THE EXCURSION PROBABILITY OF GAUSSIAN AND ASYMPTOTICALLY
GAUSSIAN RANDOM FIELDS
By
Dan Cheng

A DISSERTATION
Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of
Statistics – Doctor of Philosophy

2013

ABSTRACT
THE EXCURSION PROBABILITY OF GAUSSIAN AND
ASYMPTOTICALLY GAUSSIAN RANDOM FIELDS
By
Dan Cheng
The purpose of this thesis is to develop the asymptotic approximation to excursion probability of Gaussian and asymptotically Gaussian random fields. It is composed of two parts.
The first part is to study smooth Gaussian random fields. We extend the expected Euler characteristic approximation to a wide class of smooth Gaussian random fields with non-constant
variances. Applying similar techniques, we also find that the joint excursion probability of
vector-valued smooth Gaussian random fields can be approximated via the expected Euler
characteristic of related excursion sets. As useful applications, the excursion probabilities
over random intervals and infinite intervals are also investigated. The second part focuses on
non-smooth Gaussian and asymptotically Gaussian random fields. We study the excursion
probability of Gaussian random fields on the sphere and obtain an asymptotics based on
the Pickands’ constant. Using double sum method, we also derive the approximation, which
involves the generalized Pickands’ constant, to excursion probability of anisotropic Gaussian
and asymptotically Gaussian random fields.

Copyright by
DAN CHENG
2013

ACKNOWLEDGMENTS

I would like to express my sincere gratitude to my advisor Professor Yimin Xiao for his
excellent guidance and continuous support during my Ph.D. study and research. He guided
me not only to write this thesis but to set a career path to the future. He is extraordinarily
kind to students. Besides academic communications, we chatted and shared life experience
like friends. His enthusiasm on mathematics encourages me to keep working in the research.
I also wish to thank Professor V. S. Mandrekar, Professor Lifeng Wang and Professor
Xiaodong Wang for serving on my dissertation committee. I am grateful to the Department
of Statistics and Probability and the Graduate School who provided me the assistantships,
Dissertation Continuation Fellowship and Dissertation Completion Fellowship for working on
the dissertation. This dissertation is also supported in part by the NSF Grant DMS-1006903.
Finally, I would like to thank my family for their love and support that enable me to
pursue my career goal.

iv

TABLE OF CONTENTS

Chapter 1 Introduction and Review of Existing Literature . . . . . . . . .
1.1 Gaussian Random Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Excursion Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 2 Smooth Gaussian Random Fields with
2.1 Gaussian Fields with Stationary Increments . . .
2.2 The Mean Euler Characteristic . . . . . . . . . .
2.3 Excursion Probability . . . . . . . . . . . . . . . .
2.4 Further Remarks and Examples . . . . . . . . . .
2.5 Some Auxiliary Facts . . . . . . . . . . . . . . . .

Stationary
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .

Increments
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .

Chapter 3 Smooth Gaussian Random Fields with Non-constant Variances
3.1 Gaussian Fields on Rectangles . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Applications for Gaussian Fields with a Unique Maximum Point of the Variance
3.3 Gaussian Fields on Manifolds without Boundary . . . . . . . . . . . . . . . .
3.4 Gaussian Fields on Convex Sets with Smooth Boundary . . . . . . . . . . . .
3.5 Gaussian Fields on Convex Sets with Piecewise Smooth Boundary . . . . . .

8
8
18
26
51
59
62
62
75
80
83
91

Chapter 4
4.1
4.2
4.3

The Expected Euler Characteristic of Non-centered Stationary
Gaussian Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Preliminary Gaussian Computations . . . . . . . . . . . . . . . . . . . . . .
Stationary Gaussian Fields on Rectangles . . . . . . . . . . . . . . . . . . . .
Isotropic Gaussian Random Fields on Sphere . . . . . . . . . . . . . . . . . .

1
1
2

103
104
108
115

Chapter 5
5.1
5.2

Excursion Probability of Smooth Gaussian Processes over Random Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
Stationary Gaussian Processes . . . . . . . . . . . . . . . . . . . . . . . . . . 124
Gaussian Processes with Increasing Variance . . . . . . . . . . . . . . . . . . 142

Chapter 6
6.1
6.2
6.3

Ruin Probability of a Certain Class
cesses . . . . . . . . . . . . . . . . . . .
Self-similar Processes . . . . . . . . . . . . . .
Integrated Fractional Brownian Motion . . . .
More General Gaussian Processes . . . . . . .

of Smooth Gaussian Pro. . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . .

Chapter 7 Excursion Probability of Gaussian Random
7.1 Notations . . . . . . . . . . . . . . . . . . . . . . . . .
7.2 Non-smooth Gaussian Fields on Sphere . . . . . . . . .
7.2.1 Locally Isotropic Gaussian Fields on Sphere . .

v

Fields on Sphere
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .

.
.
.
.

152
152
155
157
169
170
170
170

7.3

7.2.2 Standardized Spherical Fractional Brownian Motion
Smooth Isotropic Gaussian Fields on Sphere . . . . . . . .
7.3.1 Preliminaries . . . . . . . . . . . . . . . . . . . . .
7.3.2 Excursion Probability . . . . . . . . . . . . . . . .

.
.
.
.

.
.
.
.

.
.
.
.

.
.
.
.

.
.
.
.

.
.
.
.

.
.
.
.

.
.
.
.

.
.
.
.

.
.
.
.

176
178
179
182

Excursion Probability of Anisotropic Gaussian and Asymptotically Gaussian Random Fields . . . . . . . . . . . . . . . . . . . .
Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Asymptotically Gaussian Random Fields . . . . . . . . . . . . . . . . . . . .
Proof of Theorem 8.1.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Example: Standardized Fractional Brownian Sheet . . . . . . . . . . . . . .
Example: Standardized Random String Processes . . . . . . . . . . . . . . .

188
188
190
205
209
212

Chapter 8
8.1
8.2
8.3
8.4
8.5

Chapter 9 Vector-valued Smooth Gaussian Random Fields . . . . . . . . . 215
9.1 Joint Excursion Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
9.2 Vector-valued Gaussian Processes . . . . . . . . . . . . . . . . . . . . . . . . 239
BIBLIOGRAPHY

. . . . . . . . . . . . .

vi

. . . . . . . . . . . . . . . . . 250

Chapter 1
Introduction and Review of Existing
Literature

1.1

Gaussian Random Fields

A real-valued random field is simply a stochastic process defined over a parameter space
T , which could be a subset of RN or even a manifold, etc. The following is the rigorous
definition [cf. Adler and Taylor (2007)].
Definition 1.1.1 Let (Ω, F, P) be a complete probability space and T a topological space.
Then a measurable mapping X : Ω → RT (the space of all real-valued functions on T ) is
called a real-valued random field. Measurable mappings from Ω to (RT )d , d ≥ 1, are called
vector-valued random fields.
Thus, X is a real-valued function X(ω, t), where ω ∈ Ω and t ∈ T . For convenience,
usually, we abbreviate X(ω, t) as X(t) or X.
We define a real-valued Gaussian (random) field to be a real-valued random field X on
a parameter space T for which the finite dimensional distributions of (X(t1 ), . . . , X(tn )) are
multivariate Gaussian ( i.e., multivariate Normal) for each 1 ≤ n < ∞ and each (t1 , . . . , tn ) ∈
T n . The functions m(t) = E{X(t)} and C(t, s) = E{(X(t) − m(t))(X(s) − m(s))} are called
respectively the mean and covariance functions of X. If m(t) ≡ 0, we call X a centered
1

Gaussian field. A vector-valued Gaussian field X taking values in Rd is the random field for
which ξ, X(t) is a real-valued Gaussian field for every ξ ∈ Rd .
The following result is Theorem 1.4.1 in Adler and Taylor (2007), which gives a sufficient
condition such that a Gaussian field X is continuous and bounded.
Theorem 1.1.2 Let {X(t) : t ∈ T } be a centered Gaussian field, where T is a compact set
of RN . If there exist positive constants K, α and η such that

E{|X(t) − X(s)|2 } ≤ K| log t − s |−1−α ,

∀ t − s ≤ η,

then X is continuous and bounded on T with probability one.
Note that the sufficient condition in the above theorem only depends on the covariance
function of X. This is a huge advantage for studying centered Gaussian random fields: all of
their properties only depend on the covariance structure. Similar sufficient conditions for the
differentiability of Gaussian fields can also be obtained, see Chapter 1 in Adler and Taylor
(2007) for more details.

1.2

Excursion Probability

The excursion probability above level u > 0 is defined as P{supt∈T X(t) ≥ u}. Due to
the wide applications in statistics and many other related areas, computing the excursion
probability becomes a classical and very important problem in probability theory. However,
usually, the exact probability is unable to obtain, instead, we try to find the asymptotic
approximation as u tends to infinity.
There is a classical result of Landau and Shepp (1970) and Marcus and Shepp (1972) that
2

gives a logarithmic asymptotics for the excursion probability of a general centered Gaussian
process. If we assume that X(t) is a.s. bounded, then they showed that

lim u−2 log P sup X(t) ≥ u = −

u→∞

t∈T

1
,
2
2σT

(1.2.1)

2
where σT = supt∈T Var(X(t)).

We present here a non-asymptotic result due to Borell (1975) and Tsirelson, Ibragimov
and Sudakov (TIS) (1976).
Theorem 1.2.1 (Borell-TIS inequality). Let {X(t) : t ∈ T } be a centered Gaussian
field, a.s. bounded, where T is a compact subset of RN . Then E{supt∈T X(t)} < ∞ and for
all u > 0,
P sup X(t) − E sup X(t) ≥ u ≤ e
t∈T

2
−u2 /(2σT )

.

t∈T

It is evident to check that the Borell-TIS inequality implies (1.2.1). There are also several
non-asymptotic bounds for the excursion probability of general (only assume continuity and
boundedness a.s.) Gaussian fields, see Chapter 4 in Adler and Taylor (2007) for more details.
If assume X to be stationary or locally stationary, then there is a famous approximation
obtained by the double sum method. This technique was developed by Pickands (1969a,
1969b) for Gaussian processes, extended to Gaussian fields by Qualls and Watanabe (1973),
and surveyed and developed in a monograph of Piterbarg (1996a).
Theorem 1.2.2 Let T be a bounded Jordan measurable set in RN such that dim(T ) = N ,
and let {X(t) : t ∈ T } be a centered Gaussian field with covariance function C(·, ·) satisfying

C(t, s) = 1 − t − s α (1 + o(1))

3

as t − s → 0.

Then as u → ∞,

P sup X(t) ≥ u = Hα Vol(T )u2N/α Ψ(u)(1 + o(1)),

(1.2.2)

t∈T
2
∞
where Hα is the Pickankds’ constant and Ψ(u) = (2π)−1/2 u e−x /2 dx.

This result was developed further by Chan and Lai (2006) for Gaussian fields with a
wider class of covariance structures. The coefficient Hα Vol(T ) above was generalized as
T

Hα (t)dt, where Hα (·) is a function on T . Moreover, the result in Chan and Lai (2006) is

applicable to certain asymptotically Gaussian random fields.
In Chapter 7, we investigate Gaussian random fields on the sphere and obtain Theorem
7.2.4, which is similar to Theorem 1.2.2. In Chapter 8, we extend the result in Chan and Lai
(2006) to anisotropic and asymptotically anisotropic Gaussian random fields, see Theorem
8.1.1 and Theorem 8.2.6.
Can we get more accurate approximation to the excursion probability of “nicer” Gaussian
random fields? The answer is yes. Sun (1993) used the tube method to find the approximation for Gaussian fields with finite Karhunen-Lo`ve expansion. Also, many authors applied
e
the Rice method to get accurate approximations for smooth Gaussian fields, see Piterbarg
(1996a), Adler (2000) and Aza¨ and Wschebor (2005, 2008, 2009), etc. Later on, these
ıs
approximations were conjectured by statisticians that they should have close connection to
the geometry of the excursion set Au = {t ∈ T : X(t) ≥ u}. Taylor, Takemura and Adler
(2005) showed the rigorous proof that the expected Euler characteristic of the excursion set,
denoted by E{ϕ(Au )}, can approximate the excursion probability very accurately. Their
result is stated as follows.
Theorem 1.2.3 Let X = {X(t) : t ∈ T } be a unit-variance smooth Gaussian random field
4

parameterized on a manifold T . Under certain conditions on the regularity of X and topology
of T , the following approximation holds:

2
P sup X(t) ≥ u = E{ϕ(Au )}(1 + o e−αu )),

as u → ∞,

(1.2.3)

t∈T

where α is some positive constant.
Moreover, E{ϕ(Au )} can be computed by the Kac-Rice formula, see Adler and Taylor
(2007),
dim(T )

E{ϕ(Au )} = C0 Ψ(u) +

2
Cj uj−1 e−u /2 ,

(1.2.4)

j=1

where Cj , j = 0, 1, . . . , dim(T ), are constants depending on X and T . Here is a simple
example. Let X be a smooth isotropic Gaussian field with unit variance and T = [0, L]N ,
then
N

E{ϕ(Au )} = Ψ(u) +
j=1

N
j

Lj λj/2

(2π)(j+1)/2

2
Hj−1 (u)e−u /2 ,

where λ = Var( ∂X (t)) and Hj−1 (u) are Hermite polynomials of order j − 1. It is worth
∂ti
mentioning here that if X is not centered or not stationary, then E{ϕ(Au )} becomes complicated to compute. In the recent monograph Adler and Taylor (2007), the authors only
considered centered Gaussian random fields with constant variance. In Chapter 4 here,
we study non-centered stationary Gaussian fields and derive exact formulae for computing
E{ϕ(Au )}.
Comparing (1.2.3) and (1.2.4) with (1.2.2), we see that the approximation in (1.2.2) only
2
uses one of the terms, which involves uN −1 e−u /2 , in E{ϕ(Au )}. Also, we note that the

error term in (1.2.2) is only o(1), and the expected Euler characteristic approximation in
(1.2.3) is much more accurate since the error is exponentially smaller than the major term
5

E{ϕ(Au )}.
The requirement of “constant variance” on the Gaussian random fields in Theorem 1.2.3
is too restrictive for many applications. However, the original proof in Taylor, Takemura
and Adler (2005) relies on this requirement heavily. If the constant variance condition is not
satisfied, little had been known on whether the approximation (1.2.3) still holds. In a recent
paper Aza¨ and Wschebor (2008, Theorem 5), the authors proved (1.2.3) for a special case
ıs
when the variance of the Gaussian field attains its maximum only in the interior of T . But
this special case excludes many important Gaussian fields in which we are interested.
As a major contribution in this thesis, we shall use the Rice method to show (1.2.3) for
more general smooth Gaussian fields without constant-variance. In Chapter 2, we study
smooth Gaussian random fields with stationary increments and obtain the desired results in
Theorem 2.3.7 and Theorem 2.3.8. Meanwhile, we provide a specific formula for computing
E{ϕ(Au )} in Theorem 2.2.2. To develop the theory further, we show in Chapter 3 that the
expected Euler characteristic approximation also holds for a large class of smooth Gaussian
random fields with non-constant variances. When computing E{ϕ(Au )}, we also find that
it can be simplified in certain sense depending on the variance function of X.
As useful applications, we study the excursion probabilities of Gaussian processes over
random intervals and infinite intervals in Chapter 5 and Chapter 6. The approximations we
derived are also more accurate than the existing ones, since the errors are super-exponentially
small.
Lastly, Chapter 9 is on a new topic: the excursion probability for vector-valued Gaussian
random fields. There has been little research on this. The only exceptions are Piterbarg and
Stamatovic (2005) and Debicki et al. (2010) who obtained some logarithmic asymptotics,
and Ladneva and Piterbarg (2000) and Anshin (2006) who obtained certain asymptotics for
6

non-smooth vector-valued Gaussian random fields with special covariance functions.
Let {(X(t), Y (s)) : t ∈ T, s ∈ S} be an R2 -valued, centered, unit-variance Gaussian
random field, where T and S are rectangles in RN . Define the excursion set

Au (X, T ) × Au (Y, S) = {(t, s) ∈ T × S : X(t) ≥ u, Y (s) ≥ u}.

We show in Theorem 9.1.9 that under certain smoothness and regularity conditions, as
u → ∞,

P sup X(t) ≥ u, sup Y (s) ≥ u
t∈T

s∈S

= E{ϕ(Au (X, T ) × Au (Y, S))} + o exp

−

u2
− αu2
1 + ρ(T, S)

.

where ρ(T, S) = supt∈T,s∈S E{X(t)Y (s)}.
Let {(X(t), Y (t)) : t ∈ T } be an R2 -valued, centered, unit-variance Gaussian process,
where T = [a, b] is a finite interval in R. Define the excursion set

Au (T, X ∧ Y ) = {t ∈ T : (X ∧ Y )(t) ≥ u}.

We show in Theorem 9.2.5 that under certain smoothness and regularity conditions, as
u → ∞,

P{∃t ∈ T such that X(t) ≥ u, Y (t) ≥ u} = P sup(X ∧ Y )(t) ≥ u
t∈T

= E{ϕ(Au (T, X ∧ Y ))} + o exp

where ρ(T ) = supt∈T E{X(t)Y (t)}.

7

−

u2
1 + ρ(T )

− αu2

,

Chapter 2
Smooth Gaussian Random Fields with
Stationary Increments

2.1

Gaussian Fields with Stationary Increments

Let X = {X(t) : t ∈ RN } be a real-valued centered Gaussian random field with stationary
increments. We assume that X has continuous covariance function C(t, s) = E{X(t)X(s)}
and X(0) = 0. Then it is known [cf. Yaglom (1957)] that

C(t, s) =

RN

(ei t,λ − 1)(e−i s,λ − 1) F (dλ) + t, Θs

(2.1.1)

where x, y is the ordinary inner product in RN , Θ is an N × N non-negative definite (or
positive semidefinite) matrix and F is a non-negative symmetric measure on RN \{0} which
satisfies
λ 2
F (dλ) < ∞.
2
RN 1 + λ

(2.1.2)

Similarly to stationary random fields, the measure F and its density (if it exists) f (λ) are
called the spectral measure and spectral density of X, respectively.

8

By (2.1.1) we see that X has the following stochastic integral representation

X(t) =

RN

(ei t,λ − 1)W (dλ) + Y, t ,

(2.1.3)

where Y is an N -dimensional Gaussian random vector and W is a complex-valued Gaussian
random measure (independent of Y) with F as its control measure. It is known that many
probabilistic, analytic and geometric properties of a Gaussian field with stationary increments
can be described in terms of its spectral measure F and, on the other hand, various interesting
Gaussian random fields can be constructed by choosing their spectral measures appropriately.
See Xiao (2009), Xue and Xiao (2011) and the references therein for more information.
For simplicity we assume that Y = 0. It follows from (2.1.1) that the variogram ν of X
is given by
ν(h) := E(X(t + h) − X(t))2 = 2

RN

(1 − cos h, λ )F (dλ).

(2.1.4)

Mean-square directional derivatives and sample path differentiability of Gaussian random
fields have been well studied. See, for example, Adler (1981), Adler and Taylor (2007),
Potthoff (2010), Xue and Xiao (2011). In particular, general sufficient conditions for a
Gaussian random field to have a modification whose sample functions are in C k are given by
Adler and Taylor (2007). For a Gaussian random field X = {X(t) : t ∈ RN } with stationary
increments, Xue and Xiao (2011) provided conditions for its sample path differentiability in
terms of the spectral density function f (λ). Similar arguments can be applied to give the
spectral condition for the sample functions of X to be in C k (RN ).
Definition 2.1.1 [Adler and Taylor (2007, p.22)]. Let

t, v1 , . . . , vk ∈ RN ;

v = (v1 , . . . , vk ) ∈ ⊗k RN .
9

We say X has a kth-order L2 partial derivative at t, in the direction v, which we denote by
Dv 2 X(t), if the limit
L

Dv 2 X(t)
L

:=

lim

k
i=1 hi

h1 ,...,hk →0

k
i=1 hi vi )

exists in L2 , where GX (t,

k

1

hi vi

i=1

k

k

(−1)k− i=1 si X t +

=

i=1

hi vi

is the symmetrized difference

k

GX t,

GX t,

si hi vi .

(2.1.5)

i=1

s∈{0,1}k

Remark 2.1.2 Recall the fact that a sequence of random variables ξn converges in L2 if and
only if E{ξn ξm } converges to a constant as n, m → ∞. It follows immediately that Dv 2 X(t)
L

exists in L2 if and only if

lim

ˆ
ˆ
h1 ,...,hk ,h1 ,...,hk →0

1
k
ˆ
i=1 hi hi

k

k

ˆ
hi vi

hi vi GX t,

E GX t,
i=1

(2.1.6)

i=1

exists.
Let e1 , e2 , . . . , eN be the standard orthonormal basis of RN . If the direction v consists
of ki many ei , 1 ≤ i ≤ N , and k =

N
i=1 ki ,

then we write Dv 2 X(t) simply as
L

∂ k X(t)
kN
k1
∂t1 ···∂tN

.

Lemma 2.1.3 Let X = {X(t) : t ∈ RN } be a real-valued centered Gaussian random field
with stationary increments and let k =
if

∂ 2k ν(0)
2k
2k
∂t1 1 ···∂tN N

N
i=1 ki .

Then

∂ k X(t)
kN
k
∂t11 ···∂tN

exists in L2 if and only

exists.

Proof To simplify the notations, we only show the proof for k = 2 and the proof for general

10

k will be similar. By the definition of the symmetric difference GX in (2.1.5),
1
ˆ
ˆ
E{GX (t, h1 ei + h2 ej )GX (t, h1 ei + h2 ej )}
ˆ 1 h2
ˆ
h1 h2 h
1
=
E{[X(t + h1 ei + h2 ej ) − X(t + h1 ei ) − X(t + h2 ej ) + X(t)]
ˆ ˆ
h1 h2 h1 h2

(2.1.7)

ˆ
ˆ
ˆ
ˆ
× [X(t + h1 ei + h2 ej ) − X(t + h1 ei ) − X(t + h2 ej ) + X(t)]}.
Expanding the product above and applying the variogram ν defined in (2.1.4), we obtain
that (2.1.7) becomes
−1
ˆ
ˆ
ˆ
{ν(h1 ei + h2 ej − h1 ei − h2 ej ) − ν(h1 ei + h2 ej − h1 ei )
ˆ ˆ
2h1 h2 h1 h2
ˆ
ˆ
ˆ
− ν(h1 ei + h2 ej − h2 ej ) + ν(h1 ei + h2 ej ) − ν(h1 ei − h1 ei − h2 ej )
ˆ
ˆ
ˆ
ˆ
+ ν(h1 ei − h1 ei ) + ν(h1 ei − h2 ej ) − ν(h1 ei ) − ν(h2 ej − h1 ei − h2 ej )

(2.1.8)

ˆ
ˆ
ˆ
ˆ
+ ν(h2 ej − h1 ei ) + ν(h2 ej − h2 ej ) − ν(h2 ej ) + ν(−h1 ei − h2 ej )
ˆ
ˆ
− ν(−h1 ei ) − ν(−h2 ej ) + ν(0)}
=

−1
ˆ
ˆ
Gν (0, h1 ei + h2 ej + (−h1 )ei + (−h2 )ej ).
ˆ
ˆ
2h1 h2 (−h1 )(−h2 )

ˆ ˆ
Note that as h1 , h2 , h1 , h2 → 0, the limit (if it exists) of the last term in (2.1.8) is just
∂ 4 ν(0)

− 2 2 , together with Remark 2.1.2, we obtain the desired result.
∂ti ∂tj
Proposition 2.1.4 Let X = {X(t) : t ∈ RN } be a real-valued centered Gaussian random
field with stationary increments and let ki (1 ≤ i ≤ N ) be non-negative integers. If there is
a constant ε > 0 such that
N
λ >1 i=1

|λi |2ki +ε F (dλ) < ∞,

11

(2.1.9)

then X has a modification X such that the partial derivative
N
i=1 ki .

RN almost surely, where k =

∂ k X(t)
kN
k
∂t11 ···∂tN

is continuous on

Moreover, ∀T > 0 and η ∈ (0, ε ∧ 1), there exists a

constant κ such that
∂ k X(t)
E

Proof

kN
k
∂t11 · · · ∂tN

−

2

∂ k X(s)
kN
k
∂s11 · · · ∂sN

≤ κ t − s η,

∀t, s ∈ [−T, T ]N .

Applying the dominated convergence theorem,
∂ 2k ν(0)

2k
2k
∂t1 1 · · · ∂tN N

=

RN

2k
2k
λ1 1 · · · λN N F (dλ)

=
λ ≤1

2k
2k
λ1 1 · · · λN N F (dλ) +

λ 2 F (dλ) +

≤
λ ≤1

λ >1

λ >1

2k
2k
λ1 1 · · · λN N F (dλ)

(2.1.10)

2k
2k
λ1 1 · · · λN N F (dλ) < ∞,

where the last inequality is due to the requirement (2.1.2) and condition (2.1.9). By Lemma
2.1.3, the partial derivative

∂ k X(t)
kN
k1
∂t1 ···∂tN

exists in L2 .

Next, we show that for any η ∈ (0, ε ∧ 1), there exists a constant κ such that
∂ k X(t)
E

kN
k
∂t11 · · · ∂tN

−

∂ k X(s)
kN
k
∂s11 · · · ∂sN

2

≤ κ t − s η,

∀t, s ∈ [−T, T ]N .

(2.1.11)

Recall that

C(t, s) =
=

RN
RN

ei t,λ − 1 e−i s,λ − 1 F (dλ)
(2.1.12)
(cos t − s, λ − cos t, λ − cos s, λ + 1)F (dλ),

12

taking the derivative gives
∂ 2k C(t, s)

=

kN k
k
∂t11 · · · ∂tN ∂s11 · · · ∂skN

2k
2k
λ1 1 · · · λN N cos t − s, λ F (dλ).

RN

It follows that
∂ k X(t)
E

k

k

N
∂t11 · · · ∂tN

−

2

N
∂t11 · · · ∂tN

=2

2

∂ k X(s)

+E

k

k

k

k

N
∂s11 · · · ∂sN

∂ k X(t)

=E

2

∂ k X(s)

k

k

N
∂s11 · · · ∂sN

− 2E

∂ k X(t)
k

k

∂ k X(s)
k

k

N
N
∂t11 · · · ∂tN ∂s11 · · · ∂sN

2k
2k
λ1 1 · · · λN N 1 − cos t − s, λ F (dλ).

RN

Let s0 = t, s1 = (s1 , t2 , . . . , tN ), s2 = (s1 , s2 , t3 . . . , tN ), ..., sN −1 = (s1 , . . . , sN −1 , tN ) and
ˆ
ˆ
ˆ
ˆ
sN = s. Let h = s − t := (h1 , . . . , hN ). Then, by Jensen’s inequality,
ˆ
∂ k X(t)
E

k

k

N
∂t11 · · · ∂tN

−

E
j=1

k

k

N
∂s11 · · · ∂sN

N

≤N

2

∂ k X(s)

∂ k X(ˆj )
s
kj

k

kj+1

−

k

kj−1

kj

k

N
∂s11 · · · ∂sj−1 ∂tj · · · ∂tN

N

N

= 2N

k

N
∂s11 · · · ∂sj ∂tj+1 · · · ∂tN

2

∂ k X(ˆj−1 )
s

N
j=1 R

|λi |2ki F (dλ)

1 − cos(hj λj )
i=1

N

(2.1.13)

N

≤ 2N
j=1

λ ≤1

|λi |2ki F (dλ)

1 − cos(hj λj )
i=1

N

N

+ 2N
j=1

λ >1

|λi |2ki F (dλ)

1 − cos(hj λj )
i=1

:= I1 + I2 .

13

Combining the result in (2.1.10) with the elementary inequality 1 − cos x ≤ x2 yields
N

|hj |2

I1 ≤ 2N
j=1

λ ≤1

λ 2 F (dλ) ≤ c1 t − s 2

(2.1.14)

for some positive constant c1 .
√
To bound the jth integral in I2 , we note that, when λ > 1, either |λj | > 1/ N or there
√
is j0 = j such that λj0 > 1/ N . We break the integral according to these two possibilities.
N
λ >1

|λi |2ki F (dλ)

1 − cos(hj λj )
i=1

N

≤

√
|λj |>1/ N

|λi |2ki F (dλ)

1 − cos(hj λj )
i=1

(2.1.15)
N

+

√
j0 =j |λj |≤1,|λj0 |>1/ N

|λi |2ki F (dλ)

1 − cos(hj λj )
i=1

:= I3 + I4 .
Combining condition (2.1.9) with the elementary inequality 1 − cos x ≤ x2 yields
N
1 − cos(hj λj )
ε
I3 ≤
|λj |
|λi |2ki F (dλ)
√
|λj |ε
1/ N <|λj |≤1/|hj |
i=1
N

1
ε
|λi |2ki F (dλ)
+
ε |λj |
|λj |>1/|hj | |λj |
i=1

(2.1.16)

≤ c2 |hj |ε
for some positive constant c2 . Similarly, it is evident to check that I4 ≤ c3 |hj |2 for some
positive constant c3 . Therefore, the H¨der condition for L2 partial derivative in (2.1.11)
o
holds, and then the desired result follows from Kolmogorov’s continuity theorem.

14

For simplicity we will not distinguish X from its modification X. As a consequence of
Proposition 2.1.4, we see that, if X = {X(t) : t ∈ RN } has a spectral density f (λ) which
satisfies
f (λ) = O

1
λ

as λ → ∞,

N +2k+H

(2.1.17)

for some integer k ≥ 1 and H ∈ (0, 1), then the sample functions of X are in C k (RN ) a.s.
When X(·) ∈ C 2 (RN ) almost surely, we write
Denote by

2 X(t)

X(t) and

∂X(t)
∂ti

∂ 2 X(t)

= Xi (t) and ∂t ∂t = Xij (t).
i j

the column vector (X1 (t), . . . , XN (t))T and the N ×N matrix

(Xij (t))i,j=1,...,N , respectively. It follows from (2.1.1) that for every t ∈ RN ,

λij :=

RN

λi λj F (dλ) =

∂ 2 C(t, s)
= E{Xi (t)Xj (t)}.
∂ti ∂sj s=t

(2.1.18)

Define the N × N matrix Λ = (λij )i,j=1,...,N , then (2.1.18) shows that Λ = Cov( X(t)) for
all t. In particular, the distribution of

λij (t) :=

RN

X(t) is independent of t. Let

λi λj cos t, λ F (dλ),

Λ(t) := (λij (t))i,j=1,...,N .

Then we have

λij (t) − λij =

RN

λi λj (cos t, λ − 1) F (λ) =

∂ 2 C(t, s)
= E{X(t)Xij (t)},
∂ti ∂tj s=t

or equivalently, Λ(t) − Λ = E{X(t) 2 X(t)}.
Let T =

N
i=1 [ai , bi ]

be a closed rectangle on RN , where ai < bi for all 1 ≤ i ≤ N and

0 ∈ T (the case of 0 ∈ T will be discussed in Remark 2.4.1). In addition to the stationary
/
increments, we will make use of the following conditions on X:

15

(H1). X(·) ∈ C 2 (T ) almost surely and its second derivatives satisfy the uniform mean-square
H¨lder condition: there exist constants L, η > 0 such that
o

E(Xij (t) − Xij (s))2 ≤ L t − s 2η ,

∀t, s ∈ T, i, j = 1, . . . , N.

(2.1.19)

(H2). For every t ∈ T , the matrix Λ − Λ(t) is non-degenerate.
(H3). For every pair (t, s) ∈ T 2 with t = s, the Gaussian random vector

(X(t), X(t), Xij (t), X(s), X(s), Xij (s), 1 ≤ i ≤ j ≤ N )

is non-degenerate.
(H3 ). For every t ∈ T , (X(t), X(t), Xij (t), 1 ≤ i ≤ j ≤ N ) is non-degenerate.
Clearly, by Proposition 2.1.4, condition (H1) is satisfied if (2.1.17) holds for k = 2. Also note
that (H3) implies (H3 ). We shall use conditions (H1), (H2) and (H3) to prove Theorems
2.3.7 and 2.3.8. Condition (H3 ) will be used for computing E{ϕ(Au )} in Theorem 2.2.2.
The following lemma shows that for Gaussian fields with stationary increments, (H2) is
equivalent to Λ − Λ(t) being positive definite.
Lemma 2.1.5 For every t = 0, Λ − Λ(t) is non-negative definite. Hence, under (H2),
Λ − Λ(t) is positive definite.

Proof

Let t = 0 be fixed. For any (a1 , . . . , aN ) ∈ RN \{0},
N

N

ai aj (λij − λij (t)) =
i,j=1

RN

2

ai λ i
i=1

16

(1 − cos t, λ ) F (λ).

(2.1.20)

Since (

N
2
i=1 ai λi ) (1 − cos

t, λ ) ≥ 0 for all λ ∈ RN , (2.1.20) is always non-negative, which

implies Λ − Λ(t) is non-negative definite. If (H2) is satisfied, then all the eigenvalues of
Λ − Λ(t) are positive. This completes the proof.
It follows from (2.1.20) that, if the spectral measure F is carried by a set of positive Lebesgue
measure (i.e., there is a set B ⊂ RN with positive Lebesgue measure such that F (B) > 0),
then (H2) holds. Hence, (H2) is in fact a very mild condition for smooth Gaussian fields
with stationary increments.
Lemma 2.1.5 and the following two lemmas indicate some significant properties of Gaussian fields with stationary increments. They will play important roles in later sections.
Lemma 2.1.6 For each t, Xi (t) and Xjk (t) are independent for all i, j, k; and E{Xij (t)Xkl (t)}
is symmetric in i, j, k, l.

Proof

By (2.1.1), one can verify that for t, s ∈ RN ,
∂ 3 C(t, s)
=
λi λj λk sin t − s, λ F (dλ),
∂ti ∂sj ∂sk
RN
∂ 4 C(t, s)
E{Xij (t)Xkl (s)} =
λi λj λk λl cos t − s, λ F (dλ).
=
∂ti ∂tj ∂sk ∂sl
RN
E{Xi (t)Xjk (s)} =

Letting s = t we obtain the desired results.
It follows immediately from Lemma 2.1.6 that the following result holds.
Lemma 2.1.7 Let A = (aij )1≤i,j≤N be a symmetric matrix, then

St (i, j, k, l) = E{(A 2 X(t)A)ij (A 2 X(t)A)kl }

is a symmetric function of i, j, k, l.
17

(2.1.21)

2.2

The Mean Euler Characteristic

The rectangle T =

N
i=1 [ai , bi ]

can be decomposed into several faces of lower dimensions.

We use the same notations as in Adler and Taylor (2007, p.134).
A face J of dimension k, is defined by fixing a subset σ(J) ⊂ {1, . . . , N } of size k and a
subset ε(J) = {εj , j ∈ σ(J)} ⊂ {0, 1}N −k of size N − k, so that
/

J = {t = (t1 , . . . , tN ) ∈ T : aj < tj < bj if j ∈ σ(J),
tj = (1 − εj )aj + εj bj if j ∈ σ(J)}.
/
Denote by ∂k T the collection of all k-dimensional faces in T , then the interior of T is given
◦

by T = ∂N T and the boundary of T is given by ∂T = ∪N −1 ∪J∈∂ T J. For J ∈ ∂k T , denote
k=0
k
by

X|J (t) and

2X

|J (t)

the column vector (Xi1 (t), . . . , Xik (t))T ,...,i ∈σ(J) and the k × k
i1
k

matrix (Xmn (t))m,n∈σ(J) , respectively.
If X(·) ∈ C 2 (RN ) and it is a Morse function a.s. [cf. Definition 9.3.1 in Adler and Taylor
(2007)], then according to Corollary 9.3.5 or page 211-212 in Adler and Taylor (2007), the
Euler characteristic of the excursion set Au = {t ∈ T : X(t) ≥ u} is given by
N

k

(−1)k

ϕ(Au ) =
k=0 J∈∂k T

(−1)i µi (J)

(2.2.1)

i=0

with
µi (J) := #{t ∈ J : X(t) ≥ u, X|J (t) = 0, index( 2 X|J (t)) = i,
ε∗ Xj (t)
j

(2.2.2)

≥ 0 for all j ∈ σ(J)},
/

where ε∗ = 2εj − 1 and the index of a matrix is defined as the number of its negative
j

18

eigenvalues. We also define

µi (J) := #{t ∈ J : X(t) ≥ u, X|J (t) = 0, index( 2 X|J (t)) = i}.

