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-)[x + +
2 +
53 +
54....]
= x - 5 + 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-)[x + 0 +
2 +
53 +
54...]
= x - 4 + 3 +
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-)[x + +
53 +
54...]
= x - 5 + 23 -
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 + 3 +
2 >
5
If =0.5,
For x > 6.625, he would deviate in the first stage of
punishment.
Or, x - 5 + 23 -
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,
= 0.5)?
A game is a subgame perfect equilibrium (SPE) if and only if
x - 5 + 3<
5
Given, = 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!
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