On Covering Monotonic Paths with Simple Random Walk

TL;DR

This paper investigates the probability that a simple random walk covers a monotonic path in high dimensions, identifying the path that maximizes this probability and providing bounds on its decay rate.

Contribution

It establishes that the monotonic increasing path maximizes covering probability and derives exponential bounds for the decay rate in dimensions four and higher.

Findings

Monotonic increasing paths maximize covering probability.

Exponential upper bounds on decay rate for dimensions ≥ 4.

Covering probability decreases exponentially with path length.

Abstract

In this paper we study the probability that a $d$ dimensional simple random walk (or the first $L$ steps of it) covers each point in a nearest neighbor path connecting 0 and the boundary of an $L_1$ ball. We show that among all such paths, the one that maximizes the covering probability is the monotonic increasing one that stays within distance 1 from the diagonal. As a result, we can obtain an exponential upper bound on the decaying rate of covering probability of any such path when $d\ge 4$.

Equations

$$A\subseteq \mathrm{Trace}(X_0,X_1,\cdots ,X_L):=\{x\in {\mathbb{Z}}^d:\exists 0\le i\le L,X_i=x\}.$$

$${\phi }_l(x,y)=(l_0+y,x-l_0).$$

$$P\left(A_{0}\cup B_{0}\subseteq{\rm Trace}\big{(}\{X_{n}\}_{n=0}^{L}\big{)}\right)\geq P\left(A_{0}\cup\varphi_{l}(B_{0})\subseteq{\rm Trace}\big{(}\{X_{n}\}_{n=0}^{L}\big{)}\right).$$

$$\mathcal{P}=\big{(}P_{0},P_{1},\cdots,P_{K}\big{)}$$

$$\mathrm{Trace}(\mathcal{P})\subseteq \mathrm{Trace}(X_0,X_1,\cdots ,X_L).$$

$$P\big{(}{\rm Trace}(\mathcal{P})\in{\rm Trace}(X_{0},\cdots,X_{L})\big{)}\leq P\big{(}\overset{\nearrow}{\mathcal{P}}\in{\rm Trace}(X_{0},\cdots,X_{L})\big{)}$$

$$\overset{\nearrow}{\mathcal{P}}=\Big{(}{\rm arc}_{1}[0:d-1],{\rm arc}_{2}[0:d-1],\cdots,{\rm arc}_{[N/d]}[0:d-1],{\rm arc}_{[N/d]+1}[0:N-d[N/d]]\Big{)}$$

$${\mathrm{arc}}_1[0:d-1]=\left(0,e_1,e_1+e_2,\cdots ,\sum _{i=1}^{d-1}{e}_i\right)$$

$$P\left({\rm Trace}(\mathcal{P})\subseteq{\rm Trace}\big{(}\{X_{n}\}_{n=0}^{\infty}\big{)}\right)\leq P_{d}^{[N/d]}.$$

$$P\left({\rm Trace}(\mathcal{P})\subseteq{\rm Trace}\big{(}\{X_{n}\}_{n=0}^{\infty}\big{)}\right)\leq\left(\frac{C}{d}\right)^{N}.$$

$$P\left({\rm Trace}(\mathcal{P})\subseteq{\rm Trace}\big{(}\{X_{n}\}_{n=0}^{\infty}\big{)}\right)\leq\hat{P}_{d}^{N^{1+\epsilon}}.$$

$$P\big{(}{\rm Trace}(\mathcal{P})\subseteq{\rm Trace}(X_{0},\cdots,X_{L})\big{)}\geq P\big{(}\overset{\rightarrow}{\mathcal{P}}\subseteq{\rm Trace}(X_{0},\cdots,X_{L})\big{)}$$

$$\overset{\rightarrow}{\mathcal{P}}=\Big{(}(0,0,\cdots,0),(1,0,\cdots,0),\cdots,(N-1,0,\cdots,0)\Big{)}.$$

$$\vec{V}=V_1\otimes V_2\otimes \cdots \otimes V_m,V_k\subseteq \Omega ,$$

$$\vec{D}\cdot \vec{V}=\bigcup _{k=1}^m{\delta }_k{V}_k\subseteq -\Omega \cup \Omega .$$

$$-A^c\cup A\subseteq \vec{D}\cdot \vec{V},$$

$$\mathcal{C}\left(\vec{V},A\right)=\left\{\vec{D}:-A^c\cup A\subseteq \vec{D}\cdot \vec{V}\right\}$$

$$\left|\mathcal{C}\left(\vec{V},\Omega \right)\right|\ge \left|\mathcal{C}\left(\vec{V},A\right)\right|$$

$$\left|\mathcal{C}\left(\vec{V},\Omega \right)\right|=\left|\mathcal{C}\left(\vec{V},\text{O}\right)\right|.$$

$$\left|\mathcal{C}\left(\vec{V},\Omega \right)\right|=0=\left|\mathcal{C}\left(\vec{V},A\right)\right|$$

$$\left|\mathcal{C}\left(\vec{V},\Omega \right)\right|=1>0=\left|\mathcal{C}\left(\vec{V},A\right)\right|$$

$$\left|\mathcal{C}\left(\vec{V},\Omega \right)\right|=\left|\mathcal{C}\left(\vec{V},A\right)\right|=2^{n_e(\vec{V})}\left(2^{m-n_e(\vec{V})}-1\right),$$

$$\vec{V}[1:m_0]=V_1\otimes V_2\otimes \cdots \otimes V_{m_0}$$

$$\vec{D}[1:m_0]=({\delta }_1,\cdots ,{\delta }_{m_0}).$$

$$\mathcal{C}\left(\vec{V},A\right)$$

$$\mathcal{P}\left(\vec{V},A\right)=\left\{\vec{D}:-A^c\cup A⊈\vec{D}[1:m_0]\cdot \vec{V}[1:m_0],-A^c\cup A\subseteq \vec{D}\cdot \vec{V}\right\}.$$

$$\left|\left\{\vec{D}:-A^c\cup A\subseteq \vec{D}[1:m_0]\cdot \vec{V}[1:m_0]\right\}\right|=2\left|\mathcal{C}\left(\vec{V}[1:m_0],A\right)\right|,$$

$$\left|\mathcal{C}\left(\vec{V},A\right)\right|=2\left|\mathcal{C}\left(\vec{V}[1:m_0],A\right)\right|+\left|\mathcal{P}\left(\vec{V},A\right)\right|.$$

$${\vec{D}}_1[1:m_0]\ne {\vec{D}}_2[1:m_0].$$

$$V_{m_0+1}\cup D_1[1:m_0]\cdot \vec{V}[1:m_0]\supseteq -A^c\cup A$$

Full text

1. 1 Introduction
2. 2 Combinatorial Inequalities
3. 3 Proof of Theorem 1.1
4. 4 Path Maximizing Covering Probability
5. 5 Proof of Theorem 1.5
6. 6 Discussions
7. A
8. B

1. Introduction

In this paper, we study the probability that a finite subset, especially the trace of a nearest neighbor path in $\mathbb{Z}^{d}$ is completely covered by the trace of a $d$ dimensional simple random walk.

For any finite subset $A\subset\mathbb{Z}^{d}$ and a $d$ dimensional simple random walk $\{X_{n}\}_{n=0}^{\infty}$ starting at 0, we say that $A$ is completely covered by the first $L$ steps of the random walk if

[TABLE]

For simplicity we state our first result for $d=2$. For integer $l_{0}\geq 0$ and the subspace of reflection $l:x=y+l_{0}$, we can define $\varphi_{l}:\mathbb{Z}^{2}\to\mathbb{Z}^{2}$ as the reflection mapping around $l$. I.e., for any $(x,y)\in\mathbb{Z}^{2}$,

[TABLE]

Suppose two disjoint finite sets $A_{0},B_{0}\subset\mathbb{Z}^{2}\cap\{x,y:x\leq y+l_{0}\}$ both stay on the left of $l$. We then have the following theorem which states that the covering probability cannot get larger when we reflect one of them to the other side of the line:

Theorem 1.1**.**

For any integer $L\geq 0$,

[TABLE]

Remark 1.2*.*

By taking the union over all the $L$’s, one can immediately see the theorem also holds for $L=\infty$.

Remark 1.3*.*

One would think (like the authors first did) that Theorem 1.1 should follow from repeated use of the reflection principle. Two problems arise when one explores this idea. The first is that reflecting a path does not conserve the hitting order within the sets, which makes it hard to determine the times of reflection. The second is that even if we consider the sets before and after reflection with the same hitting order we can get a contradiction to the monotonicity of cover probabilities with the specified order. See Figure 1 for an example. Here after hitting the reflected first vertex the path covers, the vertices we would like to hit in order 3, 4 and 5 block the way to the vertex we would like to visit second.

With Theorem 1.1, we could consider the problem of covering a nearest neighbor path in $\mathbb{Z}^{d}$. For any integer $N\geq 1$, let $\partial B_{1}(0,N)$ be the boundary of the $L_{1}$ ball in $\mathbb{Z}^{d}$ with radius $N$. We say that a nearest neighbor path

[TABLE]

connects 0 and $\partial B_{1}(0,N)$ if $P_{0}=0$ and $\inf\{n:\|P_{n}\|_{1}=N\}=K$. And we say that a path $\mathcal{P}$ is covered by the first $L$ steps of $\{X_{n}\}_{n=0}^{\infty}$ if

[TABLE]

Then we are able to use Theorem 1.1 to show that the covering probability of any such path can be bounded by that of the diagonal.

Theorem 1.4**.**

For each integers $L\geq N\geq 1$, let $\mathcal{P}$ be any nearest neighbor path in $\mathbb{Z}^{d}$ connecting 0 and $\partial B_{1}(0,N)$. $X_{n},n\geq 0$ be a $d$ dimensional simple random walk starting at 0. Then

[TABLE]

where

[TABLE]

where

[TABLE]

and ${\rm arc}_{k}=(k-1)\sum_{i=1}^{d}e_{i}+{\rm arc}_{1}$.

The following main theorem gives an upper bound of the covering probability over all nearest neighbor paths connecting 0 and $\partial B_{1}(0,N)$.

