Question

6. Consider infinitely repeated prisoner’s dilemma

	C	D
C	2,2	-1,3
D	3.-1	0,0

(a) For each δ < 1 find the shortest length of punishment that is needed in order to sustain cooperation, or show that the cooperation cannot be sustained.

(b) Using the one-shot deviation principle for each δ argue whether tit-for-tat strategy is an SPE or not. The tit-for-tat strategy is to cooperate in period 1. In period t do whatever your opponent did in period t − 1.

Answer 1

Due to lack of information in the question, we take that tit-for-tat strategy is being followed here.
By symmetry, same will be the result for P2.
From P1’s point of view
Starting point is (C,C)
Deviation action will be (D,C)
Punishment action will be (D,D)
Show that cooperation cant be sustained in the LR using one-shot deviation
Let p be the discount factor (p<1)
If (C,C) is played by both players indefinitely
Player 1 gets a payoff of 2+2p+2p2+2p3+…….. = 2/(1-p)
If P1 deviates at one point of time and then continues with tit-for-tat strategy, the actions will be
(D,C),(C,D),(D,C),(C,D),……………
So, deviation payoff should be 3+(-1)p+3p2+(-1)p3+3p4+(-1)p5+….. = 3/(1-p2) - 1p/(1-p2) = (3-p)/(1-p2)

Cooperation can be sustained iff
2/(1-p) >= (3-p)/(1-p2)
Solving gives
p>1
Which is a contradiction.

So cooperation cannot be sustained indefinitely.

If we take the length of punishment to be of 1 time period, if P1 deviates, payoff will be 3+0p+2p2+2p3+2p4+……

Comparing payoff for cooperation and deviation, we get p>0.5

The results can be summarised as follows
Length of punishment
Min value of p

1
0.5

2
0.366

3
0.343

4
0.336

And so on….