Information transmission and voting

Abstract

I analyze an individual’s incentive to disclose hard evidence in the context of committee voting. A committee consists of three members: one left-leaning, one right-leaning, and the third ex ante unbiased. They decide on whether to pursue a left or right policy by majority rule. One of them has private information about the merits of the policies and can privately send verifiable messages to the others. If the informed member is unbiased, he withholds information to neutralize the other two’s votes when preferences are sufficiently diverse. If the informed member is biased, then the others can better infer his information, knowing that any information favoring his agenda will be shared. In the latter case, because more information is effectively shared, higher social welfare results.

Appendix

1.1 The “no-signaling-what-you-don’t-know” condition

At each information set, the player about to move must have a belief about which history in the information set has been reached. In particular, I need to specify (i) the informed player’s belief before he takes a disclosure action, and (ii) all three players’ beliefs before they vote. Since the informed player’s information sets before he takes a disclosure action are singletons, I need to specify only the players’ beliefs before they vote. The set of histories before the players vote is:

$$\begin{aligned} H = M_i \times B_i \times B_{j_1} \times B_{j_2}. \end{aligned}$$

For each $\ell \in N$, the set of player $\ell$’s information sets before he votes is $H_\ell = M_\ell \times B_\ell$, with $h_\ell$ being an element. Let $\mu _\ell \left( \tilde{m}_i,\tilde{b}_i,\tilde{b}_{j_1},\tilde{b}_{j_2}\mid h_\ell \right)$ be the probability that player $\ell$ assigns to the event that $\tilde{m}_i$ occurs in the communication stage and that the bias vector is weakly below $\left( \tilde{b}_i,\tilde{b}_{j_1},\tilde{b}_{j_2} \right)$, conditional on $h_\ell$. The system of beliefs is denoted by $\mu =\{ \{\mu _{\ell }(\cdot \mid h_\ell )\}_{h_\ell \in H_\ell }\}_{\ell \in N}$.

I first characterize the informed player’s beliefs $\mu _i(\cdot \mid h_i)_{h_i \in H_i}$. Since player i’s unreached information sets can be reached only by his own deviations, his beliefs at those sets can be derived by Bayes’ rule (assuming perfect recall). Given $h_i=(m_i,b_i)$, player i knows exactly what happened in the communication stage. Hence, his belief is given by

$$\begin{aligned} \mu _i \left( \tilde{m}_i, \tilde{b}_i, \tilde{b}_{j_1}, \tilde{b}_{j_2} \mid m_i,b_i \right) = {\left\{ \begin{array}{ll} G_{j_1} \left( \tilde{b}_{j_1} \right) G_{j_2} \left( \tilde{b}_{j_2} \right) &{}\quad \mathrm{if }\;\; b_i \leqslant \tilde{b}_i \text{ and } \tilde{m}_i = m_i,\\ 0 &{}\quad \text{ otherwise }. \end{array}\right. } \end{aligned}$$

(B-1)

Here, recall that $G_\ell (\cdot )$ is the cdf of $b_\ell$. For an uninformed player j, if he sees a message $m_j$ that is on path, he can calculate the probability that $m_i$ occurs in the communication stage, conditional on $m_j$, denoted by $\mu _j\left( m_i \mid m_j\right)$.¹⁶ Hence, $\mu _{j} \left( \tilde{m}_i,\tilde{b}_i, \tilde{b}_{j}, \tilde{b}_{j^\prime } \mid m_{j},b_{j} \right)$ is well defined according to Bayes’ rule. If player j receives an out-of-equilibrium message from player i, Bayes’ rule does not impose any restriction on player j’s beliefs $\mu _{j}$. For example, if player i never shares any signal with player j on path, then player j could believe that player i’s and $j^\prime$’s biases are perfectly correlated if he receives a signal from player i. However, those beliefs violate the “no-signaling-what-you-don’t-know” condition since biases are drawn independently after the communication stage. Player j’s belief regarding player i’s disclosure behavior should be independent of player j’s belief regarding the other players’ bias realizations. Moreover, both beliefs should be independent of player j’s bias realization. This motivates the following restrictions imposed on the uninformed players’ beliefs, where $u_j(\cdot \mid m_j)$ are extended to include those nodes $m_j$ that are off-path:

• (i) For $j\in \{j_1,j_2\}$, there exists a belief system $\{\mu _j(\cdot \mid m_j):\mu _j(\cdot \mid m_j) \in \Delta (M_i)\}_{m_j \in M_j}$ such that $\mu _j(\cdot \mid m_j)$ assigns positive probability only to nodes contained in $m_j$. That is,

  $$\begin{aligned} \sum _{m_i\in m_j}\mu _{j}(m_i \mid m_j)=1, \quad \forall m_j \in M_j. \end{aligned}$$

  (B-2)

  The beliefs $\{\mu _j(\cdot \mid m_j)\}_{m_j \in M_j}$ are defined by Bayes’ rule whenever possible.

• (ii) For $j\in \{j_1,j_2\}$ and $h_j = (m_j,b_j) \in H_j$, the beliefs at $h_j$ are defined as follows:

  $$\begin{aligned}&\mu _j \left( \tilde{m}_i, \tilde{b}_i, \tilde{b}_j, \tilde{b}_{j^\prime } \mid m_j,b_j \right) \nonumber \\&\quad = {\left\{ \begin{array}{ll} \mu _j \left( \tilde{m}_i \mid m_j \right) G_i\left( \tilde{b}_i \right) G_{j^\prime }\left( \tilde{b}_{j^\prime } \right) &{}\quad \mathrm{if }\;\; b_j \leqslant \tilde{b}_j \text{ and } \tilde{m}_i \in m_j,\\ 0 &{}\quad \text{ otherwise }, \end{array}\right. } \end{aligned}$$

  (B-3)

  where $j^\prime \in \{j_1,j_2\}$ and $j^\prime \ne j$.

