# Coordination and Private Information Revelation
## 6 Extensions
### Two-Sided Private Information  
Consider the case where both the agents have private information about their own types, with agent A's (respectively B's) type being denoted by \(\theta_{A}\) (respectively \(\theta_{B}\)). For simplicity, let \(\theta_{A}\) and \(\theta_{B}\) be identically and independently distributed with distribution function \(F(\theta)\) (assumed to be strictly monotonic) with support \([\theta_{l},\theta_{h}]\). While agent \(i,i\in\{A,B\}\), knows her own type, she only knows the distribution of agent \(j\
eq i\), i.e., \(F(\theta)\). Consider a simple modification of the earlier game whereby, in Stage 1, both the agents simultaneously choose whether to reveal their type, or to reveal nothing. Thus, agent \(\theta_{A}\) chooses an element from \(\{\theta_{A}\}\cup\)[Not Reveal her Type], whereas agent \(\theta_{B}\) chooses an element from \(\{\theta_{A}\}\cup\)[Not Reveal her Type]. Let \(\pi_{i}(i=X,j=Y)\) denote the payoff of agent \(i\) in case in Stage 1 she selects \(X\) and agent \(j\) selects \(Y\), where \(X,Y\in\ M_{i}=[\theta_{l},\theta_{h}]\cup\{\)Not Reveal \(\theta_{i}\}\). We show that even with two-sided private information, a version of the earlier non-revelation result in Proposition 1 goes through. **Proposition 6**. : _Consider any perfect Bayesian equilibria where, in Stage 2, an agent uses cut-off strategies, in case she has not revealed her type, and completely mixed strategies, in case she has. In any such equilibrium, all \(\theta\) types lower than a cut-off \(\hat{\theta}\) are indifferent between revealing and not revealing her type. All types higher than this cut-off type prefer not to reveal. Thus, there exists a PBEM where there is no information revelation._  
Proof. : _Step (i)._ We first prove that in case agent \(j\) reveals her type, at most one type of agent \(i\), \(i\
eq j\), will reveal her type. We know from our earlier results that if both the agents reveal their types, then the expected payoff for agent \(i\) in the completely mixed-strategy equilibrium is \(\frac{\theta_{i}+c}{2c}\). Suppose that agent \(i\) decides not to reveal her type, given that \(j\) has revealed her type. Clearly, the mixed-strategy equilibrium in the second stage game is identical to that under the unique PBEM characterized in Propositions 2 and 3. Thus the payoff of agent \(i\) is:  
\[\pi_{i}(i=\text{Not reveal }\theta_{i}|j=\text{Reveal }\theta_{j})=\left\{ \begin{array}{l}\theta_{i}+(1-\frac{c+\hat{\theta}}{2c}).c,\text{ if }\theta_{i}>\hat{\theta},\\ \frac{c+\hat{\theta}}{2c}.c,\text{ otherwise.}\end{array} \right.\]  
The difference between the expected equilibrium payoff to \(i\) from not revealing and revealing, given that \(j\) reveals, is strictly positive for all values of \(\theta\) as shown below:  
\[\pi_{i}(i=\text{Not Reveal }\theta_{i}|j=\text{Reveal }\theta_{j}) - \pi_{i}(i=\text{Reveal }\theta_{i}|j=\text{Reveal }\theta_{j})=\] \[\left\{ \begin{array}{l}\frac{\theta_{i}-\hat{\theta}}{2}>0, \text{ if }\theta_{i}>\hat{\theta},\\ \frac{\hat{\theta}-\theta_{i}}{2}>0,\text{ otherwise.}\end{array} \right.\]  
Given that \(j\) reveals, \(i\) would therefore prefer to not reveal and adopt for \(\theta_{i}>\hat{\theta}\) and not reveal and switch for \(\theta<\hat{\theta}\). Only the type \(\hat{\theta}\) is indifferent. _Step (ii)_ We argue that in case agent \(j\) does not reveal her type, all types lower than \(\hat{\theta}\) are indifferent between revelation and non-revelation, whereas all types higher than this cut-off strictly prefer non-revelation to revelation. In this case, if agent \(i\) does not reveal, the coordination cut-off \(\hat{\theta}\) ensures that she is indifferent between adopting and switching, so that  
\[\hat{\theta}+F(\hat{\theta})c=(1-F(\hat{\theta}))c. \tag{20}\]  
As \(F(\theta)\) is strictly monotonic, \(\hat{\theta}\) exists in the interior of the type space and is unique.