BIASED RANDOM WALK IN I.I.D. POSITIVE RANDOM CONDUCTANCES ON Zfribergh/brwrc.pdf · We study the biased random walk i.i.d. positive random con-ductances on Zd. Our main result is

BIASED RANDOM WALK IN I.I.D. POSITIVE RANDOM

CONDUCTANCES ON Zd

ALEXANDER FRIBERGH

Abstract. We study the biased random walk i.i.d. positive random con-ductances on Z

d. Our main result is that the random walk is ballisticif, and only if, the conductances have finite mean. Moreover, in the sub-ballistic regime we obtain the polynomial order of the fluctuations of therandom walk. This extends results obtained by L. Shen in [15], who provedpositivity of the speed in the uniformly elliptic setting.

1. Introduction

One of the most fundamental questions in random walks in random mediais understanding the long-term behavior of the random walk. Many sys-tems of random walks in random environments (RWRE) exhibit anomalousbehaviors. One of the main reasons for such behaviors is trapping, a phe-nomenon observed by physicists long ago and which is now a central topic inRWRE. The importance of trapping in RWRE motivated the introduction ofthe Bouchaud trap model (BTM). This is an idealized model that receiveda lot of mathematical attention. A review of the main results known can befound in [2], which conjectures that the results obtained in the BTM shouldbe true in a wide variety of models.

One of the most characteristic behavior associated to trapping is the ex-istence of a zero asymptotic speed in models with directional transience.Models of biased random walks in random conductances are among the mostnatural examples of directionally transient random walks in random envi-ronments. So far, mathematically, two models of biased random walks havebeen studied: one is on a supercritical percolation cluster and the other is inenvironments assumed to be uniformly elliptic.

In the case of biased random walks on a percolation cluster, it was shownin [3] and in [16] that there exists a zero-speed regime. More recently in [8] acharacterization of the zero-speed regime has been achieved. This confirmedthe predictions of the physicists [10] that trapping occurs in the model.

2000 Mathematics Subject Classification. primary 60K37; secondary 60J45.Key words and phrases. Random walk in random conductances, Heavy-tailed random

variables.1

2 A. FRIBERGH

In the case of a biased random walk in random conductances which isuniformly elliptic, it has been shown in [15] that the walk is directionallytransient and has always positive speed and verifies an annealed central limittheorem. These results are coherent with the conjecture that a directionallytransient random walk in random environment which is uniformly ellipticshould have positive speed (see [17]). Hence, trapping does not seem toappear under uniform ellipticity conditions.

The results on these two models do not bring any understanding on the be-havior of the random walk in positive conductances that might be arbitrarilyclose to zero. In such a model, we truly lose the uniform elliptic assumption,as opposed to the biased random walk on the percolation cluster where thewalk is still uniformly elliptic on the graph where the walk is restricted.

Our purpose in this paper is to understand the ballistic-regime of a biasedrandom walk in positive i.i.d. conductances and how trapping arises in sucha model.

2. Model

We introduce P[ · ] = P⊗E(Zd)∗ , where P∗ is the law of a positive random

variable c∗ ∈ (0,∞). This measure characterizes gives a random environmentusually denoted ω.

In order to define the random walk, we introduce a bias ℓ = λ~ℓ of strength

λ > 0 and a direction ~ℓ which is in the unit sphere with respect to theEuclidian metric of R

d. In an environment ω, we consider the Markov chainof law P ω

x on Zd with transition probabilities pω(x, y) for x, y ∈ Z

d definedby

(1) X0 = x, P ωx -a.s.,

(2) pω(x, y) =cω(x, y)

∑

z∼x cω(x, z),

where x ∼ y means that x and y are adjacent in Zd and also we set

(2.1) for all x ∼ y ∈ Zd, cω(x, y) = cω

∗ ([x, y])e(y+x)·ℓ.

This Markov chain is reversible with invariant measure given by

πω(x) =∑

y∼x

cω(x, y).

The random variable cω(x, y) is called the conductance between x and yin the configuration ω. This is natural because of the links existing betweenreversible Markov chains and electrical networks. We will be making exten-sive use of this relation and we refer the reader to [5] and [12] for a furtherbackground on this relation. Moreover for an edge e = [x, y] ∈ E(Zd), wedenote cω(e) = cω(x, y).

3

Finally the annealed law of the biased random walk will be the semi-directproduct Pp = P[ · ] × P ω

0 [ · ].In the case where c∗ ∈ (1/K,K) for some K < ∞, the walk uniformly

elliptic and this model is the one previously studied in [15].

3. Results

Firstly, we prove that the walk is directionally transient.

Proposition 3.1. We have

lim Xn · ~ℓ = ∞, P − a.s.

This proposition is a consequence of Proposition 7.1.Our main theorem is

Theorem 3.1. For d ≥ 2, we have

limXn

n= v, P − a.s.,

where

(1) if E∗[c∗] < ∞, then v · ~ℓ > 0,

(2) if E∗[c∗] = ∞, then v = ~0.

If lim ln P∗[c∗>n]ln n

= −γ with γ < 1 then

limln Xn · ~ℓ

ln n= γ, P − a.s..

This theorem follows from Proposition 9.1, Proposition 10.1, Lemma 10.2and Lemma 10.4.

This result proves that trapping phenomena may occur in an elliptic regime,that is, when all transition probabilities are positive.

Let us rapidly discuss the different main ways the walk may be trapped(see Figure 1):

(1) an edge with high conductance surrounded by normal conductances,(2) a normal edge surrounded by very small conductances,

Let us discuss how the first types of traps function. Assume we have anedge e of conductance c∗(e) surrounded by edges of fixed conductances, say1. A simple computation shows that the walk will need a time of the orderof c∗(e) to leave the endpoints of e. Hence, if the expectation of c∗ is infinite,then the annealed exit time of the e is infinite. Heuristically, the walk getstrapped strongly on this one edge. This phenomenon is enough to explainthe zero-speed regime.

At first glance it is surprising that in Theorem 3.1 there is only a conditionon the tail of c∗ at infinity. Indeed, if the tail of c∗ at 0 is sufficiently big, more

4 A. FRIBERGH

1

1 1

1

1

1

M >> 1 1

ǫ << 1

ǫǫ

ǫ

ǫ

ǫ

Figure 1. The two main types of traps

precisely such that E[mini=1,...,4d−2 1/c(i)∗ ] = ∞ for c

(i)∗ i.i.d. chosen under P∗,

then the annealed exit time of an edge is infinite. This condition does notappear in Theorem 3.1, because the walk is unlikely to reach such an edgebecause it needs to cross and edge with extremely low conductance. Eventhough this type of trapping is not strong enough to create a zero-speedregime, there is evidence that the central limit theorem may not hold if thetail at 0 of c∗ is big enough (see Remark 10.1).

One may try to create traps similar to those encountered in the biasedrandom walk on the percolation cluster: a long dead-end in the direction ofthe drift. If the bias is high enough, the walk will stay long enough in thosedead-ends to force zero-speed. In our context, we are not allowed to use zeroconductances, but we may use extremely low conductances forcing the walkto exit the dead-end at the same place it entered. This type of traps is veryinefficient. Indeed, most edges forming a dead-end have to verify c∗(e) < εto be able to contain the walk for a long period, and this for any fixed ε > 0.The probabilistic cost of creating such a trap is way too high.

Hence, small conductances cannot create zero-speed but high conductancescan. To conclude, we give an idealized version of the two most importanttypes of traps in this model:

X1 = Geom(1/c∗ ∧ 1) or X2 =

0 with probability c′∗ ∧ 1

Geom(c′∗ ∧ 1) else.,

where c∗ is chosen according to the law P∗ and c′∗ has the law of maxi=1,...,4d−2 c(i)∗

where c(i)∗ are i.i.d. chosen under the law P∗. Intuitively, anything related to

trapping with biased random walks in an elliptic setting should be understoodusing those idealized traps.

5

4. Notations

Let us denote by (ei)i=1...d an orthonormal basis of Zd such that ei · ~ℓ ≥

e2 · ~ℓ ≥ · · · ≥ ed · ~ℓ ≥ 0, in particular we have e1 · ~ℓ ≥ 1/√

d. Moreover we

complete f1 := ~ℓ into an orthonormal basis (fi)1≤i≤d of Rd.

We set

H+(k) = x ∈ Zd, x · ~ℓ > k and H−(k) = x ∈ Z

d, x · ~ℓ ≤ k,and

H+x = H+(x · ~ℓ) and H−

x = H−(x · ~ℓ).Let us introduce dω(x, y) the graph distance in ω between x and y. Define

for x ∈ ω and r > 0

Bω(x, r) = y ∈ ω, dω(x, y) ≤ r.Given a set V of vertices of Z

d we denote by |V | its cardinality, by E(V ) =[x, y] ∈ E(Zd) | x, y ∈ V its edges and

∂V = x ∈ V | y ∈ Zd \ V, x ∼ y, ∂EV = [x, y] ∈ E(Zd) | x ∈ V, y /∈ V .

Given a set E of edges of Zd, we denote V (E) = x ∈ Z

d | x is an endpoint of e ∈E its vertices.

Denote for any L,L′ ≥ 1

B(L,L′) =

z ∈ Zd |

∣

∣

∣z · ~ℓ

∣

∣

∣< L, |z · fi| < L′ for i ∈ [2, d]

,

and∂+B(L,L′) = z ∈ ∂B(L,L′) | z · ~ℓ ≥ L.

We introduce the following notations. For any set of vertices A of a certaingraph on which a random walk Xn is defined, we denote

TA = infn ≥ 0, Xn ∈ A, T+A = infn ≥ 1, Xn ∈ A and T ex

A = infn ≥ 0, Xn /∈ A.This allows us to define the hitting time of “level”n by

∆n = TH(n).

Finally θn will denote the time shift by n units of time.In this paper constants are denote by c or C without emphasizing their

dependence on d and the law P∗. Moreover the value of those constants maychange from line to line.

Let us explain the organization of the paper. We begin by studying exitprobabilities of large boxes, the main point of Section 5 is to prove Theo-rem 5.1 which is a property similar to Sznitman’s conditions (T ) and (T )γ,see [17]. This property is one of the key estimates for studying directionallytransient RWRE. It allows us to define regeneration times, similar to the onesintroduced in [18], and study them; this is done in Section 7. The construc-tion of regeneration times in this model turns out to be difficult because we

6 A. FRIBERGH

lack uniform ellipticity as explained at the beginning of Section 6. The lawof large numbers in the positive speed regime is obtained in Section 8. Thezero-speed regime is studied in Section 10.

5. Exit probability of large boxes

Our first goal is to obtain estimates on the exit probabilities of large boxes.In particular we aim at showing

Theorem 5.1. For α > d + 3

P[T∂B(L,Lα) 6= T∂+B(L,Lα)] ≤ e−cL.

After this section α will be fixed, greater than d + 3.We will adapt a strategy of proof first used in [8]. For the most part, the

technical details and notations are simpler in our context. We will go overthe parts of the proof which can be simplified, but we will eventually referthe reader to [8] for the conclusion of the proof which is exactly similar inboth cases. The notations have been chosen so that the reader can follow theneeded proof in [8] word for word.

Firstly, let us describe the strategy we will follow.The fundamental idea is to partition the space into a good part where the

walk is well-behaved and a bad part consisting of small connected componentswhere we have very little control over the random walk.

Once this is done, we may study the behavior of the random walk at timeswhere it is in the good part of the space, in which it can easily be controlled.We will refer to this object as the modified walk.

The strategy is two-fold. Firstly, we show that the modified walk behavesnicely, i.e. verifies Theorem 5.1. This is essentially achieved using a com-bination of spectral gap estimates and the Carne-Varopoulos formula [4].Secondly, we need to show that information on the exit probabilities for thismodified random walk allows us to derive interesting statements on the ac-tual random walk. This is a natural thing to expect, since the bad parts ofthe environment are small.

A more detailed discussion of the strategy of proof can be found at thebeginning of Section 7 in [8].

5.1. Bad areas. We say that an edge e is K-normal if c∗(e) ∈ [1/K,K],where K will be taken to be very large in the sequel. If an edge is notK-normal, we will say it is K-abnormal which occurs with arbitrarily smallprobability ε(K) := P∗[c∗ /∈ [1/K,K]].

In relation to this, we will say that a vertex x is K-open if for all y ∼ x theedge [x, y] is K-normal. If a vertex is not K-open, we will say it is K-closed.Finally a vertex x ∈ Z

d is K-good, if there exists an infinite directed K-open

7

path starting at x, that is: we have x = x0, x1, x2, x3, . . . with x0 = x suchthat for all i ≥ 0

(1) if x2i+1 − x2i = e1 then x2i+2 − x2i+1 ∈ ν \ −e1, . . . ,−ed,(2) xi is K-open.

If a vertex is not K-good it is said to be K-bad. By taking K large enough,the probability that a vertex is K-good goes to 1, since c∗ ∈ (0,∞).

