Open Access

Higher-order expansions for distributions of extremes from general error distribution

Journal of Inequalities and Applications20142014:213

https://doi.org/10.1186/1029-242X-2014-213

Received: 2 January 2014

Accepted: 14 May 2014

Published: 27 May 2014

Abstract

In this short note, with optimal normalizing constants, the higher-order expansion for a distribution of normalized partial maximum from the general error distribution is derived, by which one deduces the associate convergence rate of the distribution of the extreme to the Gumbel extreme value distribution.

MSC:62E20, 60E05, 60F15, 60G15.

Keywords

expansion extreme general error distribution

1 Introduction

Let { X n , n 1 } be a sequence of independent and identically distributed random variables with marginal cumulative distribution function (cdf) F v following the general error distribution ( F v GED ( v ) for short), and let M n = max 1 k n X k denote the partial maximum of { X n , n 1 } . The probability density function (pdf) of GED ( v ) is given by
f v ( x ) = v exp ( ( 1 / 2 ) | x / λ | v ) λ 2 1 + 1 / v Γ ( 1 / v ) , x R ,

where v > 0 is the shape parameter, λ = [ 2 2 / v Γ ( 1 / v ) / Γ ( 3 / v ) ] 1 / 2 and Γ ( ) denotes the Gamma function (Nelson [1]). Note that GED ( 2 ) reduces to the standard normal distribution.

Recently, several contributions investigated asymptotic behaviors of normalized maxima from the GED ( v ) . It is well known that the limiting distribution of extremes from the GED ( 2 ) , i.e., the normal distribution, is a Gumbel extreme value distribution, see Leadbetter et al. [2] and Resnick [3]. Peng et al. [4] established the Mills type ratio of GED ( v ) and proved that there exist normalizing constants a n > 0 and b n R such that
lim n P ( M n a n x + b n ) = lim n F v n ( a n x + b n ) = Λ ( x ) = exp ( exp ( x ) ) , x R ,

i.e., F v is in the domain of attraction of Λ, which we denote by F v D ( Λ ) . For the uniform convergence rate of normalized maxima from the GED ( v ) , Hall [5] established the optimal uniform convergence rate as v = 2 , i.e., the normal case; Peng et al. [6] extended the result to the case of v > 1 . Both studies show that the optimal convergence rate of extremes from the GED ( v ) is proportional to 1 / log n .

For more informative studies of extremes from the GED, Nair [7] considered higher-order expansions for distribution and moments of normalized maxima from the GED ( 2 ) under optimal normalizing constants. Let Φ ( x ) denote the distribution function of the standard normal distribution GED ( 2 ) , Nair [7] proved that
b ˜ n 2 [ b ˜ n 2 ( Φ n ( a ˜ n x + b ˜ n ) Λ ( x ) ) k ˜ ( x ) Λ ( x ) ] ( w ˜ ( x ) + 1 2 k ˜ 2 ( x ) ) Λ ( x )
(1.1)
as n , where the optimal normalizing constants a ˜ n and b ˜ n are given by
1 Φ ( b ˜ n ) = n 1 , a ˜ n = b ˜ n 1 .
Here, k ˜ ( x ) and w ˜ ( x ) are, respectively, of the following form:
k ˜ ( x ) = 2 1 ( x 2 + 2 x ) e x
and
w ˜ ( x ) = 8 1 ( x 4 + 4 x 3 + 8 x 2 + 16 x ) e x .

In this short note, the aim is to establish a higher-order expansion for the distribution of normalized maxima from the GED ( v ) for v > 0 . For some recent related work on uniform convergence rates and higher-order expansions of extremes for given distributions, see Liao and Peng [8] for the log-normal distribution, and Liao et al. [9, 10] for skew distributions.

