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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).

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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.

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  • $\begingroup$ 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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