# Moderate Deviations for the Capacity of the Random Walk range in dimension
four

Arka Adhikari  Arka Adhikari Department of Mathematics, Stanford University,
Stanford, CA, USA<EMAIL_ADDRESS>and  Izumi Okada  Izumi Okada Department
of Mathematics and Informatics, Faculty of Science, Chiba University, Chiba
263- 8522, Japan<EMAIL_ADDRESS>
###### Abstract.

In this paper, we find a natural four dimensional analog of the moderate
deviation results for the capacity of the random walk, which corresponds to
Bass, Chen and Rosen [6] concerning the volume of the random walk range for
$d=2$. We find that the deviation statistics of the capacity of the random
walk can be related to the following constant of generalized Gagliardo-
Nirenberg inequalities,

$\inf_{f:\|\nabla f\|_{L^{2}}<\infty}\frac{\|f\|^{1/2}_{L^{2}}\|\nabla
f\|^{1/2}_{L^{2}}}{[\int_{(\mathbb{R}^{4})^{2}}f^{2}(x)G(x-y)f^{2}(y)\text{d}x\text{d}y]^{1/4}}.$

###### Key words and phrases:

moderate deviation, random walk, Brownian motion, capacity

###### 2010 Mathematics Subject Classification:

60F15,60G50

Research supported by NSF grant DMS 2102842

Research supported in part by JSPS KAKENHI Grant-in-Aid for Early-Career
Scientists (No. JP20K14329) (I.O.).

###### Contents

  1. 1 Introduction
    1. 1.1 Main results
    2. 1.2 Strategy
  2. 2 Proof of Theorem 1.1 and Corollary 1.2
    1. 2.1 Reduction to the study of mutual capacity
    2. 2.2 A priori Estimates on ${\rm Cap}({\mathcal{R}}_{n})$
  3. 3  Theorem 2.1: Large Deviations of the Cross Term
  4. 4 Controlling The Third Order Intersections: Proof of Theorem 3.3
    1. 4.1 Proof of Lemma 4.4
  5. 5  The Leading Term of $\chi$: Proof of Theorem 3.1
    1. 5.1 Introduction of the Auxiliary $TL^{\prime}$
    2. 5.2 Large Deviation Upper Bounds
      1. 5.2.1 A proof of Claim 5.4
      2. 5.2.2 The proof of Lemma 5.7
    3. 5.3 Lower Bound for the Large Deviation of $TL$
  6. A Green’s Function Estimates
    1. A.1 Discretization and analysis of the functional $H(f_{1},f_{2})$
    2. A.2 Comparison of the second moments of $\mathcal{G}_{n}$ and $\mathcal{G}_{n,\epsilon}$
    3. A.3 Additional Green’s function computations
    4. A.4 Generalized Gagliardo-Nirenberg constant

##  1\. Introduction

In this paper, we study the moderate deviation results for the capacity of the
random walk for $d=4$. Given an arbitrary set $A$ in $\mathbb{Z}^{d}$, the
capacity of $A$ is defined as follows: Let $\tau_{A}$ denote the first
positive hitting time of a finite set $A$ by a simple random walk
$(S_{m})_{m\geq 0}$ on $\mathbb{Z}^{d}$ and recall that the corresponding
(Newtonian) capacity is given for $d\geq 3$, by

$\displaystyle{\rm Cap}(A):=\sum_{x\in
A}P^{x}(\tau_{A}=\infty)=\lim_{|z|\to\infty}\frac{P^{z}(\tau_{A}<\infty)}{G_{D}(z)}.$

Here, $G_{D}$ is the Green’s function for the random walk on the lattice.

There has been much significant interest in studying the capacity of the range
of random walk in $d$-dimensions. As revealed in many other works,
understanding the capacity of the range of the random walk relates to
questions regarding the volume of a random walk or the intersection of random
walk ranges. This, in turn, has a multitude of applications in various fields.
For instance, random walk intersection estimates appear in the study of
quantum field theories [21], conformal field theories [15], and in the study
of the self-avoiding walk [9]. For a more detailed discussion, one can see the
references in [3].

In this direction, there are many works in the mathematical literature
studying the capacity. We let $R_{n}$ denote the capacity of the set of the
random walk range up to the time $n$,
$\mathcal{R}_{n}:=\\{S_{1},\ldots,S_{n}\\}$. Jain and Orey [17] proved a
strong law of large numbers,

$\lim_{n\to\infty}\frac{R_{n}}{n}=\alpha_{d},\quad\text{for }d\geq 3$

for some constant $\alpha_{d}$ depending on the dimension. More recently,
Chang showed a law of distributional convergence for $\frac{R_{n}}{\sqrt{n}}$.
If one defines Brownian capacity as,

$\text{Cap}_{B}(D):=\bigg{(}\inf\left\\{\iint
G(x-y)\mu(\text{d}x)\mu(\text{d}y):\mu(D)=1\right\\}\bigg{)}^{-1},$

and $G$ is the Green’s function for the Brownian motion, then, when $d=3$,
Chang [10] has shown that

$\frac{R_{n}}{\sqrt{n}}\to_{d}\text{Cap}_{B}(B([0,1])).$

Here, $B([0,1])$ is the image of a Brownian motion from time $0$ to $1$.

In addition, the paper [2] provides lower and upper bounds for the large
deviation of the capacity of the range of a random walk in various dimensions,
though without obtaining the optimal constant. The works [3, 4] also
established a law of large numbers and a central limit theorem for the
capacity of the range of a random walk in $\mathbb{Z}^{4}$. As a consequence
of these results, one conjectures a curious link between the behavior of the
capacity in $d$ dimensions and the self-intersection of random walks in $d-2$
dimensions. However, as of yet, no deeper mechanism found to explain these
parallels.

More recently, Dembo and the second author found such a parallel when they
wanted to understand the more detailed question of a law of iterated
logarithms for the capacity. In four dimensions, the main result of [14] was
the following. Let $\overline{R}_{n}$ be the centered capacity
$\overline{R}_{n}=R_{n}-\mathbb{E}[R_{n}]$. Then, the following estimates were
shown,

$\limsup_{n\to\infty}\frac{\overline{R}_{n}}{\frac{\pi^{2}}{8}\frac{n\log(\log(\log
n))}{(\log
n)^{2}}}=1,\quad\liminf_{n\to\infty}\frac{\overline{R}_{n}}{c_{*}\frac{n\log(\log
n)}{(\log n)^{2}}}=-1,$

for some constant $c_{*}>0$. Via subadditivity arguments, the upper tail of
the law of iterated logarithms can reduce to the computation of an explicit
limit. Namely, one can explicitly compute the asymptotics of
$\lim_{n=km\to\infty}k\mathbb{E}[R_{m}]-\mathbb{E}[R_{km}]$ for appropriately
chosen $k$ and $m$.

By contrast, the constant associated with the lower tail of the large
deviation is a far more delicate question. In the paper of Dembo and the
second author, it was only shown that the $\liminf$ exists; the value of the
constant depends on quite precise large deviation statistics of the capacity.
However, rather than being merely a technical question, the exact value of the
constant can reveal deep connections to other fields.

Indeed, much like how Chen and coauthors showed that the precise value of the
large deviation constant for the intersection of random walks was related to
the Gagliardo-Nirenberg inequality, we demonstrate here that the constant for
the large deviation of the lower tail of the capacity of the random walk range
is related to the generalized Gagliardo-Nirenberg inequality. This generalized
Gagliardo-Nirenberg inequality was key in the study of the polaron and many
other physical processes of interest [16, 20]. If we look at [16, Theorem
2.3], this inequality is derived from the Hardy-Littlewood-Sobolev inequality
and is used to study the Hartree equation. Hence, we find a new relationship
between the capacity of the random walk and the field of analysis.
Furthermore, the value of the large deviation constant for the capacity of the
random walk range should give great information on the corresponding large
deviation statistics of the capacity of the Wiener sausage.

###  1.1. Main results

In our main result, we find that the moderate deviation of $R_{n}$ for $d=4$
is related to best constant of the generalized Gagliardo-Nirenberg inequality.
Namely, it is the smallest constant $\tilde{\kappa}(4,2)$ such that the
following inequality should hold among $g$ with $\|\nabla g\|_{L^{2}}<\infty$:

$\displaystyle\left[\int_{(\mathbb{R}^{4})^{2}}g^{2}(x)G(x-y)g^{2}(y)\text{d}x\text{d}y\right]^{1/4}\leq\tilde{\kappa}(4,2)\|g\|^{1/2}_{L^{2}}\|\nabla
g\|^{1/2}_{L^{2}},$

where $G(x,y)=2^{-1}\pi^{-2}|x-y|^{-2}$ for $d=4$.

######  Theorem 1.1.

Assume $b_{n}\to\infty$ and $b_{n}=O(\log\log n)$. For $d=4$,

$\displaystyle\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(\overline{R}_{n}\leq-\frac{\lambda
n}{(\log n)^{2}}b_{n}\right)=-\frac{1}{2}I_{4}(\lambda),$

where

$\displaystyle
I_{4}(\lambda)=\frac{64}{\pi^{4}}\tilde{\kappa}(4,2)^{-4}\lambda.$

######  Corollary 1.2.

For $d=4$, almost surely,

$\liminf_{n\to\infty}\frac{\overline{R}_{n}}{h(n)}=-1,$

where

$\displaystyle h(n):=\frac{32}{\pi^{4}}\tilde{\kappa}(4,2)^{-4}\frac{n\log\log
n}{(\log n)^{2}}.$

###  1.2. Strategy

As mentioned before, to find the exact value of the constant associated with
the lower tail of the law of the iterated logarithms, one would need to first
prove a form of the large deviation principle. To do this, one would need to
have control over exponential moments of the quantity in question. Now, one
can find some control over such moments in the works of [14]. However, if one
exactly wants the constant, then these estimates have to be optimal. Even with
the rather technical bounds of [14], there were still multiple times when one
could not precisely track the exponential factor associated with the high
moments. While this is perfectly fine for proving that some law of iterated
logarithm holds, it is impossible to deduce anything about the value of the
lower tail of the law of the iterated logarithm.

Inspired by the connection between the capacity and the self-intersection, one
might try to see if there are any parallels one can draw from the proof of the
large deviation principle for the self-intersection in 2-dimensions. Indeed,
Bass, Chen, Kumagai and Rosen [6, 8] were able to establish an exact form for
the constant associated with the large deviation principle for the self-
intersection of random walks.

As observed in [6, 8], a vital tool in both these analyses is a splitting
formula. The self-intersection of a random walk can be written as the sum of
two self-intersections of the first and second half of the walks and the
mutual intersection of the first and second half. The large deviation behavior
when $d=2$ is largely determined by this mutual intersection. For the
capacity, one can perform a similar splitting with the quantity $\chi$ like in
the work [3].

For two arbitrary sets $A$ and $B$, $\chi$ is defined as,

(1.1) $\displaystyle\chi(A,B):=$ $\displaystyle\sum_{y\in A}\sum_{z\in
B}\mathbb{P}(R^{\prime}_{y}\cap(A\cup
B)=\emptyset)G_{D}(y-z)\mathbb{P}(R^{\prime}_{z}\cap B=\emptyset)+$
$\displaystyle\sum_{y\in A}\sum_{z\in B}\mathbb{P}(R^{\prime}_{y}\cap
A)=\emptyset)G_{D}(y-z)\mathbb{P}(R^{\prime}_{z}\cap(A\cup B)=\emptyset),$

where $R^{\prime}_{y}$ is an independent random walk starting at the point
$y$. To show the result, we will substitute two independent simple random walk
ranges until time $n$, $\mathcal{S}^{1}$ and $\mathcal{S}^{2}$, into $A$ and
$B$. The large deviation behavior should also be determined by this ‘mutual
capacity’, $\chi$. However, after this step, if one tries to imitate the
strategy of Bass, Chen, and Rosen [6] to analyze $\chi$, fundamental
difficulties arise at the very beginning that prevent one from proceeding
forward.

First of all, observe that each line of $\chi$, due to the probability term
$\mathbb{P}(R^{\prime}_{y}\cap(A\cup B)=\emptyset)$, is asymmetric in $A$ and
$B$. Furthermore, the same probability term couples the first and second parts
of the random walk. In general, many formulas that one would like to apply to
compute moments, such as the Feynman-Kac formula for lower bounds on the
asymptotic moments, would first require one to separate the two halves of the
random walk from each other. Usually, such a separation can be justified by
applying the Cauchy-Schwartz inequality, and, as in the works of [13] for the
cross term occurring when studying the moderate deviations of the range of a
random walk, one will not incur too much loss by performing this procedure.
This is no longer the case when one deals with an asymmetric cross-term like
$\chi$. Indeed, the key first step in trying to determine the exact constant
for the moderate deviations would be to try to identify a symmetric main term
contribution for $\chi$.

The first guess that one might have would be to show that the terms
$\mathbb{P}(R^{\prime}_{y}\cap(A\cup B)=\emptyset)$ could be replaced by the
expected value $\frac{\pi^{2}}{8\log n}$. This replacement was performed in
the papers [4, 14] in order to prove that a LIL exists or establish a CLT,
respectively. However, the moment estimates required to prove such results are
insufficiently strong to demonstrate a large deviation principle or determine
an exact constant. Indeed, the paper [2] remarked that it is possible that in
the large deviation regime, it would be more effective for the random walk to
reorganize itself into configurations such that
$\mathbb{P}(R^{\prime}_{y}\cap(A\cup B)=\emptyset)$ is far away from its
expected value of $\frac{\pi^{2}}{8\log n}$.

Indeed, since we cannot replace these probability terms with their
expectation, we have to determine the main and error terms via manipulations
that preserve the structure of these probability terms. Indeed, our main term
can be guessed to be of the form,

$\sum_{y\in A}\sum_{z\in B}\mathbb{P}(R^{\prime}_{y}\cap
A=\emptyset)G_{D}(y-z)\mathbb{P}(R^{\prime}_{z}\cap B=\emptyset).$

By decomposing $G_{D}=\tilde{G}_{D}*\tilde{G}_{D}$, the convolutional square
root of $G_{D}$, we see that we indeed have a decomposition that could split
the two sets $A$ and $B$ from each other. Namely, the quantity above can be
written as,

$\sum_{a\in\mathbb{R}^{4}}\sum_{y\in A}\mathbb{P}(R^{\prime}_{y}\cap
A=\emptyset)\tilde{G}_{D}(y-a)\sum_{z\in B}\mathbb{P}(R^{\prime}_{z}\cap
B=\emptyset)\tilde{G}_{D}(z-a).$

This term will indeed be symmetric, and one has more tools for computing the
exact value of the asymptotic moments. The full analysis of this term is given
in section 5. This main term will lead to the corresponding error term,

$\sum_{x^{1}\in\mathcal{S}^{1}}\sum_{x^{2}\in\mathcal{S}^{2}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2}=\emptyset,R^{\prime}_{x^{2}}\cap\mathcal{S}^{1}\neq\emptyset).$

The main observation is that this error term should approximately be of order
$\frac{n}{(\log n)^{3}}$. This is one $\log n$ factor less than the expected
order of the main term. One still needs to determine the value of high moments
of this error term; however, one no longer needs to care about the exact
values. Indeed, one only needs to derive an upper bound for the high moments
of this error term. Section 3 will justify the splitting of $\chi$ into its
main and error terms, while section 4 will analyze the error term. The
analysis of this error term involved multiple steps; the first step was to
represent the cumbersome
$\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2}=\emptyset,R^{\prime}_{x^{2}}\cap\mathcal{S}^{1}\neq\emptyset)$
into another term that is fit for moment computation. Afterward, we had to
carefully exploit a version of monotonicity for the non-intersection
probability $\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}=\emptyset)$ that
would allow us to justify the replacement of
$\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}=\emptyset)$ with its
expectation.

##  2\. Proof of Theorem 1.1 and Corollary 1.2

In this section, we show our main results, that is, Theorem 1.1 and Corollary
1.2. Recall that $R_{n}:={\rm Cap}(\mathcal{R}_{n})$ for
$\mathcal{R}_{n}:=\\{S_{1},\ldots,S_{n}\\}$. In the proof, we write
$f(n)\lesssim g(n)$ if there exists a constant $c>0$ such that $f(n)\leq
cg(n)$ for all $n$, and $f(n)\gtrsim g(n)$ if $g(n)\lesssim f(n)$.

###  2.1. Reduction to the study of mutual capacity

In order to determine the exact moderate deviation asymptotic for
$R_{n}-\mathbb{E}[R_{n}]$, it suffices to derive a moderate deviation for the
term $\chi$. For two random walks $\mathcal{S}^{1}$ and $\mathcal{S}^{2}$,
recall the cross-term in (1.1)

$\displaystyle\chi(\mathcal{S}^{1},\mathcal{S}^{2}):=$
$\displaystyle\sum_{x^{1}\in\mathcal{S}^{1}}\sum_{x^{2}\in\mathcal{S}^{2}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap(\mathcal{S}^{1}\cup\mathcal{S}^{2})=\emptyset)$
$\displaystyle+\sum_{x^{1}\in\mathcal{S}^{1}}\sum_{x^{2}\in\mathcal{S}^{2}}\mathbb{P}(R^{\prime}_{x^{1}}\cap(\mathcal{S}^{1}\cup\mathcal{S}^{2})=\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{1}=\emptyset).$

######  Theorem 2.1.

Consider $\chi(\mathcal{S}^{1},\mathcal{S}^{2})$ with
$\mathcal{S}^{1},\mathcal{S}^{2}$ independent random walks of duration $n$ and
let $b_{n}=O(\log\log n)$ with $\lim_{n\to\infty}b_{n}=\infty$. Then,

(2.1)
$\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(\chi\geq\lambda\frac{nb_{n}}{(\log
n)^{2}}\right)=-\frac{32}{\pi^{4}}\tilde{\kappa}(4,2)^{-4}\lambda.$

We will show it in Section 3. We will give the proof of Theorem 1.1 assuming
the above result.

###### Proof of Theorem 1.1.

Splitting the Walk

For simplicity in the presentation of the argument, we will perform
computations when $n$ is a multiple of a large power of $2$. For a complete
formalization of the argument, one can consider a continuous time random walk
rather than a discrete time random walk as in [6, Chapter 6] to derive large
deviation estimates, but the essential difference in the proofs are minimal.

First, fix a large integer $L$; we first subdivide our random walk
$\mathcal{S}$ of duration $n$ into $2^{L}$ parts over various iterations. Set
$m_{l}=n/2^{l}$ and let $\mathcal{S}^{(k),m_{l}}$ denote
$\mathcal{S}[(k-1)m_{l},km_{l}]$; namely, it is the $k$th portion of the
random walk once divided into $2^{l}$ equal parts. With this notation in hand,
we can define the cross-term,

$\Lambda_{l}=\sum_{j=1}^{2^{l}-1}\chi(\mathcal{S}^{(2j-1),m_{l}},\mathcal{S}^{(2j),m_{l}}).$

We also have the following decomposition of ${\rm Cap}({\mathcal{R}}_{n})$

$\displaystyle{\rm Cap}(\mathcal{S})=\sum_{i=1}^{2^{L}}{\rm
Cap}(\mathcal{S}^{(i),m_{L}})-\sum_{l=1}^{L}\Lambda_{l}+\epsilon_{L}.$

The error $\epsilon_{L}$ has the moment bound
$\mathbb{E}[\epsilon_{L}^{2}]=O((\log n)^{2})$ from [3, Proposition 2.3]. It
is actually better to deal with a slightly modified cross-term. Consider two
random walks, $\mathcal{S}^{1},\mathcal{S}^{2}$ of the same length $n$.
Define, as in equation (5.1) which will appear in the sequel, the modified
cross term:

$TL(\mathcal{S}^{1},\mathcal{S}^{2})=\sum_{x^{1}\in\mathcal{S}^{1}}\sum_{x^{2}\in\mathcal{S}^{2}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2}=\emptyset).$

The results of Theorem 3.3 show that for any $\epsilon>0$, we have that,

$\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(|\chi(\mathcal{S}^{(2j-1),m_{l}},\mathcal{S}^{(2j),m_{l}})-2TL(\mathcal{S}^{(2j-1),m_{l}},\mathcal{S}^{(2j),m_{l}})|\geq\epsilon\frac{nb_{n}}{(\log
n)^{2}}\right)=-\infty.$

Accordingly, it is natural to consider the modified term,

$\tilde{\Lambda}_{l}:=2\sum_{j=1}^{2^{l}-1}TL(\mathcal{S}^{(2j-1),m_{l}},\mathcal{S}^{(2j),m_{l}}).$

Furthermore, the moment bound on $\epsilon_{L}$ combined with Markov’s
inequality shows that

$\frac{1}{b_{n}}\log\mathbb{P}\left(\epsilon_{L}\geq\epsilon\frac{n}{(\log
n)^{2}}\right)\leq\frac{-\log n+6\log\log n+\log\epsilon}{b_{n}}.$

Thus,

$\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(\epsilon_{L}\geq\epsilon\frac{n}{(\log
n)^{2}}\right)=-\infty.$

Combining these facts, we see that if we fix $L$ and take $n\to\infty$, we
have that

$\displaystyle\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(-{\rm
Cap}(\mathcal{S})+\mathbb{E}[{\rm
Cap}(\mathcal{S})]\geq\lambda\frac{b_{n}n}{(\log n)^{2}}\right)$
$\displaystyle=\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(-\sum_{i=1}^{2^{L}}({\rm
Cap}(\mathcal{S}^{(i),m_{L}})-\mathbb{E}[{\rm
Cap}(\mathcal{S}^{(i),m_{L}})])+\sum_{l=1}^{L}(\tilde{\Lambda}_{l}-\mathbb{E}[\tilde{\Lambda}_{l}])\geq\lambda\frac{b_{n}n}{(\log
n)^{2}}\right).$

Note that in the previous expression, we used the fact that
$\mathbb{E}[\epsilon_{L}]$ and
$\mathbb{E}[|\Lambda_{l}-\tilde{\Lambda}_{l}|]$, would not contribute to the
expectations.

Our goal now is to show the following:

(2.2)
$\displaystyle\lim_{L\to\infty}\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(\sum_{i=1}^{2^{L}}(-{\rm
Cap}(\mathcal{S}^{(i),m_{L}})+\mathbb{E}[{\rm
Cap}(\mathcal{S}^{(i),m_{L}})])+\sum_{l=1}^{L}(\tilde{\Lambda}_{l}-\mathbb{E}[\tilde{\Lambda}_{l}])\geq\lambda\frac{b_{n}n}{(\log
n)^{2}}\right)$ $\displaystyle=-\frac{1}{2}I_{4}(\lambda).$

We will start with showing the upper bound of (2.2).

