Evidence and fully revealing deliberation with non-consequentialist jurors

Abstract

We analyze a model of binary choice by a committee, when information is hard and pre-voting deliberation is allowed. Each member has, independently of the others, a positive probability of getting a private signal about the true state; with the remaining probability the member is uninformed. Hard information means that lying is disallowed during deliberation—informed members can reveal publicly or hide their signals, while uninformed voters have to disclose their ignorance. We allow non-consequentialist members whose thresholds for switching to the non-status-quo action vary with the number of informative signals. We show that in general, committee members will never reveal information fully during deliberation, even when we rule out partisan types who want the same action in all states. In particular, unanimity rule performs no worse than other rules.

Appendix

Proof of Theorem 1

We prove the result by contradiction. Assume that a fully revealing equilibrium $({\varvec{\mu }}, {\varvec{\nu }})$ exists. Notice that in any fully revealing equilibrium, sincere voting strategies imply that given ${\mathbf {m}}={\mathbf {s}}$, it must be that $v(b, s, {\mathbf {m}})=C$ if and only if $({\mathbf {s}}_{-i},s_{i})\in {\mathbf {S}}_{b_{i}}(C)$ for every $i\in N$ and $b_{i}\in B$. In equilibrium, for every $i\in N$, every possible $(b_{i},s_{i})\in B\times S$, i’s incentive compatibility condition must hold at the communication stage:

$$\begin{aligned} {\mathrm{EU}}(m_{i}=s_{i},b_{i},s_{i})-{\mathrm{EU}}(m_{i}=0,b_{i},s_{i})\ge 0, \end{aligned}$$

where

$$\begin{aligned}&{\mathrm{EU}}(m_{i},b_{i},s_{i})\\&\quad =\sum _{{\mathbf {b}}_{-i}\in B^{2n}}\sum _{{\mathbf {s}}_{-i}\in S^{2n}}P({\mathbf {b}}_{-i}, {\mathbf {s}}_{-i}|b_{i}, s_{i})[Pr(A|{\mathbf {b}}, {\mathbf {s}},m_{i})u(A, b_{i},{\mathbf {s}})+Pr(C|{\mathbf {b}}, {\mathbf {s}},m_{i})u(C, b_{i},{\mathbf {s}})]. \end{aligned}$$

Here, $P({\mathbf {b}}_{-i}, {\mathbf {s}}_{-i}|b_{i}, s_{i})$ is the probability of situation $({\mathbf {b}},{\mathbf {s}})=(({\mathbf {b}}_{-i},b_{i}),({\mathbf {s}}_{-i},s_{i}))$ being obtained conditional on the pair of i’s bias and signal being $(b_{i}, s_{i})$, and $Pr(z|{\mathbf {b}},{\mathbf {s}},m_{i})$ is the probability of $z\in \{A,C\}$ being the committee’s final outcome given the bias profile ${\mathbf {b}}$, the state profile ${\mathbf {s}}$, and the message profile $(m_{i},{\mathbf {m}}_{-{\mathbf {i}}})=(m_{i},{\mathbf {s}}_{-{\mathbf {i}}})$.

Hence, i’s incentive compatibility condition holds when:

$$\begin{aligned}&{\mathrm{EU}}(m_{i}=s_{i},b_{i},s_{i})-{\mathrm{EU}}(m_{i}=0,b_{i},s_{i})\\&\quad =\sum _{{\mathbf {b}}_{-i}\in B^{2n}}\sum _{{\mathbf {s}}_{-i}\in S^{2n}}P({\mathbf {b}}_{-i}, {\mathbf {s}}_{-i}|b_{i}, s_{i})[Pr(A|{\mathbf {b}}, {\mathbf {s}},s_{i})u(A, b_{i},{\mathbf {s}})+Pr(C|{\mathbf {b}}, {\mathbf {s}},s_{i})u(C, b_{i},{\mathbf {s}})\\&\qquad -Pr(A|{\mathbf {b}}, {\mathbf {s}},0)u(A, b_{i},{\mathbf {s}})-Pr(C|{\mathbf {b}}, {\mathbf {s}},0)u(C, b_{i},{\mathbf {s}})]\ge 0. \end{aligned}$$

