Welfare comparison of electoral systems under power sharing

Abstract

We generalize Börgers’ (Am Econ Rev 94:57–66, 2004) results to a broad class of power sharing electoral systems. We show that voluntary voting under a power sharing regime Pareto dominates both random decision making and compulsory voting. We also show, however, that voluntary voting is not socially optimal, as individuals vote too frequently.

Proofs

Throughout this appendix, write individual i’s expected benefit from abstaining when each of the N other individuals votes with probability p as⁸

$$\begin{aligned} \overline{B}(p, N) = \sum _{M=0}^N \left( {\begin{array}{c}N\\ M\end{array}}\right) p^M (1-p)^{N-M} \sum _{v=0}^{M} \left( {\begin{array}{c}M\\ v\end{array}}\right) \left( \frac{1}{2}\right) ^M G(m(v,M-v)) \text {.} \end{aligned}$$

(2)

Similarly, the expected gross benefit (i.e., benefit without subtracting the voting cost) from voting for the preferred alternative when each of the N other individuals votes with probability p can be written as

$$\begin{aligned} B^*(p, N) = \sum _{M=0}^N \left( {\begin{array}{c}N\\ M\end{array}}\right) p^M (1-p)^{N-M} \sum _{v=0}^{M} \left( {\begin{array}{c}M\\ v\end{array}}\right) \left( \frac{1}{2}\right) ^M G\left( \frac{v+1}{M+1} \right) \text {.} \end{aligned}$$

(3)

We will first prove two Lemmas, and then use them to prove the Propositions.

1.1 Two useful lemmas

Lemma 2

$\overline{B}(p, N) = 1/2$ for all $p \in [0,1]$ and $N \ge 0$.

Proof

First note that $\overline{B}(p,N)$ does not depend on the alternative. Moreover, Eq. (2) indicates that $\overline{B}(p,N)$ is the expectation of G to a supporter of either alternative. Since $G(m_A) + G(m_B)=1$ for any $m_A$ (and $m_B=1-m_A$), the expectation of the sum of G (across alternatives) must be 1. This implies $\overline{B}(p,N)=1/2$.⁹ $\square$

Lemma 3

$B^*$ is weakly decreasing in p for all $N \ge 0$.

Proof

Using Eq. (3), we can write

$$\begin{aligned} B^*(p, N) = \sum _{M=0}^N \left( {\begin{array}{c}N\\ M\end{array}}\right) p^M (1-p)^{N-M} B^*(1, M) \text {.} \end{aligned}$$

In other words, $B^*(p, N)$ is the expectation of $B^*(1,M)$ over M according to the binomial distribution with N trials and success probability p. Recall that a binomial distribution with success probability p and n trials first order stochastically dominates that with success probability $p'<p$ and n trials. Therefore, to prove that $B^*$ is decreasing in p, it suffices to show that $B^*(1,\cdot )$ is decreasing (with respect to M).

Using Pascal’s formula, we can write

$$\begin{aligned} B^*(1,M+1)&= \sum _{v=0}^{M+1} \left( {\begin{array}{c}M+1\\ v\end{array}}\right) \left( \frac{1}{2}\right) ^{M+1} G\left( \frac{v+1}{M+2}\right) \\&= \sum _{v=0}^M \left( {\begin{array}{c}M\\ v\end{array}}\right) \left( \frac{1}{2}\right) ^M \left[ \frac{1}{2}G\left( \frac{v+1}{M+2}\right) + \frac{1}{2}G\left( \frac{v+2}{M+2}\right) \right] \text {.} \\ \text {Meanwhile,}\nonumber \\ B^*(1, M)&= \sum _{v=0}^{M} \left( {\begin{array}{c}M\\ v\end{array}}\right) \left( \frac{1}{2}\right) ^{M} G\left( \frac{v+1}{M+1}\right) \text {.} \end{aligned}$$

$\square$

Thus we can write

$$\begin{aligned} B^*(1, M) - B^*(1, M+1) = \sum _{v=0}^{M} \left( {\begin{array}{c}M\\ v\end{array}}\right) \left( \frac{1}{2}\right) ^{M} h_M(v) \text {,} \end{aligned}$$

where $h_M: \{0, \ldots , M\} \rightarrow [0,1]$ is defined by

$$\begin{aligned} h_M(v) = G\left( \frac{v+1}{M+1}\right) - \left[ \frac{1}{2}G\left( \frac{v+1}{M+2}\right) + \frac{1}{2} G\left( \frac{v+2}{M+2}\right) \right] \text {.} \end{aligned}$$

This $h_M$ function has the following properties:

Fact 1 For all $v=0,\ldots , M$, $h_M(M-v-1) = -h_M(v)$.