Upper Bound in (2.2): It is manifest that $\mathbb{E}[\tilde{\Lambda}_{l}]$ is
a positive number. Thus, if we only care about obtaining upper bounds on the
probability found in equation (2.2), we can drop the term
$-\mathbb{E}[\tilde{\Lambda}_{l}]$ in the computation for the upper bound. We
have,

(2.3) $\displaystyle\mathbb{P}\left(\sum_{i=1}^{2^{L}}(-{\rm
Cap}(\mathcal{S}^{(i),m_{L}})+\mathbb{E}[{\rm
Cap}(\mathcal{S}^{(i),m_{L}})])+\sum_{l=1}^{L}\tilde{\Lambda}_{l}\geq\lambda\frac{b_{n}n}{(\log
n)^{2}}\right)$ $\displaystyle\leq$
$\displaystyle\mathbb{P}\left(\sum_{i=1}^{2^{L}}(\mathbb{E}[{\rm
Cap}(\mathcal{S}^{(i),m_{L}})]-{\rm
Cap}(\mathcal{S}^{(i),m_{L}}))\geq\epsilon\frac{\lambda n}{(\log
n)^{2}}b_{n}\right)$ $\displaystyle+$
$\displaystyle\sum_{l=1}^{L}\mathbb{P}\left(\tilde{\Lambda}_{l}\geq(1-\epsilon)2^{-l}\frac{\lambda
n}{(\log n)^{2}}b_{n}\right).$

By using Lemma 2.2 and [13, Theorem 1.2.2], we can derive that

(2.4)
$\limsup_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(\sum_{i=1}^{2^{L}}(\mathbb{E}[{\rm
Cap}(\mathcal{S}^{(i),m_{L}})]-{\rm
Cap}(\mathcal{S}^{(i),m_{L}}))\geq\epsilon\frac{\lambda n}{(\log
n)^{2}}b_{n}\right)\leq-2^{L}C\epsilon.$

Now recall that $\Lambda_{l}$ is a sum of i.i.d. random variables. We can
apply our Theorem 3.1 along with [13, Theorem 1.2.2] to assert that

(2.5)
$\limsup_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(\tilde{\Lambda}_{l}\geq(1-\epsilon)2^{-l}\frac{\lambda
n}{(\log n)^{2}}b_{n}\right)\leq-\frac{1}{2}I_{4}(\lambda-\epsilon).$

If we combine equations (2.5) and (2.4) in equation (2.3), we see that,

$\displaystyle\limsup_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(\sum_{i=1}^{2^{L}}(-{\rm
Cap}(\mathcal{S}^{(i),m_{L}})+\mathbb{E}[{\rm
Cap}(\mathcal{S}^{(i),m_{L}})])+\sum_{l=1}^{L}\tilde{\Lambda}_{l}\geq\frac{\lambda
nb_{n}}{(\log n)^{2}}\right)$
$\displaystyle\leq-\min\left(2^{L}C\epsilon,\frac{1}{2}I_{4}(\lambda-\epsilon)\right).$

If we first take $L$ to $\infty$ and then $\epsilon\to 0$, we derive the
desired upper bound on the probability.

Lower bound in (2.2):

First consider the quantity $SL_{n}$ as in equation (3.4) given by,

$SL_{n}=\sum_{x^{1}\in\mathcal{S}}\sum_{x^{2}\in\mathcal{S}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}=\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}=\emptyset).$

We have that,

$\displaystyle\sum_{i=1}^{2^{L}}(-{\rm
Cap}(\mathcal{S}^{(i),m_{L}})+\mathbb{E}[{\rm
Cap}(\mathcal{S}^{(i),m_{L}})])+\sum_{l=1}^{L}(\tilde{\Lambda}_{l}-\mathbb{E}[\tilde{\Lambda}_{l}])$
$\displaystyle\geq SL_{n}+\sum_{i=1}^{2^{L}}(-{\rm
Cap}(\mathcal{S}^{(i),m_{L}})+\mathbb{E}[{\rm
Cap}(\mathcal{S}^{(i),m_{L}})])-\sum_{l=1}^{L}\mathbb{E}[\tilde{\Lambda}_{l}]$
$\displaystyle-\sum_{i=1}^{2^{L}}\sum_{x^{1},x^{2}\in\mathcal{S}^{(i),m_{L}}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}=\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}=\emptyset).$

Noting that
$\sum_{l=1}^{L}\mathbb{E}[\tilde{\Lambda}_{l}]=O\left(\frac{n}{(\log
n)^{2}}\right)$, the term $\sum_{l=1}^{L}\mathbb{E}[\tilde{\Lambda}_{l}]$ will
not contribute to the large deviation statistics to the order we are concerned
with. Thus, we have that

(2.6) $\displaystyle\mathbb{P}\left(\sum_{i=1}^{2^{L}}(-{\rm
Cap}(\mathcal{S}^{(i),m_{L}})+\mathbb{E}[{\rm
Cap}(\mathcal{S}^{(i),m_{L}})])+\sum_{l=1}^{L}\tilde{\Lambda}_{l}\geq\frac{\lambda
nb_{n}}{(\log n)^{2}}\right)$
$\displaystyle\geq\mathbb{P}\left(SL_{n}\geq\frac{(\lambda+\epsilon)nb_{n}}{(\log
n)^{2}}\right)-\mathbb{P}\left(\sum_{i=1}^{2^{L}}({\rm
Cap}(\mathcal{S}^{(i),m_{l}})-\mathbb{E}[{\rm
Cap}(\mathcal{S}^{(i),m_{l}})])\geq\frac{\epsilon nb_{n}}{2(\log
n)^{2}}\right)$
$\displaystyle-\mathbb{P}\left(\sum_{i=1}^{2^{L}}\sum_{x^{1},x^{2}\in\mathcal{S}^{(i),m_{L}}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}=\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}=\emptyset)\geq\frac{\epsilon
nb_{n}}{2(\log n)^{2}}\right).$

Now, we note that the negative quantities on the right hand side are the sum
of i.i.d random variables; the term on the last line are also of the form
$SL_{n2^{-L}}$. By using Lemma 2.2 and the result for $SL_{n}$ from Corollary
3.2 as well as [13, Theorem 1.2.2], we have that the probabilities in the last
two lines are bounded by $\exp[b_{n}(-2^{L}C\epsilon)]$ for some constant $C$.

Furthermore, the result for $SL_{n}$ from Theorem 3.2 also gives us that
$\frac{1}{b_{n}}\log P(SL_{n}\geq\frac{(\lambda+\epsilon)nb_{n}}{(\log
n)^{2}})=-\frac{1}{2}I_{4}(\lambda+\epsilon)$. Given $\epsilon$, if we first
choose $L$ such that $-2^{L}C\epsilon\ll-\frac{1}{2}I_{4}(\lambda+\epsilon)$,
we see that,

$\displaystyle\liminf_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(\sum_{i=1}^{2^{L}}(-{\rm
Cap}(\mathcal{S}^{(i),m_{l}})+\mathbb{E}[{\rm
Cap}(\mathcal{S}^{(i),m_{l}})])+\sum_{l=1}^{L}\tilde{\Lambda}_{l}\geq\frac{\lambda
nb_{n}}{(\log n)^{2}}\right)$ $\displaystyle\geq$
$\displaystyle-\frac{1}{2}I_{4}(\lambda+\epsilon).$

We can then take the limit as $L$ to $\infty$ and then $\epsilon\to 0$ to show
equation (2.2). This completes the proof of the result.

∎

We can quickly derive our corollary for the exact constant of the LIL for the
lower tail of $R_{n}-\mathbb{E}[R_{n}]$.

###### Proof of Corollary 1.2 .

This will follow by carefully applying the Borel-Cantelli lemma. The large
deviation estimates of Theorem 1.1 are used to derive the appropriate
convergence of divergence conditions. The details are the same as those found
in [13, Theorem 8.6.2]. ∎

###  2.2. A priori Estimates on ${\rm Cap}({\mathcal{R}}_{n})$

In this section, we will prove the following large deviation principle on
$R_{n}$. The following lemma will give a sufficient a-priori large deviation
estimate to bound the second term of the second line of (2.6).

######  Lemma 2.2.

Let $b_{n}=\text{O}(\log\log n)$ with $\lim_{n\to\infty}b_{n}=\infty$. There
exists some constant $C$ such that,

(2.7)
$\limsup_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(|\mathbb{E}[R_{n}]-R_{n}|\geq\frac{\lambda
n}{(\log n)^{2}}b_{n}\right)\leq-C\lambda.$

###### Proof.

We will consider proving this when $n$ is a power of $2$. By changing the
constant $C$ that appears on the right hand side of (2.7), one can use our
subdivision formula of $R_{n}$ in order to obtain estimates on general $n$ via
a binary decomposition in terms of powers of $n$.

Now, assume $n$ is power of $2$ and let $L=4\log_{2}n=4\log(\log n)$. We can
decompose $r_{n}$ iteratively $L$ times to notice that,

$R_{n}=\sum_{i=1}^{2^{L}}{\rm
Cap}(\mathcal{S}^{(i),m_{L}})-\sum_{l=1}^{L}\Lambda_{l}+\epsilon_{L},$

where we use the notation from the proof of Theorem 1.1. This time
$\epsilon_{L}$ can be shown to be of $O((\log n)^{10})$. (There will at most
$1+2+4+\ldots+2^{L}=O((\log n)^{4})$ many error terms of the form $\epsilon$
in the decomposition. Each of these error terms has moment $O((\log
n)^{2}))$.) By applying Chebyshev’s inequality, we see that the error term
$\epsilon_{L}$ provides no change to the probability at the scale $b_{n}$.
Thus, we freely drop this error term $\epsilon_{L}$ in what follows.

Bounding Upper tails of $R_{n}-\mathbb{E}[R_{n}]$

If one wants to bound the probability
$\mathbb{P}(R_{n}-\mathbb{E}[R_{n}]\geq\frac{\lambda nb_{n}}{(\log n)^{2}})$
from above, then since all the terms $\Lambda_{l}$ are positive and
$\sum_{i=1}^{L}\mathbb{E}[\Lambda_{l}]=O(\frac{n}{(\log n)^{2}})$, it suffices
to bound the probability,

$\mathbb{P}\left(\sum_{i=1}^{2^{L}}{\rm
Cap}(\mathcal{S}^{(i),m_{L}})-\mathbb{E}[{\rm
Cap}(\mathcal{S}^{(i),m_{L}})]\geq\frac{\lambda nb_{n}}{(\log n)^{2}}\right).$

Now, the sequence ${\rm Cap}(\mathcal{S}^{(i),m_{L}})-\mathbb{E}[{\rm
Cap}(\mathcal{S}^{(i),m_{L}})]$ are the sequence of i.i.d. random variables
with the property that
$\mathbb{E}\exp\left[\frac{\theta}{n}|R_{n}-\mathbb{E}[R_{n}]|\right]<\infty$.
(This is due the the fact that $R_{n}\leq n$). We can apply [6, Lemma 4.4] to
assert that there is some constant $\theta>0$ such that

$\limsup_{n\to\infty}\mathbb{E}\left[\exp\left[\frac{\theta}{2^{L/2}}\left|\sum_{i=1}^{2^{L}}\frac{2^{L}}{n}({\rm
Cap}(\mathcal{S}^{(i),m_{L}})-\mathbb{E}[{\rm
Cap}(\mathcal{S}^{(i),m_{L}})])\right|\right]\right]<\infty.$

Since $2^{L/2}\geq(\log n)^{2}$ by choice, this implies that,

$\limsup_{n\to\infty}\mathbb{E}\left[\exp\left[\theta\frac{(\log
n)^{2}}{n}\left|\sum_{i=1}^{2^{L}}({\rm
Cap}(\mathcal{S}^{(i),m_{L}})-\mathbb{E}[{\rm
Cap}(\mathcal{S}^{(i),m_{L}})])\right|\right]\right]<\infty.$

By Chebyshev’s inequality, this shows that there is some constant $C$ such
that,

(2.8)
$\limsup_{n\to\infty}\frac{1}{b_{n}}\mathbb{P}\bigg{(}\left|\sum_{i=1}^{2^{L}}({\rm
Cap}(\mathcal{S}^{(i),m_{L}})-\mathbb{E}[{\rm
Cap}(\mathcal{S}^{(i),m_{L}})])\right|\geq\frac{\lambda nb_{n}}{(\log
n)^{2}}\bigg{)}\leq-C\lambda.$

Upper Bounds on the lower tail
$\mathbb{P}(R_{n}-\mathbb{E}[R_{n}]<-\frac{\lambda nb_{n}}{(\log n)^{2}})$

Due to our control on $\left|\sum_{i=1}^{2^{L}}({\rm
Cap}(\mathcal{S}^{(i),m_{L}})-\mathbb{E}[{\rm
Cap}(\mathcal{S}^{(i),m_{L}})])\right|$ from equation (2.8). It suffices to
bound $\mathbb{P}\left(\sum_{l=1}^{L}\Lambda_{l}\geq\frac{\lambda
nb_{n}}{(\log n)^{2}}\right)$.

If we define,

$\displaystyle\alpha_{l,j}:=$
$\displaystyle\sum_{a=(2j-2)m_{l}+1}^{(2j-1)m_{l}}\sum_{b=(2j-1)m_{l}+1}^{(2j)m_{l}}\mathbb{P}(R^{\prime}_{\mathcal{S}_{a}}\cap\mathcal{S}^{(2j-1),m_{l}}=\emptyset)G_{D}(\mathcal{S}_{a}-\mathcal{S}_{b})\mathbb{P}(R^{\prime}_{\mathcal{S}_{b}}\cap\mathcal{S}^{(2j),m_{l}}=\emptyset),$

notice that we can bound,

$\chi(\mathcal{S}^{(2j-1),m_{l}},\mathcal{S}^{(2j),m_{l}})\leq\alpha_{l,j}.$

Finally, in order to show
$\limsup_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}(\sum_{l=1}^{L}\Lambda_{l}\geq\frac{\lambda{n}b_{n}}{(\log
n)^{2}})\leq-C\Lambda$ for some constant $C$. It suffices to prove that the
exponential moment,

(2.9) $\limsup_{n\to\infty}\mathbb{E}\left[\exp\left[\frac{\theta(\log
n)^{2}}{n}\sum_{l=1}^{L}\sum_{k=1}^{2^{l-1}}\alpha_{l,k}\right]\right]<\infty.$

By the consequence of Lemma 5.6, there is a parameter $\theta>0$ such that
each $\alpha_{l,k}$ has the exponential moment,

$\limsup_{n\to\infty}\mathbb{E}\left[\exp\left[\theta\frac{(\log(2^{-l}n))^{2}}{2^{-l}n}\alpha_{l,k}\right]\right]<\infty.$

Thus, we can follow the argument of [6, Theorem 5.4] from equation (5.30)
onwards to prove that the desired result (2.9). This completes the proof of
the lemma. ∎

##  3\.  Theorem 2.1: Large Deviations of the Cross Term

In this section, we provide a decomposition for $\chi$ that will give us a
proof of Theorem 2.1. Analyzing $\chi$ is not directly tractable due to the
lack of symmetry in each individual product. To deal with this issue, we can
write this in terms of the following difference,

(3.1) $\displaystyle\chi$
$\displaystyle=2\sum_{x^{1}\in\mathcal{S}^{1}}\sum_{x^{2}\in\mathcal{S}^{2}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2}=\emptyset)$
$\displaystyle-\sum_{x^{1}\in\mathcal{S}^{1}}\sum_{x^{2}\in\mathcal{S}^{2}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2}=\emptyset,R^{\prime}_{x^{2}}\cap\mathcal{S}^{1}\neq\emptyset)$
$\displaystyle-\sum_{x^{1}\in\mathcal{S}^{1}}\sum_{x^{2}\in\mathcal{S}^{2}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}=\emptyset,R^{\prime}_{x^{1}}\cap\mathcal{S}^{2}\neq\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2}=\emptyset).$

Now, in order to obtain asymptotics on $\chi$, our goal is two-fold.

  1. (1)

An upper bound on $\chi$ is found by merely considering the top line

(3.2)
$TL_{n}:=\sum_{x^{1}\in\mathcal{S}^{1}}\sum_{x^{2}\in\mathcal{S}^{2}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2}=\emptyset).$

Thus, one can obtain upper bounds on the moderate deviation statistics of
$TL_{n}$ merely from analyzing moderate deviation statistics of $TL_{n}$.

  2. (2)

Obtaining lower bounds on the moderate deviation statistics of $TL_{n}$ needs
more steps. First, one needs to show that the second line of (3.1), which we
denote as $\chi^{\prime}$ is sub-leading relative to the first line. (The
analysis of the third line would be similar to that of the second.)

(3.3)
$\chi^{\prime}:=\sum_{x^{1}\in\mathcal{S}^{1}}\sum_{x^{2}\in\mathcal{S}^{2}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2}=\emptyset,R^{\prime}_{x^{2}}\cap\mathcal{S}^{1}\neq\emptyset).$

Once this is established, lower bounds on the large deviation statistics of
$TL_{n}$ will be the same as those of $\chi$. Furthermore, observe that only
an upper bound on $\chi^{\prime}$ is necessary.

We will have two intermediate goals,

######  Theorem 3.1.

Recall $TL_{n}$ as in equation (3.2). Fix $b_{n}=O(\log\log n)$ with
$\lim_{n\to\infty}b_{n}=\infty$. We have that,

$\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(TL_{n}\geq\lambda\frac{b_{n}n}{(\log
n)^{2}}\right)=-\frac{64}{\pi^{4}}\tilde{\kappa}(4,2)^{-4}\lambda.$

We remark that by following the same proof as for the above statement, we
could obtain the following statement when dealing with the same random walk,

######  Corollary 3.2.

Let $\mathcal{S}$ be a random walk and $b_{n}=O(\log\log n)$ with
$\lim_{n\to\infty}b_{n}=\infty$. Define $SL_{n}$ as,

(3.4)
$SL_{n}:=\sum_{x^{1}\in\mathcal{S}}\sum_{x^{2}\in\mathcal{S}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}=\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}=\emptyset).$

Then, we have that,

$\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(SL_{n}\geq\lambda\frac{b_{n}n}{(\log
n)^{2}}\right)=-\frac{32}{\pi^{4}}\tilde{\kappa}(4,2)^{-4}\lambda.$

######  Theorem 3.3.

Recall $\chi^{\prime}$ as in (3.3). Fix $b_{n}=O(\log\log n)$ with
$\lim_{n\to\infty}b_{n}=\infty$. For any $\epsilon>0$, we have that,

$\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(\chi^{\prime}\geq\epsilon\frac{b_{n}n}{(\log
n)^{2}}\right)=-\infty.$

It is clear that Theorem 2.1 is a consequence of Theorems 3.1 and 3.3 along
with the decomposition (3.1). The next few sections will be devoted to proving
these theorems.

##  4\. Controlling The Third Order Intersections: Proof of Theorem 3.3

Via a path decomposition, one can see that we have the following expression
for the more complicated probability term,

(4.1)
$\displaystyle\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2}=\emptyset,R^{\prime}_{\mathcal{S}^{2}_{j}}\cap\mathcal{S}^{1}\neq\emptyset)=\sum_{x^{1}\in\mathcal{S}^{2}}G_{\mathcal{S}^{2}}(x^{2},x^{1})\mathbb{P}(R^{\prime}_{x^{1}}\cap(\mathcal{S}^{1}\cup\mathcal{S}^{2})=\emptyset).$

For any $A$, let

$G_{A}(a,b)=\sum_{t=0}^{\infty}\mathbb{P}(R^{\prime}_{a}[t]=b,R^{\prime}_{a}[0,t]\cap
A=\emptyset).$

Here $G_{\mathcal{S}^{2}}(a,b)$ is a restricted Green’s functions by
$\mathcal{S}^{2}$. It computes the total sum of all probabilities of random
walk paths from $a$ to $b$ that do not intersect $\mathcal{S}^{2}$. Namely,

$G_{\mathcal{S}^{2}}(a,b)=\sum_{t=0}^{\infty}\mathbb{P}(R^{\prime}_{a}[t]=b,R^{\prime}_{a}[0,t]\cap\mathcal{S}^{2}=\emptyset).$

The equality in (4.1) comes from a path decomposition. Namely, since
$R^{\prime}_{x^{2}}\cap\mathcal{S}^{1}\neq\emptyset$, then the random walk
$R^{\prime}_{x^{2}}$ must intersect $\mathcal{S}^{1}$ at some final point
$x^{1}$. After this point, the random walk starting from $\mathcal{S}^{1}_{k}$
must not intersect either one of $\mathcal{S}^{1}$ or $\mathcal{S}^{2}$. Then,
we can further bound

$\chi^{\prime}\leq\sum_{x^{1}_{1},x^{1}_{2}\in\mathcal{S}^{1}}\sum_{x^{2}\in\mathcal{S}^{2}}\mathbb{P}(R^{\prime}_{x^{1}_{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}_{1}-x^{2})G_{\mathcal{S}^{2}}(x^{2},x^{1}_{2})\mathbb{P}(R^{\prime}_{x^{1}_{2}}\cap\mathcal{S}^{1}=\emptyset).$

Though we have simplified the probability term in question, we are still not
ready to analyze this due to the appearance of the term $G_{\mathcal{S}^{2}}$.
We have to introduce a more sophisticated analysis in order to deal with this
term. First we fix two parameters $\beta<\alpha<1$, the specific value will be
chosen later in accordance with what is appropriate for later upper bounds. It
is important to decompose the walk $\mathcal{S}^{2}$ into appropriate
intervals of size $n^{\beta}$. We define,

$\mathcal{S}^{2}_{\beta,i}:=\mathcal{S}^{2}[(i-1)n^{\beta},in^{\beta}].$

Namely, $\mathcal{S}^{2}_{\beta,i}$ contains the range of the $i$th
$n^{\beta}$ block of the random walk. We would also define,

$\mathcal{S}^{2}_{\beta,\alpha,i}:=\mathcal{S}^{1}[(i-1/2)n^{\beta},(i+1/2)n^{\beta}].$

By adding back points when necessarily, and also using the fact that if the
set $A\subset B$ that $G_{B}(x,y)\leq G_{A}(x,y)$ for all points, we see that
we have,

$\displaystyle\chi^{\prime}\leq\chi^{\prime}_{\beta,\alpha}$
$\displaystyle:=\sum_{x^{1}_{1},x^{1}_{2}\in\mathcal{S}^{1}}\sum_{j=1}^{\lceil
n^{1-\beta}\rceil}\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,j}}\mathbb{P}(R^{\prime}_{x^{1}_{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}_{1}-x^{2})$
$\displaystyle\hskip 56.9055pt\times
G_{\mathcal{S}^{2}_{\beta,\alpha,j}}(x^{2},x^{1}_{2})\mathbb{P}(R^{\prime}_{x^{1}_{2}}\cap\mathcal{S}^{1}=\emptyset).$

One way to express the modified Green’s function
$G_{\mathcal{S}^{2}_{\beta,\alpha,i}}$ is as follows. Note that these Green’s
functions satisfy the following system of equations:

$G_{D}(x^{2}-x^{1}_{2})=\sum_{\tilde{x}\in\mathcal{S}^{2}_{\beta,\alpha,i}}^{n}G_{D}(x^{2}-\tilde{x})G_{\mathcal{S}^{2}_{\beta,\alpha,i}}(\tilde{x},x^{1}_{2}).$

If we define the matrix with size $|\mathcal{S}^{2}_{\beta,\alpha,i}|$

(4.2)
$[\mathcal{G}^{\mathcal{S}^{2}_{\beta,\alpha,i}}]_{a,b}=G_{D}(a-b),\quad\text{
for }a,b\in\mathcal{S}^{2}_{\beta,\alpha,i},$

we see that,

(4.3) $\begin{bmatrix}G_{\mathcal{S}^{2}_{\beta,\alpha,i}}(a_{1},x^{1}_{2})\\\
G_{\mathcal{S}^{2}_{\beta,\alpha,i}}(a_{2},x^{1}_{2})\\\ \vdots\\\
G_{\mathcal{S}^{2}}(a_{|\mathcal{S}^{2}_{\beta,\alpha,i}|},x^{1}_{2})\end{bmatrix}=(\mathcal{G}^{\mathcal{S}^{2}_{\beta,\alpha,i}})^{-1}\begin{bmatrix}G_{D}(a_{1},\mathcal{S}^{1}_{k})\\\
G_{D}(a_{2},\mathcal{S}^{1}_{k})\\\ \vdots\\\
G_{D}(a_{|\mathcal{S}^{2}_{\beta,\alpha,i}|},\mathcal{S}^{1}_{k})\end{bmatrix},$

where $a$ varies over all the points in $\mathcal{S}^{2}_{\beta,\alpha,i}$.

