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I'm doing this finitely repeated Prisoner's dilemma with switching costs but I have trouble showing the fact that $\varepsilon$ had to be $1 < \varepsilon < 2$. I do see why and that it is a subgame-perfect Nash equilibrium, but how do I show it.

My approach: Consider deviations by player 1, given that player 2 sticks to his strategies in his subgames following histories that end in each of the four outcomes of the game. (By symmetry the below would also account for deviations by player 2 if player 1 sticks to his strategies).

We know that this game is played $T$ times. Let $m$ denote the number of cooperation stages (i.e. number of times the strategy profile $(C,C)$ has been played), let $n$ denote the number of punishment stages (number of times the strategy profile $(D,D)$ has been played), and let $T = m + n +1$. If the players cooperate $T$ times they get $3$ in payoffs. Thus, we can write $$3T=3(m+n+1)$$ If player 1 deviates at any stage of the game he'll get a payoff of $$3 \cdot m + (4 - \varepsilon) + 2 \cdot n$$ For the deviation to be profitable for player 1 the deviation payoff has to be greater than or equal to the cooperative payoff. That is, $$3 \cdot m + (4 - \varepsilon) + 2 \cdot n \geq 3m + 3n + 3$$ We can simplify this $$n \leq 1 - \varepsilon$$

And this is where I'm stuck. What am I supposed to interpret from this? That when $\varepsilon \in (1,2)$ then the number of punishment stages is negative?

Another question I have is: Assume that the question above got answered. Is that enough? Or do I need to show something more? Because I'm well aware that playing $(C,C)$ $T$ times is a SPNE and playing $(D,D)$ $T$ times is also a SPNE and either way there is no profitable deviations.

So, to summarize: I would like to know if I'm approaching this problem in the right way. I would like to get some help on how to move forward from where I got stuck. And if the above would be enough to show what the problem is asking for.

Many thanks in advance!

[Update]---------------- Below is my updated attempt on a solution with some inspiration from M.J. Osbourne and the feedback I got from @Herr K.. enter image description here enter image description here enter image description here

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  • $\begingroup$ This doesn't change your conclusion qualitatively, but I'd suggest being more explicit about the steam of payoffs after a deviation. For example, suppose player 1 deviates at time $t$, and the outcome at $t-1$ was $(C,C)$. Then the continuation payoff from $t$ onwards would be $4-\epsilon+2(T-t)$. Compared to the continuation payoff from not deviating at $t$, which is $3(T-t+1)$, deviation is not profitable for any $t\le T$ as long as $\epsilon >1$. $\endgroup$ – Herr K. Jan 18 at 18:00
  • $\begingroup$ Ah I think I understand. If let's say I would apply the same method you used to for example ($C,D$). If we suppose player 1 deviates at time $t$, and the outcome at time $t-1$ was ($C,D$). Then the continuation payoff from $t$ onwards would be $2-\varepsilon + 2(T-t)$. If he doesn't deviate it would be a stream of 0's from time from time $t$, which is $0(T-t+1)$. Thus, we have that deviation is profitable if $2-\varepsilon 2(T-t) \geq 0T \quad \Rightarrow \quad 0 \leq 2 - \varepsilon + 2T - 2t$. That is, a deviation will be profitable for any $t \leq T$ as long as $\varepsilon < 2$. $\endgroup$ – Justin Malik Jan 20 at 16:33
  • $\begingroup$ Well, if the outcome at $t-1$ is $(C,D)$, then the given strategy prescribes playing $D$ at $t$ for player 1. So playing $C$ is deviating, whereas playing $D$ is not. Also, since we only need to consider one-shot deviations, if player 1 deviates to playing $C$ at $t$, he should revert back to the original strategy from $t+1$ onwards, i.e. playing $D$ from $t+1$ till $T$. $\endgroup$ – Herr K. Jan 20 at 22:11
  • $\begingroup$ It might be a dumb question but doesn’t the outcome $(C,D)$ at $t-1$ count as a deviation? If the play was $(C,D)$ at time $t-1$ and again at time $t$ wouldn’t it be two deviations then? $\endgroup$ – Justin Malik Jan 20 at 22:26
  • $\begingroup$ The one stage deviation is for every subgame. So in any subgame that comes after the outcome $(C,D)$, we verify that there is no profitable one-shot deviation within that subgame, even though the subgame in question is clearly off-equilibrium path. $\endgroup$ – Herr K. Jan 20 at 23:22
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A couple hints.

  1. Regarding the lower bound on $\epsilon$: What happens if deviation occurs at stage $T$? In other words, there is no opportunity for your so-called "punishment stages".
  2. Regarding the upper bound on $\epsilon$: Suppose player 2 deviates at stage $T-1$ but player 1 does not. What must be true about $\epsilon$ in order for player 1 to have an incentive to switch to $D$ at the very last stage?

In general, to show that the given strategy profile is subgame perfect, you need to argue that playing $C$ after $(C,C)$ and $D$ after either $(C,D),(D,C),(D,D)$ are optimal.

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  • $\begingroup$ Correct me if I'm wrong. First, in the last stage, $T$, if either player deviates it will yield the deviating player $4-\varepsilon$ which is less that 3 for $\varepsilon > 1$. Thus, it will not be a profitable deviation in the last stage. Second, if player 2 deviates at $T-1$ then he'll gain $4-\epsilon$ and $2$ at $T$ i.e. the total payoff of the last two stages for player 2 is $4-\varepsilon + 2$. On the other hand, if player 2 deviates such that $(C,D)$ is played at time $T-1$, then player 1 would want to deviate in the last stage, $T$. Thus, he'll gain $0+2-\varepsilon$. $\endgroup$ – Justin Malik Jan 17 at 19:48
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    $\begingroup$ @JustinMalik: Yes, you are on the right track. To ensure that player 1 switches to $D$ after player 2's deviation at $T-1$, you'd need $0+2-\varepsilon > 0+0$, the payoff from not switching. Of course, to conclude that the given strategy profile is subgame perfect, you'd still have to argue that any one-shot deviation at earlier stages, either on or off equilibrium path, is not profitable. $\endgroup$ – Herr K. Jan 18 at 0:33
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    $\begingroup$ Thank you so much. I think I understand now. I've updated the post. Would you have a look at the solution I proposed to see if I indeed understood it the right way? $\endgroup$ – Justin Malik Jan 18 at 16:06

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