(2.2.3)

2
2
2
Let σt = Var(X(t)) and let σT = supt∈T σt be the maximum variance. For Gaussian
2
fields with stationary increments, it follows from (2.1.4) that ν(t) = σt . For t ∈ J ∈ ∂k T ,

where k ≥ 1, let

ΛJ = (λij )i,j∈σ(J) ,

ΛJ (t) = (λij (t))i,j∈σ(J) ,

2
θt = Var(X(t)| X|J (t)),

2
γt = Var(X(t)| X(t)),

(2.2.4)

{J1 , . . . , JN −k } = {1, . . . , N }\σ(J),
E(J) = {(tJ1 , . . . , tJ

N −k

) ∈ RN −k : tj ε∗ ≥ 0, j = J1 , . . . , JN −k }.
j

Then for all t ∈ J,

ΛJ = Cov( X|J (t)),

ΛJ (t) − ΛJ = E{X(t) 2 X|J (t)}.

(2.2.5)

2
2
2
2
Note that θt ≥ γt for all t ∈ T and θt = γt if t ∈ ∂N T . For {t} ∈ ∂0 T , then

X|J (t) is

2
2
not defined, in this case we set θt as σt by convention. Let Cj (t) be the (1, j + 1) entry

of (Cov(X(t), X(t)))−1 , i.e. Cj (t) = M1,j+1 /detCov(X(t), X(t)), where M1,j+1 is the
cofactor of the (1, j + 1) entry, E{X(t)Xj (t)}, in the covariance matrix Cov(X(t), X(t)).
k
2
2
Denote by Hk (x) the Hermite polynomial of order k, i.e., Hk (x) = (−1)k ex /2 d k (e−x /2 ).
dx

Then the following identity holds [cf. Adler and Taylor (2007, p.289)]:
∞
u

2
2
Hk (x)e−x /2 dx = Hk−1 (u)e−u /2 ,

19

(2.2.6)

where u > 0 and k ≥ 1. For a matrix A, |A| denotes its determinant. Let R+ = [0, ∞),
2
∞
R− = (−∞, 0] and Ψ(u) = (2π)−1/2 u e−x /2 dx.

The following lemma is an analogue of Lemma 11.7.1 in Adler and Taylor (2007). It
provides a key step for computing the mean Euler characteristic in Theorem 2.2.2, meanwhile,
it has close connection with Theorem 2.3.7.
Lemma 2.2.1 Let X = {X(t) : t ∈ RN } be a centered Gaussian random field with stationary increments satisfying (H1), (H2) and (H3 ). Then for each J ∈ ∂k T with k ≥ 1,
k

(−1)i µi (J)

E

=

i=0

Proof

(−1)k

u −u2 /(2θ2 )
|ΛJ − ΛJ (t)|
t dt. (2.2.7)
Hk−1
e
k
θt
θt
(2π)(k+1)/2 |ΛJ |1/2 J

Let Di be the collection of all k × k matrices with index i. Recall the definition

of µi (J) in (2.2.3), thanks to (H1) and (H3 ), we can apply the Kac-Rice metatheorem [cf.
Theorem 11.2.1 or Corollary 11.2.2 in Adler and Taylor (2007)] to get that the left hand side
of (2.2.7) becomes
k

J

p X (t) (0)dt
|J

(−1)i E{|det 2 X|J (t)|1{ 2 X (t)∈D } 1{X(t)≥u} | X|J (t) = 0}.
i
|J
i=0
(2.2.8)

Note that on the event Di , the matrix

2X

|J (t)

has i negative eigenvalues, which implies

(−1)i |det 2 X|J (t)| = det 2 X|J (t). Also, ∪k { 2 X|J (t) ∈ Di } = Ω a.s., hence (2.2.8)
i=0

20

equals

J

p X (t) (0)dt E{det 2 X|J (t)1{X(t)≥u} | X|J (t) = 0}
|J
2
2
e−x /(2θt )

=
J

(2π)(k+1)/2 |ΛJ |1/2 θt

(2.2.9)

∞

dt
u

dx E{det

2X

|J (t)|X(t)

= x,

X|J (t) = 0}.

Now we turn to computing E{det 2 X|J (t)|X(t) = x, X|J (t) = 0}. By Lemma 2.1.5,
under (H2), Λ − Λ(t) and hence ΛJ − ΛJ (t) are positive definite for every t ∈ J. Thus there
exists a k × k positive definite matrix Qt such that

Qt (ΛJ − ΛJ (t))Qt = Ik ,

(2.2.10)

where Ik is the k × k identity matrix. By (2.2.5),

E{X(t)(Qt 2 X|J (t)Qt )ij } = −(Qt (ΛJ − ΛJ (t))Qt )ij = −δij ,

where δij is the Kronecker delta function. One can write

E{det(Qt 2 X|J (t)Qt )|X(t) = x,

X|J (t) = 0} = E{det∆(t, x)},

(2.2.11)

where ∆(t, x) = (∆ij (t, x))i,j∈σ(J) with all elements ∆ij (t, x) being Gaussian variables. To
study ∆(t, x), we only need to find its mean and covariance. Note that

21

X(t) and

2 X(t)

are independent by Lemma 2.1.6, then we apply Lemma 2.5.1 to obtain

E{∆ij (t, x)} = E{(Qt 2 X|J (t)Qt )ij |X(t) = x,

X|J (t) = 0}

= (E{X(t)(Qt 2 X|J (t)Qt )ij }, 0, . . . , 0)(Cov(X(t), X|J (t)))−1 (x, 0, . . . , 0)T

(2.2.12)

x
= (−δij , 0, . . . , 0)(Cov(X(t), X|J (t)))−1 (x, 0, . . . , 0)T = − 2 δij ,
θ
t

where the last equality comes from the fact that the (1, 1) entry of (Cov(X(t), X|J (t)))−1
2
is detCov( X|J (t))/detCov(X(t), X|J (t)) = 1/θt . For the covariance, applying Lemma

2.5.1 again gives

E{(∆ij (t, x) − E{∆ij (t, x)})(∆kl (t, x) − E{∆kl (t, x)})}
= E{(Qt 2 X|J (t)Qt )ij (Qt 2 X|J (t)Qt )kl } − (E{X(t)(Qt 2 X|J (t)Qt )ij }, 0, . . . , 0)
· (Cov(X(t), X|J (t)))−1 (E{X(t)(Qt 2 X|J (t)Qt )kl }, 0, . . . , 0)T
= St (i, j, k, l) − (−δij , 0, . . . , 0)(Cov(X(t), X|J (t)))−1 (−δkl , 0, . . . , 0)T
= St (i, j, k, l) −

δij δkl
2
θt

,

where St is a symmetric function of i, j, k, l by applying Lemma 2.1.7 with A replaced by
Qt . Therefore (2.2.11) becomes

E

1
det(θt Qt ( 2 X|J (t))Qt ) X(t) = x,
k
θt

X|J (t) = 0

1
x
= k E det ∆(t) − Ik
θt
θt

where ∆(t) = (∆ij (t))i,j∈σ(J) and all ∆ij (t) are Gaussian variables satisfying

E{∆ij (t)} = 0,

2
E{∆ij (t)∆kl (t)} = θt St (i, j, k, l) − δij δkl .

22

,

−k
By Corollary 11.6.3 in Adler and Taylor (2007), (2.2.11) is equal to (−1)k θt Hk (x/θt ), hence

E{det 2 X|J (t)|X(t) = x,

X|J (t) = 0}

= E{det(Q−1 Qt 2 X|J (t)Qt Q−1 )|X(t) = x,
t
t

X|J (t) = 0}

= |ΛJ − ΛJ (t)|E{det(Qt 2 X|J (t)Qt )|X(t) = x,
=

X|J (t) = 0}

(−1)k
x
|ΛJ − ΛJ (t)|Hk
.
k
θt
θt

Plugging this into (2.2.9) and applying (2.2.6), we obtain the desired result.

Theorem 2.2.2 Let X = {X(t) : t ∈ RN } be a centered Gaussian random field with
stationary increments such that (H1), (H2) and (H3 ) are fulfilled. Then
N

P(X(t) ≥ u,

E{ϕ(Au )} =
{t}∈∂0 T

k=1 J∈∂k T
∞

×

dt
J

× Hk

dx

···

u

E(J)

(2π)k/2 |ΛJ |1/2

|ΛJ − ΛJ (t)|
k
N −k
γt

dyJ1 · · · dyJ

(2.2.13)

x
+ γt CJ1 (t)yJ1 + · · · + γt CJ
(t)yJ
N −k
N −k
γt

× pX(t),X (t),...,X
J
J
1

Proof

1

X(t) ∈ E({t})) +

N −k

(t) (x, yJ1 , . . . , yJN −k |

X|J (t) = 0).

According to Corollary 11.3.2 in Adler and Taylor (2007), (H1) and (H3 ) imply

that X is a Morse function a.s. It follows from (2.2.1) that
N

k

(−1)k E

E{ϕ(Au )} =

(−1)i µi (J) .

(2.2.14)

i=0

k=0 J∈∂k T

If J ∈ ∂0 T , say J = {t}, it turns out E{µ0 (J)} = P(X(t) ≥ u, X(t) ∈ E({t})). If J ∈ ∂k T
with k ≥ 1, we apply the Kac-Rice metatheorem to obtain that the expectation on the right
23

hand side of (2.2.14) becomes
k

J

p X (t) (0)dt
|J

(−1)i E{|det 2 X|J (t)|1{ 2 X (t)∈D } 1{(X (t),...,X
J1
JN −k (t))∈E(J)}
i
|J
i=0
× 1{X(t)≥u} | X|J (t) = 0}

=

∞

1
(2π)k/2 |ΛJ |1/2 J

dt

dx

···

u

E(J)

dyJ1 · · · dyJ

N −k

× E{det 2 X|J (t)|X(t) = x, XJ1 (t) = yJ1 , . . . , XJ

N −k

× pX(t),X (t),...,X
J
J
1

N −k

(t) (x, yJ1 , . . . , yJN −k |

(t) = yJ

N −k

, X|J (t) = 0}

X|J (t) = 0).
(2.2.15)

For fixed t, let Qt be the positive definite matrix in (2.2.10). Then, similarly to the proof in
Lemma 2.2.1, we can write

E{det(Qt 2 X|J (t)Qt )|X(t) = x, XJ1 (t) = yJ1 , . . . , XJ
= yJ
, X|J (t) = 0}
N −k
N −k

as E{det∆(t, x)}, where ∆(t, x) is a matrix consisting of Gaussian entries ∆ij (t, x) with
mean
E{(Qt 2 X|J (t)Qt )ij |X(t) = x, XJ1 (t) = yJ1 , . . . , XJ
= yJ
, X|J (t) = 0}
N −k
N −k
= (−δij , 0, . . . , 0)(Cov(X(t), XJ1 (t), . . . , XJ
· (x, yJ1 , . . . , yJ

N −k

N −k

(t), X|J (t)))−1

, 0, . . . , 0)T

δij
2
2
(t)yJ
),
= − 2 (x + γt CJ1 (t)yJ1 + · · · + γt CJ
N −k
N −k
γt
(2.2.16)

24

and covariance

E{(∆ij (t, x) − E{∆ij (t, x)})(∆kl (t, x) − E{∆kl (t, x)})} = St (i, j, k, l) −

δij δkl
2
γt

.

Following the same procedure in the proof of Lemma 2.2.1, we obtain that the last conditional
expectation in (2.2.15) is equal to
x
(−1)k |ΛJ − ΛJ (t)|
Hk
+ γt CJ1 (t)yJ1 + · · · + γt CJ
(t)yJ
.
k
N −k
N −k
γt
γt

(2.2.17)

Plug this into (2.2.15) and (2.2.14), yielding the desired result.

Remark 2.2.3 Usually, for nonstationary (including constant-variance) Gaussian field X
on RN , its mean Euler characteristic involves at least the third-order derivatives of the
covariance function. For Gaussian random fields with stationary increments, as shown in
Lemma 2.1.6, E{Xij (t)Xk (t)} = 0 and E{Xij (t)Xkl (t)} is symmetric in i, j, k, l, so the
mean Euler characteristic becomes relatively simpler, contains only up to the second-order
derivatives of the covariance function. In various practical applications, (2.2.13) could be
simplified with only an exponentially smaller difference, see the discussions in Section 2.4.

25

2.3

Excursion Probability
N
k=0 ∂k T

As in Section 3.1, we decompose T into several faces as T =

=

N
k=0

J∈∂k T

J.

For each J ∈ ∂k T , define the number of extended outward maxima above level u as
E
Mu (J) := #{t ∈ J : X(t) ≥ u,

X|J (t) = 0, index( 2 X|J (t)) = k,

/
ε∗ Xj (t) ≥ 0 for all j ∈ σ(J)}.
j
E
In fact, Mu (J) is the same as µk (J) defined in (2.2.2) with i = k. We will make use of the

following lemma.
Lemma 2.3.1 Let X = {X(t) : t ∈ RN } be a Gaussian random field satisfying (H1) and
(H3 ), then for any u > 0,
N
E
{Mu (J) ≥ 1} a.s.

sup X(t) ≥ u =
t∈T

Proof

k=0 J∈∂k T

E
By the definition of Mu (J), it is clear that
N
E
{Mu (J) ≥ 1} a.s.

sup X(t) ≥ u ⊃
t∈T

k=0 J∈∂k T

Suppose supt∈T X(t) ≥ u, since X(t) ∈ C 2 (RN ) a.s., there exists t0 ∈ T such that X(t0 ) =
supt∈T X(t). Without loss of generality, assume t0 ∈ J ∈ ∂k T . Note that t0 is a local
maximum restricted on J, thus

X|J (t0 ) = 0 and

2X

|J (t0 )

is non-positive definite. Due

to (H1) and (H3 ), we apply Lemma 11.2.11 in Adler and Taylor (2007) to obtain that
almost surely, det( 2 X|J (t0 )) = 0 and hence index( 2 X|J (t0 )) = k. If ε∗ Xj (t0 ) < 0 for
j
some j ∈ σ(J), then we can find t1 ∈ T such that X(t1 ) > X(t0 ), which contradicts
/
26

E
X(t0 ) = supt∈T X(t). Hence ε∗ Xj (t0 ) ≥ 0 for all j ∈ σ(J). These indicate Mu (J) ≥ 1,
/
j

therefore
N
E
{Mu (J) ≥ 1} a.s.,

sup X(t) ≥ u ⊂
t∈T

k=0 J∈∂k T

completing the proof.
It follows from Lemma 2.3.1 that
N

N
E
P{Mu (J)

P sup X(t) ≥ u ≤
t∈T

E
E{Mu (J)}.

≥ 1} ≤

k=0 J∈∂k T

(2.3.1)

k=0 J∈∂k T

On the other hand, by the Bonferroni inequality,
N
E
P{Mu (J) ≥ 1} −

P sup X(t) ≥ u ≥
t∈T

k=0 J∈∂k T

E
E
P{Mu (J) ≥ 1, Mu (J ) ≥ 1}.
J=J

E
E
Let pi = P{Mu (J) = i}, then P{Mu (J) ≥ 1} =

∞
i=1 pi

E
and E{Mu (J)} =

∞
i=1 ipi ,

follows that
∞
E
E
E{Mu (J)} − P{Mu (J) ≥ 1} =

(i − 1)pi
i=2
∞

≤
i=2

1
i(i − 1)
E
E
pi = E{Mu (J)(Mu (J) − 1)}.
2
2

27

it

E
E
E
E
Together with the obvious bound P{Mu (J) ≥ 1, Mu (J ) ≥ 1} ≤ E{Mu (J)Mu (J )}, we

obtain the following lower bound for the excursion probability,
N

P sup X(t) ≥ u ≥
t∈T

k=0 J∈∂k T

1
E
E
E
E{Mu (J)} − E{Mu (J)(Mu (J) − 1)}
2
(2.3.2)

E
E
E{Mu (J)Mu (J )}.

−
J=J

Define the number of local maxima above level u as

Mu (J) := #{t ∈ J : X(t) ≥ u,

X|J (t) = 0, index( 2 X|J (t)) = k},

E
then obviously Mu (J) ≥ Mu (J), and Mu (J) is the same as µk (J) defined in (2.2.3) with

i = k. It follows similarly that
N

E{Mu (J)} ≥ P sup X(t) ≥ u
t∈T

k=0 J∈∂k T
N

≥
k=0 J∈∂k T

(2.3.3)
1
E{Mu (J)} − E{Mu (J)(Mu (J) − 1)} −
2

E{Mu (J)Mu (J )}.
J=J

We will use (2.3.1) and (2.3.2) to estimate the excursion probability for the general
case, see Theorem 2.3.8. Inequalities in (2.3.3) provide another method to approximate the
excursion probability in some special cases, see Theorem 2.3.7. The advantage of (2.3.3)
is that the principal term induced by
compared with the one induced by

N
k=0
N
k=0

J∈∂k T
J∈∂k T

E{Mu (J)} is much easier to compute

E
E{Mu (J)}.

The following two lemmas provide the estimations for the principal terms in approximating the excursion probability.

28

Lemma 2.3.2 Let X be a Gaussian field as in Theorem 2.2.2. Then for each J ∈ ∂k T with
k ≥ 1, there exists some constant α > 0 such that

E{Mu (J)} =

2
u −u2 /(2θ2 )
|ΛJ − ΛJ (t)|
t dt(1 + o(e−αu )).
Hk−1
e
k
(k+1)/2 |Λ |1/2 J
θt
θt
(2π)
J

1

(2.3.4)

Proof

Following the notations in the proof of Lemma 2.2.1, we obtain similarly that

E{Mu (J)} =

p X (t) (0)dt E{|det 2 X|J (t)|1{ 2 X (t)∈D } 1{X(t)≥u} | X|J (t) = 0}
|J
k
|J
J
∞

dx

dt

=
J

u

2
2
(−1)k e−x /(2θt )

(2π)(k+1)/2 |ΛJ |1/2 θt

× E{det 2 X|J (t)1{ 2 X (t)∈D } |X(t) = x,
k
|J

X|J (t) = 0}.
(2.3.5)

Recall

2X

|J (t)

= Q−1 Qt 2 X|J (t)Qt Q−1 and we can write (2.2.12) as
t
t

E{Qt 2 X|J (t)Qt |X(t) = x,

x
X|J (t) = 0} = − 2 Ik .
θt

Make change of variables
x
V (t) = Qt 2 X|J (t)Qt + 2 Ik ,
θt
where V (t) = (Vij (t))1≤i,j≤k . Then (V (t)|X(t) = x,

X|J (t) = 0) is a Gaussian matrix

whose mean is 0 and covariance is the same as that of (Qt 2 X|J (t)Qt |X(t) = x,
0). Denote the density of Gaussian vectors ((Vij (t))1≤i≤j≤k |X(t) = x,

29

X|J (t) =

X|J (t) = 0) by

ht (v), v = (vij )1≤i≤j≤k ∈ Rk(k+1)/2 , then
E{det(Qt 2 X|J (t)Qt )1{ 2 X (t)∈D } |X(t) = x,
k
|J

X|J (t) = 0}

= E{det(Qt 2 X|J (t)Qt )1{Q 2 X (t)Q ∈D } |X(t) = x,
t
t k
|J
x
=
det (vij ) − 2 Ik ht (v) dv,
θt
v:(vij )− x Ik ∈Dk

X|J (t) = 0}

(2.3.6)

2
θt

2
where (vij ) is the abbreviation of matrix (vij )1≤i,j≤k . Since {θt : t ∈ T } is bounded, there

exists a constant c > 0 such that

x
(vij ) − 2 Ik ∈ Dk ,
θt

k
2
vij

∀ (vij ) :=

1/2

<

i,j=1

x
.
c

Thus we can write (2.3.6) as

Rk(k+1)/2

x
det (vij ) − 2 Ik ht (v)dv −
det (vij ) −
θt
v:(vij )− x Ik ∈Dk
2 /
θ

x
I ht (v) dv
2 k
θt

t

= E{det(Qt 2 X|J (t)Qt )|X(t) = x,

X|J (t) = 0} + Z(t, x),
(2.3.7)

where Z(t, x) is the second integral in the first line of (2.3.7) and it satisfies

|Z(t, x)| ≤

x
det (vij ) − 2 Ik
θt
(vij ) ≥ x
c

ht (v)dv.

Denote by G(t) the covariance matrix of ((Vij (t))1≤i≤j≤k |X(t) = x,

X|J (t) = 0), then

by Lemma 2.5.2 in the Appendix, the eigenvalues of G(t) and hence those of (G(t))−1
are bounded for all t ∈ T . It follows that there exists some constant α > 0 such that

30

ht (v) = o(e

−α (vij ) 2
)

2

and hence |Z(t, x)| = o(e−αx ) for some constant α > 0 uniformly

for all t ∈ T . Combine this with (2.3.5), (2.3.6), (2.3.7) and the proof of Lemma 2.2.1,
yielding the result.

Lemma 2.3.3 Let X be a Gaussian field as in Theorem 2.2.2. Then for each J ∈ ∂k T with
k ≥ 1, there exists some constant α > 0 such that

E
E{Mu (J)} =

∞

1

dx · · ·
dyJ1 · · · dyJ
N −k
(2π)k/2 |ΛJ |1/2 J
u
E(J)
x
|Λ − Λ (t)|
Hk
+ γt CJ1 (t)yJ1 + · · · + γt CJ
(t)yJ
× J kJ
N −k
N −k
γt
γt
dt

× pX(t),X (t),...,X
J
J
1

N −k

(t) (x, yJ1 , . . . , yJN −k |

2

X|J (t) = 0)(1 + o(e−αu )).
(2.3.8)

Proof

Under the notations in the proof of Theorem 2.2.2, applying the Kac-Rice formula,

E
we see that E{Mu (J)} equals

p X (t) (0)dt E{|det 2 X|J (t)|1{ 2 X (t)∈D } 1{X(t)≥u}
|J
k
|J
J
× 1{(X (t),··· ,X
J
J
1

=

N −k

(−1)k
(2π)k/2 |ΛJ |1/2 J

(t))∈E(J)} |

X|J (t) = 0}

∞

dt

···

dx
u

E(J)

dyJ1 · · · dyJ

N −k

E{det 2 X|J (t)1{ 2 X (t)∈D } |X(t) = x, XJ1 (t) = yJ1 , · · · , XJ
(t) = yJ
,
N −k
N −k
k
|J
X|J (t) = 0}pX(t),X (t),··· ,X
J
J
1

N −k

(t) (x, yJ1 , · · ·

31

, yJ

N −k

| X|J (t) = 0).

Recall

2X

|J (t)

= Q−1 Qt 2 X|J (t)Qt Q−1 and we can write (2.2.16) as
t
t

E{Qt 2 X|J (t)Qt |X(t) = x, XJ1 (t) = yJ1 , · · · , XJ
(t) = yJ
, X|J (t) = 0}
N −k
N −k
=−

x
Ik .
(t)yJ
+ CJ1 (t)yJ1 + · · · + CJ
2
N −k
N −k
γt

Make change of variables
x
W (t) = Qt 2 X|J (t)Qt + 2 Ik ,
γt
where W (t) = (Wij (t))1≤i,j≤k . Denote the density of

((Wij (t))1≤i≤j≤k |X(t) = x, XJ1 (t) = yJ1 , · · · , XJ
(t) = yJ
, X|J (t) = 0)
N −k
N −k

(w), w = (wij )1≤i≤j≤k ∈ Rk(k+1)/2 . Similarly to the proof in Lemma
by ft,yJ ,··· ,yJ
1
N −k
2.3.2, to estimate

X(t)=x, X|J (t)=0,
E det 2 X|J (t)1{ 2 X (t)∈D } X (t)=y ,··· ,X
,
J1
J1
JN −k (t)=yJN −k
k
|J

32

we will get an expression similar to (2.3.7) with Z(t, x) replaced by Z(t, x, yJ1 , · · · , yJ
).
N −k
Then, similarly, we have

I(t, x) :=

···
E(J)

dyJ1 · · · dyJ

p
(x, yJ1 , · · ·
N −k X(t),XJ1 (t),··· ,XJN −k (t)

, yJ

|

, yJ

|

N −k

| X|J (t) = 0)|Z(t, x, yJ1 , · · · , yJ
)|
N −k
≤

···
E(J)

dyJ1 · · · dyJ

| X|J (t) = 0)|

p
(x, yJ1 , · · ·
N −k X(t),XJ1 (t),··· ,XJN −k (t)

x
det (wij ) − 2 Ik
γt
(wij ) ≥ x
c

≤ pX(t) (x| X|J (t) = 0)

N −k

(w)dw
ft,yJ ,··· ,yJ
1
N −k

x
det (wij ) − 2 Ik
γt
(wij ) ≥ x
c

ft (w)dw,

where the last inequality comes from replacing the integral region E(J) by RN −k , and ft (w)
is the density of ((Wij (t))1≤i≤j≤k |X(t) = x,

X|J (t) = 0). Hence by the same discussions

2
−αu2 −u2 /(2σT )

in the proof of Lemma 2.3.2, I(t, x) = o(e

) uniformly for all t ∈ T and some

constant α > 0. Combining the proofs of Lemma 2.3.2 and Theorem 2.2.2, we obtain the
result.
We call a function h(u) super-exponentially small (when compared with P(supt∈T X(t) ≥
u)), if there exists a constant α > 0 such that h(u) = o(e

2
−αu2 −u2 /(2σT )

) as u → ∞.

The following lemma is Lemma 4 in Piterbarg (1996b). It shows that the factorial
moments are usually super-exponentially small.
Lemma 2.3.4 Let {X(t) : t ∈ RN } be a centered Gaussian field satisfying (H1) and (H3).
Then for any ε > 0, there exists ε1 > 0 such that for any J ∈ ∂k T and u large enouth,

E{Mu (J)(Mu (J) − 1)} ≤ e

2
−u2 /(2βJ +ε)

33

+e

2
−u2 /(2σJ −ε1 )

,

2
2
where βJ = supt∈J supe∈Sk−1 Var(X(t)| X|J (t), 2 X|J (t)e) and σJ = supt∈J Var(X(t)).

Here Sk−1 is the (k − 1)-dimensional unit sphere.
Corollary 2.3.5 Let X = {X(t) : t ∈ RN } be a centered Gaussian random field with stationary increments satisfying (H1), (H2) and (H3). Then for all J ∈ ∂k T , E{Mu (J)(Mu (J)−
E
E
1)} and E{Mu (J)(Mu (J) − 1)} are super-exponentially small.

E
Proof Since Mu (J) ≤ Mu (J), we only need to show that E{Mu (J)(Mu (J) − 1)} is super-

exponentially small. If k = 0, then Mu (J) is either 0 or 1 and hence E{Mu (J)(Mu (J)−1)} =
2
2
0. If k ≥ 1, then, thanks to Lemma 2.3.4, it suffices to show that βJ is strictly less than σT .
2
Clearly, Var(X(t)| X|J (t), 2 X|J (t)e) ≤ σT . Applying Lemma 2.5.1 yields that

2
Var(X(t)| X|J (t), 2 X|J (t)e) = σT ⇒ E{X(t)( 2 X|J (t)e)} = 0.

Note that the right hand side above is equivalent to (ΛJ (t) − ΛJ )e = 0. By (H2), ΛJ (t) − ΛJ
is negative definite, which implies (ΛJ (t) − ΛJ )e = 0 for all e ∈ Sk−1 , so that

2
sup Var(X(t)| |J X(t), 2 X(t)e) < σT .
|J

e∈Sk−1

2
2
Therefore βJ < σT by continuity.

The following lemma shows that the cross terms in (2.3.2) and (2.3.3) are super-exponentially
small if the two faces are not adjacent. For the case when the faces are adjacent, the proof
is more technical, see the proofs in Theorems 2.3.7 and 2.3.8.
Lemma 2.3.6 Let X = {X(t) : t ∈ RN } be a centered Gaussian random field with stationary increments satisfying (H1) and (H3). Let J and J be two faces of T such that their
34

distance is positive, i.e., inf t∈J,s∈J

s − t > δ0 for some δ0 > 0, then E{Mu (J)Mu (J )} is

super-exponentially small.

Proof

We first consider the case when dim(J) = k ≥ 1 and dim(J ) = k ≥ 1. By the

Kac-Rice metatheorem for higher moments (the proof is the same as that of Theorem 11.5.1
in Adler and Taylor (2007)),

E{Mu (J)Mu (J )} =

dt
J

J

ds E{|det 2 X|J (t)||det 2 X|J (s)|1{X(t)≥u,X(s)≥u}

× 1{ 2 X (t)∈D , 2 X (s)∈D } |X(t) = x, X(s) = y, X|J (t) = 0,
k
|J
|J
k
X|J (s) = 0}pX(t),X(s), X (t), X (s) (x, y, 0, 0)
|J
|J
∞

∞

≤

J

J

dx

ds

dt

u

u

dy E{|det 2 X|J (t)||det 2 X|J (s)|

|X(t) = x, X(s) = y, X|J (t) = 0, X|J (s) = 0}pX(t),X(s) (x, y)
× p X (t), X (s) (0, 0|X(t) = x, X(s) = y).
|J
|J
(2.3.9)
Note that the following two inequalities hold: for constants ai and bj ,
k

k

|ai |
i=1

|bj | ≤
j=1

1
k+k

k

k

|ai |k+k +
i=1

|bj |k+k

;

j=1

and for any Gaussian variable ξ and positive integer l,

E|ξ|l ≤ E(|Eξ| + |ξ − Eξ|)l ≤ 2l (|Eξ|l + E|ξ − Eξ|l ) ≤ 2l (|Eξ|l + Cl (Var(ξ))l/2 ),

35

where the constant Cl depends only on l. Combining these two inequalities with Lemma
2.5.1, we get that there exist some positive constants C1 and N1 such that for large x and y,

sup
t∈J,s∈J

E{|det 2 X|J (t)||det 2 X|J (s)||X(t) = x, X(s) = y,
X|J (t) = 0, X|J (s) = 0} ≤ C1 xN1 y N1 .
(2.3.10)

Also, there exists a positive constant C2 such that

sup

p X (t), X (s) (0, 0|X(t) = x, X(s) = y)
|J
|J

t∈J,s∈J

≤

sup (2π)−(k+k )/2 [detCov( X|J (t), X|J (s)|X(t) = x, X(s) = y)]−1/2 ≤ C2 .

t∈J,s∈J

(2.3.11)

Let ρ(δ0 ) = sup s−t >δ
0

E{X(t)X(s)}
σt σs

which is strictly less than 1 due to (H3), then ∀ε > 0,

there exists a positive constant C3 such that for all t ∈ J, s ∈ J and u large enough,
∞
u

∞
u

≤

xN1 y N1 pX(t),X(s) (x, y)dxdy = E{[X(t)X(s)]N1 1{X(t)≥u,X(s)≥u} }

E{[X(t) + X(s)]2N1 1{X(t)+X(s)≥2u} }

≤ C3 exp

εu2

u2
.
−
2
(1 + ρ(δ0 ))σT

(2.3.12)

E
E
Combine (2.3.9) with (2.3.10), (2.3.11) and (2.3.12), yielding that E{Mu (J)Mu (J )} is

super-exponentially small.

36

When only one of the faces, say J, is a singleton, then let J = {t0 } and we have
∞

E{Mu (J)Mu (J )} ≤

ds
J

∞

dx
u

× E{|det

u
2X

|J

dy pX(t ),X(s), X (s) (x, y, 0)
0
|J

(2.3.13)

(s)||X(t0 ) = x, X(s) = y, X|J (s) = 0}.

Following the previous discussions yields that E{Mu (J)Mu (J )} is super-exponentially small.
Finally, if both J and J are singletons, then E{Mu (J)Mu (J )} becomes the joint probability of two Gaussian variables exceeding level u and hence is trivial.

Theorem 2.3.7 Let X = {X(t) : t ∈ RN } be a centered Gaussian random field with
stationary increments such that (H1), (H2) and (H3) are fulfilled. Suppose that for any face
J,
2
{t ∈ J : ν(t) = σT , νj (t) = 0 for some j ∈ σ(J)} = ∅.
/

(2.3.14)

Then there exists some constant α > 0 such that
N

2
−αu2 −u2 /(2σT )

P sup X(t) ≥ u =
t∈T

E{Mu (J)} + o(e

)

k=0 J∈∂k T

=
{t}∈∂0 T

×

u
Ψ
+
σt

N

1

k=1 J∈∂k T

(2π)(k+1)/2 |Λ

1/2
J|

2 2
2
|ΛJ − ΛJ (t)|
u −u2 /(2θ2 )
t dt + o(e−αu −u /(2σT ) ).
Hk−1
e
k
θt
θt
J

(2.3.15)

Proof Since the second equality in (2.3.15) follows from Lemma 2.3.2 directly, we only need
to prove the first one. By (2.3.3) and Corollary 2.3.5, it suffices to show that the last term
in (2.3.3) is super-exponentially small. Thanks to Lemma 2.3.6, we only need to consider

37

¯ ¯
the case when the distance of J and J is 0, or I := J ∩ J = ∅. Without loss of generality,
assume

σ(J) = {1, . . . , m, m + 1, . . . , k}, σ(J ) = {1, . . . , m, k + 1, . . . , k + k − m},

(2.3.16)

where 0 ≤ m ≤ k ≤ k ≤ N and k ≥ 1. If k = 0, we consider σ(J) = ∅ by convention.
Under such assumption, J ∈ ∂k T , J ∈ ∂k T and dim(I) = m.
2
Case 1: k = 0, i.e. J is a singleton, say J = {t0 }. If ν(t0 ) < σT , then by (2.3.13),

it is trivial to show that E{Mu (J)Mu (J )} is super-exponentially small. Now we consider
2
the case ν(t0 ) = σT . Due to (2.3.14), E{X(t0 )X1 (t0 )} = 0 and hence by continuity, there

exists δ > 0 such that E{X(s)X1 (s)} = 0 for all s − t0 ≤ δ. It follows from (2.3.13) that
E{Mu (J)Mu (J )} is bounded from above by
∞

∞

dx

ds

u

u

s∈J : s−t0 >δ

dy E{|det 2 X|J (s)||X(t0 ) = x, X(s) = y, X|J (s) = 0}
× pX(t ),X(s), X (s) (x, y, 0)
0
|J

∞

ds

+
s∈J : s−t0 ≤δ

u

dy E{|det 2 X|J (s)||X(s) = y, X|J (s) = 0}pX(s), X (s) (y, 0)
|J

:= I1 + I2 .
Following the proof of Lemma 2.3.6 yields that I1 is super-exponentially small. We apply
Lemma 2.5.1 to obtain that there exists ε0 > 0 such that

sup
s∈J : s−t0 ≤δ

Var(X(s)| X|J (s)) ≤

sup
s∈J : s−t0 ≤δ

2
Var(X(s)|X1 (s)) ≤ σT − ε0 .

Then I2 and hence E{Mu (J)Mu (J )} are super-exponentially small.

