Let’s follow up yesterday’s post with an older paper on evolutionary games by the always-lucid Sergiu Hart. As noted in the last post, there are many evolutionary dynamics for which the rest points of the evolutionary game played by completely myopic agents and the Nash equilibria of the equivalent static game played by strategic games coincide, which is really quite phenomenal (and since you know there are payoff-suboptimal Nash equilibria, results of this kind have, since Maynard Smith (1973), fundamentally changed our understanding of biology). Nash equilibria is a low bar, however. Since Kuhn (1953), we have also known that every finite game has a backward induction equilibrium, what we now call the subgame perfect equilibrium, in pure strategies. When does the invariant limit distribution of an extensive form evolutionary game coincide with the backward induction equilibrium? (A quick mathematical note: an evolutionary system with mutation, allowing any strategy to “mutate” on some agent with some probability in each state, means that by pure luck the system can move from any state to any other state. We also allow evolutionary systems to have selection, meaning that with some probability in each state an agent switches from his current strategy to one with a higher payoff. This process defines a Markov chain, and since the game is finite and the mutations allow us to reach any state, it is a finite irreducible Markov chain. Such Markov chains have a unique invariant distribution in the limit.)

In general, we can have limit distributions of evolutionary processes that are not the backward induction equilibrium. Consider the following three step game. Agent 1 chooses C or B, then if C was chosen, agent 2 (in the agent-normal form) chooses C2 or B2, then if C and B2 were chosen, agent 3 chooses C3 or B3. The payoff to each agent when B is chosen is (4,0,0), when C and C2 are chosen is (5,9,0), when C, B2 and C3 are chosen is (0,0,0), and when C, B2 and B3 are chosen is (0,10,1). You can see that (C,C2,C3) and (B,B2,B3) are both Nash, but only (B,B2,B3) is subgame perfect, and hence the backward induction equilibrium. Is (B,B2,B3) the limit distribution of the evolutionary game? In the backward induction equilibrium, agent 1 chooses B at the first node, and hence nodes 2 and 3 are never reached, meaning only mutation, and not selection, affect the distribution of strategies at those nodes. Since the Markov chain is ergodic, with probability 1 the proportion of agents at node 2 playing B2 will fall below .2; when that happens, selection at node 1 will push agents toward C instead of B. When this happens, now both nodes 2 and 3 are reached with positive probability. If less than .9 of the agents in 3 are playing B3, then selection will push agents at node 2 toward C2. Selection can therefore push the percentage of agents playing B2 down to 0, and hence (C,C2,C3) can be part of the limit invariant distribution even though it is not the backward induction solution.

So is backward induction unjustifiable from an evolutionary perspective? No! Hart shows that if the number of agents goes to infinity as the probability of mutation goes to zero, then the backward induction solution, when it is unique, is also the only element in the limit invariant distribution of the evolutionary game. How does letting the number of agents go to infinity help? Let Bi be an element of the backward induction equilibrium at node i somewhere in the game tree. Bi must be a best reply in the subgame beginning with i if Bj is played in all descendant nodes by a sufficiently high proportion of the population, so if Bi is not a best reply (and hence selection does not push us toward Bi) it must be that Bj is not being played further down the game tree. If Bi is a best reply in the subgame beginning with i, then most of the population will play Bi because of selection pressures.

Now here’s the trick. Consider the problematic case in the example, when node i is not being reached in a hypothesized limit distribution (if i is reached, then since the probability of mutation goes to zero, selection is much stronger than mutation, and hence non best replies will go away in the limit). Imagine that there is another node g preceding i which is also not reached, and that i is only reached when some strategy outside the backward induction equilibrium is played in g. When g and i are not reached, there is no selection pressure, and hence no reason that the backward induction equilibrium node will be played. Large populations help here. With some small probability, an agent in g mutates such that he plays the node which reaches i. This still has no effect unless there is also a mutation in the node before g that causes g to be reached. The larger the population, the lower the probability the specific individual who mutated in g mutates back before *any* individual in the node before g mutates. Hence larger populations make it more likely that rare mutations in unreached nodes will “coordinate” in the way needed for selection to take over.

Final GEB version (IDEAS version). Big up to Sergiu for posting final pdfs of all of his papers on his personal website.