Representative Voting Games

Abstract

We propose the stationary Markov perfect equilibria of representative voting games as a benchmark to evaluate the outcomes of dynamic elections, in which the evolution of voters’ political power is endogenous. We show that the equilibria of dynamic elections can achieve this benchmark if politicians are sufficiently office motivated. For arbitrary equilibria of the electoral model, we characterize the faithfulness of politicians’ choices to the policy objectives of representative voters through a delegated best-response property. Finally, we provide conditions under which general dynamic electoral environments admit representative voters in each state.

Appendix

Proof of Theorem 1

Let ${\tilde{\pi }}$ be a strategy profile in the representative voting game. For all states s, types t and policies x, let ${\tilde{V}}_{t}(s,x)$ denote the payoff from policy x in state s to a representative voter of type t, which solves the recursive equation

$$\begin{aligned} {\tilde{V}}_{t}(s,x)= & {} u_{t}(s,x)+\delta p(s|s,x){\tilde{V}}_{t}(s,x)+\delta \sum _{s'\not =s}p(s'|s,x)\int _{x'} {\tilde{V}}_{t}(s',x'){\tilde{\pi }}_{\kappa (s')}(dx'|s'),\nonumber \\&\end{aligned}$$

(2)

and let ${\tilde{V}}_t(s)=\int _x {\tilde{V}}_t(s,x){\tilde{\pi }}_{\kappa (s)}(dx|s)$. If furthermore the profile ${\tilde{\pi }}$ is a stationary Markov perfect equilibrium of the representative voting game, then, for all states s, ${\tilde{\pi }}_{\kappa (s)}(\cdot |s)$ puts probability one on solutions to

$$\begin{aligned} \max _{x\in Y(s)}{\tilde{V}}_{\kappa (s)}(s,x), \end{aligned}$$

(3)

so that

$$\begin{aligned} {\tilde{V}}_{\kappa (s)}(s)\geqslant & {} {\tilde{V}}_{\kappa (s)}(s,x) \end{aligned}$$

(4)

for all policies x, with equality if and only if x is a best-response for $\kappa (s)$ against ${\tilde{\pi }}$ in the representative voting game. To define the Markov electoral strategy $\sigma =(\pi ,\rho )$, we specify that for all s and all t, $\pi _{t}(\cdot |s)={\tilde{\pi }}_{\kappa (s)}(\cdot |s)$. Therefore, because ${\tilde{\pi }}_{\kappa (s)}(Y^c(s)|s)=0$ for all s, it follows that $V_{t}^F(s,t')={\tilde{V}}_{t}(s)$ and that $V^B_t(s,t',x)=V^C_t(s,t',x)$ for all states s, types t and $t'$ and policies x. In particular, any voting strategy $\rho$ is optimal.

Assume further that the equilibrium profile ${\tilde{\pi }}$ is in pure strategies. We specify the voting strategy such that, for all states s and types t, $\rho (s,t,x)=1$ if and only if $x\in \text{ supp }({\tilde{\pi }}_{\kappa (s)}(\cdot |s))$. To verify the optimality of the voting strategy, fix any state s, incumbent type t and policy x. We have that

$$\begin{aligned} V^I_{\kappa (s)}(s,t,x)-V^C_{\kappa (s)}(s,t,x)= & {} p(s|s,x)\left[ u_{\kappa (s)}(s,x)+\delta V^I_{\kappa (s)}(s,t,x)-V^F_{\kappa (s)}(s,t) \right] \\= & {} p(s|s,x)\left[ {\tilde{V}}_{\kappa (s)}(s,x)-{\tilde{V}}_{\kappa (s)}(s) \right] \\\leqslant & {} 0, \end{aligned}$$

with equality whenever $x\in \text{ supp }({\tilde{\pi }}_{\kappa (s)}(\cdot |s))$, as desired. The second equality follows from (8) and the inequality follows from (4). To verify the optimality of policy strategies, normalize stage utilities such that ${\underline{u}}\leqslant u_t(s,x)\leqslant {\overline{u}}$ for all states s, types t and policies x, and assume that $\delta b>{\overline{u}}-{\underline{u}}$. Fix any state s and type t. An office holder of type t obtains a payoff of at least $\frac{{\underline{u}}+b}{1-\delta }$ if she implements policy $x\in \text{ supp }({\tilde{\pi }}_{\kappa (s)}(\cdot |s))$, while her payoff to implementing any policy $x'\notin \text{ supp }({\tilde{\pi }}_{\kappa (s)}(\cdot |s))$ is at most $\frac{{\overline{u}}}{1-\delta }+b$, so that choosing policy x is optimal, as desired.

