Voting as a signaling device

Abstract

In this paper, citizens vote in order to influence the election outcome and in order to signal their unobserved characteristics to others. The model is one of rational voting and generates the following predictions: (i) The paradox of not voting does not arise, because the benefit of voting does not vanish with population size. (ii) Turnout in elections is positively related to the importance of social interactions. (iii) Voting may exhibit bandwagon effects and small changes in the electoral incentives may generate large changes in turnout due to signaling effects. (iv) Signaling incentives increase the sensitivity of turnout to voting incentives in communities with low opportunity cost of social interaction, while the opposite is true for communities with high cost of social interaction. Therefore, the model predicts less volatile turnout for the latter type of communities.

Appendix

1.1 First-stage equilibrium

Proof of Proposition 1

(i) When we plot $\mathcal B _H (c_H)$ on $c_{\min }/\mu < c_H < c_{\max }$, the intersection $c_H^*$ with the $45^{\circ }$ line would be an interior equilibrium satisfying (6). By the continuity of $\mathcal B _H (c_H)$ on the interval $[c_{\min }/\mu , c_{\max }]$, if $\mathcal B _H (c_{\min }/\mu )>c_{\min }/\mu$ (i.e., the starting point is above the $45^{\circ }$ line) and $\mathcal B _H(c_{\max })<c_{\max }$ (i.e., the ending point is below the $45^{\circ }$ line), then at least one such intersection exists. Moreover, since at least one intersection is such that $\mathcal B _H(c_H)$ cuts the $45^{\circ }$ line from above, a stable interior equilibrium exists if these two conditions are satisfied. In an interior equilibrium with signaling, the second period benefit is $\frac{1}{2} w_2 p(n-1)[\lambda _H \alpha _H+(1-\lambda _H)\alpha _L-d]$. With the cutoff points $c_H=c_{\min }/\mu$ and $c_L=c_{\min }$, $p=qF(c_{\min }/\mu )$ and $\lambda _H=1$. With the cutoff points $c_H=c_{\max }$ and $c_L=\mu c_{\max },$ $p=q+(1-q)F(\mu c_{\max })$ and $\lambda _H=\frac{q}{q+(1-q)F(\mu c_{\max })}$. Replacing $p$ and $\lambda _H$ in (5) and rearranging, one finds that $\mathcal B _H (c_{\min }/\mu )>c_{\min }/\mu$ and $\mathcal B _H(c_{\max })<c_{\max }$ are equivalent to inequalities (7) and (8).

In addition, we have to make sure that this intersection $c_H^*$ gives an equilibrium with signaling. This is the case if the two conditions $\lambda _H \alpha _H + (1-\lambda _H) \alpha _L \ge d$ and $(1-\lambda _L) \alpha _H + \lambda _L \alpha _L \le d$ hold for all $c_H \in [c_{\min }/\mu , c_{\max }]$ (i.e., for all possible intersection points). These two conditions are equivalent to (9) holding for all $c_H \in [c_{\min }/\mu , c_{\max }]$.

(ii) The lhs of inequality (7) is always positive. Hence, this inequality is satisfied for low enough $c_{\min }$. The lhs of inequality (8) is bounded above by $\alpha _H \big \{\frac{w_1}{2} + \frac{1}{2} w_2 (n-1)q(\alpha _H -d) \big \}$. Hence, this inequality is satisfied for high enough $c_{\max }$.

The lhs of inequality (9) is lower than $(1-q)/q$ since $F(\mu c_H)<F(c_H)$ for all $c_H \in [c_{\min }/\mu , c_{\max }]$. Similarly, the rhs of inequality (9) is greater than $(1-q)/q$ since $1-F(\mu c_H)>1-F(c_H)$ for all $c_H \in [c_{\min }/\mu , c_{\max }]$. Then, for instance, if $d$ is such that $\frac{\alpha _H-d}{d-\alpha _L}=\frac{1-q}{q}$ (equivalently $q\alpha _H+(1-q)\alpha _L=d$), both conditions are satisfied. Hence, there is a neighborhood of values of $d$ around $q\alpha _H+(1-q)\alpha _L$ in which both conditions are satisfied. Note that this neighborhood for $d$ is consistent with the fact that inequalities (7) and (8) hold for some parameter values, since the latter inequalities are satisfied by appropriate choice of $c_{\min }$ and $c_{\max }$, irrespective of $d$. $\square$

1.2 Comparative statics

In this subsection of the Appendix, for the ease of exposition, we write $c_H$ and $c_L$ to denote the equilibrium cutoffs instead of $c_H^*$ and $c_L^*$.

