# On the extreme order statistics for stationary Gaussian sequences subject to
random missing observations ††thanks: Research supported by Innovation of
Jiaxing City: a program to support the talented persons and Project of new
economy research center of Jiaxing City (No. WYZB202254).

Yuan Fang1,2, Zhongquan Tan2111 E-mail address<EMAIL_ADDRESS>

1\. Department of Mathematic, Zhejiang Normal University, Jinhua 321004, PR
China
2\. College of Data Science, Jiaxing University, Jiaxing 314001, PR China

> Abstract: Let $\mathbf{X}=\\{X_{n}\\}_{n\geq 1}$ be a sequence of stationary
> Gaussian variables and suppose that only some of the random variables from
> $\mathbf{X}$ can be observed. In this paper, by studying the limiting
> properties of multidimensional exceedance point processes for $\mathbf{X}$,
> we derived the joint limit distribution of extreme order statistics for the
> Gaussian sequence $\mathbf{X}$ and its observed ones. The joint limit
> distribution of the locations and heights of the maxima for the Gaussian
> sequence $\mathbf{X}$ and its observed ones are also obtained.
>
> Key Words: extreme order statistics; exceedance point processes; locations;
> Gaussian sequences; missing observations
>
> AMS Classification: Primary 60G70; secondary 60G15

##  1 Introduction

Let $\mathbf{X}=\\{X_{n}\\}_{n\geq 1}$ be a sequence of stationary random
variables and suppose that only some of the random variables from
$\mathbf{X}=\\{X_{n}\\}_{n\geq 1}$ can be observed. Let
$\boldsymbol{\varepsilon}=\\{\varepsilon_{n},n\geq 1\\}$ be a sequence of
Bernoulli random variables which indicates which variables are observed and is
independent of $\mathbf{X}=\\{X_{n}\\}_{n\geq 1}$. Let
$S_{n}=\sum_{k=1}^{n}\varepsilon_{k}$ be the number of observed random
variables from the set $\left\\{X_{1},X_{2},\cdots,X_{n}\right\\}$ and assume
that

$\frac{S_{n}}{n}\overset{p}{\longrightarrow}\lambda,\ \ \ \
n\rightarrow\infty,\ $

for some random or nonrandom variable $\lambda$.

In application fields, it is very interesting to investigate the asymptotic
relation between the maxima of the sequence and the maxima of the observed
ones. More precisely, we are interested in the following asymptotic

$\displaystyle\lim_{n\to\infty}P\left\\{\widetilde{M}_{n}\leq
u_{n}\left(x\right),M_{n}\leq u_{n}\left(y\right)\right\\}=H(x,y,\lambda),$
(1)

where $M_{n}=\max\left\\{X_{k},1\leq k\leq n\right\\}$,
$\widetilde{M}_{n}=\max\left\\{X_{k},\varepsilon_{k}=1,1\leq k\leq n\right\\}$
and $H$ is a nondegenerate distribution function.

If $\lambda$ is a constant, Mladenović and Piterbarg (2006) first proved that
(1) holds with

$H(x,y,\lambda)=G^{\lambda}\left(x\right)G^{1-\lambda}\left(y\right)$

for any $x<y\in\mathbb{R}$ under a type of long dependent condition
$D\left(u_{n},v_{n}\right)$ and a local dependent condition
${D}^{\prime}(u_{n})$. From then on, the same problem was extended to some
other cases such as the Gaussian cases in Cao and Peng (2011) and Peng et al.
(2010); the autoregressive process in Glavas̆ et al. (2017); the linear
process in Glavas̆ and Mladenović (2020); nonstationary random fields in Panga
and Pereira (2018); and the almost sure limit theorem in Tong and Peng (2011)
and Tan and Wang (2012).

When $\lambda$ is a random variable, Krajka (2011) showed that (1) still holds
with

$\displaystyle
H\left(\lambda,x,y\right)=E\left[G^{\lambda}\left(x\right)G^{1-\lambda}\left(y\right)\right].$

Hashorva et al. (2013) investigated this problem for Gaussian sequence
$\mathbf{X}=\\{X_{n}\\}_{n\geq 1}$ with correlation function
$r_{n}=Cov\left(X_{1},X_{n+1}\right)$ satisfying the condition

$\displaystyle\lim_{n\to\infty}r_{n}\ln{n}=\gamma\in[0,\infty)$

and proved that (1) still holds with

$\displaystyle
H\left(\lambda,x,y\right)=E\left(\int_{-\infty}^{+\infty}\exp\left(-\lambda
g(x,z)-\left(1-\lambda\right)g(y,z)\right)\right)d\Phi\left(z\right),$

where $g(x,z)=\exp(-x-\gamma+\sqrt{2\gamma}z)$ and $\Phi(\cdot)$ denotes the
standard normal distribution function.

In practice, we may be interested in the asymptotic relation between the
extreme order statistics of some sequences and their observed ones. It is
well-known that the related limit properties of extreme order statistics can
be derived directly from the associated exceedance point process (cf.
Leadbetter et al. (1983)). Peng and Tong (2010) and Peng et al. (2019) studied
the exceedance point process formed by stationary Gaussian sequences and their
observed ones, but it seems that the asymptotic relation between the extreme
order statistics of the sequence and its observed ones can not be derived from
their main results, since they only considered the one-dimensional exceedance
point process. For more related studies on exceedance point process, we refer
to Hsing et al. (1988), Ferreira (1999), Temido (2000), Novak (2002), Ferreira
and Pereira (2012), Hashorva et al. (2014), Xiao et al. (2019), Liu and Tan
(2022) and the references therein. As far as we know, the study on the
exceedance point process formed by stationary Gaussian sequences and their
observed ones is far from complete.

In this paper, we continue to study the joint limiting properties of
exceedance point processes formed by a stationary Gaussian sequence and its
observed ones. Applying the joint limiting properties of exceedance point
processes, we derived the joint asymptotic distribution between the extreme
order statistics of the sequence and its observed one. Some limit results on
the locations and heights of maxima for the stationary Gaussian sequence and
its observed ones are also derived.

##  2 Main results

We first give the definition of the exceedance point process formed by
exceedances of some levels by a sequence of stationary Gaussian variables
$\mathbf{X}=\\{X_{n}\\}_{n\geq 1}$. The levels used in this paper are defined
by $u_{n}(x)=a_{n}^{-1}x+b_{n}$ for any $x\in\mathbb{R}$, where

$a_{n}=(2\ln n)^{1/2}\ \mbox{and}\ \ b_{n}=a_{n}-\frac{\ln\ln
n+\ln(4\pi)}{2a_{n}}.$

Definition 2.1. For a Borel set $B\subset\left(0,1\right]$, the exceedance
point processes formed by $\mathbf{X}=\\{X_{n}\\}_{n\geq 1}$ are defined as

$\displaystyle N_{n}^{x}\left(B\right)=\sum_{\frac{j}{n}\in
B}I_{\left\\{X_{j}>u_{n}(x)\right\\}},$

where $I_{\\{\cdot\\}}$ denotes the indicator function.

Definition 2.2. For a Borel set $B\subset\left(0,1\right]$, the exceedance
point processes for observed and missed random variables from
$\mathbf{X}=\\{X_{n}\\}_{n\geq 1}$ are defined as

$\displaystyle\widetilde{N}_{n}^{x}\left(B\right)=\sum_{\frac{j}{n}\in
B}I_{\left\\{X_{j}>u_{n}(x),\varepsilon_{j}=1\right\\}}\ \ \mbox{and}\ \
\widehat{N}_{n}^{x}\left(B\right)=\sum_{\frac{j}{n}\in
B}I_{\left\\{X_{j}>u_{n}(x),\varepsilon_{j}=0\right\\}},$

respectively.

Now, we state our main results.

Theorem 2.1. Let $\mathbf{X}=\\{X_{\mathbf{n}}\\}_{\mathbf{n}\geq\mathbf{1}}$
be a sequence of stationary standard (mean 0, variance 1) Gaussian variables
with covariance functions $r_{n}=Cov\left(X_{1},X_{n+1}\right)$ satisfying

$\displaystyle\lim_{n\to\infty}r_{n}\ln{n}=\gamma\in[0,\infty).$ (2)

Suppose that $\boldsymbol{\varepsilon}=\\{\varepsilon_{n},n\geq 1\\}$ is a
sequence of Bernoulli random variables which indicates which variables in
$\mathbf{X}=\\{X_{\mathbf{n}}\\}_{\mathbf{n}\geq\mathbf{1}}$ are observed and
is independent of $\mathbf{X}=\\{X_{n}\\}_{n\geq 1}$. Let
$S_{n}=\sum_{k=1}^{n}\varepsilon_{k}$ be the number of observed random
variables from the set $\left\\{X_{1},X_{2},\cdots,X_{n}\right\\}$ and satisfy

$\frac{S_{n}}{n}\overset{p}{\longrightarrow}\lambda,\ \ as\ \
n\rightarrow\infty,\ $ (3)

where $\lambda\in[0,1]$ a.s. is a random variable. Then, for some positive
integers $k,l$, the point process
$(\widetilde{N}_{n}^{x_{1}},\ldots,\widetilde{N}_{n}^{x_{k}},\widehat{N}_{n}^{y_{1}},\ldots,\widehat{N}_{n}^{y_{l}})$
converges in distribution to the point process
$(\widetilde{N}^{x_{1}},\ldots,\widetilde{N}^{x_{k}},\widehat{N}^{y_{1}},\ldots,\widehat{N}^{y_{l}})$
on $\left(0,1\right]^{k+l}$, where $\widetilde{N}^{x}$ and $\widehat{N}^{y}$,
condition on $(\lambda,\xi)$, are Poisson processes with conditional intensity
$\lambda g(x,\xi)$ and $(1-\lambda)g(y,\xi)$, respectively. Here $\xi$ is a
standard Gaussian random variable independent of $\lambda$ and
$g(x,\xi)=\exp(-x-\gamma+\sqrt{2\gamma}\xi)$.

Remark 2.1. The point processes $\widetilde{N}_{n}^{x}$ and
$\widehat{N}_{n}^{y}$ are asymptotically dependent, but, condition on
$(\lambda,\xi)$, they are asymptotically independent. For the weakly dependent
case, i.e., the case $\gamma=0$, the dependence is captured by the random
variable $\lambda$, while for the strongly dependent case, i.e., the case
$\gamma>0$, the dependence is captured by both $\lambda$ and $\xi$.

Let $B_{i}\subset\left(0,1\right]$ be Borel sets whose boundary have zero
Lebesgue measure. For application, we are interested in the joint distribution
of
$(\widetilde{N}^{x_{1}}(B_{1}),\ldots,\widetilde{N}^{x_{k}}(B_{k}),\widehat{N}^{y_{1}}(B_{k+1}),\ldots,\widehat{N}^{y_{l}}(B_{k+l}))$.
Since it’s expression is very complicated, we only give a simple case here.

