Participation in fraudulent elections

Abstract

I analyze a costly voting model of elections in which the incumbent can stuff the ballot box to investigate how electoral fraud affects the decisions of voters to participate. I find that two stable equilibria may exist: an abstention equilibrium, where none of the voters vote and the incumbent always wins, and a more efficient coordination equilibrium, where a substantial share of a challenger’s supporters vote and the candidate preferred by the majority is likely to win. I further show that because the higher capability of the incumbent to stuff a ballot box discourages the participation of his own supporters and creates participation incentives for the challenger’s supporters, higher fraud does not always benefit the incumbent, even when costless. The model may help to explain two empirical observations related to fraudulent elections: a positive relationship between fraud and the margin of victory and a negative relationship between fraud and voter turnout.

Appendices

Appendix 1

Proof of Proposition 1

1. Suppose coordination equilibrium exists for some $N_0$, $\beta _0$, and $F_0$.

a) Suppose, one decreases population size to $N_1=N_0-t$, where $t\ge 2$ is an even integer. Denote:

$$\begin{aligned} \varPi _0(p)= & {} {N_0-1\atopwithdelims ()N_0/2}(\beta _0 p)^{N_0/2}(1-\beta _0 p)^{N_0/2-1},\\ \varPi _1(p)= & {} {N_0-t-1\atopwithdelims ()N_0/2-t/2}(\beta _0 p)^{N_0/2-t/2}(1-\beta _0 p)^{N_0/2-t/2-1}. \end{aligned}$$

Lemma 1

$\varPi _0(p)<\varPi _1(p)$ for all p.

Proof

See Appendix 2. $\square$

Lemma 1 immediately implies the result: if coordination equilibrium exists for $N_0$ then $\exists \bar{p}\in (0,1]$ such that ${F_0}^{-1}(\bar{p})\le \varPi _0(\bar{p})<\varPi _1(\bar{p})$ and thus, equilibrium exists for $N_1$.

b) Assume by contrast, that equilibrium does not exist for some $\beta _1>\beta _0$. Denote:

$$\begin{aligned} \varPi _0(p)= & {} {N_0-1\atopwithdelims ()N_0/2}(\beta _0 p)^{N_0/2}(1-\beta _0 p)^{N_0/2-1},\\ \varPi _1(p)= & {} {N_0-1\atopwithdelims ()N_0/2}(\beta _1 p)^{N_0/2}(1-\beta _1 p)^{N_0/2-1}. \end{aligned}$$

Note that $\varPi _1(p)=\varPi _0\left( \frac{\beta _1}{\beta _0}p\right)$, and, since $\beta _1>\beta _0$, $\varPi _1(p)\ge \varPi _0(p)$ for all $0< p\le \frac{N_0}{2\beta _1(N_0-1)}$. Further note that

$$\begin{aligned} \max _p\varPi _1(p)=\max _p\varPi _0(p)=\!{N_0-1\atopwithdelims ()N_0/2}\!\left( \frac{N_0}{2(N_0-1)}\right) ^{N_0/2}\!\left( 1-\frac{N_0}{2(N_0-1)}\right) ^{N_0/2-1}\!, \end{aligned}$$

and denote the latter value as $\varPi _{max}$. If there is no coordination equilibrium for $\beta _1$, then $\varPi _0\le \varPi _1(p)<F_0^{-1}(p)$ for all $0< p\le \frac{N_0}{2\beta _1(N_0-1)}$. Since $F_0^{-1}(p)$ is increasing function, for all $\frac{N_0}{2\beta _1(N_0-1)}<p\le 1$ it must be $\varPi _0(p)\le \varPi _{max}<F_0^{-1}(p)$. Hence, for any $0<p\le 1$, $\varPi _0(p)<F_0^{-1}(p)$, implying that coordination equilibrium does not exist for $\beta _0$ either, which contradicts the initial assumption.

c) If coordination equilibrium exists for some cost distribution ${F_0}$, then $\exists \bar{p}\in (0,1]$ such that ${F_0}^{-1}(\bar{p})\le \varPi (\bar{p})$. Any F which is first order stochastically dominated by $F_0$ satisfies $F(c)\ge F_0(c)$ for all c. Hence, $F^{-1}(p)\le {F_0}^{-1}(p)$ for all $p\in [0,1]$, including $\bar{p}$. From $F^{-1}(\bar{p})\le {F_0}^{-1}(\bar{p})\le \varPi (\bar{p})$ existence of equilibrium for cost distribution F follows immediately.