In order to derive the higher-order expansions of extremes from the GED ( v ) , we cite some results from Peng et al. [4, 6]. The following Mills ratio of the GED ( v ) is due to Peng et al. [4]:
1 F v ( x ) f v ( x ) 2 λ v v x 1 v as  x ,
(1.2)
which deduces the following distributional tail representation of GED ( v ) :
1 F v ( x ) = c ( x ) exp ( λ x g ( t ) f ( t ) d t )
for large x > 0 , where
c ( x ) exp ( 1 / 2 ) 2 1 / v Γ ( 1 / v ) as  x
and
f ( t ) = 2 v 1 λ v t 1 v , g ( t ) = 1 + 2 ( v 1 ) v 1 λ v t v .
(1.3)
Noting that f ( t ) 0 and g ( t ) 1 , we may choose normalizing constants a n and b n satisfying the following equations:
1 F v ( b n ) = n 1 , a n = f ( b n ) .
(1.4)
Under these normalizing constants, we have
lim n F v n ( a n x + b n ) = Λ ( x ) .

This paper is organized as follows. Section 2 provides the main results. Some auxiliary results and the proofs of the main results are given in Section 3.

2 Main result

In this section, we provide asymptotic expansions of a distribution for the partial maximum of the GED with normalizing constants a n and b n given by (1.4).

Theorem 1 Let F v ( x ) denote the cdf of GED ( v ) with v > 0 . Then:
  1. (i)
    For v 1 , with normalizing constants a n and b n given by (1.4), we have
    b n v [ b n v ( F v n ( a n x + b n ) Λ ( x ) ) k v ( x ) Λ ( x ) ] ( w v ( x ) + k v 2 ( x ) 2 ) Λ ( x )
    (2.1)
     
as n , where k v ( x ) and w v ( x ) are, respectively, given by
k v ( x ) = ( 1 v 1 ) λ v ( x 2 + 2 x ) e x
and
w v ( x ) = ( v 1 1 ) λ 2 v [ 4 x + 2 x 2 + 2 3 ( 2 v 1 ) x 3 + 1 2 ( 1 v 1 ) x 4 ] e x .
  1. (ii)
    For v = 1 , with normalizing constants a n = 2 1 / 2 and b n = 2 1 / 2 ( log n log 2 ) , we have
    e 2 b n [ e 2 b n ( F 1 n ( a n x + b n ) Λ ( x ) ) k 1 ( x ) Λ ( x ) ] ( w 1 ( x ) + k 1 2 ( x ) 2 ) Λ ( x )
    (2.2)
     
as n , where k 1 ( x ) and w 1 ( x ) are, respectively, given by
k 1 ( x ) = 1 4 e 2 x , w 1 ( x ) = 1 12 e 3 x .

Remark 1 The main result coincides with (1.1) as the GED reduces to the standard normal distribution GED ( 2 ) .

Remark 2 From (1.2) and (1.4), it is easy to check that b n v = O ( log n ) . Hence, for v 1 , Theorem 1(i) shows that the convergence rate of F v n ( a n x + b n ) to its ultimate extreme value distribution Λ ( x ) is proportional to 1 / log n , while for the case of v = 1 , Theorem 1(ii) shows that the convergence rate is proportional to 1 / n .

3 The proofs

In order to prove the main results, we need some auxiliary lemmas. The first lemma deals with a decomposition of the distributional tail representation of GED ( v ) .

Lemma 1 Let F v ( x ) and f v ( x ) , respectively, denote the cdf and pdf of GED ( v ) with v 1 ; for large x > 0 , we have
1 F v ( x ) = exp ( 1 / 2 ) 2 1 / v Γ ( 1 / v ) [ 1 + 2 ( v 1 1 ) λ v x v + 4 ( v 1 1 ) ( v 1 2 ) λ 2 v x 2 v + O ( x 3 v ) ] exp ( λ x g ( t ) f ( t ) d t )
(3.1)

with f ( t ) and g ( t ) given by (1.3).

