# The expected Euler characteristic approximation to excursion probabilities
of Gaussian vector fields

Dan Cheng
Arizona State University Partially supported by NSF Grant DMS-1902432 and
Simons Foundation Collaboration Grant 854127. Yimin Xiao
Michigan State University Partially supported by the NSF grants DMS-1855185
and DMS-2153846.

###### Abstract

Let $\\{(X(t),Y(s)):t\in T,s\in S\\}$ be an ${\mathbb{R}}^{2}$-valued,
centered, unit-variance smooth Gaussian vector field, where $T$ and $S$ are
compact rectangles in ${\mathbb{R}}^{N}$. It is shown that, as $u\to\infty$,
the joint excursion probability ${\mathbb{P}}\\{\sup_{t\in T}X(t)\geq
u,\sup_{s\in S}Y(s)\geq u\\}$ can be approximated by
${\mathbb{E}}\\{\chi(A_{u})\\}$, the expected Euler characteristic of the
excursion set $A_{u}=\\{(t,s)\in T\times S:X(t)\geq u,Y(s)\geq u\\}$, such
that the error is super-exponentially small. This verifies the expected Euler
characteristic heuristic (cf. Taylor, Takemura and Alder (2005), Alder and
Taylor (2007)) for a large class of smooth Gaussian vector fields.

Keywords: Gaussian vector fields, excursion probability, excursion set, Euler
characteristic, correlation, asymptotics, EEC, super-exponentially small.

Mathematics Subject Classification: 60G15, 60G60, 60G70.

##  1 Introduction

For a real-valued Gaussian random field $\\{Z(t),t\in{\mathbb{R}}^{N}\\}$ and
a compact rectangle $T\subset{\mathbb{R}}^{N}$, the excursion probability
${\mathbb{P}}\\{\sup_{t\in T}Z(t)\geq u\\}$ is a classical and very important
problem in both probability and statistics due to the vast applications in
many areas such as $p$-value computations, risk control and extreme event
analysis, etc. Various methods for precise approximations of
${\mathbb{P}}\\{\sup_{t\in T}Z(t)\geq u\\}$ have been developed. These include
the double sum method, the tube method, the Euler characteristic method, and
the Rice method. We refer to the monographs Piterbarg [12], Adler and Taylor
[2], Azaïs and Wschebor [6] and the references therein for comprehensive
accounts. However, extreme value theory of multivariate random fields (or
random vector fields) is still under-developed and only a few authors have
studied the joint excursion probability of multivariate random fields.
Piterbarg and Stamatovic [14] and Debicki et al. [8] established large
deviation results for the excursion probability in multivariate case. Anshin
[3] obtained precise asymptotics for a special class of nonstationary
bivariate Gaussian processes, under quite restrictive conditions. Hashorva and
Ji [10] and Debicki et al. [9] derived precise asymptotics for the excursion
probability of certain multivariate Gaussian processes defined on the real
line $\mathbb{R}$ with specific cross dependence structures. Zhou and Xiao
[20] studied the excursion probability of a class of non-smooth bivariate
Gaussian random fields by applying the double sum method. Their main results
show explicitly that the excursion probabilities of bivariate Gaussian random
fields depend not only on the smoothness parameters of the coordinate fields
but also on their maximum cross-correlation.

In statistical applications, such joint excursion probability is the critical
tool for constructing the simultaneous confidence region in a continuous-
domain approach [15]. In particular, motivated by the expected Euler
characteristic (EEC) approximation to excursion probabilities of real-valued
Gaussian random fields [17, 2], we study in this work that the EEC
approximation holds in general to the joint excursion probability of Gaussian
vector fields.

Let $\\{(X(t),Y(s)):t\in T,s\in S\\}$ be an ${\mathbb{R}}^{2}$-valued,
centered, unit-variance smooth Gaussian vector field, where $T$ and $S$ are
compact rectangles in ${\mathbb{R}}^{N}$. Let $A_{u}=\\{(t,s)\in T\times
S:X(t)\geq u,Y(s)\geq u\\}$ be the excursion set where both components $X$ and
$Y$ exceeding the level $u$. Our main objective is to show that, as
$u\to\infty$, the joint excursion probability ${\mathbb{P}}\\{\sup_{t\in
T}X(t)\geq u,\sup_{s\in S}Y(s)\geq u\\}$ can be approximated by
${\mathbb{E}}\\{\chi(A_{u})\\}$, the EEC of $A_{u}$, such that the error is
super-exponentially small; see Theorem 3.1 below for precise description. This
approximation result shows that the maximum correlation between $X(t)$ and
$Y(s)$, denoted by $R$ (see (2.1) below), plays an important role in both
${\mathbb{E}}\\{\chi(A_{u})\\}$ and the super-exponentially small error.
Moreover, as we will see in the proof of Theorem 3.1 (cf. $\mathcal{M}_{0}$ in
(5.23), $\mathcal{M}_{1}$ in (6.2), and $\mathcal{M}_{2}$ in (6.9)), the
points where $R$ is attained make the major contribution for
${\mathbb{E}}\\{\chi(A_{u})\\}$. Based on this observation, we also establish
two simpler approximations in Theorem 3.2 under the boundary condition (5.34)
on nonzero derivatives of the correlation function over boundary points where
$R$ is attained and in Theorem 3.3 under the condition that there is only a
unique point attaining $R$, respectively.

In general, the EEC approximation ${\mathbb{E}}\\{\chi(A_{u})\\}$ can be
expressed by the Kac-Rice formula as an integral; see (3.3) in Theorem 3.1. In
[17, 2], the authors derived a nice expression for
${\mathbb{E}}\\{\chi(A_{u})\\}$ called Gaussian kinematic formula, since they
assumed that the real-valued Gaussian field has unit variance, which is an
important condition to simplify the integration formula of
${\mathbb{E}}\\{\chi(A_{u})\\}$. However, in our case here, the integration
formula of ${\mathbb{E}}\\{\chi(A_{u})\\}$ (see (3.3)) mainly depends on the
conditional correlation of $X(t)$ and $Y(s)$, which varies over $T\times S$.
It turns to be very difficult to get an explicit expression for
${\mathbb{E}}\\{\chi(A_{u})\\}$. Instead, one can apply the Laplace method to
extract the term with the largest order of $u$ from the integral such that the
remaining error is $o(1/u){\mathbb{E}}\\{\chi(A_{u})\\}$. To explain this, we
show several examples on specific calculations in Section 8. For an intuitive
understanding on the EEC approximation, we may roughly treat the main term
${\mathbb{E}}\\{\chi(A_{u})\\}$ as $g(u)e^{-u^{2}/(1+R)}$ (by approximating
the integral in (3.3)); and the error term $o(e^{-u^{2}/(1+R)-\alpha u^{2}})$
is super-exponentially small, where $g(u)$ is a polynomial in $u$ and
$\alpha>0$ is some constant.

The paper is organized as follows. We introduce first the notations and
assumptions in Section 2, and then state our main results Theorems 3.1, 3.2
and 3.3 in Section 3. The proofs are then provided in three steps: (i) sketch
the main ideas in Section 4; (ii) study the super-exponentially small errors
between the joint excursion probability and EEC in Sections 5 and 6; and (iii)
provide final proofs for the main results in Section 7. Finally, we show in
Section 8 several examples on evaluating the EEC and hence approximating the
joint excursion probability explicitly.

##  2 Notations and assumptions

Let $\\{(X(t),Y(s)):t\in T,s\in S\\}$ be an ${\mathbb{R}}^{2}$-valued,
centered, unit-variance Gaussian vector field, where $T$ and $S$ are compact
rectangles in ${\mathbb{R}}^{N}$. Let

$\begin{split}r(t,s)={\mathbb{E}}\\{X(t)Y(s)\\},\quad R=\sup_{t\in T,\,s\in
S}r(t,s).\end{split}$ (2.1)

For a function $f(\cdot)\in C^{2}({\mathbb{R}}^{N})$ and
$t\in{\mathbb{R}}^{N}$, let

$\begin{split}f_{i}(t)&=\frac{\partial f(t)}{\partial t_{i}},\quad
f_{ij}(t)=\frac{\partial^{2}f(t)}{\partial t_{i}\partial t_{j}},\quad\forall
i,j=1,\ldots,N;\\\ \nabla
f(t)&=(f_{1}(t),\ldots,f_{N}(t))^{T},\quad\nabla^{2}f(t)=\left(f_{ij}(t)\right)_{i,j=1,\ldots,N}.\end{split}$
(2.2)

For a symmetric matrix $B$, denote by $B\prec 0$ and $B\preceq 0$ if all the
eigenvalues of $B$ are negative (i.e., $B$ is negative definite) and
nonpositive (i.e., $B$ is negative semi-definite), respectively. For two
functions $f(x)$ and $g(x)$, we write $f(x)\sim g(x)$ as $x\to\infty$ if
$\lim_{x\to\infty}f(x)/g(x)=1$.

Denote by $T=\prod^{N}_{i=1}[a_{i},b_{i}]$ and
$S=\prod^{N}_{i=1}[a^{\prime}_{i},b^{\prime}_{i}]$, where
$-\infty<a_{i}<b_{i}<\infty$ and
$-\infty<a^{\prime}_{i}<b^{\prime}_{i}<\infty$. Following the notations in
Adler and Taylor [2, p.134], we show below that $T$ and $S$ can be decomposed
into unions of their interiors and the lower dimension faces. Based on these
decompositions, the Euler characteristic of the excursion set $A_{u}$ can be
represented (see Section 3).

A face $K$ of dimension $k$ is defined by fixing a subset
$\sigma(K)\subset\\{1,\ldots,N\\}$ of size $k$ and a subset
$\varepsilon(K)=\\{\varepsilon_{j},j\notin\sigma(K)\\}\subset\\{0,1\\}^{N-k}$
of size $N-k$ so that

$\begin{split}K=\\{t=(t_{1},\ldots,t_{N})\in T:\,&a_{j}<t_{j}<b_{j}\ {\rm if}\
j\in\sigma(K),\\\ &t_{j}=(1-\varepsilon_{j})a_{j}+\varepsilon_{j}b_{j}\ {\rm
if}\ j\notin\sigma(K)\\}.\end{split}$

Denote by $\partial_{k}T$ the collection of all $k$-dimensional faces in $T$.
Then the interior of $T$ is denoted by $\overset{\circ}{T}=\partial_{N}T$ and
the boundary of $T$ is given by $\partial
T=\cup^{N-1}_{k=0}\cup_{K\in\partial_{k}T}K$. For each $t\in
K\in\partial_{k}T$ and $s\in L\in\partial_{l}S$, let

$\begin{split}\nabla
X_{|K}(t)&=(X_{i_{1}}(t),\ldots,X_{i_{k}}(t))^{T}_{i_{1},\ldots,i_{k}\in\sigma(K)},\quad\nabla^{2}X_{|K}(t)=(X_{mn}(t))_{m,n\in\sigma(K)},\\\
\nabla
Y_{|L}(s)&=(Y_{i_{1}}(s),\ldots,Y_{i_{l}}(s))^{T}_{i_{1},\ldots,i_{l}\in\sigma(L)},\quad\nabla^{2}Y_{|L}(s)=(Y_{mn}(s))_{m,n\in\sigma(L)}.\end{split}$

We can decompose $T$ and $S$ into

$T=\bigcup_{k=0}^{N}\partial_{k}T=\bigcup_{k=0}^{N}\bigcup_{K\in\partial_{k}T}K,\qquad
S=\bigcup_{l=0}^{N}\partial_{l}S=\bigcup_{l=0}^{N}\bigcup_{L\in\partial_{l}S}L,$

respectively. For each $K\in\partial_{k}T$ and $L\in\partial_{l}S$, we define
the _number of extended outward maxima above $u$_ as

$\begin{split}M_{u}^{E}(X,K)&:=\\#\\{t\in K:X(t)\geq u,\nabla
X_{|K}(t)=0,\nabla^{2}X_{|K}(t)\prec 0,\varepsilon^{*}_{j}X_{j}(t)\geq
0,\forall j\notin\sigma(K)\\},\\\ M_{u}^{E}(Y,L)&:=\\#\\{s\in L:Y(s)\geq
u,\nabla Y_{|L}(s)=0,\nabla^{2}Y_{|L}(s)\prec
0,\varepsilon^{*}_{j}Y_{j}(s)\geq 0,\forall j\notin\sigma(L)\\},\end{split}$

where $\varepsilon^{*}_{j}=2\varepsilon_{j}-1$, and define the _number of
local maxima above $u$_ as

$\begin{split}M_{u}(X,K)&:=\\#\\{t\in K:X(t)\geq u,\nabla
X_{|K}(t)=0,\nabla^{2}X_{|K}(t)\prec 0\\},\\\ M_{u}(Y,L)&:=\\#\\{s\in
L:Y(s)\geq u,\nabla Y_{|L}(s)=0,\nabla^{2}Y_{|L}(s)\prec 0\\}.\end{split}$

Clearly, $M_{u}^{E}(X,K)\leq M_{u}(X,K)$ and $M_{u}^{E}(Y,L)\leq M_{u}(Y,L)$.

We shall make use of the following smoothness condition (H1) and regularity
conditions (H2) and (H3).

  * (H1)

$X,Y\in C^{2}({\mathbb{R}}^{N})$ almost surely and their second derivatives
satisfy the _uniform mean-square Hölder condition_ : there exist constants
$C,\delta>0$ such that

$\begin{split}{\mathbb{E}}(X_{ij}(t)-X_{ij}(t^{\prime}))^{2}&\leq
C\|t-t^{\prime}\|^{2\delta},\quad\forall t,t^{\prime}\in T,\
i,j=1,\ldots,N,\\\ {\mathbb{E}}(Y_{ij}(s)-Y_{ij}(s^{\prime}))^{2}&\leq
C\|s-s^{\prime}\|^{2\delta},\quad\forall s,s^{\prime}\in S,\
i,j=1,\ldots,N.\end{split}$

  * (H2)

For every $(t,t^{\prime},s)\in T^{2}\times S$ with $t\neq t^{\prime}$, the
Gaussian vector

$\begin{split}\big{(}&X(t),\nabla X(t),X_{ij}(t),X(t^{\prime}),\nabla
X(t^{\prime}),X_{ij}(t^{\prime}),\\\ &Y(s),\nabla Y(s),Y_{ij}(s),1\leq i\leq
j\leq N\big{)}\end{split}$

is non-degenerate; and for every $(s,s^{\prime},t)\in S^{2}\times T$ with
$s\neq s^{\prime}$, the Gaussian vector

$\begin{split}\big{(}&Y(s),\nabla Y(s),Y_{ij}(s),Y(s^{\prime}),\nabla
Y(s^{\prime}),Y_{ij}(s^{\prime}),\\\ &X(t),\nabla X(t),X_{ij}(t),1\leq i\leq
j\leq N\big{)}\end{split}$

is non-degenerate.

  * (H3)

For every $(t,s)\in\partial_{k}T\times S$, $0\leq k\leq N-2$, such that
$r(t,s)=R$, and that the index set
$\mathcal{I}^{R}_{X}(t,s)=\\{\ell:\frac{\partial r}{\partial
t_{\ell}}(t,s)=0\\}$ contains at least two indices, the Hessian matrix

$\left(\frac{\partial^{2}r}{\partial t_{i}\partial
t_{j}}(t,s)\right)_{i,j\in\mathcal{I}^{R}_{X}(t,s)}\preceq 0.$ (2.3)

For every $(t,s)\in T\times\partial_{l}S$, $0\leq l\leq N-2$, such that
$r(t,s)=R$, and that the index set
$\mathcal{I}^{R}_{Y}(t,s)=\\{\ell:\frac{\partial r}{\partial
s_{\ell}}(t,s)=0\\}$ contains at least two indices, the Hessian matrix

$\left(\frac{\partial^{2}r}{\partial s_{m}\partial
s_{n}}(t,s)\right)_{m,n\in\mathcal{I}^{R}_{Y}(t,s)}\preceq 0.$

Although (H3) looks technical, it is in fact a mild condition imposed only on
the lower-dimension boundary points $(t,s)$ with $r(t,s)=R$. Roughly speaking,
it shows that the correlation function should have a negative semi-definite
Hessian matrix on boundary critical points where the maximum correlation $R$
is attained. Since $r(t,s)=R$ implies $\frac{\partial r}{\partial
t_{\ell}}(t,s)=0$ for all $\ell\in\sigma(K)$, we have
$\mathcal{I}^{R}_{X}(t,s)\supset\sigma(K)$. Similarly,
$\mathcal{I}^{R}_{Y}(t,s)\supset\sigma(L)$. We show below that, for $k=N-1$ or
$k=N$, the property (2.3) is always satisfied.

(i) If $k=N$, then $t$ becomes a maximum point of $r$ (as a function of $t$)
in the interior of $T$ and
$\mathcal{I}^{R}_{X}(t,s)=\sigma(K)=\\{1,\cdots,N\\}$, implying (2.3).

(ii) For $k=N-1$, we distinguish two cases. If
$\mathcal{I}^{R}_{X}(t,s)=\sigma(K)$, then $t$ becomes a maximum point of $r$
restricted on $K$, hence (2.3) holds. If
$\mathcal{I}^{R}_{X}(t,s)=\\{1,\cdots,N\\}$, let $s$ be fixed, it follows from
Taylor’s formula that

$r(t^{\prime},s)=r(t,s)+(t^{\prime}-t)^{T}\nabla^{2}r(t,s)(t^{\prime}-t)+o(\|t^{\prime}-t\|^{2}),\quad
t^{\prime}\in T,$

where $\nabla^{2}r(t,s)$ is the Hessian with respect to $t$. Notice that
$\\{(t^{\prime}-t)/\|t^{\prime}-t\|:t^{\prime}\in T\\}$ contains all
directions in ${\mathbb{R}}^{N}$ since $t\in K\in\partial_{N-1}T$, together
with the fact $r(t,s)=R$, we see that $\nabla^{2}r(t,s)$ cannot have any
positive eigenvalue and hence (2.3) holds.

It is also evident from the 1D Taylor’s formula that (2.3) holds if
$\mathcal{I}^{R}_{X}(t,s)$ contains only one index. Combining these facts,
together with the observations,

$\begin{split}\frac{\partial r}{\partial
t_{i}}(t,s)&={\mathbb{E}}\\{X_{i}(t)Y(s)\\},\quad\frac{\partial^{2}r}{\partial
t_{i}\partial t_{j}}(t,s)={\mathbb{E}}\\{X_{ij}(t)Y(s)\\},\\\ \frac{\partial
r}{\partial
s_{i}}(t,s)&={\mathbb{E}}\\{X(t)Y_{i}(s)\\},\quad\frac{\partial^{2}r}{\partial
s_{i}\partial s_{j}}(t,s)={\mathbb{E}}\\{X(t)Y_{ij}(s)\\},\end{split}$

we obtain the following result.

