Let $\left\{ X_{n}\right\} _{n\geq1}$ be a sequence of IID random-variables s.t $\mathbb{E}\left[X_{i}\right]=0$ and $\left|X_{i}\right|\leq K$ . Let $S_{n}=\sum_{i=1}^{n}X_{i}$ , I want to show $S_{n}$ visits the interval $\left[-K,K\right]$ an infinite amount of times w.p $1$. Obviously it would suffice to show that $S_{n}$ visits $0$ infinitely w.p $1$. I would really love a hint about how to approach this question. This is kind of similar of a random-walk but it's not defined just on the integers and we don't know the transition probabilities so I'm a bit lost.

Proof Attempt

If $S_{n}\equiv0$ then we are obviously done so assume that is not the case. We now start by showing $S_{n}$ must change signs with probability one. Suppose by contradiction that it does not, then wlog $\mathbb{P}\left(S_{n}>0\ \text{for all }n\right)=1$ and so $\mathbb{P}\left(\frac{1}{n}S_{n}>0\ \text{for all }n\right)$ and thus $\mathbb{P}\left(\lim\frac{1}{n}S_{n}>0\right)=1$.But by the SLLN we know that $\mathbb{P}\left(\lim\frac{1}{n}S_{n}=\mathbb{E}\left[X_{1}\right]=0\right)=1$ in contradiction. So with probability $1$ there exists $1<N\in\mathbb{N}$ such that $S_{N}<0$ and $S_{N+1}>0$. Note that If $S_{N}<-K$ then $S_{N+1}=S_{N}+X_{N+1}\leq S_{N}+K<0$ in contradiction. Thus $S_{N}\in\left[-K,K\right]$ and also $S_{N+1}\in\left[-K,K\right]$. This shows that $S_{n}$ visits $\left[-K,K\right]$ with probability $1$. Let $T=\min\left\{ n>1\ |\ S_{n}\in\left[-K,K\right]\right\}$ , we have shown that $\mathbb{P}\left(T<\infty\right)=1$ and so by the strong Markov property the sequence $X_{T+1},X_{T+2},...$ is IID with the same joint distribution as $X_{1},X_{2},...$ and so $S_{n}^{'}=\sum_{k=1}^{n}S_{T+k}$ has the same distribution as $S_{n}$ , thus $S_{n}^{'}$ visits $\left[-K,K\right]$ with probability $1$. This can be applied repeatedly after each visit in order to conclude that $S_{n}$ visits $\left[-K,K\right]$ infinitely many times with probability $1$.

I would appreciate if someone could check this for correctness.

  • $\begingroup$ Your approach doesn't work. Note that a sequence of positive numbers can have $0$ as its limit. Also, you can't assume wlog that $S_n > 0$ for all $n$. $\endgroup$ – Dominik Jan 31 '16 at 22:40
  • $\begingroup$ @Dominik If I assume by contradiction that $S_{n}$ doesn't change sign and it's not fixed at $0$ then it's either always positive or always negative, I just chose one of those cases, the other case would be proven in the same way. You are right though about the fact that a sequence of strictly positive numbers can have limit equaling zero. If I showed that $S_{n}\geq c>0$ instead and that would have sorted it out. What about the logic that followed later, if I do show that $S_{n}$ changes sign is the rest correct? $\endgroup$ – Serpahimz Jan 31 '16 at 23:07
  • $\begingroup$ I don't think it's correct. You can't just switch to a new sequence $S_n'$, you need to prove that your original sequence $S_n$ hits $[-K, K]$ infinitely often. The problem in your argumentation is that if $S_{n_0}'$ hits the interval, it's not necessary that $S_{n_0}$ hits it aswell. $\endgroup$ – Dominik Jan 31 '16 at 23:25
  • $\begingroup$ $S_{n}^{'}$ is just a tail of $S_{n}$ that starts after the first hit so if $S_{n}^{'}$ hits the interval again so does the original $S_{n}$, this is why it seems logical to me to just restart and reapply the same argument after each hit. Where is the flaw in the logic? $\endgroup$ – Serpahimz Jan 31 '16 at 23:29
  • $\begingroup$ If both $S_n$ and $S_n'$ have the value $K$ at the first hit, then $S_n$ is not in the interval at the time of the first hit of $S_n'$. $\endgroup$ – Dominik Feb 1 '16 at 7:29

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