THEORY PROBAB. APPL.c
2006 Socie y o Indus ial and Applied Ma hema ics
Vol. 50, No. 2, pp. 337–346 T ansla ed om Russian Jou nal
CONVERGENCE RATE OF
THE DEPENDENT BOOTSTRAPPED MEANS∗
A. VOLODIN†,M.ORD
´
O˜
NEZ CABRERA‡,AND T. C. HU§
(T ansla ed by A. Volodin)
Abs ac . In his pape , a Baum–Ka z, E d¨os, Hsu–Robbins, Spi ze ype comple e con e gence
esul is ob ained o he dependen boo s apped means.
Key wo ds. boo s apped means, dependen boo s ap, a e o con e gence, exponen ial in-
equali ies, s ong law o la ge numbe s
DOI. 10.1137/S0040585X97981688
1. In oduc ion. The main ocus o he p esen in es iga ion is o ob ain he con-
e gence a es in he o m o a Baum–Ka z, E d¨os, Hsu–Robbins, Spi ze ype comple e
con e gence esul o he dependen boo s apped means om a sequence o andom a i-
ables.
The wo k on he consis ency o boo s ap es ima o s has ecei ed much a en ion in
ecen yea s due o a g owing demand o he p ocedu e in bo h heo e ical and p ac ical
applica ions. I is impo an o no e ha exponen ial inequali ies a e o p ac ical use in
es ablishing he s ong asymp o ic alidi y o he boo s apped mean.
We begin wi h a b ie discussion o esul s in he li e a u e pe aining o a sequence o
independen and iden ically dis ibu ed (i.i.d.) andom a iables and classical (independen )
boo s ap o he mean. Le {X, Xn,n1}be a sequence o i.i.d. andom a iables defined
on a p obabili y space (Ω,F,P). Fo ω∈Ω and n1, le Pn(ω)=n−1n
i=1 δXi(ω)de-
no e he empi ical measu e and le {
X(ω)
n,j ,1jm(n)}be i.i.d. andom a iables wi h
law Pn(ω), whe e {m(n),n1}is a sequence o posi i e in ege s. In o he wo ds, he an-
dom a iables {
X(ω)
n,j ,1jm(n)} esul by sampling m(n) imes wi h eplacemen om
he nobse a ions X1(ω),... ,X
n(ω) such ha o each o he m(n) selec ions, each Xj(ω)
has p obabili y n−1o being chosen.
Fo each n1, {
X(ω)
n,j ,1jm(n)}is he so-called [7] boo s ap sample om
X1,... ,X
nwi h boo s ap sample size m(n). Le Xn(ω)=n−1n
j=1 Xj(ω) deno e he
sample mean o {Xj(ω), 1 jn},n1.
Bickel and F eedman [4] showed ha when Xis nondegene a e and EX2<∞, o
almos e e y ω∈Ω he cen al limi heo em (CLT)
n1/21
n
n

j=1 
X(ω)
n,j −Xn(ω)d
−→ N(0,σ2)
∗Recei ed by he edi o s Feb ua y 14, 2004. This wo k was suppo ed by Na ional Sciences and
Enginee ing Resea ch Council o Canada g an s BFM 2000-0344-C0201, FQM 127, and NSC91-2118-
M-007-008.
h p://www.siam.o g/jou nals/ p/50-2/98168.h ml
†Depa men o Ma hema ics and S a is ics, Uni e si y o Regina, Regina, Saska chewan, S4S
0A2, Canada ([email p o ec ed]).
‡Depa men o Ma hema ical Analysis, Uni e si y o Se ille, Se ille 41080, Spain (cab e a@
us.se).
§Depa men o Ma hema ics, Tsing Hua Uni e si y, Hsinchu 30043, Taiwan ( chu@ma h.
n hu.edu. w).
337
338 A. VOLODIN, M. ORD ´
O˜
NEZ CABRERA, AND T. C. HU
is alid. He e and in wha ollows, σ2=Va X. No e ha by he Gli enko–Can elli heo em,
Pn(ω) is close o L(X) o almos e e y ω∈Ω and all la ge n, and by he classical L´e y CLT,
n1/21
n
n

j=1
Xj−EXd
→N(0,σ2).
I ollows ha o almos e e y ω∈Ω, he boo s ap s a is ic
n1/21
n
n

