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Annealed transition density of simple random walk
on a high-dimensional loop-erased random walk

D. A. Croydon, D. Shiraishi, S. Watanabe Research Institute for Mathematical Sciences, Kyoto University, Japan, [email protected] School of Informatics, Kyoto University, Kyoto, Japan, [email protected] School of Informatics, Kyoto University, Kyoto, Japan, [email protected]
Abstract

We derive sub-Gaussian bounds for the annealed transition density of the simple random walk on a high-dimensional loop-erased random walk. The walk dimension that appears in these is the exponent governing the space-time scaling of the process with respect to the extrinsic Euclidean distance, which contrasts with the exponent given by the intrinsic graph distance that appears in the corresponding quenched bounds. We prove our main result using novel pointwise Gaussian estimates on the distribution of the high-dimensional loop-erased random walk.
AMS 2020 MSC: 60K37 (primary), 05C81, 60J35, 60K35, 82B41.
Keywords and phrases: heat kernel estimates, loop-erased random walk, random walk in random environment.

1 Introduction

The main result of this article provides annealed heat kernel estimates for the random walk on the trace of a loop-erased random walk in high dimensions. Whilst this is something of a toy model, the statement reveals an interesting difference between the quenched (typical) heat kernel estimates, which are of Gaussian form with respect to the intrinsic graph metric, and the annealed (averaged) heat kernel estimates, which are of sub-Gaussian form with respect to the extrinsic Euclidean metric. (We will present more detailed background concerning this terminology below.) Establishing such a discrepancy rigourously was motivated by a conjecture made in [6, Remark 1.5] for the two-dimensional uniform spanning tree, and naturally leads one to consider to what extent the behaviour is typical for random walks on random graphs embedded into an underlying space.

Before describing further our heat kernel estimates, we would like to present the key input into their proof, which is a time-averaged Gaussian bound on the distribution of the loop-erased random walk. In particular, throughout the course of the article, we let (Ln)n0(L_{n})_{n\geq 0} be the loop-erasure of the discrete-time simple random walk (Sn)n0(S_{n})_{n\geq 0} on d\mathbb{Z}^{d}, where d5d\geq 5, started from the origin. (See Section 2 for a precise definition of this process, which was originally introduced by Lawler in [19].) In the dimensions we are considering, it is known that the loop-erased random walk (LERW) behaves similarly to the original simple random walk (SRW), in that both have a Brownian motion scaling limit; for the SRW, this follows from the classic invariance principle of Donsker [13], whereas for the LERW, it was proved in [19]. Taking into account the diffusive scaling, such results readily imply convergence of the distributions of n1/2Snn^{-1/2}S_{n} and n1/2Lnn^{-1/2}L_{n} as nn\rightarrow\infty. For our heat kernel estimates, we will require understanding of the distribution of LERW on a finer, pointwise scale. Now, one can check that SRW satisfies pointwise Gaussian bounds of the form

cnd/2e|x|2cn𝟏{nx1}𝐏(Sn=x)+𝐏(Sn+1=x)2c1nd/2ec|x|2n,xd,n1,cn^{-d/2}e^{-\frac{|x|^{2}}{cn}}\mathbf{1}_{\{n\geq\|x\|_{1}\}}\leq\frac{\mathbf{P}(S_{n}=x)+\mathbf{P}(S_{n+1}=x)}{2}\leq c^{-1}n^{-d/2}e^{-\frac{c|x|^{2}}{n}},\qquad\forall x\in\mathbb{Z}^{d},\>n\geq 1, (1)

where cc is a constant and we write x1\|x\|_{1} for the 1\ell_{1}-norm of xx, see [3, Theorem 6.28], for example. (The averaging over two time steps is necessary for parity reasons.) Of course, one can not expect the same bounds for a LERW. Indeed, the ‘on-diagonal’ part of the distribution 𝐏(Ln=0)\mathbf{P}(L_{n}=0) is equal to zero for n1n\geq 1. What we are able to establish is the following, which demonstrates that if one averages over longer time intervals, then one still sees Gaussian behaviour.

Theorem 1.1.

The loop-erased random walk (Ln)n0(L_{n})_{n\geq 0} on d\mathbb{Z}^{d}, d5d\geq 5, started from the origin satisfies the following bounds: for all xd\{0}x\in\mathbb{Z}^{d}\backslash\{0\}, n1n\geq 1,

1nm=n2n1𝐏(Lm=x)c1nd/2ec2|x|2n,\frac{1}{n}\sum_{m=n}^{2n-1}\mathbf{P}(L_{m}=x)\leq c_{1}n^{-d/2}e^{-\frac{c_{2}|x|^{2}}{n}},

and for all xd\{0}x\in\mathbb{Z}^{d}\backslash\{0\}, n|x|n\geq|x|,

1nm=c3nc4n𝐏(Lm=x)c5nd/2ec6|x|2n,\frac{1}{n}\sum_{m=\lceil c_{3}n\rceil}^{\lfloor c_{4}n\rfloor}\mathbf{P}(L_{m}=x)\geq c_{5}n^{-d/2}e^{-\frac{c_{6}|x|^{2}}{n}},

where c1,,c6c_{1},\dots,c_{6} are constants.

Remark 1.2.

Another random path known to scale to Brownian motion in dimensions 5\geq 5 is the self-avoiding walk (SAW) [14, 15]. A local central limit theorem for a ‘spread out’ version of the SAW was established in [23]. For a precise statement, one should refer to [23, Theorem 1.3], but roughly it was shown that if cn(x)c_{n}(x) is the number of nn-step self-avoiding walk paths from 0 to xx and cn:=xdcn(x)c_{n}:=\sum_{x\in\mathbb{Z}^{d}}c_{n}(x), then, for appropriate constants c1c_{1} and c2c_{2}, as nn\rightarrow\infty,

1(2rn+1)dyB(nx,rn)cn(y)cnc1nd/2ec2|x|22,\frac{1}{(2r_{n}+1)^{d}}\sum_{y\in B_{\infty}(\sqrt{n}x,r_{n})}\frac{c_{n}(y)}{c_{n}}\sim c_{1}n^{-d/2}e^{-\frac{c_{2}|x|^{2}}{2}},

where B(x,r)B_{\infty}(x,r) is an \ell_{\infty}-ball and rnr_{n} is a divergent sequence satisfying rn=o(n1/2)r_{n}=o(n^{1/2}). (We note that the term zcnz_{c}^{n} in [23] can be replaced by 1/cn1/c_{n} by applying [24, Theorem 1.1].) In particular, the left-hand side here can be thought of as the probability a self-avoiding walk of length nn is close to xx, with the ball over which the spatial averaging is conducted being microscopic compared with the diffusive scaling of the functional scaling limit. For the LERW of this article, one might ask if it was similarly possible to reduce the intervals over which time averaging is conducted in Theorem 1.1 to ones of the form {n,n+1,,n+tn}\{n,n+1,\dots,n+t_{n}\} where tn=o(n)t_{n}=o(n). And, for both the LERW and the SAW, one could ask for what ranges of the variables a pointwise space-time bound as at (1) holds. See [2] for a result in this direction concerning the weakly self-avoiding walk.

We now introduce our model of random walk in a random environment. To this end, given a realisation of (Ln)n0(L_{n})_{n\geq 0}, we define a graph 𝒢\mathcal{G} to have vertex set

V(𝒢):={Ln:n0},V(\mathcal{G}):=\left\{L_{n}:\>n\geq 0\right\},

and edge set

E(𝒢):={{Ln,Ln+1}:n0}.E(\mathcal{G}):=\left\{\{L_{n},L_{n+1}\}:\>n\geq 0\right\}.

We then let (Xt𝒢)t0(X^{\mathcal{G}}_{t})_{t\geq 0} be the continuous-time random walk on 𝒢\mathcal{G} that has unit mean exponential holding times at each site and jumps from its current location to a neighbour with equal probability being placed on each of the possibilities. Moreover, we will always suppose that X0𝒢=L0=0X_{0}^{\mathcal{G}}=L_{0}=0. Clearly, since 𝒢\mathcal{G} equipped with its shortest-path graph metric d𝒢d_{\mathcal{G}} is isometric to +\mathbb{Z}_{+}, it is straightforward to deduce that, for each realisation of LL,

(n1/2d𝒢(0,Xnt𝒢))t0(|Bt|)t0,\left(n^{-1/2}d_{\mathcal{G}}\left(0,X_{nt}^{\mathcal{G}}\right)\right)_{t\geq 0}\rightarrow\left(|B_{t}|\right)_{t\geq 0}, (2)

in distribution, where (Bt)t0(B_{t})_{t\geq 0} is a one-dimensional Brownian motion started from 0, and we can also give one-dimensional Gaussian bounds for the transition probabilities of X𝒢X^{\mathcal{G}} on 𝒢\mathcal{G} (see Lemma 4.1 below). In this article, though, we are interested in the annealed situation, when we average out over realisations of 𝒢\mathcal{G}. Precisely, we define the annealed law of X𝒢X^{\mathcal{G}} to be the probability measure on the Skorohod space D(+,d)D(\mathbb{R}_{+},\mathbb{R}^{d}) given by

(X𝒢)=P𝒢(X𝒢)𝐏(d𝒢),\mathbb{P}\left(X^{\mathcal{G}}\in\cdot\right)=\int{P}^{\mathcal{G}}\left(X^{\mathcal{G}}\in\cdot\right)\mathbf{P}(d\mathcal{G}),

where 𝐏\mathbf{P} is the probability measure on the underlying probability space upon which LL is built, and P𝒢P^{\mathcal{G}} is the law of X𝒢X^{\mathcal{G}} on the particular realisation of 𝒢\mathcal{G} (i.e. the quenched law of 𝒢\mathcal{G}). We are able to prove the following. Note that we use the notation xy:=max{x,y}x\vee y:=\max\{x,y\} and xy:=min{x,y}x\wedge y:=\min\{x,y\}.

Theorem 1.3.

For any ε>0\varepsilon>0, there exist constants c1,c2,c3,c4(0,)c_{1},c_{2},c_{3},c_{4}\in(0,\infty) such that, for every xdx\in\mathbb{Z}^{d} and tε|x|t\geq\varepsilon|x|,

(Xt𝒢=x)c1(1|x|2d)(1t1/2)exp(c2(|x|41t)1/3)\mathbb{P}\left(X_{t}^{\mathcal{G}}=x\right)\leq c_{1}\left(1\wedge|x|^{2-d}\right)\left(1\wedge t^{-1/2}\right)\exp\left(-c_{2}\left(\frac{|x|^{4}}{1\vee t}\right)^{1/3}\right)

and also

(Xt𝒢=x)c3(1|x|2d)(1t1/2)exp(c4(|x|41t)1/3).\mathbb{P}\left(X_{t}^{\mathcal{G}}=x\right)\geq c_{3}\left(1\wedge|x|^{2-d}\right)\left(1\wedge t^{-1/2}\right)\exp\left(-c_{4}\left(\frac{|x|^{4}}{1\vee t}\right)^{1/3}\right).

Remark 1.4.

Essentially similar arguments would apply if X𝒢X^{\mathcal{G}} was replaced by the discrete-time simple random walk on 𝒢\mathcal{G}. In that case, one would need to assume tx1t\geq\|x\|_{1} for the lower bound, since the probability being estimated would be zero for t<x1t<\|x\|_{1}. In the continuous setting, the behaviour of (Xt𝒢=x)\mathbb{P}(X_{t}^{\mathcal{G}}=x) takes on a different form for t<ε|x|t<\varepsilon|x|, as the probability will be determined by certain rare events (cf. the discussion of Poisson bounds in [3, Section 5.1]; see also Lemma 4.1 below).

To put this result into context, it helps to briefly recall what kind of behaviour has been observed for anomalous random walks and diffusions in other settings. In particular, for many random walks or diffusions on fractal-like sets (either deterministic or random), it has been shown that the associated transition density (pt(x,y))(p_{t}(x,y)) satisfies, within appropriate ranges of the variables, upper and lower bounds of the form

c1tds/2exp(c2(d(x,y)dwt)1dw1),c_{1}t^{-d_{s}/2}\exp\left(-c_{2}\left(\frac{d(x,y)^{d_{w}}}{t}\right)^{\frac{1}{d_{w}-1}}\right), (3)

where d(x,y)d(x,y) is some metric on the space in question. (See [3, 18] for overviews of work in this area.) The exponent dsd_{s} is typically called the spectral dimension, since it is related to growth rate of the spectrum of the generator of the stochastic process. The exponent dwd_{w}, which is usually called the walk dimension (with respect to the metric dd), gives the space-time scaling.

Now, in our setting, we can clearly write

(Xt𝒢=x)=(Xt𝒢=xx𝒢)𝐏(x𝒢),\mathbb{P}\left(X_{t}^{\mathcal{G}}=x\right)=\mathbb{P}\left(X_{t}^{\mathcal{G}}=x\>\vline\>x\in\mathcal{G}\right)\mathbf{P}\left(x\in\mathcal{G}\right),

and, moreover, using simple facts about the intersection properties of SRW in high dimensions, one can check that 𝐏(x𝒢)1|x|2d\mathbf{P}(x\in\mathcal{G})\asymp 1\wedge|x|^{2-d} (where we use the notation \asymp to mean that the left-hand side is bounded above and below by constant multiples of the right-hand side). Hence, Theorem 1.3 gives that (Xt𝒢=xx𝒢)\mathbb{P}(X_{t}^{\mathcal{G}}=x\>\vline\>x\in\mathcal{G}) satisfies upper and lower bounds of exactly the sub-Gaussian form at (3), with ds=1d_{s}=1, dw=4d_{w}=4 and dd being the Euclidean metric. We can understand that ds=1d_{s}=1 results from one-dimensional nature of the graph 𝒢\mathcal{G} with respect to its intrinsic metric d𝒢d_{\mathcal{G}}. Moreover, the exponent dw=4d_{w}=4 gives the space-time scaling of the process X𝒢X^{\mathcal{G}} with respect to the Euclidean metric. Indeed, it is not difficult to combine (2) and the Brownian motion scaling limit of LL (from [19]) to check that, up to a deterministic constant factor in the time scaling,

(n1/4Xnt𝒢)t0(W|Bt|)t0,\left(n^{-1/4}X_{nt}^{\mathcal{G}}\right)_{t\geq 0}\rightarrow\left(W_{|B_{t}|}\right)_{t\geq 0},

where (Wt)t0(W_{t})_{t\geq 0} is a dd-dimensional Brownian motion, independent of the one-dimensional Brownian motion BB, with WW and BB each started from the origin. (The limit process is a version of the dd-dimensional iterated Brownian motion studied in [12], for example. Cf. the result in [10] for random walk on the range of a random walk.) We note that the exponent ds=1d_{s}=1 matches the quenched spectral dimension, whereas the dw=4d_{w}=4 is the multiple of the ‘2’ of the quenched bound, which is the walk dimension of X𝒢X^{\mathcal{G}} with respect to the intrinsic metric d𝒢d_{\mathcal{G}}, and the ‘2’ that gives the space-time scaling of LL. We highlight that the annealed bound is not obtained by simply replacing d𝒢(0,x)d_{\mathcal{G}}(0,x) by |x|2|x|^{2} in the quenched bound, though, as doing that does not result in an expression of the form at (3).

As remarked at the start of the introduction, it was conjectured in [6] that a similar combination of the various exponents will appear in sub-Gaussian annealed heat kernel bounds for the random walk on the two-dimensional uniform spanning tree. In that case, the spectral dimension of the quenched and annealed bounds is known to be 16/1316/13, the intrinsic walk dimension is 13/513/5 and the exponent governing the embedding is given by the growth exponent of the two-dimensional LERW, i.e. 5/45/4, giving an extrinsic walk dimension of 13/413/4. We are able to check the corresponding result for our simpler model using the simple observation that

(Xt𝒢=x)=m0P𝒢(X𝒢=Lm)𝐏(Lm=x),\mathbb{P}\left(X_{t}^{\mathcal{G}}=x\right)=\sum_{m\geq 0}P^{\mathcal{G}}(X^{\mathcal{G}}=L_{m})\mathbf{P}(L_{m}=x), (4)

and then combining the estimate on the distribution of the LERW from Theorem 1.1 with the deterministic Gaussian bounds on P𝒢(X𝒢=Lm)P^{\mathcal{G}}(X^{\mathcal{G}}=L_{m}) of Lemma 4.1 below. It is clear that a similar argument would apply for other random walks on random paths, including ones with different exponents, whenever one can suitably estimate the term corresponding to 𝐏(Lm=x)\mathbf{P}(L_{m}=x). LERW in dimensions 22, 33 and 44 would represent worthwhile examples to consider here. For more general random graphs, one could give a decomposition that is similar to (4), but to proceed from that point one would need both good quenched heat kernel estimates with respect to some distance in place of the Gaussian bounds on P𝒢(X𝒢=Lm)P^{\mathcal{G}}(X^{\mathcal{G}}=L_{m}) and good estimates on how the distance that appears in the quenched bounds behaves. Pursuing this program would be of interest for various random graphs that do not exhibit homogenisation. Apart from the uniform spanning tree/forest (in two and other dimensions, see [1, 5, 6] for some relevant background on the associated random walk), one might consider the situation for random walk on: the trace of a simple random walk on d\mathbb{Z}^{d}, the scaling limit of which was derived for d4d\geq 4 in [10, 11]; the high-dimensional branching random walk, for which a scaling limit was proved in [8, 9]; and on high-dimensional critical percolation clusters, for which a number of related random walk estimates have previously been established [16, 17], for example.

The remainder of the article is organised as follows. In Section 2, we define the LERW (Ln)n0(L_{n})_{n\geq 0} properly, and present some preliminary estimates that will be useful for understanding this process. After this, in Section 3, we study the LERW in more detail, proving Theorem 1.1. Finally, in Section 4, we derive our heat kernel estimates for X𝒢X^{\mathcal{G}}, proving Theorem 1.3.

2 Preliminaries

In this section, we will start by presenting some notation, and then precisely define the loop-erased random walk that is the main object of interest in this article. We also give some estimates on hitting probabilities for annuli by a simple random walk that will be useful in Section 3.

2.1 Subsets

Throughout this paper, we always assume that d5d\geq 5. If AdA\subseteq\mathbb{Z}^{d}, we set

A\displaystyle\partial A ={xdA:yA such that |xy|=1},\displaystyle=\left\{x\in\mathbb{Z}^{d}\setminus A\>:\>\exists y\in A\text{ such that }|x-y|=1\right\},
iA\displaystyle\partial_{i}A ={xA:ydA such that |xy|=1}\displaystyle=\left\{x\in A\>:\>\exists y\in\mathbb{Z}^{d}\setminus A\text{ such that }|x-y|=1\right\}

for the outer and inner boundary of AA, respectively. We write A¯=AA\overline{A}=A\cup\partial A for the discrete closure of AA. We often use AcA^{c} to denote dA\mathbb{Z}^{d}\setminus A.

For xdx\in\mathbb{R}^{d} and r>0r>0, we let

B(x,r)={yd:|xy|r},B(x,r)={yd:xyr}\displaystyle B(x,r)=\left\{y\in\mathbb{Z}^{d}\>:\>|x-y|\leq r\right\},\qquad B_{\infty}(x,r)=\left\{y\in\mathbb{Z}^{d}\>:\>\|x-y\|_{\infty}\leq r\right\}

be the discrete ball of radius rr centered at xx with respect to the Euclidean distance |||\cdot| and \ell_{\infty}-norm \|\cdot\|_{\infty} in d\mathbb{R}^{d}, respectively. We often call B(x,r)B_{\infty}(x,r) the box of side length 2r2r centered at xx.

If AA and BB are two subsets of d\mathbb{Z}^{d}, we define the distance between them by setting

dist(A,B)=infxA,yB|xy|.\text{dist}(A,B)=\inf_{x\in A,y\in B}|x-y|.

2.2 Paths and loop-erasure

Let m0m\geq 0 be an integer. If λ=[λ0,λ1,,λm]d\lambda=[\lambda_{0},\lambda_{1},\dots,\lambda_{m}]\subset\mathbb{Z}^{d} is a sequence of points in d\mathbb{Z}^{d} satisfying |λiλi1|=1|\lambda_{i}-\lambda_{i-1}|=1 for 1im1\leq i\leq m, we call λ\lambda a path of length mm. The length of λ\lambda is denoted by len(λ)\text{len}(\lambda). We write λ[i,j]=[λi,λi+1,,λj]\lambda[i,j]=[\lambda_{i},\lambda_{i+1},\dots,\lambda_{j}] for 0ijlen(λ)0\leq i\leq j\leq\text{len}(\lambda). If λ\lambda satisfies λiλj\lambda_{i}\neq\lambda_{j} for all iji\neq j, then λ\lambda is called a simple path. If λ=[λ0,λ1,]d\lambda=[\lambda_{0},\lambda_{1},\dots]\subset\mathbb{Z}^{d} is an infinite sequence of points in d\mathbb{Z}^{d} such that [λ0,λ1,,λm][\lambda_{0},\lambda_{1},\dots,\lambda_{m}] is a path for each m0m\geq 0, we call λ\lambda an infinite path. We write λ[i,)=[λi,λi+1,]\lambda[i,\infty)=[\lambda_{i},\lambda_{i+1},\dots] for i0i\geq 0.

If λ\lambda is a path in d\mathbb{Z}^{d} (either finite or infinite) and AdA\subseteq\mathbb{Z}^{d}, we define

τAλ=inf{j0:λjA}\tau^{\lambda}_{A}=\inf\left\{j\geq 0\>:\>\lambda_{j}\in A\right\} (5)

to be the first time that λ\lambda hits AA, with the convention that inf=+\inf\emptyset=+\infty. If A={x}A=\{x\}, then we set τxλ=τ{x}λ\tau^{\lambda}_{x}=\tau^{\lambda}_{\{x\}}.

Given a path λ=[λ0,λ1,,λm]d\lambda=[\lambda_{0},\lambda_{1},\dots,\lambda_{m}]\subset\mathbb{Z}^{d} with len(λ)=m\text{len}(\lambda)=m, we define its (chronological) loop-erasure LE(λ)\text{LE}(\lambda) as follows. Let σ0=max{k:λk=λ0}\sigma_{0}=\max\{k\>:\>\lambda_{k}=\lambda_{0}\} and also, for i1i\geq 1,

σi=max{k:λk=λσi1+1}.\sigma_{i}=\max\left\{k\>:\>\lambda_{k}=\lambda_{\sigma_{i-1}+1}\right\}. (6)

We note that these quantities are well-defined up to the index j=min{i:λσi=λm}j=\min\{i\>:\>\lambda_{\sigma_{i}}=\lambda_{m}\}, and we use them to define the loop-erasure of λ\lambda by setting

LE(λ)=[λσ0,λσ1,,λσj].\text{LE}(\lambda)=\left[\lambda_{\sigma_{0}},\lambda_{\sigma_{1}},\dots,\lambda_{\sigma_{j}}\right].

It follows by construction that LE(λ)\text{LE}(\lambda) is a simple path satisfying LE(λ)λ\text{LE}(\lambda)\subseteq\lambda, LE(λ)0=λ0\text{LE}(\lambda)_{0}=\lambda_{0} and LE(λ)j=λm\text{LE}(\lambda)_{j}=\lambda_{m}. If λ=[λ0,λ1,]d\lambda=[\lambda_{0},\lambda_{1},\dots]\subseteq\mathbb{Z}^{d} is an infinite path such that {k:λk=λi}\{k\>:\>\lambda_{k}=\lambda_{i}\} is finite for each i0i\geq 0, then its loop-erasure LE(λ)\text{LE}(\lambda) can be defined similarly.

2.3 Random walk and loop-erased random walk

As in the introduction, let S=(Sn)n0S=(S_{n})_{n\geq 0} be the discrete-time simple random walk (SRW) on d\mathbb{Z}^{d}. For xdx\in\mathbb{Z}^{d}, the law of SS started from S0=xS_{0}=x will be denoted by 𝐏x\mathbf{P}^{x}. The expectation with respect to 𝐏x\mathbf{P}^{x} will be denoted by 𝐄x\mathbf{E}^{x}. We will also define 𝐏=𝐏0\mathbf{P}=\mathbf{P}^{0} and 𝐄=𝐄0\mathbf{E}=\mathbf{E}^{0}.

When d5d\geq 5, it is well known that SS is transient. Hence for every xdx\in\mathbb{Z}^{d} and 𝐏x\mathbf{P}^{x}-almost-every realisation of SS, it is possible to define the loop-erasure of S[0,)S[0,\infty), and we will denote this infinite path by

L=LE(S[0,)).L=\text{LE}\left(S[0,\infty)\right).

