Lies and consequences

Abstract

I study a strategic-communication game between an informed sender and an uninformed receiver with partially aligned preferences. The receiver is endowed with the ability to probabilistically detect if the sender is lying. Specifically, if the sender is making a false claim about her type, with some commonly known probability p the receiver additionally observes a private signal indicating that the sender is lying. The main result is that the receiver’s stochastic lie-detection ability makes fully revealing equilibria—the best outcome for the receiver—possible, even for small p (less than $\frac{1}{2}$). Additionally, if the language consists of precise messages, fully revealing equilibria exist only for $p=1$ and for a set of intermediate values of p that is bounded away from 0 and 1, making the maximal ex-ante expected equilibrium utility of the receiver non-monotone in p. If vague messages are allowed, full revelation can be supported for all large enough p, overturning the non-monotonicity and improving communication outcomes relative to the precise-language case.

A Proofs

Proof of Lemma 2

Assume there exists a FRE where the receiver’s actions off the equilibrium path are given by $Y(\cdot ,-1)$. In each of the following cases for the possible values of b, I show that a sender’s type has a profitable deviation to lying, which is a contradiction.

First, if $b\ge 1$, if type $t=0$ deviates to the message $m=1$, her expected utility would be

$$\begin{aligned} pU^S(Y(1,-1),0)+(1-p)U^S(1,0)&=-p\ell (b-Y(1,-1))-(1-p)\ell (b-1)\\&\ge -p\ell (b)-(1-p)\ell (b-1)\\&>-p\ell (b)-(1-p)\ell (b)\\&=-\ell (b)=U^S(0,0), \end{aligned}$$

which is the sender’s equilibrium utility. Note that the non-strict inequality above follows from the fact that $Y(1,-1)\in [0,1]$.

Second, if $b\in [1/2,1)$, if type $t=0$ deviates to the message $m=b$, her expected utility would be

$$\begin{aligned} pU^S(Y(b,-1),0)+(1-p)U^S(b,0)=-p\ell (b-Y(b,-1))\ge -p\ell (b)>-\ell (b), \end{aligned}$$

where the first inequality follows from the fact that $|b-Y(b,-1)|\le b$ because $Y(b,-1)\in [0,1]$.

Third, if $b\in (1/4,1/2)$ and $Y(1/2,-1)\in [0,2b]$, if type $t=0$ deviates to the message $m=1/2$, her expected utility would be

$$\begin{aligned} pU^S(Y(1/2,-1),0)+(1-p)U^S(1/2,0)&=-p\ell (Y(1/2,-1)-b)\\&\quad -(1-p)\ell (1/2-b)\\&>-p\ell (b)-(1-p)\ell (b)\\&=-\ell (b)=U^S(0,0). \end{aligned}$$

Finally, if $b\in (1/4,1/2)$ and $Y(1/2,-1)\in (2b,1]$, if type ${\hat{t}}\equiv Y(1/2,-1)-2b$ deviates to the message $m=1/2$, her expected utility would be

$$\begin{aligned} pU^S(Y(1/2,-1),{\hat{t}})+(1-p)U^S(1/2,{\hat{t}})&=-p\ell (b)\\&\quad -(1-p)\ell (Y(1/2,-1)-b-1/2)\\&>-p\ell (b)-(1-p)\ell (b)\\&=-\ell (b)=U^S({\hat{t}}, {\hat{t}}), \end{aligned}$$

where the inequality follows from the fact that $|Y(1/2,-1)-b-1/2|<1/4<b$. $\square$

Proof of Proposition 2

By Lemma 1, in all FRE we have $M(t)=t$ for all $t\in \mathcal {T}$ and $Y(m,0)=m$ for all $m\in \mathcal {M}$. In equilibrium, a sender of type t’s utility is $U^S(t,t)=-b^2$. If she deviates to some false report $m\ne t$, it is

$$\begin{aligned}&pU^S(Y(m,-1),t)+(1-p)U^S(m,t)\\&\quad =-p(t+b-Y(m,-1))^2-(1-p)(t+b-m)^2. \end{aligned}$$

Denote this value by $\mathbb {E}\left[ U^S_m(t)\right]$. Notice that $\mathbb {E}\left[ U^S_m(t)\right]$ is strictly concave in t.

Fix p and a family of off-equilibrium actions $\{Y(m,-1)\}_{m\in \mathcal {M}}$ forming a part of a FRE. By Lemma 2, we have $b\le 1/4$. Also, we can assume that for each message m there is a unique type t(m) for whom the expected utility of falsely reporting m is larger than the expected utility of any other type when falsely reporting m.³⁴ In other words

$$\begin{aligned} \{t(m)\}=\arg \max _{t\in [0,1]}\mathbb {E}\left[ U^S_m(t)\right] . \end{aligned}$$

It is easy to check that $t(m)=\max \{pY(m,-1)+(1-p)m-b;0\}$. Then, whenever $t(m)=pY(m,-1)+(1-p)m-b\ge 0$, $\mathbb {E}\left[ U^S_m(t(m))\right]$ can be computed to satisfy

$$\begin{aligned} \mathbb {E}\left[ U^S_m(t(m))\right] =-p(1-p)(Y(m,-1)-m)^2 \end{aligned}$$

(1)

whenever $p\in (0,1)$.³⁵ Notice that in this case, $\mathbb {E}\left[ U^S_m(t(m))\right]$ is decreasing in the distance $|Y(m,-1)-m|$. Thus, in order to check that there are no incentives for deviation, it would suffice to check incentive compatibility for the message m for which that distance is the smallest. If $pY(m,-1)+(1-p)m-b<0$ (implying $t(m)=0$), we have

$$\begin{aligned} \mathbb {E}\left[ U^S_m(t(m))\right] \le -p(1-p)(Y(m,-1)-m)^2. \end{aligned}$$

