Sojourns of Stationary Gaussian Processes over a Random Interval

Krzysztof Dȩbicki Krzysztof Dȩbicki, Mathematical Institute, University of Wrocław, pl. Grunwaldzki 2/4, 50-384 Wrocław, Poland [email protected] and Xiaofan Peng Xiaofan Peng, School of Mathematical Sciences, University of Electronic Science and Technology of China, Chengdu 610054, China [email protected]

Abstract: We investigate asymptotics of the tail distribution of sojourn time

\int_{0}^{T}\mathbb{I}(X(t)>u)dt,

as $u\to\infty$ , where $X$ is a centered stationary Gaussian process and $T$ is an independent of $X$ nonnegative random variable. The heaviness of the tail distribution of $T$ impacts the form of the asymptotics, leading to four scenarios: the case of integrable $T$ , the case of regularly varying $T$ with index $\lambda=1$ and index $\lambda\in(0,1)$ and the case of slowly varying tail distribution of $T$ . The derived findings are illustrated by the analysis of the class of fractional Ornstein-Uhlenbeck processes.

Key Words: exact asymptotics; regularly varying function; sojourn time; stationary Gaussian process.

AMS Classification: Primary 60G15; secondary 60G70

1. Introduction

For given stochastic process $X(t),t\geq 0$ , by

\displaystyle L_{u}[a,b]:=\int_{a}^{b}\mathbb{I}_{u}\left(X(t)\right)\text{\rm d}t,

with $\mathbb{I}_{u}\left(x\right):=\mathbb{I}\left(x>u\right)$ , we define the sojourn time spent above a fixed level $u$ by process $X$ on interval $[a,b]$ . The interest in analysis of distributional properties of $L_{u}[a,b]$ stems both from theoretical questions related to the research on the level sets of stochastic processes and from its importance in applied probability, as e.g. in finance or insurance theory, where $L_{u}[0,T]$ , $T>0$ may be interpreted as the total time in ruin up to time $T$ for the risk process modeled by $X$ ; see e.g. [1, 2].

In the case of $X$ being a Gaussian process, the asymptotics of the tail distribution of $L_{u}[0,T]$ , as $u\to\infty$ , was analyzed extensively in a series of papers by Berman, e.g. [3, 4]; see also the seminal monograph [5] and recent refinements [6, 7].

The aim of this paper is to get the exact asymptotics of tail distribution of $L_{u}[0,T]$ for a class of centered stationary Gaussian processes over an independent of $X$ random time $T$ . The motivation to consider extremal behaviour of a stochastic process over a random time interval stems from its relevance in such problems as ruin of time-changed risk processes [8, 9], resetting models [10] or hybrid queueing models [11]. We also refer to related problems on extremes of conditionally Gaussian processes and Gaussian processes with random variance [12, 13]. Using the fact that

\mathbb{P}\left\{L_{u}[0,T]>0\right\}=\mathbb{P}\left\{\sup_{t\in[0,T]}X(t)>u\right\},

the findings of this contribution also extend results obtained in [14, 15, 16].

It appears that the form of the derived exact asymptotics strongly depends on the heaviness of the tail distribution of $T$ , leading to four scenarios: the case of finite ${\mathbb{E}}T$ (scenario D1), the case of $T$ having regularly varying tail distribution with index $\lambda=1$ (scenario D2), $\lambda\in(0,1)$ (scenario D3) and the case of slowly varying tail distribution of $T$ (scenario D4); see Section 3.

Brief organisation of the rest of the paper: In Section 2 we formalize the analyzed model and introduce notation. In Section 3 we derive the tail asymptotic behavior of the sojourn time for a class of centered stationary Gaussian processes $X$ over random interval $[0,T]$ under introduced in Section 2 scenarios D1-D4, respectively. Section 4 contains some examples illustrating the main findings of this contribution. All the proofs are displayed in Section 5, whereas few technical results are included in Section 6.

2. Notation and model description

Let $X(t),t\geq 0$ be a centered stationary Gaussian process with a.s. continuous trajectories, unit variance function and covariance function $r$ satisfying

A1:

$1-r(t)$ is regularly varying at $t=0$ with index $\alpha\in(0,2]$ ;
A2:

$r(t)<1$ for all $t>0$ ;
A3:

$\lim_{t\to\infty}r(t)\log(t)=0$ .

Assumptions A1-A3 cover wide range of investigated in the literature stationary Gaussian processes, where A3 is referred to as Berman’s condition (see, e.g., [5]); see also Section 4.

Let function $v(\cdot)$ be such that $\lim_{u\to\infty}v(u)=\infty$ and

(2.1)

\displaystyle\lim_{u\to\infty}u^{2}(1-r(1/v(u)))=1.

By [5], $v(\cdot)$ exists and is regularly varying at infinity with index $2/\alpha$ .

We are interested in the asymptotics of

\mathbb{P}\left\{L_{u}^{*}[0,T]>x\right\},

as $u\to\infty$ , where

(2.2)

\displaystyle L_{u}^{*}[0,T]:=v(u)L_{u}[0,T]

and $T$ is an independent of $X$ nonnegative random variable with distribution function $F_{T}(\cdot)$ which belongs to one of the following distribution classes:

D1:

$T$ is integrable;
D2:

$T$ has regularly varying tail distribution with index $\lambda=1$ ;
D3:

$T$ has regularly varying tail distribution with index $\lambda\in(0,1)$ ;
D4:

$T$ has slowly varying tail distribution.

Define for any $x\geq 0$

(2.3)

\displaystyle\mathcal{B}_{\alpha}(x)=\lim_{S\to\infty}S^{-1}\mathcal{B}_{\alpha}(S,x),

with

(2.4)

\displaystyle\mathcal{B}_{\alpha}(S,x)=\int_{\mathbb{R}}\mathbb{P}\left\{\int_{0}^{S}\mathbb{I}_{0}\left(W_{\alpha}(s)+z\right)\text{\rm d}s>x\right\}e^{-z}\text{\rm d}z,\quad W_{\alpha}(t)=\sqrt{2}B_{\alpha}(t)-\left\lvert t\right\rvert^{\alpha},

where $B_{\alpha}$ is a standard fractional Brownian motion (fBm) with Hurst index $\alpha/2\in(0,1]$ . By Theorem 2.1 in [6], we know that $\mathcal{B}_{\alpha}(x)$ is positive and finite for any $x\geq 0$ . Let $\mathcal{E}$ be a unit exponential random variable independent of $W_{\alpha}$ and set

\displaystyle\mathcal{G}_{\alpha}(x)=\mathbb{P}\left\{\int_{\mathbb{R}}\mathbb{I}_{0}\left(W_{\alpha}(s)+\mathcal{E}\right)\text{\rm d}s\leq x\right\}.

As shown in [7], $\mathcal{G}_{\alpha}$ is continuous on $\mathbb{R}^{+}$ , and thus by Remark 2.2 ii) in [6]

\mathcal{B}_{\alpha}(x)=\int_{x}^{\infty}\frac{1}{y}\text{\rm d}\mathcal{G}_{\alpha}(y)

holds for all $x\in\mathbb{R}^{+}$ . We note that $\mathcal{B}_{\alpha}(0)$ is equal to the classical Pickands constant; see e.g. [17] or Section 10 in [5]. Let

(2.5)

\displaystyle m(u)=\big{(}\mathcal{B}_{\alpha}(0)v(u)\Psi(u)\big{)}^{-1},

where $\Psi(u)$ is the survival function of an $N(0,1)$ random variable. Then, by Theorem 10.5.1 in [5],

(2.6)

\displaystyle\mathbb{P}\left\{\sup_{t\in[0,1]}X(t)>u\right\}\sim m^{-1}(u),\quad u\to\infty.

In our notation $\sim$ stands for asymptotic equivalence of two functions as the argument tends to 0 or to $\infty$ respectively.

3. Main results

In this section we find the exact asymptotics of

(3.1)

\displaystyle\mathbb{P}\left\{L_{u}^{*}[0,T]>x\right\}

as $u\to\infty$ , under scenarios D1-D4, respectively. All the proofs are postponed to Section 5.

3.1. Scenario D1

We begin with the case when $T$ is integrable. It appears that under this scenario the main contribution to the asymptotics of (3.1) comes from Gaussian process $X$ , whereas $T$ contributes only by its average behavior.

Theorem 3.1.

Let $X(t),t\geq 0$ be a centered stationary Gaussian process with unit variance and covariance function satisfying A1-A2. Suppose that $T$ is an independent of $X$ nonnegative random variable that satisfies D1. Then for any $x\geq 0$

(3.2)

\displaystyle\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}\sim\mathcal{B}_{\alpha}(x)\mathbb{E}\left\{T\right\}v(u)\Psi(u),\quad u\to\infty.

We can rewrite the result in Theorem 3.1 as

\displaystyle\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}\sim\mathbb{E}\left\{T\right\}\mathbb{P}\left\{L_{u}^{\ast}[0,1]>x\right\},\quad u\to\infty.

3.2. Scenario D2

Under this scenario the asymptotics of (3.1) is similar to the one obtained for case D1 with the exception that $T$ contributes to (3.1) by its integrated tail distribution rather than by its mean.

Theorem 3.2.

Let $X(t),t\geq 0$ be a centered stationary Gaussian process with unit variance and covariance function satisfying A1-A3. Suppose that $T$ is an independent of $X$ nonnegative random variable that satisfies D2. Then for any $x\geq 0$

(3.3)

\displaystyle\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}\sim\mathcal{B}_{\alpha}(x)l(m(u))v(u)\Psi(u),\quad u\to\infty,

where $l(u)=\int_{0}^{u}\mathbb{P}\left\{T>t\right\}\text{\rm d}t$ .

Remark 3.3.

We note that if $T$ satisfies D2 and is integrable, then (3.3) coincides with (3.2).

3.3. Scenario D3

This scenario leads to the asymptotics of (3.1) which depends only of the heaviness of the tail distribution of $T$ .

The following continuous distribution function

(3.4)

\displaystyle\mathcal{F}_{\alpha}(x):=\mathcal{B}_{\alpha}^{-1}(0)\int_{0}^{x}\frac{1}{y}\text{\rm d}\mathcal{G}_{\alpha}(x),\quad x\geq 0

plays an important role in further analysis. $\overline{\mathcal{F}_{\alpha}^{*k}}(x)$ denotes the tail distribution of the $k$ -th convolution of $\mathcal{F}_{\alpha}$ at $x\geq 0$ .

Theorem 3.4.

(3.5)

\displaystyle\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}\sim\lambda\sum_{k=1}^{\infty}\frac{\Gamma(k-\lambda)}{k!}\overline{\mathcal{F}_{\alpha}^{*k}}(x)\mathbb{P}\left\{T>m(u)\right\},\quad u\to\infty.

Remark 3.5.

