Confidence and competence in communication

Abstract

This paper studies information transmission between an uninformed decision maker (receiver) and an informed agent (sender) who have asymmetric beliefs (“confidence”) on the sender’s ability (“competence”) to observe the state of nature. We find that even when the material payoffs of the players are perfectly aligned, the sender’s over- and underconfidence on his information give rise to information loss in communication, although they do not by themselves completely eliminate information transmission in equilibrium. However, an underconfident sender may prefer no communication to informative communication. We also show that when the sender is biased, overconfidence can lead to more information transmission and welfare improvement.

Appendix: Proofs

1.1 Proposition 1

Proof

Part (i): This part of the proposition for the case where $\hat{\sigma }\notin [0,1]$ and $c\ge p$ is implied by Theorem 1 of CS, since $y^\mathrm{S}(\sigma )>y^\mathrm{R}(\sigma )$ for all $\sigma$.

For $c<p$, consider a partition satisfying (6) and (7). The roots are complex and any equilibrium partition must be an increasing sequence $0=a_{0}<a_{1}<\cdots <a_{N}=1$ given by

$$\begin{aligned} a_{i}=\hat{\sigma }+A_{N}\cos (i\rho +\omega _{N})\text {,} \end{aligned}$$

(12)

where $\cos \rho =(2c-p)/p$ and $\hat{\sigma }$ is the particular solution. The constants $A_{N}$ and $\omega _{N}$ are determined by (7), that is,

$$\begin{aligned}&\displaystyle \hat{\sigma }+A_{N}\cos (\omega _{N}) = 0 \end{aligned}$$

(13)

$$\begin{aligned}&\displaystyle \hat{\sigma }+A_{N}\cos (\omega _{N}+N\rho ) = 1\text {.} \end{aligned}$$

(14)

Combining (13) and (14) we have

$$\begin{aligned} \frac{\cos (\omega _{N})}{\cos (\omega _{N}+N\rho )}=-\frac{\hat{\sigma }}{1- \hat{\sigma }}\text {.} \end{aligned}$$

(15)

In order for (12) to be increasing in $i$ we must have

$$\begin{aligned} \omega _{N}\in \left( \pi -\rho /2,2\pi +\rho /2-N\rho \right) \end{aligned}$$

(16)

since, as the cosine function is strictly increasing between $\pi$ and $2\pi$ and symmetric around the extrema, for $a_{0}<a_{1}$ we need $\pi <\rho /2+\omega _{N}$; and for $a_{N-1}<a_{N}$ we need $2\pi >(N-1)\rho +\rho /2+\omega _{N}$. Any equilibrium must satisfy (15) and (16). Since (16) is empty for large $N$, the number of intervals that can be supported in equilibrium is finite.

The range of the left-hand side of (15) that also satisfies (16), when (16) is nonempty, is given by

$$\begin{aligned} \frac{\cos (\omega _{N})}{\cos (\omega _{N}+N\rho )}\in \left( \frac{\cos (\pi -\rho /2)}{\cos (\pi -\rho /2+N\rho )},\frac{\cos (2\pi +\rho /2-N\rho ) }{\cos (2\pi +\rho /2)}\right) . \end{aligned}$$

The infimum of the range is negative and nondecreasing in $N$, and the supremum is nonincreasing in $N$. Thus if there exists $\omega _{N}$ that satisfies (15) and (16) for $N$, then there also exists $\omega _{N-1}$ that satisfies (15) and (16) for $N-1$, which implies the equilibrium with $N-1$ intervals.

