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?
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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.

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  • $\begingroup$ Clear, concise and helpful answer, many thanks. $\endgroup$ – vince Oct 11 '13 at 15:59
  • $\begingroup$ You are welcome. $\endgroup$ – Did Oct 11 '13 at 16:50

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