Electoral Institutions with impressionable voters

Abstract

We use a model of impressionable voters to study multi-candidate elections under different electoral rules. Instead of maximizing expected utility, voters cast their ballots based on impressions. We show that, under each rule, there is a monotone relationship between voter preferences and vote measures. The nature of this relationship, however, varies by electoral rule. Vote measures are biased upwards for socially preferred candidates under plurality rule, but biased downwards under negative plurality. There is no such bias under approval voting or Borda count. Voters always elect the socially preferred candidate in two-way races for any electoral rule. In multi-candidate elections, however, the ability to elect a Condorcet winner varies by rule. The results show that multi-candidate elections can perform well even if voters follow simple behavioral rules. The relative performance of specific electoral institutions, however, depends on the assumed behavioral model of voting.

Appendix

1.1 Proof of Proposition 1

The proof for the characterization of the stationary distributions under approval voting, and negative plurality is similar to that under plurality rule (Andonie and Diermeier 2019, Lemma 1). First, consider approval voting. The equation for the updating of propensity $p_{\theta }^{t}(i)$ for candidate $i\in N$ can be written concisely as

$$\begin{aligned} p_{\theta }^{t+1}(i)=(1-\lambda _{p})p_{\theta }^{t}(i)+\lambda _{p}\mathbf {1 }_{\{\pi _{\theta }^{t}(i)\ge a_{\theta }^{t}\}}. \end{aligned}$$

This implies

$$\begin{aligned} E[p_{\theta }^{t+1}(i)]= \,& {} (1-\lambda _{p})E[p_{\theta }^{t}(i)]+\lambda _{p}E[{\mathbf {1}}_{\{\pi _{\theta }^{t}(i)\ge a_{\theta }^{t}\}}] \\=\, & {} (1-\lambda _{p})E[p_{\theta }^{t}(i)]+\lambda _{p}q_{\theta }^{t}(i). \end{aligned}$$

In a stationary distribution, the distributions of propensities $p_{\theta }^{t}$ and aspirations $a_{\theta }^{t}$ do not depend on time t, and therefore, dropping the index t in the previous equation, we obtain

$$\begin{aligned} E[p_{\theta }(i)]=q_{\theta }(i). \end{aligned}$$

Because there is a continuum of voters of type $\theta$, this implies that the measure of voters of type $\theta$ that vote for (or, approve) candidate i is: $S_{\theta }(i)=E[p_{\theta }(i)]=q_{\theta }(i)$.

In regard to abstentions, the equation for the updating of abstention propensity $p_{\theta }(A)$ is: $p_{\theta }^{t+1}(A)=(1-\lambda _{p})p_{\theta }^{t}(A)+\lambda _{p}{\mathbf {1}}_{\{I_{\theta }^{t}=0\}}$. Following the same steps as above, we obtain in a stationary distribution $S_{\theta }(A)=\prod _{j\in N}(1-q_{\theta }(j))$.

Now, consider negative plurality. The equation for the updating of propensity $p_{\theta }^{t}(i)$ for a candidate $i\in N$ can be written concisely as

$$\begin{aligned} p_{\theta }^{t+1}(i)=(1-\lambda _{p})p_{\theta }^{t}(i)+\lambda _{p}\left( 1-\frac{ 1}{1+J_{\theta }^{t}(-i)}{\mathbf {1}}_{\{\pi _{\theta }^{t}(i)<a_{\theta }^{t}\}}\right) . \end{aligned}$$

This implies

$$\begin{aligned} E[p_{\theta }^{t+1}(i)]= \,& {} (1-\lambda _{p})E[p_{\theta }^{t}(i)]+\lambda _{p}\left( 1-E\left[ \frac{1}{1+J_{\theta }^{t}(-i)}{\mathbf {1}}_{\{\pi _{\theta }^{t}(i)<a_{\theta }^{t}\}}\right] \right) \\=\, & {} (1-\lambda _{p})E[p_{\theta }^{t}(i)]+\lambda _{p}\left( 1-(1-q_{\theta }^{t}(i))E\left[ \frac{1}{1+J_{\theta }^{t}(-i)}|\pi _{\theta }^{t}(i)<a_{\theta }^{t}\right] \right) . \end{aligned}$$

Similarly, as above, in a stationary distribution we obtain

$$\begin{aligned} E[p_{\theta }(i)]=1-(1-q_{\theta }(i))E\left[ \frac{1}{1+J_{\theta }(-i)}|\pi _{\theta }(i)<a_{\theta }\right] . \end{aligned}$$

In a stationary distribution, aspirations $a_{\theta }$ are non-random and are equal to $a_{\theta }=\sum _{j\in N}\frac{v_{\theta }^{j}}{n}$. This permits us to drop the conditional event $\pi _{\theta }(i)<a_{\theta }$ in the expectation term above (see also Andonie and Diermeier 2019, Proof of Lemma 1), and therefore to obtain

$$\begin{aligned} S_{\theta }(i)=E[p_{\theta }(i)]=1-(1-q_{\theta }(i))E\left[ \frac{1}{1+J_{\theta }(-i)}\right] . \end{aligned}$$

Finally, the equation for updating the abstention propensity $p_{\theta }^t(A)$, under negative plurality, is: $p_{\theta }^{t+1}(A)=(1-\lambda _p)p_{\theta }^{t}(A)+\lambda _p{\mathbf {1}}_{\{J_{\theta }^t=0\}}$. Taking expectation in this equation and, dropping the time index t, we obtain that in a stationary distribution the measure of voters of type $\theta$ that abstain is: $S_{\theta }(A)=\prod _{j\in N}q_{\theta }(j)$.

1.2 Proof of Proposition 2

In the plurality rule, by Proposition 1, the measure of votes candidate i obtains in a stationary distribution is $S_{\theta }(i)=q_{\theta }(i)E[ \frac{1}{1+I_{\theta }(-i)}]$. Since the expectation term can be decomposed as

$$\begin{aligned} E\left[ \frac{1}{1+I_{\theta }(-i)}\right] =q_{\theta }(j)E\left[ \frac{1}{2+I_{\theta }(-i,j)}\right] +(1-q_{\theta }(j))E\left[ \frac{1}{1+I_{\theta }(-i,j)}\right] , \end{aligned}$$

it follows that we can write $S_{\theta }(i)$ as:

$$\begin{aligned} S_{\theta }(i)=q_{\theta }(i)q_{\theta }(j)E\left[ \frac{1}{2+I_{\theta }(-i,j)} \right] +q_{\theta }(i)(1-q_{\theta }(j))E\left[ \frac{1}{1+I_{\theta }(-i,j)}\right] . \end{aligned}$$

Therefore the difference in measures of votes of candidates i and j is:

$$\begin{aligned} S_{\theta }(i)-S_{\theta }(j)=(q_{\theta }(i)-q_{\theta }(j))E\left[ \frac{1}{ 1+I_{\theta }(-i,j)}\right] . \end{aligned}$$

In the approval rule, by Proposition 1, the measure of votes candidate i obtains in a stationary distribution is $S_{\theta }(i)=q_{\theta }(i)$. Therefore the difference in measures of votes of candidates i and j is:

$$\begin{aligned} S_{\theta }(i)-S_{\theta }(j)=q_{\theta }(i)-q_{\theta }(j). \end{aligned}$$