The analysis of the matrix inverse will depend on the distance between the
points in $x^{1}_{2}$ and the set $\mathcal{S}^{2}_{\beta,\alpha,i}$. Observe
that if $x^{1}_{2}$ were far away from the set
$\mathcal{S}^{2}_{\beta,\alpha,i}$, then the terms in the vector on the right
hand side of equation (4.3) would approximately be constant. Furthermore, it
is also rather unlikely that the $x^{1}_{2}$ would be close to the set
$\mathcal{S}^{2}_{\beta,\alpha,i}$. Following this intuition, we divide the
the points $x^{1}_{2}$ into two categories,

  1. (1)

In category 1, the point $x^{1}_{2}$ is of distance at least
$\sqrt{n}^{1-\delta}$ away from all the points in
$\mathcal{S}^{2}_{\beta,\alpha,j}$.

  2. (2)

In category 2, the point $x^{1}_{2}$ is not of distance at least
$\sqrt{n}^{1-\delta}$ away from some point in
$\mathcal{S}^{2}_{\beta,\alpha,j}$

If we are in the first category, we have a superior analysis, as will be
illustrated by the following manipulations. Indeed, assume that the point
$x^{1}_{2}$ is of distance at least $\sqrt{n}^{1-\delta}$ from all points in
$\mathcal{S}^{2}_{\beta,\alpha,j}$. Now, let $a$ and $b$ be two points in
$\mathcal{S}^{1}_{\beta,\alpha,j}$. Then, we must have

(4.4)
$\displaystyle\left|G_{D}(a-x^{1}_{2})-G_{D}(b-x^{1}_{2})\right|\lesssim\left|\frac{1}{\|a-x^{1}_{2}\|^{2}}-\frac{1}{\|b-x^{1}_{2}\|^{2}}\right|+\frac{1}{\|a-x^{1}_{2}\|^{4}}+\frac{1}{\|b-x^{1}_{2}\|^{4}}$
$\displaystyle\lesssim\frac{\|a-b\|}{\|a-x^{1}_{2}\|^{3}}+2\left(\frac{1}{\sqrt{n}^{1-\delta}}\right)^{4}\lesssim\frac{n^{\alpha}}{\left(\sqrt{n}\right)^{3-3\delta}}.$

Note that here, we have applied the estimates of [19, Theorem 4.3.1]. Then we
use the fact that $|\mathcal{S}^{1}_{a}-\mathcal{S}^{1}_{b}|$ is less than
$n^{\alpha}$ in the neighborhood $\mathcal{S}^{2}(\alpha,j)$ and our
assumption that
$||\mathcal{S}^{2}_{j}-\mathcal{S}^{1}_{a}||\geq\left(\sqrt{n}\right)^{1-\delta}$.
In this case, we can further write the inverse formula as in equation (4.3)
as,

(4.5) $\displaystyle\begin{bmatrix}G_{\mathcal{S}^{2}}(a_{1},x^{1}_{2})\\\
\vdots\\\
G_{\mathcal{S}^{2}}(a_{|\mathcal{S}^{2}_{\beta,\alpha,j}|},x^{1}_{2})\end{bmatrix}=(\mathcal{G}^{\mathcal{S}^{2}_{\beta,\alpha,j}})^{-1}\begin{bmatrix}G_{D}(a_{1}-x^{1}_{2})\\\
\vdots\\\
G_{D}(a_{|\mathcal{S}^{2}_{\beta,\alpha,j}|}-x^{1}_{2})\end{bmatrix}$
$\displaystyle=(\mathcal{G}^{\mathcal{S}^{2}_{\beta,\alpha,j}})^{-1}\begin{bmatrix}1\\\
\vdots\\\ 1\end{bmatrix}\times
G_{D}(a_{1}-x^{1}_{2})-(\mathcal{G}^{\mathcal{S}^{2}_{\alpha,\beta,j}})^{-1}\begin{bmatrix}G_{D}(a_{1}-x^{1}_{2})-G_{D}(a_{1}-x^{1}_{2})\\\
\vdots\\\
G_{D}(a_{|\mathcal{S}^{2}_{\beta,\alpha,j}|}-x^{1}_{2})-G_{D}(a_{1}-x^{1}_{2})\end{bmatrix}$
$\displaystyle=\begin{bmatrix}\mathbb{P}(R^{\prime}_{a_{1}}\cap\mathcal{S}^{2}_{\beta,\alpha,j}=\emptyset)\\\
\vdots\\\
\mathbb{P}(R^{\prime}_{a_{|\mathcal{S}^{2}_{\beta,\alpha,j}|}}\cap\mathcal{S}^{2}_{\beta,\alpha,j}=\emptyset)\end{bmatrix}\times
G_{D}(a_{1}-x^{1}_{2})$ $\displaystyle\hskip
56.9055pt-(\mathcal{G}^{\mathcal{S}^{2}_{\beta,\alpha,j}})^{-1}\begin{bmatrix}G_{D}(a_{1}-x^{1}_{2})-G_{D}(a_{1}-x^{1}_{2})\\\
\vdots\\\
G_{D}(a_{|\mathcal{S}^{2}_{\beta,\alpha,j}|}-x^{1}_{2})-G_{D}(a_{1}-x^{1}_{2})\end{bmatrix}.$

For simplicity, we will write the entries of the second vector above as
$E_{a,x^{1}_{2}}$ when dealing with the coordinate corresponding to point $a$
in $\mathcal{S}^{2}_{\beta,\alpha,j}$. Thus, we have the following
representation of $\chi^{\prime}_{\beta,\alpha}$. In what follows, we let
$\mathcal{I}(x^{1}_{2},j)$ be the indicator function of the event

$\mathcal{I}(x^{1}_{2},j):=\mathbbm{1}[\text{dist}(x^{1}_{2},\mathcal{S}^{2}_{\beta,j})\geq\sqrt{n}^{1-\delta}].$

For each set $\mathcal{S}^{2}_{\beta,j}$ choose a point $\tilde{x}^{2}_{j}$.
This point will serve as the central point used in the decomposition used in
equation (4.5):

(4.6) $\displaystyle\chi^{\prime}_{\beta,\alpha}$
$\displaystyle=\sum_{x^{1}_{1},x^{1}_{2}\in\mathcal{S}^{1}}\sum_{j=1}^{\lceil
n^{1-\beta}\rceil}\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,j}}\mathcal{I}(x^{1}_{2},j)\mathcal{I}(x^{1}_{1},j)\mathbb{P}(R^{\prime}_{x^{1}_{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}_{1}-x^{2})$
$\displaystyle\hskip
56.9055pt\times\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2}_{\beta,\alpha,j}=\emptyset)G_{D}(\tilde{x}^{2}_{j}-x^{1}_{2})\mathbb{P}(R^{\prime}_{x^{1}_{2}}\cap\mathcal{S}^{1}=\emptyset)$
$\displaystyle+\sum_{x^{1}_{1},x^{1}_{2}\in\mathcal{S}^{1}}\sum_{j=1}^{\lceil
n^{1-\beta}\rceil}\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,j}}\mathcal{I}(x^{1}_{2},j)\mathcal{I}(x^{1}_{1},j)\mathbb{P}(R^{\prime}_{x^{1}_{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}_{1}-\tilde{x}^{2}_{j})$
$\displaystyle\hskip 56.9055pt\times
E_{x^{2},x^{1}_{2}}\mathbb{P}(R^{\prime}_{x^{1}_{2}}\cap\mathcal{S}^{1}=\emptyset)$
$\displaystyle+\sum_{x^{1}_{1},x^{1}_{2}\in\mathcal{S}^{2}}\sum_{j=1}^{\lceil
n^{1-\beta}\rceil}\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,j}}\mathcal{I}(x^{1}_{2},j)\mathcal{I}(x^{1}_{1},j)\mathbb{P}(R^{\prime}_{x^{1}_{1}}\cap\mathcal{S}^{1}=\emptyset)$
$\displaystyle\hskip
56.9055pt\times[G_{D}(x^{1}_{1}-x^{2})-G_{D}(x^{1}_{1}-\tilde{x}^{2}_{j})]E_{x^{2},x^{1}_{2}}\mathbb{P}(R^{\prime}_{x^{1}_{2}}\cap\mathcal{S}^{1}=\emptyset)$
$\displaystyle+\sum_{x^{1}_{1},x^{1}_{2}\in\mathcal{S}^{2}}\sum_{j=1}^{\lceil
n^{1-\beta}\rceil}\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,j}}[1-\mathcal{I}(x^{1}_{2},j)\mathcal{I}(x^{1}_{1},j)]\mathbb{P}(R^{\prime}_{x^{1}_{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}_{1}-x^{2})$
$\displaystyle\hskip 56.9055pt\times
G_{\mathcal{S}^{2}_{\beta,\alpha,j}}(x^{2},x^{1}_{2})\mathbb{P}(R^{\prime}_{x^{1}_{2}}\cap\mathcal{S}^{1}=\emptyset)$
$\displaystyle=:$ $\displaystyle
MT_{n}+\mathcal{E}_{1}+\mathcal{E}_{2}+\mathcal{E}_{3}.$

The analysis of $\chi^{\prime}_{\beta,\alpha}$ now devolves into the following
three lemmas.

######  Lemma 4.1.

Fix a number $m$. There exists a constant depending only on $m$(and not on
$n$) such that, if we define,

(4.7) $\displaystyle MT_{n}:$
$\displaystyle=\sum_{x^{1}_{1},x^{1}_{2}\in\mathcal{S}^{1}}\sum_{j=1}^{\lceil
n^{1-\beta}\rceil}\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,j}}\mathcal{I}(x^{1}_{2},j)\mathcal{I}(x^{1}_{1},j)\mathbb{P}(R^{\prime}_{x^{1}_{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}_{1}-x^{2})$
$\displaystyle\hskip
56.9055pt\times\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2}_{\beta,\alpha,j}=\emptyset)G_{D}(\tilde{x}^{2}_{j}-x^{1}_{2})\mathbb{P}(R^{\prime}_{x^{1}_{2}}\cap\mathcal{S}^{1}=\emptyset),$

then,

$\mathbb{E}[(MT_{n})^{m}]\leq C_{m}\frac{n^{m}(\log\log n)^{2m}}{(\log
n)^{3m}}.$

By Markov’s inequality, we obtain the following as a consequence,

######  Corollary 4.2.

Recall $MT_{n}$ and fix any $\epsilon\geq 0$. If $b_{n}=O(\log\log n)$ with
$\lim_{n\to\infty}b_{n}=\infty$, we see that as $\lim_{n\to\infty}$, we have,

$\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(MT_{n}\geq\epsilon\frac{b_{n}n}{(\log
n)^{2}}\right)=-\infty.$

###### Proof.

By applying Markov’s inequality to (4.7) for some fixed power
$\mathbb{E}[(MT_{n})^{m}]$ , we can derive that,

$\displaystyle\limsup_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(MT_{n}\geq\epsilon\frac{b_{n}n}{(\log
n)^{2}}\right)$
$\displaystyle\leq\limsup_{n\to\infty}-\frac{m}{b_{n}}\left[\log\epsilon-\frac{\log
C_{m}}{m}+\log\log n-2\log\log\log n\right]\leq-m.$

The quantity above will go to $\infty$ as one takes $m$ to $\infty$.

∎

If one could freely replace the probability of non-intersection of the random
walks appearing in the above expression of (4.7) with $O\left(\frac{1}{\log
n}\right)$, then, Lemma 4.1 is a consequence of the following lemma and
repeating the proof of Claim 5.8.

######  Lemma 4.3.

Consider the following quantity,

$\displaystyle MT_{n}^{\prime}:$
$\displaystyle=\sum_{x^{1}_{1},x^{1}_{2}\in\mathcal{S}^{1}}\sum_{x^{2}\in\mathcal{S}^{2}}G_{D}(x^{1}_{1}-x^{2})G_{D}(x^{2}-x^{1}_{2}).$

There exists some constant $C_{m}$(not depending on $n$) such that

$\mathbb{E}[(MT_{n})^{m}]\leq C_{m}n^{m}(\log\log n)^{2m}.$

###### Proof.

Let $\alpha\geq 10m$. First, we show that for any $y_{i}\in{\mathbb{Z}}^{d}$
with $\inf_{1\leq i\leq 2m}|y_{i}|\geq n^{1/2}(\log n)^{-\alpha}$,

(4.8)
$\displaystyle\mathbb{E}\left[\sum_{k_{1},\ldots,k_{2m}=1}^{n}\prod_{i=1}^{2m}G_{D}(\mathcal{S}^{1}_{k_{i}}-y_{i})\right]\leq
C_{m,\alpha}(\log\log n)^{2m}.$

Let $A_{i}=\\{|\mathcal{S}^{1}_{k_{i-1}}-y_{i}|\geq n^{1/2}(\log
n)^{-2\alpha}\\}$. Indeed,

$\displaystyle\sum_{1\leq k_{1}\leq\ldots\leq k_{2m}\leq
n}\prod_{i=1}^{2m}G_{D}(\mathcal{S}^{1}_{k_{i}}-y_{i})$ $\displaystyle=$
$\displaystyle\sum_{1\leq k_{1}\leq\ldots\leq k_{2m}\leq
n}G_{D}(\mathcal{S}^{1}_{k_{2m}}-\mathcal{S}^{1}_{k_{2m-1}}+\mathcal{S}^{1}_{k_{2m-1}}-y_{2m})\prod_{i=1}^{2m-1}G_{D}(\mathcal{S}^{1}_{k_{i}}-y_{i})$
$\displaystyle=$ $\displaystyle\sum_{1\leq k_{1}\leq\ldots\leq k_{2m}\leq
n}G_{D}(\mathcal{S}^{1}_{k_{2m}}-\mathcal{S}^{1}_{k_{2m-1}}+\mathcal{S}^{1}_{k_{2m-1}}-y_{2m})1_{A_{2m}}\prod_{i=1}^{2m-1}G_{D}(\mathcal{S}^{1}_{k_{i}}-y_{i})$
$\displaystyle+$ $\displaystyle\sum_{1\leq k_{1}\leq\ldots\leq k_{2m}\leq
n}G_{D}(\mathcal{S}^{1}_{k_{2m}}-\mathcal{S}^{1}_{k_{2m-1}}+\mathcal{S}^{1}_{k_{2m-1}}-y_{2m})1_{A_{2m}^{c}}\prod_{i=1}^{2m-1}G_{D}(\mathcal{S}^{1}_{k_{i}}-y_{i}).$

Note that for $i\geq 1$, recall [18, Thm. 1.2.1] that for some $C$ finite and
any $x\in{\mathbb{Z}}^{4}$,

(4.9) $\displaystyle\mathbb{P}(\mathcal{S}^{1}_{i}=x)$ $\displaystyle\leq
Ci^{-2}\Big{[}e^{-2|x|^{2}/i}+(|x|^{2}\vee i)^{-1}\Big{]}.$

Hence if $|y|\geq n^{1/2}(\log n)^{-2\alpha}$,

$\displaystyle\mathbb{E}\left[\sum_{1\leq k\leq
n}G_{D}(\mathcal{S}^{1}_{k}-y)\right]\lesssim\log\log n.$

Then,

$\displaystyle\mathbb{E}\left[\sum_{1\leq k_{1}\leq\ldots\leq k_{2m}\leq
n}G_{D}(\mathcal{S}^{1}_{k_{2m}}-\mathcal{S}^{1}_{k_{2m-1}}+\mathcal{S}^{1}_{k_{2m-1}}-y_{2m})1_{A_{2m}}\prod_{i=1}^{2m-1}G_{D}(\mathcal{S}^{1}_{k_{i}}-y_{i})\right]$
$\displaystyle\lesssim$ $\displaystyle(\log\log n)\mathbb{E}\left[\sum_{1\leq
k_{1}\leq\ldots\leq k_{2m-1}\leq
n}\prod_{i=1}^{2m-1}G_{D}(\mathcal{S}^{1}_{k_{i}}-y_{i})\right]$

and by Lemma A.15,

$\displaystyle\mathbb{E}\left[\sum_{1\leq k_{1}\leq\ldots\leq k_{2m}\leq
n}G_{D}(\mathcal{S}^{1}_{k_{2m}}-\mathcal{S}^{1}_{k_{2m-1}}+\mathcal{S}^{1}_{k_{2m-1}}-y_{2m})1_{A_{2m}^{c}}\prod_{i=1}^{2m-1}G_{D}(\mathcal{S}^{1}_{k_{i}}-y_{i})\right]$
$\displaystyle\lesssim$ $\displaystyle(\log n)\mathbb{E}\left[\sum_{1\leq
k_{1}\leq\ldots\leq k_{2m-1}\leq
n}1_{A_{2m}^{c}}\prod_{i=1}^{2m-1}G_{D}(\mathcal{S}^{1}_{k_{i}}-y_{i})\right]$
$\displaystyle\leq$ $\displaystyle C_{m}(\log n)^{2m}(\log n)^{-4\alpha}.$

Hence, we obtain (4.8).

In addition,

$\displaystyle\max_{y_{1},\ldots,y_{2m}}\mathbb{E}\left[\sum_{k_{1},\ldots,k_{2m}=1}^{n}\prod_{i=1}^{2m}G_{D}(\mathcal{S}^{1}_{k_{i}}-y_{i})\right]\leq
C_{m}(\log n)^{2m}.$

Moreover, if $\inf\\{i_{1},\ldots i_{m}\\}\geq n(\log n)^{-\alpha}$,

$\displaystyle\mathbb{E}[1_{D}]:=\mathbb{E}\left[\cup_{i=1}^{m}1\\{|\mathcal{S}^{2}_{i}|\leq
n^{1/2}(\log n)^{-\alpha}\\}\right]\leq C_{m}(\log n)^{-4\alpha}$

and

$\displaystyle\sum_{\inf\\{i_{1},\ldots i_{m}\\}\leq n(\log n)^{-\alpha}}1\leq
C_{m}n^{m}(\log n)^{-\alpha}.$

Hence,

$\displaystyle\mathbb{E}\left[\sum_{i_{1},\ldots
i_{m}=1}^{n}\sum_{k_{1},\ldots,k_{2m}=1}^{n}\prod_{j=1}^{2m}G_{D}(\mathcal{S}^{1}_{k_{j}}-\mathcal{S}^{2}_{i_{\lceil
j/2\rceil}})\right]$ $\displaystyle\leq$ $\displaystyle C_{m}n^{m}(\log
n)^{m}(\log n)^{-\alpha}+\mathbb{E}\left[\sum_{\inf\\{i_{1},\ldots
i_{m}\\}\geq n(\log
n)^{-\alpha}}\sum_{k_{1},\ldots,k_{2m}=1}^{n}\prod_{j=1}^{2m}G_{D}(\mathcal{S}^{1}_{k_{j}}-\mathcal{S}^{2}_{i_{\lceil
j/2\rceil}})\right]$ $\displaystyle\leq$ $\displaystyle C_{m}n^{m}(\log
n)^{m}(\log n)^{-\alpha}+\mathbb{E}\left[\sum_{\inf\\{i_{1},\ldots
i_{m}\\}\geq n(\log
n)^{-\alpha}}\sum_{k_{1},\ldots,k_{2m}=1}^{n}1_{D^{c}}\prod_{j=1}^{2m}G_{D}(\mathcal{S}^{1}_{k_{j}}-\mathcal{S}^{2}_{i_{\lceil
j/2\rceil}})\right]$ $\displaystyle\leq$ $\displaystyle C_{m}n^{m}(\log\log
n)^{2m}.$

Therefore, we obtain the desired result. ∎

The other terms of equation (4.6) are of much smaller order.

######  Lemma 4.4.

Consider the second summand on the right hand side of equation (4.6). Namely,
let

(4.10) $\displaystyle\mathcal{E}_{1}$
$\displaystyle:=\sum_{x^{1}_{1},x^{1}_{2}\in\mathcal{S}^{1}}\sum_{j=1}^{\lceil
n^{1-\beta}\rceil}\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,j}}\mathcal{I}(x^{1}_{2},j)\mathcal{I}(x^{1}_{1},j)\mathbb{P}(R^{\prime}_{x^{1}_{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}_{1}-\tilde{x}^{2}_{j})$
$\displaystyle\hskip 56.9055pt\times
E_{x^{2},x^{1}_{2}}\mathbb{P}(R^{\prime}_{x^{1}_{2}}\cap\mathcal{S}^{1}=\emptyset).$

We have that,

(4.11) $\mathbb{E}[\mathcal{E}_{1}]\lesssim
n^{\frac{3}{2}\alpha+\frac{1}{2}\beta+\frac{3}{2}\delta+\frac{1}{2}}.$

We also have a similar estimate for the third summand on the right hand side
of equation (4.6). Namely, we have that,

(4.12) $\displaystyle\mathcal{E}_{2}:=$
$\displaystyle\sum_{x^{1}_{1},x^{1}_{2}\in\mathcal{S}^{2}}\sum_{j=1}^{\lceil
n^{1-\beta}\rceil}\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,j}}\mathcal{I}(x^{1}_{2},j)\mathcal{I}(x^{1}_{1},j)\mathbb{P}(R^{\prime}_{x^{1}_{1}}\cap\mathcal{S}^{1}=\emptyset)$
$\displaystyle\hskip
56.9055pt\times[G_{D}(x^{1}_{1}-x^{2})-G_{D}(x^{1}_{1}-\tilde{x}^{2}_{j})]E_{x^{2},x^{1}_{2}}\mathbb{P}(R^{\prime}_{x^{1}_{2}}\cap\mathcal{S}^{1}=\emptyset),$

will satisfy,

(4.13) $\mathbb{E}[\mathcal{E}_{2}]\lesssim
n^{\frac{3}{2}\alpha+\frac{3}{2}\beta+2\delta}.$

Lemma 4.4 will be shown later in this section. As before, Markov’s inequality
will give. one can derive that,

######  Corollary 4.5.

Recall the terms $\mathcal{E}_{1}$ and $\mathcal{E}_{2}$ from equations (4.10)
and (4.12). Fix any $\epsilon>0$ and set $b_{n}=O(\log\log n)$ with
$\lim_{n\to\infty}b_{n}=\infty$. We have that,

(4.14)
$\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}(\max(\mathcal{E}_{1},\mathcal{E}_{2})\geq\epsilon\frac{nb_{n}}{(\log
n)^{2}})=-\infty.$

###### Proof.

By Markov’s inequality applied to (4.11) and (4.13), we have that

$\frac{1}{b_{n}}\log\mathbb{P}(\mathcal{E}_{1}\geq\epsilon\frac{nb_{n}}{(\log
n)^{2}})\leq\frac{-\log\epsilon-(\frac{1}{2}-\frac{3}{2}\alpha-\frac{1}{2}\beta-\frac{3}{2}\delta)\log
n+2\log\log n}{b_{n}},$

and,

$\frac{1}{b_{n}}\log\mathbb{P}(\mathcal{E}_{2}\geq\epsilon\frac{nb_{n}}{(\log
n)^{2}})\leq\frac{-\log\epsilon-(1-\frac{3}{2}\alpha-\frac{3}{2}\beta-2\delta)\log
n+2\log\log n}{b_{n}}.$

The desired conclusion (4.14) follows from taking the limit $n\to\infty$. ∎

Finally, the last error term, the fourth summand of (4.6) will also be of
smaller order.

