Pre-communication in a coordination game with incomplete information

Abstract

We study several pre-communication protocols in a coordination game with incomplete information. Under decentralized decision making, we show that informative communication can be sustained in equilibrium, yet miscoordination arises with positive probabilities. Moreover, the equilibrium takes a partitional structure and messages are rank ordered, with higher messages becoming increasingly imprecise. Compared to centralized decision making (a mediator without commitment), decentralization leads to more informative communication when the miscoordination cost is high, and performs better when the miscoordination cost is intermediate. We also study the case in which the mediator is able to commit to a decision rule beforehand.

Appendix

1.1 Proof of Lemma 1

Proof

Suppose $a_{N-1}\ge c$, consider the marginal type of agent 1 with $\theta _{1}=a_{N-1}$. Since $a_{N-1}\ge c$, the marginal type has a dominant strategy, which is always playing his favored action A regardless of the messages sent in the first stage. We can compute the marginal type’s expected payoffs by sending message $m_{N-1}$ and by sending message $m_{N}$ as follows:

$$\begin{aligned} \pi _{1}(a_{N-1},m_{N})&=(1-a_{N-1})(a_{N-1}-c)+a_{N-1}^{2},\\ \pi _{1}(a_{N-1},m_{N-1})&=(1-x_{N-1})(a_{N-1}-c)+x_{N-1}a_{N-1}. \end{aligned}$$

It can be readily verified that $\pi _{1}(a_{N-1},m_{N})>\pi _{1}(a_{N-1} ,m_{N-1})$, as $x_{N-1}<a_{N-1}$. Therefore, type $a_{N-1}$ cannot be indifferent between sending messages $m_{N-1}$ and $m_{N}$, and thus equilibria with $a_{N-1}\ge c$ do not exist. $\square$

1.2 Proof of Lemma 2

Proof

Given that the marginal type $a_{n}$ is indifferent between sending messages $m_{n}$ and $m_{n+1}$, we only need to establish the following single crossing property

$$\begin{aligned} \frac{\partial [\pi (m_{n+1},\theta )-\pi (m_{n},\theta )]}{\partial \theta }\ge 0\text {, for }\theta \in [a_{n},a_{n+1}]. \end{aligned}$$

That is, a higher type agent has a weakly stronger incentive to send a higher message. By symmetry, we shall prove the inequality holds for agent 1 without any loss of generality.

Consider a type $\theta$ of agent 1 with $\theta \in [a_{n},a_{n+1}]$. If $\theta \in [a_{n},x_{n+1}]$, when he sends message $m_{n+1}$, his expected payoff is

$$\begin{aligned} \pi (m_{n+1},\theta )=a_{n}\theta -(a_{n+1}-a_{n})\left( \frac{x_{n+1}-a_{n} }{a_{n+1}-a_{n}}c\right) . \end{aligned}$$

Note that this type of agent 1 will yield if agent 2 sends message $m_{n+1}$. If $\theta \in [x_{n+1},a_{n+1}]$ and agent 1 sends message $m_{n+1},$ his expected payoff is

$$\begin{aligned} \pi (m_{n+1},\theta )&=a_{n}\theta +(x_{n+1}-a_{n})\theta +\left( a_{n+1}-x_{n+1}\right) \left( \theta -c\right) \\&=a_{n+1}\theta -\left( a_{n+1}-x_{n+1}\right) c. \end{aligned}$$

Note that this type of agent 1 will stick if agent 2 sends message $m_{n+1}$. On the other hand, if $\ \theta \in [a_{n},a_{n+1}]$ and agent 1 sends message $m_{n}$, his expected payoff is

$$\begin{aligned} \pi (m_{n},\theta )&=a_{n-1}\theta +(x_{n}-a_{n-1})\theta +\left( a_{n} -x_{n}\right) \left( \theta -c\right) \\&=a_{n}\theta -\left( a_{n}-x_{n}\right) c \end{aligned}$$

Note that this type of agent 1 will stick if agent 2 sends message $m_{n}$.

