# No unique stationary distribution?

I found the following Markov chain on wikipedia. $$P = \begin{bmatrix} 0.9 & 0.075 & 0.025\\ 0.15 & 0.8 & 0.05\\ 0.25 & 0.25 & 0.5 \end{bmatrix}$$

I tried to find the stationary distribution of the chain by trying to solve $$\pi P=\pi$$ but I cannot get a unique solution. How is this possible? Clearly, a unique solution has to exist! Right? Any help is appreciated.

Solving $\pi P = \pi$ will not result in a unique solution. Essentially, you are solving $\pi(P-I)= \vec{0}$ where $P-I$ should have at least one zero eigenvalue (in your case, it does). So you get an uncountable infinity of solutions, since if $\vec{x}$ is a solution, $k\vec{x}$ is also a solution for any $k \in \mathbb{R}$.

The way you get a solution to be unique is to consider only probability measures, i.e. by requiring that all components of $\pi$ sum to 1.

Multiplying out the following $$[\pi_1,\pi_2,\pi_3]P=[\pi_1,\pi_2,\pi_3]$$ gives three equations to solve. $$0.9\pi_1+0.15\pi_2+0.25\pi_3=\pi_1\\0.075\pi_1+0.8\pi_2+0.25\pi_3=\pi_2\\0.025\pi_1+0.05\pi_2+0.5\pi_3=\pi_3$$ Solving these, I get $$\pi_1=2\pi_2,\pi_2=5\pi_3$$Normalisation$$\implies \pi=\left[\frac58,\frac5{16},\frac1{16}\right]$$

Iterate the Markov chain:

$$\pi_{t+1} = \pi_t P = \pi_{t-1} P^2 = \pi_0 P^t$$

The stationary distribution will be

$$\pi_{\infty} = \pi_0 P^{\infty}$$

where

$$P^{\infty} = \lim_{n\to \infty} P^n = \lim U^\top D^n U = U^\top (\lim D^n) U$$

(where $U^\top D^n U$ is the eigenvectors/values decomposition of $P$). Due to the Perron-Frobenius theorem a Markovian matrix has 1 as its largest eigenvalue. Therefore $\lim D^n$ exists and is finite.