38

2
Case 2: k ≥ 1. For all t ∈ I with ν(t) = σT , by assumption (2.3.14), E{X(t)Xi (t)} = 0,

∀ i = m + 1, . . . , k + k − m. Note that I is a compact set, by Lemma 2.5.1 and the uniform
continuity of conditional variance, there exist ε1 , δ1 > 0 such that

sup
t∈B,s∈B

2
Var(X(t)|Xm+1 (t), . . . , Xk (t), Xk+1 (s), . . . , Xk+k −m (s)) ≤ σT − ε1 ,

(2.3.17)

where B = {t ∈ J : dist(t, I) ≤ δ1 } and B = {s ∈ J : dist(s, I) ≤ δ1 }. It follows from
(2.3.9) that E{Mu (J)Mu (J )} is bounded by
∞

dtds
(J×J )\(B×B )

∞

dx
u

u

dy pX(t),X(s), X (t), X (s) (x, y, 0, 0)
|J
|J

× E{|det 2 X|J (t)||det 2 X|J (s)||X(t) = x, X(s) = y, X|J (t) = 0, X|J (s) = 0}
∞

dtds

+
B×B

u

dx pX(t) (x| X|J (t) = 0, X|J (s) = 0)p X (t), X (s) (0, 0)
|J
|J

× E{|det 2 X|J (t)||det 2 X|J (s)||X(t) = x,

X|J (t) = 0, X|J (s) = 0}

:= I3 + I4 .
Note that

(J × J )\(B × B ) = (J\B) × B

B × (J\B)

(J\B) × (J\B) .

(2.3.18)

Since each product set on the right hand side of (2.3.18) consists of two sets with positive
distance, following the proof of Lemma 2.3.6 yields that I3 is super-exponentially small.
For I4 , taking into account (2.3.17), one has

sup
t∈B,s∈B

2
Var X(t)| X|J (t), X|J (s) ≤ σT − ε1 .

39

(2.3.19)

To estimate

p X (t), X (s) (0, 0) = (2π)−(k+k )/2 (detCov( X|J (t), X|J (s)))−1/2 ,
|J
|J

(2.3.20)

we write the determinant on the right hand side of (2.3.20) as

detCov(Xm+1 (t), . . . , Xk (t), Xk+1 (s), . . . , Xk+k −1 (s)|X1 (t), . . . , Xm (t), X1 (s), . . . , Xm (s))
× detCov(X1 (t), . . . , Xm (t), X1 (s), . . . , Xm (s)),
(2.3.21)

where the first determinant in (2.3.21) is bounded away from zero due to (H3). By (H1), as
shown in Piterbarg (1996b), applying Taylor’s formula, we can write

X(s) =

X(t) +

2 X(t)(s − t)T

+ s − t 1+η Yt,s ,

(2.3.22)

1
N
where Yt,s = (Yt,s , . . . , Yt,s )T is a Gaussian vector field with bounded variance uniformly for

all t ∈ J, s ∈ J . Hence as s − t → 0, the second determinant in (2.3.21) becomes

detCov(X1 (t), . . . , Xm (t), X1 (t) +
Xm (t) +

1
X1 (t), s − t + s − t 1+η Yt,s , . . . ,

m
Xm (t), s − t + s − t 1+η Yt,s )

= detCov(X1 (t), . . . , Xm (t),

1
X1 (t), s − t + s − t 1+η Yt,s , . . . ,

m
Xm (t), s − t + s − t 1+η Yt,s )

= s − t 2m detCov(X1 (t), . . . , Xm (t),

X1 (t), et,s , . . . ,

Xm (t), et,s )(1 + o(1)),
(2.3.23)

40

where et,s = (s − t)T / s − t and due to (H3), the last determinant in (2.3.23) is bounded
away from zero uniformly for all t ∈ J and s ∈ J . It then follows from (2.3.21) and (2.3.23)
that

detCov( X|J (t), X|J (s)) ≥ C1 s − t 2m

(2.3.24)

for some constant C1 > 0. Similarly to (2.3.10), there exist constants C2 , N1 > 0 such that

sup
t∈J,s∈J

E{|det 2 X|J (t)||det 2 X|J (s)||X(t) = x,

X|J (t) = 0, X|J (s) = 0}
(2.3.25)

≤ C2 (1 + xN1 ).
Combining (2.3.19) with (2.3.20), (2.3.24) and (2.3.25), and noting that m < k implies
1/ s − t m is integrable on J × J , we conclude that I4 and hence E{Mu (J)Mu (J )} are
finite and super-exponentially small.

Theorem 2.3.8 Let X = {X(t) : t ∈ RN } be a centered Gaussian random field with
stationary increments such that (H1), (H2) and (H3) are fulfilled. Then there exists some
constant α > 0 such that
N
E
E{Mu (J)} + o(e

P sup X(t) ≥ u =
t∈T

2
−αu2 −u2 /(2σT )

k=0 J∈∂k T

)
(2.3.26)

2
−αu2 −u2 /(2σT )

= E{ϕ(Au )} + o(e

),

where E{ϕ(Au )} is formulated in Theorem 2.2.2.
It is worth mentioning here that the main idea for the proof of Theorem 2.3.8 comes
from Aza¨ and Delmas (2002) (especially Theorem 4). Before showing the proof, we list the
ıs
41

following two lemmas.
Lemma 2.3.9 Under (H2), there exists a constant α0 > 0 such that

e, (Λ − Λ(t))e ≥ α0 ,

∀ t ∈ T, e ∈ SN −1 .

Proof Let MN ×N be the set of all N ×N matrices. Define a mapping φ : RN ×MN ×N → R
by (ξ, A) → ξ, Aξ , then φ is continuous. Since Λ−Λ(t) is positive definite, φ(e, Λ−Λ(t)) > 0
for each t ∈ T and e ∈ SN −1 . On the other hand, {(e, Λ − Λ(t)) : t ∈ T, e ∈ SN −1 } is a
compact subset of RN × MN ×N and φ is continuous, completing the proof.

Lemma 2.3.10 Let {ξ1 (t) : t ∈ T1 } and {ξ2 (t) : t ∈ T2 } be two Gaussian random fields.
Let
2
σi (t) = Var(ξi (t)),

ρ(t, s) =

E{ξ1 (t)ξ2 (s)}
,
σ1 (t)σ2 (s)

σ i = sup σi (t),
t∈Ti

ρ=

sup
t∈T1 ,s∈T2

σ i = inf σi (t),

ρ(t, s),

t∈Ti

ρ=

inf

t∈T1 ,s∈T2

ρ(t, s),

and assume 0 < σ i ≤ σ i < ∞, where i = 1, 2. If 0 < ρ ≤ ρ < 1, then for any N1 , N2 > 0,
there exists some α > 0 such that as u → ∞,

sup
t∈T1 ,s∈T2

2 2
2
E{(1 + |ξ1 (t)|N1 + |ξ2 (s)|N2 )1{ξ (t)≥u,ξ (s)<0} } = o(e−αu −u /(2σ1 ) ).
1
2

Similarly, if −1 < ρ ≤ ρ < 0, then

sup
t∈T1 ,s∈T2

2 2
2
E{(1 + |ξ1 (t)|N1 + |ξ2 (s)|N2 )1{ξ (t)≥u,ξ (s)>0} } = o(e−αu −u /(2σ1 ) ).
1
2

42

Proof

We only prove the first case, since the second case follows from the first one. By

elementary computation on the joint density of ξ1 (t) and ξ2 (s), we obtain

sup
t∈T1 ,s∈T2

≤

E{(1 + |ξ1 (t)|N1 + |ξ2 (s)|N2 )1{ξ (t)≥u,ξ (s)<0} }
1
2
∞

1

2πσ 1 σ 2 (1 − ρ2 )1/2 u
0
−∞

(1 + |x1 |N1

= o exp

−

+ |x2

exp

|N2 ) exp

x2
− 12
2σ 1

dx1

σ 2 ρx1 2
1
x2 −
− 2
dx2
σ1
2σ 2 (1 − ρ2 )

σ 2 ρ 2 u2
u2
− 2 2
+ εu2
2
2 )σ 2
2σ 1 2σ 2 (1 − ρ 1

,

as u → ∞, for any ε > 0.

Proof of Theorem 2.3.8 Note that the second equality in (2.3.26) follows from Theorem
2.2.2 and Lemma 2.3.3, and similarly to the proof in Theorem 2.3.7, we only need to show
E
E
that E{Mu (J)Mu (J )} is super-exponentially small when J and J are neighboring. Let

¯ ¯
I := J ∩ J = ∅. We follow the assumptions in (2.3.16) and assume also that all elements in
ε(J) and ε(J ) are 1, which implies E(J) = RN −k and E(J ) = RN −k .
+
+

43

We first consider the case k ≥ 1. By the Kac-Rice metatheorem, E{Mu (J)Mu (J )} is
bounded from above by
∞

dt
J

ds
J

∞

dx
u

∞

dy
u

0

∞

dzk+1 · · ·

0

∞

dzk+k −m

0

∞

dwm+1 · · ·

0

dwk

E{ det 2 X|J (t)||det 2 X|J (s)||X(t) = x, X(s) = y, X|J (t) = 0, Xk+1 (t) = zk+1 ,
. . . , Xk+k −m (t) = zk+k −m , X|J (s) = 0, Xm+1 (s) = wm+1 , . . . , Xk (s) = wk }
× pt,s (x, y, 0, zk+1 , . . . , zk+k −m , 0, wm+1 , . . . , wk )
:=

A(t, s) dtds,
J×J

(2.3.27)

where pt,s (x, y, 0, zk+1 , . . . , zk+k −m , 0, wm+1 , . . . , wk ) is the density of

(X(t), X(s), X|J (t), Xk+1 (t), . . . , Xk+k −m (t), X|J (s), Xm+1 (s), . . . , Xk (s))

evaluated at (x, y, 0, zk+1 , . . . , zk+k −m , 0, wm+1 , . . . , wk ).
Let {e1 , e2 , . . . , eN } be the standard orthonormal basis of RN . For t ∈ J and s ∈ J , let
et,s = (s − t)T / s − t and let αi (t, s) = ei , (Λ − Λ(t))et,s , then
N

(Λ − Λ(t))et,s =

N

ei , (Λ − Λ(t))et,s ei =
i=1

αi (t, s)ei .

(2.3.28)

i=1

By Lemma 2.3.9, there exists some α0 > 0 such that

et,s , (Λ − Λ(t))et,s ≥ α0

44

(2.3.29)

for all t and s. Under the assumptions (2.3.16) and that all elements in ε(J) and ε(J ) are
1, we have the following representation,

t = (t1 , . . . , tm , tm+1 , . . . , tk , bk+1 , . . . , bk+k −m , 0, . . . , 0),
s = (s1 , . . . , sm , bm+1 , . . . , bk , sk+1 , . . . , sk+k −m , 0, . . . , 0),
where ti ∈ (ai , bi ) for all i ∈ σ(J) and sj ∈ (aj , bj ) for all j ∈ σ(J ). Therefore,

ei , et,s ≥ 0,

∀ m + 1 ≤ i ≤ k,

ei , et,s ≤ 0,

∀ k + 1 ≤ i ≤ k + k − m,

ei , et,s = 0,

∀ k + k − m < i ≤ N.

(2.3.30)

Let

Di = {(t, s) ∈ J × J : αi (t, s) ≥ βi },
Di = {(t, s) ∈ J × J : αi (t, s) ≤ −βi },

if m + 1 ≤ i ≤ k,
if k + 1 ≤ i ≤ k + k − m,

(2.3.31)

m

D0 =

αi (t, s) ei , et,s ≥ β0 ,

(t, s) ∈ J × J :
i=1

where β0 , β1 , . . . , βk+k −m are positive constants such that β0 +

k+k −m
i=m+1 βi

< α0 . It follows

from (2.3.30) and (2.3.31) that, if (t, s) does not belong to any of D0 , Dm , . . . , Dk+k −m ,
then by (2.3.28),
k+k −m

N

(Λ − Λ(t))et,s , et,s =

αi (t, s) ei , et,s ≤ β0 +
i=1

βi < α0 ,
i=m+1

45

which contradicts (2.3.29). Thus D0 ∪ ∪k+k −m Di is a covering of J × J , by (2.3.27),
i=m+1
k+k −m
E
E
E{Mu (J)Mu (J

We first show that

)} ≤

A(t, s) dtds +
D0

D0 A(t, s) dtds

A(t, s) dtds.
i=m+1

Di

is super-exponentially small. Similarly to the proof

of Theorem 2.3.7, applying (2.3.20), (2.3.24) and (2.3.25), we obtain

A(t, s) dtds
D0
∞

≤

dtds
u

D0

dx p X (t), X (s) (0, 0)pX(t) (x| X|J (t) = 0, X|J (s) = 0)
|J
|J

× E{|det 2 X|J (t)||det 2 X|J (s)||X(t) = x,
∞

≤ C1

dtds
D0

u

X|J (t) = 0, X|J (s) = 0}

dx(1 + xN1 ) s − t −m pX(t) (x| X|J (t) = 0, X|J (s) = 0),
(2.3.32)

for some positive constants C1 and N1 . Due to Lemma 2.3.6, we only need to consider the
case when s − t is small. It follows from Taylor’s formula (2.3.22) that as s − t → 0,

Var(X(t)| X|J (t), X|J (s)) ≤ Var(X(t)|X1 (t), . . . , Xm (t), X1 (s), . . . , Xm (s))
1
X1 (t), s − t + s − t 1+η Yt,s , . . . ,

= Var(X(t)|X1 (t), . . . , Xm (t), X1 (t) +
Xm (t) +

m
Xm (t), s − t + s − t 1+η Yt,s )

= Var(X(t)|X1 (t), . . . , Xm (t),

1
X1 (t), et,s + s − t η Yt,s , . . . ,

m
Xm (t), et,s + s − t η Yt,s )

≤ Var(X(t)|

1
X1 (t), et,s + s − t η Yt,s , . . . ,

= Var(X(t)|

X1 (t), et,s , . . . ,

m
Xm (t), et,s + s − t η Yt,s )

Xm (t), et,s ) + o(1).

46

(2.3.33)

By Lemma 2.5.2, the eigenvalues of [Cov(

X1 (t), et,s , . . . ,

Xm (t), et,s )]−1 are bounded

Xi (t), et,s } = −αi (t, s). Applying these facts and

uniformly in t and s. Note that E{X(t)

Lemma 2.5.1 to the last line of (2.3.33), we see that there exist constants C2 > 0 and ε0 > 0
such that for s − t sufficiently small,
m

Var(X(t)| X|J (t), X|J (s)) ≤

2
σT

2
2
αi (t, s) + o(1) < σT − ε0 ,

− C2

(2.3.34)

i=1

where the last inequality is due to the fact that (t, s) ∈ D0 implies
m

m
2
αi (t, s)

2
αi (t, s)|

≥

ei , et,s

i=1

i=1

|2

1
≥
m

m

2

αi (t, s) et,s , ei
i=1

2
β0
≥
.
m

Plugging (2.3.34) into (2.3.32) and noting that 1/ s−t m is integrable on J ×J , we conclude
that

D0 A(t, s) dtds

is finite and super-exponentially small.

Next we show that

Di

follows from (2.3.27) that

Di

dx
u

Di

A(t, s) dtds is bounded by

∞

∞

dtds

A(t, s) dtds is super-exponentially small for i = m + 1, . . . , k. It

0

dwi pX(t), X (t),X (s), X (s) (x, 0, wi , 0)
i
|J
|J

× E{|det 2 X|J (t)||det 2 X|J (s)||X(t) = x,

X|J (t) = 0, Xi (s) = wi , X|J (s) = 0}.
(2.3.35)

We can write

pX(t),X (s) (x, wi |Xi (t) = 0) =
i

1
2πσ1 (t)σ2 (t, s)(1 − ρ2 (t, s))1/2

× exp

−

w2
1
x2
2ρ(t, s)xwi
+ 2 i −
2 (t, s)) σ 2 (t)
2(1 − ρ
σ2 (t, s) σ1 (t)σ2 (t, s)
1

47

,

where
2
σ1 (t) = Var(X(t)|Xi (t) = 0),

ρ(t, s) =

2
σ2 (t, s) = Var(Xi (s)|Xi (t) = 0) =

E{X(t)Xi (s)|Xi (t) = 0}
,
σ1 (t)σ2 (t, s)

detCov(Xi (s), Xi (t))
,
λii

and ρ2 (t, s) < 1 due to (H3). Therefore,

pX(t), X (t),X (s), X (s) (x, 0, wi , 0)
i
|J
|J
= p X (s),X (t),...,X (t),X (t),...,X (t) (0|X(t) = x, Xi (s) = wi , Xi (t) = 0)
1
i−1
i+1
k
|J
× pX(t),X (s) (x, wi |Xi (t) = 0)pX (t) (0)
i
i
≤ C3 exp

(2.3.36)

2
wi
1
x2
2ρ(t, s)xwi
−
+ 2
−
2 (t, s)) σ 2 (t)
2(1 − ρ
σ2 (t, s) σ1 (t)σ2 (t, s)
1

× (detCov(X(t), X|J (t), Xi (s), X|J (s)))−1/2
for some positive constant C3 . Also, by similar arguments in the proof of Theorem 2.3.7,
there exist positive constants C4 , C5 , C6 , C7 , N2 and N3 such that

detCov( X|J (t), Xi (s), X|J (s)) ≥ C4 s − t 2(m+1) ,

(2.3.37)

2
C5 s − t 2 ≤ σ2 (t, s) ≤ C6 s − t 2 ,

(2.3.38)

48

and
E{|det 2 X|J (t)||det 2 X|J (s)||X(t) = x,

X|J (t) = 0, Xi (s) = wi , X|J (s) = 0}

= E{|det 2 X|J (t)||det 2 X|J (s)||X(t) = x,

X|J (t) = 0,

Xi (t), et,s = wi / s − t + o(1), X|J (s) = 0}
≤ C7 (xN2 + (wi / s − t )N3 + 1).
(2.3.39)

Combining (2.3.35) with (2.3.36), (2.3.37) and (2.3.39), and making change of variable w =
wi / s − t , we obtain that for some positive constant C8 ,

A(t, s) dtds
Di

dx

Di

× exp

−

−

dwi (xN2 + (wi / s − t )N3 + 1)

w2
1
2ρ(t, s)xwi
+ 2 i −
2
2(1 − ρ2 (t, s)) σ1 (t) σ2 (t, s) σ1 (t)σ2 (t, s)
−m

Di

× exp

0

u
x2

dtds s − t

= C8

∞

∞

dtds s − t −m−1

≤ C8

∞

∞

dx
u

(2.3.40)

dw(xN2 + wN3 + 1)

0

1
x2
w2
2ρ(t, s)xw
+ 2
−
2 (t, s)) σ 2 (t)
2(1 − ρ
σ2 (t, s) σ1 (t)σ2 (t, s)
1

,

where σ2 (t, s) = σ2 (t, s)/ s − t is bounded by (2.3.38). Applying Taylor’s formula (2.3.22)
to Xi (s) and noting that E{X(t)

ρ(t, s) =
=

Xi (t), et,s } = −αi (t, s), we obtain

1
1
E{X(t)Xi (s)} −
E{X(t)Xi (t)}E{Xi (s)Xi (t)}
σ1 (t)σ2 (t, s)
λii
s−t
σ1 (t)σ2 (t, s)

i
− αi (t, s) + s − t η E{X(t)Yt,s }

−

s−t η
i
E{X(t)Xi (t)}E{Xi (t)Yt,s } .
λii

49

(2.3.41)

By (2.3.38) and the fact that (t, s) ∈ Di implies αi (t, s) ≥ βi > 0 for i = m + 1, . . . , k,
we conclude that ρ(t, s) ≤ −δ0 for some δ0 > 0 uniformly for t ∈ J, s ∈ J with s − t
sufficiently small. Then applying Lemma 2.3.10 to (2.3.40) yields that

Di

A(t, s) dtds is

super-exponentially small.
It is similar to prove that

Di

A(t, s) dtds is super-exponentially small for i = k +

1, . . . , k + k − m. In fact, in such case,
∞

dtds
Di

Di

A(t, s) dtds is bounded by

∞

dx
u

0

dzi pX(t), X (t),X (t), X (s) (x, 0, zi , 0)
i
|J
|J

× E{|det 2 X|J (t)||det 2 X|J (s)||X(t) = x,

X|J (t) = 0, Xi (t) = zi , X|J (s) = 0}.

We can follow the proof in the previous stage by exchanging the positions of Xi (s) and Xi (t)
and replacing wi with zi . The details are omitted since the procedure is very similar.
If k = 0, then m = 0 and σ(J ) = {1, . . . , k }. Since J becomes a singleton, we may let
J = {t0 }. By the Kac-Rice metatheorem, E{Mu (J)Mu (J )} is bounded by
∞

ds
J

∞

dx
u

∞

dy
u

0

∞

dz1 · · ·

0

dzk pt0 ,s (x, y, z1 , . . . , zk , 0)

× E{|det 2 X|J (s)||X(t0 ) = x, X(s) = y, X1 (t0 ) = z1 , . . . , Xk (t0 ) = zk , X|J (s) = 0}
:=
J

A(t0 , s) ds,

where pt0 ,s (x, y, z1 , . . . , zk , 0) is the density of (X(t0 ), X(s), X1 (t0 ), . . . , Xk (t0 ), X|J (s))
evaluated at (x, y, z1 , . . . , zk , 0). Similarly, J could be covered by ∪k Di with Di = {s ∈
i=1

50

J : αi (t0 , s) ≤ −βi } for some positive constants βi , 1 ≤ i ≤ k . On the other hand,
∞
Di

A(t0 , s) ds ≤

ds
Di

∞

dx
u

0

dzi pX(t ),X (t ), X (s) (x, zi , 0)
0
i 0
|J

× E{|det 2 X|J (s)||X(t0 ) = x, Xi (t0 ) = zi , X|J (s) = 0}.
E
E
By similar discussions, we obtain that E{Mu (J)Mu (J )} is super-exponentially small and

hence complete the proof.

2.4

Further Remarks and Examples

Remark 2.4.1 (The case when T contains the origin). We now show that Theorem 2.3.7
and Theorem 2.3.8 still hold when T contains the origin. In such case, (H3) is actually
not satisfied since X(0) = 0 is degenerate. However, we may construct a small open cube
2
T0 containing 0 such that supt∈T0 σt is sufficiently small, then according to the Borell-TIS

inequality, P{supt∈T0 X(t) ≥ u} is super-exponentially small. Let T = T \T0 , then

P sup X(t) ≥ u ≤ P sup X(t) ≥ u ≤ P sup X(t) ≥ u + P sup X(t) ≥ u . (2.4.1)
t∈T

t∈T

t∈T0

t∈T

To estimate P{supt∈T X(t) ≥ u}, similarly to the rectangle T , we decompose T into several
faces by lower dimensions such that T = ∪N ∂k T = ∪N ∪L∈∂ T L. Then we can get the
k=0
k=0
k
bounds similar to (2.3.3) with T replaced with T and J replaced with L. Following the proof
of Theorem 2.3.7 yields
N

P sup X(t) ≥ u =
t∈T

2
−αu2 −u2 /(2σT )

E{Mu (L)} + o(e
k=0 L∈∂ T
k

51

).

2
Due to the fact that supt∈T0 σt is sufficiently small, E{Mu (L)} are super-exponentially small

¯
¯
for all faces L such that L ⊂ ∂k T0 with 0 ≤ k ≤ N − 1 (note that T0 is a closed rectangle).
The same reason yields that for 1 ≤ k ≤ N , L ∈ ∂k T , J ∈ ∂k T such that L ⊂ J, the
difference between E{Mu (L)} and E{Mu (J)} is super-exponentially small. Hence we obtain

P sup X(t) ≥ u =
t∈T

Ψ
{t}∈∂0 T

×

u
+
σt

N

1

k=1 J∈∂k T

(2π)(k+1)/2 |ΛJ |1/2

2 2
2
|ΛJ − ΛJ (t)|
u −u2 /(2θ2 )
t dt + o(e−αu −u /(2σT ) ).
Hk−1
e
k
θt
θt
J

(2.4.2)
2
2
Here, by convention, if θt = 0, we regard e−u /(2θt ) as 0. Combining (2.4.1) with (2.4.2),

we conclude that Theorem 2.3.7 still holds when T contains the origin. The arguments for
Theorem 2.3.8 are similar.
Example 2.4.2 (Refinements of Theorem 2.3.7). Let Gaussian field X be as in Theorem
2
2
2.3.7. Suppose that ν(t0 ) = σT for some t0 ∈ J ∈ ∂k T (k ≥ 0) and ν(t) < σT for all t = t0 .
2
2
(i). If k = 0, then, due to (2.3.14), supt∈T \{t } θt ≤ σT −ε0 for some ε0 > 0. This implies
0

that E{Mu (J )} are super-exponentially small for all faces J other than {t0 }. Therefore,

P sup X(t) ≥ u = Ψ
t∈T

u
σT

+ o(e

2
−u2 /(2σT )+αu2

),

as u → ∞.

(2.4.3)

2
For example, let Y be a stationary Gaussian field with covariance ρ(t) = e− t and define

X(t) = Y (t) − Y (0), then X is a smooth Gaussian field with stationary increments satisfying
conditions (H1)-(H3). Let T = [0, 1]N , then we can apply (2.4.3) to approximate the
excursion probability of X with t0 = (1, . . . , 1).

52

(ii). If k ≥ 1, then similarly, E{Mu (J )} are super-exponentially small for all faces
J = J. It follows from Theorem 2.3.7 that

P sup X(t) ≥ u =
t∈T

uk−1

|ΛJ − ΛJ (t)| −u2 /(2θ2 )
t dt(1 + o(1)).
e
2k−1
(2π)(k+1)/2 |ΛJ |1/2 J
θt

2
Let τ (t) = θt , then ∀i ∈ σ(J), τi (t0 ) = 0, since t0 is a local maximum point of τ restricted

on J. Assume additionally that the Hessian matrix

ΘJ (t0 ) := (τij (t0 ))i,j∈σ(J)

(2.4.4)

2
is negative definite, then the Hessian matrix of 1/(2θt ) at t0 restricted on J,

ΘJ (t0 ) = −

1
2τ 2 (t

0)

(τij (t0 ))i,j∈σ(J) = −

1
Θ (t ),
4 J 0
2σT

2k−1 and h(t) = 1/(2θ 2 ), applying Lemma
is positive definite. Let g(t) = |ΛJ − ΛJ (t)|/θt
t

2.5.3 with T replaced with J gives us that as u → ∞,

P sup X(t) ≥ u =
t∈T

uk−1 |ΛJ − ΛJ (t0 )|

(2π)k/2

2k−1 uk |Θ (t )|1/2
(2π)(k+1)/2 |ΛJ |1/2 θt
J 0

e

2
−u2 /(2θt )
0 (1 + o(1))

0

2k/2 |Λ

(2.4.5)

u
J − ΛJ (t0 )|
=
Ψ
(1 + o(1)).
1/2 | − Θ (t )|1/2
σT
|ΛJ |
J 0

Example 2.4.2 (Continued: the cosine field). We consider the cosine random field on R2 :
2

1
Z(t) = √
(ξi cos ti + ξi sin ti ),
2 i=1

53

t = (t1 , t2 ) ∈ R2 ,

where ξ1 , ξ1 , ξ2 , ξ2 are independent, standard Gaussian variables. Z is a well-known centered,
unit-variance and smooth stationary Gaussian field [cf. Adler and Taylor (2007, p.382)].
Note that Z is periodic and Z(t) = −Z11 (t) − Z22 (t). To avoid such degeneracy, let X(t) =
ξ0 + Z(t) − Z(0), where t ∈ T ⊂ [0, 2π)2 and ξ0 is a standard Gaussian variable independent
of Z. Then X is a centered and smooth Gaussian field with stationary increments. The
variance and covariance of X are given respectively by

2
ν(t) = σt = 3 − cos t1 − cos t2 ,

1
C(t, s) = 2 +
2

(2.4.6)

2

[cos(ti − si ) − cos ti − cos si ].
i=1

Therefore, X satisfies conditions (H1), (H2) and (H3) [though X12 (t) ≡ 0, it can be shown
that this does not affect the validity of Theorems 2.3.7 and 2.3.8]. Taking the partial derivatives of C gives us that
1
E{X(t) X(t)} = (sin t1 , sin t2 )T ,
2

1
Λ = Cov( X(t)) = I2 ,
2
1
Λ − Λ(t) = −E{X(t) 2 X(t)} = [I2 − diag(cos t1 , cos t2 )],
2

(2.4.7)

where I2 is the 2 × 2 unit matrix and diag denotes the diagonal matrix.
(i).

Let T = [0, π/2]2 . Then by (2.4.6), ν attains its maximum 3 only at the cor-

ner (π/2, π/2), where both partial derivatives of ν are positive. Applying the result (i) in
√
2
Example 2.4.2, we obtain P{supt∈T X(t) ≥ u} = Ψ(u/ 3)(1 + o(e−αu )).
(ii). Let T = [0, 3π/2] × [0, π/2]. Then ν attains its maximum 4 only at the boundary
point t∗ = (π, π/2), where ν2 (t∗ ) > 0 so that the condition (2.3.14) is satisfied. In this case,
1
t∗ ∈ J = (0, 3π/2) × {π/2}. By (2.4.7), we obtain ΛJ = 2 and ΛJ − ΛJ (t∗ ) = 1 (1 − cos t∗ ) =
1
2

54

1. On the other hand, for t ∈ J, by Lemma 2.5.1 and (2.4.7),

2
τ (t) = θt = Var(X(t)|X1 (t)) = 3 − cos t1 − cos t2 −

1 2
sin t1 ,
2

(2.4.8)

therefore ΘJ (t∗ ) = τ11 (t∗ ) = −2. Plugging these into (2.4.5) with k = 1 gives us that
√
P{supt∈T X(t) ≥ u} = 2Ψ(u/2)(1 + o(1)).
(iii). Let T = [0, 3π/2]2 . Then ν attains its maximum 5 only at the interior point
t∗ = (π, π). In this case, t∗ ∈ J = (0, 3π/2)2 . By (2.4.7), we obtain ΛJ = 1 I2 and
2
ΛJ − ΛJ (t∗ ) = I2 . On the other hand, for t ∈ J, by Lemma 2.5.1 and (2.4.7),

2
τ (t) = θt = Var(X(t)|X1 (t), X2 (t)) = 3 − cos t1 − cos t2 −

1 2
1
sin t1 − sin2 t2 ,
2
2

(2.4.9)

therefore ΘJ (t∗ ) = (τij (t∗ ))i,j=1,2 = −2I2 . Plugging these into (2.4.5) with k = 2 gives us
√
that P{supt∈T X(t) ≥ u} = 2Ψ(u/ 5)(1 + o(1)).
Example 2.4.3 (Refinements of Theorem 2.3.8). Let X be a Gaussian field as in Theorem
2
2.3.8. Suppose t0 ∈ J ∈ ∂k T is the only point in T such that ν(t0 ) = σT . Assume

σ(J) = {1, . . . , k}, all elements in ε(J) are 1, νk (t0 ) = 0 for all k + 1 ≤ k ≤ N . Then by
Theorem 2.3.8,
N

P sup X(t) ≥ u =
t∈T

2
−αu2 −u2 /(2σT )

E
E{Mu (J)} +

E
E{Mu (J )} + o(e

).

¯ ¯
k =k+1 J ∈∂ T,J ∩J=∅
k

(2.4.10)

55

E
Lemma 2.3.3 indicates E{Mu (J)} = (−1)k E{

E
E{Mu (J)} = (−1)k

k
i
−αx2 )),
i=0 (−1) µi (J)}(1 + o(e

therefore

p X (t) (0)dt E{det 2 X|J (t)1
{(X
|J

N −k }
k+1 (t),...,XN (t))∈R+

J

2

× 1{X(t)≥u} | X|J (t) = 0}(1 + o(e−αx ))
2
2
(−1)k e−x /(2θt )

∞

=

dx
u

dt
J

(2π)(k+1)/2 |Λ

1/2 θ
t
J|

E{det 2 X|J (t)1
{(X

|X(t) = x,
∞

N −k }
k+1 (t),...,XN (t))∈R+
2

X|J (t) = 0}(1 + o(e−αu ))

2

AJ (x)dx(1 + o(e−αu )),

:=
u

(2.4.11)
and similarly,

E
E{Mu (J

2
2
(−1)k e−x /(2θt )

∞

dt

dx

)} =

J

u

(2π)(k +1)/2 |Λ

×1

{(X (t),...,X
J1
J

N −k

1/2 θ
t
J |

E{det 2 X|J (t)

|X(t)
(t))∈RN −k }
+

= x,

2

X|J (t) = 0}(1 + o(e−αu )).

(i). First we consider the case k ≥ 1. We shall follow the notations τ (t), ΘJ (t) and
2
ΘJ (t) in Example 2.4.2. Let h(t) = 1/(2θt ) and

gx (t) =

(−1)k
(2π)(k+1)/2 |ΛJ |1/2 θt

E{det 2 X|J (t)1
{(X

N −k }
k+1 (t),...,XN (t))∈R+

|X(t) = x,

X|J (t) = 0}.

Note that supt∈T |gx (t)| = o(xN1 ) for some N1 > 0 as x → ∞, which implies that the growth
2
of gx (t) can be dominated by the exponential decay e−x h(t) , hence both Lemma 2.5.3 and

2.5.4 are still applicable. Applying Lemma 2.5.3 with T replaced by J and u replaced by x2 ,

56

we obtain that as x → ∞,

AJ (x) =

(2π)k/2
xk (detΘJ (t0 ))1/2

2
−x2 /(2σT )

gx (t0 )e

(1 + o(1)).

(2.4.12)

On the other hand, it follows from (2.2.17) that

gx (t) =

1
(2π)(k+1)/2 |Λ
×

1/2 θ
t
J|

···

RN −k
+

dyk+1 · · · dyN

|ΛJ − ΛJ (t)|
x
Hk
+ γt Ck+1 (t)yk+1 + · · · + γt CN (t)yN
k
γt
γt

× pX

k+1 (t),...,XN (t)

Note that X(t0 ) and

gx (t0 ) =

(yk+1 , . . . , yN |X(t) = x,

X|J (t) = 0).