Theorem 1.5**.**

There is $P_{d}\in(0,1)$ such that for any nearest neighbor path $\mathcal{P}=(P_{0},P_{1},\cdots,P_{K})$ connecting 0 and $\partial B_{1}(0,N)$ and $\{X_{n}\}_{n=0}^{\infty}$ which is a $d$ dimensional simple random walk starting at 0 with $d\geq 4$, we always have

[TABLE]

Here $P_{d}$ equals to the probability that $\{X_{n}\}_{n=0}^{\infty}$ ever returns to the $d$ dimensional diagonal line.

Note that in Theorem 1.5 the upper bound is not very sharp since we only look at returning to the exact diagonal line for $[N/d]$ times, which may cover at most $1/d$ of the total points in $\overset{\nearrow}{\mathcal{P}}$. Although for any fixed $d$, we still have an exponential decay with respect to $N$, when $d\to\infty$, such exponential decaying speed, which is lower bounded by $\left(\frac{1}{2d}\right)^{1/d}$, goes to one. Fortunately, in Appendix A we are able to show that $\lim_{d\rightarrow\infty}2dP_{d}=1$, and then further find an upper bound on the asymptotic of the probability that a $d$ dimensional simple random walk starting from some point in Trace\big{(}\overset{\nearrow}{\mathcal{P}}\big{)} will ever return to Trace\big{(}\overset{\nearrow}{\mathcal{P}}\big{)}. Note that we now need to return at least $N$ time to cover all the points in $\overset{\nearrow}{\mathcal{P}}$. We state this result as an additional theorem which is stronger than Theorem 1.5 since the proof is much more elaborate and for many uses Theorem 1.5 is sufficient.

Theorem 1.6**.**

There is a $C\in(0,\infty)$ such that for any $d\geq 4$ and any nearest neighbor path $\mathcal{P}=(P_{0},P_{1},\cdots,P_{K})\subset\mathbb{Z}^{d}$ connecting 0 and $\partial B_{1}(0,N)$ and $\{X_{n}\}_{n=0}^{\infty}$ which is a $d$ dimensional simple random walk starting at 0, we always have

[TABLE]

The proof of Theorem 1.6 can be found at the end of Section A.1.

Remark 1.7*.*

Actually, any $C>3/2$ will serve as a good upper bound for sufficiently large $d$. See Remark A.6 in Appendix A for details.

Remark 1.8*.*

Note that we do not present a proof of Theorem 1.6 for $d=3$. With Theorem 1.4 at hand, it is possible to prove this case since the cover probability of the diagonal path is bounded by the probability a random walk returns n times to the diagonal before leaving the ball $B_{1}(n)$. By using concentration inequalities after truncation of the excursion times of 2 dimensional random walks (not simple). However this yields an extremely non sharp bound with a slower order of decaying rate.

We conjecture that the upper bound in Theorem 1.5 can be improved for paths of length $N^{1+\epsilon}$, for small enough $\epsilon>0$. If this conjecture holds this will yield non trivial bounds for the chemical distance within a 3 dimension random walk trace:

Conjecture 1.9**.**

There is an $\epsilon>0$ and a $\hat{P}_{d}\in(0,1)$ such that for any nearest neighbor path $\mathcal{P}$, connecting [math] and $\partial B_{1}(N)$, such that $|{\mathcal{P}}|\leq N^{1+\epsilon}$ and $\{X_{n}\}_{n=0}^{\infty}$ which is a $d$ dimensional simple random walk starting at 0 with $d\geq 4$, we always have

[TABLE]

Remark 1.10*.*

Note that the probability to cover a space filling curve in $B_{1}(0,N)$ decays asymptotically slower that $c^{N^{d}}$. Sznitman [5, Section 2] showed that the probability a random walk path covers $B_{1}(0,N)$ completely can be bounded below by $ce^{-cN^{d-1}\log N}$.

A natural approach towards Conjecture 1.9 is to try applying the same reflection process in this paper but also consider the repetition of visits rather than just looking at the trace of the path. However, it is shown in Section 6 that once we consider repetition, the diagonal line (with repetition) no longer maximizes the covering probability.

And for the family of monotonic nearest neighbor paths starting at 0, we also conjecture that the cover probability is minimized when the path goes straightly along a coordinate axis. I.e.,

Conjecture 1.11**.**

For each integers $L\geq N\geq 1$, let $\mathcal{P}$ be any nearest neighbor monotonic path in $\mathbb{Z}^{d}$ with length $N$. $X_{n},n\geq 0$ be a d dimensional simple random walk starting at 0. Then

[TABLE]

where

[TABLE]

Remark 1.12*.*

Note that the constants we get in Theorem 1.5 are not sharp. In fact, the upper bound we obtain for the covering of the diagonal path is of order $(1/2d)^{N}$. If we use the same argument as in Theorem 1.5 for the straight line we will get a bound of $[1/2(d-1)]^{N}$, since a return to the straight line is equivalent to a $d-1$ dimensional random walk returning to the origin. Thus we get that the bound we obtain is larger for the path that we conjecture minimizes the cover probability.

The structure of this paper is as follows: in Section 2 we prove a combinatorial inequality, which can be found later equivalent to finding a one-to-one mapping between paths of random walks within each of some equivalence classes. In Section 3 we use this combinatorial inequality to prove Theorem 1.1. With Theorem 1.1, we construct a finite sequence of paths with non-decreasing covering probabilities in Section 4 to show that the covering probability is maximized by the path that goes along the diagonal, see Theorem 1.4 and 4.1. The proof of Theorem 1.5 is completed is Section 5, while in Section 6 we discuss the two conjectures and show numerical simulations. In Appendix A we prove that $\lim_{d\rightarrow\infty}2dP_{d}=1$ and then show that the probability a simple random walk returns to $\overset{\nearrow}{\mathcal{P}}$ also has an upper bound of $O(d^{-1})$, which implies Theorem 1.6. In Appendix B we prove that the monotonicity fails when considering covering probability with repetitions.

2. Combinatorial Inequalities

In this section, we discuss a combinatorial inequality problem, which can be found equivalent to finding a one-to-one mapping between paths of random walks. For $n\in\mathbb{Z}^{+}$ and $\Omega$ be a set of $n$ integer numbers, say $\Omega=\{1,2,\cdots,n\}$ and any $A\subset\Omega$, abbreviate $-A=\{-x:x\in A\}$ and $A^{c}=\Omega\setminus A$. For any $m\in\mathbb{Z}^{+}$, consider a collection of arcs which is a “vector” of subsets

[TABLE]

where each $V_{k}$ is called an arc, and an $m$ dimensional vector $\vec{D}=(\delta_{1},\cdots,\delta_{m})\in\{-1,1\}^{m}$ which is called a configuration. Then we can introduce the inner product

[TABLE]

Moreover, for any subset $A\subseteq\Omega$, we say a configuration $\vec{D}$ of $\vec{V}$ covers the reflection $A$ if

[TABLE]

and let

[TABLE]

be the subset of all such configurations. Then to show that a simple random walk has higher probability to cover the reflection of a finite set which bring it closer to the origin, we first need to prove that

Lemma 2.1**.**

For any $m,n\in\mathbb{Z}^{+}$, and any collection of arcs $\vec{V}$

[TABLE]

for all $A\subseteq\Omega$.

Proof.

First by symmetry one can immediately see that

[TABLE]

So we will concentrate when $A\not=\O$ and prove the inequality by induction on $m$ and $n$. To show the basis of induction, it is easy to see that for any $n$ and $m=1$

[TABLE]

if $V_{1}\not=\Omega$ and

[TABLE]

if $V_{1}=\Omega$. Then for any $m$ and $n=1$, by definition we must have $V_{i}=\{1\}$ or $\O$ for each $i$, and we always have $A=\Omega$. Thus

[TABLE]

where $n_{e}(\vec{V})$ is the number of empty sets in $V_{1},\cdots,V_{m}$. With the method of induction, suppose the desired inequality is true for all $n<n_{0}$ and all $n=n_{0},m\leq m_{0}$. Then for $n=n_{0},m=m_{0}+1$, we can separate the last arc $V_{m_{0}+1}$ and the rest of the arcs and look at the truncated system at $m_{0}$. I.e.,

[TABLE]

and

[TABLE]

We have for any $A$

[TABLE]

where

[TABLE]

Noting that

[TABLE]

and that the two events in (2.3) are disjoint,

[TABLE]

In order to study the cardinality of $\mathcal{P}$, we first show that

Lemma 2.2**.**

For any two different $\vec{D}_{1}$ and $\vec{D}_{2}$ in $\mathcal{P}\left(\vec{V},A\right)$, we must have

[TABLE]

Proof.

The proof is straightforward. Suppose $\vec{D}_{1}[1:m_{0}]=\vec{D}_{2}[1:m_{0}]=D^{\prime}$, then their $m_{0}+1$st coordinates must be different. Thus

[TABLE]

and

[TABLE]

The first equality above implies that

[TABLE]

while the second implies that

[TABLE]

Combining the two inequalities gives us

[TABLE]

which contradicts with the definition of $\mathcal{P}\left(\vec{V},A\right)$. Thus the proof is complete. ∎

With Lemma 2.2, we now know that there is one-to-one mapping between each configuration in $\mathcal{P}\left(\vec{V},A\right)$ and its first $m_{0}$ coordinates. And note that

[TABLE]

We have

[TABLE]

where

[TABLE]

and

[TABLE]

Moreover, note that $\mathcal{C}\left(\vec{V}[1:m_{0}],A\right)$, $\mathcal{C}_{1}\left(\vec{V}[1:m_{0}],A\right)$ and $\mathcal{C}_{2}\left(\vec{V}[1:m_{0}],A\right)$ are disjoint and that

[TABLE]

where

[TABLE]

with each $V^{\prime}_{k}=V_{k}\cap V_{m_{0}+1}^{c}\subseteq\Omega^{\prime}=V_{m_{0}+1}^{c}$, $A^{\prime}=A\cap V_{m_{0}+1}^{c}$, and

[TABLE]

In words, in order to be in one of the 3 disjoint subsets above, we must guarantee that all points in $\Omega^{\prime}=V_{m_{0}+1}^{c}$ under reflection of $A$ are covered by the configuration $\vec{D}^{\prime}$ of $\vec{V}[1:m_{0}]$. To verify (2.6), one can note that for any

[TABLE]

we have

[TABLE]

which implies

[TABLE]

In (2.7) we have the right hand side equals to

[TABLE]

and the left hand side equals to

[TABLE]

Noting that for each $k$

[TABLE]

we have

[TABLE]

which shows that $\vec{D}^{\prime}$ also in $\mathcal{C}^{\prime}\left(\vec{V}^{\prime},A^{\prime}\right)$.