Since $Pr(C|{\mathbf {b}}, {\mathbf {s}},m_{i})=1-Pr(A|{\mathbf {b}}, {\mathbf {s}},m_{i})$, we simplify as follows:

$$\begin{aligned}&{\mathrm{EU}}(m_{i}=s_{i},b_{i},s_{i})-{\mathrm{EU}}(m_{i}=0,b_{i},s_{i})\\&\quad =\sum _{{\mathbf {b}}_{-i}\in B^{2n}}\sum _{{\mathbf {s}}_{-i}\in S^{2n}}P({\mathbf {b}}_{-i}, {\mathbf {s}}_{-i}|b_{i}, s_{i})[Pr(A|{\mathbf {b}}, {\mathbf {s}},s_{i})u(A, b_{i},{\mathbf {s}})+(1-Pr(A|{\mathbf {b}}, {\mathbf {s}},s_{i}))u(C, b_{i},{\mathbf {s}})\\&\qquad -Pr(A|{\mathbf {b}}, {\mathbf {s}},0)u(A, b_{i},{\mathbf {s}})-(1-Pr(A|{\mathbf {b}}, {\mathbf {s}},0))u(C, b_{i},{\mathbf {s}})]\\&\quad =\sum _{{\mathbf {b}}_{-i}\in B^{2n}}\sum _{{\mathbf {s}}_{-i}\in S^{2n}}P({\mathbf {b}}_{-i}, {\mathbf {s}}_{-i}|b_{i}, s_{i})\{[Pr(A|{\mathbf {b}},{\mathbf {s}},s_{i})-Pr(A|{\mathbf {b}},{\mathbf {s}},0)][u(A,b_{i},{\mathbf {s}})-u(C,b_{i}, {\mathbf {s}})]\}\\&\quad \ge 0. \end{aligned}$$

For any voter $i\in N$ who has observed a signal $s_{i}$, we define the following function for simplicity:

$$\begin{aligned} \Phi _i (s_{i},0;{\mathbf {b}}, {\mathbf {s}}) \equiv [Pr(A|{\mathbf {b}},{\mathbf {s}},s_{i})-Pr(A|{\mathbf {b}},{\mathbf {s}},0)]\times [u(A,b_{i},{\mathbf {s}})-u(C,b_{i}, {\mathbf {s}})]. \end{aligned}$$

(1)

Therefore, rewriting i’s incentive compatible condition, a fully revealing equilibrium exists if and only if for every i and every $(b_{i},s_{i})\in B\times S$, the following condition holds:

$$\begin{aligned} \sum _{{\mathbf {b}}_{-i}\in B^{2n}}\sum _{{\mathbf {s}}_{-i}\in S^{2n}} P({\mathbf {b}}_{-i}, {\mathbf {s}}_{-i}|b_{i},s_{i}) \Phi _{i} (s_{i},0;{\mathbf {b}}, {\mathbf {s}})\ge 0. \end{aligned}$$

For any ${\mathbf {b}}\in {\mathbf {B}}$ and any $k=0,1,2,\ldots ,r$, define the set ${\mathbf {S}}^{r,k}$ as the set of states where there are k innocent signals among r informative ones, that is:

$$\begin{aligned} {\mathbf {S}}^{r,k}:=\{{\mathbf {s}}\in {\mathbf {S}}^{r} \mid \# \{i| s_i=I\}=k\}. \end{aligned}$$

We prove the result by dividing all voting rules into two cases.

Case 1 $2\le K\le 2n+1$.

Fix majority rule for convenience, i.e., $K=n+1$, although the analysis is the same for any K-rule when $2\le K\le 2n+1$.

Consider an extremely harsh type juror i with bias $\underline{b}\in B$ who prefers C as long as there is at least one guilty signal G, i.e., ${\mathbf {S}}_{\underline{b}}^{r}(A)={\mathbf {S}}^{r, r}$ for any r. Consider the case where i has observed an innocent signal. By Full Support, such a situation occurs with positive probability. And we prove the result by showing that for such a voter i with $(b_{i},s_{i})=(\underline{b},I)$, she strictly prefers concealing her information by pretending to be an uninformed voter in the communication stage.

At any bias profile ${\varvec{b}}$, for any signal $s_i$ observed by voter i, we define

$$\begin{aligned} {\mathbf {Z}}_{i}({\mathbf {b}}, s_i) \equiv \{{\mathbf {s}}_{-{\mathbf {i}}}\in S^{2n}|Pr(A|{\mathbf {b}},{\mathbf {s}},I)\ne Pr(A|{\mathbf {b}},{\mathbf {s}},0), P({\mathbf {b}}_{-i}, {\mathbf {s}}_{-i}|b_{i}, s_{i})>0\} \end{aligned}$$

as the set of signals of other voters that have positive probability and make player i pivotal.

Using ${\mathbf {Z}}_{i}({\mathbf {b}}, s_i)$, the harsh type’s IC condition reduces to:

$$\begin{aligned} {\mathrm{EU}}(I,\underline{b},I)-{\mathrm{EU}}(0,\underline{b},I)= \sum _{{\mathbf {b}}_{-i}, {\mathbf {s}}_{-i}} P({\mathbf {b}}_{-i},{\mathbf {s}}_{-i}|\underline{b},I)\Phi _{i}(I,0;{\mathbf {b}}_{-i}, {\mathbf {s}}_{-i}) \ge 0, \end{aligned}$$

(2)

where the index ${\varvec{s}}_{-i}$ is summed over ${\mathbf {Z}}_{i}(({\mathbf {b}}_{-i},\underline{b}), I)$.