To ease the notation, we will not always mention the K-dependenceThe first of these results is stated in terms of the width of a subset A ⊆ Z

d,which we define to be

W (A) = max1≤i≤d

(

maxy∈A

y · ei − miny∈A

y · ei

)

.

Let us denote BADK(x) the connected component of K-bad vertices con-taining x, in case x is good then BADK(x) = ∅.Lemma 5.1. There exists K0 such that, for any K ≥ K0 and for any x ∈ Z

d,we have that the cluster BADK(x) is finite Pp[·]-a.s. and

Pp[W (BADK(x)) ≥ n] ≤ C exp(−ξ1(K)n),

where ξ1(K) → ∞ as K tends to infinity.

Proof. We call two vertices 2-connected if ||u − v||1 = 2, so that we maydefine the even bad points BADe

K(x) of x as the 2-connected component ofbad vertices containing x. Any element of BAD(x) is a neighbor of BADe

K(x)so that W (BADK(x)) ≤ W (BADe

K(x)) + 2.Consider now the percolation model on the even lattice v ∈ Z

d, ||v||1 is evenwhere the bond between y and y + e1 + ei is even-open if and only if in theoriginal model the vertices y, y + e1 and y + e1 + ei are open. This model isa 3-dependent oriented percolation model which has a measure we denote byPp,orient.

Fix p′ close to 1. For K large enough, the probability that a vertex isK-open can be made arbitrarily close to 1 so by Theorem 0.0 in [13] the lawPp,orient dominates an i.i.d. bond percolation with parameter p′.

Let us first describe how to finish the proof for d = 2. Consider the outeredge-boundary ∂EBADe

K(x) of BADeK(x) (represented dually in Figure 2): by

an argument similar to that of [6] (p.1026) we see that nր + nց = nւ + nտ,where nր, for example, is the number of edges labelled ր in ∂EBADe

K(x).We note that any տ edge of ∂EBADe

K(x) has one endpoint, y, which isbad, and one, y + e1 + e2, which is good. This implies that y is even-closed.

Hence, if nտ ≥ nւ/2, then at least one sixth of the edges of ∂EBADeK(x)

have an endpoint which is an even-closed vertex.Otherwise, let us assume that nւ ≥ 2nտ. We may notice (see Figure

2) any ւ edge followed (in the sense of the arrows) by an ց edge can be

8 A. FRIBERGH

e1

e2

~ℓ

Figure 2. The outer edge-boundary ∂EBADeK(x) of

BADeK(x) when d = 2.

mapped in an injective manner to an տ edge. This means that at least halfof the ւ edges are not followed by an ց edge. So at least one sixth of theedges of ∂EBADe

K(x) are ւ edges that are not followed by a ց arrow. Inthis case, we note that there is one endpoint y of ւ which is bad and thaty + 2e1 is good, and hence y is even-closed.

Finally, at least one sixth of the edges of ∂EBADeK(x) are adjacent to an

even-closed vertex. The outer boundary is a minimal cutset, as describedin [1]. The number of such boundaries of size n is bounded (by Corollary 9in [1]) by exp(Cn). Hence, if p′ is close enough to 1, we obtain the desiredexponential tail for W (BADe

K(x)) under Pp′ , and hence under Pp,orient.To extend this to general dimensions, we note that there exists i0 ∈ [2, d]

such that a proportion at least 1/(d − 1) of the ∂EBADeK(x) are edges of

the form [y, y + ei0 ]. We may then apply the previous reasoning in the planey + Ze1 + Zei0 to show that at least a proportion 1/(6(d − 1)) of the edges∂EBADe

K(x) are adjacent to an even-closed vertex, and thus infer the lemma.

Let us define BADK = ∪x∈ZBADK(x) which is a union of finite sets. Alsowe set GOODK = Z

d \ BADK . We may notice that

(5.1) for any x ∈ BADK , ∂BADK(x) ⊂ GOODK ,

since BADK(x) is a connected component of bad points.

9

For technical reasons, we will need a slightly stronger result. We say thata vertex x ∈ Z

d is K-super-open if all nearest neighbors of x are K-open. Ifa vertex is not super-open, it is say to be K-weakly-closed. We extend thenotion of good and bad points to super-good and weakly bad points.

Remark 5.1. Any neighbor of a super-good point is good.

Let us denote BAD+K(x) the connected component of K-weakly-bad ver-

tices containing x. Using those new definitions, we can mimic the proof ofLemma 5.1 to obtain the following result

Lemma 5.2. There exists K0 such that, for any K ≥ K0 and for any x ∈ Zd,

we have that the cluster BAD+K(x) is finite Pp[·]-a.s. and

Pp[W (BAD+K(x)) ≥ n] ≤ C exp(−ξ2(K)n),


In the sequel K will always be large enough so that BADK(x) and BAD+K(x)

are finite for any x ∈ Zd.

5.2. A graph transformation to erase big traps. Given a certain con-figuration ω we construct a graph ωK in which traps are erased and such thatthe random walk induced outside of large traps by the random walk in ω hasthe same law as the random walk in ωK .

We denote ωK the graph obtained from ω by the following transformation.The vertices of ωK are the vertices of GOODK , and the edges of ωK are

(1) [x, y], x, y ∈ GOODK , with x /∈ ∂BADK or y /∈ ∂BADK and haveconductance cωK ([x, y]) := cω([x, y]) ,

(2) [x, y], x, y ∈ ∂BADK (including loops) which have conductance

cωK ([x, y]) : = πω(x)P ωx [X1 ∈ BADK ∪ ∂BADK , T+

y = T+∂BADK

]

= πω(y)P ωy [X1 ∈ BADK ∪ ∂BADK , T+

x = T+∂BADK

],

the last equality being a consequence of reversibility and ensures symmetryfor the conductances.

We call the walk induced by Xn on GOODK , the walk Yn defined to beYn = Xρn

where

ρ0 = TGOODKand ρi+1 = T+

GOODK θρi

.

From [8] Proposition 7.2, we have the two following properties

Proposition 5.1. The reversible walk defined by the conductances ωK verifiesthe two following properties

(1) It is reversible with respect to πω(·).(2) If started at x ∈ ωK, it has the same law as the walk induced by Xn

on GOODK started at x.

10 A. FRIBERGH

and

Lemma 5.3. For x, y ∈ GOODK which are nearest neighbors in Zd, we have

cωK ([x, y]) ≥ cω([x, y]).

Hence, we may notice that

Remark 5.2. We may notice that for any x ∈ GOODK, we have

c

Ke−2λx·~ℓ ≤ πωK (x) ≤ CKe−2λx·~ℓ,

and for any y ∈ GOODK adjacent, in Zd, to x

c

Ke−2λx·~ℓ ≤ cωK ([x, y]) ≤ CKe−2λx·~ℓ.

5.3. Spectral gap estimate in ωK. The following arguments are heavilyinspired from [16] and uses spectral gap estimates. After showing that thespectral gap in ωK , we can deduce that the walk is likely to exit the boxquickly in ωK . Finally, we need to argue that exiting the box quickly, weshould exit it in the direction of the drift. This allows us to obtain Theo-rem 5.1, once we have argued that the exit probabilities in ω and ωK arestrongly related. This paper only contains the first step of this reasoning, thefollowing ones being treated in [8].

For technical reasons, we introduce the notation

B(L,Lα) = x ∈ Zd, −L ≤ x · ~ℓ ≤ 2L and |x · fi| ≤ Lα for i ≥ 2.

Let us introduce the principal Dirichlet eigenvalue of I−P ωK , in B(L,Lα)∩ωK

(5.2)

ΛωK(B(L,Lα)) =

infEGOODK(f, f), f|(B(L,Lα)∩ωK)c = 0, ||f ||L2(π(ωK)) = 1,

when B(L,Lα) ∩ ωK 6= ∅,∞, by convention when B(L,Lα) ∩ ωK = ∅,

where the Dirichlet form is defined for f, g ∈ L2(πωK ) by

EGOODK(f, g) = (f, (I − P ωK )g)πωK

=1

2

∑

x,y neighbors in ωK

(f(y) − f(x))(g(y) − g(x))cωK ([x, y]).

We have

Lemma 5.4. W have that for ω such that B(L,Lα) ∩ ωK 6= ∅,ΛωK

(B(L,Lα)) ≥ cL−(d+1).

11

Proof. By definition of a good point, from any vertex x ∈ ωK , there exists adirected open path x = px(0), px(1) . . . , px(lx) in ωK (which are neighbors inZ

d) such that px(i) ∈ B(L,Lα) for i < lx and px(lx) /∈ B(L,Lα) . This allowsus to say that

(5.3) maxi≤lx

πωK(x)

cωK([px(i + 1), px(i)])

≤ C maxi≤lx

πω(x)

πω(px(i))≤ C,

where we used Lemma 5.3 and Remark 5.2. Moreover lx ≤ CL.We use a classical argument of Saloff-Coste [14], we write for ||f ||L2(πωK ) =

1

1 =∑

x

f 2(x)πωK(x) =

∑

x

[

∑

i

f(px(i + 1)) − f(px(i))]2

πωK(x)

≤∑

x

lx

[

∑

i

(f(px(i + 1)) − f(px(i)))2]

πωK(x).

Now by (5.3), we obtain

1 ≤ C∑


(f(z) − f(y))2cωK([x, y]) × max

b∈E(Zd)

∑

x∈ωK∩B(L,Lα),b∈px

lx

where b ∈ px means that b = [px(i), px(i + 1)] for some i. Using that

(1) lx ≤ CL for any x ∈ ωK ,(2) b = [x, y] ∈ ωK can only be crossed by paths “pz” if b ∈ E(Zd) and

z ∈ BZd(x,CL),

we havemax

b

∑

x∈B(n,nα),b∈px

lx ≤ CLd+1,

and1 ≤ CLd+1

∑


(f(z) − f(y))2cωK([x, y]),

so that using (5.2)

ΛωK(B(n, nα)) ≥ cL−(d+1).

We explained how to obtain Theorem 5.1 at the beginning of subsection 5.3.The proof of Theorem 5.1 is almost completely similar to the end of the proofof Theorem 1.4 in [8]. The reader may read subsection 7.4, 7.5 and 7.6 of [8]for the complete details.

To ease this task, the notations have been chosen so that only two minorchanges have to be made: in our case K∞ = Z

d and I = Ω.The proof in [8] uses some reference to previous results, for the reader’s con-

venience we specify the correspondence. The following results in [8]: Lemma

12 A. FRIBERGH

7.5, Proposition 7.2, Lemma 7.9 and the second part of Lemma 7.6 correspondrespectively to Lemma 5.1, Proposition 5.1, Lemma 5.4 and (5.1).

A final remark is that any inequality needed on πωK can be found in Re-mark 5.2.

6. Construction of K-open ladder points

The construction of a regeneration structure is rather involved in this

model. We call a new maximum of the random walk in the direction ~ℓ,a ladder-point. The standard way of constructing regeneration times is toconsider successive ladder points and arguing, using some type of uniformellipticity assumption, that there is a positive probability of never backtrack-ing again. Such a ladder point creates a separation between the past and thefuture of the random walk leading to interesting independence properties.

A major issue in our case we do not have any type of uniform elliptic-ity. Moreover, ladder points are conditioned parts of the environment andit seems intuitive that the edge that led us to a ladder point has uncharac-teristically high conductance. Those high conductnces may hinder the walkfrom never backtracking and creating a regeneration time. In some sense,we need to show that the environment seen from the particle at a ladder-point is relatively normal. More precisely we will prove that we encounteropen ladder-points and find tail estimates on the location of the first openladder-point.

We define the following random variable

M(K) = infi ≥ 0, Xi is K-open and for j < i − 2, Xj · ~ℓ < Xi−2 · ~ℓXi = Xi−1 + e1 = Xi−2 + 2e1 ≤ ∞.

We will drop the dependence on K outside of major statements of defini-tions.

We need three preparatory lemmas before turning to the study of M(K).For this, we introduce

∂+i B(n, nα) = x ∈ B(n, nα), x ∼ y with y ∈ ∂+B(n, nα),

and

A(n) = T∂B(n,nα) ≥ T∂+i B(n,nα).

It follows from Theorem 5.1 that

Lemma 6.1. We have

P[A(n)c] ≤ Ce−cn.

We say that a vertex x ∈ B(n, nα) is K-n-closed, if there exists a nearestneighbor y ∈ B(n, nα) of x such that c∗([x, y]) /∈ [1/K,K].

13

Let us denote Kx(n) the K-n-closed connected component of x. Thisallows us to introduce

(6.1) B(n) = for all x ∈ ∂+i B(n, nα),

∣

∣Kx(n)∣

∣ ≤ ln n.It is convenient to set Kx(n) = x when Kx(n) is empty.

Lemma 6.2. For any M < ∞, we can find K large enough such that

P[B(n)c] ≤ Cn−M .

Proof. Obviously, for any x ∈ ∂+B(n, nα)

Kx(n) ⊂ CLOSEDK(x),

where CLOSEDK(x) is the closed connected component of K-closed pointcontaining x.

