Assume the frequency dependent selection model for a population with two types of individuals: $x=(x_1, x_2)$ such that $x_1 + x_2 = 1$. Obtain all the stable distribution for the sytem defined by the following fitness functions:
For all of the functions in question, $x=(0, 1)$ and $x=(1, 0)$ are equilibria. There is a 3rd potential equilibria given by $f_1(x) = f_2(x)$. This is bookwork: https://vknight.org/gt/chapters/11/#Frequency-dependent-selection
$f_1(x)=x_1 - x_2\qquad f_2(x)=x_2 - 2 x_1$
$f_1(x)=f_2(x)\Rightarrow x_1 - x_2 = x_2 - 2x_1 \Rightarrow 3x_1 = 2x_2$ which gives (using the fact that $x_1 + x_2=1$ single solution: $(x_1, x_2)=(2/5, 3/5)$
In [1]:
import sympy as sym
x_1 = sym.symbols("x_1")
sym.solveset(3 * x_1 - 2 * (1 - x_1), x_1)
Out[1]:
B. $f_1(x)=x_1x_2 - x_2\qquad f_2(x)=x_2 - x_1 + 1/2$
$f_1(x)=f_2(x)\Rightarrow x_1x_2 - x_2 = x_2 - x_1 + 1/2$ setting $x=x_1$ so that $1 - x = x_2$ gives: $x - x ^ 2 - 1 + x = 1 - x - x + 1/2$ which corresponds to:
$$-x ^ 2 + 4 x - 5/2=0$$This has solution $x=2 \pm \sqrt{6}/2$, thus $(x_1, x_2) = (2 - \sqrt{6}/2, -1 + \sqrt{6}/2)$ is the only set of solutions for which $1 \geq x_1 \geq 0$ and $1\geq x_2 \geq 0$.
In [2]:
x = sym.symbols("x", positive=True)
res = sym.solveset(- x ** 2 + 4 * x - sym.S(5) / 2, x)
res
Out[2]:
In [3]:
for sol in list(res):
print(sol, float(sol), float(1 - sol))
C. $f_1(x)=x_1 ^ 2 \qquad f_2(x)=x_2^2$
$f_1(x)=f_2(x)\Rightarrow x_1 = \pm x_2$ which has a single solution $(x_1, x_2)=(1/2, 1/2)$.
2. For the following games, obtain all the stable distributions for the evolutionary game:
For bother of these we aim to find $x$ such that $(1, 0)Ax^T=(0, 1)Ax^T$.
$A = \begin{pmatrix}2 & 4 \\ 5 & 3\end{pmatrix}$
$(1, 0)Ax^T=(0, 1)Ax^T\Rightarrow 2x+4(1-x)=5x+3(1-x)\Rightarrow x=1/4$
$A = \begin{pmatrix}1 & 0 \\ 0 & 1\end{pmatrix}$
$(1, 0)Ax^T=(0, 1)Ax^T\Rightarrow x=1-x\Rightarrow x=1/2$
3. Define:
mutated population.
Bookwork: https://vknight.org/gt/chapters/11/#Mutated-population
Evolutionary stable strategies
Bookwork: https://vknight.org/gt/chapters/11/#Evolutionary-Stable-Strategies
4. State and prove the general condition for ESS theorem.
Bookwork: https://vknight.org/gt/chapters/11/#General-condition-for-ESS
5. Using the general condition for ESS theorem identify what strategies are evolutionarily stable for the games of question 2.
For both these games we start by identifying the Nash equilibria:
$(A, A^T) = (\begin{pmatrix}2 & \underline{4} \\ \underline{5} & 3\end{pmatrix}, \begin{pmatrix}2 & \underline{5} \\ \underline{4} & 3\end{pmatrix})$
The pure NE are given by $\{((1, 0), (0, 1)), ((0, 1), (1, 0))\}$, these are not symmetric so do not correspond to ESS. The mixed nash equilibrium corresonds to the stationary point calculated previously: $x=1/4$.
For a general $y$, we have $u(x, x)=u(y,x)$ (by the defnining calculation of $x$). Thus, we look at the second condition:
$$u(x, y)=1/4(2y+4(1-y)) + 3/4(5y+3(1-y))=1/4(4-2y) + 1/4(6y+9)=1/4(4y+13)$$
$$u(y, y)=y(2y+4(1-y)) + (1-y)(5y+3(1-y))=y(4-2y) + (1-y)(2y+3)=-4y^2+3y+3$$
thus:
$$u(x, y) - u(y, y) = 4y^2 - 2y + 1/4=\frac{(4y-1)^2}{4}$$
which is $>0$ for $y\ne 1/4$, thus $x=(1/4, 3/4)$ is an ESS.
$(A, A^T) = (\begin{pmatrix}\underline{1} & 0 \\ 0 & \underline{1}\end{pmatrix}, \begin{pmatrix}\underline{1} & 0 \\ 0 & \underline{1}\end{pmatrix})$
The pure NE are given by $\{((1, 0), (1, 0)), ((0, 1), (0, 1))\}$, these are symmetric, for both we have $u(x, x)=1$ which will be $>u(y, x)$ for all $y\ne x$ as $x$ is a best response to itself. Thus, these are both ESS.
We now consider $x=1/2$. For a general $y$, we have $u(x, x)=u(y,x)$ (by the defnining calculation of $x$). Thus, we look at the second condition:
$$u(x, y)=1/2(y + 1 - y)=1/2$$
$$u(y, y)=y^2 + (1-y)^2=2y^2-2y+1$$
thus:
$$u(x, y) - u(y, y) = -2y^2 + 2 y - 1/2=-\frac{(2y-1)^2}{2}$$
which is $<0$ for $y\ne 1/2$, thus $x=(1/2, 1/2)$ is not an ESS.