J. Math. Anal. Appl. 395 (2012) 643–653
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Journal of Mathematical Analysis and
Applications
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Convergence rates of limit distribution of maxima of lognormal samples
Xin Liao, Zuoxiang Peng ∗
School of Mathematics and Statistics, Southwest University, 400715 Chongqing, China
article
abstract
info
Article history:
Received 31 January 2012
Available online 7 June 2012
Submitted by Vladimir Pozdnyakov
Let Mn denote the partial maximum of an independent and identically distributed
lognormal random sequence. In this paper, we derive the exact uniform convergence
rate of the distribution of the normalized maximum (Mn − bn ) /an to its extreme value
distribution, where the constants an and bn are chosen by an optimal way.
© 2012 Elsevier Inc. All rights reserved.
Keywords:
Extreme value distribution
Logarithmic normal distribution
Maximum
Uniform convergence rate
1. Introduction and main results
Let (ξn , n ≥ 1) be a sequence of independent and identically distributed standardized normal random variables with
distribution function Φ (x). Let Mn = max(ξk , 1 ≤ k ≤ n) denote the partial maximum. It is well-known that
lim P (Mn ≤ an x + bn ) = lim Φ n (an x + bn ) = exp −e−x = Λ(x)

n→∞

n→∞
(1.1)
with the normalized constants an and bn given by
an = (2 log n)−1/2 ,
an
1
b n = a−
(log log n + log 4π ) .
n −
2
The pointwise convergence rate of (1.1) is
e−x exp −e−x (log log n)2

Φ (an x + bn ) − Λ(x) ∼
n

16
log n
(1.2)
for large n. For more details, see [1,2]. Hall [1] investigated this problem further and proved
C1
log n
< sup |P (Mn ≤ an x + bn ) − Λ(x)| <
x∈R
C2
log n
(1.3)
for some absolute constants 0 < C1 < C2 if we choose an and bn by the following equations:
2π b2n exp(b2n ) = n2 ,
1
an = b −
n .
(1.4)
For the extreme value distributions and their associated uniform convergence rates of some given distributions, see [3,4]
respectively for exponential and mixed exponential distributions, [5,6] for the general error distribution, [7,8] for the shorttailed symmetric distribution.
∗
Corresponding author.
E-mail address: pzx@swu.edu.cn (Z. Peng).
0022-247X/$ – see front matter © 2012 Elsevier Inc. All rights reserved.
doi:10.1016/j.jmaa.2012.05.077
644
X. Liao, Z. Peng / J. Math. Anal. Appl. 395 (2012) 643–653
It is popular in theoretical analysis and wide applications that the normal distribution is carried over to logarithmic
normal one. Let ξ ∼ N (0, 1), we say that X follows the logarithmic normal distribution if X is defined by X = eξ , abbreviated
as lognormal distribution. In this paper, we are interested in the uniform convergence rate of extremes from samples with
common distribution F following the lognormal distribution. The probability density function of the lognormal distribution
is given by
x −1
(log x)2
F ′ ( x) = √
exp −
2
2π


,
x > 0.
In the sequel, let (Xn , n ≥ 1) be a sequence of independent random variables with common distribution F following the
lognormal distribution, and let Mn denote its partial maximum. For the limiting distribution of the maximum of lognormal,
by Lemma 2 in Section 2 and the arguments similar to the proof of Theorem 1.5.3 in [2], we can derive
lim P (Mn ≤ αn x + βn ) = lim F n (αn x + βn ) = Λ(x)
n→∞
n→∞
with normalized constants αn and βn given by
exp (2 log n)1/2

αn =

(2 log n)1/2
,
(1.5)
and
1/2
βn = exp (2 log n)




1−
log 4π + log log n
2 (2 log n)1/2

.
(1.6)
As mentioned by Hall [1] and Leadbetter et al. [2] that the choices of normalized constants may influence the convergence
rate of extremes, and the distribution tail representation may help us find the optimal normalized constants. For the
distributional tail representation of lognormal distribution, Lemma 2(ii) in Section 2 shows that
x
 
