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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?

My attempt is:

First, I find the Nash equilibrium of the game (so we know where the player would deviate if not following the proposed strategy):

For the row player we see that Row T and M are dominated by B, so we leave row B and delete the former 2 rows. Then for the column player, we see that the columns L and C are dominated by R, so we leave R and delete the former 2 rows. So our Nash Equilibrium is (B,R) with payoff $(0,0)$.

By a definition in my textbook:

enter image description here

So the expected value of staying with the strategy $(M,c)=(4,4)$ is :

$4+\delta 4+\delta^2 4+....=4\sum^{\infty}_{t=1}\delta^{t-1}= 4/(1-\delta) = 4+4\delta/(1-\delta)$

Now, if the players deviate to $(0,0)$, then they would get $5$ insted of $4$ in the immediate stafe of the deviation, followed by his continuation payoff:

$v_i'=5+0\delta+0\delta^2_+...=5$

For the player to stay and not deviate, the payoff for the first strategy should be higher than the latter strategy (where they deviate):

$$4+4\delta/(1-\delta)\geq 5 \Leftrightarrow \delta \geq 1/5$$

So, for $\delta \geq 1/5$, the players would not deviate.

Would this reasoning/solution be correct?

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    $\begingroup$ Yes, this looks correct. $\endgroup$ – Bayesian Sep 21 '20 at 20:04

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