# On Berman functions

Krzysztof Dȩbicki  Krzysztof Dȩbicki, Mathematical Institute, University of
Wrocław, pl. Grunwaldzki 2/4, 50-384 Wrocław, Poland
<EMAIL_ADDRESS>,  Enkelejd Hashorva  Enkelejd Hashorva,
Department of Actuarial Science, University of Lausanne,
Chamberonne 1015 Lausanne, Switzerland<EMAIL_ADDRESS>and
Zbigniew Michna  Zbigniew Michna, Department of Operations Research and
Business Intelligence, Wrocław University of Science and Technology, Poland
<EMAIL_ADDRESS>
> Abstract: Let $Z(t)=\exp\left(\sqrt{2}B_{H}(t)-\left\lvert
> t\right\rvert^{2H}\right),t\in\mathbb{R}$ with $B_{H}(t),t\in\mathbb{R}$ a
> standard fractional Brownian motion (fBm) with Hurst parameter $H\in(0,1]$
> and define for $x$ non-negative the Berman function
>
>
> $\displaystyle\mathcal{B}_{Z}(x)=\mathbb{E}\left\\{\frac{\mathbb{I}{\left\\{\epsilon_{0}(RZ)>x\right\\}}}{\epsilon_{0}(RZ)}\right\\}\in(0,\infty),$
>
> where the random variable $R$ independent of $Z$ has survival function
> $1/x,x\geq 1$ and
>
> $\epsilon_{0}(RZ)=\int_{\mathbb{R}}\mathbb{I}{\left\\{RZ(t)>1\right\\}}dt.$
>
> In this paper we consider a general random field (rf) $Z$ that is a spectral
> rf of some stationary max-stable rf $X$ and derive the properties of the
> corresponding Berman functions. In particular, we show that Berman functions
> can be approximated by the corresponding discrete ones and derive
> interesting representations of those functions which are of interest for
> Monte Carlo simulations, which are presented in this article.

Key Words: Berman functions; Pickands constants; max-stable random fields;
simulations

AMS Classification: Primary 60G15; secondary 60G70

##  1\. Introduction

In the study of sojourns of rf’s in a series of papers by S. Berman, see e.g.,
[1, 2] a key random variable (rv) and a related constant appear. Specifically,
let $Z(t)=\exp(\sqrt{2}B_{H}(t)-\left\lvert
t\right\rvert^{2H}),t\in\mathbb{R},$ with $B_{H}$ a fractional Brownian motion
(fBm) with Hurst parameter $H\in(0,1]$, that is a centered Gaussian process
with stationary increments, $Var(B_{H}(t))=\left\lvert
t\right\rvert^{2H},t\in\mathbb{R}$ and continuous sample paths. In view of [2,
Thm 3.3.1, Eq. (3.3.6)] the following rv (hereafter
$\mathbb{I}{\left\\{\cdot\right\\}}$ is the indicator function)

$\epsilon_{0}(RZ)=\int_{\mathbb{R}}\mathbb{I}{\left\\{RZ(t)>1\right\\}}dt$

plays a crucial role in the analysis of extremes of Gaussian processes.
Throughout this paper $R$ is a 1-Pareto rv ($\ln R$ is unit exponential)
independent of any other random element.
The distribution function of $\epsilon_{0}(RZ)$ is known only for
$H\in\\{1/2,1\\}$. For $H=1$ as shown in [2, Eq. (3.3.23)] $\epsilon_{0}(RZ)$
has probability density function (pdf) $x^{2}e^{-x^{2}/2}/(2\sqrt{\pi}),x>0$,
whereas for $H=1/2$ its pdf is calculated in [2, Eq. (5.6.9)].
The so-called Berman function defined for all $x\geq 0$ (see [2, Eq. (3.0.2)])
given by

(1.1) $\displaystyle\mathcal{B}_{Z}(x)$ $\displaystyle=$
$\displaystyle\mathbb{E}\left\\{\frac{\mathbb{I}{\left\\{\epsilon_{0}(RZ)>x\right\\}}}{\epsilon_{0}(RZ)}\right\\}\in(0,\infty)$

is defined in [2, Thm 3.3.1, Eq. (3.3.6)].

An important property of the Berman function is that for $x=0$ it equals the
Pickands constant, see [2, Thm 10.5.1] i.e.,
$\mathcal{B}_{Z}(0)=\mathcal{H}_{Z},$ where $\mathcal{H}_{Z}$ is the so called
generalised Pickands constant

$\displaystyle\mathcal{H}_{Z}=\lim_{T\to\infty}\frac{1}{T}\mathbb{E}\left\\{\sup_{t\in[0,T]}Z(t)\right\\}.$

This fact is crucial since $\mathcal{B}_{Z}(0)$ is the first known expression
of $\mathcal{H}_{Z}$ in terms of an expectation, which is of particular
usefulness for simulation purposes, see [3, 4, 5] for details on classical
Pickands constants.
Besides, Berman’s representation of Pickands constant yields tight lower
bounds for $\mathcal{H}_{Z}$, see [6, Thm 1.1]. As shown in [6] for all $x\geq
0$

$\mathcal{B}_{Z}(x)=\lim_{T\to\infty}\frac{1}{T}\mathcal{B}([0,T],x),\quad\mathcal{B}_{Z}([0,T],x)\coloneqq\int_{0}^{\infty}\mathbb{P}\left\\{\int_{0}^{T}\mathbb{I}(Z(t)>s)dt>x\right\\}ds.$

Motivated by the above definition, in this contribution we shall introduce the
Berman functions for given $\delta\geq 0$ with respect to some non-negative rf
$Z(t),t\in\mathbb{R}^{d},d\geq 1$ with càdlàg sample paths (see e.g., [7, 8]
for the definition and properties of generalised càdlàg functions) such that

(1.2) $\displaystyle\mathbb{E}\\{Z(t)\\}=1,\quad t\in\mathbb{R}^{d}.$

Specifically, for given non-negative $\delta,x$ define

$\mathcal{B}_{Z}^{\delta}(x)\coloneqq\lim_{T\rightarrow\infty}\frac{1}{T^{d}}\mathcal{B}_{Z}^{\delta}([0,T]^{d}\cap\delta\mathbb{Z}^{d},x),$

where

$\mathcal{B}_{Z}^{\delta}([0,T]^{d}\cap\delta\mathbb{Z}^{d},x)\coloneqq\int_{0}^{\infty}\mathbb{P}\left\\{\int_{[0,T]^{d}\cap\delta\mathbb{Z}^{d}}\mathbb{I}(Z(t)>s)\lambda_{\delta}(dt)>x\right\\}ds.$

Here $\lambda_{0}(dt)=\lambda(dt)$ is the Lebesgue measure on
$\mathbb{R}^{d}$, $0\mathbb{Z}^{d}=\mathbb{R}^{d}$ and
$\lambda_{\delta}(dt)/\delta^{d}$ is the counting measure on
$\delta\mathbb{Z}^{d}$ if $\delta>0$. Hence
$\mathcal{B}_{Z}^{\delta}(x),\delta>0$ is the discrete counterpart of
$\mathcal{B}_{Z}(x)$ and $\mathcal{B}_{Z}^{0}(x)=\mathcal{B}_{Z}(x)$.

In general, in order to be well-defined for the function
$\mathcal{B}_{Z}^{\delta}(x),x\geq 0$ some further restriction on the rf $Z$
are needed. A very tractable case for which we can utilise results from the
theory of max-stable stationary rf’s is when $Z$ is the spectral rf of a
stationary max-stable rf $X(t),t\in\mathbb{R}^{d}$, see (2.1) below.
An interesting special case is when $\ln Z(t)$ is a Gaussian rf with trend
equal the half of its variance function having further stationary increments.
We shall show in 4.3 that for such $Z$ the corresponding Berman function
$\mathcal{B}_{Z}(x)$ appears in the tail asymptotic of the sojourn of a
related Gaussian rf.

Organisation of the rest of the paper. In Section 2 we first present in 2.1 a
formula for Berman functions and then in 2.3 and 3.1 we show some continuity
properties of those functions. In Equation 2.15 and 2.4 we present two
representations for Berman functions and discuss conditions for their
positivity. Section 3 is dedicated to the approximation of Berman functions
focusing on the Gaussian case. All the proofs are postponed to Section 4.

##  2\. Main Results

Let the rf $Z(t),t\in\mathbb{R}^{d}$ be as above defined in the non-atomic
complete probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let further
$X(t),t\in\mathbb{R}^{d}$ be a max-stable stationary rf, which has spectral rf
$Z$ in its de Haan representation (see e.g., [10, 11])

(2.1) $\displaystyle X(t)=\max_{i\geq 1}\Gamma_{i}^{{-1}}Z^{(i)}(t),\quad
t\in\mathbb{R}^{d}.$

Here $\Gamma_{i}=\sum_{k=1}^{i}\mathcal{V}_{k}$ with $\mathcal{V}_{k},k\geq 1$
mutually independent unit exponential rv’s being independent of
$\\{Z^{(i)}\\}_{i=1}^{\infty}$ which are independent copies of $Z$. For
simplicity we shall assume that the marginal distributions of the rf $X$ are
unit Fréchet (equal to $e^{-1/x},x>0$) which in turn implies
$\mathbb{E}\\{Z(t)\\}=1$ for all $t\in\mathbb{R}^{d}$.

Suppose further that for all $T>0$

(2.2)
$\displaystyle\mathbb{E}\left\\{\sup_{t\in[0,T]^{d}}Z(t)\right\\}<\infty$

and $Z$ has almost surely sample paths on the space $D$ of non-negative càdlàg
functions $f:\mathbb{R}^{d}\mapsto[0,\infty)$ equipped with Skorohod’s
$J_{1}$-topology. We shall denote by $\mathcal{D}=\sigma(\pi_{t},t\in T_{0})$
the $\sigma$-field generated by the projection maps
$\pi_{t}:\pi_{t}f=f(t),f\in D$ with $T_{0}$ a countable dense subset of
$\mathbb{R}^{d}$. In view of [12, Thm 6.9] with $\alpha=1,L=B^{-1}$, see also
[13, Eq. (5.2)] the stationarity of $X$ is equivalent with

(2.3)
$\displaystyle\mathbb{E}\\{Z(h)F(Z)\\}=\mathbb{E}\\{Z(0)F(B^{h}Z)\\},\quad\forall
h\in\mathbb{R}^{d}$

valid for every measurable functional $F:D\to[0,\infty]$ such that
$F(cf)=F(f)$ for all $f\in D,c>0$. Here we use the standard notation
$B^{h}Z(\cdot)=Z(\cdot-h),h\in\mathbb{R}^{d}$.

We shall suppose next without loss of generality (see [14, Lem 7.1]) that

(2.4)
$\displaystyle\mathbb{P}\left\\{\sup_{t\in\mathbb{R}^{d}}Z(t)>0\right\\}=1.$

Under the assumption that $X$ is stationary $\mathcal{B}_{Z}^{\delta}(x)$ is
well-defined for all $\delta,x$ non-negative as we shall show below. We note
first that, see e.g., [6, 15]

$\lim_{T\to\infty}{\frac{1}{T^{d}}}\mathcal{B}_{Z}^{\delta}([0,T]^{d}\cap\delta\mathbb{Z}^{\delta},0)=\mathcal{B}_{Z}^{\delta}(0)=\mathcal{H}_{Z}^{\delta}\in(0,\infty),$

where $\mathcal{H}_{Z}^{\delta}$ is the discrete counterpart of the classical
Pickands constant $\mathcal{H}_{Z}=\mathcal{H}_{Z}^{0}$. Hence for any $x>0$
we have

$\mathcal{B}_{Z}^{\delta}(x)\leq\mathcal{H}_{Z}^{\delta}<\infty.$

Set below for $\delta>0$

$S_{\delta}=S_{\delta}(Z)=\int_{\delta\mathbb{Z}^{d}}Z(t)\lambda_{\delta}(dt)=\delta^{d}\sum_{t\in\delta\mathbb{Z}^{d}}Z(t)$

and let $S_{0}=S_{0}(Z)=\int_{\mathbb{R}^{d}}Z(t)\lambda(dt)$. In view of
(2.4) we have that $S_{0}>0$ almost surely. Since we do not consider the case
$\delta>0$ and $\delta=0$ simultaneously, we can assume that $S_{\delta}>0$
almost surely (we can construct a spectral rf $Z$ for $X$ that guarantees
this, see [14, Lem 7.3]).

In view of [15, Cor 2.1] if $\mathbb{P}\\{S_{0}=\infty\\}=1$, then
$\mathcal{H}_{Z}=0$ implying

$\mathcal{B}_{Z}^{\delta}(x){=}\mathcal{H}_{Z}=0,\quad\forall\delta,x\geq 0.$

The next result states the existence and the positivity of Berman functions
presenting further a tractable formula that is useful for simulations of those
functions.

######  Theorem 2.1.

If $\mathbb{P}\\{S_{0}=\infty\\}<1$, then for any $\delta,x$ non-negative
constants we have

(2.5) $\displaystyle\mathcal{B}_{Z}^{\delta}(x)$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathbb{E}\left\\{\frac{Z(0)}{S_{\delta}}\mathbb{I}\Bigl{(}\int_{\delta\mathbb{Z}^{d}}\mathbb{I}(Z(t)>s)\lambda_{\delta}(dt)>x\Bigr{)}\right\\}\lambda(ds)<\infty.$

Moreover, (2.5) holds substituting $S_{\delta}$ by $S_{\eta}$, where $\eta>0$
if $\delta=0$ and $\eta=k\delta,k\in\mathbb{N}$ if $\delta>0$, provided that

(2.6)
$\displaystyle\\{{S_{0}(Z)}<\infty\\}\subset\\{S_{\eta}(B^{r}Z)\in(0,\infty)\\},\quad\forall
r\in\delta\mathbb{Z}^{d}$

almost surely.

######  Remark 2.2.

  1. (i)

If $x=0$, then we retrieve the results of [15, Prop 2.1].

  2. (ii)

As shown in [15] condition (2.6) holds in the particular case that
$Z(t)>0,t\in\mathbb{R}^{d}$ almost surely.