1 − F (x) = c (x) exp −
e
g (t )
f (t )

dt
for sufficiently large x, where c (x) → (2π e)−1/2 as x → ∞, f (x) = x/ log x and g (x) = 1 + (log x)−2 . Noting that f ′ (x) → 0
and g (x) → 1 as x → ∞. By Proposition 1.1(a) and Corollary 1.7 of [9], we can choose the norming constants an and bn in
such a way that bn is the solution of the equation
2π (log bn )2 exp (log bn )2 = n2


(1.7)
and an satisfies
an = f (bn ) =
bn
log bn
,
(1.8)
then
lim P (Mn ≤ an x + bn ) = lim F n (an x + bn ) = Λ(x).
n→∞
n→∞
(1.9)
The aim of this paper is to prove that the uniform convergence rate of (1.9) is proportional to 1/ (log n)1/2 . However, for
F n (αn x + βn ) the convergence rate is no better than (log log n)2 / (log n)1/2 even though αn /an → 1 and (βn − bn )/an → 0
as n → ∞. The main results are stated as follows:
Theorem 1. Let {Xn , n ≥ 1} denote a sequence of independent identically distributed random variables with common
distribution F which is lognormal distribution.
(i) For norming constants an and bn given respectively by (1.7) and (1.8), then there exist absolute constants 0 < C1 < C2 such
that
C1
(log n)
for n ≥ 2.
1/2


< sup F n (an x + bn ) − Λ(x) <
x∈R
C2
(log n)1/2
(ii) For norming constants αn and βn respectively given by (1.4) and (1.6), we have
e−x exp −e−x (log log n)2

F (αn x + βn ) − Λ(x) ∼ −
n
for large n.
8

(2 log n)1/2
X. Liao, Z. Peng / J. Math. Anal. Appl. 395 (2012) 643–653
645
2. Auxiliary lemmas
In this section we provide several key properties of lognormal distribution which are needed for the proofs of our main
results. The first one is the distributional tail decomposition of lognormal distribution, which is stated as follows.
Lemma 1. Let F denote the lognormal distribution function. For x > 1, we have
1
1 − F (x) = √
(log x)
2π
−1

exp −
(log x)2

2
− γ (x)
(2.1)



(log x)2 
1 − (log x)−2 + s(x),
(log x)−1 exp −
1
= √
2π
(2.2)
2
where
∞

1
0 < γ (x) = √
2π


1
(log t )2 1
dt < √
(log t )−2 exp −
2
x
2π
t


(log x)2
(log x)−3 exp −
(2.3)


(log x)2
.
(log x)−5 exp −
(2.4)
2
and
∞

3
0 < s(x) = √
2π


3
(log t )2 1
dt < √
(log t )−4 exp −
2
x
2π
t
2
Proof. By integration by parts we have
1
(log x)2
1 − F (x) = √
(log x)−1 exp −
2
2π

1
= √
2π
(log x)
2

(log x)−1 exp −

1
∞

−√
2π

2
x


(log t )2 1
dt
(log t )−2 exp −
2
t
− γ (x),
which is (2.1). Similarly,
1
γ (x) = √
2π


(log x)2
− s(x).
(log x)−3 exp −
2
Putting it into (2.1), we obtain (2.2), where
3
s(x) = √
2π
∞

(log t )
−4


(log t )2 1
exp −
dt
2
x
t




 ∞
15
3
(log x)2
(log t )2 1
−5
−6
−√
dt
= √ (log x) exp −
(log t ) exp −
2
2
t
2π
2π x


3
(log x)2
< √ (log x)−5 exp −
.
2
2π
The proof is complete.
By Lemma 1, we can derive the Mills ratio and the distributional tail representation of lognormal distribution, which
are helpful, just as mentioned in Section 1, to find the optimal normalized constants an and bn given by (1.7) and (1.8).
The following result is about the Mills-type ratio and distributional tail representation of lognormal distribution, which is a
special case of tail behavior and Mills ratio of logarithmic general error distribution studied by Liao and Peng [10].
Lemma 2. Let F (x) denote the distribution of lognormal random variable with density F ′ (x), then
(i) for x > 1 we have
xF ′ (x)

log x
1−
1
(log x)2

< 1 − F (x) <
x
log x
F ′ (x)
and
1 − F (x)
F ′ (x)
as x → ∞;
∼
x
log x
(2.5)
646
X. Liao, Z. Peng / J. Math. Anal. Appl. 395 (2012) 643–653
(ii) for large x we have
x
 
