Feddersen and Pesendorfer meet Ellsberg

Abstract

The Condorcet Jury Theorem formalises the “wisdom of crowds”: binary decisions made by majority vote are asymptotically correct as the number of voters tends to infinity. This classical result assumes like-minded, expected utility maximising voters who all share a common prior belief about the right decision. Ellis (Theor Econo 11(3): 865–895, 2016) shows that when voters have ambiguous prior beliefs—a (closed, convex) set of priors—and follow maxmin expected utility (MEU), such wisdom requires that voters’ beliefs satisfy a “disjoint posteriors” condition: different private signals lead to posterior sets with disjoint interiors. Both the original theorem and Ellis’s generalisation assume symmetric penalties for wrong decisions. If, as in the jury context, errors attract asymmetric penalties then it is natural to consider voting rules that raise the hurdle for the decision carrying the heavier penalty for error (such as conviction in jury trials). In a classical model, Feddersen and Pesendorfer (Am Politi Sci Rev 92(1):23–35, 1998) have shown that, paradoxically, raising this hurdle may actually increase the likelihood of the more serious error. In particular, crowds are not wise under the unanimity rule: the probability of the more serious error does not vanish as the crowd size tends to infinity. We show that this “Jury Paradox” persists in the presence of ambiguity, whether or not juror beliefs satisfy Ellis’s “disjoint posteriors” condition. We also characterise the strictly mixed equilibria of this model and study their properties. Such equilibria cannot exist in the absence of ambiguity but may exist for arbitrarily large jury size when ambiguity is present. In addition to uninformative strictly mixed equilibria, analogous to those exhibited by Ellis (Theor Econo 11(3): 865–895, 2016), there may also exist strictly mixed equilibria which are informative about voter signals.

Appendices

Appendix A

In this Appendix, we describe the derivation of Fig. 1.

Let us start by observing that (5) may be written as follows:

$$\begin{aligned} V\left( \sigma _{t}^{i},\sigma ;\pi _{t}\right) \ =\ \pi _{t}+\sigma _{t}^{i}\left( \left( 1-\pi _{t}\right) \rho _{b}-\left( 1+c\right) \pi _{t}\rho _{a}\right) . \end{aligned}$$

Recalling (7), it follows that $V\left( \sigma _{t}^{i},\sigma ;\pi _{t}\right)$ is strictly decreasing in $\sigma _{t}^{i}$ iff $\pi _{t}>\pi ^{*}\left( \sigma \right)$; strictly increasing in $\sigma _{t}^{i}$ iff $\pi _{t}<\pi ^{*}\left( \sigma \right)$; and constant in $\sigma _{t}^{i}$ iff $\pi _{t}=\pi ^{*}\left( \sigma \right)$. Thus, if all the posteriors in $\Pi _{t}$ are (strictly) on the same side of $\pi ^{*}\left( \sigma \right)$, then voter i’s response follows the same logic as in Feddersen and Pesendorfer (1998): recall footnote 12. In particular, if $\pi ^{*}\left( \sigma \right) <{\underline{\pi }}_{t}$ then the function

$$\begin{aligned} V^{*}_{t} \left( \sigma _{t}^{i},\sigma \right) \ \equiv \ \min _{\pi _{t}\in \Pi _{t}}\ V\left( \sigma _{t}^{i},\sigma ;\pi _{t}\right) \end{aligned}$$

(15)

is the lower envelope of functions that are strictly decreasing in $\sigma _{t}^{i}$ so $V^{*}_{t}$ itself is strictly decreasing in $\sigma _{t}^{i}$. It follows that $\sigma _{t}^{i}=0$ is the unique best response to $\sigma$ (conditional on $t_{i}=t$) when $\pi ^{*}\left( \sigma \right) < {\underline{\pi }}_{t}$. Conversely, if $\pi ^{*}\left( \sigma \right) > {\overline{\pi }}_{t}$ then (15) is strictly increasing in $\sigma _{t}^{i}$ so $\sigma _{t}^{i}=1$ is the unique best response to $\sigma$ (again, conditional on $t_{i}=t$). This gives the green and blue regions (respectively) in Fig. 1.

It remains to consider the case in which ${\underline{\pi }}_{t}\le \pi ^{*}\left( \sigma \right) \le {\overline{\pi }}_{t}$: the case described by the pink rectangle in Fig. 1 and its (brown and purple) boundaries. To analyse this case, it is convenient to re-write (5) as follows:

$$\begin{aligned} V\left( \sigma _{t}^{i},\sigma ;\pi _{t}\right) \ =\ \sigma _{t}^{i}\rho _{b}+\pi _{t}\left( 1-\left( 1+c\right) \rho _{a}\sigma _{t}^{i}-\rho _{b}\sigma _{t}^{i}\right) . \end{aligned}$$

Recalling (8), we observe that

$$\begin{aligned} 1-\left( 1+c\right) \rho _{a}\sigma ^{*}\left( \sigma \right) -\rho _{b}\sigma ^{*}\left( \sigma \right) \ =\ 0 \end{aligned}$$

and hence

$$\begin{aligned} V^{*}_{t} \left( \sigma _{t}^{i},\sigma \right) \ =\ \left\{ \begin{array}{cc} V\left( \sigma _{t}^{i},\sigma ;{\underline{\pi }}_{t}\right) &{} \text {if } \sigma _{t}^{i}\le \sigma ^{*}\left( \sigma \right) \\ &{} \\ V\left( \sigma _{t}^{i},\sigma ;{\overline{\pi }}_{t}\right) &{} \text {if } \sigma _{t}^{i}\ge \sigma ^{*}\left( \sigma \right) \end{array} \right. \end{aligned}$$

