www.elsevier.com/locate/spa
The logarithmic average of sample extremes is
asymptotically normal
Istvan Berkes
a;1, Lajos Horvath
b;∗aMathematical Institute of the Hungarian Academy of Sciences, P.O. Box 127, H-1364 Budapest,
Hungary
bDepartment of Mathematics, University of Utah, 155 South 1440 East, Salt Lake City,
UT 84112-0090, USA
Received 15 November 1999; received in revised form 18 May 2000; accepted 22 May 2000
Abstract
We obtain a strong approximation for the logarithmic average of sample extremes. The central limit theorem and laws of the iterated logarithm are immediate consequences. c 2001 Elsevier Science B.V. All rights reserved.
MSC:primary 60F17; secondary 60F05
Keywords:Extreme value; Domain of attraction; Wiener process; Strong approximation; Laws of the iterated logarithm
1. Introduction and results
Let X1; X2; : : : be independent identically distributed random variables with
distribu-tion funcdistribu-tion F. Several authors studied the asymptotic properties of
Tn=
X
16i6n
1
ih
X
16i6n
Xj−a∗(i)
!,
b∗(i)
!
:
Brosamler (1988), Schatte (1988) and Lacey and Philipp (1990) obtained the rst laws of large numbers for Tn=logn in the case when the Xi’s have nite second moments.
For extensions for the non-i.i.d. case and unboundedh, see Berkes and Dehling (1993), Berkes et al. (1998a) and Ibragimov and Lifshits (1998, 1999). If h is the indicator of (−∞; x] Weigl (1989) and Csorg˝o and Horvath (1992) established the asymptotic nor-mality of (Tn−ETn)=(logn)1=2. For renements we refer to Horvath and Khoshnevisan
(1996), Berkes et al. (1998b) and Berkes and Horvath (1997). A detailed survey on almost sure limit theorems can be found in Berkes (1998).
∗Corresponding author. Tel.:+1-801-581-8159; fax: +1-801-581-4148.
E-mail address:horvath@math.utah.edu (L. Horvath).
1Research supported by Hungarian National Foundation for Scientic Research Grant T 19346. 0304-4149/01/$ - see front matter c 2001 Elsevier Science B.V. All rights reserved.
In this paper we investigate the asymptotic properties of logarithmic averages of maxima. Let
k=
max
16i6k Xi−a(k)
b(k)
and dene
S(n) = X
16i6n
1
ih(i):
Throughout this paper we assume that there is a non-degenerate distribution function
H such that
lim
n→∞P{n6t}=H(t): (1.1)
Fisher and Tippett (1928) and Gnedenko (1943) showed that all distribution functions
H arising as limits in (1.1) must be one of the following types:
(x) =
0 if x60;
exp(−x−) if x ¿0;
(x) =
exp(−(−x)) if x60;
1 if x ¿0
and
(x) = exp(−e−x); −∞¡ x ¡∞;
where ¿0: We say that F is in the domain of attraction ofH (F ∈D(H)), if (1.1) holds. If F ∈D(
) then we may choose
a(k) = 0; b(k) = inf{x: 1−F(x)61=k}: (1.2)
If F ∈D( ), then
a(k) =a+= sup{x: F(x)¡1}¡∞;
b(k) = sup{x: 1−F(a+−x)61=k}
(1.3)
and if F∈D(), then we use
