Intentional Vagueness

Abstract

This paper analyzes communication with a language that is vague in the sense that identical messages do not always result in identical interpretations. It is shown that strategic agents frequently add to this vagueness by being intentionally vague, i.e. they deliberately choose less precise messages than they have to among the ones available to them in equilibrium. Having to communicate with a vague language can be welfare enhancing because it mitigates conflict. In equilibria that satisfy a dynamic stability condition intentional vagueness increases with the degree of conflict between sender and receiver.

Appendices

Appendix 1: Proofs

We start with some preliminaries. As in Sect. 2.4, we define an expectation function α, which gives the expected value of the sender’s type if she sends message m ₀ when t = 0 and message m ₁ when t = 1:

$$ \alpha \left(q,m_{0},m_{1},\theta ,\sigma \right) \equiv \frac{\theta \cdot \phi _{m_{1},\sigma^{2}}\left(q\right) }{\left( 1-\theta \right) \cdot \phi _{m_{0},\sigma^{2}}\left(q\right) +\theta \cdot \phi _{m_{1},\sigma^{2}}\left(q\right) }. $$

This function has 180° rotational symmetry; the following lemma describes this symmetry in the special case where m ₁ = 1.

Lemma 1:

Suppose m ₀ ≠ 1. Then the action function α satisfies the following symmetry property:

$$ \alpha \left(q^{\ast }+x,m_{0},1,\theta ,\sigma \right) -\frac{1}{2}=\frac{1 }{2}-\alpha \left(q^{\ast }-x,m_{0},1,\theta ,\sigma \right) \quad \forall x\in {\mathbb{R}}. $$

where

$$ q^{\ast }=\sigma^{2}\frac{\ln \left(\frac{1-\theta }{\theta }\right) }{ 1-m_{0}}+\frac{1+m_{0}}{2}. $$

Proof

Notice that the condition

$$ \alpha \left(q^{\ast }+x,m_{0},1,\theta ,\sigma \right) -\frac{1}{2}=\frac{1 }{2}-\alpha \left(q^{\ast }-x,m_{0},\theta ,\sigma \right) \quad \forall x\in {\mathbb{R}} $$

is equivalent to

$$ \begin{aligned} &\frac{\theta \cdot \phi _{1,\sigma^{2}}\left(q\right) \left( q^{\ast }+x\right) }{\theta \cdot \phi _{1,\sigma^{2}}\left(q^{\ast }+x\right) +\left(1-\theta \right) \cdot \phi _{m_{0},\sigma^{2}}\left(q^{\ast }+x\right) } \\ &\quad =1-\frac{\theta \cdot \phi _{1,\sigma^{2}}\left( q^{\ast }-x\right) }{\theta \cdot \phi _{1,\sigma^{2}}(q^{\ast }-x)+(1-\theta )\cdot \phi _{m_{0},\sigma^{2}}(q^{\ast }-x)}\quad \forall x\in {\mathbb{R}} \\ &\quad \Leftrightarrow \frac{1}{1+\frac{1-\theta }{\theta }\frac{\phi _{m_{0},\sigma^{2}}\left(q^{\ast }+x\right) }{\phi _{1,\sigma^{2}}\left(q^{\ast }+x\right) }}=\frac{1}{1+\frac{\theta }{1-\theta }\frac{\phi _{1,\sigma^{2}}\left(q^{\ast }-x\right) }{\phi _{m_{0},\sigma^{2}}\left(q^{\ast }-x\right) }}\quad \forall x\in {\mathbb{R}} \\ &\quad \Leftrightarrow \frac{1-\theta }{\theta }\frac{\phi _{m_{0},\sigma^{2}}\left(q^{\ast }+x\right) }{\phi _{1,\sigma^{2}}\left(q^{\ast }+x\right) }=\frac{\theta }{1-\theta }\frac{\phi _{1,\sigma^{2}}\left(q^{\ast }-x\right) }{\phi _{m_{0},\sigma^{2}}\left(q^{\ast }-x\right) } \quad \forall x\in {\mathbb{R}} \\ &\quad\Leftrightarrow \frac{1-\theta }{\theta }\frac{e^{-\frac{\left(q^{\ast }+x-m_{0}\right)^{2}}{2\sigma^{2}}}}{e^{-\frac{\left(q^{\ast }+x-1\right)^{2}}{2\sigma^{2}}}}=\frac{\theta }{1-\theta }\frac{e^{-\frac{\left(q^{\ast }-x-1\right)^{2}}{2\sigma^{2}}}}{e^{-\frac{\left(q^{\ast }-x-m_{0}\right)^{2}}{2\sigma^{2}}}}\quad \forall x\in {\mathbb{R}}\\ &\quad \Leftrightarrow \hbox{ln} \left(\frac{1-\theta }{\theta }\right) +\frac{-\left(q^{\ast }+x-m_{0}\right)^{2}+\left(q^{\ast }+x-1\right)^{2}}{2\sigma^{2}} \\ &\quad = \hbox{ln} \left(\frac{\theta }{1-\theta }\right) +\frac{ -\left(q^{\ast }-x-1\right)^{2}+\left(q^{\ast }-x-m_{0}\right)^{2}}{ 2\sigma^{2}}\quad \forall x\in {\mathbb{R}} \\& \quad \Leftrightarrow \hbox{ln} \left(\frac{1-\theta }{\theta }\right) +\frac{ 2q^{\ast }m_{0}-m_{0}^{2}-2q^{\ast }+1}{2\sigma^{2}} \\& \quad =\hbox{ln} \left(\frac{\theta }{1-\theta }\right) +\frac{ 2q^{\ast }-1-2q^{\ast }m_{0}+m_{0}^{2}}{2\sigma^{2}} \\& \quad \Leftrightarrow q^{\ast }=\sigma^{2}\frac{\ln \left( \frac{1-\theta }{ \theta }\right) }{1-m_{0}}+\frac{1+m_{0}}{2}. \\ \end{aligned} $$

$\square$

We now present some results that describe several properties of informative equilibria. Recall that we assume without loss of generality that m ₀ ≤ m ₁; further, in an informative equilibrium, m ₀ ≠ m ₁, so m ₀ < m ₁. Lemma 2 states that, in any such equilibrium, the receiver’s chosen action is a strictly monotone function of q.

Lemma 2

In an informative equilibrium, the receiver’s action a is a strictly increasing function of the interpretation q.

Proof

In any equilibrium, the receiver’s action function satisfies

$$ {\bf a}\left(q\right) =\alpha \left(q,m_{0},m_{1},\theta ,\sigma \right) =\frac{\theta }{\left(1-\theta \right) \frac{\phi _{m_{0},\sigma^{2}}\left(q\right) }{\phi _{m_{1},\sigma^{2}}\left( q\right) }+\theta }. $$

For the normal distribution, the likelihood ratio $\frac{\phi _{m_{0},\sigma^{2}}\left(q\right) }{\phi _{m_{1},\sigma^{2}}\left( q\right) }$ is strictly decreasing in q for m ₀ < m ₁. $\square$

Since the sender, for given t, wants a higher action than the receiver, it follows that the type-1 will always choose an extremal message. Formally,

Lemma 3

In an informative equilibrium m ₁ = 1.