j=1 
X(ω)
n,j −Xn(ω)
is close in dis ibu ion o ha o
n1/21
n
n

j=1
Xj−EXd
−→ N(0,σ2)
o all la ge n. This is he basic idea behind he boo s ap. See he pionee ing wo k o
E on [7], whe e his nice idea is made explici and whe e i is subs an ia ed wi h se e al
impo an examples.
S ong laws o la ge numbe s we e p o ed by A h eya [2] and Cs¨o g˝o [6] o boo s apped
means. A enal-Gu i´e ez, Ma ´an, and Cues a-Albe os [1] analyzed he esul s o [2] and [6].
Then, by aking in o accoun he diffe en g ow h a es o he esampling size m(n), hey
ga e new and simple p oo s o hose esul s. They also p o ided examples ha show ha
he sizes o esampling equi ed by hei esul s o ensu e almos su e (a.s.) con e gence a e
no a om op imal.
Ano he e e ence which is impo an o his pape is he wo k o Mikosch [10]. He
es ablished a se ies o use ul exponen ial inequali ies ha a e an impo an ool o de i ing
esul s on he consis ency o he boo s apped mean. Based on hese exponen ial inequal-
i ies, he Baum–Ka z, E d¨os, Hsu–Robbins, Spi ze ype comple e con e gence esul o
he boo s apped means and a momen esul o he sup emum o no med boo s apped
sums we e es ablished in [9]. I is impo an o no e ha in [9] no assump ions we e made
conce ning ei he he ma ginal o he join dis ibu ions o he andom a iables om which
boo s ap esamples a e wi hd awn. We ollow he same app oach in his pape .
The no ion o he dependen boo s ap p ocedu e was in oduced in [12], whe e some
impo an p ope ies we e also es ablished.
The main goal o he p esen pape is o ex end and gene alize he esul s o [9] on
he s ong law o la ge numbe s o he case o he dependen boo s ap p ocedu e. The
main ools a e he ex ensions and gene aliza ions o he esul o [10] (sec ion 4) and [12]
(sec ion 2).
2. Dependen boo s ap. The esul s om his sec ion a e modifica ions, gene al-
iza ions, and ex ensions o he esul s o [12] and [13] o he dependen boo s ap om he
sequence o unnecessa y i.i.d. andom a iables. We men ion ha Smi h and Taylo [12], [13]
conside only he i.i.d. case. We p esen his case as a simple e e ence since i plays a ole
in he ollowing.
Le {Xn,n1}be a sequence o andom a iables (which a e no necessa ily indepen-
den o iden ically dis ibu ed) defined on a p obabili y space (Ω,F,P). Le {m(n),n1}
and {k(n),n1}be wo sequences o posi i e in ege s such ha o all n1, m(n)
nk(n).Fo ω∈Ω and n1, he dependen boo s ap is defined as he sample o size m(n),
CONVERGENCE RATE OF THE DEPENDENT BOOTSTRAPPED MEANS 339
deno ed {
X(ω)
n,j ,1jm(n)}, d awn wi hou eplacemen om he collec ion o nk(n)
i ems made up o k(n) copies, each o he sample obse a ions X1(ω),... ,X
n(ω).
This dependen boo s ap p ocedu e is p oposed as a p ocedu e o educe a ia ion o
es ima o s and o ob ain be e confidence in e als. The dependen boo s ap p ocedu e is
p oposed as a p ocedu e o educe a ia ion o es ima o s and o ob ain be e confidence
in e als. We e e o [13], whe e his ac is p o en and simula ed confidence in e als a e
used o examine possible gains in co e age p obabili ies and in e al leng hs.
The fi s p oposi ion gi es us he join dis ibu ion o he dependen boo s ap andom
a iables. We need he ollowing no a ion.
Fo ω∈Ω, n1,and a eal numbe x, deno e
τ(x)=
n

j=1
IXj(ω)x,
whe e I(·) is he indica o unc ion. Hence, τ(x) is he andom a iable ha coun s he
numbe o obse a ions less han o equal o x.
Fo a fini e sequence {x1,x
2,... ,x
m}o eal numbe s, deno e by {x(1),x
(2),... ,x
(m)}
i s nondec easing ea angemen , ha is, x(1) x(2) ··· x(m), and o any 1 jm
he e exis s 1 imsuch ha xi=x(j).
P oposi ion 1. Fo ω∈Ω, n1, and a sequence {x1,x
2,... ,x
m}o eal numbe s we
ha e he ollowing:
1) I k(n)τ(x(j))j o all 1jm(n), hen
P
X(ω)
n,1x1,... , 
X(ω)
n,m(n)xm(n)=
m(n)

