Abstract
We consider a symmetric simple random walk $\{W_i\}$ on $\mathbb{Z}^d, d=1,2$, in which the walker may choose to stand still for a limited time. The time horizon is $n$, the maximum consecutive time steps which can be spent standing still is $m_n$ and the goal is to maximize $\mathbb{P}(W_n=0)$. We show that for $d=1$, if $m_n \gg (\log n)^{2+\gamma}$ for some $\gamma>0$, there is a strategy for each $n$ yielding $\mathbb{P}(W_n = 0) \to 1$. For $d=2$, if $m_n \gg n^\epsilon$ for some $\epsilon>0$ then there are strategies yielding $\liminf_n \mathbb{P}(W_n=0)>0$.
Highlights
We consider a process {Wi, 0 ≤ i ≤ n} on Zd (d = 1 or 2) in which W0 = 0 and each step Wi −Wi−1 either is 0 or is a step ±ei of symmetric simple random walk (SSRW), with ei being the ith unit coordinate vector
We show that for d = 1, if mn2+γ for some γ > 0, there is a strategy for each n yielding P (Wn = 0) → 1
The choice between standing still and SSRW step is determined by a strategy
Summary
The choice between standing still and SSRW step is determined by a strategy. Tn = {(i, wi), 0 ≤ i ≤ j} : 0 ≤ j ≤ n − 1, |wi − wi−1| ≤ 1 for all i of all space-time trajectories of length less than n; here the values 0 and 1 for δn correspond to standing still and taking a SSRW step, respectively.
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