Competitive disclosure of correlated information

Abstract

We analyze a model of competition in Bayesian persuasion in which two senders vie for the patronage of a receiver by disclosing information about the qualities of their respective proposals, which are positively correlated. The information externality—the news disclosed by one sender contains information about the other sender’s proposal—generates two effects on the incentives for information disclosure. The first effect, which we call the underdog-handicap effect, arises because the receiver is endogenously biased toward choosing the ex ante stronger sender. The second effect, which we call the good-news curse, arises because a sender’s favorable signal realization implies that the rival is more likely to generate a strong competing signal realization. While the underdog-handicap effect encourages more aggressive disclosure, the good-news curse can lower disclosure incentives. If the senders’ ex ante expected qualities are different, and the qualities of their two proposals are highly correlated, then the underdog-handicap effect dominates. Furthermore, as the correlation approaches its maximum possible value, the competition becomes so intense that both senders engage in full disclosure in the unique limit equilibrium.

Appendix

Proof of Lemma 2

Suppose $p_{1},p_{2}\in ( 0,1)$. By a direct application of Bayes’ rule,

$$\begin{aligned} \Pr \left( U_{1}=u_{h}|p_{1},p_{2}\right)&=\frac{1}{1+\frac{\left( \left( 1-\mu _{1}\right) \left( 1-\mu _{2}\right) +\rho \right) \frac{1-p_{2}}{p_{2}}\frac{\mu _{2}}{1-\mu _{2}}+\left( \mu _{2}\left( 1-\mu _{1}\right) -\rho \right) }{\left( \mu _{1}\left( 1-\mu _{2}\right) -\rho \right) \frac{1-p_{2}}{p_{2}}\frac{\mu _{2}}{1-\mu _{2}}+\left( \mu _{1}\mu _{2}+\rho \right) }\frac{1-p_{1}}{p_{1}}\frac{\mu _{1}}{1-\mu _{1}} },\text { and}\\ \Pr \left( U_{2}=u_{h}|p_{1},p_{2}\right)&=\frac{1}{1+\frac{\left( \left( 1-\mu _{1}\right) \left( 1-\mu _{2}\right) +\rho \right) \frac{1-p_{1}}{p_{1}}\frac{\mu _{1}}{1-\mu _{1}}+\left( \mu _{1}\left( 1-\mu _{2}\right) -\rho \right) }{\left( \mu _{2}\left( 1-\mu _{1}\right) -\rho \right) \frac{1-p_{1}}{p_{1}}\frac{\mu _{1}}{1-\mu _{1}}+\left( \mu _{1}\mu _{2}+\rho \right) }\frac{1-p_{2}}{p_{2}}\frac{\mu _{2}}{1-\mu _{2}} }. \end{aligned}$$

It is straightforward to show that $\Pr ( U_{1}=u_{h}|p_{1},p_{2}) \ge \Pr ( U_{2}=u_{h}|p_{1},p_{2})$ holds if and only if $p_{2}\le \frac{1}{1+k}$, where

$$\begin{aligned} k\equiv \frac{\mu _{1}}{1-\mu _{1}}\times \frac{1-\mu _{2}}{\mu _{2}}\times \frac{\left( 1-\mu _{1}\right) \mu _{2}-\rho }{\mu _{1}\left( 1-\mu _{2}\right) -\rho }>0. \end{aligned}$$

Observe that $k<1$ if and only if $\rho >0$. Also, define a function $\delta :[ 0,1] \rightarrow [ 0,1]$ by

$$\begin{aligned} \delta \left( p\right) \equiv \left\{ \begin{array} [c]{cc} 0 &{}\quad \text {if }\,\,p=0\\ \left( 1+k\frac{1-p}{p}\right) ^{-1} &{}\quad \text {if }\,\,p\in (0,1] \end{array} \right. . \end{aligned}$$

With these definitions, if $p_{1}\in ( 0,1)$ or $p_{2}\in ( 0,1)$, then $\Pr ( U_{1}=u_{h}|p_{1},p_{2}) \ge \Pr ( U_{2}=u_{h}|p_{1},p_{2})$ holds if and only if $p_{2}\le \delta ( p_{1})$.

It remains to consider $p_{1},p_{2}\in \{ 0,1\} \times \{ 0,1\}$. If $p_{1}=0$, then $\Pr ( U_{1}=u_{h}|p_{1},p_{2}) =0\ge \Pr ( U_{2}=u_{h}|p_{1},p_{2})$ holds if and only if $p_{2}=0$. If $p_{1}=1$, then $\Pr ( U_{1}=u_{h}|p_{1},p_{2}) =1\ge \Pr ( U_{2}=u_{h}|p_{1},p_{2})$ holds for all $p_{2} \in [ 0,1]$. Finally, if $p_{2}=1$, then $\Pr ( U_{1} =u_{h}|p_{1},p_{2}) \ge 1=\Pr ( U_{2}=u_{h}|p_{1},p_{2})$ holds if and only if $p_{1}=1$. In sum, for all $p_{1},p_{2}\in [ 0,1]$, $\Pr ( U_{1}=u_{h}|p_{1},p_{2}) \ge \Pr ( U_{2}=u_{h}|p_{1},p_{2})$ holds if and only if $p_{2}\le \delta ( p_{1})$.

The function $\delta$ is increasing and concave because

$$\begin{aligned} \frac{\partial \delta \left( p\right) }{\partial p}=\frac{k}{\left( k\left( 1-p\right) +p\right) ^{2}}>0\text {; and }\frac{\partial ^{2}\delta \left( p\right) }{\partial p^{2}}=\frac{-2k\left( 1-k\right) }{\left( k\left( 1-p\right) +p\right) ^{3}}<0. \end{aligned}$$

Furthermore, the function $\delta$ is increasing in $\rho$:

$$\begin{aligned} \frac{\partial \delta \left( p\right) }{\partial \rho }=\frac{\mu _{1}}{1-\mu _{1}}\frac{1-\mu _{2}}{\mu _{2}}\frac{\mu _{1}-\mu _{2}}{\left( \mu _{1}\left( 1-\mu _{2}\right) -\rho \right) ^{2}}\frac{p\left( 1-p\right) }{\left( k\left( 1-p\right) +p\right) ^{2}}>0. \end{aligned}$$

Finally, if $\rho =0$, then $k=1$ and $\delta ( p) =p$. As $\rho \rightarrow ( 1-\mu _{1}) \mu _{2}$, $k\rightarrow 0$ and $\delta ( p)$ converges pointwise to $1_{\{ p>0\} }$. $\square$

Proof of Theorem 1

The sufficiency is straightforward. Suppose sender i’s payoff function $\varPi _{i}( p_{i};G_{j})$ satisfies the linear structure. If it has a jump at $p_{i}=1$, then any strategy with support $[ 0,p_{i}] \cup \{ 1\}$ is optimal for sender i. If it does not have a jump at $p_{i}=1$, then any strategy with support $[ 0,p_{i}]$ is optimal for sender i.

To see the necessity, suppose that $( G_{1},G_{2})$ is a pair of equilibrium strategies. We first make a few preliminary observations. For notational simplicity, let ${\tilde{\rho }}\equiv \frac{\rho }{\mu _{1}\mu _{2}( 1-\mu _{1}) ( 1-\mu _{2}) }$. Then

$$\begin{aligned} \varPi _{i}\left( p_{i};G_{j}\right)= & {} \frac{G_{j}\left( \delta _{i}\left( p_{i}\right) \right) +\lim _{p\rightarrow p_{i}^{-}}G_{j}\left( \delta _{i}\left( p\right) \right) }{2}\nonumber \\&+\,{\tilde{\rho }}\left( p_{i}-\mu _{i}\right) \left( \int _{0}^{\delta _{i}\left( p_{i}\right) }\left( s-\mu _{j}\right) \mathrm{d}G_{j}\left( s\right) \right) . \end{aligned}$$

(8)

If $G_{j}$ is differentiable at $p_{i}$, then

$$\begin{aligned} \frac{\mathrm{d}\varPi _{i}\left( p_{i};G_{j}\right) }{\mathrm{d}p_{i}}= & {} \frac{\mathrm{d}G_{j}\left( \delta _{i}\left( p_{i}\right) \right) }{\mathrm{d}p_{i}}\left( 1+\tilde{\rho }\left( p_{i}-\mu _{i}\right) \left( \delta _{i}\left( p_{i}\right) -\mu _{j}\right) \right) \nonumber \\&+\,\underset{\le 0}{\underbrace{{\tilde{\rho }}\int _{0}^{\delta _{i}\left( p_{i}\right) }\left( s-\mu _{j}\right) \mathrm{d}G_{j}\left( s\right) }}. \end{aligned}$$

(9)

$\square$

Lemma 3

Suppose $( q_{0},q_{1}) \cap {\text {*}}{supp}G_{j}=\emptyset$. (i) $\varPi _{i}( p;G_{j})$ is linear and weakly decreasing on $( \delta _{j}( q_{0}) ,\delta _{j}( q_{1}) )$. (ii) If $G_{j}( \delta _{i}( q_{1}) ) <G_{j}( \delta _{i}( q_{2}) )$ for $q_{2}>q_{1}$, and $G_{j}( \cdot )$ is continuous at $\delta _{i}( q_{2})$, then $\frac{\varPi _{i}( q_{1};G_{j}) -\varPi _{i}( q_{0};G_{j}) }{q_{1}-q_{0}}<\frac{\varPi _{i}( q_{2};G_{j}) -\varPi _{i}( q_{0};G_{j}) }{q_{2}-q_{0}}$. Consequently, ${\text {*}}{con}[ \varPi _{i}|G_{j}] ( q_{1}) >\varPi _{i}( q_{1} ;G_{j})$.