1 − F (x) = c (x) exp −
e
g (t )
f (t )

dt
(2.6)
with c (x) → (2π e)−1/2 , g (x) = 1 + (log x)−2 → 1 as x → ∞ and the auxiliary function f (x) = x(log x)−1 satisfies
f ′ (x) → 0 as x → ∞.
Let the norming constants an and bn are defined by (1.7) and (1.8) respectively. Let
a∗n = an rn ,
b∗n = bn + δn an
(2.7)
with rn → 1 and δn → 0 as n → ∞. Then a∗n /an → 1, and b∗n − bn /an → 0 which implies F n (a∗n x + b∗n ) → Λ(x). Since
a∗n x + b∗n → ∞, for large n we have the following expansion.


Lemma 3. Let a∗n and b∗n be defined by (2.7). For fixed x ∈ R and sufficiently large n,
F n a∗n x + b∗n − Λ(x) = Λ(x)e−x



1
1
2
(rn − 1) x + δn − an b−
n (rn x + δn )
2


 −1  2 
2
2
+ O (rn − 1) + δn + an bn
.
(2.8)
Proof. Note by (1.7) that log bn ∼ (2 log n)1/2 , which implies that
1
an b −
n =
1
∼ (2 log n)−1/2 → 0
log bn
as n → ∞ by virtue of (1.8). Notice
1
−1
log 1 + an b−
n (rn x + δn ) = an bn (rn x + δn ) −



 

1
1 3
1 2
,
an b −
(rn x + δn )2 + O an b−
n
n
2
implying
1
an b −
n

 −1
1
log 1 + an b−
n (rn x + δn ) = rn x + δn −


1
2
1
2
an b −
n (rn x + δn ) + O

1
an b −
n
2 
and
2
1
log 1 + an b−
n (rn x + δn )





 
1 2
1 3
= an b −
,
(rn x + δn )2 + O an b−
n
n
so we have

1
exp − an b−
n

−1
1
log 1 + an b−
n (rn x + δn ) −



2
1
1
log 1 + an b−
n (rn x + δn )
2





1
2
−1 2
2
2
1
= e−x 1 − (rn − 1) x − δn + an b−
r
x
+
δ
+
O
a
b
+
δ
+
r
−
1
(
)
(
)
n
n
n
n
n
n
n
2
and
−1
1
−1
1 + an b −
n log 1 + an bn (rn x + δn )



 


1 3
1 2
= 1 − an b −
.
(rn x + δn ) + O an b−
n
n
Hence,
−1
1
−1
1 + an b −
n log 1 + an bn (rn x + δn )







 1 
2
1 −1
−1
−1
× exp − an b−
log
1
+
a
b
r
x
+
δ
−
log
1
+
a
b
r
x
+
δ
(
)
(
)
n n
n
n
n n
n
n
n
2




 −1 2
1 3
= 1 − an b n
e −x
(rn x + δn ) + O an b−
n




1
1
2
−1 2
2
2
× 1 − (rn − 1) x − δn + an b−
r
x
+
δ
+
O
a
b
+
δ
+
r
−
1
(
)
(
)
n
n
n n
n
n
n
2



 −1 2
1
1
2
2
2
r
x
+
δ
+
O
a
b
+
δ
+
r
−
1
.
= e−x 1 − (rn − 1) x − δn + an b−
(
)
(
)
n
n
n
n
n
n
n
2
X. Liao, Z. Peng / J. Math. Anal. Appl. 395 (2012) 643–653
647
Thus for large n we have