######  Proposition 2.1.

Under the condition $({\bf H}3)$, we have that, for every $(t,s)\in T\times S$
such that $r(t,s)=R$, the matrices

$({\mathbb{E}}\\{X_{ij}(t)Y(s)\\})_{i,j\in\mathcal{I}^{R}_{X}(t,s)}\preceq
0\quad{\rm
and}\quad({\mathbb{E}}\\{X(t)Y_{kl}(s)\\})_{k,l\in\mathcal{I}^{R}_{Y}(t,s)}\preceq
0,$

where the index sets $\mathcal{I}^{R}_{X}(t,s)$ and $\mathcal{I}^{R}_{Y}(t,s)$
are defined respectively as

$\begin{split}\mathcal{I}^{R}_{X}(t,s)=\\{\ell:{\mathbb{E}}\\{X_{\ell}(t)Y(s)\\}=0\\}\quad\text{\rm
and
}\quad\mathcal{I}^{R}_{Y}(t,s)=\\{\ell:{\mathbb{E}}\\{X(t)Y_{\ell}(s)\\}=0\\}.\end{split}$

##  3 Main results

Here, we shall state our main results Theorems 3.1, 3.2 and 3.3, whose proofs
will be given in Section 7. Define respectively the excursion sets of $X$, $Y$
and $(X,Y)$ above level $u$ by

$\begin{split}A_{u}(X,T)&=\\{t\in T:X(t)\geq u\\},\\\ A_{u}(Y,S)&=\\{s\in
S:Y(s)\geq u\\}\quad{\text{\rm}and}\\\ A_{u}:=A_{u}(X,T)\times
A_{u}(Y,S)&=\\{(t,s)\in T\times S:X(t)\geq u,Y(s)\geq u\\}.\end{split}$

Let the _number of extended outward critical points of index $i$ above level
$u$_ be

$\begin{split}\mu_{i}(X,K)&:=\\#\\{t\in K:X(t)\geq u,\nabla
X_{|K}(t)=0,\text{index}(\nabla^{2}X_{|K}(t))=i,\\\
&\qquad\qquad\qquad\varepsilon^{*}_{j}X_{j}(t)\geq 0\ {\rm for\ all}\
j\notin\sigma(K)\\},\\\ \mu_{i}(Y,L)&:=\\#\\{s\in L:Y(s)\geq u,\nabla
Y_{|L}(s)=0,\text{index}(\nabla^{2}Y_{|L}(s))=i,\\\
&\qquad\qquad\qquad\varepsilon^{*}_{j}Y_{j}(s)\geq 0\ {\rm for\ all}\
j\notin\sigma(L)\\}.\end{split}$

Recall that $\varepsilon^{*}_{j}=2\varepsilon_{j}-1$ and the index of a matrix
is defined as the number of its negative eigenvalues. It follows from (H1),
(H2) and the Morse theorem (see Corollary 9.3.5 or pages 211–212 in Adler and
Taylor [2]) that the Euler characteristic of the excursion set can be
represented as

$\begin{split}\chi(A_{u}(X,T))&=\sum^{N}_{k=0}\sum_{K\in\partial_{k}T}(-1)^{k}\sum^{k}_{i=0}(-1)^{i}\mu_{i}(X,K),\\\
\chi(A_{u}(Y,S))&=\sum^{N}_{l=0}\sum_{L\in\partial_{l}S}(-1)^{l}\sum^{l}_{i=0}(-1)^{i}\mu_{i}(Y,L).\end{split}$
(3.1)

Since for two sets $D_{1}$ and $D_{2}$, $\chi(D_{1}\times
D_{2})=\chi(D_{1})\chi(D_{2})$, we have

$\begin{split}\chi(A_{u})&=\chi(A_{u}(X,T)\times
A_{u}(Y,S))=\chi(A_{u}(X,T))\times\chi(A_{u}(Y,S))\\\
&=\sum^{N}_{k,l=0}\sum_{K\in\partial_{k}T,L\in\partial_{l}S}(-1)^{k+l}\bigg{(}\sum^{k}_{i=0}(-1)^{i}\mu_{i}(X,K)\bigg{)}\bigg{(}\sum^{l}_{j=0}(-1)^{j}\mu_{j}(Y,L)\bigg{)}.\end{split}$
(3.2)

Now we state the following general result on the EEC approximation of the
excursion probability.

######  Theorem 3.1.

Let $\\{(X(t),Y(s)):t\in T,s\in S\\}$ be an ${\mathbb{R}}^{2}$-valued,
centered, unit-variance Gaussian vector field satisfying $({\bf H}1)$, $({\bf
H}2)$ and $({\bf H}3)$. Then there exists a constant $\alpha>0$ such that as
$u\to\infty$,

$\begin{split}&\quad{\mathbb{P}}\left\\{\sup_{t\in T}X(t)\geq u,\sup_{s\in
S}Y(s)\geq u\right\\}\\\
&=\sum^{N}_{k,l=0}\sum_{K\in\partial_{k}T,L\in\partial_{l}S}(-1)^{k+l}\int_{K}\int_{L}dtds\,p_{\nabla
X_{|K}(t),\nabla Y_{|L}(s)}(0,0){\mathbb{E}}\big{\\{}{\rm
det}\nabla^{2}X_{|K}(t){\rm det}\nabla^{2}Y_{|L}(s)\\\
&\quad\times\mathbbm{1}_{\\{X(t)\geq u,\ \varepsilon^{*}_{\ell}X_{\ell}(t)\geq
0\ {\rm for\ all}\ \ell\notin\sigma(K)\\}}\mathbbm{1}_{\\{Y(s)\geq u,\
\varepsilon^{*}_{\ell}Y_{\ell}(s)\geq 0\ {\rm for\ all}\
\ell\notin\sigma(L)\\}}\big{|}\nabla X_{|K}(t)=\nabla Y_{|L}(s)=0\big{\\}}\\\
&\quad+o\left(\exp\left\\{-\frac{u^{2}}{1+R}-\alpha u^{2}\right\\}\right)\\\
&={\mathbb{E}}\\{\chi(A_{u})\\}+o\left(\exp\left\\{-\frac{u^{2}}{1+R}-\alpha
u^{2}\right\\}\right).\end{split}$ (3.3)

In general, the EEC approximation ${\mathbb{E}}\\{\chi(A_{u})\\}$ is hard to
compute, since the conditional expectation in (3.3) involves the joint tail
estimate and hence the conditional correlation on $X(t)$ and $Y(s)$, which
varies over $T\times S$. However, one can apply the Laplace method to extract
the term with the largest order of $u$ from ${\mathbb{E}}\\{\chi(A_{u})\\}$
such that the remaining error is $o(1/u){\mathbb{E}}\\{\chi(A_{u})\\}$; see
Section 8 for examples on this.

Note that, in (3.3), if $k=0$, then all terms involving $\nabla X_{|K}(t)$ and
$\nabla^{2}X_{|K}(t)$ in (3.3) vanish. In particular, if $k=l=0$, then the
integral in (3.3) becomes a joint probability. We follow such notation in the
results in Theorems 3.2 and 3.3 below as well.

It can be seen from the proof of Theorem 3.1 that those points attaining the
maximal correlation $R$ make the major contribution for
${\mathbb{E}}\\{\chi(A_{u})\\}$. Therefore, in many cases, the general EEC
approximation ${\mathbb{E}}\\{\chi(A_{u})\\}$ can be simplified. The result
below is based on the boundary condition (5.34) (which implies $({\bf H}3)$)
on nonzero derivatives of the correlation function over boundary points where
$R$ is attained.

######  Theorem 3.2.

Let $\\{(X(t),Y(s)):t\in T,s\in S\\}$ be an ${\mathbb{R}}^{2}$-valued,
centered, unit-variance Gaussian vector field satisfying $({\bf H}1)$, $({\bf
H}2)$ and the boundary condition (5.34). Then there exists a constant
$\alpha>0$ such that as $u\to\infty$,

$\begin{split}&\quad{\mathbb{P}}\left\\{\sup_{t\in T}X(t)\geq u,\sup_{s\in
S}Y(s)\geq u\right\\}\\\
&=\sum^{N}_{k,l=0}\sum_{K\in\partial_{k}T,L\in\partial_{l}S}(-1)^{k+l}\int_{K}\int_{L}dtds\,p_{\nabla
X_{|K}(t),\nabla Y_{|L}(s)}(0,0){\mathbb{E}}\big{\\{}{\rm
det}\nabla^{2}X_{|K}(t){\rm det}\nabla^{2}Y_{|L}(s)\\\
&\quad\times\mathbbm{1}_{\\{X(t)\geq u,\ Y(s)\geq u\\}}\big{|}\nabla
X_{|K}(t)=\nabla
Y_{|L}(s)=0\big{\\}}+o\left(\exp\left\\{-\frac{u^{2}}{1+R}-\alpha
u^{2}\right\\}\right).\end{split}$

The following result is the asymptotic approximation for the special case when
the correlation attains its maximum $R$ only at a unique point.

######  Theorem 3.3.

Let $\\{(X(t),Y(s)):t\in T,s\in S\\}$ be an ${\mathbb{R}}^{2}$-valued,
centered, unit-variance Gaussian vector field satisfying $({\bf H}1)$, $({\bf
H}2)$ and $({\bf H}3)$. Suppose that the correlation attains its maximum $R$
only at a single point $(t^{*},s^{*})\in K\times L$, where $K\in\partial_{k}T$
and $L\in\partial_{l}S$ with $k,l\geq 0$. Then there exists a constant
$\alpha>0$ such that as $u\to\infty$,

$\begin{split}&\quad{\mathbb{P}}\left\\{\sup_{t\in T}X(t)\geq u,\sup_{s\in
S}Y(s)\geq u\right\\}\\\ &=\sum_{J}\sum_{F}(-1)^{{\rm dim}(J)+{\rm
dim}(F)}\int_{J}\int_{F}dtds\,p_{\nabla X_{|J}(t),\nabla Y_{|F}(s)}(0,0)\\\
&\quad\times{\mathbb{E}}\big{\\{}{\rm det}\nabla^{2}X_{|J}(t){\rm
det}\nabla^{2}Y_{|F}(s)\mathbbm{1}_{\\{X(t)\geq u,\
\varepsilon^{*}_{\ell}X_{\ell}(t)\geq 0\ {\rm for\ all}\
\ell\in\mathcal{I}^{R}_{X}(t^{*},s^{*})\setminus\sigma(J)\\}}\\\
&\quad\times\mathbbm{1}_{\\{Y(s)\geq u,\ \varepsilon^{*}_{\ell}Y_{\ell}(s)\geq
0\ {\rm for\ all}\
\ell\in\mathcal{I}^{R}_{Y}(t^{*},s^{*})\setminus\sigma(F)\\}}\big{|}\nabla
X_{|J}(t)=\nabla Y_{|F}(s)=0\big{\\}}\\\
&\quad+o\left(\exp\left\\{-\frac{u^{2}}{1+R}-\alpha
u^{2}\right\\}\right)\end{split}$

where the sums are taken over all faces $J$ of $T$ such that $t^{*}\in\bar{J}$
and $\sigma(J)\subset\mathcal{I}^{R}_{X}(t^{*},s^{*})$, and all faces $F$ of
$S$ such that $s^{*}\in\bar{F}$ and
$\sigma(F)\subset\mathcal{I}^{R}_{Y}(t^{*},s^{*})$.

##  4 Plan of the proofs

Note that, for a smooth real-valued function $f$, $\sup_{t\in T}f(t)\geq u$ if
and only if there exists at least one extended outward local maximum above $u$
on some face of $T$. Thus, under conditions $({\bf H}1)$ and $({\bf H}2)$, the
following relation holds for each $u\in{\mathbb{R}}$:

$\begin{split}&\quad\left\\{\sup_{t\in T}X(t)\geq u,\sup_{s\in S}Y(s)\geq
u\right\\}\\\
&=\bigcup_{k,l=0}^{N}\bigcup_{K\in\partial_{k}T,\,L\in\partial_{l}S}\\{M_{u}^{E}(X,K)\geq
1,M_{u}^{E}(Y,L)\geq 1\\}\quad{\rm a.s.}\end{split}$ (4.1)

Therefore, we obtain the following upper bound for the joint excursion
probability:

$\begin{split}&\quad{\mathbb{P}}\left\\{\sup_{t\in T}X(t)\geq u,\sup_{s\in
S}Y(s)\geq u\right\\}\\\
&\leq\sum_{k,l=0}^{N}\sum_{K\in\partial_{k}T,\,L\in\partial_{l}S}{\mathbb{P}}\\{M_{u}^{E}(X,K)\geq
1,M_{u}^{E}(Y,L)\geq 1\\}\\\
&\leq\sum_{k,l=0}^{N}\sum_{K\in\partial_{k}T,\,L\in\partial_{l}S}{\mathbb{E}}\\{M_{u}^{E}(X,K)M_{u}^{E}(Y,L)\\}.\end{split}$
(4.2)

On the other hand, notice that

$\begin{split}&\quad{\mathbb{E}}\\{M_{u}^{E}(X,K)M_{u}^{E}(Y,L)\\}-{\mathbb{P}}\\{M_{u}^{E}(X,K)\geq
1,M_{u}^{E}(Y,L)\geq 1\\}\\\
&=\sum_{i,j=1}^{\infty}(ij-1){\mathbb{P}}\\{M_{u}^{E}(X,K)=i,M_{u}^{E}(Y,L)=j\\}\\\
&\leq\sum_{i,j=1}^{\infty}[i(i-1)j+j(j-1)i]{\mathbb{P}}\\{M_{u}^{E}(X,K)=i,M_{u}^{E}(Y,L)=j\\}\\\
&={\mathbb{E}}\\{M_{u}^{E}(X,K)[M_{u}^{E}(X,K)-1]M_{u}^{E}(Y,L)\\}+{\mathbb{E}}\\{M_{u}^{E}(Y,L)[M_{u}^{E}(Y,L)-1]M_{u}^{E}(X,K)\\}\end{split}$

and

$\begin{split}&\quad{\mathbb{P}}\\{M_{u}^{E}(X,K)\geq 1,M_{u}^{E}(Y,L)\geq
1,M_{u}^{E}(X,K^{\prime})\geq 1,M_{u}^{E}(Y,L^{\prime})\geq 1\\}\\\
&\leq{\mathbb{P}}\\{M_{u}^{E}(X,K)\geq 1,M_{u}^{E}(Y,L)\geq
1,M_{u}^{E}(Y,L^{\prime})\geq 1\\}\\\
&\leq{\mathbb{E}}\\{M_{u}^{E}(X,K)M_{u}^{E}(Y,L)M_{u}^{E}(Y,L^{\prime})\\}.\end{split}$

Combining these two inequalities with (4.1) and applying the Bonferroni
inequality, we obtain the following lower bound for the joint excursion
probability:

$\begin{split}&\quad{\mathbb{P}}\left\\{\sup_{t\in T}X(t)\geq u,\sup_{s\in
S}Y(s)\geq u\right\\}\\\
&\geq\sum_{k,l=0}^{N}\sum_{K\in\partial_{k}T,L\in\partial_{l}S}\Big{\\{}{\mathbb{E}}\\{M_{u}^{E}(X,K)M_{u}^{E}(Y,L)\\}-\\\
&\quad{\mathbb{E}}\\{M_{u}^{E}(X,K)[M_{u}^{E}(X,K)-1]M_{u}^{E}(Y,L)\\}-{\mathbb{E}}\\{M_{u}^{E}(Y,L)[M_{u}^{E}(Y,L)-1]M_{u}^{E}(X,K)\\}\Big{\\}}\\\
&\quad-\sum_{k,k^{\prime},l=0}^{N}\sum_{\begin{subarray}{c}K\in\partial_{k}T,L\in\partial_{l}S\\\
K^{\prime}\in\partial_{k^{\prime}}T,K\neq
K^{\prime}\end{subarray}}{\mathbb{E}}\\{M_{u}^{E}(X,K)M_{u}^{E}(X,K^{\prime})M_{u}^{E}(Y,L)\\}\\\
&\quad-
C_{N}\sum_{k,l,l^{\prime}=0}^{N}\sum_{\begin{subarray}{c}K\in\partial_{k}T,L\in\partial_{l}S\\\
L^{\prime}\in\partial_{l^{\prime}}S,L\neq
L^{\prime}\end{subarray}}{\mathbb{E}}\\{M_{u}^{E}(X,K)M_{u}^{E}(Y,L)M_{u}^{E}(Y,L^{\prime})\\},\end{split}$
(4.3)

where $C_{N}$ is a constant depending only on $N$.

Remark 4.1  Note that, following the same arguments above, we have that the
expectations on the number of extended outward maxima $M_{u}^{E}(\cdot)$ in
both (4.2) and (4.3) can be replaced by the expectations on the number of
local maxima $M_{u}(\cdot)$. ∎

We call a function $h(u)$ _super-exponentially small_ [when compared with the
joint excursion probability ${\mathbb{P}}\\{\sup_{t\in T}X(t)\geq u,\sup_{s\in
S}Y(s)\geq u\\}$], if there exists a constant $\alpha>0$ such that
$h(u)=o(e^{-\alpha u^{2}-u^{2}/{(1+R)}})$ as $u\to\infty$. The main idea for
proving the EEC approximation Theorem 3.1 consists of the following two steps:
(i) show that, except for the upper bound in (4.2), all terms in the lower
bound in (4.3) are super-exponentially small; and (ii) prove that the
difference between the upper bound in (4.2) and
${\mathbb{E}}\\{\chi(A_{u})\\}$ is also super-exponentially small. The ideas
for proving Theorems 3.2 and 3.3 are similar.

##  5 Estimation of super-exponentially small terms in the lower bound

###  5.1 Auxiliary results on multivariate Gaussian tails

######  Lemma 5.1.

Let $\\{(\xi_{1}(x_{1}),\xi_{2}(x_{2}),\xi_{3}(x_{3})):(x_{1},x_{2},x_{3})\in
D_{1}\times D_{2}\times D_{3}\\}$ be an ${\mathbb{R}}^{3}$-valued, $C^{2}$,
centered, unit-variance, non-degenerate Gaussian vector field, where $D_{i}$,
$i=1,2,3$, are compact sets in ${\mathbb{R}}^{N}$. Let $R_{ij}=\sup_{x_{i}\in
D_{i},x_{j}\in D_{j}}{\mathbb{E}}\\{\xi_{i}(x_{i})\xi_{j}(x_{j})\\}$, where
$i,j=1,2,3$ and $i<j$. If $R_{12}\leq\min\\{R_{13},R_{23}\\}$, then there
exists a constant $\alpha>0$ such that for every integer $m\geq 0$, as
$u\to\infty$,

$\begin{split}&\sup_{x_{1}\in D_{1},x_{2}\in D_{2},x_{3}\in
D_{3}}{\mathbb{E}}\\{|\xi_{1}(x_{1})\xi_{2}(x_{2})\xi_{3}(x_{3})|^{m}\mathbbm{1}_{\\{\xi_{1}(x_{1})\geq
u,\xi_{2}(x_{2})\geq u,\xi_{3}(x_{3})\geq u\\}}\\}\\\
&\quad=o\bigg{(}\exp\bigg{\\{}-\alpha
u^{2}-\frac{u^{2}}{1+R_{12}}\bigg{\\}}\bigg{)}.\end{split}$ (5.1)

###### Proof.

Due to the exponential decay of Gaussian tails, it suffices to prove that
there exists $\alpha^{\prime}>0$ such that as $u\to\infty$,

$\begin{split}&\sup_{x_{1}\in D_{1},x_{2}\in D_{2},x_{3}\in
D_{3}}{\mathbb{P}}\\{\xi_{1}(x_{1})\geq u,\xi_{2}(x_{2})\geq
u,\xi_{3}(x_{3})\geq
u\\}\\}=o\left(\exp\left\\{-\alpha^{\prime}u^{2}-\frac{u^{2}}{1+R_{12}}\right\\}\right).\end{split}$
(5.2)

Note that,

${\mathbb{P}}\\{\xi_{1}(x_{1})\geq u,\xi_{2}(x_{2})\geq u,\xi_{3}(x_{3})\geq
u\\}\\}\leq{\mathbb{P}}\\{(\xi_{1}(x_{1})+\xi_{2}(x_{2}))/2\geq
u,\xi_{3}(x_{3})\geq u\\}\\},$

where $(\xi_{1}(x_{1})+\xi_{2}(x_{2}))/2$ is a centered Gaussian variable with
variance bounded by

$\sup_{x_{1}\in D_{1},x_{2}\in D_{2}}{\rm
Var}((\xi_{1}(x_{1})+\xi_{2}(x_{2}))/2)=\frac{1+R_{12}}{2}.$

It is known that (see for example Tong [18]), for a centered nondegenerate
bivariate Gaussian vector $(Z_{1},Z_{2})$ with ${\rm Var}(Z_{1})=\sigma^{2}$,
there exists $\alpha^{\prime}>0$ such that as $u\to\infty$,

${\mathbb{P}}\\{Z_{1}\geq u,Z_{2}\geq
u\\}=o\left(\exp\left\\{-\alpha^{\prime}u^{2}-\frac{u^{2}}{2\sigma^{2}}\right\\}\right).$

Combining these yields (5.2) and hence (5.1). ∎

######  Lemma 5.2.

Let $\big{\\{}(\xi_{1}(x_{1}),\ldots,\xi_{n}(x_{n}):x_{i}\in
D_{i},i=1,\ldots,n\big{\\}}$ be an ${\mathbb{R}}^{n}$-valued, $C^{2}$,
centered, unit-variance, non-degenerate Gaussian vector vector, where $D_{1}$,
…, $D_{n}$ ($n\geq 3$) are compact sets in ${\mathbb{R}}^{N}$. Let
$R_{12}=\sup_{x_{1}\in D_{1},x_{2}\in
D_{2}}{\mathbb{E}}\\{\xi_{1}(x_{1})\xi_{2}(x_{2})\\}$. If

$\begin{split}\\{&(x_{1},\ldots,x_{n})\in D_{1}\times\cdots\times D_{n}:\\\
&\quad{\mathbb{E}}\\{\xi_{1}(x_{1})\xi_{2}(x_{2})\\}=R_{12},\
{\mathbb{E}}\\{(\xi_{1}(x_{1})+\xi_{2}(x_{2}))\xi_{i}(x_{i})\\}=0,\ \forall
i=3,\ldots,n\\}=\emptyset,\end{split}$ (5.3)

then there exists $\alpha>0$ such that as $u\to\infty$,

$\begin{split}&\sup_{x_{i}\in
D_{i},i=1,\ldots,n}{\mathbb{E}}\\{|\xi_{1}(x_{1})\xi_{2}(x_{2})|^{m}\mathbbm{1}_{\\{\xi_{1}(x_{1})\geq
u,\xi_{2}(x_{2})\geq u\\}}|\xi_{3}(x_{3})=\cdots=\xi_{n}(x_{n})=0\\}\\\
&\quad=o\bigg{(}\exp\bigg{\\{}-\alpha
u^{2}-\frac{u^{2}}{1+R_{12}}\bigg{\\}}\bigg{)},\end{split}$

where $m\geq 0$ is any fixed integer.