2. Suppose, condition (7) holds for some $N_0$, $\beta _0$, and $F_0$, and let $p^*$ and $p^t$ be the equilibrium participation level and the enforcement threshold under these parameters.

a) Suppose, one decreases population size to $N_1=N_0-t$, and let $\tilde{p}$ and $\tilde{p}^t$ be the equilibrium participation level and the threshold under $N_1$. Denote:

$$\begin{aligned} \varPi _0(p)= & {} {N_0-1\atopwithdelims ()N_0/2}(\beta _0 p)^{N_0/2}(1-\beta _0 p)^{N_0/2-1},\\ \varPi _1(p)= & {} {N_0-t-1\atopwithdelims ()N_0/2-t/2}(\beta _0 p)^{N_0/2-t/2}(1-\beta _0 p)^{N_0/2-t/2-1}. \end{aligned}$$

According to Lemma 1, $\varPi _0(p)<\varPi _1(p)$ for all p. Because $p^*$ and $\tilde{p}$ are intersections of increasing function $F_0^{-1}(p)$ with $\varPi _0(p)$ and $\varPi _1(p)$ respectively in their decreasing parts, $p^*<\tilde{p}$. Similarly, since $p^t$ and $\tilde{p}^t$ are intersections of $F_0^{-1}(p)$ with $\varPi _0(p)$ and $\varPi _1(p)$ respectively in their increasing parts, $p^t>\tilde{p}^t$. Thus, equilibrium participation is decreasing and participation threshold is increasing in population size.

(b) Let $\tilde{p}$ is the equilibrium participation level and $\tilde{p}^t$ is the enforcement threshold under $\beta _1>\beta _0$. Then, $p^*$ is an argument of an intersection point between decreasing part of $\varPi _0(p)={N_0-1\atopwithdelims ()N_0/2}(\beta _0p)^{N_0/2}(1-\beta _0p)^{N_0/2-1}$, , i.e. for $p>\frac{1}{2\beta _0}$, and increasing function $F_0^{-1}(p)$. Likewise, $\tilde{p}$ is an argument of an intersection point between decreasing part of $\varPi _1(p)={N_0-1\atopwithdelims ()N_0/2}(\beta _1p)^{N_0/2}(1-\beta _1p)^{N_0/2-1}$, i.e. for $p>\frac{1}{2\beta _1}$, and the same $F_0^{-1}(p)$. Thus, to prove that $p^*>\tilde{p}$ it is sufficient to show that $\varPi _0(p)>\varPi _1(p)$ for $p>\frac{1}{\beta _0}$. The latter inequality follows from the fact that $\varPi _1(p)=\varPi _0(\frac{\beta _1}{\beta _0}p)$, i.e. $\varPi _1(p)$ is a horizontal stretch (in fact, compression, since $\beta _1>\beta _0)$ of $\varPi _0(p)$.

Similarly, $p^t$ is an argument of an intersection point between non-decreasing function $F_0^{-1}(p)$ and increasing part of $\varPi _0(p)$, i.e. for $p<\frac{1}{2\beta _0}$. Likewise, $\tilde{p}^t$ is an argument of an intersection point between the same $F^{-1}(p)$ and increasing part of $\varPi _1(p)$, i.e. for $p<\frac{1}{2\beta _1}$. Since $\varPi _0(p)<\varPi _1(p)$ for $p<\frac{1}{2\beta _1}$, it follows that $p^t>\tilde{p}^t$.

(c) Let $\tilde{p}$ is the equilibrium participation level and $\tilde{p}^t$ is the enforcement threshold under some cost distribution G which is first-order stochastically dominated by $F_0$, and assume by contrast that $p^*\ge \tilde{p}$. Then $F_0^{-1}(p^*)\ge F_0^{-1}(\tilde{p})\ge G^{-1}(\tilde{p})$. Because $p^*$ satisfies $\varPi (p^*)=F_0^{-1}(p^*)$ and $\tilde{p}$ satisfies $\varPi (\tilde{p})=G^{-1}(\tilde{p})$ it must be that $\varPi (p^* )\ge \varPi (\tilde{p})$. Recall that $\varPi$ is a decreasing function for any $p\ge \frac{N_0/2}{\beta (N_0-1)}$ and any equilibrium participation $p\ge \frac{N_0/2}{\beta (N_0-1)}$, implying that $p^*\le \tilde{p}$, which contradicts the initial assumption. Thus, $p^*\le \tilde{p}$. Similarly, since $p^t$ and $\tilde{p}^t$ are intersections of the increasing part of $\varPi (p)$ with $F_0^{-1}(p)$ and $G^{-1}(p)$ respectively, and $F_0^{-1}(p)\ge G^{-1}(p)$ for all p, $p^t\ge \tilde{p}^t$. $\square$