It is clear that since $U^S(t,t)$ does not depend on t, to verify that there are no incentives for deviation from truthfulness/full revelation it suffices to check that it is incentive compatible for t(m) to report truthfully rather than deviate to m for each $m\in \mathcal {M}=[0,1]$. In other words, we need to show

$$\begin{aligned} \sup _m \mathbb {E}\left[ U^S_m(t(m))\right] \le -b^2. \end{aligned}$$

Consider the following off-equilibrium actions:

$$\begin{aligned} Y(m,-1)=\left\{ \begin{array}{cl} 1&{} \text{ if } m<1/2,\\ 0&{} \text{ if } m\ge 1/2.\\ \end{array} \right. \end{aligned}$$

Clearly, these actions can be rationalized if the sender believes that she is certainly facing type $t=1$ (or $t=0$) after observing $(m,-1)$ for $m<1/2$ (or $m\ge 1/2$).

Note that under these beliefs we have

$$\begin{aligned} \sup _m \mathbb {E}\left[ U^S_m(t(m))\right] \le \sup _m -p(1-p)(Y(m,-1)-m)^2=-p(1-p)\frac{1}{4}. \end{aligned}$$

Thus, if $-p(1-p)\frac{1}{4}\le -b^2$, this would guarantee that the proposed off-equilibrium actions can support a FRE. In the case $b\le 1/4$, the inequality is true if and only if

$$\begin{aligned} p\in \left[ \frac{1}{2}-\frac{\sqrt{1-16b^2}}{2},\frac{1}{2}+\frac{\sqrt{1-16b^2}}{2}\right] . \end{aligned}$$

(2)

Other values of p could support a FRE only if the inequality

$$\begin{aligned} \sup _m \mathbb {E}\left[ U^S_m(t(m))\right] \le -p(1-p)\frac{1}{4} \end{aligned}$$

holds strictly. This is possible only if $\mathbb {E}\left[ U^S_{m}(t(m))\right] \le -p(1-p)(Y(m,-1)-m)^2$ holds strictly for some open neighborhood of values around $m=1/2$. In other words, we need $pY(m,-1)+(1-p)(m)-b<0$ and $t(m)=0$ for those values of m. Adding the inequality necessary for equilibrium compliance, the following system of inequalities is derived

$$\begin{aligned} \begin{aligned}&pY(m,-1)+(1-p)m-b< 0;\\&\mathbb {E}\left[ U^S_m(0)\right] =pU^S(Y(m,-1),0)+(1-p)U^S(m,0)\le -b^2. \end{aligned} \end{aligned}$$

(3)

The first inequality implies $Y(m,-1)< b$ because we are in the case $b\le 1/4$ and the inequalities need to be satisfied for all m in some neighborhood of 1 / 2. But the left-hand side of the second inequality is increasing over these values of $Y(m,-1)$ so, to maximize compliance, we may assume $Y(m,-1)=0$. It is then easy to check that the first inequality is satisfied for $m< \frac{b}{1-p}$ and the second—for $m\ge 2b$. Notice that it is possible for both of these to be satisfied only if $p\ge \frac{1}{2}$.

Since we already know that $p\in \left[ \frac{1}{2}-\frac{\sqrt{1-16b^2}}{2},\frac{1}{2}+\frac{\sqrt{1-16b^2}}{2}\right]$ can support a FRE, I assume $p>\frac{1}{2}+\frac{\sqrt{1-16b^2}}{2}$ for the rest of the proof. This implies $\frac{b}{1-p}>\frac{1}{2}$. To see that, note that since $p>\frac{1}{2}+\frac{\sqrt{1-16b^2}}{2}$, it suffices to show that $\frac{1}{2}-\frac{\sqrt{1-16b^2}}{2}\le 2b$. Re-arranging, we can verify that this is equivalent to $b\le \frac{1}{4}$. Therefore we have $2b\le \frac{1}{2}<\frac{b}{1-p}$. So for any message $m\in \left[ 2b,\frac{b}{1-p}\right)$, the (off-equilibrium) action $Y(m,-1)=0$ guarantees that no type would want to deviate to that message.

Now consider compliance for messages $m\ge \frac{b}{1-p}$. We want

$$\begin{aligned} \sup _{m\ge \frac{b}{1-p}}\mathbb {E}\left[ U^S_m(t(m))\right] \le -b^2. \end{aligned}$$

It is clear that $t(m)=pY(m,-1)+(1-p)m-b\ge 0$ for all $m\ge \frac{b}{1-p}$ regardless of the value of $Y(m,-1)$. Then, by (1)

$$\begin{aligned} \sup _{m\ge \frac{b}{1-p}}\mathbb {E}\left[ U^S_m(t(m))\right] =\sup _{m\ge \frac{b}{1-p}}-p(1-p)(Y(m,-1)-m)^2 \end{aligned}$$

for all $m\ge \frac{b}{1-p}$. This can be minimized by, for example, setting $Y(m,-1)=0$ for all $m\ge \frac{b}{1-p}$ so that

$$\begin{aligned} \sup _{m\ge \frac{b}{1-p}}\mathbb {E}\left[ U^S_m(t(m))\right] =-p(1-p)\left( \frac{b}{1-p}\right) ^2. \end{aligned}$$

In order to have equilibrium compliance, we must have $-p(1-p)\left( \frac{b}{1-p}\right) ^2\le -b^2$. It is easy to check that this is satisfied for all $p\in [1/2,1]$. Thus, in the case we are considering (i.e. $p>\frac{1}{2}+\frac{\sqrt{1-16b^2}}{2}$), no sender type would deviate to a message $m\ge \frac{b}{1-p}$ if $Y(m,-1)=0$ for all such m.

