Discipline by turnout

Abstract

This paper explores the possibility that voter turnout induces subsequent performance from the elected official, in a two-period signaling-game model of political agency. An election is held in each period to delegate a policy decision to a politician whose policy preferences are private information. A representative voter decides, in each election, whether to vote for a politician or abstain, and voting incurs a cost which is private information. With ex-ante identical politicians, turning out in the first election is statically not optimal for the voter. However, she may still do so to signal her willingness—low cost—to punish a wrong policy in the following election.

Appendix

For precision, the following standard notations are used in this appendix.

• $h^{3(t-1)}$ is an arbitrary history in the tth election; so $h^0$ is the first election, and $h^3$ is the second election.

• $h^{3(t-1)+1}:=(h^{3(t-1)},v^t)$ is an arbitrary history after voting in the tth election.

• $h^{3(t-1)+2}:=(h^{3(t-1)},v^t,i)$ with $v^t\ne -i$ is an arbitrary history when politician i decides $e^t$ as the elected politician in period t.

The voter’s pure strategy is $v:=(v(\cdot |c))_{c}$, where $v(\cdot |c)$ maps $h^{3(t-1)}$ into $\{-1,0,1\}$. Similarly, $e_i:=(e_i(\cdot |\theta _i))_{\theta _i}$ is politician i’s pure strategy, where $e_i(\cdot |\theta _i)$ maps $(h^{3(t-1)},v^t,i)$ into $\{0,1\}$. $s:=(v,e_{-1},e_1)$ is a strategy profile.

Proof of Lemma 2

Suppose $(s^*,\mu ^*)$ is an equilibrium in which the dissonant politician chooses the right policy in some ${h^2}'=({h^1}',i)$. Doing so must strictly increase her reelection prospect by Lemma 1 (a), in which case it must be optimal also for the congruent politician, because she prefers that policy (part (a)). Then $\mu ^*(\theta _i=1|{h^1}',i,e^1=1)=1/2$. By Lemma 1 (b), abstention is uniquely optimal for the voter at any cost (part (b)).

Notice that the probability of reelection after choosing the right policy is 1/2 as we have just shown. Lemma 1 (a) requires a strictly positive probability of punishment, i.e., the probability of reelection after choosing the wrong policy needs to be no larger than $1/2-R/\beta (E+R)$ (part (c)). $\square$

Proof of Proposition 1

For necessities, suppose $(s^*,\mu ^*)$ is such an equilibrium, and let $h^*$ be the history of the equilibrium path. Lemma 2 (c) requires punishment to be optimal for some voter. Since it is optimal for the low cost voter whenever it is so for the high cost voter, there are two possibilities. First, suppose that even the high cost voter finds punishment optimal. By Lemma 1 (b), this implies

$$\begin{aligned}{} & {} \mu ^*(\theta _i=1|{h^1}^*,i,e^1=0)\le \frac{1}{2}-\frac{2\overline{c}}{B}\nonumber \\ \Leftrightarrow \;\frac{2\overline{c}}{B}\le \frac{1}{2}{} & {} -\mu ^*(\theta _i=1|{h^1}^*,i,e^1=0)\le \frac{1}{2}\;\Rightarrow \;\overline{c}\le \frac{B}{4}. \end{aligned}$$

(7)

Secondly, suppose that only the low cost voter finds punishment optimal. Since the congruent incumbent chooses the right policy whenever the dissonant incumbent does so (Lemma 2 (a)), $\mu ^*(\theta _i=1|{h^1}^*,i,e^1=1)=1/2$. By Lemma 2 (b), the probability of reelection after choosing the right policy is 1/2. If, instead, the incumbent chooses the wrong policy, she is reelected if and only if the voting cost is high and she wins the coin toss. After the voter’s choice in the first election, the posterior belief assigns a probability $\mu ^*(c=\overline{c}|{h^1}^*)/2$ to that event. Since both voter types abstain in the first election, abstention delivers no information about the voting cost, and hence $\mu ^*(c={\overline{c}}|{h^1}^*)={\overline{p}}$. By Lemma 1 (a), we must have in this case

$$\begin{aligned} \frac{1}{2}-\frac{{\overline{p}}}{2}\ge \frac{R}{\beta (E+R)}\;\Leftrightarrow \;{\overline{p}} \le 1-\frac{2R}{\beta (E+R)}. \end{aligned}$$

(8)

For sufficiency, suppose first ${\overline{c}}\le B/4$, and consider a strategy profile $s'$ such that

• In the first election, the voter always abstains;

• Every first-period incumbent chooses the right policy;

• In the second election, the voter votes against the incumbent following a wrong policy and abstain otherwise, regardless of the voting cost;

• Every second-period incumbent chooses her preferred policy.