Proof of Proposition 2

Welfare as a function of some strategy $\tilde{c}$ is expressed as

$$\begin{aligned} W=N\left( (1-\beta )(1-v_B)+\beta {v_B}+\beta \varPi (F(\tilde{c}))F(\tilde{c})-\beta \int _0^{\tilde{c}}c\, \mathrm {d} F(c)\right) \end{aligned}$$

In abstention equilibrium, where $\tilde{c}=0$ and $v_B=0$, social welfare is then simply $N(1-\beta )$. Consider the difference between welfare in coordination equilibrium and welfare in abstention equilibrium:

$$\begin{aligned} \varDelta W=N\left( (2\beta -1)v_B+\beta \varPi (F(c^*))F(c^*)-\beta \int _0^{c^*}c\, \mathrm {d} F(c)\right) . \end{aligned}$$

Since $\int _0^{c^*}c\, \mathrm {d} F(c)<\varPi (F(c^*))$, $\varDelta W>N\left( (2\beta -1)v_B-\varPi (F(c^*))(1-\beta F(c^*)\right)$.

Note that the right hand side of the last inequality is non-positive for $\beta <1/2$, and strictly positive for $\beta =1$. To see the latter, note that condition (7) implies $F(c^*)>\frac{N}{2\beta (N-1)}$ and re-write the last inequality for $\beta =1$:

$$\begin{aligned} \varDelta W>N\left( v_B-\varPi (F(c^*))(1-F(c^*)\right) >N\left( v_B-\frac{N/2-1}{N-1}\varPi (F(c^*))\right) . \end{aligned}$$

Further, using the first element of $v_B$:

$$\begin{aligned} \varDelta W> & {} N\left( {N-1\atopwithdelims ()N/2+1}(F(c^*))^{N/2+1}(1-F(c^*))^{N/2-2}-\frac{N/2-1}{N-1}\varPi (F(c^*))\right) \\= & {} N\frac{N/2-1}{N-1}\varPi (F(c^*))\left( \frac{N-1}{N/2+1}\frac{F(c^*)}{1-F(c^*)}-1\right) \\> & {} N\frac{N/2-1}{N-1}\varPi (F(c^*))\left( \frac{N-1}{N/2+1}\frac{N}{N-2}-1\right) . \end{aligned}$$

Since the latter expression is strictly positive for $N>2$, from the Intermediate Value Theorem it immediately follows that for any even $N>2$ there exists some $\beta _0$ such that $\varDelta W>0$ for any $\beta >\beta _0$. $\square$

Proof of Proposition 3

Consider social welfare as a function of some strategy $\tilde{c}$:

$$\begin{aligned} W=N\left( (1-\beta )(1-v_B)+\beta {v_B}+\beta \varPi (F(\tilde{c}))F(\tilde{c})-\beta \int _0^{\tilde{c}}c\, \mathrm {d} F(c)\right) \end{aligned}$$

It is easy to check that the welfare function $W(F(\tilde{c}))$ is concave in $F(\tilde{c})$. Hence, the first-order condition with respect to $F(\tilde{c})$ defines the efficiency condition:

$$\begin{aligned} N\left( (2\beta -1)\frac{\partial v_B}{\partial F(\tilde{c})}+\beta \varPi (F(\tilde{c})\left( 1+\frac{N/2-\beta NF(\tilde{c})+\beta F(\tilde{c})}{1-\beta F(\tilde{c})}\right) -\beta \tilde{c}\right) =0. \end{aligned}$$

Lemma 2

$\frac{\partial v_B}{\partial F(\tilde{c})}=\frac{\beta (N/2-1)}{1-\beta F(\tilde{c})}\varPi (F(\tilde{c}))$ for all $\tilde{c}$.