Taking $x=0$ in (3.5) and using

\lambda\sum_{k=1}^{\infty}\frac{\Gamma(k-\lambda)}{k!}=\lambda\sum_{k=1}^{\infty}\frac{1}{k!}\int_{0}^{\infty}l^{k-\lambda-1}e^{-l}\text{\rm d}l=\lambda\int_{0}^{\infty}(1-e^{-l})l^{-\lambda-1}\text{\rm d}l=\Gamma(1-\lambda),

we recover Theorem 3.2 in [16].

3.4. Scenario D4

Suppose now that $T$ has slowly varying tail distribution at $\infty$ . As shown in the following theorem, similarly to case D3, the asymptotics of (3.1) depends only on the asymptotic behavior of the tail distribution of $T$ but in contrast to scenario D3 doesn’t depend on $x$ .

Theorem 3.6.

\displaystyle\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}\sim\mathbb{P}\left\{T>m(u)\right\},\quad u\to\infty.

4. Examples

In this section we illustrate the results derived in Section 3 by two classes of stationary Gaussian processes: fractional Ornstein-Uhlenbeck processes and increments of fractional Brownian motions.

4.1. Fractional Ornstein-Uhlenbeck processes

Suppose that $X$ is a centered stationary Gaussian process with covariance $r(t)=e^{-t^{\alpha}},t\geq 0$ , for $\alpha\in(0,2]$ . We call $X$ a fractional Urnstein-Uhlenbeck process with index $\alpha$ . If $\alpha=1$ , then $X$ is the classical Ornstein-Uhlenbeck process.

It is straightforward to check that A1-A3 are satisfied. Thus, the following proposition holds due to Theorems 3.1-3.6.

Proposition 4.1.

Suppose that $X$ is a fractional Ornstein-Uhlenbeck process with index $\alpha\in(0,2]$ , and $T$ is an independent of $X$ nonnegative random variable. Then for any $x\geq 0$ , as $u\to\infty$ ,
(i) If $T\in$ D1, then $\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}\sim\mathcal{B}_{\alpha}(x)\mathbb{E}\left\{T\right\}(2\pi)^{-1/2}u^{2/\alpha-1}e^{-u^{2}/2}$ .
(ii) If $T\in$ D2, then $\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}\sim\mathcal{B}_{\alpha}(x)(2\pi)^{-1/2}u^{2/\alpha-1}e^{-u^{2}/2}\int_{0}^{\sqrt{2\pi}\mathcal{B}_{\alpha}^{-1}(0)u^{1-2/\alpha}e^{u^{2}/2}}\mathbb{P}\left\{T>t\right\}\text{\rm d}t.$
(iii) If $T\in$ D3, then $\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}\sim\lambda\sum_{k=1}^{\infty}\frac{\Gamma(k-\lambda)}{k!}\overline{\mathcal{F}_{\alpha}^{*k}}(x)\mathbb{P}\left\{T>\sqrt{2\pi}\mathcal{B}_{\alpha}^{-1}(0)u^{1-2/\alpha}e^{u^{2}/2}\right\}$ .
(iv) If $T\in$ D4, then $\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}\sim\mathbb{P}\left\{T>\sqrt{2\pi}\mathcal{B}_{\alpha}^{-1}(0)u^{1-2/\alpha}e^{u^{2}/2}\right\}$ .

4.2. Increments of fractional Brownian motion

For a standard fBm $B_{\alpha}(t),t\geq 0$ with Hurst index $\alpha/2\in(0,1)$ and $a>0$ , define

\displaystyle X_{\alpha,a}(t):=\frac{B_{\alpha}(t+a)-B_{\alpha}(t)}{a^{\alpha/2}},\quad t\geq 0.

One can check that $X_{\alpha,a}$ is a centered stationary Gaussian process with unit variance and covariance function

r(t)=\frac{(a+t)^{\alpha}+\left\lvert a-t\right\rvert^{\alpha}-2t^{\alpha}}{2a^{\alpha}},\quad t\geq 0.

and $1-r(t)\sim a^{-\alpha}t^{\alpha},$ $t\to 0,$ which verifies assumption A1. Similarly for $t>a$

\displaystyle\left\lvert r(t)\right\rvert\leq\frac{\alpha\left\lvert 1-\alpha\right\rvert(t-a)^{\alpha-2}}{2a^{\alpha-2}},

which confirms assumption A3. Thus the following proposition holds.

Proposition 4.2.

Suppose that $X_{\alpha,a}(t),t\geq 0$ with $\alpha\in(0,2)$ , $a>0$ is independent of a nonnegative random variable $T$ . Then for any $x\geq 0$ , as $u\to\infty$ ,
(i) If $T\in$ D1, then $\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}\sim\mathcal{B}_{\alpha}(x)\mathbb{E}\left\{T\right\}(2\pi)^{-1/2}a^{-1}u^{2/\alpha-1}e^{-u^{2}/2}$ .
(ii) If $T\in$ D2, then

\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}\sim\mathcal{B}_{\alpha}(x)(2\pi)^{-1/2}a^{-1}u^{2/\alpha-1}e^{-u^{2}/2}\int_{0}^{\sqrt{2\pi}a\mathcal{B}_{\alpha}^{-1}(0)u^{1-2/\alpha}e^{u^{2}/2}}\mathbb{P}\left\{T>t\right\}\text{\rm d}t.

(iii) If $T\in$ D3, then $\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}\sim\lambda\sum_{k=1}^{\infty}\frac{\Gamma(k-\lambda)}{k!}\overline{\mathcal{F}_{\alpha}^{*k}}(x)\mathbb{P}\left\{T>\sqrt{2\pi}a\mathcal{B}_{\alpha}^{-1}(0)u^{1-2/\alpha}e^{u^{2}/2}\right\};$
(iv) If $T\in$ D4, then $\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}\sim\mathbb{P}\left\{T>\sqrt{2\pi}a\mathcal{B}_{\alpha}^{-1}(0)u^{1-2/\alpha}e^{u^{2}/2}\right\}$ .

5. Proofs

In this section we give detailed proofs of all the theorems presented in Section 3. We first give a simple extension of Theorem 7.4.1 of [5].

Lemma 5.1.

Let $X(t),t\geq 0$ be a centered stationary Gaussian process with unit variance and covariance function satisfying A1 and A3. If $L_{u}^{\ast}$ , $m(u)$ and $\mathcal{F}_{\alpha}$ are defined in (2.2), (2.5) and (3.4), respectively, then for any $s\geq 0$ and $0<l_{0}<l_{1}<\infty$ we have

(5.1)

\displaystyle\lim_{u\to\infty}\sup_{\tau\in[l_{0},l_{1}]}\left\lvert\mathbb{E}\left\{e^{-sL_{u}^{\ast}[0,\tau m(u)]}\right\}-e^{-\tau\int_{0}^{\infty}(1-e^{-sx})\text{\rm d}\mathcal{F}_{\alpha}(x)}\right\rvert=0.

Proof of Lemma 5.1 For any $\tau>0$ , the point convergence follows from Berman’s proof of Theorem 7.4.1 in [5]. The uniformity of the convergence on $[l_{0},l_{1}]$ follows by monotonicity of $\mathbb{E}\left\{e^{-sL_{u}^{\ast}[0,\tau m(u)]}\right\}$ and by continuity of $e^{-\tau\int_{0}^{\infty}(1-e^{-sx})\text{\rm d}\mathcal{F}_{\alpha}(x)}$ as function of $\tau$ . $\Box$

Define a compound Poisson process

(5.2)

\displaystyle Y(t)=\sum_{i=1}^{N(t)}\xi_{i},

where $\{N(t):t\geq 0\}$ is a Poisson process with unit intensity, and $\{\xi_{i}:i\geq 1\}$ are independent and identically distributed random variables, with distribution function $\mathcal{F}_{\alpha}$ , which are also independent of $N$ . The following corollary of Lemma 5.1 will play an important role in the proof of Theorem 3.4.

Corollary 5.1.

If $X$ is the Gaussian process given as in Lemma 5.1 and $Y$ is defined in (5.2), then for any $x\geq 0$ and $0<l_{0}<l_{1}<\infty$ we have

(5.3)

\displaystyle\lim_{u\to\infty}\sup_{l\in[l_{0},l_{1}]}\left\lvert\mathbb{P}\left\{L_{u}^{\ast}[0,lm(u)]>x\right\}-\mathbb{P}\left\{Y(l)>x\right\}\right\rvert=0.

Proof of Corollary 5.1 For arbitrary $l>0$ , by (5.1),

\displaystyle\mathbb{P}\left\{L_{u}^{\ast}[0,lm(u)]>x\right\}\rightarrow\mathbb{P}\left\{Y(l)>x\right\},\quad u\to\infty

holds for any $x>0$ . Further,

\displaystyle\left\lvert\mathbb{P}\left\{Y(l_{1})>x\right\}-\mathbb{P}\left\{Y(l_{0})>x\right\}\right\rvert\leq\mathbb{P}\left\{Y(\left\lvert l_{1}-l_{0}\right\rvert)>0\right\}=1-e^{-\left\lvert l_{1}-l_{0}\right\rvert}\leq\left\lvert l_{1}-l_{0}\right\rvert,

which implies that for any $x>0$ , $\mathbb{P}\left\{Y(l)>x\right\}$ is continuous in $l$ . Finally, the uniform convergence follows by the same argument as stated in Lemma 5.1. For $x=0$ in (5.3), we refer to Lemma 4.3 in [16]. This completers the proof. $\Box$

5.1. Proof of Theorem 3.1

By (2.6), for arbitrary $\varepsilon>0$ , there exists large enough $u$ such that

\mathbb{P}\left\{\sup_{t\in[0,1]}X(t)>u\right\}<(1+\varepsilon)m^{-1}(u),

which together with the stationarity of process $X$ implies that for any $x\geq 0$ and $t>0$

$\displaystyle\frac{\mathbb{P}\left\{L_{u}^{\ast}[0,t]>x\right\}}{v(u)\Psi(u)}$	$\displaystyle\leq$	$\displaystyle\frac{\mathbb{P}\left\{\sup_{s\in[0,t]}X(s)>u\right\}}{v(u)\Psi(u)}$
	$\displaystyle\leq$	$\displaystyle(t+1)\frac{\mathbb{P}\left\{\sup_{s\in[0,1]}X(s)>u\right\}}{v(u)\Psi(u)}$
	$\displaystyle\leq$	$\displaystyle(t+1)(1+\varepsilon)\mathcal{B}_{\alpha}(0).$

Consequently, for nonnegative random variable $T$ with distribution function $F_{T}$ satisfying D1, by dominated convergence theorem and Remark 2.2 i) in [6] we have

$\displaystyle\lim_{u\to\infty}\frac{\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}}{v(u)\Psi(u)}$	$\displaystyle=$	$\displaystyle\lim_{u\to\infty}\int_{0}^{\infty}\frac{\mathbb{P}\left\{L_{u}^{\ast}[0,t]>x\right\}}{v(u)\Psi(u)}\text{\rm d}F_{T}(t)$
	$\displaystyle=$	$\displaystyle\mathcal{B}_{\alpha}(x)\int_{0}^{\infty}t\text{\rm d}F_{T}(t)$
	$\displaystyle=$	$\displaystyle\mathcal{B}_{\alpha}(x)\mathbb{E}\left\{T\right\}.$

This completes the proof. $\Box$

5.2. Proof of Theorem 3.2

Let $A(u)$ satisfy

\displaystyle\lim_{u\to\infty}A(u)v(u)=\infty\quad\textrm{and}\quad\lim_{u\to\infty}A(u)=0.