Now consider the compliance for messages $m<2b$. We want

$$\begin{aligned} \sup _{m<2b}\mathbb {E}\left[ U^S_m(t(m))\right] \le -b^2. \end{aligned}$$

Fix any $m^*\in (b,2b)$. If $t(m^*)=0>pY(m^*,-1)+(1-p)m^*-b$, which is possible only if $Y(m^*,-1)<b$, then

$$\begin{aligned} \mathbb {E}\left[ U^S_{m^*}(t(m^*))\right]&=pU^S(Y(m^*,-1),0)+(1-p)U^S(m^*,0)\\&\ge pU^S(0,0)+(1-p)U^S(m^*,0) >-b^2, \end{aligned}$$

where the second inequality follows from our analysis of the system (3) and the fact that $m<2b$. So in order to ensure equilibrium compliance for all messages $m<2b$, we must have $t(m)=pY(m,-1)+(1-p)m-b$ (at least for all $m\in (b,2b)$)³⁶. We noticed above that in such cases $\mathbb {E}\left[ U^S_m(t(m))\right]$ is decreasing in the distance $|Y(m,-1)-m|$ so we can set $Y(m,-1)=1$ for all $m<2b$. Then, by (1), we have

$$\begin{aligned} \sup _{m<2b}\mathbb {E}\left[ U^S_m(t(m))\right] =-p(1-p)(1-2b)^2. \end{aligned}$$

We must have $-p(1-p)(1-2b)^2\le -b^2$. Solving, we get

$$\begin{aligned} p\in \left[ \frac{1}{2}-\frac{\sqrt{1-4\frac{b^2}{(1-2b)^2}}}{2};\frac{1}{2}+\frac{\sqrt{1-4\frac{b^2}{(1-2b)^2}}}{2}\right] . \end{aligned}$$

Since we are in the case $p>\frac{1}{2}+\frac{\sqrt{1-16b^2}}{2}$, it can be checked that for $b\le 1/4$ we have

$$\begin{aligned} \frac{1}{2}+\frac{\sqrt{1-16b^2}}{2}\le \frac{1}{2}+\frac{\sqrt{1-4\frac{b^2}{(1-2b)^2}}}{2}. \end{aligned}$$

So for all $p\in \left[ \frac{1}{2}+\frac{\sqrt{1-16b^2}}{2},\frac{1}{2}+\frac{\sqrt{1-4\frac{b^2}{(1-2b)^2}}}{2}\right]$, the following off-equilibrium actions guarantee FRE compliance:

$$\begin{aligned} Y(m,-1)=\left\{ \begin{array}{cl} 1&{} \text{ if } m<2b,\\ 0&{} \text{ if } m\ge 2b.\\ \end{array} \right. \end{aligned}$$

We conclude that a full-revelation equilibrium can be supported if and only if

$$\begin{aligned} p\in \left[ \frac{1}{2}-\frac{\sqrt{1-16b^2}}{2};\frac{1}{2}+\frac{\sqrt{1-4\frac{b^2}{(1-2b)^2}}}{2}\right] . \end{aligned}$$

$\square$

Proof of Proposition 3

Let p be given. Let $Y(m,0)=m$ for all $m\in \mathcal {M}$ and also

$$\begin{aligned} Y(m,-1)=\left\{ \begin{array}{cl} 1&{} \text{ if } m<1/2,\\ 0&{} \text{ if } m\ge 1/2.\\ \end{array} \right. \end{aligned}$$

I will show that given the receiver’s actions above, whenever p is in some open neighborhood of 1 / 2, it is optimal for the sender to choose $M(t)=t$ for all $t\in \mathcal {T}$. The pair of strategies would form a FRE.

For some message m, consider the maximum possible utility that can be achieved by a sender’s type (falsely) sending that message:

$$\begin{aligned} \max _{t\in \mathbb {R}}\left[ -p\ell (t+b-Y(m,-1))-(1-p)\ell (t+b-m)\right] . \end{aligned}$$

The value of this optimization program is decreasing in $|m-Y(m,-1)|$. To see that, first note that if $\ell$ is strictly convex, the objective function is strictly concave so the maximizer $t^*$ is unique and $t^*+b$ is strictly between m and $Y(m,-1)$. So a small decrease in the distance between m and $Y(m,-1)$ clearly increases the value of the objective function evaluated at $t^*$.

If $\ell$ is piecewise linear, the optimum is either a corner solution (i.e. $t^*+b$ equals either m or $Y(m,-1)$) for $p\ne 1/2$ or any $t^*$ such that $t^*+b$ is in the closed interval defined by m and $Y(m,-1)$ for $p=1/2$. Either way, the value of the objective function is $\min \{p,1-p\}(-\ell (m-Y(m,-1)))$, which is decreasing in $|m-Y(m,-1)|$.

Denote the expected utility of type t sending a false message³⁷m by $\mathbb {E}\left[ U^S_m(t)\right]$. Note that

$$\begin{aligned} \sup _{t\in [0,1]}\mathbb {E}\left[ U^S_m(t)\right] \le \max _{t\in \mathbb {R}}\left[ -p\ell (t+b-Y(m,-1))-(1-p)\ell (t+b-m)\right] \end{aligned}$$

and therefore

$$\begin{aligned} \sup _{m\in [0,1]}\sup _{t\in [0,1]}\mathbb {E}\left[ U^S_m(t)\right]&\le \sup _{m\in [0,1]}\max _{t\in \mathbb {R}}\left[ -p\ell (t+b-Y(m,-1))\right. \\&\left. \quad -(1-p)\ell (t+b-m)\right] \\&=\max _{t\in \mathbb {R}}\left[ -p\ell (t+b)-(1-p)\ell (t+b-1/2)\right] , \end{aligned}$$

where the equality follows from the fact that the program’s value is decreasing in $|m-Y(m,-1)|$.

