Question

2. (Level A) Suppose the following Prisoner's Dilemma is repeated infinitely 112 C D Dlx 011 Let uj be the payoff to player i in period t. Player i (i-1,2) maximizes her average discounted sum of payoffs, given by t=1 where δ-1 is the common discount factor of both players Suppose the players try to sustain (C, C) in each period by the 2-Period Limited Retaliation Strategy (2-LRS). That is, each player plays the following strategy . Play C in the first period » In any other period T if both players played C, then play C in period T+1 if at least one player played D * then play D for the next 2 periods, i.e. in periods T+1 and T +2 (The punishment phase.) * revert to C in period T +3 (a) What are the 2 possible outcomes (i.e. action profiles) in a period in which both players are following the 2-LRS? (Hint: one is (C, C).) (b) Suppose Player 1 is contemplating a one-shot deviation from the strategy profile in which both players follow the 2-LRS Calculate the average discounted sum of payoffs for Player 1 if he deviates in (i) Non-punishment phase, i.e when (C, C) is supposed to be played (ii) 1st period of the punishment phase (ii) 2nd period of the punishment phase (c) With reference to your answer to Part (c), explain whether any player would wish to one-shot deviate in the punishment phase. (d) For what values of z would the strategy profile of a pair of 2-LRS be subgame perfect (keeping in mind that -)?

Answer 1

In this question, the Prisoner's Dilemma is played infinite many times.

(a). Here, we have to check the game for two possible outcomes when the 2-LRS strategy is used by both players.

Case 1. When both players cooperate (C), at time T, both get the outcome (C,C). As reward, at time T+1, they again play C, resulting in the outcome (C,C). This goes on for all time periods and the outcome remains (C,C) for both players.

However, this isn't an interesting equilibrium if x > 5. Suppose I am Player 1 and I know that Player 2 will play C, I can play D and get payoff x (greater than 5) while Player 2 gets nothing. Of course, following this, Player 2 will punish me for the next two phases, as stated in the rules of 2-LRS. This brings us to Case 2.

Case 2. When (C,D) is played, Player 2 gets a payoff of x and Player 1 gets 0. The game now enters punishment stage. The following actions are played.

T: (C,D) -> (0,x)
T+1: (D,D) -> (1,1)
Now, since, to Player 2, it seems like Player 1 deviated, s/he will play D, because it is the punishment stage.
T+2: (D,D) -> (1,1)
Here, the outcome is (D,D).

(b). Now, Player 1 is considering a one-shot deviation. This means that he/she won't follow the rules for the 2-LRS for one period/cycle of the game.

(i) If this happens in non-punishment phase,
T: (D,C) -> a one-shot deviation
T+1: (D,D) -> according to 2-LRS, this continues for two periods
T+2: (D,D)
T+3: (C,C) -> continues

Since payoff in case of one-shot deviation is x, and for two stages it is 1, until it becomes 5 again infinitely, we can say
Corresponding payoff = (1-$\delta$)[x + $\delta$ + $\delta$² + 5$\delta$³ + 5$\delta$⁴....] = x - 5$\delta$ + $\delta$³

(ii) 1st period of punishment phase,
T: (C,D)
T+1: (D,C) -> one shot deviation
T+2: (D,D)
T+3: (C,C) -> continues
So in one period, the other player receives x instead of 1, and we receive 0 instead of 1.
Now, corresponding payoff = (1-$\delta$)[x + 0 + $\delta$² + 5$\delta$³ + 5$\delta$⁴...] = x - 4$\delta$ + $\delta$³ + $\delta$²

(iii) 2nd period of punishment phase,
T: (C,D)
T+1: (D,D)
T+2: (D,C) -> one shot deviation
T+3: (C,C) -> continues
Now, corresponding payoff = (1-$\delta$)[x + $\delta$ + 5$\delta$³ + 5$\delta$⁴...] = x - 5$\delta$ + 2$\delta$³ - $\delta$²

(c) Why would a player wish to one-shot deviate in the punishment phase?

A player would wish to deviate in punishment phase if payoff from deviation > 5
That is, x - 4$\delta$ + $\delta$³ + $\delta$² > 5
If $\delta$=0.5,
For x > 6.625, he would deviate in the first stage of punishment.
Or, x - 5$\delta$ + 2$\delta$³ - $\delta$² > 5
Similarly, for x > 7.5, he would deviate in the second punishment stage.

(d) For what values of x would the game be subgame perfect (given, $\delta$ = 0.5)?

A game is a subgame perfect equilibrium (SPE) if and only if
x - 5$\delta$ + $\delta$³< 5
Given, $\delta$ = 1/2
x < 5 + 2.5 - (1/8)
x < 7.375

The game is an SPE if and only if x < 7.375

Hope this helped!