# Equilibrium probabilites for a three-state markov process

• Apr 12th 2011, 12:30 PM
MathSucker
Equilibrium probabilites for a three-state markov process
I can find the equilibrium probabilities for a 2 x 2 matrix:

$\displaystyle (\pi_0, \pi_1) = (\pi_0 , 1 - \pi_0 )$\left( \begin{array}{cc} 0.2 & 0.8 \\ 0.4 & 0.6 \end{array} \right)$$

$\displaystyle \pi_0(0.2)+(1- \pi_0)0.4= \pi_0$

$\displaystyle \pi_0 = 0.4/1.2=1/3$ and $\displaystyle \pi_1 = 0.8/1.2=2/3$

But how do I find the probabilities for a 3 x 3 matrix?

Can somebody start me off on:

$\displaystyle $\left( \begin{array}{ccc} 0 & 1 & 0\\ 0 & 0.4 & 0.6\\ 0.4 & 0 & 0.6 \end{array} \right)$$
• Apr 12th 2011, 12:37 PM
Moo
Hello,

The same as for 2x2 :

$\displaystyle (\pi_0,\pi_1,1-\pi_1-\pi_0)=(\pi_0,\pi_1,1-\pi_1-\pi_0)$\left( \begin{array}{ccc} 0 & 1 & 0\\ 0 & 0.4 & 0.6\\ 0.4 & 0 & 0.6 \end{array} \right)$$

As a note, this is, by definition, the stationary/invariant distribution. In order to say that it's the equilibrium distribution, you need to show that all the states are positive recurrent and that the chain is irreducible and aperiodic.