Tit for tat invasion of a reactive resident ($w_T-w_M=ϵ[R-g(z)]$)

For a reactive resident $(p,q)$ with $|p-q|<1$, let $z=q/(1-p+q)$ and $g(z)=R{z}^2+(S+T)z(1-z)+P(1-z{)}^2$. Its self-payoff and both ordered cross-payoffs against Tit for tat equal $g(z)$. At introduced Tit for tat frequency $ϵ$, its fitness difference is therefore the displayed expression. If $2R\ge S+T$, it is positive for $z<1$ because $R-g(z)=(1-z)[R-P+(R+P-S-T)z]$. Fully cooperative reached histories tie; the deterministic opposite-response strategy needs a separate cycle calculation. Initial advantage is order frequency rather than a positive invasion exponent at zero.

In an $L$-round iterated prisoner's dilemma, Tit for tat against always defect gets $S$ once and $P$ thereafter; the defector gets $T$ once and $P$ thereafter. Their self-payoffs are $R,P$. At introduced frequency $ϵ$, subtracting mixed-population payoffs gives $ϵ[(R-P)-(T+S-2P)/L]-(P-S)/L$. The displayed positive threshold applies when its denominator is positive and the threshold is less than one. Taking $L\to \infty$ first removes this finite entry cost, so the horizon and rare-mutant limits cannot be casually interchanged.