j=1
k(n)τ(x(j))−(j−1)
k(n)n−(j−1) .
2) I k(n)τ(x(j))<j o a leas one 1jm(n), hen he abo e p obabili y is ze o.
P oo . Le πbe he eo de ing o {1,2,... ,m(n)}such ha π(j)=i o xi=x(j).
Then
P
X(ω)
n,1x1,... , 
X(ω)
n,m(n)xm(n)=P
X(ω)
n,π(1) x(1),... , 
X(ω)
n,π(m(n)) x(m(n))
=P
X(ω)
n,π(1) x(1)P
X(ω)
n,π(2) x(2) |
X(ω)
n,π(1) x(1)×···
×P
X(ω)
n,π(m(n)) x(m(n)) |
X(ω)
n,π(1) x(1),... , 
X(ω)
n,π(m(n)−1) x(m(n)−1)
=
m(n)

j=1
k(n)τ(x(j))−(j−1)
k(n)n−(j−1)
i k(n)τ(x(j))j o all 1 jm(n).
The second pa o he p oposi ion is ob ious.
O cou se, he dependen boo s ap andom a iables {
X(ω)
n,j ,1jm(n)}a e de-
penden . They obey he so-called nega i ely dependen p ope y; his p ope y will be
es ablished in P oposi ion 2. The concep o nega i ely dependen andom a iables was
in oduced by Lehmann [8] as ollows.
340 A. VOLODIN, M. ORD ´
O˜
NEZ CABRERA, AND T. C. HU
The andom a iables Y1,Y
2,... a e said o be nega i ely dependen i o each n2
he ollowing wo inequali ies hold:
P{Y1y1,... ,Y
nyn}
n

i=1
P{Yiyi}
and
P{Y1>y
1,... ,Y
n>y
n}
n

i=1
P{Yi>y
i},
o any sequence {y1,... ,y
n}o eal numbe s.
P oposi ion 2. Fo ω∈Ωand n1 he dependen boo s ap andom a iables {
X(ω)
n,j ,
1jm(n)}a e nega i ely dependen and exchangeable.
P oo . Fo he nega i e dependence p ope y we will p o e only he fi s inequali y. The
p oo o he second one is comple ely he same.
Le {x1,x
2,... ,x
m(n)}be a sequence o eal numbe s. I is in e es ing o conside only
he case k(n)τ(x(j))j o all 1 jm(n). By P oposi ion 1
P
X(ω)
n,1x1,... , 
X(ω)
n,m(n)xm(n)=
m(n)

j=1
k(n)τ(x(j))−(j−1)
k(n)n−(j−1)

m(n)

j=1
k(n)τ(x(j))
k(n)n=
m(n)

j=1
P
X(ω)
n,j xj.
The exchangeabili y is ob ious by P oposi ion 1.
3. A ew echnical lemmas. In his sec ion we p esen a ew echnical esul s
ha we will use in p oo s o he main esul s o he pape . Some o he lemmas a e only
gene aliza ions and ex ensions o well-known esul s. Fo exposi o y pu poses we ou line
hei p oo s.
Fo simplici y, by he log- unc ion in his sec ion we mean he na u al loga i hm unc ion.
The esul s can be easily gene alized on any o he loga i hm unc ion wi h base g ea e
han one.
The fi s lemma is well known (c ., o example, [5]) and i ial. So, we omi he p oo .
Lemma 1. Le {Yn,n1}be a sequence o nega i ely dependen andom a iables.
1) I { n,n1}is a sequence o measu able eal unc ions all o which a e mono one
inc easing (o all mono one dec easing), hen { n(Yn),n1}is a sequence o nega i ely
dependen andom a iables.
2) Fo any n1, En
1Yjn
1EYj,p o ided he expec a ions a e fini e.
Un o una ely, i is no possible o find he in e se unc ion o he unc ion φ( )=
1/β/log , >0, 0 <β<e, in he closed o m. Bu he ollowing lemma gi es a good
“app oxima ion” o he in e se unc ion.
Lemma 2. Le φ( )= 1/β/log and ψ( )= βlogβ , e,0<β<e. Then
1
β1−β
eβ
ψφ( )1
ββ
.
P oo . No e ha
ψ(φ( )) =
ββ1−βlog log
log β
and 1 −β
e1−βlog log
log 1
o e, which can be es ablished by diffe en ia ion.
CONVERGENCE RATE OF THE DEPENDENT BOOTSTRAPPED MEANS 341
The main idea o Lemma 2 is ha o a posi i e andom a iable Y, he assump ions
Eφ−1(Y)<∞and Eψ(Y)<∞a e equi alen .
The ollowing lemma can be ound in [11, Theo em 2]. No e ha he e is no indepen-
dence assump ion.
Lemma 3. Le φ( ), >0, be a con inuous unc ion ha is posi i e,s ic ly inc easing,
and sa is ying he condi ion φ( )→∞as →∞. Pu bn=φ(n), n1. Mo eo e ,le
{Yn,n1}be a sequence o iden ically dis ibu ed andom a iables. I
∞

