Tail empirical process and weighted extreme value
index estimator for randomly right-censored data
Brahim Brahimi, Djamel Meraghni, Abdelhakim Necir∗, Louiza Soltane

arXiv:1801.00572v4 [] 3 Feb 2018

Laboratory of Applied Mathematics, Mohamed Khider University, Biskra, Algeria

Abstract
A tail empirical process for heavy-tailed and right-censored data is introduced and its Gaussian approximation is established. In this context, a (weighted) new Hill-type estimator for
positive extreme value index is proposed and its consistency and asymptotic normality are
proved by means of the aforementioned process in the framework of second-order conditions
of regular variation. In a comparative simulation study, the newly defined estimator is seen
to perform better than the already existing ones in terms of both bias and mean squared
error. As a real data example, we apply our estimation procedure to evaluate the tail index
of the survival time of Australian male Aids patients. It is noteworthy that our approach
may also serve to develop other statistics related to the distribution tail such as second-order
parameter and reduced-bias tail index estimators. Furthermore, the proposed tail empirical
process provides a goodness-of-fit test for Pareto-like models under censorship.
Keywords: Extreme value index; Heavy tails; Random censoring; Tail empirical process.
AMS 2010 Subject Classification: 60F17, 62G30, 62G32, 62P05.

* Corresponding

author: necirabdelhakim@yahoo.fr

E-mail addresses:
brah.brahim@gmail.com (B. Brahimi)
djmeraghni@yahoo.com (D. Meraghni)
louiza stat@yahoo.com (L. Soltane)
1

2

1. Introduction
Let X1 , ..., Xn be n ≥ 1 independent copies of a non-negative continuous random variable

(rv) X, defined over some probability space (Ω, A, P) , with cumulative distribution function

(cdf) F. These rv’s are censored to the right by a sequence of independent copies Y1 , ..., Yn of a
non-negative continuous rv Y, independent of X and having a cdf G. At each stage 1 ≤ j ≤ n,
we only can observe the rv’s Zj := min (Xj , Yj ) and δj := 1 {Xj ≤ Yj } , with 1 {·} denoting

the indicator function. The latter rv indicates whether there has been censorship or not.
If we denote by H the cdf of the observed Z ′ s, then, in virtue of the independence of X
and Y, we have 1 − H = (1 − F ) (1 − G) . Throughout the paper, we will use the notation

S(x) := S(∞) − S(x), for any S. Assume further that F and G are heavy-tailed or, in other

words, that F and G are regularly varying at infinity with negative indices −1/γ1 and −1/γ2
respectively, notation: F ∈ RV (−1/γ1 ) and G ∈ RV (−1/γ2 ) . That is
F (tx)
G (tx)
→ x−1/γ1 and
→ x−1/γ2 ,
F (t)
G (t)

(1.1)

as t → ∞, for any x > 0. This class of distributions includes models such as Pareto, Burr,

Fréchet, α−stable (0 < α < 2) , t-Student and log-gamma, known to be very appropriate

for fitting large insurance claims, large fluctuations of prices, financial log-returns, ... (see,
e.g., Resnick, 2006). The regular variation of F and G implies that H ∈ RV (−1/γ) , where

γ := γ1 γ2 / (γ1 + γ2 ) . Since weak approximations of extreme value theory based statistics
are achieved in the second-order framework (see de Haan and Stadtmüller, 1996), then it

seems quite natural to suppose that cdf H satisfies the well-known second-order condition
of regular variation: for any x > 0
xν/γ − 1
H (tx) /H (t) − x−1/γ
→ x−1/γ
,
A (t)
νγ
as t → ∞, where ν ≤ 0 is the second-order parameter and A is a function tending to 0, not

changing sign near infinity and having a regularly varying absolute value at infinity with

index ν/γ. If ν = 0, interpret xν/γ − 1 / (νγ) as log x. Let us denote this assumption by

H ∈ 2RV (−1/γ,ν) (A) . For the use of second-order conditions in exploring the estimators

asymptotic behaviors, see, for instance, Theorem 3.2.6 in de Haan and Ferreira (2006), page

74. In the last decade, several authors showed an increasing interest in the issue of estimating the extreme-value index (EVI) when the data are subject to random censoring. In
this context, Beirlant et al. (2007) proposed estimators for the EVI and high quantiles and
discussed their asymptotic properties, when the observations are censored by a deterministic

3

threshold, while Einmahl et al. (2008) adapted various classical EVI estimators to the case
where the threshold of censorship is random, and proposed a unified method to establish their
asymptotic normality. Here, we remind the adjustment they made to Hill estimator (Hill,
1975) so as to estimate the tail index γ1 under random censorship. Let {(Zi , δi ) , 1 ≤ i ≤ n}
be a sample from the couple of rv’s (Z, δ) and Z1,n ≤ ... ≤ Zn,n represent the order statistics

pertaining to (Z1 , ..., Zn ) . If we denote the concomitant of the ith order statistic by δ[i:n]

(i.e. δ[i:n] = δj if Zi,n = Zj ), then the adapted Hill estimator of γ1 , defined by Einmahl et al.
(EF G)

(2008), is given by the formula b
γ1
k

(H)

γb

:= b
γ (H) /b
p, where

k

1X
Zn−i+1:n
1X
:=
log
δ[n−i+1:n] ,
and pb :=
k i=1
Zn−k:n
k i=1

(1.2)

for a suitable sample fraction k = kn , are respectively Hill’s estimator (Hill, 1975) of γ and an
estimator of the proportion p := γ/γ1 of the observed extreme values. By following similar
procedures, Ndao et al. (2014, 2016) addressed the nonparametric estimation of the conditional EVI and large quantiles for heavy-tailed distributions which was generalized, a couple
of years later, by Stupfler (2016) to all three extreme domains of attraction namely, Féchet,
Gumbel and Weibull types of distributions. In his working paper Stupfler (2017) considered
the dependent random right-censoring scheme and develop an interesting new topic. For
their part, Worms and Worms (2014) used Kaplan-Meier integration and the synthetic data
approach of Leurgans (1987), to respectively introduce two Hill-type estimators
(W W 1)
γ1
b

and

(W W 2)
γ1
b

where

:=

:=

k
X
1 − Fn (Zn−i+1:n )
i=1

1 − Fn (Zn−k:n)

log

Zn−i+1:n
,
Zn−i:n

k
X
1 − Fn (Zn−i+1:n ) δ[n−i+1:n]
i=1

1 − Fn (Zn−k:n )

Fn (x) = 1 −

Y 

Zi:n ≤x

i

1
1−
n−i+1

log

Zn−i+1:n
,
Zn−k:n

δ[i:n]

,

(1.3)

is the famous Kaplan-Meier estimator of cdf F (Kaplan and Meier, 1958). In their simulation
study, the authors pointed out that, for weak censoring (γ1 < γ2 ) , their estimators perform
(EF G)

better than b
γ1

, in terms of bias and mean squared error. However, in the strong censoring
(W W 1)

case (γ1 > γ2 ) , they noticed that the results become unsatisfactory for both γb1
(W W 2)

γ1
b

and

. With additional assumptions on γ1 and γ2 , they only established the consistency

of their estimators, without any indication on the asymptotic normality. Though these
conditions are legitimate from a theoretical viewpoint, they may be considered as constraints

4

in case studies. Very recently, Beirlant et al. (2018) used Worms’s estimators to derive new
reduced-bias ones, constructed bootstrap confidence intervals for γ1 and applied their results
to long-tailed car insurance portfolio. Likewise, Beirlant et al. (2016) considered maximum
likelihood inference, on the basis of the relative excesses {Zn−i+1:n /Zn−k:n}ki=1 , to propose

a new reduced-bias estimator for the tail index of models belonging to Hall’s class (Hall,
1982). The problem is that, even if this family includes a great number of usual heavytailed distributions, it represents a restriction to the larger class of regularly varying cdf’s,
in particular those with null second-order parameter (ν = 0) such as the log-gamma model.
As we can see, the only EVI estimator that does not impose any restrictive assumptions
on the model is the one introduced by Einmahl et al. (2008). For this reason, we intend to
(EF G)

construct a new weighted estimator to the index γ1 that enjoys the benefits of b
γ1

.

1.1. Constructing a new estimator for γ1 . First, we introduce two very crucial subdistribution functions H (i) (z) := P {Z1 ≤ z, δ1 = i} , i = 0, 1, for z > 0, so that one has

H (z) = H (0) (z) + H (1) (z) . The empirical counterparts are, respectively, defined by
n

Hn(0)

n

1X
1X
1 {Zi ≤ z} (1 − δi ) , Hn(1) (z) :=
1 {Zi ≤ z} δi ,
(z) :=
n i=1
n i=1

1 Pn
(0)
(1)
i=1 1 {Zi ≤ z} = Hn (z) + Hn (z) . From Lemma 4.1 of Brahimi et al.
n
(1)
(2015), under the first-order conditions (1.1) , we have H (t) /H (t) → p, as t → ∞, which

and Hn (z) :=

implies that H

(1)

∈ RV (−1/γ) too. Then it is natural to also assume that H

second-order condition of regular variation, in the sense that H

(1)

(1)

satisfies the

∈ 2RV (−1/γ,ν1 ) (A1 ) . From

Theorem 1.2.2 in de Haan and Ferreira (2006), the assumption H ∈ RV (−1/γ) implies that
R ∞ −1
x H (x) dx/H (t) → γ, as t → ∞, which, by integration by parts, gives
t
Z ∞
1
log (z/t) dH (z) → γ, as t → ∞.
(1.4)
H (t) t
In other words, we have
I (t) :=

H (t)
H

(1)

(t)

!

1
H (t)

Taking t = tn = Zn−k:n and replacing H
(1)
Hn

(1)

Z

t

∞

log (z/t) dH (z) → γ/p = γ1 .

and H by their respective empirical counterparts

and H n yield that I (t) becomes, in terms of n,
!
Z ∞
1
H n (Zn−k:n)
log (z/Zn−k:n ) dHn (z) .
(1)
H n (Zn−k:n) H n (Zn−k:n ) Zn−k:n

(1.5)

5
(1)

We have H n (Zn−k:n ) = k/n and H n (Zn−k:n ) = n−1
that
(1)

H n (Zn−k:n )
= pb and
H n (Zn−k:n)

Z

Pk

i=1 δ[n−i+1:n] ,

∞

log (z/Zn−k:n)

Zn−k:n

then it is readily checked

dHn (z)
= γb(H) .
H n (Zn−k:n)

(EF G)

Substituting this in (1.5) , leads to the definition of b
γ1
. By incorporating the quantity
R∞
(1)
H (tz) /H (tz) inside the integral 1 log (z) dH (tz) /H (t) , we get from Lemma 7.1 (for
r = 1)

1
H (t)

Z

∞

log (z) dH (tz)
H

1

(1)

→ γ1 , as t → ∞.

(tz) /H (tz)

But one has to be careful, because a division by zero may occur in the estimation procedure.
(1)

Indeed, we have for instance H n (Zn:n ) = δ[n:n] which may be 0 or 1. To avoid this boring
situation, we add to the denominator a suitable non-null sequence tending to zero and being
√
in agreement with the normalizing constant k corresponding to the limit distributions of

tail indices estimators. For convenience, to have Gaussian approximations of order OP k −1/2

(tending to zero in probability), we choose the sequence k −1 and we show in Lemma 7.1,
that for a given sequence tn → ∞, we have
1
H (tn )

Z

1

∞

log (z) dH (tn z)
H

(1)

(tn z) /H (tn z) + k −1

Thus, by letting tn = Zn−k:n and by replacing H

(1)

→ γ1 , as n → ∞.
(1)

and H by H n and H n respectively, the

left-hand side becomes
1
H n (Zn−k:n)

Z

Zn:n

H n (z) log (z/Zn−k:n ) dHn (z)

Zn−k:n

H n (z) + k −1 H n (z)

(1)

.