  3. (iii)

One example for $Z$, see for instance [15] is taking

$Z(t)=\exp(V(t)-\sigma^{2}_{V}(t)/2)),\quad t\in\mathbb{R}^{d},$

where $V(t),t\in\mathbb{R}^{d}$ is a centered Gaussian rf with almost surely
continuous trajectories and stationary increments,
$\sigma^{2}_{V}(t)=Var(V(t))$ and $\sigma_{V}(0)=0$. For this case
$Z(t)>0,t\in\mathbb{R}^{d}$ almost surely, condition (2.6) is satisfied and
(2.5) reads

(2.7) $\displaystyle\mathcal{B}_{Z}^{\delta}(x)$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathbb{E}\left\\{\frac{1}{S_{\delta}}\mathbb{I}\Bigl{(}\int_{\delta\mathbb{Z}^{d}}\mathbb{I}(V(t)-\sigma^{2}_{V}(t)/2>\ln
s)\lambda_{\delta}(dt)>x\Bigr{)}\right\\}\lambda(ds)<\infty.$

######  Corollary 2.3.

If $Z$ has almost surely continuous trajectories, then for all $x_{0}\geq 0$

(2.8) $\displaystyle\lim_{x\rightarrow
x_{0}}\mathcal{B}_{Z}^{0}(x)=\mathcal{B}_{Z}^{0}(x_{0}).$

Define next a probability measure $\mu$ on $\mathcal{D}$ by

(2.9) $\displaystyle\mu(A)=\mathbb{E}\\{Z(0)\mathbb{I}{\left\\{Z/Z(0)\in
A\right\\}}\\},\quad A\in\mathcal{D}.$

Let $\Theta$ be a rf with law $\mu$. By the definition, $\Theta$ has also
càdlàg sample paths and since $D$ is Polish, in view of [16, Lem p. 1276] we
can assume that $\Theta$ is defined in the same probability space as $Z$.
Recall that $\lambda_{\delta}(dt)/\delta^{d}$ is the counting measure on
$\delta\mathbb{Z}^{d}$ if $\delta>0$ and $\lambda_{0}$ is the Lebesgue measure
on $\mathbb{R}^{d}$. Since we can rewrite (2.5) as

(2.10) $\displaystyle\mathcal{B}_{Z}^{\delta}(x)$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathbb{E}\left\\{\frac{1}{S_{\delta}(\Theta)}\mathbb{I}\Bigl{(}\int_{\delta\mathbb{Z}^{d}}\mathbb{I}(\Theta(t)>s)\lambda_{\delta}(dt)>x\Bigr{)}\right\\}\lambda(ds)<\infty,$

where

$S_{{\delta}}(\Theta)=\int_{\delta\mathbb{Z}^{d}}\Theta(t)\lambda_{\delta}(dt)$

and the law of $\Theta$ is uniquely determined by the law of the max-stable
stationary rf $X$ and does not depend on the particular choice of $Z$, see
[12, Lem A.1], hence if $Z_{*}$ is another spectral rf for $X$, then

(2.11)
$\displaystyle\mathcal{B}_{Z}^{\delta}(x)=\mathcal{B}_{Z_{*}}^{\delta}(x)$

for all $\delta\geq 0$. Assume next that $\mathbb{P}\\{S_{0}<\infty\\}=1$ and
let

(2.12) $\displaystyle Z_{*}(t)=(p(T))^{-1}B^{T}{\bar{Q}_{\delta}(t)},\quad
t\in\delta\mathbb{Z}^{d}$

be a spectral rf of the max-stable rf
$X_{\delta}(t)=X(t),t\in\delta\mathbb{Z}^{d}$, where $\bar{Q}_{\delta}$ is
independent of a rv $T$, which has pdf $p(s)>0,\,s\in\delta\mathbb{Z}^{d}$. We
choose $p$ to be continuous when $\delta=0$. In view of [17, Thm 2.3] one
possible construction is

$\bar{Q}_{\delta}(t)=c\frac{\Theta(t)}{S_{\delta}(\Theta)},\quad
t\in\delta\mathbb{Z}^{d},$

with $c=1$ if $\delta=0$ and $c=\delta^{d}$ otherwise. Set below
$Q_{\delta}=\bar{Q}_{\delta}/c$.

######  Lemma 2.4.

  1. (i)

If $\mathbb{P}\\{S_{0}<\infty\\}=1$, then for $Q_{\delta}$ as above and all
$\delta,x$ non-negative we have

(2.13) $\displaystyle\mathcal{B}_{Z}^{\delta}(x)$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathbb{P}\left\\{\int_{\delta\mathbb{Z}^{d}}\mathbb{I}(Q_{\delta}(t)>s)\lambda_{\delta}(dt)>x\right\\}\lambda(ds)<\infty.$

  2. (ii)

If $\mathbb{P}\\{S_{0}<\infty\\}>0$, then with $V(t)=Z(t)\lvert S_{0}<\infty$
for all $\delta,x$ non-negative we have

(2.14) $\displaystyle\mathcal{B}_{Z}^{\delta}(x)$ $\displaystyle=$
$\displaystyle\mathbb{P}\\{S_{0}(\Theta)<\infty\\}\mathcal{B}_{V}^{\delta}(x)<\infty.$

Let in the following $Y(t)=R\Theta(t)$ with $R$ a $1$-Pareto rv with survival
function $1/x,x\geq 1$ independent of $\Theta$ and set hereafter

$\epsilon_{\delta}(Y)=\int_{\delta\mathbb{Z}^{d}}\mathbb{I}{\left\\{Y(t)>1\right\\}}\lambda_{\delta}(dt).$

Recall that when $\delta=0$ we interpret $\delta\mathbb{Z}^{d}$ as
$\mathbb{R}^{d}$. We establish below the Berman representation (1.1) for the
general setup of this paper.

######  Theorem 2.5.

If $\mathbb{P}\\{S_{0}=\infty\\}<1$, then for all $\delta,x$ non-negative

(2.15)
$\displaystyle\mathcal{B}_{Z}^{\delta}(x)=\mathbb{E}\left\\{\frac{\mathbb{I}{\left\\{\epsilon_{\delta}(Y)>x\right\\}}}{\epsilon_{\delta}(Y)}\right\\}<\infty.$

######  Corollary 2.6.

Under the conditions of 2.5 we have that $\epsilon_{0}(Y)$ has a continuous
distribution if $Z$ has almost surely continuous trajectories. Moreover,
$\mathcal{B}_{Z}^{\delta}(x)>0$ for all $x\geq 0$ such that
$\mathbb{P}\\{\epsilon_{\delta}(Y)>x\\}>0$.

######  Proposition 2.7.

For all $\delta\geq 0$ and $x>0$ we have

(2.16)
$\displaystyle\frac{\mathbb{P}\\{\epsilon_{\delta}(Y)>x\\}^{2}}{\mathbb{E}\\{\epsilon_{\delta}(Y)\\}}\leq\mathcal{B}_{Z}^{\delta}(x)\leq
x^{-1}\mathbb{P}\\{\epsilon_{\delta}(Y)\geq x\\}.$

######  Remark 2.8.

  1. (i)

If $x=0$, the lower bound in (2.16) holds with 1 in the numerator, see [6,
14].

  2. (ii)

If $\mathbb{E}\\{\epsilon^{p}_{\delta}(Y)\\}$ is finite for some $p>0$, then
combination of the upper bound in (2.16) with the Markov inequality gives the
following upper bound

(2.17) $\displaystyle\mathcal{B}_{Z}^{\delta}(x)\leq
x^{-p-1}\mathbb{E}\\{\epsilon^{p}_{\delta}(Y)\\},\quad x>0.$

  3. (iii)

If $\mathbb{E}\\{\epsilon_{\delta}(Y)\\}<\infty$ and
$\int_{0}^{\infty}e^{sx}(\mathcal{B}_{Z}^{\delta}(x))^{1/2}dx<\infty$, then it
follows that for all $s>0$

$\mathbb{E}\\{e^{s\epsilon_{\delta}(Y)}\\}\leq
1+s(\mathbb{E}\\{\epsilon_{\delta}(Y)\\})^{1/2}\int_{0}^{\infty}e^{sx}(\mathcal{B}_{Z}^{\delta}(x))^{1/2}dx.$

  4. (iv)

Since $Y=R\Theta$, we can calculate in case of known $\Theta$ the expectation
of $\epsilon_{\delta}(Y)$ as follows

$\displaystyle\mathbb{E}\\{\epsilon_{\delta}(Y)\\}$ $\displaystyle=$
$\displaystyle\int_{\delta\mathbb{Z}^{d}}\mathbb{P}\\{R\Theta(t)>1\\}\lambda_{\delta}(dt)=\int_{1}^{\infty}\int_{\delta\mathbb{Z}^{d}}\mathbb{P}\\{\Theta(t)>1/r\\}\lambda_{\delta}(dt)r^{-2}dr.$

If $Z(t)=\exp(V(t)-\sigma^{2}_{V}(t)/2)),t\in\mathbb{R}^{d}$ is as in Remark
2.2, Item iii, then in view of [6, Lem 5.4], [14, Eq. (5.3)] we have

(2.18)
$\displaystyle\mathbb{E}\\{\epsilon_{\delta}(Y)\\}=\int_{\delta\mathbb{Z}^{d}}\int_{1}^{\infty}{\Psi}\left(\frac{\sigma_{V}(t)}{2}-\frac{\ln
r}{\sigma(t)}\right)r^{-2}\lambda_{\delta}(dt)dr=2\int_{t\in\delta\mathbb{Z}^{d}}{\Psi}(\sigma_{V}(t)/2)\lambda_{\delta}(dt),$

where $\Psi$ is the survival function of an $N(0,1)$ rv.

##  3\. Approximation of $\mathcal{B}_{Z}^{\delta}(x)$ and its behaviour for
large $x$

We show first that $\mathcal{B}_{Z}=\mathcal{B}_{Z}^{0}$ can be approximated
by considering $\mathcal{B}_{Z}^{\delta}(x)$ and letting $\delta\downarrow 0$.

######  Proposition 3.1.

For all $x\geq 0$ we have that

$\lim_{\delta\downarrow 0}\mathcal{B}_{Z}^{\delta}(x)=\mathcal{B}_{Z}^{0}(x).$

We note in passing that for $x=0$ we retrieve the approximation for Pickands
constants derived in [15]. An approximation of $\mathcal{B}_{Z}^{\delta}(x)$
can be obtained by letting $T\to\infty$ and calculating the limit of

$\frac{\mathcal{B}^{\delta}_{Z}([0,T]^{d}\cap\delta\mathbb{Z}^{d},x)}{T^{d}}.$

For such an approximation we shall discuss the rate of convergence to
$\mathcal{B}_{Z}^{\delta}(x)$ assuming further that

$Z(t)=\exp\left(V(t)-\frac{\sigma^{2}_{V}(t)}{2}\right),\quad
t\in\mathbb{R}^{d}$

is as in Remark 2.2, Item iii.

A1 $\sigma^{2}_{V}(t)$ is a continuous and strictly increasing function, and
there exists $\alpha_{0}\in(0,2]$ and $A_{0}\in(0,\infty)$ such that

$\limsup_{\lVert t\rVert\to 0}\frac{\sigma^{2}_{V}(t)}{\lVert
t\rVert^{\alpha_{0}}}\leq A_{0},$

where $\lVert\cdot\rVert$ is the Euclidean norm.
A2 There exists $\alpha_{\infty}\in(0,2]$ such that

$\liminf_{\lVert t\rVert\to\infty}\frac{\sigma^{2}_{V}(t)}{\lVert
t\rVert^{\alpha_{\infty}}}>0.$

The following theorem constitutes the main finding of this section.

######  Theorem 3.2.

Under A1-A2 we have for all $\delta,x$ non-negative and $\lambda\in(0,1)$

(3.1)
$\displaystyle\lim_{T\to\infty}\left|\mathcal{B}_{Z}^{\delta}(x)-\frac{\mathcal{B}_{Z}^{\delta}([0,T]^{d}\cap\delta\mathbb{Z}^{d},x)}{T^{d}}\right|T^{\lambda}=0.$

######  Remark 3.3.

  1. (i)

For $x=0$ the rate of convergence in (3.1) agrees with the findings in [18].

  2. (ii)

The range of the parameter $\lambda\in(0,1)$ in Theorem 3.2 cannot be extended
to $\lambda\geq 1$. Indeed, following [19], for $V(t)=\sqrt{2}B_{1}(t)$,
$\delta=0$, $T>x$ and $d=1$ we have

$\mathcal{B}_{Z}([0,T],x)=2\Psi(x/\sqrt{2})+\sqrt{2}(T-x)\varphi(x/\sqrt{2})$

implying

(3.2) $\displaystyle\mathcal{B}_{Z}(x)=\sqrt{2}\varphi(x/\sqrt{2}),$

where $\varphi(\cdot)$ is the pdf of an $N(0,1)$ rv. Consequently, we have

$\displaystyle\lim_{T\to\infty}\left|\mathcal{B}_{Z}(x)-\frac{\mathcal{B}_{Z}([0,T],x)}{T}\right|T={2\Psi(x/\sqrt{2})}>0.$

In the rest of this section we focus on $d=1$ log-Gaussian case. In view of
(3.2) for some finite positive constant $C$

$\ln(\mathcal{B}_{Z}^{\delta}(x))\sim-{C}\sigma^{2}_{V}(x),\quad x\to\infty.$

The next result gives logarithmic bounds for $\mathcal{B}_{Z}^{\delta}(x)$ as
$x\to\infty$ that supports this hypothesis.