 ∗
1
∗ 2
log an x + bn
exp −
log an x + bn
√
2
2π




−1
2
1 
1
−1
−1
= √
log bn 1 + an bn (rn x + δn )
exp −
log bn 1 + an bn (rn x + δn )
2
2π



−1
1
−1
= n− 1 1 + an b −
n log 1 + an bn (rn x + δn )



 −1  −1
 1 
2
−1
−1
log 1 + an bn (rn x + δn ) −
× exp − an bn
log 1 + an bn (rn x + δn )
1


∗
∗
−1
2

1
1
2
= n−1 e−x 1 − (rn − 1) x − δn + an b−
n (rn x + δn ) + O
2

1
an b−
n
2
+ δn2 + (rn − 1)2

.
(2.9)
Similarly, one can get
log a∗n x + b∗n


−2



 
1 2
1 4
+ O an b −
= an b−
n
n
(2.10)
and

1
s(x) = O n−1 an b−
n

4 
(2.11)
by virtue of (2.4).
Using (2.9)–(2.11) together with (2.2) and (2.4) we have
F n a∗n x + b∗n − Λ(x)






 −1 2 
1
1
2
2
2
= 1 − n−1 e−x 1 − (rn − 1) x − δn + an b−
r
x
+
δ
+
O
r
−
1
+
δ
+
a
b
(
)
(
)
n
n
n
n n
n
n
2
×



−1 2
1 − an b n
+O

 
−1 4
an bn



  n
1 4
− Λ(x)
+ O n−1 an b−
n



 −1 2 
1
2
−1
1
2
r
−
1
+
δ
+
a
b
,
r
x
+
δ
+
O
= Λ(x)e−x (rn − 1) x + δn − an b−
(
)
(
)
n
n n
n
n
n
n
2
which completes the proof.
3. Proofs of the main results
The aim of this section is to prove the main results.
1
Proof of Theorem 1. (i) Letting rn = 1, δn = 0 in (2.7) and noting an b−
∼ 1/ (2 log n)1/2 , by Lemma 3 we can prove that
n
there exists an absolute constant C1 > 0 such that
sup F n (an x + bn ) − Λ(x) >


x∈R
C1
(log n)1/2
for n ≥ 2. In order to obtain the upper bound, we need to prove the following.
1
sup |F n (an x + bn ) − Λ(x)| < D1 an b−
n ,
(3.1)
1
sup |F n (an x + bn ) − Λ(x)| < D2 an b−
n ,
(3.2)
dn ≤x<∞
0≤x<dn
sup
−cn ≤x<0
sup
1
|F n (an + bn ) − Λ(x)| < D3 an b−
n
−∞<x<−cn
(3.3)
1
|F n (an x + bn ) − Λ(x)| < D4 an b−
n
(3.4)
for n ≥ n0 as there must exist absolute constant C2 > C1 such that n0 = sup(k : C2 /(log k)1/2 ≥ 1), where Di > 0 for
i = 1, 2, 3, 4 are absolute constants, and cn and dn are respectively defined by
cn =: log log log bn > 0,
dn =: − log log
log bn
log bn − 1
>0
648
X. Liao, Z. Peng / J. Math. Anal. Appl. 395 (2012) 643–653
for n ≥ n0 . Note that x ≥ −cn implies

1−
bn + an x ≥ bn − an cn = bn

log log log bn
log bn
> 0,
n ≥ n0 .
For the rest of the proof, let Cj , j = 1, 2, . . . , 12 stand for the absolute positive constants.
For x ≥ −cn , let Ψn (x) = 1 − F (an x + bn ), then
n log F (an x + bn ) = −nΨn (x) − Rn (x),
(3.5)
where
0 < Rn (x) = −n (Ψn (x) + log (1 − Ψn (x))) <
nΨn2 (x)
2 (1 − Ψn (x))
since
−x−
x2
2(1 − x)
< log(1 − x) < −x for 0 < x < 1.
(3.6)
We deduce from (2.1) that for x ≥ −cn ,
Ψn (x) ≤ Ψn (−cn ) = 1 − F (bn − an cn )
 