Proof Using integration by parts we have
1 F v ( x ) = v 2 1 + 1 / v Γ ( 1 / v ) x / λ exp ( t v 2 ) d t = f v ( x ) 2 λ v v x 1 v [ 1 + 2 ( v 1 1 ) λ v x v + 4 ( v 1 1 ) ( v 1 2 ) λ 2 v x 2 v + 8 ( v 1 1 ) ( v 1 2 ) ( v 1 3 ) λ 3 v x 3 v ] + 16 v 2 1 + 1 / v Γ ( 1 / v ) ( v 1 1 ) ( v 1 2 ) ( v 1 3 ) ( v 1 4 ) x / λ exp ( t v 2 ) t 4 v d t .
(3.2)
An application of L’Hospital’s rule shows that
lim x x / λ exp ( t v 2 ) t 4 v d t exp ( x v 2 λ v ) x 1 4 v = 0 .
(3.3)
Combining the latter with (1.2), (3.2), and (3.3), for large x we have
1 F v ( x ) = f v ( x ) 2 λ v v x 1 v [ 1 + 2 ( v 1 1 ) λ v x v + 4 ( v 1 1 ) ( v 1 2 ) λ 2 v x 2 v + O ( x 3 v ) ] = exp ( 1 / 2 ) 2 1 / v Γ ( 1 / v ) [ 1 + 2 ( v 1 1 ) λ v x v + 4 ( v 1 1 ) ( v 1 2 ) λ 2 v x 2 v + O ( x 3 v ) ] exp ( λ x g ( t ) f ( t ) d t ) ,

which is the desired result. □

Lemma 2 Let h v ( b n ; x ) = n log F v ( a n x + b n ) + e x with normalizing constants a n and b n given by (1.4), then for v 1 we have
lim n b n v ( b n v h v ( b n ; x ) k v ( x ) ) = w v ( x ) ,
(3.4)

where k v ( x ) and w v ( x ) are given by Theorem  1.

