On the speed of a planar random walk avoiding its past convex hull

Abstract

We consider a random walk inR²which takes steps uniformly distributed on the unit circle centered around the walker’s current position but avoids the convex hull of its past positions. This model has been introduced and studied by Angel, Benjamini and Virág. We show a large deviation estimate for the distance of the walker from the origin, which implies that the walker has positive lim inf speed.

2005 Elsevier SAS. All rights reserved.

Résumé

On considère une marche aléatoire surR²avec des pas distribués uniformément sur le cercle unité centré sur la position courante de la marche mais n’entrant pas dans l’adhérence convexe de ses positions précédentes. Ce modèle a été introduit et étudié par Angel, Benjamini et Virag. On démontre une estimée de grandes déviations pour la norme de la marche, qui implique que le limite inférieure de la vitesse de la marche est positive.

2005 Elsevier SAS. All rights reserved.

MSC: primary 60K35; secondary 60G50, 52A22

Keywords: Convex hull; Large deviations; Law of large numbers; Random walk; Self avoiding; Speed

1. Introduction

Angel, Benjamini and Virág introduced and studied in [1] the following model of a random walk(X_n)_n0inR², which they called the rancher. The walker starts at the originX₀=0. Suppose it has already takennsteps (n0) and is currently atX_n. Then its next positionX_n+1is uniformly distributed on the unit circle centered aroundX_n but conditioned so that the straight line segmentX_n, X_n+1fromX_ntoX_n+1does not intersect the convex hullK_n of the past positions{X₀, X₁, . . . , X_n}, see Fig. 1.

* Current address: Mathematisches Institut, Universität Tübingen, Auf de Morgenstelle 10, 72076 Tübingen, Germany.

E-mail address: martin.zerner@uni-tuebingen.de (M.P.W. Zerner).

0246-0203/$ – see front matter 2005 Elsevier SAS. All rights reserved.

doi:10.1016/j.anihpb.2004.08.001

Fig. 1. Three steps of the walk.X₃is uniformly distributed on the bold arc of the circle with radius 1, centered inX₂.

Note that(X_n)_n0 is not Markovian since in general one needs to know the whole history of the process in order to determine the transition probabilities for the next step. This makes this model more difficult to analyze than a Markovian random walk, a property it shares with many other self-interacting processes, see [1] and also [2] for references. We do not claim that this model is of particular importance or that its definition is very natural.

However, in the class of all self-interacting processes it seems to be among the few models which are at least partially analyzable and still have some striking properties.

Several such properties have been conjectured by Angel, Benjamini and Virág in [1]. Based on simulations and heuristics, the authors of [1] believe e.g. that the direction of the walk converges, i.e. thatXn/XnconvergesP-a.s.

to a random direction, see [1, Conjecture 1]. Moreover, if we denote byw_nas a measure for transversal fluctuations the maximal distance of a point inK_nfrom the lineX₀, X_n, then Angel, Benjamini and Virág conjecture thatw_n grows liken^3/4, see [1, Conjecture 2].

As far as results are concerned, the only major rigorous result which has been proved so far for this model to the best of our knowledge is that the walk has positive lim sup speed, i.e. there is a constantc >0 such thatP-a.s.

lim supX_n/n > casn→ ∞, see [1, Theorem 1]. Here(Ω,F, P )is the underlying probability space.

The purpose of the present paper is to improve this result by showing the following.

Theorem 1. There is a constantc₁>0 such that lim sup

n→∞

1 nlogP

X_nc₁n

<0 (1)

and consequently, lim inf

n→∞

X_n

n c₁ P-a.s. (2)

In particular, (2) proves [1, Conjecture 4]. We expect but were not able to prove that the speed limXn/nexists and isP-a.s. constant, as conjectured in [1, Conjecture 5].

Let us now describe how the present article is organized. The next section introduces general notation and gives a short overview of the proof. In Section 3 we introduce some sub- and supermartingales, which enable us in Section 4 to bound exponential moments of the time it takes the diameter of the convex hullK_nto increase. From this we deduce in Section 5 estimates for the diameter ofK_nsimilar to the ones claimed in Theorem 1 forX_n and show how this implies Theorem 1.

