Jury voting without objective probability

Abstract

Unlike in the standard jury voting experiment, the voting environment in practice has no explicit signal structure. Voters then need to conceptualize the information structure in order to update their beliefs based on “pivotal reasoning”. This paper investigates whether voters can play a strategic voting under a “detail-free” environment. We obtain non-parametric predictions in terms of the differences in voting behaviors under majority and unanimity rule. Our experimental results suggest that voters can still play the strategic voting as in the existing experiments.

Appendix

1.1 Remark 1

In Feddersen and Pesendorfer (1998), there is a binary signal $s\in \{0,1\}$ and the probability of $s=\omega$ is $\lambda \in (\frac{1}{2},1)$. Let $\mu$ be the prior probability of $\omega =1$. Since the signal is binary, there are only two possible p in their model: one is the probability of $\omega =1$ conditional on $s=1$, i.e., $p_{1}=\frac{\lambda \mu }{\lambda \mu +(1-\lambda )(1-\mu )}$ and the other is the probability of $\omega =1$ conditional on $s=0$, $p_{0}=\frac{(1-\lambda )\mu }{(1-\lambda )\mu +\lambda (1-\mu )}$.

Since their model has only two possible values of p, whereas our setting allows any value in [0, 1], discretize $P=[0,1]$ to $\{p_{0},p_{1}\}$ and let $f(p|\omega )$ be a probability mass function conditional on $\omega$. The degenerated version of our model is then equivalent to their binary signal model if $f(p_{1}|1)=f(p_{0}|0)=\lambda$ and $f(p_{0}|1)=f(p_{1}|0)=1-\lambda$.

1.2 Proof of Lemma 1

Suppose not. Then, there exists voter i such that $a_{i}=A$ for $p_{i}'$, while $a_{i}=R$ for $p_{i}''>p_{i}'$. However, note that

$$\begin{aligned} \psi _{\sigma }(k,p_{i})=\frac{\Pr (piv|\omega =1;k,\sigma )p_{i}}{\Pr (piv|\omega =1;k,\sigma )p_{i}+\Pr (piv|\omega =0;k,\sigma )(1-p_{i})}. \end{aligned}$$

Obviously, $\psi _{\sigma }(k,p_{i}'')>\psi _{\sigma }(k,p_{i}')$ given any strategy profile $\sigma$. Note that voter i’s expected payoff from $a_{i}$ given $\sigma$ and $p_{i}$ is

$$\begin{aligned} E[U(a_{i},\sigma _{-i}(p_{-i}))|p_{i}]={\left\{ \begin{array}{ll} -\frac{1}{2}(1-\psi _{\sigma }(k,p_{i})) &{}\quad \text{ if }\, a_{i}=A\\ -\frac{1}{2}\psi _{\sigma }(k,p_{i}) &{}\quad \text{ if }\, a_{i}=R \end{array}\right. }. \end{aligned}$$

Thus, whenever she prefers A to R given $p_{i}'$, she also prefers A to R given $p_{i}''$, a contradiction.

1.3 Proof of Proposition 1

Let $\psi _{\hat{p}}(k,p_{i})$ be $\psi _{\sigma }(k,p_{i})$ in which $\sigma$ is the symmetric cutoff strategy profile with $\hat{p}$.

First of all, given a strategy profile $\sigma$, voter i’s belief that voter j chooses A conditional on $\omega$ is

$$\begin{aligned} Q_{j,\omega }(\sigma )=\int _{p_{j}\in \{p_{j}:\sigma _{j}(p_{j})=A\}}f(p_{j}|\omega )dp_{j}. \end{aligned}$$

By the definition of symmetric equilibrium, $Q_{j,\omega }(\sigma )$ does not depend on j. Thus, let $Q_{\omega }(\sigma )=Q_{j,\omega }(\sigma )$.

Since each group consists of five voters in the experiment, voter i is pivotal when $k-1$ out of four voters choose A. Then, given $\sigma$, $\omega$ and k, we can compute the probability of being pivotal conditional on $\omega$. That is,

$$\begin{aligned} \Pr (piv|\omega ;k,\sigma )=\left( {\begin{array}{c}4\\ k-1\end{array}}\right) Q_{\omega }(\sigma )^{k-1}(1-Q_{\omega }(\sigma ))^{4-(k-1)}. \end{aligned}$$

Then, the probability of $\omega =1$ conditional on being pivotal given $p_{i}$ is

$$\begin{aligned} \psi _{\sigma }(k,p_{i})=\frac{\Pr (piv|\omega =1;\,k,\sigma )p_{i}}{\Pr (piv|\omega =1;\,k,\sigma )p_{i} +\Pr (piv|\omega =0;\,k,\sigma )(1-p_{i})}. \end{aligned}$$