###### Proof.

Let $\overline{\xi}(x_{1},x_{2})=[\xi_{1}(x_{1})+\xi_{2}(x_{2})]/2$. Then

$\begin{split}{\mathbb{E}}&\\{|\xi_{1}(x_{1})\xi_{2}(x_{2})|^{m}\mathbbm{1}_{\\{\xi_{1}(x_{1})\geq
u,\xi_{2}(x_{2})\geq u\\}}|\xi_{3}(x_{3})=\cdots=\xi_{n}(x_{n})=0\\}\\\
&\leq{\mathbb{E}}\\{\overline{\xi}(x_{1},x_{2})^{2m}\mathbbm{1}_{\\{\overline{\xi}(x_{1},x_{2})\geq
u\\}}|\xi_{3}(x_{3})=\cdots=\xi_{n}(x_{n})=0\\}.\end{split}$

Note that
$(\overline{\xi}(x_{1},x_{2})|\xi_{3}(x_{3})=\cdots=\xi_{n}(x_{n})=0)$ is a
centered Gaussian variable with variance

$\begin{split}{\rm
Var}(\overline{\xi}(x_{1},x_{2})|\xi_{3}(x_{3})=\cdots=\xi_{n}(x_{n})=0)&\leq{\rm
Var}(\overline{\xi}(x_{1},x_{2}))=\frac{1+{\mathbb{E}}\\{\xi_{1}(x_{1})\xi_{2}(x_{2})\\}}{2}\leq\frac{1+R_{12}}{2},\end{split}$

where the first inequality becomes equality if and only if
$\overline{\xi}(x_{1},x_{2})$ is independent of each $\xi_{i}(x_{i})$, $i\geq
3$. The desired result follows from the continuity of the conditional variance
in $x_{i}$ and the compactness of $D_{i}$, $i=1,\ldots,n$. ∎

###  5.2 Non-adjacent faces

For two sets $D,D^{\prime}\subset{\mathbb{R}}^{N}$, let
$d(D,D^{\prime})=\inf\\{\|t-t^{\prime}\|:t\in D,t^{\prime}\in D^{\prime}\\}$
denote their distance. The following result shows that the last two sums
involving the joint moment of two non-adjacent faces in (4.3) are super-
exponentially small.

######  Lemma 5.3.

Let $\\{(X(t),Y(s)):t\in T,s\in S\\}$ be an ${\mathbb{R}}^{2}$-valued,
centered, unit-variance Gaussian vector field satisfying $({\bf H}1)$ and
$({\bf H}2)$. Then there exists $\alpha>0$ such that as $u\to\infty$,

$\begin{split}{\mathbb{E}}\\{M_{u}(X,K)M_{u}(X,K^{\prime})M_{u}(Y,L)\\}&=o\Big{(}\exp\Big{\\{}-\frac{u^{2}}{1+R}-\alpha
u^{2}\Big{\\}}\Big{)},\\\
{\mathbb{E}}\\{M_{u}(X,K)M_{u}(Y,L)M_{u}(Y,L^{\prime})\\}&=o\Big{(}\exp\Big{\\{}-\frac{u^{2}}{1+R}-\alpha
u^{2}\Big{\\}}\Big{)},\end{split}$ (5.4)

where $K$ and $K^{\prime}$ are different faces of $T$ with
$d(K,K^{\prime})>0$, $L$ and $L^{\prime}$ are different faces of $S$ with
$d(L,L^{\prime})>0$.

###### Proof.

We only prove the first line in (5.4), since the proof for the second line is
similar. Consider first the case when ${\rm dim}(K)=k\geq 1$, ${\rm
dim}(K^{\prime})=k^{\prime}\geq 1$ and ${\rm dim}(L)=l\geq 1$. By the Kac-Rice
metatheorem for high moments [2],

$\begin{split}&\quad{\mathbb{E}}\\{M_{u}(X,K)M_{u}(X,K^{\prime})M_{u}(Y,L)\\}\\\
&=\int_{K}dt\int_{K^{\prime}}dt^{\prime}\int_{L}ds\,{\mathbb{E}}\big{\\{}|{\rm
det}\nabla^{2}X_{|K}(t)||{\rm det}\nabla^{2}X_{|K^{\prime}}(t^{\prime})||{\rm
det}\nabla^{2}Y_{|L}(s)|\\\ &\quad\times\mathbbm{1}_{\\{X(t)\geq
u,X(t^{\prime})\geq u,Y(s)\geq u\\}}\mathbbm{1}_{\\{\nabla^{2}X_{|K}(t)\prec
0,\,\nabla^{2}X_{|K^{\prime}}(t^{\prime})\prec 0,\,\nabla^{2}Y_{|L}(s)\prec
0\\}}\big{|}\\\ &\quad\nabla X_{|K}(t)=0,\nabla
X_{|K^{\prime}}(t^{\prime})=0,\nabla Y_{|L}(s)=0\big{\\}}p_{\nabla
X_{|K}(t),\nabla X_{|K^{\prime}}(t^{\prime}),\nabla Y_{|L}(s)}(0,0,0)\\\
&\leq\int_{K}dt\int_{K^{\prime}}dt^{\prime}\int_{L}ds\int_{u}^{\infty}dx\int_{u}^{\infty}dx^{\prime}\int_{u}^{\infty}dy\,p_{X(t),X(t^{\prime}),Y(s)}(x,x^{\prime},y)\\\
&\quad\times{\mathbb{E}}\big{\\{}|{\rm det}\nabla^{2}X_{|K}(t)||{\rm
det}\nabla^{2}X_{|K^{\prime}}(t^{\prime})||{\rm
det}\nabla^{2}Y_{|L}(s)|\big{|}\\\ &\quad
X(t)=x,X(t^{\prime})=x^{\prime},Y(s)=y,\nabla X_{|K}(t)=0,\nabla
X_{|K^{\prime}}(t^{\prime})=0,\nabla Y_{|L}(s)=0\\}\\\ &\quad\times p_{\nabla
X_{|K}(t),\nabla X_{|K^{\prime}}(t^{\prime}),\nabla
Y_{|L}(s)}(0,0,0|X(t)=x,X(t^{\prime})=x^{\prime},Y(s)=y).\end{split}$ (5.5)

Notice that the following two inequalities hold: for constants $a_{i_{1}}$,
$b_{i_{2}}$ and $c_{i_{3}}$,

$\begin{split}&\prod_{i_{1}=1}^{k}|a_{i_{1}}|\prod_{i_{2}=1}^{k^{\prime}}|b_{i_{2}}|\prod_{i_{3}=1}^{l}|c_{i_{3}}|\leq\frac{\sum_{i_{1}=1}^{k}|a_{i_{1}}|^{k+k^{\prime}+l}+\sum_{i_{2}=1}^{k^{\prime}}|b_{i_{2}}|^{k+k^{\prime}+l}+\sum_{i_{3}=1}^{l}|c_{i_{3}}|^{k+k^{\prime}+l}}{k+k^{\prime}+l};\end{split}$

and for any Gaussian variable $\xi$ and positive integer $m$, by Jensen’s
inequality,

$\begin{split}{\mathbb{E}}|\xi|^{m}\leq{\mathbb{E}}(|{\mathbb{E}}\xi|+|\xi-{\mathbb{E}}\xi|)^{m}&\leq
2^{m-1}(|{\mathbb{E}}\xi|^{m}+{\mathbb{E}}|\xi-{\mathbb{E}}\xi|^{m})=2^{m-1}(|{\mathbb{E}}\xi|^{m}+B_{m}({\rm
Var}(\xi))^{m/2}),\end{split}$

where $B_{m}$ is some constant depending only on $m$. Combining these two
inequalities with the well-known conditional formula for Gaussian variables,
we obtain that there exist positive constants $C_{1}$ and $N_{1}$ such that
for large $x$, $x^{\prime}$ and $y$,

$\begin{split}\sup_{t\in K,t^{\prime}\in K^{\prime},s\in
L}&{\mathbb{E}}\big{\\{}|{\rm det}\nabla^{2}X_{|K}(t)||{\rm
det}\nabla^{2}X_{|K^{\prime}}(t^{\prime})||{\rm
det}\nabla^{2}Y_{|L}(s)|\big{|}X(t)=x,X(t^{\prime})=x^{\prime},\\\
&Y(s)=y,\nabla X_{|K}(t)=0,\nabla X_{|K^{\prime}}(t^{\prime})=0,\nabla
Y_{|L}(s)=0\big{\\}}\leq C_{1}+(xx^{\prime}y)^{N_{1}}.\end{split}$ (5.6)

Further, there exists $C_{2}>0$ such that

$\begin{split}&\quad\sup_{t\in K,t^{\prime}\in K^{\prime},s\in L}p_{\nabla
X_{|K}(t),\nabla X_{|K^{\prime}}(t^{\prime}),\nabla
Y_{|L}(s)}(0,0,0|X(t)=x,X(t^{\prime})=x^{\prime},Y(s)=y)\\\ &\leq\sup_{t\in
K,t^{\prime}\in K^{\prime},s\in L}(2\pi)^{-(k+k^{\prime}+l)/2}[{\rm
detCov}(\nabla X_{|K}(t),\nabla X_{|K^{\prime}}(t^{\prime}),\nabla
Y_{|L}(s)|\\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad
X(t)=x,X(t^{\prime})=x^{\prime},Y(s)=y)]^{-1/2}\leq C_{2}.\end{split}$ (5.7)

Plugging (5.6) and (5.7) into (5.5), we obtain that there exists $C_{3}={\rm
Vol}(K){\rm Vol}(K^{\prime}){\rm Vol}(L)$ such that

$\begin{split}&{\mathbb{E}}\\{M_{u}(X,K)M_{u}(X,K^{\prime})M_{u}(Y,L)\\}\\\
&\quad\leq C_{3}C_{2}\sup_{t\in K,t^{\prime}\in K^{\prime},s\in
L}{\mathbb{E}}\\{(C_{1}+|X(t)X(t^{\prime})Y(s)|^{N_{1}})\mathbbm{1}_{\\{X(t)\geq
u,X(t^{\prime})\geq u,Y(s)\geq u\\}}\\}.\end{split}$ (5.8)

The desired result then follows from Lemma 5.1. The case when one of the
dimensions of $K$, $K^{\prime}$ and $L$ is zero can be proved similarly. ∎

###  5.3 Factorial moments

The following result shows that the factorial moments in (4.3) are super-
exponentially small.

######  Lemma 5.4.

Let $\\{(X(t),Y(s)):t\in T,s\in S\\}$ be an ${\mathbb{R}}^{2}$-valued,
centered, unit-variance Gaussian vector field satisfying $({\bf H}1)$, $({\bf
H}2)$ and $({\bf H}3)$. Then there exists a constant $\alpha>0$ such that for
all $K\in\partial_{k}T$ and $L\in\partial_{l}T$ with $k,l\geq 0$, as
$u\to\infty$,

$\begin{split}{\mathbb{E}}\\{M_{u}(X,K)[M_{u}(X,K)-1]M_{u}(Y,L)\\}&=o\Big{(}\exp\Big{\\{}-\frac{u^{2}}{1+R}-\alpha
u^{2}\Big{\\}}\Big{)},\\\
{\mathbb{E}}\\{M_{u}(X,K)M_{u}(Y,L)[M_{u}(Y,L)-1]\\}&=o\Big{(}\exp\Big{\\{}-\frac{u^{2}}{1+R}-\alpha
u^{2}\Big{\\}}\Big{)}.\end{split}$ (5.9)

###### Proof.

We only prove the first line in (5.9), since the proof for the second line is
similar. Note that, if $k=0$, then $M_{u}(X,K)[M_{u}(X,K)-1]\equiv 0$ and
hence the desired result holds. Without loss of generality, we assume $k\geq
1$ or even $k=N$ for simplifying notations. We first focus on the estimation
when $K$ is replaced by a small $N$-dimensional subset $J\subset K$.

Case (i): $\bm{l=0}$. The face $L$ becomes a single point, say $L=\\{s\\}$.
Applying the Kac-Rice metatheorem for high moments [2], we have the following
upper bounds (removing one restriction on $u$ and another restriction on the
negative definiteness of Hessian matrices),

$\begin{split}&\quad{\mathbb{E}}\\{M_{u}(X,J)[M_{u}(X,J)-1]M_{u}(Y,L)\\}\\\
&\leq\int_{J}dt\int_{J}dt^{\prime}\,{\mathbb{E}}\big{\\{}|{\rm
det}\nabla^{2}X(t)||{\rm det}\nabla^{2}X(t^{\prime})|\mathbbm{1}_{\\{X(t)\geq
u,Y(s)\geq u\\}}\big{|}\nabla X(t)=\nabla X(t^{\prime})=0\big{\\}}\\\
&\quad\times p_{\nabla X(t),\nabla X(t^{\prime})}(0,0)\\\
&\leq\int_{J}dt\int_{J}dt^{\prime}\int_{u}^{\infty}dx\,p_{\frac{X(t)+Y(s)}{2}}(x|\nabla
X(t)=\nabla X(t^{\prime})=0)p_{\nabla X(t),\nabla X(t^{\prime})}(0,0)\\\
&\quad\times{\mathbb{E}}\big{\\{}|{\rm det}\nabla^{2}X(t)||{\rm
det}\nabla^{2}X(t^{\prime})|\big{|}(X(t)+Y(s))/2=x,\nabla X(t)=\nabla
X(t^{\prime})=0\big{\\}},\end{split}$ (5.10)

where the last inequality is due to the fact $\mathbbm{1}_{\\{X(t)\geq
u,Y(s)\geq u\\}}\leq\mathbbm{1}_{\\{[X(t)+Y(s)]/2\geq u\\}}$. Following the
same arguments for proving Lemma 3 in Piterbarg  [13], we obtain from (5.10)
that, for any $\varepsilon>0$, there exists $\delta>0$ such that for $J$ with
${\rm diam}(J)=\sup_{t,t^{\prime}\in J}\|t-t^{\prime}\|\leq\delta$ and $u$
large enough,

${\mathbb{E}}\\{M_{u}(X,J)[M_{u}(X,J)-1]M_{u}(Y,L)\\}\leq\exp\left\\{-\frac{u^{2}}{2\beta(J,L)+\varepsilon}\right\\},$
(5.11)

where

$\begin{split}\beta(J,L)=\sup_{t\in J,s\in L,e\in\mathbb{S}^{N-1}}{\rm
Var}((X(t)+Y(s))/2|\nabla X(t)=0,\nabla^{2}X(t)e=0),\end{split}$ (5.12)

and $\mathbb{S}^{N-1}$ the $(N-1)$-dimensional unit sphere in
${\mathbb{R}}^{N}$.