Now assume that the equilibrium profile ${\tilde{\pi }}$ is in mixed strategies. For all states s, types t and policies $x\notin \text{ supp }({\tilde{\pi }}_{\kappa (s)}(\cdot |s))$, we specify the voting strategy such that $\rho (s,t,x)=0$. In particular, note that because ${\tilde{\pi }}_{\kappa (s)}(Y(s)|s)=1$ for all s, we have that, for any s and t, $\rho (s,t,x)=0$ for all $x\in Y^c(s)\cup Y^{d}(s)$. To construct the voting strategy for policies $x\in \text{ supp }({\tilde{\pi }}_{\kappa (s)}(\cdot |s))$, we assume that $\delta b$ is large enough that

$$\begin{aligned} \frac{{\overline{u}}-{\underline{u}}}{1-\delta }< & {} \frac{\delta b}{1-\delta \epsilon }\min \left\{ \epsilon , 1-\epsilon \right\} , \end{aligned}$$

(5)

where we use our normalization of stage utilities. Note that given any $0<\epsilon <1$, we can set b large enough that (5) is satisfied. Our goal is to define a continuous mapping $f:[\epsilon ,1]\rightarrow [\epsilon , 1]$, the fixed point $R^*$ of which will allow us to construct the voting strategy $\rho (s,t,x)$ for $x\in \text{ supp }({\tilde{\pi }}_{\kappa (s)}(\cdot |s))$ such that, in each state, the ex ante probability of reelection of all politician types is $R^*$, that is, for which $\int _x \rho (s,t,x){\tilde{\pi }}_{\kappa (s)}(dx|s)=R^*$ for all s and t. To this end, let $R\in [\epsilon ,1]$, and fix state s and type t.

Given any $r\in [0,1]$, define ${\tilde{w}}_{s,t}(x, r)$ as the payoff to a type t politician in s if she chooses policy x and is reelected with probability r, given that she will be reelected with ex ante probability R in all future periods (because no politician commits to policies under $\pi$, this mimics type t’s equilibrium payoff). That is,

$$\begin{aligned} {\tilde{w}}_{s,t}(x,r)= & {} u_t(s,x)+b +\delta \sum _{s'}p(s'|s,x)\left[ {\tilde{V}}_t(s')+\frac{rb}{1-\delta R} \right] , \end{aligned}$$

and note that ${\tilde{w}}_{s,t}(x,r)$ is continuous. For the restricted domain $r\in [\epsilon , 1]$, define ${\underline{x}}_{s,t}(r)=\text{ argmin}_{x\in \text{ supp }({\tilde{\pi }}_{\kappa (s)}(\cdot |s))}{\tilde{w}}_{s,t}(x,r)$ and ${\underline{w}}_{s,t}(r)={\tilde{w}}_{s,t}({\underline{x}}_{s,t}(r),r)$, and note that ${\underline{w}}_{s,t}(r)$ is continuous (by the Maximum Theorem). Finally, for all $x\in \text{ supp }({\tilde{\pi }}_{\kappa (s)}(\cdot |s))$ and $r\in [\epsilon , 1]$, define $r_{s,t}(x,r)$ as the solution $r'\in (0,r]$ to ${\underline{w}}_{s,t}(r)={\tilde{w}}_{s,t}(x,r')$. To show that $r_{s,t}(x,r)$ is well-defined, first note that ${\tilde{w}}_{s,t}(x,r)$ is strictly increasing in r because $\delta >0$, so that $r_{s,t}(x,r)\leqslant r$ follows from the fact that ${\underline{w}}_{s,t}(r)\leqslant {\tilde{w}}_{s,t}(x,r)$. Second, note that

