3. Extensive Form Games
3.6 Repeated Games and Reputation (Chapter 22)<\/p>\n\n\n\n
In many applications, players face the same interaction repeatedly<\/em><\/p>\n\n\n\n Question<\/mark>: How does this affect our predictions of play?<\/p>\n\n\n\n Repeated games<\/em> provide a framework for studying long-term relationships<\/mark><\/strong><\/p>\n\n\n\n Suppose that two players play the following normal-form game once:<\/p>\n\n\n\n At X, player 1’s Best Response is A, the BR in Y is B, the BR in Z is A pure strategy Nash Equilibrium: (A, Z), (B, Y)<\/mark><\/p>\n\n\n\n <\/p>\n\n\n\n What if the players interact over two periods? <\/p>\n\n\n\n We look for a pure-strategy SPE of this repeated game<\/p>\n\n\n\n * SPE(Subgame Perfect Equilibrium)<\/p>\n\n\n\n To this end, recall that SPE requires NE plays in every subgame. Therefore, for a strategy profile s* to be a SPE, (A, Z) or (B, Y) must be played at least in period 2 regardless of the previous history<\/em>.<\/p>\n\n\n\n The problem of identifying SPE then boils down to<\/p>\n\n\n\n (1) what actions the players choose in period 1; Putting aside (2) for a moment, we can easily find the next 4 SPE:<\/p>\n\n\n\n {((A, Z), (A, Z)), ((A, Z),(B, Y)), ((B, Y),(A, Z)), (B, Y),(B, Y))}<\/p>\n\n\n\n More generally, we have<\/p>\n\n\n\n THEOREM 3.6<\/mark>. Consider any repeated games. Any sequence of stage-Nash profiles is supported as the outcome of a subgame perfect equilibrium.<\/p>\n\n\n\n We look for a pure-strategy SPE of this repeated game<\/p>\n\n\n\n To this end, recall that SPE requires NE plays in every subgame. Therefore, for a strategy profile s* to be a SPE, (A, Z) or (B, Y) must be played at least in period 2 regardless of the previous history.<\/em><\/p>\n\n\n\n The problem of identifying SPE then boils down to<\/p>\n\n\n\n (1) what actions the players choose in period 1; Taking into account (2), we can enlarge the set of SPE outcomes<\/em>. To see this, let’s consider the following strategy profile:<\/p>\n\n\n\n Play (A, X<\/mark><\/strong>)) in period 1. If no one deviated, play (A, Z) in period 2. If someone deviated, then play (B, Y) in period 2.<\/p>\n\n\n\n (A, X) is a highest payoff to player 1, second highest payoff to player 2<\/mark><\/p>\n\n\n\n To demonstrate that this is another SPE, we have to check the players’ incentives<\/p>\n\n\n\n For player 1, 4 + 1 > 0 + 2 For player 2, 3 + 4 > 4 + 1 Ensuring that there is no incentive to deviate<\/mark><\/p>\n\n\n\n By relating the second-period actions to past outcomes, even non-stage-Nash profiles<\/em> can be supported as a SPE<\/p>\n\n\n\n Intuition<\/mark><\/strong> behind the strategy above:<\/p>\n\n\n\n By playing X, P2 can build a reputation<\/em> for cooperating and is rewarded in period 2 Suppose that two players play the following stage game twice<\/p>\n\n\n\n In contrast with the previous example, the stage game has a unique NE<\/mark><\/strong>: (D, D) EXAMPLE 3.20 (EXERCISE 4 ON PAGE 308)<\/mark>. If its stage game has exactly one Nash equilibrium, how many subgame perfect equilibria does a two-period, repeated game have? Explain. Would your answer change if there were T periods, where T is any finite integer?<\/p>\n\n\n\n <\/p>\n\n\n\n1 \\ 2<\/td> X<\/td> Y<\/td> Z<\/td><\/tr> A<\/td> 4, 3<\/td> 0, 0<\/td> 1, 4<\/td><\/tr> B<\/td> 0, 0<\/td> 2, 1<\/td> 0, 0<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
At A, player 2’s Best Response is Z, the BR in B is Y<\/mark><\/p>\n\n\n\n
Each player has total 7 information sets (one set in period 1 + six sets in period 2)
For this reason, a player strategy in repeated games is history-dependent<\/mark><\/strong><\/p>\n\n\n\n
(2) whether or not choices in period l can influence their play in period 2<\/p>\n\n\n\n
(2) whether or not choices in period 1 can influence their play in period 2<\/p>\n\n\n\n
((A, X), (A, Z) > (B, X), (B, Y)), payoff is smaller when debated, making this deviation non-profitable<\/mark><\/p>\n\n\n\n
((A, X), (A, Z) > (A, Z), (B, Y)), No incentive to deviate<\/mark><\/p>\n\n\n\n
By deviating, P2 is branded a “cheater” and is punished in period 2<\/p>\n\n\n\nRepeated Prisoners’ Dilemma<\/h3>\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
In period 2, (D, D) must be played regardless of the outcome in the 1st period
Choices in period 1 cannot influence payoffs in period 2<\/mark><\/strong>
Backward induction implies that both players choose D in period 1
Therefore, the unique SPE is that both players always<\/em> play action D<\/p>\n\n\n\n