The speed of a biased walk on a Galton–Watson tree without leaves is monotonic with respect to progeny distributions for high values of bias

www.imstat.org/aihp 2015, Vol. 51, No. 1, 304–318

DOI:10.1214/13-AIHP573

© Association des Publications de l’Institut Henri Poincaré, 2015

The speed of a biased walk on a Galton–Watson tree without leaves is monotonic with respect to progeny distributions for

high values of bias

Behzad Mehrdad, Sanchayan Sen and Lingjiong Zhu

Courant Institute of Mathematical Sciences, New York University, 251 Mercer Street, New York, NY 10012, USA.

E-mail:mehrdad@cims.nyu.edu;sen@cims.nyu.edu;ling@cims.nyu.edu Received 13 December 2012; revised 6 June 2013; accepted 11 June 2013

Abstract. Consider biased random walks on two Galton–Watson trees without leaves having progeny distributionsP₁andP₂ (GW(P₁)and GW(P₂)) whereP₁andP₂are supported on positive integers andP₁dominatesP₂stochastically. We prove that the speed of the walk on GW(P₁)is bigger than the same on GW(P₂)when the bias is larger than a threshold depending onP₁ andP₂. This partially answers a question raised by Ben Arous, Fribergh and Sidoravicius (Comm. Pure Appl. Math.67(2014) 519–530).

Résumé. Nous considérons des marches aléatoires biaisées sur deux arbres de Galton–Watson sans feuilles GW(P₁)et GW(P₂) ayant des lois de reproduction respectivementP₁etP₂, deux lois supportées par les entiers positifs telles queP₁domine stochasti- quementP₂. Nous prouvons que la vitesse de la marche sur GW(P₁)est supérieure ou égale à celle sur GW(P₂)si le biais est plus grand qu’un seuil dépendant deP₁etP₂. Ceci répond partiellement à une question posée par Ben Arous, Fribergh et Sidoravicius (Comm. Pure Appl. Math.67(2014) 519–530).

MSC:60K37; 60J80; 60G50

Keywords:Random walk in random environment; Galton–Watson tree; Speed; Stochastic domination

1. Introduction and main results

1.1. Introduction

Consider a supercritical Galton–Watson tree, i.e., a random rooted tree, where the offspring size of all individuals are i.i.d. copies of an integer random variableZ, which satisfiesP (Z=k)=p_k,k=0,1, . . .and

k≥1kp_k>1. The tree has no leaves ifp0=0. We shall use|x|to denote the distance of a vertexx from the root. Moreoverx_∗ will denote the ancestor ofx for any vertexxdifferent from the root andx_i will denote theith child ofx. Given a treeT andβ >0, we defineβ-biased random walk(Xn)n≥0onT as follows. Transitions to each of the children of the root are equally likely. If the vertexx haskchildren andxis not the root then the transition probabilities are given by

P (X_n+1=x_∗|X_n=x)= 1 1+βk, P (X_n+1=x_i|X_n=x)= β

1+βk, i=1,2, . . . , k.

We start the walk from the root of the tree and denote byP^ωthe law of(X_n)_n≥0on a treeω. We define the averaged law as the semi-direct productP=P×P^ωwhereP is the Galton–Watson measure (associated with offspring distribution P) on the space of rooted trees conditioned on non-extinction.

Lyons [6] proved that ifβ >_E[¹_Z], then the random walk is transient, i.e., limn→∞|X_n| = ∞. Lyons, Pemantle and Peres [7] showed thatPalmost surely, the speed

v(β, P ):= lim

n→∞

|Xn|

n (1.1)

exists and is a non-random constant. A lot of work has been done on the behavior of the speed as a function ofβ. It was conjectured in [7] thatv(β, P )increases inβ on(_E¹_[Z],∞)when the tree has no leaves, i.e.,P{0} =0. The conjecture has been open for a long time until proven recently in [3] for large values ofβ.

