# Theoretical basis behind calculation of steady state probability distribution of 2-state Markov chain from its transition matrix

I am studying stochastic processes and have stumbled on a result that is puzzling me. I have searched elsewhere for an answer without luck so hoping some proper mathematicians here can explain the result for me.

Given a two-state Markov process with probability transition matrix $$\begin{array}{c|c} &\begin{matrix}0&1\end{matrix}\\ \hline \begin{matrix}0\\ 1\end{matrix} &\pmatrix{a&b\\ c&d} \end{array}$$

I have found that the simplest way to calculate its steady-state probability distribution is :

state 0: $c \over {b + c}$

state 1: $b \over {b + c}$

This result holds for all examples I have tried, but I have been unable to explain it from theory, so cannot prove it. My questions are:

1. what is the theoretical explanation for this result?
2. does it extend to any $n\times n$ transition matrix?

This is specific to two-state chains. Note that $a=1-b$ and $d=1-c$ for the matrix to be a probability transition matrix hence $(b,c)$ determines entirely the probability transition matrix and the distribution $(\pi_0,\pi_1)$ is stationary if and only if $$\pi_0=a\pi_0+c\pi_1,\qquad\pi_1=b\pi_0+d\pi_1.$$ This system is equivalent to the single equation $$b\pi_0=c\pi_1.$$ The normalization $\pi_0+\pi_1=1$ yields the values in the question.