random walk on Z towards the origin

Consider a random walk on $$\mathbb{Z}$$ with rate $$a>0$$ (begin no origin). The r.w. jumps one step towards the origin with probability $$p$$ or one step away from the origin with probability $$1 −p$$. If the walk is at the origin then it jumps left or right with probability $$1/2$$.

Question. Is true that if $$p > 1/2$$, $$P(\{\omega: \exists N(\omega)\in [1,\infty), |S_t|\leq N(\omega) \forall t>0\})=1$$?

I think. I thought about making a coupling

• There obviously is a chance of $(1-p)^{N+1} \gt 0$ that the walk will exit the interval in the first $N+1$ steps and it's equally obvious the chance of exiting can only get larger as the number of steps increases. This would seem to answer your question in the negative, depending on what you mean by "a.s. there exists $N$" and by "all times." Could you explain your meaning?
– whuber
May 15 '19 at 16:14
• a.s. there exists $N$ means that $P(\{\omega\in\Omega :\limsup_{n\rightarrow\infty}S_n<\infty \})=1$ and $P(\{\omega\in\Omega :\liminf_{n\rightarrow\infty}S_n>-\infty \})=1$. May 16 '19 at 0:18
• There are paths $\omega$ such that there is $N_p(\omega)$ one such $|S_n|<N_p(\omega)$. All times means that such a trajectory the random walk will be confined to $[-N,N]$ for $t\geq0$. May 16 '19 at 0:51
• I'm not sure that the proof you are providing actually addresses the question. You seem to be proving that on average the sum of steps approaches 0. At best it seems like this shows that the average grows sub-linearly. May 16 '19 at 2:22
• The statements you quote make no sense. In the first one, for instance, you assert the existence of an "$N$" according to a predicate in which the symbol "$N$" never even appears!
– whuber
May 16 '19 at 12:21

I'm not sure that this is true for $$p<1$$.
Give me an $$N$$. I suggest if you wait $$N+1$$ steps then there is probability $$(1-p)^{N+1} > 0$$ that the random walker has exceeded the $$N$$. It is therefore not a.s. perhaps you meant a.a.s. or I'm not understanding a.s. as almost surely?