# Finite hitting time implies hits at any finite time?

I was wondering about the following problem:

Assume we have a state space $S:=\mathbb{Z}$ and a Markov chain, such that we can go from any state $x$ to some state $y$ with positive probabilities, i.e. $p_t(x,y)>0$ for any $t >0$ and $x,y \in S.$

Let $T_0^x$ be the hitting time to go from state $x$ to $0$.

If we know that $P(\lim_{x \rightarrow -\infty} T_0^x<\infty)=1.$ So we can almost surely go to state $0$ from -infinity. Does this imply that $\liminf_{x \rightarrow -\infty} p_t(x,0)>0,$ for all $t>0$ i.e. does this imply that we go to state $0$ in every finite time step?

The thing is that we a priori only know that we can go to $0$ in some finite time, but not necessarily in any finite time (with some positive probability.)

• How do you define $T_0^x$ exactly? Usually the starting state is not part of the random variable, but part of the probability measure. So you could talk about the hitting time of $0$, $T_0 = \inf\{t : X_t = 0\}$, and then ask about the probability of hitting 0 when starting at x, $P_x(T_0 < \infty)$. But then it does not make sense to have the $x$ limit inside the measure. Dec 15 '15 at 21:22
• Indeed, for pretty much this reason, "starting at $-\infty$" seems to be ill defined. Dec 15 '15 at 21:23
• Are you thinking of discrete time or continuous time here? Dec 15 '15 at 21:31
• Then isn't continuous-time simple random walk a counterexample? It's recurrent, so $P_x(T_0 < \infty) = 1$ for every $x$, but it's easy to verify that for any fixed $t$, $p_t(x,0) \to 0$ as $x \to \pm \infty$. Dec 15 '15 at 21:33

It's not clear what condition you are imposing, but I think the following example indicates that the answer is no, there are positive times so that you don't have to get to $0$ with uniformly bounded probability:
Arrange the states in a tree with root $0$ with disjoint paths of length $n$ connected to the root for each $n\in \mathbb{N}$. On a path of length $n$, let the rate of movement toward the root be $n$. If $0 \lt t\lt 1$ and we start at the leaf of a path of length $n$, the distribution at time $t$ will be concentrated about $t$ of the way along the path. So, the limit inferior of the probability of reaching $0$ by time $t$ is $0$.