In other words, ${\underline{\pi }}_{t}$ is the uniquely most “pessimistic” posterior in $\Pi _{t}$ when $\sigma _{t}^{i}<\sigma ^{*}\left( \sigma \right)$ and ${\overline{\pi }}_{t}$ is the uniquely most “pessimistic” posterior when $\sigma _{t}^{i}>\sigma ^{*}\left( \sigma \right)$. Moreover, by choosing $\sigma _{t}^{i}=\sigma ^{*}\left( \sigma \right)$ voter i can perfectly hedge against uncertainty since $V\left( \sigma ^{*}\left( \sigma \right) ,\sigma ;\pi _{t}\right)$ is independent of $\pi _{t}$. The latter hedging possibility may provide strict incentives to randomise, as noted by Ellis (2016). The t-conditional best response(s) to any $\sigma$ satisfying ${\underline{\pi }}_{t}\le \pi ^{*}\left( \sigma \right) \le {\overline{\pi }}_{t}$ may now be characterised as follows:

Fig. 4

Best responses when $\underline{\pi }_{t}\le \pi ^{*}\left( \sigma \right) \le \overline{\pi }_{t}$ and $\sigma ^{*}\left( \sigma \right) \ge 1$. In case (a), $\underline{\pi }_{t}<\pi ^{*}\left( \sigma \right)$ so $\sigma _{t}^{i}=1$ is optimal. In case (b), $\underline{\pi }_{t}=\pi ^{*}\left( \sigma \right)$ so any $\sigma _{t}^{i}\in \left[ 0,1\right]$ is optimal

Fig. 5

Best responses when $\underline{\pi }_{t}\le \pi ^{*}\left( \sigma \right) \le \overline{\pi }_{t}$ and $\sigma ^{*}\left( \sigma \right) <1$. In case (i), $\underline{\pi }_{t}<\pi ^{*}\left( \sigma \right)$ so $\sigma _{t}^{i}=\sigma ^{*}\left( \sigma \right)$ is optimal. In case (ii), $\underline{\pi }_{t}=\pi ^{*}\left( \sigma \right)$ so any $\sigma _{t}^{i}\in \left[ 0,\sigma ^{*}\left( \sigma \right) \right]$ is optimal

• If $\sigma ^{*}\left( \sigma \right) \ge 1$ then $V^{*}_{t} \left( \sigma _{t}^{i},\sigma \right) =V\left( \sigma _{t}^{i},\sigma ;{\underline{\pi }}_{t}\right)$ for any $\sigma _{t}^{i}\in \left[ 0,1\right]$, so $\sigma _{t}^{i}=1$ is uniquely optimal if ${\underline{\pi }}_{t}<\pi ^{*}\left( \sigma \right)$, and any $\sigma _{t}^{i}\in \left[ 0,1\right]$ is optimal if ${\underline{\pi }}_{t}=\pi ^{*}\left( \sigma \right)$, see Fig. 4.

• If $\sigma ^{*}\left( \sigma \right) <1$ then $V^{*}_{t} \left( \sigma _{t}^{i},\sigma \right)$ is non-decreasing in $\sigma _{t}^{i}$ when $\sigma _{t}^{i}<\sigma ^{*}\left( \sigma \right)$, since ${\underline{\pi }}_{t}\le \pi ^{*}\left( \sigma \right)$, and non-increasing in $\sigma _{t}^{i}$ when $\sigma _{t}^{i}>\sigma ^{*}\left( \sigma \right)$, since ${\overline{\pi }}_{t}\ge \pi ^{*}\left( \sigma \right)$. Hence: (i) $\sigma _{t}^{i}=\sigma ^{*}\left( \sigma \right)$ is uniquely optimal if ${\underline{\pi }}_{t}<\pi ^{*}\left( \sigma \right) <{\overline{\pi }}_{t}$; (ii) any $\sigma _{t}^{i}\in \left[ 0,\sigma ^{*}\left( \sigma \right) \right]$ is optimal if ${\underline{\pi }}_{t}=\pi ^{*}\left( \sigma \right)$; and (iii) any $\sigma _{t}^{i}\in \left[ \sigma ^{*}\left( \sigma \right) ,1\right]$ is optimal if ${\overline{\pi }}_{t}=\pi ^{*}\left( \sigma \right)$. Fig. 5 illustrates cases (i) and (ii).²⁹ Case (iii) is symmetric to (ii).

Appendix B

In this Appendix, we prove Proposition 4.2.

We start with an important preliminary result. To facilitate its statement, let us define

$$\begin{aligned} \Gamma =\left\{ \left( \sigma _{1},\sigma _{2}\right) \in \left[ 0,1\right] ^{2}\ |\ \sigma _{1}\le \sigma _{2}\ \ \text {and}\ \ \sigma _{2}>0\right\} \end{aligned}$$

Set $\Gamma$ contains all the strategies with $\sigma _{1}\le \sigma _{2}$ excluding $\sigma =\left( 0,0\right)$.