1.2 Proofs for section 3

Proof of Lemmas 1 and 2

Throughout this proof, the dependence on $\sigma$ is dropped when no confusion arises. It is shown in Sect. 3.2 that players employ threshold voting strategy. Given the information structure, the highest (resp. lowest) threshold a voter ever employs is 1 / 6 (resp. $-1/6$). Let $p_\ell (m_\ell ) = Pr [b_\ell < - E [s\mid m_\ell ,piv] ]$ denote the probability that player $\ell$ votes for L given $m_\ell$ where the expectation is taken over $\ell$’s bias. Conditional on s, the probability that $m_i = \varnothing$ is $Pr [m_i =\varnothing \mid s] = h$. The probability that i is pivotal conditional on $m_i = \varnothing$ and s is:

$$\begin{aligned} Pr [piv \mid m_i = \varnothing , s] = p_{j_1}(\varnothing ) (1-p_{j_2}(\varnothing ))+ p_{j_2}(\varnothing ) (1-p_{j_1}(\varnothing )). \end{aligned}$$

The probability that i receives no signal and is pivotal conditional on state s is:

$$\begin{aligned} Pr [m_i=\varnothing ,piv \mid s]= h \left( p_{j_1}(\varnothing ) (1-p_{j_2}(\varnothing ))+ p_{j_1}(\varnothing ) (1-p_{j_1}(\varnothing )) \right) , \end{aligned}$$

which is independent of s. The probability distribution over state conditional on $m_i=\varnothing$ and being pivotal is given by

$$\begin{aligned} g(s \mid m_i = \varnothing , piv) = \frac{ Pr [m_i=\varnothing , piv \mid s]f(s)}{\int _{-\frac{1}{2}}^{\frac{1}{2}} Pr [m_i=\varnothing ,piv \mid s] f(s) \mathrm{d} s} = f(s) =1 . \end{aligned}$$

This implies that $- E [s \mid m_i=\varnothing ,piv]=0$. When $m_i \in \{\theta _r\} \times D$, player i knows the thresholds utilized by the uninformed players. The probability of $m_i$ and i being pivotal conditional on s, $Pr [m_i,piv \mid s]$, is proportional to $1/2+s$. Therefore, I have

$$\begin{aligned} g(s \mid m_i , piv) = \frac{1}{2}+s, \quad - E [s \mid m_i,piv]=-\frac{1}{6}. \end{aligned}$$

Analogously, when an uninformed player j receives signal $\theta _r$, the probability of $m_j$ and j being pivotal conditional on s, $Pr [m_j=\theta _r,piv \mid s]$, is proportional to $1/2+s$. Therefore, j uses $-1/6$ as voting threshold as well.¹⁷ Similarly, the threshold that player i (resp. player j) employs after $m_i \in \{\theta _l\} \times D$ (resp. $m_j =\theta _l$) is 1 / 6. $\square$

1.2.1 Voting thresholds of the uninformed when $m_j=\varnothing$

Here, I show how to calculate $- E [s \mid m_{j_1}=\varnothing ,piv,\sigma ]$ while restricting attention to strategy profile $\sigma$ satisfying Lemmas 1 and 2. When $j_1$ receives no signal, either player i receives no signal or player i chooses not to disclose his signal to $j_1$. Let $\theta$ be the signal that player i obtains. I have:

$$\begin{aligned} Pr [m_{j_1}=\varnothing , piv \mid s,\sigma ]&= \sum _{\theta \in \{\theta _r,\theta _l,\varnothing \}} Pr [m_{j_1}=\varnothing , piv, \theta \mid s,\sigma ] \\&=\sum _{\theta \in \{\theta _r,\theta _l,\varnothing \}} Pr [ \theta \mid s,\sigma ] Pr [m_{j_1}=\varnothing , piv \mid \theta ,s,\sigma ] \\&= \sum _{\theta \in \{\theta _r,\theta _l,\varnothing \}} Pr [ \theta \mid s] Pr [m_{j_1}=\varnothing , piv \mid \theta ,\sigma ]. \end{aligned}$$

According to how the signal is drawn, I have

$$\begin{aligned}&Pr [ \theta =\theta _r \mid s] = (1-h) \left( \frac{1}{2}+s\right) , \quad Pr [ \theta =\theta _l \mid s] = (1-h) \left( \frac{1}{2}-s\right) , \\&\quad Pr [ \theta =\varnothing \mid s] = h. \end{aligned}$$

Next, I show how to calculate $Pr [ m_{j_1}=\varnothing , piv \mid \theta ,\sigma ]$. To illustrate, take $\theta =\varnothing$ as an example. I have

$$\begin{aligned} Pr [ m_{j_1}=\varnothing , piv \mid \theta =\varnothing ,\sigma ] = p_{i}(\varnothing ) (1-p_{j_2}(\varnothing ))+ p_{j_2}(\varnothing ) (1-p_{i}(\varnothing )). \end{aligned}$$

I want to show that this probability is strictly positive when $\beta \in [0,2/3)$. For this probability to be zero, it must be the case that both i and $j_2$ vote for the same policy with probability one. For any pair $(i,j_2)$, this is not possible. Therefore, the probability $Pr [ m_{j_1}=\varnothing , piv \mid s,\sigma ]$ is strictly positive for any $s\in [-1/2,1/2]$. The density function of state s conditional on $j_1$ receiving no signal and being pivotal is:

$$\begin{aligned} g(s \mid m_{j_1}=\varnothing ,piv,\sigma ) = \frac{ Pr [ m_{j_1}=\varnothing , piv \mid s,\sigma ] f(s)}{\int _{-\frac{1}{2}}^{\frac{1}{2}} Pr [ m_{j_1}=\varnothing ,piv \mid s,\sigma ] f(s) \mathrm{d} s}. \end{aligned}$$