Combining the above results, we have

$$\begin{aligned} \frac{\partial [\pi (m_{n+1},\theta )-\pi (m_{n},\theta )]}{\partial \theta }&=0\quad \text { for }\theta \in [a_{n},x_{n+1}],\nonumber \\ \frac{\partial [\pi (m_{n+1},\theta )-\pi (m_{n},\theta )]}{\partial \theta }&=a_{n+1}-a_{n}>0\quad \text { for }\theta \in [x_{n+1},a_{n+1}]. \end{aligned}$$

(12)

Thus, we have shown that $\frac{\partial [\pi (m_{n+1},\theta )-\pi (m_{n},\theta )]}{\partial \theta }\ge 0$.

In addition, similarly we can show

$$\begin{aligned} \frac{\partial [\pi (m_{n+1},\theta )-\pi (m_{n},\theta )]}{\partial \theta }=0\quad \text { for }\theta \in [x_{n},a_{n}]. \end{aligned}$$

(13)

Equation (12) and Equation (13) together imply that all types within $[x_{n},x_{n+1}]$ are indifferent between sending messages $m_{n}$ and $m_{n+1}$. $\square$

1.3 Proof of Lemma 3

Proof

(1) (Existence) Fix c and $N\ge 2$. We want to show that the difference equation system has a solution. To that end, we first express all the message cutoffs $\{a_{n}\}$ and the action cutoffs $\{x_{n}\}$ as functions of $x_{1}$ (and only $x_{1}$ and c). Specifically, given $x_{1}$ ($\Delta x_{1}\equiv x_{1}$), by applying (5) ($\Delta y_{n}=\frac{c+x_{n}}{c-x_{n}}\Delta x_{n}$) and (6) ($\Delta x_{n+1}=\Delta y_{n}$) recursively, all the $\{ \Delta x_{n}\}$, $\{x_{n}\}$, $\{ \Delta y_{n}\}$, and $\{a_{n}\}$ are determined. Define $x_{n}=f_{n} (x_{1})$ and $a_{n}=g_{n}(x_{1})$. Specifically,

$$\begin{aligned} x_{n+1}&=f_{n+1}(x_{1})= {\displaystyle \sum _{k=1}^{n}} [\Delta x_{k}(x_{1})+\Delta y_{k}(x_{1})]+\Delta x_{n+1}(x_{1}),\\ a_{n}&=g_{n}(x_{1})= {\displaystyle \sum _{k=1}^{n}} [\Delta x_{k}(x_{1})+\Delta y_{k}(x_{1})]. \end{aligned}$$

Note that $x_{n}<c$ (for all $n\le N$) is necessary for $\Delta y_{n}$ to be well defined ($\Delta y_{n}\ge 0$). Define $\widehat{x}_{1}\in [0,c)$ such that $f_{N}(\widehat{x}_{1})=c$ and $f_{n}(\widehat{x}_{1})<c$ for all $n<N$. Such an $\widehat{x}_{1}$ exists because, for $x_{1}>0$ but sufficiently small, $x_{n}<c$ for any $n\le N$ and $x_{n}$ strictly increases in n.

Now we restrict the domain of $x_{1}$ to $[0,\widehat{x}_{1})$. For any $x_{1}\in [0,\widehat{x}_{1})$, by applying (5) and (6) recursively, we can show that $\Delta x_{n}\ge 0$ and $\Delta y_{n}\ge 0$ for all $n\le N$. Moreover, all the $\Delta x_{n}$ and $\Delta y_{n}$ are continuous in $x_{1}$, and thus both $f_{N}(x_{1})$ and $g_{N}(x_{1})$ are continuous in $x_{1}$. Note that $x_{1}$ is a solution to the difference equation system if $g_{N}(x_{1})=1$.