Lemma 6.1

Let $\Omega \left( x\right) =\left\{ \sigma \in \Gamma \ |\ \pi ^{*}\left( \sigma \right) =x\right\}$. Then: (i) $\Omega \left( x\right) =\emptyset$ iff

$$\begin{aligned} x\notin \left[ \frac{1}{2+c},\ \frac{r^{N}}{r^{N}+\left( 1-r\right) ^{N}\left( 1+c\right) }\right] \text {;} \end{aligned}$$

(ii) $\Omega \left( x\right) =\left\{ \sigma \in \Gamma \ |\ \sigma _{1}=0\right\}$ if

$$\begin{aligned} x=\frac{r^{N}}{r^{N}+\left( 1-r\right) ^{N}\left( 1+c\right) }\text {;} \end{aligned}$$

and (iii) $\Omega \left( x\right) =\left\{ \sigma \in \Gamma \ |\ \sigma _{2}=\lambda \left( x\right) \sigma _{1}\right\}$ otherwise, where

$$\begin{aligned} \lambda \left( x\right) =\frac{r-\left( 1-r\right) \ell \left( x\right) ^{ \frac{1}{N}}}{r\ell \left( x\right) ^{\frac{1}{N}}-\left( 1-r\right) }\in \left[ 1,\infty \right) \ \ \ \text {and}\ \ \ \ell \left( x\right) =\frac{1-x }{x\left( 1+c\right) }. \end{aligned}$$

Proof

Since $\sigma \in \Gamma$, we have

$$\begin{aligned} \pi ^{*}\left( \sigma \right) =x\ \ \ \Leftrightarrow \ \ \ \frac{\rho _{a}}{\rho _{b}}\ =\ \ell \left( x\right) \end{aligned}$$

(16)

from (7).³⁰ The likelihood ratio $\rho _{a}/\rho _{b}$ is equal to

$$\begin{aligned} \left[ \frac{ry+\left( 1-r\right) \left( 1-y\right) }{\left( 1-r\right) y+r\left( 1-y\right) }\right] ^{N}, \end{aligned}$$

where $y=\sigma _{1}/\left( \sigma _{1}+\sigma _{2}\right) \in \left[ 0, \frac{1}{2}\right]$. It follows that

$$\begin{aligned} \left( \frac{1-r}{r}\right) ^{N}\le \frac{\rho _{a}}{\rho _{b}}\le 1. \end{aligned}$$

Using these inequalities and (16), we deduce that $\Omega \left( x\right) \ne \emptyset$ iff

$$\begin{aligned} \frac{1}{2+c}\le x\le \frac{r^{N}}{r^{N}+\left( 1-r\right) ^{N}\left( 1+c\right) }. \end{aligned}$$

This proves (i). When

$$\begin{aligned} x=\frac{r^{N}}{r^{N}+\left( 1-r\right) ^{N}\left( 1+c\right) } \end{aligned}$$

we have

$$\begin{aligned} \ell \left( x\right) =\left( \frac{1-r}{r}\right) ^{N}. \end{aligned}$$

Using (3)–(4), it is clear that $\sigma \in \Gamma$ satisfies

$$\begin{aligned} \frac{\rho _{a}}{\rho _{b}}=\left( \frac{1-r}{r}\right) ^{N} \end{aligned}$$

iff $\sigma _{1}=0$. This proves (ii). Finally, if

$$\begin{aligned} \frac{1}{2+c}\le x<\frac{r^{N}}{r^{N}+\left( 1-r\right) ^{N}\left( 1+c\right) }, \end{aligned}$$

then (3)–(4) and (16) imply

$$\begin{aligned} \pi ^{*}\left( \sigma \right) =x\ \ \ \Leftrightarrow \ \ \ \sigma _{2}=\lambda \left( x\right) \sigma _{1} \end{aligned}$$

with $\lambda \left( x\right) \in \left[ 1,\infty \right)$. This establishes (iii) and completes the proof of Lemma 6.1. $\square$

Fig. 6

Note that the points in $\Omega \left( x\right)$ form a line, and this line gets steeper as x increases

Figure 6 illustrates $\Omega \left( x\right)$.

Fig. 7

Strictly mixed responsive equilibria when $\overline{\pi }_{2}>\underline{\pi }_{1}$

Now consider Proposition 4.2. It is obvious from Fig. 1 that ${\overline{\pi }}_{2}\ge {\underline{\pi }}_{1}$ is necessary for the existence of a responsive, strictly mixed equilibrium; otherwise, it would not be possible for both $\left( {\underline{\pi }}_{1}, \overline{\pi }_{1}\right)$ and $\left( {\underline{\pi }}_{2},{\overline{\pi }} _{2}\right)$ to both lie in the rectangular region between the green and blue triangles in Fig. 1. We assume ${\overline{\pi }}_{2}\ge {\underline{\pi }}_{1}$ henceforth.

Suppose ${\hat{\sigma }}\in$ $\Gamma ^{\prime }=\left\{ \sigma \in \Gamma \ \left| \ 0<\sigma _{1}<\sigma _{2}<1\right. \right\}$ (i.e., ${\hat{\sigma }}$ is strictly mixed and responsive). By inspection of Fig. 1 it is evident that for ${\hat{\sigma }}$ to be an equilibrium, we must have $\overline{\pi }_{2}=\pi ^{*}\left( {\hat{\sigma }}\right)$ or $\underline{\pi }_{1}=\pi ^{*}\left( {\hat{\sigma }}\right)$ or both. If both conditions hold, then ${\overline{\pi }}_{2}={\underline{\pi }}_{1}$. Let us temporarily assume that ${\overline{\pi }}_{2}>\underline{\pi }_{1}$. The case ${\overline{\pi }}_{2}={\underline{\pi }}_{1}$ will be considered later.