If i prefers A, suppose it is at state ${\mathbf {s}}$; by construction of $\underline{b}$ there exists no guilty signal G, i.e., ${\mathbf {s}}\in {\mathbf {S}}^{r, r}$. By Presumption of Innocence, for any b, $(0,0,\dots ,0)\in {\mathbf {S}}_{b}(A)$, then by Signal Monotonicity, for any b and any state ${\mathbf {s}}'\in \mathbf {S}^{1,1}$, we have ${\mathbf {s}}'\subset {\mathbf {S}}_{b}(A)$. Recursively, for any b and any state ${\mathbf {s}}\in {\mathbf {S}}^{r, r}$, we have ${\mathbf {s}}\subset {\mathbf {S}}_{b}(A)$. Hence, ${\mathbf {s}}$ is a state where all individuals prefer A, hence i cannot be pivotal at ${\mathbf {s}}$. Therefore, the harsh type can be pivotal only in a situation $({\mathbf {b}},{\mathbf {s}})$ where i prefers C; so there exists at least one guilty signal G and $u(A,b_{i},{\mathbf {s}})-u(C,b_{i},{\mathbf {s}})<0$. If i is pivotal at the deliberation stage, then by Signal Monotonicity, it must be that sending a message 0 instead of I alters the majority vote and decision from A to C. That is, $Pr(A|{\mathbf {b}},{\mathbf {s}},I)=1$ and $Pr(A|{\mathbf {b}},{\mathbf {s}},0)=0$. Then (1) implies that $\Phi _{i}(I,0;{\mathbf {b}}_{-i}, {\mathbf {s}}_{-i})<0$. And by Full Support, for any ${\mathbf {b}}_{-i}\in B^{2n}$ and ${\mathbf {s}}_{-i}\in {\mathbf {Z}}_{i}(({\mathbf {b}}_{-i},\underline{b}), I)$, we have $P({\mathbf {b}}_{-i},{\mathbf {s}}_{-i}|\underline{b},I)>0$. Hence if the set ${\mathbf {Z}}_{i}(({\mathbf {b}}_{-i},\underline{b}), I)$ is non-empty, it follows that (2) is violated.

So all that remains is to show this non-emptiness when $K=n+1$. Consider a type of juror who has bias $b'$ such that

$$\begin{aligned} {\mathbf {S}}_{b'}^{r}(A) ={\left\{ \begin{array}{ll} {\mathbf {S}}^{r,r}\bigcup {\mathbf {S}}^{r,r-1} &{} \text { if } 2\le r\le 2n+1 \\ {\mathbf {S}}^{1,1} &{} \text { if } r=1,\\ {\mathbf {S}}^{0,0} &{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

That is, a juror with bias $b'$ prefers C if there is at least one guilty signal and all other signals are 0, or if there are at least two guilty signals. Notice that this type of bias $b'$ is allowed in B such that it occurs in the committee with positive probability.

Consider a realized bias profile where there are $n+1$ jurors who have bias $b'$, n jurors who have bias $\underline{b}$ including i, and the realized state is $(I,G,0,\dots ,0)$ including $s_{i}=I$. By Full Support, this situation occurs with positive probability. Then if i reports $m_{i}=I$, jurors with bias $b'$ will vote for A and $Pr(A|{\mathbf {b}},{\mathbf {s}},I)=1$. But if i reports $m_{i}=0$, then jurors with bias $b'$ will vote for C and $Pr(A|{\mathbf {b}},{\mathbf {s}},0)=0$. This bias profile lies in ${\mathbf {Z}}_{i}(({\mathbf {b}}_{-i},\underline{b}), I)$ and thereby gives non-emptiness.

Thus i’s incentive compatibility condition is violated when she is of type $\underline{b}$ and sees signal $s_i = I$, contradicting the existence of a fully revealing equilibrium. This proves necessity for all majority and indeed for all K-rules when $2\le K\le 2n+1$.

Case 2 $K=1$.

That is, as long as there is at least one juror votes for C, then the final outcome of the committee decision process is C. Consider an extremely lenient type juror j who prefers C if and only if all the realized signals are G, i.e., ${\mathbf {S}}_{\overline{b}}(C)={\mathbf {S}}^{2n+1,0}$. Consider the case where j has observed a guilty signal. By Full Support, such a situation occurs with positive probability. And we prove the result for $K=1$ by showing that for such voter j with $(b_{j}, s_{j})=(\overline{b},G)$, he strictly prefers concealing his private signal in the communication stage.