Corollary 2.1. Suppose that the conditions of Theorem 2.1 hold. Let
$B\subset\left(0,1\right]$ be a Borel set whose boundary has zero Lebesgue
measure.
If $x>y$, then, for $0\leq k_{1}\leq k_{3}$, $0\leq k_{2}\leq k_{4}$, we have

$\displaystyle
P(\widetilde{N}^{x}(B)=k_{1},\widehat{N}^{x}(B)=k_{2},\widetilde{N}^{y}(B)=k_{3},\widehat{N}^{y}(B)=k_{4})$
$\displaystyle=E\int_{-\infty}^{+\infty}\bigg{\\{}\frac{[\lambda
m(B)g(x,z)]^{k_{1}}}{k_{1}!}\frac{[\lambda
m(B)(g(y,z)-g(x,z))]^{k_{3}-k_{1}}}{(k_{3}-k_{1})!}\exp(-\lambda m(B)g(y,z))$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\times\frac{[(1-\lambda)m(B)g(x,z)]^{k_{2}}}{k_{2}!}\frac{[(1-\lambda)m(B)(g(y,z)-g(x,z))]^{k_{4}-k_{2}}}{(k_{4}-k_{2})!}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\times\exp(-(1-\lambda)m(B)g(y,z))\bigg{\\}}\mathrm{d}\Phi(z);$ (4)

if $x\leq y$, for $k_{1}\geq k_{3}\geq 0$, $k_{2}\geq k_{4}\geq 0$, we have

$\displaystyle
P(\widetilde{N}^{x}(B)=k_{1},\widehat{N}^{x}(B)=k_{2},\widetilde{N}^{y}(B)=k_{3},\widehat{N}^{y}(B)=k_{4})$
$\displaystyle=E\int_{-\infty}^{+\infty}\bigg{\\{}\frac{[\lambda
m(B)g(y,z)]^{k_{3}}}{k_{3}!}\frac{[\lambda
m(B)(g(x,z)-g(y,z))]^{k_{1}-k_{3}}}{(k_{1}-k_{3})!}\exp(-\lambda m(B)g(x,z))$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\times\frac{[(1-\lambda)m(B)g(y,z)]^{k_{4}}}{k_{4}!}\frac{[(1-\lambda)m(B)(g(x,z)-g(y,z))]^{k_{2}-k_{4}}}{(k_{2}-k_{4})!}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\times\exp(-(1-\lambda)m(B)g(x,z))\bigg{\\}}\mathrm{d}\Phi(z),$ (5)

where $m(B)$ denotes the Lebesgue’s measure of set $B$.

Noting the $N_{n}^{x}=\widetilde{N}_{n}^{x}+\widehat{N}_{n}^{x}$, by the
continuous mapping theorem, we have the following result which will be used to
derive the joint limit distribution for extreme order statistics for observed
and missed variables.

Corollary 2.2. Under the conditions of Theorem 2.1, (i). the point process
$(\widetilde{N}_{n}^{x},\widehat{N}_{n}^{y},N_{n}^{z})$ converges in
distribution to the point process
$(\widetilde{N}^{x},\widehat{N}^{y},\widetilde{N}^{z}+\widehat{N}^{z})$ on
$\left(0,1\right]^{3}$; (ii). the point process
$(\widetilde{N}_{n}^{x},N_{n}^{y})$ converges in distribution to the point
process $(\widetilde{N}^{x},\widetilde{N}^{y}+\widehat{N}^{y})$ on
$\left(0,1\right]^{2}$; (iii). the point process
$(\widehat{N}_{n}^{x},N_{n}^{y})$ converges in distribution to the point
process $(\widehat{N}^{x},\widetilde{N}^{y}+\widehat{N}^{y})$ on
$\left(0,1\right]^{2}$, where $\widetilde{N}^{x}+\widehat{N}^{x}$ is a Cox
process (see Page 135 Leadbetter et al. (1983) for the definition), i.e., a
conditional Poisson process with conditional intensity $g(x,\xi)$.

Applying Corollary 2.2, we derived the following useful results for the
extreme order statistics. Let ${M}_{n}^{(m)}$ be the $m$-th maximum of
$\left\\{X_{j}:j\in\mathbb{N}_{n}\right\\}$ with
$\mathbb{N}_{n}=\\{1,2,\ldots,n\\}$. Define

$\widetilde{M}_{n}^{(k)}=\left\\{\begin{matrix}k\mbox{-th}\
\max\\{X_{j}:\varepsilon_{j}=1,j\in\mathbb{N}_{n}\\},&\mbox{if}\
\sum_{j\in\mathbb{N}_{n}}\varepsilon_{j}>k,&\\\ &&\\\ -\infty,\ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \mbox{otherwise},\end{matrix}\right.$

and

$\widehat{M}_{n}^{(l)}=\left\\{\begin{matrix}l\mbox{-th}\
\max\\{X_{j}:\varepsilon_{j}=0,j\in\mathbb{N}_{n}\\},&\mbox{if}\
\sum_{j\in\mathbb{N}_{n}}\varepsilon_{j}<n-l,&\\\ &&\\\ -\infty,\ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \mbox{otherwise}.\end{matrix}\right.$

Theorem 2.2. Suppose that the conditions of Theorem 2.1 hold.
(i). For any $x,y\in\mathbb{R}$,

$\displaystyle\lim_{n\to\infty}P\left\\{\widetilde{M}_{n}^{(k)}\leq
u_{n}\left(x\right),\widehat{M}_{n}^{(l)}\leq u_{n}\left(y\right)\right\\}$
$\displaystyle=\sum_{s=0}^{k-1}\sum_{t=0}^{l-1}E\int_{-\infty}^{+\infty}\left\\{\frac{[\lambda
g(x,z)]^{s}}{s!}\exp(-\lambda
g(x,z))\frac{[(1-\lambda)g(y,z)]^{t}}{t!}\exp(-(1-\lambda)g(y,z))\right\\}\mathrm{d}\Phi(z).$
(6)

(ii). For $x<y\in\mathbb{R}$,

$\displaystyle\lim_{n\to\infty}P\left\\{\widetilde{M}_{n}^{(k)}\leq
u_{n}\left(x\right),{M}_{n}^{(m)}\leq u_{n}\left(y\right)\right\\}$
$\displaystyle=\sum_{t=0}^{m-1}\sum_{i=0}^{m-t-1}\sum_{j=i}^{k-1}E\int_{-\infty}^{+\infty}\left\\{\frac{[(1-\lambda)g(y,z)]^{t}}{t!}\exp(-(1-\lambda)g(y,z))\right.$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \times\left.\frac{[\lambda
g(y,z)]^{i}}{i!}\frac{[\lambda(g(x,z)-g(y,z))]^{j-i}}{(j-i)!}\exp(-\lambda
g(x,z))\right\\}\mathrm{d}\Phi(z);$ (7)

for $x\geq y\in\mathbb{R}$,

$\displaystyle\lim_{n\to\infty}P\left\\{\widetilde{M}_{n}^{(k)}\leq
u_{n}\left(x\right),{M}_{n}^{(m)}\leq u_{n}\left(y\right)\right\\}$
$\displaystyle=\sum_{t=0}^{m-1}\sum_{i=0}^{m-t-1}\sum_{j=0}^{\min\\{i,k-1\\}}E\int_{-\infty}^{+\infty}\left\\{\frac{[(1-\lambda)g(y,z)]^{t}}{t!}\exp(-g(y,z))\right.$
$\displaystyle\ \ \ \ \ \ \ \ \ \times\left.\frac{[\lambda
g(x,z)]^{j}}{j!}\frac{[\lambda(g(y,z)-g(x,z))]^{i-j}}{(i-j)!}\right\\}\mathrm{d}\Phi(z).$
(8)

(iii). For $x<y\in\mathbb{R}$,

$\displaystyle\lim_{n\to\infty}P(\widehat{M}_{n}^{(l)}\leq
u_{n}(x),M_{n}^{(m)}\leq u_{n}(y))$
$\displaystyle=\sum_{t=0}^{m-1}\sum_{i=0}^{m-t-1}\sum_{j=i}^{l-1}E\int_{-\infty}^{+\infty}\left\\{\frac{[\lambda
g(y,z)]^{t}}{t!}\exp(-\lambda g(y,z))\right.$ $\displaystyle\ \ \ \ \ \ \ \ \
\ \ \ \ \
\times\left.\frac{[(1-\lambda)g(y,z)]^{i}}{i!}\frac{[(1-\lambda)(g(x,z)-g(y,z))]^{j-i}}{(j-i)!}\exp(-(1-\lambda)g(x,z))\right\\}\mathrm{d}\Phi(z);$
(9)

for $x\geq y\in\mathbb{R}$,

$\displaystyle\lim_{n\to\infty}P(\widehat{M}_{n}^{(l)}\leq
u_{n}(x),M_{n}^{(m)}\leq u_{n}(y))$
$\displaystyle=\sum_{t=0}^{m-1}\sum_{i=0}^{m-t-1}\sum_{j=0}^{\min\\{i,l-1\\}}E\int_{-\infty}^{+\infty}\left\\{\frac{[\lambda
g(y,z)]^{t}}{t!}\exp(-g(y,z))\right.$ $\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \
\ \ \ \ \ \
\times\left.\frac{[(1-\lambda)g(x,z)]^{j}}{j!}\frac{[(1-\lambda)(g(y,z)-g(x,z))]^{i-j}}{(i-j)!}\right\\}\mathrm{d}\Phi(z).$
(10)

Remark 2.2. The results presented in Theorem 2.2 seem to be proven for the
first time, which extend some results of Mladenović and Piterbarg (2006), Peng
and Tong (2010), Krajka (2011) and Hashorva et al. (2013) under the Gaussian
setting.

Let $L_{n}$, $\widetilde{L}_{n}$ and $\widehat{L}_{n}$ be the location of
$M_{n}$, $\widetilde{M}_{n}$ and $\widehat{M}_{n}$, respectively. The
following theorem derives the joint limit distribution for the locations and
heights for the maxima from the Gaussian sequence and its observed ones, which
extends Theorem 2.2 of Peng et al. (2019).