Fig. 8

Strategy $\sigma$ satisfies (I$^{\prime }$) iff it lies in the interior of the red triangle at the intersection of the green and blue curves

If $\pi ^{*}\left( {\hat{\sigma }}\right) =\underline{\pi }_{1}<{\overline{\pi }}_{2}$, then ${\hat{\sigma }}$ will be an equilibrium iff ${\hat{\sigma }} _{2}=\sigma ^{*}\left( {\hat{\sigma }}\right)$. See Fig. 7(I). Similarly, if $\pi ^{*}\left( {\hat{\sigma }}\right) ={\overline{\pi }}_{2}> {\underline{\pi }}_{1}$, then ${\hat{\sigma }}$ will be an equilibrium iff ${\hat{\sigma }}_{1}={\hat{\sigma }}^{*}\left( \sigma \right)$. See Fig. 7(II). If $\pi ^{*}\left( {\hat{\sigma }}\right) =x$, then the condition ${\hat{\sigma }}_{t}=\sigma ^{*}\left( {\hat{\sigma }}\right)$ may be expressed, using (8), as follows:

$$\begin{aligned} \left[ \left( 1-r\right) {\hat{\sigma }}_{1}+r{\hat{\sigma }}_{2}\right] ^{N}{\hat{\sigma }}_{t}=x. \end{aligned}$$

Thus, when ${\overline{\pi }}_{2}>{\underline{\pi }}_{1}$, there exists a strictly mixed responsive equilibrium iff at least one of the following two conditions is met:

(I$^{\prime }$):

There exists $\sigma \in \Omega \left( \underline{\pi }_{1}\right) \cap \Gamma ^{\prime }$ satisfying

$$\begin{aligned} \left[ \left( 1-r\right) \sigma _{1}+r\sigma _{2}\right] ^{N}\sigma _{2}= {\underline{\pi }}_{1}. \end{aligned}$$

(CI)

(II$^{\prime }$):

There exists $\sigma \in \Omega \left( \overline{\pi }_{2}\right) \cap \Gamma ^{\prime }$ satisfying

$$\begin{aligned} \left[ \left( 1-r\right) \sigma _{1}+r\sigma _{2}\right] ^{N}\sigma _{1}= {\overline{\pi }}_{2}. \end{aligned}$$

(CII)

Consider condition (I$^{\prime }$). Note first, using Fig. 6, that $\Omega \left( {\underline{\pi }}_{1}\right) \cap \Gamma ^{\prime }\ne \emptyset$ iff

$$\begin{aligned} \frac{1}{2+c}<{\underline{\pi }}_{1}<\frac{r^{N}}{r^{N}+\left( 1-r\right) ^{N}\left( 1+c\right) }. \end{aligned}$$

(17)

Now consider Fig. 8. The set $\Gamma ^{\prime }$ is the interior of the red triangle. The line corresponding to $\Omega \left( {\underline{\pi }}_{1}\right)$ is indicated, under the assumption that ${\underline{\pi }}_{1}$ satisfies (17). The other locus in Fig. 8 describes the set of solutions to (CI); it is easily verified that the implicit function defined by (CI) is convex with a strictly negative slope, and that

$$\begin{aligned} \sigma _{1}=\sigma _{2}={\underline{\pi }}_{1}^{\frac{1}{N+1}} \end{aligned}$$

lies on the graph of this function. Condition (I$^{\prime }$) requires that these two loci intersect in $\Gamma ^{\prime }$. Such an intersection will obviously be unique if it exists. An intersection in $\Gamma ^{\prime }$ will obtain iff the $\Omega \left( {\underline{\pi }}_{1}\right)$ locus hits the top of the red triangle strictly to the right of the locus of solutions to (CI), as illustrated in Fig. 8.

Recall from Fig. 6 that the locus $\Omega \left( {\underline{\pi }} _{1}\right)$ hits the top of the red triangle at $\lambda \left( {\underline{\pi }}_{1}\right) ^{-1}$. Given $x\in \left( 0,1\right)$, let $h_{I}\left( x\right)$ denote the value of $\sigma _{1}$ that solves

$$\begin{aligned} \left[ \left( 1-r\right) \sigma _{1}+r\right] ^{N}=x \end{aligned}$$

(18)

(noting that this solution might be negative).³¹ Hence, condition (I$^{\prime }$) is satisfied iff (17) and

$$\begin{aligned} h_{I}\left( {\underline{\pi }}_{1}\right) <\frac{1}{\lambda \left( {\underline{\pi }}_{1}\right) }. \end{aligned}$$

(19)

Using the fact that $h_{I}\left( x\right)$ is strictly increasing, we may re-write condition (19) as

$$\begin{aligned}&{\underline{\pi }}_{1}<h_{I}^{-1}\left( \lambda \left( \underline{\pi } _{1}\right) ^{-1}\right) =\left[ \left( 1-r\right) \lambda \left( {\underline{\pi }}_{1}\right) ^{-1}+r\right] ^{N} \\&\quad \Leftrightarrow \ \ \ {\underline{\pi }}_{1}^{1/N}<\left( 1-r\right) \lambda \left( {\underline{\pi }}_{1}\right) ^{-1}+r. \end{aligned}$$

Using the definition of $\lambda$ from Lemma 6.1, this may be expressed:

$$\begin{aligned} {\underline{\pi }}_{1}^{1/N}<\frac{2r-1}{r-\left( 1-r\right) \ell \left( {\underline{\pi }}_{1}\right) ^{1/N}} \end{aligned}$$

$$\begin{aligned} \Leftrightarrow \ \ \ r{\underline{\pi }}_{1}^{1/N}-\left( 1-r\right) \left( \frac{1-{\underline{\pi }}_{1}}{1+c}\right) ^{1/N}<2r-1, \end{aligned}$$

