The “invisible hand” of vote markets

Abstract

This paper studies electoral competition between two non-ideological parties when voters are free to trade votes for money. We find that allowing for vote trading has significant policy consequences, even if trade does not actually take place in equilibrium. In particular, the parties’ equilibrium platforms are found to converge (hence, there is no reason for vote trading) to the ideal policy of the mid-range voter, instead of converging to the peak of the median voter (as they do when vote trading is forbidden). That is, a market for votes may not change the outcome only by redistributing the political power among voters when the parties’ policy proposals are fixed (e.g., Casella et al. in J Polit Econ 120:593–658, 2012, etc.), but also by acting as an invisible hand—modifying parties’ incentives when platform choice is endogenous.

Appendix

Calculations showing that the equilibrium bids ${\bar{b}}^{\alpha }$ , ${\bar{b}}^{\beta }$ are positive and budget feasible:

In expression (3) the numerator is negative because $\beta +\alpha (n-1)-n<0$ and the remaining terms are positive for $0\le \alpha <\beta \le 1$ and $n>2$. The denominator is negative because $2\beta +\alpha (n-2)-(\beta +1)n=\beta (2-n)+n(\alpha -1)-2\alpha <0$ for $0\le \alpha <\beta \le 1$ and $n>2$. Hence, ${\bar{b}}^{\alpha }$ is positive. Moreover ${\bar{b}}^{\alpha }<1$, as each term $\tfrac{(n-2)}{n}$, $\tfrac{(\alpha +\beta (n-1))^{2}}{ (2\beta +\alpha (n-2)-(\beta +1)n)^{2}}$, $\tfrac{2(\beta -\alpha ))(\beta +\alpha (n-1)-n)}{2\beta +\alpha (n-2)-(\beta +1)n}$ is less than one for $0\le \alpha <\beta \le 1$ and $n>2$.

In expression (4) the numerator is negative because $(\alpha -\beta )<0$ and the remaining terms are positive for $0\le \alpha <\beta \le 1$ and $n>2$. The denominator is negative because $2\beta +\alpha (n-2)-(\beta +1)n=\beta (2-n)+n(\alpha -1)-2\alpha <0$ for $0\le \alpha <\beta \le 1$ and $n>2$. Hence, ${\bar{b}}^{\beta }$ is positive. Moreover ${\bar{b}}^{\beta }<1$, as each term $\tfrac{(n-2)}{n}$, $\tfrac{2(\alpha -\beta )(\alpha +\beta (n-1)) }{2\beta +\alpha (n-2)-(\beta +1)n}$, $\tfrac{(\beta +\alpha (n-1)-n)^{2}}{ (2\beta +\alpha (n-2)-(\beta +1)n)^{2}}$ is less than one for $0\le \alpha <\beta \le 1$ and $n>2$.

Calculations showing that the voter with $y_{i}=0$ prefers refraining from vote trading to selling her vote:

In the full-trade equilibrium, substituting for the posited bid of the other buyer, the utility of the voter with $y_{i}=0$ from refraining from vote trading and voting for ${\mathcal {A}}$ is $u_{i}(b_{i}=0,q_{i}=0;{\mathcal {A}} )=-(-\frac{1}{n}\alpha -\frac{n-1}{n}\beta )^{2}+1$ and from selling her vote is $u_{i}(b_{i}=0,q_{i}=1)=-\beta ^{2}+1+\tfrac{2(n-2)}{n(n-1)}\tfrac{ (\alpha -\beta )(\alpha +\beta (n-1))(\beta +\alpha (n-1)-n)^{2}}{(2\beta +\alpha (n-2)-(\beta +1)n)^{3}}$. Their difference is $u_{i}(b_{i}=0,q_{i}=0; {\mathcal {A}})-u_{i}(b_{i}=0,q_{i}=1)=\frac{1}{n^{2}}(\beta -\alpha )\left( \alpha +\beta (2n-1)+\frac{2n(n-2)(\alpha +\beta (n-1))(\beta +\alpha (n-1)-n)^{2}}{(n-1)(2\beta +\alpha (n-2)-(1+\beta )n)^{3}}\right) =$

$\frac{1}{n^{2}}(\beta -\alpha )\left( \beta n+(\alpha +\beta (n-1))\left( 1+\frac{ 2n(n-2)(\beta +\alpha (n-1)-n)^{2}}{(n-1)(2\beta +\alpha (n-2)-(1+\beta )n)^{3}}\right) \right) >0$ for $0\le \alpha <\beta \le 1$ and $n>2$, because the term $\frac{n(n-2)(\beta +\alpha (n-1)-n)^{2}}{(n-1)(2\beta +\alpha (n-2)-(1+\beta )n)^{3}}$ is negative and its absolute value is less than one. Thus, the absolute value of the term $1+\frac{2n(n-2)(\beta +\alpha (n-1)-n)^{2}}{(n-1)(2\beta +\alpha (n-2)-(1+\beta )n)^{3}}$ is less than one for $0\le \alpha <\beta \le 1$ and $n>2$, which yields $\beta n+(\alpha +\beta (n-1))\left( 1+\frac{2n(n-2)(\beta +\alpha (n-1)-n)^{2}}{(n-1)(2\beta +\alpha (n-2)-(1+\beta )n)^{3}}\right) >0$.

