{"id":1133,"date":"2023-06-01T05:29:25","date_gmt":"2023-05-31T20:29:25","guid":{"rendered":"https:\/\/saraheee.com\/?p=1133"},"modified":"2023-06-27T21:28:02","modified_gmt":"2023-06-27T12:28:02","slug":"game-theory-7-chap10-infinitely-repeated-games","status":"publish","type":"post","link":"https:\/\/saraheee.com\/ko\/2023\/06\/game-theory-7-chap10-infinitely-repeated-games\/","title":{"rendered":"Game Theory 7 \u2013 chap10. Infinitely repeated games"},"content":{"rendered":"

Infinitely Repeated Prisoners’ Dilemma<\/h3>\n\n\n\n

Imagine that the prisoners’ dilemma is played infinitely many times<\/p>\n\n\n\n

In order to introduce discounting<\/em> of future payoffs, we denote by \\(\\delta \\in\\) [0, 1] the players’ common discount factor<\/p>\n\n\n\n

Suppose that the player obtains a payoff of \\(\\nu\\) every period<\/p>\n\n\n\n

Then the sum of the discounted payoff stream, or simply the discounted payoff<\/mark><\/strong>, is<\/p>\n\n\n\n

\\(\\nu + \\nu \\delta + \\nu \\delta^{2} + \\nu \\delta^{3} + \\cdots = \\frac{\\nu}{1-\\delta}\\)<\/p>\n\n\n\n

A more convenient way to express payoffs in repeated games is<\/p>\n\n\n\n

\\(\\frac{\\nu}{1-\\delta} \\cdot (1 – \\delta) = \\nu\\)<\/p>\n\n\n\n

This is referred to as the average discounted payoff<\/mark><\/strong> and denoted \\(\\pi\\)<\/p>\n\n\n\n

e.g., \\(10 + 5 \\delta + 10 \\delta^{2} + 5 \\delta^{3} + \\cdots = \\frac{10}{1-\\delta^{2}} + \\frac{5\\delta}{1-\\delta^{2}}\\)<\/mark><\/p>\n\n\n\n

average discounted payoff: \\((\\frac{10}{1-\\delta^{2}} + \\frac{5\\delta}{1-\\delta^{2}}) \\cdot (1-\\delta)\\) = \\(\\frac{10+5\\delta}{1+\\delta}\\)<\/mark><\/p>\n\n\n\n

To formalize the idea of reputation in infinitely repeated games, we consider the following simple strategy:<\/p>\n\n\n\n

Cooperate so long as no one has ever defected; otherwise defect<\/em><\/p>\n\n\n\n

Hence (D, D) is used in the punishment<\/mark><\/strong> phase and (C, C) in the cooperation<\/mark><\/strong> phase.
This kind of strategies is dubbed a grim-trigger<\/em> strategy<\/p>\n\n\n\n

To see whether this strategy constitutes a SPE, we utilize the symmetric payoff structure and focus on player 1’s incentives to deviate<\/p>\n\n\n\n

e.g., symmetric: \\(s_{1} = s_{2}^{\\prime}, s_{2} = s_{1}^{\\prime}\\) in \\(u_{1}(s_{1}, s_{2}) = u_{2}(s_{1}^{\\prime}, s_{2}^{\\prime})\\)<\/mark><\/p>\n\n\n\n

1 \\ 2<\/td>C<\/td>D<\/td><\/tr>
C<\/td>2, 2<\/td>0, 3<\/td><\/tr>
D<\/td>3, 0<\/td>1, 1<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

\\(u_{1}(C, D) = u_{2}(D, C) = 0\\)<\/mark><\/p>\n\n\n\n

(1) Cooperation Phase<\/h5>\n\n\n\n

Eqbm: (C, C), (C, C), (C, C), … \u2192 \\(\\pi_{1}\\) = 2<\/p>\n\n\n\n

Deviation: (D, C), (D, D), (D, D), … \u2192 \\(\\pi_{1}\\) = \\((1-\\delta)(3+\\sum_{t=1}^{\\infty}\\delta t)\\)<\/mark><\/p>\n\n\n\n

= \\((3 + \\delta + \\delta^{2} + \\cdots) \\cdot (1-\\delta) = (3 + \\frac{\\delta}{1-\\delta}) \\cdot (1-\\delta) = 3(1-\\delta) + \\delta = 3 – 2\\delta\\)<\/mark><\/p>\n\n\n\n

2 \u2265 3 – 2 \\(\\delta\\) \u21d4 \\(\\delta\\) \u2265 1\/2
So, if delta is greater than 1\/2, it becomes unprofitable to deviate<\/mark><\/p>\n\n\n\n

(2) Punishment Phase<\/h5>\n\n\n\n

Eqbm: (D, D), (D, D), (D, D), … \u2192 \\(\\pi_{1}\\) = 2<\/p>\n\n\n\n

Deviation: (C, D), (D, D), (D, D), … \u2192 \\(\\pi_{1}\\) = \\((1-\\delta)(0+\\sum_{t=1}^{\\infty}\\delta^{t} \\cdot 1) = \\delta\\)<\/p>\n\n\n\n

Hence defecting forever is the best response for player 1<\/p>\n\n\n\n

Therefore, the grim-trigger strategy can be supported as a SPE if \\(\\delta\\) \u2265 \\(\\frac{1}{2}\\)<\/p>\n\n\n\n

Remark: The one-shot deviation principle<\/mark><\/strong> states that a player has no profitable deviation in any subgames if and only if she has no profitable one-shot deviation. Therefore, to determine whether a player’s behavior is optimal, it is enough to check whether the player cannot benefit from deviating only in the current period.<\/p>\n\n\n\n

Another Prisoners’ Dilemma<\/h3>\n\n\n\n
1 \\ 2<\/td>C<\/td>D<\/td><\/tr>
C<\/td>4, 4<\/td>-2, 6<\/td><\/tr>
D<\/td>6, -2<\/td>0, 0<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

(1) Cooperation<\/mark><\/strong> phase<\/p>\n\n\n\n

Eqbm: \\(\\pi_{1}\\) = 4.
Deviation: \\(\\pi_{1} = 6(1-\\delta)\\), thus, 4 \u2265 6(1-\\(\\delta\\)) \u21d4 6\\(\\delta\\) \u2265 2 \u21d4 \\(\\delta\\) \u2265 1\/3<\/mark><\/p>\n\n\n\n

(2) Punishment<\/mark><\/strong> phase<\/p>\n\n\n\n

Eqbm: \\(\\pi_{1}\\) = 0.
Deviation: \\(\\pi_{1} = -2(1-\\delta)\\), thus, -2(1-\\(\\delta\\)) < 0 \u21d4 \\(\\delta\\) <<\/mark> 1 (Always true)<\/mark><\/p>\n\n\n\n

So, \\(\\delta\\) \u2265 1\/3 (C, C) can be supported as an equilibrium<\/mark><\/p>\n\n\n\n

<\/p>\n\n\n\n