It is optimal for the sender to always report truthfully if

$$\begin{aligned} \sup _{m\in [0,1]}\sup _{t\in [0,1]}\mathbb {E}\left[ U^S_m(t)\right] \le -\ell (b), \end{aligned}$$

where $-\ell (b)$ is the sender’s utility from reporting truthfully. So, in order to show what we need for $p=1/2$, it suffices to show the following inequality:

$$\begin{aligned} \max _{t\in \mathbb {R}} \left\{ -\frac{1}{2}\left[ \ell (t+b)+\ell (t+b-1/2)\right] \right\} \le -\ell (b). \end{aligned}$$

Notice that for the type $t^*$ that maximizes the left-hand side’s objective function, we again have that $t^*+b$ is between 0 and 1 / 2. So using Jensen’s inequality, the following holds for the left-hand side

$$\begin{aligned}&\frac{1}{2}(-\ell (t^*+b))+\frac{1}{2}(-\ell (1/2-t^*-b))\\&\quad \le -\ell \left( \frac{1}{2}(t^*+b)+\frac{1}{2}(1/2-t^*-b)\right) =-\ell (1/4)\le -\ell (b), \end{aligned}$$

which is what we wanted to show. Additionally, notice that if this inequality is strict (i.e. if $b<1/4$) by continuity of the left-hand side in p, there would be an open set of values around $p=1/2$ that also satisfy the inequality, proving the positive measure of the parameters supporting a FRE.

To complete the proof, we need to show that there does not exist a FRE for all $p\in (0,\varepsilon )\cup (1-\varepsilon ,1)$ for some $\varepsilon >0$. Let $m=b$ and consider the off-equilibrium actions $\{Y(m,-1)\}_{m\in \mathcal {M}}$ for some candidate FRE. If $Y(m,-1)\le b$, it is easy to see that type $t=0$ prefers sending the message m over truth-telling regardless of the values of p. If $Y(m,-1)>b$, we have

$$\begin{aligned} \mathbb {E}\left[ U^S_m(Y(m,-1)-b)\right] =-(1-p)\ell (Y(m,-1)-m)\ge -(1-p)\ell (1-b). \end{aligned}$$

We have $\lim _{p\rightarrow 1}\left[ -(1-p)\ell (1-b)\right] =0$. It is then clear that there exists $\varepsilon >0$ such that for all $p>1-\varepsilon$, we have

$$\begin{aligned} -(1-p)\ell (1-b)>-\ell (b) \end{aligned}$$

and so the inequality $\mathbb {E}\left[ U^S_m(Y(m,-1)-b)\right] > -\ell (b)$ holds for $p>1-\varepsilon$. Therefore for all $p>1-\varepsilon$, there does not exist $Y(m,-1)$ for which all sender types prefer truth-telling over sending the false message $m=b$. Thus, no off-equilibrium can be part of a FRE and there does not exist one.

For small values of p, we start by considering type $t=0$’s possible deviation to the message $m=b$:

$$\begin{aligned} \mathbb {E}\left[ U^S_b(0)\right] =-p\ell (Y(b,-1)-b). \end{aligned}$$

The rest of the proof proceeds analogously. $\square$

Proof of Proposition 4

I first show that if a FRE exists for p with $\mathcal {M}$, then it also exists for the same p with $\mathcal {M}^V$. Let $P^*$ denote the set of values of p, for which there exists a FRE $\{M,Y\}$ in the case without vague messages. We need to show that for all $p\in P^*$ there exists a fully revealing equilibrium for ${\mathcal {M}}^V$. Fix such a p and consider the messaging profile $M^V(t)=\{t\}$ and the action function

$$\begin{aligned} Y^V(\{t\},0)&=t,\\ Y^V(\{t\},-1)&=Y(t,-1),\\ Y^V(m,0)&=0 \text{ if } m\in \mathcal {M}^V{\setminus } M(\mathcal {T}),\\ Y^V(m,-1)&=0 \text{ if } 0\notin m\in \mathcal {M}^V{\setminus } M(\mathcal {T}),\\ Y^V(m,-1)&=1 \text{ if } 0\in m\in \mathcal {M}^V{\setminus } M(\mathcal {T}). \end{aligned}$$

I claim that $\{M^V,Y^V\}$ is a FRE. No sender type wants to deviate to a different equilibrium message since $\{M,Y\}$ is also a FRE. No sender type t strictly prefers the certain action $y=0$ (induced by deviating to an off-equilibrium truthful message or an off-equilibrium false message $m\not \ni 0$) to the corresponding equilibrium action $y=t$. So it remains to be shown that no sender type t prefers the lottery formed by the actions $y=0$ with probability $1-p$ and $y=1$ with probability p (induced by deviating to an off-equilibrium false message $m\ni 0$) to the equilibrium action $y=t$. Let $t^*\in [0,1]$ be a maximizer of the concave function $-p\ell (1-t-b)-(1-p)\ell (t+b)$, which is the utility that a type t derives from inducing that lottery.