$$\begin{aligned} {\underline{w}}_{s,t}(r)-{\tilde{w}}_{s,t}(x,0)\geqslant & {} {\underline{w}}_{s,t}(\epsilon )-{\tilde{w}}_{s,t}(x,0) \\\geqslant & {} \frac{{\underline{u}}}{1-\delta } +b\left[ 1+\frac{\delta \epsilon }{1-\delta \epsilon }\right] -\left[ \frac{{\overline{u}}}{1-\delta }+b \right] \\= & {} \frac{\delta b \epsilon }{1-\delta \epsilon }-\frac{{\overline{u}} -{\underline{u}}}{1-\delta } \\> & {} 0, \end{aligned}$$

where the final inequality follows from (5), so that $r_{s,t}(x,r)>0$ for all $r\in [\epsilon , 1]$. Finally, note that $r_{s,t}(x,r)$ is continuous.

Towards defining the mapping f, we first define a collection $\{R'_{s,t}(R) \}_{s,t}$, where each $R'_{s,t}(R)\in [\epsilon ,1]$. Fix state s and type t. If $\int _x r_{s,t}(x,1){\tilde{\pi }}_{\kappa (s)}(dx|s)\leqslant R$, then we set

$$\begin{aligned} R'_{s,t}(R)= & {} \int _x r_{s,t}(x,1){\tilde{\pi }}_{\kappa (s)}(dx|s). \end{aligned}$$

To ensure that $R'_{s,t}(R)$ is well-defined in this case, we need to verify that $R'_{s,t}(R)\geqslant \epsilon$, for which it is sufficient to show that $r_{s,t}(x,1)\geqslant \epsilon$ for all $x\in \text{ supp }({\tilde{\pi }}_{\kappa (s)}(\cdot |s))$. To see this, note that

$$\begin{aligned} {\underline{w}}_{s,t}(1)-{\tilde{w}}_{s,t}(x,\epsilon )\geqslant & {} \frac{{\underline{u}}}{1-\delta }+b\left[ 1+\frac{\delta }{1-\delta R}\right] -\left[ \frac{{\overline{u}}}{1-\delta }+ b\left[ 1+\frac{\delta \epsilon }{1-\delta R}\right] \right] \\\geqslant & {} \delta b\left[ \frac{1-\epsilon }{1-\delta \epsilon } \right] -\frac{{\overline{u}}-{\underline{u}}}{1-\delta } \\> & {} 0, \end{aligned}$$

yielding the desired contradiction, where the final inequality follows from (5). If instead $\int _x r_{s,t}(x,1){\tilde{\pi }}_{\kappa (s)}(dx|s)> R$, then we set $R'_{s,t}(R)=R$. Furthermore, in this case there exists $r^*\in [\epsilon ,1)$ such that

$$\begin{aligned} \int _x r_{s,t}(x,r^*){\tilde{\pi }}_{\kappa (s)}(dx|s)= R. \end{aligned}$$

To see this, note that by our previous results

$$\begin{aligned} \int _x r_{s,t}(x,\epsilon ){\tilde{\pi }}_{\kappa (s)}(dx|s)\leqslant & {} \epsilon \\\leqslant & {} R, \end{aligned}$$

so that the claim follows from the continuity of $r_{s,t}(x,r)$. Finally, we define the continuous mapping $f:[\epsilon ,1]\rightarrow [\epsilon , 1]$ such that $f(R)=\inf _{s,t}R'_{s,t}(R)$ for all $R\in [\epsilon , 1]$. This mapping has a fixed point $R^*=f(R^*)$ by Brouwer’s fixed point theorem, and, for all states s and types t,

$$\begin{aligned} R'_{s,t}(R^*)\leqslant R^*\leqslant R'_{s,t}(R^*), \end{aligned}$$

where the first inequality follows by construction of $R'_{s,t}(R^*)$ and the second inequality follows from the fact that $R^*$ is a fixed point of f. Therefore, $R_{s,t}(R^*)=R^*$ for all s and t.