j=n
1
bj
=On
bnand Eφ−1(Y1)<∞,
whe e φ−1is he in e se o φ, hen
1
bn
n

j=1
Yj=0 a.s.
In he ollowing lemma i is also impo an o no e ha he e is no independence con-
di ion.
Lemma 4. Le {Xn,n1}be a sequence o iden ically dis ibu ed andom a iables
such ha
E|X1|αlog |X1|α/2<∞
o some 0<α<2. Then
log n
n2/α
n

j=1
X2
j−→ 0a.s.
P oo . In o de o apply Lemma 3, pu Yn=Xn2,bn=n2/α/log n,n1, and β=α/2
( hen 0 <β<1). I we conside φ( )= 1/β/log , e, hen bn=φ(n) and acco ding
o Lemma 2 wi h ψ( )= β(log )β, he condi ions Eφ−1(Y1)<∞and Eψ(Y1)<∞a e
equi alen . No e ha
Eψ(Y1)=2
α/2E|X1|αlog |X1|α/2<∞.
The las hing we need o p o e is ha ∞
j=n1/bj=O(n/bn). We ha e
∞

j=n
1
bj
=
∞

j=n
log j
j1/β =
∞

m=1
n(m+1)−1

k=nm
log k
k1/β 
∞

m=1
nlog(mn)
(nm)1/β .
Since he sequence {log k/k1/β ,ke1/β}is s ic ly dec easing he las sum is no g ea e
han
nlog n
n1/β
∞

m=1
1 + log m/ log 2
m1/β =Cnlog n
n1/β =Cn
bn
.
By Lemma 3,
log n
n2/α
n

j=1
X2
j→0 a.s.
Lemma 4 is p o ed.
The ollowing wo lemmas deal wi h he con e gence o maximums o andom a iables.
Again, no assump ion o independence is made.
Lemma 5. Le {Xn,n1}be a sequence o posi i e andom a iables and le {bn,
n1}be a nondec easing sequence o posi i e cons an s such ha bn→∞. Then he
assump ions Xn/bn→0a.s. and max1jnXj/bn→0a.s. a e equi alen .

342 A. VOLODIN, M. ORD ´
O˜
NEZ CABRERA, AND T. C. HU
P oo . Le Xn/bn→0 a.s. Fo a bi a y nk2,
1
bn
max
1jnXj1
bn
max
1jk−1Xj+1
bn
max
kjnXj1
bn
max
1jk−1Xj+ max
kjn
Xj
bj
.
Since {bn,n1}is nondec easing he las exp ession is no g ea e han
1
bn
max
1jk−1Xj+ sup
jk
Xj
bj
−→ 0,
whe e fi s n→∞and hen k→∞. The e e se implica ion is ob ious.
The ollowing lemma in his sec ion is a gene aliza ion o he co olla y o Theo em 3
o [3].
Lemma 6. Le ψ( ), 0, be a s ic ly inc easing unc ion and le {bn,n1}
be a nondec easing sequence o posi i e numbe s such ha ψ(bn)Cn,n1, whe e he
cons an Cdoes no depend on n. Mo eo e ,le {Xn,n1}be a sequence o posi i e
iden ically dis ibu ed andom a iables such ha Eψ(X1/ε)<∞ o all ε>0. Then
1
bn
max
1jnXj→0a.s.
P oo . Fo any ε>0
∞

