Evolutionary Game Theory
Evolutionary Game Theory
Evolutionary game theory applies game-theoretic concepts to biological and social evolution, replacing the rationality assumption with selection dynamics. Strategies that yield higher fitness spread in the population; equilibrium corresponds to evolutionary stability rather than optimality. The evolutionarily stable strategy (ESS) concept, introduced by Maynard Smith and Price (1973), refines Nash equilibrium by requiring resistance to invasion by any mutant strategy — a refinement with both biological and economic applications.
Evolutionarily Stable Strategy
A strategy $p^*\in\Delta(S)$ is an evolutionarily stable strategy (ESS) if, for every alternative strategy $q\neq p^*$, there exists $\bar{\epsilon}>0$ such that for all $\epsilon\in(0,\bar{\epsilon})$: $$u(p^*,(1-\epsilon)p^*+\epsilon q)>u(q,(1-\epsilon)p^*+\epsilon q).$$ Equivalently, $p^*$ is ESS iff (i) $u(p^*,p^*)\geq u(q,p^*)$ for all $q$ (Nash condition), and (ii) if $u(p^*,p^*)=u(q,p^*)$ then $u(p^*,q)>u(q,q)$ (stability condition). ESS $\Rightarrow$ Nash but not conversely.
Replicator Dynamics
In a symmetric game with $n$ strategies, the replicator dynamics describes frequency evolution under selection. If $x_i(t)$ is the proportion using strategy $i$, the continuous-time replicator equation is: $$\dot{x}_i = x_i[u(e_i,x)-\bar{u}(x)]$$ where $u(e_i,x)=\sum_j x_j a_{ij}$ is $i$'s payoff against the current population, $\bar{u}=\sum_i x_i u(e_i,x)$ is the mean payoff. Strategies above average grow; below-average shrink. Nash equilibria are stationary points; ESS corresponds to asymptotically stable fixed points.
Example 1
Find the ESS of the Hawk-Dove game. Payoffs: (H,H)=((V-C)/2,(V-C)/2), (H,D)=(V,0), (D,H)=(0,V/2), (D,D)=(V/2,V/2), with $V Solution: Mixed ESS: try $p^*=P(H)=V/C$. Check Nash: $u(H,p^*)=p^*(V-C)/2+(1-p^*)V$; $u(D,p^*)=p^*\cdot 0+(1-p^*)V/2$. Set equal: $p(V-C)/2+(1-p)V=(1-p)V/2$; $p(V-C)/2=V/2-V+(1-p)V/2$... simplifying: $V/C\cdot(V-C)/2+(1-V/C)V=(1-V/C)V/2$. Direct: equal payoffs at $p^*=V/C$. Stability: $u(H,q)-u(D,q)=(q-V/C)(V-C)/2$... sign analysis confirms $p^*=V/C$ is ESS. Biologically: frequency of Hawks in population stabilizes at $V/C$.
Example 2
Apply replicator dynamics to the Rock-Paper-Scissors game (payoffs: win=1, lose=-1, draw=0). Show that all interior trajectories orbit the Nash equilibrium.
Solution: The unique Nash equilibrium is $x^*=(1/3,1/3,1/3)$. RPS is zero-sum with antisymmetric payoff matrix $A=\begin{pmatrix}0&-1&1\\1&0&-1\\-1&1&0\end{pmatrix}$. The replicator dynamics on the simplex: using the conserved quantity $H(x)=x_1 x_2 x_3$ (Zeeman's Liapunov function for this case), $\dot{H}/H = \sum_i \dot{x}_i/x_i = \sum_i [u(e_i,x)-\bar{u}]=0$ (zero-sum $\Rightarrow\sum_i x_i u(e_i,x)=\bar{u}$ always). Hence $H$ is constant along trajectories — orbits are closed curves on the simplex around $(1/3,1/3,1/3)$.
Practice
- Classify all Nash equilibria of the Hawk-Dove game as ESS, non-ESS Nash, or neither.
- Prove that every ESS is a Nash equilibrium, but give an example of a Nash equilibrium that is not an ESS.
- In the Prisoner's Dilemma, show that Defect is the unique ESS under replicator dynamics. Under what mutations/structures can Cooperate invade?
- Find all ESS of the coordination game $(A,A)=(1,1),(A,B)=(0,0),(B,A)=(0,0),(B,B)=(1,1)$. Show both pure strategies are ESS but the mixed equilibrium is not.
Show Answer Key
1. Hawk-Dove: payoffs $(V-C)/2$ (H vs H), $V$ (H vs D), $0$ (D vs H), $V/2$ (D vs D). If $V>C$: Hawk is the unique ESS. If $V 2. Every ESS $s^*$ satisfies: (1) $u(s^*,s^*)\ge u(s,s^*)$ for all $s$ (NE condition), and (2) if $u(s,s^*)=u(s^*,s^*)$, then $u(s^*,s)>u(s,s)$ (stability). So every ESS is a NE. Counter-example: in Hawk-Dove with $V 3. In the one-shot PD, Defect strictly dominates Cooperate. Against any mutant strategy, Defect earns $\ge$ the mutant. Against a Defect population, Defect earns $P$ while any mutant earns $\le P$. So Defect is the unique ESS. Cooperate can invade only with spatial structure (clusters of cooperators), kin selection, group selection, or direct/indirect reciprocity in repeated interactions. 4. ESS requires $u(A,A)>u(B,A)$ or [$u(A,A)=u(B,A)$ and $u(A,B)>u(B,B)$]. For pure A: $u(A,A)=1>u(B,A)=0$ ✓, so A is ESS. For pure B: $u(B,B)=1>u(A,B)=0$ ✓, so B is ESS. The mixed NE $p=1/2$: $u(1/2,1/2)=1/2$ and $u(s,1/2)=1/2$ for any $s$, so need $u(1/2,s)>u(s,s)$. But $u(s,s)=s^2+(1-s)^2\ge1/2=u(1/2,s)$, with equality only at $s=1/2$. Since $u(s,s)>u(1/2,s)$ for $s\neq1/2$, the mixed equilibrium is NOT an ESS.