Suppose $x_{1}=0$. Then, by the recursive structure, $x_{n}=a_{n}=0$ for any $n\le N$. Therefore, we have $g_{N}(0)=0$. Now suppose $x_{1}\rightarrow \widehat{x}_{1}$, and hence $x_{N}=f_{N}(\widehat{x}_{1})\rightarrow c$. Then we have

$$\begin{aligned} \lim _{x_{1}\rightarrow \widehat{x}_{1}}\Delta y_{N}=\lim _{x_{1}\rightarrow \widehat{x}_{1}}\frac{c+x_{N}}{c-x_{N}}\Delta x_{N}=\infty . \end{aligned}$$

Therefore, we have $\lim _{x_{1}\rightarrow \widehat{x}_{1}}g_{N}(x_{1})=\infty$. Given the boundary conditions just derived, the continuity of $g_{N} (\cdot )$ (in domain $[0,\widehat{x}_{1})$) means that there is an $x_{1} \in (0,\widehat{x}_{1})$ such that $g_{N}(x_{1})=1$. That is, the difference equation system has a solution.

(2) (Uniqueness) To prove the uniqueness, fix $N\ge 2$. Suppose there are two different solutions to the difference equation system: (a, x) and $(a^{\prime },x^{\prime })$. WLOG, suppose $a_{N-1}^{\prime }>a_{N-1}$. By Eq. (2), $a_{N-1}^{\prime }>a_{N-1}$ implies that $x_{N}^{\prime }>x_{N}$. Therefore, $\Delta y_{N}^{\prime }<\Delta y_{N}$. By (5), $x_{N}^{\prime }>x_{N}$ implies that $\frac{\Delta y_{N}^{\prime }}{\Delta x_{N}^{\prime }}>\frac{\Delta y_{N}}{\Delta x_{N}}$. Since $\Delta y_{N}^{\prime }<\Delta y_{N}$, it further implies that $\Delta x_{N}^{\prime }<\Delta x_{N}$. Now by (6), it means that $\Delta y_{N-1}^{\prime }<\Delta y_{N-1}$. Moreover, from the facts that $a_{N-1} ^{\prime }>a_{N-1}$ and $\Delta y_{N-1}^{\prime }<\Delta y_{N-1}$, we get $x_{N-1}^{\prime }>x_{N-1}$.

Now suppose $a_{n}^{\prime }>a_{n}$ and $\Delta y_{n}^{\prime }<\Delta y_{n}$, and thus $x_{n}^{\prime }>x_{n}$, we want to show $a_{n-1}^{\prime }>a_{n-1}$ and $\Delta y_{n-1}^{\prime }<\Delta y_{n-1}$, and thus $x_{n-1}^{\prime }>x_{n-1}$. By (5), $x_{n}^{\prime }>x_{n}$ implies that $\frac{\Delta y_{n}^{\prime }}{\Delta x_{n}^{\prime }}>\frac{\Delta y_{n} }{\Delta x_{n}}$. Since $\Delta y_{n}^{\prime }<\Delta y_{n}$, it further implies that $\Delta x_{n}^{\prime }<\Delta x_{n}$. Now by (6), it means that $\Delta y_{n-1}^{\prime }<\Delta y_{n-1}$. By the facts that $x_{n}^{\prime }>x_{n}$ and $\Delta x_{n}^{\prime }<\Delta x_{n}$, we get $a_{n-1}^{\prime }>a_{n-1}$. Finally, the facts that $\Delta y_{n-1}^{\prime }<\Delta y_{n-1}$ and $a_{n-1}^{\prime }>a_{n-1}$ imply that $x_{n-1}^{\prime }>x_{n-1}$.