This may be rewritten into
n
k

Z

∞

1 (Zn−k:n < z < Zn:n )
0

H n (z) log (z/Zn−k:n ) dHn (z)
(1)

H n (z) + k −1 H n (z)

n

1X
H n (Zi:n ) log (Zi:n /Zn−k:n)
=
1 (Zn−k:n < Zi:n < Zn:n ) (1)
,
k i=1
H (Zi:n ) + k −1 H n (Zi:n )
n

which equals
n−1
n−1
1 X (n − i) log (Zi:n /Zn−k:n)
1 X H n (Zi:n ) log (Zi:n /Zn−k:n)
P
=
.
k i=n−k+1 H (1) (Zi:n ) + k −1 H n (Zi:n )
k i=n−k+1 nj=i+1 δ[j:n] + (n − i) /k
n

(1.6)

6

Finally, changing i by n − i and j by n − j + 1 yields a new (random) weighted estimator
for the EVI γ1 as follows:

k−1

1 X i log (Zn−i:n /Zn−k:n)
.
γ1 :=
b
P
k i=1 ij=1 δ[n−j+1:n] + i/k

(1.7)

For the purpose of establishing the consistency and asymptotic normality of γb1 , we next

introduce a tail empirical process for censored data. For x ≥ 1, let us define
Z
n ∞
H n (z) dHn (z)
∆n (x) :=
.
(1.8)
(1)
k xZn−k:n H (z) + k −1 H n (z)
n
R∞
By integrating by parts, we show easily that γb1 = 1 x−1 ∆n (x) dx, thereby, motivated by

the tail product-limit process for truncated data, recently given in Benchaira et al. (2016a),

we defined this tail empirical process by
√

Dn (x) := k ∆n (x) − p−1 x−1/γ , x ≥ 1,
so that

√

k (b
γ1 − γ1 ) =

Z

(1.9)

∞

x−1 Dn (x) dx.

1
(1)

In the non censoring case (p = 1) , we have X ≡ Z, F n = H n = H n , with Fn being the
usual empirical cdf. In this case, we have
∆n (x) =
thus


n
F n (xXn−k:n ) + OP k −1 ,
k


√ n

k
F n (xXn−k:n ) − x−1/γ + OP k −1/2 ,
(1.10)
k
corresponds (asymptotically) to the tail empirical process for complete data (see, e.g., page
Dn (x) =

161 in de Haan and Ferreira, 2006). Note that one may think that it would have been
more natural to simply use Kaplan-Meier estimator of F, given in (1.3) , to define a tail
empirical process in the censoring case. This was done in Brahimi et al. (2016), but the
asymptotic properties were only established under the condition p > 1/2, which would
constitute a restriction for applications. Indeed, there exist real datasets used in case studies
with proportions p estimated at less than a half. We can cite, amongst others, the aids
survival data to which Einmahl et al. (2008) applied their methodology and approximately
got 0.28 for p (it is exacly the same value that we ourselves will find later on in Section
4) and the car liability insurance data studied in Beirlant et al. (2016) and Beirlant et al.
(2018) where the estimated value of p was 0.40.

7

The rest of the paper is organized as follows. In Section 2, we provide our main result, namely
two weak approximations leading to consistency and asymptotic normality of b
γ1 , whose

proofs are postponed to Section 5. The finite sample behavior of the proposed estimator

is checked by simulation in Section 3, where a comparison with the already existing ones
is made as well. Section 4 is devoted to an application to the survival time of Australian
male Aids patients. Finally, some results that are instrumental to the proofs are given in
the Appendix.
2. Main results
In the sequel, the functions f ← (s) := inf {x : f (x) ≥ s} , 0 < s < 1 and Uf (t) := f ← (1 − 1/t) ,
t > 1, respectively stand for the quantile and tail quantile functions pertaining to cdf f. For

further use, we set h = hn := UH (n/k) and define
Z
n ∞
H (w) dH (w)
, x ≥ 1.
rn (x) :=
(1)
k hx H (w) + k −1 H (w)

(2.11)

Let us now state our first result in which we provide Gaussian approximations both to
√
k (∆n (x) − rn (x)) and Dn (x) .
Theorem 2.1. Assume that F ∈ RV (−1/γ1 ) and G ∈ RV (−1/γ2 ) and let k = kn be an

integer sequence such that k → ∞ and k/n → 0. Then there exists a sequence of Brownian
bridges {Bn (s) ; 0 ≤ s ≤ 1} defined on the probability space (Ω, A, P) , such that, for every
0 ≤ τ < 1/8, we have

τ /γ

sup x
x≥1

(∆n (x) − rn (x)) − k

−1/2 −1

p

r

 P
n
Ln x−1/γ → 0,
k

(2.12)

as n → ∞, where {Ln (w) ; 0 < w ≤ 1} is a centred Gaussian process defined by

  
  Z w



k
k
k
k
∗
∗
−1
−1
∗
−1
Ln (w) := Bn
Bn
−
s
w − p wBn
s − p Bn p s
ds,
n
n
n
n
0

with Bn∗ (s) := Bn (ps) − Bn (1 − (1 − p) s) , for 0 ≤ ps < 1. If, in addition, we assume that
H ∈ 2RV (−1/γ,ν) (A) and H
then
τ /γ

sup x
x≥1

as n → ∞, where
−1 −1/γ

Rn (x) := p x

√

k



Dn (x) − p

−1

r

(1)

∈ 2RV (−1/γ,ν1 ) (A1 ) ,


n
P
Ln x−1/γ − Rn (x) → 0,
k

(2.13)



xν/γ − 1
ν∗ + xν∗ /γ − 1
A (h) +
A∗ (h) − A2 (h) + o x−τ /γ ,
νγ
γν∗ (1 − ν∗ )

8

with ν∗ := max (ν, ν1 ) , A∗ := 1 {ν > ν1 } A − 1 {ν ≤ ν1 } A1 and A2 := H
√
√
that kAi (h) = O (1) , i = 1, 2 and kA (h) = O (1) .

(1)

/H − p, provided

In the following theorem, we establish the consistency and asymptotic normality of our new
estimator b
γ1 .

Theorem 2.2. Assume that F ∈ RV (−1/γ1 ) and G ∈ RV (−1/γ2 ) and let k = kn be an integer
P

sequence such that k → ∞ and k/n → 0, then γb1 → γ1 , as n → ∞. If, in addition, we
√
√
(1)
assume that H ∈ 2RV (−1/γ,ν) (A) and H ∈ 2RV (−1/γ,ν1 ) (A1 ) , so that kA (h) , kA1 (h)
√
and kA2 (h) be asymptotically bounded, then
r
r Z 1
 
 
√ −1
k
n
n ∗ k
−1 ∗
−1
−1
s Bn
s ds − p
B
kγ1 (b
γ 1 − γ 1 ) − p µn =
k 0
n
k n n
r Z 1
  


k
n
k
∗
−1
−1
+
Bn
s
s − p Bn p s
log (s) ds + oP (1) ,
k 0
n
n
where

√

√
kA (h)
2 − ν∗ √
kA
(h)
−
γ
kA2 (h) .
(2.14)
+
∗
1−ν
(1 − ν∗ )2
√
√
√
Whenever kA (h) , kA1 (h) and kA2 (h) respectively converge to finite real numbers λ,
µn :=

λ1 and λ2 , then

√

D

k (b
γ1 − γ1 ) → N p−1 µ, (9 − 8p) γ12 /p

where



,

µ := λ/ (1 − ν) + (2 − ν∗ ) (1 − ν∗ )−2 λ∗ − γλ2 ,
with λ∗ := 1 {ν > ν1 } λ − 1 {ν ≤ ν1 } λ1 .
Remark 2.1. It is to be noted that for distributions in Hall’s class of models, (Hall, 1982),
√
√
the three assumptions kAi (h) = O (1) , i = 1, 2 and kA (h) = O (1) may be gathered in
√
a single one, namely k (k/n)−ν = O (1) , which is already used by Beirlant et al. (2016) in
√
∗
Theorem 1, written as k (k/n)β = O (1) , where 0 < β ∗ = −ν. Indeed, let us assume that

both F and G belong to this family, which contains the most popular heavy-tailed cdf ’s, such

as Burr, Fréchet, Generalized Pareto, Generalized Extreme Value, t-Student, ... Explicitly,
there exist constants γ1 , γ2 > 0, ηF , ηG < 0, cF , cG > 0 and dF , dG 6= 0, such that, as x → ∞

and


F (x) = cF x−1/γ1 1 + dF xηF /γ1 (1 + o (1)) ,


G (x) = cG x−1/γ2 1 + dG xηG /γ2 (1 + o (1)) .

9

This implies that we have, as x → ∞


H (x) = cx−1/γ 1 + dxν/γ (1 + o (1)) ,

and
H

(1)


(x) = c1 x−1/γ 1 + d1 xν1 /γ (1 + o (1)) ,

where c := cF cG , c1 := pc, ν1 = ν := max (ηF , ηG ) ,


ηF −1


if ηF > ηG ,
d
if
η
>
η
,


F
G
 F
 pηF −1 dF
1
d :=
d1 :=
d
if ηF < ηG ,
dG
if ηF < ηG ,
1−(1−p)ηG G




 d + d if η = η = ν,

η1 −1
(dF + dG ) if ηF = ηG = ν.
F
G
F
G
pηF +(1−p)ηG −1

It is easy to check that both H and H

(1)

satisfy the second-order condition of regular variation

with convergence rates A (t) ∼ νγdtν/γ and A1 (t) ∼ νγd1 tν/γ , respectively. Moreover, we
have H

(1)

(x) /H (x) = p + p (d1 − d) xν/γ (1 + o (1)) , and therefore A2 (t) ∼ p (d1 − d) tν/γ .

In this context, UH (t) ∼ c−1 tγ , as t → ∞, thus h = UH (n/k) ∼ c−1 (k/n)−γ , as n → ∞.
√
√
Consequently, by replacing t by c−1 (k/n)−γ , we get that kA1 (h) , kA2 (h) and A (h) are
√
indeed all of convergence rate k (k/n)−ν .
Remark 2.2. We emphasize that the asymptotic variance of γb1 is (9 − 8p) times that of
(EF G)

γ1
b

. This factor corresponds to earlier findings for weighted/kernel estimators of the ex-

treme value index in non-censored data cases, see for instance Csörgő et al. (1985) (complete

data) and Benchaira et al. (2016b) (truncated data).
3. Simulation study
Now, we carry out an extensive simulation study to illustrate the behavior of the proposed
estimator γb1 and compare its performance, in terms of (absolute) bias and root of the mean
(EF G)

squared error (RMSE), with those of γb1

and γ
b(W W 1) . To this end, we consider Burr’s,

Fréchet’s and the log-gamma models respectively defined, for x > 0, by:

λ
β
• Burr (β, τ, λ) : F (x) = 1 −
, for β, τ, λ > 0, with γ = 1/(τ λ).
β + xτ

• Fréchet (γ) : F (x) = exp −x−1/γ , for γ > 0.
• log-gamma (γ, β) : F (x) = (β γ Γ (γ))−1
and β > 0.

Rx
0

t−1 (log t)γ−1 exp (−β −1 ln t) dt, for γ > 0

We select two values, 33% and 70%, for the proportion p of observed upper statistics and
we make several combinations of the parameters of each model. For each case, we generate

10

200 samples of size n = 200 and we take our overall results (given in the form of graphical
representations) by averaging over all 200 independent replications. We consider three censoring schemes, namely Burr censored by Burr, Fréchet censored by Fréchet and log-gamma
censored by log-Gamma, that we illustrate by Figures 3.1, 3.2 and 3.3 respectively. In each
(EF G)

figure, we represent the biases and the RMSE’s of all three estimators γb1 , b
γ1

and b
γ (W W 1)

as functions of the number k of the largest order statistics. Our overall conclusion is that,
from the top panels of all three figures, we see that the newly proposed estimator γb1 and
(EF G)

the adapted Hill one b
γ1
γ
b

(W W 1)

perform almost equally well and better than Worms’s estimator

in the strong censoring case. However, the bottom panels of Figure 3.1 and Figure

3.2 show that, for the weak censoring scenario, the latter has a slight edge (especially for
small values of k) over the other two which still behave almost similarly. This agrees with

the conclusion of ? and means that γb(W W 1) is not reliable enough in the strong censoring

situation. We notice, from Figure 3.3, that when considering the log-gamma model with
(EF G)

outperforms the remaining two. But, with weak cen-

(EF G)
γ1
b

while γ
b(W W 1) does not even work in this case, which

strong censoring, the estimator b
γ1
soring, γb1 is clearly better than
is something of very striking.