Fig. 2. General notation.

2. Notation and outline of proof

We denote byd_n the diameter ofK_n. Since(K_n)_n is an increasing sequence of sets,(d_n)_n is non-decreasing.

The ladder timesτ_iat which the process(d_n)_n0strictly increases are defined recursively by τ₀:=0 and τ_i+1:=inf{n > τ_i: d_n> d_τi} (i0).

It follows from [1] thatP-a.s.τi <∞ for all i0. In Section 4 we will reprove this fact by showing that the differences

i:=τi+1−τi (i0),

between two successive finite ladder times have some finite exponential moments. Note thatτ₁=1 sinced₀=0 andd₁=1 and observe that theτ_i’s are stopping times with respect to the canonical filtration(Fn)_n0generated by(X_n)_n0. Since the diameter of a bounded convex set is the distance between two of its extremal points there is for alli1 withτi<∞a (P-a.s. unique) 0k(i) < τi such thatdτ_i= Xτ_i−Xk(i), see Fig. 2. Forx∈R²and r >0 we denote byB(x, r)the closed disk with centerxand radiusr. Ifτi<∞then

σ_i+1:=inf

n0|X_n∈/B(X_τi, d_τi)∩B(X_k(i), d_τi)

is the exit time of the walk from the large lens shaped region shown in Fig. 2, which we shall refer to as the lens created at timeτ_i. Observe thatK_τi is contained in the lens created at timeτ_i. Moreover,

τ_i+1σ_i+1 (3)

since ifσi+1<∞,Xσ_i+1 has a distance from eitherXτ_i orX_k(i)greater thandτ_i.

We have now introduced enough notation to be able to outline the idea of the proof of Theorem 1.

Outline of Proof. To show that(Xn)nis growing at a positive rate we shall first show that(dn)nis doing so. To this end we shall bound exponential moments of thei’s. This is the heart of the proof and done as follows.

Suppose thatd_nis already quite large and that we have just reached a new ladder timeτ_i, at which the diameter has increased. Due to (3) the diameter can stay constant only as long as the walk is hiding in the lens created at timeτ_i. However, there are two mechanisms, corresponding to Lemmas 4 and 5, which ensure that the walk cannot hide in the lens for too long.

The first one, described in Lemma 4, works while the walk is still inside a small ball aroundX_τi, see Fig. 2. The radius of this ball is chosen to be proportional todτ_i. This situation is depicted in an idealized form in Fig. 3. In this

Fig. 3. Idealized picture, assumingdτ_i= ∞. Whenever the walker crosses a wedge for the first time, which occurs at a positive rate, it has in the next step a positive drift towards the boundary of the lens.

regime the following happens: Whenever the walk crosses for the first time any of the wedge-shaped lines shown in Fig. 3, it has a drift away fromX_k(i) towards the boundary of the lens. To see this consider the two boundary lines of the convex hull which are emanating fromX_τi+nin Fig. 3. Suppose we are to draw in Fig. 3 the line going fromX_τi+n to the left. In the picture this line has slope 0, i.e. it is parallel toX_τi, X_k(i), but of course it could have a different slope. However, this slope cannot be much smaller than 0 becauseX_τi+n is close toX_τi but far away fromX_k(i). On the other hand, the line emanating fromXτ_i+nto the right must have in Fig. 3 a slope of at least 1.

Together, these two lines create a drift downwards and more importantly to the right towards the boundary of the lens. (If the left line has a positive slope then the drift to the right is even stronger.)

So every time the walker crosses for the first time such a helpful wedge-shaped line it will be pushed towards the next wedge-shaped line. If the little ball aroundXτ_i was infinitely large this would result in crossings of the wedge- shaped lines at a positive rate, which would propel the walk out of the lens very fast, i.e. this mechanism alone would give a finite exponential moment of_i, see Lemma 4. However, since the ball’s size is finite and proportional tod_τi, the walk can escape from this little ball before leaving the lense with a probability exponentially small ind_τi. (This will happen a finite number of times.)