Theorem 2.3. Suppose that the conditions of Theorem 2.1 hold.
(i). For any $0<s,t<1$ and $x,y$, we have

$\displaystyle\lim_{n\rightarrow\infty}P\left(\frac{1}{n}\widetilde{L}_{n}\leq
s,\frac{1}{n}\widehat{L}_{n}\leq t,\widetilde{M}_{n}\leq
u_{n}(x),\widehat{M}_{n}\leq u_{n}(y)\right)$
$\displaystyle=stE\int_{-\infty}^{+\infty}\exp(-\lambda
g(x,z))\exp(-(1-\lambda)g(y,z))\mathrm{d}\Phi(z).$ (11)

(ii). For any $0<s,t<1$ and $x\leq y$, we have

$\displaystyle\lim_{n\rightarrow\infty}P\left(\frac{1}{n}\widetilde{L}_{n}\leq
s,\frac{1}{n}L_{n}\leq t,\widetilde{M}_{n}\leq u_{n}(x),M_{n}\leq
u_{n}(y)\right)$ $\displaystyle=stE\int_{-\infty}^{+\infty}\exp(-\lambda
g(x,z))\exp(-(1-\lambda)g(y,z))-\lambda\exp(-g(x,z))\mathrm{d}\Phi(z)$
$\displaystyle\ \ \
+\min\\{s,t\\}E\int_{-\infty}^{+\infty}\lambda\exp(-g(x,z))\mathrm{d}\Phi(z).$
(12)

(iii). For any $0<s,t<1$ and $x\leq y$, we have

$\displaystyle\lim_{n\rightarrow\infty}P\left(\frac{1}{n}\widehat{L}_{n}\leq
s,\frac{1}{n}L_{n}\leq t,\widehat{M}_{n}\leq u_{n}(x),M_{n}\leq
u_{n}(y)\right)$
$\displaystyle=stE\int_{-\infty}^{+\infty}\exp(-(1-\lambda)g(x,z))\exp(-\lambda
g(y,z))-(1-\lambda)\exp(-g(x,z))\mathrm{d}\Phi(z)$ $\displaystyle\ \ \
+\min\\{s,t\\}E\int_{-\infty}^{+\infty}(1-\lambda)\exp(-g(x,z))\mathrm{d}\Phi(z).$
(13)

Corollary 2.3. Suppose that the conditions of Theorem 2.1 hold.
(i). For any $0<s,t<1$, we have

$\displaystyle\lim_{n\rightarrow\infty}P\left(\frac{1}{n}\widetilde{L}_{n}\leq
s,\frac{1}{n}\widehat{L}_{n}\leq t\right)=st.$ (14)

(ii). For any $0<s,t<1$, we have

$\displaystyle\lim_{n\rightarrow\infty}P\left(\frac{1}{n}\widetilde{L}_{n}\leq
s,\frac{1}{n}L_{n}\leq t\right)=st[1-E(\lambda)]+\min\\{s,t\\}E(\lambda).$
(15)

(iii). For any $0<s,t<1$ and $x\leq y$, we have

$\displaystyle\lim_{n\rightarrow\infty}P\left(\frac{1}{n}\widehat{L}_{n}\leq
s,\frac{1}{n}L_{n}\leq t\right)=stE(\lambda)+\min\\{s,t\\}[1-E(\lambda)].$
(16)

Remark 2.3. Corollary 2.3 shows that $\widetilde{L}_{n}$ and $\widehat{L}_{n}$
are asymptotically independent, but $L_{n}$ is asymptotically dependent of
$\widetilde{L}_{n}$ and $\widehat{L}_{n}$.

##  3 Auxiliary results

In this section, we state and prove several lemmas which will be used in the
proofs of our main results. We first borrow some notation from Krajka (2011).
Let $\boldsymbol{\alpha}=\\{\alpha_{n},n\geq 1\\}$ be nonrandom sequence
taking values in $\\{0,1\\}$. For a random variable $\lambda$ such that
$0\leq\lambda\leq 1$ a.s., set

$B_{r,t}=\begin{Bmatrix}&w:\lambda(w)\in\left\\{\begin{matrix}\left[0,\frac{1}{2^{t}}\right],&r=0;&\\\
\left(\frac{r}{2^{t}},\frac{r+1}{2^{t}}\right],&0<r\leq
2^{t}-1\end{matrix}\right.&\end{Bmatrix},$
$B_{r,t,\boldsymbol{\alpha},n}=\\{w:\varepsilon_{j}(w)=\alpha_{j},1\leq j\leq
n\\}\cap B_{r,t}.$

For the arbitrary random or nonrandom sequence
$\boldsymbol{\beta}=\\{\beta_{n},n\geq 1\\}$ of 0 and 1 and subset
$I\subset\mathbb{N}_{n}$, define

$\widetilde{M}_{n}(X,I,\boldsymbol{\beta})=\left\\{\begin{matrix}\max\\{X_{j}:\beta_{j}=1,j\in
I\\},&\mbox{if}\ \max_{j\in I}\beta_{j}>0,&\\\ &&\\\ -\infty,\ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \mbox{otherwise},\end{matrix}\right.$

and

$\widehat{M}_{n}(X,I,\boldsymbol{\beta})=\left\\{\begin{matrix}\max\\{X_{j}:\beta_{j}=0,j\in
I\\},&\mbox{if}\ \min_{j\in I}\beta_{j}=0,&\\\ &&\\\ -\infty,\ \ \ \ \ \ \ \ \
\ \ \ \ \ \ \mbox{otherwise}.\end{matrix}\right.$

For any integer $k>0$, let $c_{i},d_{i},i=1,2,\ldots,k$ be constants such that
$0<c_{1}<d_{1}\leq c_{2}<\cdots\leq c_{k}<d_{k}\leq 1$. Let
$A_{i}=\left\\{\lfloor nc_{i}\rfloor+1,\lfloor nc_{i}\rfloor+2,\cdots,\lfloor
nd_{i}\rfloor\right\\}$, $1\leq i\leq k$, where $\lfloor x\rfloor$ denotes the
integral part of $x$. Let also $x_{i}$ and $y_{i}$, $i=1,2,\ldots,k$, be two
sequences of real numbers.

Lemma 3.1. Let $\\{Z_{j}\\}_{j\geq 1}$ be a sequence of independent standard
Gaussian random variables. Under the assumptions of Theorem 2.1, we have

$\displaystyle\bigg{|}P\left(\bigcap_{i=1}^{k}\left\\{\widetilde{M}_{n}(X,A_{i},\boldsymbol{\alpha})\leq
u_{n}(x_{i}),\widehat{M}_{n}(X,A_{i},\boldsymbol{\alpha})\leq
u_{n}(y_{i})\right\\}\right)$ $\displaystyle\ \ \ \ \ \ \ \ \ \
-\int_{-\infty}^{+\infty}\prod_{i=1}^{k}P\left\\{\widetilde{M}_{n}(Z,A_{i},\boldsymbol{\alpha})\leq
u_{n}(x_{i},z),\widehat{M}_{n}(Z,A_{i},\boldsymbol{\alpha})\leq
u_{n}(y_{i},z)\right\\}\mathrm{d}\Phi(z)\bigg{|}\rightarrow 0,$ (17)

as $n\rightarrow\infty$, where
$u_{n}(x,z)=(1-\rho_{n})^{-\frac{1}{2}}(u_{n}(x)-\rho_{n}^{\frac{1}{2}}z)$
with $\rho_{n}=\gamma/\ln{n}$.

Proof: Define
$Y_{j}=(1-\rho_{n})^{\frac{1}{2}}Z_{j}+\rho_{n}^{\frac{1}{2}}\xi$, where $\xi$
is a standard Gaussian random variable. By Normal Comparision Lemma (see e.g.,
Leadbetter et al. (1983)), we have

$\displaystyle\bigg{|}P\left(\bigcap_{i=1}^{k}\left\\{\widetilde{M}_{n}(X,A_{i},\boldsymbol{\alpha})\leq
u_{n}(x_{i}),\widehat{M}_{n}(X,A_{i},\boldsymbol{\alpha})\leq
u_{n}(y_{i})\right\\}\right)$ $\displaystyle\ \ \ \
-P\left(\bigcap_{i=1}^{k}\left\\{\widetilde{M}_{n}(Y,A_{i},\boldsymbol{\alpha})\leq
u_{n}(x_{i}),\widehat{M}_{n}(Y,A_{i},\boldsymbol{\alpha})\leq
u_{n}(y_{i})\right\\}\right)\bigg{|}$ $\displaystyle\leq Kn\sum_{i=1}^{\lfloor
nd_{k}\rfloor}|r_{j}-\rho_{n}|\exp\left(-\frac{(\min\\{u_{n}(x_{i}),u_{n}(y_{i}),i=1,2,\ldots,k\\})^{2}}{1+\omega_{j}}\right),$
(18)

where $\omega_{j}=\max\\{|r_{j}|,\rho_{n}\\}$ and $K$ is some constant,
depending only on $\gamma$. It is directly to check that

$\displaystyle
P\left(\bigcap_{i=1}^{k}\left\\{\widetilde{M}_{n}(Y,A_{i},\boldsymbol{\alpha})\leq
u_{n}(x_{i}),\widehat{M}_{n}(Y,A_{i},\boldsymbol{\alpha})\leq
u_{n}(y_{i})\right\\}\right)$
$\displaystyle=\int_{-\infty}^{+\infty}P\left(\bigcap_{i=1}^{k}\left\\{\widetilde{M}_{n}(Z,A_{i},\boldsymbol{\alpha})\leq
u_{n}(x_{i},z),\widehat{M}_{n}(Z,A_{i},\boldsymbol{\alpha})\leq
u_{n}(y_{i},z)\right\\}\right)\mathrm{d}\Phi(z)$
$\displaystyle=\int_{-\infty}^{+\infty}\prod_{i=1}^{k}P\left\\{\widetilde{M}_{n}(Z,A_{i},\boldsymbol{\alpha})\leq
u_{n}(x_{i},z),\widehat{M}_{n}(Z,A_{i},\boldsymbol{\alpha})\leq
u_{n}(y_{i},z)\right\\}\mathrm{d}\Phi(z).$ (19)

Applying Lemma 6.4.1 of Leadbetter et al. (1983), (3) and (3), we obtain the
desired result.

The following lemma is from Krajka (2011).

Lemma 3.2. Let $d(X,Y)$ stand for Ky Fan metric,
$d(X,Y)=\inf\\{\varepsilon:P[|X-Y|>\varepsilon]<\varepsilon\\}$.

(a). For arbitrary positive integers $s$, $m$, we have

$d\left(\frac{S_{ms}-S_{m(s-1)}}{m},\lambda\right)\leq(2s-1)\left[d\left(\frac{S_{ms}}{ms},\lambda\right)+d\left(\frac{S_{m(s-1)}}{m(s-1)},\lambda\right)\right].$

(b). If $\\{X_{n},n\geq 1\\}$ and $\\{Y_{n},n\geq 1\\}$ are such that
$|X_{n}-Y_{n}|<1$ a.s. then

$\displaystyle E|X_{n}-Y_{n}|\leq 2d(X_{n},Y_{n}).$

Lemma 3.3. Under the conditions of Theorem 2.1, we have

$\displaystyle
P\left(\bigcap_{i=1}^{k}\left\\{\widetilde{M}_{n}(X,A_{i},\boldsymbol{\varepsilon})\leq
u_{n}(x_{i}),\widehat{M}_{n}(X,A_{i},\boldsymbol{\varepsilon})\leq
u_{n}(y_{i})\right\\}\right)$ $\displaystyle\longrightarrow
E\int_{-\infty}^{+\infty}\prod_{i=1}^{k}\exp\left[-(d_{i}-c_{i})(\lambda
g(x_{i},z)+(1-\lambda)g(y_{i},z))\right]\mathrm{d}\Phi(z),$ (20)

as $n\rightarrow\infty$.

