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Assume $(X_n)_{n\geq1} \subseteq \mathbb {Z}$ and $(Y_n)_{n\geq 1} \subseteq \mathbb {Z}$ to be iid, $X_i \sim Y_i$ and such that $S_n=\sum_{i=1}^n(X_i-Y_i)$ is a strongly aperiodic, recurrent random walk. Let $T \subseteq \mathbb{N}$ be a discrete random variable.

With this information can we then say anything about $P(\exists m, \;S_m =T)$?

Thank you.

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    did: I've corrected it, I meant what you said. Joriki: That is what I think too, I just want to be sure.2012-10-28

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Let $U=\{x\in\mathbb Z\mid\exists n\geqslant0,\mathbb P(S_n=x)\ne0\}$ denote the union of the supports of the random variables $S_n$. Let $V=\{S_n\mid n\geqslant0\}$ hence the event $A=[\exists m, S_m=T]$ is $A=[T\in V]$. Note that $U$ is a deterministic subset of $\mathbb Z$ and $V$ is a random subset of $\mathbb Z$.

Since $(S_n)$ is irreducible on $U$ and recurrent, $V=U$ almost surely. In particular, $A=[T\in U]$, up to a negligible event. If $(S_n)$ is furthermore aperiodic, by hypothesis $\mathbb Z=U-U$. The increments of $(S_n)$ are symmetrically distributed, hence in our case, $U-U=U$.

Finally, $A=[T\in \mathbb Z]=\Omega$, up to a negligible event, that is, $\mathbb P(\exists m, S_m=T)=1$.

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    Oh yea, sorry. Random walks was a bit new to me, now it makes sense. Thanks for your answer!2012-11-03