######  Proposition 3.4.

Suppose that $d=1$ and $V$ satisfies A1-A2. Then

$\liminf_{x\to\infty}\frac{\ln(\mathcal{B}_{Z}^{\delta}(x))}{\sigma^{2}_{V}(x/2)}\geq-1$

and

$\limsup_{x\to\infty}\frac{\ln(\mathcal{B}_{Z}^{\delta}(x))}{\sigma^{2}_{V}(x/2)}\leq-\frac{3-2\sqrt{2}}{2}.$

######  Remark 3.5.

  1. (i)

If we suppose additionally that $\sigma_{V}^{2}$ is regularly varying at
$\infty$ with parameter $\alpha>0$, then it follows from 3.4 that

$-\frac{1}{2^{\alpha}}\leq\liminf_{x\to\infty}\frac{\ln(\mathcal{B}_{Z}^{\delta}(x))}{\sigma^{2}_{V}(x)}\leq\limsup_{x\to\infty}\frac{\ln(\mathcal{B}_{Z}^{\delta}(x))}{\sigma^{2}_{V}(x)}\leq-\frac{3-2\sqrt{2}}{2^{\alpha+1}}.$

  2. (ii)

If follows from the proof of 3.4 that under A1-A2

$-\frac{1}{2}\leq\liminf_{x\to\infty}\frac{\ln(\mathbb{P}\\{\epsilon_{\delta}(Y)>x\\})}{\sigma^{2}_{V}(x/2)}\leq\limsup_{x\to\infty}\frac{\ln(\mathbb{P}\\{\epsilon_{\delta}(Y)>x\\})}{\sigma^{2}_{V}(x/2)}\leq-\frac{3-2\sqrt{2}}{2}.$

######  Example 3.6.

Let $V(t)=\sqrt{2}B_{H}(t)$, with $H\leq 1$, i.e.,
$\sigma^{2}_{V}(t)=2t^{2H}$. Then
$\mathbb{E}\\{\epsilon_{0}(Y)\\}=\frac{4^{1/(2H)+0.5}}{\sqrt{\pi}\Gamma(1/(2H)+0.5)}$,
see [6]. For $\delta>0$ we use (2.18) to compute
$\mathbb{E}\\{\epsilon_{\delta}(Y)\\}$, see Tab. 1. The graph of
$\mathbb{E}\\{\epsilon_{\delta}(Y)\\}$ as a function of $\delta$ and the upper
bound (2.17) with $p=1$ for Berman constants as a function of $x\in[1,10]$ are
presented on Fig. 1.

$\mathbb{E}\\{\epsilon_{\delta}(Y)\\}\,\,\backslash\,\,H$ | $0.1$ | $0.2$ | $0.3$ | $0.4$ | $0.5$ | $0.6$ | $0.7$ | $0.8$ | $0.9$ | $1$
---|---|---|---|---|---|---|---|---|---|---
$\delta=0$ | $60480$ | $72.216267$ | $12.309822$ | $5.866446$ | $4$ | $3.198992$ | $2.777685$ | $2.527405$ | $2.366354$ | $2.256758$
$\delta=1$ | $48824.040913$ | $72.594979$ | $12.632020$ | $6.140195$ | $4.232120$ | $3.395236$ | $2.942920$ | $2.665777$ | $2.481422$ | $2.351603$
$\delta=5$ | $57667.986631$ | $74.736598$ | $14.803952$ | $8.344951$ | $6.474827$ | $5.685059$ | $5.295399$ | $5.104008$ | $5.026130$ | $5.004070$
$\delta=10$ | $59291.12614$ | $77.87128$ | $18.22637$ | $12.07630$ | $10.54057$ | $10.09794$ | $10.00788$ | $10.00016$ | $10$ | $10$

Table 1. Values of $\mathbb{E}\\{\epsilon_{\delta}(Y)\\}$ for
$\delta=\\{0,1,5,10\\}$ and $V(t)=\sqrt{2}B_{H}(t)$.

Figure 1. $\mathbb{E}\\{\epsilon_{\delta}(Y)\\}$ as a function of
$\delta\in[0,2]$ and $H=\\{0.5,0.9\\}$ and the upper bound (2.17) with $p=1$
for Berman constants as a function of $x\in[1,10]$ for $H=0.5$, $\delta=1$ and
$H=0.9$, $\delta=10$ where $V(t)=\sqrt{2}B_{H}(t)$.

We simulated Berman constant $\mathcal{B}_{Z}(x)$ using estimator (2.15) for
different $x$ and $H$ see Tab. 2. In our simulation we generated $N=20000$
trajectories by means of Davies-Harte algorithm on the interval $[-64,64]$
with the step $e=1/2^{9}=0.001953125$. Since the sample paths of fractional
Brownian motion are very torn by the negative correlation of increments for
$H<0.5$ we cannot trust the simulation for $H$ close to 0 and we estimated
Berman constant for $H\geq 0.4$ (see the half width of 95% confidence interval
in Tab. 2). Let us note that the estimator (2.15) for $x=0$ is different from
the estimator of Pickands constant in [4]. Compare our simulation for $x=0$
with the results of [4] for Pickands constant.

$x\backslash H$ | $0.4$ | $0.5$ | $0.6$ | $0.7$ | $0.8$ | $0.9$ | $0.999$
---|---|---|---|---|---|---|---
$0$ | $1.016664$ | $1.038244$ | $0.9336774$ | $0.8059135$ | $0.7146539$ | $0.6482967$ | $0.5644909$
| $\pm 0.08504517$ | $\pm 0.08105526$ | $\pm 0.08679748$ | $\pm 0.04206874$ | $\pm 0.01835475$ | $\pm 0.01264331$ | $\pm 0.00675525$
$0.05$ | $0.7319381$ | $0.7662378$ | $0.77076$ | $0.7419288$ | $0.6972516$ | $0.641973$ | $0.5660913$
| $\pm 0.02164056$ | $\pm 0.02001129$ | $\pm 0.0188357$ | $\pm 0.01654269$ | $\pm 0.01375789$ | $\pm 0.01041449$ | $\pm 0.006183973$
$0.1$ | $0.6285574$ | $0.6975815$ | $0.7083258$ | $0.6965521$ | $0.670074$ | $0.6275897$ | $0.5638993$
| $\pm 0.01509329$ | $\pm 0.01508846$ | $\pm 0.01388002$ | $\pm 0.01262112$ | $\pm 0.01129947$ | $\pm 0.008966844$ | $\pm 0.005561737$
$0.2$ | $0.5210044$ | $0.5855574$ | $0.6169511$ | $0.6341507$ | $0.6143437$ | $0.5951561$ | $0.5601536$
| $\pm 0.01018901$ | $\pm 0.01005615$ | $\pm 0.009730012$ | $\pm 0.009250254$ | $\pm 0.008171438$ | $\pm 0.007041064$ | $\pm 0.004994959$
$0.4$ | $0.3996749$ | $0.4631784$ | $0.5050474$ | $0.5339913$ | $0.5503154$ | $0.5457671$ | $0.5397937$
| $\pm 0.00644895$ | $\pm 0.006531524$ | $\pm 0.006421914$ | $\pm 0.006178146$ | $\pm 0.005904937$ | $\pm 0.005230998$ | $\pm 0.004207075$
$0.5$ | $0.3523235$ | $0.4228112$ | $0.4642559$ | $0.4931276$ | $0.5079599$ | $0.5182281$ | $0.5306746$
| $\pm 0.005388992$ | $\pm 0.005689292$ | $\pm 0.005635201$ | $\pm 0.005447861$ | $\pm 0.005075071$ | $\pm 0.004643272$ | $\pm 0.003954409$
$1$ | $0.2390332$ | $0.2833468$ | $0.3098104$ | $0.3418117$ | $0.3673532$ | $0.3978701$ | $0.4387592$
| $\pm 0.003261513$ | $\pm 0.00342165$ | $\pm 0.003471659$ | $\pm 0.003439146$ | $\pm 0.003371701$ | $\pm 0.003232641$ | $\pm 0.002982276$
$2$ | $0.1373955$ | $0.1493499$ | $0.1606709$ | $0.1731576$ | $0.1824689$ | $0.1975992$ | $0.2061034$
| $\pm 0.001904439$ | $\pm 0.002068819$ | $\pm 0.002168721$ | $\pm 0.002275878$ | $\pm 0.002382227$ | $\pm 0.002493246$ | $\pm 0.00260955$
$5$ | $0.04407294$ | $0.0366231$ | $0.02843742$ | $0.02065171$ | $0.01204482$ | $0.004497369$ | $0.001105259$
| $\pm 0.0008552848$ | $\pm 0.0008681481$ | $\pm 0.0008307949$ | $\pm 0.0007567421$ | $\pm 0.0006104972$ | $\pm 0.0003907966$ | $\pm 0.0001991857$
$6$ | $0.03259252$ | $0.02488601$ | $0.01711836$ | $0.01001073$ | $0.003932873$ | $0.0006936917$ | 6.375708e-05
| $\pm 0.0007073306$ | $\pm 0.0006875452$ | $\pm 0.000618167$ | $\pm 0.0005032682$ | $\pm 0.0003303923$ | $\pm 0.0001440762$ |  $\pm 4.42043\\!\\!$ e-05

Table 2. Estimation of $\mathcal{B}_{Z}(x)$ for fractional Brownian motion
$B_{H}$ and the half width of 95% confidence interval.

######  Example 3.7.

Let $X(t)$, $t\in\mathbb{R}$ be a stationary Ornstein-Uhlenbeck process, i.e.,
a centered Gaussian process with zero mean and covariance
$\mathbb{E}\\{X(t)X(s)\\}=\exp(-|t-s|),s,t\in\mathbb{R}$. Then the random
process

$V(t)=\left\\{\begin{array}[]{ll}\sqrt{2}\int_{0}^{t}X(s)ds&\mbox{if $t\geq
0$}\\\ &\\\ -\sqrt{2}\int_{t}^{0}X(s)ds&\mbox{if $t<0$}\end{array}\right.$

is Gaussian with stationary increments and variance
$\sigma^{2}_{V}(t)=4(|t|+e^{-|t|}-1$). Using (2.15) we simulated the Berman
constant for $\delta=0$ and different $x$, see Tab. 3 and for $x=0$ and
$\delta=\\{0,0.1,0.2,0.5,1,2,5,10\\}$, see Tab. 4. We generated $N=20000$
trajectories with the step $e=10^{-5}$ on the interval $[-15,15]$. In Fig. 2
we graphed $\mathcal{B}_{Z}(x)$ and
$\frac{\ln(\mathcal{B}_{Z}(x))}{\sigma_{V}^{2}(x/2)}$ as function of $x$ and
we get that this ratio is asymptotically around $-0.4$. Note that according to
Remark 3.5 it should be between $-0.5$ and $-0.04289322$.

Using (2.18) we computed $\mathbb{E}\\{\epsilon_{\delta}(Y)\\}$ and
$1/\mathbb{E}\\{\epsilon_{\delta}(Y)\\}$ for
$\delta=\\{0,0.1,0.2,0.5,1,2,5,10\\}$, see Tab. 5. The graph of the lower
bound of $\mathcal{B}^{\delta}_{Z}(0)$ for the integrated Ornstein-Uhlenbeck
process that is $1/\mathbb{E}\\{\epsilon_{\delta}(Y)\\}$ as a function of
$\delta\in[0,10]$ is given in Fig. 2. The value of $\mathcal{B}_{Z}(0)$
constant for the integrated Ornstein-Uhlenbeck process with the same
parameters as here was simulated in [20] resulting in the value $0.528$.

$x$ | $0$ | $0.5$ | $1$ | $1.5$ | $2$ | $2.5$ | $3$
---|---|---|---|---|---|---|---
$\mathcal{B}_{Z}(x)$ | $0.5267956$ | $0.452556$ | $0.3482289$ | $0.2621687$ | $0.1900299$ | $0.1376086$ | $0.09881259$
| $\pm 0.01817717$ | $\pm 0.004676632$ | $\pm 0.003180162$ | $\pm 0.002588018$ | $\pm 0.002216284$ | $\pm 0.001910763$ | $\pm 0.00163841$
$x$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$
$\mathcal{B}_{Z}(x)$ | $0.05088893$ | $0.02715927$ | $0.01433577$ | $0.007437053$ | $0.003796336$ | $0.001998398$ | $0.001205136$
| $\pm 0.00116684$ | $\pm 0.0008278098$ | $\pm 0.0005788133$ | $\pm 0.0003983809$ | $\pm 0.0002730899$ | $\pm 0.0001906838$ | $\pm 0.0001414664$
$x$ | $11$ | $12$ | $13$ | $14$ | $15$ | $16$ | $17$
$\mathcal{B}_{Z}(x)$ | $0.000631948$ | $0.0003812784$ | $0.0001845301$ | 0.00010499 | 9.130422e-05 | 2.426165e-05 | 2.103512e-05
|  $\pm 9.837308\\!\\!$ e-05 |  $\pm 7.355149\\!\\!$ e-05 |  $\pm 4.89203\\!\\!$ e-05 |  $\pm 3.593126\\!\\!$ e-05 |  $\pm 3.276786\\!\\!$ e-05 |  $\pm 1.594309\\!\\!$ e-05 |  $\pm 1.463212e\\!\\!$ e-05

Table 3. Estimation of $\mathcal{B}_{Z}(x)$ for integrated Ornstein-Uhlenbeck
process and the half width of 95% confidence interval..

$\delta$ | $0$ | $0.1$ | $0.2$ | $0.5$ | $1$ | $2$ | $5$ | $10$
---|---|---|---|---|---|---|---|---
$\mathcal{B}^{\delta}_{Z}(0)$ | $0.5267956$ | $0.5131973$ | $0.5126575$ | $0.4934096$ | $0.4484668$ | $0.3843544$ | $0.1908583$ | $0.09984$
| $\pm 0.01817717$ | $\pm 0.007850686$ | $\pm 0.007194036$ | $\pm 0.005746635$ | $\pm 0.00402201$ | $\pm 0.001995524$ | $\pm 0.0004065713$ |  $\pm 3.913749\\!\\!$ e-05

Table 4. Estimation of $\mathcal{B}^{\delta}_{Z}(0)$ for integrated Ornstein-
Uhlenbeck process and the half width of 95% confidence interval.