Proof: Let $t$ be a fixed integer and
$m_{i}=\left\lfloor\frac{nd_{i}-nc_{i}}{t}\right\rfloor,i=1,2,\ldots,k$. Put

$K_{s,i}=\\{j:\lfloor nc_{i}\rfloor+(s-1)m_{i}+1\leq j\leq\lfloor
nc_{i}\rfloor+sm_{i}\\},\ \ \ \ 1\leq s\leq t,$

and

$I_{s,i}=\\{j:\lfloor nc_{i}\rfloor+(s-1)m_{i}+1\leq j\leq\lfloor
nc_{i}\rfloor+sm_{i}-l\\},\ \ J_{s,i}=K_{s,i}\setminus I_{s,i}.$

Obviously,

$\displaystyle\bigg{|}P\left(\bigcap_{i=1}^{k}\left\\{\widetilde{M}_{n}(X,A_{i},\boldsymbol{\varepsilon})\leq
u_{n}(x_{i}),\widehat{M}_{n}(X,A_{i},\boldsymbol{\varepsilon})\leq
u_{n}(y_{i})\right\\}\right)$ $\displaystyle\ \ \ \ \
-E\int_{-\infty}^{+\infty}\prod_{i=1}^{k}\exp\left[-(d_{i}-c_{i})(\lambda
g(x_{i},z)+(1-\lambda)g(y_{i},z))\right]\mathrm{d}\Phi(z)\bigg{|}$
$\displaystyle\leq\sum_{r=0}^{2^{t}-1}\sum_{\boldsymbol{\alpha}\in\\{0,1\\}^{n}}E\bigg{|}P\left(\bigcap_{i=1}^{k}\left\\{\widetilde{M}_{n}(X,A_{i},\boldsymbol{\alpha})\leq
u_{n}(x_{i}),\widehat{M}_{n}(X,A_{i},\boldsymbol{\alpha})\leq
u_{n}(y_{i})\right\\}\right)$ $\displaystyle\ \ \ \ \
-\int_{-\infty}^{+\infty}\prod_{i=1}^{k}\exp\left[-(d_{i}-c_{i})(\lambda
g(x_{i},z)+(1-\lambda)g(y_{i},z))\right]\mathrm{d}\Phi(z)\bigg{|}I_{\\{B_{r,t,\boldsymbol{\alpha},n}\\}}$
$\displaystyle\leq\sum_{r=0}^{2^{t}-1}\sum_{\boldsymbol{\alpha}\in\\{0,1\\}^{n}}E\bigg{|}P\left(\bigcap_{i=1}^{k}\left\\{\widetilde{M}_{n}(X,A_{i},\boldsymbol{\alpha})\leq
u_{n}(x_{i}),\widehat{M}_{n}(X,A_{i},\boldsymbol{\alpha})\leq
u_{n}(y_{i})\right\\}\right)$ $\displaystyle\ \ \ \ \
-\int_{-\infty}^{+\infty}\prod_{i=1}^{k}P\left\\{\widetilde{M}_{n}(Z,A_{i},\boldsymbol{\alpha})\leq
u_{n}(x_{i},z),\widehat{M}_{n}(Z,A_{i},\boldsymbol{\alpha})\leq
u_{n}(y_{i},z)\right\\}\mathrm{d}\Phi(z)\bigg{|}I_{\\{B_{r,t,\boldsymbol{\alpha},n}\\}}$
$\displaystyle+\sum_{r=0}^{2^{t}-1}\sum_{\boldsymbol{\alpha}\in\\{0,1\\}^{n}}E\bigg{|}\int_{-\infty}^{+\infty}\prod_{i=1}^{k}\exp\left[-(d_{i}-c_{i})(\lambda
g(x_{i},z)+(1-\lambda)g(y_{i},z))\right]\mathrm{d}\Phi(z)$ $\displaystyle\ \ \
\
-\int_{-\infty}^{+\infty}\prod_{i=1}^{k}P\left(\widetilde{M}_{n}(Z,A_{i},\boldsymbol{\alpha})\leq
u_{n}(x_{i},z),\widehat{M}_{n}(Z,A_{i},\boldsymbol{\alpha})\leq
u_{n}(y_{i},z)\right)\mathrm{d}\Phi(z)\bigg{|}I_{\\{B_{r,t,\boldsymbol{\alpha},n}\\}}$
$\displaystyle=\Sigma_{1}+\Sigma_{2}.$ (21)

The first term $\Sigma_{1}$ tends to $0$ by lemma 3.1. For the second term, it
follows from the inequality

$\displaystyle\bigg{|}\prod_{s=1}^{k}a_{s}-\prod_{s=1}^{k}b_{s}\bigg{|}\leq\sum_{s=1}^{k}\bigg{|}a_{s}-b_{s}\bigg{|}$
(22)

for all $a_{s}$, $b_{s}\in[0,1]$, that

$\displaystyle\Sigma_{2}$ $\displaystyle\leq$
$\displaystyle\sum_{r=0}^{2^{t}-1}\sum_{\boldsymbol{\alpha}\in\\{0,1\\}^{n}}E\int_{-\infty}^{+\infty}\sum_{i=1}^{k}\bigg{|}\exp\left[-(d_{i}-c_{i})(\lambda
g(x_{i},z)+(1-\lambda)g(y_{i},z))\right]$ (23) $\displaystyle\ \ \ \ \ \
-P\left(\widetilde{M}_{n}(Z,A_{i},\boldsymbol{\alpha})\leq
u_{n}(x_{i},z),\widehat{M}_{n}(Z,A_{i},\boldsymbol{\alpha})\leq
u_{n}(y_{i},z)\right)\bigg{|}\mathrm{d}\Phi(z)I_{\\{B_{r,t,\boldsymbol{\alpha},n}\\}}$
$\displaystyle\leq$
$\displaystyle\sum_{r=0}^{2^{t}-1}\sum_{\boldsymbol{\alpha}\in\\{0,1\\}^{n}}E\int_{-\infty}^{+\infty}\sum_{i=1}^{k}\bigg{|}\exp\left[-(d_{i}-c_{i})(\lambda
g(x_{i},z)+(1-\lambda)g(y_{i},z))\right]$ $\displaystyle\ \ \ \ \ \ \
-\left(1-\frac{(d_{i}-c_{i})(\lambda
g(x_{i},z)+(1-\lambda)g(y_{i},z))}{t}\right)^{t}\bigg{|}\mathrm{d}\Phi(z)I_{\\{B_{r,t,\boldsymbol{\alpha},n}\\}}$
$\displaystyle+$
$\displaystyle\sum_{r=0}^{2^{t}-1}\sum_{\boldsymbol{\alpha}\in\\{0,1\\}^{n}}E\int_{-\infty}^{+\infty}\sum_{i=1}^{k}\bigg{|}P\left(\widetilde{M}_{n}(Z,A_{i},\boldsymbol{\alpha})\leq
u_{n}(x_{i},z),\widehat{M}_{n}(Z,A_{i},\boldsymbol{\alpha})\leq
u_{n}(y_{i},z)\right)$ $\displaystyle\ \ \ \ \ \ \ \
-\left(1-\frac{(d_{i}-c_{i})(\lambda
g(x_{i},z)+(1-\lambda)g(y_{i},z))}{t}\right)^{t}\bigg{|}\mathrm{d}\Phi(z)I_{\\{B_{r,t,\boldsymbol{\alpha},n}\\}}$
$\displaystyle=$ $\displaystyle\Sigma_{21}+\Sigma_{22}.$

It is easy to see that $\Sigma_{21}$ tends to $0$, as $t\rightarrow\infty$.
Hence, we only need to estimate the second term $\Sigma_{22}$. We have

$\displaystyle\Sigma_{22}$ $\displaystyle\leq$
$\displaystyle\sum_{r=0}^{2^{t}-1}\sum_{\boldsymbol{\alpha}\in\\{0,1\\}^{n}}E\int_{-\infty}^{+\infty}\sum_{i=1}^{k}\bigg{|}P\left(\widetilde{M}_{n}(Z,A_{i},\boldsymbol{\alpha})\leq
u_{n}(x_{i},z),\widehat{M}_{n}(Z,A_{i},\boldsymbol{\alpha})\leq
u_{n}(y_{i},z)\right)$ (24)
$\displaystyle-\prod_{s=1}^{t}P\left(\widetilde{M}_{n}(Z,K_{s,i},\boldsymbol{\alpha})\leq
u_{n}(x_{i},z),\widehat{M}_{n}(Z,K_{s,i},\boldsymbol{\alpha})\leq
u_{n}(y_{i},z)\right)\bigg{|}\mathrm{d}\Phi(z)I_{\\{B_{r,t,\boldsymbol{\alpha},n}\\}}$
$\displaystyle+$
$\displaystyle\sum_{r=0}^{2^{t}-1}\sum_{\boldsymbol{\alpha}\in\\{0,1\\}^{n}}E\int_{-\infty}^{+\infty}\sum_{i=1}^{k}\bigg{|}\prod_{s=1}^{t}P\left(\widetilde{M}_{n}(Z,K_{s,i},\boldsymbol{\alpha})\leq
u_{n}(x_{i},z),\widehat{M}_{n}(Z,K_{s,i},\boldsymbol{\alpha})\leq
u_{n}(y_{i},z)\right)$
$\displaystyle-\prod_{s=1}^{t}\left[1-\left(\frac{r}{2^{t}}m_{i}(1-\Phi(u_{n}(x_{i},z)))+(1-\frac{r}{2^{t}})m_{i}(1-\Phi(u_{n}(y_{i},z)))\right)\right]\bigg{|}\mathrm{d}\Phi(z)I_{\\{B_{r,t,\boldsymbol{\alpha},n}\\}}$
$\displaystyle+$
$\displaystyle\sum_{r=0}^{2^{t}-1}\sum_{\boldsymbol{\alpha}\in\\{0,1\\}^{n}}E\int_{-\infty}^{+\infty}\sum_{i=1}^{k}\bigg{|}\prod_{s=1}^{t}\left[1-\left(\frac{r}{2^{t}}m_{i}(1-\Phi(u_{n}(x_{i},z)))+(1-\frac{r}{2^{t}})m_{i}(1-\Phi(u_{n}(y_{i},z)))\right)\right]$
$\displaystyle-\left(1-\frac{(d_{i}-c_{i})(\lambda
g(x_{i},z)+(1-\lambda)g(y_{i},z))}{t}\right)^{t}\bigg{|}\mathrm{d}\Phi(z)I_{\\{B_{r,t,\boldsymbol{\alpha},n}\\}}$
$\displaystyle=$ $\displaystyle\Sigma_{221}+\Sigma_{222}+\Sigma_{223}.$

By the independence of $\\{Z_{i}\\}_{i\geq 1}$, it is easy to check that

$\displaystyle\Sigma_{221}\leq\int_{-\infty}^{+\infty}\sum_{i=1}^{k}2m_{i}(1-\Phi(\min\\{u_{n}(x_{i},z),u_{n}(y_{i},z)\\}))d\Phi(z).$

