^{2}F}{\partial {\tilde{k}}_{e}\partial \sigma _{c}}={\left\{ \begin{array}{ll} \frac{1-4{\tilde{k}}_{e}^{2}}{2{\tilde{k}}_{e}^{2}\left( 1-2\sigma _{c}\right) ^{2}}<0 &{} \forall \sigma _{c}\le 1-{\tilde{k}}_{e}\\ \frac{1}{2y^{2}}>0 &{} \forall \sigma _{c}>1-{\tilde{k}}_{e} \end{array}\right. } \end{aligned}$$

Therefore, the objective function F is submodular in $\sigma _{c}$ and ${\tilde{k}}_{e}$ in the range $\sigma _{c}\le 1-{\tilde{k}}_{e}$ and is supermodular otherwise.

Now that $F'$ is continuous at $1-{\tilde{k}}_{e}$ and $F'\left( 1-{\tilde{k}}_{e}\right)$ is increasing in ${\tilde{k}}_{e}$. Moreover, if $\frac{\theta }{w}\le \frac{1}{2}$, then $F'\left( 1-{\tilde{k}}_{e}\right) <0$. In this case, the optimal ethical rule satisfies $\sigma _{c}^{*}\le 1-{\tilde{k}}_{e}$. We know from above that the objective function is submodular for $\sigma _{c}\le 1-{\tilde{k}}_{e}$. Therefore by monotone comparative statics, we get that $\sigma _{c}^{*}$ is always decreasing in ${\tilde{k}}_{e}$, i.e., there is always more misaligned voting in response to greater extremist popularity. Now, if $\frac{\theta }{w}>\frac{1}{2}$, then there exists some threshold ${\bar{k}}$ defined by $F'(1-{\bar{k}})=0$ such that if ${\tilde{k}}_{e}>(<){\bar{k}}$, then $\sigma _{c}^{*}>(<)1-{\tilde{k}}_{e}$. Given our observation about the super/submodularity of the objective function and monotone comparative statics, it follows that for ${\tilde{k}}_{e}\le {\bar{k}}$, we have that $\sigma _{c}^{*}$ is decreasing in ${\tilde{k}}_{e}$ but for ${\tilde{k}}_{e}>{\bar{k}}$, $\sigma _{c}^{*}$ is increasing. That is we get non-monotonicity in misaligned voting. $\square$

Proposition 8

Proof

We will argue that to achieve the same probability of victory, it is less costly for the ethical voters to misalign their votes for the center-right candidate than for the center-left candidate. To show this, it is sufficient to demonstrate that for any realizations of the fraction of ethical voters, $q_{c}$, the aggregate expressive benefit is higher conditional on the center-right candidate to attain vote share $y>G\left( \frac{1}{2}\right)$ than conditional on the center-left candidate to attain the same vote share. Let $q_{c}>0$ be arbitrary, and let $K_{l}(y)$ and $K_{r}(y)$ denote the expressive benefit to the ethical voter for achieving vote share $y>G\left( \frac{1}{2}\right)$ for the center-left and center-right candidates, respectively. Let $c_{l}$ and $c_{r}$ be cutoffs that solve, respectively,

$$\begin{aligned} y&=q_{c}G(c_{l})+(1-q_{c})G\left( \frac{1}{2}\right) \\ y&=q_{c}\left( 1-G(c_{r})\right) +(1-q_{c})\left( 1-G\left( \frac{1}{2}\right) \right) \end{aligned}$$

In other words, $c_{l}$ and $c_{r}$ are the ethical voting rules that achieves vote share y for the center-left and center-right candidates respectively. Now, based on the formulation of $B_{E}(\sigma _{c})$, we can compute the marginal change to expressive benefit of an increase in vote share for the center-left candidate, which is

$$\begin{aligned} \frac{dK_{l}}{dy}=\frac{dK_{l}}{dc_{l}}\cdot \frac{dc_{l}}{dy} =\theta (1-2c_{l})g(c_{l})\cdot \frac{1}{q_{c}g(c_{l})}=\frac{\theta }{q_{c}}(1-2c_{l}) \end{aligned}$$