Using lemma 5.1 in [11], we may notice that there are at most an expo-nential number of lattice animals. Hence, for any x ∈ ∂+B(n, nα)

P[|CLOSEDK(x)| ≥ ln n] ≤ C(C1ε(K))ln n = Cn−ξ2(K),

where ξ2(K) tends to infinity K goes to infinity. So, the right hand side canbe made lower than n−M by choosing K large enough.

Lemma 6.3. Take G 6= ∅ be a finite connected subset of Zd. Assume that

each edge e of Zd is assigned a positive conductance c(e) and that there exist

x ∈ ∂G and y ∈ G such that x ∼ y such that c([x, y]) ≥ c1c(e) for anye ∈ ∂EG. We have

Py[Tx ≤ T∂G] ≥ c1

4d|G|−1 ,

where Py is the law of the random walk in G ∪ ∂G started at y arising fromthe conductances (c(e))e∈E(Zd).

Proof. We will be using comparisons to electrical networks and we refer thereader to Chapter 2 of [12] for further background on this topic.

Let us first notice that a walk started at y ∈ G will reach ∂G beforeZ

d \ (G ∪ ∂G), so this lemma is actually a result on a finite graph G ∪ ∂G.To simplify the proof, we will consider the graph G where all edges em-

anating from x that are not [x, y] will be assigned conductance 0, whichcorresponds to reflecting the walk on those edges. It is plain to see that

Py[Tx ≤ T∂G] ≥ P G∪∂Gy [Tx ≤ T∂G],

where P G∪∂Gy is the law of the random walk started at y in the conductances

of the graph G ∪ ∂G.Hence, it is enough to prove our statement in the finite graph G∪ ∂G. We

may see that

(6.2) P G∪∂Gy [Tx ≤ T∂G] = u(y),

14 A. FRIBERGH

where u(·) is the voltage function verifying u(x) = 1 and u(z) = 0 for z ∈∂G\x. Let us denote i(·) the associate intensity. Since y is the only vertexadjacent to x in G∪ ∂G, we know that the current flowing into the circuit atx passes through the edges [x, y], so

1

RG∪∂G(x, ∂G \ x)= i([x, y]),

where RG∪∂G(x, ∂G\x) is the effective conductance between x and ∂G\xin G ∪ ∂G. By Ohm’s law we may deduce that

u(x) − u(y) = rG∪∂G([x, y])i([x, y]) =rG∪∂G([x, y])

RG∪∂G(x, ∂G \ x).

Now, since x is the only vertex adjacent to y, we can see by an elec-

trical network reduction of resistances in series that RG∪∂G(x, ∂G \ x) =

rG∪∂G([x, y]) + RG∪∂G\x(y, ∂G \ x). This means that

u(y) =RG∪∂G\x(y, ∂G \ x)

RG∪∂G\x(y, ∂G \ x) + rω([x, y])=

(

1+rG∪∂G([x, y])

RG∪∂G\x(y, ∂G \ x))−1

.

We consider the graph G∪∂G\x, by Rayleigh’s monotonocity principle,we know that collapsing all vertices G∪∂G\x, y into one vertex δ decreasesall effective conductances. In this new graph, y is connected to δ by at most∣

∣

∣G ∪ ∂G

∣

∣

∣edges of resistances at least c1r

G∪∂G([x, y]) by our assumptions on

the graph. By network reduction of conductances in parallel, this means

RG∪∂G\x(y, ∂G \ x) ≥ c1∣

∣

∣G ∪ ∂G

∣

∣

∣

rG∪∂G([x, y]).

The two last equations imply, with (6.2), that

u(y) ≥ c1

4d|G|−1 ,

which concludes the proof.

6.0.1. Successive attempts to find an open ladder point. We will show thatan open ladder point can occur shortly after we exit a box Bn. After exitingmany such boxes, Bn, B2n, . . . we will eventually see an open ladder pointwith high probability.

Let us denote for k ≤ n

(6.3) R(K)(nk) = M(K) > T∂Bnk+ 2,

where we usedBn := B(n, nα)

Moreover we have

15

Lemma 6.4. For any ε1 > 0 and M < ∞, we can find K0 = K0(ε1,M)large enough such that the following holds: for any n ≥ n0(ε1,M) K ≥ K0

and any k ∈ [2, n],

P[R(K)(kn)] ≤ (1 − cn−ε1)P[R(K)((k − 1)n)] + Cn−M ,

where the constants depend on K.

Proof. We set

(6.4) K(n) = KXT∂+i

Bn

(n) ⊆ Bn,

where we recall that the notation Kx(n) was defined above (6.1). In caseT∂+

i Bn= ∞, we simply set K(n) = ∅ and ∂K(n) = ∅.

We introduce

C(n) = x is open, for x ∈ ∂K(n) ∩H+(n),as well as

D(n) = T∂K(n) θT∂iBn= T∂K(n)∩H+(n) θT∂iBn

and

E(n) = XT∂B(n,nα)+2 = XT∂B(n,nα)+1 + e1 = XT∂B(n,nα)+ 2e1

and XT∂B(n,nα)+2, XT∂B(n,nα)+1 are open.Let us consider an event in A(n)∩C(n)∩D(n). It verifies all the following

conditions

(1) on A(n), we have T∂+i Bn

≤ T∂Bn, so that

T∂B(n,nα) ≥ T∂K(n) θT∂iBn,

(2) on D(n), by (6.4) we have

T∂K(n) θT∂iBn= T∂K(n)∩H+(n) θT∂iBn

,

(3) on C(n), we have ∂K(n) ∩H+(n) is open.

Hence, on A(n) ∩ C(n) ∩ D(n) ∩ E(n), we see that XT∂B(n,nα)is a new

maximum of the trajectory in the direction ~ℓ, XT∂B(n,nα)+1 = XT∂B(n,nα)+ e1

and XT∂B(n,nα)+2 = XT∂B(n,nα)+ 2e1 is a K-open point. This means that

A(n) ∩ C(n) ∩ D(n) ∩ E(n) ⊂ M ≤ T∂Bn+ 2.

The situation is illustrated in Figure 3.We have

P[R(kn)](6.5)

≤P[R((k − 1)n), (A(kn) ∩ C(kn) ∩ D(kn) ∩ E(kn))c]

≤P[A(kn)c] + P[B(kn)c]

16 A. FRIBERGH

0

K(kn)

B(nk, (nk)α)

~ℓ

XT∂B(nk,(nk)α)

Figure 3. A simple way to find an open ladder points

. . . + P[R((k − 1)n), A(kn), B(kn), C(kn)c]

. . . + P[R((k − 1)n), A(kn), B(kn), C(kn), D(kn)c]

. . . + P[R((k − 1)n), A(kn), B(kn), C(kn), D(kn), E(kn)c]

The first term is controlled by Theorem 5.1

(6.6) P[A(kn)c] ≤ C exp(−ckn) ≤ C exp(−cn),

and for any M < ∞, by Lemma 6.2, we can choose K large enough such that

(6.7) P[B(kn)c] ≤ n−M

Step 1 : Control of the third term

For k ≤ n, on A(kn) ∩ B(kn), we see that

(6.8) |K(kn)| ≤ ln(kn) ≤ 2 ln n,

in particular

P[R((k − 1)n), A(kn), B(kn), C(kn)c]

=∑

F⊂Zd,|F |≤2 ln n

P[R((k − 1)n), A(kn), B(kn),K(kn) = F,C(kn)c]

=∑


E[P ω[R((k − 1)n), A(kn), B(kn),K(kn) = F ]

× 1some x ∈ ∂F ∩H+(kn) is closed].We recall that K(kn) was defined at (6.4). We may now see that

(1) On the one hand, the random variable P ω[R((k−1)n), A(kn), B(kn),K(kn) =F ] is measurable with respect to σc∗([x, y]), with x, y /∈ H+

nk,

17

(2) on the other hand, the event some x ∈ ∂F ∩ H+(kn) is closed ismeasurable with respect to σc∗([x, y]), with x ∈ H+

nk.Hence, the random variables P ω[R((k − 1)n), A(kn), B(kn),K(kn) = F ]

and 1some x ∈ ∂F ∩H+(kn) is closed are P-independent. This yields

P[R((k − 1)n), A(kn), B(kn), C(kn)c]

=∑


P[R((k − 1)n), A(kn), B(kn),K(kn) = F ]

× P[some x ∈ ∂F ∩H+(kn) is closed]

≤∑


P[R((k − 1)n),K(kn) = F ]

× (1 − P[all x ∈ ∂F ∩H+(kn) are open]).

Now, we know by the Harris-inequality [9] that for F ⊂ Zd, with |F | ≤

2 ln n

P[all x ∈ ∂F ∩H(kn) are open] ≥ P[x is open]|F |

≥ (1 − ε(K))2 ln n = n2 ln(1−ε(K)).

By choosing K large enough, we can assume that 2 ln(1− ε(K)) ≥ −ε1/2.This means that the two previous equations imply that(6.9)P[R((k−1)n), A(kn), B(kn), C(kn)c] ≤ (1−n−ε1/2)P[R((k−1)n), A(kn), B(kn)].

Step 2 : Control of the fourth termWe wish to decompose P[R((k − 1)n), A(kn), B(kn), C(kn), D(kn)c] ac-

cording to all possible values of XT∂iBnkand K(kn). For this we notice that

(1) on A(kn), we have XT∂iBnk∈ ∂+

i Bnk and by the definition of K(kn)

(see (6.4)), XT∂iBnk∈ K(kn),

(2) moreover, on A(kn) ∩ B(kn), we have (6.8).

Hence,

P[R((k − 1)n), A(kn), B(kn), C(kn), D(kn)c]

≤∑

y,F

P[R((k − 1)n), XT∂iBnk= y,K(kn) = F,C(kn), D(kn)c],

where∑

y,F stands for∑

y∈∂+i Bnk

∑

F⊂Zd,|F |≤2 ln n,y∈F

Let us notice that, for a fixed ω, the events R((k−1)n), XT∂iBnk= y and

K(kn) = F are P ω-measurable with respect to Xi, i ≤ T∂iBnk. Thus, we

18 A. FRIBERGH

may use the Markov property at T∂iBnkto see that

P[R((k − 1)n), A(kn), B(kn), C(kn), D(kn)c]

(6.10)

≤∑

y,F

E[P ω[R((k − 1)n), XT∂iBnk= y,K(kn) = F ]

× P ωy [T∂F < T∂F∩H+(kn)]1x is open, for x ∈ ∂F ∩H+(kn)].

We claim that if K(kn) = F and x is open, for x ∈ ∂F ∩H+(kn) then∂EF is composed of normal edges. Indeed,

(1) notice that the definition of K(kn) at (6.4) (which is a K-(kn)-closedcomponent) implies that e is normal for all e ∈ ∂EK(kn) when e hasno endpoint in H+(kn).

(2) Moreover, if for any x ∈ ∂F ∩H+(kn) the vertex x is open, then anyedge e ∈ ∂EF with one endpoint in H+(kn) is normal.

Now, for any y ∈ F ∩ ∂+i Bnk, there exists x ∈ H+(kn) adjacent to y. Since

F ⊂ Bnk, we can see that for any z ∈ ∂F we have (x − z) · ~ℓ ≥ −1 . Usingthis, along with the conclusion of the previous paragraph, we see with (2.1)that

for all e ∈ ∂EF , cω(e) ≤ K2e2λcω([x, y]).

We can apply Lemma 6.3 to F , and we see that if K(kn) = F andx is open, for x ∈ ∂F ∩H+(kn), then we obtain

P ωy [T∂F < T∂F∩H(kn)] ≤ P ω

y [T∂F < Tx] ≤ (1 − c |F |−1) ≤ (1 − c ln−1 n),

since |F | ≤ 2 ln n.This turns (6.10) into

P[R((k − 1)n), A(kn), B(kn), C(kn), D(kn)c](6.11)

≤∑

y,F

E[P ω[R((k − 1)n), XT∂iBnk= y,K(kn) = F ](1 − c ln−1 n)

≤(1 − c ln n−1)P[R((k − 1)n), A(kn), B(kn), C(kn)].

Step 3: Control the fifth termOn A(kn)∩B(kn)∩C(kn)∩D(kn), we know that X∂B(n,nα) ∈ ∂+B(n, nα)

is an open point. So introducing

R′((k − 1)n) = R((k − 1)n) ∩ A(kn) ∩ B(kn) ∩ C(kn) ∩ D(kn)

we see

P[R((k − 1)n), A(kn), B(kn), C(kn), D(kn), E(kn)c]

19

≤∑

x∈∂+Bnk

P[R′((k − 1)n), XBnk= x, x is open, E(kn)c]

≤∑

x∈∂+Bnk

E[P ω[R′((k − 1)n), XBnk= x]1x is open,

(1x + e1 or x + 2e1 is not x-open+ P ω

x [X1 6= x + e1 or X2 6= x + 2e1]1x + e1, x + 2e1 are x-open)]where a vertex is said to be x-open if it is open in ωx coinciding with ω onall edges but those that are adjacent to x which are normal in ωx.