(20)

where the final equivalence uses the definition of $\ell$ from Lemma 6.1. It is obvious that the left-hand side of (20) is strictly increasing in ${\underline{\pi }}_{1}$ so there is some $\beta _{1}\left( N\right)$ such that (20) is equivalent to ${\underline{\pi }}_{1}<\beta _{1}\left( N\right)$, where the notation emphasises the dependence of this upper bound on N. Letting

$$\begin{aligned} \alpha _{1}\left( N\right) =\min \left\{ \beta _{1}\left( N\right) ,\ \frac{ r^{N}}{r^{N}+\left( 1-r\right) ^{N}\left( 1+c\right) }\right\} , \end{aligned}$$

we have therefore shown that (I$^{\prime }$) holds iff (I). (We will verify the stated properties of $\alpha _{1}$ shortly.)

Next, consider condition (II$^{\prime }$). We have $\Omega \left( {\overline{\pi }}_{2}\right) \cap \Gamma ^{\prime }\ne \emptyset$ iff

$$\begin{aligned} \frac{1}{2+c}<{\overline{\pi }}_{2}<\frac{r^{N}}{r^{N}+\left( 1-r\right) ^{N}\left( 1+c\right) }. \end{aligned}$$

(21)

This time $\sigma \in \Omega \left( {\overline{\pi }}_{2}\right) \cap \Gamma ^{\prime }$ is an equilibrium if it sits at the intersection of $\Omega \left( {\overline{\pi }}_{2}\right)$ and the locus defined by (CII). Given $x\in \left( 0,1\right)$, let $h_{II}\left( x\right)$ denote the value of $\sigma _{1}$ that solves

$$\begin{aligned} \left[ \left( 1-r\right) \sigma _{1}+r\right] ^{N}\sigma _{1}=x. \end{aligned}$$

(22)

It is obvious that this solution exists and is unique. Moreover, $h_{II}\left( x\right) \in \left( 0,1\right)$ for any $x\in \left( 0,1\right)$. Reasoning as for condition (I$^{\prime }$), mutatis mutandis, we deduce that: condition (II$^{\prime }$) is satisfied iff (21) and

$$\begin{aligned} h_{II}\left( {\overline{\pi }}_{2}\right) <\frac{1}{\lambda \left( {\overline{\pi }}_{2}\right) }. \end{aligned}$$

(23)

Using the definition of $\lambda$ from Lemma 6.1, this may be expressed:

$$\begin{aligned} h_{II}\left( {\overline{\pi }}_{2}\right) <\frac{r\ell \left( {\overline{\pi }} _{2}\right) ^{\frac{1}{N}}-\left( 1-r\right) }{r-\left( 1-r\right) \ell \left( {\overline{\pi }}_{2}\right) ^{\frac{1}{N}}}. \end{aligned}$$

It is easy to show that the denominator and numerator on the right-hand side of this expression are both positive.³² Straightforward algebra, therefore, gives

$$\begin{aligned} \left[ \frac{rh_{II}\left( {\overline{\pi }}_{2}\right) +\left( 1-r\right) }{ \left( 1-r\right) h_{II}\left( \overline{\pi }_{2}\right) +r}\right] ^{N}<\ell \left( {\overline{\pi }}_{2}\right) . \end{aligned}$$

Using the definitions of $h_{II}\left( {\overline{\pi }}_{2}\right)$ and $\ell \left( {\overline{\pi }}_{2}\right)$, respectively, this is equivalent to

$$\begin{aligned} \frac{\left[ rh_{II}\left( {\overline{\pi }}_{2}\right) +\left( 1-r\right) \right] ^{N}}{{\overline{\pi }}_{2}/h_{II}\left( {\overline{\pi }}_{2}\right) } <\ell \left( {\overline{\pi }}_{2}\right) \end{aligned}$$

$$\begin{aligned} \Leftrightarrow \ \ \ \left[ rh_{II}\left( \overline{\pi }_{2}\right) +\left( 1-r\right) \right] ^{N}h_{II}\left( {\overline{\pi }}_{2}\right) < \frac{1-{\overline{\pi }}_{2}}{1+c}. \end{aligned}$$

(24)

We have, therefore, shown that (23) is equivalent to (24). Since $h_{II}\left( x\right)$ is strictly increasing, it is easy to see that there is some $\beta _{2}\left( N\right)$ such that (24) is equivalent to $\overline{\pi }_{2}<\beta _{2}\left( N\right)$. Letting

$$\begin{aligned} \alpha _{2}\left( N\right) =\min \left\{ \beta _{2}\left( N\right) ,\ \frac{ r^{N}}{r^{N}+\left( 1-r\right) ^{N}\left( 1+c\right) }\right\} , \end{aligned}$$

we deduce that (II$^{\prime }$) holds iff (II).