n=1
P{Xn>εb
n}
∞

n=1
PC−1ψX1
ε>n
C−1EψX1
ε<∞.
Then by he Bo el–Can elli lemma Xn/bn→0 a.s. By Lemma 5 we ob ain ha
1
bn
max
1jnXj→0 a.s.
The nex exponen ial inequali y in his sec ion is a key ool used in he p oo o he
law o la ge numbe s o he dependen boo s ap o he mean p esen ed in he heo em.
I is an analogue o he Mikosch exponen ial inequali y [10, Lemma 5.1] o he case o he
dependen boo s ap.
We need o add wo mo e no a ions o he no a ions om sec ion 2. Le {Xn,n1}
be a sequence o (no necessa ily independen o iden ically dis ibu ed) andom a iables.
Fo ω∈Ω and n1 deno e
Mn(ω)= 1
m(n)max
1jnXj(ω)−Xn(ω)and Bn(ω)= 1
nm(n)
n

j=1 Xj(ω)−Xn(ω)2,
whe e
Xn(ω)= 1
n
n

j=1
Xj(ω)
deno e he sample mean o {Xj(ω), 1 jn},n1.
Lemma 7. Le {an,n1}and {hn,n1}be wo sequences o posi i e eals. Then
o ω∈Ωand n1such ha hnMn(ω)<1and all ε>0, he ollowing inequali y holds:
P
1
m(n)
m(n)

j=1 
X(ω)
n,j −Xn(ω)εan2 exp −εhnan+h2
nBn(ω)
2(1 −hnMn(ω)).
CONVERGENCE RATE OF THE DEPENDENT BOOTSTRAPPED MEANS 343
P oo . By Ma ko ’s inequali y
P
1
m(n)
m(n)

j=1 
X(ω)
n,j −Xn(ω)εan
exp{−εhnan}Eexp hn
1
m(n)
m(n)

j=1 
X(ω)
n,j −Xn(ω)
exp{−εhnan}Eexp hn1
m(n)
m(n)

j=1 
X(ω)
n,j −Xn(ω)
+ exp{−εhnan}Eexp −hn1
m(n)
m(n)

j=1 
X(ω)
n,j −Xn(ω).
We will es ima e only he expec a ion in he fi s i em o he las exp ession; he same
bound is alid o he second expec a ion.
No e ha by P oposi ion 2 he dependen boo s ap andom a iables {
X(ω)
n,j ,1j
m(n)},n1, a e nega i ely dependen and exchangeable. Hence, by Lemma 1(1) he
andom a iables exp hn
m(n)
X(ω)
n,j −Xn(ω),1jm(n)
a e nega i ely dependen and iden ically dis ibu ed.
The e o e,
Eexp hn1
m(n)
m(n)

j=1 
X(ω)
n,j −Xn(ω)=Em(n)

j=1
exp hn
m(n)
X(ω)
n,j −Xn(ω)

m(n)

j=1
Eexp hn
m(n)
X(ω)
n,j −Xn(ω)
by Lemma 1(2). By iden ical dis ibu ion his exp ession is equal o
Eexp hn
m(n)
X(ω)
n,1−Xn(ω)m(n)
=1
n
n

i=1
exp hn
m(n)Xi(ω)−Xn(ω)m(n)
=1+ 1
n
n

i=1 h2
n
2! m(n)2Xi(ω)−Xn(ω)2+h3
n
3! m(n)3Xi(ω)−Xn(ω)3
+h4
n
4! m(n)4Xi(ω)−Xn(ω)4+···m(n)
=1+ h2
n
m(n)
n

i=1
(Xi(ω)−Xn(ω))2
nm(n)
×1
2! +hn
3!
Xi(ω)−Xn(ω)
m(n)+h2
n
4! Xi(ω)−Xn(ω)
m(n)2
+···m(n)
1+ h2
n
m(n)
Bn(ω)
21+hnMn(ω)+hnMn(ω)2+···m(n)
=1+ h2
n
2m(n)
Bn(ω)
1−hnMn(ω)m(n)
exp h2
n
2m(n)
Bn(ω)
1−hnMn(ω)m(n)
= exp h2
nBn(ω)
2(1 −hnMn(ω)).
344 A. VOLODIN, M. ORD ´
O˜
NEZ CABRERA, AND T. C. HU
Hence,
P
1
m(n)
m(n)