Specifically, when $A=\Omega$, note that for any $D^{\prime}\in\mathcal{C}^{\prime}\left(\vec{V}^{\prime},\Omega^{\prime}\right)$, $\Omega^{\prime}=\vec{D}^{\prime}\cdot\vec{V}^{\prime}\subseteq D^{\prime}\cdot\vec{V}[1:m_{0}]$. Thus

[TABLE]

which implies that

[TABLE]

Combining (2.4)-(2.8) and the induction hypothesis, we have

[TABLE]

And thus the proof of Lemma 2.1 is complete. ∎

3. Proof of Theorem 1.1

With the combinatorial inequality above, we can study the covering probability of simple random walks. Let $\mathcal{N}_{L}$ be the set of all nearest neighbor paths starting at 0 of length $L+1$ and consider 2 subsets of $\mathcal{N}_{L}$ as follows:

[TABLE]

and

[TABLE]

Noting that the probabilities a simple random walk starting at 0 taking each $\vec{x}\in\mathcal{N}_{L}$ as its first $L$ steps are the same, it suffices to show that

[TABLE]

To prove (3.1), we need to first partition $\mathcal{N}_{L}$ into disjoint subsets, each of which serves as an equivalence class under the equivalence relation on $\mathcal{N}_{L}$ describe below.

For each $\vec{x}=(x_{0},x_{1},\cdots,x_{L})\in\mathcal{N}_{L}$, let $T_{0}=0$, $T_{1}=\inf\{n:x_{n}\in l\}$, and

[TABLE]

for each integer $n\in[2,L]$ to be the time of the $n$th visit to $l$. Here we use the convention that $\inf\{\O\}=\infty$ and let $T_{L+1}=\infty$. Then for each $n=0,1,\cdots,L$, define the $n$th arc of $\vec{x}$ as

[TABLE]

where we use the convention that $\vec{x}[k:\infty)=\vec{x}[k:L]$, and $\vec{x}[0:0)=\vec{x}[\infty:\infty)=\O$. I.e., the $n$th arc is the piece of the path between the (possible) $n$th and $n+1$st visit to $l$. Since $\vec{x}$ is a nearest neighbor path, all points in each arc must be on the same side of $l$. Then for any $D=(\delta_{1},\delta_{2},\cdots,\delta_{L})\in\{-1,1\}^{L}$ we can define a mapping $\varphi_{l,D}$ on $\mathcal{N}_{L}$ so that for any $\vec{x}$

[TABLE]

In words, we keep the part of the path the same until it first visits $l$. Then for the $n$th arc, we keep it unchanged if $\delta_{n}=1$ and reflect it around $l$ if $\delta_{n}=-1$. By definition, it is easy to see that \varphi_{l,D}\big{(}\vec{x}\big{)}\in\mathcal{N}_{L}. And since

[TABLE]

$\varphi_{l,D}$ forms a bijection on $\mathcal{N}_{L}$. Now we can introduce the equivalence relation on $\mathcal{N}_{L}$ previously mentioned. For each two $\vec{x},\vec{y}\in\mathcal{N}_{L}$, we say $\vec{x}\sim\vec{y}$ if there exist a $D\in\{-1,1\}^{L}$ such that (see Figure 2)

[TABLE]

Then one can immediately check that if $\vec{x}\sim\vec{y}$ and $D$ is the configuration, then

[TABLE]

and for $D_{0}\equiv 1$,

[TABLE]

Moreover, if

[TABLE]

let

[TABLE]

we have

[TABLE]

Combining (3.2)-(3.4), we have that $\sim$ forms a equivalence relation on $\mathcal{N}_{L}$, where each path in $\mathcal{N}_{L}$ belongs to one equivalence class. Thus all the equivalence classes are disjoint from each other and there has to be a finite number of them, forming a partition of $\mathcal{N}_{L}$. We denote these equivalence classes as

[TABLE]

where each of them can be represented by its specific element $\vec{x}_{k,+},\ k=1,\cdots,J$, which is the unique path in each class that always stays on the left of $l$.

Then for each $k$, let $n_{k,1}<n_{k,2}<\cdots<n_{k,m_{k}}$ be all the $n$’s such that

[TABLE]

Note that the only case when we have \Big{|}{\rm Trace}\left({\rm arc}\big{(}\vec{x}_{k,+},n\big{)}\right)\Big{|}=0 is when $T_{n}=T_{n+1}=\infty$ and the only case when it equals to 1 is when $T_{n}=L$. Then for any $\vec{x}\in\mathcal{C}_{L,k}$ and any $D_{1},D_{2}$ such that

[TABLE]

we must have $\delta_{1,n_{k,i}}=\delta_{2,n_{k,i}}$ for all $i$’s. So we have a well defined onto mapping $f$ between $\mathcal{C}_{L,k}$ and $\{-1,1\}^{m_{k}}$ where each $\vec{x}$ such that

[TABLE]

for some $D$ is mapped to

[TABLE]

Moreover, for any two configurations $D_{1}$ and $D_{2}$ such that $\delta_{1,n_{k,i}}=\delta_{2,n_{k,i}}$ for all $i$’s, we also must have

[TABLE]

The reason of that is for all $n$ not in $\{n_{k,i}\}_{i=1}^{m_{k}}$,

[TABLE]

which means those arcs are either empty or just one point $x_{T_{n}}$ right on the diagonal, which does not change at all under any possible reflection. Thus we have proved that the mapping $f$ is a bijection between $\mathcal{C}_{L,k}$ and $\{-1,1\}^{m_{k}}$.

At this point we have the tools we need and can go back to compare the two covering probabilities. Noting that the equivalence classes in (3.5) form a partition of $\mathcal{N}_{L}$. It suffices to show that for each $k\leq J$

[TABLE]

For each class $\mathcal{C}_{L,k}$. First, if

[TABLE]

then one can immediately see

[TABLE]

Otherwise, let $\Omega_{k}=(A_{0}\cup B_{0})\cap{\rm Trace}(\vec{x}_{k,+}[0:T_{1}])^{c}\cap l^{c}$ and $A_{k}=A_{0}\cap{\rm Trace}(\vec{x}_{k,+}[0:T_{1}])^{c}\cap l^{c}$. And let $n_{k}=|\Omega_{k}|$. We can also list all points in $\Omega_{k}$ as $\omega_{1},\cdots,\omega_{n_{k}}$ and all points in $\varphi_{l}(\Omega_{k})$ as $\omega_{-1},\cdots,\omega_{-n_{k}}$, where $\varphi_{l}(\omega_{j})=\omega_{-j}$ for all $j$. Then it is easy to check that

[TABLE]

since all other points in $A_{0}\cup B_{0}$ are guaranteed to be visited by $\vec{x}_{k,+}[0:T_{1}]$ or $\vec{x}_{k,+}[T_{1},\cdots,T_{L}]$ which are both shared over all paths in this equivalence class. And we also have

[TABLE]

since $A_{k}\subseteq A_{0}$, $\Omega_{k}\cap A_{k}^{c}\subseteq B_{0}$. Finally, define

[TABLE]

and $\vec{V}_{k}=V_{k,1}\otimes V_{k,2}\otimes\cdots\otimes V_{k,m_{k}}$. Then by the constructions above we have for any $\omega_{j}\in\Omega_{k}$, $\omega_{j}\in{\rm Trace}(\vec{x})$ if and only if there exists some $i$ such that $j\in V_{k,i}$ and $f(\vec{x})[i]=1$. And similarly $\varphi_{l}(\omega_{j})\in{\rm Trace}(\vec{x})$ if and only if there exists some $i$ so that $j\in V_{k,i}$ and $f(\vec{x})[i]=-1$. In combination, for any $\omega_{j}\in\Omega_{k}\cup\varphi_{l}(\Omega_{k})$, $\omega_{j}\in{\rm Trace}(\vec{x})$ if and only if there exists some $i$ such that $j\in f(\vec{x})[i]V_{k,i}$. Then taking the intersections and letting $\bar{\Omega}_{k}=\{1,2,\cdots,n_{k}\}$ and

[TABLE]

we have

[TABLE]

and

[TABLE]

Noting that the mapping $f$ is a bijection between $\mathcal{C}_{L,k}$ and $\{-1,1\}^{m_{k}}$,

[TABLE]

and

[TABLE]

Apply Lemma 2.1 on $\bar{\Omega}_{k}$, $\vec{V}_{k}$ and $\bar{A}_{k}$. The proof of this theorem is complete. ∎

Remark 3.1*.*

With exactly the same argument, we can also have the reflection theorem on reflecting over a line $y=x+n$, $n\geq 1$ and $x=-y+n$.

4. Path Maximizing Covering Probability

To apply Theorem 1.1 specifically on covering the trace of a nearest neighbor path in ${\mathcal{P}}\subset\mathbb{Z}^{2}$ connecting 0 and $\partial B_{1}(0,N)$, we can assume without loss of generality that $P_{K}\in\partial B_{1}(0,N)$ in the first quadrant. For each such path with at least one point at $(x,y)$ such that $|x-y|\geq 2$, we can always reflect it around $l$ equals to $x=y+1$ or $y=x+1$ with the part of the path on the same side of $l$ as [math] to be $A_{0}$ and the remaining points to be $\varphi_{l}(B_{0})$ and always increase the probability of covering.

Note that, after the reflection, $A_{0}\cup B_{0}$ is the trace of another nearest neighbor path, and we can reduce the total difference

[TABLE]

by at least one in each step. After a finite number of steps, we will end up with a nearest neighbor path that stays within $\{|x-y|\leq 1\}$. And among all those such paths, applying Theorem 1.1 for reflection over $x=y$, we can show that the covering probability is maximized when we move all the “one step corners” to the same side of the diagonal, which itself gives us a monotonic path that stays within distance one above or below the diagonal. Thus we have the theorem as follows.

Theorem 4.1**.**

For each integers $L\geq N\geq 1$, let $\mathcal{P}$ be any nearest neighbor path in $\mathbb{Z}^{2}$connecting 0 and $\partial B_{1}(0,N)$. $X_{n},n\geq 0$ be a 2 dimensional simple random walk starting at 0. Then

[TABLE]

where

[TABLE]

Proof.

