What does it mean for a stationary distribution to have imaginary values?

Question

Let's say I have a transition matrix P $$ \begin{bmatrix} 0.4 & 0.6 & 0. & 0.& 0.\\ 0.5 &0. & 0.5 &0. & 0. \\ 0.6 &0. & 0. & 0.4 &0. \\ 0.7 &0. & 0. & 0. & 0.3\\ 1. & 0.& 0.& 0.& 0. \end{bmatrix} $$

The stationary distribution of this matrix, which describes a real-world process, has imaginary values. The Perron-Frobenius eigenvalue is 1. I know for advection processes or processes where stuff stays in the system forever i.e. $\lambda=1$, we deal with imaginary values because those don't disappear. However, I don't know enough about transition matrices to know if the same thing is going on here.

Also, if I wanted to use the stationary distribution as a probability distribution in order to calculate something in expectation, but I end up having imaginary values in the stationary distribution, were I to calculate my expected value as normal, can I still treat it as a proper expectation or is it now meaningless?

Thanks.

Answer 1

Here's some code to compute the eigenvector using the standard scipy linear algebra library:

import numpy as np
import scipy.linalg as la

mat = [[0.4, 0.6, 0., 0.,  0],
 [0.5, 0.,  0.5, 0., 0.],
 [0.6, 0.,  0.,  0.4, 0.],
 [0.7, 0.,  0.,  0.,  0.3],
 [1.,  0.,  0.,  0.,  0. ]]
print(la.eig(np.transpose(mat))[1][:,0])

The output:

[0.8259962 +0.j 0.49559772+0.j 0.24779886+0.j 0.09911954+0.j
 0.02973586+0.j]

The eigenvectors of the matrix are the columns of the second array returned, not the rows.

Answer 2

If the stationary distribution has imaginary values, then there is no stationary distribution. Any Markov chain's state transition matrix will have a complex eigenvector with corresponding eigenvalue 1, but not every Markov chain has a stationary distribution.