Claim (iv): $\frac{\mathrm{d}p^*}{\mathrm{d}q}>0$.

Proof

$\mathrm{d} p^*/\mathrm{d} q$ is given by

$$\begin{aligned} \frac{\mathrm{d}p^*}{\mathrm{d}q}=F(c_H)-F(\mu c_H)+\frac{\mathrm{d} c_H}{\mathrm{d} q} \left[ qf(c_H)+(1-q)\mu f(\mu c_H) \right] \end{aligned}$$

We can compute $\mathrm{d} c_H/\mathrm{d} q$ by using equations (11) and (12) where $\partial \mathcal B _H / {\partial q}$ is given by

$$\begin{aligned} \frac{\partial \mathcal B _H}{\partial q}&= \alpha _H \Big \{\frac{w_1}{2}\Pi ^{\prime }(p)[F(c_H)-F(\mu c_H)]\\&\quad + \frac{w_2}{2}(n-1)\big [F(c_H)(\alpha _H-d) - F(\mu c_H)(\alpha _L-d)\big ]\Big \} \end{aligned}$$

Notice that the first term in $\partial \mathcal B _H / {\partial q}$ is negative, while the second term is positive. If, for example, $w_2 (w_1$) is sufficiently large, one obtains $\mathrm{d} c_H/\mathrm{d} q>0 (<0)$.

However, we now show that total turnout always increases. Replacing the value of $\mathrm{d} c_H/\mathrm{d} q$ in the above expression for $\mathrm{d} p^*/\mathrm{d} q$, we find that $\mathrm{d} p^*/\mathrm{d} q$ has the same sign as

$$\begin{aligned}&\bigg [F(c_H)-F(\mu c_H) \bigg ] \bigg [1-\alpha _H \Big \{ \frac{w_1}{2} \Pi ^{\prime }(p) \big [ qf(c_H)+(1-q)\mu f(\mu c_H) \big ]\\&\qquad + \frac{w_2}{2} (n-1) \big [ qf(c_H)(\alpha _H-d)+(1-q)\mu f(\mu c_H) (\alpha _L-d) \big ] \Big \} \bigg ] \\&\qquad + \bigg [qf(c_H)+(1-q)\mu f(\mu c_H)\bigg ] \bigg [\alpha _H \Big \{\frac{w_1}{2} \Pi ^{\prime }(p) \big [F(c_H)-F(\mu c_H) \big ]\\&\qquad + \frac{w_2}{2} (n-1) \big [F(c_H)(\alpha _H-d)-F(\mu c_H)(\alpha _L-d) \big ] \Big \} \bigg ] \end{aligned}$$

By lengthy but simple algebraic manipulations, it can be shown that the above expression and hence $\mathrm{d} p^*/\mathrm{d} q$ are always positive. $\square$

Claim (v): $\frac{\mathrm{d}p^*}{\mathrm{d}\alpha _H}>0$ but $\frac{\mathrm{d}p^*}{\mathrm{d}\alpha _L}$ cannot be signed.

Proof

$\frac{\partial \mathcal B _H}{\partial \alpha _H}$ is given by

$$\begin{aligned} \frac{\partial \mathcal B _H}{\partial \alpha _H}&= \Big \{\frac{w_1}{2}\Pi (p) + \frac{w_2}{2}(n-1)[qF(c_H)(\alpha _H-d)+(1-q)F(\mu c_H)(\alpha _L-d)]\Big \} \\&\quad +\alpha _H \Big \{\frac{w_1}{2}\Pi ^{\prime } (p) (1-q)f(\mu c_H)\Big (-\frac{\alpha _L c_H}{\alpha _H^2}\Big )\\&\quad + \frac{w_2}{2}(n-1)\Big [qF(c_H)+(1-q)f(\mu c_H)(\alpha _L-d)\Big (-\frac{\alpha _L c_H}{\alpha _H^2}\Big ) \Big ]\Big \} \end{aligned}$$

The term in the first bracket is the impact through the individual’s own preference parameter, and the term $\frac{w_2}{2}(n-1)qF(c_H)$ is the impact of the partners’ parameter on the signaling benefit. The terms involving $-\frac{\alpha _L c_H}{\alpha _H^2}$ are the effects of a decrease in the cutoff $c_L$ on the electoral and signaling benefits, which occurs so as to maintain the relationship $c_L=\mu c_H$ which must always hold in an equilibrium. Since all terms are positive, one has $\mathrm{d}c_H/\mathrm{d}\alpha _H>0$.