Case (ii): $\bm{l\geq 1}$.  To simplify the notations, without loss of
generality, we assume $l=N$. Applying again the Kac-Rice metatheorem for high
moments, we have the following upper bounds,

$\begin{split}&\quad{\mathbb{E}}\\{M_{u}(X,J)[M_{u}(X,J)-1]M_{u}(Y,L)\\}\\\
&\leq\int_{J}dt\int_{J}dt^{\prime}\int_{L}ds\,{\mathbb{E}}\big{\\{}|{\rm
det}\nabla^{2}X(t)||{\rm det}\nabla^{2}X(t^{\prime})||{\rm
det}\nabla^{2}Y(s)|\mathbbm{1}_{\\{X(t)\geq u,Y(s)\geq u\\}}\big{|}\\\
&\quad\nabla X(t)=0,\nabla X(t^{\prime})=0,\nabla Y(s)=0\big{\\}}p_{\nabla
X(t),\nabla X(t^{\prime}),\nabla Y(s)}(0,0,0)\\\
&\leq\int_{J}dt\int_{J}dt^{\prime}\int_{L}ds\int_{u}^{\infty}dx\,p_{\frac{X(t)+Y(s)}{2}}(x|\nabla
X(t)=0,\nabla X(t^{\prime})=0,\nabla Y(s)=0)\\\
&\quad\times{\mathbb{E}}\big{\\{}|{\rm det}\nabla^{2}X(t)||{\rm
det}\nabla^{2}X(t^{\prime})||{\rm
det}\nabla^{2}Y(s)|\big{|}[X(t)+Y(s)]/2=x,\\\ &\quad\nabla X(t)=0,\nabla
X(t^{\prime})=0,\nabla Y(s)=0\big{\\}}p_{\nabla X(t),\nabla
X(t^{\prime}),\nabla Y(s)}(0,0,0).\end{split}$ (5.13)

Comparing (5.13) with (5.10), the only essential difference is the additional
effect of $\nabla Y(s)=0$, which however will not affect the desired super-
exponentially small estimation since $(X,Y)$ is nondegenerate under the
condition (H2). Therefore, similarly to (5.11), we have that, for any
$\varepsilon>0$, there exists $\delta>0$ such that for $J$ with ${\rm
diam}(J)\leq\delta$ and $u$ large enough,

${\mathbb{E}}\\{M_{u}(X,J)[M_{u}(X,J)-1]M_{u}(Y,L)\\}\leq\exp\left\\{-\frac{u^{2}}{2\gamma(J,L)+\varepsilon}\right\\}\leq\exp\left\\{-\frac{u^{2}}{2\beta(J,L)+\varepsilon}\right\\},$
(5.14)

where

$\begin{split}\gamma(J,L)&=\sup_{t\in J,s\in L,e\in\mathbb{S}^{N-1}}{\rm
Var}((X(t)+Y(s))/2|\nabla X(t)=\nabla
Y(s)=\nabla^{2}X(t)e=0)\leq\beta(J,L).\end{split}$

The set $K$ may be covered by congruent cubes $J_{i}$ with disjoint interiors,
edges parallel to coordinate axes and sizes small enough such that ${\rm
diam}(J_{i}\cup J_{j})\leq\delta$ for any two neighboring cubes $J_{i}$ and
$J_{j}$ (i.e., $d(J_{i},J_{j})=0$). Then

$\begin{split}&\quad{\mathbb{E}}\\{M_{u}(X,K)[M_{u}(X,K)-1]M_{u}(Y,L)\\}\\\
&\leq{\mathbb{E}}\Big{\\{}\Big{(}\sum_{i}M_{u}(X,J_{i})\Big{)}\Big{[}\sum_{j}M_{u}(X,J_{j})-1\Big{]}M_{u}(Y,L)\Big{\\}}\\\
&={\mathbb{E}}\Big{\\{}\Big{(}\sum_{i}M_{u}(X,J_{i})\sum_{j}M_{u}(X,J_{j})-\sum_{i}M_{u}(X,J_{i})\Big{)}M_{u}(Y,L)\Big{\\}}\\\
&=\sum_{i}{\mathbb{E}}\\{M_{u}(X,J_{i})^{2}M_{u}(Y,L)\\}+\sum_{i\neq
j}{\mathbb{E}}\\{M_{u}(X,J_{i})M_{u}(X,J_{j})M_{u}(Y,L)\\}\\\
&\quad-\sum_{i}{\mathbb{E}}\\{M_{u}(X,J_{i})M_{u}(Y,L)\\}\\\
&=\sum_{i}{\mathbb{E}}\\{M_{u}(X,J_{i})[M_{u}(X,J_{i})-1]M_{u}(Y,L)\\}+\sum_{i\neq
j}{\mathbb{E}}\\{M_{u}(X,J_{i})M_{u}(X,J_{j})M_{u}(Y,L)\\}.\end{split}$ (5.15)

By Lemma 5.3, there exists $\alpha^{\prime}>0$ such that for $u$ large enough,

$\begin{split}\sum_{i\neq
j:\,d(J_{i},J_{j})>0}{\mathbb{E}}\\{M_{u}(X,J_{i})M_{u}(X,J_{j})M_{u}(Y,L)\\}\leq\exp\Big{\\{}-\frac{u^{2}}{1+R}-\alpha^{\prime}u^{2}\Big{\\}}.\end{split}$
(5.16)

If $J_{i}$ and $J_{j}$ are neighboring, i.e., $d(J_{i},J_{j})=0$, we have

$\begin{split}&\quad{\mathbb{E}}\\{M_{u}(X,J_{i}\cup J_{j})[M_{u}(X,J_{i}\cup
J_{j})-1]M_{u}(Y,L)\\}\\\
&={\mathbb{E}}\\{[M_{u}(X,J_{i})+M_{u}(X,J_{j})][M_{u}(X,J_{i})+M_{u}(X,J_{j})-1]M_{u}(Y,L)\\}\\\
&=2{\mathbb{E}}\\{M_{u}(X,J_{i})M_{u}(X,J_{j})M_{u}(Y,L)\\}+{\mathbb{E}}\\{M_{u}(X,J_{i})[M_{u}(X,J_{i})-1]M_{u}(Y,L)\\}\\\
&\quad+{\mathbb{E}}\\{M_{u}(X,J_{j})[M_{u}(X,J_{j})-1]M_{u}(Y,L)\\}.\end{split}$
(5.17)

Applying (5.11) and (5.14) to the second last sum in (5.15) and (5.17), we see
that for any $\varepsilon>0$ and $u$ large enough,

$\begin{split}&\sum_{i}{\mathbb{E}}\\{M_{u}(X,J_{i})[M_{u}(X,J_{i})-1]M_{u}(Y,L)\\}\\\
&+\sum_{i\neq
j:\,d(J_{i},J_{j})=0}{\mathbb{E}}\\{M_{u}(X,J_{i})M_{u}(X,J_{j})M_{u}(Y,L)\\}\leq\exp\Big{\\{}-\frac{u^{2}}{2\beta(K,L)+\varepsilon}\Big{\\}},\end{split}$
(5.18)

where $\beta(K,L)$ is defined in (5.12) with $J$ replaced by $K$. It is
evident that

$\beta(K,L)\leq\sup_{t\in K,s\in L}{\rm Var}((X(t)+Y(s))/2)=(1+R)/2.$

Moreover, we will how below that

$\begin{split}\beta(K,L)<(1+R)/2.\end{split}$ (5.19)

By the definition, if $\beta(K,L)=(1+R)/2$, then there exist
$(t,s)\in\bar{K}\times\bar{L}$ and $e\in\mathbb{S}^{N-1}$ such that

$\begin{split}{\rm Var}((X(t)+Y(s))/2|\nabla
X(t)=0,\nabla^{2}X(t)e=0)=(1+R)/2,\end{split}$ (5.20)

implying $r(t,s)=R$ and ${\mathbb{E}}\\{[X(t)+Y(s)]\nabla
X(t)\\}={\mathbb{E}}\\{Y(s)\nabla X(t)\\}=0$. By Proposition 2.1,
${\mathbb{E}}\\{Y(s)\nabla^{2}X(t)\\}\preceq 0$. Since $X(t)$ has unit
variance, ${\mathbb{E}}\\{X(t)\nabla^{2}X(t)\\}=-{\rm Cov}(\nabla X(t))\prec
0$. Therefore, ${\mathbb{E}}\\{[X(t)+Y(s)]\nabla^{2}X(t)e\\}\neq 0$ for all
$e\in\mathbb{S}^{N-1}$. This contradicts (5.20) and hence (5.19) holds.
Applying this fact and plugging (5.16) and (5.18) into (5.15), we finish the
proof. ∎

###  5.4 Adjacent faces

The following result shows that the last two sums involving the joint moment
of two adjacent faces in (4.3) are super-exponentially small.

######  Lemma 5.5.

Let $\\{(X(t),Y(s)):t\in T,s\in S\\}$ be an ${\mathbb{R}}^{2}$-valued,
centered, unit-variance Gaussian vector field satisfying $({\bf H}1)$, $({\bf
H}2)$ and $({\bf H}3)$. Then there exists $\alpha>0$ such that as
$u\to\infty$,

$\begin{split}{\mathbb{E}}\\{M_{u}^{E}(X,K)M_{u}^{E}(X,K^{\prime})M_{u}^{E}(Y,L)\\}&=o\Big{(}\exp\Big{\\{}-\frac{u^{2}}{1+R}-\alpha
u^{2}\Big{\\}}\Big{)},\\\
{\mathbb{E}}\\{M_{u}^{E}(X,K)M_{u}^{E}(Y,L)M_{u}^{E}(Y,L^{\prime})\\}&=o\Big{(}\exp\Big{\\{}-\frac{u^{2}}{1+R}-\alpha
u^{2}\Big{\\}}\Big{)},\end{split}$ (5.21)

where $K$ and $K^{\prime}$ are different faces of $T$ with
$d(K,K^{\prime})=0$, $L$ and $L^{\prime}$ are different faces of $S$ with
$d(L,L^{\prime})=0$.

###### Proof.

We only prove the first line in (5.21), since the proof for the second line is
the same. Let $I:=\bar{K}\cap\bar{K^{\prime}}$, which is nonempty since
$d(K,K^{\prime})=0$. Without loss of generality, assume

$\begin{split}\sigma(K)&=\\{1,\ldots,m,m+1,\ldots,k\\},\\\
\sigma(K^{\prime})&=\\{1,\ldots,m,k+1,\ldots,k+k^{\prime}-m\\},\\\
\sigma(L)&=\\{1,\ldots,l\\},\end{split}$

where $0\leq m\leq k\leq k^{\prime}\leq N$ and $k^{\prime}\geq 1$. If $k=0$,
we consider $\sigma(K)=\emptyset$ by convention. Under such assumption,
$K\in\partial_{k}T$, $K^{\prime}\in\partial_{k^{\prime}}T$, $\text{dim}(I)=m$
and $L\in\partial_{l}S$. We assume also that all elements in $\varepsilon(K)$
and $\varepsilon(K^{\prime})$ are 1.

We first consider the case when $k\geq 1$ and $l\geq 1$. By the Kac-Rice
metatheorem,

$\begin{split}&\quad{\mathbb{E}}\\{M_{u}^{E}(X,K)M_{u}^{E}(X,K^{\prime})M_{u}^{E}(Y,L)\\}\\\
&\leq\int_{K}dt\int_{K^{\prime}}dt^{\prime}\int_{L}ds\int_{u}^{\infty}dx\int_{u}^{\infty}dx^{\prime}\int_{u}^{\infty}dy\int_{0}^{\infty}dz_{k+1}\cdots\int_{0}^{\infty}dz_{k+k^{\prime}-m}\\\
&\quad\int_{0}^{\infty}dw_{m+1}\cdots\int_{0}^{\infty}dw_{k}{\mathbb{E}}\big{\\{}|\text{det}\nabla^{2}X_{|K}(t)||\text{det}\nabla^{2}X_{|K^{\prime}}(t^{\prime})||\text{det}\nabla^{2}Y_{|L}(s)|\big{|}\\\
&\quad X(t)=x,X(t^{\prime})=x^{\prime},Y(s)=y,\\\ &\quad\nabla
X_{|K}(t)=0,X_{k+1}(t)=z_{k+1},\ldots,X_{k+k^{\prime}-m}(t)=z_{k+k^{\prime}-m},\\\
&\quad\nabla
X_{|K^{\prime}}(t^{\prime})=0,X_{m+1}(t^{\prime})=w_{m+1},\ldots,X_{k}(t^{\prime})=w_{k},\nabla
Y_{|L}(s)=0\big{\\}}\\\ &\quad\times
p_{t,t^{\prime},s}(x,x^{\prime},y,0,z_{k+1},\ldots,z_{k+k^{\prime}-m},0,w_{m+1},\ldots,w_{k},0)\\\
&:=\int\int\int_{K\times K^{\prime}\times
L}A(t,t^{\prime},s)\,dtdt^{\prime}ds,\end{split}$ (5.22)

where
$p_{t,t^{\prime},s}(x,x^{\prime},y,0,z_{k+1},\ldots,z_{k+k^{\prime}-m},0,w_{m+1},\ldots,w_{k},0)$
is the density of

$\begin{split}&\big{(}X(t),X(t^{\prime}),Y(s),\nabla
X_{|K}(t),X_{k+1}(t),\ldots,X_{k+k^{\prime}-m}(t),\\\ &\quad\nabla
X_{|K^{\prime}}(t^{\prime}),X_{m+1}(t^{\prime}),\ldots,X_{k}(t^{\prime}),\nabla
Y_{|L}(s)\big{)}\end{split}$

evaluated at
$(x,x^{\prime},y,0,z_{k+1},\ldots,z_{k+k^{\prime}-m},0,w_{m+1},\ldots,w_{k},0)$.
We define

$\begin{split}\mathcal{M}_{0}:=\\{(t,s)\in
I\times\bar{L}:\,&r(t,s)=R,\,{\mathbb{E}}\\{X_{i}(t)Y(s)\\}={\mathbb{E}}\\{X(t)Y_{j}(s)\\}=0,\\\
&\forall i=1,\ldots,k+k^{\prime}-m,\,j=1,\ldots,l\\},\end{split}$ (5.23)

and distinguish two cases for $\mathcal{M}_{0}$ in discussions below.

Case (i): $\bm{\mathcal{M}_{0}=\emptyset}$. Since $I$ is a compact set, by the
uniform continuity of conditional variance, there exist constants
$\varepsilon_{1},\delta_{1}>0$ such that

$\begin{split}&\sup_{t\in B(I,\delta_{1}),\,t^{\prime}\in
B^{\prime}(I,\delta_{1}),\,s\in L}{\rm Var}([X(t)+Y(s)]/2|\nabla
X_{|K}(t),\nabla X_{|K^{\prime}}(t^{\prime}),\nabla
Y_{|L}(s))\leq\frac{1+R}{2}-\varepsilon_{1},\end{split}$ (5.24)

where $B(I,\delta_{1})=\\{t\in K:d(t,I)\leq\delta_{1}\\}$ and
$B^{\prime}(I,\delta_{1})=\\{t\in K^{\prime}:d(t,I)\leq\delta_{1}\\}$.
Partitioning $K\times K^{\prime}$ into $B(I,\delta_{1})\times
B^{\prime}(I,\delta_{1})$ and $(K\times
K^{\prime})\backslash(B(I,\delta_{1})\times B^{\prime}(I,\delta_{1}))$, and
applying the Kac-Rice formula, we obtain

$\begin{split}&\quad{\mathbb{E}}\\{M_{u}(X,K)M_{u}(X,K^{\prime})M_{u}(Y,L)\\}\\\
&\leq\int_{(K\times K^{\prime})\backslash(B(I,\delta_{1})\times
B^{\prime}(I,\delta_{1}))}dtdt^{\prime}\int_{L}ds\,p_{\nabla X_{|K}(t),\nabla
X_{|K^{\prime}}(t^{\prime}),\nabla Y_{|L}(s)}(0,0,0)\\\
&\quad\times{\mathbb{E}}\big{\\{}|{\rm det}\nabla^{2}X_{|K}(t)||{\rm
det}\nabla^{2}X_{|K^{\prime}}(t^{\prime})||{\rm
det}\nabla^{2}Y_{|L}(s)|\mathbbm{1}_{\\{X(t)\geq u,X(t^{\prime})\geq
u,Y(s)\geq u\\}}\big{|}\\\ &\qquad\nabla X_{|K}(t)=0,\nabla
X_{|K^{\prime}}(t^{\prime})=0,\nabla Y_{|L}(s)=0\big{\\}}\\\
&+\int_{B(I,\delta_{1})\times
B^{\prime}(I,\delta_{1})}dtdt^{\prime}\int_{L}ds\,p_{\nabla X_{|K}(t),\nabla
X_{|K^{\prime}}(t^{\prime}),\nabla Y_{|L}(s)}(0,0,0)\\\
&\quad\times{\mathbb{E}}\big{\\{}|{\rm det}\nabla^{2}X_{|K}(t)||{\rm
det}\nabla^{2}X_{|K^{\prime}}(t^{\prime})||{\rm
det}\nabla^{2}Y_{|L}(s)|\mathbbm{1}_{\\{X(t)\geq u,Y(s)\geq u\\}}\big{|}\\\
&\qquad\nabla X_{|K}(t)=0,\nabla X_{|K^{\prime}}(t^{\prime})=0,\nabla
Y_{|L}(s)=0\big{\\}}\\\ &:=I_{1}+I_{2}.\end{split}$ (5.25)

Note that

$\begin{split}(K\times K^{\prime})\backslash(B(I,\delta_{1})\times
B^{\prime}(I,\delta_{1}))&=\Big{(}(K\backslash B(I,\delta_{1}))\times
B^{\prime}(I,\delta_{1})\Big{)}\bigcup\Big{(}B(I,\delta_{1})\times(K\backslash
B(I,\delta_{1}))\Big{)}\\\ &\quad\bigcup\Big{(}(K\backslash
B(I,\delta_{1}))\times(K\backslash B(I,\delta_{1}))\Big{)}.\end{split}$

where each product on the right hand side consists of two sets with a positive
distance. It then follows from Lemma 5.3 that $I_{1}$ is super-exponentially
small. On the other hand, since $\mathbbm{1}_{\\{X(t)\geq u,Y(s)\geq
u\\}}\leq\mathbbm{1}_{\\{[X(t)+Y(s)]/2\geq u\\}}$, one has

$\begin{split}I_{2}&\leq\int_{B(I,\delta_{1})\times
B^{\prime}(I,\delta_{1})}dtdt^{\prime}\int_{L}ds\int_{u}^{\infty}dx\,p_{\frac{X(t)+Y(s)}{2}}(x|\nabla
X_{|K}(t)=\nabla X_{|K^{\prime}}(t^{\prime})=\nabla Y_{|L}(s)=0)\\\
&\quad\times{\mathbb{E}}\big{\\{}|{\rm det}\nabla^{2}X_{|K}(t)||{\rm
det}\nabla^{2}X_{|K^{\prime}}(t^{\prime})||{\rm
det}\nabla^{2}Y_{|L}(s)|\big{|}[X(t)+Y(s)]/2=x,\\\ &\qquad\nabla
X_{|K}(t)=0,\nabla X_{|K^{\prime}}(t^{\prime})=0,\nabla
Y_{|L}(s)=0\big{\\}}p_{\nabla X_{|K}(t),\nabla
X_{|K^{\prime}}(t^{\prime}),\nabla Y_{|L}(s)}(0,0,0).\end{split}$ (5.26)

Combining this with (5.24), we obtain that $I_{2}$ and hence
${\mathbb{E}}\\{M_{u}^{E}(X,K)M_{u}^{E}(X,K^{\prime})M_{u}^{E}(Y,L)\\}$ are
super-exponentially small.

Case (ii): $\bm{\mathcal{M}_{0}\neq\emptyset}$. Let

$\begin{split}B(\mathcal{M}_{0},\delta_{2}):=\\{(t,t^{\prime},s)\in K\times
K^{\prime}\times L:d((t,s),\mathcal{M}_{0})\vee
d((t^{\prime},s),\mathcal{M}_{0})\leq\delta_{2}\\},\end{split}$

where $\delta_{2}$ is a small positive number to be specified. Note that, by
the definitions of $\mathcal{M}_{0}$ and $B(\mathcal{M}_{0},\delta_{2})$,
there exists $\varepsilon_{2}>0$ such that

$\begin{split}\sup_{(t,t^{\prime},s)\in(K\times K^{\prime}\times L)\backslash
B(\mathcal{M}_{0},\delta_{2})}&{\rm Var}([X(t)+Y(s)]/2|\nabla X_{|K}(t),\nabla
X_{|K^{\prime}}(t^{\prime}),\nabla Y_{|L}(s))\\\
&\leq\frac{1+R}{2}-\varepsilon_{2}.\end{split}$ (5.27)

Similarly to (5.26), we obtain that $\int_{(K\times K^{\prime}\times
L)\backslash B(\mathcal{M}_{0},\delta_{2})}A(t,t^{\prime},s)dtdt^{\prime}ds$
is super-exponentially small. It suffices to show below that
$\int_{B(\mathcal{M}_{0},\delta_{2})}A(t,t^{\prime},s)\,dtdt^{\prime}ds$ is
super-exponentially small.