On x + e1, x + 2e1 are x-open ∩ x is open, we see that P ωx [X1 = x +

e1, X2 = x + 2e1] ≥ c > 0 by Remark 5.2.

P[R((k − 1)n), A(kn), B(kn), C(kn), D(kn), E(kn)c]

≤∑

x∈∂+Bnk

E[P ω[R′((k − 1)n), X∂B(n,nα) = x]1x is open,

(1x + e1 or x + 2e1 is not x-open + (1 − c)1x + e1, x + 2e1 are x-open)].We may also see that R′((k − 1)n), X∂Bnk

= x, x is open is measurablewith respect to σc∗(e), e ∈ E(Bnk) or e ∈ x+ν, whereas x + e1, x + 2e1 are x-openis measurable with respect to σc∗(e), e /∈ E(Bnk) and e /∈ x + ν. So theserandom variables are independent, which yields

P[R((k − 1)n), A(kn), B(kn), C(kn), D(kn), E(kn)c](6.12)

≤P[R′((k − 1)n)](P[x + e1 or x + e2 is not x-open]

+ (1 − c)P[x + e1, x + e2 are x-open])(6.13)

≤(1 − c)P[R((k − 1)n), A(kn), B(kn), C(kn), D(kn)],

since P[x + e1, x + e2 are x-open] > 0.

Step 4: Conclusion

For any ε1 > 0, we see using (6.6), (6.7), (6.9) (6.11) and (6.12), (whichare valid K chosen larger than some K0 depending only on M < ∞), thatwe have for any k ∈ [2, n]

P[R(kn)] ≤ P[R((k − 1)n)](1 − c ln n−1n−ε1/2) + Cn−M .

We now prove the following

Lemma 6.5. For any M , there exists K0 such that, for any K ≥ K0

P[XM(K) · ~ℓ ≥ n] ≤ C(K)n−M .

20 A. FRIBERGH

Proof. For any M < ∞, by Lemma 6.4, there exists K0 such that, for anyK ≥ K0 such that

P[R(nk)] ≤ (1 − cn−1/2)P[R(n(k − 1))] + n−M .

By a simple induction, this means that for

P[R(n2)] ≤ (1 − cn−1/2)n + n−M+1 ≤ 2n−M+1.

Recalling the definition of R(n) at (6.3), we see by Borel-Cantelli’s Lemmathat M(K) < ∞

Also this implies that

P[XM(K) · ~ℓ > n2] ≤ 2n−M+1,

andP[XM(K) · ~ℓ > n] ≤ 2n−(M+1)/2,

which proves the lemma, since M is arbitrary.

6.1. Consequence of our estimates on M. Let us introduce the laddertimes

(6.14) W0 = 0 and Wk+1 = infn ≥ 0, Xn · ~ℓ > XWk· ~ℓ.

We set

(6.15) M (K)(n) = for k with Wk ≤ ∆n, XM(K)θWk+Wk

·~ℓ−XWk·~ℓ ≤ n1/2,

Lemma 6.6. For any M < ∞, there exists K0 such that, for any K ≥ K0

and for all i ≥ 1 we have

P[M (K)(n)c] ≤ n−M .

Proof. Denote

Mj(n) = for k ≤ j − 1, XM(K)θWk+Wk

· ~ℓ − XWk· ~ℓ ≤ n1/2,

andNj(n) = XMθWj

+Wj· ~ℓ − XWj

· ~ℓ > n1/2.All XWi

are different necessarily, hence on T∂B(n,nα) = T∂+B(n,nα), any ksuch that XWk

≤ ∆n verifies k ≤ n2dα. Thus

(6.16) P[M(n)c] ≤ P[Mn2dα(n)c, T∂B(n,nα) = T∂+B(n,nα)] + exp(−cn),

by Theorem 5.1.

If Mn2dα(n)c, then there is k ≤ n2dα such that XMθWk+Wk

·~ℓ−XWk·~ℓ > n1/2.

By decomposing along the smallest such k we see that

P[Mn2dα(n)c, T∂B(n,nα) = T∂+B(n,nα)] ≤n2dα∑

k=1

P[Mk(n), Nk(n), T∂B(n,nα) = T∂+B(n,nα)].

21

Now, we see that Mj(n), Nj(n), T∂B(n,nα) = T∂+B(n,nα) we have XWj−1∈

B(n, nα), so by a simple union bound argument

P[Mj(n), Nj(n), T∂B(n,nα) = T∂+B(n,nα)]

≤∑

x∈B(n,nα)

P[XWj= x,XM(K)θWj

+Wj· ~ℓ − x · ~ℓ > n1/2]

≤∑

x∈B(n,nα)

E[P ωx [XM(K) · ~ℓ − x · ~ℓ ≥ n1/2]]

≤ |B(n, nα)|P[M(K) ≥ n1/2],

by Markov’s property at Wj and translation invariance of the environment.Hence by the two last equations, Lemma 6.5 and using (6.16), we may seethat

P[M(n)c] ≤ n−M .

7. Regeneration times

We define

D = 1X0 is goodD′,

where

D′ = infn > 0, Xn · ~ℓ ≤ X0 · ~ℓ.Also we introduce

M1 = supn≤D′

Xn · ~ℓ,

and

M2(x) = sup(y − x) · ~ℓ, y ∈ Kgood(x),where(7.1)

Kgood(x) = y, where x is connected to y using a directed open path,where directed open path was defined at the beginning of Section 5

We define the configuration dependent stopping times Sk, k ≥ 0 and thelevels Mk, k ≥ 0:

S0 = 0, M0 = X0 · ~ℓ, and for k ≥ 0,(7.2)

Sk+1 = M(K) θTH+(Mk)+ TH+(Mk),

where

Mk = supXm · ~ℓ, 0 ≤ m ≤ Rk.

22 A. FRIBERGH

with

Rk =

D′ θSk+ Sk, if XSk

is good,

TH+(M2(XSk)+XSk

·~ℓ) if XSkis not good.

These definitions imply that if Si+1 < ∞, then

(7.3) XSi+1· ~ℓ − XSi

· ~ℓ ≥ 2.

If XSkis good, then Rk can be infinite so that Mk = ∞.

Finally we define the basic regeneration time

(7.4) τ1 = SN , with N = infk ≥ 1, Sk < ∞,Mk = ∞.Let us give some intuition about those definitions. Assume Sk is con-

structed, it is, by definition, an open-point, so there is a positive probabilitythat it is a good point. Then, if it is a good point, it is natural to expectthere is a lower-bounded chance of never backtracking again. Hence, there isa positive probability of creating a point separating the past and the futureof the random walk: a regeneration point called τ1.

In case this does not occur, the future of the random walk and the envi-ronment ahead of us may be conditioned. A conditioning may be inducedby the fact that Sk is not good, that conditioning is limited to the value ofconductances in the finite set Kgood(XSk

). The conditioning induced by thefact that the walk will eventually backtrack is limited to the conductances ofthe edges that the walk visits before backtracking and the walk itself beforebacktracking. We may notice that by our definitions, all those edges are in-cluded in H−(Mk), and thus the environment in H+(Mk) and the walk onceit reaches this set are unconditioned. This property will give us the opportu-nity to construct another open ladder point for a walk free of any constraintsfrom its past. Figure 4 illustrates the construction of regeneration times.

7.1. Control the variables M1 and M2.

Lemma 7.1. We have

P[M1 ≥ n | D′ < ∞] ≤ C exp(−cn).

Proof. We have

P[2k ≤ M1 < 2k+1]

≤P[T∂B(2k,2αk) 6= T∂+B(2k,2αk)]

+ P[XT∂B(2k,2αk)

∈ ∂+B(2k, 2αk), T+H−(0) θT

∂+B(2k,2αk)< TH+(2k+1) θT

∂+B(2k,2αk)],

now by a union bound on the 2αk possible positions of XT∂+B(2k,2αk)

and using

translation invariance arguments

P[2k ≤ M1 < 2k+1] ≤ 2αkP[T∂B(2k+1,2α(k+1)) 6= T∂+B(2k+1,2α(k+1))] + e−(2k),

23

0S1

S2

S3

R1

R2

open ladder points

random walk

Kgood(XS2)

Figure 4. The construction of regeneration times

as a consequence of Theorem 5.1.Hence using Theorem 5.1 again

P[2k ≤ M1 < 2k+1] ≤ c2αke−c(2k),

and since M1 < ∞ on D′ < ∞, we see that

P[M (1) ≥ n | D′ < ∞] ≤ 1

P[D′ < ∞]

∑

k, 2k≥n

P[2k ≤ M1 < 2k+1] ≤ Ce−cn.

Also

Lemma 7.2. We have

P[M2 ≥ n | 0 is not good] ≤ C exp(−cn).

Proof. Hence, by Remark 5.1, on 0 is not goodx where 0 is connected to x using a directed open path ⊂ BAD+(0),

so in particular

M2 ≤ supx∈BAD+(0)

x · ~ℓ + 1 ≤ W (BAD+(0)) + 1,

and by Lemma 5.2, we are done.

24 A. FRIBERGH

We introduce

(7.5) S(n) = for i with Si ≤ ∆n and Mi < ∞, Mi − XSi· ~ℓ ≤ n1/2.

Let us prove

Lemma 7.3. We have

P[S(n)c] ≤ exp(−n1/2).

Proof. By (7.3), we know that cardi; Si ≤ ∆n ≤ n.So by Theorem 5.1 we see that

P[S(n)c] ≤ P[TB(n,nα) 6= T∂+B(n,nα)]

(7.6)

+∑

i≤n

∑

x∈B(n,nα)

P[Mi − XSi· ~ℓ > n1/2, Mi < ∞, XSi

= x].

Now, we may see that

P[Mi − XSi· ~ℓ > n1/2, Mi < ∞, XSi

= x](7.7)

≤P

[

supn≤D′θSi

+Si

(Xn − x) · ~ℓ > n1/2,Mi < ∞, XSi= x, x is good

]

+ P[

sup(y − x) · ~ℓ, y ∈ Kgood(x) > n1/2 | x is not good]

,

where Kgood(x) was defined at (7.1).By translation invariance of P and Lemma 7.2, we have

(7.8)

P[sup(y − x) · ~ℓ, y ∈ Kgood(x) > n1/2 | x is not good] ≤ C exp(−cn1/2).

We may see that if Mi < ∞ and XSiis good then D′ θSi

+Si < ∞, hence

P[

supn≤D′θSi

+Si

(Xn − x) · ~ℓ ≥ n1/2,Mi < ∞, XSi= x, x is good

]

(7.9)

≤P[

supn≤D′θSi

+Si

(Xn − x) · ~ℓ ≥ n1/2, D′ θSi+ Si < ∞, XSi

= x]

By using Markov’s property at the time Si, we see that

P

[

supn≤D′θSi

+Si

(Xn − x) · ~ℓ ≥ n1/2, D′ θSi+ Si < ∞, XSi

= x]

(7.10)

≤E[

P ωx

[

supn≤D

Xn · ~ℓ ≥ n1/2, D′ < ∞]

]

=P

[

supn≤D′

Xn · ~ℓ ≥ n1/2 | D′ < ∞]

≤ C exp(−cn1/2),

where we used and translation invariance and Lemma 7.1. The result followsfrom putting together (7.6), (7.7), (7.8), (7.9) and (7.10).

25

7.2. Exponential tails for backtracking. We can deduce that

Lemma 7.4. We have

P[TH−(−n) < ∞] ≤ exp(−cn).

Proof. Fix n > 0. For this proof, we will use the notation

A(n) = T∂B(2n,2nα) = T∂+B(2n,2nα).

For any k ≥ n, let us denote BX(2k+1, 2(k+1)α) = z ∈ Zd, z = XT

∂B(2k,2kα)+

y with y ∈ B(2k+1, 2(k+1)α) and

C(k) = T∂+BX(2k+1,2(k+1)α) θT∂B(2k,2kα)

= T∂BX(2k+1,2(k+1)α) θT∂B(2k,2kα)

.

A simple induction shows that on ∩k∈[n,m]C(k)∩A(n), we have TH−(−2n−1) ≥TB(2m,m2αm), hence, we see that

(7.11) ∩k≥n C(k) ∩ A(n) ⊆ TH−(−2n−1) = ∞.Denote for m > n,

D(n,m) = A(n), for n ≤ k < m, C(k), C(m)c,so that

(

∩k≥nC(k) ∩ A(n))c

⊂ ∪m≥nD(n,m) ∪ A(n)c.

which implies with (7.11) that(7.12)

P[TH−(−2n) < ∞] ≤ P[A(n)c]+∑

m>n

P[D(n,m)] ≤ exp(−c2n)+∑

m≥n

P[D(n,m)],

by Theorem 5.1.We may notice that on D(n,m), we have T∂B(2m,m2mα) = T∂+B(2m,m2mα)

(note that is different from A(m)). Hence, when using Markov’s property atT∂B(2m,m2mα) the random walk is located in ∂+B(2m,m2mα), so

P[D(n,m)]

≤∑

x∈∂+B(2m,m2mα)

E[P ω[XT∂B(2m,m2mα)= x]P ω

x [Tx+∂B(2m+1,2(m+1)α) 6= Tx+∂+B(2m+1,2(m+1)α)]]

≤Cmd2dmα maxx∈∂+B(2m,m2mα)

E[P ωx [T∂B(2m+1,2(m+1)α) 6= T∂+B(2m+1,2(m+1)α)]]

≤Cmd2dmαP[A(m + 1)c] ≤ exp(−c2m).

by translation invariance and Theorem 5.1.The lemma follows from the previous and (7.12).