Therefore, the expectation of state variable s conditional on $j_1$ receiving no signal and being pivotal is:

$$\begin{aligned} - E [s \mid m_{j_1}=\varnothing ,piv,\sigma ]= -\int _{-\frac{1}{2}}^{\frac{1}{2}} s g(s \mid m_j=\varnothing ,piv,\sigma ) \mathrm{d} s. \end{aligned}$$

The probability $Pr [ m_j=\varnothing ,piv \mid s,\sigma ]$ is a sum product of $Pr [ m_j=\varnothing \mid s,\sigma ]$ and $Pr [ piv \mid m_j = \varnothing ,s,\sigma ]$. Regardless of the state, there is a strictly positive probability that player i receives no signal and hence shares no signal with player j. Therefore, $Pr [ m_j =\varnothing \mid s,\sigma ]$ is well defined and strictly positive for any s. Once I restrict attention to those strategy profiles satisfying Lemmas 1 and 2, the probability $Pr [ piv \mid m_j = \varnothing ,s,\sigma ]$ is strictly positive for any s.

1.3 Proofs for section 4

To simplify notation, let $t_1,t_3\in (-1/6,1/6)$ denote the voting thresholds employed by player 1 and 3 if they observe no signal. Due to symmetry, I focus on the disclosure behavior after the central player observes signal $\theta _r$. Let $p_1^L$ and $p_3^L$ denote probabilities with which player 1 and 3 vote for L in equilibrium.

Proof of Proposition 1

When $\beta <2/15$, an uninformed player votes for policy L with probability at least 1 / 5 regardless of the messages he receives. Because the lowest threshold an uninformed player ever employs is $-1/6$, both $p_1^L$ and $p_3^L$ are bounded from below by 1 / 5,

$$\begin{aligned} p_1^L \geqslant -\frac{1}{6}-\left( -\frac{1}{2}-\beta \right)>\frac{1}{5}, \quad p_3^L \geqslant -\frac{1}{6}-\left( -\frac{1}{2}+\beta \right) > \frac{1}{5}. \end{aligned}$$

According to Lemma 3, player 2 would like $p_3^L$ and $p_1^L$ to be as low as possible. The unique optimal disclosure strategy is to share signal $\theta _r$ with both player 1 and 3. Analogously, player 2 fully disclose signal $\theta _l$ to the uninformed players. $\square$

Proof of Proposition 2

For $\beta >\frac{2}{15}$, we know from Lemma 4 that player 2 withholds $\theta _r$ from player 1 (and withholds $\theta _l$ from player 3). Moreover, after signal $\theta _r$, player 2 would like $p_3^L$ to be as small as possible. If player 2 shares $\theta _r$ with player 3, then player 3 will use the voting threshold $-\frac{1}{6}$. If player 2 does not share $\theta _r$ with player 3, then player 3 will use the voting threshold $t_3$, which is strictly above $-\frac{1}{6}$. Therefore, sharing $\theta _r$ with player 3 is always a best response (although it might not be the unique best response, as we later show in Proposition 3). Given player 2’s disclosure strategy, we calculate player 1’s and 3’s voting thresholds $t_1,t_3$ after signal $\varnothing$, which are given in Proposition 2. This shows that the profile in Proposition 2 is an equilibrium.

We next argue that this partial disclosure equilibrium is the unique equilibrium for $\frac{2}{15}<\beta <\frac{8+h}{18}$. First, if $\beta < \frac{1}{3}$, then according to Lemma 4 the probability that player 3 votes for L, if $\theta _r$ realizes, is $\max \left\{ \frac{1}{3}-\beta ,0\right\} >0$. Therefore, player 2’s unique best response is to share $\theta _r$ with player 3 (and, similarly, to share $\theta _l$ with player 1). We thus focus on the region that $\beta \geqslant \frac{1}{3}$. We let $q_1$ and $q_3$ be the probabilities with which player 2 shares $\theta _l$ with player 1 and shares $\theta _r$ with player 3, respectively. We want to argue that if $\beta <\frac{8+h}{18}$, then $q_1=q_3=1$ is the unique equilibrium. Suppose not. Suppose that $q_3<1$. This implies that not sharing $\theta _r$ with player 3 is among player 2’s best responses. Then, it must be true that $t_3 \leqslant -\frac{1}{2}+\beta$. We now discuss two cases, $q_1<1$ and $q_1=1$, separately.

• (i) If $q_1<1$, then it must be true that $t_1 \geqslant \frac{1}{2}-\beta$. Given this condition, the voting thresholds if player 1 and 3 see no signal are:

  $$\begin{aligned} t_1=\frac{(h-1) (2 q_1-1)}{12 (h-1) q_1+18}, \quad \text {and} \quad t_3=\frac{(h-1) (2 q_3-1)}{12 (h-1) q_3+18}. \end{aligned}$$

  They satisfy the conditions $t_3 \leqslant -\frac{1}{2}+\beta$ and $t_1 \geqslant \frac{1}{2}-\beta$ only if $\beta \geqslant \frac{8+h}{18}$.