a(k) =U(logk); b(k) =U(1 + logk)−U(logk); (1.4)
whereU(t) denotes the (generalized) inverse of−log(1−F(x)). For further properties and applications of extreme values we refer to de Haan (1970) and Galambos (1978). The rst almost sure max-limit theorems were established by Fahrner and Stadtmuller (1998) and Cheng et al. (1998). They proved that if h is an almost everywhere continuous, bounded function and (1.1) holds, then
lim
n→∞
1 logn
X
16i6n
1
ih(i) =
Z ∞
−∞
h(t) dH(t) a:s:
In this paper we provide an almost sure approximation for S(n). The asymptotic vari-ance of S(n) will depend on the type of the limit distribution function in (1.1). Let
2= 2
Z ∞
0
Z ∞
0
Z ∞
0
×’(s(1−e−x)−1=)−’(s)) dtdsdx
+ 2
Z ∞
0
Z ∞
0
h(s)h(sex=)(s(ex−1)−1=)’(s) dsdx; (1.5)
if F ∈D(), where ’=′
,
2= 2
Z ∞
0 Z 0
−∞
Z 0
−∞
h(s)h(t) (t)(I{t6sex=}(1−e−x)−1=
× (s(1−e−x)−1=)− (s)) dtdsdx
+ 2
Z ∞
0 Z 0
−∞
h(s)h(se−x=) (s(ex−1)1=) (s) dsdx; (1.6)
if F ∈D( ), where = ′
and
2= 2
Z ∞
0
Z ∞
−∞
Z ∞
−∞
h(s)h(t)(t)(I{t6s+x}(s+x)−(s)) dtdsdx
+ 2
Z ∞
0
Z ∞
−∞
h(s)h(s+x)(s−log(ex−1))(s) dsdx (1.7)
if F ∈D(), where =′.
Our main result is the following strong approximation:
Theorem 1.1. We assume that his of bounded variation and has compact support.If
(1:1)holds;then there is a Wiener process{W(t); 06t6∞}and a positive numerical
sequence n such that
S(n)−ES(n)−W(n) = o(1n=2−) a:s: (1.8)
with some ¿0 and
lim
n→∞n=logn=
2; (1.9)
where 2 is dened by (1:5) – (1:7).
The weak convergence of S(nt), 06t61 and the laws of the iterated logarithm are
immediate consequences of Theorem 1.1. Namely, under the conditions of Theorem 1.1 we have
(logn)−1=2{S(nt)−ES(nt)}D[0;1] → W(t);
lim sup
n→∞
(2 lognlog log log)−1=2{S(n)−ES(n)}= a:s:
and
lim inf
n→∞
log log logn
logn
1=2
max
16k6n|S(k)−ES(k)|=
22
8
1=2
a:s:
Remark 1.2. The norming and centering sequences in (1.2) – (1.4) are used in the proof of Theorem 1.1. However, these special choices of a(k) and b(k) are unimportant. If (1.1) holds then the norming and centering sequences are equivalent with a(k) and
b(k) in (1.2) – (1.4) (cf. Bingham et al. (1987)), so Theorem 1.1 remains true for any choices of a(k) and b(k) as long as (1.1) holds.
2. Computation of the variance
In this section we show that
lim
n→∞
1
lognvarS(n) =
2; (2.1)
where2 is dened by (1.5), (1.6), or (1.7) depending on which domain of attraction
F belongs to. The proof of (2.1) will be done in two steps. First, we establish (2.1) in three special cases and then we show that (2.1) holds if F is in the domain of attraction of an extreme value distribution.
Since h is bounded, by the Cauchy–Schwartz inequality we have
varS(n) = var X
R6i6n
1
ih(i)
!
+ O((log logn)1=2) var X
R6i6n
1
ih(i)
!!1=2
;
(2.2)
where R=R(n) = (logn)2. Also,
var X
R6i6n
1
ih(i)
!
