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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.

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Notes

  1. 1.

    Battaglini et al. (2010) also tested another type of pivotal voting model called “swing voter’s curse” and confirmed their predictions.

  2. 2.

    Voter i’s cognitive ability should be a major determinant of \(p_{i}\), while voter i’s cognitive ability is determined before the experiment. Thus, the assumption that \(p_{i}\) is drawn from a distribution might not be very natural if i is interpreted as a subject’s identity. However, if i is interpreted as a seat number in the session, the assumption is natural as the subjects were randomly assigned to seats.

  3. 3.

    For more detail, see Appendix.

  4. 4.

    Since the subjects have different cognitive abilities, one might think that it is unnatural to consider an equilibrium in which the strategy profile is independent of i. However, as we mentioned in the earlier footnote, if i is interpreted as a seat number in the session instead of a subject’s identity, it is natural to consider an equilibrium in which two voters in seat i and j choose the same action if \(p_{i}=p_{j}\) as the subjects were randomly assigned to seats.

  5. 5.

    Since the proposer’s payoff is independent of a group decision, there is no reason that the quality of proposals depends on a voting rule. Then, since the subjects were assigned to four treatments randomly, \(f(p|\omega )\) should be independent of voting rules.

  6. 6.

    To obtain testable predictions at the group level, we need to specify the parameters of the model, which contradicts the purpose of our experiment, a detail-free experiment.

  7. 7.

    We appreciate one of the referees for pointing out this possibility.

  8. 8.

    For example, Caplin et al. (2011) tested a rational inattention model of individual decision making. In their experiment, the value of each choice is represented by a simple algebraic expression so that the subjects fail to choose the best choice when they are inattentive.

  9. 9.

    There are some experimental papers that investigate voting games with costly information acquisition, e.g., Großer and Seebauer (2013) and Elbittar et al. (2014). However, unlike in our experiment, the cost of information acquisition in their experiments is explicit in the sense that it directly reduces the payoff.

  10. 10.

    We can compute the average probability of being pivotal from the frequency of approval/rejection votes. In the logic treatment, the probability of being pivotal was around 16.78 % under unanimity rule while it was 37.23 % under majority rule. In the math treatment, it was 6.25 and 26.46 % respectively.

  11. 11.

    For more detail, see Persico (2004) and Elbittar et al. (2014).

  12. 12.

    Wilson (1987) emphasizes the theoretical aspect whereas our paper focuses on the empirical aspect.

References

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Correspondence to Toru Suzuki.

Additional information

We appreciate two referees for their helpful suggestions. We also thank Mikhail Anufriev, Jacob Goeree, Susumu Imai, John Wooders and participants in Australasian Theory workshop, 13th SAET meeting, and Keio University seminar for helpful comments. This research was funded by Max Planck Society.

Appendix

Appendix

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 \(f(p_{0}|1)=f(p_{1}|0)=1-\lambda \).

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.

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 \(\hat{p}(5)\).

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,

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

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Li, K.K., Suzuki, T. Jury voting without objective probability. Soc Choice Welf 46, 389–406 (2016). https://doi.org/10.1007/s00355-015-0918-z

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JEL Classification

  • D71
  • D72
  • C92