Now by induction, we have $\Delta y_{1}^{\prime }<\Delta y_{1}$, $a_{1} ^{\prime }>a_{1}$, and $x_{1}^{\prime }>x_{1}$. By (5), $x_{1}^{\prime }>x_{1}$ implies that $\frac{\Delta y_{1}^{\prime }}{\Delta x_{1}^{\prime }}>\frac{\Delta y_{1}}{\Delta x_{1}}$. Since $\Delta y_{1}^{\prime }<\Delta y_{1}$, it implies that $\Delta x_{1}^{\prime }<\Delta x_{1}$, which is equivalent to $x_{1}^{\prime }<x_{1}$. A contradiction. $\square$

1.4 Proof of Lemma 4

Proof

Part (i). Fix $N\ge 1$. First, we show that $a_{N}^{\prime }>a_{N-1}$. Suppose to the contrary, $a_{N}^{\prime }\le a_{N-1}$. Then by the same logic as in the proof for Lemma 3 (the uniqueness), we have $\Delta y_{1}\le \Delta y_{2}^{\prime }$, $a_{1}\ge a_{2}^{\prime }$, and $x_{1}\ge x_{2}^{\prime }$. By (5), $x_{1}\ge x_{2}^{\prime }$ implies that $\frac{\Delta y_{2}^{\prime }}{\Delta x_{2}^{\prime }}\le \frac{\Delta y_{1} }{\Delta x_{1}}$. Since $\Delta y_{1}\le \Delta y_{2}^{\prime }$, it implies that $\Delta x_{1}\le \Delta x_{2}^{\prime }$, which further implies that $x_{1}<x_{2}^{\prime }$. A contradiction. Therefore, $a_{N}^{\prime }>a_{N-1}$.

By a similar logic, we can show that $a_{N-1}>a_{N-1}^{\prime }$. Following the induction as in the proof for Lemma 3 (the uniqueness), we can show that $\ldots a_{n}^{\prime }<a_{n}<a_{n+1}^{\prime }<a_{n+1}\ldots$.

Part (ii). We first show that $\Delta y_{N+1}^{\prime }<\Delta y_{N}$. Suppose $\Delta y_{N+1}^{\prime }\ge \Delta y_{N}$. Then it implies that $x_{N+1} ^{\prime }\le x_{N}$. Since $a_{N}^{\prime }>a_{N-1}$ (part (i)), we have $\Delta x_{N+1}^{\prime }<\Delta x_{N+1}$. By (5), $\Delta x_{N+1}^{\prime }<\Delta x_{N+1}$ and $x_{N+1}^{\prime }\le x_{N}$ imply that $\Delta y_{N+1}^{\prime }<\Delta y_{N}$, a contradiction. Note that $\Delta y_{N+1}^{\prime }<\Delta y_{N}$ means $x_{N+1}^{\prime }>x_{N}$. Next we show $\Delta x_{N+1}^{\prime }<\Delta x_{N}$. Suppose $\Delta x_{N+1}^{\prime } \ge \Delta x_{N}$. By (5), $\Delta x_{N+1}^{\prime }\ge \Delta x_{N}$ and $x_{N+1}^{\prime }>x_{N}$ imply that $\Delta y_{N+1}^{\prime }>\Delta y_{N}$, a contradiction.

In a similar fashion, we can recursively show that $\Delta y_{n+1}^{\prime }<\Delta y_{n}$ and $\Delta x_{n+1}^{\prime }<\Delta x_{n}$.

Part (iii). By (5) and (6),

$$\begin{aligned}&[(a_{n+1}^{\prime }-a_{n}^{\prime })-(a_{n}^{\prime }-a_{n-1}^{\prime })]-[(a_{n+1}-a_{n})-(a_{n}-a_{n-1})]\\&\quad =(\Delta y_{n+1}^{\prime }-\Delta x_{n}^{\prime })-(\Delta y_{n+1}-\Delta x_{n})\\&\quad =\left( \frac{c+x_{n+1}^{\prime }}{c-x_{n+1}^{\prime }}\frac{c+x_{n}^{^{\prime }} }{c-x_{n}^{\prime }}-1\right) \Delta x_{n}^{\prime }-\left( \frac{c+x_{n+1}}{c-x_{n+1}} \frac{c+x_{n}}{c-x_{n}}-1\right) \Delta x_{n}<0, \end{aligned}$$

where the last inequality uses the earlier results that $\Delta x_{n}^{\prime }<\Delta x_{n}$ and $x_{n}^{^{\prime }}<x_{n}$. $\square$