$\delta$ | $0$ | $0.1$ | $0.2$ | $0.5$ | $1$ | $2$ | $5$ | $10$
---|---|---|---|---|---|---|---|---
$\mathbb{E}\\{\epsilon_{\delta}(Y)\\}$ | $3.234658$ | $3.245584$ | $3.248405$ | $3.268183$ | $3.339158$ | $3.626068$ | $5.482154$ | $10.05426$
$1/\mathbb{E}\\{\epsilon_{\delta}(Y)\\}$ | $0.3091517$ | $0.308111$ | $0.3078434$ | $0.3059804$ | $0.2994767$ | $0.2757808$ | $0.1824101$ | $0.09946035$

Table 5. Estimation of $\mathbb{E}\\{\epsilon_{\delta}(Y)\\}$ and
$1/\mathbb{E}\\{\epsilon_{\delta}(Y)\\}$ for
$\delta=\\{0,0.1,0.2,0.5,1,2,5,10\\}$.

Figure 2. The graphs of $\mathcal{B}_{Z}(x)$ and
$\frac{\ln(\mathcal{B}_{Z}(x))}{\sigma_{V}^{2}(x/2)}$ as function of $x$ and
the lower bound of Pickands constant as a function of $\delta$ for integrated
Ornstein-Uhlenbeck process.

##  4\. Further Results and Proofs

Let in the following $X(t),t\in\mathbb{R}^{d}$ be a max-stable stationary rf
with càdlàg sample paths and spectral rf $Z$ as in 2.1 and define $\Theta$ as
in (2.9). Define $Y(t)=R\Theta(t),t\in\mathbb{R}^{d}$ with $R$ an $1$-Pareto
rv (with survival function $1/x,x\geq 1$) independent of $\Theta$ and set
$M_{Y,\delta}=\sup_{t\in\delta\mathbb{Z}^{d}}Y(t)$. Note in passing that

$\mathbb{P}\\{M_{Y,\delta}>1\\}=1$

since $M_{Y,\delta}\geq R\Theta(0)>1$ almost surely by the assumption on $R$
and by the definition $\mathbb{P}\\{\Theta(0)=1\\}=1$.

Recall that
$S_{\delta}=S_{\delta}(Z)=\int_{\delta\mathbb{Z}^{d}}Z(t)\lambda_{\delta}(dt)$.
In view of (2.4) we have that $\mathbb{P}\\{S_{0}>0\\}=1$. In the following,
for any fixed $\delta\geq 0$ (but not simultaneously for two different
$\delta$’s) we shall assume that $S_{\delta}>0$ almost surely, i.e., $Z$ is
such that $\mathbb{P}\\{\sup_{t\in\delta\mathbb{Z}^{d}}Z(t)>0\\}=1$. Such a
choice of $Z$ is possible in view of [14, Lem 7.3].
A functional $F:D\to[0,\infty]$ is said to be shift-invariant if
$F(f(\cdot-h))=F(f(\cdot))$ for all $h\in\mathbb{R}^{d}$.

We state first two lemmas and proceed with the postponed proofs.

######  Lemma 4.1.

If $\mathbb{P}\\{S_{0}<\infty\\}=1$, then
$\mathbb{P}\\{S_{\delta}<\infty\\}=1,\delta>0$ and for all $x>0$

(4.1)
$\displaystyle\mathbb{P}\\{\epsilon_{\delta}(Y/x)<\infty\\}=1,\quad\forall
x>0,\forall\delta\geq 0.$

Moreover $\mathbb{P}\\{M_{Y,\delta}<\infty\\}=1$.

Proof of Lemma 4.1 In view of (2.4) $S_{0}>0$ almost surely. The assumption
that $S_{0}<\infty$ almost surely is in view of [21, Thm 3] equivalent with
$Z(t)\to 0$ almost surely as $\lVert t\rVert\to\infty$, with
$\lVert\cdot\rVert$ some norm on $\mathbb{R}^{d}$. Hence $S_{\delta}<\infty$
almost surely follows from [21, Thm 3]. By the definition of $\Theta$ and the
fact that $\mathbb{P}\\{S_{0}(Z)\in(0,\infty)\\}=1$ we have

(4.2) $\displaystyle 0$ $\displaystyle=$
$\displaystyle\mathbb{E}\left\\{{Z(0)}\mathbb{I}{\left\\{\limsup_{\lVert
t\rVert\to\infty}Z(t)>0\right\\}}\right\\}=\mathbb{E}\left\\{Z(0)\mathbb{I}{\left\\{\limsup_{\lVert
t\rVert\to\infty}Z(t)/Z(0)>0\right\\}}\right\\}$ $\displaystyle=$
$\displaystyle\mathbb{E}\left\\{\mathbb{I}{\left\\{\limsup_{\lVert
t\rVert\to\infty}\Theta(t)>0\right\\}}\right\\}$

implying that $\mathbb{P}\\{\lim_{\lVert t\rVert\to\infty}\Theta(t)=0\\}=1$.
Consequently, $\mathbb{P}\\{\lim_{\lVert t\rVert\to\infty}Y(t)=0\\}=1$ and
hence the claim follows.

$\Box$

Below we interpret $\infty\cdot 0$ and $0/0$ as 0. The next result is a minor
extension of [22, Lem 2.7].

######  Lemma 4.2.

If $\mathbb{P}\\{S_{0}<\infty\\}=1$, then for all measurable shift invariant
functional $F$ and all $\delta,x$ non-negative

(4.3) $\displaystyle
x\mathbb{E}\left\\{\frac{F(Y/x)}{\epsilon_{\delta}(Y)}\mathbb{I}{\left\\{M_{Y,\delta}>{\max}(x,1)\right\\}}\right\\}$
$\displaystyle=$
$\displaystyle\mathbb{E}\left\\{\frac{F(Y)}{\epsilon_{\delta}(Y)}\mathbb{I}{\left\\{M_{Y,\delta}>{\max}(1/x,1)\right\\}}\right\\}.$

Proof of Lemma 4.2 For all measurable functional $F:D\to[0,\infty]$ and all
$x>0$

(4.4) $\displaystyle x\mathbb{E}\\{F(Y)\mathbb{I}({Y(h)}>x)\\}$
$\displaystyle=$
$\displaystyle\mathbb{E}\\{F(xB^{h}Y)\mathbb{I}(x{Y(-h)}>1)\\}$

is valid for all $h\in\mathbb{R}^{d}$ with
$B^{h}Y(t)=Y(t-h),h,t\in\mathbb{R}^{d}$. Note in passing that $B^{h}Y$ can be
substituted by $Y$ in the right-hand side of (4.4) if $F$ is shift-invariant.
The identity (4.4) is shown in [8]. For the discrete setup it is shown
initially in [13, 23] and for case $d=1$ in [22].
Next, if $x\in(0,1]$, since $Y(0)=R>1$ almost surely and by the assumption on
the sample paths we have that $\mathbb{P}\\{\epsilon_{\delta}(Y/x)>0\\}=1,$
recall $\mathbb{P}\\{\Theta(0)=1\\}=1$. By 4.1
$\mathbb{P}\\{M_{Y,\delta}\in(1,\infty)\\}=1$, hence for all $x>1$ we have
further that $M_{Y,\delta}>x$ implies $\epsilon_{\delta}(Y/x)>0$.
Consequently, in view of (4.1) $\epsilon_{\delta}(Y/x)/\epsilon_{\delta}(Y/x)$
is well defined on the event $M_{Y,\delta}>x,x>1$ and also it is well-defined
for any $x\in(0,1]$.

Recall that $\lambda_{\delta}(dt)$ is the Lebesgue measure on $\mathbb{R}^{d}$
if $\delta=0$ and the counting measure multiplied by $\delta^{d}$ on
$\delta\mathbb{Z}^{d}$ if $\delta>0$. Let us remark that for any shift-
invariant functional $F$, the functional

$F^{*}(Y)=\frac{F(Y/x)\mathbb{I}{\left\\{M_{Y,\delta}>\max(x,1)\right\\}}}{\epsilon_{\delta}(Y)\epsilon_{\delta}(Y/x)}$

is shift-invariant for all $h\in\mathbb{R}^{d}$ if $\delta=0$ and any shift
$h\in\delta\mathbb{Z}^{d}$ if $\delta>0$. Thus applying the Fubini-Tonelli
theorem twice and (4.4) with functional $F^{*}$ we obtain for all $\delta\geq
0,x>0$

$\displaystyle
x\mathbb{E}\left\\{\frac{F(Y/x)}{\epsilon_{\delta}(Y)}\mathbb{I}{\left\\{M_{Y,\delta}>\max(x,1)\right\\}}\right\\}$
$\displaystyle=$ $\displaystyle
x\int_{\delta\mathbb{Z}^{d}}\mathbb{E}\left\\{\frac{F(Y/x)\mathbb{I}{\left\\{M_{Y,\delta}>\max(x,1)\right\\}}}{\epsilon_{\delta}(Y)\epsilon_{\delta}(Y/x)}\mathbb{I}{\left\\{Y(h)>x\right\\}}\right\\}\lambda_{\delta}(dh)$
$\displaystyle=$
$\displaystyle\int_{\delta\mathbb{Z}^{d}}\mathbb{E}\left\\{\frac{F(Y)\mathbb{I}{\left\\{M_{Y,\delta}>\max(1/x,1)\right\\}}}{\epsilon_{\delta}(xY)\epsilon_{\delta}(Y)}\mathbb{I}{\left\\{xY(-h)>1\right\\}}\right\\}\lambda_{\delta}(dh)$
$\displaystyle=$
$\displaystyle\mathbb{E}\left\\{\frac{F(Y)\mathbb{I}{\left\\{M_{Y,\delta}>\max(1/x,1)\right\\}}}{\epsilon_{\delta}(xY)\epsilon_{\delta}(Y)}\int_{\delta\mathbb{Z}^{d}}\mathbb{I}{\left\\{xY(h)>1\right\\}}\lambda_{\delta}(dh)\right\\}$
$\displaystyle=$
$\displaystyle\mathbb{E}\left\\{\frac{F(Y)}{\epsilon_{\delta}(Y)}\mathbb{I}{\left\\{M_{Y,\delta}>\max(1/x,1)\right\\}}\right\\},$

hence the proof follows. $\Box$

Proof of Theorem 2.1 Let $\delta\geq 0$ be fixed and consider for simplicity
$d=1$. By the assumption we have $\mathbb{E}\\{\sup_{t\in[0,T]}Z(t)\\}<\infty$
for all $T>0$. Since we assume that
$\mathbb{P}\\{\sup_{t\in\mathbb{R}}Z(t)>0\\}=1$, then
$\mathbb{P}\\{S_{0}>0\\}=1$. Using the assumption we have
$\mathbb{P}\\{S_{\eta}<\infty\\}>0$ for all $\eta\geq 0$ and thus by (2.3) we
obtain

$\displaystyle\infty$ $\displaystyle>$
$\displaystyle\mathbb{E}\left\\{\sup_{t\in[0,{{\color[rgb]{0.,0.,1.}2}}]}Z(t)\frac{S_{0}}{S_{0}}\right\\}=\int_{\mathbb{R}}\mathbb{E}\left\\{Z(h)\sup_{t\in[0,2]}Z(t)/S_{0}\right\\}\lambda(dh)$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}}\mathbb{E}\left\\{Z(0)\sup_{t\in[-h,2-h]}Z(t)/S_{0}\right\\}\lambda(dh)$
$\displaystyle=$
$\displaystyle\sum_{i\in\mathbb{Z}}\int_{i}^{i+1}\mathbb{E}\left\\{Z(0)\sup_{t\in[-h,2-h]}Z(t)/S_{0}\right\\}\lambda(dh)$
$\displaystyle{\geq}$
$\displaystyle\sum_{i\in\mathbb{Z}}\mathbb{E}\left\\{Z(0)\sup_{t\in[-i,{1}-i]}Z(t)/S_{0}\right\\}$
$\displaystyle\geq$
$\displaystyle\mathbb{E}\left\\{Z(0)\sup_{t\in\mathbb{R}}Z(t)/S_{0}\right\\}.$

Since also for any $M>0$ and $\eta>0$

$\displaystyle
0<\mathbb{E}\left\\{\sup_{t\in[0,{M}]}Z(t)\frac{S_{\eta}}{S_{\eta}}\right\\}<\infty$

we conclude as above that for all $\eta\geq 0$

(4.5)
$\displaystyle\mathbb{E}\left\\{Z(0)\sup_{t\in\mathbb{R}}Z(t)/S_{\eta}\right\\}<\infty,\quad\int_{\eta\mathbb{Z}}\mathbb{E}\left\\{Z(0)\sup_{h\leq
t\leq h+1}Z(t)/S_{\eta}\right\\}\lambda_{\eta}(dh)<\infty.$

Next, for any $x\geq 0$ and $\eta\geq 0$

$\displaystyle
T^{-1}\int_{0}^{\infty}\mathbb{P}\\{\int_{[0,T]\cap\eta\mathbb{Z}}\mathbb{I}(Z(t)>s)\lambda_{\eta}(dt)>x,S_{\eta}=\infty\\}ds$
$\displaystyle\leq$ $\displaystyle
T^{-1}\int_{0}^{\infty}\mathbb{P}\left\\{\int_{[0,T]\cap\eta\mathbb{Z}}\mathbb{I}(Z(t)>s)\lambda_{\eta}(dt)>0,S_{\eta}=\infty\right\\}ds$
$\displaystyle=$ $\displaystyle
T^{-1}\int_{0}^{\infty}\mathbb{P}\left\\{\sup_{t\in[0,T]\cap\eta\mathbb{Z}}Z(t)>s,S_{\eta}=\infty\right\\}ds$
$\displaystyle=$ $\displaystyle
T^{-1}\mathbb{E}\left\\{\sup_{t\in[0,T]\cap\eta\mathbb{Z}}Z(t),S_{\eta}=\infty\right\\}$
$\displaystyle\to$ $\displaystyle 0,\quad T\to\infty,$

where the last claim follows from [15, Cor 2.1]. We shall assume that
$S_{\delta}>0$ almost surely (this is possible as mentioned at the beginning
of this section). For notational simplicity we consider next $\delta=0$. For
any $M>0,T>2M$ by the Fubini-Tonelli theorem and (2.3)