Let us now verify the stated properties of the functions $\alpha _{1}$ and $\alpha _{2}$. Comparing (22) and (18), it is obvious that $h_{I}\left( x\right) <h_{II}\left( x\right)$ for all $x\in \left( 0,1\right)$. Thus, if ${\overline{\pi }}_{2}=z$ satisfies (23) then ${\underline{\pi }}_{1}=z$ satisfies (19). It follows that $\beta _{1}\left( N\right) \ge \beta _{2}\left( N\right)$. To see why $\beta _{2}\left( N\right) >\left( 2+c\right) ^{-1}$, set ${\overline{\pi }} _{2}=\left( 2+c\right) ^{-1}$ in (24) to obtain:

$$\begin{aligned} \left[ rh_{II}\left( \frac{1}{2+c}\right) +\left( 1-r\right) \right] ^{N}h_{II}\left( \frac{1}{2+c}\right) <\frac{1}{2+c}. \end{aligned}$$

This inequality must hold since, by the definition of $h_{II}$, we have

$$\begin{aligned}&\left[ \left( 1-r\right) h_{II}\left( \frac{1}{2+c}\right) +r\right] ^{N}h_{II}\left( \frac{1}{2+c}\right) =\frac{1}{2+c} \\&\quad \Rightarrow \ \ \ \left[ rh_{II}\left( \frac{1}{2+c}\right) +\left( 1-r\right) \right] ^{N}h_{II}\left( \frac{1}{2+c}\right) <\frac{1}{2+c}, \end{aligned}$$

where we have used $r>\frac{1}{2}$ and $h_{II}\left( \left( 2+c\right) ^{-1}\right) \in \left( 0,1\right)$. Hence, $\beta _{2}\left( N\right) >\left( 2+c\right) ^{-1}$ and we have therefore established that $\alpha _{1}\left( N\right) \ge \alpha _{2}\left( N\right) >\left( 2+c\right) ^{-1}$ for all N.

Fig. 9

The case $\overline{\pi }_{2}=\underline{\pi }_{1}$

Finally, let us return to the scenario in which $\overline{\pi }_{2}= {\underline{\pi }}_{1}$. Then $\sigma$ is a strictly mixed responsive equilibrium iff

$$\begin{aligned} \sigma \in \Omega \left( {\underline{\pi }}_{1}\right) \cap \Gamma ^{\prime } \end{aligned}$$

and $\sigma _{1}\le \sigma ^{*}\left( \sigma \right) \le \sigma _{2}$. See Fig. 9a. Note that

$$\begin{aligned} \sigma _{1}\le \sigma ^{*}\left( \sigma \right) \le \sigma _{2}\ \ \ \Leftrightarrow \ \ \ \left[ \left( 1-r\right) \sigma _{1}+r\sigma _{2} \right] ^{N}\sigma _{1}\le \underline{\pi }_{1}\le \left[ \left( 1-r\right) \sigma _{1}+r\sigma _{2}\right] ^{N}\sigma _{2}. \end{aligned}$$

In other words, $\sigma$ must lie in the region whose western boundary is the locus determined by (CI) and whose eastern boundary is the locus described by the equation

$$\begin{aligned} \left[ \left( 1-r\right) \sigma _{1}+r\sigma _{2}\right] ^{N}\sigma _{1}= {\underline{\pi }}_{1}. \qquad \qquad \qquad \qquad (*)\end{aligned}$$

See Fig. 9b. This region has a non-empty intersection with $\Omega \left( {\underline{\pi }}_{1}\right) \cap \Gamma ^{\prime }$ iff (19) holds, which is equivalent to (I).

Appendix C

In this Appendix, we prove Lemma 4.5.

First, let $\varepsilon >0$ be small enough to satisfy³³

$$\begin{aligned} r\ln \left( \frac{1}{2+c}+\varepsilon \right) <\left( 1-r\right) \ln \left( \frac{1}{2+c}-\frac{\varepsilon }{1+c}\right) . \end{aligned}$$

(25)

We claim that

$$\begin{aligned} {\underline{\pi }}_{1}=\frac{1}{2+c}+\varepsilon \end{aligned}$$

(26)

(and therefore any smaller value of ${\underline{\pi }}_{1}$) satisfies (20) when N is sufficiently large, and hence that

$$\begin{aligned} \beta _{1}\left( N\right) \ge \frac{1}{2+c}+\varepsilon \end{aligned}$$

for sufficiently large N. To verify this claim, substitute (26) into (20) to get

$$\begin{aligned} r\left( \frac{1}{2+c}+\varepsilon \right) ^{1/N}-\left( 1-r\right) \left( \frac{1}{2+c}-\frac{\varepsilon }{1+c}\right) ^{1/N}\ <\ 2r-1. \end{aligned}$$

(27)

Let $g\left( N\right)$ denote the left-hand side of (27). It is obvious that

$$\begin{aligned} \lim _{N\rightarrow \infty }\ g\left( N\right) \ =\ 2r-1. \end{aligned}$$

(28)

To prove our claim, it suffices to show that g is strictly increasing when N is sufficiently large, since g then approaches this limit from below and it follows that (27) must hold for large N.

To see that g is strictly increasing for large N, let

$$\begin{aligned} {\overline{g}}\left( x\right) =r\left( \frac{1}{2+c}+\varepsilon \right) ^{x}-\left( 1-r\right) \left( \frac{1}{2+c}-\frac{\varepsilon }{1+c}\right) ^{x}. \end{aligned}$$

Then

$$\begin{aligned} {\overline{g}}^{\prime }\left( x\right) =r\left( \frac{1}{2+c}+\varepsilon \right) ^{x}\ln \left( \frac{1}{2+c}+\varepsilon \right) -\left( 1-r\right) \left( \frac{1}{2+c}-\frac{\varepsilon }{1+c}\right) ^{x}\ln \left( \frac{1}{ 2+c}-\frac{\varepsilon }{1+c}\right) . \end{aligned}$$

Hence,

$$\begin{aligned} \lim _{x\rightarrow 0}{\overline{g}}^{\prime }\left( x\right) \ =\ r\ln \left( \frac{1}{2+c}+\varepsilon \right) -\left( 1-r\right) \ln \left( \frac{1}{2+c} -\frac{\varepsilon }{1+c}\right) \end{aligned}$$

which is strictly less than zero by (25). It follows that $g\left( N\right)$ is strictly increasing for large N.