######  Lemma 4.6.

Consider the last summand of (4.6),

(4.15)
$\displaystyle\mathcal{E}_{3}:=\sum_{x^{1}_{1},x^{1}_{2}\in\mathcal{S}^{2}}\sum_{j=1}^{\lceil
n^{1-\beta}\rceil}\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,j}}$
$\displaystyle[1-\mathcal{I}(x^{1}_{2},j)\mathcal{I}(x^{1}_{1},j)]\mathbb{P}(R^{\prime}_{x^{1}_{1}}\cap\mathcal{S}^{1}=\emptyset)$
$\displaystyle
G_{D}(x^{1}_{1}-x^{2})G_{\mathcal{S}^{2}_{\beta,\alpha,j}}(x^{2},x^{1}_{2})\mathbb{P}(R^{\prime}_{x^{1}_{2}}\cap\mathcal{S}^{1}=\emptyset).$

We have that,

$\mathbb{E}\left[\mathcal{E}_{3}\right]\lesssim n^{1-\delta^{\prime}}.$

###### Proof.

By (4.9), we remark that for any point $a$ that,

(4.16)
$\mathbb{E}\left[\sum_{x^{2}\in\mathcal{S}^{2}}G_{D}(a-x^{2})\right]=\sum_{j=0}^{n}\sum_{i=j}^{\infty}\mathbb{P}(\mathcal{S}_{i}=a)\lesssim\sum_{j=0}^{n}\sum_{i=j}^{\infty}i_{+}^{-2}\lesssim\log
n,$

where $i_{+}:=1\vee i$. By symmetry,

$\displaystyle\mathbb{E}[\mathcal{E}_{3}]\leq$
$\displaystyle\mathbb{E}\bigg{[}\sum_{i,j,k=0}^{n}[1-\mathbbm{1}[|\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j}|\geq\sqrt{n}^{1-\delta}]\mathbbm{1}[|\mathcal{S}^{1}_{k}-\mathcal{S}^{2}_{j}|\geq\sqrt{n}^{1-\delta}]]$
$\displaystyle\hskip 56.9055pt\times
G_{D}(\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j})G_{D}(\mathcal{S}^{2}_{j}-\mathcal{S}^{1}_{k})\bigg{]}$
$\displaystyle\leq$ $\displaystyle
2\mathbb{E}\bigg{[}\sum_{i,j,k=0}^{n}\mathbbm{1}[|\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j}|\leq\sqrt{n}^{1-\delta}]G_{D}(\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j})G_{D}(\mathcal{S}^{2}_{j}-\mathcal{S}^{1}_{i}-(\mathcal{S}^{1}_{k}-\mathcal{S}^{1}_{i}))\bigg{]}.$

By (4.16) and Markov’s property, it is bound by

$\displaystyle C(\log
n)\times\mathbb{E}\left[\sum_{i,j=0}^{n}\mathbbm{1}[|\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j}|\leq\sqrt{n}^{1-\delta}]G_{D}(\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j})\right].$

Now, by [14, Lemma 4.1],

$\displaystyle\mathbb{E}\left[\sum_{i,j=0}^{n^{1-\delta/2}}\mathbbm{1}[|\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j}|\leq\sqrt{n}^{1-\delta}]G_{D}(\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j})\right]\leq\mathbb{E}\left[\sum_{i,j=0}^{n^{1-\delta/2}}G_{D}(\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j})\right]\lesssim
n^{1-\delta/2}$

and

$\displaystyle\mathbb{E}\left[\sum_{i=n^{1-\delta/2}}^{n}\sum_{j=0}^{n}\mathbbm{1}[|\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j}|\leq\sqrt{n}^{1-\delta}]G_{D}(\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j})\right]$
$\displaystyle\leq$ $\displaystyle
n\max_{y\in{\mathbb{Z}}^{4}}\mathbb{E}\left[\sum_{i=n^{1-\delta/2}}^{n}\mathbbm{1}[|\mathcal{S}^{1}_{i}-y|\leq\sqrt{n}^{1-\delta}]G_{D}(\mathcal{S}^{1}_{i}-y)\right]$
$\displaystyle\lesssim$ $\displaystyle
n\max_{y\in{\mathbb{Z}}^{4}}\sum_{i=n^{1-\delta/2}}^{n}\sum_{|x-y|\leq\sqrt{n}^{1-\delta}}|x-y|^{-2}\mathbb{P}(\mathcal{S}^{1}_{i}=x).$

Then again by (4.9), it is bound by

$\displaystyle C(\log n)n^{1-\delta/2}+Cn(\log
n)\max_{y\in{\mathbb{Z}}^{4}}\sum_{i=n^{1-\delta/2}}^{n}\sum_{|x-y|\leq\sqrt{n}^{1-\delta}}|x-y|^{-2}i_{+}^{-2}\lesssim
n^{1-\delta/2}.$

Therefore, by symmetry, it completes the proof.

∎

As before, one can show the following from Markov’s inequality,

######  Corollary 4.7.

Recall the term $\mathcal{E}_{3}$ from (4.15). Fix any $\epsilon>0$ and set
$b_{n}=O(\log\log n)$ with $\lim_{n\to\infty}b_{n}=\infty$. Then, we see that,

(4.17)
$\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(\mathcal{E}_{3}\geq\epsilon\frac{nb_{n}}{n(\log
n)^{2}}\right)=-\infty.$

The proof is similar to that of Corollary 4.5 and will not be shown here.

Using the previous corollaries, one can now prove Theorem 3.3.

###### Proof of Theorem 3.3.

It is clear that,

$\displaystyle\mathbb{P}\left(\chi^{\prime}\geq\epsilon\frac{nb_{n}}{(\log
n)^{2}}\right)\leq\mathbb{P}\left(\chi^{\prime}_{\beta,\alpha}\geq\epsilon\frac{nb_{n}}{(\log
n)^{2}}\right)$
$\displaystyle\leq\mathbb{P}\left(MT_{n}\geq\frac{\epsilon}{4}\frac{b_{n}n}{(\log
n)^{2}}\right)+\mathbb{P}\left(\mathcal{E}_{1}\geq\frac{\epsilon}{4}\frac{b_{n}n}{(\log
n)^{2}}\right)+\mathbb{P}\left(\mathcal{E}_{2}\geq\frac{\epsilon}{4}\frac{b_{n}n}{(\log
n)^{2}}\right)$
$\displaystyle+\mathbb{P}\left(\mathcal{E}_{3}\geq\frac{\epsilon}{4}\frac{b_{n}n}{(\log
n)^{2}}\right).$

By computing $\frac{1}{b_{n}}\log$ to both sides, we see that the conclusion
of Theorem 3.3 is a consequence of Corollaries 4.2, 4.5, and 4.7

∎

###  4.1. Proof of Lemma 4.4

The proof of this lemma requires novel techniques beyond careful computations
of Green’s functions due to the presence of the error terms $E$ occurring from
the matrix inversion. We will present the proof here.

###### Proof of Lemma 4.4.

First, we deal with,

$\displaystyle\sum_{x^{1}_{1},x^{1}_{2}\in\mathcal{S}^{2}}\sum_{j=1}^{\lceil
n^{1-\beta}\rceil}\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,j}}\mathcal{I}(x^{1}_{2},j)\mathcal{I}(x^{1}_{1},j)\mathbb{P}(R^{\prime}_{x^{1}_{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}_{1}-\tilde{x}^{2}_{j})$
$\displaystyle\hskip 56.9055pt\times
E_{x^{2},x^{1}_{2}}\mathbb{P}(\mathcal{R^{\prime}}_{x^{1}_{2}}\cap\mathcal{S}^{1}=\emptyset).$

We have that,

$\displaystyle\sum_{x^{1}_{1},x^{1}_{2}\in\mathcal{S}^{2}}\sum_{j=1}^{\lceil
n^{1-\beta}\rceil}\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,j}}\mathcal{I}(x^{1}_{2},j)\mathcal{I}(x^{1}_{1},j)\mathbb{P}(R^{\prime}_{x^{1}_{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}_{1}-\tilde{x}^{2}_{j})$
$\displaystyle\hskip 56.9055pt\times
E_{x^{2},x^{1}_{2}}\mathbb{P}(\mathcal{R^{\prime}}_{x^{1}_{2}}\cap\mathcal{S}^{1}=\emptyset)$
$\displaystyle=\sum_{x^{1}_{1}\in\mathcal{S}^{1}}\sum_{j=1}^{\lceil
n^{1-\beta}\rceil}\mathcal{I}(x^{1}_{2},j)\mathcal{I}(x^{1}_{1},j)\mathbb{P}(R^{\prime}_{x^{1}_{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}_{1}-\tilde{x}^{2}_{j})$
$\displaystyle\times\sum_{x^{1}_{2}\in\mathcal{S}^{2}}\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,j}}E_{x^{2},x^{1}_{2}}\mathbb{P}(\mathcal{R}^{\prime}_{x^{1}_{2}}\cap\mathcal{S}^{1}=\emptyset).$

We now consider,

$\displaystyle\sum_{x^{1}_{2}\in\mathcal{S}^{2}}\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,j}}|E_{x^{2},x^{1}_{2}}|\mathbb{P}(\mathcal{R}^{\prime}_{x^{1}_{2}}\cap\mathcal{S}^{1}=\emptyset)$
$\displaystyle\leq\sum_{x^{1}_{2}\in\mathcal{S}^{2}}\sqrt{|\mathcal{S}^{2}_{\beta,j}|\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,j}}|E_{x^{2},x^{1}_{2}}|^{2}}\lesssim
n\sqrt{n^{\beta}n^{\alpha}\frac{n^{2\alpha}}{n^{3-3\delta}}}\lesssim
n^{3/2\alpha+1/2\beta+3/2\delta-1/2}.$

To get the first inequality, we used the Cauchy-Schwartz inequality on the sum
over $\mathcal{S}^{2}_{\beta,j}$. From Lemma A.16, the matrix
$\mathcal{G}^{\mathcal{S}^{2}_{\beta,\alpha,j}}$ is positive definite and has
minimum eigenvalue greater than $1$. Thus, the inverse matrix has $l^{2}\to
l^{2}$ operator norm less than $2$. Thus, we know that,

$\sqrt{\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,j}}|E_{x^{2},x^{1}_{2}}|^{2}}\leq
2\sqrt{\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,\alpha,j}}|G_{D}(a_{1}-x^{1}_{2})-G_{D}(x^{2}-x^{1}_{2})|^{2}}\lesssim\sqrt{n^{\alpha}\frac{n^{2\alpha}}{n^{3-3\delta}}}.$

In the final inequality, we used the deterministic bound (4.4) to bound the
differences of the Green’s function in the region $\mathcal{S}^{2}_{\beta,j}$.
Note, we used the fact we are in the indicator function
$\mathcal{I}(x^{1}_{2},j)$. Thus, we have that, deterministically,

$\displaystyle\bigg{|}\sum_{x^{1}_{1},x^{1}_{2}\in\mathcal{S}^{2}}\sum_{j=1}^{\lceil
n^{1-\beta}\rceil}\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,j}}\mathcal{I}(x^{1}_{2},j)\mathcal{I}(x^{1}_{1},j)\mathbb{P}(R^{\prime}_{x^{1}_{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}_{1}-\tilde{x}^{2}_{j})$
$\displaystyle\hskip 56.9055pt\times
E_{x^{2},x^{1}_{2}}\mathbb{P}(\mathcal{R^{\prime}}_{x^{1}_{2}}\cap\mathcal{S}^{1}=\emptyset)\bigg{|}$
$\displaystyle\lesssim\sum_{x^{1}_{1}\in\mathcal{S}^{1}}\sum_{j=1}^{\lceil
n^{1-\beta}\rceil}\mathcal{I}(x^{1}_{2},j)\mathcal{I}(x^{1}_{1},j)\mathbb{P}(R^{\prime}_{x^{1}_{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}_{1}-\tilde{x}^{2}_{j})n^{3/2\alpha+1/2\beta+3/2\delta-1/2}$
$\displaystyle\lesssim
n^{3/2\alpha+1/2\beta+3/2\delta-1/2}\sum_{i=1}^{n}\sum_{j=1}^{n}G_{D}(\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j}).$

Therefore,

$\displaystyle\mathbb{E}\bigg{[}\bigg{|}\sum_{x^{1}_{1},x^{1}_{2}\in\mathcal{S}^{2}}\sum_{j=1}^{\lceil
n^{1-\beta}\rceil}\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,j}}\mathcal{I}(x^{1}_{2},j)$
$\displaystyle\hskip
56.9055pt\times\mathcal{I}(x^{1}_{1},j)\mathbb{P}(R^{\prime}_{x^{1}_{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}_{1}-\tilde{x}^{2}_{j})E_{x^{2},x^{1}_{2}}\mathbb{P}(\mathcal{R^{\prime}}_{x^{1}_{2}}\cap\mathcal{S}^{1}=\emptyset)\bigg{|}\bigg{]}$
$\displaystyle\lesssim$ $\displaystyle
n^{3/2\alpha+1/2\beta+3/2\delta-1/2}\mathbb{E}\left[\sum_{i=1}^{n}\sum_{j=1}^{n}G_{D}(\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j})\right]\lesssim
n^{3/2\alpha+1/2\beta+3/2\delta+1/2}.$

The computation of the expectation in the last line above comes from Theorem
A.1. The value of this last line is approximately $n^{-1/2}$ the scale of the
main order term (provided $\alpha,\beta,\delta$ are all chosen relatively
small).

Now, the other error term in $E$ can be dealt with in a similar way. To
recall, the other error term is,

$\displaystyle\sum_{x^{1}_{1},x^{1}_{2}\in\mathcal{S}^{2}}\sum_{j=1}^{\lceil
n^{1-\beta}\rceil}\sum_{x^{2}\in\mathcal{S}^{2}_{\beta,j}}\mathcal{I}(x^{1}_{2},j)\mathcal{I}(x^{1}_{1},j)\mathbb{P}(R^{\prime}_{x^{1}_{1}}\cap\mathcal{S}^{1}=\emptyset)[G_{D}(x^{1}_{1}-x^{2})-G_{D}(x^{1}_{1}-\tilde{x}^{2}_{j})]$
$\displaystyle\times
E_{x^{2},x^{1}_{2}}\mathbb{P}(\mathcal{R^{\prime}}_{x^{1}_{2}}\cap\mathcal{S}^{1}=\emptyset).$

We first use the improved estimate,

$|G_{D}(x^{1}_{1}-x^{2})-G_{D}(x^{1}_{1}-\tilde{x}^{2}_{j})|\lesssim|G_{D}(x^{1}_{1}-\tilde{x}^{2}_{j})|\frac{n^{\beta}}{n^{1/2-\delta/2}}$

under the indicator function $\mathcal{I}(x^{1}_{1},j)$. We remark here that
the factor of $|G_{D}(x^{1}_{1}-\tilde{x}^{2}_{j})|$ is an improved error term
in the case that $|x^{1}_{1}-\tilde{x}^{2}_{j}|$ is relatively large. With
this deterministic bound in hand, bounding this error term in $E$ reduces to
the error term we just treated. ∎

##  5\.  The Leading Term of $\chi$: Proof of Theorem 3.1

In this section, we will consider the large deviation statistics of the
following quantity,

$TL_{n}:=\sum_{x^{1}\in\mathcal{S}^{1}}\sum_{x^{2}\in\mathcal{S}^{2}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2}=\emptyset).$

We will prove Theorem 3.1 by separately proving lower and upper bounds for the
asymptotic moments.

###  5.1. Introduction of the Auxiliary $TL^{\prime}$

For technical reasons, $TL_{n}$ is not the most convenient quantity to
manipulate. Instead, we consider the following auxiliary quantity. We let
$\mathcal{S}^{k,i}$ denote the portion of the random walk in between the part
of the random walk
$\mathcal{S}^{k}\left[(i-1)\frac{n}{b_{n}},i\frac{n}{b_{n}}\right]$ and

(5.1)
$TL^{\prime}_{n}:=\sum_{i,j=1}^{b_{n}}\sum_{x^{1}\in\mathcal{S}^{1,i}}\sum_{x^{2}\in\mathcal{S}^{2,j}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1,i}=\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2,j}=\emptyset).$

We have the following relationship between $TL$ and $TL^{\prime}$.

######  Proposition 5.1.

Let $b_{n}$ be a sequence satisfying $b_{n}=O(\log\log n)$ and
$\lim_{n\to\infty}b_{n}=\infty$. Fix any $\lambda\geq 0$. Then, we have that,

(5.2)
$\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(TL^{\prime}_{n}\geq\lambda\frac{nb_{n}}{(\log
n)^{2}}\right)=\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(TL_{n}\geq\lambda\frac{nb_{n}}{(\log
n)^{2}}\right).$

###### Proof.

We remark that $TL^{\prime}_{n}\geq TL_{n}$. This immediately shows that,

$\mathbb{P}\left(TL^{\prime}_{n}\geq\lambda\frac{nb_{n}}{(\log
n)^{2}}\right)\geq\mathbb{P}\left(TL_{n}\geq\lambda\frac{nb_{n}}{(\log
n)^{2}}\right).$

To derive the opposite inequality, we first observe that $TL^{\prime}-TL$ can
be bounded from above by,

(5.3) $\displaystyle TL^{\prime}_{n}-TL_{n}$ $\displaystyle\leq$
$\displaystyle\sum_{i,j=1}^{b_{n}}\sum_{x^{1}\in\mathcal{S}^{1,i}}\sum_{x^{2}\in\mathcal{S}^{2,j}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1,i}=\emptyset,R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}\neq\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2,j}=\emptyset)$
$\displaystyle+\sum_{i,j=1}^{b_{n}}\sum_{x^{1}\in\mathcal{S}^{1,i}}\sum_{x^{2}\in\mathcal{S}^{2,j}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2,j}=\emptyset,R^{\prime}_{x^{2}}\cap\mathcal{S}\neq\emptyset)$
$\displaystyle+\sum_{i_{1}\neq
i_{2},j=1}^{b_{n}}\sum_{x^{1}\in\mathcal{S}^{1,i_{1}}\cap\mathcal{S}^{1,i_{2}}}\sum_{x^{2}\in\mathcal{S}^{2,j}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2}=\emptyset)$
$\displaystyle+\sum_{i,j_{1}\neq
j_{2}=1}^{b_{n}}\sum_{x^{2}\in\mathcal{S}^{2,j_{1}}\cap\mathcal{S}^{2,j_{2}}}\sum_{x^{1}\in\mathcal{S}^{1,i}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2}=\emptyset).$

For each line on the right hand side above, we will show that

(5.4)
$\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(\cdot\geq\epsilon\frac{nb_{n}}{(\log
n)^{2}}\right)=-\infty.$

The first two lines of the right hand side of (5.3) are very similar to the
type of error terms we have dealt with in Section 4. One can follow the
analysis of said section to show the relation (5.4) for these two lines.

The last two lines will be controlled by bounding the moments and applying
Markov’s inequality. We present the analysis with the term on the third line,
since the term on the fourth line can be dealt with similarly. We first bound
all the probability terms on the line by $1$.

If we consider two independent simple random walks $\mathcal{S}$ and
$\tilde{\mathcal{S}}$ starting at $0$, by (4.9) and (4.16), for any $x$ and
$y$,

$\displaystyle\mathbb{E}\left[\sum_{i,j=0}^{n}1\\{\mathcal{S}_{i}+x=\tilde{\mathcal{S}}_{j}+y\\}\right]=$
$\displaystyle\sum_{i,j=0}^{n}\mathbb{P}(\mathcal{S}_{i+j}=y-x)$
$\displaystyle\lesssim$
$\displaystyle\sum_{i,j=0}^{n}(i+j)_{+}^{-2}\lesssim\log n.$

Thus, we see that,

$\displaystyle\mathbb{E}\left[\sum_{i_{1}\neq
i_{2},j=1}^{b_{n}}\sum_{x^{1}\in\mathcal{S}^{1,i_{1}}\cap\mathcal{S}^{1,i_{2}}}\sum_{x^{2}\in\mathcal{S}^{2,j}}\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}=\emptyset)G_{D}(x^{1}-x^{2})\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2}=\emptyset)\right]$
$\displaystyle\leq\mathbb{E}\left[\sum_{i_{1}\neq
i_{2}=1}^{b_{n}}\sum_{x^{1}\in\mathcal{S}^{1,i_{1}}\cap\mathcal{S}^{1,i_{2}}}\mathbb{E}_{2}\left[\sum_{x^{2}\in\mathcal{S}^{2}}G_{D}(x^{1}-x^{2})\right]\right]\lesssim\mathbb{E}\left[\sum_{i_{1}\neq
i_{2}=1}^{b_{n}}I_{i_{1},i_{2}}\log n\right]$ $\displaystyle\lesssim
b_{n}^{2}(\log n)^{2}.$

On the second line $\mathbb{E}_{2}$ is the expectation with respect to only
the randomness of $\mathcal{S}^{2}$. $I_{i_{1},i_{2}}$ is the number of points
of intersection between $\mathcal{S}^{1}_{i_{1}}$ and
$\mathcal{S}^{1}_{i_{2}}$. We remark that
$\mathcal{S}^{1}_{i_{2}}-\mathcal{S}^{1}(i_{2}\frac{n}{b_{n}})$ and
$\mathcal{S}^{1}_{i_{1}}-\mathcal{S}^{1}(i_{1}\frac{n}{b_{n}})$ are
independent random walks. Thus, this term will not contribute to the large
deviation statistics of $TL^{\prime}_{n}$ on the scale of $\frac{nb_{n}}{(\log
n)^{2}}$.