Theorem ([3]). The speedv(β, P )of aβ-biased random walk on a Galton–Watson tree without leaves is increasing forβ > βcfor someβc>0very large whenP{0} =0.

Very recently, Aïdékon obtained an expression for the speedv. Theorem ([1]).

v(β, P )=E[(βZ−1)Y0/(1−β+β_Z

i=0Y_i)] E[(βZ+1)Y0/(1−β+β_Z

i=0Y_i)], (1.2)

whereY_iare i.i.d.copies distributed asP_x(τ_x∗= ∞)andτ_yis the first hitting time ofy.

Using his own formula, Aïdékon (private communications) can prove the monotonicity forβ≥2 whenP{0} =0.

However, the original conjecture is still open in the sense that it is not known if the monotonicity holds for every β >1/E[Z].

In this paper we shall investigate how the speed changes when one changes the progeny distribution keeping the bias fixed.

The paper is organized as the following. In Section 1.2, we shall introduce our main results. In Section2, we shall describe in details our coupling method. Finally, in Section3, we shall provide the proofs of all the results in Section1.2.

1.2. Main results

In [3], the authors raised the following interesting question, if P1 stochastically dominates P2, does it imply that v(β, P₁)≥v(β, P₂)? We show that this is indeed the case at least when the bias is large.

Throughout this paper, when we sayP₁dominatesP₂stochastically, we also mean thatP₁=P₂. We also recall that ifP₁dominatesP₂stochastically then there is a coupling of the random variablesZ₁andZ₂having distributions P1andP2respectively such thatZ2≤Z1.

We have the following result.

Theorem 1. Assume thatP1 andP2are two probability measures on positive integers such thatP1dominates P2

stochastically.Considerβ-biased random walks onGW (P₁)andGW (P₂).Then for everyδ >0,there exists aβ₀:=

β₀(P₁, P₂, δ) >0 such that for anyβ > β₀,we havev(β, P₁) > v(β, P₂).The constantβ₀ equalsmax{β₁,²³₄ +δ} where

β₁:=c_δ·min

E[(1/Z1−1/Z2)1Z₁<Z₂]

E[1/Z₂−1/Z₁] ,E[Z₂(1/Z₂ −1/Z₁)] E[1/Z₂−1/Z₁] +1

, (1.3)

andc_δis a universal constant depending only onδ.Here,Z₁,Z₂are independent and are distributed according toP₁ andP2respectively,Z₁ andZ₂ are jointly distributed so thatZ₁≥Z₂ almost surely and their marginal distributions areP1andP2.

Remark 2. There is a universal cut-offβ₁=β₁(M)which works for allP₂supported on{1,2, . . . , M}since we have E

Z₂

1 Z₂ − 1

Z₁

≤M·E 1

Z₂− 1 Z₁

.

The other expression inside the parentheses in the definition ofβ₁in Theorem1is more useful when “the distribution ofZ1is much larger than that ofZ2”,we shall illustrate this in Corollary5.

Remark 3. SupposeP₁dominatesP₂and are both supported on positive integers.Thenv(β, P₁)≥v(β, P₂)follows trivially in the following cases.

(i) It is easy to see(via a coupling argument)that if the maximum of the support ofP2is not larger than the minimum of the support ofP₁,then for anyβ >0,we havev(β, P₁)≥v(β, P₂).

(ii) We havev(1, P1)≥v(1, P2)just by considering the expression v(1, P )=E_P

Z−1 Z+1

obtained in[7].

(iii) Note thatv(1/EP₂[Z], P2)=0, v(1/EP₂[Z], P1) >0,andv(β, Pj)is continuous inβ forj =1,2.Thus,for some smallε >0we havev(β, P1)≥v(β, P2)for0< β < ε+1/EP₂[Z].

Further(ii)and(iii)hold even when the offspring distributions are supported on non-negative integers as long as we define the speed as in(1.1)conditional on non-extinction of the trees.