4. Application to Australian Aids data
The data file consists in n = 2843 Australian patients who were diagnosed with Aids
on July 1st , 1991. The file contains the identification number, the dates of first diagnosis,
birth and death, as well as the state and the encrypted transmission category. The data
are available in the package ”MASS” of the statistical software R. Our objective is to apply
the newly proposed estimation procedure to evaluate the tail index γ1 of the survival time
of the patients. To this end, we first select the optimal number of top statistics used in
the estimate computation. By applying the adaptive algorithm of Reiss and Thomas (see,
Reiss and Thomas (2007), page 137), we find that 522 extreme observations are needed to
obtain a proportion estimate value pb = 0.28. This represents a strong censoring rate of

around 70% for which γb(W W 1) is not recommended for the estimation of the tail index γ1 ,
(EF G)

as seen in Section 3. Therefore, we only compute the other two estimates b
γ1 and γb1

for

which the abovementioned technique gives 0.20. and 0.21 respectively. Note that this latter
value differs from that found by Einmahl et al. (2008) who used a graphical approach which

is not as objective as a numerical procedure.

4
3
0

0

1

2

RMSE

3
2
1

Bias

4

5

5

11

0

100 200 300 400 500

0

k

1.5
1.0
0.0

0.5

RMSE

1.0
0.0

0.5

Bias

1.5

k

100 200 300 400 500

0

100 200 300 400 500

0

k

100 200 300 400 500
k

(EF G)

Figure 3.1. Bias (left panel) and RMSE (right panel) of b
γ1 (red) b
γ1
(W W 1)

(black) and b
γ1

(green) based on 200 samples of size 200 from a Burr

distribution censored by another Burr model with p = 0.33 (top) and p = 0.70

(bottom).

5. Proofs

5.1. Preliminaries. Let Ui := δi H (1) (Zi ) + (1 − δi ) θ + H (0) (Zi ) , i = 1, ..., n, be a se-

quence of iid rv’s uniformly distributed on (0, 1) (Einmahl and Koning, 1992), and define
the corresponding empirical cdf and empirical process by
n

√
1X
1 {Ui ≤ s} and αn (s) := n (Un (s) − s) , 0 ≤ s ≤ 1,
Un (s) :=
n i=1

(5.15)

2.0
0.0

1.0

RMSE

1.0
0.0

Bias

2.0

3.0

3.0

12

0

100 200 300 400 500

0

100 200 300 400 500
k

2.0
0.0

1.0

RMSE

1.0
0.0

Bias

2.0

k

0

100 200 300 400 500

0

100 200 300 400 500

k

k

(EF G)

Figure 3.2. Bias (left panel) and RMSE (right panel) of b
γ1 (red) b
γ1
(W W 1)

(black) and γb1

(green) based on 200 samples of size 200 from a Fréchet

distribution censored by another Fréchet model with p = 0.33 (top) and p =

0.70 (bottom).
(0)

(1)

respectively. Thereby, we may represent, almost surely, both Hn and Hn in term of Un ,
as follows

and


Hn(0) (v) = Un H (0) (v) + θ − Un (θ) , for 0 < H (0) (v) < 1 − θ,

Hn(1) (v) = Un H (1) (v) , for 0 < H (1) (v) < θ.

(5.16)

(5.17)

0.0

0.2

0.6

RMSE

0.4

Bias

0.8

13

0

100 200 300 400 500

0

k

3

0

1

2

RMSE

2
1

Bias

3

4

k

100 200 300 400 500

0

100 200 300 400 500
k

0

100 200 300 400 500
k

(EF G)

Figure 3.3. Bias (left panel) and RMSE (right panel) of b
γ1 (red) b
γ1
(W W 1)

(black) and b
γ1

(green) based on 200 samples of size 200 from a log-gamma

distribution censored by another log-gamma model with p = 0.33 (top) and

p = 0.70 (bottom).
For more details, one refers to Deheuvels and Einmahl (1996). Therefore, in view of the
previous representations, we have almost surely



√  (1)
(1)
(1)
(1)
n H n (v) − H (v) = αn (θ) − αn θ − H (v) , for 0 < H (v) < θ,

(5.18)

and




√  (0)
(0)
(0)
(0)
n H n (v) − H (v) = −αn 1 − H (v) , for 0 < H (v) < 1 − θ.

(5.19)

14

Our methodology strongly relies on the well-known Gaussian approximation, given by Csörgő et al.
(1986) in Corollary 2.1, which says that on the probability space (Ω, A, P) , there exists a
sequence of Brownian bridges {Bn (s) ; 0 ≤ s ≤ 1} such that for every 0 < λ < ∞ and

0 ≤ η < 1/4,

nη |αn (s) − Bn (s)|

sup
λ/n≤s≤1−λ/n

(s (1 − s))1/2−η

= OP (1) .

(5.20)

For the increments αn (θ) − αn (θ − s) , we will need an approximation of the same type

as (5.20) . Following similar arguments, mutatis mutandis, as those used to in the proof

of assertions (2.2) of Theorem 2.1 and (2.8) of Theorem 2.2 in Csörgő et al. (1986), Necir
(2017), recently showed, in a technical report, that for every 0 < θ < 1, 0 < λ < ∞ and
0 ≤ η < 1/4, one also has

nη |{αn (θ) − αn (θ − s)} − Bn (s)|
= OP (1) .
s1/2−η
λ/n≤s<θ
sup

(5.21)

5.2. Proof of Theorem 2.1. To get the asymptotic weak approximation given in Theorem
2.1, we will perform successive decompositions that will produce several remainder terms
Rni (x) . We show, in Lemmas 7.6, 7.8, 7.9, 7.10, 7.12, 7.13 and 7.14 of the Appendix,

that for a every fixed 0 ≤ τ < 1/8 and for all large n, we have Rni (x) = oP x−τ /γ , for
i = 1, ..., 16, uniformly over x ≥ 1. Then, once one of the remainder terms appears in the

following decompositions it is systematically replaced by oP x−τ /γ . Let us begin by setting
ϕn (u) := (u + k −1 )
Z

n
∆n (x) =
k

Z

∞
xZn−k:n

n
+√
k

Z

and rewrite both (2.11) and (1.8) into
(1)

∞

ϕn

xZn−k:n

Observe now that
n
k

−1

(

√

H n (w)
H n (w)

(1)

(1)

∞

ϕn
xZn−k:n

n
dHn (w) and rn (x) =
k

Z

(1)

∞

H (w)
H (w)

ϕn

xh

!

dH (w) .

k (∆n (x) − rn (x)) =: Tn (x) may be rewritten into

H n (w)
H n (w)

ϕn

!

!

(1)

− ϕn

H (w)
H (w)

!

H (w)
H (w)

!)

dHn (w)

n
d (Hn (w) − H (w)) +
k

Z

(1)

xh

xZn−k:n

ϕn

H (w)
H (w)

!

dH (w) .

In view of Taylor’s expansion, we write ϕn (u) = u−1 + O (k −1 ) u−2 , then by using this latter
(1)

twice, for u = H n (w) /H n (w) and u = H

(1)

(w) /H (w) , we may decompose Tn (x) into the

15

sum of
n
√
k

Z

∞

H n (w)

xZn−k:n

n
−√
k

Z

(1)

H n (w)

∞

H (w)

−

H

(1)

(w)

!

dHn (w)

n
d (Hn (w) − H (w)) +
(1)
k
H (w)
H (w)

xZn−k:n

Z

xh

H (w)

xZn−k:n

H

(1)

dH (w) ,

(w)

and three remainder terms
 n
Rn1 (x) := OP k −1 √
k
Rn2 (x) := OP
and

Z

∞

xZn−k:n

 n
k −1 √
k

Rn3 (x) := OP

Z




!2

H n (w)

 H (1) (w)
n

∞

+

H (w)

H

xZn−k:n

 n
√
k
k
−1

Z

(1)

(w)

∞

!2 

H (w)
dHn (w) ,
(1)
H (w) 

!2

d (Hn (w) − H (w))

H (w)

xZn−k:n

H

(1)

dH (w) .

(w)

Let us now also decompose Tn (x) into the sum of
n
Tn1 (x) := √
k
n
Tn2 (x) := √
k

Z

Z

∞

H n (w) − H (w)

∞

1
Z

−

(1)

H n (w)

xZn−k:n

n
Tn3 (x) := √
k

(1)

H n (w)

xZn−k:n

∞

n
Tn4 (x) := √
k

H

H (w)

xZn−k:n

and

1

Z

H

(1)

(w)

xh

xZn−k:n

(1)

(w)

!

H (w) dHn (w) ,

d (Hn (w) − H (w)) ,

H (w)
H

dHn (w) ,

(1)

dH (w) .

(w)

For the purpose of establishing Gaussian approximations to Tni (x) , we introduce the following two crucial tail empirical processes βn and βn∗ defined, for w ≥ 1, by

n  (1)
(1)
βn (w) := √ H n (Zn−k:nw) − H (Zn−k:nw) ,
k

(5.22)

and

n
βn∗ (w) := √ H n (Zn−k:n w) − H (Zn−k:nw) .
k

(5.23)

16

5.2.1. Asymptotic representations to Tni (x) in terms of βn and βn∗ . Let us now decompose
Tn1 (x) into the sum of
(1)
Tn1

and
n
Rn4 (x) := √
k

Z

n
(x) := √
k
∞

Z

∞

H n (w) − H (w)

1
(1)
Hn

xZn−k:n

H

xZn−k:n

(w)

−

(1)

1
H

(1)

(w)

(w)
!

dHn (w) ,


H n (w) − H (w) dHn (w) .

The change of variable w = Zn−k:nu and the definition (5.23) , yield that
Z ∞
βn∗ (u)
(1)
dHn (Zn−k:nu) ,
Tn1 (x) =
(1)
x
H (Zn−k:nu)
and
Rn4 (x) =

Z

∞

1
(1)
Hn

x

(Zn−k:n u)

−

1
H

(1)

(Zn−k:nu)

!

βn∗ (u) dHn (Zn−k:nu) .

(1)

The term Tn1 (x) may in turn be decomposed into
Z
1 n ∞ 1/γ ∗
(2)
Tn1 (x) :=
u βn (u) dHn (Zn−k:n u) ,
pk x
and
Rn5 (x) :=

Z

x

∞

(

1
(1)

H

(Zn−k:n u)

−

n 1 1/γ
u
kp

)

βn∗ (u) dHn (Zn−k:n u) .

Thus, we end up with
1n
Tn1 (x) =
pk

Z

∞

x


u1/γ βn∗ (u) dHn (Zn−k:nu) + oP x−τ /γ .

Now, we use similar decompositions to the second term Tn2 (x) . Observe that this latter
equals
(1)
Tn2

(x) := −

Z

x

∞

H (Zn−k:nu)
h (1)
i2 βn (u) dHn (Zn−k:n u) ,
H (Zn−k:n u)

plus a remainder term
)
Z ∞(
1
1
H (Zn−k:nu)
−
Rn6 (x) :=
h
i2 βn (u) dHn (Zn−k:nu) .
(1)
(1)
(1)
x
H n (Zn−k:n u) H (Zn−k:n u)
H (Zn−k:n u)
(1)

Likewise, Tn2 (x) may be rewritten as the sum of
Z
1 n ∞ 1/γ
(2)
u βn (u) dHn (Zn−k:n u) ,
Tn2 (x) := − 2
p k x

17

and
Rn7 (x) := −

Z

x



∞

Thereby





1 n
− 2 u1/γ βn (u) dHn (Zn−k:n u) .