X(t0 ) are independent, and Cj (t0 ) = 0 for all 1 ≤ j ≤ N . Therefore,
|ΛJ − ΛJ (t0 )|
H
(k+1)/2 |Λ |1/2 σ k+1 k
(2π)
J
T

x
σT

× P{(Xk+1 (t0 ), . . . , XN (t0 )) ∈ RN −k | X|J (t0 ) = 0}.
+
Plugging this and (2.4.12) into (2.4.11), we obtain

E
E{Mu (J)} =

u
2k/2 |ΛJ − ΛJ (t0 )|
Ψ
σT
|ΛJ |1/2 | − ΘJ (t0 )|1/2

(2.4.13)

× P{(Xk+1 (t0 ), . . . , XN (t0 )) ∈ RN −k | X|J (t0 ) = 0}(1 + o(1)).
+
¯
¯
For J ∈ ∂k T with J ∩ J = ∅, similarly, applying Lemma 2.5.4 with T replaced by J , we

57

obtain that

E
E{Mu (J )} =

2k /2 |ΛJ − ΛJ (t0 )|
u
Ψ
P{ZJ (t0 ) ∈ Rk −k }
−
1/2 | − Θ (t )|1/2
σT
|ΛJ |
J 0
× P{(XJ (t0 ), . . . , XJ
1

N −k

(t0 )) ∈ RN −k | X|J (t0 ) = 0}(1 + o(1)),
+
(2.4.14)

where ZJ (t0 ) is a centered (k − k)-dimensional Gaussian vector with covariance matrix
−(τij )i,j∈σ(J )\σ(J) . Plugging (2.4.13) and (2.4.14) into (2.4.10), we obtain the asymptotic
result.
(ii). k = 0, say J = {t0 }. Note that X(t0 ) and

E
E{Mu (J)} = Ψ

X(t0 ) are independent, therefore

u
P{ X(t0 ) ∈ RN }.
+
σT

(2.4.15)

E
¯
¯
For J ∈ ∂k T with J ∩ J = ∅, then E{Mu (J )} is given by (2.4.14) with k = 0. Plug-

ging (2.4.15) and (2.4.14) into (2.4.10), we obtain the asymptotic formula for the excursion
probability.
Example 2.4.3 (Continued: the cosine field). We consider the Gaussian field X defined
in the continued part of Example 2.4.2.
(i). Let T = [0, π]2 . Then ν attains its maximum 5 only at the corner t∗ = (π, π), where
ν(t∗ ) = 0 so that the condition (2.3.14) is not satisfied. Instead, we will use the result
(ii) in Example 2.4.3 with J = {t∗ } and k = 0. Let J = (0, π) × {π}, J = {π} × (0, π).
Combining the results in the continued part of Example 2.4.2 with (2.4.15) and (2.4.14), and

58

noting that Λ = 1 I2 implies X1 (t) and X2 (t) are independent for all t, we obtain
2
√
1
E
E{Mu (J)} = Ψ(u/ 5),
4

√
1
E
E{Mu (∂2 T )} = Ψ(u/ 5)(1 + o(1)),
2
√
√
2
E
E
E{Mu (J )} = E{Mu (J )} =
Ψ(u/ 5)(1 + o(1)).
4
√
√
Summing these up, we have P{supt∈T X(t) ≥ u} = [(3 + 2 2)/4]Ψ(u/ 5)(1 + o(1)).
(ii). Let T = [0, 3π/2] × [0, π]. Then ν attains its maximum 5 only at the boundary
point t∗ = (π, π), where ν2 (t∗ ) = 0. Applying the result (i) in Example 2.4.3 with J =
(0, 3π/2) × {π} and k = 1, we obtain
√
E
E{Mu (J)}

=

√
2
Ψ(u/ 5),
2

which implies P{supt∈T X(t) ≥ u} = [(2 +

√
E
E{Mu (∂2 T )} = Ψ(u/ 5)(1 + o(1)),
√

√
2)/2]Ψ(u/ 5)(1 + o(1)).

Remark 2.4.4 Note that we only provide the first-order approximation for the examples
in this section. However, as shown in the theory of the approximations of integrals (see
e.g. Wong (2001)), the integrals in (2.3.15) and (2.3.26) can be expanded with more terms
once the covariance function of the Gaussian field is smooth enough. Hence for the examples above, higher-order approximation is available. Since the procedure is similar and the
computation is tedious, we omit such arguments here.

2.5

Some Auxiliary Facts

The following lemma is well-known and is quoted here for reader’s convenience.
Lemma 2.5.1 Let Y and Z be two Gaussian random vectors of dimension p and q, respectively. Then Y |Z = z is still a p-dimensional Gaussian random vector having the following
59

mean and covariance:
E{Y |Z = z} = EY + E{(Y − EY )(Z − EZ)T }[Cov(Z)]−1 (z − EZ),
Cov(Y |Z = z) = Cov(Y ) − E{(Y − EY )(Z − EZ)T }[Cov(Z)]−1 E{(Z − EZ)(Y − EY )T }.
In particular, if p = q = 1 and EY = EZ = 0, then

E{Y |Z = z} =

zE(Y Z)
,
Var(Z)

Var(Y |Z = z) = Var(Y ) −

(E(Y Z))2
.
Var(Z)

Using similar arguments in the proof of Lemma 2.3.9, we can obtain the following result.
Lemma 2.5.2 Let {A(t) = (aij (t))1≤i,j≤N : t ∈ T } be a family of positive definite matrices
such that all elements aij (·) are continuous. Denote by x and x the infimum and supremum
of the eigenvalues of A(t) over t ∈ T respectively, then 0 < x ≤ x < ∞.
The following two formulas state the results on the Laplace method approximation.
Lemma 2.5.3 can be found in many books on the approximations of integrals, here we refer
to Wong (2001). Lemma 2.5.4 can be derived by following similar arguments in the proof of
Laplace method for the case of boundary point in Wong (2001).
Lemma 2.5.3 (Laplace method for interior point). Let t0 be an interior point of T . Suppose
the following conditions hold: (i) g(t) ∈ C(T ) and g(t0 ) = 0; (ii) h(t) ∈ C 2 (T ) and attains
its unique minimum at t0 ; and (iii)

g(t)e−uh(t) dt =
T

2 h(t )
0

is positive definite. Then as u → ∞,

(2π)N/2
uN/2 (det 2 h(t

0

))1/2

g(t0 )e−uh(t0 ) (1 + o(1)).

Lemma 2.5.4 (Laplace method for boundary point). Let t0 ∈ J ∈ ∂k T with 0 ≤ k ≤ N −1.
Suppose that conditions (i), (ii) and (iii) in Lemma 2.5.3 hold, and additionally
60

h(t0 ) = 0.

Then as u → ∞,

g(t)e−uh(t) dt
T

=

(2π)N/2 P{ZJ (t0 ) ∈ (−E(J))}
uN/2 (det 2 h(t

0

))1/2

g(t0 )e−uh(t0 ) (1 + o(1)),

where ZJ (t0 ) is a centered (N − k)-dimensional Gaussian vector with covariance matrix
(hij (t0 ))J1 ≤i,j≤J

N −k

, −E(J) = {x ∈ RN : −x ∈ E(J)}, and the definitions of J1 , . . . , JN −k

and E(J) are in (2.2.4).

61

Chapter 3
Smooth Gaussian Random Fields with
Non-constant Variances

3.1

Gaussian Fields on Rectangles

Let {X(t) : t ∈ RN } be a smooth centered Gaussian random field with non-constant variance
and let T =

N
i=1 [ai , bi ]

be a closed rectangle in RN . In this Chapter, we study the excursion

probability of X over T .
2
Let ν(t) := σt = Var(X(t)) and assume supt∈T ν(t) = 1. A matrix is called negative

semidefinite if all of its eigenvalues are nonpositive. In addition to conditions (H1) and (H3)
in the previous chapter, we will make use of the following condition.
(H4). ∀t ∈ J ∈ ∂k T such that ν(t) = 1 and 0 ≤ k ≤ N − 2, (E{X(t)Xij (t)})i,j∈ζ(t) is
negative definite, where ζ(t) = {n : νn (t) = 0, 1 ≤ n ≤ N }.
Proposition 3.1.1 Let X(·) ∈ C 2 (RN ) a.s. If (νij (t))i,j∈ζ(t) is negative semidefinite for
each t ∈ J ∈ ∂k T such that ν(t) = 1 and 0 ≤ k ≤ N − 2, then (H4) holds.

Proof

Since 1 νij (t) = E{X(t)Xij (t)} + E{Xi (t)Xj (t)},
2
1
(E{X(t)Xij (t)})i,j∈ζ(t) = (νij (t))i,j∈ζ(t) − (E{Xi (t)Xj (t)})i,j∈ζ(t) .
2
62

(3.1.1)

But (νij (t))i,j∈ζ(t) is negative semidefinite and (E{Xi (t)Xj (t)})i,j∈ζ(t) is positive definite,
it follows that (E{X(t)Xij (t)})i,j∈ζ(t) is negative definite and hence (H4) holds.

Remark 3.1.2 In (H4), ν(t) = 1 implies νn (t) = 0 for all n ∈ σ(J) and thus ζ(t) ⊃ σ(J).
Additionally, we only consider t ∈ J ∈ ∂k T with 0 ≤ k ≤ N − 2, this is because for
N − 1 ≤ k ≤ N , (E{X(t)Xij (t)})i,j∈ζ(t) is automatically negative definite due to ν(t) = 1,
as shown below.
(i). If k = N , then t becomes a maximum point of ν in the interior of T , and ζ(t) =
σ(J) = {1, · · · , N }, hence (νij (t))i,j∈ζ(t) is always negative semidefinite. By (3.1.1), we see
that (E{X(t)Xij (t)})i,j∈ζ(t) is negative definite.
(ii). If k = N − 1, we distinguish two cases. If ζ(t) = σ(J), then (E{X(t)Xij (t)})i,j∈ζ(t)
is negative by the same arguments in the previous step. If ζ(t) = {1, · · · , N }, it follows from
Taylor’s formula that

ν(s) = ν(t) +

ν(t)(s − t)T + (s − t) 2 ν(t)(s − t)T + o( s − t 2 )
(3.1.2)

= ν(t) + (s − t)

2 ν(t)(s − t)T

+ o( s − t

2 ),

for all s ∈ T such that s − t is small enough. Since t ∈ J ∈ ∂N −1 T , {± s−t : s ∈
s−t
T } contains all the directions e ∈ SN −1 . Note that ν(t) = 1,
any positive eigenvalue and hence (νij (t))i,j∈ζ(t) =

2 ν(t)

2 ν(t)

does not have

is negative semidefinite, then

(E{X(t)Xij (t)})i,j∈ζ(t) is negative by (3.1.1).
If D is a subset (not necessary open) of J ∈ ∂k T , we define
E
Mu (D) = #{t ∈ D\∂D : X(t) ≥ u, X|J (t) = 0, index( 2 X|J (t)) = k,

ε∗ Xj (t) > 0 for all j ∈ σ(J)}.
/
j
63

For t ∈ J ∈ ∂k T , let

Λ(t) = (E{Xi (t)Xj (t)})1≤i,j≤N ,

ΛJ (t) = (E{Xi (t)Xj (t)})i,j∈σ(J) ,

Σ(t) = (E{X(t)Xij (t)})1≤i,j≤N ,

ΣJ (t) = (E{X(t)Xij (t)})i,j∈σ(J) ,

(3.1.3)

{1, · · · , N }\σ(J) = {J1 , · · · , JN −k },
E(J) = {(tJ1 , · · · , tJ

N −k

) ∈ RN −k : tj ε∗ ≥ 0, j = J1 , · · · , JN −k }.
j
M

1,j+1
Let Cj (t) be the (1, j + 1) entry of (Cov(X(t), X(t)))−1 , i.e. Cj (t) = detCov(X(t), X(t)) ,

where M1,j+1 is the cofactor of the (1, j +1) element, E{X(t)Xj (t)}, in the covariance matrix
Cov(X(t), X(t)).
Note that the notations Λ(t) and ΛJ (t) are different from those defined in Chapter 2.
The result Lemma 3.1.3 below follows immediately from similar argumentss in the proof
of Proposition 3.1.1.
Lemma 3.1.3 If t0 ∈ J ∈ ∂k T satisfies ν(t0 ) = 1, where k ≥ 1, then E{X(t0 )Xi (t0 )} = 0
for all i ∈ σ(J) and ΣJ (t0 ) is negative definite.
Corollary 3.1.4 Let {X(t) : t ∈ RN } be a Gaussian random field satisfying (H1), (H3)
and (H4), then there exists some constant α > 0 such that
N

2 2
E
E
E{Mu (J)(Mu (J) − 1)} = o(e−αu −u /2 ).

k=0 J∈∂k T

Proof

2
Due to Lemma 2.3.4, it suffices to show that βJ < 1 for each J ∈ ∂k T , which is

¯
equivalent to Var(X(t)| X|J (t), 2 X|J (t)e) < 1 for all t ∈ J = J ∪ ∂J and e ∈ Sk−1 by
continuity.

64

(i). Suppose Var(X(t0 )| X|J (t0 ), 2 X|J (t0 )e) = 1 for some t0 ∈ J, then

1 = Var(X(t0 )| X|J (t0 ), 2 X|J (t0 )e) ≤ Var(X(t0 )| 2 X|J (t0 )e) ≤ Var(X(t0 )) ≤ 1.

Note that
Var(X(t0 )| 2 X|J (t0 )e) = Var(X(t0 ))
⇔ E{X(t0 )( 2 X|J (t0 )e)} = 0
⇔ ΣJ (t0 )e = 0.
Since t0 is a maximum point, by Lemma 3.1.3, ΣJ (t0 ) is negative definite and hence
ΣJ (t0 )e = 0 for all e ∈ Sk−1 , which is a contradiction.
(ii). Suppose Var(X(t1 )| X|J (t1 ), 2 X|J (t1 )e) = 1 for some t1 ∈ ∂J. It then follows
from similar arguments in step (i) that Var(X(t1 )| X|J (t1 )) = 1 and hence νi (t1 ) = 0 for
all i ∈ σ(J), which implies (E{X(t1 )Xij (t1 )})i,j∈σ(J) is negative definite by (H4). Thus
there will be a contradiction as in step (i), completing the proof.

Lemma 3.1.5 Let {X(t) : t ∈ RN } be a Gaussian random field satisfying (H1), (H3) and
(H4). Then there exists some constant α > 0 such that as u → ∞,
N

2 2
E
E{Mu (J)} = E{ϕ(Au )} + o(e−αu −u /2 ).

(3.1.4)

k=0 J∈∂k T

Proof

Due to (2.2.1), we only need to show that for each k ∈ {0, 1, . . . , N } and J ∈ ∂k T ,
k
E
E{Mu (J)}

=

(−1)k

2 2
(−1)i E{µi (J)} + o(e−αu −u /2 ).

i=0

65

(3.1.5)

Without loss of generality, let σ(J) = {1, · · · , k} and assume all elements in ε(J) are 1. Let

¯
O(J) = {t ∈ J : ν(t) = 1} ∪ {t ∈ ∂J : ν(t) = 1, νi (t) = 0, ∀1 ≤ i ≤ k}.

¯
Our aim is to find an open neighborhood of O(J) restricted on J, say Uδ (J) = {t ∈ J :
¯
d(t, O(J)) < δ}, such that as u → ∞,
2 2
E
E
E{Mu (J)} = E{Mu (Uδ (J))} + o(e−αu −u /2 )
k

=

2 2
(−1)i E{µi (Uδ (J))} + o(e−αu −u /2 )

(−1)k
i=0
k

= (−1)k

(3.1.6)

2 2
(−1)i E{µi (J)} + o(e−αu −u /2 ).

i=0

For n = k + 1, . . . , N , let

¯
On (J) = {t ∈ J : ν(t) = 1, νn (t) = 0} ∪ {t ∈ ∂J : ν(t) = 1, νn (t) = 0, νi (t) = 0, ∀1 ≤ i ≤ k}.

¯
¯
Firstly, we consider the subset U 1 (J) = ∩N
n=k+1 On (J) and define its open neighborhood in
1
¯
J, Uδ (J) = {t ∈ J : d(t, U 1 (J)) < δ1 }, where δ1 is a small positive number to be specified.
1

E 1
Then, by the Kac-Rice metatheorem, E{Mu (Uδ (J))} becomes
1

∞

(−1)k
U 1 (J)
δ1

(2π)k/2 (detΛJ (t))1/2

|X(t) = x,
× pX(t),X

dt
u

RN −k
+

E{det 2 X|J (t)1{ 2 X (t)∈D } |
k
|J

X|J (t) = 0, Xk+1 (t) = yk+1 , . . . , XN (t) = yN }

k+1 (t),...,XN (t)

(x, yk+1 , . . . , yN | X|J (t) = 0) dxdyk+1 · · · dyN ,

where Dk is the collection of all k × k matrices with k negative eigenvalues.

66

(3.1.7)

Due to (H4) and continuity, we can choose δ1 small enough such that ΣJ (t) are neg1
ative definite for all t ∈ Uδ (J). Write
1

2X

|J (t)

= Q−1 Qt,J 2 X|J (t)Qt,J Q−1 , where
t,J
t,J

Qt,J (−ΣJ (t))Qt,J = Ik . Let al (t) = E{Xl (t)(Qt,J 2 X|J (t)Qt,J )ij } for l = 1, · · · , N , then
ij

E{(Qt,J 2 X|J (t)Qt,J )ij |X(t) = x,

X|J (t) = 0, Xk+1 (t) = yk+1 , · · · , XN (t) = yN }

= (E{X(t)(Qt,J 2 X|J (t)Qt,J )ij }, a1 (t), · · · , aN (t))(Cov(X(t), X(t)))−1
ij
ij
· (x, 0, · · · , 0, yk+1 , · · · , yN )T
= (−δij , a1 (t), · · · , aN (t))(Cov(X(t), X(t)))−1 (x, 0, · · · , 0, yk+1 , · · · , yN )T .
ij
ij
Make change of variables W (t) = (Wij (t))1≤i,j≤k , where

x
− 2 δij +
γt

Wij (t) = (Qt,J 2 X|J (t)Qt,J )ij −

N

al (t)Cl (t)x ,
ij
l=1

i.e.,

Qt,J

2X

x
|J (t)Qt,J = W (t) − 2 Ik + x
γt

N

al (t)Cl (t)
ij
l=1

1≤i,j≤k

= W (t) − xB(t),

where B(t) = 12 Ik − (
γt

N
l
l=1 aij (t)Cl (t))1≤i,j≤k .

((Wij (t))1≤i≤j≤k |X(t) = x,

Denote the density of

X|J (t) = 0, Xk+1 (t) = yk+1 , · · · , XN (t) = yN )

by gt (w), w = (wij : 1 ≤ i, j ≤ k) ∈ Rk(k+1)/2 , then gt (w) is independent of x. Let (wij ) be

67

the abbreviation of the k × k symmetric matrix (wij )1≤i≤j≤k , then
E{det 2 X|J (t)1D ( 2 X|J (t))|X(t) = x,
k

X|J (t) = 0, Xk+1 (t) = yk+1 , · · · , XN (t) = yN }

= det(−ΣJ (t))E{det(Qt,J 2 X|J (t)Qt,J )1{Q
2 X (t)Q ∈D } |
t,J
t,J
k
|J
|X(t) = x,
= det(−ΣJ (t))

X|J (t) = 0, Xk+1 (t) = yk+1 , · · · , XN (t) = yN }

(wij ):(wij )−xB(t)∈Dk

det((wij ) − xB(t))gt (w) dw.

¯
Since νn (t) = 0 for all t ∈ U 1 (J) and n = k +1, · · · , N , we can find δ1 small enough such that
2
1
Cl (t) are close to 0 for all l = 1, · · · , N and t ∈ Uδ (J). Together with the fact {γt : t ∈ J}
1

is bounded, there exists a constant c1 > 0 such that

(wij ) − xB(t) ∈ Dk ,

x
∀ (wij ) < .
c1

It then follows from similar arguments in Lemma 2.3.2 and Lemma 2.3.3 that
k
E 1
E{Mu (Uδ (J))} = (−1)k
1

2 2
1
(−1)i E{µi (Uδ (J))} + o(e−αu −u /2 ).
1

i=0

1
¯
¯
Next, we consider the subset U 2 (J) = (∩N −1 On (J)) \ Uδ (J), and define its neighborn=k+1
1

2
1
¯
hood Uδ (J) = {t ∈ J : d(t, U 2 (J)) < δ2 } \ Uδ (J), where δ2 is a small positive number to
2
1

68

E 2
be specified. Then we can write E{Mu (Uδ (J))} as
2

U 2 (J)
δ2

p X (t) (0) dt E{|det 2 X|J (t)|1{ 2 X (t)∈D } 1{X(t)≥u}
|J
k
|J
× 1{X

k+1 (t)>0,...,XN −1 (t)>0}

−

U 2 (J)
δ2

p X (t) (0) dt E{|det
|J

2X

| X|J (t) = 0}

(3.1.8)

|J (t)|1{ 2 X (t)∈D } 1{X(t)≥u}
k
|J

× 1{X

k+1 (t)>0,...,XN −1 (t)>0,XN (t)≤0}

| X|J (t) = 0}.

The second term in (3.1.8) is bounded by
∞
U 2 (J)
δ2

0

dx

dt
u

−∞

E{|det 2 X|J (t)||X(t) = x, XN (t) = yN , X|J (t) = 0}
(3.1.9)
× pX(t),X (t), X (t) (x, yN , 0, · · · , 0)dyN .
N
|J

Note that the conditional expectation in (3.1.9) can be bounded by c2 (|x|N1 + |yN |N2 ) when
u is large, for some positive constants c2 , N1 and N2 ; and

pX(t),X (t), X (t) (x, yN , 0, · · · , 0)
N
|J
= p X (t) (0, · · · , 0|X(t) = x, XN (t) = yN )pX(t),X (t) (x, yN )
N
|J
≤ c3 pX(t),X (t (x, yN )
N
¯
for some positive constant c3 . On the other hand, U 2 (J) is a compact set and for all
¯
t ∈ U 2 (J), νN (t) = 0 which implies νN (t) > 0 due to ν(t) = 1, thus we can choose δ2
2
sufficiently small such that E{X(t)XN (t)} > δ0 for all t ∈ Uδ (J) and some δ0 > 0. Hence
2

69

(3.1.9) is super-exponentially small by Lemma 2.3.10. Similar arguments give
k
2
(−1)i E{µi (Uδ (J))}

(−1)k

2

i=0

= (−1)k

U 2 (J)
δ2

p X (t) (0) dt E{det 2 X|J (t)1{X(t)≥u}
|J
2 2
X|J (t) = 0} + o(e−αu −u /2 ).

× 1{X

|
k+1 (t)>0,...,XN −1 (t)>0}

E 1
Combining this with (3.1.8), and following the same arguments to simplify E{Mu (Uδ (J))},
1

we obtain
k
E 2
E{Mu (Uδ (J))}
2

=

(−1)k

2 2
2
(−1)i E{µi (Uδ (J))} + o(e−αu −u /2 ).
2

i=0

Continue this procedure at most finite many times, and take the union of those disjoint
1
2
¯
neighborhoods (Uδ (J), Uδ (J), . . . ), we can find Uδ (J) = {t ∈ J : d(t, O(J)) < δ} inside
1
2

the union for some δ > 0 such that the second equality in (3.1.6) holds. On the other hand,
By the Kac-Rice metatheorem,

E
E
E
E{Mu (J)} − E{Mu (Uδ (J))} = E{Mu (J \ Uδ (J))}

≤

∞

1

1

(2π)k/2 J\Uδ (J) (detΛJ (t))1/2

dt

× E{|det 2 X|J (t)||X(t) = x,

u

pX(t) (x| X|J (t) = 0)

X|J (t) = 0} dx.

But, by the definition of Uδ (J) and continuity,

sup
t∈J\Uδ (J)

Var(X(t)| X|J (t) = 0) < 1 − ε0

70

(3.1.10)

for some ε0 > 0, hence the first equality in (3.1.6) holds and the third equality in (3.1.6)
follows similarly. We finish the proof.
Following the same proof in Lemma 2.3.6, we obtain the following result.
Lemma 3.1.6 Let X = {X(t) : t ∈ RN } be a centered Gaussian random field satisfying
(H1) and (H3). Let J and J be two faces of T such that their distance is positive, i.e.,
inf t∈J,s∈J

E
E
s − t > δ0 for some δ0 > 0, then E{Mu (J)Mu (J )} is super-exponentially

small.
The next result is an extension of Theorem 2.3.7.
Lemma 3.1.7 Let X = {X(t) : t ∈ RN } be a centered Gaussian random field satisfying
¯ ¯
(H1) and (H3). Let J and J be two neighboring faces, that is J ∩ J = ∅. Suppose

¯ ¯
{t ∈ J ∩ J : ν(t) = 1, νj (t) = 0, ∀j ∈ σ(J) ∪ σ(J )} = ∅,

(3.1.11)

E
E
then E{Mu (J)Mu (J )} is super-exponentially small.

Proof

Condition (3.1.11) implies that there exists ε0 > 0 such that

sup Var(X(t)| X|J (t), X|J (t)) ≤ 1 − ε0 ,

t∈U (δ)

¯ ¯
where U (δ) = {t ∈ J ∪ J : d(t, J ∩ J ) ≤ δ} and δ is a sufficiently small positive number.
Following the same proof in Theorem 2.3.7 yields our desired result.
The next result follows from similar arguments in the proof of Theorem 2.3.8.

71

Lemma 3.1.8 Let X = {X(t) : t ∈ RN } be a centered Gaussian random field satisfying
¯ ¯
(H1), (H3) and (H4). Let J and J be two neighboring faces, that is J ∩ J = ∅. Then
E
E
E{Mu (J)Mu (J )} is super-exponentially small.

Proof

¯ ¯
Let I = J ∩ J . We follow the assumptions in (2.3.16) and assume also that all

elements in ε(J) and ε(J ) are 1, which implies E(J) = RN −k and E(J ) = RN −k .
+
+
E
E
If condition (3.1.11) is satisfied, then E{Mu (J)Mu (J )} is super-exponentially small by

Lemma 3.1.7. So we will focus on the alternative case, which is

I0 := {t ∈ I : ν(t) = 1, νj (t) = 0, ∀1 ≤ j ≤ k + k − m} = ∅.

Let B(I0 , δ) = {t ∈ J ∪ J : d(t, I0 ) < δ}, where δ is a small number to be specified.
E
E
Note that E{Mu (J)Mu (J )} can be written as

E
E
E
E
E{[Mu (J ∩ B(I0 , δ)) + Mu (J ∩ B c (I0 , δ))][Mu (J ∩ B(I0 , δ)) + Mu (J ∩ B c (I0 , δ))]}
2 2
E
E
= E{Mu (J ∩ B(I0 , δ))Mu (J ∩ B(I0 , δ))} + o(e−αu −u /2 ),

since the rest terms are super-exponentially small by the same arguments in Lemma 3.1.7.
E
E
Therefore, to prove the result, we may estimate E{Mu (J ∩ B(I0 , δ))Mu (J ∩ B(I0 , δ))}
E
E
instead of E{Mu (J)Mu (J )} itself, with only a super-exponentially small difference. By

(H4),
ΣJ∪J (t) := E{X(t)Xij (t)}i,j=1,...,k+k −m
are negative definite for all t ∈ I0 , so that by continuity (similar to Lemma 2.3.9), we can

72

choose δ small enough such that

e, −ΣJ∪J (t)e ≥ α0 ,

∀t ∈ B(I0 , δ), e ∈ Sk+k −m−1

(3.1.12)

for some constant α0 > 0.
We first consider the case k ≥ 1. By the Kac-Rice metatheorem,

E
E
E{Mu (J ∩ B(I0 , δ))Mu (J ∩ B(I0 , δ))}
∞

≤

dt
J∩B(I0 ,δ)
∞
0

dzk+1 · · ·

ds
J ∩B(I0 ,δ)
∞
0

dzk+k −m

∞

dx
u

dy
u
∞

∞
0

dwm+1 · · ·

0

dwk

E{|det 2 X|J (t)||det 2 X|J (s)||X(t) = x, X(s) = y, X|J (t) = 0, Xk+1 (t) = zk+1 ,
. . . , Xk+k −m (t) = zk+k −m , X|J (s) = 0, Xm+1 (s) = wm+1 , . . . , Xk (s) = wk }
× pt,s (x, y, 0, zk+1 , . . . , zk+k −m , 0, wm+1 , . . . , wk )
A(t, s) dtds,

:=
(J∩B(I0 ,δ))×(J ∩B(I0 ,δ))

(3.1.13)

where pt,s (x, y, 0, zk+1 , . . . , zk+k −m , 0, wm+1 , . . . , wk ) is the density of

(X(t), X(s), X|J (t), Xk+1 (t), . . . , Xk+k −m (t), X|J (s), Xm+1 (s), . . . , Xk (s))

evaluated at (x, y, 0, zk+1 , . . . , zk+k −m , 0, wm+1 , . . . , wk ).
Let {e1 , e2 , . . . , ek+k −m } be the standard orthonormal basis of Rk+k −m . For t ∈ J
and s ∈ J , let et,s be the projection of (s − t)T / s − t on span{J, J }, and let αi (t, s) =

73

ei , −ΣJ∪J (t)et,s , then
k+k −m

k+k −m

ei , −ΣJ∪J (t)et,s ei =

−ΣJ∪J (t)et,s =

αi (t, s)ei .

i=1

(3.1.14)

i=1

It follows from (3.1.12) that
et,s , −ΣJ∪J (t)et,s ≥ α0

(3.1.15)

for all t ∈ J ∩ B(I0 , δ) and s ∈ J ∩ B(I0 , δ). Let

Di = {(t, s) ∈ (J ∩ B(I0 , δ)) × (J ∩ B(I0 , δ)) : αi (t, s) ≥ βi }, if m + 1 ≤ i ≤ k,
Di = {(t, s) ∈ (J ∩ B(I0 , δ)) × (J ∩ B(I0 , δ)) : αi (t, s) ≤ −βi }, if k + 1 ≤ i ≤ k + k − m,
m

D0 =

αi (t, s) ei , et,s ≥ β0 ,

(t, s) ∈ (J ∩ B(I0 , δ)) × (J ∩ B(I0 , δ)) :
i=1

(3.1.16)

where β0 , β1 , . . . , βk+k −m are positive constants such that β0 +

k+k −m
i=m+1 βi

< α0 . As in

the proof of Theorem 2.3.8, we see that D0 ∪ ∪k+k −m Di is a covering of (J ∩ B(I0 , δ)) ×
i=m+1
(J ∩ B(I0 , δ)). By (3.1.13),
E
E
E{Mu (J ∩ B(I0 , δ))Mu (J ∩ B(I0 , δ))}
k+k −m

≤

A(t, s) dtds +
D0

A(t, s) dtds.
i=m+1

Di

Following the same arguments in the proof of Theorem 2.3.8, we obtain that
and

Di

D0 A(t, s) dtds

A(t, s) dtds are all super-exponentially small for i = m + 1, . . . , k + k − m, com-

pleting the proof.

74

Theorem 3.1.9 Let X = {X(t) : t ∈ RN } be a centered Gaussian random field satisfying
(H1), (H3) and (H4). Then there exists some constant α > 0 such that as u → ∞,

2 2
P sup X(t) ≥ u = E{ϕ(Au )} + o(e−αu −u /2 ).
t∈T

Proof

The result follows from the combination of (2.3.1), (2.3.2), Corollary 3.1.4, Lemma

3.1.5, Lemma 3.1.6, Lemma 3.1.7 and Lemma 3.1.8.

3.2

Applications for Gaussian Fields with a Unique
Maximum Point of the Variance

In this section, we consider the case when ν(t) attains its maximum 1 at a unique point
t0 ∈ J ∈ ∂k T such that νj (t0 ) = 0 for all j ∈ σ(J).
/
Lemma 3.2.1 Let X be as in Theorem 3.1.9. Suppose ν(t0 ) = 1, where t0 ∈ J ∈ ∂k T and
k ≥ 1, then as x → ∞,

E{det 2 X|J (t0 )|X(t0 ) = x,

Proof

X|J (t0 ) = 0} = |ΣJ (t0 )|xk (1 + o(1)).

Since ν(t0 ) = 1, ΣJ (t0 ) is negative definite. Let Qt0 ,J be the k × k positive definite

matrix such that Qt0 ,J (−ΣJ (t0 ))Qt0 ,J = Ik . Then we can write

2X

|J (t0 )

= Q−1 Qt0 ,J 2 X|J (t0 )Qt0 ,J Q−1 ,
t0 ,J
t0 ,J

75

and therefore,

E{det 2 X|J (t0 )|X(t0 ) = x,
= | − ΣJ (t0 )|E{det(Qt0 ,J
Since X(t0 ) and

X|J (t0 ) = 0}
2X

(3.2.1)

|J (t0 )Qt0 ,J )|X(t0 )

= x,

X|J (t0 ) = 0}.

X|J (t0 ) are independent,

E{Qt0 ,J 2 X|J (t0 )Qt0 ,J |X(t0 ) = x,

X|J (t0 ) = 0} = −xIk .

It follows that

E{det(Qt0 ,J 2 X|J (t0 )Qt0 ,J )|X(t0 ) = x,

X|J (t0 ) = 0} = E{det(∆(t0 ) − xIk )},

(3.2.2)

where ∆(t0 ) = (∆ij (t0 ))i,j∈σ(J) is a k × k Gaussian random matrix such that E{∆(t0 )} = 0
and its covariance matrix is independent of x. By the Laplace expansion of the determinant,

det(∆(t0 ) − xIk ) = (−1)k [xk − S1 (∆(t0 ))xk−1 + S2 (∆(t0 ))xk−2 + · · · + (−1)k Sk (∆(t0 ))],

where Si (∆(t0 )) is the sum of the k principle minors of order i in ∆(t0 ). Taking the
i
expectation above, we see that as x → ∞,

E{det(∆(t0 ) − xIk )} = (−1)k xk (1 + o(1)).

Combining this with (3.2.1) and (3.2.2) completes the proof.

76

2
Let τ (t) = θt , then ∀i ∈ σ(J), τi (t0 ) = 0, since t0 is a local maximum point of τ restricted

on J. Assume additionally that the Hessian matrix

ΘJ (t0 ) := (τij (t0 ))i,j∈σ(J)

2
is negative definite, then the Hessian matrix of 1/(2θt ) at t0 restricted on J,

ΘJ (t0 ) = −

1
2τ 2 (t

0)

(τij (t0 ))i,j∈σ(J) = −

1
Θ (t ),
4 J 0
2σT

(3.2.3)

2k−1 and h(t) = 1/(2θ 2 ), applying Lemma
is positive definite. Let g(t) = |ΛJ − ΛJ (t)|/θt
t

2.5.3 with T replaced with J gives us that as u → ∞,
Proposition 3.2.2 Let X be as in Theorem 3.1.9. Suppose that ν(t) attains its maximum
1 at a unique point t0 ∈ J ∈ ∂k T such that νj (t0 ) = 0 for all j ∈ σ(J). If ΘJ (t0 ) is negative
/
definite, then as u → ∞,

P sup X(t) ≥ u =
t∈T

Proof

2k/2 | − ΣJ (t0 )|
|ΛJ (t0 )|1/2 | − ΘJ (t0 )|1/2

Ψ(u)(1 + o(1)).

Since t0 is the only point attaining the maximum variance, and also, νj (t0 ) = 0 for

all j ∈ σ(J), similarly to Example 2.4.2, we obtain that as u → ∞,
/
k

P sup X(t) ≥ u =
t∈T

2 2
(−1)i E{µi (J)} + o(e−αu −u /2 )

(−1)k
i=0

77

for some α > 0. Note that
k

(−1)i E{µi (J)}

(−1)k
i=0

= (−1)k
J

=

p X (t) (0)E{det 2 X|J (t)1{X(t)≥u} | X|J (t) = 0}dt
|J

(−1)k

∞

1

(2π)k/2

T

|ΛJ

(t)|1/2

E{det 2 X|J (t)|X(t) = x,

dt
u

X|J (t) = 0}

× pX(t) (x| 2 X|J (t) = 0)dx
=

∞

(−1)k

(2π)(k+1)/2 u

1

dx
T

θt |ΛJ (t)|1/2

E{det 2 X|J (t)|X(t) = x,

2
2
X|J (t) = 0}e−x /2θt dt.

Now we apply the Laplace method in Lemma 2.5.3 with

g(t) =

1
θt |ΛJ

1
h(t) = 2 ,
2θt

(t)|1/2
u=

E{det 2 X|J (t)|X(t) = x,

X|J (t) = 0 ,
(3.2.4)

x2 ,

and obtain
k

(−1)k

(−1)i E{µi (J)}
i=0

=

(−1)k (2π)k/2
(2π)(k+1)/2 |ΛJ (t0 )|1/2 |ΘJ (t0 )|1/2
∞

×
u

=
=

E{det 2 X|J (t0 )|X(t0 ) = x,

(−1)k (2π)k/2 |ΣJ (t0 )|
(2π)k/2 |ΛJ (t0 )|1/2 |ΘJ (t0 )|1/2
2k/2 | − ΣJ (t0 )|
|ΛJ (t0 )|1/2 | − ΘJ (t0 )|1/2

2
X|J (t0 ) = 0}x−k e−x /2 (1 + o(1))dx

Ψ(u)(1 + o(1))

Ψ(u)(1 + o(1)),

where the second equality is due to Lemma 3.2.1 and the last line is due to (3.2.3).