In the negative plurality rule, by Proposition 1, the measure of votes candidate i obtains in a stationary distribution is: $S_{\theta }(i)=1-(1-q_{\theta }(i))E[\frac{1}{n-I_{\theta }(-i)}]$. Since we can decompose the expectation term as

$$\begin{aligned} E\left[ \frac{1}{n-I_{\theta }(-i)}\right] =q_{\theta }(j)E\left[ \frac{1}{n-1-I_{\theta }(-i,j) }\right] +(1-q_{\theta }(j))E\left[ \frac{1}{n-I_{\theta }(-i,j)}\right] \text {,} \end{aligned}$$

it follows that $S_{\theta }(i)$ can be alternatively written:

$$\begin{aligned} S_{\theta }(i)= & {} 1-(1-q_{\theta }(i))q_{\theta }(j)E\left[ \frac{1}{n-1-I_{\theta }(-i,j)}\right] \\&-(1-q_{\theta }(i))(1-q_{\theta }(j))E\left[ \frac{1}{n-I_{\theta }(-i,j)} \right] . \end{aligned}$$

Therefore, the difference between the measures of votes of candidates i and j is:

$$\begin{aligned} S_{\theta }(i)-S_{\theta }(j)=(q_{\theta }(i)-q_{\theta }(j))E\left[ \frac{1}{ n-1-I_{\theta }(-i,j)}\right] . \end{aligned}$$

Now let us formally prove the observation on the differences between the three electoral rules discussed after Proposition 2. Consider first the plurality rule. We note that $E[\frac{1}{1+I_{\theta }(-i,j)}]$ decreases with the probability of positive impression $q_{\theta }(l)$, for each candidate $l\ne i,j$. This can be seen by writing

$$\begin{aligned} E\left[ \frac{1}{1+I_{\theta }(-i,j)}\right]= \,& {} E\left[ \frac{1}{1+I_{\theta }(-i,j,l)} \right] \\&-q_{\theta }(l)E\left[ \frac{1}{(1+I_{\theta }(-i,j,l))(2+I_{\theta }(-i,j,l))}\right] . \end{aligned}$$

Consider now two candidates i, j at the top of the vector of fixed components $u_{\theta }$, and two candidates $i^{\prime }$, $j^{\prime }$ at the bottom of the vector $u_{\theta }$. The probabilities of having a good impression of candidates i and j are larger than the corresponding probabilities of candidates $i^{\prime }$ and $j^{\prime }$: $q_{\theta }(i),q_{\theta }(j)>q_{\theta }(i^{\prime }),q_{\theta }(j^{\prime })$. As the expectation $E[\frac{1}{1+I_{\theta }(-i,j)}]$ decreases with $q_{\theta }(l)$, for $l\ne i,j$, this implies that $E[\frac{1}{1+I_{\theta }(-i,j)} ]>E[\frac{1}{1+I_{\theta }(-i^{\prime },j^{\prime })}]$. In other words, if we control for the differences $q_{\theta }(i)-q_{\theta }(j)$, the differences in measures of votes $S_{\theta }(i)-S_{\theta }(j)$ are larger if candidates i and j are at the top of vector $u_{\theta }$, and smaller if candidates i and j are at the bottom.

A similar reasoning applies to the negative plurality rule. In the negative plurality, the expectation term $E[\frac{1}{n-1-I_{\theta }(-i,j)}]$ increases with the probability of positive impression $q_{\theta }(l)$, for each candidate $l\ne i,j$. To see this, we write:

$$\begin{aligned} E\left[ \frac{1}{n-1-I_{\theta }(-i,j)}\right]= & {} E\left[ \frac{1}{n-1-I_{\theta }(-i,j,l)}\right] \\&+q_{\theta }(l)E\left[ \frac{1}{(n-2-I_{\theta }(-i,j,l))(n-1-I_{\theta }(-i,j,l))}\right] . \end{aligned}$$

Similarly as above, if we consider two candidates i and j at the top of the vector $u_{\theta }$, and two candidates $i^{\prime }$ and $j^{\prime }$ at the bottom of the vector $u_{\theta }$, we will have that $q_{\theta }(i),q_{\theta }(j)>q_{\theta }(i^{\prime }),q_{\theta }(j^{\prime })$. But as the expectation $E[\frac{1}{n-1-I_{\theta }(-i,j)}]$ increases with $q_{\theta }(l)$, for $l\ne i,j$, this implies that $E[\frac{1}{ n-1-I_{\theta }(-i,j)}]<E[\frac{1}{n-1-I_{\theta }(-i^{\prime },j^{\prime })} ]$. Therefore, if we control for the differences $q_{\theta }(i)-q_{\theta }(j)$, the differences in measures of votes $S_{\theta }(i)-S_{\theta }(j)$ are smaller if candidates i and j are at the top of vector $u_{\theta }$, and larger if candidates i and j are at the bottom.

1.3 Proof of Proposition 4

By Proposition 1, under each of the three electoral rules, the stochastic process has a unique stationary distribution. In the stationary distribution, the aspirations of the two groups of voters are: $a_{1}=\frac{1+v}{3}$ and $a_{2}=\frac{1+v}{3}$. Given these aspirations, we can compute the probabilities of positive impression for each of the three candidates, across the two groups, as follows:

$$\begin{aligned} q_{1}(1)= & {} \Pr (\pi _{1}(1)>a_{1})=\Pr \left( 1+\alpha \epsilon _{1}(1)>\frac{1+v }{3}\right) =1-F\left( \frac{v-2}{3\alpha }\right) \\ q_{1}(2)= & {} \Pr (\pi _{1}(2)>a_{1})=\Pr \left( \alpha \epsilon _{1}(2)>\frac{1+v}{3 }\right) =1-F\left( \frac{1+v}{3\alpha }\right) \\ q_{1}(3)= & {} \Pr (\pi _{1}(3)>a_{1})=\Pr \left( v+\alpha \epsilon _{1}(3)>\frac{1+v }{3}\right) =1-F\left( \frac{1-2v}{3\alpha }\right) \\ q_{2}(1)= & {} \Pr (\pi _{2}(1)>a_{2})=\Pr \left( \alpha \epsilon _{2}(1)>\frac{1+v}{3 }\right) =1-F\left( \frac{1+v}{3\alpha }\right) \\ q_{2}(2)= & {} \Pr (\pi _{2}(2)>a_{2})=\Pr \left( 1+\alpha \epsilon _{2}(2)>\frac{1+v }{3}\right) =1-F\left( \frac{v-2}{3\alpha }\right) \\ q_{2}(3)= & {} \Pr (\pi _{2}(3)>a_{2})=\Pr \left( v+\alpha \epsilon _{2}(3)>\frac{1+v }{3}\right) =1-F\left( \frac{1-2v}{3\alpha }\right) . \end{aligned}$$

Using Proposition 2, the difference between the measures of votes of candidates 1 and 3 in the plurality rule is

$$\begin{aligned} S(1)-S(3)= & {} \frac{1}{2}(q_{1}(1)-q_{1}(3))\left( 1-\frac{1}{2}q_{1}(2)\right) +\frac{1}{2 }(q_{2}(1)-q_{2}(3))\left( 1-\frac{1}{2}q_{2}(2)\right) \\= & {} \frac{1}{2}\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) \right) \left( \frac{1}{2}+ \frac{1}{2}F\left( \frac{1+v}{3\alpha }\right) \right) \\&+\frac{1}{2}\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{1+v}{3\alpha }\right) \right) \left( \frac{1}{2}+ \frac{1}{2}F\left( \frac{v-2}{3\alpha }\right) \right) . \end{aligned}$$

It is easy to verify that, as intuition suggests, the difference $S(1)-S(3)$ decreases with v. When $v=\frac{1}{2}$ we have: $S(1)-S(3)=\frac{1}{2}(- \frac{1}{2}+F(\frac{3/2}{3\alpha }))^{2}>0$, while when $v=1$ we have $S(1)-S(3)=\frac{1}{2}(F(\frac{-1}{3\alpha })-F(\frac{2}{3\alpha }))(\frac{1}{ 2}+\frac{1}{2}F(\frac{-1}{3\alpha }))<0$. Therefore, there exists a cutoff $v_{p}^{*}$ with $v_{p}^{*}\in (\frac{1}{2},1)$ such that if $v<v_{p}^{*}$ then $S(1)-S(3)>0$, and if $v>v_{p}^{*}$ then $S(1)-S(3)<0$.