Part (ii) From (2) and (3), for $c\ge p$ and $\hat{\sigma } \in \left[ 0,1\right]$, the sender has an outward bias in Gordon’s (2010) terms, which means the sender’s ideal action given the signal is higher than that of the receiver if for $\sigma >\hat{\sigma }$ and lower for $\sigma <\hat{\sigma }$. (2) and (3) also imply the sender’s preference is continuous in $y$, and the receiver’s ideal policy is continuous in $\sigma$. Thus, we can directly apply Theorem 4 of Gordon (2010) to establish that there exists an informative equilibrium with an infinite number of intervals. Theorem 2 of Gordon (2010) proves that, more generally, if there is an infinite equilibrium then there exists an equilibrium with $N$ intervals for every positive integer $N$. Hence Part (ii) of Proposition 1 holds. $\square$

1.2 Proposition 2

Proof

Let us first show that the most informative equilibrium with an extremely overconfident sender converges to the binary partition. Solving (6) explicitly for $c>p$ with $a_{0}=0$ and $a_{N}=1$, we obtain

$$\begin{aligned} a_{i}=\hat{\sigma }+\frac{1-\hat{\sigma }(1-B^{N})}{A^{N}-B^{N}}A^{i}+\frac{-1+ \hat{\sigma }(1-A^{N})}{A^{N}-B^{N}}B^{i} \end{aligned}$$

(17)

and equivalently with $a_{0}^{\prime }=1$ and $a_{N}^{\prime }=0$

$$\begin{aligned} a_{i}^{\prime }=\hat{\sigma }+\frac{-B^{N}-\hat{\sigma }(1-B^{N})}{A^{N}-B^{N}} A^{i}+\frac{A^{N}+\hat{\sigma }(1-A^{N})}{A^{N}-B^{N}}B^{i}, \end{aligned}$$

where $A$ and $B$ are two distinct roots

$$\begin{aligned} A=\frac{2c-p+2\sqrt{c(c-p)}}{p}\quad \text { and } \quad B=\frac{2c-p-2\sqrt{c(c-p)}}{p}. \end{aligned}$$

(18)

Solving (17) for $N\rightarrow \infty$ we obtain two sequences

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l} a_{i}=\hat{\sigma }-\hat{\sigma }B^{i} &{} \text {for }a_{i}\in [0,\hat{ \sigma }), \\ a_{i}^{\prime }=\hat{\sigma }+(1-\hat{\sigma })B^{i} &{} \text {for } a_{i}^{\prime }\in (\hat{\sigma },1]\text { such that }a_{0}^{\prime }=1. \end{array} \right. \end{aligned}$$

Since $0<B<1$, both sequences converge to $\hat{\sigma }$. If we fix $c=1$ and $p\rightarrow 0$, then $B\rightarrow 0$. Hence for $b=0$ both $a_{1}$ and $a_{1}^{\prime }$ converge to $\hat{\sigma }=1/2$, which implies that the infinite partition equilibrium converges to the binary equilibrium.

For $c<p$, solving (6) with $b=0$, $a_{0}=0$, and $a_{3}=1$ (equilibrium with three intervals), we obtain $a_{1}=\frac{2c-p}{4c-p}$ and $a_{2}=\frac{2c}{4c-p}$. We have $a_{1}\le 0$ or equivalently $a_{2}\ge 1$ if $c\le \frac{p}{2}$, which implies the equilibrium with three intervals does not exist and from Proposition 1 no equilibrium with three or more intervals exists. Thus, we conclude that the binary equilibrium with partition $\{[0,1/2),[1/2,1]\}$ is the only informative equilibrium for $c\le \frac{p}{2}$. $\square$

1.3 Proposition 3

Before we prove the proposition, we provide some useful lemmas and outline how we construct the main proof. Let us call a sequence $(a_{0},a_{1},\ldots ,a_{N})$ that satisfies the arbitrage condition (6) a “solution” to (6). The monotonicity condition (M) in CS requires that, for given $b$, $c$, and $p$, if we have two solutions $a^{+}$ and $a^{++}$ with $a_{0}^{+}=a_{0}^{++}$ and $a_{1}^{+}>a_{1}^{++}$, then $a_{i}^{+}>a_{i}^{++}$ for all $i=2,3,\ldots$. In other words, (M) says that starting from $a_{0}$, all solutions to (6) must move up or down together. Also, solving (6) in terms of $a_{1}$

$$\begin{aligned} a_{i}=\hat{\sigma }+\frac{a_{1}-\hat{\sigma }(1-B)}{A-B}A^{i}-\frac{a_{1}-\hat{ \sigma }(1-A)}{A-B}B^{i}, \end{aligned}$$

where $A$ and $B$ are the distinct roots in (18). Since $\mathrm{d}a_{i}/\mathrm{d}a_{1}>0$ for all $i=2,3,\ldots ,N$ we can see that the equilibrium partition of our model satisfies (M) for $c\ge p$.

