An LIL for cover times of disks by planar random walk and Wiener sausage

0.25 MB·English

by  J. Ben Hough

5 AN LIL FOR COVER TIMES OF DISKS
0
BY PLANAR RANDOM WALK AND WIENER SAUSAGE
0
2

n J. BEN HOUGH∗ AND YUVAL PERES∗
a
J

4
2 Abstract. LetRnbetheradiusofthelargestdiskcoveredafternstepsofasimplerandom
walk. We prove that almost surely limsupn→∞(logRn)2/(lognlog3n) = 1/4, where log3
] denotes3iterationsofthelogfunction. ThisismotivatedbyaquestionofErdo˝sandTaylor.
R
We also obtain the analogous result for the Wiener sausage, refining a result of Meyre and
P
Werner.
.
h
t
a
m
[
1. Introduction

2
v
9 In this paper, we consider a planar simple random walk starting from the origin and
3 determine asharpasymptotic upper boundforthegrowthoftheradius ofthelargestdiscrete
2
disk centered at the origin that is covered after n steps of the walk. More precisely, let
9
0 D = D(0,r) Z2 be the discrete disk of radius r, S(n) a planar simple random walk
r
4 ∩
starting from the origin, and R = sup r : D S[0,n] the radius of the largest discrete
0 n { r ⊂ }
/ disk centered at the origincovered by time n. It is well known that the planar simple random
h
walk is recurrent so R as n . We prove that almost surely
t n
a → ∞ → ∞
m (logR )2 1
n
(1) limsup = ,
: lognlog n 4
v n→∞ 3
i
X where log denotes k iterations of the log function. We also consider the analogous covering
k
r problemfortheplanarWiener sausage ofradius1, andprovethatthesameasymptotic result
a
holds in this case as well. Specifically, if denotes a planar Brownian motion starting from
t
B
the origin, the Wiener sausage of radius 1 up to time t is defined to be
= D( ,1).
t s
S B
s∈[0,t]
[
We define = sup r : D(0,r) and prove that almost surely
t t
R { ⊂ S }
(log )2 1
t
(2) limsup R = .
logtlog t 4
t→∞ 3
A problem related to the asymptotic behavior of R was posed as early as 1960 by Erd˝os
n
and Taylor. In [8, p.153] they ask: “How quickly does the function f(n) need to increase
Date: October 18, 2004.
∗
The authors gratefullyacknowledgethe financialsupport fromNSF grants#DMS-0104073and#DMS-
0244479.
1
2 J. BEN HOUGH AND YUVALPERES
so that in an infinite plane random walk, with probability 1, all the lattice points within
a distance n of the origin will be entered by the walk before f(n) steps except for finitely
many values of n?” The asymptotic growth of has been studied by Meyre and Werner in
t
R
[7], who establish the existence of c [1/8,1] so that almost surely limsup (logRt)2 = c.
∈ t→∞ logtlog3t
The present paper relies heavily on the use of excursions between large concentric circles
to control covering times. This technique has been used effectively in many recent works on
the subject, see [1], [2] and [7] for example. Although it is seemingly difficult to calculate
the cover time for a disk directly, one may say very precisely how many excursions will occur
between large concentric circles prior to the covering time of the disk. The cover time may
then be determined indirectly by studying the expected duration of the requisite number of
excursions. We derive the results above by establishing sufficiently tight bounds on the tails
of the relevant distributions. Much of the work lies in estimating the number of excursions
necessary to cover a disk. Lemma 2.6 gives this estimate for the simple random walk, and is
derived using results from [1] and [2]. Lemma 3.3 gives an analogous estimate for the Wiener
sausage, and is derived using results from [1].
This article is divided into two main sections. In the first section we study the planar
simple random walk process and prove (1). The second section demonstrates that analogous
results hold for the planar Wiener sausage.