−1



1
−1
1 −1
1
< n−1 1 + an b−
exp − an b−
log 1 − an b−
n log 1 − an bn cn
n
n cn


1 2


−1
an b−
n cn
−1
−1
−1
<n
1 + an bn log 1 − an bn cn
exp cn +
1
2(1 − an b−
n cn )

  −1



log log log bn
(log log log bn )2
log 1 −
log bn
log (2 log n)
log bn
1 +
 exp  

<
log log log bn
2n
log bn
2 1−
log bn
< C1 < 1
(3.7)
for n ≥ n0 , implying
inf (1 − Ψn (x)) > 1 − C1 > 0,
x≥−cn
so,
0 < R n ( x) ≤
< √
nΨn2 (x)
2 (1 − Ψn (x))
1
an b −
n
2 2π (1 − C1 )


log 1 −
1 +
log log log bn
log bn
log bn
  −2



(log log log bn )2
(log log bn )2
log bn


 exp  
log log log bn
exp 12 (log bn )2
1−
log bn
1
< C 2 an b −
n
for n ≥ n0 . Thus
1
|exp (−Rn (x)) − 1| < Rn (x) < C2 an b−
n
(3.8)
as 1 − e−x < x for x> 0.

Let An (x) = exp −nΨn (x) + e−x and Bn (x) = exp (−Rn (x)), by (3.8) we have
 n

F (an x + bn ) − Λ(x) ≤ Λ(x)|An (x) − 1| + |Bn (x) − 1|
1
< Λ(x)|An (x) − 1| + C2 an b−
n .
(3.9)
From (2.1) and (2.3) it follows that


−1 −x
1
−1
− nΨn (x) + e−x = 1 + an b−
e Cn (x)
n log 1 + an bn x
(3.10)
X. Liao, Z. Peng / J. Math. Anal. Appl. 395 (2012) 643–653
649
with
Cn (x) = −Dn (x)En (x) + Fn (x),
(3.11)
where Dn (x), En (x) and Fn (x) are respectively defined by

Dn (x) = exp x −




2  −1 −1
1
1
1
log 1 + an b−
,
log 1 + an b−
− an b n
n x
n x
2
1
En (x) = 1 − an b−
n

2 
1
−1
1 + an b −
n log 1 + an bn x

−2
κn ,
Fn (x) = 1 + an bn log 1 + an bn x ,
−1
−1


where 0 < κn < 1.
Noting that 1 + x < ex , x ∈ R and log(1 + x) < x, x > −1, for x ≥ −cn we have
Dn (x) > 1 + x −
> 1−



2  −1 −1
1
1
1
log 1 + an b−
− an b n
log 1 + an b−
n x
n x
2

2
1
1
log 1 + an b−
.
n x
2
(3.12)

2
x
1
2
First we prove (3.3). Noting that log 1 + an b−
n x < 0 in the case of −cn ≤ x < 0 and (log(1 + x)) < x − 2(1+x)
−1 < x < 0, by (3.12) we have


1 1
1
Cn (x) < an b−
an b −
n
n
2


x−
an bn x

−1
1
2 1 + an b −
n x

as

2
−1 2
2
+ an b n

−2
1
−1
1 + an b −
n log 1 + an bn x


(3.13)
for −cn ≤ x < 0. On the other hand, since ex − 1 < xex , x ∈ R and by (3.6) we have
Dn (x) < 1 +

