Game Theory 7 – chap09. Finitely repeated game

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