2


p k
 H (1) (Zn−k:n u)



1n
Tn2 (x) = − 2
p k

H (Zn−k:nu)

Z

∞


u1/γ βn (u) dHn (Zn−k:n u) + oP x−τ /γ .

x

For the third term Tn3 (x) , we make a change of variables and an integration by parts to get
)
(
Z ∞
H (xZn−k:n ) ∗
H
(Z
u)
n−k:n
Tn3 (x) = (1)
βn (x) +
.
βn∗ (u) d
(1)
x
H (xZn−k:n )
H (Zn−k:n u)
Making use of Lemma 7.3, we get Tn3 (x) =
may in turn be rewritten as the sum of


1 ∗
βn (x) + oP x−τ /γ . The fourth term Tn4 (x)
p

n
(1)
Tn4 (x) := p−1 √ H (xh)
k

(

Zn−k:n
h

−1/γ

)

−1 ,

plus two remainder terms
n
Rn8 (x) := √
k

Z

xh

xZn−k:n

and
n
Rn9 (x) := p−1 √ H (xh)
k

(

(

H (w)
H

(1)

(w)

− p−1

H (xZn−k:n )
−
H (xh)



)

dH (w) ,

Zn−k:n
h

−1/γ )

.

√
P
Note that k (Zn−k:n /h − 1) is asymptotically Gaussian, it follows that Zn−k:n /h → 1 and
√
k (Zn−k:n/h − 1) = OP (1) (see, for instance, Theorem 2.1 (assertion 2.7) in Brahimi et al.,
n
2015). On the other hand, from proposition 7.1 (see the Appendix), we infer that H (xh) =
k
(1)
H (xh) /H (h) ≤ (1 + ǫ) x−1/γ+ǫ . Then, by using the mean value theorem, in Tn4 (x) , yields



1 −1/γ √
Zn−k:n
(1)
Tn4 (x) = − x
k
− 1 + oP x−τ /γ .
pγ
h
√
From assertion (3.22) in Brahimi et al. (2015), we have k (Zn−k:n/h − 1) = γβn∗ (1)+oP (1) ,

it follows that


1
(1)
Tn4 (x) = − x−1/γ βn∗ (1) + oP x−τ /γ = Tn4 (x) .
p
To summarize, we showed that
√

k (∆n (x) − rn (x)) =

3
X
i=1


Tni (x) + oP x−τ /γ ,

18

where

and

Z
n

1 ∞ 1/γ ∗
u βn (u) d
Tn1 (x) := −
H n (Zn−k:n u) ,
p x
k
Z ∞


n
1
1/γ
H n (Zn−k:n u) ,
u βn (u) d
Tn2 (x) := 2
p x
k

(5.24)


1 ∗
βn (x) − x−1/γ βn∗ (1) .
p
In the following Section we provide Gaussian approximations to Tni (x) , i = 1, 2, 3.
Tn3 (x) :=

5.3. Gaussian approximation to Tni (x). We show that
r
 
Z −1/γ

1 x
n ∗ k
−1
Tn1 (x) = −
Bn
s ds + oP x−τ /γ ,
s
p 0
k
n
r


Z −1/γ

n
k
1 x
−1
Bn p s ds + oP x−τ /γ ,
s
Tn2 (x) = 2
p 0
k
n
and
r  

 

k −1/γ
1 −1/γ ∗ k
1 n
∗
Bn
+ oP x−τ /γ ,
− x
x
Bn
Tn3 (x) =
p k
n
p
n
∗
where Bn is the centred Gaussian process given in Theorem 2.1. We will only give

(5.25)
(5.26)

(5.27)
details

for (5.25) since the proofs of (5.26) and (5.27) follow by using similar arguments. Let us also
(1)

introduce the following Gaussian processes that we define, for 0 < H (v) < θ, by
 (1) 


 (1) 
(0)
Bn (v) := Bn H (v) and B∗n (v) := Bn H (v) − Bn 1 − H (v) .

(5.28)

It is clear that Tn1 (x) , given in (5.24) , may be rewritten as the sum of
r

 n
Z
o
1 ∞ 1/γ n ∗ k −1/γ
n
(1)
H n (Zn−k:n u) ,
Tn1 (x) := −
d
u
Bn
u
p x
k
n
k

and two remainder terms
r

 n
Z
o
n
1 ∞ 1/γ
n ∗
∗
βn (u) −
u
B (Zn−k:n u) d
Rn10 (x) := −
H n (Zn−k:n u) ,
p x
k n
k
and
1
Rn11 (x) := −
p

Z

x

∞

u

1/γ

r 

 n
o
k −1/γ
n
n
∗
∗
Bn (Zn−k:n u) − Bn
d
u
H n (Zn−k:n u) .
k
n
k
(1)

Let us now focus on the term Tn1 (x) . Since x ≥ 1 then Hn (w) = 1, for w ≥ xZn:n , it follows
that Hn (Zn−k:nu) = 1, for u ≥ xZn:n /Zn−k:n, therefore
r

 n
Z
o
n
1 xZn:n /Zn−k:n 1/γ n ∗ k −1/γ
(1)
d
Bn
u
u
H n (Zn−k:nu) .
Tn1 (x) = −
p x
k
n
k

19

Let Qn (t) := inf {w : Hn (w) ≥ t} , 0 < t < 1, be the empirical quantile function pertaining

to cdf Hn . For convenience, we set

ℓn (s) := Qn (1 − ks/n) /Zn−k:n,

(5.29)

and make the change of variable u = ℓn (s) to get
r


Z n
1 k H n (Zn−k:n x)
n ∗ k
(1)
1/γ
−1/γ
(ℓn (s))
Tn1 (x) =
B
(ℓn (s))
ds.
p nk H n (Zn:n x)
k n n
In view of the algebraic equation an bn = (an − a) (bn − b) + (an − a) b + a (bn − b) + ab, we
(1)

decompose Tn1 (x) into the sum of
(2)
Tn1

1
(x) :=
p

Z

n
H n (Zn−k:n x)
k

s

−1

n
H n (Zn:n x)
k

r

n ∗
B
k n




k
s ds,
n

and three remainder terms
 

Z n
 rn  k
k
1 k H n (Zn:n x) 
−1/γ
1/γ
∗
∗
−1
Bn
(ℓn (s))
s
ds,
− Bn
Rn12 (x) :=
(ℓn (s)) − s
p nk H n (Zn−k:n x)
k
n
n
1
Rn13 (x) :=
p

Z

r  
 

n
k
k
−1/γ
∗
∗
Bn
(ℓn (s))
s
ds,
− Bn
s
n
k
n
n
H n (Zn−k:n x)
k
n
H n (Zn:n x)
k

and
1
Rn14 (x) :=
p

Z

−1

 rn  k 

1/γ
−1
Bn∗
s ds.
(ℓn (s)) − s
n
k
n
H n (Zn−k:n x)
k
n
H n (Zn:n x)
k

(2)

Let us also decompose Tn1 (x) into the sum of
r
 
Z −1/γ
1 x
n ∗ k
(3)
−1
Tn1 (x) :=
s
B
s ds,
p 0
k n n
1
Rn15 (x) := −
p

Z

and
1
Rn16 (x) := −
p

n
H n (Zn−k:n x)
k

s

−1

x−1/γ

Z

n
H n (Zn:n x)
k

s

−1

0

r

r

n ∗
B
k n

n ∗
B
k n






k
s ds,
n


k
s ds.
n

Finally, we end up
1
Tn1 (x) = −
p

Z

0

x−1/γ

s

−1

r

n ∗
B
k n





k
s ds + oP x−τ /γ ,
n

20

which meets with approximation (5.25) . Recall that in Theorem 2.1, we set
r
r  

 

n
n
k −1/γ
1 −1/γ ∗ k
−1
∗
−1/γ
−1
p
Bn
− x
Ln x
x
Bn
=p
k
k
n
p
n
r
r
 
 
Z x−1/γ
Z x−1/γ
n ∗ k
n
k
−1
−1
−2
−1
s
s
Bn
s ds + p
Bn
s ds,
−p
k
n
k
n
0
0
r
 P

n
−1
Ln x−1/γ − 3i=1 Tni (x) = oP x−τ /γ , therefore
this means that p
k
r
√


n
−1
Ln x−1/γ = oP xτ /γ ,
k (∆n (x) − rn (x)) − p
k

uniformly over x ≥ 1, this completes the proof of weak approximation (2.12) .
√

5.4. Gaussian approximation to Dn (x). Recall that Dn (x) = k ∆n (x) − px−1/γ ,
x ≥ 1, which may be rewritten into the sum of
√
√

In (x) := k (∆n (x) − rn (x)) and Jn (x) := k rn (x) − px−1/γ .

In view of weak approximation (2.12) , it suffices to show that Jn (x) meets the remainder
(1)

term Rn (x) . Indeed, by letting ψ (z) := H (z) /H (z) , we write
Z ∞
ψ (hz) dH (hz)
−1/2
− px−1/γ ,
k
Jn (x) = −ψ (h)
ψ
(h)
H
(h)
x
which in turn may be decomposed into the sum of


H (hx)
−1/2
−1
−1/γ
,
k
J1n (x) := p
−x
H (h)

Z ∞
ψ (hz)
H (hz)
−1/2
k
J2n (x) := −ψ (z)
−1 d
ψ (h)
H (h)
x
and
H (hx)
,
k −1/2 J3n (x) := −ψ (z) A2 (h)
H (h)
where A2 := H

(1)

/H − p. By using the uniform inequalities (for the second-order regularly

varying functions) to H (see, e.g., de Haan and Ferreira, 2006, bottom of page 161) we
write: for any 0 < ǫ < 1, there exists n0 = n0 (ǫ) , such that for all n > n0 and x ≥ 1
H (hx) /H (h) − x−1/γ
xν/γ − 1
< ǫx−1/γ+ν/γ+ǫ .
− x−1/γ
A (h)
νγ
√
Recall that, by assumption, we have kA (h) = O (1) , it follows that
J1n (x) = p−1 x−1/γ


xν/γ − 1 √
kA (h) + o x−τ /γ .
νγ

21

For the second and third terms, let us write
Z
√
J2n (x) = kA∗ (h) ψ (h)

∞

x

1 − ψ (hz) /ψ (h) H (hz)
.
d
A∗ (h)
H (h)

Making use of Lemma 7.2 and Proposition 7.1 (applied to H) with the fact that ψ (h) → p−1
√
and kA∗ (h) = O (1) , we end up, after integration, with

ν + xν∗ /γ − 1 √
kA∗ (h) + o x−τ /γ .
1/γ
pγx ν∗ (1 − ν∗ )
√
For the third term, we also assumed that kA2 (h) = O (1) , then it suffices to use on again
√

Proposition 7.1 to readily get J3n (x) := x1/γ−1 kA2 (h) + o x−τ /γ , as sought.
J2n (x) =

5.5. Proof of Theorem 2.2. Let us begin by the consistency of γb1 which may be rewritten
R∞
into 1 x−1 ∆n (x) dx and write
r Z ∞
Z ∞

n
−1
−1/2 −1
γ1 =
b
x rn (x) dx + k
p
x−1 Ln x−1/γ + ǫn ,
k 1
1
where

ǫn :=

Z

∞

1

−1

x



∆n (x) − rn (x) − k

−1/2 −1

It is clear that for a fixed 0 ≤ τ < 1/8, we have
τ /γ

|ǫn | ≤ sup x
x≥1

R∞

∆n (x) − rn (x) − k

−1/2 −1

p

p

r

r



n
−1/γ
Ln x
dx.
k


n
Ln x−1/γ
k

Z

∞

x−1−τ /γ dx.