∎

The quantity $TL^{\prime}_{n}$ is easier to deal with since we can obtain
exact moment asymptotics. Namely,

######  Proposition 5.2.

Recall $TL^{\prime}_{n}$ from equation (5.1). Let $b_{n}=O(\log\log n)$ and
$\lim_{n\to\infty}b_{n}=\infty$. Then, we have the following exact moment
asymptotics on $TL^{\prime}_{n}$.

(5.5)
$\lim_{n\to\infty}\frac{1}{b_{n}}\log\sum_{m=0}^{\infty}\frac{1}{m!}\theta^{m}\left(\frac{\sqrt{b_{n}}\log
n}{\sqrt{n}}\right)^{m}\mathbb{E}[(TL^{\prime}_{n})^{m}]^{1/2}=\tilde{\kappa}(4,2)^{4}\frac{\pi^{4}\theta^{2}}{128}.$

As a consequence of the previous two propositions, one can now prove Theorem
3.1.

###### Proof of Theorem 3.1.

By [13, Theorem 1.2.7], equation (5.5) would be equivalent to showing,

$\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(TL^{\prime}_{n}\geq\lambda\frac{nb_{n}}{(\log
n)^{2}}\right)=-\frac{64}{\pi^{4}}\tilde{\kappa}(4,2)^{-4}\lambda.$

Now, since by Proposition we have 5.1 that

$\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(TL_{n}\geq\lambda\frac{nb_{n}}{(\log
n)^{2}}\right)=\lim_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{P}\left(TL^{\prime}_{n}\geq\lambda\frac{nb_{n}}{(\log
n)^{2}}\right),$

we complete the proof of the proposition. ∎

The remainder of this section is devoted to deriving upper and lower bounds to
the quantity in equation (5.5).

###  5.2. Large Deviation Upper Bounds

In this section, we establish the upper bound found in Proposition 5.2.

######  Proposition 5.3.

Let $b_{n}$ be a sequence satisfying $b_{n}=O(\log\log n)$ and
$\lim_{n\to\infty}b_{n}=\infty$. Then, for any $\lambda>0$, we satisfy,

$\limsup_{n\to\infty}\frac{1}{b_{n}}\log\sum_{m=0}^{\infty}\frac{1}{m!}\theta^{m}\left(\frac{\sqrt{b_{n}}\log
n}{\sqrt{n}}\right)^{m}\mathbb{E}[(TL^{\prime}_{n})^{m}]^{1/2}\leq\tilde{\kappa}(4,2)^{4}\frac{\pi^{4}\theta^{2}}{128}.$

The proposition above is an immediate consequence of the following lemma and
claim.

######  Claim 5.4.

There exists some constant $C>0$ such that for all $n,m>0$, we have that,

$\mathbb{E}[(TL^{\prime}_{n})^{m}]\leq C^{m}m!\left(\frac{n}{(\log
n)^{2}}\right)^{m}.$

The proof of the above claim will be postponed to later. We now present the
second necessary lemma.

######  Lemma 5.5.

Provided one knows that there is some constant $C$ such that for all $n$ and
$m$ we have that,

(5.6) $\mathbb{E}[(TL^{\prime}_{n})^{m}]\leq C^{m}m!\left(\frac{n}{(\log
n)^{2}}\right)^{m}.$

Then, one can show that,

$\limsup_{n\to\infty}\frac{1}{b_{n}}\log\sum_{m=0}^{\infty}\frac{1}{m!}\theta^{m}\left(\frac{\sqrt{b_{n}}\log
n}{\sqrt{n}}\right)^{m}\mathbb{E}[(TL)^{m}]^{1/2}\leq\tilde{\kappa}(4,2)^{4}\frac{\pi^{4}\theta^{2}}{128}.$

Similar statements hold for $TL^{\prime}$.

###### Proof.

Let $(B_{s})_{s\geq 0}$ and $(B^{\prime}_{s})_{s\geq 0}$ be independent
Brownian motions for $d=4$. The need for the bound in (5.6) is to ensure that
one can apply dominated convergence to the terms
$\mathbb{E}\left[\left(\frac{(\log n)^{2}}{n}TL\right)^{m}\right]$ when
needed, and replace them with the term:

$\left(\frac{\pi^{4}}{64}\right)^{m}\mathbb{E}\left[\left(\int_{0}^{1}\int_{0}^{1}G(B_{t}-B^{\prime}_{s})\text{d}t\text{d}s\right)^{m}\right].$

The reason why this can be done is due to the fact that

$\frac{(\log
n)^{2}}{n}TL\to\frac{\pi^{4}}{64}\int_{0}^{1}\int_{0}^{1}G(B_{t}-B^{\prime}_{s})\text{d}t\text{d}s,$

in distribution. One can justify this convergence by showing the following
second moment estimates.

$\displaystyle\lim_{n\to\infty}\mathbb{E}\left[\left(\frac{(\log
n)^{2}}{n}TL-\frac{\pi^{4}}{64n}\sum_{i,j=1}^{n}G_{D}(\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j})\right)^{2}\right]=0,$
$\displaystyle\lim_{n\to\infty}\mathbb{E}\left[\left(\frac{1}{n}\sum_{i,j=1}^{n}G_{D}(\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j})-\int_{0}^{1}\int_{0}^{1}G(B_{t}-B_{s}^{\prime})\text{d}t\text{d}s\right)^{2}\right]=0.$

The first estimate can be derived via computations similar to [14, Lemma 4.3]
and the second is a consequence of Theorem A.1.

We can follow the proof of [13, Theorem 7.2.1] to derive the appropriate upper
bound. Finally, by Remark A.17, we see we obtain our desired constant.

∎

It is manifest that Proposition 5.3 is a consequence of Claim 5.4 and Lemma
5.5. We devote the rest of this subsection to deriving Claim 5.4.

####  5.2.1. A proof of Claim 5.4

Our first remark is that the quantity $TL^{\prime}_{n}$ is less than,

(5.7) $\displaystyle TL_{n,\alpha}:=\sum_{i,j=1}^{n}$
$\displaystyle\mathbb{P}(\mathcal{R}_{\mathcal{S}^{1}_{i}}\cap\mathcal{S}^{1}[i-n^{\alpha},i+n^{\alpha}]=\emptyset)G_{D}(\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j})$
$\displaystyle\times\mathbb{P}(\mathcal{R}_{\mathcal{S}^{2}_{j}}\cap\mathcal{S}^{2}[j-n^{\alpha},j+n^{\alpha}]=\emptyset).$

We will analyze the moments of $TL_{n,\alpha}$ via a subadditivity argument
along with a careful moment analysis. Our first subadditivity argument allows
us to reduce our moment analysis of $TL_{n,\alpha}$ to a slightly weaker
analysis.

######  Lemma 5.6.

If one knows that there exists some constant $C$ such that for all $n$ and $m$
that

(5.8) $\mathbb{E}[(TL_{n,\alpha})^{m}]\leq C^{m}(m!)^{2}\left(\frac{n}{(\log
n)^{2}}\right)^{m},$

then there is some other constant $C^{\prime}$ such that,

(5.9)
$\mathbb{E}[(TL_{n,\alpha})^{m}]\leq(C^{\prime})^{m}(m!)\left(\frac{n}{(\log
n)^{2}}\right)^{m}.$

###### Proof.

We start by starting a subadditivity argument. Let $\mathcal{S}$ be some
random walk of length $n$. Define

(5.10)
$\mathcal{G}^{a}_{\mathcal{S}}:=\sum_{i=1}^{n}\tilde{G}_{D}(\mathcal{S}_{i}-a)\mathbb{P}(\mathcal{R}^{\prime}_{\mathcal{S}_{i}}\cap\mathcal{S}[i-n^{\alpha},i+n^{\alpha}]=\emptyset).$

To match the notation of [13, Chapter 6.1], we also define

$\mathcal{F}^{a}_{\mathcal{S}[0,t]}=\sum_{i=1}^{t}\tilde{G}_{D}(\mathcal{S}_{i}-a)\mathbb{P}(\mathcal{R}^{\prime}_{\mathcal{S}_{i}}\cap\mathcal{S}[i-t^{\alpha},i+t^{\alpha}]=\emptyset).$

The main thing to observe about this function is that,

$TL_{n,\alpha}=\sum_{a\in\mathbb{Z}^{d}}\mathcal{F}^{a}_{\mathcal{S}^{1}[0,n]}\mathcal{F}^{a}_{\mathcal{S}^{2}[0,n]},$

where $\mathcal{S}^{1}[0,n]$ and $\mathcal{S}^{2}[0,n]$ are two independent
random walks.

Furthermore, it is trivially true that for times $t<s$ that

$\mathcal{F}^{a}_{\mathcal{S}[0,s]}\leq\mathcal{F}^{a}_{\mathcal{S}[0,t]}+\mathcal{G}^{a}_{\mathcal{S}[t,s]}$

and $\mathcal{G}$ has the translation symmetry,

$\mathcal{G}^{a+z}_{\mathcal{S}+z}=\mathcal{G}^{a}_{\mathcal{S}}.$

For these reasons, we can apply all the results of [13, Section 6.1]. In
particular, we can apply the argument of [13, Theorem 6.2.1].

∎

It remains to prove equation (5.8)

######  Lemma 5.7.

Equation (5.8) holds. Namely, there is a constant such that for all $n$ and
$m$ we have that,

$\mathbb{E}[(TL_{n,\alpha})^{m}]\leq C^{m}(m!)^{2}\left(\frac{n}{(\log
n)^{2}}\right)^{m}.$

Before we start proving the above lemma, we will finish the proof of Claim
5.4.

###### Proof of Claim 5.4.

Since $TL_{n}^{\prime}\leq TL_{n,\alpha}$, we have by Lemmas 5.7 and 5.6 that

$\mathbb{E}[(TL_{n}^{\prime})^{m}]\leq\mathbb{E}[(TL_{n,\alpha})^{m}]\leq
C^{m}m!n^{m}.$

This is exactly what was desired. ∎

We now return to the proof of Lemma 5.7.

####  5.2.2. The proof of Lemma 5.7

###### Proof of Lemma 5.7.

In what follows, the constant $C$ may not remain the same from line to line.
Since
$TL_{n,\alpha}\leq\sum_{i,j=1}^{n}G_{D}(\mathcal{S}^{1}(i)-\mathcal{S}^{2}(j))$,
it is clear that there is some constant $C$.

$\mathbb{E}[(TL_{n,\alpha})^{m}]\leq\mathbb{E}\left[\left(\sum_{i,j=1}^{n}G_{D}(\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j})\right)^{m}\right]\leq
C^{m}m!n^{m}.$

This latter estimate immediately follows from the large deviation statistics
of $\sum_{i,j=1}^{n}G_{D}(\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j})$ from
Theorem A.1. Now, observe that when $m\geq(\log n)^{2}$. One has that

$m!\geq m^{m}e^{-m}\geq(\log n)^{2m}e^{-m}.$

Thus, for $m\geq(\log n)^{2}$, we have that,

$C^{m}m!n^{m}\leq(eC)^{m}(m!)^{2}\left(\frac{n}{(\log n)^{2}}\right)^{m}.$

It suffices to prove an upper bound for moments when $m\leq(\log n)^{2}$.

Bounding the moments when $m\leq(\log n)^{2}$

We will show that there exists a constant $C$ such that,

$\mathbb{E}[(TL_{n,\alpha})^{m}]\leq(m!)^{2}C^{m}\frac{n^{m}}{(\log n)^{2m}}$

by induction on $m$.

To do this we first need the following claim

######  Claim 5.8.

Define

$\displaystyle\overline{TL}^{m}_{n,\alpha}:=$
$\displaystyle\sum_{\begin{subarray}{c}i_{1},i_{2},\ldots,i_{m}\\\
|i_{a}-i_{b}|\geq n^{3\alpha}\forall
a,b\end{subarray}}\sum_{\begin{subarray}{c}j_{1},j_{2},\ldots,j_{m}\\\
|j_{a}-j_{b}|\geq n^{3\alpha}\forall
a,b\end{subarray}}\prod_{k=1}^{m}\mathbb{P}(R^{\prime}_{\mathcal{S}^{1}_{i_{k}}}\cap\mathcal{S}^{1}[i_{k}-n^{\alpha},i_{k}+n^{\alpha}]=\emptyset)$
$\displaystyle\times
G_{D}(\mathcal{S}^{1}_{i_{k}}-\mathcal{S}^{2}_{j_{k}})\mathbb{P}(R^{\prime}_{\mathcal{S}^{2}_{j_{k}}}\cap\mathcal{S}^{2}[j_{k}-n^{\alpha},j_{k}+n^{\alpha}]=\emptyset).$

Then, there exists some constant $B$ such that,

$\mathbb{E}[\overline{TL}_{n,\alpha}^{m}]\leq(m!)^{2}B^{m}\frac{n^{m}}{(\log
n)^{2m}}.$

Since the points $i_{k}$ are all spaced far apart, we are able to use in some
form that the probability terms
$\mathbb{P}(R^{\prime}_{\mathcal{S}^{1}(i_{k})}\cap\mathcal{S}^{1}[i_{k}-n^{\alpha},i_{k}+n^{\alpha}]=\emptyset)$
should be rather independent of each other. We will return to the proof of the
claim later. Assuming the claim we have the following result; the moments of
$TL_{\alpha}$ can be bounded from above by

(5.11) $\displaystyle\mathbb{E}[(TL_{n,\alpha})^{m}]$
$\displaystyle\leq\mathbb{E}[\overline{TL}^{m}_{n,\alpha}]$
$\displaystyle+2m^{2}\mathbb{E}\bigg{[}\sum_{\begin{subarray}{c}i_{1},\ldots,i_{m}\\\
|i_{1}-i_{2}|\leq
n^{3\alpha}\end{subarray}}\sum_{j_{1},\ldots,j_{m}}\prod_{k=1}^{m}\mathbb{P}(R^{\prime}_{\mathcal{S}^{1}_{i_{k}}}\cap\mathcal{S}^{1}[i_{k}-n^{\alpha},i_{k}+n^{\alpha}]=\emptyset)$
$\displaystyle\times
G_{D}(\mathcal{S}^{1}_{i_{k}}-\mathcal{S}^{2}_{j_{k}})\mathbb{P}(R^{\prime}_{\mathcal{S}^{2}_{j_{k}}}\cap\mathcal{S}^{2}[j_{k}-n^{\alpha},j_{k}+n^{\alpha}]=\emptyset)\bigg{]}.$

Namely, if there is a term in the $m$th moment of $TL_{n,\alpha}$ that is not
already contained in the term $\overline{TL}^{m}_{n,\alpha}$, there must be
some pair of points $(i_{a},i_{b})$ or $(j_{a},j_{b})$ that are of a distance
closer than $n^{3\alpha}$. By symmetry, we may assume that the two points are
$i_{1}$ and $i_{2}$. There are at most $2m^{2}$ such choices of pairs
$(i_{a},i_{b})$ or $(j_{a},j_{b})$. We will now bound the moment of the second
term above.

If could only sum over the terms $i_{2},\ldots,i_{m}$ and
$j_{2},\ldots,j_{m}$, then this would be the $m-1$th moment of
$(TL_{\alpha})$. We could then apply induction to this quantity. The main idea
is that if we fix $i_{2}$ there are at most $2n^{3\alpha}$ choices of $i_{1}$.
Thus, intuitively, this term should be no more than $n^{3\alpha}$ times
$\mathbb{E}[(TL_{\alpha})^{m-1}]$. The problem is to deal with the sum over
$j_{1}$.

Observe the following,

(5.12)
$\displaystyle\mathbb{E}\bigg{[}\sum_{\begin{subarray}{c}i_{1},\ldots,i_{m}\\\
|i_{1}-i_{2}|\leq
n^{3\alpha}\end{subarray}}\sum_{j_{1},\ldots,j_{m}}\prod_{k=1}^{m}\mathbb{P}(R^{\prime}_{\mathcal{S}^{1}_{i_{k}}}\cap\mathcal{S}^{1}[i_{k}-n^{\alpha},i_{k}+n^{\alpha}]=\emptyset)$
$\displaystyle\times
G_{D}(\mathcal{S}^{1}_{i_{k}}-\mathcal{S}^{2}_{j_{k}})\mathbb{P}(R^{\prime}_{\mathcal{S}^{2}_{j_{k}}}\cap\mathcal{S}^{2}[j_{k}-n^{\alpha},j_{k}+n^{\alpha}]=\emptyset)\bigg{]}$
$\displaystyle\leq
m\mathbb{E}\bigg{[}\sum_{\begin{subarray}{c}i_{1},\ldots,i_{m}\\\
|i_{1}-i_{2}|\leq
n^{3\alpha}\end{subarray}}\sum_{\begin{subarray}{c}j_{1},\ldots,j_{m}\\\
|j_{1}-j_{2}|\leq
n^{3\alpha}\end{subarray}}\prod_{k=1}^{m}\mathbb{P}(R^{\prime}_{\mathcal{S}^{1}_{i_{k}}}\cap\mathcal{S}^{1}[i_{k}-n^{\alpha},i_{k}+n^{\alpha}]=\emptyset)$
$\displaystyle\times
G_{D}(\mathcal{S}^{1}_{i_{k}}-\mathcal{S}^{2}_{j_{k}})\mathbb{P}(R^{\prime}_{\mathcal{S}^{2}_{j_{k}}}\cap\mathcal{S}^{2}[j_{k}-n^{\alpha},j_{k}+n^{\alpha}]=\emptyset)\bigg{]}$
$\displaystyle+\mathbb{E}\bigg{[}\sum_{\begin{subarray}{c}i_{1},\ldots,i_{m}\\\
|i_{1}-i_{2}|\leq
n^{3\alpha}\end{subarray}}\sum_{\begin{subarray}{c}j_{1},\ldots,j_{m}\\\
|j_{1}-j_{k}|\geq n^{3\alpha}\forall
k\end{subarray}}\prod_{k=1}^{m}\mathbb{P}(R^{\prime}_{\mathcal{S}^{1}_{i_{k}}}\cap\mathcal{S}^{1}[i_{k}-n^{\alpha},i_{k}+n^{\alpha}]=\emptyset)$
$\displaystyle\times
G_{D}(\mathcal{S}^{1}_{i_{k}}-\mathcal{S}^{2}_{j_{k}})\mathbb{P}(R^{\prime}_{\mathcal{S}^{2}_{j_{k}}}\cap\mathcal{S}^{2}[j_{k}-n^{\alpha},j_{k}+n^{\alpha}]=\emptyset)\bigg{]}.$

If we bound the product

$\mathbb{P}(R^{\prime}_{\mathcal{S}^{1}_{i_{1}}}\cap\mathcal{S}^{1}[i_{1}-n^{\alpha},i_{k}+n^{\alpha}]=\emptyset)G_{D}(\mathcal{S}^{1}_{i_{1}}-\mathcal{S}^{2}_{j_{1}})\mathbb{P}(R^{\prime}_{\mathcal{S}^{2}_{j_{k}}}\cap\mathcal{S}^{2}[j_{k}-n^{\alpha},j_{k}+n^{\alpha}]=\emptyset)$

by $1$, we see that the first term on the right hand side above in equation
(5.12) can indeed by bounded by $m\mathbb{E}[(TL_{n,\alpha})^{m-1}]$.

To deal with the second term, we do the following. First, fix the terms
$i_{2},\ldots,i_{m}$ and $j_{2},\ldots,j_{m}$. Without loss of generality, we
can assume that we order $j_{2}\leq j_{3}\leq\ldots\leq j_{m-1}\leq j_{m}$,
and that $j_{2}\leq i_{1}\leq j_{3}$ (We can apply similar logic regardless of
the relative position of $i_{1}$ in the ordering $j_{2}\leq
j_{3}\leq\ldots\leq j_{m}$.) Notice that upon conditioning on the values of
$\mathcal{S}_{i_{2}+n^{\alpha}}$ and $\mathcal{S}_{i_{3}-n^{\alpha}}$, the
walk $\mathcal{S}[i_{2}+n^{\alpha},i_{3}-n^{\alpha}]$ becomes independent of
the rest of the walk. We exploit this fact by using that

(5.13) $\mathbb{E}_{\mathcal{S}^{2}}\left[\sum_{i_{2}+n^{\alpha}\leq j_{1}\leq
i_{3}-n^{\alpha}}G_{D}(\mathcal{S}^{1}_{i_{1}}-\mathcal{S}^{2}_{j_{1}})\bigg{|}\mathcal{S}^{2}_{i_{2}+n^{\alpha}}=x,\mathcal{S}^{2}_{i_{3}-n^{\alpha}}=y\right]\lesssim\log
n$

for any pairs of values $x$ and $y$ by Lemma A.15. The expectation above is
only taken over the random walk $\mathcal{S}^{2}$. (Note, here we are bounding
the probability term
$\mathbb{P}(\mathcal{R}_{\mathcal{S}^{2}_{j_{1}}}\cap\mathcal{S}^{2}[j_{1}-n^{\alpha},j_{1}+n^{\alpha}])$
by $1$ to simplify further computations)

As a consequence, we see that we have,

$\displaystyle\mathbb{E}\bigg{[}\sum_{\begin{subarray}{c}i_{1},\ldots,i_{m}\\\
|i_{1}-i_{2}|\leq
n^{3\alpha}\end{subarray}}\sum_{\begin{subarray}{c}j_{1},\ldots,j_{m}\\\
|j_{1}-j_{k}|\geq n^{3\alpha}\forall
k\end{subarray}}\prod_{k=1}^{m}\mathbb{P}(R^{\prime}_{\mathcal{S}^{1}_{i_{k}}}\cap\mathcal{S}^{1}[i_{k}-n^{\alpha},i_{k}+n^{\alpha}]=\emptyset)$
$\displaystyle\times
G_{D}(\mathcal{S}^{1}_{i_{k}}-\mathcal{S}^{2}_{j_{k}})\mathbb{P}(R^{\prime}_{\mathcal{S}^{2}(j_{k})}\cap\mathcal{S}^{2}[j_{k}-n^{\alpha},j_{k}+n^{\alpha}]=\emptyset)\bigg{]}$
$\displaystyle\lesssim mn^{3\alpha}(\log
n)\mathbb{E}[TL_{n,\alpha}^{m-1}]\lesssim mn^{3\alpha}(\log
n)(m-1)!^{2}\frac{n^{m-1}}{(\log n)^{2m-2}}.$

The factor of $n^{3\alpha}$ comes from the possible choices of $x_{1}$ (given
its distance from $x_{2}$) and the factor of $m$ comes from the fact that
$j_{1}$ can be located in between any of the $m$ regions $[j_{i},j_{i+1}]$ in
the ordering $j_{2}\leq j_{3}\ldots\leq j_{m}$. At the final step, we applied
the induction hypothesis.

