Cfrgtkky

Consider the Markov chain with state space $S = {1,2,3,4}$ and one-step probability transition matrix $M$:

$$M = begin{bmatrix}0 & frac12 & 0 & frac12 \ frac13 & frac13 & frac13 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1end{bmatrix}$$

This Markov chain clearly does not have unique stationary distribution, as $pi_1=begin{bmatrix}0 & 0 & 1 & 0end{bmatrix}$ and $pi_2=begin{bmatrix}0 & 0 & 0 & 1end{bmatrix}$ are two stationary distributions for the chain. What can we say about the limit distribution for this chain? The definition in Parzen's book states that "a chain is said to possess a long-run distribution if for every $j,kin S$ we have $lim_{ntoinfty}p_{j,k}(n)=pi_k$. I calculated $M^n$ for big $n$ values using MATLAB, and we get something like this:

$$M^n to begin{bmatrix}0 & 0 & frac13 & frac23 \ 0 & 0 & frac23 & frac13 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1end{bmatrix} text{as } ntoinfty$$

So by the definition stated above the chain doesn't have a limit distribution? It all seems somewhat confusing.