= 2 X
R6i¡j6n
1
ij{Eh(i)h(j)−Eh(i)Eh(j)}+ O(1): (2.3)
Elementary arguments give that
Eh(i) =
Z ∞
−∞
h(t) dHi(t); (2.4)
whereHi(t) is the distribution function ofi. LetMi=max16k6iXk,Mi; j∗=maxi¡‘6jX‘
and
A1={(t; s): a(i) +tb(i)¡ a(j) +sb(j)};
A2={(t; s): a(i) +tb(i) =a(j) +sb(j)}:
Then for any i ¡ j we have
P{i6t; j6s}
=P{Mi6(a(i) +tb(i))∧(a(j) +sb(j)); Mi; j∗6a(j) +sb(j)}
=Hi
t∧
sb(j) b(i) +
a(j)−a(i)
b(i)
Hj−i
s b(j) b(j−i) +
a(j)−a(j−i)
b(j−i)
:
Let i; j be the measure generated by the distribution of (i; j). Then for any s ¿0
P{i∈[s; s+ s)}Hj−i
s b(i) b(j−i)+
a(i)−a(j−i)
b(j−i)
6P{(i; j)∈A2; i∈[s; s+ s)}
=P{Mi=Mj; i∈[s; s+ s)}
=P{Mi; j∗6Mi; i∈[s; s+ s)}
6P{i∈[s; s+ s)}Hj−i
(s+ s) b(i)
b(j−i)+
a(i)−a(j−i)
b(j−i)
:
Since Hk is right continuous, it follows that
Z Z
A2
h(s)h(t) di; j
=
Z ∞
−∞ h(s)h
sb(j) b(i) +
a(j)−a(i)
b(i)
Hj−i
s b(i) b(j−i) +
a(i)−a(j−i)
b(j−i)
dHi(s):
On the other hand, the formula for P{i6t; j6s} shows that in the open half-plane
(A1∪A2)c, P{i6t; j6s} depends only on s and thus
Z Z
(A1∪A2)c
h(s)h(t) di; j= 0:
Therefore,
Eh(i)h(j) =
Z Z
A1∪A2
h(s)h(t) di; j
=
Z ∞
−∞ h(s)
Z ∞
−∞ h(t)I
t ¡ sb(j) b(i) +
a(j)−a(i)
b(i)
dHi(t)
×dHj−i
s b(j) b(j−i)+
a(j)−a(j−i)
b(j−i)
+
Z ∞
−∞ h(s)h
sb(j) b(i) +
a(j)−a(i)
b(i)
×Hj−i
s b(i) b(j−i)+
a(i)−a(j−i)
b(j−i)
dHi(s); (2.5)
if i ¡ j.
Lemma 2.1. If the conditions of Theorem 1:1 are satised
F(x) =
0 if x61 1−x− if x ¿1;
Proof. Elementary calculations give
Next, we show that
X
we obtain immediately (2.7). By (2.7) we have that
By (2.6) we conclude
Since
we get that
The boundedness of h gives that cij and di are uniformly bounded and thus by (2.7)
we have
Combining (2.3), (2.7), (2.8), (2.12) and (2.13) we obtain that
Using the denitions of cij and di one can easily verify that
X
Change of variables gives that
+
Z ∞
0
h(s)h(se(v−u)=)(s(e(v−u)−1)−1=) d(s)
dvdu
=
Z logn
logR
Z logn
u
G(v; u)dvdu;
where
G(v; u) =
Z ∞
0
Z ∞
0
h(s)h(t)It ¡ se(v−u)= ’(t) 1−e−(v−u)
−1=
×’(s(1−e−(v−u))−1=)−’(s)’(t)
dtds
+
Z ∞
0
h(s)h(se(v−u)=)(s(e(v−u)−1)−1=)’(s) ds
=G∗1(v−u) +G2∗(v−u) =G(v−u)
and ’(t) =′(t).