Due to $({\bf H}3)$ and Proposition 2.1, we can choose $\delta_{2}$ small
enough such that for all $(t,t^{\prime},s)\in B(\mathcal{M}_{0},\delta_{2})$,

$\begin{split}\Lambda_{K\cup
K^{\prime}}(t,s)&:=-{\mathbb{E}}\\{[X(t)+Y(s)]\nabla^{2}X_{|K\cup
K^{\prime}}(t)\\}=-({\mathbb{E}}\\{[X(t)+Y(s)]X_{ij}(t)\\})_{i,j=1,\ldots,k+k^{\prime}-m}\end{split}$

are positive definite. Let $\\{e_{1},e_{2},\ldots,e_{N}\\}$ be the standard
orthonormal basis of ${\mathbb{R}}^{N}$. For $t\in K$, $t^{\prime}\in
K^{\prime}$ and $s\in L$, let
$e_{t,t^{\prime}}=(t^{\prime}-t)/\|t^{\prime}-t\|$ and
$\alpha_{i}(t,t^{\prime},s)={\langle}e_{i},\Lambda_{K\cup
K^{\prime}}(t,s)e_{t,t^{\prime}}\rangle$. Then

$\Lambda_{K\cup
K^{\prime}}(t,s)e_{t,t^{\prime}}=\sum_{i=1}^{N}{\langle}e_{i},\Lambda_{K\cup
K^{\prime}}(t,s)e_{t,t^{\prime}}\rangle
e_{i}=\sum_{i=1}^{N}\alpha_{i}(t,t^{\prime},s)e_{i}$ (5.28)

and there exists $\alpha_{0}>0$ such that for all $(t,t^{\prime},s)\in
B(\mathcal{M}_{0},\delta_{2})$,

${\langle}e_{t,t^{\prime}},\Lambda_{K\cup
K^{\prime}}(t,s)e_{t,t^{\prime}}\rangle\geq\alpha_{0}.$ (5.29)

Since all elements in $\varepsilon(K)$ and $\varepsilon(K^{\prime})$ are 1, we
may write

$\begin{split}t&=(t_{1},\ldots,t_{m},t_{m+1},\ldots,t_{k},b_{k+1},\ldots,b_{k+k^{\prime}-m},0,\ldots,0),\\\
t^{\prime}&=(t^{\prime}_{1},\ldots,t^{\prime}_{m},b_{m+1},\ldots,b_{k},t^{\prime}_{k+1},\ldots,t^{\prime}_{k+k^{\prime}-m},0,\ldots,0),\end{split}$

where $t_{i}\in(a_{i},b_{i})$ for $i\in\sigma(K)$ and
$t^{\prime}_{j}\in(a_{j},b_{j})$ for $j\in\sigma(K^{\prime})$. Therefore,

$\begin{split}{\langle}e_{i},e_{t,t^{\prime}}\rangle&\geq 0,\quad\forall\
m+1\leq i\leq k,\\\ {\langle}e_{i},e_{t,t^{\prime}}\rangle&\leq
0,\quad\forall\ k+1\leq i\leq k+k^{\prime}-m,\\\
{\langle}e_{i},e_{t,t^{\prime}}\rangle&=0,\quad\forall\ k+k^{\prime}-m<i\leq
N.\end{split}$ (5.30)

Let

$\begin{split}D_{i}&=\\{(t,t^{\prime},s)\in
B(\mathcal{M}_{0},\delta_{2}):\alpha_{i}(t,t^{\prime},s)\geq\beta_{i}\\},\quad\text{if}\
m+1\leq i\leq k,\\\ D_{i}&=\\{(t,t^{\prime},s)\in
B(\mathcal{M}_{0},\delta_{2}):\alpha_{i}(t,t^{\prime},s)\leq-\beta_{i}\\},\quad\text{if}\
k+1\leq i\leq k+k^{\prime}-m,\\\ D_{0}&=\bigg{\\{}(t,t^{\prime},s)\in
B(\mathcal{M}_{0},\delta_{2}):\sum_{i=1}^{m}\alpha_{i}(t,t^{\prime},s){\langle}e_{i},e_{t,t^{\prime}}\rangle\geq\beta_{0}\bigg{\\}},\end{split}$
(5.31)

where $\beta_{0},\beta_{1},\ldots,\beta_{k+k^{\prime}-m}$ are positive
constants such that
$\beta_{0}+\sum_{i=m+1}^{k+k^{\prime}-m}\beta_{i}<\alpha_{0}$. It follows from
(5.30) and (5.31) that, if $(t,s)$ does not belong to any of
$D_{0},D_{m+1},\ldots,D_{k+k^{\prime}-m}$, then by (5.28),

${\langle}\Lambda_{K\cup
K^{\prime}}(t,s)e_{t,t^{\prime}},e_{t,t^{\prime}}\rangle=\sum_{i=1}^{N}\alpha_{i}(t,t^{\prime},s){\langle}e_{i},e_{t,t^{\prime}}\rangle\leq\beta_{0}+\sum_{i=m+1}^{k+k^{\prime}-m}\beta_{i}<\alpha_{0},$

which contradicts (5.29). Thus $D_{0}\cup\cup_{i=m+1}^{k+k^{\prime}-m}D_{i}$
is a covering of $B(\mathcal{M}_{0},\delta_{2})$. By (5.22),

$\begin{split}{\mathbb{E}}&\\{M_{u}^{E}(X,K)M_{u}^{E}(X,K^{\prime})M_{u}^{E}(Y,L)\\}\leq\int_{D_{0}}A(t,t^{\prime},s)\,dtdt^{\prime}ds+\sum_{i=m+1}^{k+k^{\prime}-m}\int_{D_{i}}A(t,t^{\prime},s)\,dtdt^{\prime}ds.\end{split}$

By the Kac-Rice metatheorem and the fact $\mathbbm{1}_{\\{X(t)\geq u,Y(s)\geq
u\\}}\leq\mathbbm{1}_{\\{[X(t)+Y(s)]/2\geq u\\}}$, we obtain

$\begin{split}\int_{D_{0}}&A(t,t^{\prime},s)\,dtdt^{\prime}ds\leq\int_{D_{0}}dtdt^{\prime}ds\int_{u}^{\infty}dx\,p_{\nabla
X_{|K}(t),\nabla X_{|K^{\prime}}(t^{\prime}),\nabla Y_{|L}(s)}(0,0,0)\\\
&\times p_{[X(t)+Y(s)]/2}(x|\nabla X_{|K}(t)=0,\nabla
X_{|K^{\prime}}(t^{\prime})=0,\nabla Y_{|L}(s)=0)\\\
&\times{\mathbb{E}}\big{\\{}|\text{det}\nabla^{2}X_{|K}(t)||\text{det}\nabla^{2}X_{|K^{\prime}}(t^{\prime})||\text{det}\nabla^{2}Y_{|L}(s)|\big{|}[X(t)+Y(s)]/2=x,\\\
&\qquad\quad\nabla X_{|K}(t)=\nabla X_{|K^{\prime}}(t^{\prime})=\nabla
Y_{|L}(s)=0\big{\\}},\end{split}$ (5.32)

and that for $i=m+1,\ldots,k$,

$\begin{split}&\quad\int_{D_{i}}A(t,t^{\prime},s)\,dtdt^{\prime}ds\\\
&\leq\int_{D_{i}}dtdt^{\prime}ds\int_{u}^{\infty}dx\int_{0}^{\infty}dw_{i}\,{\mathbb{E}}\big{\\{}|\text{det}\nabla^{2}X_{|K}(t)||\text{det}\nabla^{2}X_{|K^{\prime}}(t^{\prime})||\text{det}\nabla^{2}Y_{|L}(s)|\big{|}\\\
&\quad[X(t)+Y(s)]/2=x,\nabla X_{|K}(t)=0,X_{i}(t^{\prime})=w_{i},\nabla
X_{|K^{\prime}}(t^{\prime})=\nabla Y_{|L}(s)=0\big{\\}}\\\ &\quad\times
p_{[X(t)+Y(s)]/2,\nabla X_{|K}(t),X_{i}(t^{\prime}),\nabla
X_{|K^{\prime}}(t^{\prime}),\nabla Y_{|L}(s)}(x,0,w_{i},0,0).\end{split}$
(5.33)

Comparing (5.32) and (5.33) with Eqs. (4.33) and (4.36) respectively in the
proof of Theorem 4.8 in Cheng and Xiao  [7], the only essential difference is
the additional effect of $\nabla Y_{|L}(s)=0$, which however will not affect
the desired super-exponentially small estimation since $(X,Y)$ is
nondegenerate under the condition (H2). Therefore, following similar arguments
therein, we obtain that $\int_{D_{0}}A(t,t^{\prime},s)\,dtdt^{\prime}ds$ and
$\int_{D_{i}}A(t,t^{\prime},s)\,dtdt^{\prime}ds$ ($i=m+1,\ldots,k$) are super-
exponentially small.

It is similar to show that $\int_{D_{i}}A(t,t^{\prime},s)\,dtdt^{\prime}ds$
are super-exponentially small for $i=k+1,\ldots,k+k^{\prime}-m$. For the case
$k=0$ or $l=0$, the argument is even simpler when applying the Kac-Rice
formula (see for example (5.10)). Hence the details are omitted here. We have
completed the proof. ∎

Notice that, in the proof of Lemma 5.5, we have shown in (5.25) that, if
$\mathcal{M}_{0}=\emptyset$, then
${\mathbb{E}}\\{M_{u}(X,K)M_{u}(X,K^{\prime})M_{u}(Y,L)\\}$ is super-
exponentially small. Under the boundary condition (5.34) below, which
generalizes the condition $\mathcal{M}_{0}=\emptyset$ in terms of the
correlation function $r(t,s)$, we have the following result.

######  Lemma 5.6.

Let $\\{(X(t),Y(s)):t\in T,s\in S\\}$ be an ${\mathbb{R}}^{2}$-valued,
centered, unit-variance Gaussian vector field satisfying $({\bf H}1)$ and
$({\bf H}2)$. If for any faces $K_{1}\subset T$ and $L_{1}\subset S$,

$\begin{split}\Big{\\{}(t,s)\in K_{1}\times
L_{1}:\,r(t,s)=R,\prod_{i\notin\sigma(K_{1})}\frac{\partial r}{\partial
t_{i}}(t,s)\prod_{j\notin\sigma(L_{1})}\frac{\partial r}{\partial
s_{j}}(t,s)=0\Big{\\}}=\emptyset,\end{split}$ (5.34)

then there exists a constant $\alpha>0$ such that as $u\to\infty$,

$\begin{split}{\mathbb{E}}\\{M_{u}(X,K)M_{u}(X,K^{\prime})M_{u}(Y,L)\\}&=o\Big{(}\exp\Big{\\{}-\frac{u^{2}}{1+R}-\alpha
u^{2}\Big{\\}}\Big{)},\\\
{\mathbb{E}}\\{M_{u}(X,K)M_{u}(Y,L)M_{u}(Y,L^{\prime})\\}&=o\Big{(}\exp\Big{\\{}-\frac{u^{2}}{1+R}-\alpha
u^{2}\Big{\\}}\Big{)},\end{split}$

where $K$ and $K^{\prime}$ are adjacent faces of $T$, and $L$ and $L^{\prime}$
are adjacent faces of $S$.

Remark 5.7  In other words, the boundary condition (5.34) indicates that, for
any point $(t,s)\in K_{1}\times L_{1}$ attaining the maximum of correlation
$R$, there must be $\frac{\partial r}{\partial t_{i}}(t,s)\neq 0$ for all
$i\notin\sigma(K_{1})$ and $\frac{\partial r}{\partial s_{j}}(t,s)\neq 0$ for
all $j\notin\sigma(L_{1})$. In particular, as an important property, we have
that the boundary condition (5.34) implies the condition $({\bf H}3)$, as well
as $\mathcal{M}_{0}=\emptyset$, where $\mathcal{M}_{0}$ defined in (5.23). ∎

##  6 Estimation of the difference between EEC and the upper bound

In this section, we shall show that the difference between the expected number
of extended outward local maxima, i.e. the upper bound in (4.2), and the
expected Euler characteristic of the excursion set is super-exponentially
small.

######  Proposition 6.1.

Let $\\{(X(t),Y(s)):t\in T,s\in S\\}$ be an ${\mathbb{R}}^{2}$-valued,
centered, unit-variance Gaussian vector field satisfying $({\bf H}1)$, $({\bf
H}2)$ and $({\bf H}3)$. Then there exists $\alpha>0$ such that for any
$K\in\partial_{k}T$ and $L\in\partial_{l}S$ with $k,l\geq 0$, as $u\to\infty$,

$\begin{split}&\quad{\mathbb{E}}\\{M_{u}^{E}(X,K)M_{u}^{E}(Y,L)\\}\\\
&=(-1)^{k+l}\int_{K}\int_{L}{\mathbb{E}}\big{\\{}{\rm
det}\nabla^{2}X_{|K}(t){\rm det}\nabla^{2}Y_{|L}(s)\mathbbm{1}_{\\{X(t)\geq
u,\ \varepsilon^{*}_{\ell}X_{\ell}(t)\geq 0\ {\rm for\ all}\
\ell\notin\sigma(K)\\}}\\\ &\quad\times\mathbbm{1}_{\\{Y(s)\geq u,\
\varepsilon^{*}_{\ell}Y_{\ell}(s)\geq 0\ {\rm for\ all}\
\ell\notin\sigma(L)\\}}\big{|}\nabla X_{|K}(t)=\nabla Y_{|L}(s)=0\big{\\}}\\\
&\quad\times p_{\nabla X_{|K}(t),\nabla
Y_{|L}(s)}(0,0)dtds+o\left(\exp\left\\{-\frac{u^{2}}{1+R}-\alpha
u^{2}\right\\}\right)\\\
&=(-1)^{k+l}{\mathbb{E}}\bigg{\\{}\bigg{(}\sum^{k}_{i=0}(-1)^{i}\mu_{i}(X,K)\bigg{)}\bigg{(}\sum^{l}_{j=0}(-1)^{j}\mu_{j}(Y,L)\bigg{)}\bigg{\\}}+o\left(\exp\left\\{-\frac{u^{2}}{1+R}-\alpha
u^{2}\right\\}\right).\end{split}$ (6.1)

###### Proof.

The second equality in (6.1) follows from the application of the Kac-Rice
theorem below:

$\begin{split}&\quad{\mathbb{E}}\bigg{\\{}\bigg{(}\sum^{k}_{i=0}(-1)^{i}\mu_{i}(X,K)\bigg{)}\bigg{(}\sum^{l}_{j=0}(-1)^{j}\mu_{j}(Y,L)\bigg{)}\bigg{\\}}\\\
&=\sum^{k}_{i=0}(-1)^{i}\sum^{l}_{j=0}(-1)^{j}\int_{K}\int_{L}dtds\,p_{\nabla
X_{|K}(t),\nabla Y_{|L}(s)}(0,0)\\\ &\quad\times{\mathbb{E}}\big{\\{}|{\rm
det}\nabla^{2}X_{|K}(t)||{\rm
det}\nabla^{2}Y_{|L}(s)|\mathbbm{1}_{\\{\text{index}(\nabla^{2}X_{|K}(t))=i\\}}\mathbbm{1}_{\\{\text{index}(\nabla^{2}X_{|L}(t))=j\\}}\\\
&\quad\times\mathbbm{1}_{\\{X(t)\geq u,\ \varepsilon^{*}_{\ell}X_{\ell}(t)\geq
0\ {\rm for\ all}\ \ell\notin\sigma(K)\\}}\mathbbm{1}_{\\{Y(s)\geq u,\
\varepsilon^{*}_{\ell}Y_{\ell}(s)\geq 0\ {\rm for\ all}\
\ell\notin\sigma(L)\\}}\big{|}\nabla X_{|K}(t)=\nabla Y_{|L}(s)=0\big{\\}}\\\
&=\int_{K}\int_{L}dtds\,p_{\nabla X_{|K}(t),\nabla
Y_{|L}(s)}(0,0){\mathbb{E}}\big{\\{}{\rm det}\nabla^{2}X_{|K}(t){\rm
det}\nabla^{2}Y_{|L}(s)\\\ &\quad\times\mathbbm{1}_{\\{X(t)\geq u,\
\varepsilon^{*}_{\ell}X_{\ell}(t)\geq 0\ {\rm for\ all}\
\ell\notin\sigma(K)\\}}\mathbbm{1}_{\\{Y(s)\geq u,\
\varepsilon^{*}_{\ell}Y_{\ell}(s)\geq 0\ {\rm for\ all}\
\ell\notin\sigma(L)\\}}\big{|}\nabla X_{|K}(t)=\nabla
Y_{|L}(s)=0\big{\\}},\end{split}$

where the last step is due to the fact $|{\rm
det}\nabla^{2}X_{|K}(t)|\mathbbm{1}_{\\{\text{index}(\nabla^{2}X_{|K}(t))=i\\}}=(-1)^{i}{\rm
det}\nabla^{2}X_{|K}(t)$.

To prove the first approximation in (6.1) and address the main idea, we first
deal with a special case when the two faces are both the interiors and then
prove the general cases.

Case (i): $\bm{k=l=N}$. By the Kac-Rice metatheorem,

$\begin{split}&\quad{\mathbb{E}}\\{M_{u}^{E}(X,K)M_{u}^{E}(Y,L)\\}\\\
&=\int_{K}\int_{L}p_{\nabla X(t),\nabla
Y(s)}(0,0)dtds\int_{u}^{\infty}\int_{u}^{\infty}dxdy\,p_{X(t),Y(s)}(x,y|\nabla
X(t)=\nabla Y(s)=0)\\\ &\times{\mathbb{E}}\big{\\{}{\rm det}\nabla^{2}X(t){\rm
det}\nabla^{2}Y(s)\mathbbm{1}_{\\{\nabla^{2}X(t)\prec
0\\}}\mathbbm{1}_{\\{\nabla^{2}Y(s)\prec 0\\}}\big{|}X(t)=x,Y(s)=y,\nabla
X(t)=\nabla Y(s)=0\big{\\}}\\\ &:=\int_{K}\int_{L}p_{\nabla X(t),\nabla
Y(s)}(0,0)dtds\int_{u}^{\infty}\int_{u}^{\infty}A(t,s,x,y)dxdy.\end{split}$

Let

$\begin{split}\mathcal{M}_{1}&=\\{(t,s)\in\bar{K}\times\bar{L}:r(t,s)=R,\
{\mathbb{E}}\\{X(t)\nabla Y(s)\\}={\mathbb{E}}\\{Y(s)\nabla X(t)\\}=0\\},\\\
B(\mathcal{M}_{1},\delta_{1})&=\left\\{(t,s)\in K\times
L:d\left((t,s),\mathcal{M}_{1}\right)\leq\delta_{1}\right\\},\end{split}$
(6.2)

where $\delta_{1}$ is a small positive number to be specified. Then, we only
need to estimate

$\begin{split}\int_{B(\mathcal{M}_{1},\delta_{1})}p_{\nabla X(t),\nabla
Y(s)}(0,0)dtds\int_{u}^{\infty}\int_{u}^{\infty}A(t,s,x,y)dxdy,\end{split}$
(6.3)

since the integral above with $B(\mathcal{M}_{1},\delta_{1})$ replaced by
$(K\times L)\backslash B(\mathcal{M}_{1},\delta_{1})$ is super-exponentially
small due to the fact

$\sup_{(t,s)\in(K\times L)\backslash B(\mathcal{M}_{1},\delta_{1})}{\rm
Var}([X(t)+Y(s)]/2|\nabla X(t)=\nabla Y(s)=0)<\frac{1+R}{2}.$

Notice that, for all $(t,s)\in\mathcal{M}_{1}$,
${\mathbb{E}}\\{X(t)\nabla^{2}X(t)\\}\prec 0$ and
${\mathbb{E}}\\{Y(s)\nabla^{2}Y(s)\\}\prec 0$ since $X(t)$ and $Y(s)$ have
unit-variance; and by $({\bf H}3)$ and Proposition 2.1,
${\mathbb{E}}\\{X(t)\nabla^{2}Y(s)\\}\preceq 0$ and
${\mathbb{E}}\\{Y(s)\nabla^{2}X(t)\\}\preceq 0$. Thus there exists
$\delta_{1}$ small enough such that
${\mathbb{E}}\\{[X(t)+Y(s)]\nabla^{2}Y(s)\\}\prec 0$ and
${\mathbb{E}}\\{[X(t)+Y(s)]\nabla^{2}X(t)\\}\prec 0$ for all $(t,s)\in
B(\mathcal{M}_{1},\delta_{1})$. In particular, let $\lambda_{0}$ be the
largest eigenvalue of ${\mathbb{E}}\\{[X(t)+Y(s)]\nabla^{2}X(t)\\}$ over
$B(\mathcal{M}_{1},\delta_{1})$, then $\lambda_{0}<0$ by the uniform
continuity. Also note that both ${\mathbb{E}}\\{X(t)\nabla Y(s)\\}$ and
${\mathbb{E}}\\{Y(s)\nabla X(t)\\}$ tend to 0 as $\delta_{1}\to 0$. Therefore,
as $\delta_{1}\to 0$,