78

(3.2.5)

If dim(T ) = 1, then the result in Proposition 3.2.2 becomes much simpler as stated in
the following.
Corollary 3.2.3 Let T ⊂ R and let X be as in Theorem 3.1.9. Suppose ν(t) attains its
maximum 1 at a unique interior point t0 , and additionally,

Var(X (t0 )) + E{X(t0 )X (t0 )} = 0.

Then as u → ∞,

P sup X(t) > u =
t∈T

Proof

−1/2
Var(X (t0 ))
Ψ(u)(1 + o(1)).
+1
E{X(t0 )X (t0 )}

Note that, under our assumptions, k = 1,

Σ(t0 ) = E{X(t0 )X (t0 )},

Λ(t0 ) = Var(X (t0 )).

2
Also, τ (t) = θt = Var(X(t)|X (t)) implies

Θ(t0 ) = τ (t0 ) = −

2E{X(t0 )X (t0 )}(Var(X (t0 )) + E{X(t0 )X (t0 )})
.
Var(X (t0 ))

Applying Proposition 3.2.2 gives

P sup X(t) ≥ u =
t∈T

=

21/2 | − Σ(t0 )|
|Λ(t0 )|1/2 | − Θ(t0 )|1/2

Ψ(u)(1 + o(1))

1/2
E{X(t0 )X (t0 )}
Ψ(u)(1 + o(1)),
Var(X (t0 )) + E{X(t0 )X (t0 )}

completing the proof.
79

(3.2.6)

3.3

Gaussian Fields on Manifolds without Boundary

In this section, we assume that T is an N -dimensional smooth manifold without boundary
(SN for example). Let {∂/∂xi }1≤i≤N be a natural coordinate vector field and let X be a
smooth function on T . Define

X=

∂X
∂X
,..., N ,
1
∂x
∂x

2
2
and let ( ∂i X j (t)) be the abbreviation of the N × N matrix ( ∂i X j (t))i,j=1,...,N .
∂x ∂x
∂x ∂x

If X is a Morse function, then according to Corollary 9.3.5 or page 211-212 in Adler and
Taylor (2007), the Euler characteristic of the excursion set Au = {t ∈ T : X(t) ≥ u} is given
by
N

ϕ(Au ) =

(−1)N

(−1)k µk (T ),
k=0

where
µk (T ) := # t ∈ T : X(t) ≥ u, X(t) = 0, index

∂ 2X
(t) = k .
∂xi ∂xj

We also define the number of local maxima above level u as

Mu (T ) := # t ∈ T : X(t) ≥ u, X(t) = 0, index

∂ 2X
(t) = N .
∂xi ∂xj

Since T has no boundary, we have the following much simpler bounds for the excursion
probability.

1
E{Mu (T )} ≥ P sup X(t) ≥ u ≥ E{Mu (T )} − E{Mu (T )(Mu (T ) − 1)}.
2
t∈T

80

(3.3.1)

We assume again supt∈T Var(X(t)) = 1.
Lemma 3.3.1 Let T be an oriented, compact C 3 manifold without boundary. Let {X(t) :
t ∈ T } be a Gaussian random field such that X ∈ C 3 (T ) a.s. and (H3) is fulfilled. Then
there exists some constant α > 0 such that as u → ∞,

2 2
E{Mu (T )(Mu (T ) − 1)} = o(e−αu −u /2 ).

Proof

Since T is compact it has a finite atlas. Let (U, ϕ) be one of its charts and consider

X := X ◦ ϕ−1 : ϕ(U ) ⊂ RN → R.

Then it follows immediately from the definition of Mu that

Mu (X, U ) ≡ Mu (X, ϕ(U )).

Since X ∈ C 3 (M ), the condition (H1) holds for X. Applying Lemma 2.3.4 yields

2 2
E{Mu (X, ϕ(U ))(Mu (X, ϕ(U )) − 1)} = o(e−αu −u /2 ).

This verifies the desired result.

Lemma 3.3.2 Let T be an oriented, compact C 3 manifold without boundary. Let {X(t) :
t ∈ T } be a Gaussian random field such that X ∈ C 3 (T ) a.s. and (H3) is fulfilled. Then

81

there exists some constant α > 0 such that as u → ∞,

2 2
E{Mu (T )} = E{ϕ(Au )} + o(e−αu −u /2 ).

Proof

The result follows from Lemma 3.1.5 and the arguments in the proof of Lemma

3.3.1.

Theorem 3.3.3 Let T be an oriented, compact C 3 manifold without boundary. Let {X(t) :
t ∈ T } be a Gaussian random field such that X ∈ C 3 (T ) a.s. and (H3) is fulfilled. Then
there exists some constant α > 0 such that as u → ∞,

2 2
P sup X(t) ≥ u = E{ϕ(Au )} + o(e−αu −u /2 ),
t∈T

where E{ϕ(Au )} is formulated by

(−1)N
T

Proof

E det

∂ 2X
(t) 1{X(t)≥u}
∂xi ∂xj

X(t) = 0 p X(t) (0)∂x1 ∧ · · · ∧ ∂xN .

The result follows immediately from the combination of (3.3.1), Lemma 3.3.1,

Lemma 3.3.2 and the Kac-Rice metatheorem on manifolds [cf. Theorem 12.1.1 in Adler and
Taylor (2007)].

82

3.4

Gaussian Fields on Convex Sets with Smooth Boundary

Let T be a compact, convex, N -dimensional subset of RN with smooth boundary ∂T . Morse
theorem gives
N

ϕ(Au ) =

(−1)N

(−1)k µ

◦

k (T

N −1

) + (−1)N −1

k=0

(−1)k µk (∂T ),
k=0

where
◦
◦
µk (T ) = #{t ∈T : X(t) ≥ u, X(t) = 0, index( 2 X(t)) = k},

µk (∂T ) = #{t ∈ ∂T : X(t) ≥ u, X|∂T (t) = 0,

X(t), n(t) ≥ 0, index( 2 X|∂T (t)) = k},

and n(t) is the unit normal vector pointing outwards. We also define the number of extended
outward local maxima above level u as
◦
E ◦
Mu (T ) = #{t ∈T : X(t) ≥ u, X(t) = 0, index( 2 X(t)) = N },
E
Mu (∂T ) = #{t ∈ ∂T : X(t) ≥ u,

X|∂T (t) = 0,

X(t), n(t) ≥ 0,

index( 2 X|∂T (t)) = N − 1}.
◦

Note that T can be stratified into T ∪∂T . We have the following bounds for the excursion

83

probability.
◦

E
E
E{Mu (T )} + E{Mu (∂T )} ≥ P sup X(t) ≥ u
t∈T

≥

E
E
E ◦
E ◦
E{Mu (T )} + E{Mu (∂T )} − E{Mu (T )Mu (∂T )}

(3.4.1)

1
1
E ◦
E ◦
E
E
− E{Mu (T )(Mu (T ) − 1)} − E{Mu (∂T )(Mu (∂T ) − 1)}.
2
2
Since ∂T is a hypersurface. We may consider ∂T as an (N − 1)-dimensional submanifold
embedded on RN . Similarly, we have the following result.
Lemma 3.4.1 Let T be a compact, convex, N -dimensional subset of RN with smooth boundary ∂T . Let {X(t) : t ∈ T } be a Gaussian random field such that (H1) and (H3) are fulfilled.
Then there exists some constant α > 0 such that as u → ∞,
2 2
E ◦
E ◦
E{Mu (T )(Mu (T ) − 1)} = o(e−αu −u /2 ),
2 2
E
E
E{Mu (∂T )(Mu (∂T ) − 1)} = o(e−αu −u /2 ),
2 2
E ◦
E
E{Mu (T )} + E{Mu (∂T )} = E{ϕ(Au )} + o(e−αu −u /2 ).

The next lemma shows that the crossing term is also super-exponentially small.
Lemma 3.4.2 Let T be a compact, convex, N -dimensional subset of RN with smooth boundary ∂T . Let {X(t) : t ∈ T } be a Gaussian random field such that (H1) and (H3) are fulfilled.
Then there exists some constant α > 0 such that as u → ∞,

2 2
E ◦
E
E{Mu (T )Mu (∂T )} = o(e−αu −u /2 ).

84

Proof

By the Kac-Rice metatheorem,
◦

E
E
E{Mu (T )Mu (∂T )}
∞

∞

≤ ◦ dt
T

dx

ds
∂T

u

u

dy E{|det 2 X(t)||det 2 X|∂T (s)||X(t) = x,

X(s) = y, X(t) = 0, X|∂T (s) = 0}pX(t),X(s), X(t), X (s) (x, y, 0, 0).
|∂T
By similar arguments in Theorem 2.3.7, if

X(s) = 0 for all s ∈ ∂T such that ν(s) = 1, then

◦

E
E
E{Mu (T )Mu (∂T )} is super-exponentially small. Hence we will consider the alternative case

when
I0 := {s ∈ ∂T : ν(s) = 1, X(s) = 0} = ∅.
Let B(I0 , δ) = {t ∈ T : d(t, I0 ) < δ}, where δ is a small positive number to be specified.

◦

◦

E
E
E
As discussed in Lemma 3.1.8, E{Mu (T )Mu (∂T )} can be reduced to E{Mu (T

E
∩B(s0 , δ))Mu (∂T ∩ B(I0 , δ))} with only a super-exponentially small difference. Due to
◦

E
E
the compactness of T , it suffice to show that E{Mu (T ∩B(s0 , δ))Mu (∂T ∩ B(s0 , δ))} is

super-exponentially small for some s0 ∈ I0 and δ > 0, where

B(s0 , δ) = {t ∈ T : d(t, s0 ) < δ}.

Notice another fact that for all s ∈ ∂T such that ν(s) = 1,

2 ν(s)

are negative semidefinite

and hence Σ(s) = E{X(s) 2 X(s)} are negative definite. Therefore by continuity, we may
choose δ small enough such that

− Σ(t)et,s , et,s ≥ α0 ,

◦

∀t ∈T ∩B(s0 , δ), s ∈ ∂T ∩ B(s0 , δ),

85

(3.4.2)

for some positive constant α0 , where et,s = (s − t)/ s − t .
For s ∈ ∂T ∩ B(s0 , δ), let Z(s) =
the tangent space of ∂T at s, so that

X(s), n(s) , and denote by Πs the projection onto
X|∂T (s) = Πs X(s). Then

◦

E
E
E{Mu (T ∩B(s0 , δ)))Mu (∂T ∩ B(s0 , δ))}

dt

T ∩B(s0 ,δ))

∂T ∩B(s0 ,δ)

dy

dx

ds

∞

∞

∞

= ◦

0

u

u

dz E{|det 2 X(t)||det 2 X|∂T (s)|

|X(t) = x, X(s) = y, X(t) = 0, X|∂T (s) = 0, Z(s) = z}
× pX(t),X(s), X(t), X (s),Z(s) (x, y, 0, 0, z)
|∂T
∞

∞

≤ ◦

T ∩B(s0 ,δ))

dx

ds

dt
∂T ∩B(s0 ,δ)

u

0

dz E{|det 2 X(t)||det 2 X|∂T (s)||X(t) = x,

X(t) = 0, X|∂T (s) = 0, Z(s) = z}pX(t), X(t), X (s),Z(s) (x, 0, 0, z)
|∂T
A(t, s)dtds.
◦
(T ∩B(s0 ,δ))×(∂T ∩B(s0 ,δ))

:=

(3.4.3)

We can bound the integral in (3.4.3) as the following.

A(t, s)dtds
◦
(T ∩B(s0 ,δ))×(∂T ∩B(s0 ,δ))

≤

A(t, s)dtds +
D1

A(t, s)dtds, (3.4.4)
D2

where
◦

D1 = {t ∈T ∩B(s0 , δ), s ∈ ∂T ∩ B(s0 , δ) : − Σ(t)et,s , n(s) ≥ b1 },
◦

N −1

D2 = {t ∈T ∩B(s0 , δ), s ∈ ∂T ∩ B(s0 , δ) :

− Σ(t)et,s , Ei (s) et,s , Ei (s) ≥ b2 },
i=1

b1 and b2 are positive numbers such that b1 + b2 < α0 , {E1 (s), · · · , EN −1 (s)} is the orthonormal basis of the tangent space of ∂T at s. This is because, if (t, s) does not belong to
86

D1 nor D2 , then

− Σ(t)et,s , et,s
N −1

= − Σ(t)et,s , n(s) et,s , n(s) +

− Σ(t)et,s , Ei (s) et,s , Ei (s)
i=1

< b1 + b2 < α0 ,
where we use the fact that the convexity of T implies et,s , n(s) ≥ 0. But this conflicts
◦

(3.4.2), hence D1 ∪ D2 is a covering of (T ∩B(s0 , δ)) × (∂T ∩ B(s0 , δ)).
We first show that

D2 A(t, s)dtds

is super-exponentially small. By similar arguments

in the proof for Gaussian fields on rectangle, we see that there exists positive constants C1 ,
C2 and N1 such that
E{|det 2 X(t)||det 2 X|∂T (s)||X(t) = x,

X(t) = 0, X|∂T (s) = 0} ≤ C1 (xN1 + 1),

detCov( X(t), X|∂T (s)) ≥ C2 s − t 2(N −1) .
Therefore,
∞

A(t, s) ≤
u

E{|det 2 X(t)||det 2 X|∂T (s)||X(t) = x,

X(t) = 0, X|∂T (s) = 0}

× pX(t) (x| X(t) = 0, X|∂T (s) = 0)p X(t), X (s) (0, 0)dx
|∂T
≤ C3 s − t 1−N

∞
u

(1 + xN1 )pX(t) (x| X(t) = 0, X|∂T (s) = 0)dx

for some positive constant C3 .

87

(3.4.5)

On the other hand, as s − t → 0,

Var(X(t)| X(t), X|∂T (s)) = Var(X(t)| X(t), Πs X(s))
= Var(X(t)| X(t), Πs ( X(s) −

X(t))/ s − t )

= Var(X(t)| X(t), Πs ( 2 X(t)et,s )) + o(1)
≤ Var(X(t)|Πs ( 2 X(t)et,s )) + o(1)
≤ 1 − (Πs (Σ(t)et,s ))[Cov(Πs ( 2 X(t)et,s ))]−1 (Πs (Σ(t)et,s ))T + o(1),
where the third equality is due to Taylor’s formula. Note that Cov(Πs ( 2 X(t)et,s )) is
bounded away from 0 because of the regularity condition (H3). Also, by the definition of
D2 , the vectors Πs (Σ(t)et,s ) are not vanishing for all (t, s) ∈ D2 , thus there exists a constant
ε1 > 0 such that

Var(X(t)| X(t), X|∂T (s)) ≤ 1 − ε1 ,

∀(t, s) ∈ D2 .

◦
Combining this with (3.4.5), and noting that s − t 1−N is integrable on (T ∩B(s0 , δ))) ×

(∂T ∩ B(s0 , δ)), we conclude that
Now we turn to estimating

D2 A(t, s)dtds
D1 A(t, s)dtds.

is super-exponentially small.

For (t, s) ∈ D1 , we have

pX(t), X(t), X (s),Z(s) (x, 0, 0, z)
|∂T
= pΠs ( X(t)), X (s) (0, 0|X(t) = x, Z(s) = z,
|∂T
× pX(t),Z(s) (x, z|

X(t), n(s) = 0)p

X(t),n(s)

X(t), n(s) = 0)
(0)

≤ C4 (detCov(X(t), X(t), X|∂T (s), Z(s)))−1/2
× exp

−

x2
z2
2ρ(t, s)xz
1
+ 2
−
2
2(1 − ρ(t, s)2 ) σ1 (t, s) σ2 (t, s) σ1 (t, s)σ2 (t, s)
88

(3.4.6)

for some positive constant C4 , where
2
σ1 (t, s) = Var(X(t)|

X(t), n(s) ) =

detCov(X(t), X(t), n(s) )
,
Var( X(t), n(s) )

2
σ2 (t, s) = Var(Z(s)|

X(t), n(s) ) =

detCov(Z(s), X(t), n(s) )
,
Var( X(t), n(s) )

ρ(t, s) =

Recall Z(s) =

E{X(t)Z(s)| X(t), n(s) = 0}
.
σ1 (t, s)σ2 (t, s)

X(s), n(s) , similarly to the rectangle case, one can check that there exits

positive constants C5 and C6 such that

2
C5 s − t 2 ≤ σ2 (t, s) ≤ C6 s − t 2 .

(3.4.7)

Applying Taylor formula, we obtain

ρ(t, s) =
=

1
E{X(t)
E{X(t)Z(s)} −
σ1 (t, s)σ2 (t, s)
1
E{X(t)
σ1 (t, s)σ2 (t, s)
−

2 X(t)(s − t) +

X(t) +

E{X(t) X(t), n(s) }
E{
Var( X(t), n(s) )

X(t), n(s) }E{Z(s) X(t), n(s) }
Var( X(t), n(s) )

X(t), n(s)

s − t 1+η Yt,s , n(s) }

X(t) +

2 X(t)(s − t)

+ s − t 1+η Yt,s , n(s) }
=

s−t
E{X(t)
σ1 (t, s)σ2 (t, s)
−

2 X(t)e

E{X(t) X(t), n(s) }
E{
Var( X(t), n(s) )

t,s

+ s − t η Yt,s , n(s) }

X(t), n(s)

2 X(t)e
t,s

+ s − t η Yt,s , n(s) } .

By our assumption, E{X(s0 ) X(s0 )}=0, therefore if δ is sufficiently small, E{X(t) X(t)}

89

◦

gets close to 0 for t ∈T ∩B(s0 , δ). Thus, as s − t → 0,
s−t
( Σ(t)et,s , n(s) − o(1))
σ1 (t, s)σ2 (t, s)

ρ(t, s) =

s−t

≤

σ1 (t, s)(C5 s − t 2 )1/2

(−b1 − o(1))

(3.4.8)

< −ε2
for some positive constant ε2 , where the second inequality comes from (3.4.7) and the definition of D1 .
By similar arguments in the proof for Gaussian fields on rectangle, we see that there
exists positive constants C7 , C8 , N2 and N3 such that
E{|det 2 X(t)||det 2 X|∂T (s)||X(t) = x,

X(t) = 0, X|∂T (s) = 0, Z(s) = z}

≤ C7 (xN2 + (z/ s − t )N3 + 1)
and
detCov(X(t), X(t), X|∂T (s), Z(s)) ≥ C8 s − t 2N .
Combining this with (3.4.6), and making change of variable z = z/ s − t and σ2 (t, s) =
˜
˜
σ2 (t, s)/ s − t , we obtain
A(t, s) ≤ C9 s − t −N
× exp

−

∞

dx
u

dz (xN2 + (z/ s − t )N3 + 1)

0

1
x2
z2
2ρ(t, s)xz
+ 2
−
2 ) σ 2 (t, s)
2(1 − ρ(t, s)
σ2 (t, s) σ1 (t, s)σ2 (t, s)
1

≤ C9 s − t 1−N
× exp

∞

∞

∞

dx
u

d˜ (xN2 + z N3 + 1)
z
˜

0

1
x2
z2
˜
2ρ(t, s)x˜
z
−
+ 2
−
2
σ
2(1 − ρ(t, s)2 ) σ1 (t, s) σ2 (t, s) σ1 (t, s)˜2 (t, s)
˜

90

(3.4.9)

for some positive constant C9 . Applying Lemma 2.3.10 yields that

D1 A(t, s)dtds is super-

exponentially small.
Theorem 3.4.3 Let T be a compact, convex, N -dimensional subset of RN with smooth
boundary ∂T . Let {X(t) : t ∈ T } be a Gaussian random field such that (H1) and (H3) are
fulfilled. Then there exists some constant α > 0 such that as u → ∞,

2 2
P sup X(t) ≥ u = E{ϕ(Au )} + o(e−αu −u /2 ).
t∈T

Proof

The result follows immediately from applying (3.4.1), Lemma 3.4.1 and Lemma

3.4.2.

3.5

Gaussian Fields on Convex Sets with Piecewise
Smooth Boundary

Let T be an N -dimensional compact and convex set in RN . Suppose it has piecewise smooth
boundary and can be stratified as T = ∪N ∂i T , where ∂i T is the i-dimensional boundary
i=0
of T made up of the disjoint union of a finite number of i-dimensional manifolds without
boundary.
Define the support cone [cf. Adler and Taylor (2007, p.188)] of T at t as

St T := {ξ ∈ RN : ∃δ > 0, C 1 curve γ : (−δ, δ) → Rd such that
γ(0) = t, γ(0) = ξ, γ(s) ∈ T for all s ∈ [0, δ)}.
It is easy to check that St T contains the tangent space of t.
91

Define the normal cone [cf. Adler and Taylor (2007, p.189)] of T at t as

Nt T := {z ∈ RN : z, ξ ≤ 0 for all ξ ∈ St T }.

So t ∈ T is called extended outward critical point if

X(t) ∈ Nt T .

Morse theorem gives
N

k

(−1)i µi (J),

(−1)k

ϕ(Au ) =

i=0

k=0 J∈∂k T

where

µi (J) := #{t ∈ J : X(t) ≥ u,

X|J (t) = 0, index( 2 X|J (t)) = i, X(t) ∈ Nt T }

= #{t ∈ J : X(t) ≥ u, X(t) ∈ Nt T, index( 2 X|J (t)) = i},
and the last line above is due to the fact that

X(t) ∈ Nt T implies

X|J (t) = 0 for all

t ∈ J.
We will need a modified version of (H4), say (H4 ), as the following.
(H4 ). ∀t ∈ J ∈ ∂k T such that ν(t) = 1 and 0 ≤ k ≤ N − 2, (E{X(t) 2 X(t)})|L is negative
definite, where L is the largest subspace of RN such that ( ν(t))|L = 0.
Here, (E{X(t) 2 X(t)})|L and ( ν(t))|L are the projections of E{X(t) 2 X(t)} and

ν(t)

onto the subspace L, respectively.
Similar to Proposition 3.1.1, we have the following result.
Proposition 3.5.1 Let X(t) ∈ C 2 (RN ) a.s. and let L be the largest subspace of RN such
that ( ν(t))|L = 0. If ( 2 ν(t))|L is negative semidefinite for each t ∈ J ∈ ∂k T such that
ν(t) = 1 and 0 ≤ k ≤ N − 2, then (H4 ) holds.
92

For J ∈ ∂k T , we also define the number of extended outward maxima above level u as
E
Mu (J) := #{t ∈ J : X(t) ≥ u, X|J (t) = 0, index( 2 X|J (t)) = k, X(t) ∈ Nt T }

= #{t ∈ J : X(t) ≥ u, X(t) ∈ Nt T, index( 2 X|J (t)) = k}.
Then similarly to Lemma 2.3.1, one has
N
E
{Mu (J) ≥ 1} a.s.

sup X(t) ≥ u =
t∈T

k=0 J∈∂k T

Thus by similar discussions, we get
N
E
E{Mu (J)} ≥ P sup X(t) ≥ u
t∈T

k=0 J∈∂k T
N

≥
k=0 J∈∂k T

1
E
E
E
E{Mu (J)} − E{Mu (J)(Mu (J) − 1)}
2

(3.5.1)

E
E
E{Mu (J)Mu (J )}.

−
J=J

Theorem 3.5.2 Let T be a compact, convex, N -dimensional subset of RN with piecewise
smooth boundary. Let {X(t) : t ∈ T } be a Gaussian random field such that (H1), (H3) and
(H4 ) are fulfilled. Then there exists some constant α > 0 such that as u → ∞,

2 2
P sup X(t) ≥ u = E{ϕ(Au )} + o(e−αu −u /2 ).
t∈T

Proof

Similar to the arguments in proving the smooth boundary case, we only need to

E
E
show that E{Mu (J)Mu (J )} is super-exponentially small for neighboring faces J ∈ ∂k T and
E
E
J ∈ ∂k T . Moreover, similarly, it suffices to show that E{Mu (J ∩B(s0 , δ))Mu (J ∩B(s0 , δ))

93

¯ ¯
is super-exponentially small for s0 ∈ I = J ∩ J such that ν(s0 ) = 1 and ( ν(s0 ))|L = 0,
¯
¯
where L = span{Ss0 J ∪ Ss0 J }, B(s0 , δ) = {t ∈ T : d(t, s0 ) < δ} and δ is a small positive
number to be specified. Without loss of generality, we assume k ≥ k and k ≥ 1.
Denote by {E1 , · · · , Em } the normal basis on I. Since J is a k-dimensional face, there
are (k −m) many (m+1)-dimensional faces which belong to the closure of J and are adjacent
to I as well. Denote these (m + 1)-dimensional faces by {Jm+1 , · · · , Jk } ⊂ ∂m+1 T . Now, for
each i = m + 1, . . . , k, we may view I as an m-dimensional boundary of Ji , and let Ei (s0 )
¯
be the unit normal vector pointing outwards at s0 , i.e., Ei (s0 ) ∈ Ns0 Ji . In such way, we
have a smooth frame (not necessary orthogonal) {E1 (t), · · · , Em (t), Em+1 (t), · · · , Ek (t)} on
¯
J ∩ B(s0 , δ).
Similarly, since J is a k -dimensional face, there are (k − m) many (m + 1)-dimensional
faces which belong to the closure of J and are adjacent to I as well. Denote these (m + 1)dimensional faces by {Jm+1 , · · · , J } ⊂ ∂m+1 T . Now, for each i = m + 1, . . . , k , we may
k

view I as an m-dimensional boundary of Ji , and let Ei (s0 ) be the unit normal vector pointing
¯
outwards at s0 , i.e., Ei (s0 ) ∈ Ns0 Ji . In such way, we have a smooth frame (not necessary
¯
orthogonal) {E1 (s), · · · , Em (s), Em+1 (s), · · · , E (s)} on J ∩ B(s0 , δ).
k
By (H4 ), ΣL (t) := (E{X(t) 2 X(t)})L is negative definite at t = s0 . If δ is small
enough, by continuity,

− ΣL (t)et,s , et,s ≥ α0 ,

∀t ∈ J ∩ B(s0 , δ), s ∈ J ∩ B(s0 , δ),

(3.5.2)

for some positive constant α0 , where et,s is the projection of (s − t)/ s − t on L (in fact, the
projection only removes the vanishing components of (s − t)/ s − t such that the number

94

of components in et,s is the same as dim(L)). For (t, s) ∈ (J ∩ B(s0 , δ)) × (J ∩ B(s0 , δ)), let

αi (t, s) = − ΣL (t)et,s , Ei (t) ,

i = 1, . . . , m, m + 1, . . . , k,

αj (t, s) = − ΣL (t)et,s , Ej (s) ,

j = m + 1, . . . , k .

We claim that there exist positive constants b0 , bm+1 , · · · , bk , bm+1 , · · · , b , whose propk
k
erty needs to be specified later, such that D0 ∪ (∪k
i=m+1 Di ) ∪ (∪j=m+1 Dj ) is a covering of

(J ∩ B(s0 , δ)) × (J ∩ B(s0 , δ)), where

Di = {(t, s) ∈ (J ∩ B(s0 , δ)) × (J ∩ B(s0 , δ)) : αi (t, s) ≥ bi } if m + 1 ≤ i ≤ k;
Dj = {(t, s) ∈ (J ∩ B(s0 , δ)) × (J ∩ B(s0 , δ)) : αj (t, s) ≤ −bj } if m + 1 ≤ j ≤ k ;
m

D0 =

|αi (t, s)|2 ≥ b0 .

(t, s) ∈ (J ∩ B(s0 , δ)) × (J ∩ B(s0 , δ)) :
i=1

(3.5.3)

Note that as δ gets smaller, (J ∩ B(s0 , δ)) ∪ (J ∩ B(s0 , δ)) becomes more similar to two
intersecting flat faces, and et,s will be around the convex cone created by vectors

{±E1 (t), . . . , ±Em (t), Em+1 (t), . . . , Ek (t), −Em+1 (s), . . . , −Ek (s)}.

Hence due to the convexity of T , there exists ε0 > 0, whose property needs to be specified
later, such that for all (t, s) ∈ (J ∩ B(s0 , δ)) × (J ∩ B(s0 , δ)) and δ sufficiently small, the
following representation holds:
k

et,s =

k

et,s , Ei (t) Ei (t) +
i=1

et,s , Ej (s) Ej (s),
j=m+1

95

(3.5.4)

where et,s , Ei (t) ∈ R for i = 1, . . . , m, and

et,s , Ei (t) ≥ −ε0 ,
et,s , Ej (s) ≤ ε0 ,

∀i = m + 1, . . . , k,
∀j = m + 1, . . . , k .

By continuity, there exists a universal positive constant r such that

max{|αi (t, s)|, |αj (t, s)|} ≤ r,

sup
(t,s)∈J×J

for all i = m + 1, . . . , k and j = m + 1, . . . , k .
If (t, s) ∈ (J ∩ B(s0 , δ)) × (J ∩ B(s0 , δ)) does not belong to any of sets in (3.5.3), then
by (3.5.4),
k

m

− ΣL (t)et,s , et,s =

αi (t, s) et,s , Ei (t)

αi (t, s) et,s , Ei (t) +
i=m+1

i=1
k

αj (t, s) et,s , Ej (s)

+

(3.5.5)

j=m+1
k

k

max{bj , rε0 },

max{bi , rε0 } +

≤ b0 +

j=m+1

i=m+1

where the last inequality we use the facts that for i = m + 1, . . . , k, if (t, s) ∈ Di and
/
et,s , Ei (t) ≥ 0, then αi (t, s) et,s , Ei (t) ≤ bi ; and for j = m + 1, . . . , k , if (t, s) ∈ Dj and
/
et,s , Ej (s) ≤ 0, then αj (t, s) et,s , Ej (s) ≤ bj .
Now, we choose the positive constants b0 , bm+1 , · · · , bk , bm+1 , · · · , b and ε0 such that
k
the last line of (3.5.5) is strictly less than α0 . Then − ΣL (t)et,s , et,s < α0 conflicts (3.5.2).
k
This verifies our claim that D0 ∪ (∪k
i=m+1 Di ) ∪ (∪j=m+1 Dj ) is a covering of (J ∩ B(s0 , δ)) ×

(J ∩ B(s0 , δ)).
96

Due to the convexity of T , if δ is small enough, then for each (t, s) ∈ (J ∩ B(s0 , δ)) ×
(J ∩ B(s0 , δ)),

X(t) ∈ Nt T ⇔

X|J (t) = 0 and

X(t), Ej (s) ≥ 0,

∀j = m + 1, . . . , k ;

and for each s ∈ J ∩ B(s0 , δ),

X(s) ∈ Ns T ⇔

X(s), Ei (t) ≥ 0,

X|J (s) = 0 and

∀i = m + 1, . . . , k.

By the Kac-Rice metatheorem,

E
E
E{Mu (J ∩ B(s0 , δ))Mu (J ∩ B(s0 , δ))
∞

∞

≤

dx

ds

dt

dy

u
u
J ∩B(s0 ,δ)
J∩B(s0 ,δ)
∞
∞
∞
∞
dwk
dzk −m
dwm+1 · · ·
dzm+1 · · ·
0
0
0
0

E{|det 2 X|J (t) det 2 X|J (s)||X(t) = x, X(s) = y,
X|J (t) = 0,
X|J (s) = 0,

X(t), Em+1 (s) = zm+1 , · · · ,
X(s), Em+1 (t) = wm+1 , · · · ,

X(t), Ek (s) = zk ,
X(s), Ek (t) = wk }

× pt,s (x, y, 0, zm+1 , · · · , zk , 0, wm+1 , · · · , wk )
:=

A(t, s) dtds,
(J∩B(s0 ,δ))×(J ∩B(s0 ,δ))

where pt,s is the density of

(X(t), X(s), X|J (t),
X|J (s),

X(t), Em+1 (s) , · · · ,

X(s), Em+1 (t) , · · · ,
97

X(t), Ek (s) ,

X(s), Ek (t) ).

Moreover, due to the covering discussed before, this integral can be bounded as

A(t, s) dtds
(J∩B(s0 ,δ))×(J ∩B(s0 ,δ))
k

≤

k

A(t, s) dtds +
Di

i=m+1

We first show that

A(t, s) dtds +
Dj

j=m+1

D0 A(t, s) dtds

A(t, s) dtds.
D0

is super-exponentially small. By similar arguments

in the proof for Gaussian fields on rectangle, we see that there exist positive constants C1 ,
C2 and N1 such that
E{|det 2 X|J (t)||det 2 X|J (s)||X(t) = x,

X|J (t) = 0, X|J (s) = 0} ≤ C1 (xN1 + 1),

detCov( X|J (t), X|J (s)) ≥ C2 s − t 2m .
Therefore,
∞

A(t, s) ≤
u

E{|det 2 X|J (t)||det 2 X|J (s)||X(t) = x,

X|J (t) = 0, X|J (s) = 0}

× pX(t) (x| X|J (t) = 0, X|J (s) = 0)p X (t), X (s) (0, 0)dx
|J
|J
≤ C3 s − t m

∞
u

(1 + xN1 )pX(t) (x| X|J (t) = 0, X|J (s) = 0)dx
(3.5.6)

for some positive constant C3 .

98

On the other hand, let Πt be the projection onto span{E1 (t), · · · , Em (t)}, then as δ → 0,

Var(X(t)| X|J (t), X|J (s)) ≤ Var(X(t)|Πt X|J (t), Πt X|J (s))
= Var(X(t)|Πt X(t), Πt X(s)) + o(1)
= Var(X(t)|Πt X(t), Πt ( X(s) −

X(t))/ s − t ) + o(1)

= Var(X(t)|Πt X(t), Πt ( 2 X(t)et,s )) + o(1)
≤ Var(X(t)|Πt ( 2 X(t)et,s )) + o(1)
≤ 1 − (Πt (ΣL (t)et,s ))[Cov(Πt ( 2 X(t)et,s ))]−1 (Πt (ΣL (t)et,s ))T + o(1),
where the third equality is due to Taylor’s formula. Note that Cov(Πt ( 2 X(t)et,s )) is
bounded away from 0 because of the regularity condition (H3). Also, by the definition of
D0 , the vectors Πt (ΣL (t)et,s ) are not vanishing for all (t, s) ∈ D0 , thus there exists a constant
ε1 > 0 such that

Var(X(t)| X|J (t), X|J (s)) ≤ 1 − ε1 ,

∀(t, s) ∈ D0 .

Combining this with (3.5.6), and noting that s − t m is integrable on (J ∩ B(s0 , δ))) × (J ∩
B(s0 , δ)), we conclude that
Now we turn to estimating

D0 A(t, s) dtds
Di

is super-exponentially small.

A(t, s)dtds, i = m + 1, . . . , k. Let Πt be the projection

onto span{E1 (t), . . . , Em (t), Em+1 (t), . . . , Ei−1 (t), Ei+1 (t), . . . , Ek (t)}, then for (t, s) ∈ Di ,

99

we have

pX(t), X (t), X (s),
|J
|J

X(s),Ei (t)

(x, 0, 0, w)

= pΠ ( X(t)), X (s) (0, 0|X(t) = x,
t
|J
× pX(t),

X(s),Ei (t)

(x, w|

X(t), Ei (t) = 0)p

≤ C4 [detCov(X(t), X|J (t), X|J (s),
× exp

−

X(s), Ei (t) = w,

X(t), Ei (t) = 0)

X(t),Ei (t)

(0)

(3.5.7)

X(s), Ei (t) )]−1/2

1
x2
w2
2ρ(t, s)xw
+ 2
−
2 ) σ 2 (t, s)
2(1 − ρ(t, s)
σ2 (t, s) σ1 (t, s)σ2 (t, s)
1

for some positive constant C4 , where
2
σ1 (t, s) = Var(X(t)|
2
σ2 (t, s) = Var(

ρ(t, s) =

X(t), Ei (t) ) =

X(s), Ei (t) |

E{X(t)

detCov(X(t), X(t), Ei (t) )
,
Var( X(t), Ei (t) )

X(t), Ei (t) ) =

detCov(

X(s), Ei (t) , X(t), Ei (t) )
,
Var( X(t), Ei (t) )

X(s), Ei (t) | X(t), Ei (t) = 0}
.
σ1 (t, s)σ2 (t, s)

Similarly to the rectangle case, one can check that there exits positive constants C5 and C6
such that
2
C5 s − t 2 ≤ σ2 (t, s) ≤ C6 s − t 2 .

