Replicator dynamics

In class we spoke about the replicator dynamics equation: a differential equation that is a building block of evolutionary game theory.

You can see a recording of this here.

I asked you all to suggest examples of social conventions. I then picked the idea of asking “How are you?” and responding “Fine” as a socially conventionally greeting to explore from an evolutionary game theoretic point of view.

We used the following game to model this using Replicator Dynamics:

\[A = \begin{pmatrix} 2 & -1 \\ 0 & 2\\ \end{pmatrix}\]

This assumes a population of two types of individuals:

• The first type \(H\): individuals who say “How are you?” and expect “Fine” to complete the greeting. These individuals get a utility of 2 when meeting each other and a utility of -1 when meeting someone who does not follow the convention.
• The second type \(\bar H\): individuals who use a different type of greeting. These individuals get a utility of 0 when meeting someone of type \(H\) and 2 when meeting someone of their own type.

We then model a given population using a vector \(x=(x_1, x_2)\) where \(x_1\) corresponds to the proportion of individuals of the type \(H\) and \(x_2=1-x_1\) is the proportion of the type \(\bar H\).

We can then compute the average fitness of individuals of each type:

\[f_1 = 2x_1 - x_2\]

The average fitness of the individuals of the second type are:

\[f_2 = 0\times x_1 + 2x_2 = 2x_2\]

The average utility over the entire population is then given by:

\[\phi=x_1f_1+x_2f_2\]

In the notes on Replicator Dynamics you can find linear algebraic expressions of these quantities \(f\) and \(\phi\) that extend naturally to populations with more than just 2 types.

The actual Replicator Dynamics equation is then given by:

\[\frac{dx_i}{dt} = x_i(f_i-\phi)\text{ for all}i\]

In the case of our game this corresponds to:

\[\begin{align} \frac{dx_1}{dt} =& x_1 (2x_1 - x_2 - x_1(2x_1 - x_2)-2x_2)\\ \frac{dx_2}{dt} =& x_2 (2x_2 - x_2 - x_1(2x_1 - x_2)-2x_2) \end{align}\]

which can be simplified to:

\[\begin{align} \frac{dx_1}{dt} =& x_{1} \left(- x_{1} \left(2 x_{1} - x_{2}\right) + 2 x_{1} - 2 x_{2}^{2} - x_{2}\right)\\ \frac{dx_2}{dt} =& x_{2} \left(- x_{1} \left(2 x_{1} - x_{2}\right) - 2 x_{2}^{2} + 2 x_{2}\right) \end{align}\]

Substituting \(x_2=1-x_1\) and factorising we have:

\[\begin{align} \frac{dx_1}{dt} =& - x_{1} \left(x_{1} - 1\right) \left(5 x_{1} - 3\right)\\ \frac{dx_2}{dt} =& x_{1} \left(x_{1} - 1\right) \left(5 x_{1} - 3\right) \end{align}\]

We note that \(\frac{dx_1}{dt}=-\frac{dx_2}{dt}\) which is expected as we have \(x_1+x_2=1\).

And we see (setting the derivatives to be equal to 0) that there are 3 stable populations:

• \(x_1=0\): No one uses “Hello how are you” as a greeting.
• \(x_1=1\): Everyone uses “Hello how are you” as a greeting.
• \(x_1=3/5\): 60% the population uses “Hello how are you” as a greeting.

The fact that this is stable mathematically (ie the derivatives are zero) corresponds to the game theoretic fact that in these populations every type of individual has the same fitness: so no one behaviour has an evolutionary advantage.

The notebook you can find here (static html here) includes the above calculations but also includes some numerical solutions to the differential equations. This is important as it shows not just what population is stable but how a population can reach a stable population.