Whenever this occurs, we rely on the second mechanism, see Lemma 5. The lineX_τi, X_k(i), which is part of the convex hull, creates a positive drift out of the lens. Since the diameter of the lens is of orderd_τi it will take the walker a time of orderd_τi to reach the boundary of the lens. In fact we shall show, see Lemma 5, that it is exponentially costly for the walk to stay in the lens for a time longer than the order ofd_τi. This is enough to make sure that the walk is not slowed down too much by occasionally resisting the first mechanism and getting lost in the lens. 2

To make this idea precise we need to introduce more notation. The point Yi:=X_τi+X_k(i)

2 (i1)

will serve as the “center” ofK_τi and the “radius”

R_i,n:= X_τi+n−Y_i (i1, n0)

is the distance ofX_τi+nfrom this center. The orthogonal projection ofX_τi+nonto the straight line passing through X_τi andX_k(i)will be calledZ_i,n(i1, n0). The distance ofX_τi+nfrom this line is denoted by

Di,n:= Xτ_i+n−Zi,n (i1, n0).

For the following definitions we assumei, n1 andτi+n < τi+1. In particular, due to (3), we assume that at timeτi+nthe walk has not yet left the lens created at timeτi. This implies thatZi,n∈Xτ_i, Xk(i)and thatXτ_i and

Fig. 4. General notation.

X_k(i)are still boundary points ofK_τi+n, as shown in Figs. 2 and 4. Hence if we start inX_τi+nand follow the two boundary line segments emanating fromX_τi+nwe will eventually reachX_τiandX_k(i). The boundary line segment whose continuation leads first toK_τi and then toX_k(i)is calleds_1,i,n, while the other line segment starting inX_τi+n is denoted bys_2,i,n, see Fig. 4. The angle betweens_j,i,nandX_τi+n, Y_i is calledϕ_j,i,n∈ [0, π](j=1,2), see the left part of Fig. 4. Similarly, the angle betweensj,i,nandXτ_i+n, Zi,nis denoted byψj,i,n∈ [0, π](j=1,2), see the right part of Fig. 4. Occasionally, we will dropped the subscriptsiandnfromϕandψ. SinceKτ_i+nis convex,

ϕ₁+ϕ₂=ψ₁+ψ₂π. (4)

Furthermore,|ϕ₁−ψ₁|is one of the angles in a right angled triangle, namely the triangle with verticesX_τi+n, Y_i andZ_i,n. Hence,

|ϕ₁−ψ₁| = |ϕ₂−ψ₂|π/2. (5)

3. Some sub- and supermartingales

The following result shows that for everyi1, both(R_i,n)_nand(D_i,n)_n(1n < τ_i+1−τ_i)are submartingales.

Lemma 2. For alli, n1,P-a.s. on{τ_i+n < τ_i+1}, E[R_i,n+1−R_i,n|Fτi+n]sinϕ1,i,n+sinϕ2,i,n

2π sinϕ1,i,n

2π 0, (6)

E[D_i,n+1−D_i,n|Fτi+n]sinψ_1,i,n+sinψ_2,i,n

2π sinψ_1,i,n

2π 0 (7)

and

E[Di,n+1−Di,n+Ri,n+1−Ri,n|Fτ_i+n]c2 (8) for some constantc₂>0.

Fig. 5 shows examples in which the expected increments ofR_i,nandD_i,nare close to 0, thus explaining, why we are not able to bound in (6) and (7) these expected increments individually away from 0. Note however, that in both situation depicted in Fig. 5, if the expected increment ofRi,nor ofDi,nis small then the expected increment of the other quantity is large. This confirms that the expected increments ofRi,nandDi,ncannot both be small at the same time, see (8).

Fig. 5. The expected increment ofR_i,nis small in the left figure and large in the right figure. ForD_i,nit is the other way round.

