Uniform payoff bound against Tit for tat (${\sum }_{t<T}{u}_t\le RT+(T_{\mathrm{tempt}}-R)$)

In a symmetric two-action stage game write $R,P,T_{\mathrm{tempt}},S$ for mutual cooperation, mutual defection, sole defection and sole cooperation payoffs. Suppose $T_{\mathrm{tempt}}\ge R$, $P\le R$ and $T_{\mathrm{tempt}}+S\le 2R$. Against an initially cooperative Tit for tat, every action sequence has limiting upper average payoff at most $R$.

For a deviator's actions $a_t$, put $a_{-1}=C$, $h(C)=0$ and $h(D)=T_{\mathrm{tempt}}-R$. The opponent at time $t$ plays $a_{t-1}$. Check the four transitions: $C$ to $C$ and $C$ to $D$ satisfywith equality; $D$ to $C$ uses $S\le 2R-T_{\mathrm{tempt}}$, and $D$ to $D$ uses $P\le R$. A telescoping sum gives the displayed finite-horizon bound. Divide by the horizon and let it increase. Mutual Tit for tat yields $R$ every round, so it is a Nash equilibrium under the long-run average payoff criterion. The argument is pathwise and therefore covers arbitrary deviations representable by finite-state machines, as well as action sequences outside that class.