# Strong Nash Equilibria in repeated games

Suppose we have a simultaneous game, that has a strong Nash equilibrium (SNA), i.e. a weak Pareto efficient Nash equilibrium (no deviation of any subset of player brings a benefit to them).

Now suppose we play this game repeatedly. Does the repeated game has a strong Nash equilibrium, too?

I would keep the question about the payoff function of the repeated game as open. Choose whatever adopted payoff works for the answer.

The idea behind the question is as follows: Games that don't have SNA, like prisoner's dilemma, might have those in the repeated scenario, since additional effects like long term strategies come into play.

Based on this, my guess is, that playing the SNA in every game, would also give a SNA in the repeated game.

Let's first recall the definition of a strong nash equilibrium: A strong Nash equilibrium is a Nash equilibrium in which no coalition, taking the actions of its complements as given, can cooperatively deviate in a way that benefits all of its members. In particular, some of the deviating players may profit, but then others must lose. Consider now the following two player game:

1,1 0,0 0,0

0,0 3,0 0,0

0,0 0,0 0,3

The top-left entry is a strong Nash equilibrium of the one stage game, but repeating it twice is not a SNE of the two-stage game. Indeed, in the two-stage game the players can get the payoff 1.5,1.5, which they cannot in the one-stage game. Does this example answers the question?

• Player 1 and player 2 can both deviate in the stage game and mix equally over the second and third actions, each achieving 1.5. – user102087 Sep 26 '18 at 9:48
• @MJW I think if they do that, they each get 0.75 (each gets a payoff 3 with probability only 1/4). – usul Sep 26 '18 at 16:49
• You're right, I was allowing them to correlate. Oops. Eilon is correct. – user102087 Sep 26 '18 at 16:55

Here, I’m assuming that the one-shot deviation principle holds.

Fix a stage game with $$n$$ players. Suppose we have an SNA in the one shot game that yields to each player a vector of payoffs $$\mathbf{x}$$. Thus for any deviation by $$k \leq n-1$$ players that yields $$\mathbf{y}$$, there must be at least one scalar corresponding to a deviating player $$y_j$$ Such that $$y_j < x_j$$.

Consider the infinitely repeated version of this game. The discounted average payoff under this one shot equilibrium repeated will be $$\mathbf{x}$$. Suppose also that in the infinite setting the equilibrium is to play the stage game equilibrium at every history. On path there is no deviation since for any subset, there is one player who would be strictly worse off. Off path, players are willing to punish since by construction there’s no profitable deviation.

• Good answer. I need some time to process this and will return later. – Mark.Neuhaus Sep 26 '18 at 11:03
• @Mark.Neuhaus Sadly, I was incorrect, and Eilon's post is the right answer. The repeated game allows for coordination (in some sense), which I neglected. – user102087 Sep 26 '18 at 16:57