$\begin{split}&\quad{\mathbb{E}}\\{X_{ij}(t)|X(t)=x,Y(s)=y,\nabla X(t)=\nabla
Y(s)=0\\}\\\
&=({\mathbb{E}}\\{X_{ij}(t)X(t)\\},{\mathbb{E}}\\{X_{ij}(t)Y(s)\\},{\mathbb{E}}\\{X_{ij}(t)X_{1}(t)\\},\ldots,{\mathbb{E}}\\{X_{ij}(t)X_{N}(t)\\},\\\
&\qquad{\mathbb{E}}\\{X_{ij}(t)Y_{1}(s)\\},\ldots,{\mathbb{E}}\\{X_{ij}(t)Y_{N}(s)\\})\left[{\rm
Cov}(X(t),Y(s),\nabla X(t),\nabla Y(s))\right]^{-1}\\\
&\quad\cdot(x,y,0,\ldots,0,0,\ldots,0)^{T}\\\
&=(1+o(1))({\mathbb{E}}\\{X_{ij}(t)X(t)\\},{\mathbb{E}}\\{X_{ij}(t)Y(s)\\})\left(\begin{array}[]{cc}1&R\\\
R&1\end{array}\right)^{-1}\left(\begin{array}[]{c}x\\\ y\end{array}\right)\\\
&=(1+o(1))\frac{{\mathbb{E}}\\{X_{ij}(t)X(t)\\}[x-Ry]+{\mathbb{E}}\\{X_{ij}(t)Y(s)\\}[y-Rx]}{1-R^{2}};\end{split}$
(6.4)

and similarly,

$\begin{split}&\quad{\mathbb{E}}\\{Y_{ij}(s)|X(t)=x,Y(s)=y,\nabla X(t)=\nabla
Y(s)=0\\}\\\
&=(1+o(1))\frac{{\mathbb{E}}\\{Y_{ij}(s)X(t)\\}[x-Ry]+{\mathbb{E}}\\{Y_{ij}(s)Y(s)\\}[y-Rx]}{1-R^{2}}.\end{split}$
(6.5)

By (6.4) and (6.5), there exists $0<\varepsilon_{0}<1-R$ such that for
$\delta_{1}$ small enough and all $(x,y)\in[u,\infty)^{2}$ with
$(\varepsilon_{0}+R)x<y<(\varepsilon_{0}+R)^{-1}x$ (so that
$x-Ry\geq\varepsilon_{0}u$ and $y-Rx\geq\varepsilon_{0}u$),

$\begin{split}\Sigma_{1}(t,s,x,y)&:={\mathbb{E}}\\{\nabla^{2}X(t)|X(t)=x,Y(s)=y,\nabla
X(t)=\nabla Y(s)=0\\}\prec 0\text{ and}\\\
\Sigma_{2}(t,s,x,y)&:={\mathbb{E}}\\{\nabla^{2}Y(s)|X(t)=x,Y(s)=y,\nabla
X(t)=\nabla Y(s)=0\\}\prec 0.\end{split}$

Let $\Delta_{1}(t,s,x,y)=\nabla^{2}X(t)-\Sigma_{1}(t,s,x,y)$ and
$\Delta_{2}(t,s,x,y)=\nabla^{2}Y(s)-\Sigma_{2}(t,s,x,y)$. Due to the following
decomposition,

$\begin{split}\\{u\leq x,y<\infty\\}&=\\{x\geq
u,\,y\geq(\varepsilon_{0}+R)^{-1}x\\}\cup\\{y\geq
u,\,x\leq(\varepsilon_{0}+R)^{-1}y\\}\\\ &\quad\cup\\{x\geq
u,\,u\vee(\varepsilon_{0}+R)x<y<(\varepsilon_{0}+R)^{-1}x\\},\end{split}$

we can write

$\begin{split}\int_{u}^{\infty}\int_{u}^{\infty}A(t,s,x,y)dxdy&=\int_{u}^{\infty}dx\int_{(\varepsilon_{0}+R)^{-1}x}^{\infty}A(t,s,x,y)dy+\int_{u}^{\infty}dy\int_{(\varepsilon_{0}+R)^{-1}y}^{\infty}A(t,s,x,y)dx\\\
&\quad+\int_{u}^{\infty}dx\int_{u\vee(\varepsilon_{0}+R)x}^{(\varepsilon_{0}+R)^{-1}y}A(t,s,x,y)dx,\end{split}$
(6.6)

where the first two integrals on right are super-exponentially small since
$(\varepsilon_{0}+R)^{-1}>1$ and

$\mathbbm{1}_{\\{X(t)\geq
u,Y(s)\geq(\varepsilon_{0}+R)^{-1}X(t)\\}}\vee\mathbbm{1}_{\\{Y(s)\geq
u,X(t)\geq(\varepsilon_{0}+R)^{-1}Y(s)\\}}\leq\mathbbm{1}_{\\{[X(t)+Y(s)]/2\geq[1+(\varepsilon_{0}+R)^{-1}]u/2\\}}.$

For the last integral in (6.6), we have

$\begin{split}&\quad\int_{u}^{\infty}dx\int_{u\vee(\varepsilon_{0}+R)x}^{(\varepsilon_{0}+R)^{-1}x}A(t,s,x,y)dy\\\
&=\int_{u}^{\infty}dx\int_{u\vee(\varepsilon_{0}+R)x}^{(\varepsilon_{0}+R)^{-1}x}dy\,p_{X(t),Y(s)}(x,y|\nabla
X(t)=\nabla Y(s)=0)\\\ &\times{\mathbb{E}}\big{\\{}{\rm
det}(\Delta_{1}(t,s,x,y)+\Sigma_{1}(t,s,x,y)){\rm
det}(\Delta_{2}(t,s,x,y)+\Sigma_{2}(t,s,x,y))\mathbbm{1}_{\\{\Delta_{1}(t,s,x,y)+\Sigma_{1}(t,s,x,y)\prec
0\\}}\\\ &\times\mathbbm{1}_{\\{\Delta_{2}(t,s,x,y)+\Sigma_{2}(t,s,x,y)\prec
0\\}}\big{|}X(t)=x,Y(s)=y,\nabla X(t)=\nabla Y(s)=0\big{\\}}\\\
&:=\int_{u}^{\infty}dx\int_{u\vee(\varepsilon_{0}+R)x}^{(\varepsilon_{0}+R)^{-1}x}dy\,p_{X(t),Y(s)}(x,y|\nabla
X(t)=\nabla Y(s)=0)E(t,s,x,y).\end{split}$ (6.7)

Note that the following are two centered Gaussian random matrices (free of $x$
and $y$):

$\begin{split}\Omega^{X}(t,s)&=(\Omega_{ij}^{X}(t,s))_{1\leq i,j\leq
N}=(\Delta_{1}(t,s,x,y)|X(t)=x,Y(s)=y,\nabla X(t)=\nabla Y(s)=0),\\\
\Omega^{Y}(t,s)&=(\Omega_{ij}^{Y}(t,s))_{1\leq i,j\leq
N}=(\Delta_{2}(t,s,x,y)|X(t)=x,Y(s)=y,\nabla X(t)=\nabla Y(s)=0).\end{split}$

Denote the density of the Gaussian vector $((\Omega_{ij}^{X}(t,s))_{1\leq
i\leq j\leq N},(\Omega_{ij}^{Y}(t,s))_{1\leq i\leq j\leq N})$ by
$h_{t,s}(v,w)$, where $v=(v_{ij})_{1\leq i\leq j\leq N}$, $w=(w_{ij})_{1\leq
i\leq j\leq N}\in{\mathbb{R}}^{N(N+1)/2}$. Then

$\begin{split}E(t,s,x,y)&={\mathbb{E}}\big{\\{}{\rm
det}(\Omega^{X}(t,s)+\Sigma_{1}(t,s,x,y)){\rm
det}(\Omega^{Y}(t,s)+\Sigma_{2}(t,s,x,y))\\\
&\quad\times\mathbbm{1}_{\\{\Omega^{X}(t,s)+\Sigma_{1}(t,s,x,y)\prec
0\\}}\mathbbm{1}_{\\{\Omega^{Y}(t,s)+\Sigma_{2}(t,s,x,y)\prec
0\\}}\big{\\}}\\\ &=\int_{v:\,(v_{ij})+\Sigma_{1}(t,s,x,y)\prec
0}\int_{w:\,(w_{ij})+\Sigma_{2}(t,s,x,y)\prec 0}{\rm
det}((v_{ij})+\Sigma_{1}(t,s,x,y))\\\ &\quad\times{\rm
det}((w_{ij})+\Sigma_{2}(t,s,x,y))h_{t,s}(v,w)dvdw,\end{split}$ (6.8)

where $(v_{ij})$ and $(w_{ij})$ are respectively the abbreviations of the
matrices $v=(v_{ij})_{1\leq i,j\leq N}$ and $w=(w_{ij})_{1\leq i,j\leq N}$.
Recall that $x\wedge y\geq u$ and
$(\varepsilon_{0}+R)x<y<(\varepsilon_{0}+R)^{-1}x$ implies
$x-Ry\geq\varepsilon_{0}u$ and $y-Rx\geq\varepsilon_{0}u$. By (6.4), there
exists a constant $0<c<-\lambda_{0}\varepsilon_{0}/(1-R^{2})$ such that for
$\delta_{1}$ small enough and all $(t,s)\in B(\mathcal{M}_{1},\delta_{1})$,
$x\geq u$ and $u\vee(\varepsilon_{0}+R)x<y<(\varepsilon_{0}+R)^{-1}x$,

$(v_{ij})+\Sigma_{1}(t,s,x,y)\prec
0,\quad\forall\|(v_{ij})\|:=\Big{(}\sum_{i,j=1}^{N}v_{ij}^{2}\Big{)}^{1/2}<cu.$

Thus $\\{v:\,(v_{ij})+\Sigma_{1}(t,s,x,y)\not\prec
0\\}\subset\\{v:\,\|(v_{ij})\|\geq cu\\}$. This implies that the last integral
in (6.8) with the integration domain replaced by
$\\{(v,w):\,(v_{ij})+\Sigma_{1}(t,s,x,y)\not\prec
0,\,w\in{\mathbb{R}}^{N(N+1)/2}\\}$ is $o(e^{-\alpha^{\prime}u^{2}})$
uniformly for all $(t,s)\in B(\mathcal{M}_{1},\delta_{1})$, where
$\alpha^{\prime}$ is a positive constant. The same result holds when replacing
the integration domain by
$\\{(v,w):\,v\in{\mathbb{R}}^{N(N+1)/2},\,(w_{ij})+\Sigma_{2}(t,s,x,y)\not\prec
0\\}$. Therefore, we have that, uniformly for all $(t,s)\in
B(\mathcal{M}_{1},\delta_{1})$, $x\geq u$ and
$u\vee(\varepsilon_{0}+R)x<y<(\varepsilon_{0}+R)^{-1}x$,

$\begin{split}E(t,s,x,y)&=\int_{{\mathbb{R}}^{N(N+1)/2}}\int_{{\mathbb{R}}^{N(N+1)/2}}{\rm
det}((v_{ij})+\Sigma_{1}(t,s,x,y))\\\ &\quad\times{\rm
det}((w_{ij})+\Sigma_{2}(t,s,x,y))h_{t,s}(v,w)dvdw+o(e^{-\alpha^{\prime}u^{2}}).\end{split}$

Plugging this into (6.7) and (6.6), we obtain that the indicator functions
$\mathbbm{1}_{\\{\nabla^{2}X(t)\prec 0\\}}$ and
$\mathbbm{1}_{\\{\nabla^{2}Y(s)\prec 0\\}}$ in (6.3) can be removed, causing
only a super-exponentially small error. Therefore, there exists $\alpha>0$
such that for $u$ large enough,

$\begin{split}&\quad{\mathbb{E}}\\{M_{u}^{E}(X,K)M_{u}^{E}(Y,L)\\}\\\
&=\int_{K}\int_{L}p_{\nabla X(t),\nabla
Y(s)}(0,0)dtds\int_{u}^{\infty}\int_{u}^{\infty}p_{X(t),Y(s)}(x,y|\nabla
X(t)=\nabla Y(s)=0)\\\ &\quad\times{\mathbb{E}}\\{{\rm det}\nabla^{2}X(t){\rm
det}\nabla^{2}Y(s)|X(t)=x,Y(s)=y,\nabla X(t)=\nabla Y(s)=0\\}dxdy\\\
&\quad+o\Big{(}\exp\Big{\\{}-\frac{u^{2}}{1+R}-\alpha
u^{2}\Big{\\}}\Big{)}.\end{split}$

Case (ii): $\bm{k,l\geq 0}$. Note that, if $k=0$ or $l=0$, then by the Kac-
Rice formula, the terms in (6.1) involving the Hessian will vanish, making the
proof easier. Therefore, without loss of generality, let $k,l\geq 1$,
$\sigma(K)=\\{1,\cdots,k\\}$, $\sigma(L)=\\{1,\cdots,l\\}$ and assume all the
elements in $\varepsilon(K)$ and $\varepsilon(L)$ are 1. By the Kac-Rice
metatheorem,

$\begin{split}&\quad{\mathbb{E}}\\{M_{u}^{E}(X,K)M_{u}^{E}(Y,L)\\}\\\
&=(-1)^{k+l}\int_{K}\int_{L}p_{\nabla X_{|K}(t),\nabla
Y_{|L}(s)}(0,0)dtds\int_{u}^{\infty}\int_{u}^{\infty}p_{X(t),Y(s)}\big{(}x,y\big{|}\nabla
X_{|K}(t)=\nabla Y_{|L}(s)=0\big{)}\\\ &\quad{\mathbb{E}}\big{\\{}{\rm
det}\nabla^{2}X_{|K}(t){\rm
det}\nabla^{2}Y_{|L}(s)\mathbbm{1}_{\\{\nabla^{2}X_{|K}(t)\prec
0\\}}\mathbbm{1}_{\\{\nabla^{2}Y_{|L}(s)\prec
0\\}}\mathbbm{1}_{\\{X_{k+1}(t)>0,\ldots,X_{N}(t)>0\\}}\\\
&\quad\times\mathbbm{1}_{\\{Y_{l+1}(s)>0,\ldots,Y_{N}(s)>0\\}}\big{|}X(t)=x,Y(s)=y,\nabla
X_{|K}(t)=\nabla Y_{|L}(s)=0\big{\\}}dxdy\\\
&:=(-1)^{k+l}\int_{K}\int_{L}p_{\nabla X_{|K}(t),\nabla
Y_{|L}(s)}(0,0)dtds\int_{u}^{\infty}\int_{u}^{\infty}A^{\prime}(t,s,x,y)dxdy.\end{split}$

Let

$\begin{split}\mathcal{M}_{2}&=\\{(t,s)\in\bar{K}\times\bar{L}:r(t,s)=R,\
{\mathbb{E}}\\{X(t)\nabla Y_{|L}(s)\\}={\mathbb{E}}\\{Y(s)\nabla
X_{|K}(t)\\}=0\\},\\\ B(\mathcal{M}_{2},\delta_{2})&=\left\\{(t,s)\in K\times
L:d\left((t,s),\mathcal{M}_{2}\right)\leq\delta_{2}\right\\},\end{split}$
(6.9)

where $\delta_{2}$ is a small positive number to be specified. Then, we only
need to estimate

$\begin{split}\int_{B(\mathcal{M}_{2},\delta_{2})}p_{\nabla X_{|K}(t),\nabla
Y_{|L}(s)}(0,0)dtds\int_{u}^{\infty}\int_{u}^{\infty}A^{\prime}(t,s,x,y)dxdy,\end{split}$
(6.10)

since the integral above with $B(\mathcal{M}_{2},\delta_{2})$ replaced by
$(K\times L)\backslash B(\mathcal{M}_{2},\delta_{2})$ is super-exponentially
small due to the fact

$\sup_{(t,s)\in(K\times L)\backslash B(\mathcal{M}_{2},\delta_{2})}{\rm
Var}([X(t)+Y(s)]/2|\nabla X(t)=\nabla Y(s)=0)<\frac{1+R}{2}.$

On the other hand, following similar arguments in the proof for Case (i), we
verify that removing the indicator functions
$\mathbbm{1}_{\\{\nabla^{2}X_{|K}(t)\prec 0\\}}$ and
$\mathbbm{1}_{\\{\nabla^{2}Y_{|L}(s)\prec 0\\}}$ in (6.10) will only cause a
super-exponentially small error. Combining these results, we have shown that
the first approximation in (6.1) holds, completing the proof. ∎

From the proof of Proposition 6.1, we see that the same arguments can be
applied to ${\mathbb{E}}\\{M_{u}(X,K)M_{u}(Y,L)\\}$, yielding the following
result.

######  Proposition 6.2.

Let $\\{(X(t),Y(s)):t\in T,s\in S\\}$ be an ${\mathbb{R}}^{2}$-valued,
centered, unit-variance Gaussian vector field satisfying $({\bf H}1)$, $({\bf
H}2)$ and $({\bf H}3)$. Then there exists a constant $\alpha>0$ such that for
any $K\in\partial_{k}T$ and $L\in\partial_{l}S$, as $u\to\infty$,

$\begin{split}&\quad{\mathbb{E}}\\{M_{u}(X,K)M_{u}(Y,L)\\}\\\
&=(-1)^{k+l}\int_{K}\int_{L}{\mathbb{E}}\big{\\{}{\rm
det}\nabla^{2}X_{|K}(t){\rm det}\nabla^{2}Y_{|L}(s)\mathbbm{1}_{\\{X(t)\geq
u,\ Y(s)\geq u\\}}\big{|}\nabla X_{|K}(t)=\nabla Y_{|L}(s)=0\big{\\}}\\\
&\quad\times p_{\nabla X_{|K}(t),\nabla
Y_{|L}(s)}(0,0)dtds+o\left(\exp\left\\{-\frac{u^{2}}{1+R}-\alpha
u^{2}\right\\}\right)\\\
&=(-1)^{k+l}{\mathbb{E}}\bigg{\\{}\bigg{(}\sum^{k}_{i=0}(-1)^{i}\widetilde{\mu}_{i}(X,K)\bigg{)}\bigg{(}\sum^{l}_{j=0}(-1)^{j}\widetilde{\mu}_{j}(Y,L)\bigg{)}\bigg{\\}}+o\left(\exp\left\\{-\frac{u^{2}}{1+R}-\alpha
u^{2}\right\\}\right).\end{split}$

##  7 Proofs of the main results

###### Proof of Theorem 3.1.

By Lemmas 5.4, 5.3 and 5.5, together with the fact $M_{u}^{E}(X,K)\leq
M_{u}(X,K)$, we obtain that the factorial moments and the last two sums in
(4.3) are super-exponentially small. It then follows from (4.2) and (4.3)
that, there exists a constant $\alpha>0$ such that as $u\to\infty$,

$\begin{split}{\mathbb{P}}&\left\\{\sup_{t\in T}X(t)\geq u,\sup_{s\in
S}Y(s)\geq u\right\\}\\\
&=\sum_{k,l=0}^{N}\sum_{K\in\partial_{k}T,\,L\in\partial_{l}S}{\mathbb{E}}\\{M_{u}^{E}(X,K)M_{u}^{E}(Y,L)\\}+o\left(\exp\left\\{-\frac{u^{2}}{1+R}-\alpha
u^{2}\right\\}\right).\end{split}$

The desired result is thus an immediate consequence of Proposition 6.1 and
(3.2). ∎