Proof

• (i) By Eq. (9), for each $p_{i}\in \left( \delta _{j}\left( q_{0}\right) ,\delta _{j}\left( q_{1}\right) \right)$, we have $d\varPi _{i}\left( p_{i};G_{j}\right) /dp_{i}={\tilde{\rho }}\int _{0}^{\delta _{i}\left( p_{i}\right) }\left( s-\mu _{j}\right) \mathrm{d}G_{j}\left( s\right) \le 0$.

• (ii) Define $D\equiv \frac{\varPi _{i}\left( q_{2};G_{j}\right) -\varPi _{i}\left( q_{0};G_{j}\right) }{q_{2}-q_{0}}-\frac{\varPi _{i}\left( q_{1};G_{j}\right) -\varPi _{i}\left( q_{0};G_{j}\right) }{q_{1}-q_{0}}$. Then, for some $\tilde{q}\in \left[ \delta _{i}\left( q_{0}\right) ,\delta _{i}\left( q_{2}\right) \right]$,

  $$\begin{aligned} D&=\frac{G_{j}\left( \delta _{i}\left( q_{2}\right) \right) -G_{j}\left( \delta _{i}\left( q_{0}\right) \right) }{q_{2}-q_{0}} +\frac{{\tilde{\rho }}\left( q_{2}-\mu _{i}\right) \left( \int _{\delta _{i}\left( q_{0}\right) }^{\delta _{i}\left( q_{2}\right) }\left( s-\mu _{j}\right) \mathrm{d}G_{j}\left( s\right) \right) }{q_{2}-q_{0}}\\&=\frac{G_{j}\left( \delta _{i}\left( q_{2}\right) \right) -G_{j}\left( \delta _{i}\left( q_{0}\right) \right) }{q_{2}-q_{0}}\left( 1+\tilde{\rho }\left( q_{2}-\mu _{i}\right) \left( {\tilde{q}}-\mu _{j}\right) \right) \\&>\frac{G_{j}\left( \delta _{i}\left( q_{2}\right) \right) -G_{j}\left( \delta _{i}\left( q_{0}\right) \right) }{q_{2}-q_{0}}\left( 1+\tilde{\rho }\left( -\mu _{1}\left( 1-\mu _{2}\right) \right) \right) \ge 0 . \end{aligned}$$

$\square$

Define ${\bar{p}}_{i}\equiv \sup ( {\text {*}}{supp}G_{i})$ and ${\hat{p}}_{i}\equiv \sup ( {\text {*}}{supp}G_{i}\cap ( 0,1) )$.

Lemma 4

The following holds in an equilibrium. (i) ${\bar{p}}_{j}=\delta _{i}( {\bar{p}}_{i})$; (ii) $G_{i}( p)$ does not have an atom at any $p\in ( 0,1)$ and $G_{1}( 0) \times G_{2}( 0) =0$; (iii) $G_{i}( p)$ is strictly increasing in $( 0,\hat{p}_{i})$; and (iv) ${\hat{p}}_{j}=\delta _{i}( {\hat{p}}_{i})$.

Proof

Let ${\mathcal {P}}_{i}$ be the set of linear upper envelopes of $\varPi _{i}( p_{i};G_{j})$ at $p_{i}=\mu _{i}$. Then, ${\text {*}}{con}[ \varPi _{i}|G_{j}] ( p_{i}) \le L_{i}( p_{i} ;G_{j})$ for all $p_{i}\in [ 0,1]$ and $L_{i} \in {\mathcal {L}}_{i}\$by definition, and

$$\begin{aligned} {\text {*}}{supp}G_{i}\subset \left\{ p_{i}:{\bar{\varPi }}_{i}\left( p_{i};G_{j}\right) =\varPi _{i}\left( p_{i};G_{j}\right) \right\} \text { for all }{\bar{\varPi }}_{i}\in {\mathcal {P}}_{i} \end{aligned}$$

(10)

• (i) ${\bar{p}}_{j}=\delta _{i}\left( {\bar{p}}_{i}\right)$: Suppose ${\bar{p}}_{i}>\delta _{j}\left( {\bar{p}}_{j}\right)$. By (8), $\varPi _{i}\left( p_{i} ;G_{j}\right) =1$ for all $p_{i}\in \left( \delta _{j}\left( {\bar{p}} _{j}\right) ,1\right]$. Therefore by (10), $\left( \delta _{j}\left( \bar{p}_{j}\right) ,1\right] \cap {\text {*}}{supp}G_{i}=\emptyset$, a contradiction.

• (ii) No atom at any $p\in \left( 0,1\right)$ and $G_{1}\left( 0\right) \times G_{2}\left( 0\right) =0$: Suppose $G_{i}$ assigns an atom at ${\tilde{p}}\in \left( 0,1\right)$. Then, $\varPi _{j}$ exhibits an upward jump at $\delta _{i}\left( {\tilde{p}}\right)$. Therefore, there exists an $\varepsilon >0$ such that $\varPi _{j}\left( p;G_{i}\right) <{\text {*}}{con}\left[ \varPi _{j}|G_{i}\right] \$for all $p\in \left[ \delta _{i}\left( {\tilde{p}}\right) -\varepsilon ,\delta _{i}\left( \tilde{p}\right) \right]$, i.e., $\left[ \delta _{i}\left( {\tilde{p}}\right) -\varepsilon ,\delta _{i}\left( {\tilde{p}}\right) \right] \cap {\text {*}}{supp}G_{j}=\emptyset$. By Lemma 3, however, this implies that $\varPi _{i}\left( {\tilde{p}};G_{j}\right) <{\bar{\varPi }}_{i}\left( {\tilde{p}};G_{j}\right)$ for all ${\bar{\varPi }}_{i}\in {\mathcal {P}}_{i}$, a contradiction. Next, if $G_{i}$ assigns an atom at 0, then $\varPi _{j}\left( 0;G_{i}\right) <{\bar{\varPi }} _{i}\left( 0;G_{j}\right)$ for all ${\bar{\varPi }}_{i}\in {\mathcal {P}}_{i}$. Therefore, $G_{i}\left( 0\right) >0$ implies $G_{j}\left( 0\right) =0$.

• (iii) $G_{i}\left( p\right)$is strictly increasing on $\left( 0,{\hat{p}}_{i}\right)$: This immediately follows from Lemma 3.

• (iv) ${\hat{p}}_{j}=\delta _{i}\left( {\hat{p}}_{i}\right)$: Suppose ${\hat{p}}_{j}>\delta _{i}\left( {\hat{p}}_{i}\right)$. By part (i), ${\bar{p}}_{j}=\delta _{i}\left( {\bar{p}}_{i}\right)$. Therefore, $\delta _{i}\left( {\bar{p}}_{i}\right) >\delta _{i}\left( {\hat{p}}_{i}\right)$, or equivalently, ${\bar{p}}_{i}>{\hat{p}}_{i}$. Then, by the definition of $\hat{p}_{i}$, we have ${\bar{p}}_{i}={\bar{p}}_{j}=1$, and $G_{i}$ assigns an atom at ${\bar{p}}_{i}$. Also, this implies that for all $p\in \left[ {\hat{p}} _{i},1\right)$, $G_{i}\left( p\right) =G_{i}\left( {\hat{p}}_{i}\right)$. Then, Lemma 3 implies that $\varPi _{j}\left( p_{j};G_{i}\right) <{\bar{\varPi }}_{j}\left( p_{j};G_{i}\right)$ for all $p_{j}\in \left( \delta \left( {\hat{p}}_{i}\right) ,1\right)$ for some ${\bar{\varPi }}_{j}\in {\mathcal {P}}_{j}$, i.e., ${\hat{p}}_{j}\le \delta \left( \hat{p}_{i}\right)$, a contradiction.

$\square$

By (10), $p_{i}\in {\text {*}}{supp}G_{i}$ only if ${\bar{\varPi }}_{i}( p_{i};G_{j}) =\varPi _{i}( p_{i};G_{j})$. Therefore, properties (i)–(iv) together imply the linear structure of the payoff functions. $\square$

Proof of Corollary 1

Suppose $( G_{1},G_{2})$ is an equilibrium. By Theorem 1, $\varPi _{i}\left( p_{i};G_{j}\right)$ is linear on the interval $(0,{\hat{p}}_{i}]$. Therefore,

$$\begin{aligned} \frac{\mathrm{d}^{2}\varPi _{i}\left( p_{i};G_{j}\right) }{\mathrm{d}p_{i}^{2}}= & {} \left( 1+{\tilde{\rho }}\left( p_{i}-\mu _{i}\right) \left( \delta _{i}\left( p_{i}\right) -\mu _{j}\right) \right) \frac{\mathrm{d}^{2}\left( G_{j}\left( \delta _{i}\left( p_{i}\right) \right) \right) }{\mathrm{d}p_{i}^{2}}\\&+\,{\tilde{\rho }}\left( 2\left( \delta _{i}\left( p_{i}\right) -\mu _{j}\right) +\left( p_{i}-\mu _{i}\right) \delta _{i}^{\prime }\left( p_{i}\right) \right) \frac{\mathrm{d}\left( G_{j}\left( \delta _{i}\left( p_{i}\right) \right) \right) }{\mathrm{d}p_{i}}=0. \end{aligned}$$

The solution to the differential equation above is

$$\begin{aligned} G_{j}\left( p_{i}\right) =G_{j}\left( 0\right) +C_{j}\int _{0}^{\delta _{j}\left( p_{i}\right) }\exp \left( \int _{0}^{s^{\prime }}\varLambda _{j}\left( s\right) \mathrm{d}s\right) \mathrm{d}s^{\prime }, \end{aligned}$$

for some integration constant $C_{j}$ and $\varLambda _{j}( s) \equiv -\frac{\rho ( ( s-\mu _{i}) \delta _{i}^{\prime }( s) +2( \delta _{i}( s) -\mu _{j}) ) }{\mu _{1}\mu _{2}( 1-\mu _{1}) ( 1-\mu _{2}) +\rho ( s-\mu _{i}) ( \delta _{i}( s) -\mu _{j}) }$. Substituting $p_{i}={\hat{p}}_{i}$ in the solution above, we get $C_{j}=\frac{G_{j}( {\hat{p}}_{j}) -G_{j}( 0) }{\int _{0}^{\delta _{j}( {\hat{p}}_{j}) }\varPhi _{j}( s^{\prime }) \mathrm{d}s^{\prime }}$. Equation (6) is obtained by substituting the integration constant to the solution above.

