Approval quorums dominate participation quorums

Abstract

We study direct democracy with population uncertainty. Voters’ participation is often among the desiderata by the election designer. We show that with a participation quorum, i.e. a threshold on the fraction of participating voters below which the status quo is kept, the status quo may be kept in situations where the planner would prefer the reform, or the reform is passed when the planner prefers the status quo. On the other hand, using an approval quorum, i.e. a threshold on the number of voters expressing a ballot in favor of the reform below which the status quo is kept, we show that those drawbacks of participation quorums are avoided. Moreover, an electoral system with approval quorum performs better than one with participation quorum even when the planner wishes to implement the corresponding participation quorum social choice function.

Appendix

Proof of Lemma 2

Let us begin with a complete description of the best reply correspondence of a citizen of type S. For this exercise, let $\theta ^S n$ denote the expected size of the population of other citizens of type S. Let us assume that the symmetric strategy of these other citizens is $\sigma$, so that the number of actual votes in favor of S, $N^S$, is Poisson distributed with mean $\sigma \theta ^S n$. We already know that the number of actual votes in favor of R, $N^R$, is Poisson distributed with mean $\theta ^R n$. By voting for S, a citizen can be pivotal in favor of S if $N^S=N^R-1$ and $N^S+N^R\ge qn$, and she may be pivotal in favor of R if $N^S<N^R-1$ and $N^S+N^R= qn-1$. Let us refer to the former case as Piv 1, and to the latter as Piv 2. We have

$$\begin{aligned} \mathrm{Prob}(Piv 1)= & {} \sum _{k=int\left( \frac{qn}{2}\right) }^{\infty }\mathrm{Prob}(N^S=k\;\mathrm{and}\;N^R=k+1)\\= & {} \sum _{k=int\left( \frac{qn}{2}\right) }^{\infty }\frac{e^{-\sigma \theta ^S n}(\sigma \theta ^S n)^{k}}{k!}\frac{e^{-\theta ^R n}(\theta ^R n)^{k+1}}{(k+1)!} \end{aligned}$$

and

$$\begin{aligned} \mathrm{Prob}(Piv 2)= & {} \sum _{k=0}^{int\left( \frac{qn-1}{2}\right) -1}\mathrm{Prob}(N^S=k\;\mathrm{and}\;N^R=qn-k-1)\\= & {} \sum _{k=0}^{int\left( \frac{qn-1}{2}\right) -1}\frac{e^{-\sigma \theta ^S n}(\sigma \theta ^S n)^{k}}{k!}\frac{e^{-\theta ^R n}(\theta ^R n)^{qn-k-1}}{(qn-k-1)!} \end{aligned}$$

where int(x) stands for the integer value of x. We look at equilibrium for n sufficiently large. As n becomes larger, the probabilities of Piv 1 and Piv 2 tend to 0. Let $\mu _1,\mu _2$ denote the magnitude of these events, that is, the speed at which they tend to zero, that is, for $i\in \{1,2\}$,

$$\begin{aligned} \mu _i=\lim _{n\rightarrow \infty }\frac{\ln (\mathrm{Prob}(Piv i))}{n}. \end{aligned}$$

The event with the largest magnitude will necessarily be more likely than the other one for n sufficiently large (that is precisely the meaning of n being sufficiently large). Let us compute these magnitudes. Using Theorem 1 in Myerson (2000), we know that the magnitude of such an event is identical to the magnitude of the most likely subevent, that is, of the exact sequence of numbers $N^S=k$ and $N^R=k+1$ that maximizes

$$\begin{aligned} \sigma \theta ^S \psi \left( \frac{k}{\sigma \theta ^S n}\right) +\theta ^R \psi \left( \frac{k+1}{\theta ^R n}\right) \end{aligned}$$

under the constraint that $k\ge int\left( \frac{qn}{2}\right)$, and the exact sequence of numbers $N^S=k$ and $N^R=qn-k-1$ that maximizes

$$\begin{aligned} \sigma \theta ^S \psi \left( \frac{k}{\sigma \theta ^S n}\right) +\theta ^R \psi \left( \frac{qn-k-1}{\theta ^R n}\right) \end{aligned}$$

under the constraint that ${\small k \le int\left( \frac{qn-1}{2}\right) -1}$, respectively, where $\psi (x)=x(1-\ln x)-1$. Let $k_1$ and $k_2$ denote the arguments maximizing the above expressions, respectively. Simple derivation leads to

$$\begin{aligned} k_1=\max \left\{ int\left( \frac{qn}{2}\right) ,\sqrt{\frac{1}{4}+\sigma \theta ^S\theta ^R n^2}-\frac{1}{2}\right\} , \end{aligned}$$

and

$$\begin{aligned} k_2=\min \left\{ int\left( \frac{qn-1}{2}\right) -1,\frac{\sigma \theta ^S}{\sigma \theta ^S+\theta ^R}(qn-1)\right\} . \end{aligned}$$