26 A. FRIBERGH

7.3. Uniformly bounded chances of never backtracking at open points.

We denote C = x > 0ν . For any a ∈ C, we define the environment ωax hav-

ing the same conductances as in ω on any edge non adjacent to x and whereall edges adjacent to c

ωax

∗ ([x, x + e]) = a(e) for any e ∈ ν.We say that a ∈ C is K-open if a(e) ∈ [1/K,K] for any e ∈ ν. Furthermore

a vertex x ∈ Zd is called open-good if x is good in a configuration ωa

x where ais open. Note that we do not need to specify the value of the conductances ofthe edges since the event x is good is measurable with respect to (c∗(e) ∈[1/K,K])e∈Zd .

Lemma 7.5. We have

E[

maxa∈C is open

P ωa0 [D′ < ∞] | 0 is open good

]

< 1.

Proof. Fix n > 0. On the event that 0 is open good, we denote P(i) adirected path starting at 0 where all points, except maybe 0, are open. Wedenote L∂+B(n,nα) = infi,P(i) ∈ ∂+B(n, nα). Now, we see that if the twofollowing conditions are verified

(1) Xi = P(i) for i ≤ L∂+B(n,nα),(2) TH−(2) θTP(L

∂+B(n,nα))= ∞,

then D′ = ∞.We can see that 0 is open good

mina∈C open

P ωa0 [Xi = P(i) for i ≤ L∂+B(n,nα)] ≥ κ2n

0 ,

by Remark 5.2.In particular, we have

E[

mina∈C open

P ωa0 [D′ = ∞] | 0 is open good

]

≥E[

mina∈C open

P ωa0 [Xi = P(i) for i ≤ L∂+B(n,nα)]

× Pωa

0

P(L∂+B(n,nα))

[TH−(2) = ∞] | 0 is open good]

≥κn0E

[

mina∈C open

Pωa

0

P(L∂+B(n,nα))

[TH−(2) = ∞] | 0 is open good]

.

Moreover we see that

Pωa

0

P(L∂+B(n,nα))

[TH−(2) = ∞] = P ωP(L

∂+B(n,nα))[TH−(2) = ∞],

so that for any n

E[

mina∈C open


]

≥κ2n0 E

[

P ωP(L

∂+B(n,nα))[TH−(2) = ∞] | 0 is open good

]

.

27

Now,

E[

P ωP(L

∂+B(n,nα))[TH−(2) < ∞] | 0 is open good

]

≤P[0 is open good]−1E[

P ωP(L

∂+B(n,nα))[TH−(1) < ∞]

]

≤CP[TH−(−n+2) < ∞],

where we used invariance by translation.Now, by Lemma 7.4, we see that the previous quantity is less than 1/2 for

n ≥ n0. Hence combining the last two equations

E[

mina∈C open


]

≥ (1/2)κn00 > 0,

which implies the result.

7.4. Number of trials before finding an open ladder point which is a

regeneration time. Let us introduce the collection of edges with maximum

scalar product with ~ℓ

E = e ∈ ν such that e · ~ℓ = e1 · ~ℓ,and

(7.13) Bx = e ∈ E(Zd), e = [−e1, f − e1] with f any unit vector of E.Imagining the bias is oriented to the right, the set of edges to the “left”of

x is defined to be

(7.14) Lx := [y, z] ∈ E(Zd), y · ℓ ≤ x · ℓ or z · ℓ ≤ x · ℓ ∪ Bx,

and the edges to the “right”are

(7.15) Rx := [y, z] ∈ E(Zd), y · ℓ > x · ℓ and z · ℓ > x · ℓ ∪ Bx.

We recall that N was defined at (7.4). Let us prove

Lemma 7.6. We have

P[N ≥ n] ≤ exp(−cn).

The discussion related to Figure 4 becomes important for the proof of thislemma.

Proof. We have(7.16)N ≥ n ⊆ C(n) := k ≤ n, Sk < ∞, XSk

is not good or D′ Sk +Sk < ∞.By the way our regeneration times are constructed, we can see that C(n)

is P ω-measurable with respect to σXk with k ≤ Sn+1, so using Markov’sproperty at Sn+1,

P[C(n + 1)]

28 A. FRIBERGH

≤∑

x∈Zd

E[1x is not goodP ω[XSn+1 = x,C(n)]]

+ E[1x is goodP ω[XSn+1 = x,C(n)]P ωx [D′ < ∞]]

≤∑

x∈Zd

E[1x is not open goodP ω[XSn+1 = x,C(n)]]

+ E[P ω[XSn+1 = x,C(n)]1x is open good maxa open

P ωx,a

x [D′ < ∞]],

where we used the fact that XSn+1 is open. Furthermore

(1) P ω[XSn+1 = x,C(n)] is measurable with respect to σc∗(e) with e ∈Lx,

(2) x is not open good and x is open goodmaxa open Pωa

xx [D < ∞] are

measurable with respect to σc∗(e) with e /∈ Lx.So we have independence between the random variable in (1) and those in

(2). Hence

P[C(n + 1)]

≤P[C(n)](P[x is not open good] + E[

1x is open good maxa∈C open

P ωax [D < ∞]

]

)

≤P[C(n)](

1 − P[0 is open good](

1 − E[

maxa∈C open

P ωa0 [D < ∞] | 0 is open good

])

)

where we used translation invariance. It is clear that P[0 is open good] ≥P[x is good] > 0 and further we use Lemma 7.5 to see that

P[C(n + 1)] ≤ (1 − c)P[C(n)] ≤ . . . ≤ (1 − c)n,

hence the result by (7.16).

7.5. Tails of regeneration times. Now

Theorem 7.1. For any M < ∞, there exists K0 such that, for any K ≥ K0

we have τ(K)1 < ∞, P-a.s. and

P[Xτ(K)1

· ~ℓ ≥ n] ≤ C(M)n−M .

Proof. Recalling the definitions (7.2) and (6.14), we may see that TH+(Mk), k ≥0 ⊂ Wk, k ≥ 0. This means that on M(n), defined at (6.15),

for k with Sk ≤ ∆n, XSk+1· ~ℓ − XTH+(Mk)

· ~ℓ ≤ n1/2,

where we used that TH+(Mk) is a ladder-point Wi0 and that TH+(Mk) = Wi0 ≤∆n implies that Sk ≤ ∆n.

Moreover on S(n), defined at (7.5), we have

for i with Si ≤ ∆n and Mi < ∞, Mi − XSi· ~ℓ ≤ n1/2.

29

Noticing that XTH+(Mk)· ~ℓ ≤ Mk + 1, we may see that, on S(n) ∩ M(n)

XSk+1· ~ℓ − XSk

· ~ℓ ≤ 2n1/2 + 1,

for any k with Sk < ∆n and Mk < ∞. By induction, this means that ifk ≤ n1/3, Sk < ∆n and Mk < ∞, then

XSk+1· ~ℓ ≤ k(2n1/2 + 1) < n,

and so that Sk+1 ≤ ∆n, since XSk+1is a new maximum for the random walk

in the direction ~ℓ.Hence, if N ≤ n1/3 and M(n) ∩ S(n), then for n large enough

Xτ1 · ~ℓ ≤ n1/3(2n1/2 + 1) < n.

Thus

P[Xτ1 · ~ℓ ≥ n] ≤P[N ≥ n1/3] + P[M(n)c] + P[S(n)c]

≤ exp(−cn1/3) + 2n−M ≤ 3n−M ,

by Lemma 7.3, Lemma 6.6 and Lemma 7.6. This concludes the proof.

7.6. Fundamental property of regeneration times. Then let us definethe sequence τ0 = 0 < τ1 < τ2 < · · · < τk < . . ., via the following procedure

(7.17) τk+1 = τ1 + τk(Xτ1+· − Xτ1 , ω( · + Xτ1)), k ≥ 0,

meaning we look at the k + 1-th regeneration time is the k-th regenerationtime after the first one.

We set

Gk := στ1, . . . , τk; (Xτk∧m)m≥0; c∗(e), e ∈ LXτk+1,Let us introduce for any x ∈ Z

d

ax = (c∗([x − e1, x − e1 + e]))e∈E = (c∗(e))e∈Bx∈ [1/K,K]E ,

recalling notation from (7.13). For a ∈ [1/K,K]E ,

Pax = δa

(

(c∗([x − e1, x − e1 + e]))e∈E

)

⊗∫

e∈E(Zd)\Bx

⊗dP(c∗(e)),

and the associated annealed measure

Pax = Pa

x × P ωx .

We may notice that Theorem 7.1, can easily be generalized to become

Theorem 7.2. For any M < ∞, there exists K0 such that, for any K ≥ K0,

we have τ(K)1 < ∞, P

a0-a.s. for a ∈ [1/K,K]E and

maxa∈[1/K,K]E

Pa0[Xτ

(K)1

· ~ℓ ≥ n] ≤ Cn−M ,

30 A. FRIBERGH

Similarly, we can turn Theorem 5.1 into

Theorem 7.3. For α > d + 3

maxa∈[1/K,K]E

Pa0[T∂B(L,Lα) 6= T∂+B(L,Lα)] ≤ e−cL.

The fundamental properties of regeneration times are that

(1) the past and the future of the random walk that has arrived Xτkare

only linked by the conductances of the edges in aXτk,

(2) the law of the future of the random walk has the same law as a random

walk under PaXτk

0 [ · | D = ∞].

Now let us state a theorem corresponding to the previous heuristic.

Theorem 7.4. Let f , g, hk be bounded and respectively, σXn : n ≥ 0-,σc∗(e), e ∈ R0− and Gk-measurable functions. Then for a ∈ [1/K,K]E ,

Ea[f(Xτk+· − Xτk

)g tXτkhk] = E

a[hkEaXτk

0 [fg | D = ∞]].

A similar theorem was proved in [15] (as Theorems 3.3 and 3.5). In ourcontext, the random variables studied: τ1, D etc. are defined differently fromthe corresponding ones in [15]. Nevertheless, our notations were chosen sothat we may prove Theorem 7.4 simply by following word for word the proofsof Theorems 3.3 and 3.5 in [15]. The reader should start reading after Remark3.2 in [15] to have all necessary notations. To avoid any possible confusionwe point out that in [15], the measures P, Px,ω and P correspond respectivelyto the environment, the quenched random walk and the annealed measureand that ω(b) denotes the conductances of an edge b.

We bring the reader’s attention to the fact that we have not proved yetτk < ∞, P-a.s. (or P

a-a.s. for any a ∈ [1/K,K]). We only know this fork = 1, which is enough to prove Theorem 7.4 for k = 1. Using Theorem 7.4for k = 1 and Theorem 7.2, we may show that τ2 < ∞, P-a.s. (or P

a-a.s. forany a ∈ [1/K,K]) and thereafter obtain Theorem 7.4 for k = 2. Hence, wemay proceed by induction to prove Theorem 7.4 alongside the following

Proposition 7.1. For any k ≥ 1, we have τk < ∞ P-a.s. (or Pa-a.s. for any

a ∈ [1/K,K]).

We notice that this implies Proposition 3.1, which states directional tran-

sience in the direction ~ℓ for the random walk.As in [15], we may notice that a consequence of Theorem 7.4 is

Proposition 7.2. Let

Γ := N × Zd × [1/K,K]E ,

31

with its canonical product σ-algebra and let yk = (jk, zk, ak) ∈ Γ, k ≥ 0. Fora ∈ [1/K,K]E and G ⊂ Γ measurable let also

R(a; G) := Pa0[(τ1, Xτ1 , aXτ1

) ∈ G | D = ∞].

Then under P the Γ-valued random variables (with τ0 = 0),

(7.18) Yk := (Jk, Zk, Ak) := (τk+1 − τk, Xτk+1− Xτk

, aτk+1), k ≥ 0,

define a Markov chain on the state space Γ, which has transition kernel

P[Yk+1 ∈ G | Y0 = y0, . . . , Yk = yk] = R(ak; G),

and initial distribution

Λ(G) := P[(τ1, Xτ1 , aXτ1) ∈ G].