By Corollary 6.1, for any $x\geq 0$ and arbitrary $\varepsilon\in(0,1)$ , there exist $\delta>0$ and $u_{0}$ such that

\displaystyle\inf_{t\in[A(u),\delta m(u)]}\frac{\mathbb{P}\left\{L_{u}^{*}[0,t]>x\right\}}{t\mathcal{B}_{\alpha}(x)v(u)\Psi(u)}\geq 1-2\varepsilon,\quad u>u_{0}

and

\displaystyle\sup_{t\in[A(u),\delta m(u)]}\frac{\mathbb{P}\left\{L_{u}^{*}[0,t]>x\right\}}{t\mathcal{B}_{\alpha}(x)v(u)\Psi(u)}\leq 1+2\varepsilon,\quad u>u_{0}.

Therefore,

$\displaystyle\liminf_{u\to\infty}\frac{\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}}{\mathcal{B}_{\alpha}(x)l(m(u))v(u)\Psi(u)}$	$\displaystyle\geq$	$\displaystyle\liminf_{u\to\infty}\frac{\int_{A(u)}^{\delta m(u)}\mathbb{P}\left\{L_{u}^{\ast}[0,t]>x\right\}\text{\rm d}F_{T}(t)}{\mathcal{B}_{\alpha}(x)l(m(u))v(u)\Psi(u)}$
	$\displaystyle\geq$	$\displaystyle(1-2\varepsilon)\liminf_{u\to\infty}\frac{\int_{A(u)}^{\delta m(u)}t\text{\rm d}F_{T}(t)}{l(m(u))}$
	$\displaystyle=$	$\displaystyle(1-2\varepsilon)\liminf_{u\to\infty}\frac{\int_{0}^{\delta m(u)}t\text{\rm d}F_{T}(t)}{l(m(u))}$
	$\displaystyle=$	$\displaystyle(1-2\varepsilon)\liminf_{u\to\infty}\frac{\int_{0}^{\delta m(u)}\mathbb{P}\left\{T>t\right\}\text{\rm d}t-\delta m(u)\mathbb{P}\left\{T>\delta m(u)\right\}}{l(m(u))}$
	$\displaystyle=$	$\displaystyle(1-2\varepsilon),$

where the last inequality follows by Proposition 1.5.9a in [18] such that $l(u)$ is slowly varying at $\infty$ and $\lim_{u\to\infty}u\mathbb{P}\left\{T>u\right\}/l(u)=0$ .

Similarly,

	$\displaystyle\quad\limsup_{u\to\infty}\frac{\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}}{\mathcal{B}_{\alpha}(x)l(m(u))v(u)\Psi(u)}$
	$\displaystyle\leq\limsup_{u\to\infty}\frac{\mathbb{P}\left\{L_{u}^{\ast}[0,A(u)]>x\right\}\mathbb{P}\left\{T\leq A(u)\right\}+\int_{A(u)}^{\delta m(u)}\mathbb{P}\left\{L_{u}^{\ast}[0,t]>x\right\}\text{\rm d}F_{T}(t)+\mathbb{P}\left\{T>\delta m(u)\right\}}{\mathcal{B}_{\alpha}(x)l(m(u))v(u)\Psi(u)}$
	$\displaystyle\leq\limsup_{u\to\infty}\frac{A(u)\mathbb{P}\left\{T\leq A(u)\right\}}{l(m(u))}+(1+2\varepsilon)\limsup_{u\to\infty}\frac{\int_{A(u)}^{\delta m(u)}t\text{\rm d}F_{T}(t)}{l(m(u))}$
	$\displaystyle=(1+2\varepsilon),$

where the last inequality follows by (6.20) and the same reasons as above. Since $\varepsilon$ is arbitrary, letting $\varepsilon\to 0$ , we complete the proof. $\Box$

5.3. Proof of Theorem 3.4

First, note that by Raabe’s Test, the series in (3.5) converges for $\lambda\in(0,1)$ . Then by integration by parts, for any $x\geq 0$ we have

(5.4)	$\displaystyle\int_{0}^{\infty}l^{-\lambda}\text{\rm d}\mathbb{P}\left\{Y(l)>x\right\}$	$\displaystyle=$	$\displaystyle\int_{0}^{\infty}\mathbb{P}\left\{Y(l)>x\right\}\lambda l^{-\lambda-1}\text{\rm d}l-\lim_{l\to 0}l^{-\lambda}\mathbb{P}\left\{Y(l)>x\right\}$
		$\displaystyle=$	$\displaystyle\lambda\sum_{k=1}^{\infty}\overline{\mathcal{F}_{\alpha}^{*k}}(x)\frac{1}{k!}\int_{0}^{\infty}l^{k-\lambda-1}e^{-l}\text{\rm d}l-\lim_{l\to 0}l^{-\lambda}\mathbb{P}\left\{Y(l)>x\right\}$
		$\displaystyle=$	$\displaystyle\lambda\sum_{k=1}^{\infty}\frac{\Gamma(k-\lambda)}{k!}\overline{\mathcal{F}_{\alpha}^{*k}}(x)<\infty,$

where the last equality, recalling that $\lambda\in(0,1)$ , follows by

(5.5)

\displaystyle\lim_{l\to 0}l^{-\lambda}\mathbb{P}\left\{Y(l)>x\right\}\leq\lim_{l\to 0}l^{-\lambda}(1-e^{-l})=0.

Next, by a similar argument as used in the proof of Theorem 3.2 in [16], for any $0<l_{0}<l_{1}<\infty$ we have

	$\displaystyle\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}$	$\displaystyle=$	$\displaystyle\Big{(}\int_{0}^{l_{0}m(u)}+\int_{l_{0}m(u)}^{l_{1}m(u)}+\int_{l_{1}m(u)}^{\infty}\Big{)}\mathbb{P}\left\{L_{u}^{\ast}[0,l]>x\right\}\text{\rm d}F_{T}(l)$
		$\displaystyle=$	$\displaystyle I_{1}+I_{2}+I_{3},$

where

\displaystyle\limsup_{u\to\infty}\frac{I_{1}}{\mathbb{P}\left\{T>m(u)\right\}}\leq\limsup_{u\to\infty}\frac{\int_{0}^{l_{0}m(u)}\mathbb{P}\left\{\sup_{s\in[0,l]}X(s)>u\right\}\text{\rm d}F_{T}(l)}{\mathbb{P}\left\{T>m(u)\right\}}\leq\frac{\lambda}{1-\lambda}l_{0}^{1-\lambda}

and

\displaystyle\limsup_{u\to\infty}\frac{I_{3}}{\mathbb{P}\left\{T>m(u)\right\}}\leq\limsup_{u\to\infty}\frac{\mathbb{P}\left\{T>l_{1}m(u)\right\}}{\mathbb{P}\left\{T>m(u)\right\}}=l_{1}^{-\lambda}.

Further, in view of Corollary 5.1, for any given $x\geq 0$ and arbitrary $\varepsilon>0$ , we have the following upper bound

$\displaystyle I_{2}$	$\displaystyle=$	$\displaystyle\int_{l_{0}}^{l_{1}}\mathbb{P}\left\{L_{u}^{*}[0,lm(u)]>x\right\}\text{\rm d}F_{T}(lm(u))$
	$\displaystyle\leq$	$\displaystyle(1+\varepsilon)\int_{l_{0}}^{l_{1}}\mathbb{P}\left\{Y(l)>x\right\}\text{\rm d}F_{T}(lm(u))$
	$\displaystyle=$	$\displaystyle(1+\varepsilon)\Big{(}\int_{l_{0}}^{l_{1}}\mathbb{P}\left\{T>lm(u)\right\}\text{\rm d}\mathbb{P}\left\{Y(l)>x\right\}$
		$\displaystyle\qquad\qquad-\mathbb{P}\left\{Y(l_{1})>x\right\}\mathbb{P}\left\{T>l_{1}m(u)\right\}+\mathbb{P}\left\{Y(l_{0})>x\right\}\mathbb{P}\left\{T>l_{0}m(u)\right\}\Big{)},$

which holds for $u$ large enough. By Potter’s Theorem (see, e.g., [18][Theorem 1.5.6]), for sufficiently large $u$ there exists some constant $C>1$ such that

\frac{\mathbb{P}\left\{T>lm(u)\right\}}{\mathbb{P}\left\{T>m(u)\right\}}\leq Cl^{-2\lambda}

holds for all $l\in[l_{0},l_{1}]$ , and thus by dominated convergence theorem

\displaystyle\limsup_{u\to\infty}\frac{I_{2}}{\mathbb{P}\left\{T>m(u)\right\}}\leq(1+\varepsilon)\Big{(}\int_{l_{0}}^{l_{1}}l^{-\lambda}\text{\rm d}\mathbb{P}\left\{Y(l)>x\right\}-\mathbb{P}\left\{Y(l_{1})>x\right\}l_{1}^{-\lambda}+\mathbb{P}\left\{Y(l_{0})>x\right\}l_{0}^{-\lambda}\Big{)}.

Similarly, we have the lower bound

\displaystyle\liminf_{u\to\infty}\frac{I_{2}}{\mathbb{P}\left\{T>m(u)\right\}}\geq(1-\varepsilon)\Big{(}\int_{l_{0}}^{l_{1}}l^{-\lambda}\text{\rm d}\mathbb{P}\left\{Y(l)>x\right\}-\mathbb{P}\left\{Y(l_{1})>x\right\}l_{1}^{-\lambda}+\mathbb{P}\left\{Y(l_{0})>x\right\}l_{0}^{-\lambda}\Big{)}.

Finally, letting $l_{0}\to 0$ and $l_{1}\to\infty$ in the above bounds, using (5.4) and (5.5), and then by the fact that $\varepsilon>0$ was arbitrary, we complete the proof. $\Box$

5.4. Proof of Theorem 3.6

According to Remark 3.3 in [16], we know that

\displaystyle\limsup_{u\to\infty}\frac{\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}}{\mathbb{P}\left\{T>m(u)\right\}}\leq\limsup_{u\to\infty}\frac{\mathbb{P}\left\{\sup_{s\in[0,T]}X(s)>u\right\}}{\mathbb{P}\left\{T>m(u)\right\}}\leq 1.