Proof

By Lemma 2 the receiver’s action function ${\bf a}\left( q\right)$ is a strictly increasing function of q. Furthermore, ${\bf a}\left(q\right) <1$ for all ${q\in \mathbb{R}}$. Therefore $-\left(1+b-a\left(q\right) \right)^{2},$ the type-1 sender’s payoff from the interpretation q, is also a strictly increasing function of q. This and the fact that $\phi _{1,\sigma^{2}}$ strictly first-order stochastically dominates $\phi_{m_{0},\sigma^{2}}$ for any m ₀ < 1 implies that

$$ \int\limits_{-\infty }^{\infty }-\left(1+b-a\left(q\right) \right)^{2}\phi _{1,\sigma^{2}}\left(q\right) dq>\int\limits_{-\infty }^{\infty }-\left(1+b-a\left(q\right) \right)^{2}\phi _{m_{0},\sigma^{2}}\left(q\right) dq ,\quad \hbox{for all }m_{0}<1. $$

$\square$

Henceforth, then, we assume that m ₁ = 1. The optimization problem for the type-0 sender is much trickier to solve, since in a candidate equilibrium with m ₀ < m ₁, different messages not only shift, but also change the shape of the induced distribution of actions. Some care is required to ensure that sending the specified m ₀ is globally optimal. To this end we repeatedly make use of an important technical observation. As long as the action function of the receiver satisfies Eq. (2) (see the definition of equilibrium in Sect. 2.1), the expected payoff of a type-t sender is a convolution of two quasi-concave functions. Furthermore, the density of the normal distribution is log-concave. Ibragimov (1956) shows that under these conditions the convolution itself will be quasi-concave. The following lemma adapts his result to the present environment.

Lemma 4

If the receiver’s action function a is strictly increasing in the interpretation q, then for any t, the sender’s expected payoff from sending message m

$$ V^{S}\left(m,t,{\bf a}\right) \equiv \int\limits_{-\infty }^{\infty }U^{S}\left({\bf a} \left(q\right) ,t,b\right) \phi _{m,\sigma^{2}}\left(q\right) dq $$

is a strictly quasi-concave function of m, and any $m^{\ast }$ with $\frac{dV^{S}}{dm\ast }\left(m^{\ast },t\right) =0$ is the unique global maximizer for type t.

Proof

To simplify notation, we suppress reference to t,  b and a and let $U\left(q\right) \equiv U^{S}\left({\bf a}\left( q\right) ,t,b\right) ,$ so that

$$ V^{S}\left(m\right) \equiv \int\limits_{-\infty }^{\infty }U\left( q\right) {\frac{ 1}{\sigma \sqrt{2\pi }}}e^{\frac{-\left( q-m\right)^{2}}{2\sigma^{2}}}dq. $$

Note that given our monotonicity assumption on a and by virtue of the fact that for any t there is a unique a _t that solves $\max_{a}U^{S}\left(a,t,b\right) ,$ U is either (i) strictly increasing, (ii) strictly decreasing, or (iii) there exists a value q ₀ such that U is strictly increasing for q < q ₀ and strictly decreasing for q > q ₀. In cases (i) and (ii), the result follows because the normal distribution satisfies the strict-monotone-likelihood-ratio property and therefore strict first-order stochastic dominance. Otherwise, U has a unique maximizer q ₀. For this case, consider

$$ \begin{aligned} \frac{dV^{S}}{dm} &=\int\limits_{-\infty }^{\infty }U\left(q\right) {\frac{d}{dm}} \left({\frac{1}{\sigma \sqrt{2\pi }}}e^{\frac{-\left(q-m\right)^{2}}{ 2\sigma^{2}}}\right) dq \\ &=-\int\limits_{-\infty }^{\infty }U\left(q\right) {\frac{d}{dq}}\left( {\frac{1}{ \sigma \sqrt{2\pi }}}e^{\frac{-\left( q-m\right)^{2}}{2\sigma^{2}}}\right) dq \\ &=\left[ U\left(q\right) {\frac{1}{\sigma \sqrt{2\pi }}}e^{\frac{-\left(q-m\right)^{2}}{2\sigma^{2}}}\right] _{-\infty }^{\infty }+\int\limits_{-\infty }^{\infty }{\frac{1}{\sigma \sqrt{2\pi }}}e^{\frac{-\left(q-m\right)^{2}}{ 2\sigma^{2}}}\frac{dU}{dq}dq \\ &=\frac{1}{\sigma \sqrt{2\pi }}\int\limits_{-\infty }^{\infty }e^{\frac{-\left(q-m\right)^{2}}{2\sigma^{2}}}\frac{dU}{dq}dq, \\ \end{aligned} $$

using the fact $\left\vert U\right\vert$ is bounded. Define λ ≡ q − q ₀. Note that we have $\frac{dU}{dq}\left( q_{0}+\lambda \right) <0$ and $\frac{dU}{dq}\left(q_{0}-\lambda \right) >0$ for all λ > 0. Now suppose that $\frac{dV^{S}}{dm}\left(m^{\ast }\right) =0$ for some $m^{\ast }$. $\frac{dV^{S}}{dm}\left(m^{\ast }\right)$ can be re-written as

$$ \frac{dV^{S}}{dm}\left(m^{\ast }\right) ={\frac{1}{\sigma \sqrt{2\pi }}} \left\{ \int\limits_{0}^{\infty }e^{\frac{-\left(m^{\ast }-\left(q_{0}+\lambda \right) \right)^{2}}{2\sigma^{2}}}\frac{dU}{dq}\left(q_{0}+\lambda \right) d\lambda +\int\limits_{0}^{\infty }e^{\frac{-\left(m^{\ast }-\left( q_{0}-\lambda \right) \right)^{2}}{2\sigma^{2}}}\frac{dU}{dq}\left( q_{0}-\lambda \right) d\lambda \right\}. $$

Also, we have

$$ \begin{aligned} \frac{dV^{S}}{dm} \left(m^{\ast }+\delta \right) &={\frac{1}{\sigma \sqrt{ 2\pi }}}e^{\frac{-\left(2\delta m^{\ast }-2\delta q_{0}+\delta^{2}\right) }{2\sigma^{2}}} \\ &\quad\times\left(\int\limits_{0}^{\infty }e^{\frac{-\left(m^{\ast }-\left( q_{0}+\lambda \right) \right)^{2}}{2\sigma^{2}}}e^{\frac{2\delta \lambda }{2\sigma^{2}}} \frac{dU}{dq}\left(q_{0}+\lambda \right) d\lambda +\int\limits_{0}^{\infty }e^{ \frac{-\left(m^{\ast }-\left( q_{0}-\lambda \right) \right)^{2}}{2\sigma^{2}}}e^{\frac{-2\delta \lambda }{2\sigma^{2}}}\frac{dU}{dq}\left(q_{0}-\lambda \right) d\lambda \right). \end{aligned} $$