As outlined above, we first show that

Lemma 4.2**.**

For each integers $L\geq N\geq 1$, let

[TABLE]

be any nearest neighbor path in $\mathbb{Z}^{2}$connecting 0 and $\partial B_{1}(0,N)$ with length $K+1\geq N+1$ where there is an $i\leq n$ such that $|x_{i}-y_{i}|\geq 2$ and where $x_{K}\geq 0,y_{K}\geq 0$. $X_{n},\ n\geq 0$ be a 2 dimensional simple random walk starting at 0. Then there exists a nearest neighbor path $\mathcal{P}_{1}$ that keep staying within $\{|x-y|\leq 1\}$ such that

[TABLE]

Proof.

For any path $\mathcal{Q}$ with lenght $K+1$, define its total difference as

[TABLE]

For each such path in this lemma, without loss of generality we can always assume there is some $i$ such that $x_{i}-y_{i}\geq 2$. Consider the line of reflection $l:x=y+1$ (otherwise consider $l:y=x+1$). It is easy to see that there is at least one point along this path on the right side of $l$. I.e.

[TABLE]

Define $A_{0}={\rm Trace}(\mathcal{P})\cap B_{0}^{c}$, $B_{0}=\varphi_{l}(B_{0}^{\prime})\cap A_{0}^{c}$, and $\hat{B}_{0}^{\prime}=\varphi_{l}(B_{0})\subseteq B_{0}^{\prime}$. Thus we have ${\rm Trace}(\mathcal{P})=A_{0}\cup B_{0}^{\prime}$ and

[TABLE]

Applying Theorem 1.1 on $A_{0}$ and $B_{0}$ gives us

[TABLE]

Then noting that $\mathcal{P}$ is a nearest neighbor path starting at 0 with length $K+1$, let $C_{K,i}$ be the equivalence class it belongs to under the relation $\sim$, and let $\mathcal{P}^{\prime}=\vec{x}_{i}$ be the representing element of $C_{K,i}$ where all arcs are reflected to the left of $l$. Then it is easy to see that

[TABLE]

Combine (4.2)-(4.4),

[TABLE]

Then note that for any $j$ such that $x_{j}-y_{j}\geq 2$,

[TABLE]

while

[TABLE]

for all $x_{i}\geq y_{i}+2$. Since $(x_{K},y_{K})$ is in the first quadrant, if in addition we also have $x_{K}\geq y_{K}+1$, then $\varphi_{l}(x_{K},y_{K})$ remains in the first quadrant with $\|\varphi_{l}(x_{K},y_{K})\|_{1}=N$. Thus the new nearest neighbor path ${\mathcal{P}}^{\prime}$ is also one connecting 0 and $\partial B_{1}(0,N)$. Moreover,

[TABLE]

while

[TABLE]

which shows that $D_{T}(\mathcal{P}^{\prime})\leq D_{T}(\mathcal{P})-2$. Then if there is a point $(x_{j}^{\prime},y_{j}^{\prime})$ in the new path $\mathcal{P}^{\prime}$ with $|x_{j}^{\prime}-y_{j}^{\prime}|\geq 2$, we can repeat the previous process and the covering probability is non-decreasing. Noting that for each time we decrease the total difference by at least 2 while $D_{T}(\mathcal{P})$ is a finite number, after repeating a finite number of times, we must end up with a nearest neighbor path where no point satisfies $|x-y|\geq 2$. Thus, we find a nearest neighbor path $\mathcal{P}_{1}$ connecting 0 and $\partial B_{1}(0,N)$ that keep staying within $\{|x-y|\leq 1\}$ with a higher covering probability. ∎

With Lemma 4.2, note that for any nearest neighbor path connecting 0 and $\partial B_{1}(0,N)$ that keeps staying within $\{|x-y|\leq 1\}$, we can always look at the part of it after its last visit to 0 and it has a higher covering probability. And note that such part has to contain a monotonic path from 0 to $\partial B_{1}(0,N)$. Thus it is sufficient to show that $\overset{\nearrow}{\mathcal{P}}$ has the highest covering probability over all nearest neighbor monotonic paths $\mathcal{P}_{1}$ connecting 0 and $\partial B_{1}(0,N)$ that keeps staying within $\{|x-y|\leq 1\}$. We can show this by specifying what the trace of each such path looks like. Letting $N_{0}=N-[N/2]$ and $\bar{\Omega}=\{1,2,\cdots,N_{0}\}$. We have

Lemma 4.3**.**

For each nearest neighbor path $\mathcal{P}_{1}$ connecting 0 and $\partial B_{1}(0,N)$that keeps staying within $\{|x-y|\leq 1\}$, we must have that

[TABLE]

and that

[TABLE]

Proof.

The proof follows easy from the fact that $\mathcal{P}_{1}$ is a nearest neighbor path. For any such path

[TABLE]

Since $\mathcal{P}_{1}$ is a nearest neighbor path between 0 and $\partial B_{1}(0,L)$ that stays within $\{|x-y|\leq 1\}$, thus for any $n\leq N$ there must be at least one point belongs to the trace of $\mathcal{P}_{1}$ with $L_{1}$ norm $n$. Then note that in $\{|x-y|\leq 1\}$ the only integer point with $L_{1}$ norm $2j$ is $(j,j)$, while the only two points with $L_{1}$ norm $2j-1$ are $(j-1,j)$ and $(j,j-1)$. Thus the proof is complete. ∎

With the lemma above, for each such $\mathcal{P}_{1}$ define

[TABLE]

and

[TABLE]

Then we have $\bar{A}\cup\bar{B}=\bar{\Omega}$, so that

[TABLE]

Define

[TABLE]

where $l_{0}$ is the line $x=y$. Then

[TABLE]

And note that

[TABLE]

which itself gives a monotonic nearest neighbor path connecting 0 and $\partial B_{1}(0,N)$. So Theorem 1.1 finishes the proof. ∎

For fixed $N$, the inequality in Theorem 4.1 gets equal when $L=\infty$ since the 2 dimensional simple random walk is recurrent and both probabilities goes to one. However, we can easily generalize the same result to higher dimensions. This will similarly give us Theorem 1.4.

Proof of Theorem 1.4.

This theorem can be proved by reflecting only on two coordinates in $\mathbb{Z}^{d}$ while keeping all the others unchanged. For any $n\geq 0$, we look at, without loss of generality, the subspace $l:a_{1}=a_{2}+l_{0},\ l_{0}\geq 0$ when $d\geq 3$, and define reflection $\varphi_{l}$ over $l$ as follows: for each point $(a_{1},\cdots,a_{d})\in\mathbb{Z}^{d}$,

[TABLE]

Then for all paths in $\mathcal{N}_{L}$ (all nearest neighbor paths starting at [math] of length $L+1$), we can again define $T_{0}=0$, $T_{1}=\inf\{n:x_{n}\in l\}$, and

[TABLE]

for each integer $n\in[2,L]$ to be the time of the $n$th visit to subpsace $l$, and divide $\mathcal{N}_{L}$ into a partition of equivalence classes under $\varphi_{l,D}$ for all $D\in\{-1,1\}^{L}$. Then for each pair of disjoint finite subsets $A_{0},B_{0}\in\{x\leq y+l_{0}\}$, let

[TABLE]

and

[TABLE]

For each equivalence class $\mathcal{C}_{L,k}$ as above, the exact same argument as in the proof of Theorem 1.1 guarantees that

[TABLE]

So again we have

[TABLE]

Then apply (4.6) to any nearest neighbor path connecting 0 and $\partial B_{1}(0,N)$

[TABLE]

And without loss of generality we can also assume that $P_{L}\in(\mathbb{Z}^{+}\cup\{0\})^{d}$. Let the subspace of reflection be $l:a_{1}=a_{2}+1$,

[TABLE]

and

[TABLE]

Without loss of generality we can assume $B_{0}^{\prime}$ is not empty, note that ${\rm Trace}(\mathcal{P})=A_{0}\cup B_{0}^{\prime}$, and that similar to the proof of Theorem 4.1, we can again let $\mathcal{P}^{\prime}$ be the representing element in the equivalence class under $\sim$ that contains $\mathcal{P}$, which is another nearest neighbor path connecting 0 and $\partial B_{1}(0,N)$ where all the arcs are reflected to the same side of $l$ as 0. Then ${\rm Trace}(\mathcal{P}^{\prime})=A_{0}\cup B_{0}$, and ${\rm Trace}(\mathcal{P})=A_{0}\cup B_{0}^{\prime}\supseteq A_{0}\cup\hat{B}_{0}^{\prime}$. By (4.6) we have

[TABLE]

Moreover, define

[TABLE]

be the total difference of $\mathcal{P}$. Then note that for each $n$

[TABLE]

where

[TABLE]

which is a convex function of $p$. Thus, we rewrite

[TABLE]

and

[TABLE]

For each $n$ such that $P^{\prime}_{n}=(p_{n,1},\cdots,p_{n,d})\in A_{0}$, we always have

[TABLE]

Otherwise, we must have $P^{\prime}_{n}\in B_{0}$ and there must always be a $P_{n}=\varphi_{l}(P_{n}^{\prime})\in\hat{B}_{0}^{\prime}\subseteq B_{0}^{\prime}$, which implies that

[TABLE]

And since that $P_{n}\in B_{0}^{\prime}$, $p_{n,1}\geq p_{n,2}+2$, so that for $p^{\prime}_{n,1}=p_{n,2}+1$ and $p^{\prime}_{n,2}=p_{n,1}-1$, we must have

[TABLE]

which implies that $|p^{\prime}_{n,1}-p^{\prime}_{n,2}|<|p_{n,1}-p_{n,2}|$. Then combining (4.8), and that $p^{\prime}_{n,1}+p^{\prime}_{n,2}=p_{n,1}+p_{n,2}$ with the fact that $f_{n}(p)$ is convex, we also have

[TABLE]

which further implies that $D_{T}(\mathcal{P})\geq D_{T}(\mathcal{P}^{\prime})+1$. Again, noting that $D_{T}(\mathcal{P})$ itself is finite, so after at most a finite number of iterations, we will end up with a path $\mathcal{P}_{1}$ connecting 0 and $\partial B_{1}(0,N)$ within the region

[TABLE]

And it is easy to see that for each point $\vec{a}_{0}=(a_{1,0},a_{2,0},\cdots,a_{d,0})$ in region $R$ and each subspace $l:a_{i}=a_{j}$, $\vec{a}_{0}^{\prime}=\varphi_{l}(\vec{a}_{0})$ must satisfy