1.5 Proof of Lemma 5

Proof

We only need to rule out the existence of three-partition equilibria. First consider the case that $\overline{m}_{1}\ge 2c$. Then by the monotonicity of $\overline{m}_{n}$ in n, $\overline{m}_{n}\ge 2c$ for all $n=1,2,3$. This means that the mediator will always choose outcome AB for sure regardless of the messages. As a result, all the messages are outcome equivalent and the equilibrium is a babbling one. Second, consider the case that $\overline{m}_{3}\le 2c$. By a similar logic, in this case outcome AB is never chosen and either AA or BB is chosen. As a result, in this case both agents will always send the highest message $m_{3}$ (again a babbling equilibrium).

Now consider the case that $\overline{m}_{1}<2c$ and $\overline{m}_{3}>2c$. Suppose $\overline{m}_{2}<2c$. Then outcome AB is chosen if and only if both agents send message $m_{3}$. In this case all types of agent i with $\theta _{i}\le a_{2}$ will have an incentive to send message $m_{2}$, in order to increase the chance that his favored coordinated outcome is chosen. That is, the two low messages are essentially combined into a single message. Next suppose $\overline{m}_{2}>2c$. Then outcome AB is chosen if the lower message is $m_{2}$. In this case the two high messages are outcome equivalent and are essentially combined into a single message.

Now the only case left is $\overline{m}_{1}<2c$, $\overline{m}_{3}>2c$, and $\overline{m}_{2}=2c$. This means that $a_{2}>2c$ and $a_{1}<2c$. To make $m_{2}$ different from $m_{1}$ and $m_{3}$, let $p\in (0,1)$ be the probability that outcome AB is chosen when both agents send $m_{2}$ (AA and BB are chosen with the same probability $(1-p)/2$), and $q\in [0,1]$ be probability that outcome AB is chosen when one agent sends $m_{3}$ and the other sends $m_{2}$ (with probability $1-q$ the favored coordinated outcome of the agent who sends $m_{3}$ is chosen). Now consider the marginal type $a_{2}$ of agent 1. His expected payoffs when sending messages $m_{2}$ and $m_{3}$ are

$$\begin{aligned} \pi _{1}(a_{2},m_{2})&=a_{1}a_{2}+(a_{2}-a_{1})[p(a_{2}-c)+\frac{1-p}{2}a_{2}]+(1-a_{2})q(a_{2}-c),\\ \pi _{1}(a_{2},m_{3})&=a_{1}a_{2}+(a_{2}-a_{1})[q(a_{2}-c)+(1-q)a_{2} ]+(1-a_{2})(a_{2}-c). \end{aligned}$$

Taking the difference, we have

$$\begin{aligned}&\pi _{1}(a_{2},m_{3})-\pi _{1}(a_{2},m_{2})\\&\quad =(a_{2}-a_{1})[-(q-p)c+\frac{1-p}{2}a_{2}]+(1-a_{2})(1-q)(a_{2}-c). \end{aligned}$$

To show that $\pi _{1}(a_{2},m_{3})-\pi _{1}(a_{2},m_{2})>0$, it is sufficient to show that $-(q-p)c+\frac{1-p}{2}a_{2}>0$. Since $a_{2}>2c$, we have

$$\begin{aligned} -(q-p)c+\frac{1-p}{2}a_{2}>-(q-p)c+(1-p)c=(1-q)c\ge 0. \end{aligned}$$

Therefore, $\pi _{1}(a_{2},m_{3})-\pi _{1}(a_{2},m_{2})>0$. But this contradicts to the fact that type $a_{2}$ of agent 1 should be indifferent between sending messages $m_{2}$ and $m_{3}$. $\square$

1.6 Proof of Lemma 6

.

