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Consider any Markov chain on a state space with exactly $r$ states. We want to find the largest $N>0$ such that there exists states $i,j$ for every $n < N$ we have $p_{ij}^{(N)} > 0$ and $p_{ij}^{(n)} = 0$.

My Thoughts:

Therefore $N$ is a recurrent state, and $n$ is transient state. The period of states $i,j$ here would need to more than N. I don't see how knowing we have exactly $r$ states helps us identify the recurrent states. I think we want to find a transient state such that all the states before it are recurrent. Wouldn't this be the first transient state of the Markov chain? Since the Markov chain is arbitrary, I am not sure how to find $N$. I looked through the different theorems in the textbook, but not sure which one I can apply here.

user77404
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1 Answers1

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It is a theorem of Wielandt that the number you are looking for is $(r-1)^2+1$. A way to achieve this is to have $p_{12},p_{23},p_{34},\dots,p_{k-1,k},p_{k,1},p_{k,2}$ positive and all other $p_{ij}$ zero. A reference is H Wielandt, Unzerlegbare, nicht negative Matrizen, Math Z. 52 (1950) 642-648.

Perhaps a better reference is Hans Schneider, Wielandt’s proof of the exponent inequality for primitive nonnegative matrices, Linear Algebra and its Applications 353 (2002) 5–10, available here.

EDIT: I knew this sounded familiar. I won a 500-point bounty at If $P$ is a regular transition probability matrix then $P^{n^2}$ has no zero element for (an extended version) of what I've written here. A question with a bounty can't be closed as a duplicate, I think.

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Gerry Myerson
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