# Find the threshold strategy of a game with incomplete information

I have a question on how to find the equilibrium outcomes of the following game as a function of $$(\epsilon_1, \epsilon_2)$$.

The game

There are 2 players.

Let $$Y_i$$ denote the action of player $$i$$ for each $$i\in \{1,2\}$$.

For each $$i\in \{1,2\}$$, player $$i$$ chooses between action $$1$$ or $$0$$.

For each $$i\in \{1,2\}$$, if player $$i$$ chooses action $$1$$, she gets $$-\frac{1}{2}Y_j+\epsilon_i$$ as payoff, with $$j\neq i\in \{1,2\}$$

For each $$i\in \{1,2\}$$, if player $$i$$ chooses action $$0$$, she gets $$0$$ as payoff.

For each $$i\in \{1,2\}$$, $$\epsilon_i$$ is private information of player $$i$$.

Players play Bayesian Nash Equilibrium.

We assume that $$(\epsilon_1, \epsilon_2)$$ are i.i.d. uniformly distributed in $$[-1,1]$$.

Question

Show that, for each $$i\in \{1,2\}$$, player $$i$$ chooses $$1$$ if and only if $$\epsilon_i\geq \frac{1}{5}$$

My thoughts: I really don't know how to answer the question. I tried with solving the following fixed point problem: $$\begin{cases} \alpha_1=Pr\Big[(\epsilon_1-\frac{1}{2})\times \alpha_2+\epsilon_1\times (1-\alpha_2)\geq 0\Big]\\ \alpha_2=Pr\Big[(\epsilon_2-\frac{1}{2})\times \alpha_1+\epsilon_2\times (1-\alpha_1)\geq 0\Big]\\ \end{cases}$$ where $$\alpha_i$$ is the probability that player $$i$$ plays $$1$$ and inside the square brackets we have the expected profit of each player.

The system gives $$\alpha_1=\alpha_2=\frac{2}{5}$$ and it doesn't seem to give/suggest the threshold that is in the question.

You've already almost solved the problem. Averaged over $$\epsilon_i$$, the probability that player $$i$$ plays $$1$$ is indeed $$\frac25$$. But each player has the private information about their own $$\epsilon_i$$. Since player $$2$$ plays $$1$$ with probability $$\frac25$$, the expected payoff for player $$1$$, given $$\epsilon_1$$, is $$-\frac12\cdot\frac25+\epsilon_1=\epsilon_1-\frac15$$. Thus player $$1$$ will play $$1$$ if and only if $$\epsilon_1\ge\frac15$$, and likewise for player $$2$$.