According to the proof of Theorem 6.5.1 in Leadbetter et al. (1983), we have

$u_{n}(x,z)=u_{n}(x+\gamma-\sqrt{2\gamma}z)+o(a_{n}^{-1}),$

and thus,

$\lim_{n\to\infty}(n(1-\Phi(u_{n}(x,z))))=g(x,z).$

Consequently, we have

$\displaystyle\limsup_{n\rightarrow\infty}\Sigma_{221}\leq\int_{-\infty}^{+\infty}\frac{\sum_{i=1}^{k}2(d_{i}-c_{i})g(\min\\{x_{i},y_{i}\\},z)}{t}d\Phi(z).$
(25)

To bound the term $\Sigma_{222}$, note that for any $0\leq r\leq 2^{t}-1$, we
have

$\displaystyle\sum_{j\in
K_{s,i}}\alpha_{j}(\Phi(u_{n}(x_{i},z))-\Phi(u_{n}(y_{i},z)))+[1-m_{i}(1-\Phi(u_{n}(y_{i},z)))]$
$\displaystyle=\left[1-\frac{m_{i}r}{2^{t}}(1-\Phi(u_{n}(x_{i},z)))-m_{i}(1-\frac{r}{2^{t}})(1-\Phi(u_{n}(y_{i},z)))\right]$
$\displaystyle\ \ \ +\left[\frac{\sum_{j\in
K_{s,i}}\alpha_{j}}{m_{i}}-\frac{r}{2^{t}}\right]m_{i}(\Phi(u_{n}(x_{i},z))-\Phi(u_{n}(y_{i},z)))$
$\displaystyle\leq
P\left\\{\widetilde{M}_{n}(Z,K_{s,i},\boldsymbol{\alpha})\leq
u_{n}(x_{i},z),\widehat{M}_{n}(Z,K_{s,i},\boldsymbol{\alpha})\leq
u_{n}(y_{i},z)\right\\}$ $\displaystyle\leq
m_{i}\sum_{j=2}^{m_{i}}P[A_{s1},A_{sj}]+\left[1-\frac{m_{i}r}{2^{t}}(1-\Phi(u_{n}(x_{i},z)))-m_{i}(1-\frac{r}{2^{t}})(1-\Phi(u_{n}(y_{i},z)))\right]$
$\displaystyle\ \ \ +\left[\frac{\sum_{j\in
K_{s,i}}\alpha_{j}}{m_{i}}-\frac{r}{2^{t}}\right]m_{i}(\Phi(u_{n}(x_{i},z))-\Phi(u_{n}(y_{i},z))),$
(26)

where
$A_{sj}=\\{Z_{(s-1)m_{i}+j}>\min\\{u_{n}(x_{i},z),u_{n}(y_{i},z),i=1,2,\ldots,k\\}\\}$,
$j\in\\{1,2,\cdots,m_{i}\\}$. Thus, combining Lemma 3.2 with (22) and (3), we
get

$\displaystyle\Sigma_{222}$ $\displaystyle\leq$
$\displaystyle\sum_{r=0}^{2^{t}-1}\sum_{\boldsymbol{\alpha}\in\\{0,1\\}^{n}}E\int_{-\infty}^{+\infty}\sum_{i=1}^{k}\sum_{s=1}^{t}\bigg{|}\left[\frac{\sum_{j\in
K_{s,i}}\alpha_{j}}{m_{i}}-\frac{r}{2^{t}}\right]m_{i}(\Phi(u_{n}(x_{i},z))-\Phi(u_{n}(y_{i},z)))$
(27) $\displaystyle\ \ \ \
+m_{i}\sum_{j=2}^{m_{i}}P[A_{s1},A_{sj}]\bigg{|}\mathrm{d}\Phi(z)I_{\\{B_{r,t,\boldsymbol{\alpha},n}\\}}$
$\displaystyle\leq$
$\displaystyle\sum_{r=0}^{2^{t}-1}E\int_{-\infty}^{+\infty}\sum_{i=1}^{k}\sum_{s=1}^{t}m_{i}\bigg{|}\frac{\sum_{j\in
K_{s,i}}\varepsilon_{j}}{m_{i}}-\frac{r}{2^{t}}\bigg{|}\bigg{|}(\Phi(u_{n}(x_{i},z))-\Phi(u_{n}(y_{i},z)))\bigg{|}\mathrm{d}\Phi(z)I_{\\{B_{r,t}\\}}$
$\displaystyle\ \ \ \
+\int_{-\infty}^{+\infty}\sum_{i=1}^{k}\sum_{s=1}^{t}[m_{i}^{2}(1-\Phi\\{\min(u_{n}(x_{i},z),u_{n}(y_{i},z))\\})^{2}]\mathrm{d}\Phi(z)$
$\displaystyle\leq$
$\displaystyle\sum_{i=1}^{k}\sum_{s=1}^{t}\left(E\bigg{|}\frac{\sum_{j\in
K_{s,i}}\varepsilon_{j}}{m_{i}}-\lambda\bigg{|}+\sum_{r=0}^{2^{t}-1}E\bigg{|}\lambda-\frac{r}{2^{t}}\bigg{|}I_{\\{B_{r,t}\\}}\right)$
$\displaystyle\ \ \ \
\times\int_{-\infty}^{+\infty}m_{i}\bigg{|}(\Phi(u_{n}(x_{i},z))-\Phi(u_{n}(y_{i},z)))\bigg{|}\mathrm{d}\Phi(z)$
$\displaystyle\ \ \ \
+\int_{-\infty}^{+\infty}\sum_{i=1}^{k}\sum_{s=1}^{t}[m_{i}^{2}(1-\Phi\\{\min(u_{n}(x_{i},z),u_{n}(y_{i},z))\\})^{2}]\mathrm{d}\Phi(z)$
$\displaystyle\leq$
$\displaystyle\sum_{i=1}^{k}\sum_{s=1}^{t}\left\\{\left[2(2s-1)\left(d\left(\frac{S_{m_{i}s+\lfloor
nc_{i}\rfloor}}{m_{i}s+\lfloor
nc_{i}\rfloor},\lambda\right)+d\left(\frac{S_{m_{i}(s-1)+\lfloor
nc_{i}\rfloor}}{m_{i}(s-1)+\lfloor
nc_{i}\rfloor},\lambda\right)\right)+\frac{1}{2^{t}}\right]\right.$
$\displaystyle\left.\ \ \ \
\times\int_{-\infty}^{+\infty}m_{i}\bigg{|}(\Phi(u_{n}(x_{i},z))-\Phi(u_{n}(y_{i},z)))\bigg{|}\mathrm{d}\Phi(z)\right\\}$
$\displaystyle\ \ \ \
+\int_{-\infty}^{+\infty}\sum_{i=1}^{k}\sum_{s=1}^{t}[m_{i}^{2}(1-\Phi\\{\min(u_{n}(x_{i},z),u_{n}(y_{i},z))\\})^{2}]\mathrm{d}\Phi(z).$

Since $\lim_{m_{i}\to\infty}d\left(\frac{S_{m_{i}s+\lfloor
nc_{i}\rfloor}}{m_{i}s+\lfloor nc_{i}\rfloor},\lambda\right)=0$, we obtain

$\displaystyle\limsup_{n\to\infty}\Sigma_{222}$ $\displaystyle\leq$
$\displaystyle\sum_{i=1}^{k}\frac{1}{2^{t}}\int_{-\infty}^{+\infty}(d_{i}-c_{i})|g(y_{i},z)-g(x_{i},z)|\mathrm{d}\Phi(z)$
(28)
$\displaystyle+\int_{-\infty}^{+\infty}\frac{\sum_{i=1}^{k}(d_{i}-c_{i})^{2}(g(\min\\{x_{i},y_{i}\\},z))^{2}}{t}\mathrm{d}\Phi(z).$

For $\Sigma_{223}$, using (22) again, we get

$\displaystyle\Sigma_{223}$ $\displaystyle\leq$
$\displaystyle\sum_{r=0}^{2^{t}-1}E\int_{-\infty}^{+\infty}\sum_{i=1}^{k}\sum_{s=1}^{t}\bigg{|}\frac{(d_{i}-c_{i})(\lambda
g(x_{i},z)+(1-\lambda)g(y_{i},z))}{t}$
$\displaystyle-\frac{\frac{r}{2^{t}}n(d_{i}-c_{i})(1-\Phi(u_{n}(x_{i},z)))+(1-\frac{r}{2^{t}})n(d_{i}-c_{i})(1-\Phi(u_{n}(y_{i},z)))}{t}\bigg{|}\mathrm{d}\Phi(z)I_{\\{B_{r,t}\\}}.$

Thus,

$\displaystyle\limsup_{n\to\infty}\Sigma_{223}$ $\displaystyle\leq$
$\displaystyle\int_{-\infty}^{+\infty}\sum_{i=1}^{k}\sum_{s=1}^{t}\sum_{r=0}^{2^{t}-1}E\left(\frac{\bigg{|}\lambda-\frac{r}{2^{t}}\bigg{|}(d_{i}-c_{i})(g(x_{i},z)+g(y_{i},z))}{t}\right)\mathrm{d}\Phi(z)I_{\\{B_{r,t}\\}}$
(29) $\displaystyle\leq$
$\displaystyle\sum_{i=1}^{k}\int_{-\infty}^{+\infty}\frac{(d_{i}-c_{i})(g(x_{i},z)+g(y_{i},z))}{2^{t}}\mathrm{d}\Phi(z).$

From (24), (25), (28) and (29) we have

$\displaystyle\limsup_{n\to\infty}\Sigma_{22}$ $\displaystyle\leq$
$\displaystyle\int_{-\infty}^{+\infty}\frac{\sum_{i=1}^{k}2(d_{i}-c_{i})g(\min\\{x_{i},y_{i}\\},z)}{t}d\Phi(z)$
(30)
$\displaystyle+\sum_{i=1}^{k}\frac{1}{2^{t}}\int_{-\infty}^{+\infty}(d_{i}-c_{i})|g(y_{i},z)-g(x_{i},z)|\mathrm{d}\Phi(z)$
$\displaystyle+\int_{-\infty}^{+\infty}\frac{\sum_{i=1}^{k}(d_{i}-c_{i})^{2}(g(\min\\{x_{i},y_{i}\\},z))^{2}}{t}\mathrm{d}\Phi(z)$
$\displaystyle+\sum_{i=1}^{k}\int_{-\infty}^{+\infty}\frac{(d_{i}-c_{i})(g(x_{i},z)+g(y_{i},z))}{2^{t}}\mathrm{d}\Phi(z).$

Now if we take $t\rightarrow\infty$ of both sides of (30), we obtain
$\Sigma_{22}\rightarrow 0$, which completes the proof.

The following lemma is taken from Wiśniewski (1994).

Lemma 3.4. Suppose that $(M_{n1},M_{n2},\cdots,M_{nd})$ is a vector point
process of exceedances of the level $w_{nk}$ by random vector
$\\{\eta_{n1},\eta_{n2},\ldots,\eta_{nd}\\}_{n\geq 1}$, i.e.,

$M_{nk}(B)=\sum_{\frac{j}{n}\in B}I_{\\{\eta_{nk}>w_{nk}\\}},k=1,2,\cdots,d,$

for any Borel set $B\subset(0,1]$. Suppose that

(i) $M_{nk}$ converges in distribution to the point process $M_{k}$ and
$M_{k}$ is simple point process.

