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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?

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  • $\begingroup$ Yes, this is correct. $\endgroup$ – Math1000 May 6 at 22:39
  • $\begingroup$ @Math1000, Are the terms Limiting distribution and Stationary distribution properly perceived? $\endgroup$ – user366312 May 6 at 22:40
  • $\begingroup$ $P$ 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. $\endgroup$ – amd May 7 at 3:08

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