Proof of Lemma 2. We fixi, n1 and drop them as subscripts ofϕ_j,i,nandψ_j,i,n(j=1,2). Then the following statements hold on the event{τ_i +n < τ_i+1}. Consider the angle betweenY_i, X_τi+n andX_τi+n, X_τi+n+1 which includess_2,i,n. This angle is chosen uniformly at random from the interval [ϕ₂,2π−ϕ₁]. Hence we get by a change of basis argument usingY_i, X_τi+nfor the first basis vector,

E[R_i,n+1|Fτ_i+n] = 1 (2π−ϕ₁)−ϕ₂

2π−ϕ1

ϕ2

(R_i,n,0)−(cosϕ,sinϕ)dϕ

1

2π−ϕ₁−ϕ₂

2π−ϕ₁

ϕ₂

|Ri,n−cosϕ|dϕ

R_i,n+ 1

2π−ϕ₁−ϕ₂

2π−ϕ₁

ϕ₂

−cosϕdϕ

=R_i,n+sinϕ₁+sinϕ₂

2π−ϕ₁−ϕ₂ R_i,n+sinϕ₁+sinϕ₂

2π ,

which shows (6). Similarly, (7) follows from E[D_i,n+1|Fτi+n] = 1

(2π−ψ₁)−ψ₂

2π−ψ₁

ψ2

|D_i,n−cosψ|dψ.

For the proof of (8) we assume without loss of generalityϕ₁π/2. Indeed, otherwise ϕ₂π/2 because of ϕ₁+ϕ₂π, see (4), and in the following proof one only has to replace the subscript 1 by the subscript 2 and swap X_τi andX_k(i). By (6) and (7),

E[Di,n+1−Di,n+Ri,n+1−Ri,n|Fτ_i+n]sinϕ₁+sinψ₁

2π . (9)

We will show that the right side of (9) is always greater thanc₂:=(4π²)⁻¹. Assume that it is less thanc₂. Then

sinϕ₁,sinψ₁(2π )⁻¹ (10)

and henceϕ₁(π/2)sinϕ₁1/4 by concavity of sin on[0, π/2]. Similarly, (10) implies that eitherψ₁1/4 or π−ψ₁1/4. Due to|ϕ₁−ψ₁|π/2, see (5), the latter case is impossible. Therefore,

|ϕ₁−ψ₁|max

|ϕ₁|,|ψ₁|

1/4. (11)

The angleα∈ [0, π/2]betweenZ_i,n, X_τi+nandX_τi+n, X_τi is less than or equal toψ₁. Consequently, sinψ₁sinα= X_τi−Z_i,n

X_τi−X_τi+n X_τi−Y_i − Y_i−Z_i,n

d_τi . (12)

β:=1+4π√

8>30 and γ:= 1 2β < 1

60. (14)

Wheneverτ_i<∞we denote the first exit time afterτ_ifromB(X_τi, γ d_τi)by γ_i+1:=inf

n > τ_i: X_n−X_τi> γ d_τi

(∞),

see Fig. 2. Ifi1 andn0 then we shall callngood foriifn=0 or if τ_i+n < τ_i+1∧γ_i+1 and E[R_i,n+1−R_i,n|Fτi+n] 1

π√

8 P-a.s. (15)

This means,n1 is good foriif at timeτi+nthe walker has not yet left the intersection of the small ball around Xτ_i and the lens shown in Fig. 2 and, roughly speaking, feels a substantial centrifugal force pushing it away from the centerYi. Good times help the walker to leave the lens shortly afterτiand closely to the pointXτ_i.

Next we introduce an Azuma type inequality for a certain family of supermartingales, which will help make this idea more precise.

Lemma 3. There are constantsc₃>0, c₄>0 and 1c₅<∞such thatP-a.s. for alli1, n0, E[M_i,n|Fτi]c₅exp(−c₄n),

where

M_i,n:=1{τ_i+n < τ_i+1}exp

c₃ n

m=1

W_i,m and

W_i,m:=D_i,m−1−D_i,m+β(R_i,m−1−R_i,m)+41{m−1 is good fori}. Proof. Fixi1. Firstly, we shall prove that for suitablec₄>0,

E

Mi,n+1|Fτ_i+n

exp(−c₄)Mi,n P-a.s. for alln1, (16)

thus showing that(M_i,n)_n1is an exponentially fast decreasing submartingale. SinceW_i,misFτi+m-measurable (m1), we have

E

M_i,n+1|Fτ_i+n

=M_i,nE

exp(c₃W_i,n+1), τ_i+n+1< τ_i+1|Fτ_i+n

M_i,nE

exp(c₃W_i,n+1)|Fτ_i+n

.