Further, by Corollary 5.1, for arbitrary $l>0$

\displaystyle\mathbb{P}\left\{L_{u}^{\ast}[0,lm(u)]>x\right\}\rightarrow\mathbb{P}\left\{Y(l)>x\right\}

holds for any $x\geq 0$ as $u\to\infty$ . Thus, for $T$ with slowly varying tail distribution we get

	$\displaystyle\liminf_{u\to\infty}\frac{\mathbb{P}\left\{L_{u}^{\ast}[0,T]>x\right\}}{\mathbb{P}\left\{T>m(u)\right\}}$	$\displaystyle\geq$	$\displaystyle\liminf_{u\to\infty}\frac{\mathbb{P}\left\{L_{u}^{\ast}[0,lm(u)]>x\right\}\mathbb{P}\left\{T>lm(u)\right\}}{\mathbb{P}\left\{T>m(u)\right\}}$
		$\displaystyle=$	$\displaystyle\mathbb{P}\left\{Y(l)>x\right\},$

which converges to $1$ as $l\to\infty$ , since by the strong law of large numbers $Y(l)/l\to\mathcal{B}_{\alpha}^{-1}(0)>0$ . This completes the proof. $\Box$

6. Appendix

Hereafter, $C_{i},i\in\mathbb{N}$ are positive constants which may be different from line to line. All vectors are column vectors unless otherwise specified. As long as it doesn’t cause confusion we use $\boldsymbol{0}$ to denote the $2\times 1$ column vector or the $2\times 2$ matrix whose entries are all 0’s. For a given vector (matrix) $\mathbf{Q}$ , let $|\mathbf{Q}|$ denote vector (matrix) with entries equal to absolute value of respective entries of $\mathbf{Q}$ .

Lemma 6.1.

Let $X(t),t\geq 0$ be a centered stationary Gaussian process with unit variance and covariance function satisfying A1 and A3. If $v(u)$ , $\mathcal{B}_{\alpha}(S,x)$ and $m(u)$ are defined in (2.1), (2.4) and (2.5), respectively, then for any $A(u)>1$ satisfying

(6.1)

\displaystyle\limsup_{u\to\infty}\frac{u^{2}}{\log A(u)}<\infty

we have

\displaystyle\ \ \ \ \ \lim_{u\to\infty}\sup_{d\geq A(u)}\left\lvert\frac{\mathbb{P}\left\{\sup_{s_{1}\in[0,S]}X(s_{1}/v(u))>u,\sup_{s_{2}\in[0,S]}X(d+s_{2}/v(u))>u\right\}}{\Psi^{2}(u)}-\mathcal{B}_{\alpha}^{2}(S,0)\right\rvert=0.

Proof of Lemma 6.1 We borrow the argument used in the proof of Theorem 5.1 in [6]. First, for notational simplicity we define

\displaystyle\xi_{u,d}(\boldsymbol{s})=(X(s_{1}/v(u)),X(d+s_{2}/v(u)))^{T},\quad\boldsymbol{s}=(s_{1},s_{2})\in\boldsymbol{D}=[0,S]^{2},

and denote by $R_{u,d}(\boldsymbol{s},\boldsymbol{t})$ the covariance matrix function of $\xi_{u,d}$ , i.e.,

$\displaystyle R_{u,d}(\boldsymbol{s},\boldsymbol{t})$	$\displaystyle=$	$\displaystyle\mathrm{Cov}(\xi_{u,d}(\boldsymbol{s}),\xi_{u,d}(\boldsymbol{t}))$
	$\displaystyle=$	$\displaystyle\mathbb{E}\left\{\xi_{u,d}(\boldsymbol{s})\xi_{u,d}(\boldsymbol{t})^{T}\right\}$
	$\displaystyle=$	$\displaystyle\begin{pmatrix}r(\frac{\|t_{1}-s_{1}\|}{v(u)})&r(d+\frac{t_{2}-s_{1}}{v(u)})\\ r(d+\frac{s_{2}-t_{1}}{v(u)})&r(\frac{\|t_{2}-s_{2}\|}{v(u)})\end{pmatrix}\quad\boldsymbol{s},\boldsymbol{t}\in\boldsymbol{D}.$

Then, conditioning on $\xi_{u,d}(\boldsymbol{0})$ we have

\displaystyle\mathbb{P}\left\{\exists_{\boldsymbol{s}\in\boldsymbol{D}}\xi_{u,d}(\boldsymbol{s})>\boldsymbol{u}\right\}=\iint_{\mathbb{R}^{2}}\mathbb{P}\left\{\exists_{\boldsymbol{s}\in\boldsymbol{D}}\xi_{u,d}(\boldsymbol{s})>\boldsymbol{u}|\xi_{u,d}(\boldsymbol{0})=\boldsymbol{y}\right\}\phi(y_{1},y_{2};r(d))\text{\rm d}y_{1}\text{\rm d}y_{2},

where $\boldsymbol{y}=(y_{1},y_{2})^{T}$ and $\phi(y_{1},y_{2};r(d))$ is the density function of bivariate normal random variable $\xi_{u,d}(\boldsymbol{0})$ . By the change of variables $\boldsymbol{y}=\boldsymbol{u}+\boldsymbol{z}/u$ and using properties of conditional distribution of normal random variable (see e.g., Chapter 2.2 in [5]), we get

	$\displaystyle\mathbb{P}\left\{\exists_{\boldsymbol{s}\in\boldsymbol{D}}\xi_{u,d}(\boldsymbol{s})>\boldsymbol{u}\right\}$	$\displaystyle=$	$\displaystyle\frac{e^{-u^{2}}}{2\pi u^{2}}\iint_{\mathbb{R}^{2}}\mathbb{P}\left\{\exists_{\boldsymbol{s}\in\boldsymbol{D}}\chi_{u,d}(\boldsymbol{s})-\theta_{u,d}(\boldsymbol{s},\boldsymbol{z})>\boldsymbol{0}\right\}f_{u,d}(\boldsymbol{z})\text{\rm d}z_{1}\text{\rm d}z_{2}$
		$\displaystyle=$	$\displaystyle\frac{e^{-u^{2}}}{2\pi u^{2}}\iint_{\mathbb{R}^{2}}\mathcal{I}_{u,d}(\boldsymbol{z})f_{u,d}(\boldsymbol{z})\text{\rm d}z_{1}\text{\rm d}z_{2},$

where $\mathcal{I}_{u,d}(\boldsymbol{z}):=\mathbb{P}\left\{\exists_{\boldsymbol{s}\in\boldsymbol{D}}\chi_{u,d}(\boldsymbol{s})-\theta_{u,d}(\boldsymbol{s},\boldsymbol{z})>\boldsymbol{0}\right\}$ with

\chi_{u,d}(\boldsymbol{s})=u\left(\xi_{u,d}(\boldsymbol{s})-R_{u,d}(\boldsymbol{s},\boldsymbol{0})R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})\xi_{u,d}(\boldsymbol{0})\right),\quad\boldsymbol{s}\in\boldsymbol{D},

\theta_{u,d}(\boldsymbol{s},\boldsymbol{z})=u^{2}\left(\boldsymbol{1}-R_{u,d}(\boldsymbol{s},\boldsymbol{0})R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})(\boldsymbol{1}+\boldsymbol{z}/u^{2})\right),\quad\boldsymbol{s}\in\boldsymbol{D},\boldsymbol{z}\in\mathbb{R}^{2}

and

f_{u,d}(\boldsymbol{z})=\frac{1}{\sqrt{1-r^{2}(d)}}\exp\left(\frac{1}{1+r(d)}(u^{2}r(d)-z_{1}-z_{2})-\frac{(z_{1}-r(d)z_{2})^{2}}{2u^{2}(1-r^{2}(d))}-\frac{z_{2}^{2}}{2u^{2}}\right),\quad\boldsymbol{z}\in\mathbb{R}^{2}.

Consequently, in order to show the claim it suffices to prove that for $W_{\alpha}(\boldsymbol{s})=(\sqrt{2}B_{\alpha}^{(1)}(s_{1})-s_{1}^{\alpha},\sqrt{2}B_{\alpha}^{(2)}(s_{2})-s_{2}^{\alpha})^{T}$ , $\boldsymbol{s}\in\boldsymbol{D}$ , where $B_{\alpha}^{(1)}$ and $B_{\alpha}^{(2)}$ are two independent fBm’s with Hurst index $\alpha/2$ ,

(6.2)	$\displaystyle\lim_{u\to\infty}\sup_{d\geq A(u)}\left\lvert\iint_{\mathbb{R}^{2}}\mathcal{I}_{u,d}(\boldsymbol{z})f_{u,d}(\boldsymbol{z})\text{\rm d}z_{1}\text{\rm d}z_{2}-\mathcal{B}_{\alpha}^{2}(S,0)\right\rvert$
		$\displaystyle=\lim_{u\to\infty}\sup_{d\geq A(u)}\left\lvert\iint_{\mathbb{R}^{2}}\bigg{(}\mathcal{I}_{u,d}(\boldsymbol{z})f_{u,d}(\boldsymbol{z})-\mathbb{P}\left\{\exists_{\boldsymbol{s}\in\boldsymbol{D}}W_{\alpha}(\boldsymbol{s})+\boldsymbol{z}>\boldsymbol{0}\right\}e^{-z_{1}-z_{2}}\bigg{)}\text{\rm d}z_{1}\text{\rm d}z_{2}\right\rvert$
		$\displaystyle=\lim_{u\to\infty}\sup_{d\geq A(u)}\left\lvert\iint_{\mathbb{R}^{2}}\big{(}\mathcal{I}_{u,d}(\boldsymbol{z})f_{u,d}(\boldsymbol{z})-\mathcal{I}(\boldsymbol{z})e^{-z_{1}-z_{2}}\big{)}\text{\rm d}z_{1}\text{\rm d}z_{2}\right\rvert=0,$

with

(6.3)		$\displaystyle\mathcal{I}(\boldsymbol{z})$	$\displaystyle:=$	$\displaystyle\mathbb{P}\left\{\exists_{\boldsymbol{s}\in\boldsymbol{D}}W_{\alpha}(\boldsymbol{s})+\boldsymbol{z}>\boldsymbol{0}\right\}$
(6.4)			$\displaystyle=$	$\displaystyle\mathbb{P}\left\{\sup_{s_{\in}[0,S]}\sqrt{2}B_{\alpha}(s)-s^{\alpha}>-z_{1}\right\}\mathbb{P}\left\{\sup_{s\in[0,S]}\sqrt{2}B_{\alpha}(s)-s^{\alpha}>-z_{2}\right\}.$

For any $\boldsymbol{s},\boldsymbol{t}\in\boldsymbol{D}$ ,

(6.5)	$\displaystyle\mathrm{Cov}(\chi_{u,d}(\boldsymbol{s}),\chi_{u,d}(\boldsymbol{t}))$
		$\displaystyle=u^{2}\mathrm{Cov}\big{(}\xi_{u,d}(\boldsymbol{s})-R_{u,d}(\boldsymbol{s},\boldsymbol{0})R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})\xi_{u,d}(\boldsymbol{0}),\,\,\xi_{u,d}(\boldsymbol{t})-R_{u,d}(\boldsymbol{t},\boldsymbol{0})R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})\xi_{u,d}(\boldsymbol{0})\big{)}$
		$\displaystyle=u^{2}\{R_{u,d}(\boldsymbol{s},\boldsymbol{t})-R_{u,d}(\boldsymbol{s},\boldsymbol{0})R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})R_{u,d}(\boldsymbol{0},\boldsymbol{t})\}$
		$\displaystyle=u^{2}\{(R_{u,d}(\boldsymbol{s},\boldsymbol{t})-E)+(E-R_{u,d}(\boldsymbol{s},\boldsymbol{0}))+R_{u,d}(\boldsymbol{s},\boldsymbol{0})(E-R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})R_{u,d}(\boldsymbol{0},\boldsymbol{t}))\}$

where $E$ is the $2\times 2$ identity matrix.