In the approval rule, we can use Proposition 2 to compute the difference between the measures of votes of candidates 1 and 3 as follows:

$$\begin{aligned} S(1)-S(3)= & {} \frac{1}{2}(q_{1}(1)-q_{1}(3))+\frac{1}{2}(q_{2}(1)-q_{2}(3)) \\= & {} \frac{1}{2}\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) \right) +\frac{1}{2} \left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{1+v}{3\alpha }\right) \right) . \end{aligned}$$

Again, it is easy to verify that the difference $S(1)-S(3)$ decreases with v. When $v=\frac{1}{2}$, we have $S(1)-S(3)=0$. Therefore if $v<v_{a}^{*}=\frac{1}{2}$ then $S(1)-S(3)>0$, and if $v>v_{a}^{*}= \frac{1}{2}$ then $S(1)-S(3)<0$.

Finally, in the negative plurality rule, the difference in the vote measures of candidates 1 and 3, using Proposition 2, is

$$\begin{aligned} S(1)-S(3)= & {} \frac{1}{2}(q_{1}(1)-q_{1}(3))\frac{1}{2}(1+q_{1}(2))+\frac{1}{2 }(q_{2}(1)-q_{2}(3))\frac{1}{2}(1+q_{2}(2)) \\= & {} \frac{1}{2}\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) \right) \frac{1}{2} \left( 2-F\left( \frac{1+v}{3\alpha }\right) \right) \\&+\frac{1}{2}\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{1+v}{3\alpha }\right) \right) \frac{1}{2} \left( 2-F\left( \frac{v-2}{3\alpha }\right) \right) . \end{aligned}$$

Again, as in the case of the other two rules, the difference $S(1)-S(3)$ decreases with v. When $v=\frac{1}{2}$ we have $S(1)-S(3)=-\frac{1}{2}(- \frac{1}{2}+F(\frac{3/2}{3\alpha }))^{2}<0$, and when $v=0$ we have $S(1)-S(3)=\frac{1}{2}(F(\frac{1}{3\alpha })-F(\frac{-2}{3\alpha }))(1-\frac{1 }{2}F(\frac{1}{3\alpha }))>0$. Therefore, there exists a cutoff $v_{n}^{*}$ with $v_{n}^{*}\in (0,\frac{1}{2})$ such that if $v<v_{n}^{*}$ then $S(1)-S(3)>0$, and if $v>v_{n}^{*}$ then $S(1)-S(3)<0$.

1.4 Proof of Lemma 1

Under each of the three electoral rules, the stochastic process has a unique stationary distribution. In the stationary distribution, the aspirations of the three groups of voters are: $a_{1}=\frac{1+v}{3}$, $a_{2}=\frac{1+v}{3}$ and $a_{3}=\frac{w}{3}$. Then, we can compute the probabilities of having a positive impression, for each group $\theta =1,2,3$, and for each candidate $i=1,2,3$ as follows:

$$\begin{aligned} q_{1}(1)= & {} \Pr (\pi _{1}(1)>a_{1})=\Pr \left( 1+\alpha \epsilon _{1}(1)>\frac{1+v }{3}\right) =1-F\left( \frac{v-2}{3\alpha }\right) \\ q_{1}(2)= & {} \Pr (\pi _{1}(2)>a_{1})=\Pr \left( \alpha \epsilon _{1}(2)>\frac{1+v}{3 }\right) =1-F\left( \frac{1+v}{3\alpha }\right) \\ q_{1}(3)= & {} \Pr (\pi _{1}(3)>a_{1})=\Pr \left( v+\alpha \epsilon _{1}(3)>\frac{1+v }{3}\right) =1-F\left( \frac{1-2v}{3\alpha }\right) \\ q_{2}(1)= & {} \Pr (\pi _{2}(1)>a_{2})=\Pr \left( \alpha \epsilon _{2}(1)>\frac{1+v}{3 }\right) =1-F\left( \frac{1+v}{3\alpha }\right) \\ q_{2}(2)= & {} \Pr (\pi _{2}(2)>a_{2})=\Pr \left( 1+\alpha \epsilon _{2}(2)>\frac{1+v }{3}\right) =1-F\left( \frac{v-2}{3\alpha }\right) \\ q_{2}(3)= & {} \Pr (\pi _{2}(3)>a_{2})=\Pr \left( v+\alpha \epsilon _{2}(3)>\frac{1+v }{3}\right) =1-F\left( \frac{1-2v}{3\alpha }\right) \\ q_{3}(1)=q_{3}(2)= & {} \Pr (\pi _{3}(1)>a_{3})=\Pr \left( \alpha \epsilon _{3}(1)> \frac{w}{3}\right) =1-F\left( \frac{w}{3\alpha }\right) \\ q_{3}(3)= & {} \Pr (\pi _{3}(3)>a_{3})=\Pr \left( w+\alpha \epsilon _{3}(3)>\frac{w}{3 }\right) =F\left( \frac{2w}{3\alpha }\right) . \end{aligned}$$

Using Proposition 2, the difference between candidates 1’s and 3’s measures of votes under plurality rule is

$$\begin{aligned} S(1)-S(3)= \,& {} s(q_{1}(1)-q_{1}(3))\left( 1-\frac{1}{2} q_{1}(2)\right) +s(q_{2}(1)-q_{2}(3))\left( 1-\frac{1}{2}q_{2}(2)\right) \\&+(1-2s)(q_{3}(1)-q_{3}(3))\left( 1-\frac{1}{2}q_{3}(2)\right) \\=\, & {} s\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) \right) \left( \frac{1}{2}+\frac{1}{2 }F\left( \frac{1+v}{3\alpha }\right) \right) \\&+s\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{1+v}{3\alpha }\right) \right) \left( \frac{1}{2}+\frac{1}{2 }F\left( \frac{v-2}{3\alpha }\right) \right) \\&+(1-2s)\left( 1-F\left( \frac{w}{3\alpha }\right) -F\left( \frac{2w}{3\alpha }\right) \right) \left( \frac{1}{2}+\frac{1 }{2}F\left( \frac{w}{3\alpha }\right) \right) . \end{aligned}$$

therefore the cutoff $w^{p}$ in the plurality rule is characterized by:

$$\begin{aligned}&s\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) \right) \left( \frac{1}{2}+\frac{1}{2} F\left( \frac{1+v}{3\alpha }\right) \right) \\&+s\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{1+v}{3\alpha }\right) \right) \left( \frac{1}{2}+\frac{1}{2}F\left( \frac{v-2}{3\alpha }\right) \right) \\&\quad =(1-2s)\left( F\left( \frac{w^{p}}{3\alpha }\right) +F\left( \frac{2w^{p}}{3\alpha }\right) -1\right) \left( \frac{1}{2} +\frac{1}{2}F\left( \frac{w^{p}}{3\alpha }\right) \right) \text {.} \end{aligned}$$