In order to show that the expected utilities of both players are higher in an equilibrium with more intervals, CS deform the partition with $N$ intervals to that with $N+1$ intervals, continuously increasing the player’s expected utility throughout the deformation. We follow this method, but we need to proceed by two-step deformation, rather than one, because when $\hat{ \sigma }\in (0,1]$ the deformation takes place towards the opposite directions for the right-hand and left-hand sides of $\hat{\sigma }$. Intuitively, as the number of interval increases, each boundary type on the left-hand side of $\hat{\sigma }$ move to the left (except for $a_{0}=0$) while each boundary type of the right-hand side of $\hat{\sigma }$ move to the right (except for $a_{N}=1$). We need to perform a different comparative statics for each case.

Let $a(N)$ be the equilibrium partition of size $N$. We show that $a(N)$ can be deformed to $a(N+1)$ by two steps, continuously increasing the players’ expected utility in each step. Here, we consider the case where $\hat{\sigma } \in (0,1]$. We omit the case where $\hat{\sigma }\notin (0,1]$ because the proposition for this case can be proven similarly, by using the first step only.

Let the sub-partition of $a(N)$ equal or below $\hat{\sigma }$ be $\underline{ a}(N)\equiv (a_{0}(N),a_{2}(N),\ldots ,a_{K}(N))$ where $a_{0}(N)=0$. Also, suppose that $a_{K}(N)$ is closer to $\hat{\sigma }$ than $a_{K+1}(N)$ is, in other words, $\hat{\sigma }-a_{K}(N)<a_{K+1}(N)-\hat{\sigma }$. In the following, we proceed in two steps:⁹

• 1. We fix $a_{K}(N)$ and make the sub-partition $(a_{K}(N),a_{K+1}(N),\ldots ,a_{N}(N))$ deform continuously to $(a_{K}(N),a_{K+1}(N+1),a_{K+2}(N+1),\ldots ,a_{N+1}(N+1))$, increasing the expected utility.

• 2. We make the sub-partition $(a_{0}(N),a_{1}(N),\ldots ,a_{K}(N))$ deform continuously to $(a_{0}(N+1),a_{2}(N+1),\ldots ,a_{K}(N+1))$, increasing the expected utility.

• If $\hat{\sigma }-a_{K}(N)\ge a_{K+1}(N)-\hat{\sigma }$ then the first step deforms $(a_{0}(N),a_{1}(N),\ldots ,a_{K}(N),a_{K+1}(N))$ to $(a_{0}(N+1),a_{1}(N+1),\ldots ,a_{K+1}(N+1),a_{K+1}(N))$ while fixing $a_{K+1}(N)$, and the second step deforms $(a_{K+1}(N),a_{K+2}(N),\ldots ,a_{N}(N))$ to $(a_{K+2}(N+1),a_{K+3}(N+1),\ldots ,a_{N+1}(N+1))$. Except for this, the same method and result as the case where $\hat{\sigma }-a_{K}(N)<a_{K+1}(N)-\hat{ \sigma }$ apply.