Note that for δ > 0 (δ < 0) we are inflating (deflating) the negative terms and deflating (inflating) the positive terms in the integrand. Therefore $\frac{dV^{S}}{dm}\left(m^{\ast }+\delta \right) <0$ for δ > 0 and $\frac{dV^{S}}{dm}\left(m^{\ast }+\delta \right) >0$ for δ < 0; i.e. V ^S is strictly quasi-concave and $m^{\ast }$ is a global maximum. $\square$

We can now start to prove our main results. The proofs make use of the function V defined in Sect, 2.4; V gives the expected utility of the type-0 sender with bias b if she sends message m when the receiver expects her to send message m ₀ and the type-1 sender to send message 1:

$$ V\left(b,m_{0},m^{\prime }\right) \equiv \int\limits_{\infty }^{\infty }-(b-\alpha \left(q,m_{0},1,\theta ,\sigma \right) )^{2}\cdot \phi _{m^{\prime },\sigma }\left(q\right) dq. $$

Proof of Proposition 1

Let $b=\frac{1}{2}$ and suppose we have an informative equilibrium where the sender chooses ${\bf m}=\left( m_{0},1\right)$ with m ₀ < 1. From Lemma 1, we know that $\alpha \left(q,m_{0},1,\theta ,\sigma \right)$ has 180° rotational symmetry about the point $\left(q^{\ast },\frac{1}{2}\right)$ , where $q^{\ast }=\sigma^{2}\frac{\ln \left(\frac{1-\theta }{\theta }\right) }{1-m_{0}}+\frac{1+m_{0}}{2}.$ It follows that $V\left( \frac{1}{2},m_{0},q^{\ast }-k\right) =V\left(\frac{1}{2},m_{0},q^{\ast }+k\right)$. Given Lemma 4, $q^{\ast }$ is the unique maximizer of $V\left( \frac{1}{2},m_{0},m\right)$. If $\theta \leq \frac{1}{2},$ $q^{\ast}$ > m ₀, so the type-0 sender will deviate and send m = min { $q^{\ast}$, 1} instead of m ₀; in this case, then, an informative equilibrium does not exist. Suppose instead that $\theta >\frac{1}{2},$ Solving m ₀ = $q^{\ast}$, we obtain

$$ m_{0}^{\ast }=1-\sigma \sqrt{2\log \left(\frac{\theta }{1-\theta }\right) }. $$

Notice that this expression must be less than 1. If $m_{0}^{\ast }$ lies between 0 and 1, we have a unique informative equilibrium with the sender choosing ${\bf m}=\left(m_{0}^{\ast },1\right) ;$ if $m_{0}^{\ast }\leq 0,$ we have a unique informative equilibrium with the sender choosing ${\bf m}=\left( 0,1\right)$. $\square$

Proof of Proposition 2

In an informative equilibrium, the sender chooses ${\bf m}=\left(m_{0},1\right)$ for some $m_{0}\in \left[ 0,1\right) ,$ and the receiver’s action function is given by ${\bf a}\left(q\right) =\alpha \left(q,m_{0},1,\theta ,\sigma \right)$. Since $\alpha \left(q,m_{0},1,\theta ,\sigma \right)$ is strictly increasing in q and bounded below by 0, for b = 0, $-\left( b-\alpha \left(q,m_{0},1,\theta ,\sigma \right) \right)^{2}$ is a strictly decreasing function of q. This, and the fact that $\Upphi _{m^{\prime },\sigma^{2}}$ first-order stochastically dominates $\Upphi _{m,\sigma^{2}}$ for any $m^{\prime }>m$ implies that $V_{3}\left(0,m,m\right) <0$ for all $m\in \left[ 0,1\right)$. This means that whatever message the receiver expects the type-0 sender to send, she wants to send a lower message, and hence the unique informative equilibrium (with b = 0) is with ${\bf m}=\left( 0,1\right)$. Given continuity of $V_{3}\left(b,m,m\right)$ in b, then, there is a non-empty interval $\left[ 0,\underline {b}\right]$ for which there is a unique informative equilibrium which exhibits maximum differentiation.$\square$

Proof of Proposition 3

Fix some $m\in \left[ 0,1\right)$. From the proof of Proposition 2 we know that $V_{3}\left(0,m,m\right) <0.$ By similar reasoning and the fact that $\alpha \left(q,m,1,\theta ,\sigma \right)$ is bounded above by 1, we have $V_{3}\left( 1,m,m\right) >0.$ Continuity and the intermediate value theorem then imply that there exists a $b\left(m\right) \in \left( 0,1\right)$ such that

$$ V_{3}\left(b\left(m\right) ,m,m\right) =0. $$

From Lemma 4, then, it follows that ${\bf m}=\left(m,1\right)$ is an equilibrium strategy for the sender when $b=b\left(m\right)$. $\square$

Proof of Proposition 4

Differentiating V (the expected utility of the type-0 sender) with respect to the message she actually sends, and evaluating at the message m ₀ that the receiver expects her to send, we obtain

$$ V_{3}\left(b,m_{0},m_{0}\right) =\int\limits_{-\infty }^{\infty }\left( 2b-\alpha \right) \alpha \phi _{m_{0},\sigma^{2}}\left(q\right) \frac{q-m_{0}}{ \sigma^{2}}dq. $$

The derivative of this expression with respect to m ₀ evaluated at m ₀ = 1 is equal to

$$ \left. \frac{dV_{3}\left(b,m_{0},m_{0}\right) }{dm_{0}}\right\vert _{m_{0}=1}=\frac{2\left(b-\theta \right) \left(-1+\theta \right) \theta \sqrt{1/\sigma^{2}}}{\sigma } $$

This derivative is positive exactly when b < θ. Using the fact that $V_{3}\left(b,1,1\right) =0,$ this implies that when b < θ , there exists m ₀ < 1 for which $V_{3}\left( b,m_{0},m_{0}\right) <0.$ Since $V_{3}\left(b,m_{0},m_{0}\right)$ is continuous in m ₀, there are two possibilities. Either $V_{3}\left(b,m_{0},m_{0}\right) <0 \forall m_{0}\in \left( 0,1\right) ,$ in which case there is an informative equilibrium with maximal differentiation, where the sender’s strategy is ${\bf m}=\left(0,1\right)$. Or there exists an $m_{0}\in \left( 0,1\right)$ for which $V_{3}\left(b,m_{0},m_{0}\right) =0,$ in which case Lemma 4 implies that we have an informative equilibrium with intentional vagueness, where the sender’s strategy is ${\bf m}=\left(m_{0},1\right)$. $\square$

Proof of Proposition 5

Suppose that $\theta \geq \frac{1}{2}$ and b = θ (we extend the result to the case where b > θ later), and the type-0 sender sends message m ₀ < 1. We claim that she obtains at least as high an expected utility from sending message 1; combining this result with Lemma 4, we derive a contradiction. To compare $V\left(\theta,m_{0},m_{0}\right)$ with $V\left(\theta,m_{0},1\right) ,$ we show that the type-0 sender (weakly) prefers a $\frac{1}{2} - \frac{1}{2}$ gamble between $\alpha \left(1-k,m_{0},1,\theta ,\sigma \right)$ and $\alpha \left(1+k,m_{0},1,\theta ,\sigma \right)$ to a $\frac{1}{2} - \frac{1}{2}$ gamble between $\alpha \left(m_0-k,m_{0},1,\theta ,\sigma \right)$ and $\alpha \left(m_0+k,m_{0},1,\theta ,\sigma \right)$ (property *), for all k ≥ 0. It follows that $V\left(\theta,m_{0},m_{0}\right) \geq V\left(\theta,m_{0},1\right)$.