Similarly, on the state space [1/K,K]E , the random variables

(7.19) Ak = aXτk+1, k ≥ 0,

also define a Markov chain under P. With a ∈ [1/K,K]E and B ⊂ [1/K,K]E

measurable, its transition kernel is

R(a; B) := Pa0[aXτ1

∈ B | D = ∞] =∑

j∈N,z∈Zd

R(a; (j, z, B)),

and the initial distribution is

(7.20) Λ(B) := P[aXτ1∈ B] = Λ((j, z, B)).

Now let us quote Lemma 3.7 and Theorem 3.8 from [15],

Theorem 7.5. There exists a unique invariant distribution ν for the transi-tion kernel R. It verifies

supa∈[1/K,K]E

||Rm(a; ·) − ν(·)||var ≤ Ce−cm, m ≥ 0,

where ||·||var denotes the total variation distance.Further, this probability measure ν is invariant with respect to the transition

kernel R; that is, νR = ν, and the Markov chain (Ak)k≥0, defined in (7.19)with transition kernel R and initial distribution ν on the state space [1/K,K]E

is ergodic. Moreover, the initial distribution Λ(·) given in (7.20) is absolutelycontinuous with respect to ν(·).

Theorem 7.6. The distribution ν := νR is the unique invariant distributionfor the transition kernel R. It verifies

supa∈[1/M,M ]E

∣

∣

∣

∣

∣

∣Rm(a; ·) − ν(·)

∣

∣

∣

∣

∣

∣

var≤ Ce−cm, m ≥ 0.

32 A. FRIBERGH

With initial distribution equal ν, the Markov chain (Yk)k≥0 defined in (7.18)is ergodic. Moreover, the law of the Markov chain (Yk+1)k≥0 under P is abso-lutely continuous with respect to the law of the chain with initial distributionν.

The proofs in [15] carry over to our context simply, once we have shownthe following Doeblin condition: there exists c > 0 such that for any a ∈[1/K,K]E , we have

R(a,B) ≥ c ⊗E P[c∗ ∈ B | c∗ ∈ [1/K,K]].

Let us prove this condition,

R(a; B) = Pa0[aXτ1

∈ B | D = ∞]

=1

Pa0[D

′ = ∞, 0 is good]Ea[P ω

0 [aXτ1∈ B,D′ = ∞, 0 is good]].

This means, using Remark 5.2

R(a; B) ≥ cEa[P ω0 [X1 = e1, X2 = 2e1, D′ θ2 = ∞]

. . . aX2 ∈ B, e1 is open and 2e1 is good]

≥cκ20E

a[P ω2e1

[D′ = ∞], a2e1 ∈ B, e1 is open and 2e1 is good]

≥cκ20E

a[

mina open

Pωa

02e1

[D′ = ∞], a2e1 ∈ B, e1, 2e1 are open

and 2e1 is open good]

,

now seeing that

(1) mina open Pωa

02e1

[D′ = ∞] and 2e1 is open good are measurable with

respect to σc∗([y, z]) with (y − 2e1) · ~ℓ > 0, z 6= 2e1 ,(2) a2e1 ∈ B, e1, 2e1 are open is measurable with respect to σc∗([y, z]) with (y−

2e1) · ~ℓ ≤ 0 or 2e1 = zhence they are P-independent so that

R(a; B) ≥ cEa[

mina open

Pωa

02e1

[D = ∞], 2e1 is open good]

× Pa0[a2e1 ∈ B, e1, 2e1 are open]

≥ cPa0[a2e1 ∈ B, e1, 2e1 are open],

by Lemma 7.5. Now by simple combinatorics we see that

R(a; B) ≥ c ⊗E P[c∗ ∈ B | c∗ ∈ [1/K.K]],

which is the Doeblin condition we were looking for.

33

8. Integrability of regeneration times

Our aim is to show that

Theorem 8.1. If E∗[c∗] < ∞, we have,

maxa∈[1/K,K]E

Ea[∆n | D = ∞] ≤ Cn.

Firstly, we notice

Lemma 8.1. For any x ∈ GOODK(ω) we have

Eωx

[

∞∑

i=0

1Xi = x]

≤ C(K) < ∞.

Proof. We see that

Eωx

[

∞∑

i=0

1Xi = x]

=1

P ωx [T+

x = ∞]=

πω(x)

Cω(x ↔ ∞),

where Cω(x ↔ ∞) is the effective conductance between x and infinity in ω.Since x ∈ GOODK , we can upper-bound πω(x) using Remark 5.2 and wemay use Rayleigh’s monotonicity principle to see that

Cω(x ↔ ∞) ≥ c

K

∑

i≥0

cω(pi) ≥ c exp(2λx · ~ℓ),

where (pi)i≥0 is a directed path of open points starting at x. This yields theresult.

8.1. Time spent in traps. For x ∈ ∂BAD(ω), we define BADsx(K) =

∪y∼xBADK(y) the union of all bad areas adjacent to x. It follows fromLemma 5.1 that

Lemma 8.2. For x ∈ ∂BAD(ω), we have that BADsx(K) is finite P-a.s. and

Pp[W (BADsx(K)) ≥ n] ≤ C exp(−ξ1(K)n),


We have

Lemma 8.3. For any x ∈ ∂BAD(ω) we have that

Eωx [T+

GOOD(ω)] ≤ C(K) exp(3λ |∂BADsx(ω)|)

(

1 +∑

e∈E(BADs

x)

cω∗ (e)

)

.

34 A. FRIBERGH

Proof. The first remark to be made is that since x ∈ ∂BAD(ω) ⊂ GOOD(ω),hence for all y ∼ x then c∗([x, y]) ∈ [1/K,K].

Let us consider the finite network obtained by taking BADsx(ω)∪∂BADs

x(ω)and merging all points of ∂BADs

x(ω) (which contains x) to one point δ. Wedenote ωδ the resulting graph which is obviously finite by Lemma 8.2 andconnected. We may apply the mean return formula at δ to obtain that

Eωδ

δ [T+δ ] = 2

∑

e∈E(ωδc(e)

πωδ

BADsx(δ)

= 2

∑

e∈E(BADsx) c(e) + πωδ

BADsx(δ)

πωδ

BADsx(δ)

.

For y a neighbor of δ in ωδ, we have by Remark 5.2

(8.1) c exp(

2λ miny∈∂BADs

x(δ)y · ~ℓ

)

≤ cωδ([δ, y]) ≤ C exp(

2λ maxy∈∂BADs

x(δ)y · ~ℓ

)

,

so that we know that(8.2)

c exp(

2λ maxy∈∂BADs

x(δ)y ·~ℓ

)

≤ πωδ

BADsx(δ) ≤ C |∂BADs

x(ω)| exp(

2λ maxy∈∂BADs

x(δ)y ·~ℓ

)

.

Using (8.1) and (8.2) and Remark 5.2

for e ∈ E(BADsx),

c(e)

πωδ

BADsx(δ)

≤ Cc∗(e),

which means that

(8.3) Eωδ

δ [T+δ ] ≤ C

∑

e∈E(BADsx)

c∗(e) + C.

The transition probabilities of the random walk in ωδ at any point differentfrom δ are the same as that of the walk in ω. This implies that

Eωδ

δ [T+δ ] =

∑

y∼δ

P ωδ

δ [X1 = y]Eωy [T∂BADs

x](8.4)

=∑

y∈BADsx, y∼∂BADs

x

P ωδ

δ [X1 = y]Eωy [T∂BADs

x],

by (8.1) and (8.2), we have

P ωδ

δ [X1 = y] =cωδ([δ, y])

πωδ

BADsx(δ)

≥ cexp(2λ miny∈∂BADs

xy · ~ℓ)

|∂BADsx(ω)| exp(2λ maxy∈∂BADs

xy · ~ℓ)

,

and now since BADsx ∪ ∂BADs

x is connected we have

maxy∈∂BADs

x(δ)y · ~ℓ − min

y∈∂BADsx

y · ~ℓ ≤ |∂BADsx(ω)| ,

so that

P ωδ

δ [X1 = y] ≥ c |∂BADsx(ω)|−1 exp(−2λ |∂BADs

x(ω)|).

35

This, with (8.3) and (8.4), and considering the fact that ∂BADsx ⊂ GOOD

yields

maxy∈BADs

x, y∼∂BADsx

Eωy [TGOOD(ω)] ≤ C exp(3λ |∂BADs

x(ω)|)(

1 +∑

e∈E(BADsx)

c∗(e))

.

So

Eωx [T+

GOOD(ω)] ≤ C exp(3λ |∂BADsx(ω)|)

(

1 +∑

e∈E(BADsx)

c∗(e))

.

8.2. Conductances in traps. Let us understand how the conductances intraps are conditioned.

Lemma 8.4. Take n ≥ 0 and K ≥ 1, F ⊂ E(Zd) such that 0 /∈ V (F ), ande ∈ F . If E∗[c∗] < ∞ then

E[1E(BADsx(K)) = FP ω[TV (F ) ≤ ∆n]c∗(e)]

≤CE[1E(BADsx(K)) = FP ω[TV (F ) ≤ ∆n]].

If lim ln P∗[c∗>n]ln n

= −γ with γ < 1 then for any ε > 0 we have

E[1E(BADsx(K)) = Fc∗(e)γ−ε] ≤ CE[1E(BADs

x(K)) = F].We introduce the notation ωe an to signify that for e′ ∈ E(Zd) \ e,

1e′ is abnormal(ωe an) = 1e′ is abnormal(ω)

and

1e is abnormal(ωe an) = 1.

Proof. Firstly let us notice that if there exists M such that P [c∗ < M ] = 1,then we may obtain the first part of the lemma with C = M . We will nowassume that P [c∗ > M ] > 0 for any M .

Take F ⊂ E(Zd) such that x ∈ F and 0 /∈ F . For any e ∈ F .

E[1E(BADsx) = FP ω[TV (F ) ≤ ∆n]c∗(e)]

(8.5)

≤KE[1E(BADsx) = FP ω[TV (F ) ≤ ∆n]]

+ E[c∗(e)1c∗(e) > K1E(BADsx) = FP ω[TV (F ) ≤ ∆n]]

=KE[1E(BADsx) = FP ω[TV (F ) ≤ ∆n]]

+ E[c∗(e)1c∗(e) > K1E(BADsx(ω

e an)) = FP ω[TV (F ) ≤ ∆n]]

≤KE[1E(BADsx) = FP ω[TV (F ) ≤ ∆n]]

+ E[c∗(e)1E(BADsx(ω

e an)) = FP ω[TV (F ) ≤ ∆n]].

36 A. FRIBERGH

Using the fact that c∗(e) is independent of E(BADsx(ω

e an)) = F andP ω[TV (F ) ≤ ∆n] since 0 /∈ V (F ) and e ∈ F . Hence

E[c∗(e)1E(BADsx(ω

e an)) = FP ω[TV (F ) ≤ ∆n]](8.6)

≤E[c∗(e)]E[1E(BADsx(ω

e an)) = FP ω[TV (F ) ≤ ∆n]].

We can use this same independence property again to write

E[1E(BADsx(ω

e an)) = FP ω[TV (F ) ≤ ∆n]]

(8.7)

≤ 1

P∗[c∗(e) > K]E[1c∗(e) > K1E(BADs

x(ωe an)) = FP ω[TV (F ) ≤ ∆n]]

≤ 1

P∗[c∗(e) > K]E[1c∗(e) > K1E(BADs

x) = FP ω[TV (F ) ≤ ∆n]]

≤ 1

P∗[c∗(e) > K]E[1E(BADs

x) = FP ω[TV (F ) ≤ ∆n]].

Putting together (8.5), (8.6) and (8.7) proves the first part of the result.The second part can be handled using exactly the same techniques.

8.3. Proof of Theorem 8.1.

Lemma 8.5. For any K, we have

maxa∈[1/k,K]E

E[∆n | D = ∞] ≤ n maxa∈[1/k,K]E

C∑

x∈Zd

Pa[TBADs

x(K) ≤ ∆n]1/2.

Proof. If 0 ∈ GOOD(ω), then a walk started at 0 can only be in a vertex ofBAD(ω) between visits to ∂BAD(ω). Hence on 0 ∈ GOOD(ω),

∑

x∈BAD(ω)

∆n∑

i=0

1Xi = x ≤∑

x∈∂BAD(ω)

∆n∑

i=0

1Xi = xT+GOOD(ω) θi.

Hence, on 0 ∈ GOOD(ω), since Zd is partitioned in two parts GOOD(ω)

and BAD(ω) we have

τ1 ≤∑

x∈Zd

∆n∑

i=0

1Xi = x

(8.8)

≤∑

x∈GOOD(ω)

∆n∑

i=0

1Xi = x +∑

x∈BAD(ω)

τ1∑

i=0

1Xi = x

≤∑

x∈GOOD(ω)

∆n∑

i=0

1Xi = x +∑

x∈∂BAD(ω)

τ1∑

i=0


37

Hence, on 0 ∈ GOOD(ω),

∆n ≤∑

x∈GOOD(ω)

1Tx ≤ ∆n∞

∑

i=0

1Xi = x(8.9)

+∑

x∈∂BAD(ω)

1Tx ≤ ∆n∞

∑

i=0


We can use Markov’s property to say that for any x ∈ Zd, on 0 ∈

GOOD(ω)

Eω[

1Tx ≤ ∆n∞

∑

i=0

1Xi = x]

= P ω[Tx ≤ ∆n]Eωx

[

∞∑

i=0

1Xi = x]

,

and

Eω[

1Tx ≤ ∆n∞

∑

i=0

1Xi = xT+GOOD(ω) θi

]

=P ω[Tx ≤ ∆n]Eωx

[

∞∑

i=0


]

.

