# Markov chain having unique stationary distribution

Let be a finite time homogeneous Markov chain with $$P_{n \times n}$$ transition matrix. As far as I know, $$P$$ has a unique stationary distribution if and only if it has a unique recurrent class (***).

So, let’s take: $$P=\begin{pmatrix} 3/4&0&1/4 \\ 1/3&1/2&1/6 \\ 1&0&0 \end{pmatrix}$$ State $$2$$ is transient, so the chain can’t have a unique recurrent class, right?

Now, if I calculate a stationary vector $$\pi = (a, b, c)$$ by $$(a \ b \ c)P=(a \ b \ c)$$, I find only one such vector $$\pi = (34/5, 0, 1/5)$$.

Isn't there a contradiction with statement (***) ?

I'm quite worried because there's obviously something I don't understand... but what the hell am I not understanding ?

This is an absorbing Markov chain, i.e. there exists a state $$i$$ such that $$\mathbb P(X_n = i \mid X_0=1)=1$$ for all $$n>0$$. For clarity of notation I'll swap columns $$0$$ and $$2$$ to write $$P = \left( \begin{array}{ccc} \frac{1}{4} & 0 & \frac{3}{4} \\ \frac{1}{6} & \frac{1}{2} & \frac{1}{3} \\ 0 & 0 & 1 \\ \end{array} \right)=\begin{pmatrix} Q & R \\ \mathbf 0 & I\end{pmatrix}$$ where $$Q$$ is the substochastic matrix obtained from $$P$$ by considering transitions betweeen transient states, $$R$$ that obtained from $$P$$ by considering transitions from a transient state to an absorbing state, $$\mathsf 0$$ that obtained by considering transitions from absorbing states to transient states, and $$I$$ that obtained by considering transitions between absorbing states (which with reordering of states can be written as the identity matrix with dimension the number of absorbing states).
By definition, a stationary distribution $$\pi$$ for a Markov chain is a row vector $$\pi$$ satisfying $$\sum_i=1$$ with $$\pi>0$$ and $$\pi P=\pi$$. Iteratively this would imply that $$\pi P^n=\pi$$ for all positive integers $$n$$, but as state $$2$$ is absorbing and $$P_{02}$$ and $$P_{12}$$ are positive, for any distribution $$\pi$$ there exists some positive integer $$N$$ such that $$\pi P^n=\begin{pmatrix} 0 & 0 & 1\end{pmatrix}$$ for $$n\geqslant N$$, meaning there can be no stationary distribution for this Markov chain.
Edit: The above answer is considering (positive) recurrent Markov chains. As shown in this answer, $$\begin{pmatrix} 0 & 0 & 1\end{pmatrix}$$ is a valid stationary distribution for $$P$$.
• Thanks for the detailed answer. I thought that $stationary \ \pi \Longleftrightarrow \sum_i = 1, \pi P = \pi$, but you added the condition $\pi >0$. In this case, my calculated vector cannot be rated as "stationary", which resolves the question... Am I right? Jan 25 at 19:47