Proof

Using the definition of $h_M$,

$$\begin{aligned} h_M(M-v-1)&= G\left( \frac{M-v}{M+1}\right) - \left[ \frac{1}{2}G\left( \frac{M-v}{M+2}\right) + \frac{1}{2} G\left( \frac{M-v+1}{M+2}\right) \right] \\&= 1 - G\left( \frac{v+1}{M+1}\right) \\&\qquad {}- \left( \frac{1}{2}\left[ 1-G\left( \frac{v+2}{M+2}\right) \right] + \frac{1}{2}\left[ 1-G\left( \frac{v+1}{M+2}\right) \right] \right) \\&= -G\left( \frac{v+1}{M+1}\right) + \left[ \frac{1}{2}G\left( \frac{v+1}{M+2}\right) + \frac{1}{2}G\left( \frac{v+2}{M+2}\right) \right] \\&= -h_M(v) \text {.} \end{aligned}$$

$\square$

Fact 2 For $v \le M/2-1$, $h_M(v) \le 0$.

Proof

Since G is convex on [0, 1 / 2], for $v \le M/2-1$,

$$\begin{aligned} h_M(v)&= G\left( \frac{v+1}{M+1}\right) - \left[ \frac{1}{2} G\left( \frac{v+1}{M+2}\right) + \frac{1}{2} G\left( \frac{v+2}{M+2}\right) \right] \\&\le G\left( \frac{v+1}{M+1}\right) - G\left( \frac{1}{2} \times \frac{v+1}{M+2} + \frac{1}{2} \times \frac{v+2}{M+2} \right) \\&= G\left( \frac{v+1}{M+1}\right) - G\left( \frac{v+1}{M+1} + \frac{M-2v-1}{2(M+1)(M+2)} \right) \\&\le 0 \text {.} \end{aligned}$$

The last inequality follows because $M-2v-1>0$ for $v \le M/2 -1$. $\square$

Given our notations, $B^*(1, M) - B^*(1, M+1)$ can be written as

$$\begin{aligned}&\sum _{v=0}^{\lfloor \frac{M}{2} \rfloor -1} \left( {\begin{array}{c}M\\ v\end{array}}\right) \left( \frac{1}{2} \right) ^M h_M(v) \\&\qquad + \mathbbm {1}\left[ M\text { odd}\right] \left( {\begin{array}{c}M\\ \frac{M-1}{2}\end{array}}\right) \left( \frac{1}{2} \right) ^M h_M\left( \frac{M-1}{2}\right) + \sum _{v=\lceil \frac{M}{2}\rceil }^{M} \left( {\begin{array}{c}M\\ v\end{array}}\right) \left( \frac{1}{2} \right) ^M h_M(v) \text {,} \end{aligned}$$

where $\mathbbm {1}\left[ \cdot \right]$ is the indicator function, which equals 1 if its argument is true and 0 otherwise; and the notations $\lfloor x \rfloor$ and $\lceil x \rceil$ denote, respectively, the largest integer smaller than or equal to and the smallest integer greater than or equal to x (i.e., the floor and ceiling of x). Notice that $(M-1)/2=M-(M-1)/2-1$. Hence if M is odd, by Fact 1, $h_M((M-1)/2)=0$ and the second term is zero. Also, we can change the summation variable of the last summation from v to $v'=M-v-1$ (except for the $v=M$ term). Using Fact 1, then

$$\begin{aligned} B^*(1,M) - B^*(1,M+1) = \sum _{v=0}^{\lfloor \frac{M}{2} \rfloor -1} \left[ \left( {\begin{array}{c}M\\ v\end{array}}\right) - \left( {\begin{array}{c}M\\ v+1\end{array}}\right) \right] \left( \frac{1}{2}\right) ^M h_M(v) + \left( \frac{1}{2} \right) ^M h_M(M) \text {.} \end{aligned}$$

