A firm has quality $\theta \in \{0,...,100\}$ each of which is equally likely. Regardless of $\theta$, with probability $\frac{1}{10}$, the firm can only send message $m=\varnothing$, while with probability $\frac{9}{10}$ it can send one of two messages $m=\varnothing$ or $m=\theta$. Consumer observes the firm's message and forms a belief about the firm's expected quality $b=E[\theta|m]$. The firm wants to maximize $b$, i.e., it wants to maximize the consumer's expectation of its quality. Is there an equilibrium $\theta^*$ such that: (i) every firm with $\theta<\theta^*$ sends a message $m=\varnothing$, (ii) every firm with $\theta\ge\theta^*$ sends a message $m=\theta$ if it can? If so, what is $\theta^*$?
I attempted to solve this by modeling the firm values as a continuous function $\theta \in [0,1]$. I thought this would be easier than working with discrete values. However, I got an answer that differed from the provided solution. Is there an error in the following solution for continuous firm values, or was I incorrect in assuming I could map the discrete values to $[0,1]$?
Here's my solution: First, I describe the probability that a firm reporting $\varnothing$ is above the cutoff, using Bayes' theorem: $$P(\theta>\theta^*|\varnothing) = \frac{P(\varnothing |\theta>\theta^*)×P(\theta>\theta^*)}{P(∅|\theta>\theta^*)×P(\theta>\theta^*) + P(\varnothing|\theta<\theta^*)×P(\theta<\theta^*)} $$ $$P(\theta>\theta^*| \varnothing ) = \frac{p×(1-\theta^*)}{p×(1-\theta^*) + \theta^*} $$ Second, I found an equation for the customer's expectation of a firm's quality, given that the firm sends $m=\varnothing$: $$\tag{1} E[\theta|\varnothing]=P(\theta<\theta^*|\varnothing)×\frac{\theta^*}{2} + P(\theta>\theta^*|\varnothing)×\frac{1-\theta^*}{2},$$ i.e. the expected value is $\frac{1}{2}$ of $\theta^*$ times the probability that $\theta<\theta^*$, plus $\frac{1}{2}$ of $1-\theta^*$ times the probability that $\theta>\theta^*$.
Finally, I note that a firm with quality $\theta=\theta^*$ should be indifferent between sending message $m=\theta=\theta^*$ or $m=\varnothing$, so $E[\theta|\varnothing]=E[\theta|\theta^*]$. Of course, $E[\theta|\theta^*]=\theta^*$, so: $$\tag{2} \theta^*=P(\theta<\theta^*|\varnothing)×\frac{\theta^*}{2} + P(\theta>\theta^*|\varnothing)×\frac{1-\theta^*}{2}$$ Now by substituting in the value of $P(\theta<\theta^*|\varnothing)$ derived above, I can solve for $\theta^*\approx0.188$ when $p=\frac{1}{10}$. Translating back to the original problem, a firm with quality $\theta\ge19$ will disclose $\theta$, while a firm with quality $\theta\le18$ will send message $m=\varnothing$.
However, this doesn't seem accurate. If $p=1$ instead of $\frac{1}{10}$, I would expect $\theta^*=\frac{1}{2}$, but I get $\theta^*\approx0.293$. The provided solution does satisfy the $p=1,\theta^*=\frac{1}{2}$ boundary condition, so I believe that it is correct. Where is my error?
You have a tiny error when you compute $E(\theta \mid \theta > \theta^*)$ in Equation $(1)$.
Since $\theta$ is uniformly distributed, it should be $$E(\theta \mid \theta > \theta^*) = \frac{1 + \theta}{2}$$ but you have "$-$" instead of "$+$". The same typo returns in Equation (2). When you fix this, the boundary condition is satisfied.