Next, we prove that
G(y)6c1exp(−c2y) (2.16)
with some c1¿0 and c2¿0. We can assume, without loss of generality, that the
support of h is in [0; A] with some A ¿0. We write
G∗1(y) =
Z Aexp(−y=)
0
Z ∞
0
h(s)h(t)’(t)(I{t ¡ sey=}(1−e−y)−1=
×’(s(1−e−y)−1=)−’(s)) dtds
+
Z ∞
Aexp(−y=)
Z ∞
0
h(s)h(t)’(t)(I{t ¡ sey=}(1−e−y)−1=
×’(s(1−e−y)−1=)−’(s)) dtds
=G1(y) +G2(y):
Since
Z ∞
0
h(s)h(t)’(t)(I{t ¡ sey=}(1−e−y)−1=’(s(1−e−y)−1=)−’(s)) dt
621=+1 sup
s
’(s) sup s
|h(s)|
Z ∞
0
|h(t)|’(t) dt;
we get that
G1(y)6c3exp(−y=)
with somec3. For all 06t6AandAexp(−y=)6s6Awe have thatI{t ¡ sexp(y=)}=
1. Since ’ and’′ are bounded, Taylor expansion yields
(1−e−y)−1=’(s(1−e−y)−1=−’(s)
6c5|1−(1−e−y)−1=|
6c6e−y:
If c∗ is so large that h(t) = 0 if t¿c∗, then
|G∗2(y)|=
Z c∗e−y=
0
h(s)h(sey=)(s(ey−1)−1=) d(s)
6sup
t
h2(t)(c∗e−y=);
completing the proof of (2.16).
Let r=r(n) = (logn)−1=2. Using (2.16) we obtain that
sup
logR6u6(1−r)logn
Z logn−u
0
G(y) dy−
Z ∞
0
G(y) dy
6
Z ∞
rlogn
G(y) dy6(c1=c2) exp(−c2(logn)1=2)
= o(1=(logn)2)
and
Z logn
(1−r) logn
Z logn−u
0
G(y) dydu6
Z logn
(1−r) logn
Z ∞
0
G(y) dydu
= O((logn)1=2):
Thus, we have
Z logn
logR
Z logn−u
0
G(y) dydu= logn
Z ∞
0
G(y) dy+ O((logn)1=2);
which also completes the proof of Lemma 2.1.
In the following lemma, we take a specially chosen function in the domain of attraction of and show that (2.1) holds again.
Lemma 2.2. If the conditions of Theorem 1:1 are satised;
F(x) =
0 if x6−1;
1−(−x) if −1¡ x60;
1 if 0¡ x
and a(i) = 0and b(i) =i−1= ( ¿0); then (2:1)holds where 2 is dened in (1:6).
Proof. Similarly to (2.6) we have
sup
−∞¡x¡∞
|Hi(x)− (x)|= O
1
i
as i→ ∞:
Let
d=d( ) =
Z 0
−∞
and
cij=cij( )
=
Z 0
−∞ h(s)
Z 0
−∞
h(t)I{t ¡ s(j=i)−1=}d (t)
!
d (s(j=(j−i))−1=)
+
Z 0
−∞
h(s)h(s(j=i)−1=) (s(i=(j−i))−1=) d (s): (2.17)
Following the proof of Lemma 2.1 one can easily verify that
var X
R6i6n
1
ih(i)
!
−2 X
R6i6j6n
1
ij(cij−d
2)
= O((log logn)2)
and
X
R6i6j6n
1
ij(cij−d
2)
=
Z n
R
Z n
x
1
xy
Z 0
−∞ h(s)
Z 0
−∞
h(t)I{t ¡ s(y=x)−1=}d (t)
×d (s(y=(y−x))−1=)−
Z 0
−∞
h(t) d (t)
!2
+
Z 0
−∞
h(s)h(s(y=x)−1=) (s(x=(y−x))−1=) d (s)
!
dydx+ o(logn)
=
Z logn
logR
Z logn
u
G∗(v; u) dvdu+ o(logn);
where R= (logn)2 and
G∗(v; u) =
Z 0
−∞
Z 0
−∞
h(s)h(t) (t)(I{t ¡ se−(v−u)=}
×(1−e−(v−u))1= (s(1−e−(v−u))1=− (s)) dtds
+
Z 0
−∞
h(s)h(se−(v−u)=) (s(e(v−u)−1)1=) (s) ds
=G∗(v−u)
with (t) = ′(t). Repeating the proof of (2.16) we get that
which implies that
Z logn
logR
Z logn
u
G∗(v; u) dvdu=
Z logn
logR
Z logn−u
0
G∗(y) dydu
= logn
Z ∞
0
G∗(y) dy+ o(logn):
Next, we consider a function which is in the domain of attraction of .
