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Math Help - Equilibrium probabilites for a three-state markov process

  1. #1
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    Equilibrium probabilites for a three-state markov process

    I can find the equilibrium probabilities for a 2 x 2 matrix:

    (\pi_0, \pi_1) = (\pi_0 , 1 - \pi_0 )\[ \left( \begin{array}{cc}<br />
0.2 & 0.8 \\<br />
0.4 & 0.6 \end{array} \right)\]

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

    \pi_0 = 0.4/1.2=1/3 and  \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:

    \[ \left( \begin{array}{ccc}<br />
0 & 1 & 0\\<br />
0 & 0.4 & 0.6\\<br />
0.4 & 0 & 0.6 \end{array} \right)\]
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  2. #2
    Moo
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    Hello,

    The same as for 2x2 :

    (\pi_0,\pi_1,1-\pi_1-\pi_0)=(\pi_0,\pi_1,1-\pi_1-\pi_0)\[ \left( \begin{array}{ccc}<br />
0 & 1 & 0\\<br />
0 & 0.4 & 0.6\\<br />
0.4 & 0 & 0.6 \end{array} \right)\]

    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.
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