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Suppose I have a transition matrix for a Markov chain, which has states {1,2,3}, such as:

\begin{bmatrix}0&1&0\\0&0&1\\0&0&1\end{bmatrix}

I wonder if this chain is reducible or not? Moreover, from my textbook, it says the communication class will partition the states of the Markov chain. However, in this case, only state 3 communicate to itself, while other states are transient. How does the communication class partition all states, since the only communication class is {3}?

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  • $\begingroup$ Could you consider marking given answers as accepted if you are content with the answer? $\endgroup$
    – Jan
    Sep 17, 2019 at 7:17

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This case depends somewhat on precisely how you define communicating class, but the most mathematically clean way to set up the definitions is to say that $i$ communicates with $j$ if there exist nonnegative integers $m=m(i,j),n=n(i,j)$ such that $P^m_{ij}>0$ and $P^n_{ji}>0$. Then taking $m=n=0$ tells you that each state communicates with itself.

If you use this definition, then the communicating classes of this chain are just the three singleton sets. If you instead require $m,n$ to be positive, then the only communicating class is $\{ 3 \}$ and the other states are simply not in any communicating class.

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  • $\begingroup$ Ok, get it! Thanks! $\endgroup$
    – LastK7
    Sep 16, 2019 at 0:19

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