Proof

Part (i). To see $p_{BA}^{l}=0$ in the optimal mechanism, by Eq. (9), we have

$$\begin{aligned} \frac{\partial a}{\partial p_{BA}^{l}}=\frac{-\left( \frac{a^{2}}{2}+ca\right) }{(p_{BA}^{h}+p_{BA}^{l})\left( a+c\right) +\frac{1+p_{AB}^{h}-p_{BA}^{h}}{2}+\left( p_{AB}^{h}+p_{AB}^{l}\right) \left( c-a\right) }, \\ \frac{\partial a}{\partial p_{AB}^{l}}=\frac{\left( \frac{a}{2}-c\right) a}{(p_{BA}^{h}+p_{BA}^{l})\left( a+c\right) +\frac{1+p_{AB}^{h}-p_{BA}^{h} }{2}+\left( p_{AB}^{h}+p_{AB}^{l}\right) \left( c-a\right) }. \end{aligned}$$

By the above equations,

$$\begin{aligned} \frac{\partial a}{\partial p_{AB}^{l}}=\frac{c-\frac{a}{2}}{c+\frac{a}{2} }\frac{\partial a}{\partial p_{BA}^{l}}. \end{aligned}$$

Define the objective function as $F\left( p_{AB}^{l},p_{BA}^{l}\right)$. Suppose $p_{BA}^{l}>0$. We construct

$$\begin{aligned} \widetilde{p}_{BA}^{l}=p_{BA}^{l}-\Delta ,\text { \ }\widetilde{p}_{AB} ^{l}=p_{AB}^{l}+\frac{c-\frac{a}{2}}{c+\frac{a}{2}}\Delta , \end{aligned}$$

where $\Delta$ is a sufficiently small and positive number. By construction, $a\left( \widetilde{p}_{AB}^{l},\widetilde{p}_{BA}^{l}\right) =a\left( p_{AB}^{l},p_{BA}^{l}\right) .$ We thus have

$$\begin{aligned} F\left( \widetilde{p}_{AB}^{l},\widetilde{p}_{BA}^{l}\right) -F\left( p_{AB}^{l},p_{BA}^{l}\right) \propto \frac{3ca^{3}}{2\left( c+\frac{a}{2}\right) }, \end{aligned}$$

which is positive. Therefore, in the optimal mechanism $p_{BA}^{l}$ cannot be positive.

By a very similar proof, we can show that $p_{BA}^{h}=0$ in the optimal mechanism.

Part (ii). Suppose to the contrary that $c\le \frac{a}{2}$, which implies that $a-c\ge \frac{a}{2}>0$. For the marginal type a agent, the expected payoff of sending the high message and that of sending the low message are

$$\begin{aligned} \pi _{1}(a,H)&=a^{2}+\left( 1-a\right) \left( p_{AB}^{h}\left( a-c\right) +\left( 1-p_{AB}^{h}\right) \frac{a}{2}\right) ,\\ \pi _{1}(a,L)&=a\left( p_{AB}^{l}\left( a-c\right) +\left( 1-p_{AB} ^{l}\right) \frac{a}{2}\right) , \end{aligned}$$

respectively (we have used the results in part (i) that $p_{BA}^{h}=p_{BA} ^{l}=0$). Notice that in the parenthesis the payoffs are linear combinations of $a-c$ and $\frac{a}{2}$. It follows that the lower bound of $\pi _{1}(a,H)$ can be achieved at $p_{AB}^{h}=0$, while the upper bound of $\pi _{1}(a,L)$ can be achieved at $p_{AB}^{l}=1$. Therefore,

$$\begin{aligned}&\pi _{1}(a,H)-\pi _{1}(a,L)\ge a^{2}+\left( 1-a\right) \left( \frac{a}{2}\right) -a\left( a-c\right) \\&\quad =\left( 1-a\right) \left( \frac{a}{2}\right) +ac>0, \end{aligned}$$

which contradicts the fact that a is the marginal type who is indifferent between sending the low message and sending the high message.