• (ii) If $q_1=1$, then $t_1 = \frac{h-1}{6+12h}$. Substituting $t_1$ into the formula for $t_3$, we can solve for $t_3$ explicitly. It is easy to verify that $t_3$ increases in $q_3$ so the condition $t_3 \leqslant -\frac{1}{2}+\beta$ is the easiest to satisfy when $q_3=0$. Substituting $q_3=0$ into $t_3$, we show that $t_3 \leqslant -\frac{1}{2}+\beta$ is true only if $\beta \geqslant \frac{8+h}{18}$.

This completes the proof that $q_1=q_3=1$ is the unique equilibrium when $\frac{2}{15}<\beta <\frac{8+h}{18}$. $\square$

Proposition 7

When $\beta =\frac{2}{15}$, player 2 shares $\theta _r$ with player 3 and $\theta _l$ with player 1. He is indifferent between sharing $\theta _r$ with player 1 or not, and is indifferent between sharing $\theta _l$ with player 3 or not. Player 2’s most preferred equilibrium is the full disclosure one.

Proof of Proposition 7

When $\beta =\frac{2}{15}$, after $\theta _r$ realizes, the probability that player 1 votes for L (i.e., $p_1^L$) is strictly higher than $\frac{1}{5}$. Hence, player 2 prefers $p_3^L$ to be as small as possible. Player 2 hence shares $\theta _r$ with player 3, so $p_3^L$ equals $\frac{1}{5}$. This implies that player 2 is indifferent between sharing $\theta _r$ with player 1 or not. We let $q_1$ be the probability with which player 2 shares $\theta _r$ with player 1. Similarly, player 2 shares $\theta _l$ with player 1 and is indifferent between sharing $\theta _l$ with player 3 or not. We let $q_3$ be the probability that player 2 shares $\theta _l$ with player 3. Given this sharing strategy, player 1’s and player 3’s voting thresholds if they see no signal are:

$$\begin{aligned} t_1=\frac{(h-1) (q_1-1)}{h (6 q_1+9)-6 q_1+6} , \quad \text {and} \quad t_3 = \frac{(h-1) (q_3-1)}{h (6 q_3+9)-6 q_3+6}. \end{aligned}$$

Player 2’s equilibrium payoff is:

$$\begin{aligned} \frac{1}{180} \left( w \left( 68-\frac{3}{5} h (300 t_1 t_3-40 t_1+40 t_3+33)\right) -180 w^2-30\right) , \end{aligned}$$

which is maximized when $q_1$ and $q_3$ both equal one. $\square$

Proof of Proposition 3

We have argued in Lemma 4 that player 2 would like $p_3^L$ to be as small as possible if $\theta _r$ has realized. If player 2 does not share $\theta _r$ with player 3, then it must be true that player 3 votes for L with zero probability (i.e., $t_3 \leqslant -\frac{1}{2}+\beta$). Similarly, if player 2 does not share $\theta _l$ with player 1, it must be true that $t_1 \geqslant \frac{1}{2}-\beta$. Given this no-disclosure strategy and the conditions $t_3,-t_1 \leqslant -\frac{1}{2}+\beta$, player 1’s and player 3’s voting thresholds after no signal are $t_3=-t_1 =\frac{h-1}{18}$. These satisfy the conditions $t_3,-t_1 \leqslant -\frac{1}{2}+\beta$ if and only if $\beta \geqslant \frac{8+h}{18}$.

1.4 Proofs for sections 5 and 6

Proof of Lemma 5

Suppose that $\beta <1/3$. If the probabilities with which player 2 and 3 vote for L are $p_2^L$ and $p_3^L$, respectively, the expected payoff to player 1 is:

$$\begin{aligned} E [u_1 \mid \theta =\theta _l,d,v]{=} p_2^L p_3^L \left( 4 \beta +\frac{2}{3}\right) w{+}\frac{2}{9} \left( 3 \beta +2 \right) ^2 \left( p_2^L +p_3^L - 2 p_2^L p_3^L\right) w+ c^{ \prime }. \end{aligned}$$

Since the highest possible threshold that player 2 or player 3 employs is 1 / 6, both $p_2^L$ and $p_3^L$ are weakly smaller than 2 / 3. The derivatives of $E [u_1 \mid \theta =\theta _l,d,v]$ with respect to $p_2^L$ and $p_3^L$ are strictly positive:

$$\begin{aligned} \frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_2^L}= & {} \frac{2}{9} \left( (3 \beta +2)^2-p_3^L (6 \beta (3 \beta +1)+5)\right) w>0, \\ \frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_3^L}= & {} \frac{2}{9} \left( (3 \beta +2)^2-p_2^L (6 \beta (3 \beta +1)+5)\right) w>0. \end{aligned}$$

Therefore, player 1 prefers $p_2^L,p_3^L$ to be as large as possible. He fully discloses signal $\theta _l$, so player 2 and 3 both use 1 / 6 as their threshold after signal $\theta _l$. If $\beta$ lies in [1 / 3, 2 / 3], player 1 fully discloses signal $\theta _l$ so as to decrease the probability that RR occurs. $\square$

Proof of Proposition 4

Suppose that if $\theta _r$ realizes the probabilities with which player 2 and 3 vote for L are $p_2^L$ and $p_3^L$. The expected payoff of player 1 is:

$$\begin{aligned} E [u_1 \mid \theta {=}\theta _r,d,v]{=} p_2^L p_3^L \left( 4 \beta -\frac{2}{3}\right) w{+}\frac{2}{9} \left( 3 \beta +1\right) ^2 \left( p_2^L +p_3^L -2 p_2^L p_3^L\right) w+ c^{\prime \prime }. \end{aligned}$$

