Let $T = (X_n:n \in \mathbb{N})$ denote a homogeneous Markov chain with state space $E=\lbrace 1, 2, 3\rbrace$ and
$$\mathbb{P}(X_1=2\vert X_0=1) = \mathbb{P}(X_1=3\vert X_0=1)=\frac{1}{3}$$
as well as
$$\mathbb{P}(X_1=1\vert X_0=2) = \mathbb{P}(X_1=2\vert X_0=3)=1.$$
I suspect that the transition matrix is
$$T = \left( \begin{array}{ccc}
1/3 & 1/3 & 1/3 \\
1 & 0 & 0 \\
0 & 1 & 0 \end{array} \right).$$
Also, I understand that the stationary distribution is a limiting form of the Markov chain, and so from the definition of a stationary distribution we find it $\pi$, the stationary distribution, by calculating $\pi=\pi T$.
We calculate the stationary distribution by finding left eigenvectors for the eigenvalue $1$. So,
\begin{align*}
\frac{1}{3}\pi_1 +\frac{1}{3}\pi_2+\frac{1}{3}\pi_3 & =\pi_1,\\
\pi_1 & =\pi_2 \hspace{1.0cm}\textrm{and}\\
\pi_2 & =\pi_3.
\end{align*}
So $\pi_1=\pi_2=\pi_3$ and as $\pi_1+\pi_2+\pi_3=1$, then $$\pi=\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right).$$
Is this correct?
Also, positive recurrent Markov chain has a unique stationary distribution. How do I determine if this Markov chain is positive recurrent, and thus if the above stationary distribution is unique?
Best Answer
The chain is irreducible on a finite state space hence it is positive recurrent and has a unique stationary distribution $\pi$. Following the precise indications in the comments, one gets that $\pi$ is in the proportions 3:2:1, thus $\pi=(\frac12,\frac13,\frac16)$.