100

(3.5.8)

Applying Taylor formula, we obtain

ρ(t, s) =

1
E{X(t)
σ1 (t, s)σ2 (t, s)
−

=

E{X(t)

X(t), Ei (t) }E{ X(s), Ei (t)
Var( X(t), Ei (t) )

1
E{X(t)
σ1 (t, s)σ2 (t, s)
−

X(s), Ei (t) }

2 X(t)(s − t) +

X(t) +

E{X(t) X(t), Ei (t) }
E{
Var( X(t), Ei (t) )

X(t), Ei (t) }

X(t), Ei (t)

s − t 1+η Yt,s , Ei (t) }

X(t) +

2 X(t)(s − t)

+ s − t 1+η Yt,s , Ei (t) }
=

s−t
E{X(t)
σ1 (t, s)σ2 (t, s)
−

2 X(t)e
t,s

E{X(t) X(t), Ei (t) }
E{
Var( X(t), Ei (t) )

+ s − t η Yt,s , Ei (t) }

X(t), Ei (t)

2 X(t)e
t,s

+ s − t η Yt,s , Ei (t) } .

By our assumption, (E{X(s0 ) X(s0 )})|L =0, therefore E{X(t)

X(t), Ei (t) } gets close to

0 for t ∈ J ∩ B(s0 , δ) and δ small enough. Thus, as s − t → 0,

ρ(t, s) =
≤

s−t
( ΣJ (t)et,s , Ei (t) − o(1))
σ1 (t, s)σ2 (t, s)
s−t
σ1 (t, s)(C5 s − t 2 )1/2

(−bi − o(1))

(3.5.9)

< −ε2
for some positive constant ε2 , where the second inequality comes from (3.5.8) and the definition of Di .
By similar arguments in the proof for Gaussian fields on rectangle, we see that there

101

exists positive constants C7 , C8 , N2 and N3 such that
E{|det 2 X|J (t)||det 2 X|J (s)||X(t) = x, X|J (t) = 0, XJ (s) = 0,

X(s), Ei (t) = w}

≤ C7 (xN2 + (w/ s − t )N3 + 1)
and
detCov(X(t), X|J (t), X|J (s),

X(s), Ei (t) ) ≥ C8 s − t 2(m+1) .

Combining this with (3.5.7), and making change of variable w = w/ s − t and σ2 (t, s) =
˜
˜
σ2 (t, s)/ s − t , we obtain
A(t, s) ≤ C9 s − t −(m+1)
× exp

−

≤ C9 s − t
× exp

∞

dx
u

dw (xN2 + (w/ s − t )N3 + 1)

0

1
w2
2ρ(t, s)xw
x2
+ 2
−
2 ) σ 2 (t, s)
2(1 − ρ(t, s)
σ2 (t, s) σ1 (t, s)σ2 (t, s)
1

−m

−

∞

∞

∞

dx
u

dw
˜

(xN2

+ w N3
˜

(3.5.10)

+ 1)

0

1
x2
w2
˜
2ρ(t, s)xw
˜
+ 2
−
2 ) σ 2 (t, s)
σ
2(1 − ρ(t, s)
σ2 (t, s) σ1 (t, s)˜2 (t, s)
˜
1

for some positive constant C9 . Applying Lemma 2.3.10 yields that

Di

A(t, s)dtds is super-

exponentially small.
Estimating

Dj

A(t, s)dtds for j = m + 1, . . . , k is similar. And the proof for the case

E
E
when k = 0 is also similar. Now we obtain that E{Mu (J)Mu (J )} is super-exponentially

small, completing the proof.

Remark 3.5.3 Our proof in Theorem 3.5.2 only focuses on the neighborhood of s0 , and
therefore the proof is also valid for the case when T is locally convex [cf. Adler and Taylor
(2007, p.189, Definition 8.2.1)].

102

Chapter 4
The Expected Euler Characteristic of
Non-centered Stationary Gaussian
Fields
It has been shown that the expected Euler characteristic of the excursion set, denoted
by E{ϕ(Au )}, can be used to approximate the excursion probability very accurately. Now
we turn to the computation of E{ϕ(Au )}. In the monograph Adler and Taylor (2007), the
authors considered centered Gaussian fields with constant variance and they obtained very
general formulae [cf. Theorem 12.4.1 and Theorem 12.4.2 therein] for E{ϕ(Au )} involving the
so called Lipschitz-Killing curvatures. Usually, it is very hard to simplify these LipschitzKilling curvatures. As a consequence, for general centered smooth Gaussian fields with
constant variance, E{ϕ(Au )} is difficult to compute. Therefore, E{ϕ(Au )} would become
even more complicated for general smooth Gaussian fields with non-constant variances.
However, for some relatively simple models and nice parameter space T , for example
centered stationary Gaussian fields on rectangles, E{ϕ(Au )} can be simplified a lot [cf.
Theorem 11.7.2 and Corollary 11.7.3 in Adler and Taylor (2007)]. The results there rely
heavily on the zero mean function. If the Gaussian field is stationary, but the mean function
is varying, then the computation for E{ϕ(Au )} will become more complicated. In this
103

chapter, we will show the formulae of E{ϕ(Au )} for stationary Gaussian fields with varying
mean functions, and also for isotropic Gaussian fields on the sphere with varying mean
functions.

4.1

Preliminary Gaussian Computations

The following result is Lemma 11.6.1 in Adler and Taylor (2007).
Lemma 4.1.1 (Wick formula). Let Z1 , Z2 , ..., ZN be a set of real-valued random variables having a joint Gaussian distribution and zero means. Then for any integer k,

E{Z1 Z2 · · · Z2k+1 } = 0,
E{Z1 Z2 · · · Z2k } =

(4.1.1)
E{Zi1 Zi2 } · · · E{Zi2k−1 Zi2k },

where the sum is taken over the (2k)!/(k!2k ) different ways of grouping Z1 , ..., Z2k into k
pairs.
Let ∆N be a symmetric N × N matrix with elements ∆ij such that each ∆ij is a zeromean normal variable with arbitrary variance but such that the following relationship holds:

E{∆ij ∆kl } = E(i, j, k, l) − δij δkl ,

(4.1.2)

where E is a symmetric function of i, j, k, l, and δij is the Kronecker delta function. Write
|∆N | for the determinant of ∆N .

104

Let ∆N be a symmetric N × N matrix with elements ∆ij such that each ∆ij is a zeromean normal variable with arbitrary variance but such that the following relationship holds:

E{∆ij ∆kl } = E(i, j, k, l) + δij δkl ,

(4.1.3)

where E is a symmetric function of i, j, k, l, and δij is the Kronecker delta function.
Let ∆N be a symmetric N × N matrix with elements ∆ij such that each ∆ij is a zeromean normal variable with arbitrary variance but such that the following relationship holds:

E{∆ij ∆kl } = E (i, j, k, l),

(4.1.4)

where E is a symmetric function of i, j, k, l.
Lemma 4.1.2 Let BN = (Bij )1≤i,j≤N be a real symmetric N × N matrix and let n be a
positive integer. Then for ∆n satisfying (4.1.2),
n/2

E{|∆n + Bn |} =

G2k Sn−2k (Bn ),

(4.1.5)

k=0
l
where G2j = (−1)j (2j)!/(j!2j ), Sk (Bl ) is the sum of the k principle minors of order k

in Bl , and G0 = S0 (Bn ) = 1 in convention. Similarly, for ∆n satisfying (4.1.3) and ∆n
satisfying (4.1.4),
n/2

E{|∆n + Bn |} =

G2k Sn−2k (Bn ),
k=0

E{|∆n + Bn |} = |Bn |,

105

(4.1.6)

where G2j = (2j)!/(j!2j ).

Proof

We first consider the case when n is even, say n = 2l, then

|∆2l + B2l | =

η(p)(∆1i1 + B1i1 ) · · · (∆2l,i + B2l,i ),
2l
2l

(4.1.7)

P

where p = (i1 , i2 · · · , i2l ) is a permutation of (1, 2, · · · , 2l), P is the set of the (2l)! such
permutations, and η(p) equals +1 or −1 depending on the order of the permutation p. Then

E{|∆2l + B2l |} =

η(p)E{(∆1i1 + B1i1 ) · · · (∆2l,i + B2l,i )}.
2l
2l

(4.1.8)

P

It follows from Lemma 4.1.1 that for k ≤ l, E{∆1i1 · · · ∆2k+1,i
} = 0 and
2k+1

{E(1, i1 , 2, i2 ) − δ1i1 δ2i2 } × · · ·

E{∆1i1 · · · ∆2k,i } =
2k

Q2k

× {E(2k − 1, i2k−1 , 2k, i2k ) − δ2k−1,i

δ
},
2k−1 2k,i2k

where Q2k is the set of the (2k)!/(k!2k ) ways of grouping (i1 , i2 , · · · , i2k ) into pairs without
regard to order, keeping them paired with the first index. Let P be the set of all the

106

permutations of (2k + 1, . . . , 2l), then

η(p)E{∆1i1 · · · ∆2k,i }B2k+1,i
2k

2k+1

· · · B2l,i

2l

P

=

{E(1, i1 , 2, i2 ) − δ1i1 δ2i2 } × · · ·

η(p)
P

Q2k

× {E(2k − 1, i2k−1 , 2k, i2k ) − δ2k−1,i
=

=

=

(−1)k (δ1i1 δ2i2 ) · · · (δ2k−1,i

η(p)
P

δ
}
2k−1 2k,i2k

2k+1

· · · B2l,i

)B
δ
2k−1 2k,i2k 2k+1,i2k+1

· · · B2l,i

B2k+1,i

2l

2l

Q2k

(−1)k (2k)!
k!2k

η(p)B2k+1,i

2k+1

· · · B2l,i

2l

P

(−1)k (2k)!
|(Bij )2k+1≤i,j≤2l |,
k!2k

where the second equality is due to the fact that all products involving at least one E term
will cancel out because of their symmetry property, and the third equality comes from noting
that for only one permutation in P is the product of the delta functions nonzero. Thus

E{|∆2l + B2l |} = |B2l | + G2 S2l−2 (B2l ) + · · · + G2l−2 S2 (B2l ) + G2l .

Similarly, we obtain that

E{|∆2l+1 + B2l+1 |} = |B2l+1 | + G2 S2l−1 (B2l+1 ) + · · · + G2l S1 (B2l+1 ).

Then we obtain (4.1.5). The proof for (4.1.6) follows similarly.

Corollary 4.1.3 Let ∆N , ∆N , ∆N , BN , G2j , G2j and Sk (Bl ) be as in Lemma 4.1.2. Let

107

IN be the N × N unit matrix, and x ∈ R. Then
n/2

N

E{|∆N + BN − xIN |} =

(−1)N

G2k Sn−2k (BN ) xN −n ,

(−1)n
n=0

k=0

N

(4.1.9)

n/2

and similarly,

E{|∆N + BN − xIN |} =

G2k Sn−2k (BN ) xN −n ,

(−1)n

(−1)N

n=0
N

k=0

(4.1.10)

(−1)n Sn (BN )xN −n .

E{|∆N + BN − xIN |} = (−1)N

n=0

Proof

It follows from the usual Laplace expansion of the determinant that
N

(−1)n Sn (∆N + BN )xN −n .

|∆N + BN − xIN | = (−1)N

(4.1.11)

n=0

It follows from Lemma 4.1.2 that
n/2

E{Sn (∆N + BN )} =

G2k Sn−2k (BN ),

(4.1.12)

k=0

and hence we obtain (4.1.9). (4.1.10) follows similarly.

4.2

Stationary Gaussian Fields on Rectangles

Consider a centered stationary Gaussian random field Z = {Z(t), t ∈ RN }. It has representation
Z(t) =

RN

ei t,λ W (dλ)

108

and covariance
C(t) =

RN

ei t,λ ν(dλ),

where W is a complex-valued Gaussian random measure and ν is the spectral measure
satisfying ν(RN ) = C(0) = σ 2 . We introduce the second-order spectral moments

λij =

RN

λi λj ν(dλ),

and denote Λ = (λij )1≤i,j≤N . Denoting also differentiation via subscripts, so that Zi =
∂Z/∂ti , Zij = ∂ 2 Z/∂ti ∂tj , etc., we have

E{Zi (t)Zj (t)} = λij = −Cij (0) = −E{Z(t)Zij (t)}.

The covariances among the second-order derivatives can be similarly defined. However,
all we shall need is that

E0 (i, j, k, l) := E{Zij (t)Zkl (t)} =

RN

λi λj λk λl ν(dλ)

(4.2.1)

is a symmetric function of i, j, k, l. Also note that for any fixed t, Zi (t) is independent of
both Z(t) and Zkl (t).
Let X(t) = Z(t) + m(t), where m(·) ∈ C 2 (RN ) is a real-valued deterministic function.
Let T =

N
i=1 [ai , bi ],

−∞ < ai < bi < ∞.

Theorem 4.2.1 Let X(t) = Z(t) + m(t), where Z(·) ∈ C 2 (RN ) a.s. is a stationary Gaussian field and m(·) ∈ C 2 (RN ) is a real-valued deterministic function. Suppose also that Z

109

satisfies (H3 ). Then we have

E{ϕ(Au (X, T ))}
=
{t}∈∂0 T

u − m(t)
P{ X(t) ∈ E({t})}Ψ
σ
∞

×

dx exp

dt
u

J

N

|ΛJ |1/2

+
k=1 J∈∂k T

(2π)(k+1)/2 σ k+1

1
1
− ( m|J (t))T Λ−1 m|J (t) − 2 (x − m(t))2
J
2
2σ

(4.2.2)

× P{(XJ1 (t), · · · , XJ
(t)) ∈ E(J)| X|J (t) = 0}
N −k
j/2

k

×

(−1)j
j=0

x k−j
,
σ

2 m (t)Q )
J
J

G2i Sj−2i (σQJ
i=0

−1/2

where G and S are defined in Lemma 4.1.2 and QJ = ΛJ

.

If J = {t} ∈ ∂0 T , then

Proof

E{µ0 (J)} = P{X(t) ≥ u, ε∗ Xj (t) ≥ 0 for all 1 ≤ j ≤ N }
j
(4.2.3)

u − m(t)
= P{ X(t) ∈ E({t})}Ψ
,
σ
where the last equality is due to the independence of X(t) and

X(t) for each fixed t.

Now we consider J ∈ ∂k T for some k ≥ 1. Let Di be the collection of all k × k matrices
with index i. Applying Kac-Rice formula [cf. Theorem 11.2.1 or Corollary 11.3.2 in Adler
and Taylor (2007)], together with the definition (2.2.2), we obtain
k

k

(−1)i µi (J)

E
i=0

=
J

p X (t) (0)dt
|J

i=0

(−1)i E{|det 2 X|J (t)|1{ 2 X (t)∈D }
i
|J

× 1{X(t)≥u} 1{(X (t),··· ,X
J
J
1

Note that on the event Di , the matrix

2X

|J (t)

110

N −k

(t))∈E(J)} |

(4.2.4)

X|J (t) = 0}.

has i negative eigenvalues which implies

(−1)i |det 2 X|J (t)| = det 2 X|J (t). Also, ∪k { 2 X|J (t) ∈ Di } = { 2 X|J (t) ∈ Rk }, and
i=0
X(t) is independent of both X(t) and

J

=

2 X(t)

for each fixed t, thus (4.2.4) becomes

p X (t) (0)dt E{det 2 X|J (t)1{X(t)≥u} 1{(X (t),··· ,X
J
J
|J
1

(2π)(k+1)/2 |ΛJ |1/2 σ J

dt

(t))∈E(J)} | X|J (t)
1
N −k
∞
− 1 ( m|J (t))T Λ−1 m|J (t) − 12 (x−m(t))2
J
e 2σ
dx e 2
u

= 0}

× P{(XJ1 (t), · · · , XJ
(t)) ∈ E(J)| X|J (t) = 0}E{det 2 X|J (t)|X(t) = x}.
N −k
(4.2.5)
Now we turn to computing E{det 2 X|J (t)|X(t) = x}. Since ΛJ is positive definite,
there exists a unique k × k positive definite matrix QJ (called principal square root of Λ−1 ,
J
−1/2

also denoted as ΛJ

) such that
QJ ΛJ QJ = Ik ,

where Ik is the k × k identity matrix. Hence

E{Z(t)(QJ 2 Z|J (t)QJ )ij } = −(QJ ΛJ QJ )ij = −δij ,

(4.2.6)

where δij is the Kronecker delta function. One can write

E{det(QJ 2 X|J (t)QJ )|X(t) = x}
= E{det(QJ 2 Z|J (t)QJ + QJ 2 m|J (t)QJ )|X(t) = x}
= E{det(∆(x) + QJ 2 mJ (t)QJ )},
where ∆(x) = (∆ij (x))i,j∈σ(J) with all elements ∆ij (x) being Gaussian variables. To study
∆(x), we only need to find its mean and covariance. Applying Lemma 2.5.1 and (4.2.6), we

111

obtain

x
E{∆ij (x)} = E{(QJ 2 Z|J (t)QJ )ij |X(t) = x} = − 2 δij
σ
and

E{(∆ij (x) − E{∆ij (x)})(∆kl (x) − E{∆kl (x)})}
δij δkl
= E{(QJ 2 Z|J (t)QJ )ij (QJ 2 Z|J (t)QJ )kl } −
σ2
δij δkl
,
= E(i, j, k, l) −
σ2
where E is a symmetric function of i, j, k, l by Lemma 2.1.7 with A replaced by QJ . Then
we have
E{det(QJ 2 X|J (t)QJ )|X(t) = x}
1
det(σQJ ( 2 X|J (t))QJ ) X(t) = x
σk
1
x
= k E det ∆ + σQJ 2 mJ (t)QJ − Ik
σ
σ
=E

,

where ∆ = (∆ij )i,j∈σ(J) and all ∆ij are Gaussian variables satisfying

E{∆ij ∆kl } = σ 2 E(i, j, k, l) − δij δkl .

E{∆ij } = 0,

Applying Corollary 4.1.3, we get

E{det(QJ 2 X|J (t)QJ )|X(t) = x}
(−1)k
=
σk

j/2

k

(−1)j
j=0

G2i Sj−2i (σQJ
i=0

112

2 m (t)Q )
J
J

x k−j
.
σ

(4.2.7)

It follows from (4.2.7) that

E{det 2 X|J (t)|X(t) = x}
= E{det(Q−1 QJ 2 X|J (t)QJ Q−1 )|X(t) = x}
t
t
= |ΛJ |E{det(QJ 2 X|J (t)QJ )|X(t) = x}
(−1)k
=
|ΛJ |
σk

j/2

k

G2i Sj−2i (σQJ 2 mJ (t)QJ )

(−1)j
i=0

j=0

x k−j
.
σ

Plugging this into (4.2.5), we obtain
k

(−1)i µi (J)

E
i=0

=

∞

(−1)k |ΛJ |1/2
(2π)(k+1)/2 σ k+1 J

dt
u

× P{(XJ1 (t), · · · , XJ

N −k

(4.2.8)
(t)) ∈ E(J)| X|J (t) = 0}

j/2

k

×

− 1 ( m|J (t))T Λ−1 m|J (t) − 12 (x−m(t))2
J
dx e 2
e 2σ

(−1)j
j=0

G2i Sj−2i (σQJ 2 mJ (t)QJ )
i=0

x k−j
.
σ

Combining (4.2.3), (4.2.8) and the definition (2.2.1) yields the desired result.

Corollary 4.2.2 Let Z be an isotropic Gaussian random field with Var(Z1 (t)) = γ 2 , then

113

under the conditions in Theorem 4.2.1, we have

E{ϕ(Au (X, T ))}
=
{t}∈∂0 T

∞

×

dt
J

N

u − m(t)
P{ X(t) ∈ E({t})}Ψ
σ
dx exp
u

−

1
2γ 2

γk

+
k=1 J∈∂k T

(2π)(k+1)/2 σ k+1

1
m|J (t) 2 − 2 (x − m(t))2
2σ

(4.2.9)

× P{(XJ1 (t), · · · , XJ
(t)) ∈ E(J)}
N −k
j/2

k

×

j=0

x k−j
.
σ

G2i Sj−2i (σγ −2 2 mJ (t))

(−1)j
i=0

Proof The result follows by applying Theorem 4.2.1 and noting that ΛJ = γ 2 Ik and hence
QJ = γ −1 Ik .

Corollary 4.2.3 Under the conditions in Theorem 4.2.1, assume that t0 , an interior point
2 m(t )
0

in T , is the unique maximal point of m(t) and that

is nondegenerate. Then as

u → ∞,

E{ϕ(Au (X, T ))} =

Proof

|Λ|1/2 uN/2
σN | −

2 m(t )|1/2
0

Ψ

u − m(t0 )
(1 + o(1)).
σ

By Theorem 4.2.1,

E{ϕ(Au (X, T ))} =

∞

|Λ|1/2

(2π)(N +1)/2 σ N +1 u
− 1 (x−m(t))2
× e 2σ2

x
σ

114

1
T −1 m(t)
e− 2 ( m(t)) Λ

dx
J
N

dt(1 + o(1)).

(4.2.10)

Applying Laplace method, we obtain that as x → ∞,
1

2

1
T −1 m(t) −
(x−m(t))
dt
e 2σ2
e− 2 ( m(t)) Λ
J

=

(2π)N/2 σ N
xN/2 | −

2 m(t )|1/2
0

(4.2.11)

− 1 (x−m(t0 ))2
e 2σ2
(1 + o(1)).

Thus as u → ∞,

E{ϕ(Au (X, T ))} =
=

4.3

∞

|Λ|1/2
(2π)1/2 σ N +1 | −

2 m(t )|1/2
0

|Λ|1/2 uN/2
σN | −

2 m(t )|1/2
0

Ψ

− 1 (x−m(t0 ))2

xN/2 e 2σ2

dx(1 + o(1))

u

u − m(t0 )
(1 + o(1)).
σ

Isotropic Gaussian Random Fields on Sphere

We consider isotropic Gaussian random fields on N -dimensional unit sphere SN . For x =
(x1 , · · · , xN +1 ) ∈ SN , we shall use the spherical coordinates as follows.

x1 = cos θ1 ,
x2 = sin θ1 cos θ2 ,
x3 = sin θ1 sin θ2 cos θ3 ,
(4.3.1)

.
.
.
xN = sin θ1 sin θ2 · · · sin θN −1 cos θN ,
xN +1 = sin θ1 sin θ2 · · · sin θN −1 sin θN ,

115

where 0 ≤ θi ≤ π for 1 ≤ i ≤ N − 1 and 0 ≤ θN < 2π. Let θ = (θ1 , · · · , θN ). Accordingly,
let y = (y1 , · · · , yN +1 ) be another point in SN , we use ϕ = (ϕ1 , · · · , ϕN ) to denote its
corresponding spherical coordinates.
Let · , ·, · be the Euclidean norm and the inner product respectively. Denote by d(·, ·)
the distance function in SN , i.e., d(x, y) = arccos x, y , ∀x, y ∈ SN .
The following theorem by Schoenberg (1942) characterizes the covariance function of
isotropic Gaussian field on sphere.
Theorem 4.3.1 A real continuous function C(d) is a valid covariance on unit sphere SN
for every dimension N ≥ 1 if and only if it has the form
∞

bn cosn d,

C(d) =

d ∈ [0, π],

n=0

where bn ≥ 0 and

∞
n=0 bn

< ∞.

Recall d(x, y) = arccos x, y , ∀x, y ∈ SN . It follows from the above theorem that a
function C(x, y), which is the covariance function of an isotropic Gaussian field on SN for
every dimension N ≥ 1, has the form
∞

bn x, y n ,

C(x, y) =

∀x, y ∈ SN ,

(4.3.2)

n=0

where where bn ≥ 0 and

∞
n=0 bn

< ∞.

Let ϕ(Au (X, SN )) be the Euler characteristic of excursion set Au (X, SN ) = {x ∈ SN :
X(x) ≥ u}. Then according to Corollary 9.3.5 in Adler and Taylor (2007),
N

ϕ(Au

(X, SN ))

=

(−1)N

(−1)i µi (SN )
i=0

116

(4.3.3)

with

µi (SN ) := #{x ∈ SN : X(x) ≥ u, X(x) = 0, index( 2 X(x)) = i},

where

X and

2X

(4.3.4)

are the gradient and Hessian on manifold respectively.

Let X(x) = Z(x) + m(x), x ∈ SN , where Z is a centered, unit-variance smooth Gaussian
random field on SN with covariance function C(·, ·) and m(·) ∈ C 2 (SN ) is a real-valued
deterministic function. Under the spherical coordinate, let X(θ) = X(x), Z(θ) = Z(x),
m(θ) = m(x), C(θ, ϕ) = C(x, y).
Lemma 4.3.2 Let h(x, y) = x, y n , x, y ∈ SN , where n is a nonnegative integer, and let
h(θ, ϕ) be its spherical version. Then h(θ, θ) = 1 and
∂ 3 h(θ, ϕ)
∂h(θ, ϕ)
|θ=ϕ =
|
= 0,
∂θi
∂θi ∂ϕj ∂ϕk θ=ϕ
∂ 2 h(θ, ϕ)
∂ 2 h(θ, ϕ)
|θ=ϕ = −
|
= nδij ,
∂θi ∂ϕj
∂θi ∂θj θ=ϕ

(4.3.5)

∂ 4 h(θ, ϕ)
|
= n(n − 1)(δij δkl + δik δjl + δil δjk ) + nδij δkl .
∂θi ∂θj ∂ϕk ∂ϕl θ=ϕ
Let Θ = {θ ∈ RN : 0 ≤ θN < 2π, 0 ≤ θi ≤ π, ∀1 ≤ i ≤ N − 1} and let dσ(θ) be the
volume element on the sphere, i.e.,
N −1

sinN −i θi dθ,

dσ(θ) =

∀θ ∈ Θ.

i=1

Then we can state our result as follows.
Theorem 4.3.3 Let X = {X(x) = Z(x) + m(x) : x ∈ SN }, where Z is a Gaussian field
on SN satisfying (H3 ) and m(·) ∈ C 2 (SN ) is a real-valued deterministic function. Suppose
117

also that X has the covariance function C(·, ·) such that
∞

bn x, y n ,

C(x, y) =

∀x, y ∈ SN ,

(4.3.6)

n=0

where bn ≥ 0,

∞
n=0 bn

∞
4
n=1 n bn

= 1 and

< ∞. Let β =

∞
n=1 nbn ,

m(θ) = m(x), and

let G, G and S be as in Lemma 4.1.2. Then for β > 1,

E{ϕ(Au (X, SN ))} =

∞

(β − 1)N/2
(2π)N/2

−

u

Θ
j/2

N

(−1)j

×

dw exp

dσ(θ)

m(θ)) 2

2 m(θ)

G2i Sj−2i

j=0

1
2β

βw

β2 − β

i=0

N −j

;

β2 − β

for β < 1,

E{ϕ(Au (X, SN ))} =

∞

(1 − β)N/2

dσ(θ)

(2π)N/2

Θ

(−1)j

1
2β

m(θ)) 2

2 m(θ)

G2i Sj−2i

j=0

−

u
j/2

N

×

dw exp

βw

β − β2

i=0

β − β2

N −j

;

and for β = 1,

E{ϕ(Au (X, SN ))} =

∞

1
(2π)N/2

dσ(θ)

dw exp

−

u

Θ

1
2

m(θ) 2

N

×

(−1)j Sj ( 2 m(θ))wN −j .
j=0

Remark 4.3.4 It is easy to check that the condition
C 4 (SN × SN ) and hence X(·) ∈ C 2 (SN ).

118

∞
4
n=1 n bn

< ∞ makes C(·, ·) ∈

Proof

Case 1: β > 1. Let κ =

∞
n=2 n(n − 1)bn .

Then by Lemma 4.3.2,

E{Z(θ)Z i (θ)} = E{Z i (θ)Z jk (θ)} = 0,
E{Z i (θ)Z j (θ)} = −E{Z(θ)Z ij (θ)} = βδij ,

(4.3.7)

E{Z ij (θ)Z kl (θ)} = κ(δij δkl + δik δjl + δil δjk ) + βδij δkl .
By the Kac-Rice metatheorem,

E{ϕ(Au (X, SN ))}
= (−1)N
Θ

=
=

p X(θ) (0)E{det 2 X(θ)1{X(θ)≥u} | X(θ) = 0}dσ(θ)
(4.3.8)

∞

(−1)N

dσ(θ)

Θ
N
(−1)

(2π)N/2 β N/2

u

p X(θ) (0)E{det

dσ(θ)
Θ

∞ − 1
e 2β
u

2 X(θ)|X(θ)

m(θ)) 2

= w}dw

E{det 2 X(θ)|X(θ) = w}dw.

Now we turn to computing E{det 2 X(θ)|X(θ) = w}. Note that
E{det 2 X(θ)|X(θ) = w}
= E{det( 2 Z(θ) +

2 m(θ))|X(θ)

= w}
(4.3.9)

= (β 2 − β)N/2 E{det((β 2 − β)−1/2 2 Z(θ) + (β 2 − β)−1/2 2 m(θ))|X(θ) = w}
= (β 2 − β)N/2 E{det(∆ + (β 2 − β)−1/2 2 m(θ) − β(β 2 − β)−1/2 wIN )},
where ∆ = (∆ij )1≤i,j≤N and all ∆ij are centered Gaussian variables satisfying

E{∆ij ∆kl } = (β 2 − β)−1 E{Z ij (θ)Z kl (θ)|X(θ) = w}
= (β 2 − β)−1 {κ(δij δkl + δik δjl + δil δjk ) + βδij δkl − β 2 δij δkl }
= E(i, j, k, l) − δij δkl ,
119

(4.3.10)

where E(i, j, k, l) = (β 2 − β)−1 κ(δij δkl + δik δjl + δil δjk ). Applying Corollary 4.1.3, we get

E{det 2 X(θ)|X(θ) = w}
j/2

N

=

(−1)N (β 2

(−1)j

− β)N/2

j=0

G2i Sj−2i
i=0

2 m(θ)

β2 − β

(4.3.11)

N −j

βw
β2 − β

.

Then
E{ϕ(Au (X, SN ))} =

∞

(β − 1)N/2
(2π)N/2

dσ(θ)
Θ
j/2

(−1)j
j=0

m(θ)) 2

u

N

×

1

−
dw e 2β

G2i Sj−2i
i=0

2 m(θ)

β2 − β

βw

N −j

β2 − β

Case 2: β < 1. Then (4.3.9) becomes

E{det 2 X(θ)|X(θ) = w}
= E{det( 2 Z(θ) +

2 m(θ))|X(θ)

= w}

= (β − β 2 )N/2 E{det((β − β 2 )−1/2 2 Z(θ) + (β − β 2 )−1/2 2 m(θ))|X(θ) = w}
= (β − β 2 )N/2 E{det(∆ + (β − β 2 )−1/2 2 m(θ) − β(β − β 2 )−1/2 wIN )},

where ∆ = (∆ij )1≤i,j≤N and all ∆ij are centered Gaussian variables satisfying

E{∆ij ∆kl } = (β − β 2 )−1 E{Z ij (θ)Z kl (θ)|X(θ) = w}
= (β − β 2 )−1 {κ(δij δkl + δik δjl + δil δjk ) + βδij δkl − β 2 δij δkl }
= E(i, j, k, l) + δij δkl ,

120

.

where E(i, j, k, l) = (β − β 2 )−1 κ(δij δkl + δik δjl + δil δjk ). Applying Corollary 4.1.3, we get

E{det 2 X(θ)|X(θ) = w}
j/2

N

=

(−1)N (β

(−1)j

− β 2 )N/2

2 m(θ)

G2i Sj−2i

j=0

β − β2

i=0

βw
β − β2

N −j

.

Then
E{ϕ(Au (X, SN ))} =

∞

(1 − β)N/2
(2π)N/2

dσ(θ)
Θ
j/2

(−1)j
j=0

m(θ)) 2

u

N

×

1

−
dw e 2β

2 m(θ)

G2i Sj−2i

β − β2

i=0

βw
β − β2

Case 3: β = 1. Then (4.3.9) becomes

E{det 2 X(θ)|X(θ) = w} = E{det( 2 Z(θ) +
= E{det(∆ +

2 m(θ))|X(θ)

= w}

2 m(θ) − wI )},
N

where ∆ = (∆ij )1≤i,j≤N and all ∆ij are centered Gaussian variables satisfying

E{∆ij ∆kl } = E{Z ij (θ)Z kl (θ)|X(θ) = w}
= κ(δij δkl + δik δjl + δil δjk ) + βδij δkl − β 2 δij δkl
= E (i, j, k, l),

where E (i, j, k, l) = κ(δij δkl + δik δjl + δil δjk ). Applying Corollary 4.1.3, we get
N

E{det

2 X(θ)|X(θ)

= w} =

(−1)j Sj ( 2 m(θ))wN −j .

(−1)N
j=0

121

N −j

.

Then
E{ϕ(Au (X, SN ))} =

∞

1
(2π)N/2 Θ

dσ(θ)

1

dw e− 2

m(θ) 2

u

N

×

(−1)j Sj ( 2 m(θ))wN −j .
j=0

122

Chapter 5
Excursion Probability of Smooth
Gaussian Processes over Random
Intervals
Let {X(t) : t ∈ [0, ∞)} be a Gaussian process and let T > 0 be a fixed number, the tail
probability P{sup0≤t≤T X(t) ≥ u} for large u has been extensively studied in the literature.
However, the supremum over the a fixed domain [0, T ] is not adequate in certain applications
[cf. Kozubowski et al. (2004, 2006)], instead, one needs to consider the asymptotics for
sup0≤t≤T X(t) where T is a non-negative random variable independent of X.
To study P{sup0≤t≤T X(t) ≥ u}, we have to take into account the behaviors of both
X and T . Therefore, some interesting phenomena arise due to the connection between the
Gaussian process and the random interval. Recently, M. Arendarczyk and K. Debicki (2011,
2012) considered the case when the Gaussian process X is non-smooth (i.e. the sample path
is not twice differentiable), and obtained the following result under certain conditions:

P

sup X(t) ≥ u = g1 (u)(1 + o(1)),

0≤t≤T

where g1 (u) is a function depending on X and T .

123

as u → ∞,

(5.0.1)

In the theory of approximating P{sup0≤t≤T X(t) ≥ u} for a fixed domain [0, T ], the
asymptotic results for smooth Gaussian processes [cf. Adler and Taylor (2007) and Aza¨ and
ıs
M. Wschebor (2009)] are much more accurate than those for non-smooth Gaussian processes
[cf. Piterbarg (1996a)]. More specifically, under certain smoothness condition, one can get
a higher-order approximation such that the error term decays exponentially faster than the
principle term. Motivated by this, one may expect that for smooth Gaussian processes over
random interval [0, T ], the following approximation holds under certain conditions:

2

P

sup X(t) ≥ u = g2 (u)(1 + o(e−αu )),

as u → ∞,

(5.0.2)

0≤t≤T

for some α > 0, where g2 (u) is a function depending on X and T . Obviously, compared with
(5.0.1), (5.0.2) provides a much more accurate approximation. In this chapter, we apply the
Rice method [cf. Aza¨ and Delmas (2002) and Aza¨ and M. Wschebor (2009)] to prove our
ıs
ıs
main results Theorem 5.1.6 and Theorem 5.2.5 which are of the form as (5.0.2).

5.1

Stationary Gaussian Processes

Let X = {X(t) : t ∈ R+ } be a centered stationary Gaussian process with Var(X(0)) = 1.
Define
r(t) := E{X(t)X(0)},

λ2 := Var(X (0)).