The derivatives of $E [u_1 \mid \theta =\theta _r,d,v]$ with respect to $p_2^L$ and $p_3^L$ are:

$$\begin{aligned} \frac{\partial ^{} E [u_1 \mid \theta =\theta _r,d,v]}{\partial p_2^L}= & {} \frac{2}{9} \left( 1-2 p_3^L\right) \left( 3 \beta +1\right) ^2 w+p_3^L \left( 4 \beta -\frac{2}{3}\right) w, \end{aligned}$$

(2)

$$\begin{aligned} \frac{\partial ^{} E [u_1 \mid \theta =\theta _r,d,v]}{\partial p_3^L}= & {} \frac{2}{9} \left( 1-2 p_2^L\right) \left( 3 \beta +1\right) ^2 w+p_2^L \left( 4 \beta -\frac{2}{3}\right) w. \end{aligned}$$

(3)

If an uninformed player receives no signal, his voting threshold is in the range $(-1/6,0]$, because signal $\theta _l$ is always shared as shown in Lemma 5. If an uninformed player observes $\theta _r$, his voting threshold is $-1/6$. This implies that $p_2^L \leqslant 1/2$ and $p_3^L <1/2$, for any $\beta \geqslant 1/6$. Hence, for any $\beta \geqslant 1/6$, $\frac{\partial ^{} E [u_1 \mid \theta =\theta _r,d,v]}{\partial p_2^L}>0$, so player 1 does not share $\theta _r$ with player 2. This further implies that $p_2^L <1/2$, so $\frac{\partial ^{} E [u_1 \mid \theta =\theta _r,d,v]}{\partial p_3^L} >0$. As a result, player 1 would like $p_3^L$ to be as large as possible.

We let $t_2,t_3\in (-1/6,0]$ be player 2’s player 3’s voting thresholds if they see no signal. Let $q_3$ be the probability with which player 1 shares $\theta _r$ with player 3. If $t_3 \leqslant -\frac{1}{2}+\beta$, then player 1 is indifferent between sharing $\theta _r$ with player 3 or not. If $t_3>-\frac{1}{2}+\beta$, then player 1 strictly prefers not to share $\theta _r$. As a result, $q_3=0$ is always a best response by player 1. Moreover, if $\beta \leqslant \frac{1}{3}$, then $-\frac{1}{2} +\beta \leqslant -\frac{1}{6}<t_3$, so $q_3=0$ is player 1’s unique best response. Hence, we only need to focus on the region $\beta > \frac{1}{3}$. We next examine when it is possible that some $q_3>0$ is a best response by player 1.

Suppose that player 1 chooses some $q_3>0$. It must be true that $t_3 \leqslant -\frac{1}{2}+\beta$. Then, we can solve $t_2,t_3$ as a function of $q_3$. For $\beta \geqslant 1/2$, we have:

$$\begin{aligned} \begin{aligned} t_2&= \frac{(3 \beta +1) (h-1)}{6 (-3 \beta (h-1)+5 h+1)}, \\ t_3&= -\frac{(h-1) (q_3-1) (2 (6 \beta -1) t_2-3)}{12 t_2 (6 \beta (h-1) (q_3-1)-h (q_3-7)+q_3-1)-18 ((h-1) q_3+h+1)}. \end{aligned} \end{aligned}$$

(4)

The value $t_3$ increases in $q_3$ so the inequality $t_3 \leqslant -\frac{1}{2}+\beta$ is easiest to satisfy when $q_3\rightarrow 0$. Substituting $q_3=0$ into $t_3$, we obtain that $t_3 \leqslant -\frac{1}{2}+\beta$ holds for any $\beta \geqslant 1/2$ and $0<h<1$. For $\beta < 1/2$, we have:

$$\begin{aligned} \begin{aligned} t_2&= \frac{(3 \beta +1) (h-1)}{6 (3 \beta (h+1)+2 h+1)}, \\ t_3&= -\frac{(h-1) (q_3-1) (2 (6 \beta -1) t_2-3)}{12 t_2 (6 \beta ((h-1) q_3+h{+}1){+}h (-q_3){+}h{+}q_3-1)-18 ((h-1) q_3+h+1)}. \end{aligned} \end{aligned}$$

(5)

The value $t_3$ increases in $q_3$ so the inequality $t_3 \leqslant -\frac{1}{2}+\beta$ is easiest to satisfy when $q_3\rightarrow 0$. Substituting $q_3=0$ into $t_3$, we obtain that $t_3 \leqslant -\frac{1}{2}+\beta$ is equivalent to:

$$\begin{aligned} h \leqslant \frac{81 \beta ^2+9 \sqrt{36 \beta ^6-12 \beta ^5-11 \beta ^4+8 \beta ^3+3 \beta ^2+1}-3 \beta -19}{54 \beta ^3-90 \beta ^2-12 \beta +35}. \end{aligned}$$

The right-hand side strictly increases in $\beta$. It equals 0 when $\beta =\frac{1}{3}$, and equals 1 when $\beta =\frac{1}{2}$. Let $\hat{\beta }(h)$ be the inverse function of the right-hand side function. Hence, it is possible to have some $q_3>0$ as a best response by player 1 if and only if $\beta > \hat{\beta }(h)$.