[TABLE]

Thus, apply the same argument as in the proof of Theorem 4.1: reflection over $a_{2}=a_{1}$, which reflects points in $\{a_{2}=a_{1}+1\}$ to $\{a_{1}=a_{2}+1\}$, and then for reflections over $a_{3}=a_{1},\ \cdots$ and then $a_{d}=a_{1}$. We will have a sequence of paths $\mathcal{P}_{2,i},\ i=2,\cdots d$ in $R$ with covering probabilities that never decrease. Moreover, by the definition of our reflections, for each $n\leq K$ and $2\leq j\leq d$ we have that $\{p_{2,i,n,1}\}_{i=2}^{d}$ is nondecreasing while $\{p_{2,i,n,j}\}_{i=2}^{d}$ is nonincreasing, and that

[TABLE]

Thus for $\mathcal{P}_{2}=\mathcal{P}_{2,d}$, we must have

[TABLE]

for all $n\leq K$. Then we reflect $\mathcal{P}_{2}$ over $a_{3}=a_{2}$, $a_{4}=a_{2},\ \cdots$, and $a_{d}=a_{2}$ which also gives us a sequence of paths $\mathcal{P}_{3,i},\ i=3,\cdots d$ in $R$ with covering probabilities that never decrease. Letting $\mathcal{P}_{3}=\mathcal{P}_{3,d}$, similarly we must have

[TABLE]

Moreover recalling the formulas of reflections within $R$ in (4.9), all reflections over $a_{i}=a_{2}$, $i\geq 3$ will not change $\max_{2\leq j\leq d}p_{2,n,j}$ for any $n$. Thus, we still have (4.10) holds for $\mathcal{P}_{3}$. Repeating this process and we will have a sequence $\mathcal{P}_{4},\mathcal{P}_{5},\cdots,\mathcal{P}_{d}$ with covering probabilities that never decrease, where each of them stays within $R$. And finally for $\mathcal{P}_{d}$, we must have

[TABLE]

for all $i\leq d-1,\ n\leq N.$ Noting again that $\mathcal{P}_{d}$ is a nearest neighbor path, then

[TABLE]

and the proof of this Theorem is complete. ∎

5. Proof of Theorem 1.5

With Theorem 1.4, the proof of Theorem 1.5 follows immediately from the fact that the simple random walk on $\mathbb{R}^{d},\ d\geq 4$ returns to the one dimensional line $x_{1}=x_{2}=\cdots=x_{d}$ with probability less than 1. Note that for any nearest neighbor path $\mathcal{P}=(P_{0},P_{1},\cdots,P_{N})$ and $\{X_{n}\}_{n=0}^{\infty}$ connecting [math] and $\partial B_{1}(0,N)$ which is a $d$ dimensional simple random walk, letting

[TABLE]

be the points in $\overset{\nearrow}{\mathcal{P}}$ on the diagonal, we always have by Theorem 1.4,

[TABLE]

Moreover, let $\{\tau_{n}\}_{n=1}^{\infty}$ be sequence of stopping times of all visits to diagonal line $\ell:x_{1}=x_{2}=\cdots=x_{d}$. Then

[TABLE]

To bound the probability on the right hand side of (5.1). Define a new Markov process $\{Y_{n}\}_{n=0}^{\infty}\in\mathbb{Z}^{d}$, where

[TABLE]

Note that we can also write $\tau_{n}=\inf\{n>\tau_{n-1},Y_{n}=0\}$ and that $Y_{n}$ itself is a $d-1$ dimensional random walk with generator

[TABLE]

With $d-1\geq 3$, we have $P(\tau_{n}<\infty|\tau_{n-1}<\infty)=P_{d}=1-G_{Y}(0)^{-1}<1$. And thus the proof of Theorem 1.5 complete. ∎

6. Discussions

In this section we discuss the conjectures and show numerical simulations.

6.1. Covering Probabilities with Repetitions

In the proof of Theorem 1.4, note that each time we apply Theorem 1.1 and get a new path $\mathcal{P}^{\prime}$ with higher covering probability, we always have

[TABLE]

and

[TABLE]

where $B_{0}=\varphi_{l}(B_{0}^{\prime})\cap A_{0}^{c}\subseteq\varphi_{l}(B_{0}^{\prime})$. This, together with the fact that $A_{0}$ is disjoint with both $B_{0}$ and $B_{0}^{\prime}$, implies that

[TABLE]

In words, although the length of the path remains the same after reflection, the size of its trace may decrease. In fact, for any simple path connecting 0 and $\partial B_{1}(0,N)$, at the end of our sequence of reflections, we will always end up with a (generally non-simple) path whose trace is of size $N+1$.

One natural approach towards a sharper upper bound is taking the repetitions of visits in a non-simple path into consideration. For any path

[TABLE]

starting at 0 which may not be simple, and any point $P\in{\rm Trace}(\mathcal{P})$, we can define the $k$th visit to $P$ as $T_{1,P}=\inf\{n:P_{n}=P\}$ and

[TABLE]

with convention $\inf\{\O\}=\infty$. Then we can define the repetition of $P\in{\rm Trace}(\mathcal{P})$ in the path $\mathcal{P}$ as

[TABLE]

and denote the collection of all such repetitions as $N_{\mathcal{P}}=\{n_{P,\mathcal{P}}:P\in{\rm Trace}(\mathcal{P})\}$. It is to easy to see that $n_{P,\mathcal{P}}\equiv 1$ for all $P\in{\rm Trace}(\mathcal{P})$ when $\mathcal{P}$ is a simple path, and that

[TABLE]

For $d$ dimensional simple random walk $\{X_{n}\}_{n=0}^{\infty}$ starting at 0 and any point $P\in\mathbb{Z}^{d}$, we can again define the stopping times $\tau_{0,P}=0$, $\tau_{1,P}=\inf\{n:X_{n}=P\}$ and

[TABLE]

with convention $\inf\{\O\}=\infty$. Then we have

Definition 6.1**.**

For each nearest neighbor path $\mathcal{P}$, and $d$ dimensional simple random walk $\{X_{n}\}_{n=0}^{\infty}$, we say that $\{X_{n}\}_{n=0}^{L}$ covers $\mathcal{P}$ up to its repetitions if

[TABLE]

And we denote such event by ${\rm Trace}(\mathcal{P})\otimes N_{\mathcal{P}}\subseteq\{X_{n}\}_{n=0}^{L}$.

Our hope was, for any nearest neighbor path $\mathcal{P}$ and subspace of reflection like $l:x_{i}=x_{j}+l_{0}$, the probability of a simple random walk $\{X_{n}\}_{n=0}^{L}$ starting at 0 covering $\mathcal{P}$ up to its repetitions may be upper bounded by that of covering the path $\mathcal{P}^{\prime}$ up to its repetitions, where $\mathcal{P}^{\prime}$ is the representing element in the equivalence class in $\mathcal{N}_{K}$ containing $\mathcal{P}$ under the reflection $\varphi_{l}$. In words, $\mathcal{P}^{\prime}$ is the path we get by making all the arcs in $\mathcal{P}$ reflected to the same side as 0.

Note that although $\mathcal{P}^{\prime}$ may not be simple and the size of its trace could decrease, this will at the same time increase the repetition on those points which are symmetric to the disappeared ones correspondingly. In fact, under Definition 6.1, the total number of points our random walk needs to (re-)visit is always

[TABLE]

So if our previous guess were true, then we will be able to follow the same process as in Section 3 and 4 and end with the same path along the diagonal, but this time with a higher probability of being covered up to its repetitions. Noting that the path we have in Theorem 1.4 will visit points exactly on the diagonal line $x_{1}=x_{2}=\cdots=x_{d}$ for $O(K)$ times, the same construction of a $d-1$ dimensional random walk as in Section 5 will give the sharp upper bound we need in Conjecture 1.9.

Unfortunately, here we present the counterexample and numerical simulations showing that Theorem 4.1 and 1.4 no longer holds for of certain non-monotonic paths. The idea of constructing those examples can be seen in the following preliminary model: Let $l$ be the line of reflection and suppose we have one point $x$ on the same side of $l$ as 0. Then suppose there is a equivalence class $\mathcal{C}_{L,k}$ with its representative element $\vec{x}_{k}$ having $2n$ arcs each visiting $x$ once. Then we look at the covering probability of $\{x,\varphi_{l}(x)\}\otimes(n,n)$ and that of its reflection $\{x\}\otimes 2n$. For the first one, we only need to choose $n$ of $2n$ arcs in $\vec{x}_{k}$ and reflect them to the other side while keeping the rest unchanged. So we have \big{(}^{2n}_{n}\big{)} configurations. However, for the second covering probability which one may hope to be higher, the only configuration that may give us the covering up to this repetition is $\vec{x}_{k}$ itself. Thus, at least in this equivalence class, the number of configurations covering $\{x,\varphi_{l}(x)\}\otimes(n,n)$ is larger than that of configurations covering $\{x\}\otimes 2n$.

With this idea in mind we give the following counterexample on actual 3 dimensional paths which shows precisely and rigorously that the covering probability is not increased after reflection.

Counterexample 1. Consider the following points $o=(0,0,0),\ y=(1,0,0)$, $z=(0,1,0),\ w=(1,1,0)\in\mathbb{Z}^{3}$, and paths

[TABLE]

and

[TABLE]

which is the representative element of the equivalence class containing $\mathcal{P}$, under reflection over $l:x_{2}=x_{1}$. Let $\{X_{n}\}_{n=0}^{\infty}$ be a simple random walk starting at 0. Moreover, we use the notation $A=\mathbb{Z}^{3}\setminus\{y,z,w\}$ and define stopping times $\tau_{a}=\inf\{n\geq 1:\ X_{n}=a\}$ for all $a\in\mathbb{Z}^{3}$, and $\tau_{A}=\inf\{n\geq 1:\ X_{n}\notin A\}$. Thus we have

Proposition 6.2**.**

For the paths $\mathcal{P}$ and $\mathcal{P}^{\prime}$ defined above,

[TABLE]

which is larger than

[TABLE]

The proof of Proposition 6.2 is basically a standard application of Green’s functions for finite subsets. So we leave the detailed calculations in Appendix B. For anyone who believes in law of larger numbers, we recommend them to look at the following numerical simulation which shows the empirical probabilities (which almost exactly agree with Proposition 6.2) of covering both paths with half a million independent paths of 3-dimensional simple random walks run up to $L=40000$.