Lemma 1

If $a(N)$ and $a(N+1)$ are two equilibrium partitions for the same values of $b$, $c$, and $p$, then $a_{i-1}(N)<a_{i}(N+1)<a_{i}(N).$

Proof

See Lemma 3 (p. 1446) in CS. The proof follows directly from (M). $\square$

The first step of deformation is carried out as follows. Let $(a_{K}^{x},a_{K+1}^{x},\ldots ,a_{i}^{x},\ldots ,a_{N+1}^{x})$ be the sub-partition that satisfies (6) for all $i=K+1,K+2,\ldots ,N$ with $a_{K}^{x}=a_{K}(N)$, $a_{N}^{x}=x$ and $a_{N+1}^{x}=1$. If $x=a_{N-1}(N)$ then $a_{K+1}^{x}=a_{K}^{x}=a_{K}(N).$ If $x=a_{N}(N+1)$ then we have $(a_{K}(N),a_{K+1}(N+1),\ldots ,a_{N}(N+1))$, where (6) is satisfied for all $i=K+2,K+3,\ldots ,N$. We will show that, if $x\in [a_{N-1}(N), a_{N}(N+1)]$, which is again a non-degenerate interval by Lemma 1, then the sender’s expected utility is strictly increasing in $x$.

In the second step, let $(a_{0}^{z}, a_{1}^{z},\ldots ,a_{i}^{z},\ldots ,a_{K}^{z})$ be the sub-partition that satisfies (6) for $i=1,2,\ldots ,K-1,$ with $a_{0}^{z}=0$ and $a_{K}^{z}=z$. If $z=a_{K}(N)$ then $a_{i}^{z}=a_{i}(N)$ for all $i=0,1,\ldots ,K$. If $z=a_{K}(N+1)$ then $a_{i}^{z}=a_{i}(N+1)$ for all $i=0,1,\ldots ,K$. We show that when $z\in [a_{K}(N+1),a_{K}(N)]$, which is again a non-degenerate interval by Lemma 1, the sender’s expected utility is strictly decreasing in $z$.

Lemma 2

Suppose that $(a_{0},a_{1},\ldots ,a_{i},\ldots ,a_{N})$ is a solution to (6). Then for all $i=1,2,\ldots ,N-1$ if $a_{i}>(<)\hat{ \sigma }$ then $a_{i}-a_{i-1}<a_{i+1}-a_{i}(a_{i}-a_{i-1}>a_{i+1}-a_{i})$. If $a_{i}=\hat{\sigma }$ then $a_{i}-a_{i-1}=a_{i+1}-a_{i}.$

Proof

Rearranging (6) we have

$$\begin{aligned} (a_{i+1}-a_{i})-(a_{i}-a_{i-1})=\frac{4ca_{i}+4b-2c}{p}-4a_{i}+2. \end{aligned}$$

(19)

The left-hand side $(a_{i+1}-a_{i})-(a_{i}-a_{i-1})=0$ if

$$\begin{aligned} \frac{4ca_{i}+4b-2c}{p}-4a_{i}+2&= 0\Rightarrow \\ 4a_{i}(c-p)&= -4b+2c-2p\Rightarrow \\ a_{i}&= \frac{1}{2}-\frac{b}{c-p}\equiv \hat{\sigma }. \end{aligned}$$

Since the right-hand side of (19) is increasing in $a_{i}$ for $c>p$. Thus if $a_{i}>\hat{\sigma }$ then $(a_{i+1}-a_{i})-(a_{i}-a_{i-1})>0,$ and if $a_{i}<\hat{\sigma }$ then $(a_{i+1}-a_{i})-(a_{i}-a_{i-1})<0$.

$\square$

The above lemma says that an interval $[a_{i},a_{i+1})$ is longer (shorter) than the previous interval $[a_{i-1},a_{i})$ when $a_{i}>(<)\hat{\sigma }$. The following Lemma is similar but cannot be implied by Lemma 2. Since by definition $a_{K}^{x}$ and $a_{K+1}^{z}$ are fixed throughout the respective deformation, (6) is not satisfied at $a_{i}=a_{K+1}^{x}$ for $x\in (a_{N-1}(N), a_{N}(N+1))$ or $a_{i}=a_{K}^{z}$ for $z\in (a_{K}(N+1),a_{K}(N))$.