We will impose conditions (H1), (H3) and the following regularity condition (A1) on X.
(A1). For any fixed δ > 0, supt≥δ r(t) < 1.

124

The number of maxima above level u over [0, T ] becomes

Mu (0, T ) = #{t ∈ (0, T ) : X(t) ≥ u, X (t) = 0, X (t) < 0}.

(5.1.1)

Note that for each fixed t, X(t) and X (t) are independent, by (2.3.1), we have the following
upper bound for the excursion probability,

P

sup X(t) ≥ u
0≤t≤T

≤ P{X(0) ≥ u, X (0) ≤ 0} + P{X(T ) ≥ u, X (T ) ≥ 0} + E{Mu (0, T )}

(5.1.2)

= Ψ(u) + E{Mu (0, T )}.
Similarly, by (2.3.2), the lower bound becomes

P

sup X(t) ≥ u
0≤t≤T

1
≥ Ψ(u) + E{Mu (0, T )} − E{Mu (0, T )(Mu (0, T ) − 1)}
2

(5.1.3)

− P{X(0) ≥ u, X (0) ≤ 0, X(T ) ≥ u, X (T ) ≥ 0}
− E{Mu (0, T )1{X(0)≥u,X (0)≤0} } − E{Mu (0, T )1{X(T )≥u,X (T )≥0} }.
Lemma 5.1.1 Let X be a centered stationary Gaussian process satisfying (H1) and (H3).
Then there exists some universal α > 0 such that for all T > 0, as u → ∞,

E{Mu (0, T )} =

2
λT −u2 /2
e
(1 + o(e−αu )).
2π

125

Proof

Due to (H1) and (H3), one can use the Kac-Rice metatheorem and get
T

E{Mu (0, T )} =

0

pX (t) (0)E{|X (t)|1{X(t)≥u,X (t)<0} |X (t) = 0}dt

∞
2
1
dt
E{X (t)1{X (t)<0} |X(t) = x, X (t) = 0}e−x /2 dx
0 2πλ
u
∞
2
T
=−
dt
E{X (0)1{X (0)<0} |X(0) = x}e−x /2 dx,
2πλ
u
T

=−

(5.1.4)

where the second equality is due to the independence of X(t) and X (t), the last equality is
due to the stationarity of X. Note that E{X(0)X (0)} = −λ2 , by Lemma 2.5.1,

E{X (0)|X(0) = x} = −λ2 x.

Make change of variable V = X (0) + λ2 x, then V |X(0) = x is a Gaussian variable with
mean 0 and variance κ2 = Var(X (0)|X(0)). Let us denote its density by g(v), then

E{X (0)1{X (0)<0} |X(0) = x}
= E{X (0)|X(0) = x} − E{X (0)1{X (0)≥0} |X(0) = x}
= −λ2 x −

v≥λ2 x

(5.1.5)

(v − λ2 x)g(v)dv.

But the last integral in (5.1.5) is non-negative and bounded by
2

4 2

−v
v
κ −λ x
√
vg(v)dv =
e 2κ2 dv = √ e 2κ2 .
2π
v≥λ2 x
v≥λ2 x 2πκ

Since λ2 and κ2 are both constants not depending on T , plugging (5.1.5) into (5.1.4), we
obtain the desired result by choosing some α ∈ (0, λ4 /(2κ2 )).

126

Lemma 5.1.2 Let X be a centered stationary Gaussian process satisfying (H1), (H3) and
(A1). For t2 > t1 > 0, let

f (t1 , t2 ) := min 1,

inf

t1 ≤t≤t2

detCov(X(0), X (0), X(t), X (t)) .

(5.1.6)

Then for any ε > 0, there exist positive constants C, δ and ε1 such that for all (0, T ) ⊂ R
and u large enough,

E{Mu (0, T )(Mu (0, T ) − 1)}
≤ CT exp

−

u2
2β 2 + ε

+ T 2 (f (δ, T ))−5/2 exp

−

u2
,
2 − ε1

where C, δ and ε1 do not depend on T and β 2 = Var(X(0)|X (0)) < 1.

Remark 5.1.3 Note that X is stationary, hence for any fixed t ≥ 0, X (t) is independent
of both X(t) and X (t), thus

0
r(t)
r (t)
 1


 0
λ2
−r (t) −r (t)

detCov(X(0), X (0), X(t), X (t)) = det 

 r(t) −r (t)
1
0


r (t) −r (t)
0
λ2













= [λ4 − (r (t))2 ][1 − r2 (t)] + (r (t))2 [(r (t))2 − 2r(t)r (t) − 2λ2 ].

Proof

Let b > a ≥ 0 and b − a ≤ 2δ for some δ > 0. Due to (H1) and (H3), one can use

the Kac-Rice formula for factorial moments [cf. Theorem 11.5.1 in Adler and Taylor (2007)],

127

thus

E{Mu (a, b)(Mu (a, b) − 1)}
b

b

ds pX (t),X (s) (0, 0)

dt

=

a

a

× E{|X (t)X (s)|1{X(t)≥u,X (t)<0} 1{X(s)≥u,X (s)<0} |X (t) = X (s) = 0}
b

≤
a

∞

b

dt

ds
a

(5.1.7)

u

dx pX(t) (x|X (t) = X (s) = 0)pX (t),X (s) (0, 0)

× E{|X (t)X (s)||X(t) = x, X (t) = X (s) = 0}.
Let E(t, s) = E{|X (t)X (s)||X(t) = x, X (t) = X (s) = 0}. By Taylor’s formula,

X (s) = X (t) + X (t)(s − t) + |s − t|1+η Yt,s ,

(5.1.8)

where Yt,s is a centered Gaussian variable. In particular, for s > t,
s
X (s) − X (t) − X (t)(s − t)
t (X (v) − X (t))dv ,
Yt,s =
=
|s − t|1+η
(s − t)1+η

and thus by (H1), Var(Yt,s ) ≤ L2 . Due to (5.1.8), we have

E(t, s) = E{|X (t)X (s)||X(t) = x, X (t) = 0, X (t)(s − t) = −|s − t|1+η Yt,s }
= |s − t|η E{|Yt,s X (s)||X(t) = x, X (t) = 0, X (t)(s − t) = −|s − t|1+η Yt,s }.
(5.1.9)

128

By stationarity and (H1),

Var(X (t)|X(t) = x, X (t) = X (s) = 0) ≤ Var(X (t)) ≤ C1 ,
(5.1.10)
Var(Yt,s |X(t) = x, X (t) = X (s) = 0) ≤ Var(Yt,s ) ≤

L2 ,

where C1 is a positive constant. On the other hand, for |s − t| small enough,

|E{X (s)|X(t) = x, X (t) = X (s) = 0}|
= |E{X (s)|X(t) = x, X (t) = 0, X (t) + |s − t|η Yt,s = 0}|
(5.1.11)
= |E{X (s)|X(t) = x, X (t) = 0, X (t) = 0}|(1 + o(1))
≤ C2 |x|,
and similarly,
|E{Yt,s |X(t) = x, X (t) = X (s) = 0}| ≤ C3 |x|,

(5.1.12)

for some C2 , C3 > 0. Note that for any Gaussian variables ξ1 and ξ2 ,

2
2
E|ξ1 ξ2 | ≤ Eξ1 + Eξ2 = (Eξ1 )2 + Var(ξ1 ) + (Eξ2 )2 + Var(ξ2 ).

(5.1.13)

Applying (5.1.13) and plugging (5.1.10), (5.1.11) and (5.1.12) into (5.1.9), we obtain that
there exists some C4 > 0 such that

E(t, s) ≤ C4 |s − t|η (1 + x2 ).

129

(5.1.14)

By Taylor’s formula (5.1.8),

Var(X(t)|X (t), X (s))
= Var(X(t)|X (t), X (t) + X (t)(s − t) + |s − t|1+η Yt,s )
= Var(X(t)|X (t), X (t) ± |s − t|η Yt,s )

(5.1.15)

= Var(X(t)|X (t), X (t))(1 + o(1))
= Var(X(0)|X (0))(1 + o(1)),

where the last equality is due to the fact that X (t) is independent of both X(t) and X (t).
Hence for any ε > 0, if |b − a| is small enough and u is large enough, then
∞

(1 + x2 )p

u

−

X(t) (x|X

(t) = X (s) = 0)dx ≤ e

u2
2β 2 +ε .

(5.1.16)

Note that
pX (t),X (s) (0, 0) ≤
2π

1
detCov(X (t), X (s))

and by Taylor’s formula (5.1.8),

detCov(X (t), X (s)) = detCov(X (t), X (t) + X (t)(s − t) ± |s − t|1+η Yt,s )
= |s − t|2 detCov(X (t), X (t) ± |s − t|η Yt,s )

(5.1.17)

= |s − t|2 detCov(X (t), X (t))(1 + o(1)),
as |s − t| → 0 uniformly. Thus there exists some C5 > 0 such that for |s − t| sufficiently
small,
C5
pX (t),X (s) (0, 0) ≤
.
|s − t|

130

(5.1.18)

Plugging (5.1.14), (5.1.16) and (5.1.18) into (5.1.7) we obtain that for any ε > 0, if δ is small
enough, then there exists C6 > 0 such that for large u,
−

E{Mu (a, b)(Mu (a, b) − 1)} ≤ C4 C5 e
≤ C6

u2
2β 2 +ε

b

b

|s − t|η−1 dtds

a

a
u2
− 2
(b − a)1+η e 2β +ε
−

≤ C6 (b − a)e

(5.1.19)

u2
2β 2 +ε .

The set [0, T ] may be covered by congruent intervals Ii = [ai , ai+1 ] with the same length
δ and disjoint interiors. Then

E{Mu (0, T )(Mu (0, T ) − 1)} ≤ E
i

=E
i

=

E{Mu (Ii
i

Mu (Ij ) −

Mu (Ii )
j

i

Mu (Ii )
i

)2 } +

(5.1.20)

E{Mu (Ii )Mu (Ij )} −
i=j

i

E{Mu (Ii )}
i

E{Mu (Ii )(Mu (Ii ) − 1)} +

=

(Mu (Ii ) − 1)

Mu (Ii )

E{Mu (Ii )Mu (Ij )}.
i=j

If Ii and Ij are neighboring, say j = i + 1, we have

E{Mu (Ii ∪ Ii+1 )(Mu (Ii ∪ Ii+1 ) − 1)}
= E{(Mu (Ii ) + Mu (Ii+1 ))(Mu (Ii ) + Mu (Ii+1 ) − 1)}
= 2E{Mu (Ii )Mu (Ii+1 )} + E{Mu (Ii )(Mu (Ii ) − 1)} + E{Mu (Ii+1 )(Mu (Ii+1 ) − 1)}.
(5.1.21)

131

It follows from (5.1.19) and (5.1.21) that for any ε > 0, if δ is small enough and u is large
enough, then
1
E{Mu (Ii )Mu (Ii+1 )} ≤ E{Mu (Ii ∪ Ii+1 )(Mu (Ii ∪ Ii+1 ) − 1)}
2
1
= E{Mu (ai , ai+2 )(Mu (ai , ai+2 ) − 1)}
2

(5.1.22)

u2

− 2
C
≤ 6 (ai+2 − ai )e 2β +ε
2

and hence

E{Mu (Ii )Mu (Ii+1 )} ≤ 2C6 T e

E{Mu (Ii )(Mu (Ii ) − 1)} +
i

−

u2
2β 2 +ε .

(5.1.23)

i=j,Ii ∩Ii+1 =∅

Next we consider the case when Ii = [ai , ai+1 ] and Ij = [aj , aj+1 ] are non-neighboring,
which implies aj − ai+1 ≥ δ. By the Kac-Rice formula for higher moments [cf. the proof is
the same as that of Theorem 11.5.1 in Adler and Taylor (2007)],

E{Mu (Ii )Mu (Ij )}
=

ai+1

dt

ai

aj+1
aj

ds pX (t),X (s) (0, 0)

× E{|X (t)X (s)|1{X(t)≥u,X (t)<0} 1{X(s)≥u,X (s)<0} |X (t) = X (s) = 0}
≤

ai+1
ai

dt

aj+1
aj

∞

ds

∞

dx
u

u

dy pX (t),X (s) (0, 0|X(t) = x, X(s) = y)pX(t),X(s) (x, y)

× E{|X (t)X (s)||X(t) = x, X(s) = y, X (t) = X (s) = 0}.
(5.1.24)

132

By Lemma 2.5.1 and the stationarity of X, there exists some C7 > 0 such that

|E{X (t)|X(t) = x, X(s) = y, X (t) = X (s) = 0}|
≤

C7 (|x| + |y|)
detCov(X(t), X(s), X (t), X (s))

and

|E{X (s)|X(t) = x, X(s) = y, X (t) = X (s) = 0}|
≤

C7 (|x| + |y|)
.
detCov(X(t), X(s), X (t), X (s))

Together with (5.1.13), similarly to (5.1.14), we obtain that

E{|X (t)X (s)||X(t) = x, X(s) = y, X (t) = X (s) = 0}
x2 + y 2
≤ C8 1 +
,
[detCov(X(t), X(s), X (t), X (s))]2

(5.1.25)

for some C8 > 0. On the other hand,

pX (t),X (s) (0, 0|X(t) = x, X(s) = y)
≤

1
2π

1
=
2π
≤

detCov(X (t), X (s)|X(t), X(s))
detCov(X(t), X(s))
detCov(X(t), X(s), X (t), X (s))
C9

detCov(X(t), X(s), X (t), X (s))

133

,

(5.1.26)

for some C9 > 0. Plugging (5.1.25) and (5.1.26) into (5.1.24), we obtain that for large u,

E{Mu (Ii )Mu (Ij )}
≤ 2(ai+1 − ai )(aj+1 − aj

)(f (δ, T ))−5/2

∞
u

∞
u

(5.1.27)
(x2

+ y 2 )p

X(t),X(s) (x, y)dxdy.

Let R(δ) := supt≥δ r(t) which is strictly less than 1 by (H3), then for sufficiently large u,
∞

∞

sup
|s−t|≥δ u

u

(x2 + y 2 )pX(t),X(s) (x, y)dxdy

≤ sup E{(X(t)X(s))2 1{X(t)≥u,X(s)≥u} }
|s−t|≥δ

(5.1.28)

≤ sup E{(X(t) + X(s))4 )1{X(t)+X(s)≥2u} }
|s−t|≥δ

2
≤ u4 e−u /(1+R(δ)) .

Let ε1 > 1 − R(δ), by (5.1.27) and (5.1.28), we obtain that for sufficiently large u,

2
E{Mu (Ii )Mu (Ij )} ≤ T 2 (f (δ, T ))−5/2 e−u /(2−ε1 ) .

(5.1.29)

i=j,Ii ∩Ij =∅

Combining (5.1.20) and (5.1.23) with (5.1.29), we obtain the desired result.

Lemma 5.1.4 Let X be a centered stationary Gaussian process satisfying (H1), (H3) and
(A1). Then there exist some universal positive constants δ and α such that for all T > 0
and u large enough,

2 2
E{Mu (0, T )1{X(0)≥u,X (0)≤0} } ≤ T [f (δ, T )]−3/2 e−αu −u /2 ,

2 2
E{Mu (0, T )1{X(T )≥u,X (T )≥0} } ≤ T [f (δ, T )]−3/2 e−αu −u /2 ,

134

where f (δ, T ) is defined in (5.1.6).

Proof We shall only prove the first inequality, since the proof for the second one is similar.
By Kac-Rice formula,
E{Mu (0, T )1{X(0)≥u,X (0)≤0} }
∞

T

=

dt
0

∞

dx

0

dy

u

−∞

u

dz E{|X (t)|1{X (t)<0} |X(t) = x, X(0) = y,

X (t) = 0, X (0) = z}pX(t),X(0),X (t),X (0) (x, y, 0, z)
∞

∞

T

≤

−∞

u

u

0

0

dy

dx

dt

dz E{|X (t)||X(t) = x, X(0) = y, X (t) = 0, X (0) = z}

× pX(t),X(0),X (t),X (0) (x, y, 0, z)
T

Au (t)dt.

:=
0

Let δ be a positive constant to be specified. We first consider
T
δ

∞

T

Au (t)dt ≤

dt
δ

∞

dx
u

u

dy E{|X (t)||X(t) = x, X(0) = y, X (t) = 0}
(5.1.30)

× pX (t) (0|X(t) = x, X(0) = y)pX(t),X(0) (x, y).
Note that E|ξ| ≤ E|ξ − Eξ| + |Eξ| ≤

Var(ξ) + |Eξ| for any random variable ξ,

Var(X (t)|X(t), X(0), X (t)) ≤ Var(X (t)) = Var(X (0)).

135

By Lemma 2.5.1 and the stationarity of X, there exists some C1 > 0 such that

sup |E{X (t)|X(t) = x, X(0) = y, X (t) = 0}|
δ≤t≤T

C1 (|x| + |y|)
δ≤t≤T detCov(X(t), X(0), X (t))
C1 (|x| + |y|)Var(X (0)|X(t), X(0), X (t))
= sup
detCov(X(t), X(0), X (t), X (0))
δ≤t≤T
≤ sup

≤ C1 λ2 [f (δ, T )]−1 (|x| + |y|).
It follows that there exists some C2 > 0 such that

E{|X (t)||X(t) = x, X(0) = y} ≤ C2 (1 + [f (δ, T )]−1 )(|x| + |y|).

(5.1.31)

Similarly, there exists some C3 > 0 such that

sup pX (t) (0|X(t) = x, X(0) = y)

δ≤t≤T

≤ sup
δ≤t≤T

(5.1.32)

1
2πVar(X (t)|X(t), X(0))

≤ C3

[f (δ, T )]−1/2 .

On the other hand, for sufficiently large u,
∞

sup

∞

dx

t≥δ u

u

dy(x + y)pX(t),X(0) (x, y)

= sup E{(X(t) + X(0))1{X(t)≥u,X(0)≥u} }
t≥δ

2
≤ sup E{(X(t) + X(0))1{X(t)+X(0)≥2u} } ≤ ue−u /(1+R(δ)) .
t≥δ

136

(5.1.33)

Plugging (5.1.31), (5.1.32) and (5.1.33) into (5.1.30), we obtain that for all T > 0, if u is
sufficiently large, then
T

Au (t)dt ≤

u2
−3/2 e− 2−ε1 ,
T [f (δ, T )]

δ

where ε1 > 1 − R(δ).
Next we consider
δ
0

∞

δ

Au (t)dt ≤

dt
0

0

dy
u

dzE{|X (t)||X(0) = y, X (0) = z, X (t) = 0}
−∞

× pX(0),X (0),X (t) (y, z, 0).
Note that

pX(0),X (0),X (t) (y, z, 0)
= pX(0) (y|X (0) = z, X (t) = 0)pX (0) (z|X (t) = 0)pX (t) (0)
−

≤ (2π)−3/2 [detCov(X(0), X (0), X (t))]−1/2 e

(y−µt,z )2
2
− z2
2
2σt
e 2γt ,

where

µt,z = E{X(0)|X (0) = z, X (t) = 0},
2
σt = Var(X(0)|X (0), X (t)),
2
γt = Var(X (0)|X (t)).

By (H1) and Taylor’s formula, we can write

X (t) = X (0) + X (0)t + Y0,t t1+η ,

137

where Y0,t is a centered Gaussian variable. We find that for t ∈ (0, δ) with δ sufficiently
small,

E{X(0)X (t)} = t(E{X(0)X (0)} + tη E{X(0)Y0,t })
= t(−λ2 + tη E{X(0)Y0,t }) ≤ 0.
Since z < 0, it follows from Lemma 2.5.1 that µt,z ≤ 0. If δ is sufficiently small, we also
have, similarly to (5.1.15),

2
σt = Var(X(0)|X (0))(1 + o(1)) < 1 − ε0 ,

and similarly to (5.1.17),
2
C4 t2 ≤ γt ≤ C5 t2 ,

detCov(X(0), X (0), X (t)) ≥ C6 t2 ,
where ε0 , C4 , C5 and C6 are some positive constants. Together with the fact that

E{|X (t)||X(0) = y, X (0) = z, X (t) = 0}
= E{|X (t)||X(0) = y, X (t) − tX (t) + t1+η Yt,0 = z, X (t) = 0}
= E{|X (t)||X(0) = y, X (t) − tη Yt,0 = z/t, X (t) = 0}
≤ C7 (y + |z/t| + 1)

138

for some C7 > 0, where the first equality is due to Taylor’s formula, we obtain that for δ
sufficiently small and u sufficiently large,

δ
0

−1/2

Au (t)dt ≤ (2π)−3/2 C6

−1/2

≤ (2π)−3/2 C6

≤ δe

−

(y−µt,z )2
2
− z2
2
2σt
e 2γt

−

y2
z2
2 − 2C t2
2σt
5
e

0
∞
1
dz(y + |z/t| + 1)e
dy
dt
−∞
u
0 t
δ

∞

δ

1
dt
0 t
u

0
−∞
0

−

dz(y + |z| + 1)e

dy

dt
0

dz(y + |z/t| + 1)e

dy

∞

δ

−1/2

= (2π)−3/2 C6

−

y2
z2
2
2σt − 2C5
e

−∞

u

u2
2(1−ε0 ) .

(5.1.34)

Combining (5.1.30) with (5.1.34), we obtain that there exist some universal δ, α > 0 such
that for all T > 0 and u large enough,

2 2
E{Mu (0, T )1{X(0)≥u,X (0)≤0} } ≤ T [f (δ, T )]−3/2 e−αu −u /2 .

This completes the proof.

Lemma 5.1.5 Let X be a centered stationary Gaussian process satisfying (H1), (H3) and
(A1). Then there exists some universal α > 0 such that for all T > 0 and u large enough,

2 2
P{X(0) ≥ u, X (0) ≤ 0, X(T ) ≥ u, X (T ) ≥ 0} ≤ e−αu −u /2 .

139

Proof

Let δ > 0, then similarly to (5.1.33), we obtain that for sufficiently large u,

sup P{X(0) ≥ u, X (0) ≤ 0, X(T ) ≥ u, X (T ) ≥ 0}

T ≥δ

(5.1.35)

≤ sup P{X(0) ≥ u, X(T ) ≥ u} ≤

2
e−u /(1+R(δ)) .

T ≥δ

For T ∈ (0, δ), by Taylor’s formula,

X (T ) = X (0) + X (0)T + Y0,T T 1+η ,

it follows that

P{X(0) ≥ u, X (0) ≤ 0, X(T ) ≥ u, X (T ) ≥ 0}
≤ P{X(0) ≥ u, X (0) ≤ 0, X (T ) ≥ 0}
(5.1.36)
= P{X(0) ≥ u, X (0) ≤ 0, X (0) + X (0)T + Y0,T T 1+η ≥ 0}
≤ P{X(0) ≥ u, X (0) + Y0,T T η ≥ 0}.
Let ξ(T ) = X (0) + Y0,T T η , κ2 (T ) = Var(ξ(T )) and ρ(T ) = E{X(0)ξ(T )}/κ(T ), T ∈ (0, δ).
Since E{X(0)X (0)} = −λ2 , if δ is sufficiently small, sup0≤T ≤δ ρ(T ) < −ε0 for some ε0 > 0.
Let

κ = sup κ(T ),

κ=

0≤T ≤δ

ρ = sup ρ(T ),

ρ=

0≤T ≤δ

140

inf

κ(T ),

inf

ρ(T ),

0≤T ≤δ

0≤T ≤δ

then 0 < κ ≤ κ < ∞ and −1 < ρ ≤ ρ < −ε0 . We obtain that as u → ∞,

sup P{X(0) ≥ u, X (0) + Y0,T T η ≥ 0}

0≤T ≤δ

1

= sup
0≤T ≤δ

× exp

∞

2πκ(T )(1 − ρ2 (T ))1/2 u
−

∞

dx1

0

dx2

x2
1
2ρ(T )x1 x2
x2 + 2 2 −
1
2 (T ))
κ(T )
2(1 − ρ
κ (T )
1

∞

dx1 exp{−x2 /2}
1
2 (T ))1/2 u
0≤T ≤δ 2πκ(T )(1 − ρ
∞
(x − κ(T )ρ(T )x1 )2
dx2
×
exp − 22
2κ (T )(1 − ρ2 (T ))
0
∞
∞
1
(x2 − κρx1 )2
2 /2}
≤
dx1 exp{−x1
exp −
2κ2 (1 − ρ2 )
2πκ(1 − ρ2 )1/2 u
0
κ2 ρ2 u2
u2
− 2
+ εu2
,
= o exp −
2
2κ (1 − ρ2 )
= sup

(5.1.37)

dx2

for any ε > 0. Combining (5.1.35) and (5.1.36) with (5.1.37) yields the result.

Theorem 5.1.6 Let {X(t) : t ∈ R+ } be a centered stationary Gaussian process satisfying
(H1), (H3) and (A1), and let T be a non-negative random variable independent of X. If
ET 2 (f (δ, T ))−5/2 < ∞ for any fixed δ > 0, then there exists α > 0 such that as u → ∞,

sup X(t) ≥ u = Ψ(u) +

P

0≤t≤T

Proof

2 2
λET −u2 /2
e
+ o(e−αu −u /2 ).
2π

Let FT be the cumulative distribution function of T . Note that
∞

P

sup X(t) ≥ u =
0≤t≤T

0

P

sup X(t) ≥ u FT (dT ),

0≤t≤T

141

(5.1.38)

combining (5.1.2), (5.1.3), Lemma 5.1.1, Lemma 5.1.2 and Lemma 5.1.4 with Lemma 5.1.5,
we obtain the result.

Example 5.1.7 Let X be a centered stationary Gaussian process with covariance function
2

t
r(t) = e− 2 . Then Var(X (0)) = Var(X(0)) = 1, E{X (t)X(0)} = −E{X(t)X (0)} =
t2

r (t) = −te− 2 and E{X (t)X (0)} = −r (t), thus

2
detCov(X(t), X (t), X(0), X (0)) = (1 − e−t )2 ,

(5.1.39)

which is increasing in t > 0. Hence if ET 2 < ∞, then

P

sup X(t) ≥ u = Ψ(u) +
0≤t≤T

5.2

2 2
ET −u2 /2
e
+ o(e−αu −u /2 ).
2π

Gaussian Processes with Increasing Variance

In this section, we consider a Gaussian process {X(t) : t ∈ R+ } with increasing variance at
infinity. Let

2
σt = Var(X(t)),

λ2 = Var(X (t)),
t

2
θt = Var(X(t)|X (t)).

(5.2.1)

Let T be a non-negative random variable satisfying

P{T ≥ t} = exp{−βtα (1 + o(1))} as t → ∞,

142

(5.2.2)

where α, β > 0. We write T ∈ E(α, β) if T satisfies (5.2.2). Note that (5.2.2) implies that
the corresponding cumulative distribution function FT (t) is continuous when t is sufficiently
large.
In additional to conditions (H1) and (H3), we will impose the following two conditions
(A2) and (A3) on X.
(A2). There exist α∞ > 0 and D1 > D2 > 0 such that as t → ∞,

2
σt = D1 tα∞ (1 + o(1)),

2
θt = D2 tα∞ (1 + o(1)).

(A3). There exists N1 > 0 such that as t → ∞,

max{λ2 , Var(X (t)), (detCov(X(t), X (t)))−1 } = O(tN1 ).
t

We will make use of the following inequality to estimate the excursion probability over
each fixed interval [0, T ]:

P{X(T ) ≥ u} ≤ P

sup X(t) ≥ u ≤ P{X(T ) ≥ u} + P{X(0) ≥ u} + E{Mu (0, T )}.
0≤t≤T

(5.2.3)
The following result is Lemma 6.2 in Arendarczyk and Decicki (2011), which is analogous
to the Laplace method.
Lemma 5.2.1 Let α1 , α2 , β1 , β2 > 0 and a(u) = u(1−δ)α1 /(α1 +α2 ) , A(u) = u(1+δ)α1 /(α1 +α2 ) ,
where 0 < δ < α2 /α1 . Then as u → ∞,
A(u)

exp
a(u)

β uα1
− 1 α − β2 xα2 dx = exp{−β3 uα3 (1 + o(1))},
x 1
143

where

α3 =

α1 α2
,
α1 + α2
α /(α1 +α2 ) α1 /(α1 +α2 )
β2

β3 = β1 2

α1 α2 /(α1 +α2 )
α2 α1 /(α1 +α2 )
.
+
α2
α1

Now we prove a lemma similar to Lemma 2.1 in Arendarczyk and Decicki (2011).
Lemma 5.2.2 Let X ∈ E(α1 , β1 ), Y ∈ E(α2 , β2 ) be independent non-negative random variables. Then XY ∈ E(α, β) with

α=

α1 α2
,
α1 + α2
α /(α1 +α2 ) α1 /(α1 +α2 )
β2

β = β1 2

Proof

α2 α1 /(α1 +α2 )
α1 α2 /(α1 +α2 )
+
.
α2
α1

Let a(u) = u(1−δ)α1 /(α1 +α2 ) , A(u) = u(1+δ)α1 /(α1 +α2 ) , where 0 < δ < α2 /α1 .

Then
∞

P{XY ≥ u} =

0

P{X ≥ u/y}dFY (y),
a(u)

A(u)

P{X ≥ u/y}dFY (y) +

=
0

a(u)

P{X ≥ u/y}dFY (y)

∞

+
A(u)

P{X ≥ u/y}dFY (y)

= I1 (u) + I2 (u) + I3 (u).
For any ε > 0 and u large enough, we see that

I1 (u) ≤ P{X ≥ u/a(u)} ≤ exp{−(β1 − ε)[u/a(u)]α1 }
2

≤ exp{−(β1 − ε)uα1 α2 /(α1 +α2 )+δα1 /(α1 +α2 ) } = o(exp{−uα3 +ε0 })

144

and
I3 (u) ≤ P{Y ≥ A(u)} ≤ exp{−(β2 − ε)(A(u))α2 }
≤ exp{−(β2 − ε)u(1+δ)α1 α2 /(α1 +α2 ) } = o(exp{−uα3 +ε0 }).
Next we estimate I2 . Note that both u/a(u) and u/A(u) tend to ∞, hence for any ε > 0
and u large enough, we have
A(u)

I2 (u) ≥
=

a(u)
A(u)

exp{−(β1 + ε)(u/y)α1 }dFY (y)

∂
exp{−(β1 + ε)(u/y)α1 }P{Y ≥ y}dy
a(u) ∂y
+ exp{−(β1 + ε)[u/a(u)]α1 }P{Y ≥ a(u)}
− exp{−(β1 + ε)[u/A(u)]α1 }P{Y ≥ A(u)}
A(u)

≥
a(u)

exp{−(β1 + ε)(1 + ε)(u/y)α1 } exp{−(β2 + ε)uα2 }dy

+ exp{−(β1 + ε)[u/a(u)]α1 } exp{−(β2 + ε)(a(u))α2 }
− exp{−(β1 + ε)[u/A(u)]α1 } exp{−(β2 − ε)(A(u))α2 }
= I 2 (u, ε) + R1 (u, ε) − R2 (u, ε),
and similarly,
A(u)

I2 (u) ≤

a(u)

exp{−(β1 − ε)(1 − ε)(u/y)α1 } exp{−(β2 − ε)uα2 }dy

+ exp{−(β1 − ε)[u/a(u)]α1 } exp{−(β2 − ε)(a(u))α2 }
− exp{−(β1 − ε)[u/A(u)]α1 } exp{−(β2 + ε)(A(u))α2 }
= I 2 (u, ε) + R1 (u, ε) − R2 (u, ε).

145

Applying Lemma 5.2.1, we obtain that for any ε > 0, as u → ∞

I 2 (u, ε) = exp{−β 3 (ε)uα3 (1 + o(1))},

I 2 (u, ε) = exp{−β 3 (ε)uα3 (1 + o(1))},

where

α3 =

α1 α2
,
α1 + α2

β 3 (ε) = [(β1 + ε)(1 + ε)]α2 /(α1 +α2 ) (β2 + ε)α1 /(α1 +α2 )
×

α1 α2 /(α1 +α2 )
α2 α1 /(α1 +α2 )
+
,
α2
α1

β 3 (ε) = [(β1 − ε)(1 − ε)]α2 /(α1 +α2 ) (β2 − ε)α1 /(α1 +α2 )
×

α2 α1 /(α1 +α2 )
α1 α2 /(α1 +α2 )
+
.
α2
α1

Together with the fact that there exists some ε0 > 0 such that all I1 (u), I3 (u), R1 (u, ε),
R2 (u, ε), R1 (u, ε), R2 (u, ε) are o(exp{−uα3 +ε0 }), we obtain the desired result.

Lemma 5.2.3 Let X be a Gaussian process satisfying (A2) and (A3) and let T ∈ E(α, β)
be a non-negative random variable independent of X. Then X(T ) ∈ E(α, β1 ) with

α=

2α
,
α + α∞

1 α/(α+α∞ )
β1 = β α∞ /(α+α∞ )
2D1

Proof

α α∞ /(α+α∞ )
α∞ α/(α+α∞ )
+
.
α∞
α

(5.2.4)

Let N be the standard Normal random variable and let ν(·) be the standard

deviation function of X, i.e. ν(t) = σt . Note that

P{X(T ) ≥ u} = P{ν(T )N ≥ u}.
146

(5.2.5)

On the other hand, as u → ∞,

P{ν(T ) ≥ u} = P{T ≥ ν −1 (u)} = exp{−β(ν −1 (u))α (1 + o(1))}

and
−1/α∞ 2/α∞
u
(1 + o(1)),

ν −1 (u) = D1
thus

−α/α∞ 2α/α∞
u
(1 + o(1))},

P{ν(T ) ≥ u} = exp{−βD1
−α/α∞

i.e., ν(T ) ∈ E(2α/α∞ , βD1

). Note that N ∈ E(2, 1/2), applying Lemma 5.2.2 in

(5.2.5), we conclude the result.

Lemma 5.2.4 Let X be a Gaussian process satisfying (H1), (H3), (A2) and (A3), and let
T ∈ E(α, β) be a non-negative random variable independent of X. Then for any ε > 0,
∞
0

E{Mu (0, T )}FT (dT ) = o(exp{−(β2 − ε)uα })

as u → ∞,

where

α=
β2 =

2α
,
α + α∞
β α∞ /(α+α∞ )

1 α/(α+α∞ )
2D2

α α∞ /(α+α∞ )
α∞ α/(α+α∞ )
> β1 .
+
α∞
α

147

(5.2.6)

Proof

By the Kac-Rice formula,

E{Mu (0, T ))}
∞

T

dt

=

u
∞

0
T

≤

dt
0

dx E{|X (t)|1{X (t)<0} |X(t) = x, X (t) = 0}pX(t),X (t) (x, 0)
dx E{|X (t)||X(t) = x, X (t) = 0}pX(t) (x|X (t) = 0)pX (t) (0)

u

∞
2
2
1
1
dt
dx E{|X (t)||X(t) = x, X (t) = 0} e−x /2θt
θt
0 2πλt
u
T

=

Note that E|ξ| ≤ E|ξ − Eξ| + |Eξ| ≤

Var(ξ) + |Eξ| for any random variable ξ, and

Var(X (t)|X(t), X (t)) ≤ Var(X (t)),

E{X (t)|X(t) = x, X (t) = 0}
=

E{X (t)X(t)}λ2 − E{X (t)X (t)}E{X (t)X(t)}
t
x.
detCov(X(t), X (t))

thus

E{|X (t)||X(t) = x, X (t) = 0}
≤

Var(X (t)) +

|E{X (t)X(t)}λ2 − E{X (t)X (t)}E{X (t)X(t)}|
t
x,
detCov(X(t), X (t))

Now let

h(t)

1
λt θt

|E{X (t)X(t)}λ2 − E{X (t)X (t)}E{X (t)X(t)}|
t
Var(X (t)) +
.
detCov(X(t), X (t))

By (A3), there is some N2 > 0 such that

h(t) = o(tN2 ) as t → ∞.

