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: https://cardiff.cloud.panopto.eu/Panopto/Pages/Viewer.aspx?id=5cc70927-0793-4dd5-9690-b11a00c682d0.

After this we discussed potential reasons for the emergence of the social convention of walking on a particular site of the road (on the left in the UK for example).

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

\[A = \begin{pmatrix} 1 & -1 \\ -1 & 1\\ \end{pmatrix}\]This game is meant to model the interaction of individuals in a given population who interact (by walking past each other). If both these individuals walk according to the same convention then they get a utility of 1 but if not they get a utility of -1.

We then model a given population using a vector \(x=(x_1, x_2)\) where \(x_1\)
corresponds to the **proportion** of individuals walking according to the
first convention (say: the left) and \(x_2=1-x_1\) is the **proportion**
walking
according to the second convention.

We can then compute the average utility of an individual who walks using the
first convention (we can refer to this as the **first type** and to utility as
**fitness**). They will interact with another individual of the first type
\(x_1\) of the time getting a fitness of \(1\) and an individual of the
second type \(x_2\) of the time getting a fitness of \(-1\). The average
utility is then:

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

\[f_2 = - x_1 + x_2\]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:

In the case of our game this corresponds to:

\[\begin{align} \frac{dx_1}{dt} =& x_1 (x_1 - x_2 - x_1(x_1 - x_2)+x_2(x_2 - x_1))\\ \frac{dx_2}{dt} =& x_2 (-x_1 + x_2 - x_1(x_1 - x_2)+x_2(x_2 - x_1)) \end{align}\]which can be simplified to:

\[\begin{align} \frac{dx_1}{dt} =& x_1 ((x_1 - x_2) - (x_1 - x_2) ^2)\\ \frac{dx_2}{dt} =& x_2 ((x_2 - x_1) - (x_1 - x_2) ^2) \end{align}\]Substituting \(x_2=1-x_1\) we have:

\[\begin{align} \frac{dx_1}{dt} =& x_1 (2x_1 - 1)2(1-x_1))\\ \frac{dx_2}{dt} =& -x_1 (2x_1 - 1)2(1-x_1)) \end{align}\]And we see (setting the derivatives to be equal to 0) that there are 3 stable populations:

- \(x_1=0\): Everyone drives on the right.
- \(x_1=1\): Everyone drives on the left.
- \(x_1=1/2\): Half the population drives on the left and half on the right.

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.**

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