Start by considering values of $k \in \left[0,\frac{1-m_{0}}{2}\right]$. For these values, $m_0+k \leq \frac{1+m_0}{2} \leq 1-k.$ It is easy to show that α satisfies the following properties:

• 1. $\alpha \left(\frac{1+m_0}{2},m_{0},1,\theta ,\sigma \right)=\theta;$

• 2. $\alpha \left(q,m_{0},1,\theta ,\sigma \right)$ is strictly increasing in q;

• 3. $\alpha \left(q,m_{0},1,\theta ,\sigma \right)$ has a unique point of inflection at

  $$ q^*=\sigma^{2}\frac{\ln \left(\frac{1-\theta }{\theta }\right) }{1-m_{0}}+\frac{1+m_{0}}{2}< \frac{1+m_0}{2} $$

  and is bounded above and below; therefore, α is strictly convex in q for q < $q^{\ast}$ and strictly concave in q for q > $q^{\ast}$; and

• 4. $\alpha(q^*,m_0,1,\theta,\sigma)=\frac{1}{2}$.

It follows that

$$ \begin{aligned} \theta -\alpha \left(m_{0}+k,m_{0},1,\theta ,\sigma \right) &\geq \alpha \left(1-k,m_{0},1,\theta ,\sigma \right) -\theta \geq 0 \quad \hbox{and} \\ \theta -\alpha \left(m_{0}-k,m_{0},1,\theta ,\sigma \right) &\geq \alpha \left(1+k,m_{0},1,\theta ,\sigma \right) -\theta \geq 0, \end{aligned} $$

for all $k\in \left[0,\frac{1-m_{0}}{2}\right],$ and thus property * is satisfied.

Now consider any k with $k>\frac{1-m_{0}}{2}.$ As before, we compare a $\frac{1}{2} - \frac{1}{2}$ gamble between interpretations q = m ₀ + k and q = m ₀ − k with a $\frac{1}{2} - \frac{1}{2}$ gamble between interpretations q = 1 + k and q = 1 − k. The first pair of gambles induces actions α(m ₀ + k, m ₀, 1, θ, σ) and α(m ₀ − k, m ₀, 1, θ, σ); let:

$$ \begin{aligned} a&=\alpha(m_0+k,m_0,1,\theta,\sigma)-\theta \\ b&=\theta - \alpha(m_0-k,m_0,1,\theta,\sigma). \end{aligned} $$

Similarly, the second pair of gambles induces actions α(1 + k, m ₀, 1, θ, σ) and α(1 − k, m ₀, 1, θ, σ); let:

$$ \begin{aligned} c&=\alpha(1+k,m_0,1,\theta,\sigma)-\theta \\ d&=\theta - \alpha(1-k,m_0,1,\theta,\sigma). \end{aligned} $$

For values of $k>\frac{1-m_{0}}{2}$ in this range, we have c ≥ a ≥ 0 and d ≥ a ≥ 0. We now show that a + b ≥ c + d; substituting in the expressions for a, b, c, and d above, this inequality becomes:

$$ \begin{aligned} & \alpha \left(m_{0}+k,m_{0},1,\theta ,\sigma \right) -\alpha \left(m_{0}-k,m_{0},1,\theta ,\sigma \right) \\ &\quad\geq \alpha \left(1+k,m_{0},1,\theta ,\sigma \right) -\alpha \left(1-k,m_{0},1,\theta ,\sigma \right). \end{aligned} $$

Simplifying the action function, we obtain

$$ \alpha \left(q,m_{0},1,\theta ,\sigma \right) =\frac{1}{1+e^{\frac{\left(1-m_{0}\right) \left(1+m_{0}-2q\right) }{2\sigma^{2}}}\left(\frac{1}{ \theta }-1\right) }. $$

Hence

$$ \begin{aligned} &\alpha \left(m_{0}+k,m_{0},1,\theta ,\sigma \right) -\alpha \left( m_{0}-k,m_{0},1,\theta ,\sigma \right) \\ &=\frac{1}{1+e^{\frac{\left(1-m_{0}\right) \left(1+m_{0}-2\left( m_{0}+k\right) \right) }{2\sigma^{2}}}\left(\frac{1}{\theta }-1\right) }- \frac{1}{1+e^{\frac{\left(1-m_{0}\right) \left( 1+m_{0}-2\left(m_{0}-k\right) \right) }{2\sigma^{2}}}\left( \frac{1}{\theta }-1\right) } \\ &=\frac{\theta \left(1-\theta \right) \left(e^{\frac{\left( 1-m_{0}\right) \left(1+m_{0}-2\left(m_{0}-k\right) \right) }{2\sigma^{2}} }-e^{\frac{\left(1-m_{0}\right) \left( 1+m_{0}-2\left(m_{0}+k\right) \right) }{2\sigma^{2}}}\right) }{\left(\theta +e^{\frac{\left(1-m_{0}\right) \left( 1+m_{0}-2\left(m_{0}+k\right) \right) }{2\sigma^{2}} }\left( 1-\theta \right) \right) \left(\theta +e^{\frac{\left( 1-m_{0}\right) \left(1+m_{0}-2\left(m_{0}-k\right) \right) }{2\sigma^{2}} }\left(1-\theta \right) \right) } \\ &=\frac{\theta \left(1-\theta \right) \left( 1-\frac{e^{\frac{\left(1-m_{0}\right) \left(1-m_{0}-2k\right) }{2\sigma^{2}}}}{e^{\frac{\left(1-m_{0}\right) \left( 1-m_{0}+2k\right) }{2\sigma^{2}}}}\right) }{\left(\theta +e^{\frac{\left(1-m_{0}\right) \left(1-m_{0}-2k\right) }{2\sigma^{2}}}\left(1-\theta \right) \right) \left(\frac{\theta }{e^{\frac{\left(1-m_{0}\right) \left(1-m_{0}+2k\right) }{2\sigma^{2}}}}+\left(1-\theta \right) \right) } \\ &=\frac{\theta \left(1-\theta \right) \left(1-e^{\frac{-2k\left( 1-m_{0}\right) }{\sigma^{2}}}\right) }{\left(\theta +e^{\frac{\left(1-m_{0}\right) \left(1-m_{0}-2k\right) }{2\sigma^{2}}}\left(1-\theta \right) \right) \left(\theta e^{\frac{-\left(1-m_{0}\right) \left(1-m_{0}+2k\right) }{2\sigma^{2}}}+\left(1-\theta \right) \right) } \\ \end{aligned} $$