Similar calculations for the center-right candidates gets that $\frac{dK_{r}}{dy}=\frac{\theta }{q_{c}}(2c_{r}-1)$. Now, by the premise of the Proposition, it must that that $c_{l}>\frac{1}{2}$, $c_{r}<\frac{1}{2}$, and $c_{l}-\frac{1}{2}>\frac{1}{2}-c_{r}$. It follows that $\frac{dK_{l}}{dy}<\frac{dK_{r}}{dy}$. Since this is true for all y, taking integral over y we get that $K_{l}(y)<K_{r}(y)$. And since this relationship holds for all realizations of $q_{c}$, it follows that there is greater expressive benefit associated with the center-right candidate achieving a certain winning probability than with the center-left candidate achieving the same winning probability. $\square$

1.1 Expressions for $P(\sigma _c)$, $C_E(\sigma _c)$ and $C_S$ under Turnout Costs

• $P(\sigma _C)$

  Given the results in Proposition 6, the fraction of ethical voters that vote for the center-right given $\sigma _c$ is $1-\sigma _c-\sigma _c^2$. Also, note that among non-ethical centrists, $\frac{3}{8}$ will vote for the center-right, $\frac{3}{8}$ for the center-left, and $\frac{1}{4}$ abstains. Hence, the center-right candidate wins the election if:

  $$\begin{aligned}&(1-q_c)\frac{3}{8} + q_c ( 1-\sigma _c-\sigma _c^2) \ge {\tilde{k}}_e \\&q_c \ge \frac{{\tilde{k}}_e - \frac{3}{8}}{\frac{5}{8}-\sigma _c-\sigma _c^2} \end{aligned}$$

  Hence,

  $$\begin{aligned} P(\sigma _C) = \max \left\{ 0, 1 - \frac{{\tilde{k}}_e - \frac{3}{8}}{\frac{5}{8}-\sigma _c-\sigma _c^2}\right\} \end{aligned}$$

• $C_E(\sigma _c)$

  $$\begin{aligned} C_{E}(L,R)&= {} \int _{(x,t)\in L}\theta (1-x-t)dxdt+\int _{(x,t)\in R}\theta (x-t)dxdt\\ C_{E}(\sigma _c) & = {} \theta \int _0^{\sigma _c} \int _0^{1-x} (1 - x -t) dt dx + \theta \int _{\sigma _c}^1 \int _0^{x+\sigma _c} (x-t) dt dx\\& = {} \theta \int _0^{\sigma _c} \frac{(1-x)^2}{2} dx + \theta \int _{\sigma _c}^1 \frac{x^2 - (1-2\sigma _c)^2}{2} dx\\& = {} \frac{\theta }{2} \left( \sigma _c - \sigma _c^2 + \frac{\sigma _c^3}{3}\right) + \frac{\theta }{2}\left( \frac{1}{3} - (1-2\sigma _c)^2 - \frac{\sigma _c^3}{3} - (1-2\sigma _c)^2\sigma _c\right) \end{aligned}$$

• $C_S$

  Notice that $C_S = C_E(\frac{1}{2})$. Hence

  $$\begin{aligned} C_S & = {} \frac{\theta }{2} \left( \frac{1}{2} - \left( {\frac{1}{2}}\right) ^2 + \frac{\left( {\frac{1}{2}}\right) ^3}{3}\right) + \frac{\theta }{2}\left( \frac{1}{3} - (1-2\frac{1}{2})^2 - \frac{\left( {\frac{1}{2}}\right) ^3}{3} - (1-2\frac{1}{2})^2\frac{1}{2}\right) \\ & = {} \frac{\theta }{2} \left( \frac{7}{24} \right) + \frac{\theta }{2} \left( \frac{7}{24} \right) = \frac{7}{24} \theta \end{aligned}$$

Proposition 9

Proof

First, note that if $\sigma _r=\frac{1}{2}$ (i.e., all center-right voters vote for the center-right), the center-left candidate never wins the election. Hence, the probability of a centrist victory is given by $P(\sigma _l, \frac{1}{2}) = 1 - \frac{\tilde{k_e} - \frac{1}{2}}{\frac{1}{2} - \sigma _l}$. This expression is the same as the one in our baseline framework. Hence, the maximization problem faced by the ethical center-left voters is the same as the one centrists faced in our baseline model, which means that $\sigma _c^*$ is a best-response for center-left ethical voters.