Returning to equation (5.12), we see that,

$\displaystyle\text{L.H.S. of }\eqref{eq:divmore}$ $\displaystyle\lesssim
mn^{6\alpha}C^{m-1}(m-1)!^{2}\frac{n^{m-1}}{(\log n)^{2m-2}}+m(\log
n)(m-1)!^{2}C^{m-1}\frac{n^{m-1}}{(\log n)^{m-1}}.$

Substituting this back into equation (5.11), we have,

$\mathbb{E}[(TL_{n,\alpha})^{m}]\leq(m!)^{2}B^{m}\frac{n^{m}}{(\log
n)^{m}}+Km^{3}n^{6\alpha}C^{m-1}(m-1)!\frac{n^{m-1}}{(\log n)^{2m-2}}.$

Notice that the right hand side is less than $C^{m}(m!)^{2}\frac{n^{m}}{(\log
n)^{2m}}$ provided,

$1\geq\left(\frac{B}{C}\right)^{m}+KC^{-1}n^{6\alpha-1}(\log
n)^{3}\geq\left(\frac{B}{C}\right)^{m}+KC^{-1}mn^{6\alpha-1}(\log n)^{2}.$

Provided $C$ is chosen large relative to $B$ and the universal constant $K$,
there is a value of $C$ such that the above inequality will be satisfied for
all $n$ and $m\leq(\log n)^{2}$. This completes the induction provided the
Claim 5.8 holds.

∎

Now we start complete the proof of Claim 5.8.

###### Proof of Claim 5.8.

Without loss of generality,we may order the times as $i_{1}\leq i_{2}\leq
i_{3}\ldots\leq i_{m}$. Our first step is to condition on the values of the
random walk at specific points as $\mathcal{S}^{1}_{i_{k}}=x^{c}_{k}$,
$\mathcal{S}_{i_{k}+n^{\alpha}}=x^{r}_{k}$,
$\mathcal{S}_{i_{k}-n^{\alpha}}=x^{l}_{k}$ and
$\mathcal{S}^{2}_{j_{k}}=y^{c}_{k}$,
$\mathcal{S}^{2}_{j_{k}+n^{\alpha}}=y^{r}_{k}$ ,
$\mathcal{S}^{2}_{j_{k}-n^{\alpha}}=y^{l}_{k}$. With the endpoints of the
neighborhoods $\mathcal{S}^{1}[i_{k}-n^{\alpha},i_{k}+n^{\alpha}]$ specified,
the neighborhoods involved in the probability terms above become independent
of each other. This is the key observation used to simplify the computations
that proceed. In what follows we let $p_{t}(x)$ denote the probability that a
SRW transitions to the point $x$ by time $t$.

To simplify what proceeds, we introduce the following notation,

(5.14)
$NI(\mathcal{S},i,x^{c},x^{r},x^{l}):=\mathbb{E}[\mathbb{P}(R^{\prime}_{\mathcal{S}_{i}}\cap\mathcal{S}[i-n^{\alpha},i+n^{\alpha}]=\emptyset)|\mathcal{S}_{i}=x^{c},\mathcal{S}_{i+n^{\alpha}}=x^{r},\mathcal{S}_{i-n^{\alpha}}=x^{l}].$

This finds the expected value of the probability that an independent random
walk $R^{\prime}_{\mathcal{S}_{i}}$ starting at $\mathcal{S}_{i}$ does not
intersect the portion of the random walk
$\mathcal{S}[i-n^{\alpha},i+n^{\alpha}]$ conditioned on the random walk being
at points $x^{c}$ at time $i$, $x^{r}$ at time $i+n^{\alpha}$ and $x^{l}$ at
time $i-n^{\alpha}$. If it is not necessary to condition $x^{r}$ and $x^{l}$,
we will slightly abuse notation and denote this by dropping the appropriate
argument on the right hand side. Note that $c$ as a superscript is used as a
shorthand for ‘center’ while $r$ and $l$ are ‘right’ and ‘left’ respectively.
We can write the expectation of $\mathbb{E}[\overline{TL}_{n,\alpha}^{m}]$ as,

(5.15) $\displaystyle(m!)\sum_{\sigma\in S_{n}}\sum_{\begin{subarray}{c}1\leq
i_{1}\leq i_{2}-n^{3\alpha}\leq\ldots\\\ \leq i_{m}-(m-1)n^{3^{\alpha}}\leq
n-(m-1)n^{3\alpha}\end{subarray}}\sum_{\begin{subarray}{c}1\leq
j_{\sigma(1)}\leq j_{\sigma(2)}-n^{\alpha}\ldots\\\ \leq
j_{\sigma(m)}-(m-1)n^{3\alpha}\leq
n-(m-1)n^{3\alpha}\end{subarray}}\sum_{\begin{subarray}{c}x^{r}_{k},x^{l}_{k},x^{c}_{k},\\\
y^{r}_{k},y^{l}_{k},y^{c}_{k}\in\mathbb{Z}^{4},\forall k\end{subarray}}$
$\displaystyle
p_{i_{1}}(x^{c}_{k})p_{n^{\alpha}}(x^{r}_{k}-x^{c}_{k})NI(\mathcal{S}^{1},i_{1},x_{1}^{c},x_{1}^{r})$
$\displaystyle\times
p_{j_{\sigma}(1)}(y^{c}_{\sigma(1)})p_{n^{\alpha}}(y^{r}_{\sigma(1)}-y^{c}_{\sigma(1)})NI(\mathcal{S}^{2},j_{\sigma(1)},y^{c}_{\sigma(1)},y^{r}_{\sigma(1)})$
$\displaystyle\prod_{i=2}^{k-1}NI(\mathcal{S}^{1},i_{k},x^{c}_{k},x^{r}_{k},x^{l}_{k})p_{n^{\alpha}}(x^{c}_{k}-x^{r}_{k})p_{n^{\alpha}}(x^{l}_{k}-x^{c}_{k})p_{i_{k+1}-i_{k}-2n^{\alpha}}(x^{l}_{k+1}-x^{r}_{k})$
$\displaystyle\times
NI(\mathcal{S}^{2},j_{\sigma(k)},y^{c}_{\sigma(k)},y^{r}_{\sigma(k)},y^{l}_{\sigma(k)})p_{n^{\alpha}}(y^{c}_{\sigma(k)}-y^{r}_{\sigma(k)})p_{n^{\alpha}}(y^{l}_{\sigma(k)}-y^{c}_{\sigma(k)})$
$\displaystyle
p_{y_{\sigma(k+1)}-y_{\sigma(k)}-2n^{\alpha}}(y^{l}_{\sigma(k+1)}-y^{r}_{\sigma(k)})$
$\displaystyle
NI(\mathcal{S}^{1},i_{m},x_{m}^{c},x_{m}^{l})p_{n^{\alpha}}(x_{m}^{c}-x_{m}^{l})NI(\mathcal{S}^{2},j_{\sigma(m)},y_{\sigma(m)}^{c},y_{\sigma(m)}^{r})p_{n^{\alpha}}(y_{\sigma(m)}^{c}-y_{\sigma(m)}^{l})$
$\displaystyle\prod_{i=1}^{k}G_{D}(x^{c}_{k}-y^{c}_{k}).$

The main observation to notice now is that if we were able to freely sum over
the values $x^{r}_{k},x^{l}_{k}$, then we would have that,

$\displaystyle\sum_{x^{r}_{k},x^{l}_{k}}NI(\mathcal{S}^{1},i_{k},x^{c}_{k},x^{r}_{k},x^{l}_{k})p_{n^{\alpha}}(x^{c}_{k}-x^{r}_{k})p_{n^{\alpha}}(x^{l}_{k}-x^{c}_{k})$
$\displaystyle=\frac{\pi^{2}}{8\log n^{\alpha}}(1+o(1)),$

because this just computes the averaged probability that an infinite random
walk does not intersect a the union of two independent random walks of length
$n^{\alpha}$ starting from the origin.

The only term that prevents us from freely summing over $x^{r}_{k}$ and
$x^{l}_{k}$ for all $k$ is the term
$p_{i_{k+1}-i_{k}-2n^{\alpha}}(x^{l}_{k+1}-x^{r}_{k})$. However, if we could
bound this term from above by a constant times
$p_{i_{k+1}-i_{k}}(x^{c}_{k+1}-x^{c}_{k})$, then we would be able to freely
sum over the variables $x^{r}_{k}$ and $x^{l}_{k}$ as desired. This is what we
will argue now.

It is clear that $|x^{l}_{k}-x^{c}_{k}|\leq n^{\alpha}$ and
$|x^{r}_{k}-x^{c}_{k}|\leq n^{\alpha}$. Provided that
$|x^{l}_{k+1}-x^{r}_{k}|\leq(i_{k+1}-i_{k})^{1/2+\epsilon}$ for some small
$\epsilon$, we can apply the local central limit as on the top of Lawler’s
Random Walk [19, Theorem 2.3.12, equation (2.46)] along with the fact that
$i_{k+1}-i_{k}\geq n^{3\alpha}$ to show that,

$p_{i_{k+1}-i_{k}-2n^{\alpha}}(x^{l}_{k+1}-x^{r}_{k})\leq(1+o(1))p_{i_{k+1}-i_{k}}(x^{c}_{k+1}-x^{c}_{k}).$

Otherwise, the probability that
$|x^{l}_{k+1}-x^{r}_{k}|\geq(i_{k+1}-i_{k})^{1/2+\epsilon}$ is exponentially
unlikely with probability at most $\exp[-n^{6\alpha\epsilon}]$. Thus, we
always have the bound,

$\displaystyle p_{i_{k+1}-i_{k}-2n^{\alpha}}(x^{l}_{k+1}-x^{r}_{k})$
$\displaystyle\leq$
$\displaystyle(1+o(1))p_{i_{k+1}-i_{k}}(x^{c}_{k+1}-x^{c}_{k})$
$\displaystyle+\mathbbm{1}[|x^{l}_{k+1}-x^{r}_{k}|\geq(i_{k+1}-i_{k})^{1/2+\epsilon}]p_{i_{k+1}-i_{k}-2n^{\alpha}}(x^{l}_{k+1}-x^{r}_{k}).$

Similar statements also hold for $j$ and $y$.

Furthermore, this term can be substituted into equation (5.15) by replacing
each appearance of $p_{i_{k+1}-i_{k}-2n^{\alpha}}(x^{l}_{k+1}-x^{r}_{k})$ with
the right hand side above. . We can expand each of these products to get a sum
over $2^{4m}$ terms (in each of these terms,
$p_{i_{k+1}-i_{k}-2n^{\alpha}}(x^{l}_{k+1}-x^{r}_{k})$ is replaced with either
$p_{i_{k+1}-i_{k}}(x^{c}_{k+1}-x^{c}_{k})$ or
$\mathbbm{1}[|x^{l}_{k+1}-x^{r}_{k}|\geq(i_{k+1}-i_{k})^{1/2+\epsilon}]p_{i_{k+1}-i_{k}-2n^{\alpha}}(x^{l}_{k+1}-x^{r}_{k})$.)
There is only one of these terms in which each
$p_{i_{k+1}-i_{k}-2n^{\alpha}}(x^{l}_{k+1}-x^{r}_{k})$ is replaced with
$p_{i_{k+1}-i_{k}}(x^{c}_{k+1}-x^{c}_{k})$.

We remark that if even one of the
$p_{i_{k+1}-i_{k}-2n^{\alpha}}(x^{l}_{k+1}-x^{r}_{k})$ were replaced with
$\mathbbm{1}[|x^{l}_{k+1}-x^{r}_{k}|\geq(i_{k+1}-i_{k})^{1/2+\epsilon}]p_{i_{k+1}-i_{k}-2n^{\alpha}}(x^{l}_{k+1}-x^{r}_{k})$,
then such a term would be exponentially suppressed. Indeed, we could trivially
bound all the terms of the form $G(x-y)$ and
$\mathbb{P}[\mathcal{R}^{\prime}_{\mathcal{S}(t)}\cap\mathcal{S}(t-n^{\alpha},t+n^{\alpha})]$
be $1$. Performing a trivial summation shows that this term can be no more
than $n^{2m}\exp[-n^{6\alpha\epsilon}]\ll(m!)^{2}\frac{n^{m}}{(\log n)^{2m}}$
provided $m\leq(\log n)^{2}$. Furthermore, there are no more than $2^{4m}$
such terms. Thus, these terms are clearly negligible.

Now we consider the term in which all
$p_{i_{k+1}-i_{k}-2n^{\alpha}}(x^{l}_{k+1}-x^{r}_{k})$ are replaced with
$(1+o(1))p_{i_{k+1}-i_{k}}(x^{c}_{k+1}-x^{c}_{k})$. In such a term, we can
finally sum over $x^{r}_{k},x^{l}_{k},y^{r}_{k},y^{l}_{k}$ for all $k$. Such a
term will be bounded by,

$\displaystyle\left(\frac{\pi^{2}}{8\alpha\log
n}(1+o(1))\right)^{2m}m!\sum_{\sigma\in S_{n}}\sum_{i_{1}\leq
i_{2}\leq\ldots\leq i_{m}}\sum_{j_{1}\leq j_{2}\ldots\leq
j_{m}}\sum_{x^{c}_{1},\ldots,x^{c}_{k}}\sum_{y^{c}_{1},\ldots,y^{c}_{k}}p_{i_{1}}(x^{c}_{1})p_{j_{\sigma(1)}}(y^{c}_{\sigma(1)})$
$\displaystyle\times\prod_{k=1}^{m-1}p_{i_{k+1}-i_{k}}(x_{k+1}-x_{k})p_{j_{\sigma(k+1)}-j_{\sigma(k)}}(y_{\sigma(k+1)}-y_{\sigma(k)})\prod_{k=1}^{m}G_{D}(x_{k}^{c}-y_{k}^{c}).$

However the last term computes the $m$-th moment of
$\sum_{i=1}^{n}\sum_{j=1}^{n}G_{D}(\mathcal{S}^{1}_{i}-\mathcal{S}^{2}_{j})$.
This is just $m!n^{m}$. Thus, we can bound the line above by

$m!\left(\frac{\pi^{4}n}{64\alpha(\log n)^{2}}(1+o(1))\right)^{m}.$

This completes the proof of the claim.

∎

###  5.3. Lower Bound for the Large Deviation of $TL$

Our goal in this section is to show the following statement.

In this section, our goal is to understand,

$\lim_{n\to\infty}\frac{1}{b_{n}}\log\sum_{m=0}^{\infty}\frac{1}{m!}\theta^{m}\left(\frac{b_{n}(\log
n)^{2}}{n}\right)^{m/2}\left(\mathbb{E}[(TL^{\prime}_{n})^{m}]\right)^{1/2}.$

######  Theorem 5.9.

If $b_{n}=\text{O}(\log\log n)$ and satisfies $\lim_{n\to\infty}b_{n}=\infty$,
one has that,

$\liminf_{n\to\infty}\frac{1}{b_{n}}\log\sum_{m=0}^{\infty}\frac{1}{m!}\theta^{m}\left(\frac{b_{n}(\log
n)^{2}}{n}\right)^{m/2}\left(\mathbb{E}[(TL^{\prime}_{n})^{m}]\right)^{1/2}\geq\tilde{\kappa}(4,2)^{4}\frac{\pi^{4}\theta^{2}}{128}.$

###### Proof.

First, notice that

$TL^{\prime}_{n}=\sum_{a\in\mathbb{Z}^{d}}\sum_{i=1}^{b_{n}}\sum_{x^{1}\in\mathcal{S}^{1}_{i}}\tilde{G}_{D}(x^{1}-a)\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}_{i}=\emptyset)\sum_{j=1}^{b_{n}}\sum_{x^{2}\in\mathcal{S}^{2}_{j}}\tilde{G}_{D}(x^{2}-a)\mathbb{P}(R^{\prime}_{x^{2}}\cap\mathcal{S}^{2}_{i}).$

If we let

$\mathcal{G}_{n}:=\mathbb{E}\left[\sum_{i=1}^{b_{n}}\sum_{x^{1}\in\mathcal{S}^{1}_{i}}\tilde{G}_{D}(x^{1}-a)\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}_{i}=\emptyset)\right],$

we see that,

$\mathbb{E}[(TL^{\prime}_{n})^{m}]=\sum_{a_{1},\ldots,a_{m}}\prod_{i=1}^{m}(\mathcal{G}_{n}(a_{i}))^{2}.$

If we now let $f$ be any function satisfying
$\int_{a\in\mathbb{R}^{4}}f(a)^{2}=1$ and
$C_{f}^{n}:=\sum_{a\in\mathbb{Z}^{4}}f(\sqrt{\frac{b_{n}}{n}}a)^{2}$, by the
Cauchy-Schwartz inequality, we see that we obtain that,

$\displaystyle\mathbb{E}[(TL^{\prime}_{n})^{m}]^{1/2}$
$\displaystyle=(C_{f}^{n})^{-m}\left(\sum_{a_{1},\ldots,a_{m}}\prod_{i=1}^{m}(\mathcal{G}_{n}(a_{i}))^{2}\right)^{1/2}\left(\sum_{a_{1},\ldots,a_{m}}\prod_{i=1}^{m}f\left(\sqrt{\frac{b_{n}}{n}}a_{i}\right)^{2}\right)^{1/2}$
$\displaystyle\geq(C_{f}^{n})^{-m/2}\sum_{a_{1},\ldots,a_{m}}\mathcal{G}_{n}(a_{i})f(\sqrt{\frac{b_{n}}{n}}a_{i}).$

Defining

$\mathcal{F}_{n}^{i}=(C_{f}^{n})^{-1/2}\sum_{x^{1}\in\mathcal{S}^{1}_{i}}\sum_{a\in\mathbb{Z}^{d}}\tilde{G}_{D}(x^{1}-a)f(\sqrt{\frac{b_{n}}{n}}a)\mathbb{P}(R^{\prime}_{x^{1}}\cap\mathcal{S}^{1}_{i}=\emptyset),$

we see that

$\mathbb{E}[(TL^{\prime}_{n})^{m/2}]\geq\mathbb{E}[(\sum_{i=1}^{b_{n}}\mathcal{F}_{n}^{i})^{m}].$

Furthermore, $\mathcal{F}_{n}^{i}$ is a function of only the portion
$\mathcal{S}^{1}_{i}$ of the random walk. Thus, we see that,

(5.16) $\sum_{m=0}^{\infty}\frac{1}{m!}\theta^{m}\left(\frac{b_{n}(\log
n)^{2}}{n}\right)^{m/2}\left(\mathbb{E}[(TL^{\prime}_{n})^{m}]\right)^{1/2}\geq\mathbb{E}\left[\exp\left[\theta\frac{\sqrt{b_{n}}(\log
n)}{\sqrt{n}}\sum_{i=1}^{b_{n}}\mathcal{F}^{i}_{n}\right]\right].$

Notice that for any $\epsilon>0$, we have,

$\displaystyle\frac{1}{b_{n}}\log\mathbb{E}\left[\exp\left[\theta\frac{\sqrt{b_{n}}\log
n}{\sqrt{n}}\sum_{i=1}^{b_{n}}\mathcal{F}^{i}_{n}\right]\right]\geq$
$\displaystyle\frac{1}{b_{n}}(1+\epsilon)\log\mathbb{E}\left[\frac{1}{1+\epsilon}\theta\frac{\sqrt{b_{n}}\log
n}{\sqrt{n}}\sum_{i=2}^{b_{n}}\mathcal{F}^{i}_{n}\right]$
$\displaystyle-\frac{1}{b_{n}}\epsilon\log\mathbb{E}\left[\exp\left(-\frac{1+\epsilon}{\epsilon}\theta\mathcal{F}^{1}_{n}\right)\right].$

Notice that for any fixed $\epsilon$, an upper bound on the large deviation
statistics of $\mathcal{F}^{1}_{n}$ (which can be inherited from an upper
bound large statistics on $TL^{\prime}_{\frac{n}{b_{n}}}$ as in equation
(5.16)) shows that as $n\to\infty$ the term on the right goes to $0$. Finally,
we can take $\epsilon$ to $0$ to note that the term,

$\displaystyle\liminf_{n\to\infty}\frac{1}{b_{n}}\log\mathbb{E}\left[\exp\left[\theta\frac{\sqrt{b_{n}}\log
n}{\sqrt{n}}\sum_{i=1}^{b_{n}}\mathcal{F}^{i}_{n}\right]\right]$
$\displaystyle\geq\lim_{\epsilon\to
0}\liminf_{n\to\infty}\frac{1}{b_{n}}(1+\epsilon)\log\mathbb{E}\left[\frac{1}{1+\epsilon}\theta\frac{\sqrt{b_{n}}\log
n}{\sqrt{n}}\sum_{i=2}^{b_{n}}\mathcal{F}^{i}_{n}\right].$

To find the lower bound on the term on the left hand side, it suffices to find
abound for the right hand side. We are now in a very similar situation to that
of [13, Theorem 7.1.2]. The function $\mathcal{F}^{i}_{n}$; though they are
not exactly in the same format, one can see that $\mathcal{F}^{i}_{n}$ takes
the same role as that of the term
$\sum_{x\in\mathcal{S}^{1}_{i}}f(\sqrt{\frac{b_{n}}{n}}x)$. Indeed, we can
define the operator,

$\displaystyle\mathfrak{B}_{n}\xi(x)=$
$\displaystyle\mathbb{E}\bigg{(}\exp\bigg{\\{}\frac{\sqrt{b_{n}}(\log
n)}{\sqrt{n}}\sum_{y-x\in\mathcal{S}[1,nb_{n}^{-1}]}(C^{n}_{f})^{-1/2}\sum_{a\in\mathbb{Z}^{d}}\tilde{G}_{D}(y-a)f\left(\sqrt{\frac{b_{n}}{n}}a\right)$
$\displaystyle\times\mathbb{P}\left(R^{\prime}_{y-x}\cap\mathcal{S}[1,b_{n}n^{-1}]=\emptyset\right)\bigg{\\}}\xi(x+\mathcal{S}_{nb_{n}^{-1}})\bigg{)}.$

This operator is a symmetric operator and following the proof of Lemma 7.1.3
of Chen’s book. We see that we can derive the following bound. Let $g$ be a
bounded function on $\mathbb{R}^{4}$ that is infinitely differentiable and
supported on a finite box with $\int_{\mathbb{R}^{4}}g^{2}(x)\text{d}x=1$.
There exists a constant depending only on $g$ (but not on $n$) that

$\mathbb{E}\left[\exp\left[\theta\frac{\sqrt{b_{n}}\log
n}{\sqrt{n}}\sum_{i=2}^{b_{n}}\mathcal{F}^{i}_{n}\right]\right]\geq\delta\langle\xi_{n},\mathcal{B}_{n}^{b_{n}-1}\xi_{n}\rangle\geq\delta\langle\xi_{n},\mathcal{B}_{n}\xi_{n}\rangle^{b_{n}-1}.$