Lemma 3

$a_{K+1}^{x}-a_{K}^{x}<a_{K+2}^{x}-a_{K+1}^{x}$ and $a_{K}^{z}-a_{K-1}^{z}>a_{K+1}^{z}-a_{K}^{z}$.

Proof

Suppose $a_{K+1}^{x}\le \hat{\sigma }$. By construction $\hat{\sigma } -a_{K}(N)<a_{K+1}(N)-\hat{\sigma }$. Since $a_{K}^{x}=a_{K}(N)$ and $a_{K+2}^{x}>a_{K+1}(N)$, we have $\hat{\sigma }-a_{K}^{x}<a_{K+2}^{x}-\hat{ \sigma }$ and hence $a_{K+1}^{x}-a_{K}^{x}<a_{K+2}^{x}-a_{K+1}^{x}$ as stated. Suppose $a_{K+1}^{x}>\hat{\sigma }.$ From Lemma 2, we have $a_{K+1}^{x}-\tilde{a}_{K}<a_{K+2}^{x}-a_{K+1}^{x}$ where $\tilde{a} _{K}$ is defined such that $\{a_{i-1}=\tilde{a}_{K}, a_{i}=a_{K+1}^{x},a_{i+1}=a_{K+2}^{x}\}$ satisfies (6). Since $a_{K}(N+1)<\tilde{a}_{K}<a_{K}(N)=a_{K}^{x}$ from Lemma 1, we have $a_{K+1}^{x}-a_{K}^{x}<a_{K+2}^{x}-a_{K+1}^{x}$. This proves the first part of the Lemma.

Similarly, we have $a_{K}^{z}-a_{K-1}^{z}\ge \check{a}_{K+1}-a_{K}^{z}$ where $\check{a}_{K+1}$ is defined such that $\{a_{i-1}=a_{K-1}^{z},a_{i}=a_{K}^{z},a_{i+1}=\check{a}_{K+1}\}$ satisfies (6). Lemma 1 implies $a_{K+1}^{z}=a_{K+1}(N+1)<\check{a} _{K+1}<a_{K+1}(N)$. Hence we have $a_{K}^{z}-a_{K-1}^{z}>a_{K+1}^{z}-a_{K}^{z}$. $\square$

Proof of Proposition 3

The sender’s expected utility for the first part of deformation is given by

$$\begin{aligned} EU^\mathrm{S}&\equiv -c\left[ \sum _{i=1}^{K}\,\,\int \limits _{a_{i-1}}^{a_{i}}\left( p\frac{ a_{i-1}+a_{i}}{2}+\frac{1-p}{2}-b-\theta \right) ^{2}\mathrm{d}\theta \right. \\&\left. \quad +\sum _{i=K+1}^{N+1}\,\,\int \limits _{a_{i-1}^{x}}^{a_{i}^{x}}\left( p\frac{ a_{i-1}^{x}+a_{i}^{x}}{2}+\frac{1-p}{2}-b-\theta \right) ^{2}\mathrm{d}\theta \right] \\&\quad -(1-c)\left[ \sum _{i=1}^{K}(a_{i}-a_{i-1})\int \limits _{0}^{1}\left( p\frac{ a_{i-1}+a_{i}}{2}+\frac{1-p}{2}-b-\theta \right) ^{2}\mathrm{d}\theta \right. \\&\quad +\left. \sum _{i=K+1}^{N+1}(a_{i}^{x}-a_{i-1}^{x})\int \limits _{0}^{1}\left( p\frac{ a_{i-1}^{x}+a_{i}^{x}}{2}+\frac{1-p}{2}-b-\theta \right) ^{2}\mathrm{d}\theta \right] . \end{aligned}$$