Now, suppose that $\sigma _l=\sigma _c^*$. We want to show that $\sigma _r^*=\frac{1}{2}$ is a best response for the center-right ethical voters. First, note that center-right voters would never adopt $\frac{1}{2}<\sigma _r \le 1 - \sigma _l^*$ as $P(\sigma _l^*, \frac{1}{2})\ge P(\sigma _l, \sigma _r)$. This is because the center-left candidate never wins when $\sigma _r \le 1 - \sigma _l$ but the expressive benefit is lower given that $\frac{1}{2}<\sigma _r \le 1 - \sigma _l^*$. If center-right voters choose $\sigma _r>1 - \sigma _l^*$, we have that $G_r(\sigma _l^*, \sigma _r) < G_l(1-\sigma _r, \frac{1}{2})$ as $\sigma _r$ induces the same expressive benefit to center-right voters as $\sigma _l=1-\sigma _r$ to center-left voters, but $P(\sigma _l^*, \sigma _r)=1 - \frac{\tilde{k_e} - \frac{1}{2}}{\sigma _l^* - (1-\sigma _r)}<P(1-\sigma _r, \frac{1}{2})=1 - \frac{\tilde{k_e} - \frac{1}{2}}{\frac{1}{2} - (1-\sigma _r)}$. Since $\sigma _l=\sigma _c^*$ is the unique best response for center-left voters when $\sigma _r=\frac{1}{2}$, we have $G_l(1-\sigma _r, \frac{1}{2}) < G_l(\sigma _c^*, \frac{1}{2})$. Finally, note that $G_l(\sigma _c^*, \frac{1}{2}) \le G_r(\sigma _c^*, \frac{1}{2})$ since both the center-right and center-left voters face the same probability of a centrist victory but center-left voters have a weakly lower expressive benefit as they are misaligning their vote. The discussion above means that $G_r(\sigma _l^*, \sigma _r)< G_l(1-\sigma _r, \frac{1}{2}) < G_l(\sigma _c^*, \frac{1}{2}) \le G_r(\sigma _c^*, \frac{1}{2})$ for all $\sigma _r>\frac{1}{2}$. Hence, $\sigma _r^*=\frac{1}{2}$ is a best response for center-right ethical voters. $\square$

Proposition 10

Proof

Suppose that $\sigma _r=\frac{1}{2}$, then the center-left never wins and so the probability of a centrist victory is $P(\sigma _l, \sigma _r) = 1 - \frac{\tilde{k_e} - \frac{1}{2}}{\frac{1}{2} - \sigma _l}$. Note that this is the same expression we had for our baseline framework. Hence, the best response for the center-left ethical voters, $\sigma _l^*$ is equal to $\sigma _c^*$ as in our baseline framework.

Now, suppose that $\sigma _l = \sigma _c^*=\frac{1}{2}$ (i.e., when it is optimal for the center-left to vote sincerely), it is not optimal for center-right voters to deviate from $\sigma _r=\frac{1}{2}$. In that case, center-right voters face the same problem as center-left voters, so their optimal strategy is to vote sincerely as well (i.e., $\sigma _r^*=\frac{1}{2}$).

Next, we consider the case when $\sigma _l =\sigma _c^*=0$. Before proceeding, lets compute the probabilities of a centrist victory for when $\sigma _l \in [0, 1-{\tilde{k}}_e]$. For the center-left candidate to win, it must be the case that $q_l \le \frac{\frac{1}{2}-{\tilde{k}}_e + q_r(\sigma _r - \frac{1}{2})}{\frac{1}{2}-\sigma _l}$. This is means that the probability of a center-left victory, $P_l(\sigma _l, \sigma _r)$ is given by:

$$\begin{aligned} P_l(\sigma _l, \sigma _r) = \int _{\min \left\{ \frac{{\tilde{k}}_e - \frac{1}{2}}{\sigma _r - \frac{1}{2}},1\right\} }^{1} \min \left\{ \frac{\frac{1}{2}-{\tilde{k}}_e + q_r(\sigma _r - \frac{1}{2})}{\frac{1}{2}-\sigma _l},1\right\} dq_r \end{aligned}$$

