<p>Why don't you attack the problem straighrforwardly and investigate the profitability of a deviation?</p>
<p>Suppose there was a Nash equilibrium in which both players decide to stay forever. Both players obtain payoff <span class="math-container">$\frac{1}{1-d}$</span> on path.</p>
<p>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 <span class="math-container">$10$</span>. If a later deviation to quitting or a smaller probability to quit was profitable then so would be this pure earliest deviation.</p>
<p>Hence, always staying is a Nash equilibrium when <span class="math-container">$\frac{1}{1-d}\geq 10 \iff d\geq 0.9$</span>.</p>

CC BY-SA 4.0
2021-01-11T10:37:10.157Z
2021-01-11T10:37:10.157Z
Bayesian
<p>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.</p>
<p>Utilities are discounted with the factor <span class="math-container">$d$</span>. We want to solve for the equilibrium in which both players choose to stay with probability <span class="math-container">$p$</span> at every period.</p>
<p>I hope someone could help me check my approach or offer any different approaches.</p>
<p>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 <span class="math-container">$v$</span>. For either player at any period, the game has a probability <span class="math-container">$p^2$</span> to continue on, and a probability <span class="math-container">$1-p$</span> that the player quits and receives the payment.</p>
<p>So <span class="math-container">$v=p^2(1+dv)+(1-p)10 \implies v=\frac{p^2+(1-p)10}{1-p^2d}$</span>.</p>
<p>Now, at every period, each player must be indifferent between staying and quitting.</p>
<p>For that the be the case, the payoff between quitting and staying muse be equal <span class="math-container">$10=p(1+dv)$</span>.</p>
<p>Plugging in <span class="math-container">$v$</span>, we have <span class="math-container">$10=p(1+d\frac{p^2+(1-p)10}{1-p^2d})$</span>.</p>
<p>Solve. <span class="math-container">$p=\frac{10}{1+10d}$</span>.</p>
<p>If <span class="math-container">$d\leq 0.9$</span>, both players would always play stay. Otherwise, they would mix according to <span class="math-container">$p$</span>.</p>

<game-theory><repeated-games>
Infinitely repeated game with stationary and symmetric equilibrium
Bunbury
# Infinitely repeated game with stationary and symmetric equilibrium

(Asked by: username_0 on 2021-01-11)

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 $p$.

**Tags:** game-theory, repeated-games


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## 1 answers

### Answer by username_1 on 2021-01-11. Score: 3

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 $\frac{1}{1-d}\geq 10 \iff d\geq 0.9$.


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