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How would you go about showing that two states in the same communicating class of a Markov chain must have the same period? Any help would be greatly appreciated.

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Let $P^{(k)}_{ij}$ denote the $k$-step transition probability from state $i$ to $j$. Let $d(i)$ denote $i$'s period.

Suppose $i$ and $j$ communicating and $\alpha=P^{(k)}_{ij}>0$ and $\beta=P^{(l)}_{ji}>0$ for some $k$ and $l$. Then, $P^{(k+l)}_{ii}>\alpha\beta>0$, so $k+l$ is multiple of $d(i)$. Suppose now that $m\in\{n\mid P^{(n)}_{jj}>0\}$. Then, $P^{(k+m+l)}_{ii}>0$ (you go from $i$ to $j$, go off somewhere and return to $j$, and then return to $i$). Therefore $k+l+m$, and therefore also $m$, are multiples of $d(i)$, and thus $d(j)\ge d(i)$. Now exchanging the roles of $i$ and $j$ we get that $d(i)\ge d(j)$.

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  • $\begingroup$ This is very easy to understand and very helpful, thanks so much, I understand this concept far better now. $\endgroup$
    – Freeman
    Feb 22, 2012 at 20:08

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