Since $A(u)>1$ satisfying (6.1) tends to $\infty$ as $u\to\infty$ , then by A3 we have

(6.6)

\displaystyle\lim_{u\to\infty}\sup_{d>A(u),\boldsymbol{s}\in\boldsymbol{D}}\left\lvert R_{u,d}(\boldsymbol{s},\boldsymbol{0})-E\right\rvert=\boldsymbol{0}

and

(6.7)

\displaystyle\lim_{u\to\infty}u^{2}\sup_{d\geq A(u)}\left\lvert r(d)\right\rvert\leq\lim_{u\to\infty}\frac{u^{2}}{\log A(u)}\sup_{d\geq A(u)}\left\lvert r(d)\right\rvert\log d=0.

Therefore,

(6.8)

\displaystyle\lim_{u\to\infty}\sup_{d\geq A(u)}u^{2}(E-R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0}))=\lim_{u\to\infty}\sup_{d\geq A(u)}\frac{u^{2}r(d)}{1-r^{2}(d)}\begin{pmatrix}-r(d)&1\\ 1&-r(d)\end{pmatrix}=\boldsymbol{0}.

Note that by A1, (2.1) and the Uniform Convergence Theorem (see, e.g., Theorem 1.5.2 in [18]) we get

	$\displaystyle\quad\lim_{u\to\infty}\sup_{s\in[0,S]}\left\lvert u^{2}(1-r(s/v(u)))-s^{\alpha}\right\rvert$
	$\displaystyle\leq\lim_{u\to\infty}\sup_{s\in[0,S]}\left\lvert\frac{1-r(s/v(u))}{1-r(1/v(u))}-s^{\alpha}\right\rvert+\lim_{u\to\infty}\sup_{s\in[0,S]}\left\lvert\frac{1-r(s/v(u))}{1-r(1/v(u))}\right\rvert\left\lvert u^{2}(1-r(1/v(u)))-1\right\rvert$
	$\displaystyle=0.$

Consequently,

(6.9)

\displaystyle\lim_{u\to\infty}\sup_{d\geq A(u),\boldsymbol{s},\boldsymbol{t}\in\boldsymbol{D}}\left\lvert u^{2}(E-R_{u,d}(\boldsymbol{s},\boldsymbol{t}))-\begin{pmatrix}\left\lvert t_{1}-s_{1}\right\rvert^{\alpha}&0\\ 0&\left\lvert t_{2}-s_{2}\right\rvert^{\alpha}\end{pmatrix}\right\rvert=\boldsymbol{0}.

Similarly,

	$\displaystyle\quad\lim_{u\to\infty}\sup_{d\geq A(u),\boldsymbol{s},\boldsymbol{t}\in\boldsymbol{D}}\left\lvert u^{2}R_{u,d}(\boldsymbol{s},\boldsymbol{0})\big{(}E-R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})R_{u,d}(\boldsymbol{0},\boldsymbol{t})\big{)}-\begin{pmatrix}\left\lvert t_{1}\right\rvert^{\alpha}&0\\ 0&\left\lvert t_{2}\right\rvert^{\alpha}\end{pmatrix}\right\rvert$
	$\displaystyle\leq\lim_{u\to\infty}\sup_{d\geq A(u),\boldsymbol{s},\boldsymbol{t}\in\boldsymbol{D}}\left\lvert u^{2}(E-R_{u,d}(\boldsymbol{0},\boldsymbol{t}))-\begin{pmatrix}\left\lvert t_{1}\right\rvert^{\alpha}&0\\ 0&\left\lvert t_{2}\right\rvert^{\alpha}\end{pmatrix}\right\rvert$
	$\displaystyle\quad+\lim_{u\to\infty}\sup_{d\geq A(u),\boldsymbol{s},\boldsymbol{t}\in\boldsymbol{D}}\left\lvert(R_{u,d}(\boldsymbol{s},\boldsymbol{0})-E)[u^{2}(E-R_{u,d}(\boldsymbol{0},\boldsymbol{t}))]\right\rvert$
	$\displaystyle\quad+\lim_{u\to\infty}\sup_{d\geq A(u),\boldsymbol{s},\boldsymbol{t}\in\boldsymbol{D}}\left\lvert R_{u,d}(\boldsymbol{s},\boldsymbol{0})[u^{2}(E-R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0}))]R_{u,d}(\boldsymbol{0},\boldsymbol{t})\right\rvert$
(6.10)		$\displaystyle=\boldsymbol{0},$

where in the last equality we have used (6.6), (6.8) and (6.9).

Substituting (6.9) and (LABEL:rcovas4) into (6.5) gives

(6.11)		$\displaystyle{\small\lim_{u\to\infty}\sup_{d\geq A(u),\boldsymbol{s},\boldsymbol{t}\in\boldsymbol{D}}\Bigg{\|}\mathrm{Cov}(\chi_{u,d}(\boldsymbol{s}),\chi_{u,d}(\boldsymbol{t}))-\begin{pmatrix}\|s_{1}\|^{\alpha}+\|t_{1}\|^{\alpha}-\left\lvert t_{1}-s_{1}\right\rvert^{\alpha}&0\\ 0&\|s_{2}\|^{\alpha}+\|t_{2}\|^{\alpha}-\left\lvert t_{2}-s_{2}\right\rvert^{\alpha}\end{pmatrix}\Bigg{\|}}$
(6.11)			$\displaystyle=$	$\displaystyle\boldsymbol{0}.$

Hence, the finite-dimensional distributions of $\chi_{u,d}$ converge to that of $\{\sqrt{2}B_{\alpha}(\boldsymbol{s}),\boldsymbol{s}\in\boldsymbol{D}\}$ uniformly with respect to $d\geq A(u)$ , where $B_{\alpha}(\boldsymbol{s})=(B_{\alpha}^{(1)}(s_{1}),B_{\alpha}^{(2)}(s_{2}))^{T}.$

Let $C(\boldsymbol{D})$ denote the Banach space of all continuous functions on $\boldsymbol{D}$ equipped with sup-norm, we now show that the measures on $C(\boldsymbol{D})$ induced by $\{\chi_{u,d}(\boldsymbol{s}),\boldsymbol{s}\in\boldsymbol{D},d\geq A(u)\}$ are uniformly tight for large $u$ . In fact, since $\mathbb{E}\left\{\xi_{u,d}(\boldsymbol{s})|\xi_{u,d}(\boldsymbol{0})\right\}=R_{u,d}(\boldsymbol{s},\boldsymbol{0})R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})\xi_{u,d}(\boldsymbol{0})$ then

	$\displaystyle\quad\mathbb{E}\left\{(\xi_{u,d}(\boldsymbol{s})-\xi_{u,d}(\boldsymbol{t}))^{T}\left(R_{u,d}(\boldsymbol{s},\boldsymbol{0})-R_{u,d}(\boldsymbol{t},\boldsymbol{0})\right)R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})\xi_{u,d}(\boldsymbol{0})\right\}$
	$\displaystyle=\mathbb{E}\left\{\mathbb{E}\left\{(\xi_{u,d}(\boldsymbol{s})-\xi_{u,d}(\boldsymbol{t}))^{T}\|\xi_{u,d}(\boldsymbol{0})\right\}\left(R_{u,d}(\boldsymbol{s},\boldsymbol{0})-R_{u,d}(\boldsymbol{t},\boldsymbol{0})\right)R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})\xi_{u,d}(\boldsymbol{0})\right\}$
	$\displaystyle=\mathbb{E}\left\{\xi_{u,d}^{T}(\boldsymbol{0})R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})(R_{u,d}(\boldsymbol{0},\boldsymbol{s})-R_{u,d}(\boldsymbol{0},\boldsymbol{t}))(R_{u,d}(\boldsymbol{s},\boldsymbol{0})-R_{u,d}(\boldsymbol{t},\boldsymbol{0}))R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})\xi_{u,d}(\boldsymbol{0})\right\}$

and thus

	$\displaystyle\quad\mathbb{E}\left\{\lVert\chi_{u,d}(\boldsymbol{s})-\chi_{u,d}(\boldsymbol{t})\rVert^{2}\right\}$
	$\displaystyle=u^{2}\Big{[}\mathbb{E}\left\{\lVert\xi_{u,d}(\boldsymbol{s})-\xi_{u,d}(\boldsymbol{t})\rVert^{2}\right\}-2\mathbb{E}\left\{(\xi_{u,d}(\boldsymbol{s})-\xi_{u,d}(\boldsymbol{t}))^{T}(R_{u,d}(\boldsymbol{s},\boldsymbol{0})-R_{u,d}(\boldsymbol{t},\boldsymbol{0}))R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})\xi_{u,d}(\boldsymbol{0})\right\}$
	$\displaystyle\quad+\mathbb{E}\left\{\xi_{u,d}^{T}(\boldsymbol{0})R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})(R_{u,d}(\boldsymbol{0},\boldsymbol{s})-R_{u,d}(\boldsymbol{0},\boldsymbol{t}))(R_{u,d}(\boldsymbol{s},\boldsymbol{0})-R_{u,d}(\boldsymbol{t},\boldsymbol{0}))R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})\xi_{u,d}(\boldsymbol{0})\right\}\Big{]}$
	$\displaystyle=u^{2}\Big{[}\mathbb{E}\left\{\lVert\xi_{u,d}(\boldsymbol{s})-\xi_{u,d}(\boldsymbol{t})\rVert^{2}\right\}-\mathbb{E}\left\{\lVert(R_{u,d}(\boldsymbol{s},\boldsymbol{0})-R_{u,d}(\boldsymbol{t},\boldsymbol{0}))R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})\xi_{u,d}(\boldsymbol{0})\rVert^{2}\right\}\Big{]}$
	$\displaystyle\leq u^{2}E{\lVert\xi_{u,d}(\boldsymbol{s})-\xi_{u,d}(\boldsymbol{t})\rVert^{2}}$
	$\displaystyle=2u^{2}[1-r(\left\lvert t_{1}-s_{1}\right\rvert/v(u))+1-r(\left\lvert t_{2}-s_{2}\right\rvert/v(u))].$

Moreover, by (2.1) and Potter’s Theorem (see, e.g., [18][Theorem 1.5.6]), for large enough $u$ there exists some constant $C>1$ such that

\displaystyle u^{2}(1-r(s/v(u)))=u^{2}(1-r(1/v(u)))\frac{1-r(s/v(u))}{1-r(1/v(u))}\leq C\left\lvert s\right\rvert^{\alpha/2}

holds for all $s\in[0,S]$ . Hence, for large enough $u$ we get

(6.12)

\displaystyle\sup_{d\geq A(u)}\mathbb{E}\left\{\lVert\chi_{u,d}(\boldsymbol{s})-\chi_{u,d}(\boldsymbol{t})\rVert^{2}\right\}\leq 2C(|t_{1}-s_{1}|^{\alpha/2}+|t_{2}-s_{2}|^{\alpha/2})

for any $\boldsymbol{s},\boldsymbol{t}\in\boldsymbol{D}$ , implying the uniform tightness of the measures induced by $\{\chi_{u,d}(\boldsymbol{s}),\boldsymbol{s}\in\boldsymbol{D},d\geq A(u)\}$ . This together with (6.11) implies that $\chi_{u,d}$ converges weakly, as $u\to\infty$ , to $\sqrt{2}B_{\alpha}(\boldsymbol{s}),\boldsymbol{s}\in\boldsymbol{D}$ uniformly for $d\geq A(u)$ .