Next, using Eq. (6), the Bayes-plausibility condition for sender j can be simplified as follows.

$$\begin{aligned} 1-\mu _{j}&=\int _{0}^{\delta _{i}\left( {\hat{p}}_{i}\right) }G_{j}\left( p\right) \mathrm{d}p+G_{j}\left( \delta _{i}\left( {\hat{p}}_{i}\right) \right) \left( 1-\delta _{i}\left( {\hat{p}}_{i}\right) \right) \\&=G_{j}\left( {\hat{p}}_{j}\right) -\left( G_{j}\left( {\hat{p}}_{j}\right) -G_{j}\left( 0\right) \right) \frac{\int _{0}^{{\hat{p}}_{j}}\left( \int _{\delta _{j}\left( p_{j}\right) }^{\delta _{j}\left( {\hat{p}}_{j}\right) }\varPhi _{j}\left( s\right) \mathrm{d}s\right) \mathrm{d}p_{j}}{\int _{0}^{\delta _{j}\left( {\hat{p}}_{j}\right) }\varPhi _{j}\left( s\right) \mathrm{d}s}\\&=G_{j}\left( {\hat{p}}_{j}\right) -\left( G_{j}\left( {\hat{p}}_{j}\right) -G_{j}\left( 0\right) \right) \frac{\int _{0}^{\delta _{j}\left( {\hat{p}} _{j}\right) }\delta _{i}\left( s\right) \varPhi _{j}\left( s\right) \mathrm{d}s}{\int _{0}^{\delta _{j}\left( {\hat{p}}_{j}\right) }\varPhi _{j}\left( s\right) \mathrm{d}s}\\&=G_{j}\left( {\hat{p}}_{j}\right) -C_{j}\int _{0}^{\delta _{j}\left( \hat{p}_{j}\right) }\delta _{i}\left( s\right) \varPhi _{j}\left( s\right) \mathrm{d}s. \end{aligned}$$

This gives the simplified Bayes-plausibility condition.

The slope of $\varPi _{i}( p_{i};G_{j})$ on $[ 0,{\hat{p}} _{i}]$ is

$$\begin{aligned} \frac{\mathrm{d}\varPi _{i}\left( p_{i};G_{j}\right) }{\mathrm{d}p_{i}}&=\lim _{p_{i} \rightarrow 0^{+}}\frac{\mathrm{d}G_{j}\left( \delta _{i}\left( p_{i}\right) \right) }{\mathrm{d}p_{i}}\times \left( 1+\frac{\rho }{\left( 1-\mu _{1}\right) \left( 1-\mu _{2}\right) }\right) \\&\quad +\frac{\rho }{\left( 1-\mu _{1}\right) \left( 1-\mu _{2}\right) }\mu _{j}G_{j}\left( 0\right) \\&=C_{j}\times \left( 1+\frac{\rho }{\left( 1-\mu _{1}\right) \left( 1-\mu _{2}\right) }\right) +\frac{\rho }{\left( 1-\mu _{1}\right) \left( 1-\mu _{2}\right) }\mu _{j}G_{j}\left( 0\right) . \end{aligned}$$

Lastly,

$$\begin{aligned} \lim _{p_{i}\rightarrow 1}\varPi _{i}\left( p_{i};G_{j}\right)&=G_{j}\left( {\hat{p}}_{j}\right) -{\tilde{\rho }}\left( 1-\mu _{i}\right) \left( \mu _{j}G_{j}\left( {\hat{p}}_{j}\right) -\left( G_{j}\left( {\hat{p}}_{j}\right) -\left( 1-\mu _{j}\right) \right) \right) \\&=\left( \frac{\rho +\mu _{1}\mu _{2}}{\mu _{1}\mu _{2}}\right) G_{j}\left( {\hat{p}}_{j}\right) -\frac{\rho }{\mu _{1}\mu _{2}}. \end{aligned}$$

Therefore, if $G_{j}( {\hat{p}}_{j}) <1$, then

$$\begin{aligned} \frac{\left( \mu _{1}\mu _{2}-\rho \right) +\left( \mu _{1}\mu _{2}+\rho \right) G_{j}\left( {\hat{p}}_{j}\right) }{2\mu _{1}\mu _{2}}&=G_{j}\left( 0\right) +C_{j}\left( 1+\frac{\rho }{\left( 1-\mu _{1}\right) \left( 1-\mu _{2}\right) }\right) \\&\quad +\frac{\rho }{\left( 1-\mu _{1}\right) \left( 1-\mu _{2}\right) }\mu _{j}G_{j}\left( 0\right) \\&=\frac{\left( 1-\mu _{1}\right) \left( 1-\mu _{2}\right) +\rho \mu _{j} }{\left( 1-\mu _{1}\right) \left( 1-\mu _{2}\right) }G_{j}\left( 0\right) \\&\quad +\frac{\left( 1-\mu _{1}\right) \left( 1-\mu _{2}\right) +\rho }{\left( 1-\mu _{1}\right) \left( 1-\mu _{2}\right) }C_{j}. \end{aligned}$$

Other conditions follows from Theorem 1. $\square$

Proof of Theorem 2

Equilibrium Existence: The strategy space is compact (with respect to weak*-topology). Sender i’s payoff (as a function of strategy profile) $V_{i}( G_{i},G_{j}) \equiv E_{G_{i}}[ \varPi _{i}( p;G_{j}) ] \$is linear in $G_{i}$ and hence is quasiconcave in $G_{i}$. The game is zero sum, i.e., $V_{1}( G_{1},G_{2}) +V_{2}( G_{2},G_{1}) =1$, and hence satisfies reciprocal upper semicontinuity. Therefore, if we show that the payoff function satisfies payoff security, Corollary 3.3 of Reny (1999) guarantees the existence of a pure-strategy equilibrium. Fix an arbitrary strategy profile $( G_{i},G_{j})$ and $\varepsilon >0$.

We show below that there exists a strategy ${\tilde{G}}_{i}$ of sender i that is continuous on $[ 0,1) \$and $V_{i}( {\tilde{G}}_{i} ,G_{j}) >V_{i}( G_{i},G_{j}) -\varepsilon /2$. Note that the set of discontinuous points $D\subset [ 0,1]$ of $G_{i}$ is countable. We thus can denote $D=\{ d_{1},d_{2},\cdots ,d_{l} ,\cdots \}$, where $d_{l}<d_{l+1}$. Let $t_{l}$ be the size of atom at $d_{l}$. First, suppose $\varPi _{i}( p_{i};G_{j})$ is continuous at $d_{l}$. Then, there exists an interval $( d_{l}-\varepsilon _{l} ,d_{l}+\varepsilon _{l})$ such that $\varPi _{i}( p_{i};G_{j}) >\varPi _{i}( d_{l};G_{j}) -( \frac{\varepsilon }{2+\varepsilon }) ^{l}$ for all $p_{i}\in ( d_{l}-\varepsilon _{l},d_{l} +\varepsilon _{l})$. Now replacing the atom $t_{l}$ at $d_{l}$ with a uniform distribution over the interval $( d_{l}-\varepsilon _{l} ,d_{l}+\varepsilon _{l})$ gives a new distribution $G_{i}^{\prime }$ such that $G_{i}^{\prime }$ does not have an atom at $d_{l}$ and $V_{i}( G_{i}^{\prime },G_{j}) >V_{i}( G_{i},G_{j}) -( \frac{\varepsilon }{2+\varepsilon }) ^{l}$. Next suppose $\varPi _{i}( p_{i};G_{j})$ is discontinuous at $d_{l}$. Recall that while $\varPi _{i}( p_{i};G_{j})$ may be decreasing, it cannot jump downward. Thus, it is necessary that $\lim _{p_{i}\rightarrow d_{l}^{-}}\varPi _{i}( p_{i};G_{j})<\varPi _{i}( d_{l};G_{j}) <\lim _{p_{i} \rightarrow d_{l}^{+}}\varPi _{i}( p_{i};G_{j})$. Choose a pair $\varepsilon _{l},\varepsilon _{l}^{\prime }>0$ such that (i) for all $p^{\prime }\in ( d_{l},d_{l}+\varepsilon _{l})$, $\varPi _{i}( d_{l} ;G_{j}) <\varPi _{i}( p^{\prime };G_{j})$, and (ii) $\frac{\varepsilon _{l}^{\prime }}{\varepsilon _{l}}>t_{l}( \frac{2+\varepsilon }{\varepsilon }) ^{l}-1$ and (iii) $d_{l}-\varepsilon _{l}^{\prime }>0$. Replace the atom at $d_{l}$ with a uniform distribution with density $\frac{\varepsilon _{l}}{\varepsilon _{l}^{\prime }( \varepsilon _{l}+\varepsilon _{l}^{\prime }) }t_{l}$ over the interval $( d_{l}-\varepsilon _{l}^{\prime },d_{l})$, as well as a uniform distribution with density $\frac{\varepsilon _{l}^{\prime }}{\varepsilon _{l}( \varepsilon _{l}+\varepsilon _{l}^{\prime }) }t_{l}$ over the interval $( d_{l},d_{l}+\varepsilon _{l})$.²⁵ The new strategy $G_{i}^{\prime }$ thus obtained has no atom at $d_{l}$. Moreover, as the additional mass assigned to the interval $( d_{l}-\varepsilon _{l}^{\prime },d_{l})$ is $\frac{\varepsilon _{l}}{\varepsilon _{l}^{\prime }( \varepsilon _{l}+\varepsilon _{l}^{\prime }) } t_{l}\times \varepsilon _{l}^{\prime }<( \frac{\varepsilon }{2+\varepsilon }) ^{l}$, we have $V_{i}( G_{i}^{\prime },G_{j}) >V_{i}( G_{i},G_{j}) -( \frac{\varepsilon }{2+\varepsilon }) ^{l}$.