Proof

See Appendix 2. $\square$

After applying Lemma 2, dropping constants and re-arranging the terms, the efficiency condition takes the following form:

$$\begin{aligned} \varPi (F(\tilde{c}))\frac{\beta N(1-F(\tilde{c}))+2(1-\beta )}{1-\beta F(\tilde{c})}=F^{-1}(F(\tilde{c})). \end{aligned}$$

Let $c^e$ be the efficient participation rule, i.e. the one that satisfies the efficiency condition above. Recall the equilibrium condition:

$$\begin{aligned} \varPi (F(c^*))\ge F^{-1}(F(c^*)), \end{aligned}$$

with equality for $c^*<c_{max}$, and note that $\frac{\beta N(1-F(\tilde{c}))+2(1-\beta )}{1-\beta F(\tilde{c})}>1$ for all $\tilde{c}$. Hence, it immediately follows that $c^*\le c^e$ with inequality for $c^*<c_{max}$. $\square$

Proof of Proposition 4

If all the voters abstain, an A-type voter is never pivotal, while a B-type voter is pivotal if and only if none of the non-participants’ votes is stolen. Thus, the expected benefit function of a B-type voter at point $(c_{min},c_{min})$, which corresponds to full abstention, is $\varPi _{min}(\alpha )\equiv \varPi _B(c_{min},c_{min})=(1-\alpha )^{N-1}$. Since, $\varPi _{min}(0)=1>c_{min}$, $\varPi _{min}(1)=0<c_{min}$, and $\varPi _{min}(\alpha )$ is strictly decreasing in $\alpha$, the Intermediate Value Theorem implies that there exists a unique value of $\alpha =\alpha _0<1$ such that $\varPi _B(c_{min},c_{min})=c_{min}$; for any $\alpha \ge \alpha _0$ $\varPi _B(c_{min},c_{min})<c_{min}$, i.e. deviation from abstention is never profitable for a B-type voter; and for any $\alpha <\alpha _0$ $\varPi _B(c_{min},c_{min})>c_{min}$, i.e. deviation is profitable, and thus, full abstention is not an equilibrium. $\square$

Appendix 2

Proof of Lemma 1

Consider two functions:

$$\begin{aligned} S_0(q)= & {} {N_0-1\atopwithdelims ()N_0/2}q^{N_0/2}(1-q)^{N_0/2-1},\\ S_1(q)= & {} {N_1-1\atopwithdelims ()N_1/2}q^{N_1/2}(1-q)^{N_1/2-1}, \end{aligned}$$

where $N_1=N_0-t$, $t\ge 2$, and t / 2 is integer. Denoting $N_1/2=x$ and $t/2=m$ for shorter notation:

$$\begin{aligned} S_0(q)= & {} {2x+2m-1\atopwithdelims ()x+m}q^{x+m}(1-q)^{x+m-1}.\\ S_1(q)= & {} {2x-1\atopwithdelims ()x}q^{x}(1-q)^{x-1}. \end{aligned}$$

To show that $S_0(q)<S_1(q)$ it is sufficient to show that

$$\begin{aligned} {2x+2m-1\atopwithdelims ()x+m}q^{x+m}(1-q)^{x+m-1}-{2x-1\atopwithdelims ()x}q^{x}(1-q)^{x-1}<0, \end{aligned}$$

or, equivalently, that

$$\begin{aligned} \frac{ {2x+2m-1\atopwithdelims ()x+m} }{{2x-1\atopwithdelims ()x}} q^{m}(1-q)^{m}<1, \end{aligned}$$

Since function $g(p)=q^{m}(1-q)^{m}$ achieves its maximum at $q=1/2$, to complete the proof it is sufficient to show that

$$\begin{aligned} \frac{ {2x+2m-1\atopwithdelims ()x+m} }{{2x-1\atopwithdelims ()x}} \frac{1}{4^m}<1. \end{aligned}$$

The left hand side of the last expression can be rewritten as

$$\begin{aligned} \frac{ {2x+2m-1\atopwithdelims ()x+m} }{{2x-1\atopwithdelims ()x}} \frac{1}{4^m}= & {} \frac{2x(2x+1)...(2x+2m-1)}{(x+1)..(x+m)x(x+1)..(x+m-1)} \frac{1}{4^m}\\= & {} \frac{2x(2x+1)...(2x+2m-1)}{(2x+2)..(2x+2m)2x(2x+2)..(2x+2m-2)}. \end{aligned}$$