This implies, using (8.9), that on 0 ∈ GOOD(ω)

Eω[∆n] ≤∑

x∈GOOD(ω)

P ω[Tx ≤ ∆n]Eωx

[

∞∑

i=0

1Xi = x]

+∑

x∈∂BAD(ω)


[

∞∑

i=0


]

,

Now, we have

maxa∈[1/K,K]E

Ea[∆n | D = ∞]

(8.10)

≤ mina∈[1/K,K]E

Pa[D = ∞]−1 max

a∈[1/K,K]E

ES[10 ∈ GOOD(ω)∆n](8.11)

≤C maxa∈[1/K,K]E

[

Ea[

∑

x∈GOOD(ω)


[

∞∑

i=0

1Xi = x]]

+ maxa∈[1/K,K]E

Ea[

10 /∈ BAD∑

x∈∂BAD(ω)


[

∞∑

i=0

1Xi = xT+GOOD(

38 A. FRIBERGH

by Lemma 7.5, so that using Lemma 8.1

maxa∈[1/K,K]E

Ea[∆n | D = ∞]

≤C[

maxa∈[1/K,K]E

Ea[

∑

x∈Zd

1Tx ≤ ∆n]

+ maxa∈[1/K,K]E

Ea[

∑

x∈∂BAD(ω)


[

∞∑

i=0


]]]

.

Let us for now focus on the second term. By Markov’s property for x ∈∂BAD(ω)

Eωx

[

∞∑

i=0


]

=∞

∑

i=0

Eωx [1Xi = x]Eω

x [T+GOOD(ω)]

=(

∞∑

i=0

Eωx [1Xi = x]

)

Eωx [T+

GOOD(ω)]

≤ CEωx [T+

GOOD(ω)],

where we used Lemma 8.1. Hence

maxa∈[1/K,K]E

Ea[

10 /∈ BAD∑

x∈∂BAD(ω)


[

∞∑

i=0


]]

(8.12)


Ea[

10 /∈ BAD∑

x∈∂BAD(ω)

P ω[Tx ≤ ∆n]Eωx [T+

GOOD(ω)]]

≤C∑

x∈Zd

maxa∈[1/K,K]E

Ea[

1x ∈ ∂BAD(ω)10 /∈ BADP ω[Tx ≤ ∆n]Eωx [T+

GOOD(ω)]]

≤C∑

x∈Zd

maxa∈[1/K,K]E

Ea[

1x ∈ ∂BAD(ω)10 /∈ BADP ω[TBADsx≤ ∆n]Eω

x [T+GOOD(ω)]

]

,

where we used that for x ∈ ∂BAD(ω), we have x ∈ BADsx by definition.

Now using Lemma 8.3

maxa∈[1/K,K]E

Ea[

10 /∈ BAD1x ∈ ∂BAD(ω)P ω[TBADsx≤ ∆n]Eω

x [T+GOOD(ω)]

]

(8.13)


Ea[

10 /∈ BAD1x ∈ ∂BAD(ω)P ω[TBADsx≤ ∆n] exp(3λ |∂BADs

x(ω)|)(

1 +∑

e∈E(BADsx)

c∗(e))]

39

≤∑

F⊂E(Zd)x∈V (F ),0/∈V (F )

C exp(3λ |∂V (F )|)∑

e∈E(F )

maxa∈[1/K,K]E

Ea[

1E(BADsx) = FP ω[TV (F ) ≤ ∆n]

(

1 + c∗(e))

]

now by Lemma 8.4 and using the fact that E∗[c∗] < ∞

maxa∈[1/K,K]E

Ea[

1E(BADsx) = F1x ∈ ∂BAD(ω)P ω[TV (F ) ≤ ∆n]c∗(e)

]

(8.14)


Ea[

1E(BADsx) = FP ω[TV (F ) ≤ ∆n]

]

,

hence with (8.13) we have

maxa∈[1/K,K]E

Ea[

10 /∈ BAD1x ∈ ∂BAD(ω)P ω[TBADsx≤ ∆n]Eω

x [T+GOOD(ω)]

]

(8.15)


Ea[

exp(3λ |∂BADsx|) |∂BADs

x|d P ω[TBADsx≤ ∆n]

]

,

where we used that |E(BADsx)| ≤ C |∂BADs

x|d.Using (8.12), (8.15) and (8.10) this means

maxa∈[1/K,K]E

Ea[∆n | D = ∞]


[

∑

x∈Zd

Pa[TBADs

x≤ ∆n] +

∑

x∈Zd

Ea[

exp(4λ |∂BADsx|)P ω[TBADs

x≤ ∆n]

]

]


∑

x∈Zd

Ea[


x≤ τn]

]

,

since τn ≥ ∆n by (7.3).Now, using the notation τ0 = 0,

maxa∈[1/K,K]E

∑

x∈Zd

Ea[


x≤ τn]

]

≤ maxa∈[1/K,K]E

n−1∑

i=0

Ea[

∑

x∈Zd

exp(4λ |∂BADsx|)1TBADs

x∈ [τi, τi+1]

]

,

and using Theorem 7.4

maxa∈[1/K,K]E

n−1∑

i=0

Ea[

∑

x∈Zd


x∈ [τi, τi+1]

]

,

≤ n

mina∈[1/K,K]E Pa[D = ∞]max

a∈[1/K,K]EE

a[

∑

x∈Zd


x≤ τ1

]

,

40 A. FRIBERGH

by Lemma 7.5.The result follows from the two last equations and Minkowski’s inequality

which implies

maxa∈[1/K,K]E

Ea[


x≤ τ1]

]

≤ maxa∈[1/K,K]E

Ea[

exp(6λ |∂BADsx|)

]1/2Ea

[

P ω[TBADsx≤ τ1]

2]1/2


Ea[

P ω[TBADsx≤ τ1]

]1/2= C max

a∈[1/K,K]EP

a[TBADsx≤ τ1]

1/2,

by Lemma 8.2 and the inequality x2 ≤ x for x ∈ [0, 1].

Let us estimate Pa[TBADs

x(K) ≤ τ

(K)1 ].

Lemma 8.6. We have for any x ∈ Zd then for any M < ∞, there exists K

large enough such that

maxa∈[1/K,K]E

Pa[TBADs

x(K) ≤ τ

(K)1 ] ≤ Cn−M .

Proof. Denote χ the smallest integer so that Xi, i ∈ [0, τ1] ⊂ B(χ, χα).First let us notice that

maxa∈[1/K,K]E

Pa[χ ≥ k] ≤ max

a∈[1/K,K]EP

a[Xτ1 · ~ℓ ≥ k]

(8.16)

+ maxa∈[1/K,K]E

Pa[Xτ1 · ~ℓ ≤ k, max

0≤i,j≤τ1maxl∈[2,d]

|(Xj − Xi) · fl| ≥ kα].

We can upper-bound the first term as follows

maxa∈[1/K,K]E

Pa[X

τ(K)1

· ~ℓ ≥ k] ≤ Cn−M ,

for any M by choosing K large enough by Theorem 7.2.The second term can be upper-bounded with the following reasoning: on

the event that Xτ1 · ~ℓ ≤ k and maxj 6=1 max0≤j1,j2≤τ1 |(Xj1 − Xj2) · fj| ≥ kα,Xn does not exit the box B(k, kα) through ∂+B(k, kα), this means

maxa∈[1/K,K]E

Pa[Xτ1 · ~ℓ ≤ k, max

0≤i,j≤τ1maxl∈[2,d]

|(Xj − Xi) · fl|]

≤ maxa∈[1/K,K]E

Pa[T∂B(k,kα) 6= T∂+B(k,kα)] ≤ ce−ck,

by Theorem 7.3.This turns (8.16) into

(8.17) maxa∈[1/K,K]E

Pa[χ ≥ k] ≤ Ck−M ,

for any M for K large enough.Now assume that TBADs

x≤ τ1 then either

41

(1) |∂BADsx| ≥ |x| /2

(2) or B(χ, χα) 6⊆ BZd(0, |x| /2),

but by Lemma 8.2

maxa∈[1/K,K]E

Pa[|∂BADs

x| ≥ |x| /2] ≤ C exp(−c |x|),

and

maxa∈[1/K,K]E

Pa[B(χ, χα) 6⊆ BZd(0, |x| /2)] ≤ max

a∈[1/K,K]EP

a[χ ≥ (|x| /2)1/α] ≤ C |x|−M/α ,

so

maxa∈[1/K,K]E

Pa[TBADs

x≤ τ1]

≤ maxa∈[1/K,K]E

P[B(χ, χα) 6⊆ BZd(0, |x| /2)] + maxa∈[1/K,K]E

Pa[|∂BADs

x| ≥ |x| /2]

≤C |x|−M/α ,

which proves the lemma.

We can now prove Theorem 8.1

Proof. By choosing M > 2d, applying Lemma 8.5 and Lemma 8.6, we cansee that

maxa∈[1/K,K]E

Ea[∆n] ≤ Cn

∑

x∈Zd

C |x|−M/2 < Cn,

for K large enough, which proves Theorem 8.1.

9. Law of large numbers

We can use exactly the same type of proof as in [15] to obtain

Proposition 9.1. If E∗[c∗] < ∞, then

Xn

n→ v =

EΠ[Xτ1 ]

EΠ[τ1]P-a.s. with v · ~ℓ > 0,

where

Π[·] =

∫

ν(da)Pa0[· | D = ∞] and E

Π[·] :=

∫

ν(da)Ea[· | D = ∞],

where ν is the unique invariant distribution on [1/K,K]E given in Theo-rem 7.5.

Proof. Firstly, we notice that (8.17), we have EΠ[|Xτ1 |] < ∞. Then, one may

follow the proof of Theorem 5.1 in [15] to obtain the result. For the conve-nience of the reader, we recall that the notations necessary to understand the

42 A. FRIBERGH

proof in [15] were defined in Proposition 7.2, Theorem 7.5 and Theorem 7.6of this paper. It allows us to prove that P− a.s. (or P

a-a.s. for a ∈ [1/K,K])

(9.1)Xn

n→ v =

EΠ[Xτ1 ]

EΠ[τ1],

and in a similar fashion

(9.2)∆n

n→ E

Π[τ1]

EΠ[Xτ1 · ~ℓ].

where the two previous equations remain true even if EΠ[τ1] = ∞.

We may see that (9.2) and Theorem 8.1 implies that, if E∗[c∗] < ∞, then

EΠ[τ1] < ∞. Since Xτ1

~ℓ > 2/√

d, this means that (9.1)implies that v · ~ℓ >0.

Remark 9.1. Interestingly, we do not know of any direct way of showingthat E

Π[τ1] < ∞.

10. Zero-speed regime

10.1. Characterization of the zero-speed regime. We set A to be theset of vertices: 0, e1, e1 + ei, 2e1 + ei, 2e1 + 2ei, 3e1 + 2ei, 3e1 + ei, 4e1 + ei,for all i ∈ [2, 2d − 1] and

A = any x ∈ A is 4e1-open and B = 4e1 is good,where a vertex is x-open if it is open in ωx coinciding with ω on all edges butthose that are adjacent to x which are normal in ωx.

Note that

(1) on A ∩ B, the vertex 0 is good,(2) A and B are independent of c∗([2e1, 3e1]).

Lemma 10.1. If E∗[c∗] = ∞, then mina∈[1/K,K]E Ea[τ1 | D = ∞] = ∞.

The typical configuration that will slow the walk down is depicted in Figure5: the walk is likely to reach the edge [2e1, 3e1] and then stay there for long.Moreover, we may notice that this picture is compatible with the conditioningD = ∞ when 4e1 is good.

Proof. Since, on A ∩ B, the vertex 0 is good, we have

(10.1) Ea[τ1 | D = ∞] ≥ E

a[1D = ∞τ1] ≥ cEa[1A,B1D′ = ∞τ1].

On A, we see, by Remark 5.2, that we have

P ω[X1 = e1, X2 = 2e1, X3 = e1, X4 = 2e1] ≥ κ40.

43

0

4e1

[2e1, 3e1]

normal edges

Figure 5. The typical configurations slowing the walk down

Using Remark 5.2 again, we may see that on A, we have P ω2e1

[X1 6= 3e1] ≤C/c∗([2e1, 3e1]) and P3e1 [X1 6= 2e1] ≤ C/c∗([2e1, 3e1]). This implies that

(10.2) P ω2e1

[TZd\2e1,3e1 ≥ n] ≥ (1 − C/c∗([2e1, 3e1]))n,

so

(10.3) Eω2e1

[TZd\2e1,3e1] ≥ cc∗([2e1, 3e1]).