Next, as ${\tilde{G}}_{i}$ is continuous on [0, 1) , $V_{i}( {\tilde{G}}_{i},G_{j})$ is lower semicontinuous with respect to $G_{j}$.²⁶ Consequently, there exists a neighborhood $O( G_{j})$ of $G_{j}$ such that for all $G_{j}\in O( G_{j})$, $V_{i}( \tilde{G}_{i},G_{j}) >V_{i}( {\tilde{G}}_{i},G_{j}) -\varepsilon /2$. We thus have the payoff security at $( G_{i},G_{j})$.

Equilibrium Uniqueness: Suppose there are two equilibria $( G_{1},G_{2})$ and $( G_{1}^{\prime },G_{2}^{\prime })$. Define ${\hat{p}}_{i}\equiv \sup ( {\text {*}}{supp}G_{i}\cap ( 0,1) )$ and ${\bar{p}}_{i}\equiv \sup ( {\text {*}}{supp} G_{i})$; ${\hat{p}}_{i}^{\prime }$ and ${\bar{p}}_{1}^{\prime }$ are defined accordingly for the equilibrium $( G_{1}^{\prime },G_{2}^{\prime })$. By the interchangeability property of zero-sum games, $( G_{1} ,G_{2}^{\prime })$ and $( G_{1}^{\prime },G_{2})$ are also equilibria. As a result, $G_{1}$ and $G_{1}^{\prime }$ have a common support, i.e., ${\hat{p}}_{1}={\hat{p}}_{1}^{\prime }$ and ${\bar{p}}_{1}={\bar{p}}_{1}^{\prime }$. Similarly, $G_{2}$, and $G_{2}^{\prime }$ have a common support, i.e., ${\hat{p}}_{2}={\hat{p}}_{2}^{\prime }$ and ${\bar{p}}_{2}={\bar{p}}_{2}^{\prime }$. In other words, the support of equilibrium strategies is unique. Finally, we explain why this, together with Bayes-plausibility and the necessity of the linear structure of equilibrium strategy uniquely pin down the equilibrium strategy. If $G_{i}( {\hat{p}}_{i}) =1$ (and hence $G_{j}( {\hat{p}} _{j}) =1$), then the simplified Bayes-plausibility condition in Corollary 1 implies that $G_{i}( 0) =G_{i}^{\prime }( 0)$ and $G_{j}( 0) =G_{j}^{\prime }( 0)$. Next, suppose $G_{1}( {\hat{p}}_{1}) <1$. For each $i\in \{ 1,2\}$, fixing a ${\hat{p}}_{i}$, the simplified Bayes-plausibility condition and the atom condition give a system of two equations with two unknowns ($G_{i}( 0)$ and $G_{i}( {\hat{p}}_{i})$). It is straightforward to verify that there exists a unique solution to the system, so $G_{i}( 0) =G_{i}^{\prime }( 0)$ and $G_{i}( {\hat{p}} _{i}) =G_{i}^{\prime }( {\hat{p}}_{i})$. $\square$

Proof of Theorem 3

For expositional simplicity, we refer an equilibrium such that $G_{1}( {\hat{p}}_{1}) =G_{2}( {\hat{p}}_{2}) =1$ as a type-N equilibrium, and $\max \{ G_{1}( {\hat{p}}_{1}) ,G_{2}( {\hat{p}}_{2}) \} <1$ as a type-A equilibrium.

Lemma 5

A strategy profile $( G_{1},G_{2})$ is a type-N equilibrium if and only if there exist (i) ${\hat{p}}_{1}\in [ 0,1]$ and $\hat{p}_{2}=\delta _{1}( {\hat{p}}_{1})$; and (ii) $G_{1}( 0) =0$ and $G_{2}( 0) \ge 0$ such that

$$\begin{aligned} G_{i}\left( p\right)&=G_{i}\left( 0\right) +\left( 1-G_{i}\left( 0\right) \right) \frac{\delta _{i}\left( \min \left\{ p,{\hat{p}}_{i}\right\} \right) }{\delta _{i}\left( {\hat{p}}_{i}\right) }\text {, and }\\ \mu _{i}&=E_{i}^{N}\left( {\hat{p}}_{i}\right) \equiv \left( 1-G_{i}\left( 0\right) \right) \times \int _{0}^{{\hat{p}}_{i}}\left( 1-\frac{\delta _{i}\left( s\right) }{\delta _{i}\left( {\hat{p}}_{i}\right) }\right) \mathrm{d}s. \end{aligned}$$

A strategy profile $( G_{1},G_{2})$ is a type-A equilibrium if and only if there exist (i) ${\hat{p}}_{1}\in [ 0,1]$ and $\hat{p}_{2}=\delta _{1}( {\hat{p}}_{1})$; and (ii) $G_{1}( 0) =0$ and $G_{2}( 0) \ge 0$ such that

$$\begin{aligned} G_{i}\left( p\right)&=\left\{ \begin{array} [c]{cc} G_{i}\left( 0\right) +\left( 1-G_{i}\left( 0\right) \right) \frac{\delta _{i}\left( \min \left\{ p,{\hat{p}}_{i}\right\} \right) }{2-\delta _{i}\left( {\hat{p}}_{i}\right) } &{} \text {if }p\in \left[ 0,1\right) \\ 1 &{} \text {if }p=1 \end{array} \right. \text {, and }\\ \mu _{i}&=E_{i}^{A}\left( {\hat{p}}_{i}\right) \equiv \left( 1-G_{i}\left( 0\right) \right) \left( \frac{2-\left( 2-{\hat{p}}_{i}\right) \delta _{i}\left( {\hat{p}}_{i}\right) -\int _{0}^{{\hat{p}}_{i}}\delta _{i}\left( s\right) \mathrm{d}s}{2-\delta _{i}\left( {\hat{p}}_{i}\right) }\right) . \end{aligned}$$

Proof

The linear structure of payoff functions implies that there exist ${\hat{p}} _{1}\in [ 0,1]$ and ${\hat{p}}_{2}=\delta _{1}( {\hat{p}} _{1})$ such that we can express ${\tilde{\varPi }}_{i}$ as

$$\begin{aligned} {\tilde{\varPi }}_{i}\left( p;G_{j}\right) =\left\{ \begin{array} [c]{cc} \frac{1}{2}G_{j}\left( 0\right) &{} \text {if }\,\,p=0\\ G_{j}\left( 0\right) +\left( G_{j}\left( {\hat{p}}_{j}\right) -G_{j}\left( 0\right) \right) \frac{p}{{\hat{p}}_{i}} &{} \text {if }\,\,p\in (0,{\hat{p}}_{i}]\\ G_{j}\left( {\hat{p}}_{j}\right) &{} \text {if }\,\,p\in ({\hat{p}}_{i},1)\\ \frac{1}{2}\left( 1+G_{j}\left( {\hat{p}}_{j}\right) \right) &{} \text {if }\,\,p=1 \end{array} \right. , \end{aligned}$$

(11)

for $i=1,2$ and $j\ne i$. Since $G_{j}( \delta _{i}( p) ) ={\tilde{\varPi }}_{i}( p;G_{j})$ for $p\in ( 0,\hat{p}_{i})$, we have $G_{j}( {\tilde{p}}) ={\tilde{\varPi }} _{i}( \delta _{j}( {\tilde{p}}) ;G_{j})$ for ${\tilde{p}}\in ( 0,\delta _{j}( {\hat{p}}_{i}) ) =( 0,{\hat{p}}_{j})$. Therefore,

$$\begin{aligned} G_{j}\left( p\right) =\left\{ \begin{array} [c]{cc} G_{j}\left( 0\right) +\left( G_{j}\left( {\hat{p}}_{j}\right) -G_{j}\left( 0\right) \right) \frac{\delta _{j}\left( \min \left\{ p,{\hat{p}}_{j}\right\} \right) }{\delta _{j}\left( {\hat{p}}_{j}\right) } &{} \text {if }\,\,p\in \left[ 0,1\right) \\ 1 &{} \text {if }\,\,p=1 \end{array} \right. . \end{aligned}$$