It follows that

$$\begin{aligned} \frac{\mathrm{d}EU^\mathrm{S}}{\mathrm{d}x}&\equiv \sum _{i=K+1}^{N+1}\frac{\mathrm{d}a_{i}^{x}}{\mathrm{d}x} \left\{ -c\left[ \left( p\frac{a_{i-1}^{x}+a_{i}^{x}}{2}+\frac{1-p}{2} -b-a_{i}^{x}\right) ^{2}\right. \right. \nonumber \\ \nonumber&\left. \left. \qquad \qquad \qquad \qquad -\,\left( p\frac{a_{i}^{x}+a_{i+1}^{x}}{2}+\frac{1-p}{2 }-b-a_{i}^{x}\right) ^{2}\right. \right. \\ \nonumber&\left. \qquad \qquad \qquad \qquad +\,p\int \limits _{a_{i-1}^{x}}^{a_{i}^{x}}\left( p\frac{a_{i-1}^{x}+a_{i}^{x} }{2}+\frac{1-p}{2}-b-\theta \right) \mathrm{d}\theta \right. \\ \nonumber&\left. \qquad \qquad \qquad \qquad +\,p\int \limits _{a_{i}^{x}}^{a_{i+1}^{x}}\left( p\frac{a_{i}^{x}+a_{i+1}^{x}}{2}+ \frac{1-p}{2}-b-\theta \right) \mathrm{d}\theta \right] \\ \nonumber&\quad -\,(1-c)\left[ \int \limits _{0}^{1}\left( p\frac{a_{i-1}^{x}+a_{i}^{x}}{2}+\frac{1-p }{2}-b-\theta \right) ^{2}\mathrm{d}\theta \right. \\ \nonumber&\left. \qquad \qquad \qquad \qquad -\,\int \limits _{0}^{1}\left( p\frac{ a_{i}^{x}+a_{i+1}^{x}}{2}+\frac{1-p}{2}-b-\theta \right) ^{2}\mathrm{d}\theta \right. \\ \nonumber&\qquad \qquad \qquad \qquad +\,p(a_{i}^{x}-a_{i-1}^{x})\int \limits _{0}^{1}\left( p\frac{a_{i-1}^{x}\!+\!a_{i}^{x}}{2 }\!+\!\frac{1-p}{2}\!-\!b\!-\!\theta \right) \mathrm{d}\theta \\ \nonumber&\left. \left. \qquad \qquad \qquad \qquad +\,p(a_{i+1}^{x}-a_{i}^{x})\int \limits _{0}^{1}\left( p\frac{ a_{i}^{x}+a_{i+1}^{x}}{2}\!+\!\frac{1-p}{2}\!-\!b\!-\!\theta \right) \mathrm{d}\theta \right] \right\} \\&= \sum _{i=K+1}^{N-1}\frac{\mathrm{d}a_{i}^{x}}{\mathrm{d}x}\frac{ (a_{i+1}^{x}-a_{i}^{x})(2c-p)p}{4}(a_{i-1}^{x}-2a_{i}^{x}+a_{i+1}^{x})>0. \end{aligned}$$

(20)

The inequality follows since $\frac{\mathrm{d}a_{i}^{x}}{\mathrm{d}x}>0$ and from Lemmas 2 and 3 we have $a_{i}^{x}-a_{i-1}^{x}<a_{i+1}^{x}-a_{i}^{x}\Rightarrow a_{i-1}^{x}-2a_{i}^{x}-a_{i+1}^{x}>0$ for $i=1,2,\ldots ,N-1$.

Let us look at the second part of deformation.

