# Are the terms Limiting distribution and Stationary distribution properly perceived?

Consider a Markov chain $$(X_n)_n$$ on $$S=\{1, 2\}$$ with initial distribution $$α$$ and the transition matrix

$$P = \begin{bmatrix} 2/3 & 1/3 \\ 2/3 & 1/3 \\ \end{bmatrix}$$

1. Limiting distribution = ?
2. 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?

• Yes, this is correct. – Math1000 May 6 at 22:39
• @Math1000, Are the terms Limiting distribution and Stationary distribution properly perceived? – user366312 May 6 at 22:40
• $P$ is not a distribution, so it is certainly not a “limiting distribution.” Also, see stats.stackexchange.com/q/48262 for a discussion of the difference between limiting and stationary distributions. – amd May 7 at 3:08