$\displaystyle\int_{0}^{\infty}\mathbb{P}\left\\{\int_{[0,T]}\mathbb{I}(Z(t)>s)\lambda(dt)>x\right\\}ds$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}}\mathbb{E}\left\\{\frac{Z(h)}{S_{0}}\int_{0}^{\infty}\mathbb{I}{\left\\{\int_{[0,T]}\mathbb{I}(Z(t)>s)\lambda(dt)>x\right\\}}ds\right\\}dh$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\int_{0}^{\infty}\mathbb{I}{\left\\{\int_{[0,T]}\mathbb{I}(Z(t-h)>s)\lambda(dt)>x\right\\}}ds\right\\}dh$
$\displaystyle=$
$\displaystyle\int_{\mathbb{R}}\int_{0}^{\infty}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\mathbb{I}{\left\\{\int_{h}^{T+h}\mathbb{I}(Z(t)>s)\lambda(dt)>x\right\\}}\right\\}dsdh$
$\displaystyle=$
$\displaystyle\int_{-M-T}^{-M}\int_{0}^{\infty}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\mathbb{I}{\left\\{\int_{h}^{T+h}\mathbb{I}(Z(t)>s)\lambda(dt)>x\right\\}}\right\\}dsdh$
$\displaystyle+\int_{h<-M-T\,\text{or}\,h>-M}\int_{0}^{\infty}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\mathbb{I}{\left\\{\int_{h}^{T+h}\mathbb{I}(Z(t)>s)\lambda(dt)>x\right\\}}\right\\}dsdh$
$\displaystyle=:$ $\displaystyle I_{M,T}+J_{M,T}.$

Thus we obtain

$\displaystyle\frac{I_{M,T}}{T}$ $\displaystyle=$
$\displaystyle\int_{-M/T-1}^{-M/T}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\int_{0}^{\infty}\mathbb{I}\Bigl{(}\int_{Th}^{T(1+h)}\mathbb{I}(Z(t)>s)\lambda(dt)>x\Bigr{)}ds\right\\}dh$
$\displaystyle=$
$\displaystyle\int_{-1}^{0}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\int_{0}^{\infty}\mathbb{I}\Bigl{(}\int_{Th-M}^{T(1+h)-M}\mathbb{I}(Z(t)>s)\lambda(dt)>x\Bigr{)}ds\right\\}dh$
$\displaystyle\to$
$\displaystyle\int_{-1}^{0}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\int_{0}^{\infty}\mathbb{I}\Bigl{(}\int_{\delta\mathbb{Z}}\mathbb{I}(Z(t)>s)\lambda(dt)>x\Bigr{)}ds\right\\}dy,\quad
T\to\infty$ $\displaystyle=$
$\displaystyle\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\int_{0}^{\infty}\mathbb{I}\Bigl{(}\int_{\delta\mathbb{Z}}\mathbb{I}(Z(t)>s)\lambda_{\delta}(dt)>x\Bigr{)}ds\right\\}$
$\displaystyle\leq$
$\displaystyle\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\int_{0}^{\infty}\mathbb{I}\Bigl{(}\int_{\mathbb{R}}\mathbb{I}(Z(t)>s)\lambda(dt)>0\Bigr{)}ds\right\\}$
$\displaystyle=$
$\displaystyle\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\sup_{t\in\mathbb{R}}Z(t)\right\\}=\mathcal{H}_{Z}^{0}<\infty,$

where $\mathcal{H}^{0}_{Z}$ is the Pickands constants, see [15, Prop 2.1] for
the last formula.
Let us consider the second term

$\displaystyle\frac{J_{M,T}}{T}$ $\displaystyle=$
$\displaystyle\int_{(-\infty,-1)\cup(0,\infty)}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\int_{0}^{\infty}\mathbb{I}\Bigl{(}\int_{Th-M}^{T(1+h)-M}\mathbb{I}(Z(t)>s)\lambda(dt)>x\Bigr{)}ds\right\\}dh$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{-1}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\int_{0}^{\infty}\mathbb{I}\Bigl{(}\int_{Th-M}^{T(1+h)-M}\mathbb{I}(Z(t)>s)\lambda(dt)>x\Bigr{)}ds\right\\}dh$
$\displaystyle+\int_{0}^{\infty}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\int_{0}^{\infty}\mathbb{I}\Bigl{(}\int_{Th-M}^{T(1+h)-M}\mathbb{I}(Z(t)>s)\lambda(dt)>x\Bigr{)}ds\right\\}dh$
$\displaystyle=:$ $\displaystyle K_{M,T}+L_{M,T}\,.$

Further, assuming for simplicity that $T$ is a positive integer we get

$\displaystyle K_{M,T}$ $\displaystyle\leq$
$\displaystyle\int_{-\infty}^{-1}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\int_{0}^{\infty}\mathbb{I}\Bigl{(}\int_{Th-M}^{T(1+h)-M}\mathbb{I}(Z(v)>s)\lambda(dv)>0\Bigr{)}ds\right\\}dh$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{-1}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\sup_{t\in[Th-M,\,T(1+h)-M]}Z(t)\right\\}dh$
$\displaystyle=$
$\displaystyle\frac{1}{T}\int_{-\infty}^{-T-M}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\sup_{t\in[h,\,h+T]}Z(t)\right\\}dh$
$\displaystyle\leq$
$\displaystyle\frac{1}{T}\sum_{i=1}^{T-1}\int_{-\infty}^{-T-M}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\sup_{t\in[h+i,\,h+i+1]}Z(t)\right\\}dh$
$\displaystyle=$
$\displaystyle\frac{1}{T}\sum_{i=1}^{T-1}\int_{-\infty}^{-T-M+i}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\sup_{t\in[h,\,h+1]}Z(t)\right\\}dh$
$\displaystyle\leq$
$\displaystyle\frac{1}{T}\sum_{i=1}^{T-1}\int_{-\infty}^{-M}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\sup_{t\in[h,\,h+1]}Z(t)\right\\}dh$
$\displaystyle=$
$\displaystyle\int_{-\infty}^{-M}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\sup_{t\in[h,\,h+1]}Z(t)\right\\}dh\to
0,\quad M\to\infty,$

where the last convergence follows from (4.5). The same way we show that
$L_{M,T}\to 0$ as $M\to\infty$ establishing the proof.
We prove next the second claim. In view of [15, proof of Prop 2.1] almost
surely for all $\delta,\eta\in[0,\infty)$

(4.6)
$\displaystyle\frac{1}{S_{\eta}(\Theta)}=\frac{1}{S_{\eta}(\Theta)}\frac{S_{\delta}(\Theta)}{S_{\delta}(\Theta)}\Theta(0),\quad\\{S_{\eta}(\Theta)<\infty\\}=\\{S_{\delta}(\Theta)<\infty\\}.$

Consequently, for any $\delta,\eta,x$ non-negative

$\displaystyle B_{\delta,\eta}(x)$ $\displaystyle:=$
$\displaystyle\int_{0}^{\infty}\mathbb{E}\left\\{\frac{Z(0)}{S_{\eta}}\mathbb{I}{\left\\{S_{\eta}<\infty\right\\}}\mathbb{I}\Bigl{(}\int_{\delta\mathbb{Z}}\mathbb{I}(Z(t)>s)\lambda_{\delta}(dt)>x\Bigr{)}\right\\}ds$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathbb{E}\left\\{\frac{1}{S_{\eta}(\Theta)}\mathbb{I}{\left\\{S_{\eta}(\Theta)<\infty\right\\}}\mathbb{I}\Bigl{(}\int_{\delta\mathbb{Z}}\mathbb{I}(\Theta(t)>s)\lambda_{\delta}(dt)>x\Bigr{)}\right\\}ds$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathbb{E}\left\\{\frac{1}{S_{\eta}(\Theta)}\frac{S_{\delta}(\Theta)}{S_{\delta}(\Theta)}\mathbb{I}{\left\\{S_{\delta}(\Theta)<\infty\right\\}}\mathbb{I}\Bigl{(}\int_{\delta\mathbb{Z}}\mathbb{I}(\Theta(t)>s)\lambda_{\delta}(dt)>x\Bigr{)}\right\\}ds.$

We proceed next with the case $\delta=0$, the other case follows with the same
argument where it is important that $\eta=k\delta$ for the shift
transformation. Taking $\delta=0,\eta>0$ we have

$\displaystyle B_{0,\eta}(x)$ $\displaystyle=$
$\displaystyle\int_{\mathbb{R}}\int_{0}^{\infty}\mathbb{E}\left\\{\frac{1}{S_{\eta}(\Theta)}\frac{\Theta(r)}{S_{0}(\Theta)}\mathbb{I}{\left\\{S_{0}(\Theta)<\infty\right\\}}\mathbb{I}\Bigl{(}\int_{\mathbb{R}}\mathbb{I}(\Theta(t)>s)\lambda(dt)>x\Bigr{)}\right\\}ds\lambda(dr)$
$\displaystyle=$
$\displaystyle\sum_{v\in\eta\mathbb{Z}}\int_{r+v\in[0,\eta]}\int_{0}^{\infty}\mathbb{E}\left\\{\frac{Z(0)}{S_{\eta}(Z)}\frac{Z(r)}{S_{0}(Z)}\mathbb{I}{\left\\{S_{0}(Z)<\infty\right\\}}\mathbb{I}\Bigl{(}\int_{\mathbb{R}}\mathbb{I}(Z(t)>s)\lambda(dt)>x\Bigr{)}\right\\}ds\lambda(dr)$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathbb{E}\left\\{\sum_{v\in\eta\mathbb{Z}}\int_{r+v\in[0,\eta]}\frac{Z(0)}{S_{\eta}(Z)}\frac{Z(r)}{S_{0}(Z)}\mathbb{I}{\left\\{S_{0}(Z)<\infty\right\\}}\mathbb{I}\Bigl{(}\int_{\delta\mathbb{Z}}\mathbb{I}(Z(t)>s)\lambda(dt)>x\Bigr{)}\right\\}\lambda(dr)ds$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathbb{E}\left\\{\sum_{v\in\eta\mathbb{Z}}\int_{r\in[0,\eta]}\frac{Z(0)}{S_{\eta}(Z)}\frac{Z(r-v)}{S_{0}(Z)}\mathbb{I}{\left\\{S_{0}(Z)<\infty\right\\}}\mathbb{I}\Bigl{(}\int_{\mathbb{R}}\mathbb{I}(Z(t)>s)\lambda(dt)>x\Bigr{)}\right\\}\lambda(dr)ds$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathbb{E}\left\\{\int_{r\in[0,\eta]}\frac{1}{\eta}\sum_{v\in\eta\mathbb{Z}}\frac{\eta
Z(v-r)}{S_{\eta}({B^{r}}Z)}\lambda(dr)\frac{Z(0)}{S_{0}(Z)}\mathbb{I}{\left\\{S_{0}(Z)<\infty\right\\}}\mathbb{I}\Bigl{(}\int_{\mathbb{R}}\mathbb{I}(Z(t)>s)\lambda(dt)>x\Bigr{)}\right\\}ds$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}(Z)}\mathbb{I}\Bigl{(}\int_{\mathbb{R}}\mathbb{I}(Z(t)>s)\lambda(dt)>x\Bigr{)}\right\\}ds,$

where we used (2.3) with $h=r-v$ to obtain the second last equality above and
(2.6) to get the last equality, hence the proof follows.

$\Box$

Proof of Corollary 2.3 Given $x\geq 0$ consider the representation (2.5)

$\mathcal{B}_{Z}^{0}(x)=\int_{0}^{\infty}\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\mathbb{I}\Bigl{(}\int_{\mathbb{R}^{d}}\mathbb{I}(Z(t)>s)\lambda(dt)>x\Bigr{)}\right\\}ds.$

By the monotonicity with respect to variable $x$ of the function

(4.7)
$\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\mathbb{I}\Bigl{(}\int_{\mathbb{R}^{d}}\mathbb{I}(Z(t)>s)\lambda(dt)>x\Bigr{)}\right\\}$

in order to show the continuity of $\mathcal{B}_{Z}^{0}(x)$ it suffices to
prove that

(4.8)
$\mathbb{E}\left\\{\frac{Z(0)}{S_{0}}\mathbb{I}\Bigl{(}\int_{\mathbb{R}^{d}}\mathbb{I}(Z(t)>s)\lambda(dt)=x\Bigr{)}\right\\}=0$

for almost all $s>0$. Let us define the following measurable sets

$A_{s}=\mathbb{I}\Bigl{(}\int_{\mathbb{R}^{d}}\mathbb{I}(Z(t)>s)\lambda(dt)=x\Bigr{)}\,.$

Since $Z$ has almost surely continuous trajectories we have $A_{s}\cap
A_{s^{\prime}}=\emptyset$ if $0<s<s^{\prime}$ and $x>0$. Thus there are
countably many $s>0$ such that $\mathbb{P}\\{A_{s}\\}>0$ because if there were
not countably many ones we would find countably many disjoint $A_{s}$ such
that $\sum\mathbb{P}\\{A_{s}\\}=\infty$. Thus we get (4.8) for almost all
$s>0$. The continuity at $x=0$ follows from the right continuity of (4.7).
$\Box$

Proof of Lemma 2.4 Item i: In view of (2.5) and substituting
$\Theta(t)=Q_{\delta}(t)/S_{\delta}(\Theta)$ to (2.10) we get

$\mathcal{B}_{Z}^{\delta}(x)=\int_{0}^{\infty}\mathbb{E}\left\\{\frac{1}{S_{\delta}(Q_{\delta})}\mathbb{I}\Bigl{(}\int_{\delta\mathbb{Z}^{d}}\mathbb{I}(Q_{\delta}(t)>s)\lambda_{\delta}(dt)>x\Bigr{)}\right\\}ds.$

Since $S_{\delta}(Q_{\delta})=1$ the claim follows.
Item ii: If $\mathbb{P}\\{S_{0}<\infty\\}>0$ we can define
$V(t)=Z(t)\lvert{S_{0}<\infty}$ and set (recall
$S_{0}=S_{0}(Z),\mathbb{E}\\{Z(0)\\}=1$)