For a finite length $\{X_{n}\}_{n=0}^{L}$ with $L<\infty$, although it is harder to calculate the exact covering probabilities theoretically, the following simulations on $L=4000,400$ and $40$ show that the inequality in Proposition 6.2 remains robust for fairly small $L$ (see Figure 5).

At last, it is important to note that Proposition 6.2 does not imply by any means a disproof of Conjecture 1.9. What it does show, however, is such a conjecture cannot be proved by the combinatorial argument on reflections shown in this paper.

6.2. Monotonic Path Minimizing Covering Probability

In Conjecture 1.11, we conjecture that when concentrating on monotonic paths, the covering probability is minimized when the path takes a straight line along some axis. The intuition is, while all monotonic paths connecting 0 and $\partial B_{1}(0,N)$ have the same $L_{1}$ distance, the $L_{2}$ distance is maximized along the straight line, which makes it the most difficult to cover. This conjecture is supported for small $N$. In the following example, we have $d=3$ and $N=3$. By symmetry of simple random walk, one can easily see that for each monotonic path of length $N+1=4$, starting at 0, the covering probability must equal to that of one of the following five:

[TABLE]

The following simulation (see Figure 4) shows that when $L=400$, the covering probability of path 1 is the smallest of them all. It should be easy to use the same calculation in Proposition 6.2 to show the rigorous result when $L=\infty$.

Appendix A

A.1. Introduction

In this appendix, we find the asymptotic behavior of the returning probability (to the origin and diagonal line) of a $d$ dimensional simple random walk as $d\to\infty$. The asymptotics of the return to the origin is stated in [3]. We present an alternative proof here for two reasons, the first is that we believe the proof in [3] is not completely rigurous and the second is that the technique we show here can be generalized to the diagonal return probability.

For a $d$ dimensional simple random walk $\{X_{d,n}\}_{n=0}^{\infty}$ starting at 0 and any $x\in\mathbb{Z}^{d}$, define the stopping time

[TABLE]

Then the returning probability is defined by

[TABLE]

Our main result is as follows:

Theorem A.1**.**

For $p_{d}$ defined above, we have

[TABLE]

While Theorem A.1 is stated for the leading order of $p_{d}$, one should be able to easily use the same technique in our proof to have the second and higher order asymptotics.

Also, noting that

[TABLE]

there is always a simple lower bound $p_{d}>1/(2d)$. With Theorem A.1, we actually know that such trivial lower bound is asymptotically sharp. Intuitively, when the dimension is very high, if a random walk does not return to 0 immediately in the first 2 steps, it will with high conditional probability get lost in the ocean of choices and never be able to go back to 0.

Moreover, a similar method as in Theorem A.1 may also work for other random walks. Particularly, for a specific $d-1$ dimensional one defined by

[TABLE]

where $X_{d,n}^{i}$ is the $i$th coordinate of $X_{d,n}$, we can show the same asymptotic for $\hat{X}_{d-1,n}$ which also gives us the asymptotic of the probability that a $d$ dimensional simple random walk ever returns to the diagonal line. To make the statement precise, consider the diagonal line in $\mathbb{Z}^{d}$

[TABLE]

Define the stopping time

[TABLE]

and let

[TABLE]

be the returning probability to $l_{d}$.

Theorem A.2**.**

For $P_{d}$ defined above, we have

[TABLE]

With Theorem A.2, we further look at the probability that a $d$ dimensional simple random walk starting from some point in Trace\big{(}\overset{\nearrow}{\mathcal{P}}\big{)} will ever return to Trace\big{(}\overset{\nearrow}{\mathcal{P}}\big{)}. Note that for each point

[TABLE]

we must have either $x=0$ or there must be some $1\leq k\leq d$ and $0\leq n\leq[N/d]$ such that

[TABLE]

Thus when looking at $\hat{x}=(x^{(1)}-x^{(2)},x^{(2)}-x^{(3)}\cdots,x^{(d-1)}-x^{(d)})$ we must have either $\hat{x}=0$ or $\hat{x}=e_{d-1,i}$ for some $i=1,2,\cdots,d$. In this appendix, we will use the notation $e_{d-1,0}=0$ and let $D_{d-1}=\{e_{d-1,i}:i=0,1,\cdots,d-1\}\subset\mathbb{Z}^{d}$. One can immediately see that when simple random walk $X_{d,n}$ starting from some point in Trace\big{(}\overset{\nearrow}{\mathcal{P}}\big{)} returns to Trace\big{(}\overset{\nearrow}{\mathcal{P}}\big{)}, we must have the corresponding non simple random walk $\hat{X}_{d-1,n}$ starting from $D_{d-1}$ returns to $D_{d-1}$. Thus for any simple random walk $X_{d,n}$ starting at 0, define the stopping times $T_{d,0}=0$

[TABLE]

and

[TABLE]

for all $k\geq 2$ with the convention $\inf\{n\geq\infty\}=\infty$. And for $\hat{X}_{d-1,n}$ also starting at 0, and any $0\leq i,j\leq d-1$, define the stopping time

[TABLE]

Then it is easy to see that for any $k\geq 0$

[TABLE]

With basically similar but more complicated technique as in the proof of Theorem A.2 we have

Theorem A.3**.**

There is a $C<\infty$ such that for all $d\geq 4$,

[TABLE]

With Theorem A.3, the proof of Theorem 1.6 is imminent.

Proof of Theorem 1.6.

With (A.4) and (A.5), we can immediately have

[TABLE]

while

[TABLE]

And the proof of Theorem 1.6 is complete. ∎

A.2. Useful Facts from Calculus

In this section, we list some very basic but useful facts from calculus that we are going to use later in the proof.

① For any function $f(x)\in C(\mathbb{R})$ and any $a\in\mathbb{R}$,

[TABLE]

② for any nonnegative integers $m,n$ and

[TABLE]

we have $C_{m,n}=0$ if at least one of $m$ and $n$ is odd.

③ There is a $c>0$ such that $1-\cos(x)\geq cx^{2}$ for all $x\in[-3\pi/2,3\pi/2]$.

Proof.

Consider function

[TABLE]

It is easy to see by its Taylor expansion that $f(x)\in C^{\infty}(\mathbb{R})$. And we have $f(x)>0$ for all $x\in[-3\pi/2,3\pi/2]$. Thus let $c=\min_{x\in[-3\pi/2,3\pi/2]}\{f(x)\}>0$, we have

[TABLE]

Actually, one can easily evaluate that $c=\frac{4}{9\pi^{2}}$. ∎

④ There is some $c_{1}>0$ such that within $[-c_{1},c_{1}]$,

[TABLE]

Proof.

Again consider function

[TABLE]

It is easy to see by its Taylor expansion that $g(x)\in C^{\infty}(\mathbb{R})$. So by continuity of $g$ around 0, there must be such a $c_{1}>0$. Actually one can easily check that $c_{1}=1$ satisfies the inequality here. ∎

⑤ For any $x>0$, $\log(1+x)\leq x$.

Proof.

Note that $x=\log(x+1)$ and that

[TABLE]

at $x=0$, $x$ is linear while $\log(1+x)$ is concave. ∎

⑥ With ②, we can also have that for any $n\in\mathbb{Z}^{+}$, integers $k_{1},k_{2},\cdots,k_{n}\geq 0$ and any $a_{1},a_{2},\cdots,a_{n}\in R$, suppose

[TABLE]

is a odd number. We always have

[TABLE]

Proof.

For each $i$, note that

[TABLE]

Thus we can expand $\prod_{i=1}^{n}\cos^{k_{i}}(\theta-a_{i})$ as a binomial of $\cos(\theta)$ and $\sin(\theta)$ and get

[TABLE]

Then look at each term in the summation on the right hand side of (A.7). For any $j$ one can immediately have one between $j$ and $K-j$ must be odd since $K$ is odd. Thus ② finishes the proof. ∎

A.3. Returning Probability to 0

In this section we prove Theorem A.1. It is well known (e.g. [2]) that

[TABLE]

where $G_{d}(\cdot)$ is the $d$ dimensional Green’s function. I.e.,

[TABLE]

where

[TABLE]

is the character function of $d$ dimensional simple random walk. See Section 4.1 of [2] for details. Since

[TABLE]

to prove the asymptotics we want, it suffices to show that

[TABLE]

To show (A.9), noting that

[TABLE]

we will concentrate on studying this integration. Since

[TABLE]

for all $x\not=1$, we have

[TABLE]

For the first integration in (A.11), we have for any $i=1,2,\cdots,d$, by ②

[TABLE]

So

[TABLE]

Then for the second integration in (A.11), we have

[TABLE]

Note that for any $i\not=j$,

[TABLE]

and that

[TABLE]

So

[TABLE]

Moreover, note that

[TABLE]

Thus

[TABLE]

Combining (A.11) and (A.12)-(A.14) , we have

[TABLE]

Thus to have the desired asymptotic, it is sufficient to show that

[TABLE]

To show A.16, we do not know how to do it with only calculus. However, since there are terms in the integration which look similar to sampled mean of random variables, we will construct a probability model and use large deviation technique to solve the problem.