(3)

and

$$ \begin{aligned} &\alpha \left(1+k,m_{0},1,\theta ,\sigma \right) -\alpha \left( 1-k,m_{0},1,\theta ,\sigma \right)\\ &=\frac{1}{1+e^{\frac{\left(1-m_{0}\right) \left(1+m_{0}-2\left( 1+k\right) \right) }{2\sigma^{2}}}\left(\frac{1}{\theta }-1\right) }-\frac{ 1}{1+e^{\frac{\left(1-m_{0}\right) \left(1+m_{0}-2\left( 1-k\right) \right) }{2\sigma^{2}}}\left(\frac{1}{\theta }-1\right)} \\ &=\frac{\theta \left(1-\theta \right) \left(e^{\frac{\left( 1-m_{0}\right) \left(1+m_{0}-2\left(1-k\right) \right) }{2\sigma^{2}}}-e^{ \frac{\left(1-m_{0}\right) \left( 1+m_{0}-2\left(1+k\right) \right) }{ 2\sigma^{2}}}\right) }{\left( \theta +e^{\frac{\left(1-m_{0}\right) \left(1+m_{0}-2\left( 1+k\right) \right) }{2\sigma^{2}}}\left(1-\theta \right) \right) \left(\theta +e^{\frac{\left(1-m_{0}\right) \left(1+m_{0}-2\left( 1-k\right) \right) }{2\sigma^{2}}}\left(1-\theta \right) \right) } \\ &=\frac{\theta \left(1-\theta \right) \left( 1-\frac{e^{\frac{\left(1-m_{0}\right) \left(m_{0}-1-2k\right) }{2\sigma^{2}}}}{e^{\frac{\left(1-m_{0}\right) \left( m_{0}-1+2k\right) }{2\sigma^{2}}}}\right) }{\left(\theta +e^{\frac{\left(1-m_{0}\right) \left(m_{0}-1-2k\right) }{2\sigma^{2}}}\left(1-\theta \right) \right) \left(\frac{\theta }{e^{\frac{\left(1-m_{0}\right) \left(m_{0}-1+2k\right) }{2\sigma^{2}}}}+\left(1-\theta \right) \right) } \\ &=\frac{\theta \left(1-\theta \right) \left(1-e^{\frac{-2k\left( 1-m_{0}\right) }{2\sigma^{2}}}\right) }{\left(\theta +e^{\frac{\left(1-m_{0}\right) \left(m_{0}-1-2k\right) }{2\sigma^{2}}}\left(1-\theta \right) \right) \left(\theta e^{\frac{-\left(1-m_{0}\right) \left(m_{0}-1+2k\right) }{2\sigma^{2}}}+\left(1-\theta \right) \right) }\\ \end{aligned} $$

(4)

Notice that numerators of (3) and (4) are the same, and positive; it follows that

$$ \alpha \left(m_{0}+k,m_{0},1,\theta ,\sigma \right) -\alpha \left( m_{0}-k,m_{0},1,\theta ,\sigma \right) \geq \alpha \left( 1+k,m_{0},1,\theta ,\sigma \right) -\alpha \left(1-k,m_{0},1,\theta ,\sigma \right) $$

if and only if

$$ \begin{aligned} &\left(\theta +e^{\frac{\left(1-m_{0}\right) \left( 1-m_{0}-2k\right) }{ 2\sigma^{2}}}\left(1-\theta \right) \right) \left(\theta e^{\frac{-\left(1-m_{0}\right) \left( 1-m_{0}+2k\right) }{2\sigma^{2}}}+\left(1-\theta \right) \right)\\ \leq &\left(\theta +e^{\frac{\left(1-m_{0}\right) \left( m_{0}-1-2k\right) }{2\sigma^{2}}}\left(1-\theta \right) \right) \left(\theta e^{\frac{-\left(1-m_{0}\right) \left( m_{0}-1+2k\right) }{2\sigma^{2}}}+\left(1-\theta \right) \right) \end{aligned} $$

if and only if

$$ \begin{aligned} &\theta^{2}e^{\frac{-\left(1-m_{0}\right) \left(1-m_{0}+2k\right) }{ 2\sigma^{2}}}+e^{\frac{\left(1-m_{0}\right) \left( 1-m_{0}-2k\right) }{ 2\sigma^{2}}}\left(1-2\theta +\theta^{2}\right) \\ \leq &\theta^{2}e^{\frac{\left(1-m_{0}\right) \left( 1-m_{0}-2k\right) }{ 2\sigma^{2}}}+e^{\frac{-\left(1-m_{0}\right) \left(1-m_{0}+2k\right) }{ 2\sigma^{2}}}\left(1-2\theta +\theta^{2}\right) \end{aligned} $$

if and only if

$$ \begin{aligned} &\left(e^{\frac{\left(1-m_{0}\right) \left(1-m_{0}-2k\right) }{2\sigma^{2}}}-e^{\frac{\left(1-m_{0}\right) \left( m_{0}-1-2k\right) }{2\sigma^{2} }}\right)\left(1-2\theta \right) \leq 0 \end{aligned} $$

if and only if

$$ \begin{aligned} &\left(1-2\theta \right) \leq 0 \end{aligned} $$

if and only if

$$ \begin{aligned} &\theta \geq \frac{1}{2}\quad \checkmark \end{aligned} $$

To recap, we have c ≥ a ≥ 0, d ≥ a ≥ 0, and a + b ≥ c + d. From the third inequality, we obtain

$$ \begin{aligned} & a^2+b^2 \geq a^2 + (c+d-a)^2 \\ & \quad \Rightarrow a^2 + b^2 \geq c^2 +d^2 + 2(d-a)(c-a). \end{aligned} $$

Thus, the first two inequalities give us

$$ a^2 + b^2 \geq c^2 + d^2 $$

It follows immediately that the $\frac{1}{2} - \frac{1}{2}$ gamble between α(m ₀ + k, m ₀, 1, θ, σ) and α(m ₀ − k, m ₀, 1, θ, σ) is preferred to the $\frac{1}{2} - \frac{1}{2}$ gamble between α(1 + k, m ₀, 1, θ, σ) and α(1 − k, m ₀, 1, θ, σ). Thus, when b = θ, the type-0 sender obtains at least as high expected utility from message 1 as from message m ₀.