Note that LL is a random simple path with L0=xL_{0}=x that satisfies that limm|Lm|=\lim_{m\to\infty}|L_{m}|=\infty, 𝐏x\mathbf{P}^{x}-almost surely. If LL is constructed from the SRW SS under 𝐏x\mathbf{P}^{x}, we will refer to it as the infinite loop-erased random walk (LERW) started at xx. See [22, Section 11] for further background.

2.4 Constants

We use cc, cc^{\prime}, CC, CC^{\prime}, c1c_{1}, C1,C_{1},\dots to denote universal positive finite constants depending only on dd, whose values may change between lines. If we want to emphasize that a constant depends on some parameter, we will use a subscript to indicate it. For example, c=cεc=c_{\varepsilon} means the constant cc depends on ε\varepsilon. If {an}\{a_{n}\} and {bn}\{b_{n}\} are two positive sequences, then we write an=O(bn)a_{n}=O(b_{n}) if there exists a constant C(0,)C\in(0,\infty) such that anCbna_{n}\leq Cb_{n} for all nn. When we add a subscript to this OO notation, it means that the constant CC depends on the subscript. For instance, by an=Oε(bn)a_{n}=O_{\varepsilon}(b_{n}), we mean that anCεbna_{n}\leq C_{\varepsilon}b_{n}.

2.5 Simple random walk estimates

Let SS be a simple random walk on d\mathbb{Z}^{d}, and suppose mm and nn are real numbers such that 1m<n1\leq m<n. Moreover, let A={xd:m|x|n}A=\{x\in\mathbb{Z}^{d}\>:\>m\leq|x|\leq n\}, and set τ=τAcS\tau=\tau^{S}_{A^{c}} to be the first time that SS exits AA. Then [20, Proposition 1.5.10] gives that, for all xAx\in A,

𝐏x(|Sτ|m)=|x|2dn2d+O(m1d)m2dn2d.\mathbf{P}^{x}\left(|S_{\tau}|\leq m\right)=\frac{|x|^{2-d}-n^{2-d}+O(m^{1-d})}{m^{2-d}-n^{2-d}}. (7)

Whilst this approximation is good for large mm, in this paper, we also need to consider the situation when m=1m=1 and |x||x| is large. In this case, |Sτ|m|S_{\tau}|\leq m if and only if Sτ=0S_{\tau}=0, and the estimate (7) is not useful due to the O(m1d)O(m^{1-d}) term. However, adapting the argument used to prove [20, Proposition 1.5.10], it is possible to establish that there exists a universal constant a=ad>0a=a_{d}>0 such that

𝐏x(Sτ=0)=a|x|2dan2d+O(|x|1d)G(0)an2d,\mathbf{P}^{x}\left(S_{\tau}=0\right)=\frac{a|x|^{2-d}-an^{2-d}+O(|x|^{1-d})}{G(0)-an^{2-d}}, (8)

where G(0)G(0), as defined by

G(0)=j=0𝐏(Sj=0),G(0)=\sum_{j=0}^{\infty}\mathbf{P}(S_{j}=0),

is the expected number of returns of SS to its starting point, which is finite in the dimensions we are considering.

In this article, we will also make use of another basic estimate for simple random walk on d\mathbb{Z}^{d}, which is often called the gambler’s ruin estimate. We take θd\theta\in\mathbb{R}^{d} with |θ|=1|\theta|=1 and set S^j=Sjθ\widehat{S}_{j}=S_{j}\cdot\theta. Let ηn=min{j0:S^j0 or S^jn}\eta_{n}=\min\{j\geq 0\mathrel{:}\widehat{S}_{j}\leq 0\mbox{ or }\widehat{S}_{j}\geq n\}. We denote by 𝐏^x\widehat{\mathbf{P}}^{x} the law of S^\widehat{S} with starting point xx\in\mathbb{R}. Then [22, Proposition 5.1.6] guarantees that there exist 0<α1<α2<0<\alpha_{1}<\alpha_{2}<\infty such that: for all 1mn1\leq m\leq n,

α1m+1n𝐏^m(S^ηnn)α2m+1n.\alpha_{1}\frac{m+1}{n}\leq\widehat{\mathbf{P}}^{m}(\widehat{S}_{\eta_{n}}\geq n)\leq\alpha_{2}\frac{m+1}{n}. (9)

The gambler’s ruin estimate gives upper and lower bounds on the probability that a simple random walk on d\mathbb{Z}^{d} projected onto a line escapes from one of the endpoints of a line segment.

3 Loop-erased random walk estimates

The aim of this section is to prove Theorem 1.1. Due to the diffusive scaling of the LERW, it is convenient to reparameterise the result. In particular, we will prove the following, which clearly implies Theorem 1.1. Throughout this section, for xdx\in\mathbb{Z}^{d}, we write τx=τxL\tau_{x}=\tau_{x}^{L} for the first time that the LERW LL hits xx.

Proposition 3.1.

There exist constants c1,c2(0,)c_{1},c_{2}\in(0,\infty) such that for every xd\{0}x\in\mathbb{Z}^{d}\backslash\{0\} and M>0M>0,

𝐏(τx[M|x|2,2M|x|21])c1(M|x|2)1d/2exp(c2M).\mathbf{P}\left(\tau_{x}\in\left[M|x|^{2},2M|x|^{2}-1\right]\right)\leq c_{1}\left(M|x|^{2}\right)^{1-d/2}\exp\left(-\frac{c_{2}}{M}\right).

Moreover, there exist constants c3,c4,c5,c6(0,)c_{3},c_{4},c_{5},c_{6}\in(0,\infty) such that for every xd\{0}x\in\mathbb{Z}^{d}\backslash\{0\} and M|x|1M\geq|x|^{-1},

𝐏(τx[c3M|x|2,c4M|x|2])c5(M|x|2)1d/2exp(c6M).\mathbf{P}\left(\tau_{x}\in\left[c_{3}M|x|^{2},c_{4}M|x|^{2}\right]\right)\geq c_{5}\left(M|x|^{2}\right)^{1-d/2}\exp\left(-\frac{c_{6}}{M}\right).

We will break the proof of this result into four pieces, distinguishing the cases M(0,1)M\in(0,1) and M1M\geq 1, and considering the upper and lower bounds separately. See Propositions 3.2, 3.3, 3.6 and 3.13 for the individual statements.

3.1 Upper bound for M1M\geq 1

The aim of this section is to establish the following, which is the easiest to prove of the constituent results making up Proposition 3.1.

Proposition 3.2.

There exist constants c1,c2(0,)c_{1},c_{2}\in(0,\infty) such that for every xd\{0}x\in\mathbb{Z}^{d}\backslash\{0\} and M1M\geq 1,

𝐏(τx[M|x|2,2M|x|21])c1(M|x|2)1d/2exp(c2M).\mathbf{P}\left(\tau_{x}\in\left[M|x|^{2},2M|x|^{2}-1\right]\right)\leq c_{1}\left(M|x|^{2}\right)^{1-d/2}\exp\left(-\frac{c_{2}}{M}\right).
Proof.

Recalling the definition of (σi)i0(\sigma_{i})_{i\geq 0} from (6), we have that

𝐏(τx[M|x|2,2M|x|21])\displaystyle\mathbf{P}\left(\tau_{x}\in\left[M|x|^{2},2M|x|^{2}-1\right]\right) =\displaystyle= i=M|x|22M|x|21𝐏(Sσi=x)\displaystyle\sum_{i=\lceil M|x|^{2}\rceil}^{\lfloor 2M|x|^{2}-1\rfloor}\mathbf{P}\left(S_{\sigma_{i}}={x}\right)
\displaystyle\leq 𝐄(#{iM|x|2:Sσi=x}).\displaystyle\mathbf{E}\left(\#\left\{i\geq\lceil M|x|^{2}\rceil:\>S_{\sigma_{i}}={x}\right\}\right).

Using that σii\sigma_{i}\geq i, this implies that

𝐏(τx[M|x|2,2M|x|21])\displaystyle\mathbf{P}\left(\tau_{x}\in\left[M|x|^{2},2M|x|^{2}-1\right]\right) \displaystyle\leq 𝐄(#{nM|x|2:Sn=x})\displaystyle\mathbf{E}\left(\#\left\{n\geq\lceil M|x|^{2}\rceil:\>S_{n}={x}\right\}\right)
=\displaystyle= n=M|x|2𝐏(Sn=x)\displaystyle\sum_{n=\lceil M|x|^{2}\rceil}^{\infty}\mathbf{P}(S_{n}=x)
\displaystyle\leq n=M|x|2Cnd/2\displaystyle\sum_{n=\lceil M|x|^{2}\rceil}Cn^{-d/2}
\displaystyle\leq C(M|x|2)1d/2,\displaystyle C\left(M|x|^{2}\right)^{1-d/2},

where for the second inequality, we have applied the upper bound on the transition probabilities of SS from [3, Theorem 6.28]. Since it also holds that exp(c2/M)C\exp(-c_{2}/M)\geq C uniformly over M1M\geq 1, the result follows. ∎

3.2 Upper bound for M(0,1)M\in(0,1)

We will give an upper bound on the probability that τx\tau_{x} is much smaller than |x|2|x|^{2}. More precisely, the goal of this section is to prove the following proposition. Replacing MM by 2M2M, this readily implies the relevant part of Proposition 3.1.

Proposition 3.3.

There exist constants c1,c2(0,)c_{1},c_{2}\in(0,\infty) such that for every xd\{0}x\in\mathbb{Z}^{d}\backslash\{0\} and M(0,2)M\in(0,2),

𝐏(τxM|x|2)c1(M|x|2)1d/2exp(c2M).\mathbf{P}\left(\tau_{x}\leq M|x|^{2}\right)\leq c_{1}\left(M|x|^{2}\right)^{1-d/2}\exp\left(-\frac{c_{2}}{M}\right). (10)

Before diving into the proof, we observe that it is enough to show (10) only for the case that both |x|1|x|^{-1} and MM are sufficiently small. To see this, suppose that there exist some c1,c2(0,)c_{1},c_{2}\in(0,\infty) and r0(0,1)r_{0}\in(0,1) such that the inequality (10) holds with constants c1,c2c_{1},c_{2} for all xx and MM satisfying |x|1Mr0|x|^{-1}\vee M\leq r_{0}. The remaining cases we need to consider are (i) |x|1r0|x|^{-1}\geq r_{0} and (ii) M[r0,2)M\in[r_{0},2). We first deal with case (i). If we suppose that |x|1r0|x|^{-1}\geq r_{0} and M<r02M<r_{0}^{2}, then M|x|2<1M|x|^{2}<1, and so the probability on the left-hand side of (10) is equal to 0. On the other hand, if |x|1r0|x|^{-1}\geq r_{0} and M[r02,2)M\in[r_{0}^{2},2), by choosing the constant c1c_{1} to satisfy c12d/21r02dexp{c2r02}c_{1}\geq 2^{d/2-1}r_{0}^{2-d}\exp\{c_{2}r_{0}^{-2}\}, we can ensure the right-hand side of (10) is greater than 1, and so the inequality (10) also holds in this case. Let us move to case (ii). We note that the probability on the left-hand side of (10) can be always bounded above by

𝐏(τxS<)C|x|2d\mathbf{P}\left(\tau^{S}_{x}<\infty\right)\leq C|x|^{2-d}

for some constant C(0,)C\in(0,\infty), where we have applied (8) to deduce the inequality. Thus, choosing the constant c1c_{1} so that c1C2d/21exp{c2r01}c_{1}\geq C2^{d/2-1}\exp\{c_{2}r_{0}^{-1}\}, the inequality (10) holds. Consequently, replacing the constant c1c_{1} by c12d/21r02dexp{c2r02}C2d/21exp{c2r01}c_{1}\vee 2^{d/2-1}r_{0}^{2-d}\exp\{c_{2}r_{0}^{-2}\}\vee C2^{d/2-1}\exp\{c_{2}r_{0}^{-1}\}, the inequality (10) holds for all xd{0}x\in\mathbb{Z}^{d}\setminus\{0\} and M(0,2)M\in(0,2).

We next give a brief outline of the proof of Proposition 3.3, assuming that both |x|1|x|^{-1} and MM are sufficiently small. We write Ax=B(0,|x|/4d)A_{x}=B_{\infty}(0,|x|/4\sqrt{d}) for the box of side length |x|/2d|x|/2\sqrt{d} centered at the origin, and let

tx=τAxcLt_{x}=\tau^{L}_{A_{x}^{c}}

be the first time that LL exits AxA_{x}. Note that xAxx\not\in A_{x}, and so

𝐏(τxM|x|2)𝐏(txτxM|x|2)𝐏(τx<txM|x|2)𝐏(txM|x|2).\displaystyle\mathbf{P}\left(\tau_{x}\leq M|x|^{2}\right)\leq\mathbf{P}\left(t_{x}\leq\tau_{x}\leq M|x|^{2}\right)\leq\mathbf{P}\left(\tau_{x}<\infty\>\vline\>t_{x}\leq M|x|^{2}\right)\,\mathbf{P}\left(t_{x}\leq M|x|^{2}\right). (11)

Writing

px,M=𝐏(τx<txM|x|2) and qx,M=𝐏(txM|x|2),p_{x,M}=\mathbf{P}\left(\tau_{x}<\infty\>\vline\>t_{x}\leq M|x|^{2}\right)\ \ \ \text{ and }\ \ \ q_{x,M}=\mathbf{P}\left(t_{x}\leq M|x|^{2}\right), (12)

we will show that

px,MC|x|2d,qx,MCexp{cM1}p_{x,M}\leq C|x|^{2-d},\qquad q_{x,M}\leq C\exp\{cM^{-1}\}

in Lemmas 3.4 and 3.5 below, respectively. Proposition 3.3 is then a direct consequence of these lemmas.

We start by dealing with px,Mp_{x,M}, as defined in (12).

Lemma 3.4.

There exists a constant C(0,)C\in(0,\infty) such that for all xd{0}x\in\mathbb{Z}^{d}\setminus\{0\} and M(0,2)M\in(0,2) with 𝐏(txM|x|2)>0\mathbf{P}(t_{x}\leq M|x|^{2})>0,

px,MC|x|2d.p_{x,M}\leq C|x|^{2-d}.
Proof.

Let

Λ={λ:𝐏(txM|x|2,L[0,tx]=λ)>0}\Lambda=\left\{\lambda\>:\>\mathbf{P}\left(t_{x}\leq M|x|^{2},\>L[0,t_{x}]=\lambda\right)>0\right\}

be the set of all possible paths for L[0,tx]L[0,t_{x}] satisfying txM|x|2t_{x}\leq M|x|^{2}. For λΛ\lambda\in\Lambda, we write R=RλR=R^{\lambda} for a random walk conditioned on the event that R[1,)λ=R[1,\infty)\cap\lambda=\emptyset. Note that RR is a Markov chain (see [22, Section 11.1]). We use 𝐏Ry\mathbf{P}^{y}_{R} to denote the law of RR started from R0=yR_{0}=y. Then the domain Markov property for LL (see [20, Proposition 7.3.1]) ensures that

px,M=λΛ𝐏Rλlen(λ)(xLE(R[0,)))𝐏(L[0,tx]=λ)qx,MmaxλΛ𝐏Rλlen(λ)(xR[0,)).p_{x,M}=\frac{\sum_{\lambda\in\Lambda}\mathbf{P}^{\lambda_{\text{len}(\lambda)}}_{R}\left(x\in\text{LE}\left(R[0,\infty)\right)\right)\mathbf{P}(L[0,t_{x}]=\lambda)}{q_{x,M}}\leq\max_{\lambda\in\Lambda}\mathbf{P}^{\lambda_{\text{len}(\lambda)}}_{R}\left(x\in R[0,\infty)\right).

Therefore, it suffices to show that there exists a constant C(0,)C\in(0,\infty) such that for all xd{0}x\in\mathbb{Z}^{d}\setminus\{0\}, M(0,2)M\in(0,2) with 𝐏(txM|x|2)>0\mathbf{P}(t_{x}\leq M|x|^{2})>0 and λΛ\lambda\in\Lambda,

𝐏Rλlen(λ)(xR[0,))C|x|2d.\mathbf{P}^{\lambda_{\text{len}(\lambda)}}_{R}\left(x\in R[0,\infty)\right)\leq C|x|^{2-d}.

With the above goal in mind, let us fix λΛ\lambda\in\Lambda. We set u:=τB(|x|/2)cRu:=\tau^{R}_{B(|x|/2)^{c}} for the first time that RR exits B(|x|/2)B(|x|/2). (Note that AxB(|x|/2)A_{x}\subseteq B(|x|/2) by our construction.) Using the strong Markov property for RR at time uu, we have

𝐏Rλlen(λ)(xR[0,))maxyB(|x|/2)𝐏Ry(xR[0,)).\mathbf{P}^{\lambda_{\text{len}(\lambda)}}_{R}\left(x\in R[0,\infty)\right)\leq\max_{y\in\partial B(|x|/2)}\mathbf{P}^{y}_{R}\left(x\in R[0,\infty)\right).

On the other hand, it follows from (7) that

minyB(|x|/2)𝐏y(S[0,)λ=)minyB(|x|/2)𝐏y(S[0,)Ax=)c0\min_{y\in\partial B(|x|/2)}\mathbf{P}^{y}\left(S[0,\infty)\cap\lambda=\emptyset\right)\geq\min_{y\in\partial B(|x|/2)}\mathbf{P}^{y}\left(S[0,\infty)\cap A_{x}=\emptyset\right)\geq c_{0}

for some constant c0>0c_{0}>0. Combining these estimates and using (8), we see that, for each yB(|x|/2)y\in\partial B(|x|/2),

𝐏Ry(xR[0,))𝐏y(xS[0,))𝐏y(S[0,)λ=)1c0𝐏y(xS[0,))C|x|2d\mathbf{P}^{y}_{R}\left(x\in R[0,\infty)\right)\leq\frac{\mathbf{P}^{y}\left(x\in S[0,\infty)\right)}{\mathbf{P}^{y}\left(S[0,\infty)\cap\lambda=\emptyset\right)}\leq\frac{1}{c_{0}}\mathbf{P}^{y}\left(x\in S[0,\infty)\right)\leq C|x|^{2-d}

for some constant C(0,)C\in(0,\infty). This finishes the proof. ∎

Recall that qx,Mq_{x,M} was defined at (12). We will next estimate qx,Mq_{x,M} as follows.

Lemma 3.5.

There exist constants c,C(0,)c,C\in(0,\infty) such that for all xd{0}x\in\mathbb{Z}^{d}\setminus\{0\} and M(0,2)M\in(0,2),

qx,MCexp{cM1}.q_{x,M}\leq C\exp\{-cM^{-1}\}. (13)
Proof.

As per the discussion after (10), it suffices to prove (13) only in the case that both |x|1|x|^{-1} and MM are sufficiently small. In particular, throughout the proof, we assume that

M(3200d)1.M\leq(3200d)^{-1}. (14)

Furthermore, we may assume

|x|M(4d)1,|x|M\geq(4\sqrt{d})^{-1}, (15)

since qx,M=0q_{x,M}=0 when |x|M<(4d)1|x|M<(4\sqrt{d})^{-1}. (Notice that it must hold that tx|x|(4d)1t_{x}\geq|x|(4\sqrt{d})^{-1}.)

Now, define the increasing sequence of boxes {Ai}i=1N\{A^{i}\}_{i=1}^{N}, where N=(1600dM)1N=\lfloor(1600dM)^{-1}\rfloor, by setting

Ai=B(0,400d|x|Mi)A^{i}=B_{\infty}\left(0,400\sqrt{d}\,|x|Mi\right)

for 1iN1\leq i\leq N. Observe that the particular choice of NN ensures that ANAx=B(0,|x|/4d)A^{N}\subseteq A_{x}=B_{\infty}(0,|x|/4\sqrt{d}), and the assumption (14) guarantees that

N(3200dM)1.N\geq(3200dM)^{-1}. (16)

Also, we note that dist(Ai1,Ai)\text{dist}(\partial A^{i-1},\partial A^{i}) is bigger than 400d|x|M1400\sqrt{d}\,|x|M-1, which is in turn bounded below by 9999 because of (15). As a consequence, it is reasonable to compare the number of lattice points in the set L(AiAi1)L\cap(A^{i}\setminus A^{i-1}) with |x|2M2|x|^{2}M^{2}. To be more precise, let t0=0t^{0}=0, and, for i1i\geq 1, set

ti=τ(Ai)cLt^{i}=\tau^{L}_{(A^{i})^{c}}

to be the first time that LL exits AiA^{i}. Then [7, Corollary 3.10] shows that there exists a deterministic constant c1(0,1)c_{1}\in(0,1) such that for all xd{0}x\in\mathbb{Z}^{d}\setminus\{0\} and M(0,2)M\in(0,2) satisfying (14) and (15),

𝐏(titi1c1|x|2M2L[0,ti1])c1,1iN.\mathbf{P}\left(t^{i}-t^{i-1}\geq c_{1}|x|^{2}M^{2}\>\vline\>L[0,t^{i-1}]\right)\geq c_{1},\qquad 1\leq i\leq N. (17)

With the inequality (17) and a constant a>0a>0 satisfying

22a1c1<16400dlog11c1,2\sqrt{\frac{2a}{1-c_{1}}}<\frac{1}{6400d}\log\frac{1}{1-c_{1}}, (18)

it is possible to apply [4, Lemma 1.1] to deduce the result of interest. In particular, the following table explains how the quantities of this article are substituted into [4, Lemma 1.1].

[4, Lemma 1.1] XX YiY_{i} nn pp bb xx
This article txt_{x} titi1t^{i}-t^{i-1} NN 1c11-c_{1} 2|x|2M22\,|x|^{-2}M^{-2} aM|x|2aM|x|^{2}

Then, from [4, Lemma 1.1], one has

𝐏(txaM|x|2)\displaystyle\mathbf{P}\left(t_{x}\leq aM|x|^{2}\right) exp{2M12a1c1Nlog11c1}\displaystyle{\leq}\exp\left\{2M^{-1}\sqrt{\frac{2a}{1-c_{1}}}-N\log\frac{1}{1-c_{1}}\right\}
exp{(22a1c113200dlog11c1)M1}\displaystyle{\leq}\exp\left\{\left(2\sqrt{\frac{2a}{1-c_{1}}}-\frac{1}{3200d}\log\frac{1}{1-c_{1}}\right)M^{-1}\right\}
exp{M16400dlog11c1},\displaystyle{\leq}\exp\left\{-\frac{M^{-1}}{6400d}\log\frac{1}{1-c_{1}}\right\},

where for the second and third inequalities, we apply (16) and (18), respectively. Rewriting aM=MaM=M^{\prime} completes the proof. ∎

Proof of Proposition 3.3.

Proposition 3.3 follows directly from (11) and Lemmas 3.4 and 3.5 (and the basic observation that M1d/221d/2M^{1-d/2}\geq 2^{1-d/2} for M(0,2)M\in(0,2)). ∎

3.3 Lower bound for M(0,1)M\in(0,1)

Recall that for xd{0}x\in\mathbb{Z}^{d}\setminus\{0\}, τx\tau_{x} indicates the first time that LL hits xx. The aim of this section is to bound below the probability that τx\tau_{x} is much smaller than |x|2|x|^{2}. In particular, the following is the main result of this section, which readily implies the part of the lower bound of Proposition 3.1 with |x|1M<1|x|^{-1}\leq M<1.

Proposition 3.6.