Therefore is suffices to show for the proof of (16) that on the event{τi+n < τi+1},E[exp(c₃Wi,n+1)|Fτ_i+n]<1 for some suitablec₃>0. However, since theWi,m’s(m1)are uniformly bounded by a constant this follows from the fact that on the event{τi+n < τi+1},

E[W_i,n+1|Fτi+n] =E[D_i,n−D_i,n+1+R_i,n−R_i,n+1|Fτi+n] +4

π√

8E[Ri,n−Ri,n+1|Fτ_i+n] +1{nis good fori} −c2<0

P-a.s. by virtue of definition (15) and Lemma 2 (6), (8). Using (16) for induction overnwe obtain by the tower property,

E[Mi,n|Fτ_i]exp

−c₄(n−1)

E[Mi,1|Fτ_i] P-a.s.

for alln1. SinceM_i,1is bounded by a constant this finishes the proof.

4. Exponential moments ofτi+1−τi

The next two lemmas provide tools needed to bound in Proposition 6 the timesi during which the diameter does not increase. The first lemma shows that the walk cannot spend too much time inside the lens and the small ball aroundX_τishown in Fig. 2.

Lemma 4. For alli1,n0,P-a.s.

P[τ_i+n < τ_i+1∧γ_i+1|Fτi]c₅exp(−c₄n), wherec₄andc₅are as in Lemma 3.

The second lemma gives a first crude upper bound for_i.

Lemma 5. There is a finite constantc₆such that for alli1, n0,P-a.s.

P[_i> n|Fτ_i]c₅exp

c₄(c₆d_τi−n) , wherec₄andc₅are as in Lemma 3.

In the proof of both lemmas we will use the fact that the sum in the definition ofM_i,nis in part telescopic, i.e.

n

m=1

Wi,m= −Di,n+β(Ri,0−Ri,n)+4 n

m=1

1{m−1 is good fori} (17)

sinceD_i,0=0.

Proof of Lemma 4. Forn=0 the statement is true sincec₅1. Fixn1. The statement follows from Lemma 3 and (17) once we have shown that on the event{τi+n < τi+1∧γi+1}, the right-hand side of (17) is nonnegative.

First we will show that on{τi+n < τi+1∧γi+1}, Di,n+β(Ri,n−Ri,0)2

Di,n− Xτ_i−Zi,n

. (18)

This is done by brute force. For abbreviation we setd:=dτ_i,y :=Di,nandx:= Xτ_i −Zi,n and note that on {τi+n < τi+1∧γi+1}we havex, y∈ [0, γ d]. Observe thatxandyplay the role of Cartesian coordinates ofXτ_i+n, see Fig. 6. UsingR_i,0=d/2 andR_i,n=

(d/2−x)²+y²we see that (18) is equivalent to β

(d/2−x)²+y²y−2x+βd/2.

Both sides of this inequality are nonnegative sincex is less thanγ d, which is tiny compared toβd. Taking the square and rearranging shows that (18) is equivalent to

x(4x−4y−2βd−β²x+β²d)+y(y+βd−β²y)0. (19)

Fig. 6. More realistic and detailed than Fig. 3. The darkly shaded convex hullKτ_iat timeτ_ihas been enlarged after three steps by the lightly shaded part.j=3 is good forisince it satisfies the sufficient criterionψ_1,i,3π/4, see (22), which corresponds to the fact that the dashed line, which intersects the horizontal axis at an angle ofπ/4, does not intersectK_τ^o

i+3.j=1 is also good forifor the same reason, while j=2 might be good foribut fails to satisfy the sufficient condition (22), since the corresponding dashed line starting inXτ_i+2would have intersectedK_τ^o

i+2.