To complete the proof, consider the case where b > θ. Here, the Proposition follows from the result just obtained together with the single-crossing condition

$$ \frac{\partial^{2}U^{s}(a,t,b)}{\partial t\partial a}>0 $$

and the fact that the normal distribution satisfies the strict monotone likelihood ratio property (see the proof of Proposition 9 for details). $\square$

Lemma 5

$z_1(b,m)=V_{31}(b,m,m) > 0 \forall m \in [0,1).$

Proof

Notice that $\phi _{m,\sigma^{2}}\left(m+x\right) \frac{x}{\sigma^{2}}=-\phi _{m,\sigma^{2}}\left(m-x\right) \frac{\left(-x\right) }{\sigma^{2}}.$ Furthermore, for all $m \in [0,1),$ $\frac{\theta \phi _{1,\sigma^{2}}\left(q\right) }{\theta \phi _{1,\sigma^{2}}\left(q\right) +\left(1-\theta \right) \phi _{m,\sigma^{2}}\left(q\right) }$ is a strictly increasing function of q and therefore gives greater weight to $\phi _{m,\sigma^{2}}\left(m+x\right) \frac{x}{\sigma^{2}}$ than to $\phi _{m,\sigma^{2}}\left(m-x\right) \frac{\left(-x\right) }{\sigma^{2}}$ for all x > 0. Therefore

$$ V_{31}\left(b,m,m\right) =\int\limits_{-\infty }^{\infty }2\frac{\theta \phi _{1,\sigma^{2}}\left(q\right) }{\theta \phi _{1,\sigma^{2}}\left(q\right) +\left(1-\theta \right) \phi _{m,\sigma^{2}}\left(q\right) } \phi _{m,\sigma^{2}}\left(q\right) \frac{q-m}{\sigma^{2}}>0. $$

$\square$

Proof of Proposition 6

For m ₀ < 1, define

$$ q^{\ast }\left(m_{0}\right) \equiv \sigma^{2}\frac{\log \left( \frac{ 1-\theta }{\theta }\right) }{1-m_{0}}+\frac{1+m_{0}}{2} $$

(so $\left(q^{\ast }\left(m_{0}\right) ,\frac{1}{2}\right)$ is the point of symmetry of the expectation function $\alpha \left(q,m_{0},1,\theta ,\sigma \right)$—see Lemma 1 above). If $\theta <\frac{1}{2},$ then $q^{\ast }(m_{0})>m_{0},$ in which case we claim that the existence of an informative equilibrium requires that $b<\frac{1}{2}.$ To see why, consider

$$ \begin{aligned} V_{3}\left(b,m_{0},m_{0}\right) &=\int\limits_{-\infty }^{\infty }-\left( b-\frac{ \theta \phi _{1,\sigma^{2}}\left(q\right) }{\theta \phi _{1,\sigma^{2}}\left(q\right) +\left(1-\theta \right) \phi _{m_{0},\sigma^{2}}\left(q\right) }\right)^{2}\phi _{m_{0},\sigma^{2}}\left(q\right) \frac{q-m_{0}}{\sigma^{2}}dq \\ &=\int\limits_{-\infty }^{\infty }\left(-b^{2}+2b\frac{\theta \phi _{1,\sigma^{2}}\left(q\right) }{\theta \phi _{1,\sigma^{2}}\left( q\right) +\left(1-\theta \right) \phi _{m_{0},\sigma^{2}}\left( q\right) }\right. \\ &\quad \left. -\left(\frac{\theta \phi _{1,\sigma^{2}}\left( q\right) }{\theta \phi _{1,\sigma^{2}}\left(q\right) +\left( 1-\theta \right) \phi _{m_{0},\sigma^{2}}\left(q\right) }\right)^{2}\right) \phi _{m_{0},\sigma^{2}}\left(q\right) \frac{q-m_{0}}{\sigma^{2}}dq \\ &=\int\limits_{-\infty }^{\infty }\left(2b-\frac{\theta \phi _{1,\sigma^{2}}\left(q\right) }{\theta \phi _{1,\sigma^{2}}\left( q\right) +\left(1-\theta \right) \phi _{m_{0},\sigma^{2}}\left( q\right) }\right) \\ &\quad \quad \frac{\theta \phi _{1,\sigma^{2}}\left(q\right) }{\theta \phi _{1,\sigma^{2}}\left(q\right) +\left(1-\theta \right) \phi _{m_{0},\sigma^{2}}\left(q\right) }\phi _{m_{0},\sigma^{2}}\left(q\right) \frac{q-m_{0}}{\sigma^{2}}dq. \end{aligned} $$

Setting $b=\frac{1}{2},$ we obtain

$$ \begin{aligned} V_{3}\left(\frac{1}{2},m_{0},m_{0}\right) &=\int\limits_{-\infty }^{\infty }\left(1-\frac{\theta \phi _{1,\sigma^{2}}\left(q\right) }{\theta \phi _{1,\sigma^{2}}\left(q\right) +\left(1-\theta \right) \phi _{m_{0},\sigma^{2}}\left(q\right) }\right) \\ &\quad \quad \frac{\theta \phi _{1,\sigma^{2}}\left(q\right) }{\theta \phi _{1,\sigma^{2}}\left(q\right) +\left(1-\theta \right) \phi _{m_{0},\sigma^{2}}\left(q\right) }\phi _{m_{0},\sigma^{2}}\left(q\right) \frac{q-m_{0}}{\sigma^{2}}dq. \end{aligned} $$

The quantity $\left(1-\frac{\theta \phi _{1,\sigma^{2}}\left( q\right) }{\theta \phi _{1,\sigma^{2}}\left(q\right) +\left( 1-\theta \right) \phi _{m_{0},\sigma^{2}}\left(q\right) }\right) \frac{\theta \phi _{1,\sigma^{2}}\left(q\right) }{\theta \phi _{1,\sigma^{2}}\left(q\right) +\left(1-\theta \right) \phi _{m_{0},\sigma^{2}}\left(q\right) }$ is a function of q that is symmetric about $q^{\ast }\left(m_{0}\right) ,$ strictly increasing to the left of $q^{\ast }\left(m_{0}\right)$ and strictly decreasing to the right of $q^{\ast }\left(m_{0}\right)$. So as long as $q^{\ast }\left(m_{0}\right) >m_{0},$ it assigns greater weight to $\phi _{m_{0},\sigma^{2}}\left(m_{0}+x\right) \frac{x}{\sigma^{2}}$ than to $\phi _{m_{0},\sigma^{2}}\left( m_{0}-x\right) \frac{\left(-x\right) }{\sigma^{2}}$ for all x > 0. Hence, $V_{3}\left(\frac{1}{2},m_{0},m_{0}\right) >0.$

Now consider raising b above $\frac{1}{2}.$ From Lemma 5, $V_{31}(b,m,m) > 0 \forall m \in [0,1)$ and in particular for m = m ₀. Hence $V_{3}\left(b,m_{0},m_{0}\right) >0$ for all $b>\frac{1}{2},$ which is not consistent with equilibrium. $\square$

Proof of Proposition 8

Recall that $z\left(b,1\right) =0$ and

$$ \frac{dz\left(b,1\right) }{dm}=\frac{2\left(b-\theta \right) \left(-1+\theta \right) \theta \sqrt{1/\sigma^{2}}}{\sigma }. $$

Therefore, if b > θ , by continuity of z there exists an $\underline {m}\in \left(0,1\right)$ such that for all $m^{\prime }\in \left(\underline {m},1\right]$ we have $z\left(b,m^{\prime }\right) >0.$ Hence, for b > θ pooling at message $m^{\ast }=1$ is asymptotically stable. If b < θ , then $\frac{dz\left( b,1\right) }{dm}>0,$ i.e. m = 1 is a hyperbolic source rather than a sink of the vagueness dynamic, and therefore unstable.

It remains to show that there is a (Lyapunov) stable equilibrium when b ≤ θ.