Further, by (6.8) and (6.9) we have

	$\displaystyle\lim_{u\to\infty}\sup_{d\geq A(u),\boldsymbol{s}\in\boldsymbol{D}}\left\lvert u^{2}\big{(}E-R_{u,d}(\boldsymbol{s},\boldsymbol{0})R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})\big{)}-\begin{pmatrix}\left\lvert s_{1}\right\rvert^{\alpha}&0\\ 0&\left\lvert s_{2}\right\rvert^{\alpha}\end{pmatrix}\right\rvert$
			$\displaystyle\leq\lim_{u\to\infty}\sup_{d\geq A(u),\boldsymbol{s}\in\boldsymbol{D}}\left\lvert u^{2}(E-R_{u,d}(\boldsymbol{s},\boldsymbol{0}))-\begin{pmatrix}\left\lvert s_{1}\right\rvert^{\alpha}&0\\ 0&\left\lvert s_{2}\right\rvert^{\alpha}\end{pmatrix}\right\rvert$
			$\displaystyle\quad+\lim_{u\to\infty}\sup_{d\geq A(u),\boldsymbol{s}\in\boldsymbol{D}}\left\lvert R_{u,d}(\boldsymbol{s},\boldsymbol{0})[u^{2}(E-R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0}))]\right\rvert$
			$\displaystyle=\boldsymbol{0}$

and thus for any $\boldsymbol{z}\in\mathbb{R}^{2}$

	$\displaystyle\lim_{u\to\infty}\sup_{d\geq A(u),\boldsymbol{s}\in\boldsymbol{D}}\left\lvert\theta_{u,d}(\boldsymbol{s})-(\|s_{1}\|^{\alpha}-z_{1},\|s_{2}\|^{\alpha}-z_{2})^{T}\right\rvert$
			$\displaystyle\leq\lim_{u\to\infty}\sup_{d\geq A(u),\boldsymbol{s}\in\boldsymbol{D}}\left\lvert\Bigg{[}u^{2}\big{(}E-R_{u,d}(\boldsymbol{s},\boldsymbol{0})R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})\big{)}-\begin{pmatrix}\left\lvert s_{1}\right\rvert^{\alpha}&0\\ 0&\left\lvert s_{2}\right\rvert^{\alpha}\end{pmatrix}\Bigg{]}\begin{bmatrix}1\\ 1\end{bmatrix}\right\rvert$
			$\displaystyle\quad+\quad\lim_{u\to\infty}\sup_{d\geq A(u),\boldsymbol{s}\in\boldsymbol{D}}\left\lvert\big{(}E-R_{u,d}(\boldsymbol{s},\boldsymbol{0})R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})\big{)}\boldsymbol{z}\right\rvert$
			$\displaystyle=\boldsymbol{0}.$

Therefore, for each $\boldsymbol{z}\in\mathbb{R}^{2}$ , the probability measures on $C(\boldsymbol{D})$ induced by $\{\chi_{u,d}(\boldsymbol{s})-\theta_{u,d}(\boldsymbol{s},\boldsymbol{z}),\boldsymbol{s}\in\boldsymbol{D}\}$ converge weakly, as $u\to\infty$ , to that induced by $\{W_{\alpha}(\boldsymbol{s})+\boldsymbol{z},\boldsymbol{s}\in\boldsymbol{D}\}$ uniformly with respect to $d\geq A(u)$ . Then, by the continuous mapping theorem, (6.4) and the fact that the set of discontinuity points of cumulative distribution function of $\sup_{s_{\in}[0,S]}\sqrt{2}B_{\alpha}(s)-s^{\alpha}$ consists of at most of one point (see, e.g., Theorem 7.1 in [19] or related Lemma 4.4 in [20]), we get

\displaystyle\lim_{u\to\infty}\sup_{d\geq A(u)}\left\lvert\mathcal{I}_{u,d}(\boldsymbol{z})-\mathcal{I}(\boldsymbol{z})\right\rvert=0

for almost all $\boldsymbol{z}\in\mathbb{R}^{2}$ , where $\mathcal{I}(\boldsymbol{z})$ is defined in (6.3). Further, by (6.7) we know

\lim_{u\to\infty}\sup_{d\geq A(u)}\left\lvert f_{u,d}(\boldsymbol{z})-e^{-z_{1}-z_{2}}\right\rvert=0,\quad\forall\,\boldsymbol{z}\in\mathbb{R}^{2},

and thus for almost all $\boldsymbol{z}\in\mathbb{R}^{2}$

(6.14)

\displaystyle\lim_{u\to\infty}\sup_{d\geq A(u)}\left\lvert\mathcal{I}_{u,d}(\boldsymbol{z})f_{u,d}(\boldsymbol{z})-\mathcal{I}(\boldsymbol{z})e^{-z_{1}-z_{2}}\right\rvert=0.

Therefore, to verify (6.2), we have to put the limit into integral. In the following, we look for an integrable upper bound for $\sup_{d\geq A(u)}\mathcal{I}_{u,d}(\boldsymbol{z})f_{u,d}(\boldsymbol{z})$ . We first give a lower bound for $\inf_{d\geq A(u),\boldsymbol{s}\in\boldsymbol{D}}\theta_{u,d}(\boldsymbol{s},\boldsymbol{z})$ . Let $\varepsilon(<1/2)$ be a positive constant. In view of (6), we know that, for sufficiently large $u$

\displaystyle\sup_{d\geq A(u),\boldsymbol{s}\in\boldsymbol{D}}\left\lvert E-R_{u,d}(\boldsymbol{s},\boldsymbol{0})R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0})\right\rvert\leq\begin{pmatrix}\varepsilon&\varepsilon\\ \varepsilon&\varepsilon\end{pmatrix},

and thus

$\displaystyle\inf_{d\geq A(u),\boldsymbol{s}\in\boldsymbol{D}}\theta_{u,d}(\boldsymbol{s},\boldsymbol{z})$	$\displaystyle=$	$\displaystyle\inf_{d\geq A(u),\boldsymbol{s}\in\boldsymbol{D}}\{u^{2}(E-R_{u,d}(\boldsymbol{s},\boldsymbol{0})R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0}))\boldsymbol{1}+(E-R_{u,d}(\boldsymbol{s},\boldsymbol{0})R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0}))\boldsymbol{z}-\boldsymbol{z}\}$
	$\displaystyle\geq$	$\displaystyle-\boldsymbol{1}-\boldsymbol{z}+\inf_{d\geq A(u),\boldsymbol{s}\in\boldsymbol{D}}\{(E-R_{u,d}(\boldsymbol{s},\boldsymbol{0})R_{u,d}^{-1}(\boldsymbol{0},\boldsymbol{0}))\boldsymbol{z}\}$
	$\displaystyle\geq$	$\displaystyle-\boldsymbol{1}-\boldsymbol{z}-\begin{pmatrix}\varepsilon&\varepsilon\\ \varepsilon&\varepsilon\end{pmatrix}\left\lvert\boldsymbol{z}\right\rvert:=h(\boldsymbol{z}),\quad\boldsymbol{z}\in\mathbb{R}^{2}.$

Let $\{\boldsymbol{e}_{k},k=1,2,3\}$ denotes $(1,1)^{T}$ , $(0,1)^{T}$ and $(1,0)^{T}$ , respectively. By Cauchy-Schwartz inequality and (6.12), for large enough $u$

(6.15)		$\displaystyle\sup_{d\geq A(u)}\mathbb{E}\left\{\left(\boldsymbol{e}_{k}^{T}(\chi_{u,d}(\boldsymbol{s})-\chi_{u,d}(\boldsymbol{t}))\right)^{2}\right\}$	$\displaystyle\leq$	$\displaystyle\sup_{d\geq A(u)}2\mathbb{E}\left\{\lVert\chi_{u,d}(\boldsymbol{s})-\chi_{u,d}(\boldsymbol{t})\rVert^{2}\right\}$
(6.15)			$\displaystyle\leq$	$\displaystyle 4C(\|t_{1}-s_{1}\|^{\alpha/2}+\|t_{2}-s_{2}\|^{\alpha/2}),\quad k=1,2,3$

holds for any $\boldsymbol{s},\boldsymbol{t}\in\boldsymbol{D}$ . Thus, by Sudakov-Fernique inequality (see, e.g., [21][Theorem 2.9]), we have

(6.16)

\displaystyle\sup_{d\geq A(u)}\mathbb{E}\left\{\sup_{\boldsymbol{s}\in\boldsymbol{D}}\boldsymbol{e}_{k}^{T}\chi_{u,d}(\boldsymbol{s})\right\}\leq\mathbb{E}\left\{\sup_{\boldsymbol{s}\in\boldsymbol{D}}\sum_{i=1}^{2}2\sqrt{C}B_{\alpha/2}^{(i)}(s_{i})\right\}:=C_{1}<\infty,\quad k=1,2,3,

where $B_{\alpha/2}^{(i)}$ ’s are independent fBm’s with Hurst index $\alpha/4$ . Then, for all large enough $u$ ,