Note that the numerator consists of 2m elements in the form of $(2x+i)$, $i=0..2m-1$. The denominator contains two groups of m elements each. The first group consists of elements in the form of $2x+2i$, $i=1..m$, while the second group consists of elements in the same form $2x+2i$, $i=0..m-1$. Hence, the entire expression can be rewritten as

$$\begin{aligned} \frac{\prod \limits _{i=0}^{2m-1}(2x+i)}{\prod \limits _{i=1}^{m}(2x+2i)\prod \limits _{i=0}^{m-1}(2x+2i)}=\frac{2x\left( \prod \limits _{i=1}^{m-1} (2x+2i-1)(2x+2i)\right) (2x+2m-1)}{2x\left( \prod \limits _{i=1}^{m-1}(2x+2i)(2x+2i)\right) (2x+2m)}. \end{aligned}$$

After cancelling out equal elements, each element in the numerator is smaller than the corresponding element in the denominator. Hence, the entire expression is less than 1:

$$\begin{aligned} \frac{\left( \prod \limits _{i=1}^{m-1}(2x+2i-1)\right) (2x+2m-1)}{\left( \prod \limits _{i=1}^{m-1}(2x+2i)\right) (2x+2m)}<1. \end{aligned}$$

$\square$

Proof of Lemma 2

Recall the following identity⁷:

$$\begin{aligned} 1-I_x(a,b)=(1-x)^{a+b-1}\sum \limits _{i=0}^{a-1}{a+b-1\atopwithdelims ()i}\left( {\frac{x}{1-x}}\right) ^{i}, \end{aligned}$$

where $I_x(a,b)$ is regularized incomplete beta-function. Denote $\beta F(\tilde{c})=q$ for shorter notation. Then

$$\begin{aligned} v_B=\sum \limits _{i=N/2+1}^{N-1}{N-1\atopwithdelims ()i}q^{i}(1-q)^{N-i}, \end{aligned}$$

Consider the following regularized incomplete beta-function: $I_q(N/2+1,N/2-1)$. Using the identity above:

$$\begin{aligned} 1-I_q(N/2+1,N/2-1)= & {} (1-q)^{N-1}\sum \limits _{i=0}^{N/2}{N-1\atopwithdelims ()i}\left( {\frac{q}{1-q}}\right) ^{i}\\= & {} \sum \limits _{i=0}^{N/2}{N-1\atopwithdelims ()i}q^{i}(1-q)^{N-i-1}=1-v_B. \end{aligned}$$

Hence, $v_B=I_q(N/2+1,N/2-1)$. Also recall Chebyshev’s integral:

$$\begin{aligned} \int x^a(1-x)^b\, \mathrm {d} x=B_x(a+1,b+1), \end{aligned}$$

where $B_x(a+1,b+1)$ is incomplete beta-function. Thus,

$$\begin{aligned} B_q(N/2+1,N/2-1)=\int q^{N/2}(1-q)^{N/2-2}\, \mathrm {d} q. \end{aligned}$$

By definition of regularized incomplete beta-function:

$$\begin{aligned} I_q(N/2+1,N/2-1)=\frac{B_q(N/2+1,N/2-1)}{B(N/2+1,N/2-1)}, \end{aligned}$$

where $B(N/2+1,N/2-1)$ is beta-function. Since $B(N/2+1,N/2-1)=\frac{(N/2)!(N/2-2)!}{(N-1)!}$:

$$\begin{aligned}&(N/2-1){N-1\atopwithdelims ()N/2}\int q^{N/2}(1-q)^{N/2-2}\, \mathrm {d} q\\&\quad =\frac{(N-1)!}{(N/2)!(N/2-2)!}B_q(N/2+1,N/2-1)\\&\quad =\frac{B_q(N/2+1,N/2-1)}{B(N/2+1,N/2-1)}=I_q(N/2+1,N/2-1)=v_B. \end{aligned}$$

Hence, $v_B=(N/2-1){N-1\atopwithdelims ()N/2}\int q^{N/2}(1-q)^{N/2-1}\frac{1}{1-q}\, \mathrm {d} q$. Differentiating both parts of the last identity with respect to $F(\tilde{c})$:

$$\begin{aligned} \frac{\partial v_B}{\partial F(\tilde{c})}= & {} \beta \frac{\partial v_B}{\partial q}=\beta (N/2-1)\frac{\partial }{\partial q}\left( \int {N-1\atopwithdelims ()N/2}q^{N/2}(1-q)^{N/2-1}\frac{1}{1-q}\, \mathrm {d} q\right) \\= & {} \frac{\beta (N/2-1)}{1-\beta F(\tilde{c})} \varPi (F(\tilde{c})). \end{aligned}$$

$\square$