Similarly, for the approval rule, we can use Proposition 2 to compute the difference $S(1)-S(3)$ as follows:

$$\begin{aligned} S(1)-S(3)=\, & {} s(q_{1}(1)-q_{1}(3))+s(q_{2}(1)-q_{2}(3))+(1-2s)(q_{3}(1)-q_{3}(3)) \\= \,& {} s\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) \right) +s\left( F\left( \frac{1-2v}{ 3\alpha }\right) -F\left( \frac{1+v}{3\alpha }\right) \right) \\&+(1-2s)\left( 1-F\left( \frac{w}{3\alpha }\right) -F\left( \frac{2w}{3\alpha }\right) \right) . \end{aligned}$$

Therefore the cutoff $w^{a}$ in the approval rule is characterized by:

$$\begin{aligned}&s\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) \right) +s\left( F\left( \frac{1-2v}{ 3\alpha }\right) -F\left( \frac{1+v}{3\alpha }\right) \right) \\&\quad =(1-2s)\left( F\left( \frac{w^{a}}{3\alpha }\right) +F\left( \frac{2w^{a}}{3\alpha }\right) -1\right) . \end{aligned}$$

Finally, for the negative plurality rule, using Proposition 2, the difference $S(1)-S(3)$ is

$$\begin{aligned} S(1)-S(3)= \,& {} s(q_{1}(1)-q_{1}(3))\frac{1}{2}(1+q_{1}(2))+s(q_{2}(1)-q_{2}(3)) \frac{1}{2}(1+q_{2}(2)) \\&+(1-2s)(q_{3}(1)-q_{3}(3))\frac{1}{2}(1+q_{3}(2)) \\=\, & {} s\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) \right) \left( 1-\frac{1}{2}F\left( \frac{ 1+v}{3\alpha }\right) \right) \\&+s\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{1+v}{3\alpha }\right) \right) \left( 1-\frac{1}{2}F\left( \frac{ v-2}{3\alpha }\right) \right) \\&+(1-2s)\left( 1-F\left( \frac{w}{3\alpha }\right) -F\left( \frac{2w}{3\alpha }\right) \right) \left( 1-\frac{1}{2}F\left( \frac{w}{3\alpha }\right) \right) . \end{aligned}$$

and so the cutoff $w^{n}$ is characterized by the equation

$$\begin{aligned}&s\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) \right) \left( 1-\frac{1}{2}F\left( \frac{ 1+v}{3\alpha }\right) \right) \\&\quad +s\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{1+v}{3\alpha }\right) \right) \left( 1-\frac{ 1}{2}F\left( \frac{v-2}{3\alpha }\right) \right) \\&\quad =(1-2s)\left( F\left( \frac{w^{n}}{3\alpha }\right) +F\left( \frac{2w^{n}}{3\alpha }\right) -1\right) \left( 1-\frac{1}{ 2}F\left( \frac{w^{n}}{3\alpha }\right) \right) . \end{aligned}$$

We first make a simple observation on the relationship between the value of v and the thresholds $w^{p}$, $w^{a}$, and $w^{n}$. Intuitively, when v increases in absolute value, the relative support of voters in groups 1 and 2 for candidate 1 increases. Similarly, when w increases, the relative support of voters in group 3 for candidate 3 increases. As $w^{p}$, $w^{a}$, and $w^{n}$ are defined as the threshold for w at which the relative support of groups 1 and 2 for candidate 1, and support of group 3 for candidate 3 are balanced, then all three thresholds $w^{p}$, $w^{a}$, and $w^{n}$ should decrease with v. Indeed, it is easy to prove this result also formally, by observing that the left sides of the three equations characterizing $w^{p}$, $w^{a}$, and $w^{n}$ all decrease with v, while the right sides all increase with w.

Let us next prove two preliminary results. In the first preliminary result, we show that for any $v\le -1$ we have: $w^{p}<w^{a}<w^{n}$. If $v\le -1$, then, using the equation for $w^{p}$, and because $w^{p}\ge 0$, we can write:

$$\begin{aligned}&(1-2s)\left( F\left( \frac{w^{p}}{3\alpha }\right) +F\left( \frac{2w^{p}}{3\alpha }\right) -1\right) \\&\quad =s\left( F\left( \frac{ 1-2v}{3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) \right) \frac{\frac{1}{2}+\frac{1}{2}F\left( \frac{1+v}{3\alpha }\right) }{\frac{1}{2}+\frac{1}{2}F\left( \frac{w^{p}}{3\alpha }\right) } \\&\qquad +s\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{1+v}{3\alpha }\right) \right) \frac{\frac{1}{2}+ \frac{1}{2}F\left( \frac{v-2}{3\alpha }\right) }{\frac{1}{2}+\frac{1}{2}F\left( \frac{w^{p}}{ 3\alpha }\right) } \\&\quad <s\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) \right) +s\left( F\left( \frac{1-2v}{ 3\alpha }\right) -F\left( \frac{1+v}{3\alpha }\right) \right) \\&\quad =(1-2s)\left( F\left( \frac{w^{a}}{3\alpha }\right) +F\left( \frac{2w^{a}}{3\alpha }\right) -1\right) . \end{aligned}$$

Therefore $w^{p}<w^{a}$. Similarly, using the equation for $w^{n}$ we can write:

$$\begin{aligned}&(1-2s)\left( F\left( \frac{w^{n}}{3\alpha }\right) +F\left( \frac{2w^{n}}{3\alpha }\right) -1\right) \\&\quad =s\left( F\left( \frac{ 1-2v}{3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) \right) \frac{1-\frac{1}{2}F\left( \frac{1+v}{ 3\alpha }\right) }{1-\frac{1}{2}F\left( \frac{w^{n}}{3\alpha }\right) } \\&\qquad +s\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{1+v}{3\alpha }\right) \right) \frac{1-\frac{1}{2}F\left( \frac{v-2}{3\alpha }\right) }{1-\frac{1}{2}F\left( \frac{w^{n}}{3\alpha }\right) } \\&\quad >s\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) \right) +s\left( F\left( \frac{1-2v}{ 3\alpha }\right) -F\left( \frac{1+v}{3\alpha }\right) \right) \\&\quad =(1-2s)\left( F\left( \frac{w^{a}}{3\alpha }\right) +F\left( \frac{2w^{a}}{3\alpha }\right) -1\right) \end{aligned}$$

Therefore $w^{a}<w^{n}$. Thus, when $v\le -1$ we have $w^{p}<w^{a}<w^{n}$.