Similar to $\mathcal B _H(c_H)$, we define $\mathcal B _L(c_L)$ as

$$\begin{aligned} \mathcal B _L (c_L)&\equiv B_L(c_L/\mu ,c_L) \nonumber \\&= \alpha _L\Big \{\frac{w_1}{2}\Pi (p) + \frac{w_2}{2}(n-1)\big [qF(c_L/\mu )(\alpha _H-d)+(1-q)F(c_L)(\alpha _L-d)\big ]\Big \} \end{aligned}$$

(16)

The second term in this benefit is the signaling benefit which increases after an increase in $\alpha _H$. The electoral benefit in the first term decreases if and only if $\Pi (p)$ increases. Thus, $\mathrm{d} c_L/\mathrm{d} \alpha _H$ may be negative. However, if total turnouts $p$ were to decrease, then $\Pi (p)$, and hence all terms in (16), would increase, which together with $\mathrm{d}c_H/\mathrm{d}\alpha _H>0$ yields a contradiction.

From (16), we obtain

$$\begin{aligned} \frac{\partial \mathcal B _L}{\partial \alpha _L}&= \Big \{\frac{w_1}{2}\Pi (p) + \frac{w_2}{2}(n-1)[qF(c_L/\mu )(\alpha _H-d)+(1-q)F(c_L)(\alpha _L-d)]\Big \}\nonumber \\&\quad + \alpha _L \Big \{\frac{w_1}{2}\Pi ^{\prime }(p) q f(c_L/\mu )\Big (\frac{-c_L \alpha _H}{\alpha _L^2}\Big ) \nonumber \\&+ \frac{w_2}{2}(n-1)\Big [(1-q)F(c_L)+q f(c_L/\mu )(\alpha _H-d)\Big (\frac{-c_L \alpha _H}{\alpha _L^2}\Big )\Big ]\Big \} \end{aligned}$$

(17)

The last term in (17) expresses the impact of an increase in $\alpha _L$ via the associated change in the share of high-type agents among the interaction partners and is negative. If the weight $w_2$ on signaling is large and if the density of cost $f(c_L/\mu )$ happens to be large at the original equilibrium, then (17) may turn negative, so that $\mathrm{d}c_L/\mathrm{d} \alpha _L$ need not be positive. $\square$

Proof of Proposition 3 First, we derive the thresholds $d_H$ and $d_L$ from the two conditions given at the end of Sect. 3.1 as a function of the cutoff value $c_H$ for the high types in the equilibrium under consideration. To do this, we substitute the relevant values for $\lambda _H$ and $\lambda _L$ into $\lambda _H \alpha _H + (1-\lambda _H)\alpha _L \ge d$ and $(1-\lambda _L)\alpha _H + \lambda _L\alpha _L \le d$:

$$\begin{aligned}&d_H \!=\! \lambda _H\alpha _H\!+\!(1-\lambda _H)\alpha _L\\&\quad \Rightarrow \, d_H \!=\! \frac{qF(c_H)}{qF(c_H)\!+\!(1\!-\!q)F(\mu c_H)}\alpha _H\!+\!\left( 1\!-\!\frac{qF(c_H)}{qF(c_H)\!+\!(1\!-\!q)F(\mu c_H)}\right) \alpha _L \\&\quad \Rightarrow \, d_H \!=\! \frac{\alpha _HqF(c_H)\!+\!\alpha _L(1-q) F(\mu c_H)}{qF(c_H)\!+\!(1-q)F(\mu c_H)}\quad \quad \end{aligned}$$

Similarly:

$$\begin{aligned} d_L=(1-\lambda _L)\alpha _H+\lambda _L\alpha _L \, \Rightarrow \, d_L=\frac{\alpha _Hq(1-F(c_H))+\alpha _L(1-q)(1-F(\mu c_H))}{q(1-F(c_H))+(1-q)(1-F(\mu c_H))} \end{aligned}$$

For part (i), suppose that $f(c_H)/F(c_H) \ge \mu f(\mu c_H)/F(\mu c_H)$. One has

$$\begin{aligned}&\frac{f(c_H)}{F(c_H)} > (=) \frac{\mu f(\mu c_H)}{F(\mu c_H)} \, \Leftrightarrow \, (\alpha _H-\alpha _L)f(c_H)F(\mu c_H) > (=) (\alpha _H-\alpha _L)\mu f(\mu c_H)F(c_H) \\&\quad \Leftrightarrow \, \alpha _Hf(c_H)F(\mu c_H)\!+\!\alpha _L\mu f(\mu c_H)F(c_H) \!>\! (=) \alpha _H\mu f(\mu c_H)F(c_H) \!+\!\alpha _L f(c_H)F(\mu c_H)\quad \quad \end{aligned}$$