Lemma 2.3. If the conditions of Theorem 1:1 are satised;
F(x) =
0 if x60;
1−e−x if x ¿0
and a(i) = logi and b(i) = 1; then (2:1) holds where 2 is dened in (1:7).
Proof. It is easy to see that
sup
−∞¡x¡∞
|Hi(x)−(x)|= O
1
i
as i→ ∞: (2.18)
We have the same rate of convergence in (2.6) as well as in (2.18), so following the proof of Lemma 2.1 we arrive at
var X
R6i6n
1
ih(i)
!
= 2 X
R6i6j6n
1
ij(cij−d
2) + o(logn);
where R=R(n) = (logn)2,
d=d() =
Z ∞
−∞
h(t) d(t)
and
cij=cij()
=
Z ∞
−∞ h(t)
Z ∞
−∞
h(t)I{t ¡ s+ log(j=i)}d(t) d(s+ log(j=i))
+
Z ∞
−∞
h(s)h(s+ log(j=i))(s+ log(i=(j−i)) d(s): (2.19)
Also,
X
R6i6j6n
1
ij(cij−d
2) =I
n;2+ o(logn);
where
In;2= Z n
R
Z n
x
1
xy
Z ∞
−∞ h(s)
Z ∞
−∞
h(t)I{t ¡ s+ log(y=x)}d(t)
×d(s+ log(y=x))−
Z ∞
−∞
h(s) d(t)
2
+
Z ∞
−∞
h(s)h(s+ log(y=x))
Change of variables gives that
which completes the proof of Lemma 2.3.
max Bingham et al. (1987).
According to Theorem 8:13:4 in Bingham et al. (1987), there is a slowly varying function ‘ such that
U(x+u)−U(x)
u‘(ex) →1 as x→ ∞
for anyu ¿0. Observing thatU is a monotone function, Polya’s theorem (cf. Bingham et al., 1987, p. 60) yields that
sup
The uniform convergence in (2.28) implies immediately (2.24) – (2.27).
We recall that cij are dened in (2.9), (2.17) and (2.19) and the denition depends
on the extreme value distribution we have in the limit.
Lemma 2.5. We assume that the conditions of Theorem1:1 are satised; ¿0 and
B ¿1. Then for any ¿0 there is N such that and their proofs will be omitted. Since is continuous, by (1.1) we have that
sup
−∞¡x¡∞
|Hk(x)−(x)|= o(1) as k→ ∞:
So integration by parts and the bounded variation of h(t) give
if i¿N1 andj−i¿N2. By Lemma 2.4 and the fact that h has a compact support we
Hence, integration by parts yields
max
completing the proof.
Now, we are ready to prove the main result of this section.
Theorem 2.1. If the conditions of Theorem 1:1 are satised; then (2:1) holds.