Let $X_{1},X_{2},\cdots,X_{d}$ be i.i.d. uniform random variables on $[-\pi,\pi]$, and let $Y_{i}=\cos(X_{i})$, $i=1,2,\cdots,d$, which are also i.i.d random variables. Moreover, one can immediately check that $Y_{i}\in[-1,1]$ and

[TABLE]

and

[TABLE]

Then the sampled mean

[TABLE]

and let $Z_{d}$ be the nonnegative random variable

[TABLE]

Then according to our construction and the definition of $\mathcal{E}_{d}$, we have

[TABLE]

To bound (A.17), define the event $A_{d}=\{|\bar{Y}_{d}|\leq d^{-0.4}\}$, then for any $d\geq 6$,

[TABLE]

Then define the event

[TABLE]

We can further have

[TABLE]

Note that the first term in (A.18) is already of order $o(d^{-1})$. In order to control the second term, we first have that whenever $\bar{Y}_{d}\not=0$,

[TABLE]

Then according to ③, we have

[TABLE]

which implies that

[TABLE]

On the other hand, noting that we have $d^{-0.4}$ which is of a higher order than the central limit scaling $d^{-0.5}$. Cramér’s Theorem can be used to give an exponentially small term on $P(A_{d}^{c})$. For the self-completeness of this paper we also include details of the large deviation argument used here. Note that

[TABLE]

and that $E[\bar{Y}_{d}]=E[-\bar{Y}_{d}]=0$, and that $\hbox{var}\,(\bar{Y}_{d})=\hbox{var}\,(-\bar{Y}_{d})=1/2$, we can without loss of generality concentrate on controlling the first probability. The calculations we have below will work exactly the same on the second one if we take $-\bar{Y}_{d}$ as our random variable. Note that for any $\theta>0$

[TABLE]

And by Markov inequality,

[TABLE]

for any $\theta>0$. Noting that $\bar{Y}_{d}$ is the sampled mean of i.i.d. sequence $Y_{1},\cdots,Y_{d}$, one can rewrite the inequality above as

[TABLE]

Let $\theta_{1}=\theta/d$, then the inequality above can be further simplified as

[TABLE]

which holds true for all $d$ and any $\theta_{1}>0$. Now let $\theta_{1}=d^{-0.4}$. Then we have that for all $d\geq 2$

[TABLE]

which by ④ implies

[TABLE]

Then by ⑤ we can further have

[TABLE]

Combining (A.20) and (A.22),

[TABLE]

Finally, for the last term in (A.18), noting that the upper bound of $Z_{d}$ in (A.19) and the definition of $B_{d}$, we have

[TABLE]

where $B_{2,d}(0,1/d)$ is the $L_{2}$ ball in $\mathbb{R}^{d}$ centered at 0 with radius $1/d$. For the integral in (A.24), use the $d$ dimensional spherical coordinates

[TABLE]

where $r\geq 0$, $\theta_{i}\in[0,\pi]$ for $i=1,2,\cdots,d-2$, and $\theta_{d-1}\in[0,2\pi]$. Then we have

[TABLE]

Combining (A.24) and (A.25), we have

[TABLE]

Thus in (A.23) and (A.26), we have shown that all terms in (A.18) is of order $o(d^{-1})$ and the proof of Theorem A.1 is complete. ∎

A.4. Returning Probability to the Diagonal Line

In this section we prove Theorem A.2. Recalling that

[TABLE]

we have $X_{d,n}\in l_{d}$ if and only if $X_{d,n}^{1}=X_{d,n}^{2}=\cdots=X_{d,n}^{d}$, which in turns is equivalent to $\hat{X}_{d-1,n}=0$. And for the new process $\hat{X}_{d-1,n}$, one can easily check that it also forms a $d-1$ dimensional random walk with transition probability

[TABLE]

so that $\hat{X}_{d-1,n}$ also forms a finite range symmetric random walk. Moreover, the characteristic function of $\hat{X}_{d-1,n}$ is given by

[TABLE]

And we also have

[TABLE]

together with

[TABLE]

where $\hat{G}_{d-1}(\cdot)$ is the Green’s function for $\hat{X}_{d-1,n}$. I.e.,

[TABLE]

Then again we only need to show that

[TABLE]

Moreover, using exactly the same embedded random walk argument as in Lemma 1 of [4] on $X_{d,n}$ and $\tau_{d,l_{d}}$, one can immediately have $P_{d+1}\leq P_{d}$, which is also equivalent to $\hat{G}_{d}(0)\leq\hat{G}_{d-1}(0)$. So in order to show (A.29), we can without loss of generality concentrate on even $d$’s.

Then again we have

[TABLE]

Note that by ① and ②, for any $i=1,\cdots,d-2$,

[TABLE]

[TABLE]

and

[TABLE]

It is easy to see the first 3 integration terms will be exactly the same as in Section A.3. I.e.,

[TABLE]

and

[TABLE]

Thus, we again have

[TABLE]

And we only need to show that for sufficiently large even $d$

[TABLE]

To show (A.32), we again rewrite the integral above into the expectation of some function of a sequence of i.i.d. random variables. Let $\hat{X}_{1},\hat{X}_{2},\cdots,\hat{X}_{d-1}$ be i.i.d. uniform random variables on $[-\pi,\pi]$, we can define

[TABLE]

and

[TABLE]

Then according to our construction and the definition of $\mathcal{\hat{E}}_{d}$, we have

[TABLE]

Again let event $\hat{A}_{d-1}=\{|\hat{Y}_{d-1}|\leq d^{-0.4}\}$, then for any $d\geq 6$,

[TABLE]

Then let

[TABLE]

We can further have

[TABLE]

To control the third term in (A.34), note that for any $d\geq 3$, and any $i=1,2,\cdots,d-2$, within the event $\hat{B}_{d-1}$,

[TABLE]

Thus within the event $\hat{B}_{d-1}\cap\{\hat{Y}_{d-1}\not=1\}$, we have by ③

[TABLE]

Moreover, for any $(x_{1},x_{2},\cdots,x_{d-1})\in\mathbb{R}^{d}$, we have

[TABLE]

where $\sigma(d-1,1)$ is the smallest singular value of $d-1$ by $d-1$ Jordan block with $\lambda=1$. In [1] it has been proved that

[TABLE]

Thus we have

[TABLE]

Combining (A.35) and (A.36) gives us

[TABLE]

which implies that

[TABLE]

Using exactly the same argument of spherical coordinates, we have

[TABLE]

Then for the second term $P(\hat{A}^{c}_{d-1})\max_{\omega\in\hat{B}_{d-1}^{c}}\{\hat{Z}_{d-1}(\omega)\}$, we first control the probability $P(\hat{A}^{c}_{d-1})$ for sufficiently large even number $d=2n$. Note that

[TABLE]

and that

[TABLE]

So we have for sufficiently large even number $d=2n$

[TABLE]

and

[TABLE]

Moreover note that for $d=2n$ we have

[TABLE]

where

[TABLE]

and

[TABLE]

Noting that $\hat{Y}_{1,d-1}$ and $\hat{Y}_{2,d-1}$ are again sampled means of i.i.d. random variables with expectation 0 and variance $1/2$. Although now we have $\hat{Y}_{1,d-1}$ and $\hat{Y}_{2,d-1}$ are correlated, we can still have the upper bound

[TABLE]

and

[TABLE]

Apply Cramér’s Theorem on $\hat{Y}_{1,d-1}$ and $\hat{Y}_{2,d-1}$, we again have that there is some $u,U\in(0,\infty)$ (actually we can use $u=1/16$ and $U=2$) such that

[TABLE]

Lastly for $\max_{\omega\in\hat{B}_{d-1}^{c}}\{\hat{Z}_{d-1}(\omega)\}$, note that the range of $\hat{X}_{i+1}-\hat{X}_{i}$ is $[-2\pi,2\pi]$ which is no longer a subset of $[-3\pi/2,3\pi/2]$, we will not be able to use ③ directly to find an upper bound. to overcome this issue, we have the following lemma:

Lemma A.4**.**

For any $d$, consider the following two subsets of $\mathbb{R}^{d-1}$:

[TABLE]

and

[TABLE]

where $c$ is the constant in ③. Then there is some $d_{0}<\infty$ such that for all $d\geq d_{0}$, $D_{1}(d)\subseteq D_{2}(d)$.

Proof.

Let $d_{0}$ be a positive integer such that $\sqrt{c}d_{0}^{2}>1$. Then for any $d\geq d_{0}$ and any $(x_{1},\cdots,x_{d-1})\in D_{1}(d)$. By the definition of $D_{1}(d)$ and the fact that $x_{1}\in[-\pi,\pi]$, we must have

[TABLE]

which implies that

[TABLE]

Now suppose there is a $(x_{1},\cdots,x_{d-1})\in D_{1}(d)\cap D_{2}(d)^{c}$. Let $k=\inf\{i:\ |x_{i}|>\frac{i}{\sqrt{c}d^{3}}\}$. Then (A.41) ensures that $k>1$. Then for $x_{k-1}$,

[TABLE]

Thus we must have $|x_{k-1}-x_{k}|\leq 3\pi/2$, which gives that

[TABLE]

And now we have a contradiction. ∎

Moreover, we have that for any $(x_{1},\cdots,x_{d-1})\in D_{2}(d)$,

[TABLE]

Thus there is another $d_{1}<\infty$ such that for all $d\geq d_{1}$,

[TABLE]

This means for any $d\geq d_{1}$, and any $(x_{1},\cdots,x_{d-1})\in B_{2,d}(0,1/d)^{c}$,

[TABLE]

which gives us

[TABLE]

Thus combine (A.40) and (A.44) we finally have

[TABLE]

and the proof of Theorem A.2 is complete. ∎

A.5. Proof of Theorem A.3

With the asymptotic of the return of $\{\hat{X}_{d-1,n}\}_{n=0}^{\infty}$ obtained in the previous section, we will be able to use a similar but more complicate argument to show the same asymptotic for $\{\hat{X}_{d-1,n}\}_{n=0}^{\infty}$ to return to the set $D_{d-1}$. First using again exactly the same embedded random walk argument as in Lemma 1 of [4], it is easy to note that each time $\{\hat{X}_{d-1,n}\}_{n=0}^{\infty}$ returns to $D_{d-1}$ is also a time when $\{\hat{X}_{d-2,n}\}_{n=0}^{\infty}$, the embedded Markov chain tracking the changes of the first $d-2$ coordinates of $\{\hat{X}_{d-1,n}\}_{n=0}^{\infty}$, which is also a $d-2$ dimensional version of the non simple random walk of interest, returns to $D_{d-2}$. This implies

[TABLE]

and we can without loss of generality again concentrate on even numbers of $d$’s. Then for each $i$, one can immediately have

[TABLE]

Thus, in order to prove Theorem A.3, it is sufficient to show that for all sufficiently large even $d$’s, there is a $C<\infty$ such that for any $0\leq i\leq d-1$

[TABLE]

Then for any $0\leq i\not=j\leq d-1$, by strong Markov property

[TABLE]

and

[TABLE]

In Theorem A.2, we have already proved that $\hat{G}_{d-1}(0)=1+(2d)^{-1}+o(d^{-1})$. Thus now it is sufficient to show that for any $i$

[TABLE]

For any $0\leq i,j\leq d-1$, we have

[TABLE]

Thus, we will concentrate on controlling

[TABLE]

with $0\leq i<j\leq d-1$. Using the same technique as in the proof of Theorem A.2, and noting that

[TABLE]

we first have

[TABLE]

and we call

[TABLE]

to be the tail term. For any $0\leq i\not=j\leq d-1$, let $d(i,j)$ be their distance up to mod$(d)$. I.e.,

[TABLE]

The reason we want to have the distance under mod$(d)$ is that our non simple random walk $\{\hat{X}_{d-1,n}\}_{n=0}^{\infty}$ has some “periodic boundary condition” where we need one transition to move from $e_{d-1,d-1}$ to $e_{d-1,0}$. Then we have the following lemma which implies that for all but a finite number of $(i,j)$’s, the tail term $\mathcal{\hat{E}}^{(i,j)}_{d}$ is actually all we get for $G_{d-1}^{(i,j)}$.