$\displaystyle\sup_{d\geq A(u)}\mathcal{I}_{u,d}(\boldsymbol{z})$	$\displaystyle=$	$\displaystyle\sup_{d\geq A(u)}\mathbb{P}\left\{\exists_{\boldsymbol{s}\in\boldsymbol{D}}\chi_{u,d}(\boldsymbol{s})-\theta_{u,d}(\boldsymbol{s},\boldsymbol{z})>\boldsymbol{0}\right\}$
	$\displaystyle\leq$	$\displaystyle\sup_{d\geq A(u)}\mathbb{P}\left\{\exists_{\boldsymbol{s}\in\boldsymbol{D}}\chi_{u,d}(\boldsymbol{s})>\inf_{d\geq A(u),\boldsymbol{s}\in\boldsymbol{D}}\theta_{u,d}(\boldsymbol{s},\boldsymbol{z})\right\}$
	$\displaystyle\leq$	$\displaystyle\sup_{d\geq A(u)}\mathbb{P}\left\{\sup_{\boldsymbol{s}\in\boldsymbol{D}}\boldsymbol{e}_{k}^{T}\chi_{u,d}(\boldsymbol{s})>\boldsymbol{e}_{k}^{T}h(\boldsymbol{z})\right\}$
	$\displaystyle\leq$	$\displaystyle\sup_{d\geq A(u)}\exp\left(-\frac{\left(\boldsymbol{e}_{k}^{T}h(\boldsymbol{z})-\mathbb{E}\left\{\sup_{\boldsymbol{s}\in\boldsymbol{D}}\boldsymbol{e}_{k}^{T}\chi_{u,d}(\boldsymbol{s})\right\}\right)^{2}}{2\text{Var}_{\boldsymbol{s}\in\boldsymbol{D}}\boldsymbol{e}_{k}^{T}\chi_{u,d}(\boldsymbol{s})}\right)$
	$\displaystyle\leq$	$\displaystyle\exp\left(-C_{2}\left(\boldsymbol{e}_{k}^{T}h(\boldsymbol{z})-C_{1}\right)^{2}\right),\quad\boldsymbol{z}\in\boldsymbol{Z}_{k},k=1,2,3,$

where (6) follows from Borell-TIS inequality (see, e.g., [22][Theorem 2.1.1]), the last inequality follows by (6.15)-(6.16) with $C_{2}=(16CS^{\alpha/2})^{-1}$ , and

\boldsymbol{Z}_{1}=\{(z_{1},z_{2})|z_{1}<0,z_{2}<0,(2\varepsilon-1)(z_{1}+z_{2})>2+C_{1}\},

\boldsymbol{Z}_{2}=\{(z_{1},z_{2})|z_{1}>0,z_{2}<0,(\varepsilon-1)z_{2}-\varepsilon z_{1}>1+C_{1}\},

\boldsymbol{Z}_{3}=\{(z_{1},z_{2})|z_{1}<0,z_{2}>0,(\varepsilon-1)z_{1}-\varepsilon z_{2}>1+C_{1}\}.

Therefore,

\displaystyle\sup_{d\geq A(u)}\mathcal{I}_{u,d}(\boldsymbol{z})\leq g(\boldsymbol{z}):=\left\{\begin{array}[]{ll}\exp\left(-C_{2}\left(\boldsymbol{e}_{k}^{T}h(\boldsymbol{z})-C_{1}\right)^{2}\right),&\boldsymbol{z}\in\boldsymbol{Z}_{k},k=1,2,3,\\ 1,&\quad\boldsymbol{z}\in\mathbb{R}^{2}\backslash\bigcup_{k=1}^{3}\boldsymbol{Z}_{k},\\ \end{array}\right.

holds for sufficiently large $u$ . Moreover, by (6.7)

	$\displaystyle\quad\sup_{d\geq A(u)}f_{u,d}(\boldsymbol{z})e^{z_{1}+z_{2}}$
	$\displaystyle=\sup_{d\geq A(u)}\frac{1}{\sqrt{1-r^{2}(d)}}\exp\left(\frac{u^{2}r(d)}{1+r(d)}+\frac{2u^{2}r(d)(1-r(d))(z_{1}+z_{2})-(z_{1}^{2}-2r(d)z_{1}z_{2}+z_{2}^{2})}{2u^{2}(1-r^{2}(d))}\right)$
	$\displaystyle\leq\frac{3}{2}\sup_{d\geq A(u)}\exp\left(\frac{u^{2}r(d)}{1+r(d)}+\frac{-\frac{1+r(d)}{2}(z_{2}-z_{1})^{2}-\frac{1-r(d)}{2}(z_{1}+z_{2}-2u^{2}r(d))^{2}+2u^{4}r^{2}(d)(1-r(d))}{2u^{2}(1-r^{2}(d))}\right)$
	$\displaystyle\leq\frac{3}{2}\sup_{d\geq A(u)}e^{u^{2}r(d)}\leq 2,\quad\boldsymbol{z}\in\mathbb{R}^{2}$

holds for all large enough $u$ , and thus

\displaystyle\sup_{d\geq A(u)}\mathcal{I}_{u,d}(\boldsymbol{z})f_{u,d}(\boldsymbol{z})\leq 2g(\boldsymbol{z})e^{-z_{1}-z_{2}},\quad\boldsymbol{z}\in\mathbb{R}^{2}.

We now show that $g(\boldsymbol{z})e^{-z_{1}-z_{2}}$ is integrable on $\mathbb{R}^{2}$ . In fact,

\displaystyle\iint_{\mathbb{R}^{2}}g(\boldsymbol{z})e^{-z_{1}-z_{2}}\text{\rm d}z_{1}\text{\rm d}z_{2}=\left(\iint_{\boldsymbol{Z}_{1}}+\iint_{\boldsymbol{Z}_{2}}+\iint_{\boldsymbol{Z}_{3}}+\iint_{\mathbb{R}^{2}\backslash\bigcup_{k=1}^{3}\boldsymbol{Z}_{k}}\right)g(\boldsymbol{z})e^{-z_{1}-z_{2}}\text{\rm d}z_{1}\text{\rm d}z_{2},

where

$\displaystyle\iint_{\boldsymbol{Z}_{1}}g(\boldsymbol{z})e^{-z_{1}-z_{2}}\text{\rm d}z_{1}\text{\rm d}z_{2}$	$\displaystyle\leq$	$\displaystyle\int_{-\infty}^{0}\int_{-\infty}^{0}\exp\left(-C_{2}\left((2\varepsilon-1)z_{1}+(2\varepsilon-1)z_{2}-2-C_{1}\right)^{2}-z_{1}-z_{2}\right)\text{\rm d}z_{1}\text{\rm d}z_{2}$
	$\displaystyle\leq$	$\displaystyle\left(\int_{-\infty}^{0}\exp\left(-C_{2}(2\varepsilon-1)^{2}z_{1}^{2}+\left(2C_{2}(2\varepsilon-1)(2+C_{1})-1\right)z_{1}\right)\text{\rm d}z_{1}\right)^{2}$
	$\displaystyle<$	$\displaystyle\infty,$

$\displaystyle\iint_{\boldsymbol{Z}_{2}}g(\boldsymbol{z})e^{-z_{1}-z_{2}}\text{\rm d}z_{1}\text{\rm d}z_{2}$	$\displaystyle=$	$\displaystyle\iint_{\boldsymbol{Z}_{3}}g(\boldsymbol{z})e^{-z_{1}-z_{2}}\text{\rm d}z_{1}\text{\rm d}z_{2}$
	$\displaystyle=$	$\displaystyle\int^{\infty}_{0}\left(\int_{-\infty}^{\frac{\varepsilon z_{1}+1+C_{1}}{(\varepsilon-1)}}\exp\left(-C_{2}\left((\varepsilon-1)z_{2}-\varepsilon z_{1}-1-C_{1}\right)^{2}-z_{2}\right)\text{\rm d}z_{2}\right)e^{-z_{1}}\text{\rm d}z_{1}$
	$\displaystyle=$	$\displaystyle\frac{e^{\frac{1+C_{1}}{1-\varepsilon}}}{1-\varepsilon}\int^{\infty}_{0}e^{(\frac{\varepsilon}{1-\varepsilon}-1)z_{1}}\text{\rm d}z_{1}\int_{0}^{\infty}\exp\left(-C_{2}z_{2}^{2}-\frac{z_{2}}{\varepsilon-1}\right)\text{\rm d}z_{2}<\infty,$

since $\varepsilon<1/2$ , and

	$\displaystyle\iint_{\mathbb{R}^{2}\backslash\bigcup_{k=1}^{3}\boldsymbol{Z}_{k}}g(\boldsymbol{z})e^{-z_{1}-z_{2}}\text{\rm d}z_{1}\text{\rm d}z_{2}$	$\displaystyle\leq$	$\displaystyle\left(\iint_{z_{1}<0,z_{2}<0,z_{1}+z_{2}\geq\frac{2+C_{1}}{2\varepsilon-1}}+2\int_{0}^{\infty}\int^{\infty}_{\frac{\varepsilon z_{1}+C_{1}+1}{\varepsilon-1}}\right)e^{-z_{1}-z_{2}}\text{\rm d}z_{1}\text{\rm d}z_{2}$
		$\displaystyle\leq$	$\displaystyle\left(\frac{2+C_{1}}{1-2\varepsilon}\right)^{2}e^{\frac{2+C_{1}}{1-2\varepsilon}}+2e^{\frac{1+C_{1}}{1-\varepsilon}}\int^{\infty}_{0}e^{(\frac{\varepsilon}{1-\varepsilon}-1)z_{1}}\text{\rm d}z_{1}<\infty.$

Consequently, (6.2) follows by the dominated convergence theorem and (6.14). This completes the proof. $\Box$

Lemma 6.2.

Let $X(t),t\geq 0$ be a centered stationary Gaussian process with unit variance and covariance function satisfying A1 and A3. Let $v(u)$ , $\mathcal{B}_{\alpha}(x)$ and $m(u)$ be defined in (2.1), (2.3) and (2.5) respectively. Then for $A(u)$ such that

(6.19)

\displaystyle\lim_{u\to\infty}A(u)v(u)=\infty\quad\textrm{and}\quad\lim_{u\to\infty}\frac{A(u)}{m(u)}=0

and any $x\geq 0$ we have

(6.20)

\displaystyle\mathbb{P}\left\{L_{u}^{\ast}[0,A(u)]>x\right\}\sim\mathcal{B}_{\alpha}(x)A(u)v(u)\Psi(u),\quad u\to\infty.

Proof of Lemma 6.2 We follow the argument used in the proof of Theorem 2.1 in [6]. Let $A(u)$ satisfy (6.19), for any $S>1$ define

\Delta_{k}=[kS/v(u),(k+1)S/v(u)],\quad k=0,\ldots,N_{u}

with $N_{u}=\lfloor A(u)v(u)/S\rfloor$ , i.e., the integer part of $A(u)v(u)/S$ . By stationarity of $X$ , we have for all $u$ positive and $x\geq 0$

\displaystyle I_{1}(u)\leq\mathbb{P}\left\{L_{u}^{*}[0,T]>x\right\}\leq I_{2}(u),

where

	$\displaystyle I_{1}(u)=(N_{u}-1)\mathbb{P}\left\{L_{u}^{*}\Delta_{0}>x\right\}-\sum_{0\leq i<k\leq N_{u}-1}q_{i,k}(u),$
	$\displaystyle I_{2}(u)=(N_{u}+1)\mathbb{P}\left\{L_{u}^{*}\Delta_{0}>x\right\}+\sum_{0\leq i<k\leq N_{u}}q_{i,k}(u),$

with $q_{i,k}(u)=\mathbb{P}\left\{\sup_{t\in\Delta_{i}}X(t)>u,\sup_{t\in\Delta_{k}}X(t)>u\right\}.$ By Theorem 5.1 in [6] and (2.3), we have

(6.21)

\displaystyle\lim_{S\to\infty}\lim_{u\to\infty}\frac{N_{u}\mathbb{P}\left\{L_{u}^{*}\Delta_{0}>x\right\}}{A(u)v(u)\Psi(u)}=\mathcal{B}_{\alpha}(x)

for any $x\geq 0$ . Therefore, it suffices to show that the double sum is negligible with respect to $A(u)v(u)\Psi(u)$ as $u\to\infty$ and then as $S\to\infty$ .