Sincex, y∈ [0, γ d]andβγ=1/2, see (14), the termsβ²d in the first bracket andβd in the second bracket are the dominant terms, respectively, which shows that (19) and thus (18) holds. For the proof that the right-hand side of (17) is indeed nonnegative on{τ_i+n < τ_i+1∧γ_i+1}it therefore suffices to show that,

D_i,n− X_τi−Z_i,n2

n−1

j=0

1{j is good fori}. (20)

BothD_i,j and−X_τi−Z_i,jcan increase by at most 1 ifj increases by 1. Therefore, the left-hand side of (20) is less than or equal to 2(#J_i,n)where

J_i,n:=

0j < n| ∀0m < j: D_i,m− X_τi−Z_i,mD_i,j− X_τi−Z_i,j .

Hence it suffices to show that the elements ofJ_i,nare good fori. Note thatj =0∈J_i,nis good foriby definition of being good. So fix 1j∈J_i,n. By Lemma 2 (6) it is enough to show that sinϕ_1,i,j2^−1/2, that is

ϕ_1,i,j∈ [π/4,3π/4]. (21)

On the one hand,ϕ₁−ψ₁is close toπ/2, as can be seen in Fig. 6. More precisely, sin(ϕ1−ψ1)= Y_i−Z_i,j

Y_i−X_τi+j Y_i−X_τi − X_τi−Z_i,j Y_i−X_τi + X_τi−X_τi+j d_τi/2−γ d_τi

d_τi/2+γ d_τi =1−2γ 1+2γ 1

√2 =sinπ 4.

Since 0ψ₁ϕ₁this impliesϕ₁π/4, thus proving the first part of (21). On the other hand,ϕ₁−ψ₁π/2, see (5). Hence all that remains to be shown for the completion of the proof of (21) is that

ψ1,i,jπ/4. (22)

Consider the half line (dashed in Fig. 6) starting atX_τi+j which includes an angle of π/4 withX_τi+j, Z_i,j that containss_1,i,j. We claim that this line does not intersectK_τ^o

i+j. This would imply (22). To prove this claim observe that for anyc >0 the set of possible values forX_τi+mwithD_i,m−X_τi−Z_i,m =cis a line parallel to the half line just described. Sincej∈Ji,nthe walker did not cross between timeτi and timeτi+j−1 the dashed line passing throughXτ_i+j. Consequently, it suffices to show that the dashed line does not intersectK_τ^o

i. If it did intersectK_τ^o

i

then this would force the walker on its way fromX_τitoX_τi+j to cross the dashed line strictly before timeτ_i+j, which is impossible as we just saw. 2

Proof of Lemma 5. Again, the statement is true forn=0 sincec₅1. Forn1, by the Pythagorean theorem and (13),D_i,nR_i,nd_τi on the event{τ_i+n < τ_i+1}. Thus on this event the right-hand side of (17) is greater than or equal to−d_τi+β(0−d_τi)+0. Hence due to Lemma 3 and (17),P-a.s. for alln1,

exp

−c₃(1+β)d_τi

P[τ_i+n < τ_i+1|Fτi]c₅exp(−c₄n),

which is equivalent to the claim of the lemma withc₆:=c₃(1+β)/c₄. 2 The following result is stronger than Lemma 5.

Proposition 6.τ_i<∞P-a.s. for all i0. Moreover, there are positive constantsc₇, c₈and finite constantsc₉ andc₁₀1 such that for alli0 andn0,P-a.s.,

P[_in|Fτ_i]c₉exp(−c₇n) and (23)

E

exp(c_8i)|Fτi

c₁₀. (24)

Proof. (24) is an immediate consequence of (23). For the proof of (23) choose c7:= γ c₄

c₆+γ and c9:=c5e^c4 (25)

withc₄,c₅andc₆according to Lemma 5. We only need to show that (23) holds for alli1 withτ_i<∞. Indeed, the casei=0 is trivial sinceτ₀=0, τ₁=1 and hence₀=1. Moreover, since (23) implies_i<∞P-a.s., we then haveτ_i=₀+ · · · +_i−1<∞as well.