Observe that since $\{M,Y\}$ is a FRE, no type wants to deviate to the message $\{0\}$ and therefore the expected utility of any type t from inducing the lottery over the actions 0 with probability $1-p$ and $Y^V(\{0\},-1)=Y(0,-1)$ with probability p (denoted by $\mathbb {E}[U^S_{\{0\}}(t)]$) is no greater than $-\ell (b)$. So it suffices to show that

$$\begin{aligned} -p\ell (1-t^*-b)-(1-p)\ell (t^*+b)\le \mathbb {E}[U^S_{\{0\}}(t)] \end{aligned}$$

for some $t\ne 0$. If $Y(0,-1)=1$, we are done. Assume instead $Y(0,-1)<1$. Note also that $Y(0,-1)\ge 2b$ because, otherwise, types close to $t=0$ would have a profitable deviation to $m=\{0\}$.³⁸

If $t^*>0$, consider $t^{**}=\max \{0,t^*-(1-Y(0,-1))\}$. If $t^{**}>0$, we have

$$\begin{aligned} \mathbb {E}[U^S_{\{0\}}(t^{**})]&=-p\ell (Y(0,-1)-t^{**}-b)-(1-p)\ell (t^{**}+b)\\&= -p\ell (1-t^{*}-b)-(1-p)\ell (t^{**}+b)\\&> -p\ell (1-t^{*}-b)-(1-p)\ell (t^{*}+b), \end{aligned}$$

which is what we wanted to show, and where the inequality follows from the fact that $t^{**}<t^*$.

If $t^{**}=0$, then $t^{**}\ge t^*-(1-Y(0,-1))$ and so $Y(0,-1)-t^{**}-b\le 1-t^*-b$. Since $Y(0,-1)-t^{**}-b\ge 2b-b=b>0$ and, by optimality of $t^*$, $t^*+b\in (0,1)$, we have $\ell (Y(0,-1)-t^{**}-b)\le \ell (1-t^{*}-b)$. Thus, we have

$$\begin{aligned} \mathbb {E}[U^S_{\{0\}}(t^{**})]&=-p\ell (Y(0,-1)-t^{**}-b)-(1-p)\ell (t^{**}+b)\\&\ge -p\ell (1-t^{*}-b)-(1-p)\ell (t^{*}+b). \end{aligned}$$

Finally, assume $t^*=0$. Since $b\le 1/4$ and $Y(0,-1)\in [2b,1)$ we have

$$\begin{aligned} -p\ell (1-t^{*}-b)-(1-p)\ell (t^{*}+b)< \mathbb {E}[U^S_{\{0\}}(0)]. \end{aligned}$$

Since $\mathbb {E}[U^S_{\{0\}}(0)]$ is continuous in type, there is some type $t^{**}$ close to $t=0$ which satisfies

$$\begin{aligned} -p\ell (1-t^{*}-b)-(1-p)\ell (t^{*}+b)< \mathbb {E}[U^S_{\{0\}}(t^{**})]. \end{aligned}$$

Next, I show that a FRE exists for all $p\ge \frac{1}{2}$. Assume $b<1/4$. Consider the following fully revealing strategy profile (Y, M):

$$\begin{aligned} M(0)&=[0,1],\\ M(t)&=[t,1-\varepsilon _t] \text{ for } \text{ all } t\in \mathcal {T}{\setminus }\{0\},\\ Y(M(t),0)&=t \text{ for } \text{ all } t\in \mathcal {T},\\ Y(M(t),-1)&=1-\varepsilon _t \text{ if } t\le 2b,\\ Y(M(t),-1)&=0 \text{ if } t> 2b,\\ Y(m,0)&=0 \text{ if } m\in \mathcal {M}{\setminus } M(\mathcal {T}),\\ Y(m,-1)&=0 \text{ if } 0\notin m\in \mathcal {M}{\setminus } M(\mathcal {T}),\\ Y(m,-1)&=\sup m \text{ if } 0\in m\in \mathcal {M}{\setminus } M(\mathcal {T}) \end{aligned}$$

for some $\varepsilon _t\ge 0$ such that $1-\varepsilon _t\ge \max \{t,4b\}$ and $\varepsilon _t>0$ whenever $t\le 2b$. I will show that this strategy profile is an equilibrium for all $p\ge \frac{1}{2}$.

No type has a profitable deviation to a true or false off-path message $m\notin M(\mathcal {T})$ with $0\notin m$ since that induces the action 0 for sure. Similarly, no type has a profitable deviation to a true $m\notin M(\mathcal {T})$ with $0\in m$. A sender type t deviating to a false message $m\notin M(\mathcal {T})$ with $0\in m$ induces a lottery between actions 0 (with probability $1-p$) and $\sup m$ (with probability p). Note that $t\notin [0,\sup m]$ is possible only if $t>\sup m$. Thus, both 0 and $\sup m$ are worse for t than the action she induces in equilibrium.³⁹

No type has a profitable deviation to a true on-path message $m\in M(\mathcal {T})$ since that would induce a lower action than truth-telling induces. What is left to verify is that no type has a profitable deviation to a false on-path message $m\in M(\mathcal {T})$. No type $t^*>1-\varepsilon _t$ has a profitable deviation to M(t) since both induced actions are weakly worse than the one $t^*$ induces in equilibrium.

We need to consider possible deviations only from types in $t^*\in [0,t)$. If $t\le 2b$:

$$\begin{aligned} \mathbb {E}[U^S_{M(t)}(t^*)]=-(1-p)\ell (t^*+b-t)-p\ell (1-\varepsilon _t-t^*-b). \end{aligned}$$

If $\ell$ is linear, for all $p\ge \frac{1}{2}$ we have

$$\begin{aligned} \mathbb {E}[U^S_{M(t)}(t^*)]&\le \mathbb {E}[U^S_{M(t)}(t)]\\&=-(1-p)b-p(1-\varepsilon _t-t-b)\\&\le -(1-p)b-pb\\&=-b, \end{aligned}$$

which is the equilibrium payoff and where the second inequality follows from $1-\varepsilon _t \ge 4b$ and $t\le 2b$.

