# Markov Chain Stationary distribution

I constructed a 4*4 state transition matrix from a discrete-time Markov Chain Model as follows:

A=[p0 p0 p0 p0; p*p0+(1-p) p0 p0 p0; p0 p*p0+(1-p) p0 p0; p0 p0 p*p0+(1-p) p0];

I want to find the stationary distribution probability of this matrix. How do I do that??

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What have you tried? –  draks ... Mar 14 '13 at 16:07
Is A really a transition matrix? –  Did Mar 14 '13 at 16:28
Answer by the OP to my previous comment: Yes."A" is a transition matrix derived from a discrete time markov chain model. Unfortunately, A is not a transition matrix (except if p0=1/4 and p=1). As you are aware, any line (or any column, depending on your convention) should sum to 1. They do not. –  Did Mar 14 '13 at 16:57
Actually p0=1/8. –  renata Mar 14 '13 at 18:34
renata: The. Matrix. A. Is. Not. A. Transition. Matrix. –  Did Mar 17 '13 at 9:42

## 1 Answer

You need to find the limit as you multiply $A$ by itself $n$-times and $n \rightarrow \infty,$ ie. $\lim_{n \rightarrow \infty} A^n$

I would suggest you first take $p0$ as a common factor (it is a scalar so you can just divide everything by it and place it in the front) and the resulting matrix seems to have a rather simple form, it has $1$s on the first row and the rest is a symmetrical. Try to find the $n$th power of this matrix.

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You need to find the limit as you multiply A by itself n-times and... One almost never proceeds like this. –  Did Mar 14 '13 at 16:29
@Did Well, you need a matrix $B$ such that $B A = B$. Been a while since I did Markov's chains. –  Valtteri Mar 14 '13 at 16:34
Then why send the OP, probably a newcomer to the field, to this dead-end? –  Did Mar 14 '13 at 16:36
@Did Well, because that's what I would have done first...was probably a bad idea. –  Valtteri Mar 14 '13 at 16:38
@Did Yes."A" is a transition matrix derived from a discrete time markov chain model. –  renata Mar 14 '13 at 16:52