# Repeatedly playing an equilibrium in a repeated game

What is a formal proof to this known fact about repeated games?

The situation in which, in every time step, the players play a Nash equilibrium in the basic game unconditioned on history, is a Nash equilibrium in the repeated game.

• Perhaps you ought to reformat this as a question and post your proof as an answer. Aug 26, 2015 at 13:58
• Your proof is essentially correct. It is difficult to say if it is sufficiently formal. Sufficient for who? The logic was easy to understand (this is subjective), but $X^t, Y^t$ could probably be defined as functions of strategies. Aug 26, 2015 at 14:00

Suppose all players play unconditioned Nash equilibria, except player $i$ who diverts and plays another strategy. The other strategy of player $i$ may depend on history. We have to show that player $i$'s payoff in the repeated game does not increase.

For every time-period t, define the following random variables:

• $X^{t}$ - the utility of player $i$ in time t when all players, including player $i$, play the basic Nash equilibrium.
• $Y_{h}^{t}$ (where $h$ is a history vector of length $t-1$) - the utility of player $i$ in time $t$ when player $i$ diverts and plays the alternative strategy while all other players continue to play the basic Nash equilibrium.
• $Y^{t}$ - the utility of player $i$ in time $t$ when player $i$ diverts while all other players continue to play the basic Nash equilibrium: $$Y^{t}=\sum_{h}Prob[h]\cdot Y_{h}^{t}$$

By definition of a Nash equilibrium, for every time $t$ and for every history $h$:

$$E[Y_{h}^{t}]\leq E[X^{t}]$$

Hence also:

$$E[Y^{t}]\,=\,\sum_{h}Pr[h]\cdot E[Y_{h}^{t}]\,\leq\,\sum_{h}Pr[h]\cdot E[X^{t}]\,\leq\,E[X^{t}]\cdot\sum_{h}Pr[h]\,\leq\,E[X^{t}]$$

Summing over the entire series, and using the fact that the expectation of a sum is the sum of expectations:

$$E[\sum_{t=1}^{T}Y^{t}]\leq E[\sum_{t=1}^{T}X^{t}]$$

So the player cannot gain by deviating.

Thanks to @denesp for verifying this proof!