Observe that the second critical value of $k_1$ can be approximated by its limit value $\sqrt{\sigma \theta ^S\theta ^R}n$. In the case where $\sqrt{\sigma \theta ^S\theta ^R}\le \frac{q}{2},$ $k_1$ tends towards its first critical value $\frac{qn}{2}$, assuming $\frac{qn}{2}$ is an integer. Event Piv1, that is, a tie between R and S, is more likely for values $N^S=\frac{qn}{2}-1$ and $N^R=\frac{qn}{2}.$ We compute that

$$\begin{aligned} \mu _1= & {} \lim _{n\rightarrow \infty }n^{-1}\ln \left( \frac{e^{-\sigma \theta ^S n}(\sigma \theta ^S n)^{\frac{qn}{2}-1}}{(\frac{qn}{2}-1) !} \frac{e^{-\theta ^R n}(\theta ^R n)^{\frac{qn}{2}}}{(\frac{qn}{2}) !} \right) \\= & {} \lim _{n\rightarrow \infty }n^{-1}\ln \left( \frac{e^{-(\sigma \theta ^S + \theta ^R) n}(\sigma \theta ^S \theta ^R n^2)^{\frac{qn}{2}}}{\left( (\frac{qn}{2}) !\right) ^2} \frac{q}{2\sigma \theta ^S} \right) , \end{aligned}$$

which, using the Stirling formula (according to which k ! can be approximated by $\sqrt{2\pi k}\left( \frac{k}{e} \right) ^k$), yields

$$\begin{aligned} \mu _1= & {} \lim _{n\rightarrow \infty }n^{-1}\ln \left( \frac{e^{-(\sigma \theta ^S + \theta ^R - q) n}}{\pi q n} \left( \frac{4\sigma \theta ^S \theta ^R n^2}{(qn)^2} \right) ^{\frac{qn}{2}}\frac{q}{2\sigma \theta ^S}\right) ,\\= & {} \lim _{n\rightarrow \infty }-(\sigma \theta ^S + \theta ^R - q) + \frac{q}{2}\ln \frac{4\sigma \theta ^S \theta ^R}{q^2} - \frac{\ln \pi q n}{n} + \frac{\ln \frac{q}{2\sigma \theta ^S}}{n}, \end{aligned}$$

which gives

$$\begin{aligned} \mu _1=q-q\ln q + q\ln 2\sqrt{\sigma \theta ^S\theta ^R}-(\sigma \theta ^S+\theta ^R). \end{aligned}$$

(1)

Similar computations lead to the following magnitude equations. If $\sqrt{\sigma \theta ^S\theta ^R}> \frac{q}{2}, k_1$ tends towards its second critical value, $n\sqrt{\sigma \theta ^S\theta ^R}$, and

$$\begin{aligned} \mu _1=2\sqrt{\sigma \theta ^S\theta ^R}-(\sigma \theta ^S+\theta ^R). \end{aligned}$$

(2)

If $k_2$ tends to $\frac{qn}{2}-1$, then

$$\begin{aligned} \mu _2=q-q\ln q + q\ln 2\sqrt{\sigma \theta ^S\theta ^R}-(\sigma \theta ^S+\theta ^R). \end{aligned}$$

(3)

If $k_2$ tends to $\frac{\sigma \theta ^S qn}{\sigma \theta ^S+\theta ^R}$, then

$$\begin{aligned} \mu _2=q-q\ln q + q\ln (\sigma \theta ^S+\theta ^R)-(\sigma \theta ^S+\theta ^R). \end{aligned}$$

(4)

Region 1. :

$\theta ^S<\theta ^R$ and $\theta ^S+\theta ^R<q$. In this region, $k_1$ tends to $\frac{qn}{2}$ and $k_2$ tends to $\frac{\sigma \theta ^S qn}{\sigma \theta ^S+\theta ^R}$. Consequently, $\mu _1<\mu _2$ (as the geometric mean $\sqrt{\sigma \theta ^S\theta ^R}$ is always smaller than the arithmetic mean $\frac{\sigma \theta ^S+\theta ^R}{2}$). Independently of $\sigma$, a citizen has incentive to abstain. The only equilibrium, therefore, is $\sigma =0$, and the expected outcome is S.