2. Asymptotic results for the random walk
We begin with some definitions. Recall that S(n) is a simple random walk in the plane
started at zero, and D = D(0,r) Z2 is the disk of radius r in Z2. Define random variables
r
∩
R = sup r : D S[0,n]
n r
{ ⊂ }
T = inf n : D S[0,n]
r r
{ ⊂ }
so that R is the largest disk centered at zero that is covered by time n, and T is the time
n r
necessary to cover D . The following discussion will demonstrate that
r
(logR )2 1
n
(3) limsup = a.s.
lognlog n 4
n→∞ 3
The proof has two main components, we first check that 1 is an upper bound for the limsup,
4
and then deduce that it is a lower bound as well. The proofs of a few technical estimates
will be saved for the end.
2.1. Establishing that 1 is an upper bound. For notational simplicity, define
4
(4) f(x) = exp [λlogxlog x]1/2 ,
{ 3 }
℘(x) = x(logx)3 and t = eαn where α > 1 is arbitrary. It is sufficient to show that
n
whenever λ > 1 there exists α = α(λ) > 1 so that P A i.o. = 0 where A is the event
4 { n } n
AN LIL FOR COVER TIMES OF DISKS 3
that R f(t ). We compute probabilities of events determined by A ’s by counting
excursiotnn+s1.≥Let ∂Dn be the boundary of D in Z2: n
r r
(5) ∂D = z Dc Z2 : y z = 1 for some y D .
r { ∈ r ∩ | − | ∈ r}
Definition 2.1. We say that an excursion from ∂D to ∂D occurs between times s < t if
r p
S(s) ∂D , S(t) ∂D and the times s and t satisfy:
r p
∈ ∈
1. t = min n : n > s and S(n) ∂D
p
{ ∈ }
2. s = min n : n > τ and S(n) ∂D where τ = max n : n < t and S(n) ∂D (we
r p
{ ∈ } { ∈ }
take the max of the empty set to be 0).
The key result linking excursions to our covering problem is that if N is the number of
r
excursions between ∂D and ∂D that occur before T , and φ = (logr)2 then
2r ℘(r) r r log r
2
N 2
r
(6)
φ → 3
r
in probability as r ([1, Lemma 5.1] and [2, Lemma 2.4, Theorem 2.5]).
→ ∞
With this result in mind, we are inspired to introduce the events A˜ that at least (2
n 3 −
ǫ )φ excursions from ∂D to ∂D occur before the walk first hits ∂D .
1 f(tn) 2f(tn) ℘(f(tn)) t1/2+ǫ2
n+1
The events A and A˜ should then be comparable since a random walk requires roughly r2
n n
steps to first hit ∂D . We now quote [3, Exercise 1.6.8] which will allow us to compute the
r
˜
probabilities of the events A . In what follows, ζ(r) gives the hitting time of ∂D .
n r
Lemma 2.1. If ρ < r < P and x ∂D ,
r
∈
log(P) log(r)+O(1/ρ)
(7) Px ζ(ρ) < ζ(P) = − ,
{ } log(P) log(ρ)
−
where the O( ) term is bounded uniformly in x, ρ and P.
·
The following corollary follows easily, see also [3, Proposition 1.6.7]
Corollary 2.2. If 0 < r < P and x ∂D then:
r
∈
1 log(P) log(r)+O(1)
(8) Px ζ(0) < ζ(P) − .
{ } ≥ 16 log(P)
(cid:18) (cid:19)
From the above lemma, we see immediately that
4 J. BEN HOUGH AND YUVALPERES
logt1/2+ǫ2 log(℘(f(t )))]+O(1) ⌊(2/3−ǫ1)φf(tn)⌋
P(A˜ ) = n+1 − n
n " logt1/2+ǫ2 log[2f(t )] #
n+1 − n
log[℘(f(t ))] log[2f(t )]+O(1) ⌊(2/3−ǫ1)φf(tn)⌋
n n
= 1 −
" − logt1/2+ǫ2 log[2f(t )] #
n+1 − n
3log [f(t )]+O(1) ⌊(2/3−ǫ1)φf(tn)⌋
(9) = 1 2 n .