###### Proof of Theorem 3.2.

By Remark 4, both inequalities (4.2) and (4.3) still hold with
$M_{u}^{E}(\cdot)$ replaced by $M_{u}(\cdot)$. Therefore, the corresponding
factorial moments and the last two sums in (4.3) with $M_{u}^{E}(\cdot)$
replaced by $M_{u}(\cdot)$ are super-exponentially small by Lemmas 5.4, 5.3
and 5.6. Consequently, there exists a constant $\alpha>0$ such that as
$u\to\infty$,

$\begin{split}{\mathbb{P}}&\left\\{\sup_{t\in T}X(t)\geq u,\sup_{s\in
S}Y(s)\geq u\right\\}\\\
&=\sum_{k,l=0}^{N}\sum_{K\in\partial_{k}T,\,L\in\partial_{l}S}{\mathbb{E}}\\{M_{u}(X,K)M_{u}(Y,L)\\}+o\left(\exp\left\\{-\frac{u^{2}}{1+R}-\alpha
u^{2}\right\\}\right).\end{split}$

The desired result is thus an immediate consequence of Proposition 6.2 and
(3.2). ∎

###### Proof of Theorem 3.3.

Note that, in the proof of Theorem 3.1, we have seen that the points in
$\mathcal{M}_{2}$ defined in (6.9) make major contribution to the joint
excursion probability. That is, with up to a super-exponentially small error,
we can focus only on those product faces, say $J\times F$, whose closure
$\bar{J}\times\bar{F}$ contains the unique point $(t^{*},s^{*})$ with
$r(t^{*},s^{*})=R$ and satisfying
$\sigma(J)\subset\mathcal{I}^{R}_{X}(t^{*},s^{*})$ and
$\sigma(F)\subset\mathcal{I}^{R}_{Y}(t^{*},s^{*})$ (i.e., the partial
derivatives of $r$ are 0 at $(t^{*},s^{*})$ restricted on $J$ and $F$).
Specifically, let

$\begin{split}T^{*}&=\\{J\in\partial_{k}T:t^{*}\in\bar{J},\,\sigma(J)\subset\mathcal{I}^{R}_{X}(t^{*},s^{*}),\,k=0,\ldots,N\\},\\\
S^{*}&=\\{F\in\partial_{\ell}S:s^{*}\in\bar{F},\,\sigma(F)\subset\mathcal{I}^{R}_{Y}(t^{*},s^{*}),\,\ell=0,\ldots,N\\};\end{split}$

and for each $J\in T^{*}$ and $F\in S^{*}$, let

$\begin{split}M_{u}^{E^{*}}(X,J)&:=\\#\\{t\in J:X(t)\geq u,\nabla
X_{|J}(t)=0,\nabla^{2}X_{|J}(t)\prec 0,\\\
&\qquad\qquad\qquad\varepsilon^{*}_{j}X_{j}(t)\geq 0\ {\rm for\ all}\
j\in\mathcal{I}^{R}_{X}(t^{*},s^{*})\setminus\sigma(J)\\},\\\
M_{u}^{E^{*}}(Y,F)&:=\\#\\{s\in F:Y(s)\geq u,\nabla
Y_{|F}(s)=0,\nabla^{2}Y_{|F}(s)\prec 0,\\\
&\qquad\qquad\qquad\varepsilon^{*}_{j}Y_{j}(s)\geq 0\ {\rm for\ all}\
j\in\mathcal{I}^{R}_{Y}(t^{*},s^{*})\setminus\sigma(F)\\}.\end{split}$

Note that, both inequalities (4.2) and (4.3) hold with $M_{u}^{E}(\cdot)$
replaced by $M_{u}^{E^{*}}(\cdot)$ when the corresponding face therein belongs
to $T^{*}$ or $S^{*}$, and replaced by $M_{u}(\cdot)$ otherwise. Following
similar arguments in deriving Theorems 3.1 and 3.2, we obtain that, there
exists $\alpha>0$ such that as $u\to\infty$,

$\begin{split}{\mathbb{P}}&\left\\{\sup_{t\in T}X(t)\geq u,\sup_{s\in
S}Y(s)\geq u\right\\}\\\ &=\sum_{J\in T^{*},\,F\in
S^{*}}{\mathbb{E}}\\{M_{u}^{E^{*}}(X,J)M_{u}^{E^{*}}(Y,F)\\}+o\left(\exp\left\\{-\frac{u^{2}}{1+R}-\alpha
u^{2}\right\\}\right).\end{split}$

The desired result then follows from Proposition 6.1. ∎

##  8 Examples

Throughout this section, we assume that $\\{(X(t),Y(s)):t\in T,s\in S\\}$,
where $T=S=[0,1]$, is an ${\mathbb{R}}^{2}$-valued, centered, unit-variance
Gaussian vector process satisfying $({\bf H}1)$, $({\bf H}2)$ and $({\bf
H}3)$.

###  8.1 Example with correlation attaining the maximum at a unique point

Suppose $r(t,s)$ attains the maximum $R$ only at a point $(t^{*},s^{*})$,
i.e., $r(t^{*},s^{*})=R$. Let

$\begin{split}\lambda_{1}(t)&={\rm Var}(X^{\prime}(t)),\ \lambda_{2}(s)={\rm
Var}(Y^{\prime}(s)),\ r_{1}(t,s)={\mathbb{E}}\\{X^{\prime}(t)Y(s)\\},\
r_{2}(t,s)={\mathbb{E}}\\{X(t)Y^{\prime}(s)\\},\\\
r_{11}(t,s)&={\mathbb{E}}\\{X^{\prime\prime}(t)Y(s)\\},\
r_{22}(t,s)={\mathbb{E}}\\{X(t)Y^{\prime\prime}(s)\\},\
r_{12}(t,s)={\mathbb{E}}\\{X^{\prime}(t)Y^{\prime}(s)\\}\\\
\lambda_{1}&=\lambda_{1}(t^{*}),\ \lambda_{2}=\lambda_{2}(s^{*}),\
R_{11}=r_{11}(t^{*},s^{*}),\ R_{22}=r_{22}(t^{*},s^{*}),\
R_{12}=r_{12}(t^{*},s^{*}).\end{split}$

Case 1: $\bm{(t^{*},s^{*})=(0,0)}$ and $\bm{r_{1}(0,0)r_{2}(0,0)\neq 0}$. By
Theorem 3.2,

$\begin{split}{\mathbb{P}}\left\\{\sup_{t\in T}X(t)\geq u,\sup_{s\in
S}Y(s)\geq u\right\\}&={\mathbb{P}}\\{X(0)\geq u,Y(0)\geq
u\\}+o\left(\exp\left\\{-\frac{u^{2}}{1+R}-\alpha u^{2}\right\\}\right)\\\
&=\frac{(1+R)^{2}}{2\pi\sqrt{1-R^{2}}}\frac{1}{u^{2}}e^{-\frac{u^{2}}{1+R}}(1+o(1)),\end{split}$

where the last line is due to a well-know asymptotics for
${\mathbb{P}}\\{X(0)\geq u,Y(0)\geq u\\}$, see [11].

Case 2: $\bm{(t^{*},s^{*})=(0,0)}$, $\bm{r_{1}(0,0)=0}$ and
$\bm{r_{2}(0,0)\neq 0}$. By Theorem 3.3,

$\begin{split}{\mathbb{P}}&\left\\{\sup_{t\in T}X(t)\geq u,\sup_{s\in
S}Y(s)\geq u\right\\}\\\ &={\mathbb{P}}\\{X(0)\geq u,Y(0)\geq
u,X^{\prime}(0)<0\\}+I(u)+o\left(\exp\left\\{-\frac{u^{2}}{1+R}-\alpha
u^{2}\right\\}\right),\end{split}$ (8.1)

where

$\begin{split}I(u)&=(-1)\int_{0}^{1}p_{X^{\prime}(t)}(0)dt\int_{u}^{\infty}\int_{u}^{\infty}p_{X(t),Y(0)}(x,y|X^{\prime}(t)=0)\\\
&\quad\times{\mathbb{E}}\\{X^{\prime\prime}(t)|X(t)=x,Y(0)=y,X^{\prime}(t)=0\\}dxdy.\end{split}$

Since $X^{\prime}(0)$ is independent of both $X(0)$ and $Y(0)$, we have

$\begin{split}{\mathbb{P}}\\{X(0)\geq u,Y(0)\geq
u,X^{\prime}(0)<0\\}=\frac{(1+R)^{2}}{4\pi\sqrt{1-R^{2}}}\frac{1}{u^{2}}e^{-\frac{u^{2}}{1+R}}(1+o(1)).\end{split}$
(8.2)

Let $\Sigma(t)=(\Sigma_{ij}(t))_{i,j=1,2}={\rm
Cov}((X(t),Y(0))|X^{\prime}(t)=0)$, implying $\Sigma_{11}(t)=1$,
$\Sigma_{22}(t)=1-r_{1}^{2}(t,0)/\lambda_{1}(t)$ and
$\Sigma_{12}(t)=\Sigma_{21}(t)=r(t,0)$. Then

$\begin{split}I(u)&=(-1)\int_{0}^{1}\frac{1}{\sqrt{2\pi\lambda(t)}}dt\int_{0}^{\infty}\int_{0}^{\infty}\frac{e^{-\frac{1}{2}(x+u,\,y+u)\Sigma(t)^{-1}(x+u,\,y+u)^{T}}}{2\pi\sqrt{{\rm
det}(\Sigma(t))}}\\\
&\quad\times{\mathbb{E}}\\{X^{\prime\prime}(t)|X(t)=x+u,Y(0)=y+u,X^{\prime}(t)=0\\}dxdy,\end{split}$

where the expectation can be written as $f(t)u+g(t)$ such that

$\begin{split}f(0)=(-\lambda_{1},R_{11})\left(\begin{array}[]{cc}1&R\\\
R&1\end{array}\right)^{-1}\left(\begin{array}[]{c}1\\\
1\end{array}\right)=\frac{R_{11}-\lambda_{1}}{1+R}.\end{split}$

By Theorem 7.5.3 in Tong (1990), as $u\to\infty$, the Mills ratio

$\begin{split}&\int_{0}^{\infty}\int_{0}^{\infty}e^{-\frac{1}{2}(x,y)\Sigma(t)^{-1}(x,y)^{T}-(u,u)\Sigma(t)^{-1}(x,y)^{T}}dxdy\\\
&\sim\frac{1}{u^{2}[(\Sigma(t)^{-1})_{11}+(\Sigma(t)^{-1})_{21}][(\Sigma(t)^{-1})_{12}+(\Sigma(t)^{-1})_{22}]}.\end{split}$
(8.3)

Therefore,

$\begin{split}I(u)&\sim(-1)\int_{0}^{1}\frac{1}{\sqrt{2\pi\lambda_{1}(t)}}\frac{1}{2\pi\sqrt{{\rm
det}(\Sigma(t))}}f(t)\frac{1}{u}e^{-\frac{1}{2}u^{2}(1,1)\Sigma(t)^{-1}(1,1)^{T}}\\\
&\quad\times\frac{1}{[(\Sigma(t)^{-1})_{11}+(\Sigma(t)^{-1})_{21}][(\Sigma(t)^{-1})_{12}+(\Sigma(t)^{-1})_{22}]}dt.\end{split}$

It can be checked that the function

$\begin{split}h(t):=\frac{1}{2}(1,1)\Sigma(t)^{-1}(1,1)^{T}=\frac{2-r_{1}(t,0)^{2}/\lambda_{1}(t)-2r(t,0)}{2[1-r_{1}(t,0)^{2}/\lambda_{1}(t)-r(t,0)^{2}]}\end{split}$

attains its minimum only at $0$ with $h(0)=1/(1+R)$ and
$h^{\prime\prime}(0)=R_{11}(R_{11}-\lambda_{1})/[\lambda_{1}(1+R)^{2}]$.
Applying the Laplace method (see, for example, Lemma A.3 in Cheng and Xiao
[7]), we obtain

$\begin{split}I(u)&\sim\frac{1}{2}\frac{1}{\sqrt{2\pi\lambda_{1}}}\frac{1}{2\pi\sqrt{1-R^{2}}}\frac{\lambda_{1}-R_{11}}{1+R}(1+R)^{2}\bigg{(}\frac{2\pi}{u^{2}}\frac{\lambda_{1}(1+R)^{2}}{R_{11}(R_{11}-\lambda_{1})}\bigg{)}^{1/2}\frac{1}{u}e^{-\frac{u^{2}}{1+R}}\\\
&=\frac{\sqrt{\lambda_{1}-R_{11}}}{2\sqrt{-R_{11}}}\frac{(1+R)^{2}}{2\pi\sqrt{1-R^{2}}}\frac{1}{u^{2}}e^{-\frac{u^{2}}{1+R}}.\end{split}$
(8.4)

Combining (8.1) with (8.2) and (8.4), we obtain

$\begin{split}{\mathbb{P}}\left\\{\sup_{t\in T}X(t)\geq u,\sup_{s\in
S}Y(s)\geq
u\right\\}=\left(\frac{1}{2}+\frac{\sqrt{\lambda_{1}-R_{11}}}{2\sqrt{-R_{11}}}\right)\frac{(1+R)^{2}}{2\pi\sqrt{1-R^{2}}}\frac{1}{u^{2}}e^{-\frac{u^{2}}{1+R}}(1+o(1)).\end{split}$

Case 3: $\bm{(t^{*},s^{*})=(0,0)}$ and $\bm{r_{1}(0,0)=r_{2}(0,0)\neq 0}$. By
Theorem 3.3,

$\begin{split}&{\mathbb{P}}\Big{\\{}\sup_{t\in T}X(t)\geq u,\sup_{s\in
S}Y(s)\geq u\Big{\\}}\\\ &={\mathbb{P}}\\{X(0)\geq u,Y(0)\geq
u,X^{\prime}(0)<0,Y^{\prime}(0)<0\\}\\\
&\quad+(-1)\int_{0}^{1}p_{X^{\prime}(t)}(0)dt\int_{u}^{\infty}\int_{u}^{\infty}\int_{-\infty}^{0}p_{X(t),Y(0),Y^{\prime}(0)}(x,y,z|X^{\prime}(t)=0)\\\
&\qquad\times{\mathbb{E}}\\{X^{\prime\prime}(t)|X(t)=x,Y(0)=y,Y^{\prime}(0)=z,X^{\prime}(t)=0\\}dxdydz\\\
&\quad+(-1)\int_{0}^{1}p_{Y^{\prime}(s)}(0)ds\int_{u}^{\infty}\int_{u}^{\infty}\int_{-\infty}^{0}p_{X(0),Y(s),X^{\prime}(0)}(x,y,z|Y^{\prime}(s)=0)\\\
&\qquad\times{\mathbb{E}}\\{Y^{\prime\prime}(s)|X(0)=x,Y(s)=y,X^{\prime}(0)=z,Y^{\prime}(s)=0\\}dxdydz\\\
&\quad+\int_{0}^{1}\int_{0}^{1}p_{X^{\prime}(t),Y^{\prime}(s)}(0,0)dtds\int_{u}^{\infty}\int_{u}^{\infty}p_{X(t),Y(s)}(x,y|X^{\prime}(t)=Y^{\prime}(s)=0)\\\
&\qquad\times{\mathbb{E}}\\{X^{\prime\prime}(t)Y^{\prime\prime}(s)|X(t)=x,Y(s)=y,X^{\prime}(t)=Y^{\prime}(s)=0\\}dxdy\\\
&\quad+o\Big{(}\exp\Big{\\{}-\frac{u^{2}}{1+R}-\alpha u^{2}\Big{\\}}\Big{)}\\\
&:=I_{1}(u)+I_{2}(u)+I_{3}(u)+I_{4}(u)+o\Big{(}\exp\Big{\\{}-\frac{u^{2}}{1+R}-\alpha
u^{2}\Big{\\}}\Big{)}.\end{split}$ (8.5)

Since $(X^{\prime}(0),Y^{\prime}(0))$, which has covariance matrix ${\rm
Var}(X^{\prime}(0))=\lambda_{1}$, ${\rm Var}(Y^{\prime}(0))=\lambda_{2}$ and
${\mathbb{E}}\\{X^{\prime}(0)Y^{\prime}(0)\\}=R_{12}$, is independent of
$(X(0),Y(0))$, we have

$\begin{split}I_{1}(u)={\mathbb{P}}\\{X^{\prime}(0)<0,Y^{\prime}(0)<0\\}\frac{(1+R)^{2}}{2\pi\sqrt{1-R^{2}}}\frac{1}{u^{2}}e^{-\frac{u^{2}}{1+R}}(1+o(1)).\end{split}$
(8.6)

Note that, if $R_{12}=0$, then
${\mathbb{P}}(X^{\prime}(0)<0,Y^{\prime}(0)<0)=1/4$. Similarly to (8.7), we
have

$\begin{split}I_{2}(u)&\sim\frac{\sqrt{\lambda_{1}-R_{11}}}{2\sqrt{-R_{11}}}\frac{(1+R)^{2}}{2\pi\sqrt{1-R^{2}}}\frac{1}{u^{2}}e^{-\frac{u^{2}}{1+R}}(1+o(1)),\\\
I_{3}(u)&\sim\frac{\sqrt{\lambda_{2}-R_{22}}}{2\sqrt{-R_{22}}}\frac{(1+R)^{2}}{2\pi\sqrt{1-R^{2}}}\frac{1}{u^{2}}e^{-\frac{u^{2}}{1+R}}(1+o(1)).\end{split}$
(8.7)

Let us compute $I_{4}$. Let $\Sigma(t,s)=(\Sigma_{ij}(t,s))_{i,j=1,2}={\rm
Cov}((X(t),Y(s))|X^{\prime}(t)=Y^{\prime}(s)=0)$, implying

$\begin{split}\Sigma_{11}(t,s)&=1-\frac{\lambda_{1}(t)r_{2}^{2}(t,s)}{\lambda_{1}(t)\lambda_{2}(s)-r_{12}^{2}(t,s)},\quad\Sigma_{22}(t,s)=1-\frac{\lambda_{2}(s)r_{1}^{2}(t,s)}{\lambda_{1}(t)\lambda_{2}(s)-r_{12}^{2}(t,s)},\\\
\Sigma_{12}(t,s)&=\Sigma_{21}(t,s)=r(t,s)+\frac{r_{12}(t,s)r_{1}(t,s)r_{2}(t,s)}{\lambda_{1}(t)\lambda_{2}(s)-r_{12}^{2}(t,s)}.\end{split}$

Then

$\begin{split}I_{4}(u)&=\int_{0}^{1}\int_{0}^{1}\frac{1}{2\pi\sqrt{\lambda_{1}(t)\lambda_{2}(s)-r_{12}^{2}(t,s)}}dtds\int_{0}^{\infty}\int_{0}^{\infty}\frac{e^{-\frac{1}{2}(x+u,y+u)\Sigma(t,s)^{-1}(x+u,y+u)^{T}}}{2\pi\sqrt{{\rm
det}(\Sigma(t,s))}}\\\
&\qquad\times{\mathbb{E}}\\{X^{\prime\prime}(t)Y^{\prime\prime}(s)|X(t)=x+u,Y(s)=y+u,X^{\prime}(t)=Y^{\prime}(s)=0\\}dxdy.\end{split}$
(8.8)

where the expectation is on the product of two non-centered (conditional)
Gaussian variables and hence its highest-order term in $u$ can be derived from
the product of the means of Gaussian variables. We can write
${\mathbb{E}}\\{X^{\prime\prime}(t)|X(t)=x+u,Y(s)=y+u,X^{\prime}(t)=Y^{\prime}(s)=0\\}=f(t,s)u+f_{0}(t,s,x,y)$
such that

$\begin{split}f(0,0)=(-\lambda_{1},R_{11})\left(\begin{array}[]{cc}1&R\\\
R&1\end{array}\right)^{-1}\left(\begin{array}[]{c}1\\\
1\end{array}\right)=\frac{R_{11}-\lambda_{1}}{1+R};\end{split}$

and write
${\mathbb{E}}\\{Y^{\prime\prime}(s)|X(t)=x+u,Y(s)=y+u,X^{\prime}(t)=Y^{\prime}(s)=0\\}=g(t,s)u+g_{0}(t,s,x,y)$
such that