j=1 
X(ω)
n,j −Xn(ω)εan2 exp −εhnan+h2
nBn(ω)
2(1 −hnMn(ω)).
4. Comple e con e gence a es o he dependen boo s ap o he
mean. Wi h he p elimina ies accoun ed o , he law o la ge numbe s o he dependen
boo s ap o he mean may now be es ablished. The heo em is an analogue o Theo em 2.1
o [9] o he case o he dependen boo s ap.
Theo em. Le {Xn,n1}be a sequence o (no necessa ily independen o iden ically
dis ibu ed) andom a iables and le {an,n1}be a sequence o posi i e eal numbe s. I
(i) log n
m(n)an
max
1in|Xi|−→0a.s.
and
(ii) log n
nm(n)a2
n
n

i=1
X2
i−→ 0a.s.,
hen o any eal ,e e y ε>0, and almos e e y ω∈Ω
∞

n=1
n P
1
m(n)
m(n)

j=1 
X(ω)
n,j −Xn(ω)εan<∞.
We make se e al ema ks conce ning he heo em be o e p o ing i .
Rema ks. 1. The conclusion o he heo em is o cou se s onge he la ge is aken. In
con as wi h he Baum–Ka z, E d¨os, Hsu–Robbins, Spi ze comple e con e gence heo em,
he cons an does no play a ole in any condi ion o he heo em and i can be aken
a bi a ily la ge.
2. Taking = 0, i ollows om he Bo el–Can elli lemma and he conclusion o he
heo em ha o almos e e y ω∈Ω
1
an1
m(n)
m(n)

j=1 
X(ω)
n,j −Xn(ω)−→ 0 a.s.
3. Acco ding o Lemma 5, i (log n)/(m(n)an)↓0 mono onically, hen assump ion (i)
om he heo em is equi alen o he appa en ly weake and s ic ly simple condi ion
log n
m(n)an
Xn→0 a.s.
4. Ca e ul analysis o he p oo o he heo em shows ha assump ions (i) and (ii) can
be sligh ly weakened:
(i)log n
m(n)an
max
1in|Xi−Xn|−→0 a.s.,
(ii)log n
nm(n)a2
n
n

i=1
(Xi−Xn)2−→ 0 a.s.
Igno ing he ac ha assump ions (i) and (ii) a e ob iously weake han assump-
ions (i) and (ii), i is wo h men ioning ha hey a e cumbe some and mo e difficul o
check.
P oo o he heo em. The conclusion o he heo em ob iously holds o <−1, so i
will be assumed ha −1. Using he no a ion o Lemma 7 deno e
Ω0=ω:log n
an
Mn(ω)−→ 0 and log n
a2
n
Bn(ω)→0.
I is easy o check ha condi ions (i) and (ii) imply P(Ω0)=1.
CONVERGENCE RATE OF THE DEPENDENT BOOTSTRAPPED MEANS 345
Fo fixed −1, ε>0, and ω∈Ω0, le
hn=3+
εan
log n, n 1.
Since ω∈Ω0,we ha e
hnMn(ω)=3+
ε
log n
an
Mn(ω)−→ 0 and h2
n
log nBn(ω)=3+
ε2log n
a2
n
Bn(ω)−→ 0.
Le nbe sufficien ly la ge such ha
h2
nBn(ω)log nand hnMn(ω)1
2.
Applying Lemma 7 we ob ain
n Pm(n)
j=1 
X(ω)
n,j
m(n)−Xn(ω)εn2n exp −εhnan+h2
nBn(ω)
2(1 −hnMn(ω))
2n exp −(3+ ) log n+ log n=2n exp −(2+ ) log n2n−2
since −1, and he conclusion ollows.
Co olla y. Le {Xn,n1}be a sequence o iden ically dis ibu ed (no necessa ily
independen ) andom a iables and 0<α<2.I E|X1|α|log |X1||α<∞, hen o e e y eal
,e e y ε>0, and almos e e y ω∈Ω
∞

n=1
n P
1
n1/α
n

j=1 
X(ω)
n,j −Xn(ω)ε<∞.
P oo . Conside m(n)=nand an=n(1−α)/α,n1,in he heo em. We need o check
ha assump ions (i) and (ii) a e ue.
Fo (i) we deno e bn=n1/α/log nand ψ( )= α(log )α, 1. Acco ding o Lemma
2, ψ(bn)Cn, whe e he cons an Cdoes no depend on n. Assump ion (i) ollows om
Lemma 6.
Assump ion (ii) ollows om Lemma 4 di ec ly. We should men ion ha Lemma 4
equi es an e en sligh ly weake momen assump ion han we ha e.
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