In the second preliminary result, we will show that if $v=0$ then: (a) If $s< \frac{1}{3}$ then $w^{p},w^{a},w^{n}<1$; and (b) If $s>\frac{1}{3}$ then $w^{p},w^{a},w^{n}>1$. When $v=0$, the equation characterizing $w^{p}$ in the plurality rule becomes

$$\begin{aligned}&s\left( F\left( \frac{1}{3\alpha }\right) +F\left( \frac{2}{3\alpha }\right) -1\right) \left( \frac{1}{2}+\frac{1}{2}F\left( \frac{1}{3\alpha }\right) \right) \\&\quad =(1-2s)\left( F\left( \frac{w^{p}}{3\alpha }\right) +F\left( \frac{2w^{p}}{ 3\alpha }\right) -1\right) \left( \frac{1}{2}+\frac{1}{2}F\left( \frac{w^{p}}{3\alpha }\right) \right) . \end{aligned}$$

Therefore, if $s>\frac{1}{3}$ we must have $w^{p}>1$; and if $s<\frac{1}{3}$ we must have $w^{p}<1$. In the approval rule, when $v=0$ the equation characterizing $w^{a}$ is

$$\begin{aligned} s\left( F\left( \frac{1}{3\alpha }\right) +F\left( \frac{2}{3\alpha }\right) -1\right) =(1-2s)\left( F\left( \frac{w^{a}}{ 3\alpha }\right) +F\left( \frac{2w^{a}}{3\alpha }\right) -1\right) . \end{aligned}$$

Therefore, if $s>\frac{1}{3}$ then we must have $w^{a}>1$; and if $s<\frac{1 }{3}$ then we must have $w^{a}<1$. Finally, in the negative plurality rule, the equation characterizing the threshold $w^{n}$ for $v=0$ is

$$\begin{aligned}&s\left( F\left( \frac{1}{3\alpha }\right) +F\left( \frac{2}{3\alpha }\right) -1\right) \left( 1-\frac{1}{2}F\left( \frac{1}{ 3\alpha }\right) \right) \\&\quad =(1-2s)\left( F\left( \frac{w^{n}}{3\alpha }\right) +F\left( \frac{2w^{n}}{3\alpha }\right) -1\right) \left( 1- \frac{1}{2}F\left( \frac{w^{n}}{3\alpha }\right) \right) . \end{aligned}$$

As in the other two rules, if $s>\frac{1}{3}$ we must have $w^{n}>1$; and if $s<\frac{1}{3}$ we must have $w^{n}<1$.

We can now use these two preliminary results to prove the lemma. Suppose first that $s<\frac{1}{3}$. The first preliminary result from above shows that if $v\le -1$ then $w^{p}<w^{a}<w^{n}$. When $v=0$, we can use the equation for $w^{p}$ and the second preliminary result to write:

$$\begin{aligned} (1-2s)\left( F\left( \frac{w^{p}}{3\alpha }\right) +F\left( \frac{2w^{p}}{3\alpha }\right) -1\right)= & {} s\left( F\left( \frac{1 }{3\alpha }\right) \right. \\&\left. +F\left( \frac{2}{3\alpha }\right) -1\right) \frac{\frac{1}{2}+\frac{1}{2}F\left( \frac{1}{ 3\alpha }\right) }{\frac{1}{2}+\frac{1}{2}F\left( \frac{w^{p}}{3\alpha }\right) } \\> & {} s\left( F\left( \frac{1}{3\alpha }\right) +F\left( \frac{2}{3\alpha }\right) -1\right) \\= & {} (1-2s)\left( F\left( \frac{w^{a}}{ 3\alpha }\right) +F\left( \frac{2w^{a}}{3\alpha }\right) -1\right) . \end{aligned}$$

Therefore, if $v=0$ then $w^{a}<w^{p}$. Similarly, if $v=0$ we can use the equation for $w^{n}$ and the second preliminary result to write:

$$\begin{aligned} (1-2s)\left( F\left( \frac{w^{n}}{3\alpha }\right) +F\left( \frac{2w^{n}}{3\alpha }\right) -1\right)= & {} s\left( F\left( \frac{1 }{3\alpha }\right) +F\left( \frac{2}{3\alpha }\right) -1\right) \frac{1-\frac{1}{2}F\left( \frac{1}{3\alpha }\right) }{1-\frac{1}{2}F\left( \frac{w^{n}}{3\alpha }\right) } \\< & {} s\left( F\left( \frac{1}{3\alpha }\right) +F\left( \frac{2}{3\alpha }\right) -1\right) \\= & {} (1-2s)\left( F\left( \frac{w^{a}}{ 3\alpha }\right) +F\left( \frac{2w^{a}}{3\alpha }\right) -1\right) . \end{aligned}$$

and so $w^{n}<w^{a}$. Thus, if $v=0$ then $w^{n}<w^{a}<w^{p}$.

Because for all $v\le -1$ we have $w^{p}<w^{a}<w^{n}$, and for $v=0$ we have $w^{n}<w^{a}<w^{p}$, there must exist a cutoff $v^{*}\in (-1,0)$ at which $w^{p}=w^{a}=w^{*}$. We will show that at the cutoff $v=v^{*}$ we also have $w^{n}=w^{*}$. To see this, we write the equations for thresholds $w^{p}$ and $w^{a}$ when $v=v^{*}$ as

$$\begin{aligned}&s\left( F\left( \frac{1-2v^{*}}{3\alpha }\right) -F\left( \frac{v^{*}-2}{3\alpha }\right) \right) \left( \frac{1 }{2}+\frac{1}{2}F\left( \frac{1+v^{*}}{3\alpha }\right) \right) \\&\quad +s\left( F\left( \frac{1-2v^{*}}{ 3\alpha }\right) -F\left( \frac{1+v^{*}}{3\alpha }\right) \right) \left( \frac{1}{2}+\frac{1}{2}F\left( \frac{ v^{*}-2}{3\alpha }\right) \right) \\&\quad =(1-2s)\left( F\left( \frac{w^{*}}{3\alpha }\right) +F\left( \frac{2w^{*}}{3\alpha }\right) -1\right) \left( \frac{1}{2}+\frac{1}{2}F\left( \frac{w^{*}}{3\alpha }\right) \right) . \end{aligned}$$

and

$$\begin{aligned}&s\left( F\left( \frac{1-2v^{*}}{3\alpha }\right) -F\left( \frac{v^{*}-2}{3\alpha }\right) \right) +s\left( F\left( \frac{1-2v^{*}}{3\alpha }\right) -F\left( \frac{1+v^{*}}{3\alpha }\right) \right) \\&\quad =(1-2s)\left( F\left( \frac{w^{*}}{3\alpha }\right) +F\left( \frac{2w^{*}}{3\alpha }\right) -1\right) . \end{aligned}$$

Multiplying the second equation by $\frac{3}{2}$ and then subtracting the two equations we obtain:

$$\begin{aligned}&s\left( F\left( \frac{1-2v^{*}}{3\alpha }\right) -F\left( \frac{v^{*}-2}{3\alpha }\right) \right) \left( 1- \frac{1}{2}F\left( \frac{1+v^{*}}{3\alpha }\right) \right) \\&\quad +s\left( F\left( \frac{1-2v^{*}}{3\alpha } \right) -F\left( \frac{1+v^{*}}{3\alpha }\right) \right) \left( 1-\frac{1}{2}F\left( \frac{v^{*}-2}{3\alpha }\right) \right) \\&\quad =(1-2s)\left( F\left( \frac{w^{*}}{3\alpha }\right) +F\left( \frac{2w^{*}}{3\alpha }\right) -1\right) \left( 1- \frac{1}{2}F\left( \frac{w^{*}}{3\alpha }\right) \right) . \end{aligned}$$