A few remarks are useful for our subsequent analysis. First, $\delta _{1}( p)$ is an increasing concave function, while $\delta _{2}( p)$ is an increasing convex function. Therefore, as illustrated in Fig. 5, $G_{1}( p_{1})$ is concave on the interval $(0,{\hat{p}}_{1}]$, whereas $G_{2}( p_{2})$ is convex on the interval $(0,{\hat{p}}_{2}]$. Second, ${\hat{p}}_{i}=\delta _{i}( {\hat{p}}_{j})$; and $\delta _{i}( p) =1$ if and only if $p=1$. Third, if an equilibrium is type-A, i.e., $\max \{ G_{1}( {\hat{p}}_{1}) ,G_{2}( {\hat{p}}_{2}) \} <1$, then the linear structure of payoff functions requires that for $i=1,2$ and $j\ne i$,

$$\begin{aligned} \frac{{\tilde{\varPi }}_{i}\left( {\hat{p}}_{i};G_{j}\right) -\lim _{p\rightarrow 0^{+}}{\tilde{\varPi }}_{i}\left( p;G_{j}\right) }{{\hat{p}}_{i}}={\tilde{\varPi }} _{i}\left( 1;G_{j}\right) -\lim _{p\rightarrow 0^{+}}{\tilde{\varPi }}_{i}\left( p;G_{j}\right) . \end{aligned}$$

As $\lim _{p\rightarrow 0^{+}}{\tilde{\varPi }}_{i}\left( p;G_{j}\right) =G_{j}\left( 0\right)$, ${\tilde{\varPi }}_{i}\left( 1;G_{j}\right) =\frac{1}{2}\left( 1+G_{j}\left( {\hat{p}}_{j}\right) \right)$ and $\tilde{\varPi }_{i}\left( {\hat{p}}_{i};G_{j}\right) =G_{j}\left( {\hat{p}}_{j}\right)$, the equation above can be written as

$$\begin{aligned} \frac{G_{j}\left( {\hat{p}}_{j}\right) -G_{j}\left( 0\right) }{{\hat{p}}_{i} }=\frac{1}{2}\left( 1+G_{j}\left( {\hat{p}}_{j}\right) \right) -G_{j}\left( 0\right) . \end{aligned}$$

(12)

We are left to show that $G_{1}( 0) =0$. If $G_{1}( 0) >0$, then $G_{2}( 0) =0$. Suppose that there exists a type-N equilibrium, i.e., an equilibrium such that $G_{1}( {\hat{p}} _{1}) =1$. Recall that ${\hat{p}}_{2}=\delta _{1}( {\hat{p}} _{1}) >{\hat{p}}_{1}$. As $G_{1}$ is concave on the interval $(0,\hat{p}_{1}]$, whereas $G_{2}$ is convex on the interval $(0,{\hat{p}}_{2}]$, we have that $G_{1}( p) \ge G_{2}( p)$ for all p with strict inequality on the interval $( 0,{\hat{p}}_{1})$. This implies $G_{2}$ first-order stochastically dominates $G_{1}$, contradicting that $\mu _{1}>\mu _{2}$. Next, suppose there exists a type-A equilibrium, i.e., an equilibrium such that $G_{1}( {\hat{p}}_{1}) <1$. Rearranging condition (12) and using the hypothesis that $G_{1}( 0) >0=G_{2}( 0)$, we get

$$\begin{aligned} G_{1}\left( {\hat{p}}_{1}\right) =\frac{{\hat{p}}_{2}+2\left( 1-{\hat{p}} _{2}\right) G_{1}\left( 0\right) }{2-{\hat{p}}_{2}}\text {, and }G_{2}\left( {\hat{p}}_{2}\right) =\frac{{\hat{p}}_{1}}{2-{\hat{p}}_{1}}. \end{aligned}$$

As ${\hat{p}}_{2}>{\hat{p}}_{1}$, we have $G_{1}( {\hat{p}}_{1}) >G_{2}( {\hat{p}}_{2})$. Consequently, $G_{1}( p) \ge G_{2}( p)$ with strict inequality on the interval (0, 1) . This implies $G_{2}$ first-order stochastically dominates $G_{1}$, contradicting that $\mu _{1}>\mu _{2}$. Therefore, we have the required result.

Finally, the conditions $\mu _{i}=E_{i}^{A}({\hat{p}}_{1})$ for type-A equilibrium, and $\mu _{i}=E_{i}^{N}( {\hat{p}}_{1})$ for type-N equilibrium, are obtained by the Bayes-plausibility condition for sender i. $\square$

Now we prove the comparative statics result. Take a pair of normalized quality covariances $\rho _{L}$ and $\rho _{H}>\rho _{L}$. Let $k_{l}=\frac{\mu _{1} }{1-\mu _{1}}\frac{1-\mu _{2}}{\mu _{2}}\frac{( 1-\mu _{1}) \mu _{2}-\rho _{l}}{\mu _{1}( 1-\mu _{2}) -\rho _{l}}$, $l=L,H$. It is immediate that $\rho _{L}<\rho _{H}$ is equivalent to $k_{L}>k_{H}$. Denote by $( G_{1l},G_{2l})$ the unique equilibrium under normalized covariance $\rho _{l}$, for $l\in \{ L,H\}$. Other functions such as $\delta _{il}$, ${\tilde{\varPi }}_{il}$, $E_{il}^{N},$ and $E_{il}^{A}$, as well as parameters such as ${\hat{p}}_{il}$ are also defined accordingly for $l\in \{ L,H\}$.

Case 1 $\mu _{1}\in ( 0,1-\int _{0}^{1}\delta _{1H}( p) \mathrm{d}p)$: As $\delta _{1L}( p) <\delta _{1H}( p)$ for all $p\in ( 0,1)$, the equilibria are of type-N for both $\rho _{L}$ and $\rho _{H}$, i.e., $G_{1L}( {\hat{p}}_{1L}) =G_{1H}( {\hat{p}}_{1H}) =1$. Furthermore, as $E_{1L}^{N}( \cdot ) >E_{1H}^{N}( \cdot )$ and $E_{il}^{N}( \cdot )$ is a strictly increasing function, we have ${\hat{p}}_{1L} <{\hat{p}}_{1H}$ because of the equilibrium condition ${\hat{p}}_{1l}=( E_{il}^{N}) ^{-1}( \mu _{1})$. Next, note that on the interval $[ 0,{\hat{p}}_{2L}]$, while ${\tilde{\varPi }}_{2L}( p;G_{1L}) =G_{1L}( \delta _{2L}( p) )$ is linear, ${\tilde{\varPi }}_{2H}( \delta _{1H}( \delta _{2L}( p) ) ;G_{1H}) =G_{1H}( \delta _{2L}( p) )$ is concave. Furthermore, as $G_{1L}( \delta _{2L}( 0) ) =G_{1H}( \delta _{2L}( 0) ) =0$, and $1=G_{1L}( \delta _{2L}( {\hat{p}}_{2L}) ) >G_{1H}( \delta _{2L}( {\hat{p}}_{2L}) )$, $G_{1L}( p)$ and $G_{1H}( p)$ cross exactly once on $(0,{\hat{p}}_{1L}]$. This proves $G_{1H}\succ G_{1L}$.

Next, we consider sender 2’s strategy. First ${\hat{p}}_{1L}<{\hat{p}}_{1H}$ implies ${\hat{p}}_{2L}<{\hat{p}}_{2H}$. We show below that $G_{2L}( 0) <G_{2H}( 0)$. Suppose not, then $G_{2L}( 0) \ge G_{2H}( 0)$, together with the linear structure of ${\tilde{\varPi }}_{1}$, implies ${\tilde{\varPi }}_{1}( p;G_{2L}) \ge {\tilde{\varPi }}_{1}( p;G_{2H})$ on $[ 0,{\hat{p}} _{1L}]$. Since for all $p>0$, ${\tilde{\varPi }}_{1L}( p;G_{2L} ) =G_{2L}( \delta _{1L}( p) )$ and ${\tilde{\varPi }}_{1H}( p;G_{2H}) =G_{2H}( \delta _{1H}( p) ) >G_{2H}( \delta _{1L}( p) )$, we have that $G_{2H}$ first-order stochastically dominates $G_{2L}$, a contradiction. Next, on $(0,{\hat{p}}_{1L}]$, while $G_{2L}( \delta _{1L}( p) ) ={\tilde{\varPi }}_{1L}( p;G_{2L})$ is linear, $G_{2H}( \delta _{1L}( p) ) ={\tilde{\varPi }} _{1L}( \delta _{2H}( \delta _{1L}( p) ) ;G_{2H})$ is strictly convex in p. Also $G_{2L}( \delta _{1L}( 0) ) =G_{2L}( 0) <G_{2H}( 0) =G_{2H}( \delta _{1L}( 0) )$, and $1=G_{2L}( \delta _{1L}( {\hat{p}}_{1L}) ) >G_{2H}( \delta _{1L}( {\hat{p}}_{1L}) )$. Hence, $G_{2L}( p)$ and $G_{2H}( p)$ cross exactly once on the interval $(0,{\hat{p}}_{2L}]$. We thus have $G_{2H}\succ G_{2L}$.

Case 2 $\mu _{1}\in [ 1-\int _{0}^{1}\delta _{1L}( p) \mathrm{d}p,1)$: In this case, both $G_{1H}$ and $G_{1L}$ are of type-A. We start with the proof that ${\hat{p}}_{1H}<{\hat{p}}_{1L}$ and ${\hat{p}}_{2H} <{\hat{p}}_{2L}$. Suppose that ${\hat{p}}_{2H}\ge {\hat{p}}_{2L}$. Then, the linear structure of ${\tilde{\varPi }}_{2l}$ requires that $\frac{{\tilde{\varPi }}_{2L}( {\hat{p}}_{2L};G_{1L}) }{{\hat{p}}_{2L}}<\frac{{\tilde{\varPi }}_{2H}( {\hat{p}}_{2H};G_{1H}) }{{\hat{p}}_{2H}}$. This implies that $\tilde{\varPi }_{2L}( p;G_{1L}) \le {\tilde{\varPi }}_{2H}( p;G_{1H})$ for all $p\in [ 0,{\hat{p}}_{2L}]$. Notice, however, that ${\tilde{\varPi }}_{2L}( p;G_{1L}) =G_{1L}( \delta _{2L}( p) ) >G_{1L}( \delta _{2H}( p) )$ and ${\tilde{\varPi }}_{2H}( p;G_{1H}) =G_{1H}( \delta _{2H}( p) )$. Therefore, $G_{1L}$ first-order stochastically dominates $G_{1H}$, a contradiction. Since $\delta _{1L}( p) <\delta _{1H}( p)$, ${\hat{p}}_{2H}<{\hat{p}}_{2L}$ implies ${\hat{p}} _{1H}<{\hat{p}}_{1L}$.

