3. Extensive Form Games

3.6 Repeated Games and Reputation (Chapter 22)

### Repeated Interactions

In many applications, players face the same interaction *repeatedly*

Question: How does this affect our predictions of play?

*Repeated games* provide a framework for studying **long-term relationships**

Suppose that two players play the following normal-form game once:

1 \ 2 | X | Y | Z |

A | 4, 3 | 0, 0 | 1, 4 |

B | 0, 0 | 2, 1 | 0, 0 |

At X, player 1’s Best Response is A, the BR in Y is B, the BR in Z is A

At A, player 2’s Best Response is Z, the BR in B is Y

pure strategy Nash Equilibrium: (A, Z), (B, Y)

What if the players interact over two periods?

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**

We look for a pure-strategy SPE of this repeated game

* SPE(Subgame Perfect Equilibrium)

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*.

The problem of identifying SPE then boils down to

(1) what actions the players choose in period 1;

(2) whether or not choices in period l can influence their play in period 2

Putting aside (2) for a moment, we can easily find the next 4 SPE:

{((A, Z), (A, Z)), ((A, Z),(B, Y)), ((B, Y),(A, Z)), (B, Y),(B, Y))}

More generally, we have

THEOREM 3.6. Consider any repeated games. Any sequence of stage-Nash profiles is supported as the outcome of a subgame perfect equilibrium.

We look for a pure-strategy SPE of this repeated game

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.*

The problem of identifying SPE then boils down to

(1) what actions the players choose in period 1;

(2) whether or not choices in period 1 can influence their play in period 2

Taking into account (2), we can *enlarge the set of SPE outcomes*. To see this, let’s consider the following strategy profile:

Play (A, **X**)) in period 1. If no one deviated, play (A, Z) in period 2. If someone deviated, then play (B, Y) in period 2.

(A, X) is a highest payoff to player 1, second highest payoff to player 2

To demonstrate that this is another SPE, we have to check the players’ incentives

For player 1, 4 + 1 > 0 + 2

((A, X), (A, Z) > (B, X), (B, Y)), payoff is smaller when debated, making this deviation non-profitable

For player 2, 3 + 4 > 4 + 1

((A, X), (A, Z) > (A, Z), (B, Y)), No incentive to deviate

Ensuring that there is no incentive to deviate

By relating the second-period actions to past outcomes, even *non-stage-Nash profiles* can be supported as a SPE

**Intuition** behind the strategy above:

By playing X, P2 can build a *reputation* for cooperating and is rewarded in period 2

By deviating, P2 is branded a “cheater” and is punished in period 2

### Repeated Prisoners’ Dilemma

Suppose that two players play the following stage game twice

1 \ 2 | C | D |

C | 2, 2 | 0, 3 |

D | 3, 0 | 1, 1 |

In contrast with the previous example, the stage game has a **unique NE**: (D, D)

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**

Backward induction implies that both players choose D in period 1

Therefore, the unique SPE is that both players *always* play action D

EXAMPLE 3.20 (EXERCISE 4 ON PAGE 308). 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?

- Reference: Chang-Koo Chi, (25/50) Game Theory and Applications 7 – Finitely repeated games, Jul 8, 2020, https://youtu.be/GF96ScVGZWk