There exist constants c,c,R(0,)c,c^{\prime},R\in(0,\infty) such that for every xd\{0}x\in\mathbb{Z}^{d}\backslash\{0\} and |x|1M<1|x|^{-1}\leq M<1,

𝐏(τx[R1M|x|2,RM|x|2])c(M|x|2)1d/2exp(cM).\mathbf{P}\left(\tau_{x}\in[R^{-1}M|x|^{2},RM|x|^{2}]\right)\geq c^{\prime}(M|x|^{2})^{1-d/2}\exp\left(-\frac{c}{M}\right). (19)

Before moving on to the proof, we will first show that once we prove that there exists a constant n01n_{0}\geq 1 such that (19) holds for n0|x|1M<1n_{0}|x|^{-1}\leq M<1, we obtain (19) for every xd{0}x\in\mathbb{Z}^{d}\setminus\{0\} and |x|1M<1|x|^{-1}\leq M<1 by adjusting cc, cc^{\prime} and RR as needed. Let us consider the following three events:

  • S[0,τxS]S[0,\tau^{S}_{x}] is a simple path of length M|x|2\lceil M|x|^{2}\rceil,

  • S[τxS+1,τB(0,2r)cS]S[\tau^{S}_{x}+1,\tau^{S}_{B(0,2r)^{c}}] is a simple path that does not intersect S[0,τxS]S[0,\tau^{S}_{x}],

  • S[τB(0,2r)cS,)B(0,32|x|)=S[\tau^{S}_{B(0,2r)^{c}},\infty)\cap B(0,\frac{3}{2}|x|)=\emptyset,

where we set r=|x|n0r=|x|\vee n_{0}. It is straightforward to see that τx[M|x|2,2M|x|2]\tau_{x}\in[M|x|^{2},2M|x|^{2}] holds on the intersection of these events. By constructing a simple random walk path that satisfies the first two conditions up to the first exit time from the Euclidean ball B(0,2r)B(0,2r) and then applying (7) and the strong Markov property, we have

𝐏(τx[M|x|2,2M|x|2])a(2d)2M|x|2(2d)2r,\mathbf{P}\left(\tau_{x}\in[M|x|^{2},2M|x|^{2}]\right)\geq a(2d)^{-2M|x|^{2}}(2d)^{-2r},

for some a>0a>0 that does not depend on MM or xx. Suppose 1M|x|n01\leq M|x|\leq n_{0}. If |x|<n0|x|<n_{0}, then the right-hand side is bounded below as follows:

a(2d)2M|x|2(2d)2ra(2d)(2n02+2n0)Cc(M|x|2)1d/2ecM.a(2d)^{-2M|x|^{2}}(2d)^{-2r}\geq a(2d)^{-(2n_{0}^{2}+2n_{0})}\geq C\geq c^{\prime}(M|x|^{2})^{1-d/2}e^{-\frac{c}{M}}.

On the other hand, if |x|n0|x|\geq n_{0}, then the right-hand side satisfies

a(2d)2M|x|2(2d)2rC(M|x|2)1d/2ecM|x|2c|x|C(M|x|2)1d/2ec′′/M.a(2d)^{-2M|x|^{2}}(2d)^{-2r}\geq C\left(M|x|^{2}\right)^{1-d/2}e^{-cM|x|^{2}-c^{\prime}|x|}\geq C\left(M|x|^{2}\right)^{1-d/2}e^{-c^{\prime\prime}/M}.

In particular, by replacing RR by R2R\vee 2 and adjusting cc, cc^{\prime} appropriately, the result at (19) can be extended to 1M|x|<|x|1\leq M|x|<|x|.

The structure of this section is as follows. First, we define several subsets of d\mathbb{R}^{d}. These will be used to describe a number of events involving the simple random walk SS whose loop-erasure is LL. We provide some key estimates on the probabilities of these events in Lemmas 3.9 and Lemma 3.12. Finally, applying these results, we prove Proposition 3.6 at the end of this section.

We begin by defining “a tube connecting the origin and xx”, which will consist of a number of boxes of side-length M|x|M|x|. To this end, for M(0,1)M\in(0,1), let

NM=1M+12.N_{M}=\left\lceil\frac{1}{M}+\frac{1}{2}\right\rceil.

Moreover, for x=(x1,,xd)d{0}x=(x^{1},\dots,x^{d})\in\mathbb{Z}^{d}\setminus\{0\} and M(0,1)M\in(0,1), define a sequence {bi}\{b_{i}\} of vertices of d\mathbb{R}^{d} by setting

bi=(iMx1,,iMxd)b_{i}=\left(iMx^{1},\dots,iMx^{d}\right) (20)

for i{0,1,,2NM}i\in\{0,1,\dots,2N_{M}\}. Let us consider a rotation around the origin that aligns the x1x^{1}-axis with the line through the origin and xx. We denote by BB and QQ the images of [M|x|/2,M|x|/2]d[-{M|x|}/{2},{M|x|}/{2}]^{d} and {0}×[M|x|/2,M|x|/2]d1\{0\}\times[-{M|x|}/{2},{M|x|}/{2}]^{d-1} under this rotation, respectively. For y=(y1,,yd)dy=(y^{1},\dots,y^{d})\in\mathbb{R}^{d}, we let

B~(y,r)={(y1+rz1,,yd+rzd):(z1,,zd)B},\widetilde{B}(y,r)=\left\{\left(y^{1}+rz^{1},\dots,y^{d}+rz^{d}\right)\mathrel{:}(z^{1},\dots,z^{d})\in B\right\}, (21)

be the tilted cube of side-length rM|x|rM|x| centered at yy, and we write BiB_{i} for B~(bi,1)\widetilde{B}(b_{i},1). For i=0,1,,2NMi=0,1,\dots,2N_{M} and a,ba,b\in\mathbb{R} with a<ba<b, also let

Q(y,r)={(y1+rz1,,yd+rzd):(z1,,zd)Q},Q(y,r)=\left\{(y^{1}+rz^{1},\dots,y^{d}+rz^{d})\mathrel{:}(z^{1},\dots,z^{d})\in Q\right\},
Bi[a,b]={(z1+sMx1,,zd+sMxd):s[a,b],(z1,,zd)Qi(0)},B_{i}[a,b]=\left\{\left(z^{1}+sMx^{1},\dots,z^{d}+sMx^{d}\right)\mathrel{:}s\in[a,b],\ (z^{1},\dots,z^{d})\in Q_{i}(0)\right\},

where

Qi(b)=Q(((i12+b)M|x|,,(i12+b)M|x|),1).Q_{i}(b)=Q\left(\left((i-\frac{1}{2}+b)M|x|,\dots,(i-\frac{1}{2}+b)M|x|\right),1\right).

We also set

Q~i(b)=Q(((i12+b)M|x|,,(i12+b)M|x|),12),\widetilde{Q}_{i}(b)=Q\left(\left((i-\frac{1}{2}+b)M|x|,\dots,(i-\frac{1}{2}+b)M|x|\right),\frac{1}{2}\right),

and note that, by definition, Q~i(b)Qi(b)\widetilde{Q}_{i}(b)\subseteq Q_{i}(b) for all i0i\geq 0 and bb\in\mathbb{R}. Observe that every Q~i(b)\widetilde{Q}_{i}(b) is perpendicular to the line through the origin and xx, and that QiQi(0)Q_{i}\coloneqq Q_{i}(0) is the “left face” of the cube Bi=Bi[0,1]B_{i}=B_{i}[0,1]. Finally, we write

Q~iQ~i(0),i=1,2,,2NM+1,\widetilde{Q}_{i}\coloneqq\widetilde{Q}_{i}(0),\qquad i=1,2,\dots,2N_{M}+1, (22)

and set Q0=Q~0={0}Q_{0}=\widetilde{Q}_{0}=\{0\} for convenience. See Figure 1 for a graphical representation of the situation.

Refer to caption
Figure 1: Illustration of BiB_{i}, QiQ_{i} and Q~i\widetilde{Q}_{i} for a given xx.

In this section, it will be convenient to consider SS (recall that L=LE(S[0,))L=\mathrm{LE}(S[0,\infty))) as a continuous curve in d\mathbb{R}^{d} by linear interpolating between discrete time points, and thus we may assume that S(k)S(k) is defined for all non-negative real kk. If λ\lambda is a continuous path in d\mathbb{R}^{d} and AdA\subseteq\mathbb{R}^{d}, we write

τλ(A)=inf{t0:λ(t)A},\tau^{\lambda}(A)=\inf\left\{t\geq 0\>:\>\lambda(t)\in A\right\},

and also, for xdx\in\mathbb{R}^{d}, we set τxλ=τλ({x})\tau^{\lambda}_{x}=\tau^{\lambda}(\{x\}), analogous to the notation of first hitting times for discrete paths (5).

In order to obtain the lower bound (19), we consider events under which the LERW LL, started at the origin, travels through the “tube” i=0NMBi\bigcup_{i=0}^{N_{M}}B_{i} until it hits xx. See Figure 2 for a graphical representation.

Definition 3.7.

We define the events FiF_{i}, i=0,1,,2NMi=0,1,\dots,2N_{M}, as follows. Firstly,

F0={τS(Q1)<,S(τS(Q1))Q~1,S[0,τS(Q1)]B0,S[τS(Q1(ε)),τS(Q1)]Q1(2ε)=}.F_{0}=\left\{\begin{array}[]{c}\tau^{S}(Q_{1})<\infty,\>S(\tau^{S}(Q_{1}))\in\widetilde{Q}_{1},\>S[0,\tau^{S}(Q_{1})]\subset B_{0},\\ S[\tau^{S}(Q_{1}(-\varepsilon)),\tau^{S}(Q_{1})]\cap Q_{1}(-2\varepsilon)=\emptyset\end{array}\right\}.

For i=1,2,,NM1i=1,2,\dots,N_{M}-1,

Fi={τS(Qi)<τS(Qi+1)<,S(τS(Qi+1))Q~i+1,S[τS(Qi),τS(Qi+1)]Bi[ε,1],S[τS(Qi+1(ε)),τS(Qi+1)]Qi+1(2ε)=}.F_{i}=\left\{\begin{array}[]{c}\tau^{S}(Q_{i})<\tau^{S}(Q_{i+1})<\infty,\>S(\tau^{S}(Q_{i+1}))\in\widetilde{Q}_{i+1},\>S[\tau^{S}(Q_{i}),\tau^{S}(Q_{i+1})]\subset B_{i}[-\varepsilon,1],\\ S[\tau^{S}(Q_{i+1}(-\varepsilon)),\tau^{S}(Q_{i+1})]\cap Q_{i+1}(-2\varepsilon)=\emptyset\end{array}\right\}.

Moreover,

FNM={τS(QNM)<τxS<τS(QNM+1(14))<,S(τS(QNM+1(14)))Q~NM+1(14),S[τS(QNM),τS(QNM+1)]BNM[14,54],S[τxS,τS(QNM+1(14))]LE(S[τS(QNM),τxS])=},F_{N_{M}}=\left\{\begin{array}[]{c}\tau^{S}(Q_{N_{M}})<\tau^{S}_{x}<\tau^{S}\left(Q_{N_{M}+1}\left(\frac{1}{4}\right)\right)<\infty,\\ S\left(\tau^{S}\left(Q_{N_{M}+1}\left(\frac{1}{4}\right)\right)\right)\in\widetilde{Q}_{N_{M}+1}\left(\frac{1}{4}\right),\>S[\tau^{S}(Q_{N_{M}}),\tau^{S}(Q_{N_{M}+1})]\subset B_{N_{M}}\left[-\frac{1}{4},\frac{5}{4}\right],\\ S\left[\tau^{S}_{x},\tau^{S}\left(Q_{N_{M}+1}\left(\frac{1}{4}\right)\right)\right]\cap\mathrm{LE}(S[\tau^{S}(Q_{N_{M}}),\tau^{S}_{x}])=\emptyset\end{array}\right\},
FNM+1={τS(QNM+1(14))<τS(QNM+2),τS(QNM+2)Q~NM+2,S[τS(QNM+1(14)),τS(QNM+2)]BNM+1[14ε,1]},F_{N_{M}+1}=\left\{\begin{array}[]{c}\tau^{S}\left(Q_{N_{M}+1}\left(\frac{1}{4}\right)\right)<\tau^{S}(Q_{N_{M}+2}),\>\tau^{S}(Q_{N_{M}+2})\in\widetilde{Q}_{N_{M}+2},\\ S\left[\tau^{S}\left(Q_{N_{M}+1}\left(\frac{1}{4}\right)\right),\tau^{S}(Q_{N_{M}+2})\right]\subset B_{N_{M}+1}\left[\frac{1}{4}-\varepsilon,1\right]\end{array}\right\},

and, for i=NM+1,,2NMi=N_{M}+1,\dots,2N_{M},

Fi={τS(Qi)<τS(Qi+1)<,S(τS(Qi+1))Q~i+1,S[τS(Qi),τS(Qi+1)]Bi[ε,1]}.F_{i}=\left\{\begin{array}[]{c}\tau^{S}(Q_{i})<\tau^{S}(Q_{i+1})<\infty,\>S(\tau^{S}(Q_{i+1}))\in\widetilde{Q}_{i+1},\\ S[\tau^{S}(Q_{i}),\tau^{S}(Q_{i+1})]\subset B_{i}[-\varepsilon,1]\end{array}\right\}.
Refer to caption
Figure 2: Illustration of the events FiF_{i}, i=0,1,,NMi=0,1,\dots,N_{M}.

The first three conditions of the definition of each FiF_{i}, i=0,1,,2NMi=0,1,\dots,2N_{M}, require that SS travels inside the “tube” and it exits each BiB_{i} at a point which is not close to Qi+1\partial Q_{i+1}. Furthermore, the last condition in the definition of F0F_{0}, the last two conditions in that of each FiF_{i}, i=1,2,,NM1i=1,2,\dots,N_{M}-1, and the third condition in that of FNMF_{N_{M}} control the range of backtracking of SS. Finally, the last condition in the definition of FNMF_{N_{M}} and events FiF_{i}, i=NM+1,,2NMi=N_{M}+1,\dots,2N_{M} guarantee that xx remains in LE(S[0,τS(Q2NM+1)])\mathrm{LE}(S[0,\tau^{S}(Q_{2N_{M}+1})]).

Next, we define events that provide upper and lower bounds for the length of the loop-erasure of SS in each BiB_{i}. For i{0,1,,NM1}i\in\{0,1,\dots,N_{M}-1\}, let

ξi=LE(S[0,τS(Qi+1)]),\xi_{i}=\mathrm{LE}\left(S[0,\tau^{S}(Q_{i+1})]\right), (23)

and also set ξNM=LE(S[0,τxS])\xi_{N_{M}}=\mathrm{LE}(S[0,\tau^{S}_{x}]). We further define

λi\displaystyle\lambda_{i} =LE(S[τS(Qi),τS(Qi+1)]),1iNM1,\displaystyle=\mathrm{LE}\left(S[\tau^{S}(Q_{i}),\tau^{S}(Q_{i+1})]\right),\qquad 1\leq i\leq N_{M}-1,
λNM\displaystyle\lambda_{N_{M}} =LE(S[τS(QNM),τxS]),\displaystyle=\mathrm{LE}\left(S[\tau^{S}(Q_{N_{M}}),\tau^{S}_{x}]\right),
ξ0\displaystyle\xi^{\prime}_{0} =ξ0,\displaystyle=\xi_{0},
ξi\displaystyle\xi^{\prime}_{i} =ξ0λ1λi,i1.\displaystyle=\xi_{0}\oplus\lambda_{1}\oplus\cdots\oplus\lambda_{i},\qquad i\geq 1. (24)

Since ξi\xi_{i} is not necessarily a simple curve, ξi\xi_{i} and ξi\xi^{\prime}_{i} do not coincide in general. However, the restriction on the backtracking of SS on the events FiF_{i} and a cut time argument (see Definition 3.10 and Definition 3.11 below) enable us to handle the difference between ξi\xi_{i} and ξi\xi^{\prime}_{i}. This will be discussed later, in the proof of Proposition 3.6.

We now define events upon which the length of λi\lambda_{i} is bounded above.

Definition 3.8.

For C>0C>0, the event G0(C)G_{0}(C) is given by

G0(C)={len(ξ0)CM2|x|2},G_{0}(C)=\left\{\mathrm{len}(\xi_{0})\leq CM^{2}|x|^{2}\right\},

and for i=1,2,,NMi=1,2,\dots,N_{M}, the event Gi(C)G_{i}(C) is given by

Gi(C)={len(λi)CM2|x|2}.G_{i}(C)=\left\{\mathrm{len}(\lambda_{i})\leq CM^{2}|x|^{2}\right\}.

In the following lemma, we demonstrate that GiG_{i} occurs with high conditional probability. Recall that Q~i\widetilde{Q}_{i} was defined at (22).

Lemma 3.9.

For any δ>0\delta>0, there exists a constant C+>0C_{+}>0 such that

𝐏z(Gi(C+)Fi)1δ,\mathbf{P}^{z}\left(G_{i}(C_{+})\>\vline\>F_{i}\right)\geq 1-\delta, (25)

uniformly in xd\{0}x\in\mathbb{Z}^{d}\backslash\{0\}, |x|1M<1|x|^{-1}\leq M<1, i{0,1,,NM}i\in\{0,1,\dots,N_{M}\} and zQ~iz\in\widetilde{Q}_{i}.

Proof.

For i=0,1,,NM1i=0,1,\dots,N_{M}-1, yBi[ε,1]y\in B_{i}[-\varepsilon,1] and zQ~iz\in\widetilde{Q}_{i}, we have that

𝐏z(yλiFi)𝐏z(τyS<τS(Qi+1)Fi).\mathbf{P}^{z}\left(y\in\lambda_{i}\mid F_{i}\right)\leq\mathbf{P}^{z}(\tau^{S}_{y}<\tau^{S}(Q_{i+1})\mid F_{i}).

Moreover, by (9) and translation invariance, we have that there exists some constant c>0c>0 such that

infzQ~i𝐏z(Fi)cε,\inf_{z\in\widetilde{Q}_{i}}\mathbf{P}^{z}(F_{i})\geq c\varepsilon, (26)

for all i=1,2,,NM1i=1,2,\dots,N_{M}-1. For i=0i=0, the same argument yields that 𝐏(F0)cε\mathbf{P}(F_{0})\geq c\varepsilon. Thus, it follows from (8) and (26) that

𝐏z(τyS<τS(Qi+1)Fi)𝐏z(τyS<τS(Qi+1))𝐏z(Fi)C𝐏z(τyS<)C(|yz|2d1),\mathbf{P}^{z}(\tau^{S}_{y}<\tau^{S}(Q_{i+1})\mid F_{i})\leq\frac{\mathbf{P}^{z}(\tau^{S}_{y}<\tau^{S}(Q_{i+1}))}{\mathbf{P}^{z}(F_{i})}\leq C\mathbf{P}^{z}(\tau^{S}_{y}<\infty)\leq C\left(|y-z|^{2-d}\wedge 1\right),

for some constant C>0C>0. By taking the sum over yBi[ε,1]y\in B_{i}[-\varepsilon,1], we have that

𝐄z(len(λi)Fi)=yBi[ε,1]𝐏z(yλiFi)CyBi[ε,1](|yz|2d1)CM2|x|2.\mathbf{E}^{z}(\mathrm{len}(\lambda_{i})\mid F_{i})=\sum_{y\in B_{i}[-\varepsilon,1]}\mathbf{P}^{z}(y\in\lambda_{i}\mid F_{i})\leq C\sum_{y\in B_{i}[-\varepsilon,1]}\left(|y-z|^{2-d}\wedge 1\right)\leq CM^{2}|x|^{2}. (27)

The same argument also applies to the case i=0i=0, and thus we have

𝐄0(len(ξ0)F0)CM2|x|2.\mathbf{E}^{0}(\mathrm{len}(\xi_{0})\mid F_{0})\leq CM^{2}|x|^{2}. (28)

Similarly, for the case i=NMi=N_{M}, recalling that λNM\lambda_{N_{M}} was defined at (24), we have that

𝐏z(yλNMFNM)\displaystyle{\mathbf{P}^{z}\left(y\in\lambda_{N_{M}}\mid F_{N_{M}}\right)} 𝐏z(τyS<τxSFNM)\displaystyle\leq\mathbf{P}^{z}\left(\tau^{S}_{y}<\tau^{S}_{x}\mid F_{N_{M}}\right)
𝐏z(τyS<τxS<τS(BNM[14,54]c))𝐏z(FNM)\displaystyle\leq\frac{\mathbf{P}^{z}\left(\tau^{S}_{y}<\tau^{S}_{x}<\tau^{S}(B_{N_{M}}[-\frac{1}{4},\frac{5}{4}]^{c})\right)}{\mathbf{P}^{z}\left(F_{N_{M}}\right)}
=𝐏z(τyS<τxSτS(BNM[14,54]c))𝐏y(τxS<τS(BNM[14,54]c))𝐏z(FNM),\displaystyle=\frac{\mathbf{P}^{z}\left(\tau^{S}_{y}<\tau^{S}_{x}\wedge\tau^{S}(B_{N_{M}}[-\frac{1}{4},\frac{5}{4}]^{c})\right)\mathbf{P}^{y}\left(\tau^{S}_{x}<\tau^{S}(B_{N_{M}}[-\frac{1}{4},\frac{5}{4}]^{c})\right)}{\mathbf{P}^{z}\left(F_{N_{M}}\right)}, (29)

where we used the strong Markov inequality to obtain the last inequality. We will prove that 𝐏z(FNM)C(M|x|)2d\mathbf{P}^{z}(F_{N_{M}})\geq C^{\prime}(M|x|)^{2-d} later in this subsection, see (59).

Now we bound above the sum of the numerator of (29) over yBNM[14,54]y\in B_{N_{M}}[-\frac{1}{4},\frac{5}{4}], separating into three cases depending on the location of yy.

  1. (i)

    For yB(z,118M|x|)y\in B(z,\frac{1}{18}M|x|), it follows from (8) that

    𝐏z(τyS<τS(BNM[14,54]c)τxS)\displaystyle\mathbf{P}^{z}\left(\tau^{S}_{y}<\tau^{S}\left(B_{N_{M}}\left[-\frac{1}{4},\frac{5}{4}\right]^{c}\right)\wedge\tau^{S}_{x}\right) =C(|yz|2d(M|x|/2)2d)+O(|yz|1d),\displaystyle=C(|y-z|^{2-d}-(M|x|/2)^{2-d})+O(|y-z|^{1-d}),
    𝐏y(τxS<τS(BNM[14,54]c))\displaystyle\mathbf{P}^{y}\left(\tau^{S}_{x}<\tau^{S}\left(B_{N_{M}}\left[-\frac{1}{4},\frac{5}{4}\right]^{c}\right)\right) C(M|x|)2d.\displaystyle\leq C(M|x|)^{2-d}.

    Thus we have that

    yB(z,118M|x|)\displaystyle\sum_{y\in B(z,\frac{1}{18}M|x|)} 𝐏z(τyS<τxSτS(BNM[14,54]c))𝐏y(τxS<τS(BNM[14,54]c))\displaystyle\mathbf{P}^{z}\left(\tau^{S}_{y}<\tau^{S}_{x}\wedge\tau^{S}\left(B_{N_{M}}\left[-\frac{1}{4},\frac{5}{4}\right]^{c}\right)\right)\mathbf{P}^{y}\left(\tau^{S}_{x}<\tau^{S}\left(B_{N_{M}}\left[-\frac{1}{4},\frac{5}{4}\right]^{c}\right)\right)
    C(M|x|)2dk=1118M|x||yz|=k(k2d(M|x|/2)2d+O(k1d))\displaystyle\leq C(M|x|)^{2-d}\sum_{k=1}^{\frac{1}{18}M|x|}\sum_{|y-z|=k}(k^{2-d}-(M|x|/2)^{2-d}+O(k^{1-d}))
    C(M|x|)4d.\displaystyle\leq C(M|x|)^{4-d}.
  2. (ii)

    For yB(x,118M|x|)y\in B(x,\frac{1}{18}M|x|), a similar argument to case (i) yields that

    yB(x,118M|x|)\displaystyle\sum_{y\in B(x,\frac{1}{18}M|x|)} 𝐏z(τyS<τxSτS(BNM[14,54]c))𝐏y(τxS<τS(BNM[14,54]c))\displaystyle\mathbf{P}^{z}\left(\tau^{S}_{y}<\tau^{S}_{x}\wedge\tau^{S}\left(B_{N_{M}}\left[-\frac{1}{4},\frac{5}{4}\right]^{c}\right)\right)\mathbf{P}^{y}\left(\tau^{S}_{x}<\tau^{S}\left(B_{N_{M}}\left[-\frac{1}{4},\frac{5}{4}\right]^{c}\right)\right)
    C(M|x|)4d.\displaystyle\leq C(M|x|)^{4-d}.
  3. (iii)

    For yBNM[14,54](B(z,118M|x|)B(x,118M|x|))y\in B_{N_{M}}[-\frac{1}{4},\frac{5}{4}]\setminus(B(z,\frac{1}{18}M|x|)\cup B(x,\frac{1}{18}M|x|)), we have that

    𝐏z(τyS<τxS<τS(BNM[14,54]c))\displaystyle\mathbf{P}^{z}\left(\tau^{S}_{y}<\tau^{S}_{x}<\tau^{S}\left(B_{N_{M}}\left[-\frac{1}{4},\frac{5}{4}\right]^{c}\right)\right) 𝐏z(τyS<)C(M|x|)2d,\displaystyle\leq\mathbf{P}^{z}(\tau^{S}_{y}<\infty)\leq C(M|x|)^{2-d},
    𝐏y(τxS<τS(BNM[14,54]c))\displaystyle\mathbf{P}^{y}\left(\tau^{S}_{x}<\tau^{S}\left(B_{N_{M}}\left[-\frac{1}{4},\frac{5}{4}\right]^{c}\right)\right) 𝐏y(τxS<)C(M|x|)2d,\displaystyle\leq\mathbf{P}^{y}(\tau^{S}_{x}<\infty)\leq C(M|x|)^{2-d},

    which gives

    yBNM[14,54]|yz|,|yx|118M|x|\displaystyle\sum_{\begin{subarray}{c}y\in B_{N_{M}}[-\frac{1}{4},\frac{5}{4}]\\ |y-z|,|y-x|\geq\frac{1}{18}M|x|\end{subarray}} 𝐏z(τyS<τxS<τS(BNM[14,54]c))𝐏y(τxS<τS(BNM[14,54]c))\displaystyle\mathbf{P}^{z}\left(\tau^{S}_{y}<\tau^{S}_{x}<\tau^{S}\left(B_{N_{M}}\left[-\frac{1}{4},\frac{5}{4}\right]^{c}\right)\right)\mathbf{P}^{y}\left(\tau^{S}_{x}<\tau^{S}\left(B_{N_{M}}\left[-\frac{1}{4},\frac{5}{4}\right]^{c}\right)\right)
    CyBNM[14,54]|yz|,|yx|118M|x|(M|x|)42dC(M|x|)4d.\displaystyle\leq C\sum_{\begin{subarray}{c}y\in B_{N_{M}}[-\frac{1}{4},\frac{5}{4}]\\ |y-z|,|y-x|\geq\frac{1}{18}M|x|\end{subarray}}(M|x|)^{4-2d}\leq C(M|x|)^{4-d}.

Thus, by (29), it holds that

𝐄z(len(λNM)FNM)=yBNM[14,54]𝐏z(yλNMFNM)C(M|x|)4dcC(M|x|)2dCM2|x|2.\mathbf{E}^{z}(\mathrm{len}(\lambda_{N_{M}})\mid F_{N_{M}})=\sum_{y\in B_{N_{M}}[-\frac{1}{4},\frac{5}{4}]}\mathbf{P}^{z}(y\in\lambda_{N_{M}}\mid F_{N_{M}})\leq\frac{C(M|x|)^{4-d}}{c^{\prime}\cdot C^{\prime}(M|x|)^{2-d}}\leq CM^{2}|x|^{2}. (30)

Combining (27), (28) and (30) with Markov’s inequality, it holds that

𝐏z(len(λi)C+M2|x|2Fi)C/C+,\mathbf{P}^{z}(\mathrm{len}(\lambda_{i})\geq C_{+}M^{2}|x|^{2}\mid F_{i})\leq C/C_{+},

for i=0,1,,NMi=0,1,\dots,N_{M}. By taking C+=δ1CC_{+}=\delta^{-1}C, we obtain (25). ∎

Now we will deal with events involving SS upon which the length of λi\lambda_{i} is bounded below and, at the same time, the gap between the lengths of ξi\xi_{i} and ξi\xi^{\prime}_{i} is bounded above. First, we define a special type of cut time of SS in each BiB_{i}.

Definition 3.10.

A nice cut time in BiB_{i} is a time kk satisfying the following conditions:

  • k[τS(Qi(ε/3)),τS(Qi(ε))]k\in\left[\tau^{S}(Q_{i}(\varepsilon/3)),\tau^{S}(Q_{i}(\varepsilon))\right],

  • S(k)B(S(τS(Qi)),εM|x|/2)S(k)\in B(S(\tau^{S}(Q_{i})),\varepsilon M|x|/2),

  • S[τS(Qi),k]S[k+1,τS(Qi(ε))]=S[\tau^{S}(Q_{i}),k]\cap S\left[k+1,\tau^{S}(Q_{i}(\varepsilon))\right]=\emptyset,

  • S[k+1,τS(Qi(ε))]Qi=S[k+1,\tau^{S}(Q_{i}(\varepsilon))]\cap Q_{i}=\emptyset.

Second, let BiB^{\prime}_{i} (respectively Bi′′B^{\prime\prime}_{i}) be the cube of side-length M|x|/3M|x|/3 (respectively M|x|/9M|x|/9) centered at bib_{i} whose faces are parallel to those of BiB_{i}, i.e.

Bi=B~(bi,13),Bi′′=B~(bi,19),B^{\prime}_{i}=\widetilde{B}\left(b_{i},\frac{1}{3}\right),\qquad B^{\prime\prime}_{i}=\widetilde{B}\left(b_{i},\frac{1}{9}\right),

where bib_{i} and B~(y,r)\widetilde{B}(y,r) are as defined in (20) and (21), respectively. We denote by QiLQ_{i}^{L} (respectively QiRQ_{i}^{R}) the “left (respectively right) face” of BiB^{\prime}_{i}. See Figure 3.

Let ρi=inf{nτS(Bi′′):S(k)(Bi)c}\rho_{i}=\inf\{n\geq\tau^{S}(B^{\prime\prime}_{i}):\>S(k)\in(B^{\prime}_{i})^{c}\} be the first time that SS exits BiB^{\prime}_{i} after it first entered Bi′′B^{\prime\prime}_{i}. We define a set of local cut times of SS by

Ki={τS(Bi′′)kρi:S(k)Bi′′,S[τS(Bi),k]S[k+1,ρi]=}.K_{i}=\left\{\tau^{S}(B^{\prime\prime}_{i})\leq k\leq\rho_{i}\mathrel{:}S(k)\in B^{\prime\prime}_{i},\>S[\tau^{S}(B^{\prime}_{i}),k]\cap S[k+1,\rho_{i}]=\emptyset\right\}.

Finally, we define events Hi(j)(j=1,2,3,4)H_{i}^{(j)}~{}(j=1,2,3,4) as follows.

Refer to caption
Figure 3: Illustration of BiB^{\prime}_{i} and Bi′′B^{\prime\prime}_{i}.
Definition 3.11.

For 1iNM11\leq i\leq N_{M}-1 and l>0l>0,

Hi(1)\displaystyle H_{i}^{(1)} ={S has a nice cut time in Bi}{0<τS(Qi(ε))τS(Qi)Cε2M2|x|2},\displaystyle=\left\{S\mbox{ has a nice cut time in $B_{i}$}\right\}\cap\left\{0<\tau^{S}(Q_{i}(\varepsilon))-\tau^{S}(Q_{i})\leq C\varepsilon^{2}M^{2}|x|^{2}\right\},
Hi(2)\displaystyle H_{i}^{(2)} ={τS(Bi)<τS(Qi+1),S(τS(Qi(1/3)))QiL,S[τS(Qi(ε)),τS(Qi(1/3))]Qi(ε/2)=},\displaystyle=\left\{\begin{array}[]{c}\tau^{S}(B^{\prime}_{i})<\tau^{S}(Q_{i+1}),\>S(\tau^{S}(Q_{i}(1/3)))\in Q_{i}^{L},\\ S[\tau^{S}(Q_{i}(\varepsilon)),\>\tau^{S}(Q_{i}(1/3))]\cap Q_{i}(\varepsilon/2)=\emptyset\end{array}\right\}, (33)
Hi(3)(l)\displaystyle H_{i}^{(3)}(l) ={#KilM2|x|2,S(ρi)QiR,S[τS(Bi),k]Qi[0,5/9]for allkKi},\displaystyle=\left\{\#K_{i}\geq lM^{2}|x|^{2},\>S(\rho_{i})\in Q_{i}^{R},\>S[\tau^{S}(B^{\prime}_{i}),k]\in Q_{i}[0,5/9]\ \mbox{for all}\ k\in K_{i}\right\},
Hi(4)\displaystyle H_{i}^{(4)} ={S[ρi,τS(Qi+1)]Qi[0,11/18]=},\displaystyle=\left\{S[\rho_{i},\tau^{S}(Q_{i+1})]\cap Q_{i}[0,11/18]=\emptyset\right\},

where #A\#A denotes the cardinality of set AA. Moreover, Hi(l)=Hi(1)Hi(2)Hi(3)(l)Hi(4)H_{i}(l)=H_{i}^{(1)}\cap H_{i}^{(2)}\cap H_{i}^{(3)}(l)\cap H_{i}^{(4)}.

Note that, on the event Hi(l)H_{i}(l), a local cut time kKik\in K_{i} satisfies

S[τS(Qi),k]S[k+1,τS(Qi+1)]=,S[\tau^{S}(Q_{i}),k]\cap S[k+1,\tau^{S}(Q_{i+1})]=\emptyset,

and thus len(λi)#Ki\mathrm{len}(\lambda_{i})\geq\#K_{i} holds.

Lemma 3.12.

Let Hi(l)H_{i}(l) be as defined above. Then there exists constants c>0c>0, ε>0\varepsilon>0, l>0l>0 and R>0R^{\prime}>0 such that

𝐏z(Hi(l)Fi)c,\mathbf{P}^{z}\left(H_{i}(l)\mid F_{i}\right)\geq c, (34)

uniformly in xx and MM with M|x|>RM|x|>R^{\prime}, i{1,2,,NM1}i\in\{1,2,\dots,N_{M}-1\} and zQ~iz\in\widetilde{Q}_{i}.

Proof.

By the strong Markov property, we have that

𝐏z(Hi(l)Fi)\displaystyle\mathbf{P}^{z}\left(H_{i}(l)\mid F_{i}\right)
infz1,z2,z3,z41𝐏z(Fi)j=13𝐏zj(Hi(j){S[τS(zj),τS(zj+1)]Bi[ε,1]})\displaystyle\geq\inf_{z_{1},z_{2},z_{3},z_{4}}\frac{1}{\mathbf{P}^{z}(F_{i})}\prod_{j=1}^{3}\mathbf{P}^{z_{j}}\left(H_{i}^{(j)}\cap\{S[\tau^{S}(z_{j}),\tau^{S}(z_{j+1})]\subseteq B_{i}[-\varepsilon,1]\}\right)
×𝐏z4(Hi(4){S[τS(z3),τS(z4)]Bi[ε,1],S(τS(Qi+1))Q~i+1,S[τS(Qi+1(ε)),τS(Qi+1)]Qi+1(2ε)})\displaystyle\qquad\times\mathbf{P}^{z_{4}}\left(H_{i}^{(4)}\cap\left\{\begin{array}[]{c}S[\tau^{S}(z_{3}),\tau^{S}(z_{4})]\subseteq B_{i}[-\varepsilon,1],\>S(\tau^{S}(Q_{i+1}))\in\widetilde{Q}_{i+1},\\ S[\tau^{S}(Q_{i+1}(-\varepsilon)),\>\tau^{S}(Q_{i+1})]\cap Q_{i+1}(-2\varepsilon)\end{array}\right\}\right) (37)
infz1𝐏z1(Hi(1))infz2𝐏z2(Hi(2){S[τS(z2),τS(z3)]Bi[ε,1]})infz3𝐏z3(Hi(3)(l))\displaystyle\geq\inf_{z_{1}}\mathbf{P}^{z_{1}}(H_{i}^{(1)})\inf_{z_{2}}\mathbf{P}^{z_{2}}(H_{i}^{(2)}\mid\{S[\tau^{S}(z_{2}),\tau^{S}(z_{3})]\subseteq B_{i}[-\varepsilon,1]\})\inf_{z_{3}}\mathbf{P}^{z_{3}}(H_{i}^{(3)}(l))
×infz4𝐏z4(Hi(4)S[τS(z3),τS(z4)]Bi[ε,1],S(τS(Qi+1))Q~i+1,S[τS(Qi+1(ε)),τS(Qi+1)]Qi+1(2ε)),\displaystyle\qquad\times\inf_{z_{4}}\mathbf{P}^{z_{4}}\left(H_{i}^{(4)}\>\vline\>\begin{array}[]{c}S[\tau^{S}(z_{3}),\tau^{S}(z_{4})]\subseteq B_{i}[-\varepsilon,1],\>S(\tau^{S}(Q_{i+1}))\in\widetilde{Q}_{i+1},\\ S[\tau^{S}(Q_{i+1}(-\varepsilon)),\>\tau^{S}(Q_{i+1})]\cap Q_{i+1}(-2\varepsilon)\end{array}\right), (40)

where the infima are taken over z1Q~iz_{1}\in\widetilde{Q}_{i}, z2B~(z1,2ε)z_{2}\in\partial\widetilde{B}(z_{1},2\varepsilon), z3QiLz_{3}\in Q_{i}^{L} and z4QiRz_{4}\in Q_{i}^{R} (see (21) for the definition of B~(y,r)\widetilde{B}(y,r)).

First we estimate the conditional probability of Hi(1)H_{i}^{(1)}. Recall that B(x,r)B(x,r) denotes the Euclidean ball of radius rr with center point xx. We consider the event of SS up to the first exit time of the small box B~(z1,2ε)\widetilde{B}(z_{1},2\varepsilon). Let k1=τS(B(z,ε2M|x|)c)k_{1}=\tau^{S}(B(z,\frac{\varepsilon}{2}M|x|)^{c}) and k2=τS(B~(z1,2ε)c)k_{2}=\tau^{S}(\widetilde{B}(z_{1},2\varepsilon)^{c}). Then

𝐏z1(Hi(1))\displaystyle\mathbf{P}^{z_{1}}(H_{i}^{(1)}) \displaystyle\geq 𝐏z1(S has a nice cut time k[k1,k2], 0<k2τS(Qi)Cε2M2|x|2)\displaystyle\mathbf{P}^{z_{1}}\left(S\mbox{ has a nice cut time }k\in[k_{1},k_{2}],\>0<k_{2}-\tau^{S}(Q_{i})\leq C^{\prime}\varepsilon^{2}M^{2}|x|^{2}\right) (44)
\displaystyle\geq 𝐏z1(S has a nice cut time k[k1,k2],#S[τS(Qi),k1]C1ε2M2|x|2, 0<k2τS(Qi)Cε2M2|x|2)\displaystyle\mathbf{P}^{z_{1}}\left(\begin{array}[]{c}S\mbox{ has a nice cut time }k\in[k_{1},k_{2}],\\ \#S[\tau^{S}(Q_{i}),k_{1}]\geq C^{\prime-1}\varepsilon^{2}M^{2}|x|^{2},\>0<k_{2}-\tau^{S}(Q_{i})\leq C^{\prime}\varepsilon^{2}M^{2}|x|^{2}\end{array}\right)
\displaystyle\geq 𝐏z1(S has a nice cut time k[C1ε2M2|x|2,Cε2M2|x|2])\displaystyle\mathbf{P}^{z_{1}}\left(S\mbox{ has a nice cut time }k\in[C^{\prime-1}\varepsilon^{2}M^{2}|x|^{2},C^{\prime}\varepsilon^{2}M^{2}|x|^{2}]\right)
𝐏z1(#S[τS(Qi),k1]C1ε2M2|x|2)\displaystyle\qquad-\mathbf{P}^{z_{1}}\left(\#S[\tau^{S}(Q_{i}),k_{1}]\geq C^{\prime-1}\varepsilon^{2}M^{2}|x|^{2}\right)
𝐏z1(0<k2τS(Qi)Cε2M2|x|2).\displaystyle\qquad-\mathbf{P}^{z_{1}}\left(0<k_{2}-\tau^{S}(Q_{i})\leq C^{\prime}\varepsilon^{2}M^{2}|x|^{2}\right).

If we take C>1C^{\prime}>1 sufficiently large, then the second and third terms on the right-hand side of (44) are bounded below by some small constant, while it follows from [21, equation (1)] that the first term is bounded below by some universal constant. Thus, we have

𝐏z1(Hi(1))c1\mathbf{P}^{z_{1}}(H_{i}^{(1)})\geq c_{1} (45)

for some constant c1>0c_{1}>0.

Second we consider the conditional probability of Hi(2)H_{i}^{(2)}. By (7), there exists some universal constant C>0C>0 such that

𝐏z2(S[τS(B(z,εM|x|/2)c),τS(Bi)](B(z,ε2M|x|/2)))Cεd2,\mathbf{P}^{z_{2}}\left(S\left[\tau^{S}(B(z,\varepsilon M|x|/2)^{c}),\tau^{S}(B^{\prime}_{i})\right]\cap(B(z,\varepsilon^{2}M|x|/2))\neq\emptyset\right)\leq C\varepsilon^{d-2},

for M|x|εdM|x|\geq\varepsilon^{-d}. It follows from (9) that

c2ε\displaystyle c_{2}\varepsilon 𝐏z2(τS(Bi)<τS(Qi+1),S[τS(z2),τS(z3)]Bi[ε,1])\displaystyle\leq\mathbf{P}^{z_{2}}\left(\tau^{S}(B^{\prime}_{i})<\tau^{S}(Q_{i+1}),\ S[\tau^{S}(z_{2}),\tau^{S}(z_{3})]\subseteq B_{i}[-\varepsilon,1]\right)
𝐏z2(S[τS(z2),τS(z3)]Bi[ε,1])\displaystyle\leq\mathbf{P}^{z_{2}}\left(S[\tau^{S}(z_{2}),\tau^{S}(z_{3})]\subseteq B_{i}[-\varepsilon,1]\right)
c3ε,\displaystyle\leq c_{3}\varepsilon,

uniformly in z2B(z1,εM|x|/2)z_{2}\in B(z_{1},\varepsilon M|x|/2). Thus we have that

𝐏z2(Hi(2)S[τS(z2),τS(z3)]Bi[ε,1])c2εCεd2c3εc,\mathbf{P}^{z_{2}}\left(H_{i}^{(2)}\>\vline\>S[\tau^{S}(z_{2}),\tau^{S}(z_{3})]\subseteq B_{i}[-\varepsilon,1]\right)\geq\frac{c_{2}\varepsilon-C\varepsilon^{d-2}}{c_{3}\varepsilon}\geq c, (46)

for some constant c>0c>0 and sufficiently small ε\varepsilon.

Again by (9), it holds that

𝐏z4\displaystyle\mathbf{P}^{z_{4}} (Hi(4)S[τS(z4),τS(Qi+1)]Bi[ε,1],S(τS(Qi+1))Q~i+1,S[τS(Qi+1(ε)),τS(Qi+1)]Qi+1(2ε))c\displaystyle\left(H_{i}^{(4)}\>\vline\>\begin{array}[]{c}S[\tau^{S}(z_{4}),\tau^{S}(Q_{i+1})]\subseteq B_{i}[-\varepsilon,1],\>S(\tau^{S}(Q_{i+1}))\in\widetilde{Q}_{i+1},\\ S[\tau^{S}(Q_{i+1}(-\varepsilon)),\tau^{S}(Q_{i+1})]\cap Q_{i+1}(-2\varepsilon)\end{array}\right)\geq c (49)

for some constant c>0c>0.

We will next derive a lower bound for 𝐏z3(Hi(3))\mathbf{P}^{z_{3}}(H_{i}^{(3)}) by applying the second moment method. We consider the ball B(y,M|x|/18)\mathcal{B}\coloneqq B\left(y,{M|x|}/{18}\right) and two independent simple random walks R1R^{1} and R2R^{2} with starting point yy. Let wj=Rj(τRj(c))w_{j}=R^{j}(\tau^{R^{j}}(\mathcal{B}^{c})) for j=1,2j=1,2. We define two events of R1R^{1} and R2R^{2} as follows:

I1\displaystyle I_{1} ={R1[1,τR1(c)]R2[1,τR2(c)]=},\displaystyle=\left\{R^{1}[1,\tau^{R^{1}}(\mathcal{B}^{c})]\cap R^{2}[1,\tau^{R^{2}}(\mathcal{B}^{c})]=\emptyset\right\},
I2\displaystyle I_{2} ={dist({w1},R2[1,τR2(c)])dist({w2},R1[1,τR1(c)])M|x|/36}.\displaystyle=\left\{\mathrm{dist}(\{w_{1}\},R^{2}[1,\tau^{R^{2}}(\mathcal{B}^{c})])\vee\mathrm{dist}(\{w_{2}\},R^{1}[1,\tau^{R^{1}}(\mathcal{B}^{c})])\geq M|x|/36\right\}.

Let us denote by PP the joint distribution of R1R^{1} and R2R^{2}. By [7, Lemma 3.2], we have

𝐏(I2I1)c4,\mathbf{P}(I_{2}\mid I_{1})\geq c_{4}, (50)

while it follows from [20, equation (3.2)] that 𝐏(I1)c4\mathbf{P}(I_{1})\geq c_{4} for some constant c4>0c_{4}>0. On I2I_{2}, without loss of generality, we suppose that |w1z3||w2z3||w_{1}-z_{3}|\leq|w_{2}-z_{3}|. Let

J1=\displaystyle J_{1}= {τz3R1<τR1(Bic),dist(R1(k),l1)M|x|/200 for all k[τR1(c),τz3R1]}\displaystyle\left\{\tau^{R^{1}}_{z_{3}}<\tau^{R^{1}}({B^{\prime}_{i}}^{c}),\ \mathrm{dist}(R^{1}(k),l^{1})\leq M|x|/200\mbox{ for all }k\in[\tau^{R^{1}}(\mathcal{B}^{c}),\tau^{R^{1}}_{z_{3}}]\right\}
J2=\displaystyle J_{2}= {R2(τR2(Bic))QiR,dist(R2(k),l2)M|x|/200 for all k[τR2(c),τR2(Bic)]},\displaystyle\left\{R^{2}(\tau^{R_{2}}({B_{i}}^{c}))\in Q_{i}^{R},\ \mathrm{dist}(R^{2}(k),l^{2})\leq M|x|/200\mbox{ for all }k\in[\tau^{R^{2}}(\mathcal{B}^{c}),\tau^{R^{2}}({B^{\prime}_{i}}^{c})]\right\},

where l1l^{1} (respectively l2l^{2}) is the line segment between the points w1w_{1} and z3z_{3} (or between w2w_{2} and R2(τR2(Bic))R^{2}(\tau^{R^{2}}({B^{\prime}_{i}}^{c})), respectively). (See Figure 4 for a depiction of I1I2J1J2I_{1}\cap I_{2}\cap J_{1}\cap J_{2}.)

Refer to caption
Figure 4: Illustration of the event I1I2J1J2I_{1}\cap I_{2}\cap J_{1}\cap J_{2}.

Since dist(w2,Bic)\mathrm{dist}(w_{2},{B^{\prime}_{i}}^{c}) is comparable to M|x|M|x|, we have that

𝐏(J2)c,\mathbf{P}(J_{2})\geq c^{\prime},

for some constant c>0c^{\prime}>0. Moreover, by the strong Markov property and (8),

𝐏(J1)\displaystyle\mathbf{P}(J_{1}) 𝐏(dist(R1(k),l1)M|x|/200 for all k[τR1(c),τR1(B(z3,M|x|/400))])\displaystyle\geq\mathbf{P}\left(\mathrm{dist}(R^{1}(k),l^{1})\leq M|x|/200\mbox{ for all }k\in[\tau^{R^{1}}(\mathcal{B}^{c}),\tau^{R^{1}}(B(z_{3},M|x|/400))]\right)
×𝐏R1(τR1(B(z3,M|x|/400)))(τz3R1<τR1(Bic))×𝐏z3(R1(τR1(B(z3,M|x|/200)))Bi)\displaystyle\>\>\>\>\>\>\times\mathbf{P}^{R^{1}(\tau^{R^{1}}(B(z_{3},M|x|/400)))}\left(\tau^{R^{1}}_{z_{3}}<\tau^{R^{1}}({B^{\prime}_{i}}^{c})\right)\times\mathbf{P}^{z_{3}}\left(R^{1}(\tau^{R^{1}}(B(z_{3},M|x|/200)))\not\in B^{\prime}_{i}\right)
c|w1z3|2d,\displaystyle\geq c|w_{1}-z_{3}|^{2-d}, (51)

for some constant c>0c>0. By the strong Markov property, we bound from below the expectation of #Ki\#K_{i} on the event A{S(ρi)QiR,S[τS(Bi),k]Qi[0,5/9]for allkKi}A\coloneqq\{S(\rho_{i})\in Q_{i}^{R},S[\tau^{S}(B^{\prime}_{i}),k]\in Q_{i}[0,5/9]\ \mbox{for all}\ k\in K_{i}\} by

𝐄z3(#Ki𝟏A)\displaystyle\mathbf{E}^{z_{3}}\left(\#K_{i}\mathbf{1}_{A}\right) yBi′′𝐏(I1I2J1J2)\displaystyle\geq\sum_{y\in B^{\prime\prime}_{i}}\mathbf{P}\left(I_{1}\cap I_{2}\cap J_{1}\cap J_{2}\right)
=yBi′′𝐏(I1)𝐏(I2I1)𝐏(J1)𝐏(J2)\displaystyle=\sum_{y\in B^{\prime\prime}_{i}}\mathbf{P}(I_{1})\mathbf{P}(I_{2}\mid I_{1})\mathbf{P}(J_{1})\mathbf{P}(J_{2})
yBi′′c42c|yz3|c\displaystyle\geq\sum_{y\in B^{\prime\prime}_{i}}c_{4}^{2}\cdot c|y-z_{3}|\cdot c^{\prime}
cM2|x|2.\displaystyle\geq cM^{2}|x|^{2}.\hskip 240.0pt (52)

On the other hand, the first and second moment of KiK_{i} is bounded above as follows. Since |yz3|19M|x||y-z_{3}|\geq\frac{1}{9}M|x| for yBi′′y\in B^{\prime\prime}_{i} and zQiLz\in Q_{i}^{L},

𝐄z3(#Ki)\displaystyle\mathbf{E}^{z_{3}}(\#K_{i}) 𝐄z3(yBi′′𝟏(τyS<))yBi′′𝐏z3({τyS<})yBi′′C|yz3|2dCM2|x|2,\displaystyle\leq\mathbf{E}^{z_{3}}\left(\sum_{y\in B^{\prime\prime}_{i}}\mathbf{1}(\tau^{S}_{y}<\infty)\right)\leq\sum_{y\in B^{\prime\prime}_{i}}\mathbf{P}^{z_{3}}\left(\{\tau^{S}_{y}<\infty\}\right)\leq\sum_{y\in B^{\prime\prime}_{i}}C|y-z_{3}|^{2-d}\leq CM^{2}|x|^{2}, (53)
𝐄z3((#Ki)2)\displaystyle\mathbf{E}^{z_{3}}((\#K_{i})^{2}) 𝐄z3((yBi′′𝟏(τyS<))2)\displaystyle\leq\mathbf{E}^{z_{3}}\left(\left(\sum_{y\in B^{\prime\prime}_{i}}\mathbf{1}(\tau^{S}_{y}<\infty)\right)^{2}\right)
CM2|x|2+yBi′′yBi′′(𝐏z3(τyS<τyS<)+𝐏z3(τyS<τyS<))\displaystyle\leq CM^{2}|x|^{2}+\sum_{y\in B^{\prime\prime}_{i}}\sum_{y^{\prime}\in B^{\prime\prime}_{i}}\left(\mathbf{P}^{z_{3}}(\tau^{S}_{y}<\tau^{S}_{y^{\prime}}<\infty)+\mathbf{P}^{z_{3}}(\tau^{S}_{y^{\prime}}<\tau^{S}_{y}<\infty)\right)
CM2|x|2+yBi′′yBi′′(𝐏z3(τyS<)+𝐏z3(τyS<))𝐏y(τyS<)\displaystyle\leq CM^{2}|x|^{2}+\sum_{y\in B^{\prime\prime}_{i}}\sum_{y\in B^{\prime\prime}_{i}}\left(\mathbf{P}^{z_{3}}(\tau^{S}_{y}<\infty)+\mathbf{P}^{z_{3}}(\tau^{S}_{y^{\prime}}<\infty)\right)\mathbf{P}^{y}(\tau^{S}_{y^{\prime}}<\infty)
CM2|x|2+yBi′′k=1118M|x|yBi′′|yy|=k(|yz3|2d+|yz3|2d)k2d\displaystyle\leq CM^{2}|x|^{2}+\sum_{y\in B^{\prime\prime}_{i}}\sum_{k=1}^{\frac{1}{18}M|x|}\sum_{\begin{subarray}{c}y\in B^{\prime\prime}_{i}\\ |y-y^{\prime}|=k\end{subarray}}\left(|y-z_{3}|^{2-d}+|y^{\prime}-z_{3}|^{2-d}\right)k^{2-d}
+yBi′′k118M|x|+1yBi′′|yy|=k(|yz3|2d+|yz3|2d)(M|x|/9)2d\displaystyle\qquad+\sum_{y\in B^{\prime\prime}_{i}}\sum_{k\geq\frac{1}{18}M|x|+1}\sum_{\begin{subarray}{c}y\in B^{\prime\prime}_{i}\\ |y-y^{\prime}|=k\end{subarray}}\left(|y-z_{3}|^{2-d}+|y^{\prime}-z_{3}|^{2-d}\right)(M|x|/9)^{2-d}
CM4|x|4,\displaystyle\leq C^{\prime}M^{4}|x|^{4}, (54)

where CC and CC^{\prime} depend only on dd. Now, for 0θ10\leq\theta\leq 1, we have that

𝐄z3\displaystyle\mathbf{E}^{z_{3}} (#Ki𝟏A)θ𝐄z3(#Ki𝟏A)+𝐄z3(#Ki𝟏A𝟏(#Ki𝟏A>θ𝐄z3(#Ki𝟏A))).\displaystyle\left(\#K_{i}\mathbf{1}_{A}\right)\leq\theta\mathbf{E}^{z_{3}}\left(\#K_{i}\mathbf{1}_{A}\right)+\mathbf{E}^{z_{3}}\left(\#K_{i}\mathbf{1}_{A}\mathbf{1}\left(\#K_{i}\mathbf{1}_{A}>\theta\mathbf{E}^{z_{3}}(\#K_{i}\mathbf{1}_{A})\right)\right).

From this, since #Ki0\#K_{i}\geq 0, the Cauchy-Schwarz inequality yields

𝐏z3({#Ki>θ𝐄z3(#Ki𝟏A)}A)\displaystyle\mathbf{P}^{z_{3}}(\{\#K_{i}>\theta\mathbf{E}^{z_{3}}(\#K_{i}\mathbf{1}_{A})\}\cap A) 𝐏z3({#Ki𝟏A>θ𝐄z3(#Ki𝟏A)}A)\displaystyle\geq\mathbf{P}^{z_{3}}(\{\#K_{i}\mathbf{1}_{A}>\theta\mathbf{E}^{z_{3}}(\#K_{i}\mathbf{1}_{A})\}\cap A)
(1θ)2𝐄z3(#Ki𝟏A)2𝐄z3((#Ki)2).\displaystyle\geq(1-\theta)^{2}\frac{\mathbf{E}^{z_{3}}(\#K_{i}\mathbf{1}_{A})^{2}}{\mathbf{E}^{z_{3}}((\#K_{i})^{2})}.

By substituting (52), (53) and (54) into the above estimate, we obtain that

𝐏z3(Hi(3)(l))𝐏z3(#KilC𝐄z3(#Ki))(1lC)2𝐄z3(#Ki𝟏A)2𝐄z3((#Ki)2)c.\mathbf{P}^{z_{3}}(H_{i}^{(3)}(l))\geq\mathbf{P}^{z_{3}}\left(\#K_{i}\geq\frac{l}{C}\mathbf{E}^{z_{3}}(\#K_{i})\right)\geq\left(1-\frac{l}{C}\right)^{2}\frac{\mathbf{E}^{z_{3}}(\#K_{i}\mathbf{1}_{A})^{2}}{\mathbf{E}^{z_{3}}((\#K_{i})^{2})}\geq c. (55)

Finally, substituting (45), (46), (49) and (55) into (40) gives (34). ∎

We are now ready to prove Proposition 3.6. Recall that FiF_{i}, GiG_{i} and HiH_{i} were defined in Definitions 3.7, 3.8 and 3.11, respectively. Let

UNM={S[τS(Q2NM+1),]B(0,NMM|x|)=},U_{N_{M}}=\left\{S[\tau^{S}(Q_{2N_{M}+1}),\infty]\cap B(0,N_{M}\cdot M|x|)=\emptyset\right\},

and

Θ=Θ(C,l)=(i=02NMFi)(i=0NMGi(C))(i=1NM1Hi(l))UNM.\Theta=\Theta(C,l)=\left(\bigcap_{i=0}^{2N_{M}}F_{i}\right)\cap\left(\bigcap_{i=0}^{N_{M}}G_{i}(C)\right)\cap\left(\bigcap_{i=1}^{N_{M}-1}H_{i}(l)\right)\cap U_{N_{M}}.
Proof of Proposition 3.6.

We will first demonstrate that the bound τx[R1M|x|2,RM|x|]\tau_{x}\in[R^{-1}M|x|^{2},RM|x|], as appears in the probability on the left-hand side of (19), holds on Θ(C,l)\Theta(C,l). Suppose that Θ(C,l)\Theta(C,l) occurs. By the definition of FiF_{i}, i=0,1,,2NMi=0,1,\dots,2N_{M},

LE(S[0,τxS])S[τxS+1,]=\mathrm{LE}(S[0,\tau^{S}_{x}])\cap S[\tau^{S}_{x}+1,\infty]=\emptyset

holds. Thus xLx\in L and τx=len(S[0,τxS])\tau_{x}=\mathrm{len}(S[0,\tau^{S}_{x}]). Let kik_{i} be a nice cut time of SS in BiB_{i} (see Definition 3.10), and recall that ξi\xi_{i} and ξi\xi^{\prime}_{i} are as defined in (23) and (24), respectively. Let

si\displaystyle s_{i} =inf{n0:ξi1(n)S[τS(Qi),τS(Qi+1)]},\displaystyle=\inf\left\{n\geq 0\mathrel{:}\xi_{i-1}(n)\in S[\tau^{S}(Q_{i}),\tau^{S}(Q_{i+1})]\right\},
ti\displaystyle t_{i} =sup{n[τS(Qi),τS(Qi+1)]:S(n)=ξi1(si)}.\displaystyle=\sup\left\{n\in[\tau^{S}(Q_{i}),\tau^{S}(Q_{i+1})]\mathrel{:}S(n)=\xi_{i-1}(s_{i})\right\}.

Then we have that

λi=LE(S[τS(Qi),ki])LE(S[ki,τS(Qi+1)]),\lambda_{i}=\mathrm{LE}\left(S[\tau^{S}(Q_{i}),k_{i}]\right)\oplus\mathrm{LE}\left(S[k_{i},\tau^{S}(Q_{i+1})]\right), (56)

and also

ξi=ξi1[0,si]LE(S[ti,ki])LE(S[ki,τS(Qi+1)])ξi1S[ti,ki]λi,\xi_{i}=\xi_{i-1}[0,s_{i}]\oplus\mathrm{LE}\left(S[t_{i},k_{i}]\right)\oplus\mathrm{LE}\left(S[k_{i},\tau^{S}(Q_{i+1})]\right)\subseteq\xi_{i-1}\cup S[t_{i},k_{i}]\cup\lambda_{i},

for i=1,2,,NM1i=1,2,\dots,N_{M}-1, where we have applied (56) for the inclusion. Furthermore,

ξNM=ξNM1[0,sNM]LE(S[tNM,τxS])ξNM1λNM.\xi_{N_{M}}=\xi_{N_{M}-1}[0,s_{N_{M}}]\oplus\mathrm{LE}(S[t_{N_{M}},\tau^{S}_{x}])\subseteq\xi_{N_{M}-1}\cup\lambda_{N_{M}}.

Thus, by induction, it follows that

i=1NM1LE(S[ki,τS(Qi+1)])ξNMξ0i=1NM1(S[τS(Qi),τS(Qi(ε))]λi)λNM.\bigcup_{i=1}^{N_{M}-1}\mathrm{LE}\left(S[k_{i},\tau^{S}(Q_{i+1})]\right)\subseteq\xi_{N_{M}}\subseteq\xi_{0}\cup\bigcup_{i=1}^{N_{M}-1}\left(S[\tau^{S}(Q_{i}),\tau^{S}(Q_{i}(\varepsilon))]\cup\lambda_{i}\right)\cup\lambda_{N_{M}}.

Note that, on Hi(l)H_{i}(l), kKik^{\prime}\in K_{i} is a cut time of the path S[ki,τS(Qi+1)]S[k_{i},\tau^{S}(Q_{i+1})], and thus S(k)LE(S[ki,τS(Qi+1)])S(k^{\prime})\in LE(S[k_{i},\tau^{S}(Q_{i+1})]). By the definition of Gi(C)G_{i}(C) and Hi(l)H_{i}(l) (see Definitions 3.8 and 3.11, respectively), we have that

len(ξNM)\displaystyle\mathrm{len}(\xi_{N_{M}}) i=1NM1#Ki12lM|x|2,\displaystyle\geq\sum_{i=1}^{N_{M}-1}\#K_{i}\geq\frac{1}{2}lM|x|^{2},
len(ξNM)\displaystyle\mathrm{len}(\xi_{N_{M}}) len(ξ0)+i=1NM1(#S[τS(Qi),τS(Qi(ε))]+len(λi))+len(λNM)\displaystyle\leq\mathrm{len}(\xi_{0})+\sum_{i=1}^{N_{M}-1}\left(\#S[\tau^{S}(Q_{i}),\tau^{S}(Q_{i}(\varepsilon))]+\mathrm{len}(\lambda_{i})\right)+\mathrm{len}(\lambda_{N_{M}})
3(C+ε2)M|x|2,\displaystyle\leq 3(C+\varepsilon^{2})M|x|^{2},

on Θ(C,l)\Theta(C,l). Hence choosing RR suitably large gives the desired bound on Θ(C,l)\Theta(C,l).

Consequently, to complete the proof, it will be enough to show that 𝐏(Θ)\mathbf{P}(\Theta) is bounded below by the right-hand side of (19). By Lemma 3.12, there exist constants c5,ε,l>0c_{5},\varepsilon,l>0 such that infzQ~i𝐏z(Hi(l)Fi)c5\inf_{z\in\widetilde{Q}_{i}}\mathbf{P}^{z}(H_{i}(l)\mid F_{i})\geq c_{5} for i=1,2,,NMi=1,2,\dots,N_{M}. Moreover, by Lemma 3.9, there exists a constant C>0C>0 such that infzQ~i𝐏z(Gi(C)Fi)1c5/2\inf_{z\in\widetilde{Q}_{i}}\mathbf{P}^{z}(G_{i}(C)\mid F_{i})\geq 1-c_{5}/2 for i=0,1,,NMi=0,1,\dots,N_{M}. Thus we have

infzQ~i𝐏z(Gi(C)Hi(l)Fi)c52,i{1,2,,NM1},\inf_{z\in\widetilde{Q}_{i}}\mathbf{P}^{z}(G_{i}(C)\cap H_{i}(l)\mid F_{i})\geq\frac{c_{5}}{2},\qquad i\in\{1,2,\dots,N_{M}-1\}, (57)
infzQ~i𝐏z(Gi(C)Fi)1c52,i=0,NM.\inf_{z\in\widetilde{Q}_{i}}\mathbf{P}^{z}(G_{i}(C)\mid F_{i})\geq 1-\frac{c_{5}}{2},\qquad i=0,N_{M}. (58)

As already noted in the proof of Lemma 3.9, we also have that

infzQ~i𝐏z(Fi)c6ε,\inf_{z\in\widetilde{Q}_{i}}\mathbf{P}^{z}(F_{i})\geq c_{6}\varepsilon, (59)

for all i=1,2,,NM1i=1,2,\dots,N_{M}-1, and a similar bound holds for 𝐏(F0)\mathbf{P}(F_{0}). And, repeating a similar argument to the lower bound for 𝐏z3(Hi(3))\mathbf{P}^{z_{3}}(H_{i}^{(3)}) from the proof of Lemma 3.12, from (50) to (51) we have that

infzQ~NM𝐏z(FNM)c6M2d|x|2d,\inf_{z\in\widetilde{Q}_{N_{M}}}\mathbf{P}^{z}(F_{N_{M}})\geq c_{6}M^{2-d}|x|^{2-d},

where c6>0c_{6}>0 is adjusted if necessary. By combining these estimates on 𝐏(Fi)\mathbf{P}(F_{i}) with (57) and (58), we obtain that

infzQ~i𝐏z(FiGi(C)Hi(l))c5c6ε2,i{1,2,,NM1},\inf_{z\in\widetilde{Q}_{i}}\mathbf{P}^{z}(F_{i}\cap G_{i}(C)\cap H_{i}(l))\geq\frac{c_{5}c_{6}\varepsilon}{2},\qquad i\in\{1,2,\dots,N_{M}-1\},
𝐏(F0G0(C))(1c52)c6,\mathbf{P}(F_{0}\cap G_{0}(C))\geq\left(1-\frac{c_{5}}{2}\right)c_{6},
infzQ~NM𝐏z(FNMGNM(C))(1c52)c6M2d|x|2d.\inf_{z\in\widetilde{Q}_{N_{M}}}\mathbf{P}^{z}(F_{N_{M}}\cap G_{N_{M}}(C))\geq\left(1-\frac{c_{5}}{2}\right)c_{6}M^{2-d}|x|^{2-d}.

Furthermore, similarly to the case with i=1,2,,NM1i=1,2,\dots,N_{M}-1, it holds that

infzQ~i𝐏z(Fi)c6ε,i{NM+1,NM+2,,2NM},\inf_{z\in\widetilde{Q}_{i}}\mathbf{P}^{z}(F_{i})\geq c_{6}\varepsilon,\qquad i\in\{N_{M}+1,N_{M}+2,\dots,2N_{M}\},

and it follows from (7) that

infzQ~2NM+1𝐏z(UNM)c\inf_{z\in\widetilde{Q}_{2N_{M}+1}}\mathbf{P}^{z}(U_{N_{M}})\geq c

for some constant c>0c>0. Finally, by the strong Markov property, we have that

𝐏(Θ(C,l))\displaystyle\mathbf{P}(\Theta(C,l))
𝐏(F0G0(C))i=1NM1infzQ~i𝐏z(FiGi(C)Hi(l))\displaystyle\geq\mathbf{P}(F_{0}\cap G_{0}(C))\prod_{i=1}^{N_{M}-1}\inf_{z\in\widetilde{Q}_{i}}\mathbf{P}^{z}\left(F_{i}\cap G_{i}(C)\cap H_{i}(l)\right)
×infzQ~NM𝐏z(FNMGNM(C))×i=NM+12NMinfzQ~i𝐏z(Fi)×infzQ~2NM+1𝐏z(UNM)\displaystyle\qquad\qquad\times\inf_{z\in\widetilde{Q}_{N_{M}}}\mathbf{P}^{z}(F_{N_{M}}\cap G_{N_{M}}(C))\times\prod_{i=N_{M}+1}^{2N_{M}}\inf_{z\in\widetilde{Q}_{i}}\mathbf{P}^{z}(F_{i})\times\inf_{z\in\widetilde{Q}_{2N_{M}+1}}\mathbf{P}^{z}(U_{N_{M}})
CM2d|x|2decM\displaystyle\geq CM^{2-d}|x|^{2-d}e^{-\frac{c}{M}}
CM1d/2|x|2decM,\displaystyle\geq CM^{1-d/2}|x|^{2-d}e^{-\frac{c}{M}},

where the third inequality holds simply because M1M\leq 1. ∎

3.4 Lower bound for M1M\geq 1

We now turn to the proof of the lower bound of Proposition 3.1 with M1M\geq 1. In particular, we will establish the following.

Proposition 3.13.

There exist constants c,c,R(0,)c,c^{\prime},R\in(0,\infty) such that for every xd\{0}x\in\mathbb{Z}^{d}\backslash\{0\} and M1M\geq 1,

𝐏(τx[R1M|x|2,RM|x|2])c(M|x|2)1d/2exp(cM).\mathbf{P}\left(\tau_{x}\in[R^{-1}M|x|^{2},RM|x|^{2}]\right)\geq c^{\prime}(M|x|^{2})^{1-d/2}\exp\left(-\frac{c}{M}\right).

As in the previous section, the basic strategy involves the construction of a set of particular realisations of LL that we can show occurs with suitably high probability. To do this, we will use a certain reversibility property of simple random walk, as is set out in the next lemma. In the statement of this, for a finite path λ=[λ(0),λ(1),,λ(m)]\lambda=[\lambda(0),\lambda(1),\cdots,\lambda(m)], we write λR=[λ(m),λ(m1),,λ(0)]\lambda^{R}=[\lambda(m),\lambda(m-1),\cdots,\lambda(0)] for its time reversal.

Lemma 3.14.

Let x,zdx,z\in\mathbb{Z}^{d}, xzx\neq z, and write SxS^{x}, SzS^{z} for independent simple random walks in d\mathbb{Z}^{d} started at xx, zz, respectively. Moreover, write τxz:=inf{j:Sjz=x}\tau_{x}^{z}:=\inf\{j:\>S^{z}_{j}=x\}, τzx:=inf{j:Sjx=z}\tau_{z}^{x}:=\inf\{j:\>S^{x}_{j}=z\},

σ1=sup{jτzx:Sjx=x},u=inf{jτzx:Sjx=x},σ2=sup{j<u:Sjx=z}.\sigma_{1}=\sup\{j\leq\tau^{x}_{z}\>:\>S^{x}_{j}=x\},\quad u=\inf\{j\geq\tau_{z}^{x}\>:\>S^{x}_{j}=x\},\quad\sigma_{2}=\sup\{j<u\>:\>S^{x}_{j}=z\}.

It then holds that

{λ:𝐏((Sz[0,τxz])R=λτxz<)>0}={λ:𝐏(Sx[σ1,σ2]=λτxz<)>0},\left\{\lambda\>:\>\mathbf{P}\left(\left(S^{z}[0,\tau^{z}_{x}]\right)^{R}=\lambda\>\vline\>\tau^{z}_{x}<\infty\right)>0\right\}=\left\{\lambda\>:\>\mathbf{P}\left(S^{x}[\sigma_{1},\sigma_{2}]=\lambda\>\vline\>\tau^{z}_{x}<\infty\right)>0\right\}, (60)

and, denoting the set above Λ\Lambda,

𝐏((Sz[0,τxz])R=λτxz<)=𝐏(Sx[σ1,σ2]=λτzx<),λΛ.\mathbf{P}\left(\left(S^{z}[0,\tau^{z}_{x}]\right)^{R}=\lambda\>\vline\>\tau^{z}_{x}<\infty\right)=\mathbf{P}\left(S^{x}[\sigma_{1},\sigma_{2}]=\lambda\>\vline\>\tau^{x}_{z}<\infty\right),\qquad\forall\lambda\in\Lambda. (61)
Proof.

Since (60) is easy to see, we only check (61). Take λ=[λ(0),λ(1),,λ(m)]Λ\lambda=[\lambda(0),\lambda(1),\cdots,\lambda(m)]\in\Lambda. Note that 𝐏(τxz<)=𝐏(τzx<)\mathbf{P}(\tau^{z}_{x}<\infty)=\mathbf{P}(\tau^{x}_{z}<\infty) by symmetry. It follows that

𝐏((Sz[0,τxz])R=λ,τxz<)=𝐏(Sz[0,τxz]=λR,τxz<)=𝐏(Sz[0,τxz]=λR)=(2d)m.\mathbf{P}\left(\left(S^{z}[0,\tau^{z}_{x}]\right)^{R}=\lambda,\ \tau^{z}_{x}<\infty\right)=\mathbf{P}\left(S^{z}[0,\tau^{z}_{x}]=\lambda^{R},\ \tau^{z}_{x}<\infty\right)=\mathbf{P}\left(S^{z}[0,\tau^{z}_{x}]=\lambda^{R}\right)=(2d)^{-m}.

