# Clarification on the definition of a closed communicating class

If I have a transition matrix defined as $p_{ij}=1$ for $j=i+1$ and $p_{ij}=0$ otherwise, where the state-space is countably infinite, what would this be? It doesn't look like a communicating class because no 2 states communicate but then the "communication" relation partitions the state-space, hence the entire state-space must be one big communicating class here? Then the whole thing is closed? But that shouldn't be right...

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## 1 Answer

Correct me if you use the wrong definition of the communication, I base on the book 'Markov Chains and Stochastic Stability'. There $i,j$ communicate iff either $i=j$ or $p^m_{ij}>0$ and $p^n_{ji}>0$ for some $0<m,n<\infty$. In your case any state communicates only with itseld, so you have countably many communicating classes. I.e. in terms of relations, $[i] = i$ for all $i=1,2,3,...$

To be precise, in your question the statement

no 2 states communicate

is incorrect since any state communicate with itself by definition.

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Ah, thanks! :-) I didn't realize that. – doob Oct 12 '11 at 12:34
You are welcome – Ilya Oct 12 '11 at 13:11