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November 2nd, 2014, 02:16 PM  #1 
Newbie Joined: Nov 2014 From: NUM Posts: 1 Thanks: 0  Markov chain, decomposition of communicating class
Let $(X_n)_{n\in\mathbb{N}_0}$ be a Markov chain with discrete state space $E$ and transition matrix $P$. Let $C\subseteq E$ be a communicating class. Prove or disprove the following statement. $C$ is periodic with period $p>1$ if $C$ can be decomposed into a disjoint union of sets $C_0\cup\ldots\cup C_{p1}$ in such a way that if $i\in C_n$ and $j\in C_m$ with $m\neq n+1\text{ mod }p$, then $p_{ij}=0$. Distinguish between the cases when (a) $E$ is finite, (b) $E$ is countably infinite.  I tried to prove that as good as I could: Let $E$ be finite. Then $1\leqslant p<\infty$. Because there cannot be infinite many disjoint sets and only finite states. If $p=1$ then there is no such decomposition at all. So this case can be neglected. If $p>1$ then for any $i\in C$ $i$ is in exactly one $C_k$. Because of the decomposition of $C$ it then needs $\ell\cdot p, \ell\geq 1$ steps to return from $i$ to $i$, so here is $R(i)=\left\{n > 0: p_{ii}^{(n)}>0\right\}=\left\{kp: k\geq 1\right\}$ because of the decomposition of $C$. Thus $p(i)=gcd(R(i))=p>1$. So here the statement holds.  Now let $E$ be countably infinite. I think here the problem is that then it is possible that $p$ can be very large, i.e. "$p\to\infty$". For example one could look at the natural numbers and set $C_k:=\left\{k\right\}, k\in\mathbb{N}$, and assume that for the state $i$ in $\lim_{k\to\infty}C_k$ it is $i\rightarrow 1$. Then there is a decomposition as stated in the task, but "$p=\infty$". I am not sure if this then fullfills what is meant witht $p(C)>1$, what I would rather understand as $p(C)<1<\infty$. Maybe anybody can tell me if parts of this proof are right. 

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chain, class, communicating, decomposition, markov, periodicity 
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