Fixi1 andn0. We distinguish three cases:

{n < γ d_τi},

γ d_τin < (c₆+γ )d_τi

and

(c₆+γ )d_τin . Note that these events are elements ofFτ_i and partitionΩ. By definition ofγ_i+1,

γ_i+1τ_i+ γ d_τi (26)

since the walker takes steps of length one. Therefore, on{n < γ dτ_i}, P[_i> n|Fτ_i] ⁽²⁵⁾= P[τ_i+n < τ_i+1∧γ_i+1|Fτ_i]

Lemma 4

c₅exp(−c₄n)⁽²⁵⁾ c₉exp(−c₇n).

On{γ d_τin < (c₆+γ )d_τi}, P[_i> n|Fτ_i] P

τ_i+ γ d_τi< τ_i+1|Fτ_i

⁽²⁶⁾

P

τ_i+ γ d_τi −1< τ_i+1∧γ_i+1|Fτ_i

Lemma 4

c₅exp

−c₄(γd_τi −1)⁽²⁵⁾

c₉exp(−c₇n).

Finally, on{(c₆+γ )d_τin}, P[_i> n|Fτi]^{Lemma 5} c₅exp

c₄(c₆d_τi−n)⁽²⁵⁾

c₉exp(−c₇n), where the last inequality can easily be checked. 2

E

exp(dm−dn)

c12(n+1)exp

c11(m−n) .

For the proof of this lemma and of Theorem 1 we need the following definition: Given n 0 let i_n:=sup{i0|τ_in}. Note that

d_τin=d_n and i_nτ_inn < τ_in+1. (27)

Proof of Lemma 7. The casen=mis trivial. So let 0m < nand set f (m, n):=n−m

2n >0 and g(m, n):=c8f (m, n)

2 lnc₁₀ >0, (28)

wherec₈andc₁₀are according to Proposition 6. A simple union bound yields E[e^dm−dn]I+II+III, where

I:=P

τ_im+1−mf (m, n)n , II:=P

τ_im+1−m < f (m, n)n, i_n< i_m+ g(m, n)n and III:=E

exp(d_m−d_n), i_ni_m+ g(m, n)n ,

see also Fig. 7. Here term I corresponds to the situation in which after timemthe diameter does not increase for an untypical long while. Term II handles the case in which the diameter does increase shortly after timem, as it should, but not often enough in the remaining time untiln. The third term III considers the original random variable on the typical event that the number of times at which the diameter increases is at least proportional tonwith a constant of proportionality not too small.

It suffices to show that each of these three terms decays asn→ ∞in the way claimed in Lemma 7 with constants c₁₁andc₁₂independent ofnandm. As for the first term,

I P

_im

f (m, n)n(27)

= m

i=0

P

i_m=i, _if (m, n)n

(23) c₉(m+1)e^{−c7f (m,n)n}c₉(n+1)e^c7(m−n)/2,

which is an upper bound like the one requested in Lemma 7. The second term is estimated as follows.

II⁽²⁷⁾= P

τ_im+1< m+f (m, n)n, n < τ_in+1τ_im+g(m,n)n

(28) P

τ_im+g(m,n)n−τ_im+1n−m−f (m, n)n=f (m, n)n

E

exp c₈

τi_m+g(m,n)n−τi_m+1−f (m, n)n

Fig. 8. The eventAioccurs if the first trial point sampled lies on the bold arc.

= e^{−c8f (m,n)n}

k1

E exp

c₈(τ_{k+g(m,n)n−1}−τ_k)

, i_m+1=k

= e^{−c8f (m,n)n}

k1

E

_g(m,n)n−2

i=0

exp(c_8k+i), i_m+1=k

. (29)

Note that{im+1=k}is the event thatτk is the first time after timemat which the diameter increases. Therefore,

{i_m+1=k} ∈Fτ_k. (30)

Moreover, the increments_k+i are measurable with respect toFτ_k+i+1. Consequently, by conditioning in (29) on

Fτk+g(m,n)n−2 and applying Proposition 6 (24) withi=k+ g(m, n)n −2 we conclude

IIe^{−c8f (m,n)n}c₁₀

k1

E

_g(m,n)n−3

i=0

exp

c₈(_k+i)

, i_m+1=k

.