We have argued that for $\beta > \hat{\beta }(h)$, it is possible to have some $q_3>0$ in equilibrium other than $q_3=0$. However, the value $t_2$ is independent of $q_3$ and the condition $t_3\leqslant -\frac{1}{2}+\beta$ implies that, no matter what $q_3$ is, player 3 must vote for R for sure if no signal or signal $\theta _r$ realizes. This implies that when multiple equilibria exist, they are all outcome-equivalent to the equilibrium in which player 1 does not share $\theta _r$ with player 3. $\square$

Proof of Proposition 5

The signs of $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_2^L}$ and $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_3^L}$ are the same as those of $A-p_3^L$ and $A-p_2^L$, respectively, where $A:=\frac{(3 \beta +1)^2}{18 \beta ^2-6 \beta +5}$ is strictly positive and increasing for $\beta \in \left[ 0,1/6\right)$. Since the voting threshold by player 3 or 2 is at least $-1/6$, it is always true that $p_3^L \geqslant \frac{1}{3}-\beta$ and that $p_2^L \geqslant \frac{1}{3}$. We let $\beta ^{*}$ be the root of $A=\frac{1}{3}-\beta$, and $\beta ^{**}$ the root of $A=\frac{1}{3}$. We have $\left( \beta ^*, \beta ^{**} \right) \approx \left( 0.05, 0.08\right)$. For any $\beta < \beta ^{*}$, it holds that $A-p_3^L<0$, so player 1 shares $\theta _r$ with player 2. We let $\beta$. For any $\beta < \beta ^{**}$, it holds that $A-p_2^L<0$, so player 1 shares $\theta _r$ with player 3. We are ready to state the equilibrium for any $\beta < \beta ^{**}$.

• (i) For any $\beta <\beta ^{*}$, there is a unique equilibrium in which player 1 shares $\theta _r$ with players 2 and 3.

• (ii) For any $\beta =\beta ^{*}$, player 1 shares $\theta _r$ with player 3, so $p_3^L =\frac{1}{3}-\beta$. As a result, $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_2^L}=0$ so player 1 is indifferent between sharing $\theta _r$ with player 2 or not.

• (iii) For any $\beta \in \left( \beta ^{*},\beta ^{**} \right)$, there is a unique equilibrium. Player 1 shares $\theta _r$ with player 3 so $p_3^L =\frac{1}{3}-\beta$. As a result, $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_2^L}=0$ so player 1 does not share $\theta _r$ with player 2.

From now on, we focus on the parameter region in which $\beta \in \left[ \beta ^{**}, \frac{1}{6} \right)$. We let $q_2, q_3$ be the probabilities with which player 1 shares signal $\theta _r$ with player 2 and 3. We let $t_2,t_3$ be the voting thresholds by player 2 and 3 if they see no signal. We first argue that there exists an equilibrium in which $q_2=0$, so it must be true that $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_2^L}\geqslant 0$.

• (i) If $q_3=1$ in the equilibrium, then $t_2$ and $t_3$ are given by:

  $$\begin{aligned} t_2 = \frac{\left( 9 \beta ^2+2\right) (h-1)}{3 \left( 18 \beta ^2 (h+1)+5 h+4\right) } \quad \text {and} \quad t_3 =0. \end{aligned}$$

  It is a best response to choose $q_3=1$ if and only if $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_3^L}\leqslant 0$. The equilibrium conditions $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_2^L}\geqslant 0$ and $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_3^L}\leqslant 0$ are satisfied if and only if:

  $$\begin{aligned} h \geqslant h_1(\beta ) :=\frac{4 \left( 9 \beta ^2+2\right) (3 \beta (3 \beta +8)-2)}{6 \beta (18 (\beta -3) \beta (3 \beta -1)-49)+65}. \end{aligned}$$

• (ii) If $q_3 \in (0,1)$ in the equilibrium, then $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_3^L}= 0$ must hold. This implies that:

  $$\begin{aligned} t_2 = \frac{3 (6 \beta -1)}{2 (6 \beta (3 \beta -1)+5)}. \end{aligned}$$

  Substituting this value into the formula for $t_2$ and $t_3$, we can solve $t_3, q_3$ as a function of $h, \beta$. We are left with the constraints that $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_2^L}\geqslant 0$ and that $q_3 \in (0,1)$. They are satisfied if and only if:

  $$\begin{aligned} h_2 (\beta )< h < h_1 (\beta ), \end{aligned}$$

  where $h_2(\beta )$ solves the equation that $q_3=0$.

• (iii) If $q_3 =0$ in the equilibrium, then $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_3^L} \geqslant 0$ must hold. We can solve $t_2$ and $t_3$ as a function of $h, \beta$. Substituting $t_2,t_3$ into the equilibrium conditions $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_2^L}\geqslant 0$ and $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_3^L}\geqslant 0$, we show that they are satisfied if and only if:

  $$\begin{aligned} h \leqslant h_2(\beta ). \end{aligned}$$

We next examine the equilibrium in which $q_2=1$. This implies that $p_2^L = 1/3$.

• (i) For any $\beta > \beta ^{**}$, we have $A-1/3>0$. This implies that $q_3=0$ for any $\beta > \beta ^{**}$. Then $t_2$ and $t_3$ are given by:

  $$\begin{aligned} t_2 = 0 \quad \text {and} \quad t_3 = -\frac{(3 \beta +4) (h-1)}{6 (3 \beta (h-1)-5 h-4)}. \end{aligned}$$

  Substituting $t_2,t_3$ into the equilibrium conditions $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_2^L}\leqslant 0 \leqslant \frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_3^L}$, we obtain that these conditions are satisfied if and only if:

  $$\begin{aligned} h \geqslant \frac{2 (3 \beta +4) (3 \beta (3 \beta (6 \beta -1)+13)-2)}{(6 \beta -1) (6 \beta (3 \beta (3 \beta -5)+11)-65)}. \end{aligned}$$

  The right-hand side increases in $\beta$ for $\beta \in \left[ \beta ^{**},1/6 \right)$, and equals $\frac{4}{13}$ when $\beta =\beta ^{**}$. Therefore, it is possible to have $q_2=1$ and $q_3=0$ in an equilibrium only if $h \geqslant \frac{4}{13}$.

