# Thread: Markov Chains - Steady State

1. ## Markov 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

2. 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:

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

is a steady state solution.

RonL