Consider a Markov chain $(X_n)_n$ on $S=\{1, 2\}$ with initial distribution $α$ and the transition matrix
$P = \begin{bmatrix} 2/3 & 1/3 \\ 2/3 & 1/3 \\ \end{bmatrix}$
- Limiting distribution = ?
- Stationary distribution = ?
My Solution:
$\underline {\text{Limiting Distribution}}$
$P^2 = \begin{bmatrix} 2/3 & 1/3 \\ 2/3 & 1/3 \\ \end{bmatrix} \begin{bmatrix} 2/3 & 1/3 \\ 2/3 & 1/3 \\ \end{bmatrix} = \begin{bmatrix} 2/3 & 1/3 \\ 2/3 & 1/3 \\ \end{bmatrix}$
So, the limiting distribution of $P$ is $P$ itself.
$\underline {\text{Stationary Distribution}}$
Let the stationary distribution $\pi = \begin{bmatrix} p & 1-p \end{bmatrix}$.
So,
$\pi P = \pi $
$\Rightarrow \pi (P-1) = 0$
$\Rightarrow \begin{bmatrix} p & 1-p \end{bmatrix} \left(\begin{bmatrix} 2/3 & 1/3 \\ 2/3 & 1/3 \\ \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}\right) = 0$
$\Rightarrow \begin{bmatrix} p & 1-p \end{bmatrix} \begin{bmatrix} -1/3 & 1/3 \\ 2/3 & -2/3 \\ \end{bmatrix} = 0$
$\Rightarrow \begin{bmatrix} \frac{-p}{3}+\frac{2}{3}+\frac{-2p}{3} & \frac{p}{3} + \frac{-2}{3} + \frac{2p}{3} \end{bmatrix} = 0$
$\Rightarrow p = 2/3$
So,
$\pi = \begin{bmatrix} \frac{2}{3} & \frac{1}{3} \end{bmatrix}$
Are the terms Limiting distribution and Stationary distribution properly perceived in this solution?