The term is

$\displaystyle\langle\xi_{n},\mathcal{B}_{n}\xi_{n}\rangle$
$\displaystyle=(1+o(1))\int_{\mathbb{R}^{4}}\text{d}xg(x)$
$\displaystyle\times\mathbb{E}\bigg{(}\exp\bigg{[}\theta\frac{\sqrt{b_{n}}\log
n}{\sqrt{n}}\sum_{y\in\mathcal{S}[1,nb_{n}^{-1}]}(C_{f}^{n})^{-1/2}\sum_{a\in\mathbb{Z}^{d}}\tilde{G}_{D}(y+x-a)f\left(\sqrt{\frac{b_{n}}{n}}a\right)$
$\displaystyle\times\mathbb{P}(R^{\prime}_{y}\cap\mathcal{S}[1,nb_{n}^{-1}]=\emptyset)\bigg{]}g\left(x+\sqrt{\frac{b_{n}}{n}}\mathcal{S}_{nb_{n}^{-1}}\right)\bigg{)}$
$\displaystyle\rightarrow\int_{\mathbb{R}^{2}}\text{d}xg(x)\mathbb{E}\left(\exp\left\\{\int_{0}^{1}\frac{\pi^{2}}{8}(f*\tilde{G}_{D})(x+W(t))\text{d}t\right\\}g(x+W(1))\right).$

Let us explain some steps in the derivation above. First, we remark that the
term inside the exponential has finite expectation. Secondly, we also have the
second moment comparison estimate

(5.17) $\displaystyle\mathbb{E}\bigg{[}\frac{\sqrt{b_{n}}\log
n}{\sqrt{n}}\sum_{y\in
S[1,nb_{n}^{-1}]}(C_{f}^{n})^{-1/2}\sum_{a\in\mathbb{Z}^{d}}\tilde{G}_{D}(y+x-a)f\left(\sqrt{\frac{b_{n}}{n}}a\right)\mathbb{P}(\mathcal{R}^{\prime}_{y}\cap\mathcal{S}[1,nb_{n}^{-1}]=\emptyset)$
$\displaystyle-\frac{\pi^{2}}{8}\frac{\sqrt{b_{n}}}{\sqrt{n}}(C_{f}^{n})^{-1/2}\sum_{a\in\mathbb{Z}^{d}}\tilde{G}_{D}(y+x-a)f\left(\sqrt{\frac{b_{n}}{n}}a\right)\bigg{]}^{2}\to
0.$

As before, this follows from computations similar to those found in the proof
of Claim 5.8 to allow us to replace $\mathbb{P}(\mathcal{R}^{\prime}_{y}\cap
S[1,nb_{n}^{-1}]=\emptyset)$ with $\frac{\pi^{2}}{8\log n}$. Combining these
observations, we see that as $n\to\infty$

(5.18)
$\displaystyle\int_{\mathbb{R}^{4}}\text{d}xg(x)\mathbb{E}\bigg{(}\exp\bigg{[}\theta\frac{\sqrt{b_{n}}\log
n}{\sqrt{n}}\sum_{y\in\mathcal{S}[1,nb_{n}^{-1}]}(C_{f}^{n})^{-1/2}\sum_{a\in\mathbb{Z}^{d}}\tilde{G}_{D}(y+x-a)f\left(\sqrt{\frac{b_{n}}{n}}a\right)$
$\displaystyle\times\mathbb{P}(\mathcal{R}^{\prime}_{y}\cap\mathcal{S}[1,nb_{n}^{-1}]=\emptyset)\bigg{]}g\left(x+\sqrt{\frac{b_{n}}{n}}\mathcal{S}_{nb_{n}^{-1}}\right)\bigg{)}\to$
$\displaystyle\int_{\mathbb{R}^{4}}\text{d}xg(x)\mathbb{E}\bigg{(}\exp\bigg{[}\theta\frac{\pi^{2}}{8}\frac{\sqrt{b_{n}}\log
n}{\sqrt{n}}\sum_{y\in\mathcal{S}[1,nb_{n}^{-1}]}(C_{f}^{n})^{-1/2}\sum_{a\in\mathbb{Z}^{d}}\tilde{G}_{D}(y+x-a)f\left(\sqrt{\frac{b_{n}}{n}}a\right)\bigg{]}$
$\displaystyle\times
g\left(x+\sqrt{\frac{b_{n}}{n}}\mathcal{S}_{nb_{n}^{-1}}\right)\bigg{)}.$

The last line will go to the corresponding Brownian motion quantity by the
invariance principle. Finally, by Remark A.17, we see we obtain our desired
constant.

∎

##  Appendix A Green’s Function Estimates

In this section, we will study the weak convergence of an appropriately scaled
version of $\mathfrak{G}_{n}$ to $\mathcal{G}_{B}[0,1]$. We consider the
following quantity:

(A.1)
$\mathcal{G}_{n}:=\frac{1}{n^{2}}\sum_{i=1}^{n}\sum_{j=1}^{n}(\sqrt{n})^{d-2}G_{D}\left(\sqrt{n}\left(\frac{\mathcal{S}^{1}_{i}}{\sqrt{n}}-\frac{\mathcal{S}^{2}_{j}}{\sqrt{n}}\right)\right)$

and show that that it will converge weakly following Brownian quantity,

(A.2)
$\mathcal{G}_{B}([0,1])=\int_{0}^{1}\int_{0}^{1}G(B_{1}(t)-B_{2}(s))\text{d}t\text{d}s.$

Our main result will be the following,

######  Theorem A.1.

As $n\to\infty$, we have that for any integer $m\geq 0$ that

(A.3)
$\lim_{n\to\infty}\mathbb{E}[(\mathcal{G}_{n})^{m}]=\mathbb{E}[(\mathcal{G}_{B}[0,1])^{m}].$

One can imagine that such a convergence is a consequence of the invariance
principle for the random walk. However, we need to be more careful regarding
the behavior of the Green’s function near the origin. For this reason, we will
consider the following smoothed out versions of our quantities of interest.
Note that

(A.4) $\displaystyle\mathcal{G}_{n,\epsilon}:=$
$\displaystyle\frac{1}{n^{2}}\sum_{i=1}^{n}\sum_{j=1}^{n}\frac{1}{n^{2}}\sum_{z_{1}\in\frac{1}{\sqrt{n}}\mathbb{Z}^{4}}\frac{1}{n^{2}}\sum_{z_{2}\in\frac{1}{\sqrt{n}}\mathbb{Z}^{4}}P_{\epsilon}\left(\frac{1}{\sqrt{n}}\mathcal{S}^{1}_{i}-z_{1}\right)$
$\displaystyle\hskip 28.45274pt\times
n^{2/2}G_{D}(\sqrt{n}z_{1}-\sqrt{n}z_{2})P_{\epsilon}\left(\frac{1}{\sqrt{n}}\mathcal{S}^{2}_{j}-z_{2}\right),$

where $P_{\epsilon}$ is the probability density function for a Gaussian random
variable with variance $2\epsilon$,
$P_{\epsilon}(z)\propto\frac{1}{\epsilon^{2}}\exp[-|z|^{2}/\epsilon]$, and
$C_{\epsilon}$ is the normalization factor
$C_{\epsilon}=\sum_{x\in\frac{1}{\sqrt{n}}\mathbb{Z}^{4}}P_{\epsilon}(x)$.
This will, in turn, approach the following function weakly:

$\mathcal{G}_{B,\epsilon}:=\int_{0}^{1}\int_{0}^{1}\int_{\mathbb{R}^{4}}\int_{\mathbb{R}^{4}}P_{\epsilon}(B_{1}(t)-z_{1})G(z_{1}-z_{2})P_{\epsilon}(B_{2}(s)-z_{2})\text{d}t\text{d}s\text{d}z_{1}\text{d}z_{2}.$

All moments of $\mathcal{G}_{B,\epsilon}$ will match those of
$\mathcal{G}_{B}$ as $\epsilon$ approaches 0. The details are in Corollary
A.11. Using this fact, our main Theorem A.1 will be a consequence of the
following two results,

######  Theorem A.2.

Fix $\epsilon$, then as $n\to\infty$, we have that for all $m$ that,

(A.5)
$\lim_{n\to\infty}\mathbb{E}[(\mathcal{G}_{n,\epsilon})^{m}]=\mathbb{E}[(\mathcal{G}_{B,\epsilon})^{m}].$

and

######  Theorem A.3.

We have the following limiting relation between the moments of
$\mathcal{G}_{\epsilon}$ and $\mathcal{G}$.

$\lim_{\epsilon\to
0}\limsup_{n\to\infty}\mathbb{E}[(\mathcal{G}_{n,\epsilon}-\mathcal{G}_{n})^{2}]=0.$

Given Theorems A.2 and A.3, the proof of Theorem A.1 follows similarly from
the proof of [13, Theorem 5.3.2]. At this point, we start with proving
Theorems A.2

To start our analysis, we consider continuity properties of the following
function:
$\tilde{G}_{\epsilon}(a):=\int_{e\in\mathbb{R}^{4}}P_{\epsilon}(a-e)\tilde{G}(e)\text{d}e$
and the discrete analog
$\tilde{G}_{D,\epsilon}(\sqrt{n}a):=\frac{1}{n^{2}}\sum_{e\in\frac{1}{\sqrt{n}}\mathbb{Z}^{4}}P_{\epsilon}(a-e)\tilde{G}_{D}(\sqrt{n}e)$.
(Note: $\tilde{G}_{D,\epsilon}$ has an implicit dependence in $n$, we will
drop this from the notation in what follows since the context is clear.) One
can see that these functions have the property that,

(A.6)
$\displaystyle\int\tilde{G}_{\epsilon}(a-e)\tilde{G}_{\epsilon}(b-e)\text{d}e=\int_{\mathbb{R}^{4}}P_{\epsilon}(a-z_{1})G(z_{1}-z_{2})P_{\epsilon}(b-z_{2})\text{d}z_{1}\text{d}z_{2},$
$\displaystyle\frac{1}{n^{2}}\sum_{e\in\frac{1}{\sqrt{n}}\mathbb{Z}^{4}}n^{3/2}\tilde{G}_{D,\epsilon}(\sqrt{n}a-\sqrt{n}e)n^{3/2}\tilde{G}_{D,\epsilon}(\sqrt{n}b-\sqrt{n}e)$
$\displaystyle=\frac{1}{n^{4}}\sum_{z_{1},z_{2}\in\frac{1}{\sqrt{n}}\mathbb{Z}^{4}}P_{\epsilon}(a-z_{1})n^{2/2}G_{D}(\sqrt{n}z_{1}-\sqrt{n}z_{2})P_{\epsilon}(b-z_{2}).$

######  Lemma A.4.

There is some constant, depending on $\epsilon$, such that the following
estimates hold uniformly in $a$,

$|\tilde{G}_{\epsilon}(a)|\lesssim
1,\quad|\tilde{G}_{\epsilon}(a+\kappa)-\tilde{G}_{\epsilon}(a)|\lesssim\kappa.$

###### Proof.

First see that,

$\displaystyle\tilde{G}_{\epsilon}(a)$ $\displaystyle=\int_{|e|\leq
1}P_{\epsilon}(a-e)\tilde{G}(e)\text{d}e+\int_{|e|\geq
1}P_{\epsilon}(a-e)\tilde{G}(e)\text{d}e$
$\displaystyle\leq||P_{\epsilon}||_{\infty}\int_{|e|\leq
1}\tilde{G}(e)\text{d}e+\left[\int_{\mathbb{R}^{4}}P^{2}_{\epsilon}(a-e)\text{d}e\right]^{1/2}\left[\int_{|e|\geq
1}\tilde{G}^{2}(e)\text{d}e\right]^{1/2}.$

By applying a similar inequality, we also have that,

$\displaystyle\int_{\mathbb{R}^{4}}|\nabla
P_{\epsilon}(a-e)|\tilde{G}(e)\text{d}e$ $\displaystyle\leq\int_{|e|\leq
1}||\nabla P_{\epsilon}(a-e)||\tilde{G}(e)\text{d}e+\int_{|e|\geq 1}||\nabla
P_{\epsilon}(a-e)||\tilde{G}(e)\text{d}e$ $\displaystyle\leq||\nabla
P_{\epsilon}||_{\infty}\int_{|e|\leq
1}\tilde{G}(e)\text{d}e+\left[\int_{\mathbb{R}^{4}}||\nabla
P_{\epsilon}(a-e)||^{2}\text{d}e\right]^{1/2}\left[\int_{|e|\geq
1}\tilde{G}^{2}(e)\text{d}e\right].$

Thus, we see that,

$\displaystyle|\tilde{G}_{\epsilon}(a+\kappa)-\tilde{G}_{\epsilon}(a)|\leq\int_{\mathbb{R}^{4}}|P_{\epsilon}(a+\kappa-e)-P_{\epsilon}(a-e)|\tilde{G}(e)\text{d}e$
$\displaystyle\leq\int_{0}^{\kappa}\text{d}t\int_{\mathbb{R}^{4}}\left|\left\langle\nabla
P_{\epsilon}(a+t-e),\frac{\kappa}{||\kappa||}\right\rangle\right|\tilde{G}(e)\text{d}e$
$\displaystyle\leq\int_{0}^{\kappa}\text{d}t\int_{\mathbb{R}^{4}}||\nabla
P_{\epsilon}(a+t-e)||\tilde{G}(e)\text{d}e\lesssim\kappa$

and we obtain the desired result. ∎

We can now give a brief overview of Theorem A.2.

###### Proof of Theorem A.2.

Consider the following function of $f_{1}$ and $f_{2}$

(A.7)
$H(f_{1},f_{2}):=\int_{\mathbb{R}^{4}}\text{d}e\prod_{i=1}^{2}\left[\int_{0}^{1}\text{d}t\tilde{G}_{\epsilon}(f_{i}(t)-e)\right].$

As a consequence of the above lemma, we know that $H(f_{1},f_{2})$ is
continuous in $f_{1}$ and $f_{2}$. Thus, we can apply the invariance
principle, we can deduce that,

(A.8)
$\displaystyle\int_{\mathbb{R}^{4}}\text{d}e\prod_{i=1}^{2}\left[\frac{1}{n}\sum_{t=1}^{n}\tilde{G}_{\epsilon}\left(\frac{\mathcal{S}^{i}_{t}}{\sqrt{n}}-e\right)\right]\rightarrow_{d}\int_{\mathbb{R}^{4}}\text{d}e\prod_{i=1}^{2}\left[\int_{0}^{1}\text{d}t\tilde{G}_{\epsilon}(B_{i}(t)-e)\right]$
$\displaystyle\hskip
28.45274pt=\int_{0}^{1}\int_{0}^{1}\int_{\mathbb{R}^{4}}\int_{\mathbb{R}^{4}}P_{\epsilon}(B_{1}(t)-z_{1})G(z_{1}-z_{2})P_{\epsilon}(B_{2}(s)-z_{2})\text{d}t\text{d}s\text{d}z_{1}\text{d}z_{2}.$

Now, we have the bound the difference between the left hand side of the
quantity above in (A.8) to the quantity that appears in the right hand side of
(A.4). Namely, we will get weak convergence of $\mathcal{G}_{B,\epsilon}$,
provided we have the following discretization estimate.

######  Lemma A.5.

Fix $\epsilon>0$. If uniformly in $a_{1}$ and $a_{2}$, we have that as
$n\to\infty$,

(A.9) $\displaystyle\left|\int_{\mathbb{R}^{4}}\tilde{G}_{\epsilon}(\lfloor
a_{1}\rfloor_{n}-e)\tilde{G}_{\epsilon}(\lfloor
a_{2}\rfloor_{n}-e)\text{d}e-\frac{1}{n^{2}}\sum_{e\in\frac{1}{\sqrt{n}}\mathbb{Z}^{4}}\prod_{i=1}^{2}n^{3/2}\tilde{G}_{D,\epsilon}(\sqrt{n}(a_{i}-e))\right|=o(1).$

Provided the above lemma is true, we apply it with values
$a_{1}=\mathcal{S}^{1}_{t_{1}}$ and $a_{2}=\mathcal{S}^{2}_{t_{2}})$, sum, and
then average to see that,

(A.10)
$\displaystyle\left|\frac{1}{n^{2}}\sum_{t_{1},t_{2}=1}^{n}\int_{\mathbb{R}^{4}}\tilde{G}_{\epsilon}\left(\frac{\mathcal{S}^{1}_{t_{1}}}{\sqrt{n}}-e\right)\tilde{G}_{\epsilon}\left(\frac{\mathcal{S}^{2}_{t_{2}}}{\sqrt{n}}-e\right)\text{d}e-\mathcal{G}_{n,\epsilon}\right|=o(1).$

In the next subsection, we will provide technical estimates to derive Lemma
A.5.

∎

###  A.1. Discretization and analysis of the functional $H(f_{1},f_{2})$

We start with a few intermediate lemmas. The first lemma allows us to reduce
the domain of integration of $\tilde{G}_{\epsilon}(a)$ from all of
$\mathbb{R}^{4}$, to an integration over a region of finite support.

######  Lemma A.6.

Fix some $\delta_{2}>\delta_{1}>0$. Let $\chi$ be a smooth positive function
supported on $[-n^{\delta_{2}},n^{\delta_{2}}]^{4}$ bounded by $1$ and such
that $\chi$ is 1 on $[-n^{\delta_{1}},n^{\delta_{1}}]^{4}$. Then, uniformly
for $a,b\in\mathbb{R}^{4}$, there is some constant, possibly depending on
$\epsilon$, such that the following estimate holds,

(A.11)
$\int_{\mathbb{R}^{4}}\tilde{G}_{\epsilon}(a-e)\tilde{G}_{\epsilon}(b-e)\text{d}e-\int_{\mathbb{R}^{4}}\chi(a-e)\tilde{G}_{\epsilon}(a-e)\tilde{G}_{\epsilon}(b-e)\chi(b-e)\text{d}e\lesssim
n^{-2\delta_{1}/4}.$

As intermediate estimates, we also have, for sufficiently large $n$ depending
on $\delta_{1}$ and $\epsilon$, that

(A.12) $\int_{\mathbb{R}^{4}}\tilde{G}_{\epsilon}^{2}(e)\text{d}e\lesssim
1,\quad\int_{|e|\geq
n^{\delta_{1}}}\tilde{G}_{\epsilon}^{2}(e)\text{d}e\lesssim
n^{-2\delta_{1}/4}.$

######  Remark A.7.

By similar methods, we would also have a corresponding equation for
$\tilde{G}_{\epsilon,D}$. Namely, we would have,

$\displaystyle\frac{1}{n^{2}}\sum_{e\in\frac{1}{\sqrt{n}}\mathbb{Z}^{4}}n^{3/2}\tilde{G}_{\epsilon,D}(\sqrt{n}a-\sqrt{n}e)n^{3/2}\tilde{G}_{\epsilon,D}(\sqrt{n}b-\sqrt{n}e)$
$\displaystyle-\frac{1}{n^{2}}\sum_{e\in\frac{1}{\sqrt{n}}\mathbb{Z}^{4}}\chi(a-e)n^{3/2}\tilde{G}_{\epsilon,D}(\sqrt{n}a-\sqrt{n}e)n^{3/2}\tilde{G}_{\epsilon,D}(\sqrt{n}b-\sqrt{n}e)\chi(b-e)\lesssim
n^{-3\delta_{1}/2}.$

###### Proof.

By the Cauchy-Schwartz inequality, we see that we have,

(A.13)
$\displaystyle\left|\int_{\mathbb{R}^{4}}\tilde{G}_{\epsilon}(a-e)\tilde{G}_{\epsilon}(b-e)\text{d}e-\int_{\mathbb{R}^{4}}\chi(a-e)\tilde{G}_{\epsilon}(a-e)\tilde{G}_{\epsilon}(b-e)\chi(b-e)\text{d}e\right|$
$\displaystyle\leq\int_{\mathbb{R}^{4}}\tilde{G}_{\epsilon}(a-e)(1-\chi(a-e))\tilde{G}_{\epsilon}(b-e)\text{d}e+\int_{\mathbb{R}^{4}}\tilde{G}_{\epsilon}(a-e)\chi(a-e)\tilde{G}_{\epsilon}(b-e)(1-\chi(b-e))\text{d}e$
$\displaystyle\leq
2\left[\int_{\mathbb{R}^{4}}(\tilde{G}_{\epsilon}(a-e)(1-\chi(a-e)))^{2}\text{d}e\right]^{1/2}\left[\int_{\mathbb{R}^{4}}\tilde{G}^{2}_{\epsilon}(a-e)\text{d}e\right]^{1/2}$
$\displaystyle\leq
2\left[\int_{\mathbb{R}^{4}}(\tilde{G}_{\epsilon}(e))^{2}\text{d}e\right]^{1/2}\left[\int_{|e|\geq
n^{\delta_{1}}}\tilde{G}^{2}_{\epsilon}(e)\text{d}e\right]^{1/2}.$

All that remains is to estimate the integrals
$\int_{\mathbb{R}^{4}}\tilde{G}^{2}_{\epsilon}(e)\text{d}e$ and $\int_{|e|\geq
n^{\delta_{1}}}\tilde{G}^{2}_{\epsilon}(e)\text{d}e.$ We have that,

(A.14)
$\displaystyle\int_{\mathbb{R}^{4}}\tilde{G}^{2}_{\epsilon}(e)\text{d}e$
$\displaystyle=\int_{z_{1},a\in\mathbb{R}^{4}}\text{d}aG(a)\int
P_{\epsilon}(z_{1})P_{\epsilon}(z_{1}+a)\text{d}z_{1}$
$\displaystyle\lesssim\int_{a\in\mathbb{R}^{4}}\text{d}aG(a)\exp[-|a|^{2}/(2\epsilon)]\leq\infty.$

We use the fact that
$G(a)\exp[-|a|^{2}/2]\propto\frac{1}{|a|^{d-2}}\exp[-|a|^{2}/2]$ is integrable
both near the origin and at $\infty$.

