A solvable replicator model

We present a soluble variant of the replicator model well established in theoretical biology and game theory. By using methods of statistical physics we derive an analytical solution to our model which becomes exact in the long time limit. We apply our model to the iterated prisoner’s dilemma game and compare our results to numerical simulations. The replicator model describes the evolution of self-replicating entities, replicators, in different areas of biological sciences, for example sociobiology, ecology, prebiotic evolution and genetics [1, 2]. While it finds increasing interest, in particular in the context of dynamical game theory, most investigations still essentially rely on numerical simulations [3, 5, 6]. Here we introduce a diluted version of the replicator dynamics and present an analytic solution to our model. As the original replicator model, our model describes the time evolution of a population by a set of so-called game dynamical equations. The population consists of n species, each adopting a certain strategy i. The state of the population at time step (generation) t is characterized by a vector x = (x 0, . . . , x n−1) with x i being the fraction of the population which belongs to species i (x i > 0, ∑n−1 i=0 x t i = 1). A pay-off matrix A = (Aij )n−1 ij=0 encodes the interactions between the species. In the original replicator model each species interacts with all members of the population in every time step. In our diluted annealed [7] model we draw in each generation one specific species j (t) randomly with equal probability out of the set {0, . . . , n−1}. Then all members of the population interact with j (t). At the next time step a new species is drawn and so on. The pay-off of species i, Aij(t)x j (t), is viewed as a measure of its reproductive success (fitness) [8]. The offspring inherits its species strategy, and hence the growth rate of a species is set proportional to its fitness. Accordingly, the time evolution of the entire population is governed by xt+1 i = x i Aij (t)x t j (t) ∑n−1 i=0 x t i Aij (t)x t j (t) i = 0, . . . , n− 1. (1) Since the iterated prisoner’s dilemma has become the leading paradigm for the emergence of cooperation in biological societies, we are in particular interested in the † Present address: Institut für Theoretische Physik III, Heinrich-Heine-Universität zu Düsseldorf, Universitätsstaße 1, D-40225 Düsseldorf, Germany. 0305-4470/98/150301+04$19.50 c © 1998 IOP Publishing Ltd L301 L302 Letter to the Editor situation where the members of the population play this game. The prisoner’s dilemma (PD) is a two-player game [9, 10]. The players have to opt simultaneously for one of the two strategies C (cooperate) or D (defect). If both players cooperate they receive a pay-off of R points. The pay-off for joint defection is P points. An unilaterally defecting player obtains T points while his opponent ends up with S points. Because T > R > P > S and R > (T + S)/2 (we use T = 5, R = 3, P = 1 and S = 0) a rational player will choose D since this yields the higher pay-off no matter whether the opponent opts for C or D. However, if there is a sufficiently high probability that the players will meet again there is no longer a single best strategy for the iterated prisoner’s dilemma (IPD). We consider a simple version of the IPD where strategies are entirely specified by the outcome of the previous round. Such strategies are said to have one-step memory and can be described by a quadruple p := (p 1, p 2, p 3, p 4), where the components denote the probabilities of cooperating after the outcome of the previous round was R, S, T or P , respectively. If all components are either 0 or 1 such a strategy is deterministic. There are 16 deterministic strategies with one-step memory (which we label by i = 0, . . . , 15, with the ith quadruple being the binary representation for i). We take uncertainty into account by introducing a small probability to misimplement a move and replacing 1 by 1 − and 0 by in the deterministic strategies [11]. If > 0 (we use = 10−5) the game between two strategies i and j can be modelled as a Markov process with fully ranked transition matrix  p 1p j 1 p i 1(1− p 1) (1− p 1)pj 1 (1− p 1)(1− p 1) p 2p j 3 p i 2(1− p 3) (1− p 2)pj 3 (1− p 2)(1− p 3) p 3p j 2 p i 3(1− p 2) (1− p 3)pj 2 (1− p 3)(1− p 2) p 4p j 4 p i 4(1− p 4) (1− p 4)pj 4 (1− p 4)(1− p 4)  . The stationary distribution π : = (π1, π2, π3, π4) corresponds to the left-hand eigenvector of the transition matrix with eigenvalue 1. The mean pay-off for i against j can be calculated as Aij = Rπ1 + Sπ2 + T π3 + Pπ4. The diluted model can be solved by iteration of equation (1) which yields x i = x0 i ∏t−1 τ=0Aij(τ) ∑n−1 i=0 x 0 i ∏t−1 τ=0Aij(τ) where x0 i is the initial frequency of species i. Next we average over all possible realizations of the sequence {j (τ )}t−1 τ=0 and obtain 〈xt i 〉 = 1 n n−1 ∑ j (0)=0 . . . 1 n n−1 ∑ j (t−1)=0 x0 i ∏t−1 τ=0Aij(τ) ∑n−1 i=0 x 0 i ∏t−1 τ=0Aij(τ) . (2) To perform the average it is convenient to introduce new summation variables mi = ∑t−1 τ=0 δij (τ ), which count how often j (τ ) is equal to i for τ = 0, . . . , t − 1. With these new variables equation (2) reads 〈xt i 〉 = ∑