I have an exercise in Steven Tadelis Game theory Introduction book (10.2) :

Grim Trigger: Consider the infinitely repeated game with discount factor $δ < 1$ of the following variant of the Prisoner’s Dilemma:

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a) For which values of the discount factor δ can the players support the pair of actions (M, C) played in every period?

b)For which values of the discount factor δ can the players support the pair of actions (T,L) played in every period? Why is your answer different from that for (a)?

I have answered (a) in this question

Now I would like to check if my (b) is correct:

In order to find which strategies the players would deviate to, we find the nash equilibrium of the game, which is $(0,0)$. So, this means that player 1 plays $B$ all the time and player 2 plays $R$ all the time , given that they would deviate to any other strategy than (T,L). So, the payoff for both of them given that they stay on $(6,6)$ is :

$6+\delta 6+\delta^2 6+...=6+\delta\sum^{\infty}_{t=1}\delta^{t-1}=6+6\delta/(1-\delta)$

And if they deviate from (6,6), they will have a payoff of $8$ the first timestep they move from (6,6) and then a payoff of $0$'s afterwards(since the opposing player(s) would play the nash equilibrium strategy, as they would not like to play the same strategy as (6,6) and be losing $-2$ all the time ) :


So, in order not to deviate from (6,6), the payoff should be greater than the deviating strategy, so we calculate for which $\delta$ values they don't deviate:

$6+6\delta/(1-\delta)\geq 8 \Leftrightarrow \delta \geq 1/4$

So, for $\delta \geq 1/4$ they stay on the (T,L) strategy forever.

The reason the answer is different from (a) (my previous question) is that the strategies they deviate from are different. So, to not deviate from 6 to 8, they would need a bigger incentive to stay on track. And for them to not deviate from 4 to 5, they wouldn't be as "motivated" to move from 4 to 5 since the difference is lower than 8-6=1, hence they need a lower incentive to stay "on track".

Would my reasoning be correct?

  • $\begingroup$ Looks correct to me. $\endgroup$ – Bayesian Dec 15 '20 at 20:32

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