$\begin{split}g(0,0)=(R_{22},-\lambda_{2})\left(\begin{array}[]{cc}1&R\\\
R&1\end{array}\right)^{-1}\left(\begin{array}[]{c}1\\\
1\end{array}\right)=\frac{R_{22}-\lambda_{2}}{1+R}.\end{split}$

Therefore, in the expectation in (8.8), the highest-order term in $u$
evaluated at $(0,0)$ is given by
$[(\lambda_{1}-R_{11})(\lambda_{2}-R_{22})/(1+R)^{2}]u^{2}$. Note that the
Mills ratio in (8.3) with $\Sigma(t)$ replaced by $\Sigma(t,s)$ is
asymptotically $(1+R)^{2}/u^{2}$ at $(t,s)=(0,0)$. Plugging these into (8.8)
yields

$\begin{split}I_{4}(u)&\sim\int_{0}^{1}\int_{0}^{1}\frac{1}{2\pi\sqrt{\lambda_{1}(t)\lambda_{2}(s)-r_{12}^{2}(t,s)}}\frac{1}{2\pi\sqrt{{\rm
det}(\Sigma(t,s))}}f(t,s)g(t,s)u^{2}\\\
&\qquad\times\frac{1}{u^{2}[(\Sigma(t,s)^{-1})_{11}+(\Sigma(t,s)^{-1})_{21}]^{2}}e^{-\frac{1}{2}u^{2}(1,1)\Sigma(t,s)^{-1}(1,1)^{T}}dtds.\end{split}$

Since

$\begin{split}h(t):=\frac{1}{2}(1,1)\Sigma(t,s)^{-1}(1,1)^{T}=\frac{1}{2}\frac{\Sigma_{11}(t,s)+\Sigma_{22}(t,s)-2\Sigma_{12}(t,s)}{\Sigma_{11}(t,s)\Sigma_{22}(t,s)-\Sigma_{12}^{2}(t,s)}\end{split}$

attains its minimum only at $(t,s)=(0,0)$ with $h(0)=1/(1+R)$ and

$\begin{split}&\nabla^{2}h(0,0)=\frac{1}{(1+R)^{2}(\lambda_{1}\lambda_{2}-R_{12}^{2})}\left(\begin{array}[]{cc}(\lambda_{1}-R_{11})(R_{12}^{2}-\lambda_{2}R_{11})&R_{12}(\lambda_{1}-R_{11})(R_{22}-\lambda_{2})\\\
R_{12}(\lambda_{1}-R_{11})(R_{22}-\lambda_{2})&(\lambda_{2}-R_{22})(R_{12}^{2}-\lambda_{1}R_{22})\end{array}\right),\\\
&{\rm
det}(\nabla^{2}h(0,0))=\frac{(\lambda_{1}-R_{11})(\lambda_{2}-R_{22})(R_{11}R_{22}-R_{12}^{2})}{(1+R)^{4}(\lambda_{1}\lambda_{2}-R_{12}^{2})}.\end{split}$

Applying the Laplace method (see Lemma A.3 in [7]) yields

$\begin{split}I_{4}(u)&\sim{\mathbb{P}}(Z_{1}>0,Z_{2}>0)\frac{1}{2\pi\sqrt{\lambda_{1}\lambda_{2}-R_{12}^{2}}}\frac{1}{2\pi\sqrt{1-R^{2}}}\frac{(\lambda_{1}-R_{11})(\lambda_{2}-R_{22})}{(1+R)^{2}}u^{2}\\\
&\qquad\times\frac{(1+R)^{2}}{u^{2}}\frac{2\pi}{u^{2}}\bigg{(}\frac{(1+R)^{4}(\lambda_{1}\lambda_{2}-R_{12}^{2})}{(\lambda_{1}-R_{11})(\lambda_{2}-R_{22})(R_{11}R_{22}-R_{12}^{2})}\bigg{)}^{1/2}e^{-\frac{u^{2}}{1+R}}\\\
&={\mathbb{P}}(Z_{1}>0,Z_{2}>0)\frac{\sqrt{(\lambda_{1}-R_{11})(\lambda_{2}-R_{22})}}{\sqrt{R_{11}R_{22}-R_{12}^{2}}}\frac{(1+R)^{2}}{2\pi\sqrt{1-R^{2}}}\frac{1}{u^{2}}e^{-\frac{u^{2}}{1+R}},\end{split}$
(8.9)

where $(Z_{1},Z_{2})$ is a centered bivariate Gaussian variable with
covariance $\nabla^{2}h(0,0)$. Plugging (8.6), (8.7) and (8.9) into (8.5), we
obtain

$\begin{split}&{\mathbb{P}}\Big{\\{}\sup_{t\in T}X(t)\geq u,\sup_{s\in
S}Y(s)\geq
u\Big{\\}}=\Bigg{[}{\mathbb{P}}(X^{\prime}(0)<0,Y^{\prime}(0)<0)+\frac{\sqrt{\lambda_{1}-R_{11}}}{2\sqrt{-R_{11}}}+\frac{\sqrt{\lambda_{2}-R_{22}}}{2\sqrt{-R_{22}}}\\\
&\quad+{\mathbb{P}}(Z_{1}>0,Z_{2}>0)\frac{\sqrt{(\lambda_{1}-R_{11})(\lambda_{2}-R_{22})}}{\sqrt{R_{11}R_{22}-R_{12}^{2}}}\Bigg{]}\frac{(1+R)^{2}}{2\pi\sqrt{1-R^{2}}}\frac{1}{u^{2}}e^{-\frac{u^{2}}{1+R}}(1+o(1)),\end{split}$

where the two probabilities on right become $1/4$ when $R_{12}=0$.

Case 4: $\bm{(t^{*},s^{*})=(t^{*},0)}$, where $\bm{t^{*}\in(0,1)}$ and
$\bm{r_{2}(t^{*},0)\neq 0}$. By Theorem 3.3 and similar arguments in Case 2,
we obtain

$\begin{split}{\mathbb{P}}\Big{\\{}\sup_{t\in T}X(t)\geq u,\sup_{s\in
S}Y(s)\geq
u\Big{\\}}=\frac{\sqrt{\lambda_{1}-R_{11}}}{\sqrt{-R_{11}}}\frac{(1+R)^{2}}{2\pi\sqrt{1-R^{2}}}\frac{1}{u^{2}}e^{-\frac{u^{2}}{1+R}}(1+o(1)).\end{split}$

Case 5: $\bm{(t^{*},s^{*})=(t^{*},0)}$, where $\bm{t^{*}\in(0,1)}$ and
$\bm{r_{2}(t^{*},0)=0}$. By Theorem 3.3 and similar arguments in Case 3, we
obtain

$\begin{split}{\mathbb{P}}\Big{\\{}\sup_{t\in T}X(t)\geq u,\sup_{s\in
S}Y(s)\geq
u\Big{\\}}&=\Bigg{[}\frac{\sqrt{\lambda_{1}-R_{11}}}{\sqrt{-R_{11}}}+\frac{\sqrt{(\lambda_{1}-R_{11})(\lambda_{2}-R_{22})}}{2\sqrt{R_{11}R_{22}-R_{12}^{2}}}\Bigg{]}\\\
&\quad\times\frac{(1+R)^{2}}{2\pi\sqrt{1-R^{2}}}\frac{1}{u^{2}}e^{-\frac{u^{2}}{1+R}}(1+o(1)).\end{split}$

Case 6: $\bm{(t^{*},s^{*})\in(0,1)^{2}}$. By Theorem 3.3 and similar arguments
in Case 3, we obtain

$\begin{split}{\mathbb{P}}\Big{\\{}\sup_{t\in T}X(t)\geq u,\sup_{s\in
S}Y(s)\geq
u\Big{\\}}=\frac{\sqrt{(\lambda_{1}-R_{11})(\lambda_{2}-R_{22})}}{\sqrt{R_{11}R_{22}-R_{12}^{2}}}\frac{(1+R)^{2}}{2\pi\sqrt{1-R^{2}}}\frac{1}{u^{2}}e^{-\frac{u^{2}}{1+R}}(1+o(1)).\end{split}$

###  8.2 Examples with correlation attaining the maximum on a line

Here we consider the bivariate Gaussian random fields in Zhou and Xiao [20],
where the smooth case was not studied since the double sum method therein is
not applicable. Let $X(t)$ and $Y(s)$ be smooth stationary Gaussian processes
with covariances satisfying

$\begin{split}{\mathbb{E}}\\{X(0)X(t)\\}&=1-\frac{\lambda_{1}}{2}|t|^{2}(1+o(1)),\quad{\rm
as}\ |t|\to 0,\\\
{\mathbb{E}}\\{Y(0)Y(s)\\}&=1-\frac{\lambda_{2}}{2}|s|^{2}(1+o(1)),\quad{\rm
as}\ |s|\to 0,\end{split}$

which implies ${\rm
Var}(X^{\prime}(t))=-{\mathbb{E}}\\{X(t)X^{\prime\prime}(t)\\}=\lambda_{1}$
and ${\rm
Var}(Y^{\prime}(s))=-{\mathbb{E}}\\{Y(s)Y^{\prime\prime}(s)\\}=\lambda_{2}$.
Assume that the correlation of $X$ and $Y$ satisfies

$r(t,s)={\mathbb{E}}\\{X(t)Y(s)\\}=\rho(|t-s|),\quad\forall t,s\in[0,1],$

where $\rho$ is a real function. Suppose $\rho$ attains its maximum $R$ only
at 0 with $\rho^{\prime}(0)=0$ and $\rho^{\prime\prime}(0)<0$. This indicates
that the maximum correlation $R$ is only achieved on the diagonal line
$\\{t=s:0\leq t,s\leq 1\\}$. By Theorem 3.1, we have

$\begin{split}&\quad{\mathbb{P}}\Big{\\{}\sup_{t\in[0,1]}X(t)\geq
u,\sup_{s\in[0,1]}Y(s)\geq u\Big{\\}}\\\
&=\int_{0}^{1}\int_{0}^{1}p_{X^{\prime}(t),Y^{\prime}(s)}(0,0)dtds\int_{u}^{\infty}\int_{u}^{\infty}p_{X(t),Y(s)}(x,y|X^{\prime}(t)=Y^{\prime}(s)=0)\\\
&\qquad\times{\mathbb{E}}\\{X^{\prime\prime}(t)Y^{\prime\prime}(s)|X(t)=x,Y(s)=y,X^{\prime}(t)=Y^{\prime}(s)=0\\}dxdy\\\
&\quad+{\mathbb{P}}\\{X(0)\geq u,Y(0)\geq
u,X^{\prime}(0)<0,Y^{\prime}(0)<0\\}\\\ &\quad+{\mathbb{P}}\\{X(1)\geq
u,Y(1)\geq
u,X^{\prime}(1)>0,Y^{\prime}(1)>0\\}+o\Big{(}\exp\Big{\\{}-\frac{u^{2}}{1+R}-\alpha
u^{2}\Big{\\}}\Big{)}\\\ &=I(u)(1+o(1)),\end{split}$

where $I(u)$ denotes the integral term in the second and third lines. We shall
derive below the asymptotics of $I(u)$ which gives the highest-order term in
$u$. By the stationarity and change of variables (using $z=s$ and $w=t-s$ for
$0<s<t<1$ and the symmetry property),

$\begin{split}I(u)&=\int_{0}^{1}\int_{0}^{1}p_{X^{\prime}(0),Y^{\prime}(|t-s|)}(0,0)dtds\int_{u}^{\infty}\int_{u}^{\infty}p_{X(0),Y(|t-s|)}(x,y|X^{\prime}(0)=Y^{\prime}(|t-s|)=0)\\\
&\qquad\times{\mathbb{E}}\\{X^{\prime\prime}(0)Y^{\prime\prime}(|t-s|)|X(0)=x,Y(|t-s|)=y,X^{\prime}(0)=Y^{\prime}(|t-s|)=0\\}dxdy\\\
&=2\int_{0}^{1}(1-t)p_{X^{\prime}(0),Y^{\prime}(t)}(0,0)dt\int_{u}^{\infty}\int_{u}^{\infty}p_{X(0),Y(t)}(x,y|X^{\prime}(0)=Y^{\prime}(t)=0)\\\
&\qquad\times{\mathbb{E}}\\{X^{\prime\prime}(0)Y^{\prime\prime}(t)|X(0)=x,Y(t)=y,X^{\prime}(0)=Y^{\prime}(t)=0\\}dxdy\\\
&:=2I_{0}(u).\end{split}$

Let $\Sigma(t)=(\Sigma_{ij}(t))_{i,j=1,2}={\rm
Cov}((X(0),Y(t))|X^{\prime}(0)=Y^{\prime}(t)=0)$, implying

$\begin{split}\Sigma_{11}(t)&=1-\frac{\lambda_{1}\rho^{\prime}(t)^{2}}{\lambda_{1}\lambda_{2}-\rho^{\prime\prime}(t)^{2}},\quad\Sigma_{22}(t)=1-\frac{\lambda_{2}\rho^{\prime}(t)^{2}}{\lambda_{1}\lambda_{2}-\rho^{\prime\prime}(t)^{2}},\\\
\Sigma_{12}(t)&=\Sigma_{21}(t)=\rho(t)+\frac{\rho^{\prime\prime}(t)\rho^{\prime}(t)^{2}}{\lambda_{1}\lambda_{2}-\rho^{\prime\prime}(t)^{2}}.\end{split}$

Then

$\begin{split}I_{0}(u)&=\int_{0}^{1}\frac{1-t}{2\pi\sqrt{\lambda_{1}\lambda_{2}-\rho^{\prime\prime}(t)^{2}}}dt\int_{0}^{\infty}\int_{0}^{\infty}\frac{1}{2\pi\sqrt{{\rm
det}(\Sigma(t))}}e^{-\frac{1}{2}(x+u,y+u)\Sigma(t)^{-1}(x+u,y+u)^{T}}\\\
&\qquad\times{\mathbb{E}}\\{X^{\prime\prime}(0)Y^{\prime\prime}(t)|X(0)=x+u,Y(t)=y+u,X^{\prime}(0)=Y^{\prime}(t)=0\\}dxdy.\end{split}$
(8.10)

We have
${\mathbb{E}}\\{X^{\prime\prime}(0)|X(0)=x+u,Y(t)=y+u,X^{\prime}(0)=Y^{\prime}(t)=0\\}=f(t)u+f_{0}(t,x,y)$
with

$\begin{split}f(0)=(-\lambda_{1},\rho^{\prime\prime}(0))\left(\begin{array}[]{cc}1&R\\\
R&1\end{array}\right)^{-1}\left(\begin{array}[]{c}1\\\
1\end{array}\right)=\frac{\rho^{\prime\prime}(0)-\lambda_{1}}{1+R};\end{split}$

and
${\mathbb{E}}\\{Y^{\prime\prime}(t)|X(0)=x+u,Y(t)=y+u,X^{\prime}(0)=Y^{\prime}(t)=0\\}=g(t)u+g_{0}(t,x,y)$
with

$\begin{split}g(0)=(\rho^{\prime\prime}(0),-\lambda_{2})\left(\begin{array}[]{cc}1&R\\\
R&1\end{array}\right)^{-1}\left(\begin{array}[]{c}1\\\
1\end{array}\right)=\frac{\rho^{\prime\prime}(0)-\lambda_{2}}{1+R}.\end{split}$

Therefore, in the expectation in (8.10), the highest-order term in $u$
evaluated at $t=0$ is given by
$[(\lambda_{1}-\rho^{\prime\prime}(0))(\lambda_{2}-\rho^{\prime\prime}(0))/(1+R)^{2}]u^{2}$.
Note that the Mills ratio in (8.3) is asymptotically $(1+R)^{2}/u^{2}$ at
$t=0$. Plugging these into (8.10) yields

$\begin{split}I_{0}(u)&\sim\int_{0}^{1}\frac{1-t}{2\pi\sqrt{\lambda_{1}\lambda_{2}-\rho^{\prime\prime}(t)^{2}}}\frac{1}{2\pi\sqrt{{\rm
det}(\Sigma(t))}}f(t)g(t)\\\
&\qquad\times\frac{1}{[(\Sigma(t)^{-1})_{11}+(\Sigma(t)^{-1})_{12}]^{2}}e^{-\frac{1}{2}u^{2}(1,1)\Sigma(t)^{-1}(1,1)^{T}}dt.\end{split}$

Since

$\begin{split}h(t):=\frac{1}{2}(1,1)\Sigma(t)^{-1}(1,1)^{T}=\frac{1}{2}\frac{\Sigma_{11}(t)+\Sigma_{22}(t)-2\Sigma_{12}(t)}{\Sigma_{11}(t)\Sigma_{22}(t)-\Sigma_{12}(t)^{2}}\end{split}$

attains its minimum only at $t=0$ with $h(0)=1/(1+R)$ and

$\begin{split}h^{\prime\prime}(0)=\frac{-\rho^{\prime\prime}(0)(\lambda_{1}-\rho^{\prime\prime}(0))(\lambda_{2}-\rho^{\prime\prime}(0))}{(1+R)^{2}[\lambda_{1}\lambda_{2}-\rho^{\prime\prime}(0)^{2}]}.\end{split}$

Applying the Laplace method (see Lemma A.3 in [7]) yields

$\begin{split}I_{0}(u)&\sim\frac{1}{2}\frac{1}{2\pi\sqrt{\lambda_{1}\lambda_{2}-\rho^{\prime\prime}(0)^{2}}}\frac{1}{2\pi\sqrt{1-R^{2}}}\frac{[\lambda_{1}-\rho^{\prime\prime}(0)][\lambda_{2}-\rho^{\prime\prime}(0)]}{(1+R)^{2}}u^{2}\\\
&\qquad\times\frac{(1+R)^{2}}{u^{2}}\bigg{(}-\frac{2\pi}{u^{2}}\frac{(1+R)^{2}[\lambda_{1}\lambda_{2}-\rho^{\prime\prime}(0)^{2}]}{\rho^{\prime\prime}(0)(\lambda_{1}-\rho^{\prime\prime}(0))(\lambda_{2}-\rho^{\prime\prime}(0))}\bigg{)}^{1/2}e^{-\frac{u^{2}}{1+R}}\\\
&=\frac{1}{2}\frac{\sqrt{(\lambda_{1}-\rho^{\prime\prime}(0))(\lambda_{2}-\rho^{\prime\prime}(0))}(1+R)}{(2\pi)^{3/2}\sqrt{1-R^{2}}\sqrt{-\rho^{\prime\prime}(0)}}\frac{1}{u}e^{-\frac{u^{2}}{1+R}}.\end{split}$

Thus we obtain

$\begin{split}{\mathbb{P}}&\Big{\\{}\sup_{t\in[0,1]}X(t)\geq
u,\sup_{s\in[0,1]}Y(s)\geq u\Big{\\}}\\\
&=\frac{1}{(2\pi)^{3/2}}\sqrt{\frac{(\lambda_{1}-\rho^{\prime\prime}(0))(\lambda_{2}-\rho^{\prime\prime}(0))(1+R)}{-\rho^{\prime\prime}(0)(1-R)}}\frac{1}{u}e^{-\frac{u^{2}}{1+R}}(1+o(1)).\end{split}$

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> Dan Cheng
> School of Mathematical and Statistical Sciences
> Arizona State University
> 900 S Palm Walk
> Tempe, AZ 85281, USA
> E-mail<EMAIL_ADDRESS>>
> Yimin Xiao
> Department of Statistics and Probability
> Michigan State University
> 619 Red Cedar Road, C-413 Wells Hall
> East Lansing, MI 48824, USA
> E-mail<EMAIL_ADDRESS>