1 2
an b −
n x
1
2 1 + an b −
n x

 exp

1 2
an b −
n x
1
2 1 + an b−
n x

 ,
so,

Cn (x) > an bn
−1
− 

x2
 exp
1
2 1 + an b −
n x

1 2
an b−
n x
2 1 + an bn−1 x


−1

+ an b n x −
1
an b −
n x
2

1
2 1 + an b −
n x

 .
Combining (3.13) with (3.14), for −cn ≤ x < 0 we have

|Cn (x)| < an bn
−1
+

1
 1 an b−
x−
n
2
x2
1
2 1 + an b −
n x


< an bn
−1
+
1
8
2 1 + an b n x
 exp
x2
1
2 1 − an b −
n cn

−1


1 2
an b −
n cn
2
1 2
an b −
n x

2
1
−1
1 − an bn cn
− an b n x +

1
an b −
n x
1
2 1 + an b −
n x




−2
1
1
−1
+ an b −
1 + an b−
n
n log 1 − an bn cn

1 2
an b −
n cn
1
2 1 − an b −
n cn


2

−1
1
2 1 + an b−
n x

 exp

1 2
an b −
n x

1+


−2
1
1
−1
+ an b −
1 + an b −
n
n log 1 + an bn x


1
an b −
n cn
−1

+ an bn cn +
2
1
2 1 − an b −
n cn



1
< an b −
C 3 + C 4 x2
n
for n ≥ n0 . So for −cn ≤ x < 0 we have


−1 −x
1
−1
| − nΨn (x) + e−x | = 1 + an b−
e |Cn (x)|
n log 1 + an bn x


−1 −x −1 

−1
−1
< 1 + an bn log 1 + an bn x
e an bn C3 + C4 x2


−1 −x −1 

1
−1
< 1 + an b −
e an bn C3 + C4 x2
n log 1 − an bn cn


−1 cn −1 

1
−1
< 1 + an b −
e an bn C3 + C4 cn2
n log 1 − an bn cn


(3.14)
650
X. Liao, Z. Peng / J. Math. Anal. Appl. 395 (2012) 643–653


−1


log log bn
log log bn (log log log bn )2
1
−1
C
= 1 + an b−
log
1
−
a
b
c
+
C
3
n n
n
4
n
log bn


< 1 +
log 1 −
log log log bn
log bn
 −1

log bn

log log bn
C3
log bn
log bn
+ C4
(log log bn )3

log bn
< C5
for n ≥ n0 . Hence noting |ex − 1| < |x| (ex + 1) , x ∈ R, we have


Λ(x)|An (x) − 1| = Λ(x)| exp −nΨn (x) + e−x − 1|




< Λ(x)| − nΨn (x) + e−x | exp −nΨn (x) + e−x + 1




−1 −x −1 

1
−1
e an bn C3 + C4 x2
< eC5 + 1 Λ(x) 1 + an b−
n log 1 − an bn cn
 2

−1 




x
2
1
−1
−1
exp
−
C
+
C
x
1
+
a
b
log
1
−
a
b
c
< eC5 + 1 e−1 an b−
3
4
n
n
n
n
n
n
2
1
< C 6 an b −
n

 2
for n ≥ n0 , since max−cn ≤x≤0 C3 + C4 x2 e−x /2 ≤ C3 + C4 . Combining above inequality with (3.9) we have


1
sup F n (an x + bn ) − Λ(x) < (C2 + C6 ) an b−
n
−cn ≤x<0
for n > n0 .
Second, consider the case in which 0 ≤ x < dn . By (3.6), (3.10) and (3.12) we can derive

1
C n ( x ) < an b −
n
1
2
1 2
−1
1
−1
an b −
1 + an b−
n x + an b n
n log 1 + an bn x


−2

1
+ an b −
x
.
n
(3.15)
On the other hand, note that x − x2 /2 < log(1 + x), x > 0 and ex − 1 < xex , x ∈ R, implies

