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I approached this question in this way: $(P_1,P_2), (R_1,R_2), (S_1,S_2)$ are the Nash Equilibria of the Stage 1 game. For the given strategy to be sustained as SPNE, there should be no way unilateral deviations increase payoffs of the individuals. When $(Q_1,Q_2)$ is played in stage 1, the payoff at the end of stage 2 would be $4+2=6$ each. Now if player 1 deviates to a higher payoff, by playing $P1$, the total payoff he will get after two stages would be $x+0$ Deviation would be futile if $x \leq 6$. Also, it is given tha $x>4$ So, one value of $x$ which can sustain the above strategy as SPNE is $5$. Please correct me where I am wrong. Thanks!


1 Answer 1


In a finitely repeated game with a unique NE, the only SPNE is the repetition of the unique NE. The reason is that by backward induction the NE will be played in the last period and, hence, also in the penultimate period, and so on.

Here, there are multiple NE so that the different NE can be used to reward and punish previous behavior. Your idea of how to construct such a SPNE is correct. The reward payoff in the second (so the final) period is 2 and the punishment payoff is 0 for each player. So we only have our SPNE if $4+2 \geq x+0 \iff x\leq 6$. Otherwise, the reward is not profitable enough (or the punishment NE not harsh enough).

If $x<4$, $(Q,Q)$ is a NE, and repeating this NE twice would be a SPNE.

  • $\begingroup$ Thank You for the answer. Does the fact that x needs to be strictly less than 6 ensure that punishment is 'harsh'? $\endgroup$
    – S.Rana
    Apr 8, 2019 at 9:04
  • $\begingroup$ I apologize, it only has to be weakly larger. If $x=6$, a unilateral deviation from the strategy profile you outlined would still not be (strictly) profitable. $\endgroup$
    – Bayesian
    Apr 8, 2019 at 9:08
  • $\begingroup$ Got it. Thank you! $\endgroup$
    – S.Rana
    Apr 8, 2019 at 9:25

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