Lemma A.5**.**

For any $k\in\mathbb{Z}^{+}$ and any $0\leq i\not=j\leq d-1$ such that $d(i,j)>k$,

[TABLE]

Proof.

By symmetry we can without generality assume that $j>i$. Recalling that

[TABLE]

we have

[TABLE]

where we use the convention that $\theta_{0}=\theta_{d}=0$. For each term in the summation above, it is easy to see that there is some nonnegative integers $k_{0},\cdots,k_{d-1}$ with $\sum_{h=0}^{d-1}k_{h}=k$ such that we can rewrite the term as

[TABLE]

Thus, it is sufficient to show that for any nonnegative integers $k_{0},\cdots,k_{d-1}$ with $\sum_{h=0}^{d-1}k_{h}=k$

[TABLE]

First, if $i=0$ then we have $j>k$ and $d-j>k$. Thus we can separate the product in (A.52) as

[TABLE]

where

[TABLE]

Thus $\Pi[0:j-1]$ is a product of $j$ terms while $\Pi[j:d-1]$ is a product of $d-j$ terms. Note that

[TABLE]

If $k_{j-1}+k_{j}$ is an even number, integrate over $\theta_{j}$ and ⑥ gives us (A.53). If $k_{j-1}+k_{j}$ is odd, without loss of generality we can assume $k_{j-1}$ is odd. Noting that

[TABLE]

by the pigeon hole principle we must have at least one of those $k_{h}$’s to be zero, which is even. Thus, let $h_{0}=\sup_{h\leq j-1}\{k_{h}{\rm\ is\ even}\}$. Then $h_{0}\in[0,j-2]$, where we use the standard convention that $\sup\{\O\}=-\infty$ and $\inf\{\O\}=\infty$. By definition $k_{h_{0}+1}$ is odd, and thus

[TABLE]

Note that $k_{h_{0}}+k_{h_{0}+1}$ is odd, so we integrate over $\theta_{h_{0}+1}$ and ⑥ again gives us (A.53).

Symmetrically, if we have $k_{j}$ is odd, then we can look at $h_{1}=\inf_{h\geq j}\{k_{h}{\rm\ is\ even}\}$ and have $h_{1}\in[j+1,d-1]$. This in turn implies that $k_{h_{1}}+k_{h_{1}-1}$ is odd, so we integrate over $\theta_{h_{1}}$ and use ⑥. We use the same argument in the following discussions.

Similarly if $i>0$, with $d(i,j)>k$ implying $j-i>k$ as well as $d+i-j>k$, we can also have

[TABLE]

where

[TABLE]

And again note that

[TABLE]

and that

[TABLE]

So if either $k_{i-1}+k_{i}$ or $k_{j-1}+k_{j}$ is a even number,⑥ again gives us (A.53).

Now suppose both $k_{i-1}+k_{i}$ and $k_{j-1}+k_{j}$ are odd. If either $k_{i}$ or $k_{j-1}$ is odd, we can without loss of generality assume the odd one is $k_{j-1}$. Note that

[TABLE]

Let $h_{0}=\sup_{h\leq j-1}\{k_{h}{\rm\ is\ even}\}$. Then $h_{0}\in[i,j-2]$. Then again we have that $k_{h_{0}}+k_{h_{0}+1}$ is odd, so we can integrate over $\theta_{h_{0}+1}$ and ⑥ again gives us (A.53).

Otherwise, we must have both $k_{i-1}$ and $k_{j}$ are odd numbers. Again note that

[TABLE]

At least one of the $k_{h}$’s above must be 0, and let’s say again without loss of generality it is in $[0,i-1]$. Once more let $h_{0}=\sup_{h\leq i-1}\{k_{h}{\rm\ is\ even}\}$. Then $h_{0}\in[0,i-2]$, and $k_{h_{0}}+k_{h_{0}+1}$ is odd so we can once again integrate over $\theta_{h_{0}+1}$ to use ⑥ to gives us (A.53). Combining all the possible situations together, the proof of this lemma is complete. ∎

With Lemma A.5, one can immediately see that for any $0\leq i\leq d-1$ and any $j$ such that $d(i,j)\geq 6$,

[TABLE]

which immediately implies that

[TABLE]

Then recalling that in the proof Theorem A.2 we have $\hat{X}_{1},\hat{X}_{2},\cdots,\hat{X}_{d-1}$ be i.i.d. uniform random variables on $[-\pi,\pi]$ and

[TABLE]

And we define

[TABLE]

Then again we have for any $i,j$

[TABLE]

Recall the event $\hat{A}_{d-1}=\{|\hat{Y}_{d-1}|\leq d^{-0.4}\}$, then for any $d\geq 6$,

[TABLE]

Then recall

[TABLE]

We can similarly have

[TABLE]

Noting that in (A.35)-(A.43) and (A.44), we find upper bounds for $\hat{Z}_{d-1}{\mathbbm{1}}_{\hat{B}_{d-1}}$ and $\max_{\omega\in\hat{B}_{d-1}^{c}}\{\bar{Z}_{d-1}(\omega)\}$ using $1/(1-\hat{Y}_{d-1})$ which is also an upper bound for the smaller corresponding terms with $\bar{Z}_{d-1}$. Thus (A.39) and (A.45) give us the second and third term in (A.56) is also $o(d^{-2.4})$. Which implies there is a $C_{1}<\infty$ such that for sufficiently large even number $d$,

[TABLE]

whenever $d(i,j)\geq 6$, and that

[TABLE]

for all $0\leq i,j\leq d-1$. Combining the observation here with Lemma A.5, we have for sufficiently large $d$ any $i$

[TABLE]

Note that for any $n>0$, $|\{j:d(i,j)=n\}|\leq 2$. So the second term in (A.57) is just a finite summation of no more than 55 terms. When $p=1$, if $d(i,j)=0$,

[TABLE]

And if $d(i,j)=1$,

[TABLE]

For $d(i,j)\geq 2$, by Lemma A.5

[TABLE]

And for $p\geq 2$ and any $i,j$

[TABLE]

Thus we have shown that all terms in this finite summation is either 0 or $O(d^{-1})$. Take $C=28$ and the proof of Theorem 1.6 is complete. ∎

Remark A.6*.*

It is clear that the upper bound $C=28$ we find here is not precise since here we only want the right order and are actually having very weak upper bounds for those 55 terms in the summation. Actually, any $C>3/2$ will be a good upper bound for sufficiently large $d$. Among the 55 terms in the summation, one can easily see that the term $j=i$, $p=2$ and the two terms with $d(i,j)=1$, $p=1$ are the only ones $\sim d^{-1}$ and each of them is $1/(2d)+o(d^{-1})$. All the other terms are either 0 or $o(d^{-1})$. The calculation is trivial calculus but very tedious, especially for someone who is reading (or writing) this not too short paper.

Appendix B

In this appendix we prove that the monotonicity fails when considering covering probability with repetitions.

Proof of Proposition 6.2: To show the first part of Equation (6.3) and (6.4), note that $\left\{{\rm Trace}(\mathcal{P})\otimes N_{\mathcal{P}}\subseteq\{X_{n}\}_{n=0}^{\infty}\right\}$ is a subset of event $\{\tau_{A}<\infty\}$. Thus by strong Markov property and symmetry of simple random walk

[TABLE]

Similarly, note that $\left\{{\rm Trace}(\mathcal{P}^{\prime})\otimes N_{\mathcal{P}^{\prime}}\subseteq\{X_{n}\}_{n=0}^{\infty}\right\}$ is a subset of $\{\tau_{A_{1}}<\infty\}$, where $A_{1}=\mathbb{Z}^{3}\setminus\{y,w\}$

[TABLE]

where $\tau_{2,y}$ is defined in (6.2). To calculate the probability we have in (LABEL:probability_11) and (B.2), one may first note that

[TABLE]

where $G(\cdot)$ is the Green’s function of 3-dimensional simple random walk. I.e.,

[TABLE]

with

[TABLE]

Thus

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

Then for $A_{2}=\mathbb{Z}^{3}\setminus\{z\}$, we have

[TABLE]

where $G_{A_{2}}(\cdot)$ is the Green’s function for set $A_{2}$, see Section 4.6 of [2] for reference. Then by Proposition 4.6.2 of [2],

[TABLE]

which gives

[TABLE]

Similarly, for $A_{3}=\mathbb{Z}^{3}\setminus\{y,z\}$ we have

[TABLE]

where

[TABLE]

and

[TABLE]

which gives

[TABLE]

Then for $A_{4}=\mathbb{Z}^{3}\setminus\{w\}$,

[TABLE]

where

[TABLE]

Thus

[TABLE]

And for $A_{5}=\mathbb{Z}^{3}\setminus\{y\}$,

[TABLE]

where

[TABLE]

Thus

[TABLE]

And for $A_{6}=\mathbb{Z}^{3}\setminus\{z,w\}$,

[TABLE]

where

[TABLE]

and

[TABLE]

Thus we have

[TABLE]

which by symmetry also equals to $P_{o}(\tau_{z}=\tau_{A})$. Finally for $P_{o}(\tau_{o}<\tau_{y})$ and $P_{o}(\tau_{o}<\tau_{y})$, using one step argument at time 0,

[TABLE]

and

[TABLE]

So again for $A_{7}=\mathbb{Z}^{3}\setminus\{2y\}$, we have

[TABLE]

where

[TABLE]

and

[TABLE]

Thus

[TABLE]

And for $P_{o}(\tau_{2y}<\tau_{y})$, recalling that $A_{5}=\mathbb{Z}^{3}\setminus\{y\}$ we have

[TABLE]

where

[TABLE]

Thus

[TABLE]

At this point, we finally have all the variables needed calculated, apply (B.3-B.17) to (LABEL:probability_11) and (B.2), the proof of Proposition 6.2 is complete.