Region 2. :

$\theta ^S\ge \theta ^R$ and $\theta ^S+\theta ^R<q$: in this region, $k_1$ tends to $\frac{qn}{2}$, and $k_2$ tends to either value. If it tends to $\frac{\sigma \theta ^S qn}{\sigma \theta ^S+\theta ^R}$, then the same reasoning as above holds, and $\sigma =0$ is the only equilibrium. If it tends to $\frac{qn}{2}-1$, then $\mu _1=\mu _2$. The probabilities of Piv1 and Piv2 tend to zero at the same speed, but that does not mean that they are equal. Actually, the most likely subevent of Piv1 and Piv2 are when $N^S=N^R-1=int(\frac{qn}{2})$ and $N^S=int(\frac{qn+1}{2})-2,N^R=int(\frac{qn}{2})+1$ respectively. In both cases, $N^R$ takes the same value, so that $\mathrm{Prob}(Piv1)>\mathrm{Prob}(Piv2)\Leftrightarrow \mathrm{Prob}(N^S=int(\frac{qn}{2}))>\mathrm{Prob}(N^S=int(\frac{qn+1}{2})-2)$. Consequently, $\mathrm{Prob}(Piv1)>\mathrm{Prob}(Piv2)\Leftrightarrow \sigma \theta ^S n>\frac{qn}{2}-1$. That shows that there is an equilibrium with $\sigma = \frac{qn-2}{2\theta ^S n}$. But this equilibrium is unstable: for any slight decrease (resp., increase) in $\sigma$, $\mu _1<\mu _2$ (resp., $\mu _1>\mu _2$) and abstaining (resp., voting for S) is a best reply. So we have two stable symmetric equilibria in this region, namely $\sigma =0$ and $\sigma =1$. In both cases, the expected outcome is S, as the number of voters in favor of R is expected to be below the quorum, and the total expected number of votes for S if all S supporters vote for S is larger than the expected number of votes for R.

Region 3. :

$\theta ^S<\theta ^R<q<\theta ^S+\theta ^R$: in this region, $k_2$ tends to $\frac{\sigma \theta ^S qn}{\sigma \theta ^S+\theta ^R}$. If $\sqrt{\sigma \theta ^S \theta ^R}<\frac{q}{2}$, then $k_1$ tends to $\frac{qn}{2}$ and $\mu _1<\mu _2$ and a citizen maximizes her utility by abstaining. If $\sqrt{\sigma \theta ^S \theta ^R}\ge \frac{q}{2}$, then $k_1$ tends to $n\sqrt{\sigma \theta ^S\theta ^R}$. We may, again, have a mixed strategy equilibrium with $\mu _1=\mu _2$ and

$$\begin{aligned} 2\sqrt{\sigma \theta ^S\theta ^R}=q-q\ln q + q\ln (\sigma \theta ^S+\theta ^R). \end{aligned}$$

But, again, such an equilibrium cannot be stable. Indeed,

$$\begin{aligned} \frac{\partial (\mu _1-\mu _2)}{\partial \sigma }=\sqrt{\frac{\theta ^S\theta ^R}{\sigma }}-\frac{q\theta ^S}{\sigma \theta ^S+\theta ^R}, \end{aligned}$$

and, as $q<2\sqrt{\sigma \theta ^S\theta ^R}$, we can deduce, by replacing q with its upper bound,

$$\begin{aligned} \frac{\partial (\mu _1-\mu _2)}{\partial \sigma }>\sqrt{\frac{\theta ^S\theta ^R}{\sigma }}\frac{\theta ^R-\sigma \theta ^S}{\sigma \theta ^S+\theta ^R}>0, \end{aligned}$$

where the last inequality comes from $\theta ^S<\theta ^R$ and $\sigma \le 1$. So, only $\sigma =0$ and $\sigma =1$ are equilibrium candidates. If $\sigma =0$, then $\mu _1<\mu _2$ and abstaining is an equilibrium in the whole region, with outcome S. If $\sigma =1$, then $\mu _1>\mu _2$ if and only if $\sqrt{\theta ^S \theta ^R}\ge \frac{q}{2}$ and $2 \sqrt{\theta ^S \theta ^R}>q-q\ln q + q\ln (\theta ^S+\theta ^R)$. In this subregion, voting for S is also an equilibrium, and the expected outcome is R. To sum up, in this region, where the planner always prefers R, there is always an equilibrium with outcome S and in one subregion it is the only equilibrium outcome.