$b=\mathbb{E}\\{Z(0)\mathbb{I}{\left\\{S_{0}<\infty\right\\}}\\}=\mathbb{P}\\{S_{0}(\Theta)<\infty\\}>0.$

For this choice of $b$ by (2.3) we have

$\mathbb{E}\\{V(t)\\}=\frac{\mathbb{E}\\{Z(t)\mathbb{I}{\left\\{S_{0}<\infty\right\\}}\\}}{\mathbb{P}\\{S_{0}(\Theta)<\infty\\}}=\frac{\mathbb{E}\\{Z(0)\mathbb{I}{\left\\{S_{0}<\infty\right\\}}\\}}{\mathbb{P}\\{S_{0}(\Theta)<\infty\\}}=1$

for all $t\in\mathbb{R}$. Clearly,
$\mathbb{P}\\{\sup_{t\in\mathbb{R}^{d}}V(t)>0\\}=1$. In view of [21] $V$ is
the spectral rf of a stationary max-stable rf $X_{*}$ with càdlàg sample paths
and moreover $S_{0}(V)=\int_{\mathbb{R}^{d}}V(t)\lambda(dt)<\infty$ almost
surely. In view of [15, proof of Prop 2.1] we have that

$\\{S_{\delta}(\Theta)<\infty\\}=\\{S_{0}(\Theta)<\infty\\}$

almost surely for all $\delta>0$. Consequently, we obtain for all $\delta>0$

$\displaystyle\mathcal{B}_{Z}^{\delta}(x)$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathbb{E}\left\\{\frac{Z(0)}{S_{\delta}}\mathbb{I}\Bigl{(}\int_{\delta\mathbb{Z}^{d}}\mathbb{I}(Z(t)>s)\lambda_{\delta}(dt)>x\Bigr{)}\mathbb{I}{\left\\{S_{\delta}<\infty\right\\}}\right\\}ds$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathbb{E}\left\\{\frac{1}{S_{\delta}(\Theta)}\mathbb{I}\Bigl{(}\int_{\delta\mathbb{Z}^{d}}\mathbb{I}(\Theta(t)>s)\lambda_{\delta}(dt)>x\Bigr{)}\mathbb{I}{\left\\{S_{\delta}(\Theta)<\infty,S_{0}(\Theta)<\infty\right\\}}\right\\}ds$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathbb{E}\left\\{\frac{Z(0)}{S_{\delta}(Z)}\mathbb{I}\Bigl{(}\int_{\delta\mathbb{Z}^{d}}\mathbb{I}(Z(t)>s)\lambda_{\delta}(dt)>x\Bigr{)}\mathbb{I}{\left\\{S_{0}(Z)<\infty\right\\}}\right\\}ds$
$\displaystyle=$ $\displaystyle
b\int_{0}^{\infty}\mathbb{E}\left\\{\frac{V(0)}{S_{\delta}(V)}\mathbb{I}\Bigl{(}\int_{\delta\mathbb{Z}^{d}}\mathbb{I}(V(t)>s)\lambda_{\delta}(dt)>x\Bigr{)}\right\\}ds$
$\displaystyle=$ $\displaystyle b\mathcal{B}_{V}^{\delta}(x)<\infty$

establishing the proof. $\Box$

Proof of Theorem 2.5 Assume first that $\mathbb{P}\\{S_{0}<\infty\\}=1$. In
view of (4.1) we have that $\epsilon_{\delta}<\infty$ almost surely, hence as
in [11, 22] where $d=1$ is considered it follows that (2.12) holds with

$\bar{Q}_{\delta}(t)=c\frac{Y(t)}{\epsilon_{\delta}(Y)M_{Y,\delta}},\quad
t\in\mathbb{R},$

with $c=1$ if $\delta=0$ and $c=\delta^{d}$ otherwise. Set below
$Q_{\delta}=\bar{Q}/c$ and for simplicity omit the subscript below writing
simply $M_{Y}$ instead of $M_{Y,\delta}$. Since $Y(t)/M_{Y}\leq 1$ almost
surely for all $t\in\delta\mathbb{Z}^{d}$ and
$\mathbb{P}\\{M_{Y}\in(1,\infty)\\}=1$, in view of 2.4 we have using further
the Fubini-Tonelli theorem and 4.2

$\displaystyle\mathcal{B}_{Z}^{\delta}(x)$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathbb{E}\left\\{\mathbb{I}\Bigl{(}\int_{\delta\mathbb{Z}^{d}}\mathbb{I}({Q}_{\delta}(v)>s)\lambda_{\delta}(dv)>x\Bigr{)}\right\\}ds$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathbb{E}\left\\{\frac{1}{\epsilon_{\delta}(Y)}\frac{1}{M_{Y}}\mathbb{I}{\left\\{\epsilon_{\delta}(Y/s)>x\right\\}}\right\\}ds$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\mathbb{E}\left\\{\frac{1}{\epsilon_{\delta}(Y)}\mathbb{I}{\left\\{M_{Y}>s\right\\}}\frac{1}{M_{Y}}\mathbb{I}{\left\\{\epsilon_{\delta}(Y/s)>x\right\\}}\right\\}ds$
$\displaystyle=:$
$\displaystyle\int_{0}^{\infty}\mathbb{E}\left\\{\frac{1}{{s}\epsilon_{\delta}(Y)}\mathbb{I}{\left\\{M_{Y}>s\right\\}}F(Y/s)\right\\}ds$
$\displaystyle=$
$\displaystyle\int_{0}^{\infty}\frac{1}{s^{2}}\mathbb{E}\left\\{\frac{\mathbb{I}{\left\\{\epsilon_{\delta}(Y)>x\right\\}}}{\epsilon_{\delta}(Y)M_{Y}}\mathbb{I}{\left\\{M_{Y}>\max(1/s,1)\right\\}}\right\\}ds$
$\displaystyle=$
$\displaystyle\mathbb{E}\left\\{\frac{\mathbb{I}{\left\\{\epsilon_{\delta}(Y)>x\right\\}}}{\epsilon_{\delta}(Y)M_{Y}}\int_{0}^{\infty}\frac{1}{s^{2}}\mathbb{I}{\left\\{M_{Y}>\max(1/s,1)\right\\}}ds\right\\}$
$\displaystyle=$
$\displaystyle\mathbb{E}\left\\{\frac{\mathbb{I}{\left\\{\epsilon_{\delta}(Y)>x\right\\}}}{\epsilon_{\delta}(Y)}\right\\}.$

The last equality follows from (recall $M_{Y}\in(1,\infty)$ almost surely)

$\int_{0}^{\infty}\frac{1}{s^{2}}\mathbb{I}{\left\\{M_{Y}>\max(1/s,1)\right\\}}ds=\int_{1}^{\infty}\frac{1}{s^{2}}ds+\int_{0}^{1}\frac{1}{s^{2}}\mathbb{I}{\left\\{M_{Y}>1/s\right\\}}ds=M_{Y}.$

In view of (4.1) for all $x$ non-negative such that
$\mathbb{P}\\{\epsilon_{\delta}(Y)>x\\}>0$ we have that
$\mathcal{B}_{Z}^{\delta}(x)\in(0,\infty)$, hence the proof follows.
Assume now that $\mathbb{P}\\{S_{0}<\infty\\}\in(0,1)$. In view of 2.4 we have

$\displaystyle\mathcal{B}_{Z}^{\delta}(x)$ $\displaystyle=$ $\displaystyle
b\mathcal{B}_{V}^{\delta}(x),$

with $V(t)=Z(t)\lvert S_{0}<\infty$, which is well-defined since
$\mathbb{P}\\{S_{0}<\infty\\}>0$ by the assumption. Since $S_{0}(V)<\infty$
almost surely and $Y_{*}(t)=Y(t)\lvert S_{0}(\Theta)<\infty,t\in\mathbb{R}$ by
the proof above

$\displaystyle\mathcal{B}_{Z}^{\delta}(x)$ $\displaystyle=$
$\displaystyle\mathbb{P}\\{S_{0}<\infty\\}\mathbb{E}\left\\{\frac{\mathbb{I}{\left\\{\epsilon_{\delta}(Y_{*})>x\right\\}}}{\epsilon_{\delta}(Y_{*})}\right\\}$
$\displaystyle=$
$\displaystyle\mathbb{E}\left\\{\frac{\mathbb{I}{\left\\{\epsilon_{\delta}(R\Theta)>x\right\\}}}{\epsilon_{\delta}(R\Theta)}\mathbb{I}{\left\\{S_{0}(\Theta)<\infty\right\\}}\right\\}.$

In view of [22, Lem 2.5, Cor 2.9] and [14, Thm 3.8] and the above

(4.9) $\displaystyle
H_{Z}^{\delta}=\mathcal{B}_{Z}^{\delta}(0)=\mathbb{E}\left\\{\frac{1}{\epsilon_{\delta}(R\Theta)}\right\\}=\mathbb{E}\left\\{\frac{1}{\epsilon_{\delta}(R\Theta)}\mathbb{I}{\left\\{S_{0}(\Theta)<\infty\right\\}}\right\\}$

and thus $\epsilon_{\delta}(R\Theta)<\infty$ implies $S_{0}<\infty$ almost
surely. Hence the proof is complete. $\Box$

Proof of Corollary 2.6 In view of (2.3), the representation (2.15) and the
finiteness of $\mathcal{B}_{Z}^{0}(x)$ for all $x\geq 0$, the monotone
convergence theorem yields for all $x_{0}\geq 0$

$\lim_{x\downarrow
x_{0}}\mathbb{E}\left\\{\frac{\mathbb{I}{\left\\{x_{0}\leq\epsilon_{0}(Y)<x\right\\}}}{\epsilon_{0}(Y)}\right\\}=\mathbb{E}\left\\{\frac{\mathbb{I}{\left\\{\epsilon_{0}(Y)=x_{0}\right\\}}}{\epsilon_{0}(Y)}\right\\}=0$

consequently, since by our assumption 4.1 implies
$\mathbb{P}\\{\epsilon_{0}(Y)\in(0,\infty)\\}=1$, then

$\mathbb{P}\\{\epsilon_{0}(Y)=x_{0}\\}=\mathbb{E}\\{\mathbb{I}{\left\\{\epsilon_{0}(Y)=x_{0}\right\\}}\\}=0$

follows establishing the claim. $\Box$

Proof of Proposition 2.7 In order to prove (2.16) note first that for any non-
negative rv $U$ with df $G$ and $x\geq 0$ such that $\mathbb{P}\\{U>x\\}>0$

$\displaystyle\frac{1}{\mathbb{P}\\{U>x\\}}\int_{x}^{\infty}\frac{1}{y}dG(y)$
$\displaystyle\geq$
$\displaystyle\frac{\mathbb{P}\\{U>x\\}}{\int_{x}^{\infty}ydG(y)}\geq\frac{\mathbb{P}\\{U>x\\}}{\mathbb{E}\\{U\\}}.$

Consequently, we obtain for all $x>0$

$\displaystyle\mathcal{B}_{Z}^{\delta}(x)$ $\displaystyle\geq$
$\displaystyle\frac{\mathbb{P}\\{\epsilon_{\delta}(Y)>x\\}^{2}}{\mathbb{E}\\{\epsilon_{\delta}(Y)\mathbb{I}\\{\epsilon_{\delta}(Y)>x\\}\\}}\geq\frac{\mathbb{P}\\{\epsilon_{\delta}(Y)>x\\}^{2}}{\mathbb{E}\\{\epsilon_{\delta}(Y)\\}}$

establishing the proof of the lower bound (2.16). The proof of the upper bound
follows from the fact that

$\mathcal{B}_{Z}^{\delta}(x)=\int_{x}^{\infty}\frac{1}{y}dF(y)\leq\frac{1}{x}\int_{x}^{\infty}dF(y)=x^{-}{\mathbb{P}\\{\epsilon_{\delta}(Y)\geq
x\\}},$

where $F$ is the distribution of $\epsilon_{\delta}(Y)$. This completes the
proof. $\Box$

Proof of Proposition 3.1 Since $\mathcal{B}_{Z}^{\delta}(0)$ is the
generalised Pickands constant $\mathcal{H}_{Z}^{\delta}$, then the claim
follows for $x=0$ from [15]. In view of (2.14) we can assume without loss of
generality that $\mathbb{P}\\{S_{0}<\infty\\}=1$. Under this assumption, from
the proof of 4.1 we have that $Y(t)\to 0$ almost surely as $\lVert
t\rVert\to\infty$. Hence for some $M$ sufficiently large $Y(t)<1$ almost
surely for all $t$ such that $\lVert t\rVert>M$. Consequently, for all
$\delta\geq 0$

$\epsilon_{\delta}(Y)=\int_{\delta
Z^{d}\cap[-M,M]^{d}}\mathbb{I}{\left\\{Y(t)>1\right\\}}\lambda_{\delta}(dt).$

Moreover, $\epsilon_{\delta}(Y)<\infty$ almost surely for all $\delta\geq 0$
implying $\epsilon_{\delta}(Y)\to\epsilon_{0}(Y)$ almost surely as
$\delta\downarrow 0$. In view of [22, Lem 2.5, Cor 2.9] and [14, Thm 3.8] for
all $\delta\geq 0$

$\mathcal{H}_{Z}^{\delta}=\mathbb{E}\\{1/\epsilon_{\delta}(Y)\\}.$

Applying [15, Thm 2] and (4.9) yields

$\mathbb{E}\\{1/\epsilon_{\delta}(Y)\\}=\mathcal{H}_{Z}^{\delta}\to\mathcal{H}_{Z}^{0}=\mathbb{E}\\{1/\epsilon_{0}(Y)\\},\quad\delta\downarrow
0.$

Hence $1/\epsilon_{\delta}(Y),\delta>0$ is uniformly integrable and hence

$\mathcal{B}_{Z}^{\delta}(x)=\mathbb{E}\left\\{\frac{\mathbb{I}{\left\\{\epsilon_{\delta}(Y)>x\right\\}}}{\epsilon_{\delta}(Y)}\right\\}\to\mathcal{B}_{Z}^{0}(x),\quad\delta\downarrow
0$

establishing the proof. $\Box$

###  4.1. Proof of Theorem 3.2

Suppose that $V(t),t\in\mathbb{R}^{d}$ is a centered Gaussian field with
stationary increments and variance function $\sigma^{2}_{V}(\cdot)$ that
satisfies A1-A2. Then, by stationarity of increments $\sigma^{2}_{V}(\cdot)$
is negative definite, which combined with Schoenberg’s theorem, implies that
for each $u>0$

$R_{u}(s,t)\coloneqq\exp\left(-\frac{1}{2u^{2}}\sigma_{V}^{2}(s-t)\right),\
s,t\in\mathbb{R}^{d}$

is positive definite, and thus a valid covariance function of some centered
stationary Gaussian rf $X_{u}(t),t\in\mathbb{R}^{d}$, where $s-t$ is meant
component-wise. The proof of 3.2 is based on the analysis of the asymptotics
of sojourn time of $X_{u}(t)$. Since the idea of the proof is the same for
continuous and discrete scenario, in order to simplify notation, we consider
next only the case $\delta=0$.