Observing that the random variable
h(j)−h
i.e. with probability not greater than i=j, we get that
+ X
R6i¡j6n; j=i61+
1
ijcov(h(i); h(j))
+ X
R6i¡j6n;1+6j=i6B
1
ijcov(h(i); h(j))
=n;1+· · ·+n;4: (2.30)
Clearly,
n;1= O(1) as n→ ∞; (2.31)
and by (2.29) we have
n;26c10
X
R6i¡j6n; j=i¿B
1
ij× i j6
c11
B logn (2.32)
and
n;36c12
X
R6i¡j6n; j=i61+
1
ij× i
jn;36c13logn: (2.33)
Let ¿0 and choose B= 1= and = in (2.30) – (2.33). Integration by parts and (1.1) yield that
lim
i→∞Eh(i) =d
and therefore by Lemma 2.5 there is N=N() such that
max
(1+)6j=i6B
cov(h(i); h(j))−(cij−d2)
6
if i¿N. Hence, ifn¿n0, then
n;4−2
X
R6i¡j6n;1+6j=i6B
1
ij(cij−d
2)
62 X
R6i¡j6n;1+6j=i6B
1
ij
6c14log(1=) logn: (2.34)
Putting together (2.2) and (2.30) – (2.34) we conclude that for any ¿0
lim sup
n→∞
1
lognvarS(n)−
1 logn
X
R6i¡j6n;1+6j=i61=
1
ij(cij−d
2)
6c15log(1=): (2.35)
3. Proof of Theorem 1.1
We can and shall assume, without loss of generality, that |h|61. For any i ¡ k we dene
Mk; i= max
i¡j6kXj and Mk=Mk;0= max16j6kXj:
Let r ¡ p ¡ q be positive integers and dene
X= X
2p¡i62q 1
ih
Mi−a(i)
b(i)
and
X′= X
2p¡i62q 1
ih
Mi;2r−a(i)
b(i)
:
Lemma 3.1. If the conditions of Theorem 1:1 are satised; then for any d¿1 we have
E|X−X′|d6(2(q−p))d2−(p−r): (3.1)
Proof. Let 2p¡ i62q. Clearly, M
i;0 6=Mi;2r implies that the maximum of X1; : : : ; Xi is taken among the rst 2r terms and thus
P{Mi6=Mi;2r}62r=i62−(p−r): (3.2)
Now by |h|61 we have
|X −X′|6 X
2p¡i62q 2
iI{Mi6=Mi;2r}
and therefore (3.2) yields
E|X−X′|6 X
2p¡i62q 2
iP{Mi6=Mi;2r}
62−(p−r) X
2p¡i62q 2
i
62(q−p)2−(p−r) (3.3)
and
|X−X′|6 X
2p¡i62q 2
i62(q−p): (3.4)
Now (3.1) follows immediately from (3.3) and (3.4).
Let
k=
X
2k¡i62k+1 1
i
h
Mi−a(i)
b(i)
−Eh
Mi−a(i)
b(i)
Lemma 3.2. If the conditions of Theorem 1:1 are satised; then for any positive
integers M and N we have
E X
M ¡k6M+N
k
!2
6c1N (3.5)
and
E X
M ¡k6M+N
k
!4
6c2N3 (3.6)
with some constants c1 and c2.
Proof. We prove only (3.6) since the proof of (3.5) is similar; in fact simpler. It is easy to see that
E X
M ¡k6M+N
k
!4
= X
M ¡k6M+N
E4k+ 6 X
M ¡i¡j6M+N
E2i2j
+ 4 X
M ¡i6=j6M+N
E3ij+ 12
X
M ¡i6=j6=k6M+N
E2ijk
+ 24 X
M ¡i¡j¡k¡‘6M+N
Eijk‘
=S(1)+· · ·+S(5):
Since |h|61 we get that |i|62 and therefore
S(1)+S(2)+S(3)6c3N2: (3.7)
Next, we prove that
S(5)6c4N3: (3.8)
First, we show that if M ¡ i ¡ j ¡ k ¡ ‘6M+N and at least one of j−i and‘−k
is larger than N1=2, then
|Eijk‘|6c5N−4: (3.9)
Assume that j−i¿N1=2 and set
∗j; i=
X
2j¡m62j+1 1
m
h
Mm;2i+1−a(m)
b(m)
−Eh
Mm;2i+1−a(m)
b(m)
:
Using Lemma 3.1 we obtain that
E|j−∗j; i|6c62−(j−i)6c72−N
1=2
:
Similar estimates hold for E|k−∗k; i|; E|‘−∗‘; i| and thus
|Eijk‘−Eij; i∗∗k; i∗‘; i|6c82−N
1=2
(3.10)
for allM ¡ i ¡ j ¡ k ¡ ‘6M+N, if j−i¿N1=2. Observing that
i and{∗j; i; ∗k; i; ∗‘; i}
equals 0 and therefore (3.9) is proven if j−i¿N1=2. The case ‘−k¿N1=2 can be treated similarly.
