Useful analogy to interpret the notion of evolutionary stable strategy (ESS)

Question

I am seeking a good analogy to understand the concept of evolutionary stable strategy (state)

Let $\pi$ denote the fitness of a population, $\pi_{ij}$ is the fitness of strategy $i$ against strategy $j$

Then, a strategy $i$ is an evolutionary stable strategy (ESS) (or state) satisfies:

$i$ is an ESS iff $\forall j \neq i$, the following holds:

1. $\pi_{ij} \geq \pi_{ji}$

2. $\pi_{ij} = \pi_{ji}$, then $\pi_{ij} > \pi_{jj}$

It is well known that (1) corresponds to the definition of Nash equilibrium point. A strategy $i$ is Nash if it yields better payoff (fitness) than any other strategy $j$ (under assumption of unilateral deviation).

What about the second condition? In most textbook, it says

"If $i$ is a Nash equilibrium but not a strict one, then $i$ can still be an ESS if it can invade mutant strategy $j$"

Hmm...invasion...mutants....unfortunately these terminologies are not very meaningful to me without the biological context for which evolutionary game theory is built upon.

Can someone provide an useful analogy or realistic case study to motivate the second condition of ESS?

Answer 1

A strategy is an ESS if it is not profitable for a small fraction of the population to deviate from the societal norm. Think of folks developing software in a research lab. Folks can either use the .NET platform on Windows or the Python platform on Linux. If most folks adopt Linux, then it doesn't make sense for a minority to play Windows. The folks adopting Windows won't be contributing, and evolutionary forces will drive them to adopt Windows.

Intuitively, a societal norm is an ESS if a small fraction of the population can deviate (be it by mistake, experimentation, etc.) and find this deviation not to be profitable. So evolutionary forces drive them back to the ESS.

Note: The definition you provided (the standard textbook definition) works for infinite populations. In finite populations, we do not have this notion of replacement. So suppose we have $n$ agents, with $a$ of them programmed to play $A$ and $b$ programmed to play $B$. If we draw an $A$ player, there are $a-1$ remaining $a$ players. So the odds of drawing another $A$ player are lower. This is because agents are not replaced. This blog entry provides a generalized notion of an ESS which converges, and discusses the convergence to a Nash equilibrium.

Answer 2

Example: A group of people is allowed to buy liquor (with assumed utility for all) if at least one of them carries an ID. (Carrying an ID is an earlier, individual, uncoordinated choice that is assumed not to hurt or benefit your utility otherwise.) In the ESS everybody will carry their ID.

NB: Your statement that (1) defines a NE is not correct. (1) is not unilateral.