$$\begin{aligned} \frac{\mathrm{d}EU^\mathrm{S}}{\mathrm{d}z}&\equiv \sum _{i=1}^{K}\frac{\mathrm{d}a_{i}^{z}}{\mathrm{d}z}\left\{ -c\left[ \left( p\frac{a_{i-1}^{z}+a_{i}^{z}}{2}+\frac{1-p}{2} -b-a_{i}^{z}\right) ^{2}\right. \right. \nonumber \\ \nonumber&\left. \left. \qquad \qquad \qquad \qquad -\,\left( p\frac{a_{i}^{z}+a_{i+1}^{z}}{2}+\frac{1-p}{2 }-b-a_{i}^{z}\right) ^{2}\right. \right. \\ \nonumber&\left. \qquad \qquad \qquad \qquad +p\int \limits _{a_{i-1}^{z}}^{a_{i}^{z}}\left( p\frac{a_{i-1}^{z}+a_{i}^{z} }{2}+\frac{1-p}{2}-b-\theta \right) \mathrm{d}\theta \right. \\ \nonumber&\left. \qquad \qquad \qquad \qquad +\,p\int \limits _{a_{i}^{z}}^{a_{i+1}^{z}}\left( p\frac{a_{i}^{z}+a_{i+1}^{z}}{2}+ \frac{1-p}{2}-b-\theta \right) \mathrm{d}\theta \right] \\ \nonumber&\quad -\,(1-c)\left[ \int \limits _{0}^{1}\left( p\frac{a_{i-1}^{z}+a_{i}^{z}}{2}+\frac{1-p }{2}-b-\theta \right) ^{2}\mathrm{d}\theta \right. \\ \nonumber&\left. \qquad \qquad \qquad \qquad -\,\int \limits _{0}^{1}\left( p\frac{ a_{i}^{z}+a_{i+1}^{z}}{2}+\frac{1-p}{2}-b-\theta \right) ^{2}\mathrm{d}\theta \right. \\ \nonumber&\qquad \qquad \qquad \qquad +\,p(a_{i}^{z}-a_{i-1}^{z})\int \limits _{0}^{1}\left( p\frac{a_{i-1}^{z}+a_{i}^{z}}{2 }+\frac{1-p}{2}-b-\theta \right) \mathrm{d}\theta \\ \nonumber&\qquad \qquad \qquad \qquad \left. \left. +\,p(a_{i+1}^{z}-a_{i}^{z})\int \limits _{0}^{1}\left( p\frac{ a_{i}^{z}+a_{i+1}^{z}}{2}\!+\!\frac{1-p}{2}\!-\!b\!-\!\theta \right) \mathrm{d}\theta \right] \right\} \\&= \sum _{i=1}^{K}\frac{\mathrm{d}a_{i}^{z}}{\mathrm{d}z}\frac{(a_{i+1}^{z}-a_{i}^{z})(2c-p)p}{4 }(a_{i-1}^{z}-2a_{i}^{z}+a_{i+1}^{z})<0. \end{aligned}$$

(21)

The inequality follows since we have $\frac{da_{i}^{z}}{dz}>0$ and Lemmas 2 and 3 imply $a_{i}^{z}-a_{i-1}^{z}>a_{i+1}^{z}-a_{i}^{z}\Rightarrow a_{i-1}^{z}-2a_{i}^{z}+a_{i+1}^{z}<0$. Therefore, $EU^\mathrm{S}$ is increasing throughout the second part of deformation for which $z$ decreases from $a_{K}(N)$ to $a_{K}(N+1)$.

Since we have completed the deformation from $a(N)$ to $a(N+1)$ by two steps while increasing the expected utility, we conclude that the sender’s expected utility is higher in an equilibrium with more intervals.

Following the above two-step deformation, the receiver’s expected utility for the first part of deformation is given by

$$\begin{aligned} EU^\mathrm{R}&\equiv -p\left[ \sum _{i=1}^{K}\int \limits _{a_{i-1}}^{a_{i}}\left( p\frac{ a_{i-1}+a_{i}}{2}+\frac{1-p}{2}-\theta \right) ^{2}\mathrm{d}\theta \right. \\&\left. \qquad \qquad +\sum _{i=K+1}^{N+1}\int \limits _{a_{i-1}^{x}}^{a_{i}^{x}}\left( p\frac{ a_{i-1}^{x}+a_{i}^{x}}{2}+\frac{1-p}{2}-\theta \right) ^{2}\mathrm{d}\theta \right] \\&-(1-p)\left[ \sum _{i=1}^{K}(a_{i}-a_{i-1})\int \limits _{0}^{1}\left( p\frac{ a_{i-1}+a_{i}}{2}+\frac{1-p}{2}-\theta \right) ^{2}\mathrm{d}\theta \right. \\&\qquad \qquad \qquad +\left. \sum _{i=K+1}^{N+1}(a_{i}^{x}-a_{i-1}^{x})\int \limits _{0}^{1}\left( p\frac{ a_{i-1}^{x}+a_{i}^{x}}{2}+\frac{1-p}{2}-\theta \right) ^{2}\mathrm{d}\theta \right] . \end{aligned}$$

