Suppose that two individuals play the prisoner's dilemma (PD) a finite number of times; and assume that they both discount the future at a constant rate. Can cooperation be sustained by a Nash equilibrium? Notice that I am not restricting attention to sub-game perfect Nash equilibria (obviously, there are no SPNE which sustain cooperation).


There is also no NE which sustains coopration for more or less the same reason as in the SPNE case.

Consider, a PD played twice. A strategy contains five actions, one for each decision node: one in the beginning (empty history) and one for each of the four period-2 histories (CC,CD,DC,DD). I claim that any strategy other than (D;D;D;D;D) is dominated.

Consider any strategy in which you play C in period 2, say (C,D,C,C,D). A deviation to (C,D,D,D,D) is profitable because defection cannot be punished after period 2 as the game ends. Behavior in period 1 cannot be conditioned on the future. Given that any equilibrium candidate has the structure (_,D,D,D,D) cooperation in period 1 is also dominated. If you are not convinced, you can write down the game with all its strategies in a big normal form matrix.

You can iterate the argument for any (commonly known) finite number of periods.

  • $\begingroup$ Thanks, I am just reading this now. When you say 'dominated', do you mean 'strictly dominated'? $\endgroup$ – afreelunch Nov 12 '20 at 19:56
  • $\begingroup$ I am seeing weak dominance, not strict dominance. But that is not enough to prove uniqueness of NE. $\endgroup$ – afreelunch Nov 12 '20 at 19:57
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    $\begingroup$ True, there are multiple NE, but in none of them there is cooperation on-path. A NE would be both players playing D in period 1 and D after (D,D) and anything after the other three histories. The off-path play in period 2 has no impact. However, if there was C onpath in the last period, a deviation to D is profitable. So on-path there is always DD in the end. Iterate this argument and ONPATH there are only Ds. I thought you meant C on path with "sustain cooperation." $\endgroup$ – Bayesian Nov 12 '20 at 20:47
  • $\begingroup$ Right, that is what I thought -- thanks for clarifying $\endgroup$ – afreelunch Nov 13 '20 at 8:36

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