1

x−1−τ /γ dx = γ/τ (finite), then by using weak approximation (2.12) given in Ther

n R ∞ −1
P
−1
orem 2.1, we end up with ǫn → 0. On the other hand p
x Ln x−1/γ is a centred
1
k
Gaussian random with variance
 r Z ∞

 2
n
2
−1
−1
−1/γ
.
σn := E p
x Ln x
k 1
Since

1

By a tedious (but elementary) computation we show that σn2 → (9p−1 − 8) γ12 . A similar

calculation may be found in the proof of Corollary 2.1 of Brahimi et al. (2015). It remains
R∞
to show that 1 x−1 rn (x) dx → γ1 as n → ∞. Indeed, we have
)
Z ∞ (Z ∞
Z ∞
1
H
(hw)
dH
(hw)
d log x.
x−1 rn (x) dx =
(1)
H (h) 1
x
1
H (hw) + k −1 H (hw)

By an integration by parts, this latter becomes
Z ∞
1
H (hx) dH (hx)
(log x) dx,
(1)
H (h) 1 H (hx) + k −1 H (hx)

22

which, from Lemma 7.1, tends to γ1 as n → ∞, as sought. Let us now consider the asymptotic
√
R∞
γ1 − γ1 ) = 1 x−1 Dn (x) dx and use the weak approximation
normality. Recall that k (b
(2.13) to get

√
where µn :=

R∞
1

r Z ∞

n
x−1 Ln x−1/γ + p−1 µn + oP (1) ,
k (b
γ1 − γ1 ) = p−1
k 1

x−1 Rn (x) dx. By using integrations by parts with elementary calculations,

we show that µn meets formula (2.14) and
r Z ∞
r
r
 
 
Z 1

n
n
n ∗ k
k
−1
−1
−1/γ
−1 ∗
γ1 s Bn
s ds − p
B
x Ln x
=
k 1
k
n
k n n
0
r


  
Z 1
n
k
k
−1
∗
−1
γ1 s
s − p Bn p s
log (s) ds,
Bn
+
k
n
n
0

with variance tending to (9p−1 − 8) γ12 , which leads to the Gaussian approximation and
√
γ1 − γ1 ) .
therefore the asymptotic normality of k (b
6. Concluding notes
In this work, we first defined a tail empirical process for Pareto-like distributions which are
randomly right-censored and established its Gaussian approximation. The latter will play
a central role, in the context of right censorship, in determining the asymptotic distributions of statistics that are functionals of the tail index estimator such as the estimators
of large quantiles, risk measures, second-order parameters of regular variation... Then, we
introduced a new Hill-type estimator for positive EVI of right-censored heavy-tailed data
whose asymptotic behavior (consistency and asymptotic normality) is assessed by using the
above-mentioned tail process. Compared to other existing estimators, the new tail index estimator performs better, as far as bias and mean squared error are concerned, at least from
a simulation viewpoint. As a case study, we provided an estimator to the survival time of
Australian male Aids patients. It is also worth mentioning, that assumption H ∈ RV (−1/γ)
also implies that

R∞
t

Γt (g, α) :=

g




dH (x)
H (x)
(log (x/t))α
H (t−)
H (t)
→ γ α , as t → ∞,
R1
α
g (x) (− log x) dx
0

where g is a suitable weight function and α is some positive real number. Thus, assertion (1.4)
becomes a special case of the limit above. Thereby, the functional Γt (g, α) can be considered
as a basic tool to constructing a whole class of estimators for distribution tail parameters for
complete data, see for instance Ciuperca and Mercadier (2010). Recently, Benchaira et al.

23

(2016a) and Haouas et al. (2017) benefited from this result to derive, respectively, a kernel
estimator of tail index and a second-order parameter estimator for random right-truncation
data. As we deal with the tail modeling of underlying cdf F, we have to work with the limit
!α


R∞
dH (x)
H (x)
log (x/tn )
g
t
(1)
H (tn −)
H (tn )
H (z) /H (z) + k −1
(c)
→ γ1α , as n → ∞,
Γtn (g, α) :=
R1
α
g (x) (− log x) dx
0
which is generalization of the result (1.6) . This latter may be readily shown by using similar

arguments as those used in Lemma 7.2. Thereby, by letting t = Zn−k:n, then by replacing
H

(1)

(1)

and H by their respective empirical counterparts H n and H n , we end up with an
(c)

estimator of Γt (g, α) given by
k−1
P

(i)

an,k (log (Zn−i:n /Zn−k:n))α
(c)
Γn,k (g, α) := i=1 R 1
,
g (x) (− log x)α dx
0
 


Pi
i
i
(i)
where an,k := g
/
δ
+
i/k
. This would have fruitful consequences
j=1 [n−j+1:n]
k
k+1
on the statistical analysis of extremes under random censoring. To finish, notice that the
Gaussian approximations corresponding to pb, γb1 and Dn (x) are jointly established in terms
of the same sequence of Brownian bridges Bn . This allows to establish the limit distributions

to the statistics of Kolmogorov-Smirnov and Cramer-von Mises type respectively defined by
KSn,k :=

and
CVn,k := (b
pb
γ)

−1

k

√

k sup ∆n (x) − pb−1 x−1/bγ ,
x≥1

Z

∞

x−1/bγ −1 ∆n (x) − pb−1 x−1/bγ

1

2

dx.

These statistics provide goodness-of-fit tests for Pareto-like distributions under right-censorship.
This matter will be addressed in our future work. In the case of complete data (p = 1) , the
latter two statistics respectively become
√

n
k sup F n (xXn−k:n ) − x−1/bγ and (b
γ )−1 k
x≥1 k

Z

∞

x−1/bγ −1
1

n
k

F n (xXn−k:n ) − x−1/bγ

2

dx.

They are used in Koning and Peng (2008), amongst others, to test the heaviness of cdf’s.
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24

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7. Appendix
A key result related to the regular variation concept, namely Potter-type inequalities (see,
e.g., Proposition B.1.10, page 369 in de Haan and Ferreira, 2006), will be applied quite
frequently. For this reason, we need to recall this very useful tool here.
Proposition 7.1. Let f ∈ RV (α) , for α ∈ R. Then, for any sufficiently small ǫ > 0, there

exists t0 = t0 (ǫ) > 0, such that for any t ≥ t0 , we have |f (tx) /f (t) − xα | ≤ ǫxα+ǫ , uniformly
on x ≥ 1.

Lemma 7.1. Let F ∈ RV (−1/γ1 ) and G ∈ RV (−1/γ2 ) . Then, for every real r ≥ 0, we have
!r
Z ∞
γ
H (z)
1
log
(z/t)
dH
(z)
=
.
(7.30)
lim
(1)
t→∞ H (t) t
pr
H (z)
Moreover, for a given sequence t = tn → ∞, as n → ∞, we have
Z ∞
log (z/t) dH (z)
1
lim
= γ1 .
(1)
n→∞ H (t) t
H (z) /H (z) + k −1
Proof. For r ≥ 0, let us set
1
Ir (t) :=
H (t)

Z

t

∞

H (z)
H

(1)

(z)

!r

log (z/t) dH (z) ,

(7.31)

27

which, by a change of variables, becomes
1
H (t)

Z

∞

H (tz)
H

1

(1)

(tz)

!r

log (z) dH (tz) .

This may be decomposed into the sum of
!r
)
Z ∞(
H (tz)
1
I1 (t) :=
− p−r log (z) dH (tz) ,
(1)
H (t) 1
H (tz)
R∞
and I2 (t) := p−r 1 log (z) dH (tz) /H (t) . We will show that I1 (t) → 0 and I2 (t) → γ/pr ,

as t → ∞. Indeed, we have

1
|I1 (t)| ≤
H (t)

Z

∞

H (tz)
H

1

(1)

(tz)

!r

−

1
log (z) dH (tz) .
pr

Recall that from Lemma 4.1 of Brahimi et al. (2015), under the first-order conditions (1.1) ,
we have H (z) /H

(1)

(z) → p−1 , as z → ∞, it follows that H/H

Proposition (7.1) to H/H

(1)

(1)

∈ RV (0) . Thus, by applying

, we write : for all large t and any sufficiently small ǫ > 0

H (tz) /H
H (t) /H

(1)
(1)

(tz)
(t)

− 1 < ǫ, for any z ≥ 1,

(7.32)

leading to
H (tz)
H

(1)

(tz)

−

H (t)
H

(1)

< ǫ (1 + ǫ) p−1 z ǫ .

(t)

Observe now that
H (tz)
H

(1)

(tz)

− p−1 <

H (tz)
H

(1)

(tz)

−

H (t)
H

(1)

(t)

+

H (t)
H

(1)

(t)

− p−1 ,

(7.33)

(1)

it follows that H (tz) /H (tz) − p−1 = o (z ǫ ) , uniformly over z ≥ 1. Using the mean-value
r

(1)
theorem yields that H (tz) /H (tz) − p−r = o (z ǫ ) as well, which implies that for all
R∞

large t, I1 (t) = o (1) 1 z ǫ log (z) d H (tz) /H (t) . Then, it suffices to show that the latter

RM
integral is finite. Indeed, let JM := 1 z ǫ log (z) d H (tz) /H (t) , which, by an integration
by parts, equals


−M H (tM) /H (t) log M +
ǫ

Z

1

M


H (tz) /H (t) d (z ǫ log z) .

Once again, we use Proposition (7.1) to H, to write that for any z ≥ 1,
H (tz) /H (t) − z −1/γ < ǫz −1/γ+ǫ ,

(7.34)

28

for sufficiently large t. It follows that

M ǫ H (tM) /H (t) log M = M ǫ (1 + o (M ǫ )) M −1/γ log M,

which tends to zero as M → ∞. By using similar arguments, we end up with
Z M
Z M

ǫ
H (tz) /H (t) d (z log z) =
(1 + o (z ǫ )) z −1/γ d (z ǫ log z) ,
1

1

which tends to γ (γǫ − 1) / (2γǫ − 1)2 (1 + o (1)) as M → ∞, hence I1 (t) = o (1) . For the
R∞
second term I2 (t) , it suffices to use (1.4) to have pr I2 (t) = 1 log (z) dH (tz) /H (t) → γ
as t → ∞, which meets (7.30) . To prove (7.31) , we first apply Taylor’s expansion. Indeed,
since 1/k → 0 as n → ∞, then



!2 
1
H (z)
H (z)
 , as n → ∞,
= (1)
+o
(1)
(1)
H (z) /H (z) + 1/k
H (z)
H (z)

thereby, we use result (7.30) twice, for r = 1 and r = 2, which completes the proof of the
lemma.



Lemma 7.2. Let ψ := H/H

(1)

. Then, for any small ǫ > 0 and x ≥ 1, we have

ψ (tx) /ψ (t) − 1 xν∗ /γ − 1
< ǫxν∗ /γ+ǫ , as t → ∞,
−
A∗ (t)
ν∗ γ

(7.35)

where ν∗ := max (ν, ν1 ) and A∗ (t) := 1 {ν > ν1 } A (t) − 1 {ν ≤ ν1 } A1 (t) .
Proof. Let us write
ψ (tx)
=
ψ (t)



H (tx)
− x−1/γ
H (t)



H
H

(1)

(1)

(t)

H

+ x−1/γ

(tx)

H

(1)

(1)

(t)

(tx)

− x1/γ

!

+ 1.

It is easy to check that


ϑt (x) + x

+

A1 (t)
A∗ (t)

A (t)
ψ (tx) /ψ (t) − 1
=
A∗ (t)
A∗ (t)

ν/γ
−1/γ x



−1
νγ



H
H

(1)

(t)

(1)

(tx)

ν1 /γ
−1
(1)
1/γ x
ϑt (x) − x
x−1/γ ,
ν1 γ

where
ϑt (x) :=

H (tx) /H (t) − x−1/γ
xν/γ − 1
− x−1/γ
A (t)
νγ

and
(1)
ϑt

(x) :=

H

(1)

(1)

xν1 /γ − 1
(t) /H (tx) − x1/γ
+ x1/γ
.
A1 (t)
ν1 γ

29

Thus, the uniform inequalities (for second-order regularly varying functions) to both tails H
and H

(1)

(see, e.g., Proposition 4 together with Remark 1 in Hua and Joe (2011)) yield that

: for any 0 < ǫ < 1 and x ≥ 1,
(1)

|ϑt (x)| < ǫx−1/γ+ν/γ+ǫ and ϑt (x) < ǫx1/γ+ν1 /γ+ǫ , as t → ∞.
On the other hand, we have 1/H
(1)

(1)

(1)

∈ RV (1/γ) , then from Proposition (7.1) , we infer that,

(t) /H (tx) = (1 + o (xǫ )) x1/γ , as t → ∞. Therefore


ν/γ

−1
A (t)
ψ (tx) /ψ (t) − 1
−1/γ+ν/γ+ǫ
−1/γ x
o x
+x
(1 + o (xǫ )) x1/γ
=
A∗ (t)
A∗ (t)
νγ


ν1 /γ

−1
A1 (t)
1/γ+ν1 /γ+ǫ
1/γ x
o x
−x
x−1/γ .
+
A∗ (t)
ν1 γ

If ν > ν1 , then A∗ (t) = A (t) and A1 (t) /A∗ (t) = A1 (t) /A (t) = O t(ν1 −ν)/γ which tends

uniformly over x ≥ 1, H

to zero as t → ∞, it follows that


ψ (tx) /ψ (t) − 1 xν/γ − 1
−
= o xν/γ+ǫ .
A∗ (t)
νγ


If ν ≤ ν1 , then A∗ (t) = −A1 (t) and A (t) /A∗ (t) = A (t) /A1 (t) = O t(ν−ν1 )/γ = O (1) ,

which goes to zero as t → ∞, then


ψ (tx) /ψ (t) − 1 xν1 /γ − 1
−
= o xν1 /γ+ǫ ,
A∗ (t)
ν1 γ

which completes the proof of (7.35) .



