# Coordination and Private Information Revelation
## 6 Extensions
### Mandatory Disclosure  
We now compare the coordination probability under the PBEM vis-a-vis that under mandatory information revelation followed by a play of completely mixed strategies.  
Fixing \(\theta\), let \(\psi_{R}(\theta)\) denote the coordination probability on either one of the two ideas, either A or B, under mandatory disclosure (full revelation) followed by completely mixed strategies in the coordination phase. Similarly, let \(\phi_{NR}\) denote the coordination probabilities under the unique PBEM. With complete revelation, the coordination probability in the completely mixed-strategy equilibrium is  
\[\psi_{R}(\theta)=(1-p_{R})q_{R}+(1-q_{R})p_{R}=\frac{c^{2}-b\theta}{2c^{2}}, \tag{21}\]  
where \(p_{R}\) and \(q_{R}\) are the probabilities with which agents A and B adopt their own ideas respectively, \(0<p_{R},q_{R}<1\). In contrast, the coordination probability under the unique PBEM is  
\[\psi_{NR}=(1-F(\hat{\theta}))(1-q_{NR})+F(\hat{\theta})q_{NR}=\frac{c^{2}-b \hat{\theta}}{2c^{2}}, \tag{22}\]  
recalling that \(q_{NR}\) is the probability with which B adopts own idea and A plays a simple cut-off strategy at \(\hat{\theta}\). Thus, the difference in expected probability of coordination is  
\[\int_{\theta_{1}}^{\theta_{h}}\psi d(F(\theta))=\int_{\theta_{1}}^{\theta_{h}} [\psi_{R}(\theta)-\psi_{NR}]d(F(\theta))=\frac{b}{2c^{2}}(\bar{\theta}-\hat{ \theta}), \tag{23}\]  
where, \(\bar{\theta}\) is the average value of \(\theta\).  
**Proposition 8**.: _The overall expected coordination probability through mandatory information disclosure is greater than or equal to that with the non-revelation equilibrium iff \(\bar{\theta}\geq\hat{\theta}\)._  
It is interesting to note, therefore, that mandatory disclosure need not improve coordination probability in the game. This follows directly from the fact that revelation makes the opponent more aggressive. Recall that \(q_{R}(\theta)\) is increasing in \(\theta\), whereas \(q_{NR}\) is independent of \(\theta\). For all types \(\theta\) higher than \(\hat{\theta}\), \(q_{NR}\) is greater than \(q_{R}(\theta)\). Consequently, agent B is more aggressive without mandated disclosure than with mandated disclosure. For all types lower than \(\hat{\theta}\), agent B is less aggressive without mandated disclosure and the converse holds.  
**Example 1**.: _If \(\theta\) is uniformly distributed over \([0,1]\), \(c=2\) and \(b=1\), with \(\hat{\theta}=\frac{c-b}{2c}=\frac{1}{4}\) and \(\bar{\theta}=\frac{1}{2}\), \(\int_{\theta_{1}}^{\theta_{h}}\psi d(F(\theta))=\frac{1}{4}>0\) leading to lower overall expected coordination probability with mandated disclosure in comparison with the no disclosure equilibrium. On the other hand, if the support of the distribution changes to \([0,\frac{1}{2}]\), mandated disclosure and no disclosure equilibria achieve the same expected probability of coordination, as \(\bar{\theta}=\hat{\theta}=\frac{1}{4}\). If we change the support of the distribution to \([0,\frac{1}{4}]\), expected coordination probability is higher with mandated disclosure than no disclosure, as the mean of the distribution falls to \(\frac{1}{8}<\hat{\theta}\)._