# Repeated Game SPNE

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!

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.
• 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. Apr 8, 2019 at 9:08