Calculations showing that an individual with $y_{i}\in (0,1)$ prefers selling her vote to refraining from vote trading and just voting for ${\mathcal {A}}$:

In the full-trade equilibrium, substituting for the posited bids of the extreme voters, the utility of an individual with $y_{i}\in (0,1)$ from selling her vote is $u_{i}(b_{i}=0,q_{i}=1)=$

$\frac{-2(\alpha -\beta )^{3}+(\alpha -\beta )^{2}(-3+2\alpha -2\beta -4y_{i}(y_{i}-1))n+2(\alpha -\beta )((\alpha -1)y_{i}(2y_{i}-1)+\beta (-2+\alpha +(3-2y_{i})y_{i}))n^{2}-(\beta (y_{i}-1)+y_{i}-\alpha y_{i})^{2}n^{3}}{n((\beta -\alpha )(n-2)+n)^{2}}$ and from refraining from vote trading and voting for ${\mathcal {A}}$ is

$u_{i}(b_{i}=0,q_{i}=0;{\mathcal {A}})=-\frac{((\alpha -\beta )^{2}-(\alpha -\beta )(\alpha -2+2y_{i})n+(\beta +(\alpha -\beta -1)y_{i})n^{2})^{2}}{ n^{2}((\beta -\alpha )(n-2)+n)^{2}}$.

Their difference is $u_{i}(b_{i}=0,q_{i}=1)-u_{i}(b_{i}=0,q_{i}=0;{\mathcal {A}} )=$

$\frac{(\alpha -\beta )(\beta +\alpha (n-1)-n)\left( -(\alpha -\beta )^{2}+(\alpha -\beta )(-1+\alpha +4y_{i})n+2(1+\beta -\alpha )y_{i}n^{2}\right) }{n^{2}((\beta -\alpha )(n-2)+n)^{2}}>0$ for $y_{i}\in (0,1)$, $0\le \alpha <\beta \le 1$ and $n>2$, because the product $(\alpha -\beta )(\beta +\alpha (n-1)-n)$ is positive and the term $(-(\alpha -\beta )^{2}+(\alpha -\beta )(-1+\alpha +4y_{i})n+2(1+\beta -\alpha )y_{i}n^{2})$, which can be written as $(\beta -\alpha )((\alpha -\beta )+(1-\alpha )n)+(\beta -\alpha )(2n-4)y_{i}n+2y_{i}n^{2}$, is also positive.

Proof of Lemma 2

Considering that $\beta =1-\alpha$ and $Q=n-2-2k$, expressions (1), (2) yield that the equilibrium bids of the two individuals with $y_{i}=\{0,1\}$ are ${\bar{b}}^{\alpha }={\bar{b}}^{\beta }=\frac{\left( 1-2\alpha \right) (n-2-2k)}{4n}$, which are positive and budget feasible for $\alpha \in [0,\frac{1}{2})$, $n>2$ and $k<\frac{n-2}{2}.$ Given the concavity of the maximization problem, neither of the two extreme voters wishes to deviate to any other bid.

Moreover, the two voters with $y_{i}=\{0,1\}$ never deviate to any other strategy. Given the posited strategies of the other players, the utility that the voter with $y_{i}=0$ derives from playing ${\bar{b}}^{\alpha }$ is $u_{i}(b_{i}={\bar{b}}^{\alpha },q_{i}=0;{\mathcal {A}})=-\left( -\frac{1}{2}\right) ^{2}+1- \frac{\left( 1-2\alpha \right) (n-2-2k)}{4n}$ and from refraining from vote trading and voting for ${\mathcal {A}}$ is $u_{i}(b_{i}=0,q_{i}=0;{\mathcal {A}} )=-\left( -\frac{k+1}{n}\alpha -\left( \frac{n-k-1}{n}\right) (1-\alpha )\right) ^{2}+1$. Their difference is $u_{i}(b_{i}={\bar{b}}^{\alpha },q_{i}=0;{\mathcal {A}} )-u_{i}(b_{i}=0,q_{i}=0;{\mathcal {A}})=\frac{(1-2\alpha )(n-2k-2)}{2n^{2}} (n-k-1-n\alpha +2\alpha +2k\alpha )>0$ for $\alpha \in [0,\frac{1}{2} )$, $n>2$ and $k<\frac{n-2}{2}$. Furthermore, the utility that the voter with $y_{i}=0$ derives from selling her vote is $u_{i}(b_{i}=0,q_{i}=1)=-\left( - \frac{k}{n}\alpha -(\frac{n-k}{n})(1-\alpha )\right) ^{2}+1+\frac{1}{n-1-2k}\frac{ \left( 1-2\alpha \right) (n-2-2k)}{4n}$ and hence $u_{i}(b_{i}={\bar{b}} ^{\alpha },q_{i}=0;{\mathcal {A}})-u_{i}(b_{i}=0,q_{i}=1)=\frac{(1-2\alpha )\left( n-2k\right) ^{2}}{4n^{2}(n-2k-1)}\left( 2n-2k-2n\alpha -1+2\alpha +4k\alpha \right) >0$ for $\alpha \in [0,\frac{1}{2})$, $n>2$ and $k< \frac{n-2}{2}$.