We can improve the thresholdβ₀of Theorem1by making stronger assumptions.

Theorem 4. Suppose P1 and P2 are two probability measures on positive integers such that for some >1, there exists a coupling of Z₁⁽¹⁾, Z₁⁽²⁾, . . . , Z⁽⁾₁ and Z₂⁽¹⁾, Z₂⁽²⁾, . . . , Z₂⁽⁾ for which min{Z₁⁽¹⁾, Z₁⁽²⁾, . . . , Z₁⁽⁾} ≥ max{Z₂⁽¹⁾, Z₂⁽²⁾, . . . , Z⁽⁾₂ }almost surely,whereZ⁽¹⁾_j , . . . , Z_j⁽⁾are i.i.d.distributed according toP_jforj=1,2.Then for anyδ >0,we havev(β, P₁)≥v(β, P₂)for anyβ >max{K·β₁^1/,²³₄ +δ}where the constantKequals ²⁷₄ ·3^5/3. Corollary 5. Assume thatP₁ andP₂ are two probability measures on positive integers such thatP₁stochastically dominatesP₂.Letm_i:=E_Pi[Z]andZ_i⁽ⁿ⁾be the number of children in thenth generation inGW (P_i),denote the law ofZ_i⁽ⁿ⁾byP_i⁽ⁿ⁾fori=1,2.Assume that there exists someθ >0such thatE[e^{θ Z1(1)}]<∞.Letf be the generating function for P₁ and α:= −logf(0)/logf(1).Further assume that m₁> m^max₂ ^{{2/α,(1/α)+1}} (if P₁{1} =0, then α= ∞and this condition is automatically satisfied).

Then,for anyβ >23/4,there exists somek=k(P1, P2, β)such thatv(β, P₁^(k)) > v(β, P₂^(k)). (We emphasize that v(β, P_i^(k))is the speed of aβ-biased random walk on a Galton–Watson tree havingP_i^(k)as its offspring distribution.)

The following corollary is the counterpart to Theorem 1.2 in [3].

Corollary 6. Assume all the assumptions in Theorem1and recall the definition ofβ₀from there.Moreover,assume that the minimum degrees of bothP₁andP₂are bigger thand (for somed≥2),i.e.,d_i:=min{k≥1, Pi(Z=k) >

0} ≥d,fori∈ {1,2}.

Then we have,v(β, P₁) > v(β, P₂)for anyβ > β₀/d.

2. Constructing the walks

Let us describe precisely the coupling we use. Let U₁ have uniform distribution on (1/(β+1),1). Let(U_i)_i≥2 be i.i.d. uniformly distributed random variables on [0,1] independent of U₁. Let {(Z_1,k , Z_2,k )}k≥1 be i.i.d. ran- dom vectors such that for each k, Z_1,k has the marginal distribution P1 and Z_2,k has the marginal distribution

P₂ and with probability 1, we have Z_2,k ≤Z_1,k. Finally let {Z_i,k}k≥1 be i.i.d. P_i for i=1,2. The sequences {Ui}i≥1,{Z1,k}k≥1,{Z2,k}k≥1,{(Z_1,k, Z_2,k )}k≥1are independent of each other.

In our proof we shall work conditional on an event which ensures that the roots are only visited once, for this reason we only need one copy ofU₁. Note that our definition ofU₁is slightly different from the one in [3].

We construct two random walksX_n⁽¹⁾ andX⁽²⁾_n (onGW (P₁)andGW (P₂)) and another walkY_n onZ_≥0 in the following way. DefineY₀:=0 and forn≥1,

Y_n:=

n

i=1

{1U_i>1/(β+1)−1U_i≤1/(β+1)}, n∈N.

We startX⁽¹⁾andX⁽²⁾at the roots and grow the treesGW (P1)andGW (P2)dynamically. For simplicity we drop the time parameternand denote the position ofX⁽ⁱ⁾_n byx⁽ⁱ⁾.

