

In [1]:

    
import numpy as np
from game_tools import NormalFormGame
from logitdyn import LogitDynamics
from __future__ import division



In [29]:

    
coo_payoffs = np.array([[4,0], [3,2]])
g_coo = NormalFormGame(coo_payoffs)

test_simulate_LLN



In [43]:

    
u = coo_payoffs
beta = 1.0
P = np.zeros((2,2))

I made a probabilistic choice matrix $P$ in a redundant way just in case.



In [44]:

    
P[0,0] = np.exp(u[0,0] * beta) / (np.exp(u[0,0] * beta) + np.exp(u[1,0] * beta))
P[0,0]









    Out[44]:





0.73105857863000479



In [45]:

    
P[1,0] = np.exp(u[1,0] * beta) / (np.exp(u[0,0] * beta) + np.exp(u[1,0] * beta))
P[1,0]









    Out[45]:





0.2689414213699951



In [46]:

    
P[0,1] = np.exp(u[0,1] * beta) / (np.exp(u[0,1] * beta) + np.exp(u[1,1] * beta))
P[0,1]









    Out[46]:





0.11920292202211755



In [47]:

    
P[1,1] = np.exp(u[1,1] * beta) / (np.exp(u[0,1] * beta) + np.exp(u[1,1] * beta))
P[1,1]









    Out[47]:





0.88079707797788243



In [48]:

    
print P









    



[[ 0.73105858  0.11920292]
 [ 0.26894142  0.88079708]]

$P[i,j]$ represents the probability that a player chooses an action $i$ provided that his opponent takes an action $j$.



In [49]:

    
Q = np.zeros((4,4))
Q[0, 0] = P[0, 0]
Q[0, 1] = 0.5 * P[1, 0]
Q[0, 2] = 0.5 * P[1, 0]
Q[0, 3] = 0
Q[1, 0] = 0.5 * P[0, 0]
Q[1, 1] = 0.5 * P[0, 1] + 0.5 * P[1, 0]
Q[1, 2] = 0
Q[1, 3] = 0.5 * P[1, 1]
Q[2, 0] = 0.5 * P[0, 0]
Q[2, 1] = 0
Q[2, 2] = 0.5 * P[1, 0] + 0.5 * P[0, 1]
Q[2, 3] = 0.5 * P[1, 1]
Q[3, 0] = 0
Q[3, 1] = 0.5 * P[0, 1]
Q[3, 2] = 0.5 * P[0, 1]
Q[3, 3] = P[1, 1]
print Q









    



[[ 0.73105858  0.13447071  0.13447071  0.        ]
 [ 0.36552929  0.19407217  0.          0.44039854]
 [ 0.36552929  0.          0.19407217  0.44039854]
 [ 0.          0.05960146  0.05960146  0.88079708]]

$Q$ is the transition probability matrix. The first row and column represent the state $(0,0)$, which means that player 1 takes action 0 and player 2 also takes action 0. The second ones represent $(0,1)$, the third ones represent $(1,0)$, and the last ones represent $(1,1)$.



In [52]:

    
from quantecon.mc_tools import MarkovChain



In [53]:

    
mc = MarkovChain(Q)



In [55]:

    
mc.stationary_distributions[0]









    Out[55]:





array([ 0.22451524,  0.08259454,  0.08259454,  0.61029569])

I take 0.61029569 as the criterion for the test.



In [56]:

    
ld = LogitDynamics(g_coo)



In [130]:

    
# New one  (using replicate)
n = 1000
seq = ld.replicate(T=100, num_reps=n)
count = 0
for i in range(n):
    if all(seq[i, :] == [1, 1]):
        count += 1
ratio = count / n
ratio









    Out[130]:





0.605



In [131]:

    
# Old one
counts = np.zeros(1000)
for i in range(1000):
    seq = ld.simulate(ts_length=100)
    count = 0
    for j in range(100):
        if all(seq[j, :] == [1, 1]):
            count += 1
    counts[i] = count
m = counts.mean() / 100
m









    Out[131]:





0.59267999999999998