If $\ell$ is strictly convex, then

$$\begin{aligned} \arg \max _{{\hat{t}}\in \mathbb {R}}\mathbb {E}[U^S_{M(t)}({\hat{t}})]+b \end{aligned}$$

is in the interval $(t,1-\varepsilon _t)$, it is strictly increasing in p, and it is at the midpoint of the interval $[t,1-\varepsilon _t]$ when $p=\frac{1}{2}$. Therefore, since the distance of the interval is at least 2b, for all $p\ge \frac{1}{2}$, we have

$$\begin{aligned} \arg \max _{{\hat{t}}\in \mathbb {R}}\mathbb {E}[U^S_{M(t)}({\hat{t}})]+b\ge t+b\Leftrightarrow \arg \max _{{\hat{t}}\in \mathbb {R}}\mathbb {E}[U^S_{M(t)}({\hat{t}})]\ge t. \end{aligned}$$

Thus, for all $t^*<t$ we have

$$\begin{aligned} \mathbb {E}[U^S_{M(t)}(t^*)]&<-(1-p)\ell (t+b-t)-p\ell (1-\varepsilon _t-t-b)\\&\le -(1-p)\ell (b)-p\ell (b)\\&=-\ell (b), \end{aligned}$$

where the second inequality follows from $1-\varepsilon _t \ge 4b$ and $t\le 2b$. Thus, for $p\ge \frac{1}{2}$ no type $t^*<t$ has a profitable deviation to the message M(t) whenever $t\le 2b$.

Now let’s consider the case $t>2b$. As above, for $t^*<t$ we have

$$\begin{aligned} \mathbb {E}[U^S_{M(t)}(t^*)]=-(1-p)\ell (t^*+b-t)-p\ell (t^*+b). \end{aligned}$$

If $\ell$ is linear, for all $p\ge \frac{1}{2}$ we have

$$\begin{aligned} \mathbb {E}[U^S_{M(t)}(t^*)]&\le \mathbb {E}[U^S_{M(t)}(0)]\\&=-(1-p)(t-b)-pb\\&<-b, \end{aligned}$$

where the second inequality follows from the fact that $t>2b$.

If $\ell$ is strictly convex instead, $\arg \max _{{\hat{t}}}\mathbb {E}[U^S_{M(t)}({\hat{t}})]+b$ is in the set (0, t), it is decreasing in p, and, for $p=\frac{1}{2}$, it is at the midpoint of [0, t]. Set $t^{**}=\max \left\{ 0,\arg \max _{{\hat{t}}}\mathbb {E}[U^S_{M(t)}({\hat{t}})]\right\}$. Thus, for $p\ge \frac{1}{2}$ we have

$$\begin{aligned} \left( t-\left( t^{**}+b\right) \right) >b \end{aligned}$$

since $t>2b$. For all $t^*<t$ we have

$$\begin{aligned} \mathbb {E}[U^S_{M(t)}(t^*)]\le \mathbb {E}[U^S_{M(t)}(t^{**})]&=-(1-p)\ell (t-(t^{**}+b))-p\ell (t^{**}+b)\\&<-(1-p)\ell (b)-p\ell (b)\\&=-\ell (b). \end{aligned}$$

Thus, no type $t^*<t$ has a profitable deviation to the message M(t) whenever $t>2b$. This completes the first part of the proof.

To establish the statement pertaining to the case of quadratic utility, note that the sufficiency follows from Proposition 2 and the first half of this proof. To demonstrate necessity, it suffices to show that there does not exist a FRE for any $p<\frac{1}{2}-\frac{\sqrt{1-16b^2}}{2}$. Toward contradiction, fix a p satisfying the inequality and let (Y, M) be a FRE for that value of p. I will show that regardless of the values of M(1 / 2) and $Y(M(1/2),-1)$ in this strategy profile, the sender type $t^*$, defined by

$$\begin{aligned} t^*:=pY(M(1/2),-1)+(1-p)(1/2)-b, \end{aligned}$$

satisfies $t^*\notin M(1/2)$ and would have an incentive to deviate to the message M(1 / 2).

As we are in a case where $p<1/2$ and $b\le 1/4$, the inequality $t^*\ge 0$ holds, regardless of the value of $Y(M(1/2),-1)$. Note also that a sufficient condition for the inequality $t^*<1/2$ to hold for all $Y(M(1/2),-1)\in [0,1]$ is $p<2b$. This is satisfied because $\frac{1}{2}-\frac{\sqrt{1-16b^2}}{2}\le 2b$ as observed in the proof of Proposition 2. Thus, $t^*\in [0,1/2)$. By Lemma 3, $\inf M(1/2)=1/2$ and so $t^*\notin M(1/2)$. Furthermore, we have

$$\begin{aligned} \mathbb {E}\left[ U^S_{M(1/2)}(t^*)\right] =&-p(pY(M(1/2),-1)+(1-p)(1/2)-Y(M(1/2),-1))^2\\&-(1-p)(pY(M(1/2),-1)+(1-p)(1/2)-1/2)^2\\ =&-p(1-p)(Y(M(1/2),-1)-1/2)^2\\ \ge&-p(1-p)(1/4)\\ >&-b^2, \end{aligned}$$

where the first inequality follows from the fact that $Y(M(1/2),-1)\in [0,1]$ and the second follows from how we established (2). Thus, $t^*$ has a profitable deviation to M(1 / 2), which is the contradiction we need to complete the proof.

$\square$

Proof of Proposition 5

Let (M, Y) be a FRE. The equilibrium payoff for all t is $-\ell (b)$. Consider any announcement $\langle M',X \rangle$. I will show that $\langle M',X \rangle$ cannot satisfy condition C1.