Before we proceed to the proof of 3.2, we need the following lemmas, where
$Z(t)=\exp\left(V(t)-\frac{\sigma^{2}_{V}(t)}{2}\right)$ is as in Remark 2.2,
Item iii.

######  Lemma 4.3.

For all $T>0$ and $x\geq 0$

  1. (i)

$\lim_{u\to\infty}\frac{\mathbb{P}\left\\{\int_{[0,T]^{d}}\mathbb{I}(X_{u}(t)>u)dt>x\right\\}}{\Psi(u)}=\mathcal{B}_{Z}([0,T]^{d},x).$

  2. (ii)

For all $x\geq 0$

$\lim_{u\to\infty}\frac{\mathbb{P}\left\\{\int_{[0,\ln(u)]^{d}}\mathbb{I}(X_{u}(t)>u)dt>x\right\\}}{(\ln(u))^{d}\Psi(u)}=\mathcal{B}_{Z}(x),\quad\lim_{T\to\infty}\frac{\mathcal{B}_{Z}([0,T]^{d},x)}{T^{d}}=\mathcal{B}_{Z}(x)\in(0,\infty).$

Proof of Lemma 4.3 Item i follows straightforwardly from [24, Lem 4.1]. The
proof of Item i follows by the application of the double sum technique applied
to the sojourn functional, as demonstrated e.g., in [24, Prop 3.1]. The claim
in Item ii follows by the same argument as its counterpart in [24, Lem 4.2].
$\Box$

The following lemma is a slight modification of [25, Lem 6.3] to the family
$X_{u},\ u>0$. Let $\mathbf{i}=(i_{1},...,i_{d})$, with
$i_{1},...,i_{d}\in\\{0,1,2,...\\}$,
$\mathcal{R}_{\mathbf{i}}:=\prod_{k=1}^{d}[i_{k}T,(i_{k}+1)T]$ and

$\displaystyle\widehat{\mathcal{K}}$ $\displaystyle:=$
$\displaystyle\\{\mathbf{i}=(i_{1},...,i_{d}):0\leq
i_{k},(i_{k}-1)T\leq\ln(u),k=1,...,d\\},$
$\displaystyle\widecheck{\mathcal{K}}$ $\displaystyle:=$
$\displaystyle\\{\mathbf{i}=(i_{1},...,i_{d}):0\leq
i_{k}T\leq\ln(u),k=1,...,d\\}.$

######  Lemma 4.4.

There exists a constant $C\in(0,\infty)$ such that for sufficiently large $u$,
for all
$\mathbf{i},\mathbf{j}\in\widehat{\mathcal{K}},\mathbf{i}\neq\mathbf{j}$ we
have

$\mathbb{P}\left\\{\max_{t\in\mathcal{R}_{\mathbf{i}}}X_{u}(t)>u,\max_{t\in\mathcal{R}_{\mathbf{j}}}X_{u}(t)>u\right\\}\leq
CT^{2d}\exp\left(-\frac{1}{8}\inf_{t\in\mathcal{R}_{\mathbf{i}},s\in\mathcal{R}_{\mathbf{j}}}\sigma_{V}^{2}(t-s)\right)\Psi(u).$

Proof of Theorem 3.2. The proof consists of two steps, where we find an
asymptotic upper and lower bound for the ratio

$\frac{\mathbb{P}\left\\{\int_{[0,\ln(u)]^{d}}\mathbb{I}(X_{u}(t)>u)dt>x\right\\}}{(\ln(u))^{d}\Psi(u)},$

as $u\to\infty$. We note that by 4.3 the limit, as $u\to\infty$, of the above
fraction is positive and finite.
Asymptotic upper bound. If $T>0$, then for sufficiently large $u$

$\displaystyle\mathbb{P}\left\\{\int_{[0,\ln(u)]^{d}}\mathbb{I}(X_{u}(t)>u)dt>x\right\\}$
$\displaystyle\leq$
$\displaystyle\mathbb{P}\left\\{\sum_{\mathbf{i}\in\widehat{\mathcal{K}}}\int_{\mathcal{R}_{\mathbf{i}}}\mathbb{I}(X_{u}(t)>u)dt>x\right\\}$
$\displaystyle\leq$
$\displaystyle\mathbb{P}\left\\{\exists_{\mathbf{i}\in\widehat{\mathcal{K}}}\int_{\mathcal{R}_{\mathbf{i}}}\mathbb{I}(X_{u}(t)>u)dt>x\right\\}+\mathbb{P}\left\\{\exists_{\mathbf{i},\mathbf{j}\in\widehat{\mathcal{K}},\mathbf{i}\neq\mathbf{j}}\max_{t\in\mathcal{R}_{\mathbf{i}}}X_{u}(t)>u,\max_{t\in\mathcal{R}_{\mathbf{j}}}X_{u}(t)>u\right\\}$
$\displaystyle\leq$
$\displaystyle\sum_{\mathbf{i}\in\widehat{\mathcal{K}}}\mathbb{P}\left\\{\int_{\mathcal{R}_{\mathbf{i}}}\mathbb{I}(X_{u}(t)>u)dt>x\right\\}+\mathbb{P}\left\\{\exists_{\mathbf{i},\mathbf{j}\in\widehat{\mathcal{K}},\mathbf{i}\neq\mathbf{j}}\max_{t\in\mathcal{R}_{\mathbf{i}}}X_{u}(t)>u,\max_{t\in\mathcal{R}_{\mathbf{j}}}X_{u}(t)>u\right\\}$
$\displaystyle\leq$
$\displaystyle\left\lceil\frac{(\ln(u))^{d}}{T^{d}}\right\rceil\mathbb{P}\left\\{\int_{[0,T]^{d}}\mathbb{I}(X_{u}(t)>u)dt>x\right\\}+\mathbb{P}\left\\{\exists_{\mathbf{i},\mathbf{j}\in\widehat{\mathcal{K}},\mathbf{i}\neq\mathbf{j}}\max_{t\in\mathcal{R}_{\mathbf{i}}}X_{u}(t)>u,\max_{t\in\mathcal{R}_{\mathbf{j}}}X_{u}(t)>u\right\\},$

where $\lceil\cdot\rceil$ is the ceiling function and the last inequality
above follows from the stationarity of $X_{u}$. Using again the stationary of
$X_{u}$, we obtain

$\displaystyle\mathbb{P}\left\\{\exists_{\mathbf{i},\mathbf{j}\in\widehat{\mathcal{K}},\mathbf{i}\neq\mathbf{j}}\max_{t\in\mathcal{R}_{\mathbf{i}}}X_{u}(t)>u,\max_{t\in\mathcal{R}_{\mathbf{j}}}X_{u}(t)>u\right\\}$
$\displaystyle\leq$
$\displaystyle\sum_{\mathbf{i},\mathbf{j}\in\widehat{\mathcal{K}},\mathbf{i}\neq\mathbf{j}}\mathbb{P}\left\\{\max_{t\in\mathcal{R}_{\mathbf{i}}}X_{u}(t)>u,\max_{t\in\mathcal{R}_{\mathbf{j}}}X_{u}(t)>u\right\\}$
$\displaystyle\leq$
$\displaystyle\left\lceil\frac{(\ln(u))^{d}}{T^{d}}\right\rceil\sum_{\mathbf{k}\in\widehat{\mathcal{K}},\mathbf{k}\neq\mathbf{0}}\mathbb{P}\left\\{\max_{t\in\mathcal{R}_{\mathbf{0}}}X_{u}(t)>u,\max_{t\in\mathcal{R}_{\mathbf{k}}}X_{u}(t)>u\right\\}$
$\displaystyle=$
$\displaystyle\left\lceil\frac{(\ln(u))^{d}}{T^{d}}\right\rceil\left(\sum_{\mathbf{k}\in\widehat{\mathcal{K}},\mathbf{k}\neq\mathbf{0},\mathcal{R}_{\mathbf{0}}\cap\mathcal{R}_{\mathbf{k}}\neq\emptyset}\mathbb{P}\left\\{\max_{t\in\mathcal{R}_{\mathbf{0}}}X_{u}(t)>u,\max_{t\in\mathcal{R}_{\mathbf{k}}}X_{u}(t)>u\right\\}\right.$
$\displaystyle+\left.\sum_{\mathbf{k}\in\widehat{\mathcal{K}},\mathbf{k}\neq\mathbf{0},\mathcal{R}_{\mathbf{0}}\cap\mathcal{R}_{\mathbf{k}}=\emptyset}\mathbb{P}\left\\{\max_{t\in\mathcal{R}_{\mathbf{0}}}X_{u}(t)>u,\max_{t\in\mathcal{R}_{\mathbf{k}}}X_{u}(t)>u\right\\}\right)$
(4.12) $\displaystyle=:$
$\displaystyle\left\lceil\frac{\ln^{d}(u)}{T^{d}}\right\rceil\left(\Sigma_{1}+\Sigma_{2}\right).$

Next, by 4.4, for sufficiently large $T,u$ and some $\rm{Const_{0}}>0$

(4.13) $\displaystyle\Sigma_{2}$ $\displaystyle\leq$ $\displaystyle
CT^{2d}\sum_{\mathbf{k}\in\widehat{\mathcal{K}},\mathbf{k}\neq\mathbf{0},\mathcal{R}_{\mathbf{0}}\cap\mathcal{R}_{\mathbf{k}}=\emptyset}\exp\left(-\frac{1}{8}\sigma_{V}^{2}(T\mathbf{k})\right)\Psi(u)$
$\displaystyle\leq$ $\displaystyle
CT^{2d}\sum_{\mathbf{k}>\mathbf{0}}\exp\left(-\rm{Const_{0}}T^{\alpha_{\infty}}\lVert\mathbf{k}\rVert^{\alpha_{\infty}}\right)\Psi(u)$
$\displaystyle\leq$
$\displaystyle{\rm{Const_{1}}}T^{2d}\exp\left(-T^{\alpha_{\infty}/2}\right)\Psi(u).$

The upper bound for $\Sigma_{1}$ follows by a similar argument as used in the
proof of [25, Lem 6.3], thus we explain only main steps of the argument. For a
while, consider the following probability

$\mathbb{P}\left\\{\max_{t\in\mathcal{R}_{\mathbf{0}}}X_{u}(t)>u,\max_{t\in\mathcal{R}_{{(1,0,...,0)}}}X_{u}(t)>u\right\\}.$

Then, for each $\varepsilon>0$ and sufficiently large $T$, $u$,

(4.14)
$\displaystyle\mathbb{P}\left\\{\max_{t\in\mathcal{R}_{\mathbf{0}}}X_{u}(t)>u,\max_{t\in\mathcal{R}_{(1,0,...,0)}}X_{u}(t)>u\right\\}$
$\displaystyle\leq$
$\displaystyle\mathbb{P}\left\\{\max_{t\in[0,T^{\varepsilon}]\times[0,T]^{d-1}}X_{u}(t)>u\right\\}+\mathbb{P}\left\\{\max_{t\in[0,T]^{d}}X_{u}(t)>u,\max_{t\in[T^{\varepsilon},T^{\varepsilon}+T]\times[0,T]^{d-1}}X_{u}(t)>u\right\\}$
$\displaystyle\leq$ $\displaystyle{\rm
Const_{2}}T^{d-1+\varepsilon}\Psi(u)+{\rm
Const_{3}}T^{2d}\exp\left(-T^{\varepsilon/2}\right),$

where the above inequality follows by 4.4 and

$\displaystyle\lim_{u\to\infty}\frac{\mathbb{P}\left\\{\max_{t\in[0,T^{\varepsilon}]\times[0,T]^{d-1}}X_{u}(t)>u\right\\}}{\Psi(u)}$
$\displaystyle\leq$ $\displaystyle\lceil
T\rceil^{(d-1)(1-\epsilon)}\lim_{u\to\infty}\frac{\mathbb{P}\left\\{\max_{t\in[0,T^{\varepsilon}]^{d}}X_{u}(t)>u\right\\}}{\Psi(u)}$
$\displaystyle=$ $\displaystyle\lceil
T\rceil^{(d-1)(1-\epsilon)}\mathcal{B}_{Z}([0,T^{\varepsilon}]^{d},0)$
$\displaystyle\leq$ $\displaystyle{\rm Const_{4}}T^{d-1+\varepsilon},$

which is a consequence of the stationarity of $X_{u}$ and statement (i) of 4.3
applied to $x=0$. Again, by the stationarity of $X_{u}$ we can obtain the
bound as in (4.14) uniformly for all the summands in $\Sigma_{1}$.

