In the previous chapter:

In this chapter we’ll take a look at what happens when games are repeatedly infinitely.

To illustrate infinitely repeated games (\(T\to\infty\)) we will consider a Prisoners dilemma as our stage game.

Let us denote \(s_{C}\) as the strategy “cooperate at every stage”. Let us denote \(s_{D}\) as the strategy “defect at every stage”.

If we assume that both players play \(s_{C}\) their utility would be:

Similarly:

It is impossible to compare these two strategies. To be able to carry out analysis of strategies in infinitely repeated games we make use of a discounting factor \(0v_1\) in stage \(k\) then:

Recalling that player 1 would receive \(v_1\) in every stage with no devitation, the biggest gain to be made from deviating is if player 1 deviates in the first stage (all future gains are more heavily discounted). Thus if we can find \(\bar\delta\) such that \(\delta>\bar\delta\) implies that \(U_1^{(1)}\leq \frac{v_1}{1-\delta}\) then player 1 has no incentive to deviate.

as \(u_1(\sigma_1’,\bar \sigma_2)>v_1>u_1^*\), taking \(\bar\delta=\frac{u_1(\sigma_1’,\bar\sigma_2)-v_1}{u_1(\sigma_1’,\bar\sigma_2)-u_1^*}\) gives the required required result for player 1 and repeating the argument for player 2 completes the proof of the fact that the prescribed strategy is a Nash equilibrium.

By construction this strategy is also a subgame perfect Nash equilibrium. Given any history both players will act in the same way and no player will have an incentive to deviate: