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arXiv:1012.4711v1 [math.PR] 21 Dec 2010

Connectivity properties of random interlacement and

intersection of random walks

Bal´ azs R´ ath∗

Art¨ em Sapozhnikov∗

December 2010

Abstract

We consider the interlacement Poisson point process on the space of doubly-infinite Zd-valued

trajectories modulo time shift, tending to infinity at positive and negative infinite times. The set of

vertices and edges visited by at least one of these trajectories is the random interlacement at level u

of Sznitman [12]. We prove that for any u > 0, almost surely, (1) any two vertices in the random

interlacement at level u are connected via at most ⌈d/2⌉ trajectories of the point process, and (2) there

are vertices in the random interlacement at level u which can only be connected via at least ⌈d/2⌉

trajectories of the point process. In particular, this implies the already known result of [12] that the

random interlacement at level u is connected.

1Introduction

The model of random interlacements was recently introduced by Sznitman in [12] in order to describe

the local picture left by the trajectory of a random walk on the discrete torus (Z/NZ)d, d ≥ 3 when

it runs up to times of order Nd, or on the discrete cylinder (Z/NZ)d× Z , d ≥ 2, when it runs up to

times of order N2d, see [11], [14]. Informally, the random interlacement Poisson point process consists of

a countable collection of doubly infinite trajectories on Zd, and the trace left by these trajectories on a

finite subset of Zd“looks like” the trace of the above mentioned random walks.

So far, research related to random interlacements mainly focused on the description of the connectivity

properties of the vacant set (which corresponds to the set of vertices not visited by the random walker).

In this paper we investigate connectivity properties of the random interlacement, giving a detailed picture

about how the collection of doubly infinite trajectories are actually interlaced. Our methods are further

developed in [9] to study properties of percolation and random walks on the random interlacement.

1.1The model

Let W be the space of doubly-infinite nearest-neighbor trajectories in Zd(d ≥ 3) which tend to infinity

at positive and negative infinite times, and let W∗be the space of equivalence classes of trajectories in

W modulo time shift. We write W for the canonical σ-algebra on W generated by the coordinates Xn,

n ∈ Z, and W∗for the largest σ-algebra on W∗for which the canonical map π∗from (W,W) to (W∗,W∗)

is measurable. Let u be a positive number. We say that a Poisson point measure µ on W∗has distribution

∗ETH Z¨ urich, Department of Mathematics, R¨ amistrasse 101, 8092 Z¨ urich.

artem.sapozhnikov@math.ethz.ch. The research of both authors has been supported by the grant ERC-2009-AdG 245728-

RWPERCRI.

0MSC2000: Primary 60K35, 82B43.

0Keywords: Random interlacement; random walk; intersection of random walks; capacity; Wiener test.

Email: balazs.rath@math.ethz.ch and

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Pois(u,W∗) if the following properties hold: For a finite subset A of Zd, let µAbe the restriction of µ to

the set of trajectories from W∗that intersect A, and let NAbe the number of trajectories in Supp(µA).

Then µA=?NA

(1) The random variable NAhas Poisson distribution with parameter ucap(A) (see (2.2) for the defini-

tion of the cap(A)).

i=1δπ∗(Xi), where Xiare doubly-infinite trajectories from W parametrized in such a way

that Xi(0) ∈ A and Xi(t) / ∈ A for all t < 0 and for all i ∈ {1,...,NA}, and

(2) Given NA, the points Xi(0), i ∈ {1,...,NA}, are independent and distributed according to the

normalized equilibrium measure on A (see (2.7) for the definition).

(3) Given NAand (Xi(0))NA

pendent, (Xi(t),t ≥ 0)NA

are distributed as independent random walks conditioned on not hitting A.

i=1, the corresponding forward and backward paths are conditionally inde-

i=1are distributed as independent simple random walks, and (Xi(t),t ≤ 0)NA

i=1

Properties (1)-(3) uniquely define Pois(u,W∗) as proved in Theorem 1.1 in [12]. In fact, Theorem 1.1

in [12] gives a coupling of the Poisson point measures µ(u) with distribution Pois(u,W∗) for all u > 0,

but we will not need such a general statement here. We also mention a couple of properties of the

distribution Pois(u,W∗), which will be useful in the proofs. Property (4) follows from the above definition

of Pois(u,W∗), and (5) is a property of Poisson point measures.

(4) Let µ1and µ2be independent Poisson point measures on W∗with distributions Pois(u1,W∗) and

Pois(u2,W∗), respectively. Then µ1+ µ2has distribution Pois(u1+ u2,W∗).

(5) Let S1,...,Skbe disjoint elements of W∗. We denote by I(Si)µ the restriction of µ to the set of

trajectories from Si. Then I(S1)µ,...,I(Sk)µ are independent Poisson point measures on W∗.

We refer the reader to [12] for more details. For a Poisson point measure µ with distribution Pois(u,W∗),

the random interlacement I at level u is defined as

?

I = I(µ) =

w∈Supp(µ)

range(w).(1.1)

1.2The result

We consider a random point measure µ on W∗distributed as Pois(u,W∗). We denote by P the law of

µ. Our main result concerns the geometric properties of the support of µ. Remember that the support of

µ consists of a countable set of doubly-infinite random walk trajectories modulo time shift. We construct

the random graph G = (V,E) as follows. The set of vertices V is the set of trajectories from Supp(µ), and

the set of edges E is the set of pairs of different trajectories from Supp(µ) that intersect. Let diam(G)

be the diameter of G. Our main result is the following theorem.

Theorem 1. For d ≥ 3, let

sd= ⌈(d − 2)/2⌉,(1.2)

where ⌈a⌉ is the smallest integer not less than a. Then

P(diam(G) = sd) = 1,

In particular, we get an alternative proof of (2.21) in [12], which states that the random interlacement I

is a connected subgraph of Zd.

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Remark 1. In dimensions 3 and 4, the result is a trivial consequence of Theorem 2.6 in [4] (see also

remark at the bottom of page 661 in [4]) which states that two independent random walks in dimension

3 or 4 intersect infinitely often with probability 1. Therefore, it remains to prove the theorem for d ≥ 5.

The structure of the proof of Theorem 1 can be non-rigorously summarized as follows: first we pick

one of the doubly infinite trajectories from Supp(µ). Denote by A(1)the set of vertices of Zdvisited by

this trajectory. The second layer A(2)consists of the vertices visited by those trajectories of Supp(µ) that

intersect A(1), and recursively let A(s)denote the set of vertices visited by the trajectories that intersect

A(s−1). We prove that P(diam(G) = sd) = 1 by showing that, almost surely, A(sd)?= I and A(sd+1)= I.

Let us recall the following well-known fact (see, e.g., Proposition 2.3 in [4]): For d ≥ 3, the probability

that a simple random walk from 0 hits x is comparable with min(1,|x|2−d). We will use this fact and the

following elementary lemma to show that A(sd)?= I.

Lemma 1. There exists a finite constant C = C(d) such that for any positive integer n and for any

z0,zn+1∈ Zd,

?

z1,...,zn∈Zd

n

?

i=0

min

?

1,|zi− zi+1|2−d?

?

≤ C|z0− zn+1|2n+2−d

= ∞

if n < sd,

otherwise.

(See, e.g. (1.38) of Proposition 1.7 in [2] for a proof of Lemma 1.) Lemma 1 gives bounds on n-fold

convolutions of the probability that a random walk from z0ever visits zn+1. We will see that P(0,x ∈ A(s))

can be estimated as a (s−1)-fold convolution of such hitting probabilities, and, therefore, we will conclude

from Lemma 1 that P(0,x ∈ A(s)) ≤ C|x|2s−d. In particular, P(0,x ∈ A(sd)) → 0 as |x| → ∞. This

contradicts A(sd)= I, since I has positive density.

In order to show that A(sd+1)= I, we argue as follows. Heuristically, A(s)is a 2s-dimensional object

as long as 2s < d. The capacity of A(s)intersected with a ball of radius R (see (2.2) for the definition

of the capacity) is comparable to R2sas long as 2s ≤ d − 2. The set A(sd)already saturates the ball in

terms of capacity, thus it is visible for an independent random walk started somewhere inside the ball of

radius R. We apply a variant of Wiener’s test (see, e.g., Proposition 2.4 in [4]) to show that any random

walk hits A(sd)almost surely.

This is the general strategy of the proof. Instead of following it directly, we benefit from property (4) of

Pois(u,W∗) by decomposing µ into a sum of sdi.i.d point measures µ(s)with distribution Pois(u/sd,W∗)

and constructing each A(s)from the “new” measure µ(s).

The paper is organized as follows. In Section 2 we collect most of the notation and facts used in the

paper. The most important of those are the definitions and properties of the Green function and the

capacity. We prove the lower bound of Theorem 1 in Section 3, and the upper bound in Section 4. The

structure of the proof of the upper bound of Theorem 1 is given at the beginning of Section 4.

2Notation and facts about Green function and capacity

In this section we collect most of the notation, definitions and facts used in the paper. For a ∈ R, we

write |a| for the absolute value of a, ⌊a⌋ for the integer part of a, and ⌈a⌉ for the smallest integer not less

than a. For x ∈ Zd, we write |x| for max(|x1|,...,|xd|). For a set S, we write |S| for the cardinality of

S. For R > 0 and x ∈ Zd, let B(x,R) = {y ∈ Zd: |x − y| ≤ R} be the ball of radius R centered at

x. We denote by I(A) the indicator of event A, and by E[X;A] the expected value of random variable

XI(A). Throughout the text, we write c and C for small positive and large finite constants, respectively,

that may depend on d and u. Their values may change from place to place.

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For x ∈ Zd, let Pxbe the law of a simple random walk X on Zdwith X(0) = x. We write g(·,·) for

the Green function of the walk:

g(x,y) =

∞

?

t=0

Px(X(t) = y), x,y ∈ Zd.

We also write g(·) for g(0,·). The Green function is symmetric and, by translation invariance, g(x,y) =

g(y − x). It follows from [5, Theorem 1.5.4] that for any d ≥ 3 there exist a positive constant cg= cg(d)

and a finite constant Cg= Cg(d) such that for all x and y in Zd,

?

Definition 2.1. Let K be a subset of Zd. The energy of a finite Borel measure ν on K is

?

cgmin1,|x − y|2−d?

≤ g(x,y) ≤ Cgmin

?

1,|x − y|2−d?

. (2.1)

E(ν) =

K

?

K

g(x,y)dν(x)dν(y) =

?

x,y∈K

g(x,y)ν(x)ν(y).

The capacity of K is

cap(K) =

?

inf

νE(ν)

?−1, (2.2)

where the infimum is over probability measures ν on K. (We assume that ∞−1= 0, i.e. the capacity of

the empty set is 0.)

The following properties of the capacity immediately follow from (2.2):

Monotonicity: for any K1⊂ K2⊂ Zd, cap(K1) ≤ cap(K2);

for any K1,K2⊂ Zd, cap(K1∪ K2) ≤ cap(K1) + cap(K2);

for any x ∈ Zd, cap({x}) = 1/g(0).

(2.3)

Subadditivity: (2.4)

Capacity of a point:(2.5)

It will be useful to have an alternative definition of the capacity in d ≥ 3.

Definition 2.2. Let K be a finite subset of Zd. The equilibrium measure of K is defined by

eK(x) = Px(X(t) / ∈ K for all t ≥ 1)I(x ∈ K), x ∈ Zd.(2.6)

The capacity of K is then equal to the total mass of the equilibrium measure of K:

cap(K) =

?

x

eK(x),

and the unique minimizer of the variational problem (2.2) is given by the normalized equilibrium measure

? eK(x) = eK(x)/cap(K).(2.7)

(See, e.g., Lemma 2.3 in [3] for a proof of this fact.)

As a simple corollary of the above definition, we get for d ≥ 3,

Px(H(K) < ∞) =

?

y∈K

g(x,y)eK(y), for x ∈ Zd.(2.8)

Here, we write H(K) for the first entrance time in K, i.e. H(K) = inf{t ≥ 0 : X(t) ∈ K}. We will

repeatedly use the following bound on the capacity of B(0,R) in d ≥ 3 (see (2.16) on page 53 in [5]):

There exist constants cb= cb(d) > 0 and Cb= Cb(d) < ∞ such that for all positive R,

cbRd−2≤ cap(B(0,R)) ≤ CbRd−2.(2.9)

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3Proof of Theorem 1: lower bound on the diameter

Remember the definition of sdin (1.2). In this section we prove that P(diam(G) ≥ sd) = 1. Since,

almost surely, diam(G) ≥ 1, we only need to consider the case d ≥ 5. For two trajectories v and w in

V , we write ρ(v,w) for the distance between v and w in G. In order to prove that the probability of the

event {diam(G) ≥ sd} is 1, we assume by contradiction that this probability is ≤ 1−δ, for some positive

δ. In other words, the probability of event

E = {ρ(v,w) ≤ sd− 1 for all v,w ∈ V }

is bounded from below by δ.

For x,y ∈ Zd, we denote by S(x,y) the subset of doubly-infinite trajectories in W∗that intersect both

vertices x and y. Remember the definition (1.1) of the random interlacement I. The next lemma gives

an estimate on the probability that E occurs and two different vertices x and y of Zdare in I:

Lemma 2. For any x,y ∈ Zd,

P({x,y ∈ I} ∩ E) ≤

sd−1

?

n=1

?

z1,...,zn∈Zd

n

?

i=0

E[µ(S(zi,zi+1))], (3.1)

where we take z0= x and zn+1= y.

We postpone the proof of Lemma 2 until the end of this section.

E[µ(S(zi,zi+1))] in (3.1) is bounded from above by 2ug(zi,zi+1). (This follows, for example, from (1.33)

in [13] applied to K = {zi} and K′= {zi+1}.) Therefore, we obtain

Each of the expectations

P({x,y ∈ I} ∩ E) ≤

sd−1

?

n=1

(2u)n+1

?

z1,...,zn∈Zd

n

?

i=0

g(zi,zi+1),

where we again assume z0 = x and zn+1 = y. Recall from (2.1) that g(x,y) ≤ Cgmin(1,|x − y|2−d).

Therefore, by Lemma 1,

sd−1

?

n=1

?

z1,...,zn∈Zd

n

?

i=0

g(zi,zi+1) ≤ C|z0− zn+1|2sd−d≤ C|z0− zn+1|−1.

In particular, P({x,y ∈ I} ∩ E) ≤ C|x − y|−1→ 0, as |x − y| → ∞. By property (1) of Pois(u,W∗), for

any R > 0,

P(I ∩ B(0,R) ?= ∅) = P?NB(0,R)≥ 1?= 1 − e−ucap(B(0,R)).

By (2.9), we can take R big enough so that

P(I ∩ B(0,R) ?= ∅) ≥ 1 −δ

3.

With this choice of R, for any z ∈ Zd, we obtain

P({I ∩ B(0,R) ?= ∅} ∩ {I ∩ B(z,R) ?= ∅} ∩ E) ≥ P(E) − 2P(I ∩ B(0,R) = ∅) ≥ δ/3.

On the other hand, for z ∈ Zdwith |z| > 3R,

P({I ∩ B(0,R) ?= ∅} ∩ {I ∩ B(z,R) ?= ∅} ∩ E) ≤

?

x∈B(0,R)

?

y∈B(z,R)

P({x,y ∈ I} ∩ E) ≤ CR2d|z|−1,

which tends to 0 as |z| tends to infinity. This is a contradiction, and we conclude that P-a.s. the diameter

of G is at least sd.

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Proof of Lemma 2. One can deduce the result almost immediately from the Palm theory for general

Poisson point processes (see, e.g. Chapter 13.1 in [1]). Remember the definition of the set S(x,y) given

before the statement of Lemma 2. Let D(x,y) be the event that S(x,y) ∩ Supp(µ) ?= ∅.

words, D(x,y) = {µ(S(x,y)) ?= 0}. For x,y,x′,y′∈ Zd, we write D(x,y) ◦ D(x′,y′) for the event that

there exist different trajectories w and w′in Supp(µ) such that w ∈ S(x,y) and w′∈ S(x′,y′). Let

?∗be the sum over all (n + 1)-tuples of pairwise different doubly-infinite trajectories modulo time-shift

sd−1

?

sd−1

?

where we take z0= x and zn+1= y. The result then follows from the Slivnyak-Mecke theorem (See, e.g.

Theorem 3.3 in [6], where it is proved for point processes in Rd, and Chapter 13.1 in [1] for the theory of

Palm distributions in general spaces.):

??∗

In other

w0,...,wn∈ Supp(µ). We have

P({x,y ∈ I} ∩ E)

≤

n=1

?

?

z1,...,zn∈Zd

P(D(z0,z1) ◦ ... ◦ D(zn,zn+1))

≤

n=1

z1,...,zn∈Zd

E

??∗

n

?

i=0

I(wi∈ S(zi,zi+1))

?

,

E

n

?

i=0

I(wi∈ S(zi,zi+1))

?

=

n

?

i=0

E[µ(S(zi,zi+1))].

4Proof of Theorem 1: upper bound on the diameter

The proof of the upper bound on the diameter of G in Theorem 1 is organized as follows. Section 4.1

contains preliminary lemmas. Lemma 4 gives some bounds on the expected capacity of a certain family

of traces of random walks. Lemma 5 provides bounds on the expected capacity of a set of vertices visited

by trajectories from Supp(µ) that intersect a given set of vertices. Both lemmas state that the capacity

of such sets of vertices is either comparable with the volume of the set (when trajectories are “well

spread-out”) or with the capacity of the ball that contains the set (when the set is “dense” in the ball).

In Lemma 6, we show that the exclusion of a (small) number of trajectories from Supp(µ) that visit a

certain ball does not decrease too much the capacity of sets in Lemma 5. This step is needed to benefit

from property (5) of Pois(u,W∗) and create some additional independence.

In Section 4.2 we use these bounds on the capacity to construct certain subsets of Supp(µ) (see (4.10)

and (4.11)) that are visible by an independent random walk started near the origin.

In Section 4.3 we construct a sequence of almost independent visible subsets of Supp(µ) and use ideas

similar in spirit to Wiener’s test to show that, almost surely, infinitely many of these sets are visited by

an independent random walk. This is done in Lemma 10.

We finish Section 4.3 by completing the proof of Theorem 1.

4.1 Bounds on the capacity of certain collection of random walk trajectories

Lemma 3. Let d ≥ 5. Let (xi)i≥1be a sequence in Zd, and let Xibe a sequence of independent simple

random walks on Zdwith Xi(0) = xi. Then for all positive integers N and n, we have

i,j=1

s,t=n+1

E

N

?

2n

?

g (Xi(s),Xj(t))

≤ C

?

Nn + N2n3−d/2?

.(4.1)

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Proof. Let X be a simple random walk with X(0) = 0, then for all y ∈ Zdand for all positive integers s,

Eg(X(s),y) ≤ Cs1−d/2.(4.2)

Indeed, by the Markov property,

Eg (X(s),y) =

∞

?

t=s

P(X(t) = y) ≤ C

∞

?

t=s

t−d/2≤ Cs1−d/2.

Here we used the fact that [10, Proposition 7.6]

sup

y∈ZdP(X(t) = y) ≤ Ct−d/2.

In order to prove (4.1), we consider separately the cases i = j and i ?= j. In the first case, the Markov

property and the fact that g(x,y) = g(x − y) imply

i=1

s,t=n+1

E

N

?

2n

?

g(Xi(s),Xi(t))

=NE

2n

?

s,t=n+1

?

g (X(|s − t|))

(4.2)

≤

CNn1 +

n

?

s=1

s1−d/2

?

(d≥5)

≤ CNn.

In the case i ?= j, an application of (4.2) gives

E

2n

?

s,t=n+1

g (Xi(s),Xj(t))

≤ n2Cn1−d/2.

This completes the proof.

Let (Xi(t) : t ≥ 0)i≥1be a sequence of nearest-neighbor trajectories on Zd, and XN= (X1,...,XN).

For positive integers N and R, we define the subset Φ(XN,R) of Zdby

Φ(XN,R) =

N?

i=1

??Xi(t) : 1 ≤ t ≤ R2/2?∩ B(Xi(0),R)?. (4.3)

Lemma 4. Let Xi be a sequence of independent simple random walks on Zdwith Xi(0) = xi. There

exists a positive constant c such that for any sequence (xi)i≥1⊂ Zdand for all positive integers N and R,

cap?Φ(XN,R)?≤NR2

Ecap?Φ(XN,R)?≥ cmin

2g(0),(4.4)

and for d ≥ 5,

?

NR2,Rd−2?

.(4.5)

Proof. The upper bound on the capacity of Φ(XN,R) follows from properties (2.4) and (2.5), and the

fact that the number of vertices in Φ(XN,R) is at most NR2/2.

We proceed with the lower bound on Ecap?Φ(XN,R)?. The following inequality follows from Kol-

?

mogorov’s maximal inequality applied coordinatewise: For each λ > 0 and n ≥ 1,

P

max

1≤t≤n|X(t)| ≥ λ

?

≤n

λ2.(4.6)

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Take positive integers N and R, random walks X1,...,XNwith Xi(0) = xi, and set

n = ⌊R2/4⌋.(4.7)

We define the random subset J of {1,...N} by

J = {i :sup

1≤t≤2n|Xi(t) − xi| ≤ R}.

We also consider the event A = {|J| ≥ N/4}. It follows from (4.6) that

E|J| ≥ N

?

1 −2n

R2

?

≥N

2.

Since |J| ≤ N, we get P(A) ≥1

By the definition (2.2) of the capacity of Φ(XN,R), we have

Ecap?Φ(XN,R)?≥ E?E(ν)−1?≥ E?E(ν)−1;A?,

where ν stands for the probability measure

3.

ν(x) =

1

|J|n

?

i∈J

2n

?

t=n+1

I(Xi(t) = x), x ∈ Zd.

The energy of ν equals

E(ν) =

1

|J|2n2

?

i,j∈J

2n

?

s,t=n+1

g(Xi(s),Xj(t)).

Therefore, in order to prove the lower bound on Ecap?Φ(XN,R)?, it suffices to show that

|J|2n2

i,j∈J

s,t=n+1

E

1

?

2n

?

g(Xi(s),Xj(t))

−1

;A

≥ cmin

?

NR2,Rd−2?

.

By the Cauchy-Schwarz inequality and the definition of the event A, we get

E

1

|J|2n2

?

i,j∈J

2n

?

s,t=n+1

g(Xi(s),Xj(t))

−1

;A

≥ (N/4)2n2P(A)2

E

?

i,j∈J

2n

?

s,t=n+1

g(Xi(s),Xj(t));A

−1

.

Since J is a subset of {1,...,N}, the right-hand side is bounded from below by

(N/4)2n2P(A)2

E

N

?

i,j=1

2n

?

s,t=n+1

g(Xi(s),Xj(t))

−1

(4.1)

≥

N2n2

144C(Nn + N2n3−d/2)

(4.7)

≥

cmin

?

NR2,Rd−2?

.

This completes the proof.

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Let A be a finite set of vertices in Zd. For a point measure ω =?

distribution Pois(u,W∗), we have NA= NA(µ).) Let X1,...,XNA(ω)be these trajectories parametrized

in such a way that Xi(0) ∈ A and Xi(t) / ∈ A for all t < 0 and for all i ∈ {1,...,NA(ω)}. We write XA(ω)

for (X1,...,XNA(ω)). We also define Ψ(ω,A,R) as Φ(XA(ω),R), i.e.,

i≥0δwiwith wi∈ W∗, we denote by

NA(ω) the number of trajectories from Supp(ω) that intersect A. (In particular, for a point measure µ with

Ψ(ω,A,R)

(4.3)

= Φ(XA(ω),R) =

NA(ω)

?

i=1

??Xi(t) : 1 ≤ t ≤ R2/2?∩ B(Xi(0),R)?.(4.8)

Lemma 5. Let d ≥ 5. Let µ be a Poisson point measure with distribution Pois(u,W∗), then for all finite

subsets A of Zdand for all positive R, one has

Ecap(Ψ(µ,A,R)) ≥ c min

?

ucap(A)R2,Rd−2?

.

Proof. Let λ = ucap(A). Properties (2) and (3) of Pois(u,W∗) and Lemma 4 imply that

Ecap(Ψ(µ,A,R)) ≥ c Emin

?

A] = λ2+ λ, and P(NA = 0) = exp(−λ). If

NAR2,Rd−2?

.

Property (1) of Pois(u,W∗) implies that ENA = λ, E[N2

λ ≤ 1/2, we estimate

?

If λ ≥ 1/2, we write

EminNAR2,Rd−2?

≥ R2P(NA≥ 1) = R2(1 − e−λ) ≥ R2λ/2.

Emin

?

NAR2,Rd−2?

≥ min

?R2λ

2

,Rd−2

?

P

?

NA≥λ

2

?

.

Remember the Paley-Zygmund inequality [8]: Let ξ be a non-negative random variable with finite second

moment. For any θ ∈ (0,1),

P(ξ ≥ θEξ) ≥ (1 − θ)2[Eξ]2

E[ξ2].(4.9)

An application of (4.9) to NAgives

P

?

NA≥λ

2

?

≥1

4

λ2

λ2+ λ≥

1

12.

This completes the proof.

Definition 4.1. For positive integers r and R with r < R, and a point measure ω =?

restriction of ω to the set of trajectories that do not intersect B(r), and ωr,Rfor the restriction of ω to

the set of trajectories that intersect B(R) but do not intersect B(r). By property (5) of Pois(u,W∗), the

measures µrand µr,Rare independent for any r > 0 and R ∈ (r,∞]. Moreover, for any r > 0, we have

µ = µr+ µr,∞.

i≥0δwiwith

wi∈ W∗, we write ωrfor the restriction of ω to the set of trajectories that intersect B(r), ωr,∞for the

Lemma 6. Let d ≥ 5. For all finite subsets A of Zdand for all positive integers r and R with r < R,

?

Ecap(Ψ(µr,∞,A,R)) ≥ c minucap(A)R2,Rd−2?

− Curd−2R2.

9

Page 10

Proof. By the subadditivity of the capacity and the fact that µ = µr+ µr,∞,

Ecap(Ψ(µr,∞,A,R)) ≥ Ecap(Ψ(µ,A,R)) − Ecap(Ψ(µr,A,R)).

We use Lemma 5 to bound Ecap(Ψ(µ,A,R)) from below. As for an upper bound on Ecap(Ψ(µr,A,R)),

note that |Supp(µr)| = µr(W∗) = NB(r)(µr) = NB(r). Therefore, by Lemma 4,

Ecap(Ψ(µr,A,R)) ≤R2ENB(r)

2g(0)

=R2ucap(B(r))

2g(0)

(2.9)

≤ Curd−2R2.

4.2Construction of visible sets

Let X be a simple random walk on Zdwith X(0) = x. We denote the corresponding probability

measure and the expectation by Pxand Ex, respectively. Let µ(2),µ(3),... be independent random point

measures with distribution Pois(u,W∗) (The parameter u is fixed here.), which are also independent of X.

The corresponding probability measures and expectations are denoted by P(2),P(3),... and E(2),E(3),...,

respectively. For s ≥ 1, we write P(s)

x

for Px⊗ P(2)⊗ ... ⊗ P(s).

Let r and R be positive integers with r < R and |x| < R. Let TB(R)be the first exit time of X from

B(R), i.e., TB(R)= inf{t ≥ 0 : X(t) / ∈ B(R)}. We denote by Y the random walk X(TB(R)+ ·). We

define the following sequence of random subsets of Zd:

A(1)(r,R) = A(1)(R) = Φ(Y,R)

(4.3)

=

?Y (t) : 1 ≤ t ≤ R2/2?∩ B(Y (0),R), (4.10)

and for s ≥ 2 (see (4.8) for notation),

A(s)(r,R) = Ψ

?

µ(s)

r,∞,A(s−1)(r,R),R

?

= Ψ

?

µ(s)

r,sR,A(s−1)(r,R),R

?

,(4.11)

where the last equality follows from the fact that A(s−1)(r,R) is a subset of B(sR) by construction.

Remark 2. Note that for each y ∈ A(s)(r,R), there exist doubly-infinite trajectories wi∈ Supp(µ(i)

2 ≤ i ≤ s, such that (1) the vertex y is visited by ws, (2) the random walk X intersects w2, and (3) for

all i ∈ {2,...,s − 1}, the trajectories wiand wi+1intersect.

r,∞),

Lemma 7. Let s be a positive integer. There exist finite constants Cs = C(u,d,s) such that for all

positive integers r and R with r < R and for all x ∈ B(R),

?

and

E(s)

x

capA(s)(r,R)

E(s)

xcapA(s)(r,R)

?

≤ CsRmin(d−2,2s), (4.12)

?

??2?

≤ CsR2min(d−2,2s).(4.13)

Proof. We fix r and R throughout the proof, and we write A(s)for A(s)(r,R). Since A(s)is a subset of

B((s + 1)R), the monotonicity of the capacity implies that

cap

?

A(s)?

≤ cap(B((s + 1)R))

(2.9)

≤ CsRd−2.

10

Page 11

Therefore, it suffices to show that the first and the second moments of cap?A(s)?are bounded from above

?

(The last inequality follows from the monotonicity of the capacity.) Remember that NA(s−1)(µ(s)) is a

Poisson random variable with parameter ucap?A(s−1)?, therefore, we have

E(s)

xcap

≤

2g(0)E(s−1)

The bound on the first moment of cap?A(s)?follows by induction. The bound on the second moment of

?

and

?

=

E(s−1)

x

u2cap

by CsR2sand CsR4s, respectively. It follows from (4.4) that

E(s)

xcapA(s)?

≤

R2

2g(0)E(s)

xNA(s−1)(µ(s)

r,∞) ≤

R2

2g(0)E(s)

xNA(s−1)(µ(s)).

?

A(s)?

R2

x

ucap

?

A(s−1)?

.

cap?A(s)?is also obtained using (4.4). In a similar fashion as above, we obtain the relations:

E(s)

x

cap

≤

?

A(s)?2?

R4

4g(0)2E(s)

x

?

NA(s−1)(µ(s)

r,∞)2?

,

E(s)

x

NA(s−1)(µ(s)

r,∞)2?

≤

E(s)

x

?

NA(s−1)(µ(s))2?

?

?

A(s−1)?2?

+ E(s−1)

x

ucap

?

A(s−1)?

.

The bound on the second moment of cap?A(s)?follows from these inequalities and from the first statement

Lemma 8. Let d ≥ 5. Let s be a positive integer. There exist positive constants cs = c(u,d,s) and

ε = ε(u,d,s) such that for all positive integers r and R with

of the lemma.

rd−2≤ εR(4.14)

and for all x ∈ B(R),

E(s)

xcap

?

A(s)(r,R)

?

≥ csRmin(d−2,2s). (4.15)

Remark 3. Remember that A(s)(r,R) is constructed as a subset of a (random) number of pieces of

random walk trajectories of lengths ⌊R2/2⌋. The expected capacity of a single random walk is comparable

with its length in dimension ≥ 5, as shown in Lemma 4. Note that min(d−2,2s) is 2s for s < ⌈(d−2)/2⌉

and d−2 for s ≥ ⌈(d−2)/2⌉. One can interpret the results of Lemma 8 as follows. If s ≤ ⌈(d−2)/2⌉, the

random walk pieces that form A(s)(r,R) are well spread-out, so that the expected capacity of A(s)(r,R)

is comparable with its volume. On the other hand, if s ≥ ⌈(d − 2)/2⌉, the set A(s)(r,R) saturates the

ball B((s+1)R) and its expected capacity becomes comparable with the capacity of the ball, which is of

order Rd−2by (2.9).

Proof. We prove (4.15) by induction on s.

It follows from (4.5) that

Excap

?

A(1)(R)

?

≥ c1R2.

Let s ≥ 2, and assume that the induction hypothesis holds:

E(s−1)

x

cap

?

A(s−1)(r,R)

?

≥ cs−1Rmin(d−2,2s−2).

11

Page 12

With this lower bound on the expected value of cap?A(s−1)(r,R)?

that

P(s−1)

x

and the corresponding upper bound

(4.13), the Paley-Zygmund inequality (4.9) yields that there exists a positive constant c = c(u,d,s) such

?

Lemma 6 implies that

?

(4.16)

≥

c min

cap

?

A(s−1)(r,R)

?

≥ cRmin(d−2,2s−2)?

≥ c.(4.16)

E(s)

xcapA(s)(r,R)

?

≥

c E(s−1)

x

min

?

ucap(A(s−1)(r,R))R2,Rd−2?

Rmin(d−2,2s−2)R2,Rd−2?

c Rmin(d−2,2s)− Curd−2R2

− Curd−2R2

?

− Curd−2R2

=

(4.14)

≥

(c/2) Rmin(d−2,2s).

(The last inequality holds if ε in (4.14) is taken small enough, since we only consider d ≥ 5 and s ≥ 2.)

In the next lemma we study the probability that a simple random walk hits A(s)(r,R). Remember

the definitions of X and µ(s), s ≥ 2 at the beginning of Section 4.2, and sdin (1.2).

Lemma 9. Let d ≥ 5. Let Z be a simple random walk on Zdwith Z(0) = z, which is independent of X

and µ(s), s ≥ 2, with law Pz. There exist positive constants c = c(u,d), and ε = ε(u,d) > 0 such that, for

all positive integers r and R with rd−2≤ εR, x ∈ B(R), and z ∈ B(R), we have

?

where H(A(s)(r,R)) is the entrance time of Z in A(s)(r,R) and TB(R2)the exit time of Z from B(R2).

Pz⊗ P(sd)

x

H(A(sd)(r,R)) < TB(R2)

?

≥ c,

Proof. We write A for A(sd)(r,R) throughout the proof. We use the identity (2.8):

Pz(H(A) < ∞) =

?

y∈A

g(z,y)eA(y),

where eAis the equilibrium measure of A (see (2.6)). We have

Pz⊗ P(sd)

x

(H(A) < ∞) = E(sd)

x

?

y∈A

g(z,y)eA(y)

.

Note that A is a subset of B((sd+ 1)R) ⊂ B(dR) by construction. Therefore, inequality (2.1) implies

that, for any y ∈ A and z ∈ B(R), g(z,y) ≥ cg(2dR)2−d. Also remember that?

Pz⊗ P(sd)

x

(H(A) < ∞) ≥ cg(2dR)2−dE(sd)

y∈AeA(y) = cap(A).

These observations give

x

[cap(A)].

It follows from the previous lemma that, for d ≥ 5, we can choose ε > 0 so that

E(sd)

x

[cap(A)] ≥ cRmin(d−2,2sd)= cRd−2.

Therefore,

Pz⊗ P(sd)

x

(H(A) < ∞) ≥ c.

12

Page 13

On the other hand, by the strong Markov property of the random walk Z,

?TB(R2)< H(A) < ∞?

Pz⊗ P(sd)

x

≤

sup

z′/ ∈B(R2)

sup

z′/ ∈B(R2)

Pz′ ⊗ P(sd)

x

(H(A) < ∞)

≤

Pz′ (H(B(dR)) < ∞).

In the second inequality we use the fact that A is a subset of B(dR). We bound the right-hand side, using

(2.1), (2.8) and (2.9):

sup

z′/ ∈B(R2)

Pz′ (H(B(dR)) < ∞) ≤ Cg(R2− dR)2−dcap(B(dR)) ≤ CR2−d.

Remember that R ≥ rd−2/ε ≥ 1/ε. Therefore, by taking ε small enough, we get

sup

z′/ ∈B(R2)

Pz′ (H(B(dR)) < ∞) ≤1

2Pz⊗ P(s)

x

?

H(A(s)) < ∞

?

.

The result follows.

4.3 Construction of recurrent sets

We will now use the result of Lemma 9 to construct a sequence of subsets A(sd)(rk,Rk) of Zdsuch

that the union of these sets ∪kA(sd)(rk,Rk) is hit by an independent random walk (infinitely often) with

probability 1. Remember the definitions of X and µ(s), s ≥ 2 at the beginning of Section 4.2.

Lemma 10. Let d ≥ 5. For z ∈ Zd, let Z be a simple random walk on Zdwith Z(0) = z, which is

independent of X and µ(s), s ≥ 2. Let Pz be its law. Let X(0) = x. There exist sequences of positive

integers rkand Rksuch that

?

where H(A(s)(r,R)) is the entrance time of Z in A(s)(r,R).

Pz⊗ P(sd)

x

limsup

k

?

H(A(sd)(rk,Rk)) < ∞

??

= 1,

Proof. Let ε be the positive number from Lemma 9. We define rkand Rkrecursively:

r0= max(|x|,|z|), R0= ⌈ε−1rd−2

0

⌉,

and, for k ≥ 1,

rk= dR2

k−1, Rk= ⌈ε−1rd−2

k

⌉.

(Any sequences that grow faster than these would do.) We consider the following sequence of (random)

subsets of Zd:

Ak= A(sd)(rk,Rk) (⊂ B(dRk)).

Note that the following properties hold:

(i) the set of vertices {X(t) : t ≤ TB(Rk)+ (R2

k/2)} is contained in B(rk+1),

(ii) rkand Rksatisfy the assumptions of Lemma 9, and

(iii) the set Akis measurable with respect to the sigma-algebra generated by {X(t) : t ≤ TB(rk+1)} and

µ(i)

rk,rk+1for 2 ≤ i ≤ sd.

13

Page 14

Property (i) follows from the fact that Rk+(R2

from the fact that rd−2

k

A(s)(r,R) in (4.10) and (4.11), set A(i−1)(rk,Rk) is contained in B(iRk). Therefore, set Akis measurable

with respect to the sigma-algebra generated by {X(t) : t ≤ TB(Rk)+ (R2

Since sdRk< rk+1and {X(t) : t ≤ TB(Rk)+ (R2

k/2) < rk+1. Property (ii) follows from our choice of ε and

≤ εRk. In order to see that property (iii) holds, note that, by the definition of

k/2)} and µ(i)

rk,iRkfor i ≤ sd.

k/2)} ⊂ B(rk+1), property (iii) follows.

Consider the events Γk = {H(Ak) < TB(R2

definition, H(Ak) is the entrance time of Z in Akand TB(R2

show that there exists a positive constant c such that for all k ≥ 1 and for any g1,...,gk−1∈ {0,1},

k)} and their indicator functions γk = I(Γk). In this

k)is the exit time of Z from B(R2

k). We will

Pz⊗ P(sd)

x

(Γk| γ1= g1,...,γk−1= gk−1) ≥ c > 0. (4.17)

The result will then follow from Borel’s lemma [7]:

Lemma 11. Consider a probability space (Ω,F,P) and a sequence of events ∆n∈ F. Let δn= I(∆n) be

the indicator function of the event ∆n. If there exists a sequence bnsuch that?

P(∆n| δ1= d1,...,δn−1= dn−1) ≥ bn> 0

nbn= ∞ and for any

di∈ {0,1}, i = 1,...,n − 1,

then

P

?

limsup

k

∆k

?

= 1.

We will now prove (4.17). We denote by E the event {γ1= g1,...,γk−1= gk−1}. By property (iii)

above and the fact that {Z(t) :t ≤ TB(rk−1)} ⊂ B(rk), the event E is measurable with respect to

the sigma-algebra Fk−1generated by {X(t) : t ≤ TB(rk)}, µ(s)

(Here, the two occurrences of TB(rk)correspond to the exit times of X and Z from B(rk), respectively,

which are, in general, different.) By property (5) of Pois(u,W∗), the sets of point measures {µ(s)

{µ(s)

rk,rk+1}s≥2are independent. Therefore, using strong Markov property for X and Z and integrating

over the µ(s)

rk,rk+1, s ≥ 2, we obtain

rkfor s ≤ sd, and {Z(t) : t ≤ TB(rk)}.

rk}s≥2and

Pz⊗ P(sd)

x

(Γk∩ E) = Ez⊗ E(sd)

x

?

Pz′ ⊗ P(sd)

x′

(Γk);E

?

,

where x′= X(TB(rk)), and z′= Z(TB(rk)). It follows from Lemma 9 that

Pz′ ⊗ P(sd)

x′

(Γk) ≥ c.

This proves (4.17) and completes the proof of the lemma.

As a corollary of Lemma 10 we obtain the following lemma. Let µ(i), i ∈ {1,...,sd−1} be independent

Poisson point measures with distribution Pois(u,W∗), where sdis defined in (1.2). Let P be their joint

law. We construct the graph G′= (V′,E′) as follows. The set of vertices V′is the set of trajectories from

∪sd−1

that intersect.

i=1Supp(µ(i)), and the set of edges E′is the set of pairs of different trajectories from ∪sd−1

i=1Supp(µ(i))

Lemma 12. Let d ≥ 5 and u > 0. Then, with the above notation,

P(diam(G′) ≤ sd) = 1.

14

Page 15

Proof. Take a positive integer r. By Definition 4.1 (see also the notation there), for each i ∈ {1,...,sd−1},

µ(i)= µ(i)

r

and µ(i)

any i ∈ {1,...,sd− 1}, let N(i)be the number of trajectories in Supp(µ(i)

is the number of doubly-infinite trajectories modulo time-shift from Supp(µ) that intersect B(r). By

property (1) of Pois(u,W∗), N(i)has the Poisson distribution with parameter ucap(B(r)). By the defi-

nition of Pois(u,W∗), we know that (recall the notation from Section 1.1), for each i ∈ {1,...,sd− 1},

µ(i)

r

j=1δπ∗(X(i)

are parametrized in such a way that X(i)

j ∈ {1,...,N(i)}, and (b) they satisfy properties (2) and (3) of Pois(u,W∗). In particular, given N(i)and

(X(i)

walks.

Property (5) of Pois(u,W∗) gives that for each i ∈ {1,...,sd− 1}, all the random walks (X(i)

0)N(i)

r,∞for k ∈ {1,...,sd−1}. Therefore, Lemma 10 and Remark 2 imply that,

given N(i)and (X(i)

j=1for all i ∈ {1,...,sd− 1}, almost surely, for each pair of different random

walks (X(i)

l

1 ≤ m ≤ sd− 1, such that X(i)

j

l

∩ wsd−1?= ∅, and wi∩ wi+1?= ∅ for i ∈ {1,...,sd− 2}.

Since this holds for any r, the result follows.

r + µ(i)

r,∞, and the measures µ(i)

r,∞are independent by property (5) of Pois(u,W∗). For

r ).In other words, N(i)

=?N(i)

j), where X(i)

1,...,X(i)

N(i)are doubly-infinite trajectories from W such that (a) they

j(0) ∈ B(r) and X(i)

j(t) / ∈ B(r) for all t < 0 and for all

j(0))N(i)

j=1, the forward trajectories (X(i)

j(t),t ≥ 0)N(i)

j=1are distributed as independent simple random

j(t),t ≥

j=1are independent from µ(k)

j(0))N(i)

j(t),t ≥ 0) and (X(k)

(t),t ≥ 0), there exist doubly-infinite trajectories wm ∈ Supp(µ(m)

∩ w1?= ∅, X(k)

r,∞),

Proof of Theorem 1: upper bound on diameter. We complete the proof of Theorem 1 by showing that

P(diam(G) ≤ sd) = 1. By Remark 1, we may and will assume that d ≥ 5. Let µ(1),...,µ(sd−1)be

independent Poisson point measures on W∗with distribution Pois(u/(sd− 1),W∗). We construct the

graph G′= (V′,E′) as follows. The set of vertices V′is the set of trajectories from ∪sd−1

and the set of edges E′is the set of pairs of different trajectories from ∪sd−1

Lemma 12 implies that the diameter of G′is at most sd. On the other hand, by property (4) of Pois(u,W∗),

graphs G and G′have the same law. This completes the proof.

i=1Supp(µ(i)),

i=1Supp(µ(i)) that intersect.

Acknowledgments. We would like to thank A.-S. Sznitman for suggesting to look for an alternative

proof of the connectivity of the random interlacement, inspiring discussions, and valuable comments. We

also thank A. Drewitz for comments on the manuscript.

References

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16