(ii) For any Borel sets $B_{k}\subset(0,1],k=1,2,\cdots,d$,
$\lim_{n\to\infty}P(M_{nk}(B_{k})=0,\ \ k=1,2\cdots,d)$ exists.
Then the vector point process $(M_{n1},M_{n2},\cdots,M_{nd})$ converges in
distribution to $(M_{1},M_{2},\cdots,M_{d})$ on $(0,1]^{d}$ as
$n\rightarrow\infty$.

##  4 Proof of main results

In this section, we give the proofs of our main results.

The proof of Theorem 2.1. Firstly, we prove that $\widetilde{N}_{n}^{x}$ and
$\widehat{N}_{n}^{y}$ converge in distribution to $\widetilde{N}^{x}$ and
$\widehat{N}^{y}$, respectively. From Kallenberg’s Theorem (see e.g.,
Kallenberg (1975), Leadbetter et al. (1983)), it suffices to show that as
$n\rightarrow\infty$,

(a) $E[\widetilde{N}_{n}^{x}(B_{1})]\rightarrow E[\widetilde{N}^{x}(B_{1})]$,
$E[\widehat{N}_{n}^{y}(B_{1})]\rightarrow E[\widehat{N}^{y}(B_{1})]$, where
$B_{1}=(c,d]$, $0<c<d\leq 1$;

(b) $P(\widetilde{N}_{n}^{x}(B_{2})=0)\rightarrow
P(\widetilde{N}^{x}(B_{2})=0)$, $P(\widehat{N}_{n}^{y}(B_{2})=0)\rightarrow
P(\widehat{N}^{y}(B_{2})=0)$, where $B_{2}=\bigcup_{i=1}^{k}(c_{i},d_{i}]$,
$0<c_{1}<d_{1}\leq c_{2}<\cdots<c_{k}<d_{k}\leq 1$.
To show assertion (a), let $A=\left\\{\lfloor nc\rfloor+1,\lfloor
nc\rfloor+2,\cdots,\lfloor nd\rfloor\right\\}$. According to the conditions of
the Theorem 2.1, we have

$\frac{S_{n}(A)}{\lfloor nd\rfloor-\lfloor
nc\rfloor}\overset{p}{\longrightarrow}\lambda,\ \ \ as\ \ n\to\infty,$

which combined with the dominated convergence theorem yields

$E\left(\frac{S_{n}(A)}{\lfloor nd\rfloor-\lfloor nc\rfloor}\right)\rightarrow
E(\lambda),\ \ \ as\ \ n\to\infty,$

i.e.,

$\displaystyle\ \ \ \frac{\sum_{j\in
A}P(\varepsilon_{j}=1)}{n(d-c)}\rightarrow E(\lambda),\ \ \ as\ \ n\to\infty.$
(31)

Thus, we also have

$\displaystyle\ \ \ \frac{\sum_{j\in
A}P(\varepsilon_{j}=0)}{n(d-c)}\rightarrow E(1-\lambda),\ \ \ as\ \
n\to\infty.$ (32)

Using the facts (31) and
$\lim_{n\to\infty}n(1-\Phi(u_{n}(x)))=\exp\left(-x\right)$, we have

$\displaystyle E[\widetilde{N}_{n}^{x}(B_{1})]$ $\displaystyle=$
$\displaystyle
E\left[\sum_{\frac{j}{n}\in(c,d]}I_{\\{X_{j}>u_{n}(x),\varepsilon_{j}=1\\}}\right]$
(33) $\displaystyle=$ $\displaystyle(1-\Phi(u_{n}(x)))\sum_{j\in
A}P(\varepsilon_{j}=1)$ $\displaystyle\rightarrow$
$\displaystyle(d-c)E(\lambda)e^{-x}.$

Noting that $\widetilde{N}^{x}$ is a conditional Poisson process with
conditional intensity $\lambda g(x,\xi)$, we get

$\displaystyle E[\widetilde{N}^{x}(B_{1})]$ $\displaystyle=$ $\displaystyle
E\\{E[\widetilde{N}^{x}(B_{1})|(\lambda,\xi)]\\}$ (34) $\displaystyle=$
$\displaystyle E((d-c)\lambda g(x,\xi))$ $\displaystyle=$
$\displaystyle(d-c)E(\lambda)E(g(x,\xi))$ $\displaystyle=$
$\displaystyle(d-c)E(\lambda)e^{-x}.$

Similarly,

$E[\widehat{N}_{n}^{y}(B_{1})]\rightarrow(d-c)E((1-\lambda))e^{-y}=E[\widehat{N}^{y}(B_{1})].$

Thus, the first assertion holds.
To prove the second assertion (b), using Lemma 3.3 for the case $x_{i}\equiv
x$ and $y_{i}\equiv y$, $i=1,2,\ldots,k$, we have

$\displaystyle\lim_{n\to\infty}P(\widetilde{N}_{n}^{x}(B_{2})=0,\widehat{N}_{n}^{y}(B_{2})=0)$
$\displaystyle\ \ \
=\lim_{n\to\infty}P\left(\bigcap_{i=1}^{k}\left\\{\widetilde{M}_{n}(X,A_{i},\boldsymbol{\varepsilon})\leq
u_{n}(x),\widehat{M}_{n}(X,A_{i},\boldsymbol{\varepsilon})\leq
u_{n}(y)\right\\}\right)$ $\displaystyle\ \ \
=E\int_{-\infty}^{+\infty}\prod_{i=1}^{k}\exp\left[-(d_{i}-c_{i})(\lambda
g(x,z)+(1-\lambda)g(y,z))\right]\mathrm{d}\Phi(z).$ (35)

On the one hand, letting $y\to\infty$ and $x\to\infty$, we obtain

$\displaystyle\lim_{n\to\infty}P(\widetilde{N}_{n}^{x}(B_{2})=0)=E\int_{-\infty}^{+\infty}\prod_{i=1}^{k}\exp\left[-(d_{i}-c_{i})(\lambda
g(x,z))\right]\mathrm{d}\Phi(z)$ (36)

and

$\displaystyle\lim_{n\to\infty}P(\widehat{N}_{n}^{y}(B_{2})=0)=E\int_{-\infty}^{+\infty}\prod_{i=1}^{k}\exp\left[-(d_{i}-c_{i})((1-\lambda)g(y,z))\right]\mathrm{d}\Phi(z),$
(37)

respectively. On the other hand, since $\widetilde{N}^{x}$ is a conditional
Poisson process with conditional intensity $\lambda g(x,\xi)$, we have

$\displaystyle P(\widetilde{N}^{x}(B_{2})=0)$ $\displaystyle=$ $\displaystyle
P\left(\bigcap_{i=1}^{k}\\{\widetilde{N}^{x}((c_{i},d_{i}])=0\\}\right)$ (38)
$\displaystyle=$
$\displaystyle\int_{0}^{1}\int_{-\infty}^{+\infty}\prod_{i=1}^{k}P(\widetilde{N}^{x}(c_{i},d_{i}]=0|\lambda=e,\xi=z)\mathrm{d}F_{\lambda}(e)\mathrm{d}\Phi(z)$
$\displaystyle=$ $\displaystyle
E\int_{-\infty}^{+\infty}\prod_{i=1}^{k}\exp\left[-(d_{i}-c_{i})(\lambda
g(x,z))\right]\mathrm{d}\Phi(z),$

where $F_{\lambda}(e)=P(\lambda\leq e)$. Hence,
$P(\widetilde{N}_{n}^{x}(B_{2})=0)\rightarrow P(\widetilde{N}^{x}(B_{2})=0)$,
as $n\rightarrow\infty$. Similarly, we have

$P(\widehat{N}_{n}^{y}(B_{2})=0)\rightarrow P(\widehat{N}^{y}(B_{2})=0),$

as $n\rightarrow\infty$, which proves (b).
Secondly, we show that for any Borel sets $C_{i}\subset(0,1]$,
$i=1,\ldots,k+l$,

$\displaystyle\lim_{n\to\infty}P\left(\widetilde{N}_{n}^{x_{1}}(C_{1})=0,\ldots,\widetilde{N}_{n}^{x_{k}}(C_{k})=0,\widehat{N}_{n}^{y_{1}}(C_{k+1})=0,\ldots,\widehat{N}_{n}^{y_{l}}(C_{k+l})=0\right)$
(39)

exists. For the simplicity, we only deal with the case $k=2$ and $l=2$, since
the general case can be done similarly. We first deal with the case $x_{1}\leq
x_{2}$ and $y_{1}\leq y_{2}$. Put $nC_{i}=\\{nx,x\in C_{i}\\}$ and let
$C_{5}=C_{2}\setminus C_{1}$, $C_{6}=C_{4}\setminus C_{3}$. We have

$\displaystyle\lim_{n\to\infty}P\left(\widetilde{N}_{n}^{x_{1}}(C_{1})=0,\widetilde{N}_{n}^{x_{2}}(C_{2})=0,\widehat{N}_{n}^{y_{1}}(C_{3})=0,,\widehat{N}_{n}^{y_{2}}(C_{4})=0\right)$
$\displaystyle\ \ \ \ \ =\lim_{n\to\infty}P\left(\widetilde{M}_{C_{1}}\leq
u_{n}(x_{1}),\widetilde{M}_{C_{2}}\leq u_{n}(x_{2}),\widehat{M}_{C_{3}}\leq
u_{n}(y_{1}),\widehat{M}_{C_{4}}\leq u_{n}(y_{2})\right)$ $\displaystyle\ \ \
\ \ =\lim_{n\to\infty}P\left(\widetilde{M}_{C_{1}}\leq
u_{n}(x_{1}),\widetilde{M}_{C_{5}}\leq u_{n}(x_{2}),\widehat{M}_{C_{3}}\leq
u_{n}(y_{1}),\widehat{M}_{C_{6}}\leq u_{n}(y_{2})\right),$

where
$\widetilde{M}_{C_{i}}=\widetilde{M}_{n}(X,nC_{i},\boldsymbol{\varepsilon})$
and $\widehat{M}_{C_{i}}=\widehat{M}_{n}(X,nC_{i},\boldsymbol{\varepsilon})$.
Note that $C_{1}\bigcap C_{5}=\varnothing$ and $C_{3}\bigcap
C_{6}=\varnothing$. It follows now from Lemma 3.3 that

$\displaystyle\lim_{n\to\infty}P\left(\widetilde{N}_{n}^{x_{1}}(C_{1})=0,\widetilde{N}_{n}^{x_{2}}(C_{2})=0,\widehat{N}_{n}^{y_{1}}(C_{3})=0,,\widehat{N}_{n}^{y_{2}}(C_{4})=0\right)$
$\displaystyle=\lim_{n\to\infty}P\left(\widetilde{M}_{C_{1}}\leq
u_{n}(x_{1}),\widetilde{M}_{C_{5}}\leq u_{n}(x_{2}),\widehat{M}_{C_{3}}\leq
u_{n}(y_{1}),\widehat{M}_{C_{6}}\leq u_{n}(y_{2})\right)$
$\displaystyle=E\int_{-\infty}^{+\infty}\exp\left(-m(C_{1})\lambda
g(x_{1},z)-m(C_{3})(1-\lambda)g(y_{1},z)\right)$ $\displaystyle\ \ \ \ \ \
\times\exp\left(-m(C_{2}\setminus C_{1})\lambda g(x_{2},z)-m(C_{4}\setminus
C_{3})(1-\lambda)g(y_{2},z)\right)\mathrm{d}\Phi(z).$ (40)

