In class today we revisited the Moran Process and specifically we looked at calculating so called fixation probabilities analytically.

You can see a recording of this here: cardiff.cloud.panopto.eu/Panopto/Pages/Viewer.aspx?id=09358805-0821-4a71-b50d-afc000a5ac93

The notions considered follow the class text closely which you can read here: nashpy.readthedocs.io/en/stable/text-book/moran-process.html.

The formula for the respective fitness values used is:

\[\begin{aligned} f_{1i} &= \frac{3(N-i)}{N - 1}=3\frac{N-i}{N-1}\\ f_{2i} &= \frac{i+2(N - i -1)}{N - 1}=\frac{2N-2-i}{N - 1}\\ \end{aligned}\]For \(N=3\) this gives (recall we’re using the Hawk Dove game)

\[f_{1}(1)=3\qquad f_{1}(2)=\frac{3}{2}\]and:

\[f_{2}(1)=\frac{3}{2}\qquad f_{2}(2)=1\]Which leads to:

\[\gamma_1=\frac{1}{2}\qquad \gamma_2=\frac{2}{3}\]Thus, applying the formula we get:

\[x_1 = \frac{1}{1 + 1/2 + 1/2\times2/3}=\frac{1}{11/6}\approx.545455\]This is not far from the approximations we simulated.

This was/is our last so called “content” based class, everything from here will be preparing for your assessment: both the individual and the group.

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