Notice that ${\hat{p}}_{2H}<{\hat{p}}_{2L}$ and the linear structure of ${\tilde{\varPi }}_{2}$ together imply ${\tilde{\varPi }}_{2L}( p;G_{1L}) >{\tilde{\varPi }}_{2H}( p;G_{1H})$ and ${\tilde{\varPi }}_{2L}( {\hat{p}}_{2L};G_{1L}) >{\tilde{\varPi }}_{2H}( {\hat{p}}_{2H} ;G_{1H})$. On the interval $[ 0,{\hat{p}}_{2H}]$, $G_{1L}( \delta _{2L}( p) ) ={\tilde{\varPi }}_{2L}( p;G_{1L})$ is linear while $G_{1H}( \delta _{2L}( p) ) ={\tilde{\varPi }}_{2H}( \delta _{1H}( \delta _{2L}( p) ) ;G_{1H})$ is strictly concave. Furthermore, $G_{1L}( \delta _{2L}( 0) ) =G_{1H}( \delta _{2L}( 0) ) =0$ and $G_{1L}( \delta _{2L}( {\hat{p}}_{2H}) ) >G_{1H}( \delta _{2L}( {\hat{p}}_{2H}) )$. Therefore, $G_{1L}( p)$ and $G_{1H}( p)$ cross exactly once on $(0,{\hat{p}}_{1H}]$. Since $G_{1H}( p) <G_{1L}( p)$ for all $p\in ( {\hat{p}}_{1H},1)$, we have $G_{1H}\succ G_{1L}$.

Next, we establish that $G_{2H}\succ G_{2L}$. To this end, we first show $G_{2H}( 0) >G_{2L}( 0)$ and $G_{2H}( \hat{p}_{2H}) <G_{2L}( {\hat{p}}_{2L})$. Recall from the lemma above that the simplified Bayes-plausibility condition for sender 2 is $\mu _{2}=( 1-G_{2l}( 0) ) B( {\hat{p}} _{1l},\delta _{1l})$, where $B( {\hat{p}}_{1l},\delta _{1l}) \equiv \frac{\int _{0}^{{\hat{p}}_{1l}}\delta _{1l}( s) \mathrm{d}s+2( 1-{\hat{p}}_{1l}) }{2-{\hat{p}}_{1l}}$. Furthermore, $B( {\hat{p}} _{1L},\delta _{1H}) <B( {\hat{p}}_{1H},\delta _{1H})$ because ${\hat{p}}_{1H}<{\hat{p}}_{1L}$ and $\frac{\partial B( {\hat{p}}_{1l} ,\delta _{1l}) }{\partial {\hat{p}}_{1l}}=\frac{-2( 1-\delta ( {\hat{p}}_{1l}) ) -\int _{0}^{{\hat{p}}_{1}}[ \delta _{1l}( {\hat{p}}_{1}) -\delta _{1l}( s) ] \mathrm{d}s}{( 2-{\hat{p}}_{1l}) ^{2}}<0$. As $B( {\hat{p}}_{1L} ,\delta _{1L}) <B( {\hat{p}}_{1L},\delta _{1H})$, $G_{2H}( 0) >G_{2L}( 0)$. Next, we show that $G_{2H}( {\hat{p}}_{2H}) <G_{2L}( {\hat{p}}_{2L})$. Observe that the simplified Bayes-plausibility condition, together with the linear structure of the equilibrium, implies that $G_{2l}( {\hat{p}} _{2l}) =1-\mu _{2}( 1+\frac{\int _{0}^{\delta _{2l}( \hat{p}_{2l}) }\delta _{1l}( s) \mathrm{d}s}{2( 1-\delta _{2l}( {\hat{p}}_{2l}) ) }) ^{-1}$. Recall that $\delta _{1l}( p) \equiv ( 1+k_{l}\frac{1-p}{p}) ^{-1}$. Evaluating the integral gives $\frac{\int _{0}^{\delta _{2l}( {\hat{p}}_{2l}) }\delta _{1l}( s) \mathrm{d}s}{2( 1-\delta _{2l}( {\hat{p}}_{2l}) ) }=\frac{\frac{k_{l}}{1-k_{l}} {\hat{p}}_{2l}+\frac{k_{l}}{( 1-k_{l}) ^{2}}( 1-( 1-k_{l}) {\hat{p}}_{2l}) \ln ( 1-( 1-k_{l}) {\hat{p}}_{2l}) }{1-{\hat{p}}_{2l}}\equiv Q( {\hat{p}}_{2l} ,k_{l})$. By direct computation, $\frac{\partial Q}{\partial \hat{p}_{2l}}=\frac{k_{l}( {\hat{p}}_{2l}( 1-k_{l}) +k_{l} \ln ( 1-( 1-k_{l}) {\hat{p}}_{2l}) ) }{( 1-{\hat{p}}_{2l}) ^{2}( k_{l}-1) ^{2}}>0$ and $\frac{\partial Q}{\partial k_{l}}=\frac{( 1-k_{l}^{2}) {\hat{p}} _{2l}+( 1-( 1-k_{l}) {\hat{p}}_{2l}+k_{l}) \ln ( 1-( 1-k_{l}) {\hat{p}}_{2l}) }{( 1-{\hat{p}} _{2l}) ( 1-k_{l}) ^{3}}>0$. We thus have $Q( \hat{p}_{2L},k_{L})>Q( {\hat{p}}_{2L},k_{H}) >Q( \hat{p}_{2H},k_{H})$, so $G_{2L}( {\hat{p}}_{2L}) >G_{2H}( {\hat{p}}_{2H})$.

The linear structure of ${\tilde{\varPi }}_{1l}$, together with the facts that ${\tilde{\varPi }}_{1L}( 0;G_{2L}) <{\tilde{\varPi }}_{1H}( 0;G_{2H})$, ${\hat{p}}_{1H}<{\hat{p}}_{1L}$, and ${\tilde{\varPi }}_{1}( {\hat{p}}_{1L};G_{2L}) >{\tilde{\varPi }}_{1}( {\hat{p}}_{1H} ;G_{2H})$ imply ${\tilde{\varPi }}_{1L}( 1;G_{2L}) >\tilde{\varPi }_{1H}( 1;G_{2H})$. Next, denote by ${\text {*}}{con}[ {\tilde{\varPi }}_{1L}( p;G_{2L}) ]$ the concave closure of ${\tilde{\varPi }}_{1L}( p;G_{2L})$. Note that both ${\text {*}}{con}[ {\tilde{\varPi }}_{1L}( p;G_{1L}) ]$ and ${\text {*}}{con}[ {\tilde{\varPi }}_{1H}( \delta _{2H}( \delta _{1L}( p) ) ;G_{2H}) ]$ are linear in $p\in [ 0,1]$. Moreover, ${\text {*}}{con}[ \tilde{\varPi }_{1H}( \delta _{2H}( \delta _{1L}( 0) ) ;G_{2H}) ] ={\text {*}}{con}[ {\tilde{\varPi }}_{1H}( 0;G_{2H}) ] =G_{2H}( 0) >G_{2L}( 0) ={\text {*}}{con}[ {\tilde{\varPi }}_{1L}( 0;G_{2L}) ]$, and ${\text {*}}{con}[ {\tilde{\varPi }}_{1H}( \delta _{2H}( \delta _{L}( 1) ) ;G_{2H}) ] ={\text {*}}{con}[ {\tilde{\varPi }}_{1H}( 1;G_{2H}) ] =\frac{1}{2}( 1+G_{2H}( {\hat{p}}_{2H}) ) <\frac{1}{2}( 1+G_{2L}( {\hat{p}}_{2L}) ) ={\text {*}}{con} [ {\tilde{\varPi }}_{1L}( 1;G_{2L}) ]$. Therefore, ${\text {*}}{con}[ {\tilde{\varPi }}_{1L}( p;G_{2L}) ]$ and ${\text {*}}{con}[ {\tilde{\varPi }}_{1H}( \delta _{2H}( \delta _{1L}( p) ) ;G_{2H}) ]$ cross exactly once. Consequently, it is necessary that $G_{2L}( p)$ and $G_{2H}( p)$ cross exactly once, and $G_{2H}\succ$ $G_{2L}$.

Case 3 $\mu _{1}\in [ 1-\int _{0}^{1}\delta _{1H}( p) \mathrm{d}p,1-\int _{0}^{1}\delta _{1L}( p) \mathrm{d}p)$: In this case, while $G_{1L}$ is of type-N, $G_{1H}$ is of type-A. An argument identical to that for Cases 1 and 2 above implies that $G_{iH}\succ G_{iL}$.