Consider the case b < θ first. Then there are two subcases: If $z\left(b,m\right) \leq 0$ for all $m\in \left[ 0,1\right) ,$ then m = 0 is stable. Otherwise, there exists $m^{\prime }\in \left[ 0,1\right)$ with $z\left(b,m^{\prime }\right) >0.$ At the same time, given the case we are considering, there is $m^{\prime \prime }\in \left(m^{\prime },1\right)$ with $z\left(b,m^{\prime \prime }\right) <0.$ Define $\overline{m}\equiv \inf \left\{ m\mid z\left( b,m\right) <0 , m>m^{\prime }\right\}$ and $\underline {m}\equiv \sup \left\{ m\mid z\left(b,m\right) >0 , m<\overline{m}\right\} .$ Note that $\underline {m}\leq \overline{m}.$ If $\underline {m}<\overline{m},$ then $z\left(b,m\right) =0$ for all m ⁰ in the open set $\left(\underline {m},\overline{m}\right) ,$ and therefore any such m ⁰ is stable. If $\underline {m}=\overline{m},$ then any open set $\left( \overline{m},\overline{m}+\epsilon \right)$ contains a subinterval on which $z\left(b,m\right) <0$ and any open set $\left( \overline{m}-\epsilon ,\overline{m}\right)$ contains a subinterval on which $z\left(b,m\right) >0,$ and therefore $\overline{m}$ is stable.

Finally, consider b = θ . If $z\left(b,m\right) \geq 0$ for all $m\in \left[ 0,1\right] ,$ then $m^{\ast }=1$ is stable. If $z\left(b,m^{\prime \prime }\right) <0$ for some $m^{\prime \prime}\in \left[ 0,1\right) ,$ then either $z\left(b,m\right) \leq 0$ for all $m\in \left[ 0,m^{\prime \prime }\right)$ or there exists $m^{\prime }<m^{\prime \prime }$ with $z\left(b,m^{\prime}\right) >0.$ In that case, the argument given for the case b < θ applies. $\square$

Lemma 6

In any pure-strategy equilibrium the receiver’s action rule is continuously differentiable. If in addition the sender’s strategy is monotone and informative, then the receiver’s action rule is strictly increasing.

Proof

If the sender uses a pure strategy, then with any equilibrium message m we can associate the set of types $\Uptheta \left( m\right)$ who use that message. Let $M^{\ast }$ denote the set of equilibrium messages. Then the receiver’s posterior belief about the sender’s type given interpretation q is

$$ \mu \left(t\mid q\right) =\frac{\phi _{m_{t},\sigma^{2}}\left( q\right) \nu \left(t\right) }{\sum_{m\in M^{\ast }}\phi _{m,\sigma^{2}}\left(q\right) \nu \left(\Uptheta \left(m\right) \right) }. $$

In equilibrium, the receiver’s action rule is given by

$$ \begin{aligned} {\bf a}\left(q\right) &=\arg \max_{a}\sum_{t\in T}-\left( a-t\right)^{2}\mu \left(t\mid q\right) \\ &=\frac{\sum_{m\in M^{\ast }}\phi _{m,\sigma^{2}}\left(q\right) \nu \left(\Uptheta \left(m\right) \right) E\left[ t\mid t\in \Uptheta \left(m\right) \right] }{\sum_{m\in M^{\ast }}\phi _{m,\sigma^{2}}\left(q\right) \nu \left(\Uptheta \left(m\right) \right) }. \end{aligned} $$

(Note that, as in the two-type case, the receiver’s best response is to choose an action equal to the expectation of t.) Continuous differentiability of the receiver’s action rule follows from continuous differentiability of $\phi _{m,\sigma^{2}}$ for any m and the fact that $\phi _{m,\sigma^{2}}$ is everywhere positive.

If the sender’s strategy is monotone and informative, more than one message is sent. Let there be k > 1 such messages. It will be convenient to reindex messages and to use m _i to denote the ith equilibrium message and $\Uptheta _{i}$ to denote the set of types who send that message. Then we can rewrite the receiver’s action rule as

$$ {\bf a}\left(q\right) =\frac{\sum_{i=1}^{k}\phi _{m_{i},\sigma^{2}}\left(q\right) \nu \left(\Uptheta _{i}\right) E\left[ t\mid t\in \Uptheta _{i}\right] }{\sum_{i=1}^{k}\phi _{m_{i},\sigma^{2}}\left(q\right) \nu \left(\Uptheta _{i}\right) }, $$

where $E\left[ t\mid t\in \Uptheta _{i+1}\right] >E\left[ t\mid t\in \Uptheta _{i}\right]$ for all $i=1,\ldots ,k-1$ (which can be satisfied because of monotonicity) and $\nu \left(\Uptheta _{i}\right) >0$ for all $i=1,\ldots ,k$. To prove that a is a strictly increasing function of q, we proceed by induction. Define $\xi \left(q\mid m_{i}\right) \equiv \frac{\phi _{m_{i},\sigma^{2}}\left(q\right) \nu \left(\Uptheta _{i}\right) }{\sum_{j=1}^{k}\phi _{m_{j},\sigma^{2}}\left(q\right) \nu \left( \Uptheta _{j}\right) },$ so that

$$ a\left(q\right) =\sum_{i=1}^{k}\xi \left(q\mid m_{i}\right) E\left[ t\mid t\in \Uptheta _{i}\right]. $$

Notice that $\sum_{i=1}^{k-1}\frac{\xi \left(q\mid m_{i}\right) }{\xi \left(q\mid m_{k}\right) }+1=\frac{1}{\xi \left(q\mid m_{k}\right) }.$ SMLRP implies that each of the fractions on the left–hand side decrease as q increases. Hence $\xi \left(q\mid m_{k}\right)$ is (strictly) increasing in q. This establishes the claim for k = 2. We will now show that if it holds for k, then it holds for k + 1. For $i=1,\ldots ,k$ define $\tilde{\xi}\left(q\mid m_{i}\right) \equiv \frac{\xi \left(q\mid m_{i}\right) }{\sum_{j=1}^{k}\xi \left(q\mid m_{j}\right) }.$ Then

$$ \begin{aligned} &\sum_{i=1}^{k+1}\xi \left(q\mid m_{i}\right) E\left[ t\mid t\in \Uptheta _{i}\right] \\ &\quad =\left(1-\xi \left(q\mid m_{k+1}\right) \right) \left\{ \sum_{i=1}^{k}\tilde{\xi}\left(q\mid m_{i}\right) E\left[ t\mid t\in \Uptheta _{i}\right] \right\} +\xi \left(q\mid m_{k+1}\right) E\left[ t\mid t\in \Uptheta _{k+1}\right]. \end{aligned} $$

The result follows because the expression in curly brackets, which is strictly smaller than $E\left[ t\mid t\in \Uptheta _{k+1}\right] ,$ by the induction hypothesis is strictly increasing in q and because $\xi \left(q\mid m_{k+1}\right)$ is strictly increasing in q. $\square$

Proof of Proposition 9

Given the receiver’s equilibrium action rule a, define type t’s payoff from sending message m as

$$ V^{S}\left(m,t,{\bf a}\right) \equiv \int\limits_{-\infty }^{\infty }U^{s}\left({\bf a}\left(q\right) ,t,b\right) \phi _{m,\sigma^{2}}\left(q\right) dq. $$