Note that if $\sigma$ is a symmetric cutoff strategy profile with $\hat{p}$ , $Q_{\omega }(\sigma )=1-F(\hat{p}|\omega )$. Since $F(\hat{p}|\omega )$ is continuous in $\hat{p}$, $\psi _{\hat{p}}(k,\hat{p})$ is also continuous in $\hat{p}$. Moreover, $\lim _{\hat{p}\rightarrow 1}\psi _{\hat{p}}(k,\hat{p})=1$ and $\lim _{\hat{p}\rightarrow 0}\psi _{\hat{p}}(k,\hat{p})=0$. Then, there exists $\hat{p}(k)\in (0,1)$ such that $\psi _{\hat{p}(k)}(k,\hat{p}(k))=\frac{1}{2}$ given k. Since $\psi _{\hat{p}(k)}(k,p)$ is strictly increasing in p, $\psi _{\hat{p}(k)}(k,p)>(<)\frac{1}{2}$ if $p>(<)\hat{p}(k)$. Thus, given voting rule k, there exists a symmetric equilibrium in which each voter uses the cutoff strategy with $\hat{p}(k)$.

Now, we claim that given any symmetric equilibrium under each voting rule, $\hat{p}(3)>\hat{p}(5)$. Note that $\Pr (piv|\omega ;5,p_{i})=[1-F(\hat{p}|\omega )]{}^{4}$. The probability of $\omega =1$ conditional on being pivotal is then

$$\begin{aligned} \psi _{\hat{p}}(5,p_{i})=\frac{[1-F(\hat{p}|1)]{}^{4}p_{i}}{[1-F(\hat{p}|1)]^{4}p_{i}+[1-F(\hat{p}|0)]^{4}(1-p_{i})}. \end{aligned}$$

By the monotone likelihood ratio property, $f(\hat{p}|0)[1-F(\hat{p}|1)]\ge f(\hat{p}|1)[1-F(\hat{p}|0)]$. It follows that $\frac{d\psi _{\hat{p}}(5,\hat{p})}{d\hat{p}}>0$. Since $\psi _{\hat{p}}(5,\hat{p})$ is strictly increasing in $\hat{p}$, $\hat{p}(5)$, which solves $\psi _{\hat{p}}(5,\hat{p})=\frac{1}{2}$, is unique. That is, there exists a unique symmetric equilibrium under unanimity rule. On the other hand, since unanimity rule requires a larger number of A when a voter is pivotal, $\psi _{\hat{p}}(3,\hat{p})<\psi _{\hat{p}}(5,\hat{p})$ given any $\hat{p}$. Then, $\hat{p}(3)$, which solves $\psi _{\hat{p}}(3,\hat{p})=\frac{1}{2}$, has to be strictly higher than $\hat{p}(5)$.

1.4 Proof of Proposition 2

Let $\mu :=\int _{p\in P}f(p,\omega =1)dp$. Since $\Pr (R|\omega ;k)=F(\hat{p}(k)|\omega )$ given equilibrium cutoff $\hat{p}(k)$, the probability of $\omega =0$ conditional on $a_{i}=R$ in the symmetric equilibrium is

$$\begin{aligned} \Pi _{R}(k)=\frac{F(\hat{p}(k)|0)(1-\mu )}{F(\hat{p}(k)|0)(1-\mu )+F(\hat{p}(k)|1)\mu }. \end{aligned}$$

From Proposition 1, $\hat{p}(3)>\hat{p}(5)$. Then, by the monotone likelihood property,

$$\begin{aligned} \frac{F(\hat{p}(5)|0)}{F(\hat{p}(5)|1)}>\frac{F(\hat{p}(3)|0)}{F(\hat{p}(3)|1)}. \end{aligned}$$

It follows that

$$\begin{aligned} \frac{F(\hat{p}(5)|0)}{F(\hat{p}(3)|0)}>\frac{F(\hat{p}(5)|0)(1-\mu )+F(\hat{p}(5)|1)\mu }{F(\hat{p}(3)|0)(1-\mu )+F(\hat{p}(3)|1)\mu } \end{aligned}$$

for any $\mu >0$. Thus,

$$\begin{aligned} \frac{F(\hat{p}(5)|0)(1-\mu )}{F(\hat{p}(5)|0)(1-\mu )+F(\hat{p}(5)|1)\mu }>\frac{F(\hat{p}(3)|0)(1-\mu )}{F(\hat{p}(3)|0)(1-\mu )+F(\hat{p}(3)|1)\mu }. \end{aligned}$$