" − logt1/2+ǫ2 log[2f(t )]#
n+1 − n
We use the following two estimates to simplify the expression for P(A˜ ).
n
Lemma 2.3. For ǫ < 1 a and δ < 1 b we have
| n| 2| n| n 2| n|
a ǫ a ǫ a δ
n n n n n n
− = +O +O .
b +δ b b b2
n n n (cid:18) n(cid:19) (cid:18) n (cid:19)
Lemma 2.4. If α2β and β ǫ are bounded, α , ǫn 0 and δ < 1 then:
n n n n n αn → | n|
[1 α +ǫ ]βn+δn = e−αnβn[1+O(α2β )+O(β ǫ )+O(α δ )].
− n n n n n n n n
Using Lemma 2.3 to simplify the fraction in (9) and then applying Lemma 2.4, we may
simplify the expression for P(A˜ ):
n
4(1 3ǫ /2)(logf(t ))2 [log f(t )]2 φ
P(A˜ ) = exp − − 1 n 1+O 2 n φ +O f(tn)
n 1+2ǫ logt (logt )2 f(tn) log(t )
(cid:20) 2 n+1 (cid:21)(cid:26) (cid:18) n+1 (cid:19) (cid:18) n+1 (cid:19)
log f(t )
(10) + O 2 n .
logt
(cid:18) n+1 (cid:19)(cid:27)
It is easy to verify that
[log f(t )]2 φ log f(t )
(11) lim 2 n φ = 0, and lim f(tn) = 0, and lim 2 n = 0
n→∞ (logtn+1)2 f(tn) n→∞ log(tn+1) n→∞ logtn+1
so ∞ P(A˜ ) converges if and only if
n=1 n
P
∞ 4(1 3ǫ /2)(logf(t ))2
1 n
(12) exp − −
1+2ǫ logt
n=1 (cid:20) 2 n+1 (cid:21)
X
converges. Now compute:
∞ 4(1 3ǫ /2)(logf(t ))2 ∞ 4(1 3ǫ /2)λαnlog(nlogα)
1 n 1
exp − − = exp − −
1+2ǫ logt 1+2ǫ αn+1
n=1 (cid:20) 2 n+1 (cid:21) n=1 (cid:20) 2 (cid:21)
X X
∞
−4λ1−3ǫ1/2
(13) = C n α 1+2ǫ2 .
n=1
X
AN LIL FOR COVER TIMES OF DISKS 5
It follows that if λ > 1 we may choose ǫ > 0 and ǫ > 0 and α > 1 so that the above sum
4 1 2
converges, and hence P A˜ i.o. = 0.
n
n o
To finish the argument, we must demonstrate that the event A i.o. A˜ i.o. has
n n
{ } − { }
probability zero. Observe that A i.o. A˜ i.o. C D where
n n 1 1
{ }−{ } ⊂ ∪
C = ζ(t1/2+ǫ2) < t i.o.
1 n n
D = N < (2/3 ǫ )φ i.o. .
1 (cid:8) f(tn) − 1(cid:9) f(tn)
(cid:8) (cid:9)
The fact that C has probability zero follows from the estimate [3, p.146]:
1
Lemma 2.5. P ζ(r) / (r2−ǫ,r2+ǫ) = O(e−rγ), for some γ = γ(ǫ) > 0.
{ ∈ }
And the following lemma shows that D is also null.
1
Lemma 2.6. P Nr / 2 ǫ, 2 +ǫ O 1 .
φr ∈ 3 − 3 ≤ (log2r)2
n o (cid:16) (cid:17)
(cid:0) (cid:1)
Indeed the claim follows from the following computation:
∞ ∞
1 4
(14) = <
(log f(t ))2 (nlogα+logλ+log αn)2 ∞
n=1 2 n n=1 3
X X
and an application of the Borel-Cantelli lemma. The proof of this lemma is rather involved,
and will be given at the end of the section.
