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In an excercise I'm given the following matrix of a markov chain $\begin{pmatrix} 0&1&0 \\ 1/2 &1/2 & 0 \\ 0 & 1/2 &1/2 \end{pmatrix}$. It has the stationary distribution $\pi=(1/3,2/3,0)$ (which I think is unique).

In the answer to the problem it says that the limiting distribution aproaches the stationary independently of the initial distribution.

The only theorem that I know that talks about this is the "ergodicity theorem"(if the chain is ergodic then $p(n)\rightarrow \pi$ indpendently of the initial distribution). However it doesn't seem to me that this chain is ergodic; since no state communicates with the last state it isn't irreducible and therefore not ergodic (I think?).

What is the argument for the limiting distribution approaching the stationary independently of the initial distribution?

user202542
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    Did you happen to find the argument for this? I have the same problem.. – Biggiez Oct 27 '18 at 12:51
  • The final state is transient, while the other two are aperiodic positive-recurrent and communicate so this finite-state Markov chain is ergodic and you can apply the theorem – Henry Jul 10 '21 at 17:23

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I think, there are two things that you should check. Firstly, you should be looking if a Markov chain is periodic or not. Secondly, check if the markov chain has only one minimal subset. If it is not periodic and it has only one minimal subset, which is the case for this problem, you can have a unique steady state, regardless of the initial distribution.

Med
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  • Thank you! What do you mean by minimal subset? Do youknow where I can find a proof of this? – user202542 Oct 23 '16 at 15:15
  • A minimal subset, is a closed subset of states of a Markov chain that contains no proper closed subset. Unfortunately, I do not know reference for the proof. – Med Oct 23 '16 at 16:46