With similar arguments we can show that the voter with $y_{i}=1$ never deviates to selling her vote or to refraining from vote trading. Hence, the posited bids of the two extreme voters are their unique best responses.

Next, we show what no individual with ideal policy $y_{i}\in (0,1)$ places a monetary bid to acquire more votes. Given the posited bids of the voters with $y_{i}=\{0,1\}$, there is no positive bid that satisfies expression (1) for an individual with $y_{i}>0$ who votes for party ${\mathcal {A}}$. Similarly, there is no positive bid that satisfies expression (2) for an individual with $y_{i}<1$ who votes for party ${\mathcal {B}}$.

Next, we consider an individual with $y_{i}\in (0,1)$ who sells her vote in this partial-trade profile of strategies and all others expect it. Substituting for the posited strategies of the other players, the utility from selling her vote is $u_{i}(b_{i}=0,q_{i}=1)=-(y_{i}-\frac{1}{2})^{2}+1+ \frac{\left( 1-2\alpha \right) }{2n}$ and from voting for party ${\mathcal {A}}$ without engaging in vote trading is $u_{i}(b_{i}=0,q_{i}=0;{\mathcal {A}} )=-(y_{i}-{\hat{z}})^{2}+1$, where ${\hat{z}}=\frac{1}{n}(2+k+\frac{1}{2} (n-3-2k))\alpha +(1-\frac{1}{n}(2+k+\frac{1}{2}(n-3-2k)))(1-\alpha )=\frac{1}{2n}\left( n+2\alpha -1\right)$. Their difference is $u_{i}(b_{i}=0,q_{i}=1)-u_{i}(b_{i}=0,q_{i}=0;{\mathcal {A}})=\frac{(1-2\alpha )(4ny_{i}-2\alpha +1)}{4n^{2}}>0$ for $\alpha \in [0,\frac{1}{2})$ and $n>2$; that is, she has no incentives to deviate to voting for party ${\mathcal {A}}$ without engaging in vote trading. With similar arguments we can establish that an individual with $y_{i}\in (0,1)$ who sells in a partial-trade profile of strategies will not deviate to voting for party ${\mathcal {B}}$ without engaging in vote trading.

Consider now an individual who refrains from vote trading and just votes for party ${\mathcal {A}}$ in this partial-trade profile of strategies and all others expect it. Substituting for the posited strategies of the other players, her utility from refraining from vote trading is $u_{i}(b_{i}=0,q_{i}=0;{\mathcal {A}})=-(y_{i}-\frac{1}{2})^{2}+1$ and from selling her vote is $u_{i}(b_{i}=0,q_{i}=1)=-(y_{i}-{\tilde{z}} )^{2}+1+\frac{1}{n-1-2k}\frac{\left( 1-2\alpha \right) (n-2-2k)}{2n}$ where ${\tilde{z}}=\frac{1}{n}(k+\frac{1}{2}(n-1-2k))\alpha +(1-\frac{1}{n}(k+\frac{1 }{2}(n-1-2k)))(1-\alpha )=\frac{1}{2n}\left( n-2\alpha +1\right)$. This individual will not deviate to selling her vote if $u_{i}(b_{i}=0,q_{i}=0;{\mathcal {A}})\ge u_{i}(b_{i}=0,q_{i}=1)\Rightarrow$ $y_{i}\le$ $\xi =\tfrac{(1-2\alpha )\left( n-1-2k\right) +2n}{4n\left( n-1-2k\right) }$. That is, an individual with $y_{i}\in (0,\xi ]$ who refrains from vote trading and votes for ${\mathcal {A}}$ in this partial-trade profile of strategies will not deviate to selling her vote.

Similarly, an individual with ideal policy $y_{i}\in [1-\xi ,1)$ who refrains from vote trading and just votes for ${\mathcal {B}}$ in this partial-trade profile of strategies will not deviate to selling her vote.