# Finding the limiting distribution of a $3 \times 3$ Markov chain

This is a question from a book.

Find $\lim_{n\rightarrow \theta}P^n$ where $$P=\begin{pmatrix}0 & 1 & 0\\ \frac{1}{6} & \frac{1}{2} & \frac{1}{3}\\ 0 & \frac{2}{3} & \frac{1}{3} \end{pmatrix}$$

I assumed that there was a typo and in fact what is being asked was $\lim_{n\rightarrow \infty}P^n$. I then proceeded to solve

$$\begin{pmatrix}\pi_{0} & \pi_{1} & \pi_{2}\end{pmatrix}\begin{pmatrix}0 & 1 & 0\\ \frac{1}{6} & \frac{1}{2} & \frac{1}{3}\\ 0 & \frac{2}{3} & \frac{1}{3} \end{pmatrix}=\begin{pmatrix}\pi_{0} & \pi_{1} & \pi_{2}\end{pmatrix}$$

which converts to the system of equations

\begin{align} 0\pi_{0} + \frac{1}{6}\pi_{1} + 0\pi_{2} &= \pi_{0} \\ 1\pi_{0} + \frac{1}{2}\pi_{1} + \frac{2}{3}\pi_{2} &= \pi_{1} \\ 0\pi_{0} + \frac{2}{3}\pi_{1} + \frac{1}{3}\pi_{2} &= \pi_{2} \\ \pi_0 + \pi_1 + \pi_2 &= 1 \end{align}

which doesn't seem to be consistent, since ignoring the final line gives $\pi_0=\pi_1=\pi_2=0$ when I row-reduce. However, the textbook does give a solution of $\pi^T=(0.4, 0.45, 0.15)$. What am I doing wrong?

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The equation $\pi P = \pi$ will always have as solution $\pi =0$. The point is it should have more solutions. –  leonbloy Mar 30 '13 at 15:33

1. Your third equation has a wrong coefficient. It should be $0\pi_{0}+\color{red}{\frac{1}{3}}\pi_{1}+\frac{1}{3}\pi_{2}=\pi_{2}$.
2. The textbook answer doesn't seem to be correct. The solution should be $(\pi_0,\,\pi_1,\,\pi_2)=(0.1,\, 0.6,\, 0.3)$.