Now, if at timen≥0,X⁽¹⁾_n andX⁽²⁾_n are at two sitesx⁽¹⁾andx⁽²⁾, neither of them visited before by the correspond- ing walks, then we assignZ_1,n+1andZ_{2,n +1}many children tox⁽¹⁾andx⁽²⁾respectively (recall thatZ_{1,n +1}≥Z_{2,n +1}).

If at timen, one of the walks, sayX⁽¹⁾is at a sitex⁽¹⁾previously visited by the walk while the other walkX⁽²⁾is at a new sitex⁽²⁾then we assignZ2,n+1many children tox⁽²⁾.

Let us now explain the rules for transition. Denote the number of offsprings ofx⁽ⁱ⁾byZ_i and letx_k⁽ⁱ⁾be thekth child ofx⁽ⁱ⁾(i=1,2).

Define

η₁:= β (β+1)Z1

, η₂:=

β β+1

1 Z2− 1

Z1

, η₃:=

1

β+1− 1 Z2β+1

1 Z2

,

η₄:=

1

β+1− 1 Z₁β+1

1

Z₁, η₅:= |η₃−η₄|.

Then wheneverZ₁≥Z₂, we move according to the rule explained below.

WhenU_n+1∈(1/(β+1),1)we have the following cases.

(1) Consider the random walkX⁽¹⁾.

• IfU_n+1∈(_β¹₊₁+(i−1)η1,_β+¹₁+iη1], thenX_n⁽¹⁾₊₁=x_Z⁽¹⁾

1+1−ifori=1,2, . . . , Z1. (2) Consider the random walkX⁽²⁾.

• IfU_n+1∈(_β¹₊₁+(i−1)η2,_β+¹₁+iη₂], then we haveX⁽²⁾_n+1=x_Z⁽²⁾

2+1−i, wherei=1,2, . . . , Z2.

• IfU_n+1∈(_β+¹₁+Z₂η₂+(i−1)η1,_β+¹₁+Z₂η₂+iη₁], then we haveX⁽²⁾_n+1=x_Z⁽²⁾

2+1−i, wherei=1,2, . . . , Z2. When Un+1∈(0,1/(β+1)) we have to consider two cases. If η3≥η4, then we use the following coupling.

Figure1gives an illustration.

(1) Consider the random walkX⁽¹⁾.

• IfU_n+1∈ [0,_Z1β¹₊₁], then we haveX_n⁽¹⁾₊₁=x_∗⁽¹⁾.

• IfU_n+1∈(_Z ¹

1β+1+(i−1)η4,_Z ¹

1β+1+iη₄], then we haveX_n⁽¹⁾₊₁=x_Z⁽¹⁾

1+1−i, wherei=1,2, . . . , Z1. (2) Consider the random walkX⁽²⁾.

• IfU_n+1∈ [0,_Z ¹

2β+1], then we haveX_n⁽²⁾₊₁=x_∗⁽²⁾.

• IfU_n+1∈(_Z ¹

2β+1+(i−1)η5,_Z ¹

2β+1+iη5], then we haveX_n⁽²⁾₊₁=x_Z⁽²⁾

2+1−i, wherei=1,2, . . . , Z2.

• If U_n+1∈(_Z ¹

2β+1 +Z₂η₅+(i−1)η4,_Z ¹

2β+1 +Z₂η₅+iη₄], then we have X_n⁽²⁾₊₁=x_Z⁽²⁾

2+1−i, where i= 1,2, . . . , Z2.

Ifη₃< η₄, then we use the following coupling. Figure2is an illustration of the following coupling.

(1) Consider the random walkX⁽¹⁾.

Fig. 1. The coupling forη₃≥η₄. In the illustration, we useZ₂;η₄etc. to denoteZ₂many subintervals with each subinterval of lengthη₄etc.

Fig. 2. The coupling forη₄> η₃.