We now discuss the limit of the unique equilibrium of the auxiliary game when $\rho$ approaches ${\bar{\rho }}\equiv \mu _{2}( 1-\mu _{1})$. Note that for a sufficiently large $\rho$, the unique equilibrium is of type-A, i.e., $\mu _{1}\in ( 1-\int _{0}^{1}\delta ( p) \mathrm{d}p,1)$, because $1-\int _{0}^{1}\delta ( p) \mathrm{d}p\rightarrow 0$ as $\rho \rightarrow {\bar{\rho }}$. Consider the limit of sender 1’s strategy. Note first that $\lim _{\rho \rightarrow {\bar{\rho }}}{\hat{p}}_{1}=0$. For otherwise, there exists an increasing sequence $\{ \rho _{N}\}$ with $\rho _{N}\rightarrow {\bar{\rho }}$ such that in the associated sequence of equilibria, we have ${\hat{p}}_{2}\rightarrow 1$. This contradicts that $\hat{p}_{2}$ is decreasing in $\rho$. The Bayes-plausibility condition with respect to $\mu _{1}$ implies that $\mu _{1}\ge ( 1-G_{1}( \hat{p}_{1}) ) ( 1-{\hat{p}}_{1})$ and $1-\mu _{1}\ge G_{1}( {\hat{p}}_{1}) ( 1-{\hat{p}}_{1})$. Rearranging the two inequalities above gives $1-\frac{\mu _{1}}{1-{\hat{p}}_{1}}\le G_{1}( {\hat{p}}_{1}) \le \frac{1-\mu _{1}}{1-{\hat{p}}_{1}}$, which implies that $G_{1}( {\hat{p}}_{1}) \rightarrow 1-\mu _{1}$. Therefore, $G_{1}( p)$ converges to $G_{1,F}( p)$ in distribution as $\rho \rightarrow {\bar{\rho }}$.

Next, consider sender 2’s limit strategy. Since ${\hat{p}}_{1}=\frac{G_{2}( {\hat{p}}_{2}) -G_{2}( 0) }{\frac{1}{2}( 1+G_{2}( {\hat{p}}_{2}) ) -G_{2}( 0) }$, and ${\hat{p}}_{1}\rightarrow 0$, it is necessary that $G_{2}( {\hat{p}} _{2}) -G_{2}( 0) \rightarrow 0$. Thus, $G_{2}( p)$ converges to $G_{2,F}( p)$ in distribution as $\rho \rightarrow {\bar{\rho }}$. $\square$

Proof of Theorem 4

Since the game is zero sum and symmetric, Theorem 2 implies that the equilibrium is unique and symmetric. Moreover, $\mu _{1}=\mu _{2}\equiv \mu$ implies that $\delta _{1}( p) =\delta _{2}( p) =p$. Therefore, $\varLambda ( s) =-\frac{3\rho ( s-\mu ) }{\mu ^{2}( 1-\mu ) ^{2}+\rho ( s-\mu ) ^{2}}$. By Lemma 4, $G( 0) =0$ in a symmetric equilibrium. By Lemma 1, in an equilibrium such that $G( {\hat{p}}) =1$, we have

$$\begin{aligned} {\hat{p}}=2\mu \text {; and }G\left( p\right) =\min \left\{ \frac{1}{2} +\frac{\sqrt{\left( 1-\mu \right) ^{2}+\rho }}{2}\frac{p-\mu }{\sqrt{\mu ^{2}\left( 1-\mu \right) ^{2}+\rho \left( p-\mu \right) ^{2}}},1\right\} . \nonumber \\ \end{aligned}$$

(13)

Also, in an equilibrium such that $G( {\hat{p}}) <1$, we have $\mu \varPi ( 1) =\frac{1}{2}$ by the linear structure of the payoff function. Then, using Lemma 4, we obtain

$$\begin{aligned} {\hat{p}}= & {} \frac{2\mu ^{2}\left( \left( 1-\mu \right) ^{2}+\rho \right) }{\mu ^{2}\left( 1-\mu \right) +\rho \left( 3\mu -1\right) };\,\,\text {and }G\left( p\right) \nonumber \\= & {} \left\{ \begin{array} [c]{cc} \frac{1}{2}+\frac{\sqrt{\left( 1-\mu \right) ^{2}+\rho }}{2}\frac{\min \left\{ p,{\hat{p}}\right\} -\mu }{\sqrt{\mu ^{2}\left( 1-\mu \right) ^{2}+\rho \left( \min \left\{ p,{\hat{p}}\right\} -\mu \right) ^{2}}} &{} \text {if }\,\,p\in [0,1)\\ 1 &{} \text {if}\,\,p=1 \end{array} \right. . \end{aligned}$$

(14)

Let $G_{{\tilde{\rho }}}( p)$ be the equilibrium strategy when $\rho ={\tilde{\rho }}$. Similarly, define ${\hat{p}}_{{\tilde{\rho }}}\equiv \sup ( {\text {*}}{supp}G_{{\tilde{\rho }}}\cap ( 0,1) )$. Take a pair $\rho ,\rho ^{\prime }\in [ 0,\mu ( 1-\mu ) ]$ with $\rho ^{\prime }>\rho$.

First, suppose $\mu \le 1/2$. By (13), for $p<2\mu$,

$$\begin{aligned} \frac{\partial }{\partial \rho }G_{\rho }\left( p\right) =\frac{\left( 1-\mu \right) ^{2}}{4\sqrt{\left( 1-\mu \right) ^{2}+\rho }}\frac{\left( \mu -p\right) p\left( 2\mu -p\right) }{\left( \mu ^{2}\left( 1-\mu \right) ^{2}+\rho \left( p-\mu \right) ^{2}\right) ^{\frac{3}{2}}}. \end{aligned}$$

(15)

Thus, $G_{\rho }( p) >G_{\rho ^{\prime }}( p)$ for all $p\in ( 0,\mu )$, and $G_{\rho }( p) <G_{\rho ^{\prime }}( p)$ for all $p\in ( \mu ,1)$. Therefore, $G_{\rho }\succ G_{\rho ^{\prime }}$.

Next, suppose $\mu >1/2$. Observe that both ${\hat{p}}$ and $G( \hat{p})$, as given by Eq. (14), are increasing in $\rho$. Moreover, for $p<{\hat{p}}$, $\frac{\partial G( p) }{\partial \rho }$ is given by Eq. (15) above. Therefore, if ${\hat{p}}_{\rho }\ge \mu$, then $G_{\rho }( p) >G_{\rho ^{\prime }}( p)$ for all $p\in ( 0,\mu )$ and $G_{\rho }( p) <G_{\rho ^{\prime }}( p)$ for all $p\in ( \mu ,1)$. On the other hand, if ${\hat{p}}_{\rho }<\mu$, then $G_{\rho }( p) >G_{\rho ^{\prime } }( p)$ for all $p\in ( 0,{\hat{p}}_{\rho })$. Thus, $G_{\rho }\succ G_{\rho ^{\prime }}$.

Finally, as $\lim _{\rho \rightarrow {\bar{\rho }}}{\hat{p}}_{\rho }>0$ for all possible values of $\mu$, the limiting strategy G(p) as $\rho \rightarrow {\bar{\rho }}$ is informative. $\square$

Proof of Theorem 5

In a game in which the normalized covariance of proposal qualities is $\rho$, denote by $\varPi _{i}^{\rho }( p_{i};G_{j})$ the payoff function of sender i, denote by $G_{i,\rho }$ the equilibrium strategy of sender i, and denote by $\delta _{i}^{\rho }( p)$ the transformation function identified in Lemma 2. Define ${\hat{p}}_{i}^{\rho }\equiv \sup ( {\text {*}}{supp}( G_{i,\rho }) \backslash \{ 1\} )$.

We first show that$\ G_{1,\rho }( {\hat{p}}_{1}^{\rho }) <1$ for $\rho$ sufficiently close to ${\bar{\rho }}$. Suppose not. Then, there exists a sequence $\{ \rho _{n}\}$ such that $\lim _{n\rightarrow \infty }\rho _{n}={\bar{\rho }}$, and for all n, $G_{1,\rho _{n}}( {\hat{p}} _{1}^{\rho _{n}}) =\varPi _{2}^{\rho _{n}}( 1;G_{1,\rho _{n}}) =1$. Then, the linear structure of $\varPi _{2}$ implies that for any $p_{2} \in ( 0,1)$, $\varPi _{2}^{\rho _{n}}( p_{2};G_{1,\rho _{n} }) /p_{2}>1$. Recall that $\delta _{2}^{\rho _{n}}( p_{2}) \rightarrow 0$ as $\rho _{n}\rightarrow {\bar{\rho }}$ and

$$\begin{aligned} \varPi _{2}^{\rho _{n}}\left( p_{2};G_{1,\rho _{n}}\right)= & {} \left( 1+\frac{\rho _{n}\left( p_{2}-\mu _{2}\right) \left( \delta _{2}^{\rho _{n}}\left( p_{2}\right) -\mu _{1}\right) }{\mu _{1}\left( 1-\mu _{1}\right) \mu _{2}\left( 1-\mu _{2}\right) }\right) G_{1,\rho _{n}}\left( \delta _{2} ^{\rho _{n}}\left( p_{2}\right) \right) \\&-\,\frac{\rho _{n}\left( p_{2}-\mu _{2}\right) }{\mu _{1}\left( 1-\mu _{1}\right) \mu _{2}\left( 1-\mu _{2}\right) }\int _{0}^{\delta _{2}^{\rho _{n} }\left( p_{2}\right) }G_{1,\rho _{n}}\left( s\right) \mathrm{d}s. \end{aligned}$$

Therefore, for an arbitrary pair $\varepsilon _{1},\varepsilon _{2}>0$ and for $p_{2}\in \left( 1-\varepsilon _{1},1\right)$, there exists an ${\bar{n}}_{1}$ such that $n>{\bar{n}}_{1}$ implies

$$\begin{aligned} \varPi _{2}^{\rho _{n}}\left( p_{2};G_{1,\rho _{n}}\right) <\varepsilon _{2}\times G_{1,\rho _{n}}\left( \delta _{2}^{\rho _{n}}\left( p_{2}\right) \right) \le \varepsilon _{2}, \end{aligned}$$

and hence $\varPi _{2}^{\rho _{n}}( p_{2};G_{1,\rho _{n}}) /p_{2}<1$, which is a contradiction.