Region 4. :

$\theta ^S>\theta ^R$, $\theta ^R<q<\theta ^S+\theta ^R$: in this region, both $k_1$ and $k_2$ can converge towards any of their respective values. By the same argument as above, we can prove that there are two equilibria, $\sigma =0$ and $\sigma =1$, but the expected outcomes associated to these equilibria are both S, as either the quorum is not reached in equilibrium (if all S supporters abstain) or S gets more votes than R (if all S supporters actually vote). There is also a mixed strategy equilibrium, which, for the same reason as above, is unstable.

Region 5. :

$\theta ^S<\theta ^R$ and $\theta ^R>q$: independently of the optimal strategy of citizens of type S, the expected outcome is R, as citizens of type R are numerous enough to reach the quorum and they are more numerous than citizens of type S.

Region 6. :

$\theta ^S>\theta ^R>q$. Let us look immediately at the two extreme equilibrium candidates, $\sigma =0$ and $\sigma =1$. In the former case, $k_1$ and $k_2$ converge towards $\frac{qn}{2}$ and 0 respectively, so that unambiguously $\mu _1<\mu _2$, and abstaining is a best reply. This is, therefore, an equilibrium, with expected outcome R, as $\theta ^R n>q n$. This is the most surprising equilibrium of this game form. The reform is passed, whereas more citizens strictly prefer S to R than R to S. If $\sigma =1$, then $k_1$ and $k_2$ converge to $n\sqrt{\theta ^S \theta ^R}$ and $\frac{qn}{2}-1$ respectively. Then, $\mu _1>\mu _2$ if and only if

$$\begin{aligned} 2\sqrt{\sigma \theta ^S\theta ^R}-q+q\ln q - q\ln (\sigma \theta ^S+\theta ^R)>0, \end{aligned}$$

but this is always the case, as the inequality holds for $\theta ^S=\theta ^R=q$ (the smallest values of these parameters in region 6) and the expression is increasing in both $\theta ^S$ and $\theta ^R$. This proves that $\sigma =1$ is an equilibrium, and the equilibrium outcome is S.$\square$

Proof of Lemma 3

The same kind of computations as in the above proof reveal that in state $i\in \{r,s\}$ if $\sqrt{\lambda ^{R|i}\lambda ^{S|i}}\ge q$, then the most likely subevent of Piv1 occurs when k tends to $\sqrt{\lambda ^{R|i}\lambda ^{S|i}}n$, and

$$\begin{aligned} \mu _{1|i}=2\sqrt{\lambda ^{R|i}\lambda ^{S|i}}-(\lambda ^{R|i}+\lambda ^{S|i}), \end{aligned}$$

(5)

whereas, if $\sqrt{\lambda ^{R|i}\lambda ^{S|i}}\le q$, then the most likely subevent of Piv1 occurs when k tends to qn, and

$$\begin{aligned} \mu _{1|i}=2q-2q\ln q + q\ln \lambda ^{R|i}\lambda ^{S|i}-(\lambda ^{R|i}+\lambda ^{S|i}). \end{aligned}$$

(6)

If $\lambda ^{S|i}\le q$, then the most likely subevent of Piv2 occurs when k tends to $\lambda ^{S|i}n$ (its most likely value), and

$$\begin{aligned} \mu _{2|i}=q-q\ln q + q\ln \lambda ^{R|i}-\lambda ^{R|i}, \end{aligned}$$

(7)

whereas, if $\lambda ^{S|i}\ge q$, then the most likely subevent of Piv2 occurs when k tends to $qn-1$, and

$$\begin{aligned} \mu _{2|i}=2q-2q\ln q + q\ln \lambda ^{R|i}\lambda ^{S|i}-(\lambda ^{R|i}+\lambda ^{S|i}). \end{aligned}$$

(8)

Having computed the magnitudes, we prove first that it is impossible that S wins in state r when $\theta ^R + \theta ^I\ge q.$ Note that this only happens if $\sigma _R<1.$ We have to distinguish between two cases.