Therefore when $v=v^{*}$ we also have $w^{n}=w^{*}$. We will next show that if $v<v^{*}$ then $w^{p}<w^{a}<w^{n}$; while if $v>v^{*}$ then $w^{n}<w^{a}<w^{p}$. Consider the two equations characterizing the thresholds $w^{p}$ and $w^{a}$, and let us divide the two equations side by side:

$$\begin{aligned}&\frac{F\left( \frac{w^{p}}{3\alpha }\right) +F\left( \frac{2w^{p}}{3\alpha }\right) -1}{F\left( \frac{w^{a} }{3\alpha }\right) +F\left( \frac{2w^{a}}{3\alpha }\right) -1}\left( \frac{1}{2}+\frac{1}{2}F\left( \frac{ w^{p}}{3\alpha }\right) \right) \\&\quad =\frac{F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) }{F\left( \frac{1-2v}{ 3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) +F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{1+v}{ 3\alpha }\right) }\left( \frac{1}{2}+\frac{1}{2}F\left( \frac{1+v}{3\alpha }\right) \right) \\&\qquad +\frac{F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{1+v}{3\alpha }\right) }{F\left( \frac{1-2v}{ 3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) +F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{1+v}{ 3\alpha }\right) }\left( \frac{1}{2}+\frac{1}{2}F\left( \frac{v-2}{3\alpha }\right) \right) . \end{aligned}$$

It can be shown that the right hand side increases in v. This can be shown by differentiating the expression and proving that its derivative is positive. At the same time, $(\frac{1}{2}+\frac{1}{2}F(\frac{w^{p}}{3\alpha } ))$ decreases in v. Therefore, in order to preserve the balance of the right and left side of the equation, it must then be that the ratio $\frac{F( \frac{w^{p}}{3\alpha })+F(\frac{2w^{p}}{3\alpha })-1}{F(\frac{w^{a}}{3\alpha })+F(\frac{2w^{a}}{3\alpha })-1}$ increases in v. When $v=v^{*}$ we have $w^{p}=w^{a}=w^{*}$, and so $\frac{F(\frac{w^{p}}{3\alpha })+F( \frac{2w^{p}}{3\alpha })-1}{F(\frac{w^{a}}{3\alpha })+F(\frac{2w^{a}}{ 3\alpha })-1}=1$. This implies

$$\begin{aligned}&\text{ if } v>v^* \text{ then: } \frac{F(\frac{w^{p}}{3\alpha })+F(\frac{2w^{p}}{ 3\alpha })-1}{F(\frac{w^{a}}{3\alpha })+F(\frac{2w^{a}}{3\alpha })-1}>1 \text{ and } \text{ so } w^{a}<w^{p}\text {, and } \\&\quad \text{ if } v<v^* \text{ then: } \frac{F(\frac{w^{p}}{3\alpha })+F(\frac{2w^{p}}{ 3\alpha })-1}{F(\frac{w^{a}}{3\alpha })+F(\frac{2w^{a}}{3\alpha })-1}<1 \text{ and } \text{ so } w^{p}<w^{a}. \end{aligned}$$

Similarly, we can divide the equations characterizing the thresholds $w^{n}$ and $w^{a}$ to obtain

$$\begin{aligned}&\frac{F\left(\frac{w^{n}}{3\alpha }\right)+F\left(\frac{2w^{n}}{3\alpha }\right)-1}{F\left(\frac{w^{a} }{3\alpha }\right)+F\left(\frac{2w^{a}}{3\alpha }\right)-1}\left(1-\frac{1}{2}F(\frac{w^{n}}{ 3\alpha })\right) \\&\quad =\frac{F\left(\frac{1-2v}{3\alpha }\right)-F\left(\frac{v-2}{3\alpha }\right)}{F\left(\frac{1-2v}{ 3\alpha }\right)-F\left(\frac{v-2}{3\alpha }\right)+F\left(\frac{1-2v}{3\alpha }\right)-F\left(\frac{1+v}{ 3\alpha }\right)}\left(1-\frac{1}{2}F\left(\frac{1+v}{3\alpha }\right)\right) \\&\qquad +\frac{F\left(\frac{1-2v}{3\alpha }\right)-F\left(\frac{1+v}{3\alpha }\right)}{F\left(\frac{1-2v}{ 3\alpha }\right)-F\left(\frac{v-2}{3\alpha }\right)+F\left(\frac{1-2v}{3\alpha }\right)-F\left(\frac{1+v}{ 3\alpha }\right)}\left(1-\frac{1}{2}F\left(\frac{v-2}{3\alpha }\right)\right). \end{aligned}$$

As above, it can be shown that the right hand side of the equation decreases in v. As the term $(1-\frac{1}{2}F(\frac{w^{n}}{3\alpha }))$ increases in v, in order to preserve the balance, it must be that the ratio: $\frac{F( \frac{w^{n}}{3\alpha })+F(\frac{2w^{n}}{3\alpha })-1}{F(\frac{w^{a}}{3\alpha })+F(\frac{2w^{a}}{3\alpha })-1}$ decreases in v. When $v=v^{*}$ we have $w^{n}=w^{a}=w^{*}$, and so $\frac{F(\frac{w^{n}}{3\alpha })+F( \frac{2w^{n}}{3\alpha })-1}{F(\frac{w^{a}}{3\alpha })+F(\frac{2w^{a}}{ 3\alpha })-1}=1$. This implies:

$$\begin{aligned}&\text{ if } v>v^* \text{ then: } \frac{F(\frac{w^{n}}{3\alpha })+F(\frac{2w^{n}}{ 3\alpha })-1}{F(\frac{w^{a}}{3\alpha })+F(\frac{2w^{a}}{3\alpha })-1}<1 \text{ and } \text{ so } w^{n}<w^{a},\text { and } \\&\quad \text{ if } v<v^* \text{ then: } \frac{F(\frac{w^{n}}{3\alpha })+F(\frac{2w^{n}}{ 3\alpha })-1}{F(\frac{w^{a}}{3\alpha })+F(\frac{2w^{a}}{3\alpha })-1}>1 \text{ and } \text{ so } w^{a}<w^{n}. \end{aligned}$$

Finally, let us prove part (2) of the lemma, where $s>\frac{1}{3}$. From the second preliminary result, we know that when $v=0$, we have $w^{p},w^{a},w^{n}>1$. As all three thresholds $w^{p}$, $w^{a}$, and $w^{n}$ decrease with v, this implies that for all $v\le 0$, we have $w^{p},w^{a},w^{n}>1$. Using the equation for threshold $w^{p}$, and because $v\le 0$ and $w^{p}>1$, we can write:

$$\begin{aligned}&(1-2s)\left( F\left( \frac{w^{p}}{3\alpha }\right) +F\left( \frac{2w^{p}}{3\alpha }\right) -1\right) \\&\quad =s\left( F\left( \frac{ 1-2v}{3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) \right) \frac{\frac{1}{2}+\frac{1}{2}F\left( \frac{1+v}{3\alpha }\right) }{\frac{1}{2}+\frac{1}{2}F\left( \frac{w^{p}}{3\alpha }\right) } \\&\qquad +s\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{1+v}{3\alpha }\right) \right) \frac{\frac{1}{2}+ \frac{1}{2}F\left( \frac{v-2}{3\alpha }\right) }{\frac{1}{2}+\frac{1}{2}F\left( \frac{w^{p}}{ 3\alpha }\right) } \\&\quad <s\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) \right) +s\left( F\left( \frac{1-2v}{ 3\alpha }\right) -F\left( \frac{1+v}{3\alpha }\right) \right) \\&\quad =(1-2s)\left( F\left( \frac{w^{a}}{3\alpha }\right) +F\left( \frac{2w^{a}}{3\alpha }\right) -1\right) . \end{aligned}$$