Let $\mu '$ be a posterior belief that follows Bayes’ rule whenever applicable. The optimality of each part of $s'$ is trivial or a straightforward application of Bayes’ rule and Lemma 1, except for the voter’s decision in the second election following a wrong policy, i.e., ${h^4}'=(h^0,v^1=0,i,e^1=0)$, on which the strategy profile assigns no probability. By Lemma 1 (b), voting against the incumbent in the second election is optimal for both voter types if and only if $\mu '(\theta _i=1|{h^4}')\le 1/2-2{\overline{c}}/B$; thus, let us restrict $\mu '(\theta _i=1|h^0,v^1=0,i,e^1=0)\le 1/2-2{\overline{c}}/B$, noting that $1/2-2{\overline{c}}/B\ge 0$ (because ${\overline{c}}\le B/4$). It follows that $(s', \mu ')$ is indeed an equilibrium. Notice that the last restriction on $\mu '$ restricts the posterior belief to put little probability on the event that the incumbent is congruent upon observing a deviation to the wrong policy.

Finally, suppose that ${\overline{p}}\le 1-2R/[\beta (E+R)]$ and, for simplicity, that ${\overline{c}}>B/4$. Consider a strategy profile as in the previous case, except that the high-cost voter always abstains in the second election. Let $s^{**}$ and $\mu ^{**}$ denote, respectively, this strategy profile and and a posterior belief that follows Bayes’ rule whenever applicable. This time, the only nontrivial decision is in the first period following turnout, i.e., ${h^3}{''}=(h^0,v^1=i,i)$. If the dissonant first-period incumbent chooses the wrong policy, she is reelected if and only if the voting cost is high and she wins the coin toss, on which event the posterior belief puts a probability $\mu ^{**}(\overline{c}|{h^3}{''})/2$. If, instead, she chooses the right policy, both types of voter will abstain in the second election, and her reelection probability is 1/2. By Lemma 1 (a), choosing the right policy is optimal if (and only if) $\mu ^{**}(\overline{c}|{h^3}{''})\le 1-2R/[\beta (E+R)]$. By restricting $\mu ^{**}$ to satisfy the last inequality, noting that $1-2R/[\beta (E+R)]>0$, it follows that $(s^{**}, \mu ^{**})$ is indeed an equilibrium. Notice that the last restriction on $\mu ^{**}$ restricts the posterior belief to put little probability on the event that the voting cost is high upon observing a deviation to turnout. $\square$

Proof of Lemma 3

Let ${\overline{c}}<B/4$, and suppose on the contrary that $(s',\mu ')$ is such an equilibrium. If congruent politician chooses the right policy in the first period following abstention, it reveals her type, i.e., $\mu '(\theta _i=1|h^0,v^1=0,i,e^1=1)=1$. In that case, the voter’s optimal action in the second election following abstention and right policy is to vote for the incumbent at any cost by Lemma 1 (b). Congruent politician’s expected payoff when choosing $e^1=1$ is $(1+\beta )(E+B)$. If, instead, she chooses the wrong policy in the first period following abstention, it conveys no information about her type, and $\mu '(\theta _i=1|h^0,v^1=0,i,e^1=0)=1/2$. In that case, the optimal choice of the voter in the second election following abstention and wrong policy is to abstain at any cost by Lemma 1 (b). Congruent politician’s expected payoff when choosing $e^1=0$ is $E+\beta (E+B)/2$, which is strictly smaller than $(1+\beta )(E+B)$. Thus congruent politician must choose $e^1=1$ following abstention. But dissonant politician’s expected payoff in that first period is $E+R$ if she chooses $e^1=0$, while it is $E+\beta (E+R)/2$, which is strictly larger than $E+R$, if she chooses $e^1=1$. This contradicts the optimality of the dissonant politician’s choice of $e^1=0$ following abstention.

Proof of Proposition 2

Let us first derive necessary and sufficient conditions for two different kinds of equilibria in which discipline requires the costly signal of turnout: (separating) the voter’s choice in the first election depends on the voting cost, i.e., $v(h^0|{\underline{c}})\ne v(h^0|{\overline{c}})$; (pooling) the voter’s choice in the first election does not depend on the voting cost, i.e., $v(h^0|{\underline{c}})=v(h^0|{\overline{c}})$. We then use those conditions to prove the statements of the proposition.

Separating: Suppose $(s^*,\mu ^*)$ is such an equilibrium. Since ${\underline{c}}<B/4$, the low-cost voter must turnout in the first election, that is, $v^*(h^0|\underline{c})\ne 0$, by Lemma 4. Any action in the first election other than $v^*(h^0|{\underline{c}})$ is a sure sign that the voting cost is high; and since ${\overline{c}}>B/4$, the high-cost voter never finds it optimal to turnout in the second election by Lemma 1 (b), in which case the left-hand side of the inequality in Lemma 1 (a) becomes 0. Thus, the unique optimal action of dissonant politician in the first period following any action other than $v^*(h^0|{\underline{c}})$ is to choose the wrong policy; and, reasoning backward, the high-cost voter must abstain in the first election, which is optimal for her if and only if ${\overline{c}}\ge B/2$ by Lemma 4.