Then

$$ \frac{\partial V^{S}\left(m,t,{\bf a}\right) }{\partial t}=\int\limits_{-\infty }^{\infty }\frac{\partial U^{s}({\bf a}(q),t,b)}{\partial t}\phi _{m,\sigma^{2}}(q)dq. $$

The sender’s payoff function satisfies the single–crossing condition

$$ \frac{\partial^{2}U^{s}(a,t,b)}{\partial t\partial a}>0. $$

This and the fact that a is a strictly increasing function of q from Lemma 6 implies that $\frac{\partial U^{s}({\bf a}(q),t,b)}{\partial t}$ is strictly increasing in q. Since $\phi_{m,\sigma^{2}}$ satisfies the strict monotone likelihood ratio property $\Upphi _{m^{\prime },\sigma^{2}}$ first–order stochastically dominates $\Upphi _{m,\sigma^{2}}$ for any $m^{\prime }>m.$ Therefore

$$ \frac{\partial^{2}\tilde{V}({\bf a},t,m)}{\partial m\partial t}>0. $$

Suppose that

$$ \frac{\partial \tilde{V}({\bf a},s,m_{s})}{\partial m}\geq 0 $$

for a type s < 1. Then

$$ \frac{\partial \tilde{V}({\bf a},\tau ,m_{s})}{\partial m}>0 $$

for any type τ > s. Using Lemma 4, this implies that either m _τ > m _s or $m_{\tau }^{\prime }=1$ for all $\tau^{\prime }\geq s.$ Similarly, when

$$ \frac{\partial \tilde{V}({\bf a},t,m_{t})}{\partial m}\leq 0 $$

for a type t > 0, we get that for any type τ < t either m _τ < m _t or $m_{\tau }^{\prime }=0$ for all $\tau^{\prime }\leq t.$ $\square$

Proof of Proposition 10

The receiver’s utility coincides with the utility of a sender whose type is less than t. Given the strict monotonicity of the receiver’s action rule from Lemma 6, single crossing and SMLRP, this type would want to send a message less than m _t . $\square$

Proof of Proposition 12

With $b> \frac{1}{2}$ beliefs that are concentrated on the lowest type are the least favorable ones for every type. Therefore, if there is a monotone informative equilibrium in the noisy-channel game with sender strategy m , receiver strategy a and belief system ν, there is an equilibrium in the channel-choice game in which the sender uses strategy m, the receiver responds with a(q) to any interpretation q that is received through the noisy channel, responds with action 0 to any interpretation that is received through the clear channel, has belief ν(q) after any interpretation that is received through the noisy channel and believes that the sender is the lowest type after every interpretation that is received through the clear channel. $\square$

Proof of Proposition 13

Recall that the receiver uses a pure strategy in any Perfect Bayesian equilibrium. Therefore in any PBE each clear message m induces exactly one action $a_m \in [0,1].$ In the common interest game there is a type t for whom a _m is the ideal action. This type strictly prefers sending message m to any noisy equilibrium message $\tilde m.$ By continuity, there is an open neighborhood ${{\mathcal{O}}}$ of t such that all types in ${{\mathcal{O}}}$ strictly prefer sending the clear message m to sending any noisy equilibrium message. $\square$

Appendix 2: Utility Loss from Intentional Vagueness in the CS Model

We observed in the introduction that there is intentional vagueness even in the CS model: Given the receiver’s interpretation of messages in an informative equilibrium, he would prefer that some subset of sender types deviate from their equilibrium strategy and send messages that are associated with lower types. For example, consider the two-step equilibrium of the uniform-quadratic version of the CS model when the sender’s bias $b=\frac{1}{8}:$ sender types $t \in [0,\frac{1}{4})$ send one message, say m ₁, while sender types $t \in [\frac{1}{4},1]$ send a different message, say m ₂; the receiver chooses action $a=\frac{1}{8}$ if he observes m ₁, and action $a=\frac{5}{8}$ if he observes m ₂. Given this equilibrium response, however, note that the receiver would be better off if types between $\frac{1}{4}$ and $\frac{3}{8}$ deviated from their equilibrium strategy and sent message m ₁ instead of m ₂, since m ₁ induces an action closer his ideal than does m ₂ as long as $t<(\frac{1}{8}+\frac{5}{8})/2=\frac{3}{8}.$

Consider an n-step equilibrium of the uniform-quadratic version of the CS model, with sender’s bias b. The boundary types are given by

$$ t_i=\frac{i}{n}+2i(i-n)b, \quad i=0,\ldots,n. $$

Following the reasoning above, it is easy to see that the receiver (assuming his equilibrium interpretation of messages remains unchanged) would prefer all and only sender types between t _i and t _i  + b ($i=1,\ldots,n-1$) to send the message corresponding to the preceding partition step. Compared with this scenario, his ex ante utility loss in the actual equilibrium is

$$ \sum^{n-1}_{i=1}\int\limits^{t_i+b}_{t_i} \left(\theta - \frac{t_i + t_{i-1}}{2} \right)^2 - \left(\theta - \frac{t_{i+1} + t_i}{2} \right)^2 d\theta = -\frac{b^2(n-1)}{n}. $$

In the most informative equilibrium, the number of steps is given by

$$ n^*=\left\lceil -\frac{1}{2}+\frac{1}{2}\sqrt{1+\frac{2}{b}} \right\rceil $$

(where $\lceil x \rceil$ is the smallest integer greater than x). Figure 9 plots the receiver’s utility loss in this equilibrium.

Fig. 9

Receiver’s ex ante utility loss from intentional vagueness in the CS model

Appendix 3: Sets of Strategies that are Closed Under Rational Behavior (Curb)

Let X _i , i = S, R, be a set of pure strategies of player i and X = X _S  × X _R a set of pure strategy profiles for sender and receiver. For player i = S, R define β _i (X _j ), i ≠ j, as the set of pure-strategy best replies to those beliefs over X _j for which a best reply is well-defined. Define β(X): = β _S (X _R ) × β _R (X _S ). Then we call X a curb set if $X \supseteq \beta(X) \neq \emptyset.$ ²³ It is a minimal curb set if it does not contain a proper subset that is a curb set. Note that, as usual, the set of all strategy profiles is a curb set.

Lemma 7

Assume that there is a strictly monotone equilibrium (σ ^e, ρ ^e). Then the singleton set {(σ ^e, ρ ^e)} is a minimal curb set and no pooling equilibrium belongs to a minimal curb set.

Proof

By Lemma 6, ρ ^e is strictly increasing. Therefore by Lemma 4, σ ^e is a unique best reply, and hence (σ ^e, ρ ^e) is strict. A strict equilibrium trivially forms a minimal curb set as a singleton.

If a pooling equilibrium did belong to a minimal curb set X, then X would have to include every sender strategy and corresponding best reply of the receiver. Hence X would have to include (σ ^e, ρ^e), which is a curb set in its own right. Therefore X could not be minimal. $\square$