Also on A, we may notice that for any neighbor of 2e1 or 3e1, there existsan open nearest-neighbor path of length at most 5 in A \ 0 to 4e1. UsingRemark 5.2, this implies that

(10.4) P ω2e1

[T4e1 θTZd\2e1,3e1

θ4 < T∂A\0 θTZd\2e1,3e1

θ4 ] ≥ κ50.

We may notice that on A, if

(1) X1 = e1, X2 = 2e1, X3 = e1, X4 = 2e1, (hence τ1 ≥ 4)(2) T4e1 θT

Zd\2e1,3e1θ4 < T∂A\0 θT

Zd\2e1,3e1θ4 ,

(3) D′ θT4e1= ∞,

then we have D′ = ∞ and τ1 ≥ TZd\2e1,3e1θ4. Using this, Markov’s propertytwice along with (10.4) and (10.3) we may see that

Ea[1A,B1D′ = ∞τ1] ≥cEa[1A,BEω

2e1[TZd\2e1,3e1]P

ω4e1

[D′ = ∞]]

(10.5)

≥cEa[1A,Bc∗([2e1, 3e1])Pω4e1

[D′ = ∞]].

44 A. FRIBERGH

We may now notice that 1A, c∗([2e1, 3e1]) and 1BP ω4e1

[D′ = ∞] arePa-independent, so that

Ea[1A,B1D′ = ∞τ1] ≥ Pa

0[A]Ea[c∗([2e1, 3e1])]Ea[1BP ω

4e1[D′ = ∞]].

We have mina∈[1/K,K]E Pa[A] ≥ c > 0, Ea[c∗([2e1, 3e1])] = E[c∗([2e1, 3e1])] =∞ by translation invariance and

Ea[1BP ω4e1

[D′ = ∞]] = E[10 is goodP ω[D′ = ∞]] > 0,

by Lemma 7.5 and the fact that P[0 is good] > 0.Hence, by (10.1) and (10.5), we have

mina∈[1/K,K]E

Ea[τ1 | 0 − regen] = ∞.

Remark 10.1. Using a reasoning similar to the previous proof but using anormal edge surrounded by edges with small conductances, we may show that

if c∗ < M < ∞ and E[mini=1,...,4d−2 1/c(i)∗ ] = ∞ then

E0[τ21 | 0 − regen] = ∞.

This implies

Proposition 10.1. If E∗[c∗] = ∞, then lim Xn/n = ~0 P-a.s.

Proof. Using Theorem 7.2

maxa∈[1/K,K]E

Ea[X

τ(K)1

· ~ℓ] < ∞.

which implies

EΠ[Xτ1 · ~ℓ] < ∞.

Because of Theorem 7.6, we may Birkhoff’s ergodic theorem (p.341 in [7])to see that

(10.6)(Xτn

− Xτ1) · ~ℓn

→ EΠ[Xτ1 · ~ℓ] < ∞.

Now, by Lemma 10.1, we see that

EΠ[τ1] = ∞,

which implies thatτn − τ1

n→ ∞.

From here, we may argue as in Theorem 5.1 in [15] to see that

limXn

n= ~0.

45

10.2. Lower-bound on the fluctuations of the random walk.

Lemma 10.2. If − lim ln P∗[c∗>n]ln n

= γ < 1, we have

lim infln ∆n

ln n≥ 1/γ, P-a.s..

Proof. Firstly, using (10.6) and the law of large numbers, we see that for csmall enough

P[τcn ≤ ∆n] → 1.

Using (10.2) and a reasoning similar to the proof of Lemma 10.1

mina∈[1/K,K]E

Pa0[τ1 ≥ n | D = ∞] ≥ cE[(1 − C/c∗([e1, 2e1]))

n]

≥ cP∗[c∗ ≥ n1/(1+ε1/2)] ≥ cn−(γ(1+ε1)),

for any ε1 > 0.

Notice that if∑cn−1

i=1 (τi+1 − τi) ≤ n1γ−ε then τi+1 − τi ≤ n

1γ−ε for any

i ≤ cn − 1. The previous two equations imply for any ε > 0 we have

P[∆n ≤ n1γ−ε] ≤ P

[

cn−1∑

i=1

(τi+1 − τi) ≤ n1γ−ε

]

+ o(1)

≤ o(1) +cn−1∏

i=1

maxa∈[1/K,K]E

Pa0[τ1 < cn

1γ−ε | D = ∞]

≤ o(1) + (1 − cn−( 1γ−ε)γ(1+ε1))cn−1,

by Theorem 7.4, for any ε1 > 0.Then taking ε1 > 0 small enough such that ( 1

γ− ε)γ(1 + ε1) < 1, we see

that

P[∆n ≤ n1γ−ε] → 1.

This being true for all ε > 0, we have the lemma.

10.3. Upper-bound on the fluctuations of the random walk.

Lemma 10.3. Assume that − lim ln P∗[c∗>n]ln n

= γ < 1. For any ε > 0, thereexists K0 such that, for any K ≥ K0

maxa∈[1/K,K]E

Pa[τ

(K)1 > n | D = ∞] ≤ C(K)n−(γ−ε)

Proof. To simplify the notations, we will do the proof for P. In a similarfashion, we could do it for any P

a for a ∈ [1/K,K]K.In this proof, we will point out the K dependence of constants. Fix ε > 0.

Denote χ the smallest integer so that Xi, i ∈ [0, τ1] ⊂ B(χ, χα). By (8.8),

46 A. FRIBERGH

on 0 − regen

τ1 =∑

x∈Zd

τ1∑

i=1

1Xi = x

≤∑

x∈GOOD(ω)

τ1∑

i=1

1Xi = x +∑

x∈∂BAD(ω)

τ1∑

i=1


Recalling the definition of χ at the beginning of the proof of Lemma 8.6.On 0 − regen

Eω[1χ ≤ nετ1]

≤Eω[

1χ ≤ nε∑

x∈Zd

τ1∑

i=1

1Xi = x]

≤∑

x∈B(nε,nCε)

Eω[

∞∑

i=1

1Xi = x]

≤∑

x∈B(nε,nCε)

[

1x ∈ GOOD(ω)Eω[

∞∑

i=1

1Xi = x]

+ 1x ∈ ∂BAD(ω)Eω[

∞∑

i=1


]]

.

Using Markov’s property and Lemma 8.1 we obtain

Eω[1χ ≤ nετ1]

≤C(K)∑

x∈B(nε,nCε)

[1x ∈ GOOD(ω) + 1x ∈ ∂BAD(ω)Eωx [T+

GOOD(ω)]],

and now, since γ < 1, we have

Eω[1χ ≤ nετ γ−ε1 ] ≤ Eω[1χ ≤ nετ1]

γ−ε

≤C(K)[

∑

x∈B(nε,nCε)

[1x ∈ GOOD(ω) + 1x ∈ ∂BAD(ω)Eωx [T+

GOOD(ω)]]]γ−ε

We may now apply Lemma 8.3

Eω[1χ ≤ nετ γ−ε1 ]

≤C(K)[

∑

x∈B(nε,nCε)

1x ∈ GOOD(ω) + 1x ∈ ∂BAD(ω)Eωx [T+

GOOD(ω)]]γ−ε

47

≤C(K)[

∑

x∈B(nε,nCε)

1x ∈ GOOD(ω)

+∑

x∈B(nε,nCε)

1x ∈ ∂BAD(ω) exp(3λ |∂BADsx(ω)|)

(

1 +∑

e∈E(BADsx(ω))

c∗(e))

]γ−ε

≤C(K)nCε maxx∈B(nε,nCε)

1x ∈ ∂BAD(ω) |E(BADsx(ω))| exp(3λ |∂BADs

x(ω)|)

×(

1 + maxe∈E(BADs

x(ω))c∗(e)

γ−ε)

≤C(K)nCε∑

x∈B(nε,nCε)


×(

1 +∑

e∈E(BADsx(ω))

c∗(e)γ−ε

)

.

Now

E[1χ ≤ nετ γ−ε1 | D = ∞]

≤C(K)E[

10-goodnCε∑

x∈B(nε,nCε)


×(

1 +∑

e∈E(BADsx(ω))

c∗(e)γ−ε

)

]

,

so using a reasoning similar to (8.13) and (8.14) with Lemma 8.4 yields

E[1χ ≤ nετ γ−ε1 | D = ∞]/(C(K)nCε)

≤∑

x∈B(nε,nCε)

E[

exp(4λ |∂BADsx(ω)|)

(

1 +∑

e∈E(BADsx(ω))

c∗(e)γ−ε

)]

≤C(K)(1 + E∗[cγ−ε∗ ])

∑

x∈B(nε,nCε)

E[|∂BADsx(ω)|d exp(3λ |∂BADs

x(ω)|)].

Now, we may see that Lemma 8.2 implies that

E[1x ∈ ∂BAD(ω) |∂BADsx(ω)|d exp(4λ |∂BADs

x(ω)|)] < C(K) < ∞.

which means that for any ε > 0

E[1χ ≤ nετ γ−ε1 | D = ∞] ≤ C(K)nCε.

From this, using Chebyshev’s inequality, we get

P[χ ≤ nε, τ1 > n | D = ∞] = P[1χ ≤ nετ1 > n | D = ∞]

≤ n−(γ−ε)E[1χ ≤ nετ γ−ε

1 | D = ∞]

≤ C(K)n−(γ−ε)nC1ε.

48 A. FRIBERGH

For any ε1 > 0, we may apply the previous equality for ε which dependsonly on γ and C1 to obtain

P[χ ≤ nε, τ1 > n | D = ∞] ≤ C(K)n−(γ−ε1).

Hence (8.17) and the previous equation imply that for any ε1 > 0 thereexists K large enough we obtain

P[τ(K)1 > n | D = ∞] ≤ P[χ > nε | D = ∞] + P[χ ≤ nε, τ

(K)1 > n | D = ∞]

≤ C(K)n−(γ−ε1),

which proves the lemma.

Hence, we have

Lemma 10.4. If lim ln P∗[c∗>n]ln n

= −γ with γ < 1 then

lim supln ∆n

ln n≤ 1

γ, P-a.s..

Proof. Fix M ≥ 1. We know, by Lemma 10.3 and Theorem 7.4, that thereexists K large enough such that for i ≤ M + 1

E[cardj ≤ n, τ(K)j − τ

(K)j−1 ≥ ni/(Mγ)] =E

[

∑

j≤n

1τ (K)j − τ

(K)j−1 ≥ ni/(Mγ)

]

≤Cnn−(i/M)(1−1/M) = Cn(1−i/M)+2/M ,

hence by Markov’s inequality for any L

P[

cardj ≤ n, τ(K)j − τ

(K)j−1 ≥ ni/(Mγ) ≥ 1

2Mn1/γ+L/M−(i+1)/(Mγ)

]

≤C(M)n(1−1/γ)(1−i/M)+(1/γ+2−L)/M .

Fix L ≥ 1/γ + 3 (which does not depend on M), we denote the event

B(n,M)

=

cardj ≤ n, τ(K)j − τ

(K)j−1 ∈ (ni/(Mγ), n(i+1)/(Mγ)] ≥ 1

2Mn1/γ+L/M−(i+1)/(Mγ)

,

and we get that for any fixed M and i ≤ M

(10.7) P[B(n, i,M)] ≤ Cn−1/M = o(1),

since γ ≤ 1.In the same way, by Lemma 10.3 and Theorem 7.4, we get that

B(n,M + 1,M) = cardj ≤ n, τ(K)j − τ

(K)j−1 ≥ n(M+1)/(Mγ) ≥ 1,

verifies

P[B(n,M + 1,M)] ≤ n−ε = o(1).

49

Also denoting B(n,M) = ∪M+1j=0 B(n, i,M) we have

P[B(n,M)] = o(1).

Now on B(n,M)c, we can give an upper bound for τ(K)n by

τ (K)n ≤

M∑

i=0

n(i+1)/(Mγ)( 1

2Mn1/γ+L/M−(i+1)/(Mγ)

)

=M + 1

2Mn1/γ+L/M .

Since by definition of the n-th regeneration time we necessarily have Xτ(K)n

·e1 ≥ n, it follows that ∆n ≤ τ

(K)n ≤ n1/γ+L/M on B(n,M)c. Hence

on B(n,M)c,ln ∆n

ln n≤ 1/γ + L/M.

Hence we have proved that for any M ≥ 1, by (10.7)

lim supln ∆n

ln n≤ 1/γ + L/M, P-a.s.,

and letting M go to infinity we get the result (we recall that L does notdepend on M).

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50 A. FRIBERGH

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BIASED RANDOM WALK IN I.I.D. POSITIVE RANDOM CONDUCTANCES ON Zfribergh/brwrc.pdf · We study the biased random walk i.i.d. positive random con-ductances on Zd. Our main result is

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