# Markov Chains - Steady State

• Jan 6th 2008, 10:33 PM
3leggeddog
"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 AM
CaptainBlack
Quote:

Originally Posted by 3leggeddog
"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

The steady state is an eigen vector of the transition matrix coresponding to
the eigen value 1. That is you want a solution of:

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

which is a non zero solution of:

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

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

Alternativly you could compute:

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

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

Or you could just show that:

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