By Fact 2, $h_M(v) \le 0$ for each v summed in the first term. Meanwhile, since $v < M/2$ for each v summed, $\left( {\begin{array}{c}M\\ v\end{array}}\right) < \left( {\begin{array}{c}M\\ v+1\end{array}}\right)$. Hence the first term must be positive. The second term is positive since

$$\begin{aligned} h_M(M)&= G\left( \frac{M+1}{M+1}\right) - \left[ \frac{1}{2}G\left( \frac{M+1}{M+2}\right) + \frac{1}{2}G\left( \frac{M+2}{M+2}\right) \right] \\&= \frac{1}{2} \left[ G(1) - G\left( \frac{M+1}{M+2}\right) \right] \\&\ge 0 \text {.} \end{aligned}$$

Therefore, $B^*(1,\cdot )$ is decreasing (in M), implying that $B^*(\cdot , N)$ is decreasing (in p).

1.2 Proof of Proposition 1

Recall that $p^*$ is a symmetric equilibrium voting probability under voluntary voting if and only if it solves

$$\begin{aligned} F\left( B^*(p,N) - \overline{B}\,(p,N) \right) = p \text {.} \end{aligned}$$

The left hand side of this equation is continuous and, by Lemmas 2 and 3, decreasing in p. The right hand side is continuous and strictly increasing in p. At $p=0$, the left hand side is weakly positive (since it is a probability) while the right hand side is zero. At $p=1$, the left hand side is weakly less than 1 (since it is a probability) while the right hand side is 1. Hence there exists a unique $p^*$ solving the equation, meaning that the voluntary voting game has a unique symmetric equilibrium voting probability.

1.3 Proof of Proposition 2

To each individual, random decision making gives the same expected interim payoff as abstaining in a voting game with $N+1$ other individuals who all vote with probability 1. The interim expected payoff under this regime is therefore $\overline{B}(1, N+1)$.

Given an equilibrium voluntary voting probability $p^*>0$, let $c^*$ be the associated equilibrium cut-off cost. Since $p^*>0$, there exists $c \in C$ such that $c< c^*$.

An individual with voting cost $c > c^*$ abstains under voluntary voting, yielding an interim expected payoff of $\overline{B}(p^*, N)$. By Lemma 2, this is the same as the interim expected payoff under the random decision making regime.

An individual with voting cost $c^*$ is indifferent between voting and abstaining under voluntary voting. Thus her interim expected payoff under voluntary voting is $\overline{B}(p^*, N)$, which by Lemma 2 is equal to the interim expected payoff under random decision making.

An individual with voting cost $c < c^*$ votes under voluntary voting, yielding an interim expected payoff of $B^*(p^*, N) - c$. Note that $c^*$ is defined such that $B^*(p^*,N)-c^* \ge \overline{B}(p^*,N)$ (the inequality may be strict if $p^*=1$). The right hand side of this inequality is equal to $\overline{B}(1,N+1)$ due to Lemma 2. Hence, $B^*(p^*,N) - c > \overline{B}(1,N+1)$ for all $c<c^*$.

Therefore, at the interim stage, all individuals are at least as well off under voluntary voting than under random decision making. Those with voting cost strictly below $c^*$ are strictly better off under voluntary voting.

1.4 Proof of Proposition 3

To an individual with voting cost c, compulsory voting gives the interim payoff of $B^*(1,N) - c$.

Given an equilibrium voluntary voting probability $p^*<1$, let $c^*$ be the associated equilibrium cut-off cost. Since $p^*<1$, there exists $c \in C$ such that $c>c^*$.

An individual with voting cost $c>c^*$ strictly prefers abstaining to voting under voluntary voting, yielding an interim expected payoff of $\overline{B}(p^*,N)>B^*(p^*,N)-c$. Under compulsory voting, she gets $B^*(1,N)-c$, which by Lemma 3 is weakly less than $B^*(p^*,N)-c$. Therefore, she is strictly worse off under compulsory voting.