and after some tedious algebra, we have that

$$\begin{aligned} P_l(\sigma _l, \sigma _r) = {\left\{ \begin{array}{ll} 0 &{} \text { if }\sigma _{r}\le {\tilde{k}}_e\\ \frac{\sigma _r - \frac{1}{2}}{2(\frac{1}{2}-\sigma _l)} - \frac{{\tilde{k}}_e - \frac{1}{2}}{\frac{1}{2} - \sigma _l}+\frac{({\tilde{k}}_e-\frac{1}{2})^2}{2(\frac{1}{2}-\sigma _l)(\sigma _r - \frac{1}{2})} &{} \text { if }{\tilde{k}}_e< \sigma _{r}< {\tilde{k}}_e + \frac{1}{2}-\sigma _l\\ 1 - \frac{{\tilde{k}}_e - \frac{\sigma _l}{2}-\frac{1}{4}}{\sigma _r - \frac{1}{2}} &{} \text { if }\sigma _{r} \ge \min \left\{ {\tilde{k}}_e + \frac{1}{2}-\sigma _l,1\right\} \end{array}\right. } \end{aligned}$$

The center-right candidate wins if $q_l>\frac{{\tilde{k}}_e - \frac{1}{2}+q_r(\sigma _r - \frac{1}{2})}{\frac{1}{2}-\sigma _l}$. Hence, the probability that the center-right wins, $P_r(\sigma _l,\sigma _r)$ is:

$$\begin{aligned} P_r(\sigma _l, \sigma _r) = \int _{0}^{\min \left\{ \frac{1-{\tilde{k}}_e - \sigma _l}{\sigma _r - \frac{1}{2}},1\right\} } 1 - \frac{{\tilde{k}}_e - \frac{1}{2}+q_r(\sigma _r - \frac{1}{2})}{\frac{1}{2}-\sigma _l} dq_r \end{aligned}$$

and after some tedious algebra, we have that

$$\begin{aligned} P_r(\sigma _l, \sigma _r) = {\left\{ \begin{array}{ll} 1 - \frac{{\tilde{k}}_e - \frac{1}{2}}{\frac{1}{2}-\sigma _l} - \frac{\sigma _r - \frac{1}{2}}{2(\frac{1}{2}-\sigma _l)} &{} \text { if }\sigma _{r}\le 1 - {\tilde{k}}_e + \frac{1}{2}-\sigma _l\\ \frac{(1-{\tilde{k}}_e-\sigma _l)^2}{2(\sigma _r - \frac{1}{2})(\frac{1}{2}-\sigma _l)}&{} \text { if }\sigma _{r}> 1 - {\tilde{k}}_e + \frac{1}{2}-\sigma _l \end{array}\right. } \end{aligned}$$

Note that the above implies that, given $\sigma _l \in [0, 1-{\tilde{k}}_e]$, center-right voters have no incentive to deviate from $\sigma _r=\frac{1}{2}$ to $\sigma _r \le \max \left\{ \tilde{k_e}, 1 - {\tilde{k}}_e + \frac{1}{2} - \sigma _l\right\}$. In particular, the center-right voters will not deviate to $\sigma _r \le {\tilde{k}}_e$ is trivial as, in that case there is no increase in the probability of a center-left victory from increasing $\sigma _r$ (and there is a cost in terms of a lower probability of a center-right victory). If $1 - {\tilde{k}}_e + \frac{1}{2} - \sigma _l>{\tilde{k}}_e$ and $\sigma _r \le 1 - {\tilde{k}}_e + \frac{1}{2} - \sigma _l$, the probability of a centrist victory is $P(\sigma _l,\sigma _r)=P_l(\sigma _l,\sigma _r)+P(\sigma _l, \sigma _r) = 1 - 2\frac{\tilde{k_e} - \frac{1}{2}}{\frac{1}{2} - \sigma _l} + \frac{({\tilde{k}}_e-\frac{1}{2})^2}{2(\frac{1}{2}-\sigma _l)(\sigma _r - \frac{1}{2})}$. This is lower than $P(\sigma _l,\frac{1}{2})=P_l(\sigma _l,\frac{1}{2})+P(\sigma _l, \frac{1}{2}) = 1 - \frac{\tilde{k_e} - \frac{1}{2}}{\frac{1}{2} - \sigma _l}$ as $-\frac{\tilde{k_e} - \frac{1}{2}}{\frac{1}{2} - \sigma _l} + \frac{({\tilde{k}}_e-\frac{1}{2})^2}{2(\frac{1}{2}-\sigma _l)(\sigma _r - \frac{1}{2})}<-\frac{\tilde{k_e} - \frac{1}{2}}{\frac{1}{2} - \sigma _l} + \frac{({\tilde{k}}_e-\frac{1}{2})^2}{2(\frac{1}{2}-\sigma _l)({\tilde{k}}_e - \frac{1}{2})}=-\frac{\tilde{k_e} - \frac{1}{2}}{2(\frac{1}{2} - \sigma _l)} <0, \forall \sigma _r>{\tilde{k}}_e$.

