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Let $\xi$ be a random variable with symmetric and non-lattice distribution. In other words, there dose not exists $\delta > 0$ such that $\mathrm P\{\xi \in \delta \mathbb Z\} = 1$.

Define a random walk $S_n = \sum_{i=1}^n \xi_i$ where $\xi_i \sim \xi$ are independent.

If we think $S_n$ as a Markov chain, is it always irreducible with the above assumptions? And how can we prove this? (If $S_n$ has uncountable states, I just want to know if there is always a way for $S_n$ to reach from one possible value to another possible value.)

NonalcoholicBeer
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    Is it a walk on the integers or the reals? If it is the reals then you need to study the density of $n \delta_1 + m \delta_2$ for integers $n,m $ with irrational $\delta_1/\delta_2$. I think it holds by some ergodic type theorem for translates by an irrational number mod 1. – Ian Jul 02 '17 at 11:37
  • @Ian I assume that $\xi \in \mathbb R$. – NonalcoholicBeer Jul 02 '17 at 11:40
  • If $\xi$ does not have a discrete probability distribution then there would be uncountably many states, so ${S_n}$ would not be a Markov chain. – Math1000 Jul 02 '17 at 12:32
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    @Math1000 The restrictions on the term "Markov chain" have varied dramatically over the years. At the very least it is a discrete time Markov process with state space $\mathbb{R}$, so we know what is meant. – Ian Jul 02 '17 at 13:25
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    @ablmf The definition of irreducible is different for uncountable state space chains. It is usually replaced with the notion of $\psi$-irreducibility where $\psi$ is some reference measure. This says that for every $x \in S$ and measurable $A \subset S$ with $\psi(A)>0$, there exists $n$ such that $P_n(x,A)>0$. Frequently the state space is a topological space and $\psi$ assigns positive measure to each nonempty open set. In this context $\psi$ can be taken to be the Lebesgue measure. Cf. the book Markov Chains and Stochastic Stability by Meyn and Tweedie. – Ian Jul 02 '17 at 13:28

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