By the same arguments as the above, we have, for $x_{1}\leq x_{2}$ and
$y_{1}>y_{2}$,

$\displaystyle\lim_{n\to\infty}P\left(\widetilde{N}_{n}^{x_{1}}(C_{1})=0,\widetilde{N}_{n}^{x_{2}}(C_{2})=0,\widehat{N}_{n}^{y_{1}}(C_{3})=0,,\widehat{N}_{n}^{y_{2}}(C_{4})=0\right)$
$\displaystyle=E\int_{-\infty}^{+\infty}\exp\left(-m(C_{1})\lambda
g(x_{1},z)-m(C_{4})(1-\lambda)g(y_{2},z)\right)$ $\displaystyle\ \ \ \ \ \
\times\exp\left(-m(C_{2}\setminus C_{1})\lambda g(x_{2},z)-m(C_{3}\setminus
C_{4})(1-\lambda)g(y_{1},z)\right)\mathrm{d}\Phi(z);$ (41)

for $x_{1}>x_{2}$ and $y_{1}\leq y_{2}$,

$\displaystyle\lim_{n\to\infty}P\left(\widetilde{N}_{n}^{x_{1}}(C_{1})=0,\widetilde{N}_{n}^{x_{2}}(C_{2})=0,\widehat{N}_{n}^{y_{1}}(C_{3})=0,,\widehat{N}_{n}^{y_{2}}(C_{4})=0\right)$
$\displaystyle=E\int_{-\infty}^{+\infty}\exp\left(-m(C_{2})\lambda
g(x_{2},z)-m(C_{3})(1-\lambda)g(y_{1},z)\right)$ $\displaystyle\ \ \ \ \ \
\times\exp\left(-m(C_{1}\setminus C_{2})\lambda g(x_{1},z)-m(C_{4}\setminus
C_{3})(1-\lambda)g(y_{2},z)\right)\mathrm{d}\Phi(z)$ (42)

and for $x_{1}>x_{2}$ and $y_{1}>y_{2}$,

$\displaystyle\lim_{n\to\infty}P\left(\widetilde{N}_{n}^{x_{1}}(C_{1})=0,\widetilde{N}_{n}^{x_{2}}(C_{2})=0,\widehat{N}_{n}^{y_{1}}(C_{3})=0,,\widehat{N}_{n}^{y_{2}}(C_{4})=0\right)$
$\displaystyle=E\int_{-\infty}^{+\infty}\exp\left(-m(C_{2})\lambda
g(x_{2},z)-m(C_{4})(1-\lambda)g(y_{2},z)\right)$ $\displaystyle\ \ \ \ \ \
\times\exp\left(-m(C_{1}\setminus C_{2})\lambda g(x_{1},z)-m(C_{3}\setminus
C_{4})(1-\lambda)g(y_{1},z)\right)\mathrm{d}\Phi(z).$ (43)

Therefore, by Lemma 3.4, we prove that
$\left(\widetilde{N}_{n}^{x_{1}},\ldots,\widetilde{N}_{n}^{x_{k}},\widehat{N}_{n}^{y_{1}},\ldots,\widehat{N}_{n}^{y_{l}}\right)$
converges in distribution to the point process
$\left(\widetilde{N}^{x_{1}},\ldots,\widetilde{N}^{x_{k}},\widehat{N}^{y_{1}},\ldots,\widehat{N}^{y_{l}}\right)$
on $\left(0,1\right]^{k+l}$.

Proof of Corollary 2.1. By Theorem 2.1, we have

$\displaystyle
P\left(\widetilde{N}^{x}(B)=k_{1},\widehat{N}^{x}(B)=k_{2},\widetilde{N}^{y}(B)=k_{3},\widehat{N}^{y}(B)=k_{4}\right)$
$\displaystyle=\int_{-\infty}^{+\infty}\int_{0}^{1}P\left(\widetilde{N}^{x}(B)=k_{1},\widetilde{N}^{y}(B)=k_{3}|\lambda=e,\xi=z\right)$
$\displaystyle\ \ \ \ \ \ \times
P\left(\widehat{N}^{x}(B)=k_{2},\widehat{N}^{y}(B)=k_{4}|\lambda=e,\xi=z\right)\mathrm{d}\Phi(z)\mathrm{d}F_{\lambda}(e).$
(44)

For $x>y$, applying Theorem 5.6.1 in Leadbetter et al. (1983), we have

$\displaystyle
P\left(\widetilde{N}^{x}(B)=k_{1},\widetilde{N}^{y}(B)=k_{3}|\lambda=e,\xi=z\right)$
$\displaystyle\ \
=\frac{[em(B)(g(y,z)-g(x,z))]^{k_{3}-k_{1}}}{(k_{3}-k_{1})!}\frac{[em(B)g(x,z)]^{k_{1}}}{k_{1}!}\exp(-em(B)g(y,z)).$
(45)

Similarly,

$\displaystyle
P\left(\widehat{N}^{x}(B)=k_{2},\widehat{N}^{y}(B)=k_{4}|\lambda=e,\xi=z\right)$
$\displaystyle\ \
=\frac{[(1-e)m(B)(g(y,z)-g(x,z))]^{k_{2}-k_{4}}}{(k_{2}-k_{4})!}\frac{[(1-e)m(B)g(x,z)]^{k_{2}}}{k_{2}!}\exp(-(1-e)m(B)g(y,z)).$

Hence, plugging (4) and(4) into (4), we obtain the desired result (2). By the
same arguments, we can prove (2) also holds.

Proof of Theorem 2.2. (i). First, note that

$\\{\widetilde{M}_{n}^{(k)}\leq
u_{n}(x)\\}=\\{\widetilde{N}_{n}^{x}((0,1])\leq k-1\\}$

and

$\\{\widehat{M}_{n}^{(l)}\leq u_{n}(y)\\}=\\{\widehat{N}_{n}^{y}((0,1])\leq
l-1\\}.$

Thus, by Theorem 2.1, we have

$\displaystyle\lim_{n\to\infty}P\left\\{\widetilde{M}_{n}^{(k)}\leq
u_{n}\left(x\right),\widehat{M}_{n}^{(l)}\leq u_{n}\left(y\right)\right\\}$
$\displaystyle=\lim_{n\to\infty}P\left(\widetilde{N}_{n}^{x}((0,1])\leq
k-1,\widehat{N}_{n}^{y}((0,1])\leq l-1\right)$
$\displaystyle=P\left(\widetilde{N}^{x}((0,1])\leq
k-1,\widehat{N}^{y}((0,1])\leq l-1\right)$
$\displaystyle=\int_{0}^{1}\int_{-\infty}^{+\infty}P\left(\widetilde{N}^{x}((0,1])\leq
k-1|\lambda=e,\xi=z\right)$ $\displaystyle\ \ \ \ \ \ \ \ \ \times
P\left(\widehat{N}^{y}((0,1])\leq
l-1|\lambda=e,\xi=z\right)\mathrm{d}F_{\lambda}(e)\mathrm{d}\Phi(z)$
$\displaystyle=\sum_{s=0}^{k-1}\sum_{t=0}^{l-1}E\int_{-\infty}^{+\infty}\left\\{\frac{[\lambda
g(x,z)]^{s}}{s!}\exp(-\lambda g(x,z))\right.$ $\displaystyle\ \ \ \ \ \ \ \ \
\ \
\left.\times\frac{[(1-\lambda)g(y,z)]^{t}}{t!}\exp(-(1-\lambda)g(y,z))\right\\}\mathrm{d}\Phi(z),$
(47)

which proves the (i).

(ii). By the same way as (i), we have

$\displaystyle\lim_{n\to\infty}P\left(\widetilde{M}_{n}^{(k)}\leq
u_{n}(x),M_{n}^{(m)}\leq u_{n}(y)\right)$ $\displaystyle\ \ \ \ \ \
=\lim_{n\to\infty}P\left(\widetilde{N}_{n}^{x}((0,1])\leq
k-1,\widetilde{N}_{n}^{y}((0,1])+\widehat{N}_{n}^{y}((0,1])\leq m-1\right)$
$\displaystyle\ \ \ \ \ \ =P\left(\widetilde{N}^{x}((0,1])\leq
k-1,\widetilde{N}^{y}((0,1])+\widehat{N}^{y}((0,1])\leq m-1\right)$
$\displaystyle\ \ \ \ \ \
=\int_{0}^{1}\int_{-\infty}^{+\infty}\sum_{t=0}^{m-1}P\left(\widehat{N}^{y}((0,1])=t|\lambda=e,\xi=z\right)$
$\displaystyle\ \ \ \ \ \ \ \ \times P\left(\widetilde{N}^{x}((0,1])\leq
k-1,\widetilde{N}^{y}((0,1])\leq
m-1-t|\lambda=e,\xi=z\right)\mathrm{d}F_{\lambda}(e)\mathrm{d}\Phi(z)$
$\displaystyle\ \ \ \ \ \
=\int_{0}^{1}\int_{-\infty}^{+\infty}\sum_{t=0}^{m-1}\frac{[(1-e)g(y,z)]^{t}\exp(-(1-e)g(y,z))}{t!}$
$\displaystyle\ \ \ \ \ \ \ \ \times P\left(\widetilde{N}^{x}((0,1])\leq
k-1,\widetilde{N}^{y}((0,1])\leq
m-1-t|\lambda=e,\xi=z\right)\mathrm{d}F_{\lambda}(e)\mathrm{d}\Phi(z).$ (48)

For $x<y$, using (4), we have

$\displaystyle P\left(\widetilde{N}^{x}((0,1])\leq
k-1,\widetilde{N}^{y}((0,1])\leq m-1-t|\lambda=e,\xi=z\right)$ $\displaystyle\
\ \ \ \
=\sum_{i=0}^{m-t-1}\sum_{j=i}^{k-1}P\left(\widetilde{N}^{x}((0,1])=j,\widetilde{N}^{y}((0,1])=i|\lambda=e,\xi=z\right)$
$\displaystyle\ \ \ \ \
=\sum_{i=0}^{m-t-1}\sum_{j=i}^{k-1}\frac{[eg(y,z)]^{i}}{i!}\frac{[e(g(x,z)-g(y,z))]^{j-i}}{(j-i)!}\exp(-eg(x,z)).$
(49)

Plugging (4) into (4), we get the desired result (2). For $x\geq y$,
similarly, we have

$\displaystyle P\left(\widetilde{N}^{x}((0,1])\leq
k-1,\widetilde{N}^{y}((0,1])\leq m-1-t|\lambda=e,\xi=z\right)$ $\displaystyle\
\ \ \ \
=\sum_{i=0}^{m-t-1}\sum_{j=0}^{\min\\{i,k-1\\}}P\left(\widetilde{N}^{x}((0,1])=j,\widetilde{N}^{y}((0,1])=i|\lambda=e,\xi=z\right)$
$\displaystyle\ \ \ \ \
=\sum_{i=0}^{m-t-1}\sum_{j=0}^{\min\\{i,k-1\\}}\frac{[eg(x,z)]^{j}}{j!}\frac{[e(g(y,z)-g(x,z))]^{i-j}}{(i-j)!}\exp(-eg(y,z)).$
(50)

Plugging (4) into (4), we get the desired result (2).