Multiplying both sides by $q(1-q)$ and adding  $\alpha _Hq^2f(c_H)F(c_H)$  and  $\alpha _L(1-q)^2\mu f(\mu c_H)F(\mu c_H)$  on both sides yields:

$$\begin{aligned}&\alpha _Hq^2f(c_H)F(c_H)+ \alpha _Hq(1-q)f(c_H)F(\mu c_H) \\&\qquad +\alpha _Lq(1-q)\mu f(\mu c_H)F(c_H) +\alpha _L(1-q)^2\mu f(\mu c_H)F(\mu c_H)\\&\quad > (=)\alpha _Hq^2f(c_H)F(c_H)+\alpha _Hq(1-q)\mu f(\mu c_H)F(c_H)\\&\qquad +\alpha _Lq(1-q) f(c_H)F(\mu c_H)+\alpha _L(1-q)^2\mu f(\mu c_H)F(\mu c_H)\\&\quad \Leftrightarrow \big [\alpha _Hqf(c_H)+\alpha _L(1-q)\mu f(\mu c_H)\big ]\big [qF(c_H)+(1-q)F(\mu c_H)\big ]\\&\quad > (=) \big [\alpha _HqF(c_H)+\alpha _L(1-q) F(\mu c_H)\big ]\big [qf(c_H)+(1-q)\mu f(\mu c_H)\big ] \\&\quad \Leftrightarrow \frac{\alpha _Hqf(c_H)+\alpha _L(1-q)\mu f(\mu c_H)}{qf(c_H)+(1-q)\mu f(\mu c_H)}~> (=)~ \frac{\alpha _HqF(c_H)+\alpha _L(1-q) F(\mu c_H)}{qF(c_H)+(1-q)F(\mu c_H)} \\&\quad \Leftrightarrow \tilde{\alpha } ~> (=) d_H \end{aligned}$$

From the lines above, we conclude more specifically that

$$\begin{aligned} \frac{f(c_H)}{F(c_H)} ~~> (=)~~ \frac{\mu f(\mu c_H)}{F(\mu c_H)} \, \Leftrightarrow \,\tilde{\alpha } ~>(=)~ d_H \end{aligned}$$

When $\tilde{\alpha }$ is greater than (resp. equal to) $d_H$, this implies that any value of $d$ that satisfies the equilibrium conditions also satisfies $d < \tilde{\alpha }$ (resp. $d \le \tilde{\alpha }$). Hence $\frac{d^2c_H^*}{\mathrm{d}w_1 \mathrm{d}w_2} > 0$ (resp. $\ge 0$).

For part (ii), note first that the above argument implies $\tilde{\alpha } < d_H$. Furthermore, substitute in the proof above the terms $1-F(c_H)$ and $1-F(\mu c_H)$ for the terms $F(c_H)$ and $F(\mu c_H)$, respectively, and iterate the same steps. Then, we obtain:

$$\begin{aligned}&\frac{f(c_H)}{1-F(c_H)} >(=) \frac{\mu f(\mu c_H)}{1-F(\mu c_H)}\\&\quad \Leftrightarrow \, \frac{\alpha _Hqf(c_H)+\alpha _L(1-q)\mu f(\mu c_H)}{qf(c_H)+(1-q)\mu f(\mu c_H)} >(=) \frac{\alpha _Hq(1-F(c_H))+\alpha _L(1-q)(1- F(\mu c_H))}{q(1-F(c_H))+(1-q)(1-F(\mu c_H))} \\&\quad \Leftrightarrow \, \tilde{\alpha } > (=)d_L\quad \end{aligned}$$

When $\tilde{\alpha }$ is greater than $d_L$, the cost of the match $d$ may satisfy $d < \tilde{\alpha }$ or not. This depends on the other parameters of the model. Hence, either  $\frac{d^2c_H^*}{\mathrm{d}w_1 \mathrm{d}w_2} \!\ge \! 0$  or  $\frac{d^2c_H^*}{\mathrm{d}w_1 \mathrm{d}w_2} \!<\! 0$. When $\tilde{\alpha }$ is equal to $d_L, d \ge \tilde{\alpha }$. Hence, $\frac{d^2c_H^*}{\mathrm{d}w_1 \mathrm{d}w_2} \le 0$.

Finally, part (iii) follows from part (ii). This is because when the initial condition of part (ii) does not hold, then $\tilde{\alpha } < d_L$, which implies that the condition $d \le \tilde{\alpha }$ is mutually exclusive with the equilibrium conditions.$\square$