1
Dn (x) < exp x − an b−
n
< 1+

1 2
an b −
n x
2
−1

exp
1
log 1 + an b−
n x

1 2
an b −
n x

2

,
so,
C n ( x ) > −1 −
>−
1 2
an b −
n x
2
1 2
an b −
n x
2

exp

exp
1 2
an b −
n x
2
1 2
an b −
n x



1
−1
+ 1 + an b −
n log 1 + an bn x

2
as x ≥ 0. Combining (3.15) with (3.16), for 0 ≤ x < dn we can get


1
|Cn (x)| < an b−
C7 + C8 x2
n
for n ≥ n0 , implying


−1 −x −1 

1
−1
| − nΨn (x) + e−x | < 1 + an b−
e an b n C 7 + C 8 x 2
n log 1 + an bn x


1
< e−x an b−
C7 + C8 x2
n


1
< an b −
C7 + 4C8 e−2
n
< C9 .
Hence by |ex − 1| < |x| (ex + 1) , x ∈ R,




Λ(x)|An (x) − 1| = Λ(x) exp −nΨn (x) + e−x − 1





< Λ(x) −nΨn (x) + e−x  exp −nΨn (x) + e−x + 1




1
< eC9 + 1 Λ(x)e−x an b−
C 7 + C 8 x2
n






1
C7 + C8 x2 exp −e−x − x
= eC9 + 1 an b−
n
(3.16)
X. Liao, Z. Peng / J. Math. Anal. Appl. 395 (2012) 643–653





1
1
C7 + C8 x2 exp −1 − x2
< eC9 + 1 an b−
n
2




1
≤ eC9 + 1 e−1 C7 + 2C8 e−1 an b−
n
651

1
= C10 an b−
n
for n ≥ n0 since 0 ≤ x < dn . Combining with (3.9) to get (3.2).
Third we prove (3.1). Obviously,
1
sup (1 − Λ(x)) ≤ 1 − Λ(dn ) = an b−
n ,
(3.17)
x≥dn
and from (3.5) we deduce that
1 − F n (an dn + bn ) < nΨn (dn ) + Rn (dn ).
(3.18)
By (2.1) we have

n
1
nΨn (dn ) < √
(log (an dn + bn ))−1 exp − (log (an dn + bn ))2
2
2π

 



1 −1
−1
< exp − an b−
log
1
+
a
b
d
n
n
n
n

−1 2 
an b n d n
< exp −dn +
2

2 
log bn


log
log
log bn −1
log bn


1
= an b−
exp 

n (log bn ) log
log bn − 1
2 log bn
1
< C11 an b−
n
(3.19)

for n ≥ n0 since the continuous function x log (x/(x − 1)) exp
x
log log x−
1
2x
2 
is decreasing as x > 2. Note that Rn (x) <
1
C 2 an b −
n by (3.7). Combining with (3.17), (3.18) and (3.19), we derive that
sup |F n (an x + bn ) − Λ(x)| ≤ 1 − F n (an dn + bn ) + 1 − Λ(dn )
x≥dn
1
< (C11 + C2 + 1) an b−
n
for n ≥ n0 which completes the proof of (3.1).
Finally, consider the case in which −∞ < x ≤ −cn . In this case,
sup F n (an x + bn ) ≤ F n (bn − an cn )
x≤−cn




2  −1 2 

−2
1
−1
−1
−1
cn
≤ exp −e
1−
log 1 − an bn cn
− an b n
1 + an bn log 1 − an bn cn
2

1
(log log bn ) cn2
(log log bn ) cn3
1

≤ an b −
+ 
n exp
2
−1
2 (log bn )
(log bn )3
2 1 − an bn cn


−2
1
log log bn 
(log log bn ) cn4
−1
−1
+ 
+
1 + an bn log 1 − an bn cn
2
1
(log bn )4
(log bn )2
8 1 − an b −
n cn
1
< C12 an b−
n
for n ≥ n0 . Thus we have
sup |F n (an x + bn ) − Λ(x)| ≤ F n (bn − an cn ) + Λ(−cn )
x<−cn
1
< (C12 + 1) an b−
n
for n ≥ n0 . This completes the proof of (3.4). The proof of Theorem 1(i) is complete.
652
X. Liao, Z. Peng / J. Math. Anal. Appl. 395 (2012) 643–653
(ii) Setting
bn = βn + θn



1/2
= exp (2 log n)
1−

Bn
2 (2 log n)1/2
+

θn
exp (2 log n)1/2

 ,
(3.20)
where βn is defined by (1.6), Bn = log 4π + log log n and θn = exp (2 log n)1/2 o 1/ (log n)1/2 . Note that

log (1 + x) = x −
x2

as x → 0.
 