Similarly, at any bias profile ${\varvec{b}}$, for any signal $s_i$ observed by voter i, we define

$$\begin{aligned} {\mathbf {Z}}_{j}({\mathbf {b}}, s_j) \equiv \{{\mathbf {s}}_{-j}\in S^{2n}|Pr(A|{\mathbf {b}},{\mathbf {s}},G)\ne Pr(A|{\mathbf {b}},{\mathbf {s}},0), P({\mathbf {b}}_{-j}, {\mathbf {s}}_{-j}|b_{j}, s_{j})>0\} \end{aligned}$$

as the set of signals of other voters that have positive probability and make player j pivotal. Using this, the lenient type’s IC condition reduces to:

$$\begin{aligned} {\mathrm{EU}}(G,\overline{b},I)-{\mathrm{EU}}(0,\overline{b},I)= \sum _{{\mathbf {b}}_{-j}, {\mathbf {s}}_{-j}} P({\mathbf {b}}_{-j},{\mathbf {s}}_{-j}|\overline{b},G)\Phi _{j}(G,0;{\mathbf {b}}_{-j}, {\mathbf {s}}_{-j}) \ge 0, \end{aligned}$$

(3)

where the index ${\varvec{s}}_{-j}$ is summed over ${\mathbf {Z}}_{j}(({\mathbf {b}}_{-j},\overline{b}), G)$.

Given that the other individuals are telling the truth, juror j infers that if he hears a communication result showing all reports from the others are G, he prefers the final outcome to be C and he can unilaterally ensure that the final outcome is C at the voting stage regardless of his speech, since $K=1$. Thus he is not pivotal at this case. Otherwise, he hears a communication result showing that not all the others have observed guilty signals, and he prefers the final outcome to be A now. Conditional on him being pivotal at the communication stage, it must be that concealing instead of reporting G will turn the final result from C to A. That is, for all ${\mathbf {b}}_{-j}\in B^{2n}$, for any ${\mathbf {s}}_{-j}\in {\mathbf {Z}}_{j}(({\mathbf {b}}_{-j}, \overline{b}), G)$, $Pr(A|{\mathbf {b}},{\mathbf {s}},G)=0$, $Pr(A|{\mathbf {b}},{\mathbf {s}},0)=1$, and $u(A, b_{j},{\mathbf {s}})-u(C, b_{j},{\mathbf {s}})>0$. Therefore, $\Phi _{j}(G,0;{\mathbf {b}}_{-j}, {\mathbf {s}}_{-j})<0$. And by Full Support, for any ${\mathbf {b}}_{-j}\in B^{2n}$ and ${\mathbf {s}}_{-j}\in {\mathbf {Z}}_{j}(({\mathbf {b}}_{-j},\overline{b}), G)$, we have $P({\mathbf {b}}_{-j},{\mathbf {s}}_{-j}|\overline{b}, G)>0$. Hence if the set ${\mathbf {Z}}_{j}(({\mathbf {b}}_{-i},\overline{b}), G)$ is non-empty, it follows that (3) is violated.

To see the non-emptiness of ${\mathbf {Z}}_{j}(({\mathbf {b}}_{-j},\overline{b}), G)$ when $K=1$, consider a type of juror who has bias $b''$ such that ${\mathbf {S}}_{b''}(C)={\mathbf {S}}^{2n+1,0}\bigcup {\mathbf {S}}^{2n,0}$. That is, jurors with bias $b''$ prefers C if there is at most one uninformative signal and all informative signals are guilty signals. Again, $b''$ is permissible; thus it occurs with positive probability.

Consider a realized bias profile where there are 2n jurors who have bias $b''$, and juror j has bias $\overline{b}$, and the realized state is $(G,G,\dots ,G,0)$ including $s_{j}=G$. By Full support, this situation happens with positive probability. Then if juror j reports $m_{j}=G$, jurors with bias $b''$ will vote for C and $Pr(A|{\mathbf {b}},{\mathbf {s}},G)=0$. But if j reports $m_{j}=0$, then jurors with bias $b''$ will vote for A and $Pr(A|{\mathbf {b}},{\mathbf {s}},0)=1$. This bias profile lies in ${\mathbf {Z}}_{j}(({\mathbf {b}}_{-j},\overline{b}), I)$ and thereby gives non-emptiness.

Thus j’s incentive compatibility condition is violated when she is of type $\overline{b}$ and sees signal $s_{j} = G$, contradicting the existence of a fully revealing equilibrium. This proves the result for $K=1$ and completes the whole proof. $\square$