Lemma 7.3. For every 0 < η < 1/2, we have
sup u(1/2−η)/γ |βn∗ (u)| = OP (1) = sup u(1/2−η)/γ |βn (u)| .
u≥1

u≥1

Proof. Let 0 < η < 1/2 and recall, from (5.23) , that

n
βn∗ (u) = √ H n (uZn−k:n) − H (uZn−k:n) .
k

By using the change of variables t = uZn−k:n/h, we write
(1/2−η)/γ

n
n
h
(1/2−η)/γ
∗
sup t(1/2−η)/γ √ H n (th) − H (th) ..
sup u
|βn (u)| = √
u≥1
ht≥Zn−k:n
k Zn−k:n
k
From Theorem 2 in Shorack and Wellner (1986) (page 4), we may write


H n (z) , z ≥ 0

n≥1


D 
= Un H (z) , z ≥ 0

n≥1

,

(7.36)

30

where Un being the uniform empirical
givenin (5.15)
 cdf 
 . Let us now introduce the uniform
√
n
k
tail empirical process ϑn (s) := k
Un
s − s , s > 0. It is easy to verify, in view of
k
n
(7.36) , that



o
n n

n
D
√ H n (th) − H (th) , t > 0
H (th) , t > 0
= ϑn
,
k
n≥1
k
n≥1
it follows that
sup u

(1/2−η)/γ

u≥1

|βn∗



D

(u)| =

h
Zn−k:n

(1/2−η)/γ

sup

t(1/2−η)/γ ϑn

ht≥Zn−k:n

n
k

H (th)



.

Since H is decreasing and continuous, then the previous supremum is equivalent to that of
over 0 < nk H (th) ≤ nk H (Zn−k:n ) , thereby the latter expression becomes

(1/2−η)/γ
(1/2−η)
n
|ϑn (s)|
h
H (th)
sup
t(1/2−η)/γ
.
sup
(1/2−η)
Zn−k:n
k
0<s≤ n H(Zn−k:n ) s
0< n H(th)≤ n H(Zn−k:n )
k

k

k

Let c > 0 be fixed and ǫ > 0 be sufficiently small. It is clear that
)
(
|ϑn (s)|
>c
P
sup
(1/2−η)
s
H(Z
)
0<s≤ n
n−k:n
k


n n
o
|ϑn (s)|
=P
H (Zn−k:n ) − 1 > ǫ + P
sup (1/2−η) > c =: Pn1 + Pn2 .
k
0<s≤1+ǫ s
n n
o
n
P
H (Zn−k:n) − 1 < ǫ occurs with a large probabilSince H (Zn−k:n) → 1, then An :=
k
k
ity, that is Pn1 ↓ 0, as n → ∞. Next we show that Pn2 is asymptotically bounded. Indeed,
in set An , we have

sup
0<s≤ n
H(Zn−k:n )
k

s−(1/2−η) |ϑn (s)| ≤

sup s−(1/2−η) |ϑn (s)| ,

0<s≤1+ǫ

which, from Proposition 3.1 in Einmahl et al. (2006), is asymptotically bounded. On the
other hand, by applying Proposition 7.1 (to H), we write:
(1 − ǫ) t−1/γ+ǫ ≤

n
H (th) ≤ (1 + ǫ) t−1/γ+ǫ ,
k

(7.37)

uniformly over v ≥ 1. Then is easy to verify that, in set An , we have

(1/2−η)
(1/2−η)/γ n
t
sup
H (th)
= OP (1) ,
k
0< n H(th)≤ n H(Zn−k:n )
k

k

this means that Pn2 = O (1) , which meets the first assertion. To show the second one, let
us set


n  (0)
(0)
βen (u) := √ H n (Zn−k:n u) − H (Zn−k:n u) .
k

(7.38)

31

r

n 
(0)
αn 1 − H (Zn−k:nu) , (almost surely). Since
From (5.23) , we have βen (u) = −
k
D

{αn (s) , 0 ≤ s ≤ 1}n≥1 = − {αn (1 − s) , 0 ≤ s ≤ 1}n≥1 ,

then

n
o
βen (u) , u ≥ 1

D

n≥1

=

r


n  (0)
αn H (Zn−k:nu) , u ≥ 1
k



.

n≥1

As we already did for βn , we write
(1/2−η)/γ

 n (0)

h
D
(1/2−η)/γ e
sup t(1/2−η)/γ ϑn
sup u
βn (u) =
H (th) .
Zn−k:n
k
u≥1
ht≥Zn−k:n

Recall that H (t) /H
that H (t) /H

(0)

(1)

(t) → p−1 and H

(1)

(t) = H (t) − H

(0)

(t) , then is easy to verify

(t) → q −1 , as t → ∞, where q = 1 − p, this implies that H

(0)

is regularly

varying with index (−1/γ) too. By using similar arguments as used for βn , we also show
supu≥1 u(1/2−η)/γ βen (u) = OP (1) . Note that βn (u) = βn∗ (u) − βen (u) , it follows that
sup u(1/2−η)/γ |βn (u)| ≤ sup u(1/2−η)/γ |βn∗ (u)| + sup u(1/2−η)/γ βen (u) .
u≥1

u≥1

u≥1

which by the first and the second results is OP (1) . This completes the proof of the Lemma.

Lemma 7.4. For every 0 ≤ τ < 1/8, we have

1
βn∗ (x) − βn∗ (x) = oP x−τ /γ =
(1)
p
H (xZn−k:n )
H (xZn−k:n )

Proof. Let 0 ≤ τ < 1/8 and write
1 ∗
∗
β
(x)
=
β (x) +
n
(1)
p n
H (xZn−k:n )
H (xZn−k:n )

(

Z

x

∞

βn∗ (u) d

(

H (Zn−k:n u)
H

1
−
(1)
H (xZn−k:n ) p
H (xZn−k:n )

(1)

)

(Zn−k:n u)

)

.

βn∗ (x) .

Let ǫ > 0 be sufficiently small and 0 < η < 1/2 so that η − 1/2 + τ < 0. By using inequality
(7.33) and Lemma 7.3 together, we get
(
)


1
H (xZn−k:n )
−
βn∗ (x) = oP x(η−1/2)/γ+ǫ = oP x−τ /γ ,
(1)
H (xZn−k:n ) p

uniformly over x ≥ 1, which meets the first assertion of this lemma. For the second assertion,

we once again make use of Lemma 7.3, to write
)
)
(
(
Z ∞
Z ∞
H
(Z
u)
H
(Z
u)
n−k:n
n−k:n
= OP (1)
.
u−(1/2−η)/γ d
βn∗ (u) d
(1)
(1)
x
x
H (Zn−k:n u)
H (Zn−k:n u)

32

By making an integration by parts, then by using inequality (7.33) , we readily show that

the latter integral equals oP x−τ /γ , as well. This completes the proof of the lemma.

Lemma 7.5. For n ≥ 1, we have
(i)

t
Un (t)
= OP (1) = sup
.
t
U1:n ≤t≤1 Un (t)
U1:n ≤t≤1

(ii)

Vn (t)
t
= OP (1) = sup
.
t
1/n≤t≤1
1/n≤t≤1 Vn (t)

sup

sup

For every 0 < η < 1/2, we have
nη |t − Vn (t)|
nη |Un (t) − t|
=
O
(1)
=
sup
P
1−η
1−η .
0≤t≤1 (t (1 − t))
1/n≤t≤1−1/n (t (1 − t))

(iii) sup

Proof. The proofs of the first two assertions may be found in Shorack and Wellner (1986)
(pages 415 and 416, inequality 2). The left result of assertion (iii) is also given in Shorack and Wellner
(1986) (page 425, assertion 15). For the right one, we have
nη |t − Vn (t)|
1−η
1/n≤t≤1−1/n (t (1 − t))

1−η 
1−η η
1 − Vn (t)
n |Vn (t) − Un (Vn (t))|
Vn (t)
,
=
sup
1−t
t
(Vn (t) (1 − Vn (t)))1−η
1/n≤t≤1−1/n
sup

which is less than or equal to

1−η
1−η

1 − Vn (t)
(Vn (t))
sup
sup
1−t
t
1/n≤t≤1−1/n
1/n≤t≤1−1/n

nη |Vn (t) − Un (Vn (t))|
×
sup
1−η .
1/n≤t≤1−1/n (Vn (t) (1 − Vn (t)))

D

Note that {1 − Vn (t) ; 0 ≤ t ≤ 1} = {Vn (1 − t) ; 0 ≤ t ≤ 1} , which may be rewritten into

{Vn (s) ; 0 ≤ s ≤ 1} . Hence, without loss of generality, we may write

1−η
1−η

1 − Vn (t)
Vn (s)
sup
=
sup
,
1−t
s
1/n≤t≤1−1/n
1/n≤s≤1−1/n

which, from (ii) , equals OP (1) . Observe now that 1/n ≤ t ≤ 1 − 1/n is equivalent to
U1:n ≤ Vn (t) ≤ Un:n , it follows that

nη |Un (s) − s|
nη |Vn (t) − Un (Vn (t))|
≤
sup
1−η
1−η .
0≤s≤1 (s (1 − s))
1/n≤t≤1−1/n (Vn (t) (1 − Vn (t)))
sup

Finally, using the left result of (iii) completes the proof.

Lemma 7.6. For every 0 ≤ τ < 1/8, we have Rni (x) = oP x−τ /γ , for i = 1, 2, 3.



33

Proof. By using representation (7.36) and (5.17) with assertion (i) of Lemma 7.5, we get
(1)

H n (w) /H (w) = OP (1) = H n (w) /H

(1)

(w) , uniformly over Z1:n ≤ w ≤ Zn:n . Since

Hn (w) = 1, for w ≥ Zn:n , it follows that for any x ≥ 1,
Rn1 (x) = OP

Z

 n
k −1 √
k

∞

H (w)
H

xZn−k:n

(1)

(w)

!2

dH (w) .

As we did in the proof of Lemma 7.5, let us consider the set An := {|Zn−k:n /h − 1| < ǫ} , in

which the previous equation becomes
Rn1 (x) = OP
Let us rewrite this latter into

 n
k −1 √
k

Rn1 (x) = OP k


−1/2

Z

Z

∞

H (w)
H

(1−ǫ)xh

∞

(1)

(w)

H (wh)
H

(1−ǫ)x

(1)

(wh)

!2

!2

d

dH (w) .

H (wh)
.
H (h)

(1)

(wh) = (1 + o (1)) p−1 , as n → ∞, uniformly

on w > x. Then by using Proposition 7.1 (for H), we obtain Rn1 (x) = OP k −1/2 x−1/γ ,

which equals oP x−τ /γ as n → ∞, uniformly on x ≥ 1. For the second term, in set An , we
From inequality (7.33) , we have H (wh) /H

write

Rn2 (x) = OP

 n
k −1 √
k

Z

∞

(1−ǫ)xh

H (w)
H

(1)

(w)

!2


d H n (w) − H (w) .

For convenience let us set hǫ := (1 − ǫ) h. By using a change of variables and integration by
parts, the latter integral becomes
!2
Z ∞


H (hǫ )
−
H n (hǫ ) − H (hǫ ) −
H n (hǫ w) − H (hǫ w) d
(1)
x
H (hǫ )

H (hǫ w)
H

(1)

(hǫ w)

!2

.