2.2. Establishing that 1 is a lower bound. This proof is nearly identical to the one just
4
given. We define the events H and H˜ so that H is the event that R > f(t ) and H˜ is
n n n tn n n
the event that there are at least (2/3+ǫ )φ excursions between ∂D and ∂D
1 f(tn) 2f(tn) ℘(f(tn))
before ζ(t1/2−ǫ2). The task now is to show that if λ < 1 then P H i.o. = 1. We will do
n 4 { n }
this by first showing that we may choose ǫ > 0 and ǫ > 0 so that P H˜ i.o. = 1 and then
1 2 n
{ }
checking that for such ǫ and ǫ the two sets are identical up to null sets.
1 2
To this end, we define sigma fields = σ S(k) : k < ζ(t1/2−ǫ2) , and check that if
n n
F { }
λ < 1 then for all sufficiently small ǫ > 0 and ǫ > 0, we have ∞ P(H˜ ) = with
4 1 2 n=1 n|Fn−1 ∞
probability one. Observe that by the strong Markov property:
P
P(H˜ ) min Px(ζ(0) < ζ(t1/2−ǫ2))P(H˜ )
n|Fn−1 ≥ x∈∂Dt1n/−21−ǫ2 n n
1 log[t1/2−ǫ2] log[t1/2−ǫ2]+O(1)
n − n−1 P(H˜ )
≥ 16 log[t1/2−ǫ2] ! n
n
1 1
(15) 1 P(H˜ ),
n
≥ 32 − α
(cid:18) (cid:19)
where the last inequality follows from Corollary 2.2.
6 J. BEN HOUGH AND YUVALPERES
Now compute:
∞ ∞ log(t1/2−ǫ2) log(℘(f(t )))+O(1) ⌊(2/3+ǫ1)φf(tn)⌋
P(H˜ ) = n − n
n " log(t1/2−ǫ2) log(2f(t )) #
Xn=1 Xn=1 n − n
∞ 3log (f(t ))+O(1) ⌊(2/3+ǫ1)φf(tn)⌋
= 1 2 n
− (1/2 ǫ )log(t ) log(2f(t ))
n=1(cid:20) − 2 n − n (cid:21)
X
∞ 4(1+3ǫ /2)[logf(t )]2 [log f(t )]2 φ
= exp − 1 n 1+O 2 n φ +O f(tn)
1 2ǫ logt (logt )2 f(tn) logt
n=1 (cid:20) − 2 n (cid:21)(cid:26) (cid:18) n (cid:19) (cid:18) n(cid:19)
X
log f(t )
(16) + O 2 n .
logt
(cid:18) n (cid:19)(cid:27)
Both error terms tend to zero as n , so it suffices to consider the sum
→ ∞
∞ 4(1+3ǫ /2)[logf(t )]2 ∞ 4(1+3ǫ /2)λαnlog αn
exp − 1 n = exp − 1 2
1 2ǫ logt 1 2ǫ αn
n=1 (cid:20) − 2 n (cid:21) n=1 (cid:20) − 2 (cid:21)
X X
∞
−4λ1+3ǫ1/2
(17) = C n 1−2ǫ2 .
n=1
X
If λ < 1, we may choose ǫ > 0 and ǫ > 0 small enough to that the above sum diverges
4 1 2
and thus the sum ∞ P(H˜ ) diverges a.s. Now applying the Borel-Cantelli lemma
n=1 n|Fn−1
for filtrations [5, Corollary 7.20] we may conclude that P H˜ i.o. = 1.
n
P { }
To complete the proof, we must show that P( H˜ i.o. H i.o ) = 0. Observe
n n
{ } − { }
that H˜ i.o. H i.o C D , where
n n 2 2
{ }−{ } ⊂ ∪
C = ζ(t1/2−ǫ2) > t i.o.