An individual with voting cost $c=c^*$ is indifferent between voting and abstaining under voluntary voting, so her interim expected payoff must be $B^*(p^*,N) - c^*$. By Lemma 3, this is weakly higher than $B^*(1,N)-c^*$, the interim expected benefit under compulsory voting.

An individual with voting cost $c<c^*$ votes under voluntary voting and obtains an interim expected payoff of $B^*(p^*,N) - c$. By Lemma 3, this is weakly higher than $B^*(1,N)-c^*$, the interim expected benefit under compulsory voting.

Therefore, at the interim stage, all individuals are at least as well off under voluntary voting than under compulsory voting. Those with voting cost strictly above $c^*$ are strictly better off under voluntary voting.

1.5 Proof of Proposition 4

Suppose the social planner chooses a symmetric cut-off cost c to maximize the representative agent’s ex-ante payoff. Then the decision problem is

$$\begin{aligned} \max _{c \in C}\quad F(c) B^*(F(c),N) + (1-F(c))\overline{B}(F(c),N) - \int _{\underline{c}}^{c} c' f(c') dc' \text {.} \end{aligned}$$

Using Lemma 2, the first order condition yields

$$\begin{aligned} f(c) \left[ B^*(F(c), N) - \overline{B}(F(c), N) - c + F(c) \frac{\partial B^*}{\partial p} \right] = 0 \text {.} \end{aligned}$$

Recall that $f(c)>0$ for all $c \in C$. By Lemma 3, $\frac{\partial B^*}{\partial p} \le 0$. Hence, at the socially optimal turnout level, we must have

$$\begin{aligned} B^*(F(c), N) - \overline{B}(F(c), N) \ge c \text {.} \end{aligned}$$

Meanwhile, the voluntary equilibrium voting probability $p^*$ satisfies

$$\begin{aligned} B^*(p^*, N) - \overline{B}(p^*, N) = F^{-1}(p^*) \text {.} \end{aligned}$$

Since $B^*$ is weakly decreasing in p, this implies that the socially optimal $p=F(c)$ is weakly lower than $p^*$.

1.6 Workings for Example 1

Using Eq. (3), one can obtain, in this example

$$\begin{aligned} B^*(p,3)&= \frac{11}{16} p^3 + \frac{15}{8} p^2(1-p) + \frac{9}{4} p(1-p)^2 + (1-p)^3 \text {.} \\ \text {Hence} \\ \frac{\partial B^*}{\partial p}&= \frac{3}{16}\left( p^2 + 4p-4 \right) \text {.} \end{aligned}$$

Thus $B^*(\cdot , 3)$ is strictly increasing on $p \in (2(\sqrt{2}-1), 1]$.

Table 2 Social planner’s problem numerical results

By Lemma 2, the three voluntary voting equilibria (in terms of the voting probability p) are solutions to

$$\begin{aligned} B^*(p,3) - \frac{1}{2} = F^{-1}(p) \text {.} \end{aligned}$$

The three equilibria are obtained numerically.

We solve the social planner’s problem in terms of the voting probability (instead of cut-off voting cost), that is

$$\begin{aligned} \max _{p} \quad \frac{1}{2} + p\left[ B^*(p,3) - \frac{1}{2} \right] - \int _0^{F^{-1}(p)} c\, dF(c) \text {.} \end{aligned}$$

Ignoring the kinks in the distribution function, the FOC is

$$\begin{aligned} B^*(p,3) - \frac{1}{2} - F^{-1}(p) + p \frac{\partial B^*}{\partial p} = 0 \text {;} \end{aligned}$$

while the second order derivative of the objective function is

$$\begin{aligned} 2 \frac{\partial B^*}{\partial p} + p \frac{\partial ^2 B^*}{\partial p^2} - \frac{1}{f(F^{-1}(p))} \text {.} \end{aligned}$$

Table 2 shows the ex-ante social welfare at the points where the FOC holds (one of them is a local minimum, as the SOC indicates), as well as at the boundary points and the kinked points of the distribution function.