• IfU_n+1∈ [0,_Z ¹

1β+1], then we haveX⁽¹⁾_n+1=x_∗⁽¹⁾.

• IfUn+1∈(_Z ¹

1β+1+(i−1)η4,_Z ¹

1β+1+iη4], then we haveX⁽¹⁾_n+1=x_Z⁽¹⁾

1+1−i, wherei=1,2, . . . , Z1−Z2.

• If U_n+1∈(_Z ¹

1β+1+(Z₁−Z₂)η₄+(i−1)η5,_Z ¹

1β+1+(Z₁−Z₂)η₄+iη₅], then X_n⁽¹⁾₊₁=x_Z⁽¹⁾

2+1−i, where i=1,2, . . . , Z2.

• IfU_n+1∈(_Z ¹

2β+1+(i−1)η3,_Z ¹

2β+1+iη₃], then we haveX⁽¹⁾_n+1=x_Z⁽¹⁾

2+1−i, wherei=1,2, . . . , Z2. (2) Consider the random walkX⁽²⁾.

• IfU_n+1∈ [0,_Z2β¹₊₁], then we haveX⁽²⁾_n+1=x_∗⁽²⁾.

• IfU_n+1∈(_Z ¹

2β+1+(i−1)η3,_Z ¹

2β+1+iη₃], then we haveX⁽²⁾_n+1=x_Z⁽²⁾

2+1−i, wherei=1,2, . . . , Z2. Finally ifZ₁< Z₂we move according to the following rule.

(1) Fori=1,2:

• IfU_n+1∈ [0,_Z ¹

iβ+1], then we haveX⁽ⁱ⁾_n+1=x_∗⁽ⁱ⁾.

• IfUn+1∈(_Z ¹

iβ+1+(j−1)_Z ^β

iβ+1,_Z ¹

iβ+1+j_Z ^β

iβ+1], then we haveX⁽ⁱ⁾_n+1=x_j⁽ⁱ⁾, wherej=1,2, . . . , Zi. It is routine to check thatX⁽ⁱ⁾is aβ-biased random walk onGW (Pi)fori=1,2.

3. Proofs

The main idea in our proof is to use a technique originally used in [3], to couple the walks on the Galton–Watson trees with a random walk onZ. We shall use a super-regeneration time which is a regeneration time for all the three walks Y,GW (P₁)andGW (P₂). Regeneration time is an often-used technique in the study of random walks in random media. (See, e.g., [10].) Informally, a regeneration time is a maximum of a random walk which is also a minimum of the future of the random walk. A timeτ is a regeneration time for theβ-biased random walk(Y_n)_n≥0onZif we have

Y_τ>max

n<τY_n and Y_τ<min

n>τY_n.

Consider the regeneration time for walks on GW (P₁)and GW (P₂)in the sense that is usually defined on trees (see [8]). As in [3] if τ is a regeneration time for (Y_n)_n≥0, then it is also a regeneration time for GW (P₁)and GW (P₂). In this respect,τ is called a super-regeneration time.

Let us consider the event that 0 is a regeneration time for(Y_n)_n≥0. Following the notation in [3], we denote this event by{0−SR}. Then, we have

p_∞:=P (0−SR)=β−1 β . Let us define the probability measureP˜ as

P (˜ ·):=P (·|0−SR).

UnderP˜, 0 is a regeneration time and letτ_ibeith non-zero regeneration time.

Then,(|Xτ_i+1−Xτ_i|, τi+1−τi)i≥1is a sequence of i.i.d. random vectors having the same distribution as(|Xτ₁|, τ1) underP˜ and as in [3], we have, for anyβ >1,

v(β, P₁)=E˜[|X⁽¹⁾_τ1 |]

˜

E[τ₁] and v(β, P₂)=E˜[|X⁽²⁾_τ1 |]

˜ E[τ₁] . Hence,v(β, P₁) > v(β, P₂)is equivalent toE˜[|X_τ⁽¹⁾₁ |]>E˜[|X⁽²⁾_τ1 |].