• (ii) For $\beta =\beta ^{**}$, the argument in the previous step shows that it is possible to have an equilibrium with $q_3=0$ for any $h \geqslant \frac{4}{13}$. It is not possible to have $q_3=1$ in an equilibrium since $p_3^L$ would be $1/3-\beta$ and $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_2^L}$ would be strictly negative. It is possible to have $q_3 \in (0,1)$ if and only if:

  $$\begin{aligned} h \left( \left( 15-8 \sqrt{2}\right) q_3+26 \sqrt{2}-39\right) +4 \left( 2 \sqrt{2}-3\right) (q_3-1)\le 0. \end{aligned}$$

  This condition holds for some $q_3 \in (0,1)$ only if $h \geqslant \frac{4}{13}$.

To sum, it is possible to have $q_2=1$ in an equilibrium only if $h \geqslant \frac{4}{13}$.

Lastly, we examine the equilibrium in which $q_2\in (0,1)$. First, it is not possible to have $q_3=1$. Suppose not. Then, $p_3^L=\frac{1}{3}-\beta$. For any $\beta \geqslant \beta ^{**}$, $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_2^L} >0$ so $q_2$ should be zero. This leads to a contradiction. Therefore, either $q_3=0$ or $q_3 \in (0,1)$.

• (i) Suppose that $q_3=0$ in equilibrium. Given that $q_2 \in (0,1)$, $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_2^L}= 0$ must hold. This implies that:

  $$\begin{aligned} t_3=\beta +\frac{3 (6 \beta -1)}{2 (6 \beta (3 \beta -1)+5)}. \end{aligned}$$

  Substituting this value into the formula for $t_2$ and $t_3$, we can solve $t_2, q_2$ as a function of $h, \beta$. We are left with the constraints that $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_3^L}\geqslant 0$ and that $q_2 \in (0,1)$. They are satisfied if and only if

  $$\begin{aligned}&\frac{2 (3 \beta -2) (3 \beta +1) (\beta (\beta (6 \beta (6 \beta +7)+1)+39)-2)}{2 \beta (\beta (3 \beta (9 \beta (2 \beta (6 \beta -13)+5)-74)-146)+138)-37} \geqslant h \\&\quad > \frac{2 (3 \beta +4) (3 \beta (3 \beta (6 \beta -1)+13)-2)}{(6 \beta -1) (6 \beta (3 \beta (3 \beta -5)+11)-65)}. \end{aligned}$$

  The left-hand side follows from the constraint that $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_3^L}\geqslant 0$, and the right-hand side follows from the constraint that $q_2<1$.

• (ii) Suppose that $q_3\in (0,1)$ in equilibrium. We let $(a_1,a_2,a_3,1-a_1-a_2-a_3)$ denote player 1’s disclosure strategy, where $a_1,a_2,a_3$ correspond to the probability to share with both players 2 and 3, that to share only with player 2, and that to share only with player 3, respectively. Then, $q_2=a_1+a_2$ and $q_3=a_1+a_3$. There are two cases. First, player 1 chooses whether to share with player 2 and 3 independently, so $a_1=q_2 q_3, a_2=q_2 (1-q_3), a_3=(1-q_2) q_3$. In the second case, this condition does hold.

  We now consider the first case. The conditions $\frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_2^L} = \frac{\partial ^{} E [u_1 \mid \theta =\theta _l,d,v]}{\partial p_3^L}=0$ imply that:

  $$\begin{aligned}&t_2 = \frac{6 \beta (-3 \beta q_2+q_2-9)-5 q_2+9}{6 (6 \beta (3 \beta -1)+5) (q_2-1)}, \quad \text {and}\nonumber \\&\quad t_3 = \frac{2-3 \beta (3 \beta (6 \beta -1)+13)}{3 (6 \beta (3 \beta -1)+5) (q_3-1)}-\frac{1}{6}. \end{aligned}$$

  Substituting these values into the formula for $t_2$ and $t_3$, we can solve $q_2, q_3$ as a function of $h, \beta$. The conditions $0<q_2,q_3<1$ are satisfied if and only if:

  $$\begin{aligned} h > \frac{2 (3 \beta -2) (3 \beta +1) (\beta (\beta (6 \beta (6 \beta +7)+1)+39)-2)}{2 \beta (\beta (3 \beta (9 \beta (2 \beta (6 \beta -13)+5)-74)-146)+138)-37}. \end{aligned}$$

  (6)

  The right-hand side increases in $\beta$ and equals $\frac{4}{13}$ when $\beta =\beta ^{**}$. For the second case, the same condition 6 is required for such an equilibrium to exist.

To sum, it is possible to have $q_2 \in (0,1)$ in equilibrium only if $h > \frac{4}{13}$. This shows that for any $\beta \in \left[ \beta ^{**}, 1/6\right)$ and $h < \frac{4}{13}$, there is an unique equilibrium in which $q_2=0$. $\square$

Proof of Proposition 6

We first consider the case in which player 2 is informed. For $\beta \geqslant \frac{h+8}{18}$, player 2’s most preferred equilibrium is the no-disclosure one characterized in Proposition 3. Player 1 and 3 vote for L and R, respectively, with probability one if they see no signal. This is as if player 2 were choosing the policy by himself. The social welfare equals:

$$\begin{aligned} \frac{1}{18} (w (-5 h-54 w+14)-9)-2 \beta ^2. \end{aligned}$$

(7)