Let us fix 0 < η < 1/2. The representation (7.36) and assertion (iii) of of Lemma 7.5 together
1−η
imply that, uniformly over w ≥ x, we have H n (hǫ w) − H (hǫ w) = OP (n−η ) H (hǫ w)
.

By using this latter with inequality (7.33) and Proposition 7.1 (to together H), we end up
with Rn2 (x) = OP (k −η ) x(η−1)/γ = oP (1) x−τ /γ . For the third term
Z ∞
 n
H (w)
−1
√
dH (w) ,
Rn3 (x) = OP k
k xZn−k:n H (1) (w)

we use similar arguments to show that equals OP k −1/2 x−1/γ = oP (1) x−τ /γ , which finishes
the proof of the lemma.

Lemma 7.7. For every 0 < η < 1/2, we have supk−1 ≤s≤1 sη+γ |ℓn (s) − s−γ | = oP (1) .



34

Proof. Recall that in (5.29) , we set ℓn (s) = Qn (1 − ks/n) /Zn−k:n, 0 < s ≤ 1. Observe that
ℓn (s) − s−γ may be rewritten into the sum of
−γ
−γ 
−γ


Vn (ks/n)
Vn (ks/n)
Vn (ks/n)
, L2,n (s) :=
−
,
L1,n (s) := ℓn (s) −
Uk+1:n
Uk+1:n
k/n

and
L3,n (s) :=



Vn (ks/n)
k/n
D

−γ

− s−γ .

First note that, for each n ≥ 1, we have Zn−k:n = Q (1 − Uk+1:n ) and
D

{Qn (1 − ks/n) , 0 < s < 1} = {Q (1 − Vn (ks/n)) , 0 < s < 1} ,
then, without loss of generality, we may assume that
ℓn (s) =

Q (1 − Vn (ks/n))
, 0 < s < 1.
Q (1 − Uk+1:n )
P

On the other hand, we have Q (1 − ·) ∈ RV (−γ) and (k/n) /Uk+1:n → 1. By making use of

Proposition 7.1 with f (s) = Q (1 − s) and t = (k/n) /Uk+1:n , we infer that, for any small

ǫ > 0, we have L1,n (s) = oP (1) (Vn (ks/n) /Uk+1:n )−γ+ǫ , uniformly over k −1 ≤ s < 1, in

other term L1,n (s) = oP (1) (Vn (ks/n) / (k/n))−γ+ǫ . By using assertion (ii) of Lemma 7.5,

we end up with


L1,n (s) = oP s−γ+ǫ , as n → ∞.

(7.39)

The second term may be rewritten into
)
−γ (
−γ

k/n
Vn (ks/n)
−1 .
L2,n (s) =
k/n
Uk+1:n
P

Since nUk+1:n /k → 1, then uniformly over k −1 ≤ s < 1, we have

L2,n (s) = oP s−γ , as n → ∞.
(7.40)

Observe now that L3,n (s) = (k/n)γ Vn (ks/n)−γ − (sk/n)−γ , which by using the mean
value theorem equals −γ (k/n)γ {V∗n (ks/n)}−γ−1 (Vn (ks/n) − sk/n) , where V∗n (ks/n) is
between Vn (ks/n) and sk/n. Once again by using assertion (ii) of Lemma 7.5, we show

that V∗n (ks/n) / (ks/n) = OP (1) uniformly over k −1 ≤ s < 1, therefore
L3,n (s) = OP (1) (n/k) s−γ−1 |Vn (ks/n) − sk/n| .
It is clear that L3,n (s) = s−γ−η OP (k −η ) nη |Vn (ks/n) − sk/n| / (sk/n)1−η . Making use of
assertion (iii) of Lemma 7.5, we get


L3,n (s) = oP s−γ−η , as n → ∞.

(7.41)

35

By combining (7.39) , (7.40) and (7.41) we show that
over k −1 ≤ s < 1, as sought.

P3

i=1

Li,n (s) = oP (s−γ−η ) uniformly



Lemma 7.8. For every 0 ≤ τ < 1/8, we have Rni (x) = oP x−τ /γ , for i = 4, 5.
Proof. Recall that

Rn4 (x) = −
First note that H

(1)

Z

(1)

∞

H n (uZn−k:n) − H
(1)
Hn

x

(uZn−k:n) H

(1)

(1)

(uZn−k:n)

(uZn−k:n)

βn∗ (u) dHn (uZn−k:n) .

(1)

> H and H n > H n . On the other hand, by using (7.36) and assertion

(i) of Lemma 7.5, we infer that H (uZn−k:n) /H n (uZn−k:n) = OP (1) , uniformly over u ≥ 1.

It follows that

Rn4 (x) = OP (1)

Z

x

(1)

∞

(1)

(1)

H n (uZn−k:n) − H (uZn−k:n)
|βn∗ (u)| dHn (uZn−k:n) .

2
H (uZn−k:n)

Observe that H n (uZn−k:n) − H

(1)

(uZn−k:n) may be rewritten into

  (0)
(0)
H n (uZn−k:n) − H (uZn−k:n) − H n (uZn−k:n) − H (uZn−k:n) .



(0)
It is clear that Rn,4 (x) = OP (1) Cn (x) + Cn (x) , where
Cn (x) :=

Z

∞

x

and
Z

H n (uZn−k:n) − H (uZn−k:n) ∗
|βn (u)| dHn (uZn−k:n) ,

2
H (uZn−k:n)
(0)

(0)

H n (uZn−k:n) − H (uZn−k:n)
Cn(0) (x) :=
|βn∗ (u)| dHn (uZn−k:n) .

2
x
H (uZn−k:n)

(0)
Next, we show that both Cn (x) and Cn (x) are equal to oP x−τ /γ . Indeed, by making use
∞

of assertion (ii) in Lemma 7.5, we write
Z
 ∞ u−(1/2−η)/γ+ǫ
−η
Cn (x) = OP n

1+η dHn (uZn−k:n) .
x
H (uZn−k:n)

In view of inequality (7.34) , we get
Z
  n 1+η ∞ (1+η)/γ−(1/2−η)/γ+ǫ
−η
Cn (x) = OP n
u
dHn (uZn−k:n ) .
k
x

By an integration by parts and once again by making use of assertion (i) of Lemma 7.5,

we end up with Cn (x) = OP (k −η ) x(η−1/2)/γ . Since k → ∞ and x(η−1/2)/γ = O x−τ /γ ,

(0)
uniformly over x ≥ 1, it follows that Cn (x) = oP x−τ /γ . Let us now consider Cn (x) . Note

36

that we also have H

(0)

(0)

> H and H n > H n and recall that from (5.16) , we have almost

surely
(0)
Hn

(v) − H

(0)


 

(0)
(0)
(v) = Un 1 − H (v) − 1 − H (v) .


(0)
Then, by using similar arguments as for Cn (x) we also readily show that Cn (x) = oP x−τ /γ ,
therefore we omit further details. Let us now take care of the second term
)
Z ∞(
1
n 1 1/γ
Rn5 (x) =
βn∗ (u) dHn (Zn−k:n u) .
u
−
(1)
k
p
x
H (Zn−k:n u)

Recall that 1/H

(1)

∈ RV (1/γ) , then by using Lemma 7.5, we infer that with a probability

tending to 1, large n and for every small ǫ > 0, we have
H
H

(1)

(1)

(Zn−k:n)

(Zn−k:nu)

− u1/γ < ǫuǫ ,

uniformly over u ≥ 1. However, by making use of inequality (7.33) , we get
H (Zn−k:n )
H

(1)

(Zn−k:n)

−

1
< ǫ.
p

n
P
On the other hand, since H (Zn−k:n) → 1 then with a probability tending to 1 and large
k
n
n, we have H (Zn−k:n ) − 1 < ǫ. By combining the last three inequalities, we end up with
k
1
H

(1)

(Zn−k:nu)

−

n 1 1/γ
u
< ǫuǫ+1/γ ,
kp

(7.42)

uniformly over u ≥ 1, this implies that
Z ∞
Rn5 (x) = oP (1)
u−(1/2−η)/γ−ǫ z ǫ+1/γ dHn (Zn−k:nu) .
x

Once again, by using assertion (i) of Lemma 7.5, yields
Z ∞
Rn5 (x) = oP (1)
u−(1/2−η)/γ−ǫ u−1/γ−1+1/γ+ǫ+ǫ du,
x


which equals oP (1) x(η−1/2)/γ+ǫ = oP x−τ /γ . This achieves the proof of the lemma.

Lemma 7.9. For every 0 ≤ τ < 1/8, we have Rni (x) = oP x−τ /γ , i = 6, 7.
Proof. Recall that

Rn6 (x) =

Z

x

∞

(

1
− u1/γ
(1)
H (uZn−k:n) p
H (Zn−k:n)

)

βn∗

(u) d



Hn (Zn−k:n u)
H (Zn−k:n )



.



37

It it clear that
|Rn6 (x)| ≤

Z

∞

x



1
− u1/γ |βn∗ (u)| d
(1)
H (uZn−k:n) p
H (Zn−k:n)

Hn (Zn−k:n u)
H (Zn−k:n )



.

Let us write
1
− u1/γ
(uZn−k:n) p
! (1)
1 H (Zn−k:n )
1
H (Zn−k:n)
−
+
(1)
(1)
H (Zn−k:n) p H (uZn−k:n) p

H (Zn−k:n )
H

(1)

=

By using inequality (7.33) and Proposition 7.1 to H

(1)

H
H

(1)

(1)

(Zn−k:n )

(uZn−k:n)

− u1/γ

!

.

, we infer that


1
− u1/γ = oP u1/γ+ǫ , uniformly over u ≥ 1.
H (uZn−k:n) p
R∞

It follows that Rn6 (x) = oP (1) x u(1/2+η)/γ+ǫ d Hn (Zn−k:n u) /H (Zn−k:n ) . By using an
H (Zn−k:n)
(1)

integration by parts then once again by making use of assertion (i) of Lemma 7.5, we show

easily that Rn6 (x) = oP x−τ /γ . By similar routine arguments, we also show Rn7 (x) =

oP x−τ /γ that we omit further details.


Lemma 7.10. For every 0 ≤ τ < 1/8, we have Rni (x) = oP x−τ /γ , i = 8, 9.
Proof. Recall that

n
Rn8 (x) = − √
k

Z

xh
xZn−k:n

(

H (w)
H

(1)

(w)

− p−1

)

dH (w) .

By making a change of variables, we get
√ Z
Rn8 (x) = − k

x

xZn−k:n /h

(

H (hw)
H

(1)

(hw)

− p−1

)

d

H (hw)
.
H (h)

In view of inequality (7.33) and Proposition 7.1 to H we get
√ Z x
Rn8 (x) = oP (1) k
w −1/γ−1+ǫ dw.
xZn−k:n /h


√ 
k 1 − (Zn−k:n /h)−1/γ+ǫ . Recall that
After integration, we obtain Rn8 (x) = oP x−1/γ+ǫ
√
k (1 − Zn−k:n /h) = OP (1) , then by using the mean theorem, we infer that

√ 
k 1 − (Zn−k:n /h)−1/γ+ǫ = OP (1) ,

38



as well. It follows that Rn8 (x) = oP x−1/γ+ǫ which also equals oP x−τ /γ . Observe now

that

H (xh) √
Rn9 (x) = p−1
k
H (h)

H (xZn−k:n )
−
H (xh)



Zn−k:n
h

−1/γ !

.

By applying Proposition 7.1 implies that, for any 0 < ǫ < 1, there exists n0 = n0 (ǫ) , such
that for all n > n0 and y ≥ 1
H (xZn−k:n )
−
H (xh)



Zn−k:n
h

−1/γ

≤ǫ



Zn−k:n
h

−1/γ+

,

it follows from the previous inequality, that

−1/γ
H (xZn−k:n )
Zn−k:n
−
= oP (1) , .
h
H (xh)



uniformly over x ≥ 1. Since H (xh) /H (h) → x−1/γ thus Rn9 (x) = oP x−1/γ = oP x−τ /γ ,

as sought.