Proof It is well known that n ( 1 F v ( a n x + b n ) ) e x as n . By 1 F v ( b n ) = n 1 , we know that b n if and only if n . The following fact holds by (1.2):
lim n 1 F v ( a n x + b n ) b n m v = 0 for  m = 1 , 2 .
(3.5)
Let
A v ( n , x ) = 1 + 2 ( 1 v ) v λ v b n v + 4 ( 1 v ) ( 1 2 v ) v 2 λ 2 v b n 2 v + O ( b n 3 v ) 1 + 2 ( 1 v ) v λ v ( a n x + b n ) v + 4 ( 1 v ) ( 1 2 v ) v 2 λ 2 v ( a n x + b n ) 2 v + O ( ( a n x + b n ) 3 v ) .
It is easy to check that lim n A v ( n , x ) = 1 and
A v ( n , x ) 1 = ( 1 + o ( 1 ) ) [ 4 ( 1 v ) v λ 2 v b n 2 v x + 16 ( 1 v ) ( 1 2 v ) v 2 λ 3 v b n 3 v x + O ( b n 3 v ) ] .
Hence,
lim n A v ( n , x ) 1 b n v = 0
(3.6)
and
lim n A v ( n , x ) 1 b n 2 v = 4 ( v 1 1 ) λ 2 v x .
(3.7)
By (3.1) we have
1 F v ( b n ) 1 F v ( a n x + b n ) e x = A v ( n , x ) exp [ 0 x ( ( v 1 ) a n b n + a n t + v a n ( b n + a n t ) v 1 2 λ v 1 ) d t ] = A v ( n , x ) { 1 + 0 x ( ( v 1 ) a n b n + a n t + v a n ( b n + a n t ) v 1 2 λ v 1 ) d t + 1 2 [ 0 x ( ( v 1 ) a n b n + a n t + v a n ( b n + a n t ) v 1 2 λ v 1 ) d t ] 2 ( 1 + o ( 1 ) ) } .
(3.8)
It follows from (3.5)-(3.8) that
lim n b n v h v ( b n ; x ) = lim n log F v ( a n x + b n ) + n 1 e x n 1 b n v = lim n [ ( 1 F v ( a n x + b n ) ) 1 2 ( 1 F v ( a n x + b n ) ) 2 ( 1 + o ( 1 ) ) n 1 b n v + ( 1 F v ( b n ) ) e x n 1 b n v ] = lim n ( 1 F v ( a n x + b n ) ) n 1 ( 1 F v ( b n ) 1 F v ( a n x + b n ) e x 1 ) b n v = e x lim n [ A v ( n , x ) [ 0 x ( ( v 1 ) a n b n + a n t + v a n ( b n + a n t ) v 1 2 λ v 1 ) d t ] ( 1 + o ( 1 ) ) b n v + A v ( n , x ) 1 b n v ] = e x lim n 0 x b n v ( ( v 1 ) a n b n + a n t + v a n ( b n + a n t ) v 1 2 λ v 1 ) d t = ( 1 v 1 ) λ v ( x 2 + 2 x ) e x = k v ( x ) ,
(3.9)
where the last step is due to the dominated convergence theorem since
lim n b n v ( v a n ( b n + a n t ) v 1 2 λ v 1 ) = 2 ( 1 v 1 ) λ v t
(3.10)
and
lim n ( v 1 ) a n b n v b n + a n t = 2 ( 1 v 1 ) λ v .
(3.11)
By arguments similar to (3.9), we have
lim n b n v ( b n v h v ( b n ; x ) k v ( x ) ) = lim n log F v ( a n x + b n ) + n 1 e x n 1 b n v k v ( x ) n 1 b n 2 v = lim n [ ( 1 F v ( a n x + b n ) ) 1 2 ( 1 F v ( a n x + b n ) ) 2 ( 1 + o ( 1 ) ) n 1 b n 2 v + n 1 e x n 1 b n v k v ( x ) n 1 b n 2 v ] = lim n ( 1 F v ( a n x + b n ) ) + n 1 e x ( 1 k v ( x ) e x b n v ) n 1 b n 2 v = lim n 1 F v ( a n x + b n ) n 1 1 F v ( b n ) 1 F v ( a n x + b n ) e x ( 1 k v ( x ) e x b n v ) 1 b n 2 v = e x lim n [ A v ( n , x ) b n 2 v ( 0 x ( ( v 1 ) a n b n + a n t + v a n ( b n + a n t ) v 1 2 λ v 1 ) d t k v ( x ) e x b n v ) k v ( x ) e x A v ( n , x ) b n v 0 x ( ( v 1 ) a n b n + a n t + v a n ( b n + a n t ) v 1 2 λ v 1 ) d t + 1 2 ( 1 + o ( 1 ) ) A v ( n , x ) b n 2 v ( 0 x ( ( v 1 ) a n b n + a n t + v a n ( b n + a n t ) v 1 2 λ v 1 ) d t ) 2 + A v ( n , x ) 1 b n 2 v ] = ( v 1 1 ) λ 2 v [ 4 x + 2 x 2 + 2 3 ( 2 v 1 ) x 3 + 1 2 ( 1 v 1 ) x 4 ] e x = w v ( x ) .

The proof is complete. □

For v = 1 , noting that the GED ( 1 ) is the Laplace distribution with pdf given by
f 1 ( x ) = 2 1 / 2 exp ( 2 1 / 2 | x | ) , x R ,
(3.12)
and the Laplace distributional tail can be written by
1 F 1 ( x ) = 2 1 / 2 f 1 ( x ) = 2 1 exp ( 2 1 / 2 ) exp ( 1 x 1 f ( t ) d t ) , x > 0 ,
(3.13)

with f ( t ) = 2 1 / 2 . For the Laplace distribution, we have the following result.