2 { n n }
D = N (2/3+ǫ )φ i.o. .
2 { f(tn) ≥ 1 f(tn) }
Lemmas 2.5 and 2.6 show that both these sets are null, so the proof is complete.
2.3. Proof of Lemma 2.6. The method of proof is to consider the simpler problem of
covering D by random sets that are covered during i.i.d. excursions and then to show
r
that our problem does not differ substantially from this situation. More precisely, define
k(0,r) = 0, s(0,r) = ζ(℘(r)) and for j 1:
≥
k(i,r) = inf j > s(i 1,r) : S(j) ∂D
2r
{ − ∈ }
s(i,r) = inf j > k(i,r) : S(j) ∂D .
℘(r)
{ ∈ }
Let A(j,r) = S[k(j,r),s(j,r)] D consist of the points in D covered by the jth excursion of
r r
∩
the random walk. Take C(j,r), E(j,r) and F(j,r) to be i.i.d. subsets of D chosen according
r
to the following distribution: pick x ∂D according to harmonic measure, start a random
2r
∈
walk at x, stop the walk when it reaches ∂D and choose the points in D that are visited
℘(r) r
AN LIL FOR COVER TIMES OF DISKS 7
by the walk. Harmonic measure, H , on ∂D is the hitting probability from infinity:
r r
H (x) = lim Py S(ζ(r)) = x .
r
|y|→∞ { }
Let U(k,r) = k A(j,r). Also, define V (k,r) = k C(j,r) and V , V similarly.
j=0 C j=0 E F
In what follows, we frequenty conserve notation by writing V (x,r) when we should more
i
S S
properly write V ( x ,r). We now prove an estimate for the i.i.d. covering process:
i
⌊ ⌋
Lemma 2.7. P V ((2 ǫ)φ ,r) = D O 1 and P V ((2 +ǫ)φ ,r) = D O 1 ,
i 3 − r r ≤ φr i 3 r 6 r ≤ φr
where i = C,E or F. (cid:16) (cid:17) (cid:16) (cid:17)
(cid:8) (cid:9) (cid:8) (cid:9)
Proof. From [2, Proposition 2.6] we have the following result:
Lemma 2.8. Suppose for some 0 < u,c,α < and all r
∞
(18) P V (uφ ,r) = D c(logr)−α.
i r r
{ 6 } ≥
Then for all v < u,
(19) lim P V (vφ ,r) = D = 0.
i r r
r→∞ { }
Applying the proof given in [2] to a subsequence of r’s, we deduce that if equation (18)
holds for infinitely many r’s then for all v < u,
(20) liminfP V (vφ ,r) = D = 0.
i r r
r→∞ { }
From (6) we know that Nr 2 in probability and quoting [2, Lemma 2.4] we have that for
φr → 3
any u > 0,
(21) lim P V (uφ ,r) = D = lim P U(uφ ,r) = D
i r r r r
r→∞ { } r→∞ { }
provided that one of the limits exists and equals 0 or 1. Combining these two facts
P V ((2/3+ǫ/2)φ ,r) = D 1.
i r r
{ } →
Inparticular, (20)doesnotholdwhenv = 2/3+ǫ, sobythereasoning immediately preceding
(20) there exists c < such that for all r:
1
∞
c 1
(22) P V ((2/3+ǫ)φ ,r) = D 1 O .