Next, let $\{ \rho _{n}\}$ be a sequence such that $\lim _{n\rightarrow \infty }\rho _{n}={\bar{\rho }}$. Since $\{ {\hat{p}}_{2} ^{\rho _{n}}\} _{n}$ is increasing, $\lim _{n\rightarrow \infty }\hat{p}_{2}^{\rho _{n}}$ exists. We show that $\lim _{n\rightarrow \infty }{\hat{p}} _{2}^{\rho _{n}}<1$ and $\lim _{n\rightarrow \infty }{\hat{p}}_{1}^{\rho _{n}}=0$. If $\lim _{n\rightarrow \infty }{\hat{p}}_{2}^{\rho _{n}}=1$, then by the argument of the preceding paragraph, for any $\varepsilon _{1}>0$ and $\varepsilon _{2}>0$; and $p_{2}>1-\varepsilon _{1}$, there exists ${\bar{n}}_{2}$ such that $n>\bar{n}_{2}$ implies ${\hat{p}}_{2}^{\rho _{n}}>p_{2}$ and $\varPi _{2}^{\rho _{n}}( p_{2};G_{1,\rho _{n}}) <\varepsilon _{2}$, and hence

$$\begin{aligned} \frac{\varPi _{2}^{\rho _{n}}( p_{2};G_{1,\rho _{n}}) -p_{2}\varPi _{2}^{\rho _{n}}( 1;G_{1,\rho _{n}}) }{1-p_{2}}<\frac{\varepsilon _{2}-p_{2}/2}{1-p_{2}}<0. \end{aligned}$$

However, the linear structure of $\varPi _{2}$ implies $\varPi _{2}^{\rho _{n}}( 0;G_{1,\rho _{n}}) =\frac{\varPi _{2}^{\rho _{n}}( p_{2};G_{1,\rho _{n} }) -p_{2}\varPi _{2}^{\rho _{n}}( 1;G_{1,\rho _{n}}) }{1-p_{2}}$. Therefore, we have $\varPi _{2}^{\rho _{n}}( 0;G_{1,\rho _{n}}) <0$, a contradiction. Furthermore, since ${\hat{p}}_{1}^{\rho _{n}}=\delta _{2} ^{\rho _{n}}( {\hat{p}}_{2})$, we have $\lim _{n\rightarrow \infty }{\hat{p}}_{1}^{\rho _{n}}=0$.

We are ready to show that $G_{1,\rho _{n}}$ converges to $G_{1,F}$ in distribution. Observe that

$$\begin{aligned} T_{1}\left( {\hat{p}}_{2}^{\rho _{n}}\right) \equiv \frac{\int _{0}^{{\hat{p}} _{2}^{\rho _{n}}}\delta _{2}^{\rho _{n}}\left( x\right) \exp \left( \int _{0}^{x}\varLambda _{1}\left( s\right) \mathrm{d}s\right) \mathrm{d}x}{\int _{0}^{{\hat{p}} _{2}^{\rho _{n}}}\exp \left( \int _{0}^{x}\varLambda _{1}\left( s\right) \mathrm{d}s\right) \mathrm{d}x}<{\hat{p}}_{1}^{\rho _{n}}. \end{aligned}$$

Therefore, the simplified Bayes-plausibility condition for sender 1 implies

$$\begin{aligned} 1-\mu _{1}\ge G_{1,\rho _{n}}\left( {\hat{p}}_{1}^{\rho _{n}}\right) -\left( G_{1,\rho _{n}}\left( {\hat{p}}_{1}^{\rho _{n}}\right) -G_{1,\rho _{n}}\left( 0\right) \right) {\hat{p}}_{1}^{\rho _{n}}, \end{aligned}$$

and hence $1-\mu _{1}\ge \lim _{n\rightarrow \infty }G_{1,\rho _{n}}( \hat{p}_{1}^{\rho _{n}})$. At the same time, the simplified Bayes-plausibility condition implies $1-\mu _{1}\le G_{1,\rho _{n}}( {\hat{p}}_{1}^{\rho _{n}}) ( 1-T_{1}( {\hat{p}}_{2}^{\rho _{n} }) ) \le G_{1,\rho _{n}}( {\hat{p}}_{1}^{\rho _{n}})$. Therefore, $\lim _{n\rightarrow \infty }G_{1,\rho _{n}}( {\hat{p}} _{1}^{\rho _{n}}) =$ $1-\mu _{1}$, and $G_{1,\rho _{n}}$ converges to $G_{1,F}$ in distribution.

We now show that $G_{2,\rho _{n}}$ converges to $G_{2,F}$ in distribution. We first show that $G_{1,\rho }( 0) =0$ for all $\rho$ sufficiently close to ${\bar{\rho }}$. Suppose not. Then, there exists a sequence $\{ \rho _{n}\}$ such that $\lim _{n\rightarrow \infty }\rho _{n}={\bar{\rho }}$ and $\ G_{1,\rho _{n}}( 0) >0$ for all n. For each $\varepsilon _{3}\in ( 0,\mu _{1}) ,\varepsilon _{4}>0$, and $p_{1}\in ( 0,\varepsilon _{3})$, there exists an ${\bar{n}}_{3}$ such that $n>{\bar{n}}_{3}$ implies that

$$\begin{aligned} \varPi _{1}^{\rho _{n}}\left( p_{1};G_{2,\rho _{n}}\right)= & {} \left( 1+\frac{\rho _{n}\left( p_{1}-\mu _{1}\right) \left( \delta _{1}^{\rho _{n}}\left( p_{1}\right) -\mu _{2}\right) }{\mu _{1}\left( 1-\mu _{1}\right) \mu _{2}\left( 1-\mu _{2}\right) }\right) G_{2,\rho _{n}}\left( \delta _{1} ^{\rho _{n}}\left( p_{1}\right) \right) \\&-\,\frac{\rho _{n}\left( p_{1}-\mu _{1}\right) }{\mu _{1}\left( 1-\mu _{1}\right) \mu _{2}\left( 1-\mu _{2}\right) }\int _{0}^{\delta _{1}^{\rho _{n} }\left( p_{1}\right) }G_{2,\rho _{n}}\left( s\right) \mathrm{d}s \end{aligned}$$

is bounded from below by $1-\varepsilon _{4}$. However, $G_{1,\rho _{n}}\left( 0\right) >0$ implies $G_{2,\rho _{n}}\left( 0\right) =0$ and hence $\varPi _{1}^{\rho _{n}}\left( 0;G_{2,\rho _{n}}\right) =0$. Furthermore, $\varPi _{1}^{\rho _{n}}\left( 1;G_{2,\rho _{n}}\right) \ge 1/2$. Therefore, $\varPi _{1}^{\rho _{n}}\left( 0;G_{2,\rho _{n}}\right) <\varPi _{1}^{\rho _{n}}\left( p_{1};G_{2,\rho _{n}}\right)$ and $\varPi _{1}^{\rho _{n}}\left( 0;G_{2,\rho _{n} }\right) <\varPi _{1}^{\rho _{n}}\left( 1;G_{2,\rho _{n}}\right)$. Therefore, $0\notin {\text {*}}{supp}G_{1,\rho _{n}}$, a contradiction to $G_{1,\rho _{n}}\left( 0\right) >0$.

It remains to show ${\hat{p}}_{2}^{\rho }\rightarrow 0$ as $\rho \rightarrow {\bar{\rho }}$. An argument similar to that for the convergence of $G_{1}$ can establish that $G_{2,\rho }$ converges to to $G_{2,F}$ in distribution. To see that $\lim _{\rho \rightarrow {\bar{\rho }}}{\hat{p}}_{2}^{\rho }=0$, suppose there exists a sequence $\{ \rho _{n}\}$ that converges to ${\bar{\rho }}$ and $\lim _{n\rightarrow \infty }{\hat{p}}_{2}^{\rho _{n}}={\tilde{p}}_{2}\in ( 0,1)$. Note that we have already established that $G_{1,\rho _{n} }( 0) =0$ for all n sufficiently large. Thus, for sufficiently large n, the atom condition at the top, as well as that $\varPi _{2}^{\rho }( 1;G_{1,\rho _{n}}) \ge 1/2$, implies $\varPi _{2}^{\rho _{n}}( {\tilde{p}}_{2};G_{1,\rho _{n}}) \ge {\tilde{p}}_{2}/2$. On the other hand, $\lim _{n\rightarrow \infty }p_{1}^{\rho _{n}}=0$ and $\lim _{n\rightarrow \infty }G_{1,\rho _{n}}( \delta _{2}^{\rho _{n}}( {\tilde{p}}_{2}) ) =0$. Therefore, for any $\varepsilon _{5}$, there exists an $\bar{n}_{4}$ such that $n>{\bar{n}}_{4}$ implies $\varPi _{2}^{\rho _{n}}( {\tilde{p}}_{2};G_{1,\rho _{n}}) \le \frac{1-{\tilde{p}}_{2}}{1-\mu _{2} }\varepsilon _{5}$, which is a contradiction to $\varPi _{2}^{\rho _{n}}( {\tilde{p}}_{2};G_{1,\rho _{n}}) \ge {\tilde{p}}_{2}/2$. $\square$