Conversely, let ${\overline{c}}\ge B/2$, and consider a strategy profile $(s',\mu ')$ in which

• In the first period,

  • The low-cost voter turns out, and the high-cost voter abstains;

  • Dissonant politician chooses the right policy if and only if the voter turns out (i.e., discipline by turnout);

  • Congruent politician always chooses the right policy;

• And in the second period $h^3=(h^0,v^1,i,e^1)$,

  • The low-cost voter votes against the incumbent whenever $e^1=0$, votes for the incumbent if $v^1=0$ and $e^1=1$, and abstains otherwise,

  • The high-cost voter always abstains;

  • Every politician chooses her preferred policy.

$\mu '$ is a posterior belief that follows Bayes’ rule whenever applicable.

The optimality of each part in the description above is trivial or a straightforward application of Bayes’ rule and Lemma 1, except for the low-cost voter’s decision in the second election following turnout and wrong policy, i.e., ${h^4}'=(h^0,v^1=i,i,e^1=0)$, on which the strategy profile does not assign probability. Voting against the incumbent is optimal for the low-cost voter unless $\mu '(\theta _i=1|{h^4}')>1/2-2{\underline{c}}/B$ by Lemma 1; thus, let $\mu '(\theta _i=1|{h^4}')\le 1/2-2{\underline{c}}/B$, noting that $1/2-2{\underline{c}}/B>0$ (because ${\underline{c}}<B/4$). It follows that $(s',\mu ')$ is indeed an equilibrium. Note that the last restriction restricts the posterior belief to put little probability on the event that the incumbent is congruent upon observing a deviation to the wrong policy.

Pooling: Suppose $(s^{**},\mu ^{**})$ is such an equilibrium. Since the low-cost voter turns out in the first election by Lemma 4, the high-cost voter must also turnout in the first election. Her expected payoff in this case is $B-C+\beta B/2$, while it is $B/2+\beta B/2$ if she abstains in the first election instead. It follows that ${\overline{c}}\le B/2$. Also, in the second election, the high-cost voter never turns out by Lemma 1 (b) (because ${\overline{c}}>B/4$) while the low-cost voter votes against the incumbent if $v^1\ne 0$ and $e^1=0$, also by Lemma 1 (b) (because ${\underline{c}}<B/4$). The first-period incumbent is reelected following $v^1\ne 0$ and $e^1=0$ if and only if the voting cost is high and she wins the coin toss. Since turnout in the first election gives no information about the voter, $\mu ^{**}(c={\overline{c}}|h^0,v^1\ne 0)={\overline{p}}$. By Lemma 1 (b), it follows that ${\overline{p}}\le 1-2R/[\beta (E+R)]$.

Conversely, let ${\overline{c}}\le B/2$ and ${\overline{p}}\le 1-2R/[\beta (E+R)]$. Consider a strategy profile $(s'',\mu '')$ that is the same as in the separation case above, except that $v''(h^0|{\overline{c}})=v''(h^0|{\underline{c}})\ne 0$. $\mu ''$ is a posterior belief that follows Bayes’ rule whenever applicable.

This time, there are two nontrivial decisions: in ${h^4}'=(h^0,v^1=i,i,e^1=0)$ as in the separation case and in the first period following abstention, i.e., ${h^1}'=(h^0,v^1=0)$. By Lemma 1 (b), voting against the incumbent in ${h^4}'$ is optimal for the low-cost voter if and only if $\mu ''(\theta _i=1|{h^4}')\le 1/2-2{\underline{c}}/B$; thus, let us restrict $\mu ''$ to be so, noting that $1/2-2{\underline{c}}/B>0$ (because ${\underline{c}}<B/4$). Also, choosing the wrong policy in $(h^0,v^1=0,i)$ is optimal for the dissonant politician if and only if $\mu ''({\overline{c}}|h^0,v^1=0)\ge 1-2R/[\beta (E+R)]$ by Lemma 1 (a); thus, let us further restrict $\mu ''$ to be so. It follows that $(s'',\mu '')$ is indeed an equilibrium. Note that the last restriction restricts the posterior belief to put a high probability on the event that the voting cost is high upon observing a deviation to abstention in the first election.

We have shown that a separating equilibrium exists if and only if ${\overline{c}}\ge B/2$, and a pooling equilibrium exists if and only if ${\overline{c}}\le B/2$ and ${\overline{p}}\le 1-2R/[\beta (E+R)]$. Thus, there is an equilibrium in which discipline requires the costly signal of turnout if and only if ${\overline{c}}\ge B/2$ or ${\overline{p}}\le 1-2R/[\beta (E+R)]$.

If ${\overline{c}}>B/2$, no pooling equilibrium exists; and the low-cost voter separates herself by turning out in the first election. Similarly, no separating equilibrium exists if $\overline{c}<B/2$, in which case the two types of the voter pool on turning out in the first election. $\square$