Let us denote by Bthe set of times before τ1when the random walk onZ_≥0takes a step back, i.e.,B= {j≤ τ1|Uj≤1/(β+1)}.

We quote the following lemma from [3].

Lemma 7 (Lemma 4.1, [3]). If{|B| =k},then{τ1≤3k+2}.

Proof of Theorem1. Consider|B| =k, i.e.,B= {i₁<· · ·< i_k}, wherek≥1 andτ₁=n. Let us make two simple observations.

(i) |X⁽¹⁾τ₁ | − |X⁽²⁾τ₁ | =2 or 0 whenk=1.

(ii) |X⁽¹⁾τ₁ | − |X⁽²⁾τ₁ | ≥ −2(k−1)whenk≥2.

We have EX˜ ⁽¹⁾_τ

1 −X⁽²⁾_τ

1

= ˜EX⁽¹⁾_τ

1 −X_τ⁽²⁾

1 ; |B| =1 +

∗

EX˜ _τ⁽¹⁾

1 −X_τ⁽²⁾

1 ;B= {i₁<· · ·< i_k}, τ₁=n

≥ ˜EX⁽¹⁾_τ

1 −X_τ⁽²⁾

1 ; |B| =1

−

∗

2(k−1)P˜X_τ⁽¹⁾

1 −X_τ⁽²⁾

1 <0;B= {i₁<· · ·< i_k}, τ₁=n ,

where

∗stands for summation over alln≥2, k≥2 and{i₁, . . . , i_k} ⊆ {1, . . . , n}for which the walkY_k does not come back to the origin.

For the first term, we have EX˜ _τ⁽¹⁾

1 −X_τ⁽²⁾

1 ; |B| =1

≥2 β

β+1 4

E 1

Z₂β+1− 1 Z₁β+1

. (3.1)

Let us explain the inequality in (3.1). Letε_i=I(U_i ≥1/(β+1))−I(U_i<1/(β+1)). When|B| =1,|X⁽¹⁾_τ1 |−|X⁽²⁾_τ1 | = 2 or 0, hence we have

EX˜ _τ⁽¹⁾

1 −X_τ⁽²⁾

1 ; |B| =1

= 2

p_∞PX⁽¹⁾_τ

1 −X⁽²⁾_τ

1 =2; |B| =1;0−SR and thus we get the lower bound in (3.1) by considering the event

A=

ε₁=ε₂=1, ε3= −1 andX⁽¹⁾

3 −X⁽²⁾

3 =2, ε4=ε₅=1, τ1=5 . For the second term, we have

P˜X⁽¹⁾_τ

1 −X_τ⁽²⁾

1 <0;B= {i1<· · ·< ik}, τ1=n

≤ 1

p_∞PX_τ⁽¹⁾

1 −X⁽²⁾_τ

1 <0;B= {i1<· · ·< ik}, τ1=n .

On{|X⁽¹⁾τ₁ | − |Xτ⁽²⁾₁ |<0}, letσ be the first time when the walk onGW (P1)goes up but the walk onGW (P2)goes down, necessarilyσ ∈B. (When {|Xτ⁽¹⁾₁ | − |Xτ⁽²⁾₁ | ≥0}, defineσ = ∞.) We introduce some notation here. Given a sequenceθ= {θn}n≥1 whereθn= ±1 we denote byτ (θ ), the first non-zero regeneration time for the walk Zn= n

i=1θi; e.g., the first non-zero regeneration time forYn,τ1equalsτ (ε)whereε= {εn}n≥1andεn=I(Un≥1/(β+ 1))−I(Un<1/(β+1)). Define

τ₁^{(j )}=τ ε^{(j )}

whereε^{(j )}:= {ε₁, . . . , ε_j−1,−1, εj+1, . . .}. LetB^{(j )}:= {i≤τ₁^{(j )}: ε_i^{(j )}= −1}. Also define

τ_{1(j )}=τ (ε_{(j )}) whereε_{(j )}:= {ε₁, . . . , ε_j−1,+1, εj+1, . . .}.

