# Moran processes - solutions¶

1. Give the transition probabilities for the Moran process with neutral drift.

2. Obtain the transition probability matrix for the Moran process with neutral drift with $N=4$ individuals.

$$M = \begin{pmatrix} 1&0&0&0&0\\ 3/16&5/8&3/16&0&0\\ 0&1/4&1/2&1/4&0\\ 0&0&3/16&5/8&3/16\\ 0&0&0&0&1\\ \end{pmatrix}$$

3. State and prove the theorem for fixation probabilities in a birth death process.

4. Extend the formulae of question 3 to the case of a Moran process on a game.

5. For the following games, obtain the fixation probabilities for $N=4$:

1. $A=\begin{pmatrix}1 & 1 \\ 1 & 1\end{pmatrix}$

This game corresponds to neutral drift, thus:

$x_1=1/N=1/4$ (for both strategies).

2. $A=\begin{pmatrix}1 & 2 \\ 3 & 1\end{pmatrix}$

Assuming $i$ individuals of the first type, for this game we have $N=4$ and $(a, b, c, d)=(1, 2, 3, 1)$ the fitness of both types is given respectively by:

$$f_{1i}=\frac{a(i-1)+b(N-i)}{N-1}=\frac{7-i}{3}$$ $$f_{2i}=\frac{c(i)+d(N-i-1)}{N-1}=\frac{2i+3}{3}$$

which gives:

$$\gamma_i=\frac{f_{2i}}{f_{1i}}=\frac{2i+3}{7-i}$$

thus:

$$x_1=\frac{1}{1+\sum_{j=1}^{3}\prod_{k=1}^j\frac{2k+3}{5-k}}=\frac{1}{1+5/6+(5/6)(7/5)+(5/6)(7/5)(9/4)}=\frac{8}{45}$$

Some code to verify this:

In [1]:
import sympy as sym
import numpy as np
def theoretic_fixation(N, game, i=1):
"""
Calculate x_i as given by the above formula
"""
f_ones = np.array([(game[0, 0] * (i - 1) + game[0, 1] * (N - i)) / (N - 1) for i in range(1, N)])
f_twos = np.array([(game[1, 0] * i + game[1, 1] * (N - i - 1)) / (N - 1) for i in range(1, N)])
gammas = f_twos / f_ones
return (1 + np.sum(np.cumprod(gammas[:i-1]))) / (1 + np.sum(np.cumprod(gammas)))

In [2]:
game = np.array([[sym.S(1), sym.S(2)], [sym.S(3), sym.S(1)]])
theoretic_fixation(N=4, game=game)

Out[2]:
8/45

6. Consider the game $A=\begin{pmatrix}r & 1 \\ 1 & 1\end{pmatrix}$ for $r>1$, and obtain $x_1$ as a function of $r$ and $N$. How does $r$ effect the chance of fixation?

Assuming $i$ individuals of the first type, for this game we have $N=4$ and $(a, b, c, d)=(r, 1, 1, 1)$ the fitness of both types is given respectively by:

$$f_{1i}=\frac{r(i-1)+(N-i)}{N-1}$$$$f_{2i}=\frac{(i)+(N-i-1)}{N-1}=1$$

which gives:

$$\gamma_i=\frac{f_{2i}}{f_{1i}}=\frac{N-1}{r(i-1)+(N-i)}$$

thus:

$$x_1=\frac{1}{1+\sum_{j=1}^{N-1}\prod_{k=1}^j\frac{N-1}{r(k-1)+(N-i)}}$$

Note that as $r$ increases the denominator of $x_1$ decreases (each term in the product decreases), thus $x_1$ increases as a function of $r$ (as expected).

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