Finally, we can use the fixed point $R^*$ of the mapping f to back out the voting strategy $\rho (s,t,x)$ for all states s, types t and policies $x\in \text{ supp }({\tilde{\pi }}_{\kappa (s)}(\cdot |s))$. Note that, by construction, $\int _x r_{s,t}(x,1){\tilde{\pi }}_{\kappa (s)}(dx|s)\geqslant R^*$ for all $x\in \text{ supp }({\tilde{\pi }}_{\kappa (s)}(\cdot |s))$, so that there exists $r^*\in [\epsilon , 1]$ such that $\int _x r_{s,t}(x,r^*){\tilde{\pi }}_{\kappa (s)}(dx|s)= R^*$. Therefore, we define $\rho (s,t,x)=r_{s,t}(x,r^*)$.

To verify the optimality of the policy strategy, note that, by construction, politician t is indifferent between all $x\in \text{ supp }({\tilde{\pi }}_{\kappa (s)}(\cdot |s))$, which all yield payoff ${\underline{w}}_{s,t}(r^*)$. Furthermore, by choosing policy $x\notin \text{ supp }({\tilde{\pi }}_{\kappa (s)}(\cdot |s))$, the politician t can obtain a payoff of at most

$$\begin{aligned} \frac{{\overline{u}}}{1-\delta }+b< & {} {\underline{w}}_{s,t}(\epsilon )\\\leqslant & {} {\underline{w}}_{s,t}(r^*), \end{aligned}$$

where the first inequality follows by (5) and the final inequality follows because, by construction, $r^*\geqslant \epsilon$. Therefore, there is no profitable deviation in s for politician t to a policy outside the support of ${\tilde{\pi }}_{\kappa (s)}(\cdot |s)$ □

Proof of Theorem 2

Fix a convergent Markov electoral equilibrium $\sigma =(\pi ,\rho )$. A first claim is that the symmetry of policy strategies and the optimality of the voting strategy $\rho$ imply that, for all states s, types t and $t'$ and policies x, $V_{\kappa (s)}^F(s,t')=V_{\kappa (s)}^F(s,t)$, $V_{\kappa (s)}^I(s,t,x)=V_{\kappa (s)}^I(s,t',x)$, $V_{\kappa (s)}^C(s,t,x)=V_{\kappa (s)}^C(s,t',x)$ and

$$\begin{aligned}&\rho (s,t,x)V_{\kappa (s)}^I(s,t,x)+(1-\rho (s,t,x))V_{\kappa (s)}^C(s,t,x) \\&\quad =\rho (s,t',x)V_{\kappa (s)}^I(s,t',x)+(1-\rho (s,t',x))V_{\kappa (s)}^C(s,t',x). \end{aligned}$$

Now suppose, towards a contradiction, that there exists a policy $x\in X(s)$ such that, for all types t,

$$\begin{aligned}&{u_{\kappa (s)}(s,x)+ \delta \left[ \rho (s,t,x)V_{\kappa (s)}^I(s,t,x)+(1-\rho (s,t,x))V_{\kappa (s)}^C(s,t,x)\right] } \nonumber \\&\quad > \int _{x'} \left[ u_{\kappa (s)}(s,x')+\delta \left[ \rho (s,t,x')V_{\kappa (s)}^I(s,t,x')+(1-\rho (s,t,x'))V_{\kappa (s)}^C(s,t,x') \right] \right] \pi _t(dx'|s)\nonumber \\&\quad = V_{\kappa (s)}^F(s,t), \end{aligned}$$

(6)

and notice that we have that $x\in Y(s)$ if and only if $\varphi (x)\in Y^c(s)$ also satisfies (6). Similarly, because the equilibrium is convergent, we have that $x\in Y(s)$ if and only if $\xi (s)\in Y^{d}(s)$ also satisfies (6). Correspondingly, in the sequel we assume that $x\in Y^c(s)$. If a politician of type $\kappa (s)$ commits to policy x in state s, it follows that