Let $\varepsilon^{\ast}(<2)$ be the positive root of equation $x^{2}-(2-\alpha)x-\frac{3}{2}\alpha=0$ and put $\beta=\inf_{t\geq 1}\{1-r(t)\},$ which by A3 is positive. Define

\displaystyle A_{0}(u)=0,\,A_{1}(u)=u^{\frac{\varepsilon^{*}-2}{\alpha}}\wedge A(u),\,A_{2}(u)=1\wedge A(u),\,A_{3}(u)=e^{\beta u^{2}/8}\wedge A(u),\,A_{4}(u)=A(u)

and

\Lambda_{l}(u)=\{(i,k):1\leq i+1<k\leq N_{u},A_{l-1}(u)<(k-i-1)S/v(u)\leq A_{l}(u),\,l=1,2,3,4\}.

Then

	$\displaystyle\sum_{0\leq i<k\leq N_{u}}q_{i,k}(u)$	$\displaystyle=$	$\displaystyle\sum_{0\leq i<N_{u}}q_{i,i+1}(u)+\sum_{l=1}^{4}\sum_{(i,k)\in\Lambda_{l}(u)}q_{i,k}(u)$
		$\displaystyle:=$	$\displaystyle Q_{0}(u)+\sum_{l=1}^{4}Q_{l}(u).$

According to (4.7)-(4.9) in [6] we know that

(6.23)

\displaystyle\limsup_{u\to\infty}\frac{Q_{2}(u)}{A(u)v(u)\Psi(u)}=0,

(6.24)

\displaystyle\lim_{S\to\infty}\limsup_{u\to\infty}\frac{Q_{1}(u)}{A(u)v(u)\Psi(u)}=0,

and

(6.25)

\displaystyle\lim_{S\to\infty}\limsup_{u\to\infty}\frac{Q_{0}(u)}{A(u)v(u)\Psi(u)}=0.

Next, by stationarity of $X$ , for sufficiently large $u$

\displaystyle\sup_{(i,k)\in\Lambda_{3}(u)}\mathbb{E}\left\{\sup_{s\in\Delta_{i},t\in\Delta_{k}}(X(s)+X(t))\right\}\leq 2\mathbb{E}\left\{\sup_{s\in[0,1]}X(s)\right\}=:C_{3}<\infty,

	$\displaystyle\sup_{(i,k)\in\Lambda_{3}(u),s\in\Delta_{i},t\in\Delta_{k}}\text{Var}(X(s)+X(t))$	$\displaystyle=$	$\displaystyle 4-2\inf_{(i,k)\in\Lambda_{3}(u),(s,t)\in\Delta_{i}\times\Delta_{k}}\{1-r(t-s)\}$
		$\displaystyle\leq$	$\displaystyle 4-2\beta.$

Then, by Borell-TIS inequality we have for large enough $u$

$\displaystyle\sup_{(i,k)\in\Lambda_{3}(u)}q_{i,k}(u)$	$\displaystyle\leq$	$\displaystyle\sup_{(i,k)\in\Lambda_{3}(u)}\mathbb{P}\left\{\sup_{s\in\Delta_{i},t\in\Delta_{k}}X(s)+X(t)>2u\right\}$
	$\displaystyle\leq$	$\displaystyle\exp\left(-\frac{(2u-C_{3})^{2}}{2(4-2\beta)}\right)$
	$\displaystyle\leq$	$\displaystyle\exp\left(-\frac{1+\beta/2}{2}(u-C_{3}/2)^{2}\right),$

and thus

(6.26)		$\displaystyle\limsup_{u\to\infty}\frac{Q_{3}(u)}{A(u)v(u)\Psi(u)}$	$\displaystyle\leq$	$\displaystyle\limsup_{u\to\infty}\frac{N_{u}A_{3}(u)v(u)}{SA(u)v(u)\Psi(u)}\exp\left(-\frac{1+\beta/2}{2}(u-C_{3}/2)^{2}\right)$
(6.26)			$\displaystyle\leq$	$\displaystyle\limsup_{u\to\infty}\frac{\sqrt{2\pi}uv(u)}{S^{2}}\exp\left(-\frac{\beta}{8}u^{2}+C_{3}u(1+\beta/2)\right)=0.$

Further, since $e^{\beta u^{2}/8}$ satisfies (6.1), then by Lemma 6.1 and stationarity of $X$ ,

Q_{4}(u)\leq 2N^{2}_{u}\Psi^{2}(u)\mathcal{B}_{\alpha}^{2}(S,0)

holds for $u$ sufficiently large. Therefore,

	$\displaystyle\limsup_{u\to\infty}\frac{Q_{4}(u)}{A(u)v(u)\Psi(u)}$	$\displaystyle\leq$	$\displaystyle\limsup_{u\to\infty}\frac{2N_{u}^{2}\Psi^{2}(u)\mathcal{B}_{\alpha}^{2}(S,0)}{A(u)v(u)\Psi(u)}$
		$\displaystyle\leq$	$\displaystyle\limsup_{u\to\infty}\frac{2\mathcal{B}_{\alpha}^{2}(S,0)}{S^{2}\mathcal{B}_{\alpha}(0)}\frac{A(u)}{m(u)}=0,$

where the last equality follows by (6.19).

Consequently, substituting (6.23)-(6) into (6) yields

\displaystyle\lim_{S\to\infty}\limsup_{u\to\infty}\frac{1}{A(u)v(u)\Psi(u)}\sum_{0\leq i<k\leq N_{u}}q_{i,k}(u)=0,

which together with (6.21) completes the proof. $\Box$

Corollary 6.1.

If $X$ , $v(u)$ , $\mathcal{B}_{\alpha}(x)$ , $m(u)$ and $A(u)$ are given as in Lemma 6.2, then for any $x\geq 0$ and $\varepsilon\in(0,1)$ there exists $\delta>0$ such that

(6.27)

\displaystyle\liminf_{u\to\infty}\inf_{t\in[A(u),\delta m(u)]}\frac{\mathbb{P}\left\{L_{u}^{*}[0,t]>x\right\}}{t\mathcal{B}_{\alpha}(x)v(u)\Psi(u)}\geq 1-\varepsilon

and

(6.28)

\displaystyle\limsup_{u\to\infty}\sup_{t\in[A(u),\delta m(u)]}\frac{\mathbb{P}\left\{L_{u}^{*}[0,t]>x\right\}}{t\mathcal{B}_{\alpha}(x)v(u)\Psi(u)}\leq 1+\varepsilon.

Proof of Corollary 6.1 Let $x\geq 0$ be fixed, recalling (5.2) we have that, for arbitrary $\varepsilon>0$ , there exists some $\delta>0$ such that

(6.29)

\displaystyle(1-\varepsilon/4)\leq\frac{\mathbb{P}\left\{Y(t)>x\right\}}{t\overline{\mathcal{F}_{\alpha}}(x)}\leq(1+\varepsilon/4),\quad t\in(0,\delta).

For such $\varepsilon$ and $\delta$ , suppose that (6.27) does not hold. Then, there exist two sequences $\{u_{n},n\in\mathbb{N}\}$ and $\{t_{n},n\in\mathbb{N}\}$ such that $u_{n}\to\infty$ as $n\to\infty$ and

(6.30)

\displaystyle\frac{\mathbb{P}\left\{L_{u_{n}}^{*}[0,t_{n}]>x\right\}}{t_{n}\mathcal{B}_{\alpha}(x)v(u_{n})\Psi(u_{n})}<1-\varepsilon,\quad t_{n}\in[A(u_{n}),\delta m(u_{n})],n\in\mathbb{N}.

Putting ${\hat{t}}_{n}=t_{n}/m(u_{n})$ , by (2.5) and (3.4), we get

(6.30^′)

\frac{\mathbb{P}\left\{L_{u_{n}}^{*}[0,{\hat{t}}_{n}m(u_{n})]>x\right\}}{{\hat{t}}_{n}\overline{\mathcal{F}_{\alpha}}(x)}<1-\varepsilon,\quad{\hat{t}}_{n}\in[A(u_{n})/m(u_{n}),\delta],n\in\mathbb{N}.

Since sequence $\{{\hat{t}}_{n},n\in\mathbb{N}\}$ is bounded, then there exists a convergent subsequence $\{{\hat{t}}_{n_{k}},k\in\mathbb{N}\}$ such that $\lim_{k\to\infty}{\hat{t}}_{n_{k}}\geq 0$ . If $\lim_{k\to\infty}{\hat{t}}_{n_{k}}>0$ , then by Corollary 5.1

\displaystyle\frac{\mathbb{P}\left\{L_{u_{n_{k}}}^{*}[0,{\hat{t}}_{n_{k}}m(u_{n_{k}})]>x\right\}}{\mathbb{P}\left\{Y({\hat{t}}_{n_{k}})>x\right\}}>1-\varepsilon/4

holds for sufficiently large $k$ , which together with (6.29) implies

\displaystyle\frac{\mathbb{P}\left\{L_{u_{n_{k}}}^{*}[0,{\hat{t}}_{n_{k}}m(u_{n_{k}})]>x\right\}}{{\hat{t}}_{n_{k}}\overline{\mathcal{F}_{\alpha}}(x)}>(1-\varepsilon/4)^{2}.

This however contradicts (6.30^′). If $\lim_{k\to\infty}{\hat{t}}_{n_{k}}=0$ , then

\displaystyle\lim_{k\to\infty}t_{n_{k}}v(u_{n_{k}})\geq\lim_{k\to\infty}A(u_{n_{k}})v(u_{n_{k}})=\infty\quad\textrm{and}\quad\lim_{k\to\infty}\frac{t_{n_{k}}}{m(u_{n_{k}})}=\lim_{k\to\infty}{\hat{t}}_{n_{k}}=0,

and thus by Lemma 6.2

\displaystyle\frac{\mathbb{P}\left\{L_{u_{n_{k}}}^{*}[0,t_{n_{k}}]>x\right\}}{t_{n_{k}}\mathcal{B}_{\alpha}(x)v(u_{n_{k}})\Psi(u_{n_{k}})}>1-\varepsilon/4

holds for sufficiently large $k$ . This contradicts (6.30). An analogous argument can be used to verify (6.28). This completes the proof. $\Box$

Acknowledgement: The authors would like to thank Enkelejd Hashorva for his numerous valuable remarks on all the steps of preparation of the manuscript. K.D. was partially supported by NCN Grant No 2018/31/B/ST1/00370 (2019-2022). X.P. thanks National Natural Science Foundation of China (11701070,71871046) for partial financial support. Financial support from the Swiss National Science Foundation Grant 200021-175752/1 is also kindly acknowledged.

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