I'm having a lot of trouble computing the equilibrium values for this game. Suppose we have a game where a player with true skill $\theta \sim U[0, 1]$ can choose to take a test or not, which costs $c$.
Let $X \in \{0, 1\}$ be an indicator for whether they take the test. Let $Y \in \{0,1\}$ be an indicator for whether they do well on the test. Suppose $$Pr(Y=1 | X=1)=\theta$$ i.e., if they take the test, they do well with probability $\theta$.
If they don't take the test, they get payoff $p = E(\theta | X=0)$; if they take it and do well they get $q = E(\theta | X=1, Y=1)$, and if they take it and do poorly, they get $r = E(\theta|X=1, Y=0)$.
We can see that they take the test if and only if $$\theta(q-c) + (1-\theta)(r-c) \geq p \iff \theta \geq \frac{p + c - r}{q-r} =k.$$ Now clearly we have that $Pr(X=0) = \min(k, 1)$ so $p = \frac{k}{2}$. However, when I calculate $q, r$, I get the same posterior expectation $\frac{1+k}{2}$ for both quantities, which makes no sense. Here's my work for $q$:
$$q = E(\theta | X=1, Y=1) = \int_{0}^{1} \theta Pr(\theta | X=1, Y=1) d\theta = $$$$\int_{0}^{1}\theta \frac{Pr(Y=1|\theta, X=1)Pr(\theta|X=1)}{Pr(Y=1|X=1)} d\theta.$$ Here we can see the problem: it appears that $$Pr(Y=1 | \theta, X=1) = \theta = Pr(Y=1 | X=1), \text{ (**)}$$ in which case the terms cancel out and we are left with $$\int_{0}^{1} \theta Pr(\theta | X=1) d\theta$$ However, if we do the same process for $r$, the same two terms (with $Y=0$ instead) cancel out, giving us the same integral, which happens to evaluate to $$\frac{1+k}{2}.$$
This doesn't seem to make sense: why would the posterior expectation for $\theta$ be the same regardless of how well you do on the test? There's clearly something wrong with my conditioning, but I can't figure out what I'm missing - any insights?
I think the issue might be in $P(Y = 1|X=1) = P(Y = 1| \theta,X=1) = \theta$.
Lets assume $P(Y = 1| \theta,X=1) = \theta$ and proceed.
By marginalization we have, $$P(Y = 1 | X = 1) = \int_0^1 P(Y = 1, \theta | X = 1) d\theta = \int_0^1 P(Y = 1 | \theta, X = 1) P(\theta | X = 1) d\theta$$
If you assume $P(Y = 1 | \theta, X = 1) = \theta$ then, $$P(Y = 1 | X = 1) = \int_0^1 \theta P(\theta|X =1 ) d\theta = E(\theta | X = 1)$$.
So now Evaluating $q$, $$q = \int_0^1 \theta \frac{P(Y = 1 | \theta, X = 1 ) P(\theta | X = 1) } {P(Y = 1, X = 1)} d\theta = \int_0^1 \theta \frac{\theta P(\theta | X = 1)} {E(\theta | X = 1)} d\theta = \frac{E(\theta^2 | X = 1)}{E(\theta|X = 1)}$$