Lemma 7.11. For every 0 < η < 1/2, we have
r

 

n
k
k
−1/γ
∗
(η−1)/2
∗
Bn
sup s
(ℓn (s))
s = oP (1) .
− Bn
k k−1 ≤s≤1
n
n
Proof. Let 0 < η < 1/2 be fixed, then from Lemma 7.7 and by using the mean value theorem,
we may write
sup s1−η (ℓn (s))−1/γ − s = oP (1) .

(7.43)

k −1 ≤s≤1

Recall now that Bn∗ (s) = Bn (ps) − Bn (1 − qs) with q = 1 − p, and write

 

k
k
−1/γ
∗
∗
(ℓn (s))
s
− Bn
Bn
n
n








k
k
k
k
−1/γ
−1/γ
≤ Bn p (ℓn (s))
− Bn p s + Bn 1 − q (ℓn (s))
− Bn 1 − q s .
n
n
n
n
From inequality (1.11) in Csörgő et al. (1986), we infer that
(
)
P

sup

|d|<1, 0≤x+d≤1

d1/2 |Bn (x + d) − Bn (x)| ≥ λ


≤ cλ−1 exp −λ2 /8 ,

(7.44)

for every 0 < λ < ∞, where c is a suitably chosen universal constant. Then, by using this
k
k
k
inequality with d = p (ℓn (s))−1/γ − p s and x = p s, together with (??) , we may readily
n
n
n
show that
r





k
k
n
−1/γ
Bn p (ℓn (s))
− Bn p s = oP s(η−1)/2 .
k
n
n

39

k
k
k
Likewise, with d = q s − q (ℓn (s))−1/γ and x = 1 − q s, we get
n
n
n
r





n
k
k
−1/γ
Bn 1 − q (ℓn (s))
− Bn 1 − q s = oP s(η−1)/2 ,
k
n
n

uniformly over k −1 ≤ s ≤ 1, which finishes the proof of the Lemma.




Lemma 7.12. For every 0 ≤ τ < 1/8, we have Rni (x) = oP x−τ /γ , i = 10, 11.

Proof. Let us first fix 0 ≤ τ < 1/8 and recall that
r
 n

Z Zn:n /Zn−k:n
o
n ∗
n
∗
1/γ
−1
Bn (uZn−k:n) d
H n (Zn−k:nu) ,
βn (u) −
u
Rn10 (x) = −p
k
k
x

e n := B∗ − Bn , as the sum of
which may be rewritten, by letting βen := βn∗ − βn and B
n
r

 n
Z Zn:n /Zn−k:n
o
n
n
1/γ
−1
βn (u) −
u
Bn (uZn−k:n) d
K1n (x) := −p
H n (Zn−k:n u) ,
k
k
x

and

K2n (x) := −p

−1

Z

Zn:n /Zn−k:n

u
x

1/γ



βen (u) −

r

 n
o
n
ne
Bn (uZn−k:n) d
H n (Zn−k:n u) .
k
k


Next we show that, Kin (x) = oP x−τ /γ , i = 1, 2. Indeed, let 0 < η < 1/8 and 0 < ǫ < 1 be
fixed and write

K1n (x) = −p−1

Since is H

(1)

Z

r

n
Bn (uZn−k:n)
k
1/γ
u
h (1)
i1/2−η
H (uZn−k:n)
i1/2−η n n
o
h (1)
H n (Zn−k:n u) .
d
× H (uZn−k:n)
k

Zn:n /Zn−k:n
x

βn (u) −

is a decreasing function, then over the interval x < u < Zn:n /Zn−k:n, we have
H

(1)

(Zn:n ) ≤ H

(1)

(uZn−k:n) ≤ H

(1)

(xZn−k:n ) ≤ H

for any x ≥ 1. Then with a large probability (1 − ǫ) p n1 ≤ H

(1)

(1)

(Zn−k:n ) ,

(uZn−k:n) ≤ (1 + ǫ) p nk , hence,

we may apply Gaussian approximations (5.21) to get
r
Z
i1/2−η n n
o
 ∞ 1/γ h (1)
n
−η
K1n (x) =
OP n
u
H (uZn−k:n)
H n (Zn−k:nu) .
d
k
k
x




(0)
e n (u) = −Bn 1 − H (0) (v) .
From their definitions above, βen (u) = −αn 1 − H (v) and B

Then by applying Gaussian approximations (5.20) we have
r
Z
o
i1/2−η n n
 ∞ 1/γ h (0)
n
−η
H (uZn−k:n)
H n (Zn−k:nu) .
OP n
u
K2n (x) =
d
k
k
x

40

By using the routine manipulations as used in Lemma 7.9, we end up with

Kin (x) = OP k −η x(2η−1/2)/γ+ǫ , i = 1, 2.

Since k → ∞ and 2η − 1/2 + τ < 0, then Kin (x) = oP x−τ /γ , i = 1, 2. Let us now consider
Rn11 (x) which may be rewritten into the sum of
r 

 n
Z
o
k −1/γ
1 ∞ 1/γ n
n
H n (Zn−k:n u) ,
Bn (uZn−k:n) − Bn p u
A1n (x) := −
d
u
p x
k
n
k

and

1
A2n (x) :=
p

Z

∞

u
x

1/γ

r 

 n
o
k
n
n e
−1/γ
en
Bn (uZn−k:n) − B
d
u
H n (Zn−k:nu) .
k
n
k

Recall that from (5.28) and write





 (1)
k −1/γ
k −1/γ
Bn p u
− Bn (uZn−k:n) = Bn p u
− Bn H (uZn−k:n) .
n
n

k
(1)
Once again, by applying this inequality (7.44) with x = p u−1/γ , y = H (uZn−k:n) and
n
(1)
k −1/γ
ρ = H (uZn−k:n) − p n u
, we may write
s
!


 (1)

k
k −1/γ
(1)
− Bn H (uZn−k:n) = OP
H (uZn−k:n) − p u−1/γ ,
Bn p u
n
n

uniformly over u ≥ 1. On the other hand, from inequality (7.42) , uniformly on u ≥ 1, we
k
k
(1)
write H (uZn−k:n) − p u−1/γ ≤ ǫp u−1/γ+ǫ , therefore
n
n
r  

 (1)


k −1/γ
n
Bn p u
− Bn H (uZn−k:n )
= oP u(ǫ−1/γ)/2 .
k
n

R∞
which implies that Ain (x) = oP (1) x u(ǫ−1/γ)/2 d nk H n (Zn−k:n u) , i = 1, 2. By using rou-

tine manipulations of assertion (i) of Lemma 7.5, with the fact (1 − ǫγ) /2 > τ, as ǫ ↓ 0, we

get Ain (x) = oP x−τ /γ , uniformly over x ≥ 1, as sought.


Lemma 7.13. For every 0 ≤ τ < 1/8, we have Rni (x) = oP x−τ /γ , i = 12, 13, 14.

Proof. Let fix 0 ≤ τ < 1/8 and recall that

 
Z n H n (Zn:n x) 
 rn  k
k
k
−1/γ
1/γ
∗
∗
−1
−1
Bn
(ℓn (s))
s
ds,
− Bn
(ℓn (s)) − s
Rn12 (x) = p
n
k
n
n
H n (Zn−k:n x)
k

where ℓn (s) is as in (5.29) . Note that H n (Zn:n x) = n−1 for x ≥ 1, therefore
r


 
Z n H n (Zn−k:n x)
k
k
n ∗ k
1/γ
−1/γ
−1
∗
−1
Bn
(ℓn (s)) − s
(ℓn (s))
s ds.
− Bn
|Rn12 (x)| ≤ p
k
n
n
k −1

41

Let 1/8 < η < 1/4 and small ǫ > 0, then from Lemma 7.7 and by using the mean value
theorem, we deduce that (ℓn (s))1/γ − s−1 = oP (s−1−η ) , uniformly over k −1 ≤ s < 1. On the

other hand, by using Lemma 7.12,
Z n H n (Zn−k:n x)
n
− 32 η+ 21
k
H n (Zn−k:n x)
s−1−η s−(η−1)/2 ds = OP (1)
.
Rn12 (x) = OP (1)
k
0

Note that H n (Zn−k:n x) = oP x−1/γ+ǫ , uniformly over x ≥ 1, it follows that Rn12 (x) =
1
3
∗
oP (1) x( 2 η− 2 )/γ+ǫ , with ǫ∗ ↓ 0. We verified that for 0 ≤ τ < 1/8 and 1/8 < η < 1/4, we have

(3η − 1) /2 + τ < 0, this means that Rn12 (x) = oP x−τ /γ . By using similar arguments, we

also end up with Rn13 (x) = oP x−τ /γ = Rn14 (x) as sought.


Lemma 7.14. For every 0 ≤ τ < 1/8, we have Rni (x) = oP x−τ /γ , i = 15, 16.
Proof. Let us fix 0 ≤ τ < 1/8 and recall that
r
 
Z n H n (Zn−k:n x)
k
n ∗ k
−1
−1
B
s ds.
Rn15 (x) = −p
s
k n n
x−1/γ

It is clear that
|Rn15 (x)| ≤ p

−1

Z

max( n
H n (Zn−k:n x),x−1/γ )
k
min( n
H n (Zn−k:n x),x−1/γ )
k

s

−1

r

n ∗
B
k n



k
s
n



ds.

From de Haan and Resnick (1998) (Propositions 2.2, 2.3, 3.1), see also de Haan and Resnick
(1993) (Proposition 4.1), there exists a Brownian bridge {B(t); t > 0} such that

o
√ nn
 √
xν/γ − 1
−1/γ
sup x
= oP (1) ,
− B x−1/γ − kA (h) x−1/γ
k
H n (Zn−k:n x) − x
k
νγ
x≥1

for every 4τ < ρ < 1/2. Note that supx≥1 xρ/γ B x−1/γ = sup0≤s≤1 s−ρ |B (s)| which is
ρ/γ

stochastically bounded (see for instance Lemma 3.2 in Einmahl et al. (2006)). On the other
√
√

hand, we have kA (h) = O (1) , it follows that kA (h) xρ/γ−1/γ xν/γ − 1 / (νγ) = O (1) ,
uniformly over x ≥ 1, therefore
sup xρ/γ
x≥1


n
H n (Zn−k:n x) − x−1/γ = OP k −1/2 = oP (1) .
k

Let 0 < ǫ < 1 be sufficiently small and set


ρ/γ n
−1/γ
Dn (ǫ; ρ) := sup x
H n (Zn−k:n x) − x
>ǫ .
k
x≥1
For a fixed c > 0, let cx := cx−τ /γ and write

P (|Rn15 (x)| > cx ) < P (|Rn15 (x)| > cx , Dn (ǫ; ρ)) + P (Dn (ǫ; ρ)) .

42

It is clear that
P (|Rn15 (x)| > cx , Dn (ǫ; ρ)) ≤ P p

−1

Z

x−1/γ +ǫx−ρ/γ

s
x−1/γ −ǫx−ρ/γ

−1

r

n ∗
B
k n



k
s
n



ds > cx

!

,

which, by using Markov inequality, is
r
 
Z x−1/γ +ǫx−ρ/γ
k
n
−1 −2
−1
∗
≤ p cx
s
E Bn
s ds.
k
n
x−1/γ −ǫx−ρ/γ
It is easy to verify that E |Bn∗ (s)| ≤ s1/2 , it follows that the previous quantity is
Z x−1/γ +ǫx−ρ/γ
n
1/2
1/2 o
−1 −2
s−1/2 ds = 2p−1 c−2 x2τ /γ x−1/γ + ǫx−ρ/γ
− x−1/γ − ǫx−ρ/γ
.
≤ p cx
x−1/γ −ǫx−ρ/γ

It is ready to check, in view for the mean value theorem, that the expression between two

brackets equals o x−ρ/(2γ) . This implies that
Z x−1/γ +ǫ

−1 −2
p cx
s−1/2 ds = o x(2τ −ρ/2)/γ = o (1) ,
x−1/γ −ǫ

because 4τ < ρ < 1/2, therefore P (|Rn14 (x)| > cx , Dn (ǫ; ρ)) = o (1) . Since P (Dn (ǫ; ρ)) =

o (1) , hence Rn15 (x) = oP x−τ /γ , uniformly over x ≥ 1. By using similar arguments, we

also show that Rn16 (x) = oP x−τ /γ , that we omit further details.