LetB(j ):= {i≤τ_{1(j )}: ε_{(j )i}= −1}. Note that, for fixedj,ε^{(j )}andε_{(j )}are functions of{U_i: i=j}, and are hence independent ofU_j.

Also note that if|X⁽¹⁾_τ1 | − |X⁽²⁾_τ1 |<0 then the event E=

i,j≤τ₁

Z_1,i< Z_2,j

i,j≤τ₁

Z_1,i < Z_2,j

i,j≤τ₁

{Z_1,i< Z_2,j}

i=j i,j≤τ1

Z_1,i < Z_2,j

is true. For everyn≥2, letE_iⁿ:=4

j=1E_j,iⁿ where E1,iⁿ:=

i,j≤n

Z1,i< Z_2,j

∩

Ui_l ∈ 1

Z_2,j β+1, 1 Z_1,iβ+1

andE_2,iⁿ ,E_3,iⁿ,E_4,iⁿ are defined similarly (i.e., use the random variablesZ_1,i , Z_2,jin the definition ofE_2,iⁿ etc.).

PX_τ⁽¹⁾

1 −X_τ⁽²⁾

1 <0;B= {i1<· · ·< i_k}, τ1=n

≤

k

=1

P

B= {i₁<· · ·< i_k}, τ₁=n, σ=i

=

k

=1

P

B⁽ⁱ⁾= {i₁<· · ·< i_k}, τ₁⁽ⁱ⁾=n, σ=i

≤

k

=1

P

B⁽ⁱ⁾= {i₁<· · ·< i_k}, τ₁⁽ⁱ⁾=n

;E_iⁿ

≤

k

=1

4n²P

B⁽ⁱ⁾= {i₁<· · ·< i_k}, τ₁⁽ⁱ⁾=n E

1

Z1β+1− 1

Z2β+1;1Z1<Z2

,

where we used independence ofε⁽ⁱ⁾andUi. Then, by Lemma7, PX_τ⁽¹⁾

1 −X_τ⁽²⁾

1 <0;B= {i₁<· · ·< i_k}, τ₁=n

≤4(3k+2)²

k

=1

P

B⁽ⁱ⁾= {i₁<· · ·< i_k}, τ₁⁽ⁱ⁾=n, U_i≤ 1 β+1

·(β+1)·E 1

Z₁β+1− 1

Z₂β+1;1Z1<Z2

=4(β+1)(3k+2)²E 1

Z₁β+1− 1

Z₂β+1;1Z₁<Z₂

k =1

P

B= {i1<· · ·< ik}, τ1=n

≤8βk(3k+2)²E

(Z2−Z1)β

(Z1β+1)(Z2β+1)1Z1<Z2

P

B= {i₁<· · ·< i_k}, τ₁=n

≤8k(3k+2)²E 1

Z₁− 1 Z₂

1Z₁<Z₂

P

B= {i₁<· · ·< i_k}, τ₁=n .

Therefore, by using the simple upper boundP (|B| =k)≤c(₄₍₁²⁷_+β))^k (Lemma 6.1 in [3]) for a universal constant cand the fact thatp_∞=(β−1)/β, we get

EX˜ ⁽¹⁾_τ

1 −X⁽²⁾_τ

1

≥2 β

β+1 4

E 1

Z₂β+1− 1 Z₁β+1

− ^∞

k=2

16

p_∞k(k−1)(3k+2)²E 1

Z₁− 1 Z₂

1Z1<Z2

P

|B| =k

≥2 β

β+1 4

E

(Z₁−Z₂)β (Z₂β+1)(Z₁β+1)