Using the fact that $r\in \left( \frac{1}{2},1\right)$, we have

$$\begin{aligned} \lim _{N\rightarrow \infty }\frac{r^{N}}{r^{N}+\left( 1-r\right) ^{N}\left( 1+c\right) }\ =\ \lim _{N\rightarrow \infty }\frac{1}{1+\left( \frac{1-r}{r} \right) ^{N}\left( 1+c\right) }=1 \end{aligned}$$

(29)

and hence

$$\begin{aligned} \alpha _{1}\left( N\right) \ge \frac{1}{2+c}+\varepsilon \end{aligned}$$

for all sufficiently large N. Since $\alpha _{1}\left( N\right) >\left( 2+c\right) ^{-1}$ for all N, there must exist some $\eta >0$ such that

$$\begin{aligned} \alpha _{1}\left( N\right) \ge \frac{1}{2+c}+\eta \end{aligned}$$

for all N.

Appendix D

In this Appendix, we prove Proposition 5.2.

By Lemmas 4.1 and 4.4, if N is sufficiently large, any responsive equilibrium with $\sigma _{2}=1$ must have $\sigma _{1}\in \left( 0,1\right)$. Hence, if $\left( \sigma _{1},1\right)$ is a responsive equilibrium and N is sufficiently large, then

$$\begin{aligned} {\underline{\pi }}_{1}\le \pi ^{*}\left( \left( \sigma _{1},1\right) \right) \le {\overline{\pi }}_{1} \end{aligned}$$

(30)

(see Fig. 1). Recalling (29), we have

$$\begin{aligned} {\overline{\pi }}_{1}<\frac{r^{N}}{r^{N}+\left( 1-r\right) ^{N}\left( 1+c\right) } \end{aligned}$$

for sufficiently large N. Combining this fact with (30) and Fig. 6 we see that

$$\begin{aligned} 0<\lambda ^{-1}\left( {\overline{\pi }}_{1}\right) \le \sigma _{1} \end{aligned}$$

in any responsive equilibrium with $\sigma _{2}=1$, provided N is sufficiently large. In such an equilibrium:³⁴

$$\begin{aligned} \Pr \left( {\mathcal {C}}|a\right) \ \ge \ \left[ r\lambda \left( {\overline{\pi }}_{1}\right) ^{-1}+\left( 1-r\right) \right] ^{N+1}\ =\ \left[ \frac{ \left( 2r-1\right) \ell \left( \overline{\pi }_{1}\right) ^{\frac{1}{N}}}{ r-\left( 1-r\right) \ell \left( {\overline{\pi }}_{1}\right) ^{\frac{1}{N}}} \right] ^{N+1}>0, \end{aligned}$$

where the final inequality uses $\lambda ^{-1}\left( {\overline{\pi }} _{1}\right) >0$. It follows that

$$\begin{aligned} \ell \left( {\overline{\pi }}_{1}\right) <1\ \ \ \Leftrightarrow \ \ \ {\overline{\pi }}_{1}>\frac{1}{2+c} \end{aligned}$$

(31)

in any responsive equilibrium with $\sigma _{2}=1$ and large N:³⁵ if $\ell \left( \overline{\pi }_{1}\right) \ge 1$, then $2r-1>0$ and

$$\begin{aligned} \left[ \frac{\left( 2r-1\right) \ell \left( \overline{\pi }_{1}\right) ^{ \frac{1}{N}}}{r-\left( 1-r\right) \ell \left( {\overline{\pi }}_{1}\right) ^{ \frac{1}{N}}}\right] ^{N+1}>0 \end{aligned}$$

would imply $r-\left( 1-r\right) \ell \left( \overline{\pi }_{1}\right) ^{ \frac{1}{N}}$ and hence

$$\begin{aligned} \Pr \left( {\mathcal {C}}|a\right) \ \ge \ \left[ \frac{\left( 2r-1\right) \ell \left( {\overline{\pi }}_{1}\right) ^{\frac{1}{N}}}{r-\left( 1-r\right) \ell \left( \overline{\pi }_{1}\right) ^{\frac{1}{N}}}\right] ^{N+1}\ge \left[ \frac{\left( 2r-1\right) \ell \left( {\overline{\pi }}_{1}\right) ^{ \frac{1}{N}}}{r\ell \left( {\overline{\pi }}_{1}\right) ^{\frac{1}{N}}-\left( 1-r\right) \ell \left( \overline{\pi }_{1}\right) ^{\frac{1}{N}}}\right] ^{N+1}=1 \end{aligned}$$

which is impossible since conviction cannot be certain, conditional on $s=a$, in an equilibrium with $\sigma _{1}<1$.