If $m\in \mathcal {M'}$ is a message sent as a part of the announcement, let $M'^{-1}(m)$ denote all types in X that can send m. If $M'^{-1}(m)$ is a singleton for all m, then if $M'^{-1}(m)=\{t\}$ it follows that $Y'(m)=t$ and so $U^S(Y'(m),t)=-\ell (b)$ for all m and $t\in X$. Therefore, the inequality in C1 cannot hold strictly.

If $M'^{-1}(m)$ is not a singleton for some m and $Y'(m)$ is the corresponding best-response action by the receiver, it has to be the case that there are types $t_0,t_1\in M'^{-1}(m)$ such that $t_0<Y'(m)<t_1$. But then

$$\begin{aligned} U^S(Y'(m),t_1)<U^S(Y(M(t_1)),t_1)=U^S(t_1,t_1) \end{aligned}$$

and so the inequality in C1 does not hold for all $t\in X$.

$\square$

Proof of Proposition 6

Consider some message-connected equilibrium with pooling. Namely, let all types between $\underline{t}$ and $\overline{t}$ (with $\overline{t}>\underline{t}$) pool on some message m. Notice that since F is atomless, the probability distributions induced by Bayes’ rule conditional on (m, 0) and on $(m,-1)$ differ only on a set of Lebesgue measure zero⁴⁰ and they, in turn, differ on a zero-measure set from the probability density function over the set of types pooling on m

$$\begin{aligned} g(t)=\frac{F'(t)}{F(\overline{t})-F(\underline{t})}. \end{aligned}$$

So we can treat $g(\cdot )$ as the pdf of the distributions that are induced by the observation of either (m, 0) or $(m,-1)$. I will show that, under either of the proposition’s premises, the receiver’s equilibrium action Y(m, v) is just $\mathbb {E}_g[t]$.

First, if the receiver’s loss function is quadratic, the receiver chooses y to maximize

$$\begin{aligned} \mathbb {E}_g[-(y-t)^2]&=-\int _{\underline{t}}^{\overline{t}}g(t)(y-t)^2dt\\&=-\mathbb {E}_g[t^2]+2y\mathbb {E}_g[t]-y^2. \end{aligned}$$

This function is strictly concave in y so we can maximize it using the first-order condition. It is $2\mathbb {E}_g[t]=2y$. Thus, $y^{\max }=\mathbb {E}_g[t]$.

Second, let’s consider the case of F being uniform. Note then that, in this case, g is (essentially) uniform. In other words, we can assume $g(t)=1/(\overline{t}-\underline{t})$. In this case, the receiver chooses y to maximize

$$\begin{aligned} -\int _{\underline{t}}^{\overline{t}}\ell ^R(|y-t|)dt. \end{aligned}$$

Since $\ell ^R$ is strictly convex, it is not hard to see that the unique maximizer here is

$$\begin{aligned} y^{\max }=\mathbb {E}_g[t]=\frac{\underline{t}+\overline{t}}{2}. \end{aligned}$$

Now consider the expected value of the expected distance between the sender’s bliss action and the equilibrium action conditional on m—i.e. $\mathbb {E}_g[|t+b-\mathbb {E}_g[t]|]$. Letting $t^*:=\max \{\underline{t},\mathbb {E}_g[t]-b\}$, we have

$$\begin{aligned} \int _{\underline{t}}^{\overline{t}}g(t)tdt=&\mathbb {E}_g[t]\\ \Leftrightarrow&b-b+\int _{\underline{t}}^{t^*}g(t)tdt+\int _{t^*}^{\overline{t}}g(t)tdt-\mathbb {E}_g[t]=0\\ \Leftrightarrow&\int _{t^*}^{\overline{t}}g(t)(t+b-\mathbb {E}_g[t])dt-b=\int _{\underline{t}}^{t^*}g(t)(\mathbb {E}_g[t]-t-b)dt. \end{aligned}$$

(I)

Analogously

$$\begin{aligned} E_g[t]&=\int _{\underline{t}}^{t^*}g(t)tdt+\int _{t^*}^{\overline{t}}g(t)tdt\\&\le t^*\int _{\underline{t}}^{t^*}g(t)dt+\int _{t^*}^{\overline{t}}g(t)tdt\\&\Rightarrow \int _{t^*}^{\overline{t}}g(t)tdt\ge \mathbb {E}_g[t]-t^*\int _{\underline{t}}^{t^*}g(t)dt. \end{aligned}$$

(II)

Thus

$$\begin{aligned} E_g[|t+b-\mathbb {E}_g[t]|]=&\int _{\underline{t}}^{\overline{t}}g(t)|t+b-\mathbb {E}_g[t]|dt\\ =&\int _{\underline{t}}^{t^*}g(t)(\mathbb {E}_g[t]-t-b)dt+\int _{t^*}^{\overline{t}}g(t)(t+b-\mathbb {E}_g[t])dt\\ =\,&2\int _{t^*}^{\overline{t}}g(t)(t+b-\mathbb {E}_g[t])dt-b\\ =\,&2\int _{t^*}^{\overline{t}}g(t)tdt+2(b-\mathbb {E}_g[t])\int _{t^*}^{\overline{t}}g(t)dt-b, \end{aligned}$$

(III)

where the third equality follows from (I). If $t^*=\underline{t}$, (III) reduces to

$$\begin{aligned} \mathbb {E}_g[|t+b-\mathbb {E}_g[t]|]&=2\int _{\underline{t}}^{\overline{t}}g(t)tdt+2(b-\mathbb {E}_g[t])\int _{\underline{t}}^{\overline{t}}g(t)dt-b\\&=2\mathbb {E}_g[t]+2(b-\mathbb {E}_g[t])-b\\&=b. \end{aligned}$$