$$\begin{aligned}&{V_{\kappa (s)}^I(s,\kappa (s),x)-V^C_{\kappa (s)}(s,\kappa (s),x)} \\&\quad =p(s|s,x)\left[ u_{\kappa (s)}(s,x)+\delta V_{\kappa (s)}^I(s,\kappa (s),x) -V_{\kappa (s)}^F(s,\kappa (s))\right] \\&\quad \geqslant \delta p(s|s,x)(1-\rho (s,\kappa (s),x))\left[ V_{\kappa (s)}^I(s,\kappa (s),x)-V^C_{\kappa (s)}(s,\kappa (s),x) \right] , \end{aligned}$$

where the inequality, which is strict because $p(s|s,x)>0$, follows from (6). Therefore, because $\delta p(s|s,x)(1-\rho (s,\kappa (s),x))<1$, we have that

$$\begin{aligned} V_{\kappa (s)}^I(s,\kappa (s),x)> V^C_{\kappa (s)}(s,\kappa (s),x), \end{aligned}$$

and hence $\rho (s,\kappa (s),x)=1$.

To complete the proof, suppose that the equilibrium $\sigma$ is reelection-balanced with ex ante reelection probability $R^*$. Therefore, the payoff to politician $\kappa (s)$ from choosing according to policy strategy $\pi _{\kappa (s)}(\cdot |s)$ in state s is

$$\begin{aligned} V_{\kappa (s)}^F(s,\kappa (s)) +b \left[ 1+\frac{\delta R^*}{1-\delta R^*}\right] . \end{aligned}$$

(7)

If instead politician $\kappa (s)$ chooses deviating policy x in state s, her payoff is

$$\begin{aligned} u_{\kappa (s)}(s,x)+\delta V^I_{\kappa (s)}(s,\kappa (s),x) +b \left[ 1+\frac{\delta }{1-\delta p(s|s,x)}\left[ \frac{1-\delta p(s|s,x)R^*}{1-\delta R^*} \right] \right] , \end{aligned}$$

which, using (6), is strictly higher than her equilibrium payoff from (7), yielding the desired contradiction. □

Proof of Corollary 1

Fix a convergent and reelection-balanced Markov electoral equilibrium $\sigma$ with reelection probability $R^*$, and consider the profile ${\tilde{\pi }}$ in the representative voting game defined such that ${\tilde{\pi }}_{\kappa (s)}(\cdot |s)=\pi _{t}(\cdot |s)$ for all s and arbitrary t. First, if either (i) the policy profile $\pi$ is pure or (ii) politicians never commit to policies, i.e., $\pi _t(Y^c(s)|s)=0$ for all s and t, then we have that, for all states s, types t and $t'$ and policies x,

$$\begin{aligned} V^F_{t}(s,t')= & {} {\tilde{V}}_t(s), \text{ and } \nonumber \\ V^I_t(s,t',x)= & {} p(s|s,x){\tilde{V}}_t(s,x)+\sum _{s'\not =s}p(s'|s,x){\tilde{V}}_t(s'), \end{aligned}$$

(8)

where payoffs ${\tilde{V}}_{t}(s,x)$ and ${\tilde{V}}_{t}(s)$ are defined as in (2) Second, by Theorem 2, and invoking the optimality of the voting strategy $\rho$, we have that $\pi _t(\cdot |s)$ puts probability one on solutions to

$$\begin{aligned} \max _{x\in Y(s)}u_{\kappa (s)}(s,x)+\delta \max \left\{ V_{\kappa (s)}^I(s,t,x),V_{\kappa (s)}^C(s,t,x)\right\} . \end{aligned}$$

(9)

Third, in both cases (i) and (ii), we have that $V_{\kappa (s)}^I(s,t,x)=V_{\kappa (s)}^C(s,t,x)$, so that any solution to (9) must also be a solution to

$$\begin{aligned} \max _{x\in Y(s)}u_{\kappa (s)}(s,x)+\delta V_{\kappa (s)}^I(s,t,x). \end{aligned}$$

(10)

Finally, that ${\tilde{\pi }}_{\kappa (s)}(\cdot |s)$ must put probability one on solutions to (3) follows by substituting (8) into (10), so that the profile ${\tilde{\pi }}$ is a stationary Markov perfect equilibrium of the representative voting game. □