(iii). The proof of this case is similar to that of case (ii), so we omit it.

Proof of Theorem 2.3. (i). Recall that $\mathbb{N}_{n}=\\{1,2,\ldots,n\\}$ and
define $\mathbb{N}_{ns}=\\{1,2,\ldots,\lfloor ns\rfloor\\}$. Note that

$\displaystyle P\left(\frac{1}{n}\widetilde{L}_{n}\leq
s,\frac{1}{n}\widehat{L}_{n}\leq t,\widetilde{M}_{n}\leq
u_{n}(x),\widehat{M}_{n}\leq u_{n}(y)\right)$
$\displaystyle=P\left(\widetilde{M}_{n}(X,\mathbb{N}_{n}\setminus\mathbb{N}_{ns},\boldsymbol{\varepsilon})\leq\widetilde{M}_{n}(X,\mathbb{N}_{ns},\boldsymbol{\varepsilon})\leq
u_{n}(x),\right.$ $\displaystyle\ \ \ \ \ \ \ \
\left.\widehat{M}_{n}(X,\mathbb{N}_{n}\setminus\mathbb{N}_{nt},\boldsymbol{\varepsilon})\leq\widehat{M}_{n}(X,\mathbb{N}_{nt},\boldsymbol{\varepsilon})\leq
u_{n}(y)\right)=:P_{n}(s,t,x,y).$ (51)

Define $\mathbb{R}_{st}=(s,t]$ for $0\leq s<t\leq 1$. By Theorem 2.1, we have

$\displaystyle
P\left(\widetilde{M}_{n}(X,\mathbb{N}_{n}\setminus\mathbb{N}_{ns},\boldsymbol{\varepsilon})\leq
u_{n}(x_{1}),\widetilde{M}_{n}(X,\mathbb{N}_{ns},\boldsymbol{\varepsilon})\leq
u_{n}(x_{2}),\right.$ $\displaystyle\ \ \ \ \ \ \ \
\left.\widehat{M}_{n}(X,\mathbb{N}_{n}\setminus\mathbb{N}_{nt},\boldsymbol{\varepsilon})\leq
u_{n}(x_{3}),\widehat{M}_{n}(X,\mathbb{N}_{nt},\boldsymbol{\varepsilon})\leq
u_{n}(x_{4})\right)$
$\displaystyle=P\left(\widetilde{N}_{n}^{x_{1}}(\mathbb{R}_{s1})=0,\widetilde{N}_{n}^{x_{2}}(\mathbb{R}_{0s})=0,\widehat{N}_{n}^{x_{3}}(\mathbb{R}_{t1})=0,\widehat{N}_{n}^{x_{4}}(\mathbb{R}_{0t})=0\right)$
$\displaystyle\rightarrow
P\left(\widetilde{N}^{x_{1}}(\mathbb{R}_{s1})=0,\widetilde{N}^{x_{2}}(\mathbb{R}_{0s})=0,\widehat{N}^{x_{3}}(\mathbb{R}_{t1})=0,\widehat{N}^{x_{4}}(\mathbb{R}_{0t})=0\right)$
$\displaystyle=\int_{0}^{1}\int_{-\infty}^{+\infty}P\left(\widetilde{N}^{x_{1}}(\mathbb{R}_{s1})=0,\widetilde{N}^{x_{2}}(\mathbb{R}_{0s})=0|\lambda=e,\xi=z\right)$
$\displaystyle\ \ \ \ \ \ \ \ \ \times
P\left(\widehat{N}^{x_{3}}(\mathbb{R}_{t1})=0,\widehat{N}^{x_{4}}(\mathbb{R}_{0t})=0|\lambda=e,\xi=z\right)\mathrm{d}F_{\lambda}(e)\mathrm{d}\Phi(z)$
$\displaystyle=\int_{0}^{1}\int_{-\infty}^{+\infty}\exp\left(-(1-s)eg(x_{1},z)-seg(x_{2},z)\right)$
$\displaystyle\ \ \ \ \ \ \ \ \ \
\times\exp\left(-(1-t)(1-e)g(x_{3},z)-t(1-e)g(x_{4},z)\right)\mathrm{d}F_{\lambda}(e)\mathrm{d}\Phi(z)$
$\displaystyle=:H(x_{1},x_{2},x_{3},x_{4}).$ (52)

Let
$h(x_{1},x_{2},x_{3},x_{4})=\frac{\partial^{4}H(x_{1},x_{2},x_{3},x_{4})}{\partial
x_{1}\partial x_{2}\partial x_{3}\partial x_{4}}$ be the joint density
function of random vector $(Y_{1},Y_{2},Y_{3},Y_{4})$, then by some direct
computations

$\displaystyle\lim_{n\rightarrow\infty}P_{n}(s,t,x,y)$ $\displaystyle=$
$\displaystyle P(Y_{1}\leq Y_{2}\leq x,Y_{3}\leq Y_{4}\leq y)$ (53)
$\displaystyle=$ $\displaystyle\iiiint\limits_{x_{1}\leq x_{2}\leq x,x_{3}\leq
x_{4}\leq
y}h(x_{1},x_{2},x_{3},x_{4})\mathrm{d}x_{1}\mathrm{d}x_{2}\mathrm{d}x_{3}\mathrm{d}x_{4}$
$\displaystyle=$ $\displaystyle stE\int_{-\infty}^{+\infty}\exp(-\lambda
g(x,z))\exp(-(1-\lambda)g(y,z))\mathrm{d}\Phi(z),$

which combining with (4) implies the desired result (2).

(ii). Note that for $x\leq y$

$\displaystyle P\left(\frac{1}{n}\widetilde{L}_{n}\leq s,\frac{1}{n}L_{n}\leq
t,\widetilde{M}_{n}\leq u_{n}(x),M_{n}\leq u_{n}(y)\right)$
$\displaystyle=P\left(\frac{1}{n}\widetilde{L}_{n}\leq s,\frac{1}{n}L_{n}\leq
t,\widetilde{M}_{n}\leq u_{n}(x),\widehat{M}_{n}\leq u_{n}(y)\right)$
$\displaystyle=P\left(\frac{1}{n}\widetilde{L}_{n}\leq s,\frac{1}{n}L_{n}\leq
t,\widetilde{M}_{n}\leq u_{n}(x),\widehat{M}_{n}\leq
u_{n}(y),\widetilde{M}_{n}\leq\widehat{M}_{n}\right)$ $\displaystyle\ \ \
+P\left(\frac{1}{n}\widetilde{L}_{n}\leq s,\frac{1}{n}L_{n}\leq
t,\widetilde{M}_{n}\leq u_{n}(x),\widehat{M}_{n}\leq
u_{n}(y),\widetilde{M}_{n}>\widehat{M}_{n}\right)$
$\displaystyle=P\left(\frac{1}{n}\widetilde{L}_{n}\leq
s,\frac{1}{n}\widehat{L}_{n}\leq t,\widetilde{M}_{n}\leq
u_{n}(x),\widehat{M}_{n}\leq
u_{n}(y),\widetilde{M}_{n}\leq\widehat{M}_{n}\right)$ $\displaystyle\ \ \
+P\left(\frac{1}{n}\widetilde{L}_{n}\leq\min\\{s,t\\},\widetilde{M}_{n}\leq
u_{n}(x),\widehat{M}_{n}\leq
u_{n}(y),\widetilde{M}_{n}>\widehat{M}_{n}\right).$ (54)

It follows from case (i) that for any $x,y\in\mathbb{R}$

$\displaystyle\lim_{n\rightarrow\infty}P\left(\frac{1}{n}\widetilde{L}_{n}\leq
s,\frac{1}{n}\widehat{L}_{n}\leq t,\widetilde{M}_{n}\leq
u_{n}(x),\widehat{M}_{n}\leq u_{n}(y)\right)$
$\displaystyle=stE\int_{-\infty}^{+\infty}\exp(-\lambda
g(x,z))\exp(-(1-\lambda)g(y,z))\mathrm{d}\Phi(z),$ (55)

which indicates that $\widetilde{L}_{n},\widehat{L}_{n}$ are asymptotically
independent of $\widetilde{M}_{n}$ and $\widehat{M}_{n}$. Thus, we have for
$x<y$

$\displaystyle\lim_{n\rightarrow\infty}P\left(\frac{1}{n}\widetilde{L}_{n}\leq
s,\frac{1}{n}\widehat{L}_{n}\leq t,\widetilde{M}_{n}\leq
u_{n}(x),\widehat{M}_{n}\leq
u_{n}(y),\widetilde{M}_{n}\leq\widehat{M}_{n}\right)$
$\displaystyle=\lim_{n\rightarrow\infty}P\left(\frac{1}{n}\widetilde{L}_{n}\leq
s,\frac{1}{n}\widehat{L}_{n}\leq
t\right)\times\lim_{n\rightarrow\infty}P\left(\widetilde{M}_{n}\leq
u_{n}(x),\widehat{M}_{n}\leq
u_{n}(y),\widetilde{M}_{n}\leq\widehat{M}_{n}\right)$
$\displaystyle=stE\int_{-\infty}^{+\infty}\exp(-\lambda
g(x,z))\exp(-(1-\lambda)g(y,z))-\lambda\exp(-g(x,z))\mathrm{d}\Phi(z)$ (56)

and

$\displaystyle\lim_{n\rightarrow\infty}P\left(\frac{1}{n}\widetilde{L}_{n}\leq\min\\{s,t\\},\widetilde{M}_{n}\leq
u_{n}(x),\widehat{M}_{n}\leq
u_{n}(y),\widetilde{M}_{n}>\widehat{M}_{n}\right)$
$\displaystyle=\lim_{n\rightarrow\infty}P\left(\frac{1}{n}\widetilde{L}_{n}\leq\min\\{s,t\\}\right)\times\lim_{n\rightarrow\infty}P\left(\widetilde{M}_{n}\leq
u_{n}(x),\widehat{M}_{n}\leq
u_{n}(y),\widetilde{M}_{n}>\widehat{M}_{n}\right)$
$\displaystyle=\min\\{s,t\\}E\int_{-\infty}^{+\infty}\lambda\exp(-g(x,z))\mathrm{d}\Phi(z).$
(57)

Plugging (4) and (4) into (4), we get the desired result (2).

(iii). The proof is the same as that of case (ii), so we omit it.

Conflict of interest: The authors have no conflict of interest.

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