Application of the bounds (4.12), (4.13), (4.14) to (LABEL:up.1) leads to the
following upper estimate

(4.15)
$\displaystyle\limsup_{u\to\infty}\frac{\mathbb{P}\left\\{\int_{[0,\ln(u)]^{d}}\mathbb{I}(X_{u}(t)>u)dt>x\right\\}}{\ln^{d}(u)\Psi(u)}$
$\displaystyle\leq$
$\displaystyle\frac{\mathcal{B}_{Z}([0,T]^{d},x)}{T^{d}}+{\rm
Const_{4}}\frac{1}{T^{d}}\left(T^{d-1+\varepsilon}+T^{2d}\exp\left(-T^{\alpha_{\infty}/2}\right)+T^{2d}\exp\left(-T^{\varepsilon/2}\right)\right),$

which is valid for all $\varepsilon>0$ and $T$ sufficiently large.
Asymptotic lower bound. Taking $T>0$, for sufficiently large $u$

$\displaystyle\mathbb{P}\left\\{\int_{[0,\ln(u)]^{d}}\mathbb{I}(X_{u}(t)>u)dt>x\right\\}$
$\displaystyle\geq\mathbb{P}\left\\{\sum_{\mathbf{i}\in\widecheck{\mathcal{K}}}\int_{\mathcal{R}_{\mathbf{i}}}\mathbb{I}(X_{u}(t)>u)dt>x\right\\}$
$\displaystyle\geq\mathbb{P}\left\\{\exists_{\mathbf{i}\in\widecheck{\mathcal{K}}}\int_{\mathcal{R}_{\mathbf{i}}}\mathbb{I}(X_{u}(t)>u)dt>x\right\\}$
$\displaystyle\geq\sum_{\mathbf{i}\in\widecheck{\mathcal{K}}}\mathbb{P}\left\\{\int_{\mathcal{R}_{\mathbf{i}}}\mathbb{I}(X_{u}(t)>u)dt>x\right\\}-\sum_{\mathbf{i},\mathbf{j}\in\widecheck{\mathcal{K}},\mathbf{i}\neq\mathbf{j}}\mathbb{P}\left\\{\max_{t\in\mathcal{R}_{\mathbf{i}}}X_{u}(t)>u,\max_{t\in\mathcal{R}_{\mathbf{j}}}X_{u}(t)>u\right\\}$
$\displaystyle\geq\left\lfloor\frac{\ln^{d}(u)}{T^{d}}\right\rfloor\mathbb{P}\left\\{\int_{[0,T]^{d}}\mathbb{I}(X_{u}(t)>u)dt>x\right\\}-\sum_{\mathbf{i},\mathbf{j}\in\widecheck{\mathcal{K}},\mathbf{i}\neq\mathbf{j}}\mathbb{P}\left\\{\max_{t\in\mathcal{R}_{\mathbf{i}}}X_{u}(t)>u,\max_{t\in\mathcal{R}_{\mathbf{j}}}X_{u}(t)>u\right\\}\,,$

where in (4.1) we used Bonferroni inequality.

Using that $\widecheck{\mathcal{K}}\subset\widehat{\mathcal{K}}$ with the
upper bound for

$\sum_{\mathbf{i},\mathbf{j}\in\widehat{\mathcal{K}},\mathbf{i}\neq\mathbf{j}}\mathbb{P}\left\\{\max_{t\in\mathcal{R}_{\mathbf{i}}}X_{u}(t)>u,\max_{t\in\mathcal{R}_{\mathbf{j}}}X_{u}(t)>u\right\\}$

derived in (4.1), we conclude that for each $T$ sufficiently large and
$\varepsilon>0$,

(4.18)
$\displaystyle\liminf_{u\to\infty}\frac{\mathbb{P}\left\\{\int_{[0,\ln(u)]^{d}}\mathbb{I}(X_{u}(t)>u)dt>x\right\\}}{(\ln(u))^{d}\Psi(u)}$
$\displaystyle\geq$
$\displaystyle\frac{\mathcal{B}_{Z}([0,T]^{d},x)}{T^{d}}-{\rm
Const_{4}}\frac{1}{T^{d}}\left(T^{d-1+\varepsilon}+T^{2d}\exp\left(-T^{\alpha_{\infty}/2}\right)+T^{2d}\exp\left(-T^{\varepsilon/2}\right)\right).$

Thus, by statement (ii) of 4.3 combined with (4.15) and (4.18), in view of the
fact that $\varepsilon$ can take any value in $(0,1)$, we arrive at

$\displaystyle\lim_{T\to\infty}\left|\mathcal{B}_{Z}(x)-\frac{\mathcal{B}_{Z}([0,T]^{d},x)}{T^{d}}\right|T^{\lambda}=0$

for all $\lambda\in(0,1)$ establishing the proof. $\Box$

Proof of Proposition 3.4 The idea of the proof is to analyze the asymptotic
upper and lower bound of

$\mathbb{P}\\{\epsilon_{\delta}(Y)>x\\}$

as $x\to\infty$ and then to apply 2.7. In order to simplify the notation, we
consider only the case $\delta=0$. Let
$Z(t)=V(t)-\sigma^{2}_{V}(t)/2,t\in\mathbb{R}$ with $V$ a centered Gaussian
process with stationary increments that satisfies A1-A2 and $\mathcal{W}$ an
independent of $V$ exponentially distributed rv with parameter $1$.
Logarithmic upper bound. Let $A\in(0,1/2)$. We begin with an observation that

(4.19) $\displaystyle\mathbb{P}\left\\{\epsilon_{\delta}(Y)>x\right\\}$
$\displaystyle=$
$\displaystyle\mathbb{P}\left\\{\int_{\mathbb{R}}\mathbb{I}\\{\mathcal{W}+V(t)-\sigma^{2}_{V}(t)/2>0\\}dt>x\right\\}$
$\displaystyle\leq$ $\displaystyle\mathbb{P}\left\\{\mathcal{W}\leq
A\sigma^{2}_{V}(x/2),\int_{\mathbb{R}}\mathbb{I}\\{A\sigma^{2}_{V}(x/2)+V(t)-\sigma^{2}_{V}(t)/2>0\\}dt>x\right\\}$
$\displaystyle+\mathbb{P}\left\\{\mathcal{W}>A\sigma^{2}_{V}(x/2)\right\\}$
$\displaystyle\leq$ $\displaystyle
e^{-A\sigma^{2}_{V}(x/2)}+\mathbb{P}\left\\{\int_{\mathbb{R}}\mathbb{I}\\{A\sigma^{2}_{V}(x/2)+V(t)-\sigma^{2}_{V}(t)/2>0\\}dt>x\right\\}$
$\displaystyle\leq$ $\displaystyle
e^{-A\sigma^{2}_{V}(x/2)}+\mathbb{P}\left\\{\sup_{t\in\left(-\infty,-x/2]\cup[x/2,\infty\right)}V(t)-\sigma^{2}_{V}(t)/2>-A\sigma^{2}_{V}(x/2)\right\\}$
$\displaystyle\leq$ $\displaystyle
e^{-A\sigma^{2}_{V}(x/2)}+2\mathbb{P}\left\\{\exists_{t\in\left[x/2,\infty\right)}V(t)>\left(\frac{1}{2}-A\right)\sigma^{2}_{V}(t)\right\\}$
(4.20) $\displaystyle=$ $\displaystyle
e^{-A\sigma^{2}_{V}(x/2)}+2\mathbb{P}\left\\{\exists_{t\in\left[x/2,\infty\right)}\frac{V(t)}{\sigma_{V}^{2}(t)}>\left(\frac{1}{2}-A\right)\right\\}\,,$

where in (4.19) we used that $\\{V(-t),t\geq 0\\}\stackrel{{\scriptstyle
d}}{{=}}\\{V(t),t\geq 0\\}$ and the assumption that $\sigma^{2}_{V}$ is
increasing. Next, by A1, for sufficiently large $x$ and $s,t\geq x/2$ such
that $|t-s|\leq 1$

$Cov\left(\frac{V(t)}{\sigma_{V}(t)},\frac{V(s)}{\sigma_{V}(s)}\right)\geq\exp\left(-|t-s|^{\alpha_{0}/2}\right)=:Cov\left(Z(t),Z(s)\right),$

where $Z$ is some centered stationary Gaussian process. Hence, by Slepian
inequality (see, e.g., Corollary 2.4 in [26])

$\displaystyle\mathbb{P}\left\\{\exists_{t\in\left[x/2,\infty\right)}\frac{V(t)}{\sigma^{2}_{V}(t)}>\left(\frac{1}{2}-A\right)\right\\}$
$\displaystyle\leq$
$\displaystyle\sum_{k=0}^{\infty}\mathbb{P}\left\\{\exists_{t\in[x/2+k,x/2+k+1]}\frac{V(t)}{\sigma^{2}_{V}(t)}>\left(\frac{1}{2}-A\right)\right\\}$
$\displaystyle\leq$
$\displaystyle\sum_{k=0}^{\infty}\mathbb{P}\left\\{\exists_{t\in[0,1]}Z(t)>\left(\frac{1}{2}-A\right){\sigma_{V}(x/2+k)}\right\\}$

and by Landau-Shepp (see, e.g., [26, Eq. (2.3)]), uniformly with respect to
$k$

$\lim_{x\to\infty}\frac{\ln\left(\mathbb{P}\left\\{\exists_{t\in[0,1]}Z(t)>\left(\frac{1}{2}-A\right){\sigma_{V}(x/2+k)}\right\\}\right)}{\sigma_{V}^{2}(x/2+k)}=-\frac{1}{2}\left(\frac{1}{2}-A\right)^{2}\,.$

The above implies that

$\lim_{x\to\infty}\frac{\ln\left(\mathbb{P}\left\\{\exists_{t\in\left[x/2,\infty\right)}\frac{V(t)}{\sigma^{2}_{V}(t)}>\left(\frac{1}{2}-A\right)\right\\}\right)}{\sigma_{V}^{2}(x/2)}\leq-\frac{1}{2}\left(\frac{1}{2}-A\right)^{2}\
.$

Thus, in order to optimize the value of $A$ in (4.20) it suffices now to solve

$\left(\frac{1}{2}-A\right)^{2}=2A$

that leads to (recall that $A<1/2$)

$A=\frac{3-2\sqrt{2}}{2}.$

Hence

$\lim_{x\to\infty}\frac{\ln\left(\mathbb{P}\\{\epsilon_{\delta}(Y)>x\\}\right)}{\sigma_{V}^{2}(x/2)}\leq-\frac{3-2\sqrt{2}}{2},$

which combined with (2.16) in 2.7 completes the proof of the logarithmic upper
bound.
Logarithmic lower bound. Taking $A>1/2$ we have

$\displaystyle\mathbb{P}\\{\epsilon_{\delta}(Y)>x\\}$ $\displaystyle=$
$\displaystyle\mathbb{P}\left\\{\int_{\mathbb{R}}\mathbb{I}\\{\mathcal{W}+V(t)-\sigma^{2}_{V}(t)/2>0\\}dt>x\right\\}$
$\displaystyle\geq$
$\displaystyle\mathbb{P}\left\\{\mathcal{W}>A\sigma^{2}_{V}(x/2),\int_{\mathbb{R}}\mathbb{I}\\{A\sigma^{2}_{V}(x/2)+V(t)-\sigma^{2}_{V}(t)/2>0\\}dt>x\right\\}$
$\displaystyle\geq$
$\displaystyle\mathbb{P}\left\\{\mathcal{W}>A\sigma^{2}_{V}(x/2)\right\\}\mathbb{P}\left\\{\inf_{t\in[-x/2,x/2]}V(t)>-(A-1/2)\sigma^{2}_{V}(x/2)\right\\}$
$\displaystyle=$ $\displaystyle
e^{-A\sigma^{2}_{V}(x/2)}\left(1-\mathbb{P}\\{\sup_{t\in[-x/2,x/2]}V(t)>(A-1/2)\sigma^{2}_{V}(x/2)\\}\right).$

Using that

(4.22)
$\displaystyle\mathbb{P}\left\\{\sup_{t\in[-x/2,x/2]}V(t)>(A-1/2)\sigma^{2}_{V}(x/2)\right\\}$
$\displaystyle\leq$ $\displaystyle 2\sum_{i\in\\{0,...,\lfloor
x/2\rfloor-1\\}}\mathbb{P}\left\\{\sup_{t\in[i,i+1]}V(t)>(A-1/2)\sigma^{2}_{V}(x)\right\\}$
$\displaystyle+2\mathbb{P}\left\\{\sup_{t\in[\lfloor
x/2\rfloor,x/2]}V(t)>(A-1/2)\sigma^{2}_{V}(x)\right\\}$

and the fact that by the stationarity of increments of $V$

$\mathbb{E}\left\\{\sup_{t\in[i,i+1]}V(t)\right\\}=\mathbb{E}\left\\{\sup_{t\in[i,i+1]}(V(t)-V(i))+V(i)\right\\}=\mathbb{E}\left\\{\sup_{t\in[0,1]}V(t)\right\\}=:\mu<\infty$

we can apply Borell inequality (e.g., [26, Thm 2.1]) uniformly for all the
summands in (4.22) to get that for sufficiently large $x$ (recall that
$\sigma^{2}_{V}$ is supposed to be increasing)

$\displaystyle\mathbb{P}\left\\{\sup_{t\in[-x/2,x/2]}V(t)>(A-1/2)\sigma^{2}_{V}(x/2)\right\\}$
$\displaystyle\leq$ $\displaystyle
4(x+1)\exp\left(-\frac{((A-1/2)\sigma^{2}_{V}(x/2)-\mu)^{2}}{2\sigma^{2}_{V}(x/2)}\right)$
$\displaystyle\leq$
$\displaystyle\exp\left(-\frac{(A-1/2)^{2}\sigma^{2}_{V}(x/2)}{4}\right)\to 0$

as $x\to\infty$. Hence we arrive at

$\liminf_{x\to\infty}\frac{\ln(\mathbb{P}\\{\epsilon_{\delta}(Y)>x\\})}{\sigma^{2}_{V}(x/2)}\geq-A,$

which combined with (2.16) in 2.7 and the fact that, by the proof of the
logarithmic upper bound $\mathbb{E}\\{\epsilon_{\delta}(Y)\\}<\infty$ implies

$\liminf_{x\to\infty}\frac{\ln(\mathcal{B}_{Z}(x))}{\sigma^{2}_{V}(x)}\geq-2A$

for all $A>1/2$. This completes the proof. $\Box$

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