Case 1: $\lambda ^{S|r}\ge \lambda ^{R|r}\ge q$. That clearly requires $\sigma _S>0.$ Given the dominant strategy of the informed independent citizens, we have $\lambda ^{S|s}>\lambda ^{S|r}\ge \lambda ^{R|r}>\lambda ^{R|s}$. Magnitude $\mu _{1|r}$ is given by Eq. (5), $\mu _{2|r}$ by Eq. (8). Proving that $\mu _{1|r}>\mu _{2|r}$ amounts to proving that

$$\begin{aligned} 2\sqrt{\lambda ^{R|r}\lambda ^{S|r}}>2q-2q\ln q + q\ln \lambda ^{R|r}\lambda ^{S|r}, \end{aligned}$$

or

$$\begin{aligned} \sqrt{\lambda ^{R|r}\lambda ^{S|r}}-q\ln \sqrt{\lambda ^{R|r}\lambda ^{S|r}}>q-q\ln q, \end{aligned}$$

which follows from the fact that function $x-q\ln x$ is increasing for $x>q$. Magnitude $\mu _{1|s}$ may be given either by Eq. (5), in which case $\mu _{1|r}>\mu _{1|s}$ follows from $\lambda ^{S|r} \lambda ^{R|r}>\lambda ^{S|s} \lambda ^{R|s}$ (remember that $\lambda ^{S|r} + \lambda ^{R|r}=\lambda ^{S|s} + \lambda ^{R|s}$), or by (6), in which case $\mu _{1|r}>\mu _{1|s}$ follows from the same argument as for $\mu _{2|r}$ above. Obviously, $\mu _{2|r}>\mu _{2|s}$, so that it is clear that $\mu _{1|r}>\mu _{2|s}$. Consequently, conditional on her vote being pivotal, an uninformed independent citizen is sure to be in state r, so that $\sigma _S>0$ is not a best reply, a contradiction.

Case 2: $\lambda ^{R|r}<q$. Again, it is crucial that $\lambda ^{S|s}=\lambda ^{S|r}+\gamma \theta ^I,$ and $\lambda ^{R|r}=\lambda ^{R|s}+\gamma \theta ^I$. If $\lambda ^{S|r}$ is such that $\sqrt{\lambda ^{S|r} \lambda ^{R|r}}\ge q$, then we are back to a case similar to the one above, and $\sigma _S>0$ cannot be a best reply. If $q\le \lambda ^{S|r} \le \frac{q^2}{\lambda ^{R|r}}$, then $\mu _{1|r}=\mu _{2|r}$ (given by Eqs. (6) and (8)). Magnitudes $\mu _{1|s}$ and $\mu _{2|s}$ are given by the same equations, so that $\mu _{1|r}>\mu _{1|s},\mu _{2|s}$ follows from $\lambda ^{S|r} \lambda ^{R|r}>\lambda ^{S|s} \lambda ^{R|s}$. If $\lambda ^{S|r}<q,$ then $\mu _{2|r}$ is given by Eq. (7); $\mu _{2|r}>\mu _{2|s}$ follows from $\lambda ^{R|s}<\lambda ^{R|r}$ and the fact that function $x-q\ln x$ is decreasing for $x<q$. The fact that Piv1 are less likely than Piv2 is immediate and comes from the fact that Piv1 occurs when the votes for R just reaches the quorum (same requirement as Piv2) and there is a tie between R and S. Consequently, conditional on her vote being pivotal, an uninformed independent voter is sure to be in state r, proving that $\sigma _R<1$ is not a best reply, a contradiction. Note that this also shows that if $\lambda ^{R|r}<q$, the best reply is $\sigma _S=0$: uninformed citizens don’t vote for S, (but may vote for R), using the quorum as a guarantee that S will win the election in s.

Second, we prove that it is impossible that R wins in state s when $\theta ^R < \theta ^I.$ That occurs if $\lambda ^{R|s}>q,\lambda ^{S|s},$ which implies that $\sigma _S<1$. Note that $\lambda ^{R|r}=\lambda ^{R|s}+\gamma \theta ^I>\lambda ^{R|s}>q$, so that clearly $\mu _{2|s}>\mu _{2|r}$. Also, $\lambda ^{R|r}>\lambda ^{R|s}>q,\lambda ^{S|s}>\lambda ^{S|r}$ makes it clear that $\mu _{1|s}>\mu _{1|r}$. Consequently (independently on how Piv1 and Piv2 are ranked), state s is infinitely more likely than r conditional on her vote being pivotal, so an uninformed independent citizen votes for S, so that $\sigma _S<1$ is not a best reply.