i r r
{ 6 } ≤ (logr)2 ≤ φ
(cid:18) r(cid:19)
For the other bound, we follow the proof of [2, Proposition 2.6] to see that since
c
(23) P V ((2/3 ǫ/2)φ ,r) = D 1
i r r
{ − 6 } ≥ (logr)α
for some c ,α > 1 and all r (note that the LHS converges to 1), we have:
1
(24) P V ((2/3 ǫ)φ ,r) = D E 1 c (logr)−α Yr
i r r 2
{ − } ≤ −
where Y is a random variable satisfying P(Y 1(log(cid:0)r)α+1) O( 1 )(cid:1)and c > 0. (Equation
r r ≤ 2 ≤ φr 2
(24) is equivalent to the last equation on p. 197 of [2]. The bound on the distribution of Y
r
8 J. BEN HOUGH AND YUVALPERES
follows from the last equation on p. 198 of [2], once one observes that Y J (1 Z )).
r ≥ i=1 − i
Therefore
P
1 1
(25) P V ((2/3 ǫ)φ ,r) = D e−c2logr[1+O((logr)1−α)]+O O .
i r r
{ − } ≤ φ ≤ φ
(cid:18) r(cid:19) (cid:18) r(cid:19)
(cid:3)
The next task is to relate the i.i.d. covering problem to the simple random walk covering
problem by constructing a coupling between the two processes. In particular, we construct
therandomsets A(i,r) basedclosely onthei.i.d.sets so thatA(i,r) equals thecorresponding
i.i.d.set withhighprobability. The idea behindtheconstruction isthefollowing. The hitting
probability on ∂D starting from any point on ∂D is nearly the same as harmonic
2r ℘(r)
measure. So for i 1 we can take A(i,r) equal to a new i.i.d. set with probability p close
≥
to 1, and equal to a random set with a biased starting position with probabilty 1 p. The
−
distribution of the set A(0,r) is substantially different. However, we can couple the random
walk to the i.i.d. set process so that A(0,r) is a subset of the first i.i.d. set generated that
contains the origin. This allows us to bound the contribution of A(0,r) from above. A
similar construction will also be used to control the second discrepancy between the coupled
processes.
We now give a precise description of the coupling. Construct A(0,r) as follows: let
mE = min j : 0 E(j,r) and let SE be the random walk which was used to generate
{ ∈ }
E(mE,r). Set tE = min j : SE(j) = 0 and tE = min j : SE(j) ∂D . Define
i { } f { ∈ ℘(r)}
A(0,r) = SE[tE,tE] D . Now, assuming that A(0,r),...,A(ℓ,r) have been constructed let
i f ∩ r
µℓ be the measure on ∂D which satisfies:
2r 2r
(26) µℓ (x) = P S(k(ℓ+1,r)) = x A(0,r),...,A(ℓ,r) .
2r { | }
That is, µℓ is the hitting probability of ∂D conditioned on the subsets of D covered by
2r 2r r
the first ℓ excursions. By [2, Lemma 2.1] we know that µℓ (x) H (x) < c1 H (x)
| 2r − 2r | (logr)2 2r
where c is independent of ℓ and the A(i,r)’s. Thus,
1
µℓ 1 c1 H
(27) νℓ = 2r − − (logr)2 2r
2r (cid:16) c1 (cid:17)
(logr)2
is a probability measure and we have µℓ = (1 c1 )H + c1 νℓ . Define i.i.d. random
2r − (logr)2 2r (logr)2 2r
variables ξ ,ξ ,ξ ,... so that:
1 2 3
c
P(ξ = 0) = 1 1
i − (logr)2
c
P(ξ = 1) = 1 .
i (logr)2
Then if ξ = 0 we set A(ℓ + 1,r) = C(ℓ + 1,r). If ξ = 1 and ξ = 1 for some s ℓ
ℓ+1 ℓ+1 s
≤
we simply pick x ∂D according to the distribution νℓ , start a random walk at x, stop
∈ 2r 2r
it when it hits ∂D and let A(ℓ + 1,r) consist of the points in D visited by the walk.