If $t^*=\mathbb {E}_g[t]-b>\underline{t}$, we must have $\mathbb {E}_g[t]-b> \underline{t}\ge 0$ and hence $\mathbb {E}_g[t]> b$. Then (III) becomes

$$\begin{aligned} \mathbb {E}_g[|t+b-\mathbb {E}_g[t]|]=\,&2\int _{t^*}^{\overline{t}}g(t)tdt+2(b-\mathbb {E}_g[t])\int _{t^*}^{\overline{t}}g(t)dt-b\\ \ge \,&{2}\left( \mathbb {E}_g[t]-t^*\int _{\underline{t}}^{t^*}g(t)dt\right) +2(b-\mathbb {E}_g[t])\int _{t^*}^{\overline{t}}g(t)dt-b\\ =\,&2\mathbb {E}_g[t]-2(\mathbb {E}_g[t]-b)(1-P)+2(b-\mathbb {E}_g[t])P-b\\ =\,&2\mathbb {E}_g[t]+2(b-\mathbb {E}_g[t])-b\\ =\,&b, \end{aligned}$$

where the inequality follows from (II) and I have denoted $P:=\int _{t^*}^{\overline{t}}g(t)dt$. Either way, we have $\mathbb {E}_g[|t+b-\mathbb {E}_g[t]|]\ge b$. In other words, the expected distance between the sender’s bliss action and the receiver’s equilibrium action conditional on m is no less than b. Thus, the expected utility of the sender conditional on sending the message m in the equilibrium is

$$\begin{aligned} \mathbb {E}\left[ -\ell \left( \left| t+b-\mathbb {E}_g[t]\right| \right) |m\right] \le -\ell \left( \mathbb {E}_g[|t+b-\mathbb {E}_g[t]|]\right) \le -\ell (b), \end{aligned}$$

where I use the fact that $\ell$ is convex and increasing in its argument’s absolute value.

Notice that if a type t separates by being the only to send a certain message m in equilibrium, we must have $Y(m,0)=Y(m,-1)=t$, and hence that type’s equilibrium utility is $U^S(y,t)=-\ell (b)$. By the Law of Iterated Expectations

$$\begin{aligned} \mathbb {E}[U^S(y,t)]=\mathbb {E}\left[ \mathbb {E}[U^S|m]\right] \le -\ell (b), \end{aligned}$$

because $\mathbb {E}[U^S|m]\le -\ell (b)$ for all equilibrium messages m. The sender’s utility in all FRE is $-\ell (b)$ and this completes the proof. $\square$

Proof of Proposition 7

Assume $p\ge \frac{\ell (b)}{\ell (b)+\varepsilon }$, and consider the following fully revealing strategy profile:

$$\begin{aligned} M(t) = t \text{ and } Y(m,v) = {\left\{ \begin{array}{ll} m &{}\text {if }v=0,\\ 0 &{}\text {if }v=-1. \end{array}\right. } \end{aligned}$$

It is clear Y is a best response to M. To establish that this is an equilibrium, we only need to verify that the sender does not have profitable deviation. As $Y(m,-1)=0$ regardless of m, a sender of type t maximizes her expected utility from a deviation by sending the message $m=t+b$. The expected utility then is:

$$\begin{aligned} -p(\ell (t+b)+\varepsilon )\le -p(\ell (b)+\varepsilon )\le -\frac{\ell (b)}{\ell (b)+\varepsilon }(\ell (b)+\varepsilon )=-\ell (b), \end{aligned}$$

which is the utility from truth-telling.

Finally, note that the receiver need not choose $Y(0,-1)=0$ in FRE: it is easy to show that as long as $\ell (b-Y(0,-1))+\varepsilon >\ell (b)$, no agent would deviate to the message $m=0$. $\square$

Proof of Proposition 8

Abusing notation, extend the domain of the function $b(\cdot )$ by setting $b(t)=b(0)$ for $t<0$ and $b(t)=b(1)$ for $t>1$. With this modification in place, showing that there exists a FRE for values of p around 1 / 2 is identical to the first part of the proof of Proposition 3.

Now I show that there does not exist a FRE for all $p\in (0,\varepsilon )\cup (1-\varepsilon ,1)$ for some $\varepsilon >0$. Let $m=b(0)$ and consider the off-equilibrium actions $\{Y(m,-1)\}_{m\in \mathcal {M}}$ for some candidate FRE. If $Y(m,-1)\le b(0)$, it is easy to see that type $t=0$ prefers sending the message m over truth-telling. If $Y(m,-1)>b(0)$, by continuity of $t+b(t)$, which maps onto $[b(0),1+b(1)]$ for $t\in [0,1]$, we can find some $t^*$ such that $t^*+b(t^*)=Y(m,-1)$. Then

$$\begin{aligned} \mathbb {E}\left[ U^S_m(t^*)\right] =-(1-p)\ell (Y(m,-1)-m). \end{aligned}$$

Since $Y(m,-1)$ is bounded, we have $\lim _{p\rightarrow 1}\left[ -(1-p)\ell (Y(m,-1)-m)\right] =0$. Since $b(\cdot )$ is a positive continuous function on the compact domain $\mathcal {T}$, it is bounded away from zero. It is clear then that for all sufficiently high p, the inequality

$$\begin{aligned} \mathbb {E}\left[ U^S_m(t^*)\right] > -\min _t\ell (b(t)) \end{aligned}$$

holds. Therefore, for sufficiently high p type $t^*$ prefers lying over truth-telling. Either way, the off-equilibrium actions cannot be part of a FRE. Therefore, there does not exist one.

For small p, note that

$$\begin{aligned} \mathbb {E}\left[ U^S_{b(0)}(0)\right] =-p\ell (b(0)-Y(b(0),-1)). \end{aligned}$$

The proof then proceeds analogously.

$\square$