On the other hand, we have

𝐏(Sx[σ1,σ2]=λ,τzx<)\displaystyle\mathbf{P}\left(S^{x}[\sigma_{1},\sigma_{2}]=\lambda,\ \tau^{x}_{z}<\infty\right) =𝐏(Sx[σ1,σ2]=λ)\displaystyle=\mathbf{P}\left(S^{x}[\sigma_{1},\sigma_{2}]=\lambda\right)
=k0𝐏(Sx[σ1,σ2]=λ,σ1=k)\displaystyle=\sum_{k\geq 0}\mathbf{P}\left(S^{x}[\sigma_{1},\sigma_{2}]=\lambda,\ \sigma_{1}=k\right)
=k0𝐏(zSx[0,k],Skx=x,Sx[k,k+m]=λ,σ2=k+m)\displaystyle=\sum_{k\geq 0}\mathbf{P}\left(z\notin S^{x}[0,k],\ S^{x}_{k}=x,\ S^{x}[k,k+m]=\lambda,\ \sigma_{2}=k+m\right)
=k0𝐏(zSx[0,k],Skx=x,Sx[k,k+m]=λ,Sk+mx=z,F),\displaystyle=\sum_{k\geq 0}\mathbf{P}\left(z\notin S^{x}[0,k],\ S^{x}_{k}=x,\ S^{x}[k,k+m]=\lambda,\ S^{x}_{k+m}=z,\ F\right),

where F:={zSx[k+m+1,u)}F:=\{z\notin S^{x}[k+m+1,u^{\prime})\} with u=inf{jk+m:Sjx=x}u^{\prime}=\inf\{j\geq k+m:\>S^{x}_{j}=x\}. Therefore, the Markov property ensures that

𝐏(zSx[0,k],Skx=x,Sx[k,k+m]=λ,Sk+mx=z,F)\displaystyle\mathbf{P}\left(z\notin S^{x}[0,k],\ S^{x}_{k}=x,\ S^{x}[k,k+m]=\lambda,\ S^{x}_{k+m}=z,\ F\right)
=𝐏(zSx[0,k],Skx=x)𝐏(Sx[0,m]=λ)𝐏(F)\displaystyle=\mathbf{P}\left(z\notin S^{x}[0,k],\ S^{x}_{k}=x\right)\mathbf{P}\left(S^{x}[0,m]=\lambda\right)\mathbf{P}(F^{\prime})
=(2d)m𝐏(zSx[0,k],Skx=x)𝐏(F),\displaystyle=(2d)^{-m}\mathbf{P}\left(z\notin S^{x}[0,k],\ S^{x}_{k}=x\right)\mathbf{P}(F^{\prime}),

where F:={zSz[1,τxz]}F^{\prime}:=\{z\not\in S^{z}[1,\tau^{z}_{x}]\}. Writing

ξx=inf{j1:Sjx=x} and p=𝐏(ξx<,zSx[0,ξx]),\xi_{x}=\inf\{j\geq 1:\>S^{x}_{j}=x\}\qquad\text{ and }\qquad p=\mathbf{P}\left(\xi_{x}<\infty,\ z\notin S^{x}[0,\xi_{x}]\right),

we note that

k0𝐏(zSx[0,k],Skx=x)=11p.\sum_{k\geq 0}\mathbf{P}\left(z\notin S^{x}[0,k],\ S^{x}_{k}=x\right)=\frac{1}{1-p}.

Moreover, by symmetry again, it holds that 𝐏(F)=1p\mathbf{P}(F^{\prime})=1-p. Hence we conclude that

𝐏(Sx[σ1,σ2]=λ,τzx<)=(2d)m,\mathbf{P}\left(S^{x}[\sigma_{1},\sigma_{2}]=\lambda,\ \tau^{x}_{z}<\infty\right)=(2d)^{-m},

which gives (61). ∎

In order to explain our application of the previous result, we need to introduce some notation. First, let xd\{0}x\in\mathbb{Z}^{d}\backslash\{0\} and M1M\geq 1. Moreover, set J=CM|x|2J=C\sqrt{M|x|^{2}} for some C1C\geq 1 that will be determined later, and, for ii\in\mathbb{Z}, write b^i=(2iJ,0,,0)d\widehat{b}_{i}=(2iJ,0,\dots,0)\in\mathbb{R}^{d} and

B^i=B(b^i,J),\widehat{B}_{i}=B_{\infty}\left(\widehat{b}_{i},J\right),

which represent adjacent cubes of side length 2J2J. We also introduce the following smaller cubes centred at b=(94J,J2,0,,0)db^{\prime}=(\frac{9}{4}J,\frac{J}{2},0,\dots,0)\in\mathbb{R}^{d},

B^=B(b,J/6),B^′′=B(b,J/18)\widehat{B}^{\prime}=B_{\infty}(b^{\prime},{J}/{6}),\qquad\widehat{B}^{\prime\prime}=B_{\infty}(b^{\prime},J/18)

Note that B^′′B^B^1\widehat{B}^{\prime\prime}\subset\widehat{B}^{\prime}\subset\widehat{B}_{1}. See Figure 5 for a sketch showing the cubes B^1\widehat{B}_{-1}, B^0\widehat{B}_{0}, B^1\widehat{B}_{1} and B^\widehat{B}^{\prime}, as well as some of the other objects that we now define. In particular, we introduce a collection of surfaces:

Q={52J}×[J,J]d1,Q^{*}=\left\{\frac{5}{2}J\right\}\times[-J,J]^{d-1},
Q={3J}×[J,J]d1,Q_{*}=\{3J\}\times[-J,J]^{d-1},
Q={3J}×[J4,J4]×[J,J]d2Q,Q=\{3J\}\times\left[-\frac{J}{4},\frac{J}{4}\right]\times[-J,J]^{d-2}\subset Q_{*},
Q~={J}×[J,J]d1,Q={(y1J/16,y2,,yd):(y1,y2,,yd)Q~},\widetilde{Q}=\left\{J\right\}\times[-J,J]^{d-1},\qquad Q^{\prime}=\left\{(y^{1}-J/16,y^{2},\cdots,y^{d})\mathrel{:}(y^{1},y^{2},\cdots,y^{d})\in\widetilde{Q}\right\},
Q~±={J}×[±J2J8,±J2+J8]×[J,J]d2Q~,\widetilde{Q}_{\pm}=\left\{J\right\}\times\left[\pm\frac{J}{2}-\frac{J}{8},\pm\frac{J}{2}+\frac{J}{8}\right]\times[-J,J]^{d-2}\subset\widetilde{Q},
Q1=B^1{3J}×d1;Q_{-1}=\widehat{B}_{-1}\cap\{-3J\}\times\mathbb{R}^{d-1};

the hyperplane 85J/36(1)\mathbb{H}^{(1)}_{85J/36}, where for aa\in\mathbb{R} and i{1,,d}i\in\{1,\dots,d\}, we denote

a(i)={(x1,,xd)d:xi=a}\mathbb{H}^{(i)}_{a}=\left\{(x_{1},\cdots,x_{d})\in\mathbb{R}^{d}\mathrel{:}x_{i}=a\right\}

(see Figure 7 below for the location of 85J/36(1)\mathbb{H}^{(1)}_{85J/36} in particular); and also the following regions:

D±=[J,4916J]×[J,J]d1[1516J,52J]×[J2J2,J2J2]×[J,J]d2,D_{\pm}=\left[-J,\frac{49}{16}J\right]\times[-J,J]^{d-1}\setminus\left[\frac{15}{16}J,\frac{5}{2}J\right]\times\left[-\frac{J}{2}\mp\frac{J}{2},\frac{J}{2}\mp\frac{J}{2}\right]\times[-J,J]^{d-2},
D~±=D±[1516J,)×d1.\widetilde{D}_{\pm}=D_{\pm}\cap\left[\frac{15}{16}J,\infty\right)\times\mathbb{R}^{d-1}.

We highlight that D+D_{+} is shown as the shaded region on Figure 5.

Refer to caption
Figure 5: Cubes and other regions appearing in the proof of Proposition 3.13.

Roughly speaking, to establish the main result of this section, we will show that, with high enough probability, the loop-erased random walk LL passes from 0 to (somewhere close to) QQ through D+D_{+}, spending a suitable time in B^′′\widehat{B}^{\prime\prime} on the way, before returning to xx through DD_{-}, and then escapes to \infty via Q1Q_{-1}. To make this precise, we will consider an event based on the simple random walk started from 0; see Figure 6. Controlling the probability of this will involve an appeal to Lemma 3.14, through which we obtain a bound that depends on three independent random walks, one started from 0 and two started from xx (see Lemma 3.15 below).

Refer to caption
Figure 6: A sketch of a realisation of S0S^{0} yielding τxM|x|2\tau_{x}\geq M|x|^{2}.

Concerning notation, as in the statement of Lemma 3.14, for each zdz\in\mathbb{Z}^{d}, we will write SzS^{z} for a simple random walk started from zz. We assume that the elements of the collection (Sz)zd(S^{z})_{z\in\mathbb{Z}^{d}} are independent. We moreover write (S~z)zd(\widetilde{S}^{z})_{z\in\mathbb{Z}^{d}} for an independent copy of (Sz)zd(S^{z})_{z\in\mathbb{Z}^{d}}. We also set

τAz:=inf{k0:SkzA},σAz=sup{k0:SkzA},\tau^{z}_{A}:=\inf\{k\geq 0\mathrel{:}S^{z}_{k}\in A\},\qquad\sigma^{z}_{A}=\sup\{k\geq 0\mathrel{:}S^{z}_{k}\in A\},

τxz=τ{x}z\tau^{z}_{x}=\tau^{z}_{\{x\}} and σxz=σ{x}z\sigma^{z}_{x}=\sigma^{z}_{\{x\}}. A particularly important point in the argument that follows is given by

ρ=SτQ00,\rho=S^{0}_{\tau^{0}_{Q}},

i.e. the location where S0S^{0} hits QQ, which is defined when τQ0<\tau_{Q}^{0}<\infty. Additionally, we introduce

τ~=inf{k0:S~kxd(B^0B^1)},\widetilde{\tau}=\inf\left\{k\geq 0\mathrel{:}\widetilde{S}^{x}_{k}\in\mathbb{R}^{d}\setminus(\widehat{B}_{0}\cup\widehat{B}_{-1})\right\},

and, to describe a collection of local cut points for a path λ\lambda,

Γ(λ[i,j])={λ(k):λ[i,k]λ[k+1,j]=}.\Gamma(\lambda[i,j])=\left\{\lambda(k)\mathrel{:}\lambda[i,k]\cap\lambda[k+1,j]=\emptyset\right\}.

The following result gives the key decomposition of the simple random walk underlying LL that we will consider later in the section. It already takes into account the time-reversal property of Lemma 3.14. We will break the complicated event that appears in the statement into several more convenient pieces below.

Lemma 3.15.

In the setting described above, 𝐏(τxM|x|2)\mathbf{P}(\tau_{x}\geq M|x|^{2}) is bounded below by the probability of the following event:

{τQ0<,S0[0,τQ0]D+,S0[0,σB^′′0]85J/36(1)=,S0[τQ0,τQ0]85J/36(1)=,#(Γ(S0[0,τQ0])B^′′)M|x|2,τρx<,Sx[0,σρx]D,(S0[0,τQ0]Sx[0,σρx])B^0=,S~x(S0[0,τQ0]Sx[0,σρx])=}.\left\{\begin{array}[]{c}\tau_{Q}^{0}<\infty,\>S^{0}[0,\tau^{0}_{Q}]\subset D_{+},\>S^{0}[0,\sigma^{0}_{\widehat{B}^{\prime\prime}}]\cap\mathbb{H}^{(1)}_{85J/36}=\emptyset,\>S^{0}[\tau^{0}_{Q^{*}},\tau^{0}_{Q}]\cap\mathbb{H}^{(1)}_{85J/36}=\emptyset,\\ \#(\Gamma(S^{0}[0,\tau^{0}_{Q}])\cap\widehat{B}^{\prime\prime})\geq M|x|^{2},\>\tau^{x}_{\rho}<\infty,\>S^{x}[0,\sigma^{x}_{\rho}]\subset D_{-},\\ (S^{0}[0,\tau^{0}_{Q}]\cap S^{x}[0,\sigma^{x}_{\rho}])\cap\widehat{B}_{0}=\emptyset,\>\widetilde{S}^{x}\cap(S^{0}[0,\tau^{0}_{Q}]\cup S^{x}[0,\sigma^{x}_{\rho}])=\emptyset\end{array}\\ \right\}.
Proof.

Clearly,

{τQ0<τx0<,S0[0,τQ0]D+,S0[0,σB^′′0]85J/36(1)=,S0[τQ0,τQ0]85J/36(1)=,#(Γ(S0[0,τQ0])B^′′)M|x|2,S0[τQ0,τx0]D,(S0[0,τQ0]S0[τQ0,τx0])B^0=,S0[τx0,)(S0[0,τQ0]S0[τQ0,τx0])=}\left\{\begin{array}[]{c}\tau_{Q}^{0}<\tau^{0}_{x}<\infty,\>S^{0}[0,\tau^{0}_{Q}]\subset D_{+},\>S^{0}[0,\sigma^{0}_{\widehat{B}^{\prime\prime}}]\cap\mathbb{H}^{(1)}_{85J/36}=\emptyset,\>S^{0}[\tau^{0}_{Q^{*}},\tau^{0}_{Q}]\cap\mathbb{H}^{(1)}_{85J/36}=\emptyset,\\ \#(\Gamma(S^{0}[0,\tau^{0}_{Q}])\cap\widehat{B}^{\prime\prime})\geq M|x|^{2},\>S^{0}[\tau^{0}_{Q},\tau^{0}_{x}]\subset D_{-},\\ (S^{0}[0,\tau^{0}_{Q}]\cap S^{0}[\tau^{0}_{Q},\tau^{0}_{x}])\cap\widehat{B}_{0}=\emptyset,\>S^{0}[\tau^{0}_{x},\infty)\cap(S^{0}[0,\tau^{0}_{Q}]\cup S^{0}[\tau^{0}_{Q},\tau^{0}_{x}])=\emptyset\end{array}\\ \right\}

is a subset of the event {τxM|x|2}\{\tau_{x}\geq M|x|^{2}\}. Now, conditioning on the value of ρ\rho and applying the strong Markov property at times τQ0\tau_{Q}^{0} and τx0\tau^{0}_{x}, we have that the probability of the above event is equal to

zQ𝐏(τQ0<,ρ=z,τxz<,S0[0,τQ0]D+,S0[0,σB^′′0]85J/36(1)=,S0[τQ0,τQ0]85J/36(1)=,#(Γ(S0[0,τQ0])B^′′)M|x|2,Sz[0,τxz]D,(S0[0,τQ0]Sz[0,τxz])B^0=,S~x(S0[0,τQ0]Sz[0,τxz])=).\sum_{z\in Q}\mathbf{P}\left(\begin{array}[]{c}\tau_{Q}^{0}<\infty,\>\rho=z,\>\tau^{z}_{x}<\infty,\\ S^{0}[0,\tau^{0}_{Q}]\subset D_{+},\>S^{0}[0,\sigma^{0}_{\widehat{B}^{\prime\prime}}]\cap\mathbb{H}^{(1)}_{85J/36}=\emptyset,\>S^{0}[\tau^{0}_{Q^{*}},\tau^{0}_{Q}]\cap\mathbb{H}^{(1)}_{85J/36}=\emptyset,\\ \#(\Gamma(S^{0}[0,\tau^{0}_{Q}])\cap\widehat{B}^{\prime\prime})\geq M|x|^{2},\>S^{z}[0,\tau^{z}_{x}]\subset D_{-},\\ (S^{0}[0,\tau^{0}_{Q}]\cap S^{z}[0,\tau^{z}_{x}])\cap\widehat{B}_{0}=\emptyset,\>\widetilde{S}^{x}\cap(S^{0}[0,\tau^{0}_{Q}]\cup S^{z}[0,\tau^{z}_{x}])=\emptyset\end{array}\\ \right).

Applying Lemma 3.14, we can replace τxz\tau^{z}_{x} and Sz[0,τxz]S^{z}[0,\tau^{z}_{x}] in the above expression by τzx\tau^{x}_{z} and Sx[σ1,σ2]S^{x}[\sigma_{1},\sigma_{2}], respectively, where σ1\sigma_{1}, σ2\sigma_{2} are defined as in the statement of that result. Since 0σ1σ2σzx0\leq\sigma_{1}\leq\sigma_{2}\leq\sigma^{x}_{z}, it holds that Sx[σ1,σ2]Sx[0,σzx]S^{x}[\sigma_{1},\sigma_{2}]\subseteq S^{x}[0,\sigma_{z}^{x}]. Consequently, we obtain that the above sum is bounded below by

zQ𝐏(τQ0<,ρ=z,τzx<,S0[0,τQ0]D+,S0[0,σB^′′0]85J/36(1)=,S0[τQ0,τQ0]85J/36(1)=,#(Γ(S0[0,τQ0])B^′′)M|x|2,Sx[0,σzx]D,(S0[0,τQ0]Sx[0,σzx])B^0=,S~x(S0[0,τQ0]Sx[0,σzx])=),\sum_{z\in Q}\mathbf{P}\left(\begin{array}[]{c}\tau_{Q}^{0}<\infty,\>\rho=z,\>\tau^{x}_{z}<\infty,\\ S^{0}[0,\tau^{0}_{Q}]\subset D_{+},\>S^{0}[0,\sigma^{0}_{\widehat{B}^{\prime\prime}}]\cap\mathbb{H}^{(1)}_{85J/36}=\emptyset,\>S^{0}[\tau^{0}_{Q^{*}},\tau^{0}_{Q}]\cap\mathbb{H}^{(1)}_{85J/36}=\emptyset,\\ \#(\Gamma(S^{0}[0,\tau^{0}_{Q}])\cap\widehat{B}^{\prime\prime})\geq M|x|^{2},\>S^{x}[0,\sigma_{z}^{x}]\subset D_{-},\\ (S^{0}[0,\tau^{0}_{Q}]\cap S^{x}[0,\sigma_{z}^{x}])\cap\widehat{B}_{0}=\emptyset,\>\widetilde{S}^{x}\cap(S^{0}[0,\tau^{0}_{Q}]\cup S^{x}[0,\sigma_{z}^{x}])=\emptyset\end{array}\\ \right),

and replacing the sum with a union inside the probability completes the proof. ∎

Now, we will rewrite the event we defined in the statement of Lemma 3.15 as the intersection of various smaller events concerning S0S^{0}, SxS^{x} and S~x\widetilde{S}^{x}. For convenience, we will write

η0SτQ~00,ηxSτQ~xx,η~S~τ~x\eta_{0}\coloneqq S^{0}_{\tau^{0}_{\widetilde{Q}}},\qquad\eta_{x}\coloneqq S^{x}_{\tau^{x}_{\widetilde{Q}}},\qquad\widetilde{\eta}\coloneqq\widetilde{S}^{x}_{\widetilde{\tau}}

in the remainder of this subsection. We moreover define the event E1E_{1} by setting

E1={τQ~0<,η0Q~+,τρx<,ηxQ~,τ~<,η~Q1,(S0[0,τQ~0]Sx[0,τQ~x])B^0=,S0[0,τQ~0]D+,Sx[0,τQ~x]D,S~x[0,τ~](S0[0,τQ~0]Sx[0,τQ~x])=}.E_{1}=\left\{\begin{array}[]{c}\tau_{\widetilde{Q}}^{0}<\infty,\>\eta_{0}\in\widetilde{Q}_{+},\>\tau^{x}_{\rho}<\infty,\>\eta_{x}\in\widetilde{Q}_{-},\>\widetilde{\tau}<\infty,\>\widetilde{\eta}\in Q_{-1},\\ (S^{0}[0,\tau^{0}_{\widetilde{Q}}]\cap S^{x}[0,\tau^{x}_{\widetilde{Q}}])\cap\widehat{B}_{0}=\emptyset,\\ S^{0}[0,\tau^{0}_{\widetilde{Q}}]\subset D_{+},\>S^{x}[0,\tau^{x}_{\widetilde{Q}}]\subset D_{-},\>\widetilde{S}^{x}[0,\widetilde{\tau}]\cap(S^{0}[0,\tau^{0}_{\widetilde{Q}}]\cup S^{x}[0,\tau^{x}_{\widetilde{Q}}])=\emptyset\end{array}\right\}.

On E1E_{1}, the paths S0S^{0}, SxS^{x} and S~x\widetilde{S}^{x} do not have an intersection and move along the different courses until they first exit the union of B^1\widehat{B}_{-1} and B^0\widehat{B}_{0}.

Next, we will define some events that restrict the behavior of S0S^{0} after τQ~0\tau^{0}_{\widetilde{Q}}. First, recall that b=(94J,J2,0,,0)db^{\prime}=(\frac{9}{4}J,\frac{J}{2},0,\cdots,0)\in\mathbb{R}^{d} and B^=B(b,J6)\widehat{B}^{\prime}=B_{\infty}(b^{\prime},\frac{J}{6}). We define the “left” and “right” faces of B~\widetilde{B}^{\prime} by

QL={2512J}×[J3,23J]×[J6,J6]d2,QR={2912J}×[J3,23J]×[J6,J6]d2.Q_{L}=\left\{\frac{25}{12}J\right\}\times\left[\frac{J}{3},\frac{2}{3}J\right]\times\left[-\frac{J}{6},\frac{J}{6}\right]^{d-2},\qquad Q_{R}=\left\{\frac{29}{12}J\right\}\times\left[\frac{J}{3},\frac{2}{3}J\right]\times\left[-\frac{J}{6},\frac{J}{6}\right]^{d-2}.

Moreover, we define a subset of D~+\widetilde{D}_{+} by setting

B^L=[1718J,7936J]×[J3,23J]×[J2,J2]d2.\widehat{B}^{\prime}_{L}=\left[\frac{17}{18}J,\frac{79}{36}J\right]\times\left[\frac{J}{3},\frac{2}{3}J\right]\times\left[-\frac{J}{2},\frac{J}{2}\right]^{d-2}.

See Figure 7. Then, writing uy=inf{nτQ~y:SnyQ}u^{y}=\inf\{n\geq\tau^{y}_{\widetilde{Q}}\mathrel{:}S^{y}_{n}\in Q^{\prime}\} and σ=inf{nτB^′′y:Sny(B^)c}\sigma^{\prime}=\inf\{n\geq\tau^{y}_{\widehat{B}^{\prime\prime}}\mathrel{:}S^{y}_{n}\in(\widehat{B}^{\prime})^{c}\}, let

F1(y)=\displaystyle F_{1}(y)= {τQLy<uy,Sy[τQ~y,τQLy]B^L},\displaystyle\left\{\tau^{y}_{Q_{L}}<u^{y},S^{y}[\tau^{y}_{\widetilde{Q}},\tau^{y}_{Q_{L}}]\subset\widehat{B}^{\prime}_{L}\right\}, (62)
F2(y)=\displaystyle F_{2}(y)= {τB^′′y<inf{nτQLy:Sny(B^L)c}<,σQR,#{k[τB^′′y,σ]:Sy[τQLy,k]Sy[k+1,σ]=,SkyB^′′}M|x|2},\displaystyle\left\{\begin{array}[]{c}\tau^{y}_{\widehat{B}^{\prime\prime}}<\inf\{n\geq\tau^{y}_{Q_{L}}\mathrel{:}S^{y}_{n}\in(\widehat{B}^{\prime}_{L})^{c}\}<\infty,\>\sigma^{\prime}\in Q_{R},\\ \#\left\{k\in[\tau^{y}_{\widehat{B}^{\prime\prime}},\sigma^{\prime}]\mathrel{:}S^{y}[\tau^{y}_{Q_{L}},k]\cap S^{y}[k+1,\sigma^{\prime}]=\emptyset,\>S^{y}_{k}\in\widehat{B}^{\prime\prime}\right\}\geq M|x|^{2}\end{array}\right\}, (65)
F3(y)=\displaystyle F_{3}(y)= {τQy<τQy<,Sy[σ,τQy]D+[8536J,)×d1},\displaystyle\left\{\tau^{y}_{Q^{*}}<\tau^{y}_{Q}<\infty,\>S^{y}[\sigma^{\prime},\tau^{y}_{Q}]\subset D_{+}\cap\left[\frac{85}{36}J,\infty\right)\times\mathbb{R}^{d-1}\right\}, (66)

and set E2=F1(0)F2(0)F3(0)E_{2}=F_{1}(0)\cap F_{2}(0)\cap F_{3}(0). In particular, on the event E2E_{2}, we have the existence of cut points of S[τQLy,σ]S[\tau^{y}_{Q_{L}},\sigma^{\prime}] contained in B^′′\widehat{B}^{\prime\prime}. Finally let

E3=\displaystyle E_{3}= {τρx<,Sx[τQ~x,σρx]D~},\displaystyle\left\{\tau^{x}_{\rho}<\infty,\>S^{x}[\tau^{x}_{\widetilde{Q}},\sigma^{x}_{\rho}]\subset\widetilde{D}_{-}\right\},
E4=\displaystyle E_{4}= {S~x[τ~,](S0[0,τQ0]Sx[0,σρx])=},\displaystyle\left\{\widetilde{S}^{x}[\widetilde{\tau},\infty]\cap(S^{0}[0,\tau^{0}_{Q}]\cup S^{x}[0,\sigma^{x}_{\rho}])=\emptyset\right\},

be events that restrict the regions where SxS^{x} and S~x\widetilde{S}^{x} can explore, respectively.

Refer to caption
Figure 7: Illustration of the sets used in controlling the number of cut points of S0S^{0} in B^′′\widehat{B}^{\prime\prime}.