Relation (3.9) implies that the contribution of those terms inS(5) where at least one of j−i and‘−k is greater thanN1=2 is O(1). On the other hand, the contribution of
the remaining terms is less thanc9N3, since the number of 4-tuples (i; j; k; ‘) satisfying
M ¡ i ¡ j ¡ k ¡ ‘6M +N, j−i6N1=2, ‘−k6N1=2 is clearly at most N3. Hence
(3.8) is proven.
Following the proof of (3.8) one can easily verify that
ES(4)6c10N3
and therefore (3.6) follows from (3.7) and (3.8).
Let us divide [1;∞) into consecutive intervals △1= [p1; q1], △′1= [p′1; q′1], △2=
[p2; q2], △′2= [p′2; q′2]; : : : ; where p1= 1, p′k=qk and pk =q′k−1. We choose these
intervals so that
| △k|= [k1=2] and | △′k|= [k1=4];
where | △ | denotes the length of the interval△. Set
k=
X
2pk¡i62qk 1
ih
Mi−a(i)
b(i)
;
∗k= X
2pk¡i62qk 1
ih
M
i;2qk−1 −a(i)
b(i)
;
k=
X
2p′k¡i62q′k 1
ih
Mi−a(i)
b(i)
and
∗k= X
2p′k¡i62q′k 1
ih M
i;2q′k−1−a(i)
b(i)
!
:
By Lemma 3.1 for any integer d¿1 we have that
E|k−∗k|d6c11(qk−pk)d2−(pk−qk−1)6c12kd=22−k
1=4
6c13k−4 (3.11)
and similarly,
E|k−∗k|d6c14k−4: (3.12)
Lemma 3.3. If the conditions of Theorem 1:1 are satised; then
X
16i6k
(i−Ei) = O(k5=8logk) a:s:
Proof. Applying (3.12) with d= 1 and using the monotone convergence theorem we get
X
16i¡∞
Also, (3.12) with d= 2 and Lemma 3.2 give
k∗k −E∗kk=kk−Ekk+ O(1)
= O(k1=8);
where k · kdenotes the L2 norm. Thus,
X
16k¡∞ E|∗
k −E∗k|2
k5=4(logk)2 ¡∞ is a:s: convergent: (3.14)
Since ∗
1; ∗2; : : :are independent random variables with zero means, (3.14) implies that
X
16k¡∞ ∗
k−E∗k
k5=8logk ¡∞ a:s:
and thus by the Kronecker lemma (cf. Chow and Teicher, 1988, p. 114) we have
X
16i6k
(∗i −E∗i) = O(k5=8logk) a:s: (3.15)
Putting together (3.13) and (3.15) we obtain Lemma 3.3.
Let
Nk=
X
16i6k
([i1=2] + [i1=4])
and
n= var
X
16k6n
1
kh
Mk−a(k)
b(k)
!
:
Lemma 3.4. If the conditions of Theorem 1:1 are satised; then
X
16i6k
E(∗
i −E∗i)2=2Nk(1 + O(k−1=8)):
Proof. Lemma 3.2 implies that vari= O(i1=4)
and (3.12) with d= 2 gives
var∗i = O(i1=4):
Hence by the independence of ∗
1; ∗2; : : :it follows that
var X
16i6k
∗i
!
= O(k5=4): (3.16)
Using (3.16), the Minkowski inequality and (3.11), (3.12) with d= 2 we see that the rst two of the quantities
var1=2 X
16i6k
i∗
!
; var1=2 X
16i6k
(∗i +∗i)
!
; var1=2 X
16i6k
(i+i)
!
dier at most by O(k5=8), while the second and third dier at most by O(1). Since the third expression in (3.17) equals 12=Nk2, we proved that
var1=2 X 16i6k
∗
i
!