Continuing in this way we obtain by induction afterg(m, n)n −1 steps, IIe^{−c8f (m,n)n}c₁₀^g(m,n)n−1e^{−c8f (m,n)n}c^g(m,n)n₁₀ ⁽²⁸⁾= e^{(c8/4)(m−n)}, which is again of the form required in Lemma 7.

In order to demonstrate that also the third term III behaves properly we will show that the incrementsd_τi+1−d_τi, i1,have a uniformly positive chance of being larger than a fixed constant, say 1/2, independently of the past. To this end, we may assume that the process(X_n)_nis generated in the following way: There are i.i.d. random variables U_n,k,n0,k0,uniformly distributed on the unit circle centered in 0 such thatX_n+1=X_n+U_n,k,wherekis the smallest integer such thatX_n, X_n+U_n,kdoes not intersectK_n. Then for anyi1, by definition ofτ_i,

{d_τi+1d_τi+1/2} ⊇ {d_τi+1d_τi+1/2} ⊇

U_τi,0·X_τi−X_k(i) d_τi 1

2

=:A_i. (31)

The reason for the second inclusion in (31) is illustrated in Fig. 8: IfAi occurs thenKτ_i andXτ_i +Uτ_i,0 are separated by a slab of width 1/2. In particular, the line connectingXτ_i andXτ_i +Uτ_i,0does not intersect Kτ_i. Henceτi+1=τi+1 andXτ_i+1=Xτ_i+1=Xτ_i+Uτ_i,0. Therefore,

dτ_i+1Xτ_i+1−Xk(i)Xτ_i−Xk(i) +1/2=dτ_i+1/2.

It follows from (31) that for all 1j₁j₂, d_τj

2−d_τj

1 1

2

j₂−1 i=j₁

1{d_τi+1 d_τi+1/2}1 2

j₂−1 i=j₁

1{A_i}. (32)

E exp(d_τim+1−d_τin), i_n(i_m+1)+ g(m, n)n −1

k1

E

exp(dτ_k−dτ_{k+g(m,n)n−1}), im+1=k

(32)

k1

E

exp

−1 2

k+g(m,n)n −2 i=k

1{A_i}), i_m+1=k

. (35)

As seen in (30),{i_m+1=k} ∈Fτk⊆Fτk. Therefore, after conditioning in (35) onFτk, we see with the help of (34) forikand (33) that the right-hand side of (35) equals

E

exp

−1 2A₁

g(m,n)n−1

,

which decays as required in Lemma 7, see (28). 2

Lemma 7 directly implies a weaker version of Theorem 1 in whichX_nis replaced byd_n. For the full statement we need the following additional argument.

Proof of Theorem 1. (2) follows from (1) by the Borel–Cantelli lemma. For the proof of (1) pickc₁₁ andc₁₂ according to Lemma 7 and choosec₁₃>0 andc₁>0 small enough such that

2c₁₃−c₁₁<0 and 2c₁−c₁₁(c₁₃−c₁) <0. (36) We denote byMn:=max{Xm |mn}the walker’s maximal distance from the origin by timen. Note thatMn

andd_nare related via

Mndn2Mn for alln0 (37)

because ofX₀=0. By a union bound for anyn0, P

X_nc₁n

P[_inc₁n] +P[M_nc₁₃n] +P[B_n], where (38) B_n:=

_in< c₁n, M_n> c₁₃n,X_nc₁n .

It suffices to show that each one of the three terms on the right-hand side of (38) decays exponentially fast inn. As for the first term,

P[_inc₁n]⁽²⁷⁾= n

i=0

P

i_n=i, _ic₁n⁽²³⁾

c₉(n+1)e^−c7c1n,

which decays exponentially fast innindeed. So does the second term in (38) since by Chebyshev’s inequality, P[Mnc₁₃n]⁽³⁷⁾ P[dn2c₁₃n]e^2c13nE[e^−dn]^{Lemma 7} c₁₂(n+1)e^{(2c13−c11)n},