0
$\begingroup$

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.

  • 0
    Since I gave some incomplete answers to your previous related question, I'll try this in a comment first :-). If your definition of recurrence is that almost surely $S_m=0$ infinitely often (if not, what's your definition of recurrence?), then the walk has infinitely many tries to get from $0$ to $T$, and since it's strongly aperiodic each try has a non-zero probability of succeeding; it follows that it reaches $T$ with probability $1$?2012-10-28
  • 0
    Not sure you meant to write *there exists a $m$ s.t. $P(S_m =T)$*. Rather it seems that you are asked about $P(\exists m,S_m=T)$.2012-10-28
  • 0
    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

1 Answers 1

2

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, $\mathbb P(\exists m, S_m=T)=1$.

  • 0
    The first version was sloppy, please see the revised one.2012-10-29
  • 0
    Thank you for your answer! Would you mind elaborating on the $\mathbb {Z}=U-U$ part? Is it a setminus? And why does the statement hold?2012-10-29
  • 0
    A setminus? Heaven's, no! Say, which definitions of the aperiodicity and the strong aperiodicity of a random walk do you use?2012-10-29
  • 0
    Oh yea, sorry. Random walks was a bit new to me, now it makes sense. Thanks for your answer!2012-11-03