Now, suppose that $\sigma _l = \sigma _c^* = 0$. In this case,

$$\begin{aligned} P_l(\sigma _l, \sigma _r) = \sigma _r - \frac{1}{2}- 2\left( {\tilde{k}}_e - \frac{1}{2}\right) +\frac{({\tilde{k}}_e-\frac{1}{2})^2}{(\sigma _r - \frac{1}{2})} \end{aligned}$$

and

$$\begin{aligned} P_r(\sigma _l, \sigma _r) = {\left\{ \begin{array}{ll} 1 - 2\left( {\tilde{k}}_e - \frac{1}{2}\right) - \left( \sigma _r - \frac{1}{2}\right) &{} \text { if }\sigma _{r}\le 1 - {\tilde{k}}_e + \frac{1}{2}\\ \frac{(1-{\tilde{k}}_e)^2}{\sigma _r - \frac{1}{2}}&{} \text { if }\sigma _{r}> 1 - {\tilde{k}}_e + \frac{1}{2} \end{array}\right. } \end{aligned}$$

As argued above, the center-right voters will not deviate from $\sigma _r=\frac{1}{2}$ to

$$\begin{aligned} \sigma _r \le \max \left\{ \tilde{k_e}, 1 - {\tilde{k}}_e + \frac{1}{2} - \sigma _l\right\} . \end{aligned}$$

For $\sigma _r \ge 1 - {\tilde{k}}_e + \frac{1}{2}$, we have that $P(0,\sigma _r) = (\sigma _r - \frac{1}{2}) - 2({\tilde{k}}_e - \frac{1}{2}) + \frac{({\tilde{k}}_e - \frac{1}{2})^2}{\sigma _r - \frac{1}{2}} + \frac{(1-{\tilde{k}}_e)^2}{\sigma _r - \frac{1}{2}} < (1 - \frac{1}{2}) - 2({\tilde{k}}_e - \frac{1}{2}) + \frac{({\tilde{k}}_e - \frac{1}{2})^2}{(1 - {\tilde{k}}_e + \frac{1}{2}) - \frac{1}{2}} + \frac{(1-{\tilde{k}}_e)^2}{(1 - {\tilde{k}}_e + \frac{1}{2}) - \frac{1}{2}} = 1 - 2({\tilde{k}}_e - \frac{1}{2}) = P(0, \frac{1}{2})$. Thus the center right ethical voters would not find such a deviation profitable. Therefore, we have that $\sigma _r^*=\frac{1}{2}$ is a best response to $\sigma _l=0$.

Now, notice that $P(\sigma _l, \sigma _r)$ is continuous in its arguments. Hence, for all $\delta$, there is $\epsilon$ such that for $\sigma _l^* = \sigma _c^*<\epsilon$, $|P(\sigma _l^*,\sigma _r) - P(0, \sigma _r)| < \delta , \ \forall \sigma _r \in [\frac{1}{2},1]$. Since the only profitable deviations imply a decrease of at least $\int _{\frac{1}{2}}^{\sigma _r} \theta (1-x)dx + \int _{\sigma _r}^{1} \theta x dx - \frac{3}{8} \theta > 0$ in $G_r(\sigma _l, \sigma _r)$, there is $\epsilon >0$ for which $\sigma _r^*=\frac{1}{2}$ and $\sigma _l^*<\epsilon$ is a consistent rule profile. $\square$