We continue by checking the local cut points that we construct on the event F2(0)F_{2}(0) are cut points of the loop-erasure of S0[0,τQ0]S^{0}[0,\tau^{0}_{Q}]. Note that on the event E2E_{2}, it follows from the definition of F2(0)F_{2}(0) and F3(0)F_{3}(0) that

S[τQ0,τQ0]85J/36(1)=,S[σ,τQ0]B^′′=.S[\tau^{0}_{Q^{*}},\tau^{0}_{Q}]\cap\mathbb{H}^{(1)}_{85J/36}=\emptyset,\qquad S[\sigma^{\prime},\tau^{0}_{Q_{*}}]\cap\widehat{B}^{\prime\prime}=\emptyset.

Moreover, on the event E1E2E_{1}\cap E_{2}, we have that

  • S0[0,τQ~0]Q=S^{0}[0,\tau^{0}_{\widetilde{Q}}]\cap Q^{*}=\emptyset,

  • S0[τQ~0,τQL0]Q=S^{0}[\tau^{0}_{\widetilde{Q}},\tau^{0}_{Q_{L}}]\cap Q^{*}=\emptyset,

  • S0[τQL0,σ]Q=S^{0}[\tau^{0}_{Q_{L}},\sigma^{\prime}]\cap Q^{*}=\emptyset.

The first and second statements follow from the definitions of E1E_{1} and F1(0)F_{1}(0), respectively, while the third statement is derived from the definitions of F2(0)F_{2}(0) and σ\sigma^{\prime} (recall the definitions of the sets defined above, which are also shown in Figure 7). From these statements, we immediately conclude that

S0[0,σ]Q=.S^{0}[0,\sigma^{\prime}]\cap Q^{*}=\emptyset.

For the rest of the path S0[0,τQ0]S^{0}[0,\tau^{0}_{Q}], the definition of F3(0)F_{3}(0) implies that

S0[σ,τQ0]B^′′=,S0[τQ0,τQ0]85/36J(1)=.S^{0}[\sigma^{\prime},\tau^{0}_{Q^{*}}]\cap\widehat{B}^{\prime\prime}=\emptyset,\qquad S^{0}[\tau^{0}_{Q^{*}},\tau^{0}_{Q}]\cap\mathbb{H}^{(1)}_{85/36J}=\emptyset.

Combining the preceding three statements, we see that, on E1E2E_{1}\cap E_{2},

S0[0,σB^′′0]85/36J(1)=,S0[τQ0,τQ0]85J/36(1)=,S^{0}[0,\sigma^{0}_{\widehat{B}^{\prime\prime}}]\cap\mathbb{H}^{(1)}_{85/36J}=\emptyset,\qquad S^{0}[\tau^{0}_{Q^{*}},\tau^{0}_{Q}]\cap\mathbb{H}^{(1)}_{85J/36}=\emptyset,

where we recall that is σB^′′0\sigma^{0}_{\widehat{B}^{\prime\prime}} be the last exit time of B^′′\widehat{B}^{\prime\prime} by S0S^{0} (we assume here that S0S^{0} is stopped at τQ0\tau^{0}_{Q}). Thus, the local cut points of the event F2(0)F_{2}(0) are indeed cut points of the loop-erasure of S0[0,τQ0]S^{0}[0,\tau^{0}_{Q}] and the probability of the event we defined in the statement of Lemma 3.15 is bounded below by

𝐏(E1E2E3E4).\mathbf{P}(E_{1}\cap E_{2}\cap E_{3}\cap E_{4}).

In what follows, we will bound below this probability below. To start with, we will prove that S0S^{0}, SxS^{x} and S~x\widetilde{S}^{x} do not have an intersection and are separated in a cube with positive probability. Let Trz=τB(0,r)zT^{z}_{r}=\tau^{z}_{B_{\infty}(0,r)}. We define the event GnG_{n} by setting

Gn={S0[0,Tn0]Sx[0,Tnx]=S0[0,Tn0]S~x[0,T~nx]=Sx[0,Tnx]S~x[1,T~nx]=},G_{n}=\left\{S^{0}[0,T^{0}_{n}]\cap S^{x}[0,T^{x}_{n}]=S^{0}[0,T^{0}_{n}]\cap\widetilde{S}^{x}[0,\widetilde{T}^{x}_{n}]=S^{x}[0,T^{x}_{n}]\cap\widetilde{S}^{x}[1,\widetilde{T}^{x}_{n}]=\emptyset\right\},

and let ZnZ_{n} be given by

min{d(STn00,Sx[0,Tnx]S~x[0,T~nx]),d(STnxx,S0[0,Tn0]S~x[0,T~nx]),d(S~T~nxx,S0[0,Tn0]Sx[0,Tnx])},\min\left\{d(S^{0}_{T^{0}_{n}},S^{x}[0,T^{x}_{n}]\cup\widetilde{S}^{x}[0,\widetilde{T}^{x}_{n}]),d(S^{x}_{T^{x}_{n}},S^{0}[0,T^{0}_{n}]\cup\widetilde{S}^{x}[0,\widetilde{T}^{x}_{n}]),d(\widetilde{S}^{x}_{\widetilde{T}^{x}_{n}},S^{0}[0,T^{0}_{n}]\cup S^{x}[0,T^{x}_{n}])\right\},

where dd here is the Euclidean distance, i.e. ZnZ_{n} is the minimum of the distance between the point from which either S0S^{0}, SxS^{x} or S~x\widetilde{S}^{x} exits B(0,n)B_{\infty}(0,n) and the union of the other two paths up to their exit times.

Lemma 3.16.

There exists c>0c>0 and n01n_{0}\geq 1 such that: for all nn0n\geq n_{0},

𝐏(Gn{Znn2})c.\mathbf{P}\left(G_{n}\cap\left\{Z_{n}\geq\frac{n}{2}\right\}\right)\geq c.
Proof.

For readability, we assume that x=(0,|x|,0,,0)x=(0,|x|,0,\cdots,0). (Other cases will follow by a small modification of the argument.) We follow the idea of [7, Lemma 3.2]. Let e1=(1,0,0,,0)de_{1}=(1,0,0,\cdots,0)\in\mathbb{Z}^{d} and e2=(0,1,0,,0)de_{2}=(0,1,0,\cdots,0)\in\mathbb{Z}^{d}. We define the event I1I_{1} by setting

I1={Si0=ie2,Six=ie1,S~ix=ie1for 1ik},I_{1}=\left\{S^{0}_{i}=ie_{2},\>S_{i}^{x}=ie_{1},\>\widetilde{S}_{i}^{x}=-ie_{1}\ \mbox{for}\ 1\leq i\leq k\right\},

where k1k\geq 1 will be fixed later. Then we have 𝐏(I1)=(2d)3k\mathbf{P}(I_{1})=(2d)^{-3k}.

We will show that the probability that S0S^{0}, SxS^{x} and S~x\widetilde{S}^{x} do not intersect before they first exit from B(0,n)B_{\infty}(0,n) conditioned on I1I_{1} is bounded above by arbitrarily small ε\varepsilon by taking kk sufficiently large. Let K(j)K(j) be the number of intersections of S0[j,Tn0]S^{0}[j,T^{0}_{n}], Sx[j,Tnx]S^{x}[j,T^{x}_{n}] and S~x[j,T~nx]\widetilde{S}^{x}[j,\widetilde{T}^{x}_{n}], i.e.

K(j)=#((S0[j,Tn0]Sx[j,Tnx])(S0[j,Tn0]S~x[j,T~nx])(Sx[j,Tnx]S~x[max{1,j},T~nx])).K(j)=\#\left((S^{0}[j,T^{0}_{n}]\cap S^{x}[j,T^{x}_{n}])\cup(S^{0}[j,T^{0}_{n}]\cap\widetilde{S}^{x}[j,\widetilde{T}^{x}_{n}])\cup(S^{x}[j,T^{x}_{n}]\cap\widetilde{S}^{x}[\max\{1,j\},\widetilde{T}^{x}_{n}])\right).

Then by the Markov inequality,

𝐏(K(0)>0I1)\displaystyle\mathbf{P}(K(0)>0\mid I_{1}) 𝐏(K(k)1I1)\displaystyle\leq\mathbf{P}(K(k)\geq 1\mid I_{1})
𝐄(K(k)I1)\displaystyle\leq\mathbf{E}(K(k)\mid I_{1})
3maxx,y:d(x,y)k𝐄(#(Sx[0,Tnx]Sy[0,Tny]))\displaystyle\leq 3\max_{x,y:\>d(x,y)\geq k}\mathbf{E}\left(\#(S^{x}[0,T^{x}_{n}]\cap S^{y}[0,T^{y}_{n}])\right)
3maxx,y:d(x,y)km=0m=0zd𝐏x(Smx=z)𝐏y(Smy=z)\displaystyle\leq 3\max_{x,y:\>d(x,y)\geq k}\sum_{m=0}^{\infty}\sum_{m^{\prime}=0}^{\infty}\sum_{z\in\mathbb{Z}^{d}}\mathbf{P}^{x}(S^{x}_{m}=z)\mathbf{P}^{y}(S^{y}_{m^{\prime}}=z)
3maxx,y:d(x,y)km=0m=0𝐏x(Sm+mx=y)\displaystyle\leq 3\max_{x,y:\>d(x,y)\geq k}\sum_{m=0}^{\infty}\sum_{m^{\prime}=0}^{\infty}\mathbf{P}^{x}(S^{x}_{m+m^{\prime}}=y)
3l=1lCld/2eck2/l\displaystyle\leq 3\sum_{l=1}^{\infty}l\cdot Cl^{-d/2}e^{-ck^{2}/l}
Cl=1l1d/2eck2/l,\displaystyle\leq C\sum_{l=1}^{\infty}l^{1-d/2}e^{-ck^{2}/l}, (67)

for some CC, c>0c>0, where we applied the Gaussian estimate for the off-diagonal heat kernel of the simple random walk on d\mathbb{Z}^{d} for the last inequality (see (1)). Since d5d\geq 5, the right-hand side of (67) converges to 0 as kk\to\infty.

Our next step is to construct subsets where each simple random walk path is constrained to move until it first exits from B(0,n)B_{\infty}(0,n). We define by

HL={n2}×[n4,n4]d1,HR={n2}×[n4,n4]d1,H_{L}=\left\{-\frac{n}{2}\right\}\times\left[-\frac{n}{4},\frac{n}{4}\right]^{d-1},\qquad H_{R}=\left\{\frac{n}{2}\right\}\times\left[-\frac{n}{4},\frac{n}{4}\right]^{d-1},

the subsets of the left and right face of B(0,n2)B_{\infty}(0,\frac{n}{2}) in the direction of x1x_{1}-axis, respectively, and by

H+=[n4,n4]×{n2}×[n4,n4]d2,H_{+}=\left[-\frac{n}{4},\frac{n}{4}\right]\times\left\{\frac{n}{2}\right\}\times\left[-\frac{n}{4},\frac{n}{4}\right]^{d-2},

the subset of the upper face of B(0,n2)B_{\infty}(0,\frac{n}{2}) in the direction of x2x_{2}-axis. Let

I20\displaystyle I^{0}_{2} ={STn/200H+,STn00n(2),S0[Tn/20,Tn0](n/3(1)n/3(1)n/3(2))=},\displaystyle=\left\{S^{0}_{T^{0}_{n/2}}\in H_{+},\>S^{0}_{T^{0}_{n}}\in\mathbb{H}^{(2)}_{n},\>S^{0}[T^{0}_{n/2},T^{0}_{n}]\cap(\mathbb{H}^{(1)}_{n/3}\cup\mathbb{H}^{(1)}_{-n/3}\cup\mathbb{H}^{(2)}_{n/3})=\emptyset\right\},
I2x\displaystyle I^{x}_{2} ={STn/2xxHR,STnxxn(1),Sx[Tn/2x,Tnx](n/3(1)n/3(2))=},\displaystyle=\left\{S^{x}_{T^{x}_{n/2}}\in H_{R},\>S^{x}_{T^{x}_{n}}\in\mathbb{H}^{(1)}_{n},\>S^{x}[T^{x}_{n/2},T^{x}_{n}]\cap(\mathbb{H}^{(1)}_{n/3}\cup\mathbb{H}^{(2)}_{n/3})=\emptyset\right\},
I~2x\displaystyle\widetilde{I}^{x}_{2} ={S~T~n/2xxHL,S~T~nxxn(1),S~x[T~n/2x,T~nx](n/3(1)n/3(2))=},\displaystyle=\left\{\widetilde{S}^{x}_{\widetilde{T}^{x}_{n/2}}\in H_{L},\>\widetilde{S}^{x}_{\widetilde{T}^{x}_{n}}\in\mathbb{H}^{(1)}_{-n},\>\widetilde{S}^{x}[\widetilde{T}^{x}_{n/2},\>\widetilde{T}^{x}_{n}]\cap(\mathbb{H}^{(1)}_{-n/3}\cup\mathbb{H}^{(2)}_{n/3})=\emptyset\right\},

and I2=I20I2xI~2xI_{2}=I_{2}^{0}\cap I_{2}^{x}\cap\widetilde{I}^{x}_{2}. It is an elementary exercise to check that there exists some ε>0\varepsilon>0 such that 𝐏(I2I1)ε\mathbf{P}(I_{2}\mid I_{1})\geq\varepsilon holds uniformly in n1n\geq 1 and xdx\in\mathbb{Z}^{d}. Note that on the event I1I2I_{1}\cap I_{2}, it holds that

STnxxn(1),S0[0,Tn0]S~x[0,T~nx](,n2]×d1,S^{x}_{T^{x}_{n}}\in\mathbb{H}^{(1)}_{n},\qquad S^{0}[0,T^{0}_{n}]\cup\widetilde{S}^{x}[0,\widetilde{T}^{x}_{n}]\subset(-\infty,\frac{n}{2}]\times\mathbb{R}^{d-1},

so that d(STnxx,S0[0,Tn0]S~x[0,T~nx])n2d(S^{x}_{T^{x}_{n}},S^{0}[0,T^{0}_{n}]\cup\widetilde{S}^{x}[0,\widetilde{T}^{x}_{n}])\geq\frac{n}{2}. Similarly, the same bound holds for d(STn00,Sx[0,Tnx]S~x[0,T~nx])d(S^{0}_{T^{0}_{n}},S^{x}[0,T^{x}_{n}]\cup\widetilde{S}^{x}[0,\widetilde{T}^{x}_{n}]) and d(S~T~nxx,S0[0,Tn0]Sx[0,Tnx])d(\widetilde{S}^{x}_{\widetilde{T}^{x}_{n}},S^{0}[0,T^{0}_{n}]\cup S^{x}[0,T^{x}_{n}]). Thus we have Znn2Z_{n}\geq\frac{n}{2} on the event in question. Finally, by taking kk large so that 𝐏(K(0)>0I1)ε2\mathbf{P}(K(0)>0\mid I_{1})\leq\frac{\varepsilon}{2}, we obtain that

𝐏(Gn{Znn2})\displaystyle\mathbf{P}\left(G_{n}\cap\left\{Z_{n}\geq\frac{n}{2}\right\}\right) 𝐏(I1I2{K(0)=0})\displaystyle\geq\mathbf{P}(I_{1}\cap I_{2}\cap\{K(0)=0\})
=𝐏(I1)(𝐏(I2I1)𝐏(K(0)>0I1))\displaystyle=\mathbf{P}(I_{1})\left(\mathbf{P}(I_{2}\mid I_{1})-\mathbf{P}(K(0)>0\mid I_{1})\right)
(2d)3kε/2,\displaystyle\geq(2d)^{-3k}\varepsilon/2,

which completes the proof. ∎

We are now ready to complete the proof of the main result of this subsection.

Proof of Proposition 3.13.

For any R>1R>1, we have that

𝐏(τx[R1M|x|2,RM|x|2])𝐏(τxM|x|2)𝐏(τx>RM|x|2)\mathbf{P}\left(\tau_{x}\in[R^{-1}M|x|^{2},RM|x|^{2}]\right)\geq\mathbf{P}\left(\tau_{x}\geq M|x|^{2}\right)-\mathbf{P}\left(\tau_{x}>RM|x|^{2}\right)

Now, by the argument used to prove Proposition 3.2, we have that

𝐏(τx>RM|x|2)C(RM|x|2)1d/2.\mathbf{P}\left(\tau_{x}>RM|x|^{2}\right)\leq C\left(RM|x|^{2}\right)^{1-d/2}.

Thus, by taking RR suitably large, and applying Lemma 3.15 and the argument above Lemma 3.16, to complete the proof it suffices to prove that

𝐏(E1E2E3E4)cJ2d\mathbf{P}\left(E_{1}\cap E_{2}\cap E_{3}\cap E_{4}\right)\geq cJ^{2-d}

for some constant cc.

First, we derive an estimate for 𝐏(E1)\mathbf{P}(E_{1}) from the result of Lemma 3.16. We take C1C\geq 1 large so that J=CM|x|23n0J=C\sqrt{M|x|^{2}}\geq 3n_{0}. We then define the event FF^{\prime} by setting

F={τQ~0<,η0Q~+,τρx<,ηxQ~,τ~<,η~Q1,S0[TJ/30,τQ~0](J/9(1)5J/36(2))=,Sx[TJ/3x,τQ~x](J/9(1)5J/36(2))=,S~x[T~J/3x,τ~]J/9(1)=}.F^{\prime}=\left\{\begin{array}[]{c}\tau_{\widetilde{Q}}^{0}<\infty,\>\eta_{0}\in\widetilde{Q}_{+},\>\tau^{x}_{\rho}<\infty,\>\eta_{x}\in\widetilde{Q}_{-},\>\widetilde{\tau}<\infty,\>\widetilde{\eta}\in Q_{-1},\\ S^{0}[T^{0}_{J/3},\tau^{0}_{\widetilde{Q}}]\cap\left(\mathbb{H}^{(1)}_{J/9}\cup\mathbb{H}^{(2)}_{5J/36}\right)=\emptyset,\>S^{x}[T^{x}_{J/3},\tau^{x}_{\widetilde{Q}}]\cap\left(\mathbb{H}^{(1)}_{J/9}\cup\mathbb{H}^{(2)}_{5J/36}\right)=\emptyset,\\ \widetilde{S}^{x}[\widetilde{T}^{x}_{J/3},\widetilde{\tau}]\cap\mathbb{H}^{(1)}_{-J/9}=\emptyset\end{array}\right\}.

Recall the definition of the events I1I_{1} and I2I_{2} and the random variable K(j)K(j) from the proof of Lemma 3.16. It is straightforward to check that if we take n=J/3n=J/3, then

I1I2{K(0)=0}FE1.I_{1}\cap I_{2}\cap\{K(0)=0\}\cap F^{\prime}\subset E_{1}.

Moreover, by the strong Markov property and the approximation to Brownian motion, there exists some constant c>0c>0 such that 𝐏(FI1I2{K(0)=0})c\mathbf{P}(F^{\prime}\mid I_{1}\cap I_{2}\cap\{K(0)=0\})\geq c, uniformly in xx and MM. By Lemma 3.16, we thus obtain that

𝐏(E1)𝐏(FI1I2{K(0)=0})𝐏(I1I2{K(0)=0})c2.\mathbf{P}(E_{1})\geq\mathbf{P}\left(F^{\prime}\mid I_{1}\cap I_{2}\cap\{K(0)=0\}\right)\mathbf{P}(I_{1}\cap I_{2}\cap\{K(0)=0\})\geq c^{2}. (68)

Next we estimate 𝐏(E2E1)\mathbf{P}(E_{2}\mid E_{1}). It follows from the strong Markov property that

𝐏(E2E1)infa1Q~+𝐏a1(F1(a1)F2(a2)F3(a3)),\mathbf{P}\left(E_{2}\mid E_{1}\right)\geq\inf_{a_{1}\in\widetilde{Q}_{+}}\mathbf{P}^{a_{1}}\left(F_{1}(a_{1})\cap F_{2}(a_{2})\cap F_{3}(a_{3})\right),

where F1(y)F_{1}(y), F2(y)F_{2}(y) and F3(y)F_{3}(y) are as defined in (62), (65) and (66), respectively. Thus it suffices to bound from below the right-hand side of the above inequality. We begin with a lower bound for 𝐏a1(F1)\mathbf{P}^{a_{1}}(F_{1}). By the gambler’s ruin estimate (9), we have that 𝐏a1(F1)c\mathbf{P}^{a_{1}}(F_{1})\geq c for some universal constant c>0c>0. Next, applying a similar argument to that used to obtain (55) in the proof of Lemma 3.12 and the strong Markov property, we obtain that 𝐏a1(F2F1)c\mathbf{P}^{a_{1}}(F_{2}\mid F_{1})\geq c. Finally, again by (9) and the strong Markov property, we have that 𝐏a1(F3F1F2)c\mathbf{P}^{a_{1}}(F_{3}\mid F_{1}\cap F_{2})\geq c. Since c>0c>0 does not depend on a1Q~+a_{1}\in\widetilde{Q}_{+}, we can conclude that

𝐏a1(E2E1)c3.\mathbf{P}^{a_{1}}(E_{2}\mid E_{1})\geq c^{3}. (69)

By the independence of S0S^{0} and SxS^{x} and the strong Markov property, we also have that

𝐏(E3E1E2)infa2Q~,zQ𝐏a2(τza2<,Sa2[0,σza2]D~).\mathbf{P}(E_{3}\mid E_{1}\cap E_{2})\geq\inf_{a_{2}\in\widetilde{Q}_{-},\>z\in Q}\mathbf{P}^{a_{2}}\left(\tau^{a_{2}}_{z}<\infty,\>S^{a_{2}}[0,\sigma_{z}^{a_{2}}]\subset\widetilde{D}_{-}\right).

Let a2Q~a_{2}\in\widetilde{Q}_{-} and zQz\in Q. Again by the strong Markov property,

𝐏a2(τza2<,Sa2[0,σza2]D~)=𝐏a2(τza2<,Sa2[0,τza2]D~)𝐏z(Sz[0,σzz]D~).\mathbf{P}^{a_{2}}\left(\tau^{a_{2}}_{z}<\infty,\>S^{a_{2}}[0,\sigma_{z}^{a_{2}}]\subset\widetilde{D}_{-}\right)=\mathbf{P}^{a_{2}}\left(\tau^{a_{2}}_{z}<\infty,\>S^{a_{2}}[0,\tau_{z}^{a_{2}}]\subset\widetilde{D}_{-}\right)\mathbf{P}^{z}\left(S^{z}[0,\sigma^{z}_{z}]\subset\widetilde{D}_{-}\right). (70)

We will give lower bounds for the two probabilities in the right-hand side. Let la2,zl_{a_{2},z} be the piecewise linear curve that runs from a2a_{2} in the direction of e2e_{2} until its second coordinate reaches 5J/25J/2, and then runs along the line from that point to zz. Similarly to (51) in Lemma 3.12, we obtain that

𝐏a2(τza2<,Sa2[0,τza2]D~)\displaystyle\mathbf{P}^{a_{2}}\left(\tau^{a_{2}}_{z}<\infty,S^{a_{2}}[0,\tau_{z}^{a_{2}}]\subset\widetilde{D}_{-}\right) 𝐏a2(τza2<,dist(Sa2(k),la2,z)J/16for allk[0,τza2])\displaystyle\geq\mathbf{P}^{a_{2}}\left(\tau^{a_{2}}_{z}<\infty,\mathrm{dist}(S^{a_{2}}(k),l_{a_{2},z})\leq J/16\ \mbox{for all}\ k\in[0,\tau^{a_{2}}_{z}]\right)
cJ2d,\displaystyle\geq cJ^{2-d}, (71)

uniformly in a2a_{2} and zz for some c>0c>0. Furthermore, we have that

𝐏z(Sz[0,σzz]D~)\displaystyle\mathbf{P}^{z}\left(S^{z}[0,\sigma^{z}_{z}]\subset\widetilde{D}_{-}\right) 1𝐏z(Sz[0,σzz]B(z,J/16)c)\displaystyle\geq 1-\mathbf{P}^{z}\left(S^{z}[0,\sigma^{z}_{z}]\cap B(z,J/16)^{c}\neq\emptyset\right)
1supwB(z,J/16)𝐏w(τzw<)\displaystyle\geq 1-\sup_{w\in\partial B(z,J/16)}\mathbf{P}^{w}(\tau^{w}_{z}<\infty)
1aG(0)J2d,\displaystyle\geq 1-\frac{a}{G(0)}J^{2-d},

where we applied (8) with nn\to\infty to the last inequality. Thus, by increasing the value of the constant C>0C>0 in J=CM|x|2J=C\sqrt{M|x|^{2}} if necessary, we have that

𝐏z(Sz[0,σzz]D~)c,\mathbf{P}^{z}\left(S^{z}[0,\sigma^{z}_{z}]\subset\widetilde{D}_{-}\right)\geq c, (72)

for some uniform constant c>0c>0. Plugging (71) and (72) into (70) yields that

𝐏(E3E1E2)cJ2d.\mathbf{P}(E_{3}\mid E_{1}\cap E_{2})\geq cJ^{2-d}. (73)

Now we will give a lower bound for 𝐏(E4E1E2E3)\mathbf{P}(E_{4}\mid E_{1}\cap E_{2}\cap E_{3}). Recall that Q1=B^1{3J}×d1Q_{-1}=\widehat{B}_{-1}\cap\{-3J\}\times\mathbb{R}^{d-1}. By the strong Markov property and the definition of the events E1E_{1}, E2E_{2} and E3E_{3}, we have that

𝐏(E4E1E2E3)infa3Q1𝐏~a3(S~a3[0,](D+D)=).\mathbf{P}(E_{4}\mid E_{1}\cap E_{2}\cap E_{3})\geq\inf_{a_{3}\in Q_{-1}}\widetilde{\mathbf{P}}^{a_{3}}(\widetilde{S}^{a_{3}}[0,\infty]\cap(D_{+}\cup D_{-})=\emptyset).