+ O x3
2
 
Taking logarithms of (3.20), we have

log bn = (2 log n)1/2 1 −

1
−
Bn
4 log n
2 (2 log n)1/2
+
θn
(2 log n)
Bn
2 (2 log n)1/2
−
1/2
exp (2 log n)1/2


2
θn
exp (2 log n)1/2



+O
B3n
(log n)2


.
Hence,


1 − (2 log n)1/2 θn2




+
8 log n
exp 2 (2 log n)1/2
(2 log n)1/2 exp (2 log n)1/2


 3 
2 (2 log n)1/2 + Bn θn
B2n
Bn


−
+
O
+
1/2
1/2
log n
exp (2 log n)
4 (2 log n)
B2n
(log bn )2 = 2 log n − Bn +
Bn θn
−
and
log log bn =
1
1
Bn
θn
B2
n

−
2
2
4 log n
32 (log n)2
(2 log n) exp (2 log n)1/2


Bn θn
B2n
θn2

+

 +O
.
−
4 (log n) exp 2 (2 log n)1/2
2 (2 log n)3/2 exp (2 log n)1/2
(log n)3/2
log 2 +
log log n −
+
1/2
Putting above expansions of (log bn )2 and log log bn into
log 2π + 2 log log bn + (log bn )2 = 2 log n,
(3.21)
we have
exp (2 log n)1/2 B2n

θn ∼

16 log n
.
Hence,
bn = exp (2 log n)1/2



1−
Bn
2 (2 log n)1/2
+
B2n

16 log n
+o
Now for αn and βn defined by (1.4) and (1.6), one can check that
rn − 1 =
αn
an
−1∼
log log n
2 (2 log n)1/2
,
and
δn =
βn − bn
an
∼−
(log log n)2
8 (2 log n)1/2
for large n. Hence by Lemma 3 we can derive
F n (αn x + βn ) − Λ(x) ∼ −
e−x Λ(x) (log log n)2
8
(2 log n)1/2
for large n, which completes the proof of Theorem 1(ii).
(log log n)2
log n

.
X. Liao, Z. Peng / J. Math. Anal. Appl. 395 (2012) 643–653
653
Acknowledgments
The authors would like to thank the Editor-in-Chief (Professor Dr. Goong Chen) and the referee for carefully reading the
paper and for their comments which greatly improved the paper. This work was supported by the National Natural Science
Foundation of China Grant No. 11171275 and the SWU Grant for Statistics PH.D.
References
[1]
[2]
[3]
[4]
[5]
[6]
[7]
[8]
[9]
[10]
P. Hall, On the rate of convergence of normal extremes, J. Appl. Probab. 16 (1979) 433–439.
M.R. Leadbetter, G. Lindgren, H. Rootzén, Extremes and Related Properties of Random Sequences and Processes, Springer-Verlag, New York, 1983.
W.J. Hall, J.A. Wellner, The rate of convergence in law of the maximum of an exponential sample, Stat. Neerl. 33 (1979) 151–154.
Z. Peng, Z. Weng, S. Nadarajah, Rates of convergence of extremes for mixed exponential distributions, Math. Comput. Simul. 81 (2010) 92–99.
Z. Peng, B. Tong, S. Nadarajah, Tail behavior of the general error distribution, Comm. Statist. Theory Methods 38 (2009) 1884–1892.
Z. Peng, S. Nadarajah, F. Lin, Convergence rate of extremes for the general error distribution, J. Appl. Probab. 47 (2010) 668–679.
F. Lin, Z. Peng, Tail behavior and extremes of short-tailed symmetric distribution, Comm. Statist. Theory Methods 39 (2010) 2811–2817.
F. Lin, X. Zhang, Z. Peng, Y. Jiang, On the rate of convergence of STSD extremes, Comm. Statist. Theory Methods 40 (2011) 1795–1806.
S.I. Resnick, Extreme Values, Regular Variation and Point Processes, Springer-Verlag, New York, 1987.
X. Liao, Z. Peng, 2012, Tail behavior and extremes of logarithmic general error distribution, (submitted for publication).