Solutions: Moran Process Exercises

import nashpy as nash
import matplotlib.pyplot as plt
import numpy as np

a_values = np.linspace(-10, 10, 500)
number_of_equilibrium = []
for a in a_values:
    M_r = np.array(
        (
            (3 * a, a),
            (2 * a, 2),
        )
    )
    M_c = M_r.T
    game = nash.Game(M_r, M_c)
    number_of_equilibrium.append(
        len(
            list(game.support_enumeration())
        )
    )

plt.figure()
plt.scatter(a_values, number_of_equilibrium)
plt.xlabel("$a$")
plt.title("number of equilibrium")
plt.axvline(2)
plt.axvline(0)
plt.text(-7.5, 2, r"$\{(r_2, c_2)\}$")
plt.text(5, 2, r"$\{(r_1, c_1)\}$")
plt.text(0.75, 1.5, r"$\{(r_1, c_1), (\sigma_1, \sigma_2),(r_2, c_2) \}$", rotation=90);

For $N=2$ :

\begin{align} f_1(a) &= \frac{(i - 1)3a+(2 - i)a}{1}\\ f_2(a) &= \frac{2ai+2(1 - i)}{1}\\ \end{align}

(10)

This gives:

\gamma_i = \frac{i(2a - 2) + 2}{2ai - 2} = \frac{2ai-2i + 2}{2ai - a}

(11)

This gives:

\rho_1 = \frac{1}{1 + \gamma_1} = \frac{1}{1 + \frac{2a}{a}}=\frac{1}{1 + 2}=3

(12)

Let us use some code to confirm these calculations:

import sympy as sym

i = sym.Symbol("i")
a = sym.Symbol("a")
f_1 = ((3 * a * (i - 1) + (2 - i) * a) / 2)
f_2 = ((2 * a * i + 2 * (1 - i)) / 2)
gamma = sym.simplify(f_2 / f_1)
gamma

rho_1_N_2 = 1 / (1 + gamma.subs({i: 1}))
rho_1_N_2

For $N=3$ :

\begin{align} f_1(a) &= \frac{(i - 1)3a+(3 - i)a}{2}\\ f_2(a) &= \frac{2ai+2(2 - i)}{2}\\ \end{align}

(13)

\gamma_i = \frac{2ai +4 - 2i}{3ai-3a+3a-ai}=\frac{2ai+4-2i}{2ai}=\frac{i(a-1) + 2}{ai}

(14)

This gives:

\begin{align} \rho_1 &= \frac{1}{1 + \gamma_1 + \gamma_1\gamma_2} = \frac{1}{1 + \frac{a + 1}{a} + \frac{a+1}{a}\frac{2a}{2a}} &= \frac{1}{\frac{a + 2a + 2}{a}}\\ &=\frac{a}{3a + 2} \end{align}

(15)

Let us use some code to confirm these calculations:

f_1 = ((3 * a * (i - 1) + (3 - i) * a) / 2)
f_2 = ((2 * a * i + 2 * (2 - i)) / 2)
gamma = sym.simplify(f_2 / f_1)
gamma

rho_1_N_3 = 1 / (1 + gamma.subs({i: 1})+ gamma.subs({i: 1}) * gamma.subs({i: 2}))
sym.simplify(rho_1_N_3)

For $N=4$ :

\begin{align} f_1(a) &= \frac{(i - 1)3a+(4 - i)a}{3}\\ f_2(a) &= \frac{2ai+2(3 - i)}{3}\\ \end{align}

(16)

\gamma_i = \frac{i(2a - 2) + 6}{2ai + a}=\frac{2(ai-i+3)}{a(2i+1)}

(17)

This gives:

\begin{align} \rho_1 &= \frac{1}{1 + \gamma_1 + \gamma_1\gamma_2 + \gamma_1\gamma_2\gamma_3}\\ & = \frac{1}{1 + \frac{2(a + 2)}{3a} + \frac{2(a + 2)}{3a}\frac{2(2a+1)}{5a} + \frac{2(a + 2)}{3a}\frac{2(2a+1)}{5a}\frac{6a}{7a}}\\ & = \frac{1}{\frac{3\cdot 5 a ^2}{3\cdot 5 a ^2} + \frac{2(a + 2)\cdot 5 a}{3\cdot 5a^2} + \frac{\frac{7 + 6}{7}(2(a + 2)2(2a+1))}{3\cdot 5 a ^ 2}}\\ & = \frac{1}{\frac{3\cdot 5\cdot 7 a ^2}{3\cdot 5 \cdot 7 a ^2} + \frac{7 \cdot 2(a + 2)\cdot 5 a}{3\cdot 5\cdot 7a^2} + \frac{13(2(a + 2)2(2a+1))}{3\cdot 5 \cdot 7a ^ 2}}\\ & = \frac{3\cdot 5 \cdot 7 a ^ 2}{3\cdot 5\cdot 7 a ^2 + 7 \cdot 2(a + 2)\cdot 5 a + 13(2(a + 2)2(2a+1))}\\ & = \frac{3\cdot 5 \cdot 7 a ^ 2}{105 a ^ 2 + 70 a ^ 2 + 140 a + 104a ^ 2 + 260 a + 104}\\ & = \frac{3\cdot 5 \cdot 7 a ^ 2}{279a^2 + 400a + 104}\\ \end{align}

(18)

Let us use some code to confirm these calculations:

f_1 = ((3 * a * (i - 1) + (4 - i) * a) / 3)
f_2 = ((2 * a * i + 2 * (3 - i)) / 3)
gamma = sym.simplify(f_2 / f_1)
gamma

rho_1_N_4 = 1 / (1 + gamma.subs({i: 1})+ gamma.subs({i: 1}) * gamma.subs({i: 2})+ gamma.subs({i: 1}) * gamma.subs({i: 2}) * gamma.subs({i: 3}))
sym.simplify(rho_1_N_4)

a. For all $N=2$ we have: