A Divergent Random Walk on Stairs

TL;DR

This paper studies a state-dependent random walk on natural numbers influenced by a sequence _n, showing that for any probability less than 1, a sequence exists making the walk diverge with at least that probability, but the case for probability 1 remains open.

Contribution

The paper proves that for any <<1, there exists a sequence _n making the walk diverge with probability at least , advancing understanding of the walk's divergence behavior.

Findings

Constructed sequences _n ensuring divergence with probability <1

Extended previous results by achieving divergence probability arbitrarily close to 1

The case for divergence probability exactly 1 remains unresolved.

Abstract

We consider a state-dependent, time-dependent, discrete random walks $X_t^{\{a_n\}}$ defined on natural numbers $\mathbb{N}$ (bent to a "stair" in $\mathbb{N}^2$) where the random walk depends on input of a positive deterministic sequence $\{a_n\}$. This walk has the peculiar property that if we set $a_n$ to be $+\infty$ for all $n$, it converges to a stationary distribution $\pi(\cdot)$; but if $a_n$ is uniformly bounded (over all $n$) by any upper bound $a \in (0,\infty)$, this walk diverges to infinity with probability 1. It is thus interesting to consider the intermediate case where $a_n<\infty$ for all $n$ but $a_n$ eventually tends to $+\infty$. (Latuszynski et al., 2013) first defined this walk and conjectured that a particular choice of sequence $\{a_n\}$ exists such that (i) $a_n \to \infty$ and, (ii) $P(X_t^{\{a_n\}} \to \infty )=1$. They managed to construct a sequence $\{a_n\}$ that satisfies (i) and $P(X_t^{\{a_n\}}\to \infty)>0$, which is weaker than (ii). In this paper, we obtain a stronger result: for any $\sigma<1$, there exists a choice of $\{a_n\}$ so that $P(X_t\to \infty)\ge \sigma$. Our result does not apply when $\sigma=1$, the original conjecture remains open. We record our method here for technical interests.