# Discount factor and deviating from strategy - Game Theory

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: 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: 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?

• Yes, this looks correct. – Bayesian Sep 21 '20 at 20:04