Lemma 3 For v = 1 , let h 1 ( b n ; x ) = n log F 1 ( a n x + b n ) + e x with normalizing constants a n = 2 1 / 2 and b n = 2 1 / 2 ( log n log 2 ) . Then
lim n e 2 b n ( e 2 b n h 1 ( b n ; x ) k 1 ( x ) ) = w 1 ( x ) ,
(3.14)

where k 1 ( x ) and w 1 ( x ) are those given by Theorem  1.

Proof Noting that for GED ( 1 ) , i.e., the Laplace distribution with pdf f 1 ( x ) = 2 1 / 2 exp ( 2 1 / 2 | x | ) , we have
lim n F 1 n ( a n x + b n ) = Λ ( x )
with normalizing constants a n = 2 1 / 2 and b n = 2 1 / 2 ( log n log 2 ) . So, by (1.4) and (3.13), we have
lim n e 2 b n h 1 ( b n ; x ) = lim n log F 1 ( a n x + b n ) + n 1 e x n 1 e 2 b n = lim n [ ( 1 F 1 ( a n x + b n ) ) 1 2 ( 1 F 1 ( a n x + b n ) ) 2 ( 1 + o ( 1 ) ) n 1 e 2 b n + ( 1 F 1 ( b n ) ) e x n 1 e 2 b n ] = lim n 1 2 ( 1 F 1 ( a n x + b n ) ) 2 ( 1 + o ( 1 ) ) n 1 e 2 b n = 1 4 e 2 x = k 1 ( x )
(3.15)
and
lim n e 2 b n ( e 2 b n h 1 ( b n ; x ) k 1 ( x ) ) = lim n log F 1 ( a n x + b n ) + n 1 e x n 1 e 2 2 b n k 1 ( x ) e 2 b n = lim n [ ( 1 F 1 ( a n x + b n ) ) 1 2 ( 1 F 1 ( a n x + b n ) ) 2 1 3 ( 1 F 1 ( a n x + b n ) ) 3 ( 1 + o ( 1 ) ) n 1 e 2 2 b n + ( 1 F 1 ( b n ) ) e x n 1 e 2 2 b n k 1 ( x ) e 2 b n ] = lim n 1 3 ( 1 F 1 ( a n x + b n ) ) 3 ( 1 + o ( 1 ) ) n 1 e 2 2 b n = 1 12 e 3 x = w 1 ( x ) .
(3.16)

The proof is complete. □

Proof of Theorem 1 By (3.9) and (3.15), we have
h v ( b n ; x ) 0 and | i = 3 h v i 3 ( b n ; x ) i ! | < exp ( | h v ( b n ; x ) | ) 1
(3.17)
as n . For the case of v 1 , by Lemma 2 and (3.17), we have
b n v [ b n v ( F v n ( a n x + b n ) Λ ( x ) ) k v ( x ) Λ ( x ) ] = b n v [ b n v ( exp ( h v ( b n ; x ) ) 1 ) k v ( x ) ] Λ ( x ) = [ b n v ( b n v h v ( b n ; x ) k v ( x ) ) + b n 2 v h v 2 ( b n ; x ) ( 1 2 + h v ( b n ; x ) i = 3 h v i 3 ( b n ; x ) i ! ) ] Λ ( x ) ( w v ( x ) + 1 2 k v 2 ( x ) ) Λ ( x )
as n . Similarly, by Lemma 3 and (3.17), we get
e 2 b n [ e 2 b n ( F 1 n ( a n x + b n ) Λ ( x ) ) k 1 ( x ) Λ ( x ) ] ( w 1 ( x ) + 1 2 k 1 2 ( x ) ) Λ ( x )

as n .

The proof is complete. □

Declarations

Acknowledgements

This work was supported by the National Natural Science Foundation of China (11171275), the Natural Science Foundation Project of CQ (cstc2012jjA00029) and the Fundamental Research Funds for the Central Universities (XDJK2013C021, XDJK2014D020).

Authors’ Affiliations

(1)
School of Mathematics and Statistics, Southwest University

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© Jia and Li; licensee Springer. 2014

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