Therefore $w^{p}<w^{a}$. Similarly, using the equation for threshold $w^{n}$, and because $v\le 0$ and $w^{p}>1$, we can write:

$$\begin{aligned}&(1-2s)\left( F\left( \frac{w^{n}}{3\alpha }\right) +F\left( \frac{2w^{n}}{3\alpha }\right) -1\right) \\&\quad =s\left( F\left( \frac{ 1-2v}{3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) \right) \frac{1-\frac{1}{2}F\left( \frac{1+v}{ 3\alpha }\right) }{1-\frac{1}{2}F\left( \frac{w^{n}}{3\alpha }\right) } \\&\qquad +s\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{1+v}{3\alpha }\right) \right) \frac{1-\frac{1}{2}F\left( \frac{v-2}{3\alpha }\right) }{1-\frac{1}{2}F\left( \frac{w^{n}}{3\alpha }\right) } \\&\quad >s\left( F\left( \frac{1-2v}{3\alpha }\right) -F\left( \frac{v-2}{3\alpha }\right) \right) +s\left( F\left( \frac{1-2v}{ 3\alpha }\right) -F\left( \frac{1+v}{3\alpha }\right) \right) \\&\quad =(1-2s)\left( F\left( \frac{w^{a}}{3\alpha }\right) +F\left( \frac{2w^{a}}{3\alpha }\right) -1\right) \end{aligned}$$

Therefore $w^{a}<w^{n}$. Thus, for all $v\le 0$ we have $w^{p}<w^{a}<w^{n}$.

1.5 Proof of Proposition 5

The proof is similar to that of Proposition 1. Using the equations for the updating of propensities, we can write:

$$\begin{aligned} E[p_{\theta }^{t+1}(i)]= & {} E[p_{\theta }^{t+1}(i)|i\in N_{pos}^{t}]\Pr (i\in N_{pos}^{t})\\&+E[p_{\theta }^{t+1}(i)|i\in N_{neg}^{t},N_{pos}^{t}\ne \emptyset ]\Pr (i\in N_{neg}^{t},N_{pos}^{t}\ne \emptyset ) \\&+E\left[ p_{\theta }^{t+1}(i)|N_{pos}^{t}=\emptyset \right] \Pr (N_{pos}^{t}=\emptyset ) \\= & {} (1-\lambda _{p})E[p_{\theta }^{t}(i)]+\lambda _{p}E\left[ \frac{n-1+n-I_{\theta }^{t}}{2}|i\in N_{pos}^{t}\right] \Pr (i\in N_{pos}^{t}) \\&+\lambda _{p}E[\frac{n-(I_{\theta }^{t}+1)}{2}|i\in N_{neg}^{t},N_{pos}^{t}\ne \emptyset ]\Pr (i\in N_{neg}^{t},N_{pos}^{t}\ne \emptyset ). \end{aligned}$$

In a stationary distribution, the distributions of propensities and aspirations, $p_{\theta }^{t}$ and $a_{\theta }^{t}$, do not depend on time. Therefore, dropping the superscript time index t, we obtain:

$$\begin{aligned} E[p_{\theta }(i)]= \,& {} E\left[ \frac{n-1+n-I_{\theta }}{2}|i\in N_{pos}\right] \Pr (i\in N_{pos}) \\&+E\left[ \frac{n-(I_{\theta }+1)}{2}|i\in N_{neg},N_{pos}\ne \emptyset \right] \Pr (i\in N_{neg},N_{pos}\ne \emptyset ). \end{aligned}$$

Computing the expression on the right hand side, we obtain:

$$\begin{aligned} S_{\theta }(i)=E[p_{\theta }(i)]=\frac{n-1}{2}(1-\prod _{j\in N}(1-q_{\theta }(j)))-\frac{1}{2}\sum _{j\in N}q_{\theta }(j)+\frac{n}{2}q_{\theta }(i). \end{aligned}$$

Finally, the equation for the updating of the abstention propensity is: $p_{\theta }^{t+1}(A)=(1-\lambda _p)p_{\theta }^t(A)+\lambda _p{\mathbf {1}} _{\{I_{\theta }^t=0\}}$. Similarly as in the other electoral rules, this implies that the measure of voters of type $\theta$ that abstain is $S_{\theta }(A)=\prod _{j\in N}(1-q_{\theta }(j))$.

1.6 Proof of Proposition 6

Consider first the first round. In the stationary distribution of the first round, the aspirations of the two groups of voters are $a_{1}=\frac{1}{3}$, and $a_{2}=\frac{2}{3}$. The probabilities of positive impressions for each candidate $i=1,2,3$ across the two groups of voters are

$$\begin{aligned} q_{1}(1)= & {} \Pr (\pi _{1}(1)>a_{1})=\Pr \left( 1+\alpha \epsilon _{1}(1)>\frac{1}{3 }\right) =F\left( \frac{2}{3\alpha }\right) \\ q_{1}(2)=q_{1}(3)= & {} \Pr (\pi _{1}(2)>a_{1})=\Pr \left( \alpha \epsilon _{1}(2)> \frac{1}{3}\right) =1-F\left( \frac{1}{3\alpha }\right) \\ q_{2}(1)= & {} \Pr (\pi _{2}(1)>a_{2})=\Pr \left( \alpha \epsilon _{2}(1)>\frac{2}{3} \right) =1-F\left( \frac{2}{3\alpha }\right) \\ q_{2}(2)=q_{2}(3)= & {} \Pr (\pi _{2}(2)>a_{2})=\Pr \left( 1+\alpha \epsilon _{2}(2)> \frac{2}{3}\right) =F\left( \frac{1}{3\alpha }\right) . \end{aligned}$$

Because $q_{\theta }(2)=q_{\theta }(3)$ for each group $\theta =1,2$, the difference in measures of votes of candidates 2 and 3 is $S^{I}(2)-S^{I}(3)=0$. On the other hand, the difference in measures of votes of candidates 1 and 2 is:

$$\begin{aligned} S^{I}(1)-S^{I}(2)= & {} s(q_{1}(1)-q_{1}(2))\left(1-\frac{1}{2} q_{1}(3)\right)\\&+(1-s)(q_{2}(1)-q_{2}(2))\left( 1-\frac{1}{2}q_{2}(3)\right) \\= & {} \left( F\left( \frac{1}{3\alpha }\right) +F\left( \frac{2}{3\alpha }\right) -1\right) \left( \frac{3}{2}s-\left( 1-\frac{1}{2 }F\left( \frac{1}{3\alpha }\right) \right) \right) . \end{aligned}$$

Therefore $S^{I}(1)-S^{I}(2)>0$ if $s>s^{*}$, where $s^{*}=\frac{2}{3 }(1-\frac{1}{2}F(\frac{1}{3\alpha }))$ with $s^{*}<\frac{1}{2}$. We can further show that the share of votes candidate 1 obtains in the first round is below 50%: $S^{I}(1)<0.5(S^{I}(1)+S^{I}(2)+S^{I}(3))$, or equivalently $S^{I}(1)<S^{I}(2)+S^{I}(3)$. Using the characterization of measures of votes in the stationary distribution and upon re-arranging, the condition can be written as:

$$\begin{aligned}&s\left\{ F\left( \frac{2}{3\alpha }\right) \left( 1-\frac{1}{3}\left( 1-F\left( \frac{1}{3\alpha }\right) \right) ^{2}\right) +\left( F\left( \frac{1}{3\alpha }\right) \right) ^{2}\right. \\&\quad \left. -\left( 1-F\left( \frac{2}{3\alpha }\right) \right) \left( 1-\frac{1}{3}\left( F\left( \frac{1}{ 3\alpha }\right) \right) ^{2}\right) -\left( 1-F\left( \frac{1}{3\alpha }\right) \right) ^{2}\right\} \\&\quad <1-\left( 1-F\left( \frac{2}{3\alpha }\right) \right) \left( 1-\frac{1}{3}\left( F\left( \frac{1}{3\alpha } \right) \right) ^{2}\right) -\left( 1-F\left( \frac{1}{3\alpha }\right) \right) ^{2}. \end{aligned}$$