To complete the proof we will show that

$$\begin{aligned} \left[ \frac{\left( 2r-1\right) \ell \left( \overline{\pi }_{1}\right) ^{ \frac{1}{N}}}{r-\left( 1-r\right) \ell \left( {\overline{\pi }}_{1}\right) ^{ \frac{1}{N}}}\right] ^{N+1}=\left[ \frac{\left( 2r-1\right) }{r-\left( 1-r\right) \ell \left( {\overline{\pi }}_{1}\right) ^{\frac{1}{N}}}\right] ^{N+1}\ell \left( {\overline{\pi }}_{1}\right) ^{\frac{N+1}{N}} \end{aligned}$$

(32)

has a strictly positive limit as $N\rightarrow \infty$ when $0<\ell \left( {\overline{\pi }}_{1}\right) <1$. Since

$$\begin{aligned} \lim _{N\rightarrow \infty }\ell \left( {\overline{\pi }}_{1}\right) ^{\frac{N+1 }{N}}\ =\ \ell \left( {\overline{\pi }}_{1}\right) >0, \end{aligned}$$

the limit of (32) is strictly positive iff

$$\begin{aligned} \lim _{N\rightarrow \infty }\left[ \frac{\left( 2r-1\right) }{r-\left( 1-r\right) \ell \left( {\overline{\pi }}_{1}\right) ^{\frac{1}{N}}}\right] ^{N+1}\ >\ 0. \end{aligned}$$

(33)

We follow the logic on p. 32 of Feddersen and Pesendorfer (1998) to establish (33) as follows:

$$\begin{aligned}&\quad \lim _{N\rightarrow \infty }\left[ \frac{\left( 2r-1\right) }{r-\left( 1-r\right) \ell \left( {\overline{\pi }}_{1}\right) ^{\frac{1}{N}}}\right] ^{N+1} \ge \lim _{N\rightarrow \infty }\left[ \frac{\left( 2r-1\right) }{r-\left( 1-r\right) \left[ 1-\left( N+1\right) ^{-1}\ln \ell \left( {\overline{\pi }}_{1}\right) \right] }\right] ^{N+1} \\&\quad =\ \lim _{N\rightarrow \infty }\left[ 1+\left( 1-r\right) \left( N+1\right) ^{-1}\right] ^{-\left( N+1\right) } \\&\quad =\ \exp \left[ -\left( \frac{1-r}{2r-1}\right) \ln \ell \left( \overline{ \pi }_{1}\right) \right] \\&\quad =\ \ell \left( {\overline{\pi }}_{1}\right) ^{-\left( 1-r\right) /\left( 2r-1\right) }, \end{aligned}$$

where the inequality and second equality use Feddersen and Pesendorfer’s (1998) expressions (6) and (5), respectively.

Appendix E

In this Appendix, we prove Proposition 5.3.

Suppose $\sigma ^{\left( k\right) }$ is a strictly mixed responsive equilibrium for a jury of size $N_{k}\in \left\{ 1,2,...\right\}$, with $N_{k}\rightarrow \infty$ as $k\rightarrow \infty$. It follows that ${\underline{\pi }}_{1}\le {\overline{\pi }}_{2}$. From the proof of Proposition 4.2, each $\sigma ^{\left( k\right) }$ satisfies

$$\begin{aligned} \left[ \left( 1-r\right) \sigma _{1}^{\left( k\right) }+r\sigma _{2}^{\left( k\right) }\right] ^{N_{k}}\sigma _{2}^{\left( k\right) }={\underline{\pi }}_{1} \end{aligned}$$

or

$$\begin{aligned} \left[ \left( 1-r\right) \sigma _{1}^{\left( k\right) }+r\sigma _{2}^{\left( k\right) }\right] ^{N_{k}}\sigma _{1}^{\left( k\right) }={\overline{\pi }}_{2} . \end{aligned}$$

Therefore, $\sigma _{t}^{\left( k\right) }\rightarrow 1$ as $k\rightarrow \infty$ for each $t\in T$, and hence

$$\begin{aligned} \underset{k\rightarrow \infty }{\lim \inf \ }\left[ \left( 1-r\right) \sigma _{1}^{\left( k\right) }+r\sigma _{2}^{\left( k\right) }\right] ^{N_{k}}\ \ge \ {\underline{\pi }}_{1}. \end{aligned}$$

Since

$$\begin{aligned} \lim _{k\rightarrow \infty }\ \left[ \left( 1-r\right) \sigma _{1}^{\left( k\right) }+r\sigma _{2}^{\left( k\right) }\right] \ =\ 1, \end{aligned}$$

we have

$$\begin{aligned} \underset{k\rightarrow \infty }{\lim \inf \ }\left[ \left( 1-r\right) \sigma _{1}^{\left( k\right) }+r\sigma _{2}^{\left( k\right) }\right] ^{N_{k}+1}\ \ge \ {\underline{\pi }}_{1}. \end{aligned}$$

Note that, for equilibrium $\sigma ^{\left( k\right) }$,

$$\begin{aligned} \Pr \left( {\mathcal {C}}|b\right) \ =\ \left[ \left( 1-r\right) \sigma _{1}^{\left( k\right) }+r\sigma _{2}^{\left( k\right) }\right] ^{N_{k}+1} \end{aligned}$$

so the probability of conviction in $\sigma ^{\left( k\right) }$ is at least

$$\begin{aligned} \left( 1-{\overline{p}}\right) \left[ \left( 1-r\right) \sigma _{1}^{\left( k\right) }+r\sigma _{2}^{\left( k\right) }\right] ^{N_{k}+1} \end{aligned}$$

for any $p\in \left[ {\underline{p}},{\overline{p}}\right]$. Thus,

$$\begin{aligned} \min _{p\in \left[ {\underline{p}},{\overline{p}}\right] }\left[ \underset{ k\rightarrow \infty }{\lim \inf \ }\gamma _{k}\left( p\right) \right] \ \ge \ \left( 1-{\overline{p}}\right) {\underline{\pi }}_{1}>0. \end{aligned}$$