Part (iii). We now show that $p_{AB}^{l}$ cannot be positive in equilibrium. To see this, by Eq. (9), we have

$$\begin{aligned} \frac{\partial a}{\partial p_{AB}^{h}}=\frac{\left( 1-a\right) \left( c+\frac{a}{2}\right) }{\left( \frac{a}{2}-c\right) a}\frac{\partial a}{\partial p_{AB}^{l}}. \end{aligned}$$

Define the objective function as $F\left( p_{AB}^{h},p_{AB}^{l}\right)$. Suppose $p_{AB}^{l}>0$. We construct

$$\begin{aligned} \widetilde{p}_{AB}^{l}=p_{AB}^{l}-\Delta ,\text { \ }\widetilde{p}_{AB} ^{h}=p_{AB}^{h}+\frac{\left( 1-a\right) \left( c+\frac{a}{2}\right) }{\left( \frac{a}{2}-c\right) a}\Delta , \end{aligned}$$

where $\Delta$ is positive and sufficiently small. By construction, $a\left( \widetilde{p}_{AB}^{h},\widetilde{p}_{AB}^{l}\right) =a\left( p_{AB} ^{l},p_{AB}^{l}\right)$. We thus have

$$\begin{aligned}&F\left( \widetilde{p}_{AB}^{h},\widetilde{p}_{AB}^{l}\right) -F\left( p_{AB}^{h},p_{AB}^{l}\right) \\&\propto a^{2}\left( c-\frac{a}{4}\right) +\left( \frac{\left( 1-a\right) \left( c+\frac{a}{2}\right) }{\left( c-\frac{a}{2}\right) a}\right) \left( 1-a\right) ^{2}\left( \frac{\left( 1+a\right) }{4}+c\right) >0, \end{aligned}$$

since $c>\frac{a}{2}$. Therefore, $p_{AB}^{l}$ cannot be positive in the optimal mechanism. $\square$

1.7 Proof of Lemma 7

Proof

Now define the LHS of (11) as G(a). It can be verified that $G^{\prime }(a)>0$, $G(0)<0$, and $G(c)>0$. Therefore, $a\in [0,c)$ for any $p_{AB}^{h}$. By previous results, $\frac{\partial a}{\partial p_{AB}^{h} }>0$, or a is increasing in $p_{AB}^{h}$.

By (11), choosing $p_{AB}^{h}$ is equivalent to choosing a. From (11), we can solve for $p_{AB}^{h}$ as a function of a. Substituting $p_{AB}^{h}$ in the objective function and simplifying, we get a new objective function F(a), with the restriction that $a\in [0,\overline{a}]$, where $\overline{a}$ is the solution to (11) when $p_{AB}^{h}=1$. Specifically,

$$\begin{aligned} F(a)=\frac{1}{4}+\frac{1-2c}{4}\frac{a(1-a)}{2c-a}. \end{aligned}$$

(14)

Observing (14), we notice that $\frac{a(1-a)}{2c-a}\ge 0$ since $2c>a$. It immediately follows that when $c>1/2$, the optimal $a=0$, which means that the optimal $p_{AB}^{h}=0$. When $c=1/2$, then any a is optimal, or any $p_{AB}^{h}\in [0,1]$ is optimal. Now consider the case that $c<1/2$. Since $a<c$, it means that $a<1/2$. For the term $\frac{a(1-a)}{2c-a}$, the numerator $a(1-a)$ is increasing in a since $a<1/2$; the denominator is decreasing in a. Therefore, $\frac{a(1-a)}{2c-a}$ is increasing in a for $a\in [0,\overline{a}]$. As a result, the optimal a is the corner solution $a=\overline{a}$, which implies that the optimal $p_{AB}^{h}=1$. $\square$