Note that the expected utility is identical to that of the sender, except that $b=0$ and $c=p$. Therefore, in order to show that the receiver’s expected utility is higher in an equilibrium with more intervals, we can directly apply the argument we have used for the sender’s expected utility. $\square$

1.4 Proposition 4

Proof

Differentiating (17) with respect to $\hat{\sigma }$, we have

$$\begin{aligned} \underset{c\rightarrow p}{\lim }\frac{\mathrm{d}a_{i}}{\mathrm{d}c}&= \frac{\overset{-}{ \overbrace{i(i-N)}}[-2bN(1-i^{2}+iN+N^{2})+(2i-N)p]}{\underset{+}{ \underbrace{3Np^{2}}}}>0\nonumber \\&\text {for }i=1,2,\ldots ,N-1. \end{aligned}$$

(22)

For the inequality to follow, the term in the square brackets must be negative

$$\begin{aligned} \underset{-}{\underbrace{-2bN(1-i^{2}+iN+N^{2})}}+(2i-N)p<0. \end{aligned}$$

If this inequality holds for $i=N-1$ (so that the second term on the left-hand side is the largest), it is also satisfied for all $i=1,2,\ldots ,N-1$. Thus substituting $i=N-1$ we obtain

$$\begin{aligned} b>\frac{(N-2)p}{2N^{2}(N+1)}. \end{aligned}$$

(23)

The assumption that the equilibrium with $N$ intervals is the most informative equilibrium implies

$$\begin{aligned} \underset{\text {Sup of bias for eqm with }N\text { intervals.}}{\underbrace{ \frac{p}{2N(N-1)}}}>b>\underset{\text {Sup of bias for eqm with }N+1\text { intervals}}{\underbrace{\frac{p}{2N(N+1)}}} \end{aligned}$$

(24)

and it is easy to check that

$$\begin{aligned} \frac{p}{2N(N+1)}>\frac{(N-2)p}{2N^{2}(N+1)}. \end{aligned}$$

This implies that in the most informative equilibrium (23) and hence (22) holds.

The receiver’s expected utility is given by

$$\begin{aligned} EU^\mathrm{R}&= -p\sum _{i=1}^{N}\int \limits _{a_{i-1}}^{a_{i}}\left( p\frac{a_{i-1}+a_{i}}{ 2}+\frac{1-p}{2}-\theta \right) ^{2}\mathrm{d}\theta \\&-(1-p)\sum _{i=1}^{N}(a_{i}-a_{i-1})\int \limits _{0}^{1}\left( p\frac{a_{i-1}+a_{i} }{2}+\frac{1-p}{2}-\theta \right) ^{2}\mathrm{d}\theta \end{aligned}$$

Fix the level of $p$ and let $a(N,\bar{c})$ be the partition with $N$ non-degenerate intervals and $p=c=\bar{c}$. Since $b>0$ any informative equilibrium has a finite number of intervals (Proposition 1). By continuity we can construct the equilibrium partition with $N$ intervals with $c=\bar{c}+\epsilon$ for small enough $\epsilon$, which we denote by $a(N,\bar{c}+\epsilon )$.

Now we can deform $a(N,\bar{c})$ into $a(N,\bar{c}+\epsilon )$, increasing the receiver’s expected utility throughout the deformation, as we have done in (20). $\quad \square$