Steady State Probabilities (Markov Chain) Python Implementation

Question

Hi I am trying to generate steady state probabilities for a transition probability matrix. Here is the code I am using:

import numpy as np

one_step_transition = np.array([[0.125     , 0.42857143, 0.75      ],
                                [0.75      , 0.14285714, 0.25      ],
                                [0.125     , 0.42857143, 0.        ]])


def steady_state_prop(p):
    dim = p.shape[0]
    q = (p-np.eye(dim))
    ones = np.ones(dim)
    q = np.c_[q,ones]
    QTQ = np.dot(q, q.T)
    bQT = np.ones(dim)
    return np.linalg.solve(QTQ,bQT)

steady_state_matrix = steady_state_prop(one_step_transition.transpose())

print (steady_state_matrix)

#result is :
#array([0.38268793, 0.39863326, 0.21867882])

#Expected Result = (0.4,0.4,0.2)

My question is why the outcome is slightly different from exact answer?

Asked By: Nadipineni Naimisha

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Source

Answer 1

The expected result is wrong. For the steady state the product of the transition matrix and the steady state must be the steady state again.

tobe = np.array(((0.4, 0.4, 0.2)))
print(tobe)
print(np.dot(one_step_transition.T, tobe))
print()

result = steady_state_prop(one_step_transition)
print(result)
print(np.dot(one_step_transition.T, result))
print()

Output is

[0.4 0.4 0.2]
[0.37142857 0.40714286 0.22142857]

[0.38268793 0.39863326 0.21867882]
[0.38268793 0.39863326 0.21867882]

So your functions seems to be correct, the result you expect is not.

Answered By: rocksportrocker

Answer 2

I have used a different approach to solve the problem:

def Markov_Steady_State_Prop(p):
    p = p - np.eye(p.shape[0])
    for ii in range(p.shape[0]):
        p[0,ii] = 1
    
    P0 = np.zeros((p.shape[0],1))    
    P0[0] = 1
    return np.matmul(np.linalg.inv(p),P0)

The results are the same as yours and I think your expected results are somehow wrong or they are the approximate version.

Answered By: Koorosh Aslansefat

Steady State Probabilities (Markov Chain) Python Implementation

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