Infinitely repeated game with stationary and symmetric equilibrium We have two players playing a repeated game. At every period, each player decides to stay or to quit. If both decide to stay, then they both receive 1. If either decides to quit, then the quitter receives 10 for the period while the other player receives 0. The game terminates once someone decides to quit.
Utilities are discounted with the factor $d$. We want to solve for the equilibrium in which both players choose to stay with probability $p$ at every period.
I hope someone could help me check my approach or offer any different approaches.
Since the game could terminate at any period, we first need to find the expected payoff for either player. Suppose the value of the game is $v$. For either player at any period, the game has a probability $p^2$ to continue on, and a probability $1-p$ that the player quits and receives the payment.
So $v=p^2(1+dv)+(1-p)10 \implies v=\frac{p^2+(1-p)10}{1-p^2d}$.
Now, at every period, each player must be indifferent between staying and quitting.
For that the be the case, the payoff between quitting and staying muse be equal $10=p(1+dv)$.
Plugging in $v$, we have $10=p(1+d\frac{p^2+(1-p)10}{1-p^2d})$.
Solve. $p=\frac{10}{1+10d}$.
If $d\leq 0.9$, both players would always play stay. Otherwise, they would mix according to $p$.
Why don't you attack the problem straighrforwardly and investigate the profitability of a deviation?
Suppose there was a Nash equilibrium in which both players decide to stay forever. Both players obtain payoff $\frac{1}{1-d}$ on path.
This really is a Nash equilibrium if anf only if there is no profitable deviation. Consider the deviation to quit in the first period, yielding the deviation payoff $10$. If a later deviation to quitting or a smaller probability to quit was profitable then so would be this pure earliest deviation.
Hence, always staying is a Nash equilibrium when $\frac{1}{1-d}\geq 10 \iff d\geq 0.9$.