From this, it is an easy application of (7) to deduce that

𝐏(E4E1E2E3)c,\mathbf{P}\left(E_{4}\mid E_{1}\cap E_{2}\cap E_{3}\right)\geq c, (74)

for some universal constant c>0c>0.

Finally by multiplying (68), (69), (73) and (74), we obtain the desired lower bound. ∎

4 Heat kernel estimates for the associated random walk

The aim of this section is to prove Theorem 1.3. As explained in the introduction, the main input concerning the loop-erased random walk will be Theorem 1.1. To estimate (Xt𝒢=x)\mathbb{P}(X_{t}^{\mathcal{G}}=x) using the decomposition at (4), we also require control over P𝒢(Xt𝒢=Lm)P^{\mathcal{G}}(X_{t}^{\mathcal{G}}=L_{m}), where in this section (Lm)m0(L_{m})_{m\geq 0} is always the infinite LERW started from 0. Since the structure of the graph 𝒢\mathcal{G} is simply that of +\mathbb{Z}_{+} equipped with nearest-neighbour bonds, we have the obvious identity

P𝒢(Xt𝒢=Lm)=qt(0,m),P^{\mathcal{G}}(X_{t}^{\mathcal{G}}=L_{m})=q_{t}(0,m),

where (qt(x,y))x,y+,t>0(q_{t}(x,y))_{x,y\in\mathbb{Z}_{+},\>t>0} gives the transition probabilities of the continuous-time simple random walk on +\mathbb{Z}_{+} with unit mean holding times. For this, we have the following estimates from [3]. (We note that although the result we will cite in [3] is stated for the simple random walk on \mathbb{Z}, it is easy to adapt to apply to the half-space +\mathbb{Z}_{+}.)

Lemma 4.1.

For any ε>0\varepsilon>0, there exist constants c1,c2,c3,c4,c5,c6(0,)c_{1},c_{2},c_{3},c_{4},c_{5},c_{6}\in(0,\infty) such that for every m+m\in\mathbb{Z}_{+} and tεmt\geq\varepsilon m,

qt(0,m)c1(1t1/2)exp(c2m21t)q_{t}(0,m)\leq c_{1}\left(1\wedge t^{-1/2}\right)\exp\left(-\frac{c_{2}m^{2}}{1\vee t}\right)

and also

qt(0,m)c3(1t1/2)exp(c4m21t).q_{t}(0,m)\geq c_{3}\left(1\wedge t^{-1/2}\right)\exp\left(-\frac{c_{4}m^{2}}{1\vee t}\right).

Moreover, for m1m\geq 1 and t<εmt<\varepsilon m, we have that

qt(0,m)c5exp(c6m(1+log(m/t))).q_{t}(0,m)\leq c_{5}\exp\left(-c_{6}m\left(1+\log(m/t)\right)\right).
Proof.

From [3, Theorem 6.28(b)], we obtain the relevant bounds for t1mt\geq 1\vee m. Moreover, the bounds for m=0m=0, t(0,1)t\in(0,1), follow from [3, Theorem 6.28(d)]. As for m1m\geq 1, t(εm,m)t\in(\varepsilon m,m), we can apply [3, Theorem 6.28(c)] to deduce that qt(0,m)q_{t}(0,m) is bounded above and below by an expression of the form:

cexp(c1m(1+log(m/t))).c\exp\left(-c^{-1}m\left(1+\log(m/t)\right)\right).

This can be bounded above and below by an expression of the form cexp(c1m)c\exp(-c^{-1}m), and that in turn by c(1t1/2)exp(c1m21t)c(1\wedge t^{-1/2})\exp(-\frac{c^{-1}m^{2}}{1\vee t}), uniformly over the range of mm and tt considered. This completes the proof of the first two inequalities in the statement of the lemma. The third inequality is given by again applying [3, Theorem 6.28(c)]. ∎

We are now ready to proceed with the proof of Theorem 1.3.

Proof of Theorem 1.3.

Clearly, if x=0x=0, then Lemma 4.1 immediately yields

(Xt𝒢=x)=qt(0,0)1t1/2,\mathbb{P}\left(X_{t}^{\mathcal{G}}=x\right)=q_{t}(0,0)\asymp 1\wedge t^{-1/2},

which gives the result in this case.

We next suppose x0x\neq 0. In this case, applying Lemma 4.1 with ε=1\varepsilon=1, we find that

(Xt𝒢=x)\displaystyle\mathbb{P}\left(X_{t}^{\mathcal{G}}=x\right) =\displaystyle= m=1(Xt𝒢=Lm)𝐏(Lm=x)\displaystyle\sum_{m=1}^{\infty}\mathbb{P}\left(X_{t}^{\mathcal{G}}=L_{m}\right)\mathbf{P}\left(L_{m}=x\right) (75)
=\displaystyle= m=1qt(0,m)𝐏(Lm=x)\displaystyle\sum_{m=1}^{\infty}q_{t}(0,m)\mathbf{P}\left(L_{m}=x\right)
\displaystyle\leq c1(1t1/2)m=1texp(c2m21t)𝐏(Lm=x)\displaystyle c_{1}\left(1\wedge t^{-1/2}\right)\sum_{m=1}^{t}\exp\left(-\frac{c_{2}m^{2}}{1\vee t}\right)\mathbf{P}\left(L_{m}=x\right)
+c3m=t+1exp(c4m(1+log(m/t)))𝐏(Lm=x).\displaystyle\qquad+c_{3}\sum_{m={t+1}}^{\infty}\exp\left(-c_{4}m\left(1+\log(m/t)\right)\right)\mathbf{P}\left(L_{m}=x\right).

Now, the second sum here is readily bounded as follows:

c3m=t+1exp(c4m(1+log(m/t)))𝐏(Lm=x)c3m=t+1exp(c4m)c3exp(c5t).c_{3}\sum_{m={t+1}}^{\infty}\exp\left(-c_{4}m\left(1+\log(m/t)\right)\right)\mathbf{P}\left(L_{m}=x\right)\leq c_{3}\sum_{m={t+1}}^{\infty}\exp\left(-c_{4}m\right)\leq c_{3}\exp\left(-c_{5}t\right).

Moreover, since we are assuming tε|x|εt\geq\varepsilon|x|\geq\varepsilon, the final expression is readily bounded above by one of the form

c6(1|x|2d)(1t1/2)exp(c7(|x|41t)1/3).c_{6}\left(1\wedge|x|^{2-d}\right)\left(1\wedge t^{-1/2}\right)\exp\left(-c_{7}\left(\frac{|x|^{4}}{1\vee t}\right)^{1/3}\right).

Thus, to complete the proof of upper bound in the statement of Theorem 1.3, it remains to derive a similar bound for the first sum on the right-hand side at (75). For this, we have that

c1(1t1/2)m=1texp(c2m21t)𝐏(Lm=x)\displaystyle c_{1}\left(1\wedge t^{-1/2}\right)\sum_{m=1}^{t}\exp\left(-\frac{c_{2}m^{2}}{1\vee t}\right)\mathbf{P}\left(L_{m}=x\right) (76)
\displaystyle\leq c1(1t1/2)k=0exp(c2(2k)21t)m=2k2k+11𝐏(Lm=x)\displaystyle c_{1}\left(1\wedge t^{-1/2}\right)\sum_{k=0}^{\infty}\exp\left(-\frac{c_{2}(2^{k})^{2}}{1\vee t}\right)\sum_{m=2^{k}}^{2^{k+1}-1}\mathbf{P}\left(L_{m}=x\right)
\displaystyle\leq c1(1t1/2)k=0exp(c2(2k)21t)(2k)1d/2exp(c3|x|22k)\displaystyle c_{1}\left(1\wedge t^{-1/2}\right)\sum_{k=0}^{\infty}\exp\left(-\frac{c_{2}(2^{k})^{2}}{1\vee t}\right)(2^{k})^{1-d/2}\exp\left(-\frac{c_{3}|x|^{2}}{2^{k}}\right)
\displaystyle\leq c1(1t1/2)m=1md/2exp(c2m21tc3|x|2m)\displaystyle c_{1}\left(1\wedge t^{-1/2}\right)\sum_{m=1}^{\infty}m^{-d/2}\exp\left(-\frac{c_{2}m^{2}}{1\vee t}-\frac{c_{3}|x|^{2}}{m}\right)
\displaystyle\leq c1(1t1/2)1ud/2exp(c2u21tc3|x|2u)𝑑u,\displaystyle c_{1}\left(1\wedge t^{-1/2}\right)\int_{1}^{\infty}u^{-d/2}\exp\left(-\frac{c_{2}u^{2}}{1\vee t}-\frac{c_{3}|x|^{2}}{u}\right)du,

where we have applied Theorem 1.1 for the second inequality. To bound the integral, we first note that, for any δ>0\delta>0, it is possible to find a constant C<C<\infty such that ad/2Ceδaa^{d/2}\leq Ce^{\delta a} for all a0a\geq 0. In particular, choosing δ=c3/2\delta=c_{3}/2, this implies that

ud/2=|x|d(u|x|2)d/2C|x|dexp(c3|x|22u).u^{-d/2}=|x|^{-d}\left(\frac{u}{|x|^{2}}\right)^{-d/2}\leq C|x|^{-d}\exp\left(\frac{c_{3}|x|^{2}}{2u}\right).

Hence, applying this estimate and the change of variable v=u/((1t)|x|2)1/3v=u/((1\vee t)|x|^{2})^{1/3}, we obtain

1ud/2exp(c2u21tc3|x|2u)𝑑u\displaystyle\int_{1}^{\infty}u^{-d/2}\exp\left(-\frac{c_{2}u^{2}}{1\vee t}-\frac{c_{3}|x|^{2}}{u}\right)du (77)
\displaystyle\leq C|x|d1exp(c2u21tc3|x|22u)𝑑u\displaystyle C|x|^{-d}\int_{1}^{\infty}\exp\left(-\frac{c_{2}u^{2}}{1\vee t}-\frac{c_{3}|x|^{2}}{2u}\right)du
\displaystyle\leq C|x|2d(|x|41t)1/30exp((c2v2+c32v)×(|x|41t)1/3)𝑑v.\displaystyle C|x|^{2-d}\left(\frac{|x|^{4}}{1\vee t}\right)^{-1/3}\int_{0}^{\infty}\exp\left(-\left(c_{2}v^{2}+\frac{c_{3}}{2v}\right)\times\left(\frac{|x|^{4}}{1\vee t}\right)^{1/3}\right)dv.

Now, let f(v):=c2v2+c32vf(v):=c_{2}v^{2}+\frac{c_{3}}{2v}, and note that this is a function that has a unique minimum v0v_{0} on (0,)(0,\infty) such that f(v0)>0f(v_{0})>0. Thus, for |x|41t|x|^{4}\geq 1\vee t, the remaining integral above is estimated as follows:

0exp((c2v2+c32v)×(|x|41t)1/3)𝑑v\displaystyle\int_{0}^{\infty}\exp\left(-\left(c_{2}v^{2}+\frac{c_{3}}{2v}\right)\times\left(\frac{|x|^{4}}{1\vee t}\right)^{1/3}\right)dv
\displaystyle\leq 0exp((f(v)f(v0)))𝑑vexp(f(v0)(|x|41t)1/3)\displaystyle\int_{0}^{\infty}\exp\left(-\left(f(v)-f(v_{0})\right)\right)dv\exp\left(-f(v_{0})\left(\frac{|x|^{4}}{1\vee t}\right)^{1/3}\right)
=\displaystyle= Cexp(c(|x|41t)1/3).\displaystyle C\exp\left(-c\left(\frac{|x|^{4}}{1\vee t}\right)^{1/3}\right).

Putting this together with (76) and (77), we deduce the desired result in the range |x|41t|x|^{4}\geq 1\vee t. If |x|4<1t|x|^{4}<1\vee t, then we follow a simpler argument to deduce:

(Xt𝒢=x)\displaystyle\mathbb{P}\left(X_{t}^{\mathcal{G}}=x\right) =\displaystyle= m=1qt(0,m)𝐏(Lm=x)\displaystyle\sum_{m=1}^{\infty}q_{t}(0,m)\mathbf{P}\left(L_{m}=x\right)
\displaystyle\leq c1(1t1/2)m=1𝐏(Lm=x)\displaystyle c_{1}\left(1\wedge t^{-1/2}\right)\sum_{m=1}^{\infty}\mathbf{P}\left(L_{m}=x\right)
=\displaystyle= c1(1t1/2)𝐏(Lm=x for some m0)\displaystyle c_{1}\left(1\wedge t^{-1/2}\right)\mathbf{P}\left(L_{m}=x\mbox{ for some }m\geq 0\right)
\displaystyle\leq c1(1t1/2)𝐏(Sm=x for some m0)\displaystyle c_{1}\left(1\wedge t^{-1/2}\right)\mathbf{P}\left(S_{m}=x\mbox{ for some }m\geq 0\right)
\displaystyle\leq c1(1t1/2)|x|2d,\displaystyle c_{1}\left(1\wedge t^{-1/2}\right)|x|^{2-d},

where we have applied Lemma 4.1 for the first inequality, and (8) for the third. This is enough to establish that the upper bound of Theorem 1.3 holds in this case as well.

For the lower bound when x0x\neq 0, we follow a similar argument to the upper bound, but with additional care about the range of summation/integration. In what follows, we set α=c4/c3\alpha=c_{4}/c_{3}, where, here and for the rest of the proof, c3c_{3}, c4c_{4} are the constants of Theorem 1.1. Clearly, we can assume that c31<c4c_{3}\leq 1<c_{4}, so that α>1\alpha>1. Recall that we are also assuming tε|x|t\geq\varepsilon|x|, and without loss of generality, we may suppose ε(0,1)\varepsilon\in(0,1). Applying the bounds of Lemma 4.1 with ε\varepsilon given by

ε:=min{ε1+α2,αε4/34c4(1+α2)},\varepsilon^{\prime}:=\min\left\{\frac{\varepsilon}{1+\alpha^{2}},\frac{\alpha\varepsilon^{4/3}}{4c_{4}(1+\alpha^{2})}\right\},

we deduce that

(Xt𝒢=x)\displaystyle\mathbb{P}\left(X_{t}^{\mathcal{G}}=x\right) =\displaystyle= m=1qt(0,m)𝐏(Lm=x)\displaystyle\sum_{m=1}^{\infty}q_{t}(0,m)\mathbf{P}\left(L_{m}=x\right)
\displaystyle\geq c(1t1/2)m=1t/εexp(Cm21t)𝐏(Lm=x)\displaystyle c\left(1\wedge t^{-1/2}\right)\sum_{m=1}^{\lfloor t/\varepsilon^{\prime}\rfloor}\exp\left(-\frac{Cm^{2}}{1\vee t}\right)\mathbf{P}\left(L_{m}=x\right)
\displaystyle\geq c(1t1/2)k=0logα(t/ε)1exp(C(αk)21t)m=αkαk+1𝐏(Lm=x),\displaystyle c\left(1\wedge t^{-1/2}\right)\sum_{k=0}^{\lfloor\log_{\alpha}(\lfloor t/\varepsilon^{\prime}\rfloor)\rfloor-1}\exp\left(-\frac{C(\alpha^{k})^{2}}{1\vee t}\right)\sum_{m=\lceil\alpha^{k}\rceil}^{\lfloor\alpha^{k+1}\rfloor}\mathbf{P}\left(L_{m}=x\right),

where for the second inequality, we have applied that

[1,t/ε]k=0logα(t/ε)1[αk,αk+1]\left[1,\lfloor t/\varepsilon^{\prime}\rfloor\right]\supseteq\bigcup_{k=0}^{\lfloor\log_{\alpha}(\lfloor t/\varepsilon^{\prime}\rfloor)\rfloor-1}\left[\alpha^{k},\alpha^{k+1}\right]

and the observation that each mm can appear in at most two of the intervals [αk,αk+1][\lceil\alpha^{k}\rceil,\lfloor\alpha^{k+1}\rfloor]. (We also note that our choice of cc ensures logα(t/ε)11\lfloor\log_{\alpha}(\lfloor t/\varepsilon^{\prime}\rfloor)\rfloor-1\geq 1, and so the sum is non-empty.) Consequently, applying Theorem 1.1 with n=αk/c3n=\alpha^{k}/c_{3}, we find that

(Xt𝒢=x)\displaystyle\mathbb{P}\left(X_{t}^{\mathcal{G}}=x\right) \displaystyle\geq c(1t1/2)k=0logα(c3|x|)logα(t/ε)1exp(C(αk)21t)(αk)1d/2exp(C|x|2αk)\displaystyle c\left(1\wedge t^{-1/2}\right)\sum_{k=0\vee\lceil\log_{\alpha}(c_{3}|x|)\rceil}^{\lfloor\log_{\alpha}(\lfloor t/\varepsilon^{\prime}\rfloor)\rfloor-1}\exp\left(-\frac{C(\alpha^{k})^{2}}{1\vee t}\right)(\alpha^{k})^{1-d/2}\exp\left(-\frac{C|x|^{2}}{\alpha^{k}}\right)
\displaystyle\geq c(1t1/2)m=1αc3|x|α1t/εmd/2exp(Cm21tC|x|2m)\displaystyle c\left(1\wedge t^{-1/2}\right)\sum_{m=1\vee\lceil\alpha c_{3}|x|\rceil}^{\lfloor\alpha^{-1}\lfloor t/\varepsilon^{\prime}\rfloor\rfloor}m^{-d/2}\exp\left(-\frac{Cm^{2}}{1\vee t}-\frac{C|x|^{2}}{m}\right)
\displaystyle\geq c(1t1/2)2c4|x|αt/(1+α2)εud/2exp(Cu21tC|x|2u)𝑑u,\displaystyle c\left(1\wedge t^{-1/2}\right)\int_{2c_{4}|x|}^{\alpha t/(1+\alpha^{2})\varepsilon^{\prime}}u^{-d/2}\exp\left(-\frac{Cu^{2}}{1\vee t}-\frac{C|x|^{2}}{u}\right)du,

where we have used that 1αc3|x|=1c4|x|=c4|x|1\vee\lceil\alpha c_{3}|x|\rceil=1\vee\lceil c_{4}|x|\rceil=\lceil c_{4}|x|\rceil to obtain the bottom limit of the integral, and the choice of ε\varepsilon^{\prime} to obtain the top one. Making the change of variable v=uε4/3/((1t)|x|2)1/3v=u\varepsilon^{4/3}/((1\vee t)|x|^{2})^{1/3} yields a lower bound for the integral of

c(((1t)|x|2)1/3)1d/22c4ε×(ε|x|/t)1/3αε7/3(1+α2)ε×(t/ε|x|)2/3vd/2exp(C(v2+1v)×(|x|41t)1/3)𝑑v,c\left(((1\vee t)|x|^{2})^{1/3}\right)^{1-d/2}\int_{2c_{4}\varepsilon\times\left(\varepsilon|x|/t\right)^{1/3}}^{\frac{\alpha\varepsilon^{7/3}}{(1+\alpha^{2})\varepsilon^{\prime}}\times\left(t/\varepsilon|x|\right)^{2/3}}v^{-d/2}\exp\left(-C\left(v^{2}+\frac{1}{v}\right)\times\left(\frac{|x|^{4}}{1\vee t}\right)^{1/3}\right)dv,

and, since tε|x|t\geq\varepsilon|x|, our choice of ε\varepsilon^{\prime} implies that this is bounded below by

c(((1t)|x|2)1/3)1d/22c4ε4c4εvd/2exp(C(v2+1v)×(|x|41t)1/3)𝑑v\displaystyle c\left(((1\vee t)|x|^{2})^{1/3}\right)^{1-d/2}\int_{2c_{4}\varepsilon}^{4c_{4}\varepsilon}v^{-d/2}\exp\left(-C\left(v^{2}+\frac{1}{v}\right)\times\left(\frac{|x|^{4}}{1\vee t}\right)^{1/3}\right)dv
\displaystyle\geq c((1t)|x|2)1/3d/6exp(C(|x|41t)1/3).\displaystyle c((1\vee t)|x|^{2})^{1/3-d/6}\exp\left(-C\left(\frac{|x|^{4}}{1\vee t}\right)^{1/3}\right).\hskip 140.0pt

Hence, if |x|41t|x|^{4}\geq 1\vee t, we can put the pieces together to find that

(Xt𝒢=x)\displaystyle\mathbb{P}\left(X_{t}^{\mathcal{G}}=x\right) \displaystyle\geq c(1t1/2)((1t)|x|2)1/3d/6(|x|41t)1/3d/6exp(C(|x|41t)1/3)\displaystyle c\left(1\wedge t^{-1/2}\right)((1\vee t)|x|^{2})^{1/3-d/6}\left(\frac{|x|^{4}}{1\vee t}\right)^{1/3-d/6}\exp\left(-C\left(\frac{|x|^{4}}{1\vee t}\right)^{1/3}\right)
=\displaystyle= c(1t1/2)|x|2dexp(C(|x|41t)1/3),\displaystyle c\left(1\wedge t^{-1/2}\right)|x|^{2-d}\exp\left(-C\left(\frac{|x|^{4}}{1\vee t}\right)^{1/3}\right),

as required. Finally, for |x|4<1t|x|^{4}<1\vee t, continuing to suppose that c4c_{4} is the constant of Theorem 1.1, we have that

(Xt𝒢=x)\displaystyle\mathbb{P}\left(X_{t}^{\mathcal{G}}=x\right) \displaystyle\geq m=14c42tqt(0,m)𝐏(Lm=x)\displaystyle\sum_{m=1}^{\lfloor\sqrt{4c_{4}^{2}t}\rfloor}q_{t}(0,m)\mathbf{P}\left(L_{m}=x\right)
\displaystyle\geq c(1t1/2)m=14c42t𝐏(Lm=x)\displaystyle c\left(1\wedge t^{-1/2}\right)\sum_{m=1}^{\lfloor\sqrt{4c_{4}^{2}t}\rfloor}\mathbf{P}\left(L_{m}=x\right)
\displaystyle\geq c(1t1/2)m=c3|x|c4|x|𝐏(Lm=x)\displaystyle c\left(1\wedge t^{-1/2}\right)\sum_{m=\lceil c_{3}\lceil|x|\rceil\rceil}^{\lfloor c_{4}\lceil|x|\rceil\rfloor}\mathbf{P}\left(L_{m}=x\right)
\displaystyle\geq c(1t1/2)|x|2d,\displaystyle c\left(1\wedge t^{-1/2}\right)|x|^{2-d},

where we have applied Lemma 4.1 with ε=1/4c42\varepsilon=1/4c_{4}^{2} for the second inequality, that c4|x|2c4|x|4c42tc_{4}\lceil|x|\rceil\leq 2c_{4}|x|\leq\sqrt{4c_{4}^{2}t} for the third, and Theorem 1.1 for the final one. ∎

Acknowledgements

DC was supported by JSPS Grant-in-Aid for Scientific Research (C) 19K03540 and the Research Institute for Mathematical Sciences, an International Joint Usage/Research Center located in Kyoto University. DS was supported by JSPS Grant-in-Aid for Scientific Research (C) 22K03336, JSPS Grant-in-Aid for Scientific Research (B) 22H01128 and 21H00989. SW is supported by JST, the Establishment of University Fellowships Towards the Creation of Science Technology Innovation, Grant Number JPMJFS2123.

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