Letting $a=F(\frac{1}{3\alpha })$ and $b=F(\frac{2}{3\alpha })$, the condition can be written as:

$$\begin{aligned}&s\left\{ b\left( 1-\frac{1}{3}(1-a)^{2}\right) +a^{2}-(1-b)\left( 1-\frac{1}{3}a^{2}\right) -(1-a)^{2} \right\} \\&\quad <1-(1-b)\left( 1-\frac{1}{3}a^{2}\right) -(1-a)^{2}, \end{aligned}$$

or, upon re-arranging:

$$\begin{aligned} s\left( \frac{5}{3}b-\frac{2}{3}a^{2}b+\frac{2}{3}ab+\frac{1}{3}a^{2}+2a-2\right) -\left( b- \frac{2}{3}a^{2}-\frac{1}{3}a^{2}b+2a-1\right) <0. \end{aligned}$$

We note that the left hand side is linear in b. Given that $1\ge b\ge a\ge \frac{1}{2}$, if the inequality holds at $b=a$ and $b=1$, then the inequality will hold for all $b\ge a$. The fact that inequality holds for $b=a$, and $b=1$ can be shown by straightforward algebra.

Consider now the second round, where candidate 1 confronts candidate 2. The difference in measures of votes of candidates 1 and 2 in the second round is:

$$\begin{aligned} S^{II}(1)-S^{II}(2)= \,& {} s(q_{1}(1)-q_{1}(2))+(1-s)(q_{2}(1)-q_{2}(2)) \\=\, & {} \left( F\left( \frac{1}{3\alpha }\right) +F\left( \frac{2}{3\alpha }\right) -1\right) (2s-1). \end{aligned}$$

As $s<\frac{1}{2}$, we have $S^{II}(1)-S^{II}(2)<0$, i.e., candidate 2 wins.

1.7 Proof of Proposition 7

In the stationary distribution of the first round, the aspirations of the three groups of voters are $a_{1}=\frac{1}{4}$, $a_{2}=\frac{1}{4}$ and $a_{3}=\frac{1}{2}$. The probabilities of positive impressions for each candidate $i=1,2,3$ are:

$$\begin{aligned} q_{1}(1)= & {} \Pr (\pi _{1}(1)>a_{1})=\Pr \left( 1+\alpha \epsilon _{1}(1)>\frac{1}{4 }\right) =F\left( \frac{3}{4\alpha }\right) \\ q_{1}(2)=q_{1}(3)=q_{1}(4)= & {} \Pr (\pi _{1}(2)>a_{1})=\Pr \left( \alpha \epsilon _{1}(2)>\frac{1}{4}\right) =1-F\left( \frac{1}{4\alpha }\right) \\ q_{2}(1)=q_{2}(3)=q_{2}(4)= & {} \Pr (\pi _{2}(1)>a_{2})=\Pr \left( \alpha \epsilon _{2}(1)>\frac{1}{4}\right) =1-F\left( \frac{1}{4\alpha }\right) \\ q_{2}(2)= & {} \Pr (\pi _{2}(2)>a_{2})=\Pr \left( 1+\alpha \epsilon _{2}(2)>\frac{1}{4 }\right) =F\left( \frac{3}{4\alpha }\right) \\ q_{3}(1)=q_{3}(2)= & {} \Pr (\pi _{3}(1)>a_{3})=\Pr \left( \alpha \epsilon _{3}(1)> \frac{1}{2}\right) =1-F\left( \frac{1}{2\alpha }\right) \\ q_{3}(3)=q_{3}(4)= & {} \Pr (\pi _{3}(3)>a_{3})=\Pr \left( 1+\alpha \epsilon _{3}(3)> \frac{1}{2}\right) =F\left( \frac{1}{2\alpha }\right) . \end{aligned}$$

Because of the symmetry between candidates 1 and 2, we have $S^{I}(1)=S^{I}(2)$. Similarly, by the symmetry of candidates 3 and 4, we have $S^{I}(3)=S^{I}(4)$. Using previous results, the difference in measures of votes of candidates 1 and 3 is:

$$\begin{aligned} S^{I}(1)-S^{I}(3)= & {} s(q_{1}(1)-q_{1}(3))E\left[ \frac{1}{1+I_{1}(-1,3)} \right] \\&+(1-2s)(q_{3}(1)-q_{3}(3))E\left[ \frac{1}{1+I_{3}(-1,3)}\right] \\= & {} s\left( F\left( \frac{1}{4\alpha }\right) +F\left( \frac{3}{4\alpha }\right) -1\right) \left( F\left( \frac{1}{4\alpha }\right) + \frac{1}{3}\left( 1-F\left( \frac{1}{4\alpha }\right) \right) ^{2}\right) \\&-(1-2s)\left( 2F\left( \frac{1}{2\alpha }\right) -1\right) \left( \frac{1}{2}+\frac{1}{3}\left( 1-F\left( \frac{1}{ 2\alpha }\right) \right) F\left( \frac{1}{2\alpha }\right) \right) \end{aligned}$$

The difference $S^{I}(1)-S^{I}(3)$ increases with s. If $s=\frac{1}{3}$, then the difference $S^{I}(1)-S^{I}(3)$ is:

$$\begin{aligned} (S^{I}(1)-S^{I}(3))_{s=\frac{1}{3}}= & {} \left( F\left( \frac{1}{4\alpha }\right) +F\left( \frac{3}{ 4\alpha }\right) -1\right) \left( F\left( \frac{1}{4\alpha }\right) +\frac{1}{3}\left( 1-F\left( \frac{1}{4\alpha }\right) \right) ^{2}\right) \\&-\left( 2F\left( \frac{1}{2\alpha }\right) -1\right) \left( \frac{1}{2}+\frac{1}{3}\left( 1-F\left( \frac{1}{2\alpha } \right) \right) F\left( \frac{1}{2\alpha }\right) \right) . \end{aligned}$$

Therefore, if the expression on the right hand is positive, then there exists a cutoff $s^{*}<\frac{1}{3}$ such that for all $s\in (s^{*}, \frac{1}{3})$ we have $S^{I}(1)-S^{I}(3)>0$.

If $\epsilon \sim U[-1,+1]$, then it can be shown by straightforward algebra that $(S^{I}(1)-S^{I}(3))_{s=\frac{1}{3}}>0$ for all $\alpha \ge 0$. If $\epsilon \sim N(0,1)$, we can show by numerical analysis that $(S^{I}(1)-S^{I}(3))_{s=\frac{1}{3}}>0$ for all $\alpha \ge 0$ as well. For other distributions F(.), a sufficient condition that ensures that $(S^{I}(1)-S^{I}(3))_{s=\frac{1}{3}}>0$ is that $\alpha <\alpha ^{*}$, where the cutoff $\alpha ^{*}$ is a function of the distribution F(). If $\alpha \rightarrow 0$ then the difference $(S^{I}(1)-S^{I}(3))_{s=\frac{1 }{3}}\rightarrow \frac{1}{2}>0$, and therefore, by continuity, $(S^{I}(1)-S^{I}(3))_{s=\frac{1}{3}}>0$ if $\alpha$ is not “too large”.