I am trying to show the following markov chain is aperiodic (it could very well not be)
$S = \mathbb{Z}$ and $h : S \rightarrow (0,1)$ with $h(i)>0 \space \forall i \in S$ and $\sum_{i \in S}h(i) = 1$
the probabilities on $S$ given by $p_{ij} = (1/4)min(1,h(j)/h(i))$ if $|i-j| \leq 2$ and $i \neq j$, and $p_{ii} = 1-p_{i,i-2}-p_{i,i-1}-p_{i,i+1}-p_{i,i+2}$, and $p_{ij} = 0$ otherwise.
($p_{ij}$ is defined as the probability to go from state $i$ to state $j$)
Now I have some reason to think this chain is aperiodic, this is my work so far:
Assume $h(i)=h(j) \space \forall i,j \in S$, given $\sum_{i\in S} h(i) =1$
$h(i) = h(j) = \lim_{n \to \infty} (1/n)\sum_{i \in S}h(i) = 0$ with $n$ being the number of elements composing S.
Therefore $h(i) = h(j) = 0\space \forall i,j \in S$ implying $\sum_{i \in S}h(i) = 0$ a contradiction, showing $h(i) \neq h(j)$ for some $i,j$.
Now what this shows is that for some states in the chain, $p_{ii} \neq 0$ meaning that state is aperiodic. I hope however the whole chain is. Fow now I have the impression that I would have some 100 consecutive states with probability 1/10000 each, making my chain periodic, or is that possibility somehow not possible (pardon my wording)?
Best Answer
This answer is based on these lecture notes.
Let $P$ be the transition matrix, with $(P)_{i,j} = p_{i,j}$. A chain is irreducible if for all $i$,$j$, there is some $t$ such that $(P^t)_{i,j} > 0$. Since $P$ is tridiagonal and all entries are positive, it is irreducible.
Now from the notes, we have the proposition (Excercise 2.9): "Suppose $P$ is irreducible and contains at least one self-loop (i.e., $(P)_{i,i} > 0$ for some $i$). Then $P$ is aperiodic."
Clearly $P$ has self-loops, since $p_{i,j} \in (0,\frac{1}{4})$ if $i\neq j$ and $\lvert i-j\rvert \le 2$, and from the definition: $$p_{i,i} = 1 - p_{i,i-1} - p_{i, i-2} - p_{i,i+1} - p_{i,i+2} > 1 - \frac{1}{4} - \frac{1}{4} - \frac{1}{4} - \frac{1}{4} = 0$$
Putting this together with the proposition gives your aperiodicity result.