=21=Nk2+ O(k5=8)
=21=Nk2(1 + O(k−1=8)); (3.18)
where the second equality follows from the observation thatn is proportional to logn
(cf. Theorem 2.1) and therefore
12=Nk2¿c14(log 2Nk)1=2¿c15k3=4
with somec14¿0 andc15¿0. Since∗1,∗2; : : : ; are independent, Lemma 3.4 follows
from (3.18).
Lemma 3.5. If the conditions of Theorem 1:1 are satised; then there is a Wiener
process {W(t);06t ¡∞} such that
X
16i6k
(i−Ei) =W(2Nk) + O(211Nk=24+) a:s:
with any ¿0.
Proof. Let
Zi=∗i −E∗i and s2k=
X
16i6k
E(∗i −Ei∗)2:
By Lemma 3.2 we have
E(k−Ek)4= O((qk−pk)3) = O(k3=2): (3.19)
Using (3.11) with d= 4, (3.19) yields that
EZk4= O(k3=2):
Also, by Lemma 3.4 we have
s2k=2Nk(1 + O(k−1=8)): (3.20)
Since N ∼c16logN (cf. Theorem 2.1) and Nk∼c17k3=2 with some 0¡ c16,c17¡∞
we get for any 5=6¡ # ¡1 that
X
16k¡∞
1
s2# k
Z
x2¿s2# k
x2dP{Zk6x}6
X
16k¡∞
1
s4# k
Z ∞
−∞
x4dP{Zk6x}
= X
16k¡∞
1
s4# k
EZk46c18 X
16k¡∞
k3=2=22#Nk
6c19 X
16k¡∞
k3=2=Nk2#
6c20 X
Thus using the invariance principle in Theorem 4:4 of Strassen (1965) we can dene a Wiener process W such that
X
16i6k
(∗i −E∗i)−W(s2k) = O(s(1+k #)=2logsk) a:s: (3.21)
Since n∼logn, (3.20) implies that
|s2k−2Nk|= O(112Nk=12)
and thus Theorem 1:2:1 of Csorg˝o and Revesz (1981) on the increments of W imply that
|W(s2k)−W(2Nk)|= o(112Nk=24+) a:s: (3.22)
Also, (3.11) with d= 1 and the monotone convergence theorem yield
X
16i¡∞
|(i−Ei)−(i∗−E∗i)|¡∞ a:s: (3.23)
Now Lemma 3.5 follows from (3.20) – (3.23).
After the preliminary results the proof of Theorem 1.1 will be very simple.
Proof of Theorem 1.1. By Lemmas 3.3 and 3.5 we have that
X
16i6Nk 1
i
h
Mi−a(i)
b(i)
−Eh
Mi−a(i)
b(i)
=W(2Nk) + o(211Nk=24+) a:s: (3.24)
with any ¿0. Now if 2Nk6n ¡2Nk+1, then the random variable
X
16i6n
1
i
h
Mi−a(i)
b(i)
−Eh
Mi−a(i)
b(i)
diers from its value at n= 2Nk by at most
O
X
2Nk6i62Nk+1 1
i
= O(Nk+1−Nk) = O(k1=2)
= O(Nk1=3) = O((logn)1=3) = O(1n=3): (3.25)
Also, Minkowski’s inequality and (3.25) imply for 2Nk6n62Nk+1
|1n=2−12Nk=2|= O
X
2Nk6i62Nk+1 1
i
= O(1n=3) (3.26)
and thus 1=26n=2Nk62 for all k large enough. By (3.26) we have that
|n−2Nk|= O(n5=6)
and therefore using Theorem 1:2:1 of Csorg˝o and Revesz (1981) we conclude
with any ¿0. Thus (3.24) implies
X
16i6n
1
i
h
Mi−a(i)
b(i)
−Eh
Mi−a(i)
b(i)
=W(n) + o(n11=24+) a:s:
with any ¿0, completing the proof of Theorem 1.1.
Acknowledgements
We are grateful to Renate Caspers and Ingo Fahrner for pointing out a missing term in the denition of 2 in the rst version of our paper.
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