$\square$

Proof of Theorem 2

Lemma 3 shows that a $\hat{q}$-approval quorum always gives the outcome that coincides with the $\hat{q}$-approval social preferences. Consequently, in state r, it also coincides with the q-participation quorum social preferences, with $\hat{q}=q$, as they are the same in that state. We simply need to show that approval quorum does better than participation quorum in state s. Let us restrict ourselves to proving that the undesired equilibria (leading to S being chosen in Region 3 and R being chosen in Region 6) highlighted in Theorem 1 still prevails under the current assumptions.

Claim 1: an equilibrium exists such that S is elected in s, whereas $\theta ^R<q$, $\theta ^R>\theta ^I$ and $\theta ^R + \theta ^I \le q.$ Let $\sigma _S^*\ge 0$ be defined by $\theta ^R + \sigma _S^* (1-\gamma )\theta ^I < q < \theta ^R + (\sigma _S^* (1-\gamma )+\gamma )\theta ^I,$ and

$$\begin{aligned} \hbox {Prob}(N^R + N^S=qn-1|R)=\hbox {Prob}(N^R + N^S=qn-1|S), \end{aligned}$$

that is,

$$\begin{aligned}&\frac{e^{-[\theta ^R + (\sigma _S^* (1-\gamma )+\gamma )\theta ^I]n}([\theta ^R + (\sigma _S^* (1-\gamma )+\gamma )\theta ^I]n)^{qn-1}}{(qn-1)!} \\= & {} \frac{e^{-[\theta ^R + (\sigma _S^* (1-\gamma ))\theta ^I]n}([\theta ^R + (\sigma _S^* (1-\gamma ))\theta ^I]n)^{qn-1}}{(qn-1)!}, \end{aligned}$$

or,

$$\begin{aligned} e^{\frac{-\gamma \theta ^I}{qn-1}}[\theta ^R + (\sigma _S^* (1-\gamma )+\gamma )\theta ^I]=[\theta ^R + (\sigma _S^* (1-\gamma ))\theta ^I] \end{aligned}$$

which yields

$$\begin{aligned} \sigma _S^*=\frac{\theta ^R}{(1-\gamma )\theta ^I}\frac{1-\gamma \theta ^Ie^{\frac{-\gamma \theta ^I}{qn-1}}}{\theta ^R e^{\frac{-\gamma \theta ^I}{qn-1}}-1}. \end{aligned}$$

We claim that informed independents playing $(R,\emptyset )$, and the uninformed independents playing $\sigma _S^*$ is an equilibrium. By definition of $\sigma _S^*$, $\hbox {Prob}(Piv2|r)=\hbox {Prob}(Piv2|s)$. Clearly, $\hbox {Prob}(Piv1|i)<\hbox {Prob}(Piv2|i)$, all $i \in \{r,s\}$, as a Piv1 event requires a tie between R and S and that the quorum be reached. When informed citizens observe the state is r, they clearly vote for R. When they observe s, given that $\hbox {Prob}(Piv1|s)<\hbox {Prob}(Piv2|s)$, they prefer abstaining. Uninformed citizens do not observe the state. Conditional on their vote being pivotal, they are sure that they face a Piv2 event, so that, as $n\rightarrow \infty$, they tend to be indifferent between voting for S and abstaining. That equilibrium is stable: if $\sigma _S<\sigma _S^*,$ (resp., $\sigma _S>\sigma _S^*$) then $\hbox {Prob}(Piv2|r)>\hbox {Prob}(Piv2|s)$ (resp., $\hbox {Prob}(Piv2|r)<\hbox {Prob}(Piv2|s)$) so that citizens prefer voting for S (resp. abstaining).

Claim 2: an equilibrium exists such that R is elected in s, whereas $\theta ^R<\theta ^I$. Assume $q<\theta ^R < \theta ^I.$ We claim that, like in the public information framework of the previous sections, we have an equilibrium where independent uninformed citizens prefer S but abstain, as voting for S could make R win the election. Let us prove that informed independents playing $(R,\emptyset )$, and the uninformed independents playing $\emptyset$ is an equilibrium. As $\lambda ^{R|r}=\lambda ^{R|s}+\gamma \theta ^I > \lambda ^{R|s}>q$ and $\lambda ^{S|r}=\lambda ^{S|s}=0,$ we have $\mu _{2|s}>\mu _{2|r}>\mu _{1|s},\mu _{1|r}.$ In state s, given that $\mu _{2|s}>\mu _{1|s},$ any independent citizen prefers to abstain. As informed citizens observe the state of nature, voting for R in r and abstaining in s is a best reply. Conditional on her vote being pivotal, an uninformed independent citizen is sure to be in state s, so that abstaining is her best reply.$\square$