We next consider the case in which player 1 is informed. When $\beta \geqslant \frac{h+8}{18}$, there is essentially a unique equilibrium characterized in Proposition 4 in which player 1 withholds $\theta _r$ and shares $\theta _l$. When $\beta \geqslant 1/2$, player 2’s and 3’s voting thresholds if they see no signal are given by (4) with $q_3$ being zero. When $\beta < 1/2$, their voting thresholds if they see no signal are given by (5) with $q_3$ being zero. Substituting these voting thresholds into the social welfare, we obtain that for $\beta \geqslant 1/2$ the social welfare is:

$$\begin{aligned} a_1 t_2^2 + a_2 t_2+ a_3, \end{aligned}$$

where $t_2$ is given by (4) and

$$\begin{aligned} a_1 =&\frac{1}{3} w (3 \beta (h-1)-5 h-1) , \\ a_2 =&\frac{1}{9} (3 \beta +1)^2 (h-1) w,\\ a_3 =&\frac{1}{108} \left( -54 \left( 4 \beta ^2+1\right) +w (9 \beta (10 \beta +1) (h-1)-61 h+115)-324 w^2\right) . \end{aligned}$$

For $\beta < 1/2$, the social welfare is:

$$\begin{aligned} b_1 t_2^2 + b_2 t_2+ b_3, \end{aligned}$$

where $t_2$ is given by (5) and

$$\begin{aligned} b_1 =&-\frac{1}{3} w (3 \beta (h+1)+2 h+1) ,\\ b_2 =&-\frac{1}{18} w (6 \beta (3 \beta (h+1)-2 h+2)-11 h+2),\\ b_3 =&\frac{1}{108} \left( -18 \beta ^2 ((h{+}5) w{+}12){+}9 \beta (7 h{-}1) w{+}w (-61 h-324 w+115)-54\right) . \end{aligned}$$

It is easy to verify that for any $\beta <1/2$, the social welfare when player 1 is informed is strictly higher than (7). For any $\beta \geqslant 1/2$, it is strictly higher than (7) if and only if

$$\begin{aligned} h < \bar{h}(\beta ) = \frac{4 (3 \beta -2) (3 \beta +1) (3 \beta +2)}{(3 \beta -7) (6 \beta (6 \beta +1)-11)}. \end{aligned}$$

$\square$

Proposition 8

(Bias-dependent sharing). Suppose that $\beta \geqslant (h+8)/18$ and an informed player chooses his disclosure strategy after learning his bias. If player 2 is informed, his most preferred equilibrium is the no-disclosure one as in Proposition 3. If player 1 is informed, the equilibrium is outcome-equivalent to that in Proposition 4. Hence, Proposition 6 still holds.

Proof of Proposition 8

We first argue that the strategy profile in Proposition 3 is an equilibrium even if player 2 learns his bias before sharing. The key observation is that if player 1 and player 3 vote differently when they observe no signal, then player 2 has no incentive to disclose any signal, even if the signal is favorable from player 2’s perspective. This is because player 2’s vote will always be the policy choice. We prove this result by showing that no player has any profitable deviation.

First, if player 2 shares no information with the other two players, player 1 always votes for L and player 3 always votes for R, as shown in the proof of Proposition 3. Second, given that players 1 and 3 always vote differently when they observe no signal, player 2’s vote is the final policy and he gets his highest possible payoff. Hence, not sharing is indeed player 2’s best response.

We next characterize the equilibrium when player 1 is informed. We let $t_2,t_3 \in (-1/6,1/6)$ be player 2’s and 3’s voting thresholds if they see no signal. If player 1 sees $\theta _l$, he prefers L for all bias realization. Therefore, he shares $\theta _l$ for sure. If player 1 sees $\theta _r$, he prefers R if and only if $b_1>-\frac{1}{6}$, so he shares $\theta _r$ if and only if $b_1 >-\frac{1}{6}$. Given this sharing behavior, we can solve for $t_2,t_3$. For $\beta \geqslant \frac{1}{2}$, we have:

$$\begin{aligned} \begin{aligned} t_2&= \frac{(3 \beta +1) (h-1)}{6 (-3 \beta (h-1)+5 h+1)}, \\ t_3&= -\frac{(3 \beta +1) (h-1) (3 \beta (h-1)-8 h-1)}{6 (3 \beta (h-1)-17 h-1) (3 \beta (h-1)-2 h-1)}. \end{aligned} \end{aligned}$$

(8)

For $\beta <\frac{1}{2}$, we have:

$$\begin{aligned} \begin{aligned} t_2&= \frac{(3 \beta +1) (h-1)}{6 (3 \beta (h+1)+2 h+1)}, \\ t_3&= -\frac{(2 \beta +1) (h-1) (6 \beta (2 h+1)+7 h+2)}{6 \left( 12 \beta ^2 (h-1) (4 h+1)-2 \beta (h (h+30)+5)-17 h (h+1)-2\right) }. \end{aligned} \end{aligned}$$

(9)

Note that the $t_2$ in (8) and (9) is the same as that in (4) and (5), respectively. Moreover, the values of $t_3$ in (4), (5), (8), and (9) are smaller than $-\frac{1}{2}+\beta$.

Lastly, we argue that the equilibrium is outcome-equivalent to that in Proposition 4. If $\theta _l$ realizes, it is shared and all three players use $\frac{1}{6}$ as their voting threshold. This is the same as before. If $\varnothing$ realizes, player 3 votes for R for sure, and player 1’s and 2’s voting thresholds are 0 and $t_2$, which are the same as before. If $\theta _r$ realizes, when player 1 votes for R, the policy will be R since player 3 votes for R for sure. When player 1 votes for L and hence withholds $\theta _r$, then player 2 is the pivotal player and his voting threshold $t_2$ is the same as before. Therefore, this equilibrium is outcome-equivalent to the one in Proposition 4. $\square$