0
$\begingroup$

Let's say a Markov chain has the state-space $S = \{A,B,C,D\}$ Transition Matrix as follows:

$ \begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \end{bmatrix} $

This is NOT irreducible as far as I am aware due to state C being transient.

To find an equilibrium distribution for the chain, would I just treat it the same way as if it were irreducible in which case I get $\pi = (1/3)(1,1,0,1)$. Is this unique? I'm not sure. Thanks in advance!

1 Answers 1

0

This Markov chain is periodic, so no stationary distribution exists. If we start in state A we see the pattern ADBADBADB... so we know every third time step we will be in either A, D or B. And if we start the chain in state C then the pattern is CADBADBADB...