℘(r) r
Finally, if ξ = 1 and ξ = 0 for all s ℓ then we pick x ∂D according to νℓ as
ℓ+1 s ≤ ∈ 2r 2r
AN LIL FOR COVER TIMES OF DISKS 9
before. But now, we let mF = min j : x F(j,r) and let SF be the random walk used to
{ ∈ }
generate F(mF,r). Set tF = min j : SF(j) = x , tF = min j : SF(j) ∂D , and take
i { } f { ∈ ℘(r)}
A(ℓ+1,r) = SF[tF,tF]. It is easy to see that the A(i,r)’s have the correct joint distributions,
i f
so this construction gives a coupling between the i.i.d. covering process and covering process
of interest.
The motivation for this construction is as follows: with overwhelming probability A(i,r)
will equal C(i,r) for all i < O(φ ) except for i = 0 and possibly some other value, say i .
r p
We will use the E(i,r) sets to control the effect of the discrepancy at i = 0 and the F(i,r)
sets to control the potential discrepancy at i = i . Our next lemma bounds the probability
p
that there are more than two discrepancies.
Lemma 2.9. If u is a constant then P uφr ξ > 1 O 1 .
i=1 i ≤ (log2r)2
n o (cid:16) (cid:17)
P
Proof. Set m = uφ and α = c1 . Then
r (logr)2
uφr
P ξ > 1 = 1 (1 α)m mα(1 α)m−1
i
− − − −
( )
i=1
X
m(m 1)
= 1 [1 mα+ − α2 +O(m3α3)]
− − 2
mα[1 (m 1)α+O(m2α2)]
− − −
m(m 1)
= − α2 +O(m3α3)
2
1
(28) O(m2α2) = O .
≤ (log r)2
(cid:18) 2 (cid:19)
(cid:3)
Next we show that if a = 64logrlog3r then with high probability A(0,r) V (a,r) and,
log2r ⊂ E
if i exists, A(i ,r) V (a,r). Intuitively, two discrepancies are worth no more than 2a
p p F
⊂
excursions, which is insignificant relative to φ .
r
Lemma 2.10. P mE > a O 1
{ } ≤ (log2r)2
(cid:16) (cid:17)
Proof.
P mE > a = (P 0 / E(1,r) )a
{ } { ∈ }
a
1 log(℘(r)) log(2r) c
2
1 − −
≤ − 16 log(℘(r))
(cid:20) (cid:18) (cid:19)(cid:21)
10 J. BEN HOUGH AND YUVALPERES
by Corollary 2.2. Thus,
a
1 log r
P mE > a 1 2
{ } ≤ − 32 logr
(cid:20) (cid:18) (cid:19)(cid:21)
(log r)(log r)
exp[ 2log r] 1+O 2 3
≤ − 3 logr
(cid:20) (cid:18) (cid:19)(cid:21)
1
(29) O ,
≤ (log r)2
(cid:18) 2 (cid:19)
(cid:3)
Lemma 2.11. P mF > a O 1
{ } ≤ (log2r)2
(cid:16) (cid:17)
Proof. Denote by τ the hitting time of the point x by a simple random walk. Then
x
a
P mF > a max Py τ > ζ(℘(r))
x
{ } ≤ (cid:20)x,y∈∂D2r { }(cid:21)
a
1
max Pz τ > ζ ℘(r)
0
≤ (cid:20)z∈∂D4r (cid:26) (cid:18)2 (cid:19)(cid:27)(cid:21)
1 log(1℘(r)) log(4n) c a
1 2 − − 2
≤ − 16 log(1℘(r))
(cid:20) (cid:18) 2 (cid:19)(cid:21)
a
1 log r
1 2
≤ − 32 logr
(cid:20) (cid:18) (cid:19)(cid:21)
1
(30) O
≤ (log r)2
(cid:18) 2 (cid:19)
(cid:3)
by the same computation as in the previous lemma.
The above lemmas provide us with sufficient control over the discrepancies between the
two covering processes to establish the desired bound. To expedite what follows, let us define
k
Y(k,r) = A(i,r) and Z(k,r) = A(i,r).
i[=0 1≤[i≤k
ξi=0