− c p_∞E

1 Z1− 1

Z2

1Z1<Z2

_∞

k=2

16k(k−1)(3k+2)² 27

4(1+β) k

≥2 β

β+1 4

E

(Z₁−Z₂)β 4Z₂Z₁β²

− cβ (β−1)E

1 Z₁− 1

Z₂

1Z₁<Z₂

∞ k=2

16k(k−1)(3k+2)² 27

4(1+β) k

≥2 β

β+1 4

E

(Z₁ −Z₂) 4Z₂Z₁β

− c·27²β 4²(β−1)(β+1)²E

1 Z1− 1

Z2

1Z1<Z2

· ^∞

k=2

16k(k−1)(3k+2)² 27

4(1+β) k−2

, (3.2)

where we used the fact thatZ₁≥Z₂ ≥1 andβ >1. Hence we conclude that for anyδ >0,E˜[|X⁽¹⁾_τ1 | − |X⁽²⁾_τ1 |]>0 if we have

β >max

cδ·E[((1/Z1)−(1/Z2))1Z1<Z2] E[(1/Z₂)−(1/Z₁)] ,23

4 +δ

, (3.3)

for some universal constantc_δ>0 that depends only onδ.

Now we derive the other lower bound in (1.3). On{|X⁽¹⁾_τ1 | − |X_τ⁽²⁾₁ |<0}, let us define the eventsEandF as E:=

for someσ₁≤τ₁,X⁽¹⁾_σ

1+1=X_σ⁽²⁾

1+1andX_j⁽¹⁾=X⁽²⁾_j for anyj≤σ₁ , F :=

for someσ₂≤τ₁, X⁽¹⁾_j =X_j⁽²⁾for anyj≤σ₂, andX_σ⁽¹⁾

2+1=X⁽²⁾_σ

2+1,butX_σ⁽¹⁾

2+1=X⁽²⁾_σ

2+1.

In other words,Eis the event that the first time the walks onGW (P₁)andGW (P₂)decouple, the walk onGW (P₂) goes up and the walk onGW (P₁)goes down. Clearly this happens at timeσ₁∈B.F is the event that the first time the walks onGW (P₁)andGW (P₂)decouple, they both go downwards but to different offsprings. This happens at time σ₂which may or may not be inB.

Next, PX⁽¹⁾_τ

1 −X_τ⁽²⁾

1 <0;B= {i₁, . . . , i_k}, τ₁=n

=PX_τ⁽¹⁾

1 −X_τ⁽²⁾

1 <0;B= {i₁, . . . , i_k}, τ₁=n;E +PX⁽¹⁾_τ

1 −X⁽²⁾_τ

1 <0;B= {i1, . . . , ik}, τ1=n;F

. (3.4)

Let us get an upper bound for the second term in (3.4).

PX⁽¹⁾_τ

1 −X_τ⁽²⁾

1 <0;B= {i₁, . . . , i_k}, τ₁=n;F

=

n

=1

PX⁽¹⁾_τ

1 −X_τ⁽²⁾

1 <0;B= {i₁, . . . , i_k}, τ₁=n, σ₂=;F

≤

n

=1

P

B= {i₁, . . . , i_k}, τ₁=n, σ₂=;F

. (3.5)

If /∈ {i1, . . . , ik}, then we have P

B= {i1, . . . , ik}, τ1=n, σ2=;F

≤P

B= {i1, . . . , ik}, τ1=n;U∈ n

m=1

1

β+1,(Z_1,m−Z_2,m)β (β+1)Z_1,m + 1

β+1

=P

B()= {i1, . . . , ik}, τ1()=n, U∈ n

m=1

1

β+1,(Z_1,m−Z_2,m)β (β+1)Z_1,m + 1

β+1

=P

B()= {i1, . . . , ik}, τ1()=n P

U∈

n

m=1

1

β+1,(Z_1,m −Z_2,m )β (β+1)Z_1,m + 1

β+1