Furthermore, we see that,

(A.15) $\int_{|e|\geq
n^{\delta_{1}}}\tilde{G}^{2}(e)\text{d}e=\int_{z_{1},z_{2}\in\mathbb{R}^{4}}\text{d}z_{1}\text{d}z_{2}P_{\epsilon}(z_{1})P_{\epsilon}(z_{2})\int_{|e|\geq
n^{\delta_{1}}}\tilde{G}(e-z_{1})\tilde{G}(z_{2}-e)\text{d}e.$

If one of $z_{1}$ or $z_{2}$ were greater in absolute value that
$n^{\delta_{1}}/2$, then we apply the trivial bound that $\int_{|e|\geq
n^{\delta_{1}}}\tilde{G}(e-z_{1})\tilde{G}(z_{2}-e)\text{d}e\leq\int_{e\in\mathbb{R}^{4}}\tilde{G}(e-z_{1})\tilde{G}(z_{2}-e)=G(z_{1}-z_{2})$.
If instead, both of $z_{1}$ and $z_{2}$ are less in absolute value that
$n^{\delta_{1}}/2$, then we have the bound,

$\displaystyle\int_{|e|\geq
n^{\delta_{1}}}\tilde{G}(e-z_{1})\tilde{G}(z_{2}-e)\text{d}e$
$\displaystyle\leq\left[\int_{|e|\geq
n^{\delta_{1}}}\tilde{G}^{2}(e-z_{1})\text{d}e\right]^{1/2}\left[\int_{|e|\geq
n^{\delta_{1}}}\tilde{G}^{2}(z_{2}-e)\text{d}e\right]^{1/2}$
$\displaystyle\leq\int_{|e|\geq
n^{\delta_{1}}/2}\tilde{G}^{2}(e)\text{d}e\lesssim\frac{1}{n^{2\delta_{1}}}.$

Returning to (A.15), we see that we can bound this by,

(A.16) $\displaystyle\int_{|e|\geq n^{\delta_{1}}}\tilde{G}^{2}(e)\text{d}e$
$\displaystyle\lesssim\frac{1}{n^{2\delta_{1}}}\int_{|z_{1}|,|z_{2}|\leq
n^{\delta_{1}}/2}P_{\epsilon}(z_{1})P_{\epsilon}(z_{2})\text{d}z_{1}\text{d}z_{2}$
$\displaystyle\hskip 28.45274pt+2\int_{|z_{1}|\geq
n^{\delta}/2,z_{2}\in\mathbb{R}^{4}}P_{\epsilon}(z_{1})P_{\epsilon}(z_{2})G(z_{1}-z_{2})\text{d}z_{1}\text{d}z_{2}$
$\displaystyle\lesssim\frac{1}{n^{2\delta_{1}}}+2\int_{a\in\mathbb{R}^{4}}\text{d}aG(a)\int_{|z_{1}|\geq
n^{\delta_{1}}/2}P_{\epsilon}(z_{1})P_{\epsilon}(a+z_{1})\text{d}z_{1}.$

To finish with our estimate, we need to have two different estimates on
$\int_{|z_{1}|\geq
n^{\delta_{1}}/2}P_{\epsilon}(z_{1})P_{\epsilon}(a+z_{1})\text{d}z_{1}$ based
on the value of $a$. If $a\geq n^{\delta_{1}}/2$, we use the trivial estimate,

$\int_{|z_{1}|\geq
n^{\delta_{1}}/2}P_{\epsilon}(z_{1})P_{\epsilon}(a+z_{1})\text{d}z_{1}\leq\int_{z_{1}\in\mathbb{R}^{4}}P_{\epsilon}(-z_{1})P_{\epsilon}(a+z_{1})\text{d}z_{1}=O(1)\exp[-|a|^{2}/(2\epsilon)].$

Otherwise, we have the estimate,

$\displaystyle\int_{|z_{1}|\geq
n^{\delta_{1}}/2}P_{\epsilon}(z_{1})P_{\epsilon}(a+z_{1})\text{d}z_{1}$
$\displaystyle\leq\left[\int_{|z_{1}|\geq
n^{\delta_{1}}/2}P^{2}_{\epsilon}(z_{1})\text{d}z_{1}\right]^{1/2}\left[\int_{|z_{1}|\geq
n^{\delta_{1}}/2}P^{2}_{\epsilon}(z_{1}+a)\text{d}z_{1}\right]^{1/2}$
$\displaystyle\lesssim\exp[-n^{2\delta_{1}}/(8\epsilon)].$

Here, we mainly used properties of the Gaussian distribution to assert that
the integral $\int_{|z_{1}|\geq
n^{\delta_{1}}/2}P^{2}_{\epsilon}(z_{1})\text{d}z_{1}\lesssim\exp[-n^{2\delta_{1}}/(4\epsilon)]$
for $n$ sufficiently large. The other integral $\int_{|z_{1}|\leq
n^{\delta_{1}}/2}P^{2}_{\epsilon}(z_{1})\text{d}z_{1}\leq\int_{z_{1}\in\mathbb{R}^{4}}P^{2}_{\epsilon}(z_{1})\text{d}z_{1}\lesssim
1$. Returning to the last line of (A.16), we see that,

(A.17) $\displaystyle 2\int_{a\in\mathbb{R}^{4}}\text{d}aG(a)\int_{|z_{1}|\geq
n^{\delta_{1}}/2}P_{\epsilon}(z_{1})P_{\epsilon}(a+z_{1})\text{d}z_{1}$
$\displaystyle\lesssim\int_{|a|\geq
n^{\delta_{1}}/2}G(a)\exp[-|a|^{2}/(2\epsilon)]\text{d}a+\int_{|a|\leq
n^{\delta_{1}}/2}G(a)\exp[-n^{2\delta}/(8\epsilon)]\text{d}a$
$\displaystyle\lesssim
n^{3\delta_{1}/2}\exp[-n^{2\delta_{1}}/(8\epsilon)]+n^{2\delta_{1}}\exp[-n^{2\delta_{1}}/(8\epsilon)].$

The first and second integrals on the second line both come from direct
integration, one you substitute $G(a)\propto\frac{1}{|a|^{3}}$. When $n$ is
sufficiently large,
$\exp[-n^{2\delta_{1}}/(8\epsilon)]\ll\frac{1}{n^{2\delta_{1}}}$. Combining
the estimates in (A.17) and (A.16), we finally have the estimate,

$\int_{|e|\geq
n^{\delta_{1}}}\tilde{G}^{2}_{\epsilon}(e)\text{d}e\lesssim\frac{1}{n^{2\delta_{1}/2}}.$

This estimate, along with equation (A.14) can be put into the last line of
equation (A.13) to derive the inequality as desired in equation (A.11).

∎

By substituting $a=f_{1}(t)$ and $b=f_{2}(s)$ in Lemma A.6 and integrating
over the two time variables $t$ and $s$, one can obtain the following as a
simple corollary.

######  Corollary A.8.

Let $\chi$ be a function as in Lemma A.6. Then, we have that,

$\displaystyle\int_{\mathbb{R}^{4}}\text{d}e\prod_{i=1}^{2}\left[\int_{0}^{1}\text{d}t\int_{\mathbb{R}^{4}}P_{\epsilon}\left(f_{i}(t)-z\right)\tilde{G}(z-e)\text{d}z\right]$
$\displaystyle-\int_{\mathbb{R}^{4}}\text{d}e\prod_{i=1}^{2}\left[\int_{0}^{1}\text{d}t\int_{\mathbb{R}^{4}}P_{\epsilon}\left(f_{i}(t)-z\right)\tilde{G}(z-e)\chi(z-e)\text{d}z\right]=o(1),$

uniformly in $f_{1},f_{2}$.

We also have similar claims when considering $\tilde{G}_{D,\epsilon}$.

After the reduction to a region of finite support, our next lemma allows us to
replace $\tilde{G}_{\epsilon}$ with an appropriate discrete form closer to one
found in the expression of the discrete computation.

######  Lemma A.9.

First, fix some $\epsilon$ not changing with $n$. We have the following
estimates uniform in $a$. Fix some $\delta_{1}>\delta_{2}>0$ sufficiently
small, then there is some constant, possibly depending on $\epsilon$, such
that,

(A.18)
$\Bigg{|}\tilde{G}_{\epsilon}(a)-\frac{1}{n^{2}}\sum_{\begin{subarray}{c}z\in\frac{1}{\sqrt{n}}\mathbb{Z}^{4}\\\
|z|\geq n^{-\delta_{1}},|a-z|\leq
n^{\delta_{2}}\end{subarray}}P_{\epsilon}(a-z)\tilde{G}(z)\Bigg{|}\lesssim
n^{-\delta_{1}}.$

From our modulus of continuity estimates for $\tilde{G}_{\epsilon}(a)$ from
Lemma A.4, we would also have the following relation as a consequence of the
equation above,

(A.19)
$\Bigg{|}\tilde{G}_{\epsilon}(a)-\frac{1}{n^{2}}\sum_{\begin{subarray}{c}z\in\frac{1}{\sqrt{n}}\mathbb{Z}^{4}\\\
|z|\geq n^{-\delta_{1}},|\lfloor a\rfloor_{n}-z|\leq
n^{\delta_{2}}\end{subarray}}P_{\epsilon}(\lfloor
a\rfloor_{n}-z)\tilde{G}(z)\Bigg{|}\lesssim n^{-\delta_{1}}.$

Here, $\lfloor a\rfloor_{n}$ denotes the element in the lattice
$\frac{1}{\sqrt{n}}\mathbb{Z}^{4}$ that is formed by considering
$\frac{1}{\sqrt{n}}(\lfloor\sqrt{n}a_{1}\rfloor,\lfloor\sqrt{n}a_{2}\rfloor,\ldots,\lfloor\sqrt{n}a_{d}\rfloor)$,
where we apply the least integer function to each coordinate.

Similarly, one can show that,

(A.20)
$\Bigg{|}n^{3/2}\tilde{G}_{D,\epsilon}(\sqrt{n}a)-\frac{1}{n^{2}}\sum_{\begin{subarray}{c}z\in\frac{1}{\sqrt{n}}\mathbb{Z}^{4}\\\
|z|\geq n^{-\delta_{1}},|a-z|\leq
n^{\delta_{2}}\end{subarray}}P_{\epsilon}(a-z)n^{3/2}\tilde{G}_{D}(\sqrt{n}a)\Bigg{|}\lesssim
n^{-\delta_{1}}.$

###### Proof.

We will only consider proving equation (A.18); the proof for (A.20) would be
simpler. First, observe that

(A.21) $\int_{|z|\leq
n^{-\delta_{1}}}P_{\epsilon}(a-z)\tilde{G}_{D}(a)\text{d}z\lesssim\int_{|z|\leq
n^{-\delta_{1}}}\frac{1}{|z|^{3}}\text{d}z\lesssim n^{-\delta_{1}}.$

Secondly, we have that,

(A.22) $\displaystyle\int_{\begin{subarray}{c}|z|\geq n^{-\delta_{1}}\\\
|a-z|\geq n^{\delta_{2}}\end{subarray}}P_{\epsilon}(a-z)\tilde{G}(z)\text{d}z$
$\displaystyle\leq\left[\int_{|a-z|\geq
n^{\delta_{2}}}P_{\epsilon}^{2}(a-z)\text{d}z\right]^{1/2}\left[\int_{|z|\geq
n^{-\delta_{1}}}\tilde{G}^{2}(z)\text{d}z\right]^{1/2}$
$\displaystyle\lesssim\exp[-O(1)n^{2\delta_{2}}]n^{d\delta_{1}/2}\lesssim
n^{-\delta_{1}}.$

To get the last line, we used properties of the Gaussian integral to say that
the first integral on the second line is $\exp[-O(1)n^{2\delta_{2}}]$, and we
used direct computation for the second integral. Combining estimates (A.21)
and (A.22), we can deduce that,

$\left|\tilde{G}_{\epsilon}(a)-\int_{\begin{subarray}{c}|z|\geq
n^{-\delta_{1}}\\\ |z-a|\leq
n^{\delta_{2}}\end{subarray}}P_{\epsilon}(a-z)\tilde{G}(z)\text{d}z\right|\lesssim
n^{-\delta_{1}}.$

Now, we compute the difference between the quantity on the right hand side
above, and the appropriate discretization. If we let $\lfloor z\rfloor_{n}$
denote the point in the lattice $\frac{1}{\sqrt{n}}\mathbb{Z}^{4}$ that is
closest to $z$, then we can observe the following,

$|\tilde{G}(z)-\tilde{G}(\lfloor z\rfloor_{n})|\lesssim
n^{5\delta_{1}-2}\tilde{G}(z),\quad\forall|z|\geq n^{-\delta_{1}}.$

This comes from the fact gradient of $|z|^{-5}$ is
$3|z|^{-5}[z_{1},z_{2},\ldots,z_{4}]=3G(z)|z|^{-2}[z_{1},z_{2},\ldots,z_{4}]$
and that $|z|^{-1}\leq n^{\delta_{1}}$.

Secondly, we also have the following estimate,

$|P_{\epsilon}(a-z)-P_{\epsilon}(a-\lfloor z\rfloor_{n})|\lesssim
n^{\delta_{2}-2}P_{\epsilon}(a-z),|a-z|\leq n^{\delta_{2}}.$

Again, this uses the fact that the gradient of the Gaussian distribution
$P_{\epsilon}(a-z)$ is given by
$P_{\epsilon}(a-z)\left[\frac{2(a_{1}-z_{1})}{\epsilon},\frac{2(a_{2}-z_{2})}{\epsilon},\ldots,\frac{2(a_{d}-z_{d})}{\epsilon}\right].$
Now, using the assumption that $|a-z|\leq n^{\delta_{2}}$, will give us our
desired estimate. By applying the triangle inequality, we ultimately see that,

$|P_{\epsilon}(a-z)\tilde{G}(z)-P_{\epsilon}(a-\lfloor
z\rfloor_{n})\tilde{G}(\lfloor z\rfloor_{n})|\lesssim
P_{\epsilon}(a-z)\tilde{G}(z).$

Thus, we see that,

$\displaystyle\int_{\begin{subarray}{c}|z|\geq n^{-\delta_{1}}\\\ |a-z|\leq
n^{\delta_{2}}\end{subarray}}P_{\epsilon}(a-z)\tilde{G}(z)\text{d}z-\sum_{\begin{subarray}{c}z\in\frac{1}{\sqrt{n}}\mathbb{Z}^{4}\\\
|z|\geq n^{-\delta_{1}}\\\ |a-z|\leq
n^{\delta_{2}}\end{subarray}}P_{\epsilon}(a-\lfloor
z\rfloor_{n})\tilde{G}(\lfloor z\rfloor_{n})$
$\displaystyle\lesssim\max[n^{5\delta_{1}-2},n^{\delta_{2}-2}]\int_{\begin{subarray}{c}|z|\geq
n^{-\delta_{1}}\\\ |a-z|\leq
n^{\delta_{2}}\end{subarray}}|P_{\epsilon}(a-z)\tilde{G}(z)-P_{\epsilon}(a-\lfloor
z\rfloor_{n})\tilde{G}(\lfloor z\rfloor_{n})|\text{d}z$
$\displaystyle\lesssim\max[n^{5\delta_{1}-2},n^{\delta_{2}-2}]\int_{\mathbb{R}^{4}}P_{\epsilon}(a-z)\tilde{G}(z)\text{d}z\lesssim\max[n^{5\delta_{1}-2},n^{\delta_{2}-2}].$

This completes the proof of the lemma. ∎

As a corollary of the lemma, we have the following estimates.

######  Corollary A.10.

First, fix some $\epsilon$ not changing with $n$. Additionally, fix parameters
$\delta_{1}>\delta_{2}$ sufficiently small. For $|a|\leq n^{\delta_{2}}$, we
have the following estimate,

$\left|\tilde{G}_{\epsilon}(a)-n^{3/2}\tilde{G}_{\epsilon,D}(\lfloor\sqrt{n}a\rfloor)\right|\lesssim\frac{n^{20\delta_{1}+2\delta_{2}}}{n}+n^{-\delta_{1}}.$

###### Proof.

By using (A.18) and (A.20), it suffices to estimate,

$\Bigg{|}\frac{1}{n^{2}}\sum_{\begin{subarray}{c}z\in\frac{1}{\sqrt{n}}\mathbb{Z}^{4}\\\
|z|\geq n^{-\delta_{1}},|a-z|\leq
n^{\delta_{2}}\end{subarray}}P_{\epsilon}(a-z)\left[n^{3/2}\tilde{G}_{D}(\sqrt{n}z)-\tilde{G}(z)\right]\Bigg{|}.$

From equation (A.27), which we can apply since if $|a|\leq n^{\delta_{2}}$,
then $|z|\leq 2n^{\delta_{2}}$ in the sum above, we can bound the quantity
above as,

$\displaystyle\lesssim\Bigg{|}\frac{1}{n^{2}}\sum_{\begin{subarray}{c}z\in\frac{1}{\sqrt{n}}\mathbb{Z}^{4}\\\
|z|\geq n^{-\delta_{1}},|a-z|\leq
n^{\delta_{2}}\end{subarray}}P_{\epsilon}(a-z)\frac{n^{20\delta_{1}}}{n|z|^{2}}\Bigg{|}\lesssim\Bigg{|}\frac{1}{n^{2}}\sum_{z\in\frac{1}{\sqrt{n}}\mathbb{Z}^{4},|z|\geq
n^{-\delta_{1}},|z|\leq
2n^{\delta_{2}}}\frac{n^{20\delta}_{1}}{n|z|^{2}}\Bigg{|}$
$\displaystyle\lesssim n^{20\delta_{1}}\int_{2n^{\delta}\geq|z|\geq
n^{-\delta}}\frac{1}{n|z|^{2}}\text{d}z\lesssim\frac{n^{20\delta_{1}+2\delta_{2}}}{n}.$

In our application of equation (A.27), we made the choice of parameter
$\epsilon=8\delta_{1}$. ∎

We finally have all results necessary to prove Lemma A.5.

###### Proof of Lemma A.5.

Fix parameters $\delta_{1},\delta_{2},\delta_{3}>0$ sufficiently small
satisfying $\frac{1}{400}>\delta_{1}>20\delta_{2}>20\delta_{3}>0$. Recalling
the function $\chi$ from A.6, we set $\chi$ to a be a smooth function
supported on the interval $[-n^{\delta_{2}},n^{\delta_{2}}]^{4}$ and $1$ on
$[-n^{\delta_{3}},n^{\delta_{3}}]^{4}$,

$\displaystyle\Bigg{|}\int_{\mathbb{R}^{4}}\chi(a-e)\tilde{G}_{\epsilon}(a-e)\tilde{G}_{\epsilon}(b-e)\chi(b-e)\text{d}e$
$\displaystyle-\frac{1}{n^{2}}\sum_{e\in\frac{1}{\sqrt{n}}\mathbb{Z}^{4}}\chi(\lfloor
a\rfloor_{n}-e)n^{3/2}\tilde{G}_{\epsilon,D}(\sqrt{n}\lfloor
a\rfloor_{n}-\sqrt{n}e)n^{3/2}\tilde{G}_{\epsilon,D}(\sqrt{n}\lfloor
b\rfloor_{n}-\sqrt{n}e)\chi(\lfloor b\rfloor_{n}-e)\Bigg{|}=o(1).$

Thus, we see that it suffices to show that,

$\displaystyle\int_{\mathbb{R}^{4}}\chi(a-e)\left|\tilde{G}_{\epsilon}(a-e)-n^{3/2}\tilde{G}_{\epsilon,D}(\sqrt{n}(\lfloor
a\rfloor_{n}-e))\right|\chi(b-e)\tilde{G}_{\epsilon}(b-e)\text{d}e=o(1),$
$\displaystyle\int_{\mathbb{R}^{4}}\chi(a-e)n^{3/2}\tilde{G}_{\epsilon,D}(\sqrt{n}(\lfloor
a\rfloor_{n}-e))\left|\tilde{G}_{\epsilon}(b-e)-n^{3/2}\tilde{G}_{\epsilon,D}(\sqrt{n}(\lfloor
b\rfloor_{n}-e))\right|\chi(b-e)\text{d}e=o(1).$

We will show the former estimate, as both are similar. By Corollary A.10, we
can bound the difference of $\tilde{G}_{\epsilon}$ and
$\tilde{G}_{\epsilon,D}$ in the region on which $\chi$ is not equal to $0$.
Thus, we have,

$\displaystyle\int_{\mathbb{R}^{4}}\chi(a-e)\left|\tilde{G}_{\epsilon}(a-e)-n^{3/2}\tilde{G}_{\epsilon,D}(\sqrt{n}(\lfloor
a\rfloor_{n}-e))\right|\chi(b-e)\tilde{G}_{\epsilon}(b-e)\text{d}e$
$\displaystyle\lesssim\int_{\mathbb{R}^{4}}\chi(a-e)n^{-\delta_{1}}\chi(b-e)\tilde{G}_{\epsilon}(b-e)\text{d}e\lesssim\left[\int_{[-n^{\delta_{2}},n^{\delta_{2}}]^{4}}n^{-2\delta_{1}}\text{d}e\right]^{1/2}\left[\int_{\mathbb{R}^{4}}\tilde{G}^{2}_{\epsilon}(b-e)\text{d}e\right]^{1/2}$
$\displaystyle\lesssim n^{4\delta_{2}/2-\delta_{1}}.$

We used the fact that $\chi$ is supported on
$[-n^{\delta_{2}},n^{\delta_{2}}]$ and the $L^{2}$ integrability of
$\tilde{G}_{\epsilon}^{2}$ from equation (A.12). Our choice of parameters
$\delta_{1}$ and $\delta_{2}$ ensures that the bound in the last inequality is
$o(1)$. This completes the proof of the lemma.

∎

###  A.2. Comparison of the second moments of $\mathcal{G}_{n}$ and
$\mathcal{G}_{n,\epsilon}$

In this subsection, we will prove Theorem A.3.

###### Proof of Theorem A.3.

Define $\hat{P}_{n,\epsilon}$ as follows,

$\hat{P}_{n,\epsilon}(l):=\sum_{b\in\mathbb{Z}^{4}}P_{\epsilon}\left(\frac{b}{\sqrt{n}}\right)\exp[2\pi\text{i}lb].$

By methods similar to the proof of Lemma A.12, we see that,

$\tilde{G}_{D,\epsilon}(\sqrt{n}b)=\int_{(-1/2,1/2]^{4}}\hat{\tilde{G}}_{D}(l)(n^{-2}\hat{P}_{n,\epsilon}(l))\text{d}l.$

Now, notice that we can write $\mathcal{G}^{2}_{n,\epsilon}$ as,

$\displaystyle\frac{1}{n^{4}}\sum_{u,u^{\prime}=1}^{n}\sum_{t,t^{\prime}=1}^{n}\frac{1}{n^{4}}\sum_{e,e^{\prime}\in\frac{1}{\sqrt{n}}\mathbb{Z}^{4}}n^{3/2}\tilde{G}_{D,\epsilon}(\mathcal{S}_{1}(u)-\sqrt{n}e)n^{3/2}\tilde{G}_{D,\epsilon}(\mathcal{S}_{1}(u^{\prime})-\sqrt{n}e^{\prime})$
$\displaystyle\hskip 28.45274pt\times
n^{3/2}\tilde{G}_{D,\epsilon}(\mathcal{S}_{2}(t)-\sqrt{n}e)n^{3/2}\tilde{G}_{D,\epsilon}(\mathcal{S}_{2}(t^{\prime})-\sqrt{n}e^{\prime}).$

By taking Fourier transforms, we see that this can be written as,

$\displaystyle\mathcal{G}^{2}_{n,\epsilon}=\frac{1}{n^{4}}\sum_{u,u^{\prime}=1}^{n}\sum_{t,t^{\prime}=1}^{n}n^{2}\int_{((-1/2,1/2]^{4})^{2}}\prod_{i=1}^{2}\hat{\tilde{G}}^{2}_{D}(\lambda_{i})n^{-d}\hat{P}^{2}_{n,\epsilon}(\lambda_{i})$
$\displaystyle\hskip
28.45274pt\times\exp[2\pi\text{i}[\lambda_{1}(\mathcal{S}_{1}(u)-\mathcal{S}_{2}(t))+\lambda_{2}(\mathcal{S}_{1}(u^{\prime})-\mathcal{S}_{2}(t^{\prime}))]]\prod\text{d}\lambda_{i}.$