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 $[-1,1]$.


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.


1 Answer 1


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 $2$.


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