"Find a steady state of the Markov process given by the following transition matrix:

(.7 .3 0 )

(.2 .4 .8)

(.1 .3 .2)

Does anyone know how to go about doing this?

Thanks,

3LD

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- Jan 6th 2008, 10:33 PM3leggeddogMarkov Chains - Steady State
"Find a steady state of the Markov process given by the following transition matrix:

(.7 .3 0 )

(.2 .4 .8)

(.1 .3 .2)

Does anyone know how to go about doing this?

Thanks,

3LD - Jan 7th 2008, 12:22 AMCaptainBlack
The steady state is an eigen vector of the transition matrix coresponding to

the eigen value 1. That is you want a solution of:

$\displaystyle \bold{A}\bold{x}=\bold{x}$

which is a non zero solution of:

$\displaystyle [\bold{A}-\bold{I}]\bold{x}=\bold{0}$

subject to the constraints that $\displaystyle \bold{x}_i \ge 0$ and $\displaystyle \sum \bold{x}_i =1$

Alternativly you could compute:

$\displaystyle \bold{x}_n=\bold{A}^n \bold{e}_1$

for larger and larger $\displaystyle n$ utill you get convergence.

Or you could just show that:

$\displaystyle \bold{x}=\left[ \begin{array}{c}0.4\\0.4\\0.2 \end{array} \right]$

is a steady state solution.

RonL