148

2
2
Let T0 be a large number such that for t > T0 , h(t) ≤ tN1 , θt is increasing and θt ≤ tα1 +1 .
2
Let A be a large number such that sup0≤t≤T0 h(t) ≤ A and sup0≤t≤T0 θt ≤ A, then for u

large enough,

E{Mu (0, T ))} ≤

T0
0

T 1
∞
∞
2
2
2
1
−x2 /2θt
+
dxe
dxe−x /2θt
h(t)dt
h(t)dt
2π
T0 2π
u
u

T 1
∞
2
2
T
−x2 /2θT
≤ 0 Ae−u /(2A) +
tN1 dt
dxe
2π
T0 2π
u
∞
2
2
T
1
1
−x2 /2θT
≤ 0 Ae−u /(2A) + √ T N1 +α1 /2+3/2
dx √
e
.
2π
2π
2πθT
u

Hence we have
∞
0

E{Mu (0, T )}dFT (T )
∞
∞
2
2
T
1
−x2 /2θT
√
T N1 +α1 +2 dFT (T )
≤ 0 Ae−u /(2A) +
e
dx,
2π
2πθT
0
u

= I1 (u) + I2 (u).

Let T be a non-negative random variable with cumulative distribution function satisfying
dFT (t) = tN1 +α1 +2 dFT (t), then T ∈ E(α, β). Let {X(t) : t ∈ R+ } be a Gaussian process
2
with Var(X(t)) = θt , then by Lemma 5.2.3,

X(T ) ∈ E(α, β2 ),

where α and β2 are as shown in (5.2.6). Note that

I2 (u) = P{X(T ) ≥ u}

and 2 > α hence I1 (u) = o(exp{−uα+δ }) for any δ ∈ (0, 2 − α). Thus both I1 (u) and I2 (u)
149

are o(exp{−(β2 − ε)uα }) for any ε > 0. The proof is completed.

Theorem 5.2.5 Let {X(t) : t ∈ R+ } be a Gaussian process satisfying (H1), (H3), (A2)
and (A3), and let T ∈ E(α, β) be a non-negative random variable independent of X. Then
X(T ) ∈ E(α, β1 ) and as u → ∞,

P

sup X(t) ≥ u = P{X(T ) ≥ u} + o(exp{−(β2 − ε)uα })

0≤t≤T

= P{X(T ) ≥ u}(1 + o(exp{−(β2 − β1 − ε)uα })

for any ε > 0, where α, β1 and β2 are as shown in (5.2.4) and (5.2.6).

Proof

Note that
∞

sup X(t) ≥ u =

P

0

0≤t≤T

P

sup X(t) ≥ u FT (dT ),

0≤t≤T

combining (5.2.3) and Lemma 5.2.3 with Lemma 5.2.4, we obtain the result.
t s
Example 5.2.6 Let X(t) = 0 0 B(v)dvds, where B(v) is the standard Brownian motion.

Then one has
t3
,
3
t4
E{X(t)X (t)} = ,
8
2
σt =

t5
,
20

λ2 =
t

Var(X (t)) = t,
E{X(t)X (t)} =

2
θt = Var(X(t)|X (t)) =

t3
,
6

E{X (t)X (t)} =

t2
,
2

t5
.
320

t s
Example 5.2.7 Let X(t) = 0 0 Z(v)dvds, where Z(v) is a continuous stationary Gaussian

150

process with covariance function R(t) such that R(0) = 1 and

R(t) = Dtα∞ −4 (1 + o(1)) as t → ∞,

where D > 0, 2 < α∞ < 4. Then

2
σt =

2D
tα∞ (1 + o(1))
α∞ (α∞ − 2)(α∞ − 3)

and
2
2
θt = σt − [E{X(t)X (t)}]2 /Var(X (t))

=

(4 − α∞ )D
tα∞ (1 + o(1)).
2α∞ (α∞ − 2)(α∞ − 3)

151

Chapter 6
Ruin Probability of a Certain Class of
Smooth Gaussian Processes
Let {X(t) : t ≥ 0} be a centered smooth Gaussian process with variance t2γ for some
γ > 2. We consider the probability P{supt≥0 (X(t) − ctβ ) ≥ u} as u → ∞, where c > 0
and β > γ. We derive some asymptotic approximations to such probability which refine the
result of H¨sler and Piterbarg (1999).
u

6.1

Self-similar Processes

Let {X(t) : t ≥ 0} be a centered smooth Gaussian process with variance t2γ for some γ > 2.
We say X is self similar if its covariance function C(t, s) satisfies

C(at, as) = a2γ C(t, s),

152

∀t, s ≥ 0, a > 0.

(6.1.1)

Let Y (t) =

X(t)
,
1+ctβ

then

P sup(X(t) − ctβ ) ≥ u = P{X(t) ≥ u + ctβ for some t ≥ 0}
t≥0

= P{X(u1/β t) ≥ u + c(u1/β t)β for some t ≥ 0}
= P{uγ/β X(t) ≥ u(1 + ctβ ) for some t ≥ 0}

(6.1.2)

X(t)
≥ u1−γ/β
1 + ctβ
t≥0

= P sup

= P sup Y (t) ≥ u1−γ/β ,
t≥0

where the third equality is due to the self-similarity (6.1.1). Note that Var(Y (t)) =

t2γ
,
(1+ctβ )2

as a function of t, attains its maximum

σ2

=

2γ/β
−2
β
γ
c(β − γ)
β−γ

at the unique point
t0 =

1/β
γ
.
c(β − γ)

Theorem 6.1.1 Let {X(t) : t ≥ 0} be a centered self-similar Gaussian process with variance
t2γ for some γ > 2. Let β > γ and c > 0. Suppose X satisfies (H1) and (H3). Then there
exists some α > 0 such that as u → ∞,

P sup(X(t) − ctβ ) ≥ u = P sup Y (t) ≥ u1−γ/β
=−

t≥0
2t0

t≥0

∞

dt

t0 /2

+ o exp

E{Y (t)|Y (t) = x, Y (t) = 0} −x2 /2θ2
t dx
e
2π detCov(Y (t), Y (t))
u1−γ/β
−

u2−2γ/β
− αu2−2γ/β
2
2σ

153

,

(6.1.3)

where Y (t) =

Proof

X(t)
1+ctβ

2
and θt = Var(Y (t)|Y (t)).

The first equality in (6.1.3) is the result in (6.1.2). Note that

Y (t) ≥ u1−γ/β

sup

P

≤ P sup Y (t) ≥ u1−γ/β
t≥0

t0 /2≤t≤2t0

≤P

Y (t) ≥ u1−γ/β + P

sup
t0 /2≤t≤2t0

+P

sup

Y (t) ≥ u1−γ/β

(6.1.4)

0≤t≤t0 /2

sup Y (t) ≥ u1−γ/β ,
t≥2t0

where the last two terms are super-exponentially small due to the Borell-TIS inequality [cf.
Theorem 2.1.1 in Adler and Taylor (2007)]. On the other hand, by Theorem 3.1.9,

Y (t) ≥ u1−γ/β

sup

P

t0 /2≤t≤2t0
2t0

=−

∞

dt

t0 /2

+ o exp

E{Y (t)|Y (t) = x, Y (t) = 0} −x2 /2θ2
t dx
e
2π detCov(Y (t), Y (t))
u1−γ/β
−

u2−2γ/β
− αu2−2γ/β
2
2σ

.

Combining this with (6.1.4) yields the desired result.

Corollary 6.1.2 Under the assumptions in Theorem 6.1.1, as u → ∞,

P sup(X(t) − ctβ ) ≥ u = P sup Y (t) ≥ u1−γ/β
t≥0

=

Var(Y (t0 ))
+1
E{Y (t0 )Y (t0 )}

t≥0

−1/2

Ψ

u1−γ/β
(1 + o(1)).
σ

Proof One can check that the second derivative of the variance function of Y , Var(Y (t)) =
t2γ
,
(1+ctβ )2

at t0 is not equal to 0. This implies that the condition in (3.2.6) holds. Applying
154

Corollary 3.2.3 and Theorem 6.1.1, we obtain the result.

6.2

Integrated Fractional Brownian Motion

In this section, we show the application to a typical example, the double integrated fractional
Brownian motion.
t s
Let X(t) = 0 0 BH (u)duds, where BH is fractional Brownian motion with Hurst index

H, i.e. Cov(BH (t)BH (s)) = 1 (t2H +s2H −|t−s|2H ). Then X satisfies (6.1.1) with γ = H+2,
2
it also satisfies (H1) and (H3), and

Var(X(t)) =

Let β > H + 2 and Y (t) =

X(t)
,
1+ctβ

t2H+4
.
2(2H + 1)(2H + 4)

(6.2.1)

we consider the probability

P sup(X(t) − ctβ ) ≥ u = P sup Y (t) ≥ u1−(H+2)/β .
t≥0

(6.2.2)

t≥0

We see that
Var(Y (t)) =

t2H+4
,
2(2H + 1)(2H + 4)(1 + ctβ )2

(6.2.3)

which attains the maximum at the unique point

t0 =

1/β
H +2
.
c(β − H − 2)

155

(6.2.4)

Note that

E(Y (t)Y (t)) =
Var(Y (t)) =

cβtβ
t2H+3
1
−
,
2(2H + 1)(1 + ctβ )2 2 (2H + 4)(1 + ctβ )

c2 β 2 t2β
t2H+2
1
cβtβ
+
,
−
(1 + ctβ )2 2H + 2 2(2H + 1)(2H + 4)(1 + ctβ )2 2(2H + 1)(1 + ctβ )

and

E{Y (t)Y (t)}
=

cβtβ
t2H+2
2H 2 + H + 1 c2 β(β + 1)t2β − cβ(β − 1)tβ
+
−
,
2H + 2
2(2H + 1)(1 + ctβ )2
(2H + 4)(1 + ctβ )2
1 + ctβ

it follows that
H2 − H
Var(Y (t0 ))
=
.
E{Y (t0 )Y (t0 )}
(β − 2)(H + 1) − 2H 2
Thus by Corollary 6.1.2,

P sup(X(t) − ctβ ) ≥ u
t≥0

= P sup Y (t) ≥ u1−(H+2)/β
t≥0

∼

−1/2
Var(Y (t0 ))
Ψ
+1
E{Y (t0 )Y (t0 )}

∼

(β − 2)(H + 1) − 2H 2 1/2
Ψ
(β − H − 2)(H + 1)

u1−(H+2)/β

(6.2.5)

Var(Y (t0 ))
u1−(H+2)/β
Var(Y (t0 ))

.

Now let H = 1/2, β = 3 and c = 1. By the discussions above,
√
P

sup(X(t) − ctβ )

≥u ∼

t≥0

4/3Ψ

144u1/6
51/3

.

(6.2.6)

However, applying the Laplace approximation of higher order to (6.1.3), we will get a more

156

accurate approximation:

P sup(X(t) − ctβ ) ≥ u
t≥0

∞
√
1
∼
πc0 +
2π u1/6

where

6.3

√

π
1
c2 + c0 2
2
x

−

exp

72x2
52/3

(6.2.7)
dx,

√
√
√
19 3055/6
8 3051/6
15
, c2 =
, c0 =
.
c0 =
5
2700
5

More General Gaussian Processes

Assume that {X(t) : t ≥ 0} is a centered smooth Gaussian process with variance t2γ for
X(u1/β t)

some γ > 2. Let Xu (t) = γ/β
, then
u
(1+ctβ )

P sup(X(t) − ctβ ) ≥ u = P{X(t) ≥ u + ctβ for some t ≥ 0}
t≥0

= P{X(u1/β t) ≥ u + c(u1/β t)β for some t ≥ 0}
=P

X(u1/β t)
uγ/β (1 + ctβ )

(6.3.1)
≥

u1−γ/β

for some t ≥ 0

= P sup Xu (t) ≥ u1−γ/β .
t≥0

Note that Var(Xu (t)) =

t2γ
,
(1+ctβ )2

σ2

=

as a function of t, attains its maximum

2γ/β
−2
γ
β
c(β − γ)
β−γ

at the unique point
t0 =

1/β
γ
.
c(β − γ)

157

We see that neither σ 2 nor t0 depend on u.
E{X(t)X(s)}
Let r(t, s) = √

Var(X(t))Var(X(s))

, we will make use of the following condition.

(A1 ). For any fixed δ > 0, R(δ) := sup|t−s|≥δ r(t, s) < 1.
Let M 1−γ/β (Xu , (t0 /2, 2t0 )) be the number of local maximum points t ∈ (t0 /2, 2t0 )
u
such that Xu (t) exceeding level u1−γ/β .
Lemma 6.3.1 Let {X(t) : t ≥ 0} be a centered Gaussian process with variance t2γ for some
γ > 2. Assume X ∈ C 2 (R+ ) a.s. and that X satisfies the regularity conditions (H3) and
(A1 ). Suppose there exist positive constants C0 , N0 and η0 such that for all t ≥ 0,
Var(X (t)) ≤ C0 (tN0 + 1),
(6.3.2)
[detCov(X(t), X (t), X

(t))]−1

≤ C0

(tN0

+ 1);

for all t = s,

E(X (t) − X (s))2 ≤ C0 [(t + s)N0 + 1](t − s)2η0 ;

(6.3.3)

and all |t − s| ≥ δ0 , where δ0 > 0 is some fixed number,

[detCov(X(t), X (t), X(s), X (s))]−1 ≤ C0 (t + s)N0 .

Then there exists some α > 0 such that as u → ∞,

E{M 1−γ/β (Xu , (t0 /2, 2t0 ))[M 1−γ/β (Xu , (t0 /2, 2t0 )) − 1]}
u
u
= o exp

−

u2−2γ/β
− αu2−2γ/β
2
2σ
158

.

(6.3.4)

Proof

By the Kac-Rice metatheorem, one has

E{M 1−γ/β (Xu , (t0 /2, 2t0 ))[M 1−γ/β (Xu , (t0 /2, 2t0 )) − 1]}
u
u
≤

2t0

dt

t0 /2

2t0
t0 /2

∞

ds

u1−γ/β

dxE{|Xu (t)Xu (s)||Xu (t) = x, Xu (t) = Xu (s) = 0}

(6.3.5)

× pXu (t) (x|Xu (t) = Xu (s) = 0)pX (t),X (s) (0, 0).
u
u
Let Eu (t, s) := E{|Xu (t)Xu (s)||Xu (t) = x, Xu (t) = Xu (s) = 0}. By Taylor’s formula,

Xu (s) = Xu (t) + Xu (t)(s − t) + |s − t|1+η Yt,s,u ,

(6.3.6)

where Yt,s,u is a centered Gaussian variable. In particular, for s > t,
s
Xu (s) − Xu (t) − Xu (t)(s − t)
t (Xu (v) − Xu (t))dv .
Yt,s,u =
=
(s − t)1+η
(s − t)1+η

Differentiating Var(X(t)) = t2γ with respective to t twice, we see that

2(Var(X (t))) + E{X(t)X (t)} = 2γ(2γ − 1)t2γ−2 .

Since |E{X(t)X (t)}| ≤ Var(X(t))Var(X (t)) = t2γ Var(X (t)), together with condition
(6.3.2), we get
Var(X (t)) ≤ C1 (tN1 + 1)
for some positive constants C1 and N1 . Combining this fact with conditions (6.3.2) and
(6.3.3), we obtain
sup

Var(Yt,s,u ) ≤ C2 uN2

t0 /2≤t<s≤2t0

159

for some positive constants C2 and N2 .
Applying (6.3.6), we have

Eu (t, s) = E{|Xu (t)Xu (s)||Xu (t) = x, Xu (t) = 0, Xu (t)(s − t) = −|s − t|1+η Yt,s,u }
= |s − t|η E{|Yt,s,u Xu (s)||Xu (t) = x, Xu (t) = 0, Xu (t)(s − t) = −|s − t|1+η Yt,s,u }.
(6.3.7)
For any Gaussian variables ξ1 , ξ2 , the following inequality holds,

2
2
E|ξ1 ξ2 | ≤ Eξ1 + Eξ2 = (Eξ1 )2 + Var(ξ1 ) + (Eξ2 )2 + Var(ξ2 ).

(6.3.8)

We have
Var(Xu (t)|Xu (t) = x, Xu (t) = Xu (s) = 0) ≤ Var(Xu (t)) ≤ C3 uN3 ,
Var(Yt,s,u |Xu (t) = x, Xu (t) = Xu (s) = 0) ≤ Var(Yt,s,u ) ≤ C2 uN2 ,
for some positive constants C3 and N3 .
On the other hand, for s > t and |s − t| → 0, there exist positive constants C4 , C5 , N4
and N5 such that for large x and u,

|E{Xu (t)|Xu (t) = x, Xu (t) = Xu (s) = 0}|
= |E{Xu (t)|Xu (t) = x, Xu (t) = 0, Xu (t) + |s − t|η Yt,s,u = 0}|
≤ |E{Xu (t)|Xu (t) = x, Xu (t) = 0, Xu (t) = 0}| + o(1)uN4 |x|
≤

C4 |x|
+ o(1)uN |x| ≤ C5 |x|uN5 ,
detCov(Xu (t), Xu (t), Xu (t))

where the last line is due to condition (6.3.2). In fact, let (ξ1 , ξ2 , ξ3 ) be a non-degenerate

160

Gaussian vector, then

detCov(ξ1 , ξ1 + ξ2 , ξ1 + ξ2 + ξ3 ) = detCov(ξ1 , ξ2 , ξ3 ).

By using this identity, we see that condition (6.3.2) implies that there exist positive constants
C0 and N0 such that for large u,

sup
t0 /2≤t<s≤2t0

[detCov(Xu (t), Xu (t), Xu (t))]−1 ≤ C0 uN0 .

Similarly we obtain that as |s − t| → 0, there exist positive constants C6 and N6 such that
for large x and u,

|E{Yt,s,u |Xu (t) = x, Xu (t) = Xu (s) = 0}| ≤ C6 |x|uN6 .

Combining these results with (6.3.7) and (6.3.8), we get

2
2
Eu (t, s) ≤ |s − t|η (C3 uN3 + C4 uN4 + C5 |x|2 u2N5 + C6 |x|2 u2N6 ).

By Taylor’s formula (6.3.6), as |s − t| → 0,

Var(Xu (t)|Xu (t), Xu (s))
= Var(Xu (t)|Xu (t), Xu (t) + Xu (t)(s − t) + |s − t|1+η Yt,s,u )
= Var(Xu (t)|Xu (t), Xu (t) ± |s − t|η Yt,s,u )
= Var(Xu (t)|Xu (t), Xu (t))(1 + o(1)).

161

(6.3.9)

Let

κ2 :=

sup
u>0,t0 /2≤t≤2t0

Var(Xu (t)|Xu (t), Xu (t)).

One can check that the second derivative of the variance function of Xu (t), Var(Xu (t)) =
t2γ
,
(1+ctβ )2

at t0 is not equal to 0. Therefore supu>0 E{Xu (t0 )Xu (t0 )} < 0 and moreover,

κ2 < σ 2 .
For any ε > 0, if |s − t| is sufficiently small, then for large u,

∞
u1−γ/β

x2 p

Xu (t) (x|Xu (t)

= Xu (s) = 0)dx ≤

2−2γ/β
−u 2
e 2κ +ε .

Note that
pX (t),X (s) (0, 0) ≤
u
u
2π

1
detCov(Xu (t), Xu (s))

,

and by the Taylor expansion, for |s − t| → 0 and large u,

detCov(Xu (t), Xu (s))
= detCov(Xu (t), Xu (t) + Xu (t)(s − t) + |s − t|1+η Yt,s,u )
= |s − t|2 detCov(Xu (t), Xu (t) + |s − t|η Yt,s,u )
= |s − t|2 detCov(Xu (t), Xu (t))(1 + uN7 o(1)),
where N7 is some positive constant. Note that

detCov(Xu (t), Xu (t)) =

detCov(Xu (t), Xu (t), Xu (t))
.
Var(Xu (t)|Xu (t), Xu (t))

162

Thus by conditions (6.3.2) and (6.3.3), there exists positive constant N8 such that for small
|s − t| and large u,

uN8
.
pX (t),X (s) (0, 0) ≤
u
u
|s − t|

Note that when u tends to infinity, the polynomials of u will be killed by the exponential
decay of u. Plugging these results into (6.3.5), we obtain that for any ε > 0, there exists
δ > 0 small enough, such that for large u,

E{M 1−γ/β (Xu , (t0 − δ, t0 + δ))[M 1−γ/β (Xu , (t0 − δ, t0 + δ)) − 1]}
u
u
−u

≤e

2−2γ/β
2κ2 +ε

≤ C7 δ exp

t0 +δ

−

t0 +δ

|s − t|η−1 dtds

t0 −δ t0 −δ
u2−2γ/β

2κ2 + ε

for some positive constant C7 .
The set [t0 /2, 2t0 ] may be covered by congruent intervals Ii = [ai , ai+1 ] with disjoint
interiors such that the lengths are less than δ/2. By similar discussions in Lemma 5.1.2, we
only need to consider non-neighboring Ii = [ai , ai+1 ] and Ij = [aj , aj+1 ], say aj −ai+1 ≥ δ/2.
Then

E{M 1−γ/β (Xu , Ii )M 1−γ/β (Xu , Ij )}
u
u
=

ai+1
ai

aj+1
aj

∞

dtds

u1−γ/β

∞
u1−γ/β

dxdyE{|Xu (t)Xu (s)||Xu (t) = x, Xu (s) = y,

Xu (t) = Xu (s) = 0}pX (t),X (s) (0, 0|Xu (t) = x, Xu (s) = y)pXu (t),Xu (s) (x, y).
u
u
Similarly to (6.3.9), by conditions (6.3.2) and (6.3.3), there exists a positive constant N9

163

such that for large u,

E{|Xu (t)Xu (s)||Xu (t) = x, Xu (s) = y, Xu (t) = Xu (s) = 0}
≤

uN9 (x2 + y 2 )
;
[detCov(Xu (t), Xu (s), Xu (t), Xu (s))]2

and also,

pX (t),X (s) (0, 0|Xu (t) = x, Xu (s) = y)
u
u
1
≤
2π detCov(Xu (t), Xu (s)|Xu (t), Xu (s))
=
≤

1
2π

detCov(Xu (t), Xu (s))
detCov(Xu (t), Xu (s), Xu (t), Xu (s))
uN9

detCov(Xu (t), Xu (s), Xu (t), Xu (s))

.

Thus by condition (6.3.4), there exists a positive constant N10 such that for large u,

E{Mu (Ii )Mu (Ij )}
≤ u2N9 (ai+1 − ai )(aj+1 − aj )

∞

∞

u1−γ/β

u1−γ/β

dxdy(x2 + y 2 )pXu (t),Xu (s) (x, y)
−5/2

×

inf
t,s∈[t0 /2,2t0 ]:|t−s|≥δ/2

≤

uN10 (a

detCov(Xu (t), Xu (s), Xu (t), Xu (s))
∞

i+1

− ai )(aj+1 − aj )

∞

u1−γ/β

u1−γ/β

164

dxdy(x2 + y 2 )pXu (t),Xu (s) (x, y).

By (A1 ), R(δ) = sup|s−t|≥δ r(t, s) is strictly less than 1 hence for u sufficiently large,
∞

∞

sup

|s−t|≥δ/2 u1−γ/β

≤

u1−γ/β

dxdy(x2 + y 2 )pXu (t),Xu (s) (x, y)

E{(Xu (t)Xu (s))2 I(Xu (t) ≥ u1−γ/β , Xu (s) ≥ u1−γ/β )}

sup
|s−t|≥δ/2

≤

E{(Xu (t) + Xu (s))4 )I(Xu (t) + Xu (s) ≥ 2u1−γ/β )}

sup
|s−t|≥δ/2

≤ u4 exp

−

u2−2γ/β
.
1 + R(δ/2)

Combining the results completes the proof.

Theorem 6.3.2 Suppose the assumptions in Lemma 6.3.1 hold. Then there exists α > 0
such that as u → ∞,

P sup(X(t) − ctβ ) ≥ u = P sup Xu (t) ≥ u1−γ/β
=−

t≥0
2t0

t≥0

∞

dt

t0 /2

+ o exp

2
E{Xu (t)|Xu (t) = x, Xu (t) = 0} −x2 /(2θu (t))
dx
e
2π detCov(Xu (t), Xu (t))
u1−γ/β

−

u2−2γ/β
− αu2−2γ/β
2σ 2

(6.3.10)

,

X(u1/β t)

2
where Xu (t) = γ/β
and θu (t) = Var(Xu (t)|Xu (t)).
β)
u
(1+ct

Proof

The first equality in (6.3.10) is the result in (6.3.1). Note that

Xu (t) ≥ u1−γ/β

sup

P

≤ P sup Xu (t) ≥ u1−γ/β
t≥0

t0 /2≤t≤2t0

≤P

sup

Xu (t) ≥ u1−γ/β + P

t0 /2≤t≤2t0

+P

sup
0≤t≤t0 /2

sup Xu (t) ≥ u1−γ/β ,
t≥2t0

165

Xu (t) ≥ u1−γ/β

(6.3.11)

where the last two terms are super-exponentially small due to the Borell-TIS inequality [cf.
Theorem 2.1.1 in Adler and Taylor (2007)]. On the other hand, by Lemma 6.3.1 and the
bounds in (2.3.3),

P

Xu (t) ≥ u1−γ/β

sup
t0 /2≤t≤2t0

u2−2γ/β
− αu2−2γ/β
2
2σ
2t0
∞
2
E{Xu (t)|Xu (t) = x, Xu (t) = 0} −x2 /(2θu (t))
=−
dt
e
dx
2π detCov(Xu (t), Xu (t))
t0 /2
u1−γ/β
u2−2γ/β
− αu2−2γ/β
,
+ o exp −
2
2σ
= E{M 1−γ/β (Xu , (t0 /2, 2t0 ))} + o exp
u

−

where the last equality comes from the combination of similar discussions in Lemma 2.3.2
and Lemma 6.3.1. Combining this with (6.3.11) yields the desired result.
Applying the Laplace method, we obtain the following result.
Corollary 6.3.3 Under the assumptions in Theorem 6.3.2, one has that as u → ∞,

P sup(X(t) − ctβ ) ≥ u = P sup Xu (t) ≥ u1−γ/β
t≥0

=

t≥0

Var(Xu (t0 ))
+1
E{Xu (t0 )Xu (t0 )}

−1/2

Ψ

u1−γ/β
(1 + o(1)).
σ

Example 6.3.4 Let β > H > 2, X(t) = tH Z(t), Xu (t) =

X(tu1/β )
uH/β (1+ctβ )

=

tH Z(tu1/β )
,
1+ctβ

where Z is a smooth stationary Gaussian process with covariance r(t) and r(0) = 1. Then
Var(X(t)) = t2H , and

P sup(X(t) − ctβ ) ≥ u = P sup Xu (t) ≥ u1−H/β .
t>0

t>0

166

Notice that Var(Xu (t)) attains it maximum

Var(Xu (t0 )) =

2H/β
−2
β
H
c(β − H)
β−H

at the unique point
t0 =

1/β
H
.
c(β − H)

By tedious computations, we get

H 1/β
HtH−1 − c(β − H)tH+β−1
1/β ) + t u
Z(tu
Z (tu1/β ),
Xu (t) =
β )2
β
(1 + ct
1 + ct

and

Xu (t) =

H(H − 1)tH−2 + c((2H − β)(H − 1) − β 2 )tH+β−2 + c2 (β − H)(1 − H)t2β+H−2
(1 + ctβ )3
× Z(tu1/β ) +

tH u2/β
2[HtH−1 (1 + ctβ ) − cβtH+β−1 ]u1/β
Z (tu1/β ) +
Z (tu1/β ).
(1 + ctβ )2
1 + ctβ

Notice that E{Z(t)Z (t)} = 0 and Var(Z (t)) = −E{Z(t)Z (t)} = r (0) for all t, we obtain

Var(Xu (t)) =

(HtH−1 − c(β − H)tH+β−1 )2
t2H u2/β
−
r (0)
(1 + ctβ )4
(1 + ctβ )2

=

t2H−2
(H − c(β − H)tβ )2
− t2 u2/β r (0) ,
(1 + ctβ )2
(1 + ctβ )2

and

E{Xu (t)Xu (t)} =

(H(H − 1) + c((2H − β)(H − 1) − β 2 )tβ + c2 (β − H)(1 − H)t2β
(1 + ctβ )2
+ t2 u2/β r (0)

t2H−2
.
(1 + ctβ )2

167

It follows that
−H(β − H) + t2 u2/β r (0)
E{Xu (t0 )Xu (t0 )}
0
=
Var(Xu (t0 )) + E{Xu (t0 )Xu (t0 )}
−H(β − H)
=1−

H 2/β−1 u2/β r (0)
c2/β (β − H)2/β+1

,

and thus
P sup(X(t) − ctβ ) > u = P sup Xu (t) > u1−H/β
t>0

t>0

∼

1/2
E{Xu (t0 )Xu (t0 )}
u1−H/β
Ψ
Var(Xu (t0 )) + E{Xu (t0 )Xu (t0 )}
Var(Xu (t0 ))1/2

=

1−

H 2/β−1 u2/β r (0) 1/2
c2/β (β − H)2/β+1

Ψ

168

u1−H/β
Var(Xu (t0 ))1/2

.

Chapter 7
Excursion Probability of Gaussian
Random Fields on Sphere
In this chapter, we consider a real-valued Gaussian random field X = {X(x) : x ∈ SN }
indexed on the N -dimensional unit sphere SN . The approximations to excursion probability
of the field P{supx∈SN X(x) ≥ u}, as u → ∞, are obtained for two cases: (i) X is locally
isotropic and the sample path is non-smooth; (ii) X is isotropic and the sample path is twice
differentiable. For the first case, it is shown that the asymptotics is similar to Pickands’
approximation on Euclidean space which involves Pickands’ constant; while for the second
case, we use the expected Euler characteristic method to obtain a more precise approximation
such that the error is super-exponentially small.

169

7.1

Notations

For x = (x1 , . . . , xN +1 ) ∈ SN , its corresponding spherical coordinate θ = (θ1 , . . . , θN ) is
defined as follows.

x1 = cos θ1 ,
x2 = sin θ1 cos θ2 ,
x3 = sin θ1 sin θ2 cos θ3 ,
.
.
.
xN = sin θ1 sin θ2 · · · sin θN −1 cos θN ,
xN +1 = sin θ1 sin θ2 · · · sin θN −1 sin θN ,
where 0 ≤ θi ≤ π for 1 ≤ i ≤ N − 1 and 0 ≤ θN < 2π.
Throughout this chapter, for two points x = (x1 , . . . , xN +1 ) and y = (y1 , . . . , yN +1 )
on SN , we always denote by θ = (θ1 , . . . , θN ) the spherical coordinate of x and by ϕ =
(ϕ1 , . . . , ϕN ) the spherical coordinate of y respectively.
Let

· , ·, · be Euclidean norm and inner product respectively. Denote by d(·, ·) the

distance function in SN , i.e., d(x, y) = arccos x, y , ∀x, y ∈ SN .

7.2

Non-smooth Gaussian Fields on Sphere

7.2.1

Locally Isotropic Gaussian Fields on Sphere

Let X = {X(x) : x ∈ SN } be a centered Gaussian random field with covariance function C
satisfying
C(x, y) = 1 − cdα (x, y)(1 + o(1)) as dα (x, y) → 0,
170

(7.2.1)

for some constants c > 0 and α ∈ (0, 2].
Covariance functions satisfying (7.2.1) behave like isotropic in local sense, hence they fall
under the general category of locally isotropic covariance. Also, there are many examples
of covariances of isotropic Gaussian fields on SN satisfying (7.2.1). For instance, C(x, y) =
α
e−cd (x,y) , where c > 0 and α ∈ (0, 1].

Recall the spherical coordinate representation, we define X(θ) := X(x) and denote by C
the covariance function of X accordingly.
Lemma 7.2.1 Let x, y ∈ SN and let x be fixed. Then as d(y, x) → 0,
N −1

d2 (y, x)

∼ (ϕ1 − θ1

)2

+ (sin2 θ1 )(ϕ2

− θ2

)2

sin2 θi (ϕN − θN )2 ,

+ ··· +
i=1

where θ = (θ1 , . . . , θN ) and ϕ = (ϕ1 , . . . , ϕN ) are the spherical coordinates of x and y
respectively.

Proof

Note that x, y ∈ SN implies x 2 = y 2 = 1, hence as d(y, x) → 0, x − y → 0

and
cos y − x ∼ 1 −

1
y − x 2 = y, x .
2

Applying the spherical coordinates, we obtain that as d(y, x) → 0, or equivalently ϕ − θ →
0 (There is an exception for θ with θN = 0, since for those ϕ such that d(y, x) → 0 and ϕN
tending to 2π, ϕ − θ does not tend to 0. In such case, we may treat θN to be 2π instead

171

of 0 and this does not affect the result thanks to the periodicity.),

d2 (y, x) = arccos2 y, x ∼ y − x 2
= (cos ϕ1 − cos θ1 )2 + (sin ϕ1 cos ϕ2 − sin θ1 cos θ2 )2 + · · ·
+ (sin ϕ1 sin ϕ2 · · · sin ϕN −1 cos ϕN − sin θ1 sin θ2 · · · sin θN −1 cos θN )2
+ (sin ϕ1 sin ϕ2 · · · sin ϕN −1 sin ϕN − sin θ1 sin θ2 · · · sin θN −1 sin θN )2
= 2 − 2 cos(ϕ1 − θ1 ) + 2(sin ϕ1 sin θ1 )[1 − cos(ϕ2 − θ2 )]
N −1

sin ϕi sin θi [1 − cos(ϕN − θN )].

+ ··· + 2
i=1

It then follows from Taylor’s expansion that
N −1

d2 (y, x)

∼ (ϕ1 − θ1

)2

+ (sin ϕ1 sin θ1 )(ϕ2 − θ2

)2

sin ϕi sin θi (ϕN − θN )2

+ ··· +
i=1
N −1

sin2 θi (ϕN − θN )2 ,

∼ (ϕ1 − θ1 )2 + (sin2 θ1 )(ϕ2 − θ2 )2 + · · · +
i=1

completing the proof.
Next, we need some existing results on the approximations to excursion probability of
Gaussian random fields over Euclidean space.
Let 0 < α ≤ 2 and let {Wt (s) : t ∈ RN , s ∈ [0, ∞)N } ba a Gaussian random field such
that
EWt (s) = − s α rt (s/ s ),
Cov(Wt (s), Wt (v)) = s α rt (s/ s ) + v α rt (v/ v )
− s − v α rt ((s − v)/ s − v ),

172

(7.2.2)

where rt (·) : SN −1 → R+ is a continuous function satisfying

sup
v∈SN −1

|rt (v) − rs (v)| → 0 as s → t.

(7.2.3)

Define
r
Hα (t) = lim K −N
K→∞

∞
0

eu P

sup

Wt (s) ≥ u du.

(7.2.4)

s∈[0,K]N

Denote by Hα the usual Pickands’ constant, that is
∞

Hα = lim K −N
K→∞

0

eu P

sup

Z(s) ≥ u du,

s∈[0,K]N

where {Z(s) : s ∈ [0, ∞)N } is a Gaussian random field such that

EZ(s) = − s α ,
Cov(Z(s), Z(v)) = s α + v α − s − v α .
r
It is clear that Hα (t) becomes Hα when rt ≡ 1.

Let D ⊂ RN be a bounded N -dimensional Jordan measurable set. Let Y = {Y (t), t ∈
RN } be a real-valued, centered Gaussian field such that the covariance function CY satisfies

CY (t, t + s) = 1 − s α rt (s/ s )(1 + o(1)) as s → 0,

(7.2.5)

¯
for some constant α ∈ (0, 2], uniformly over t ∈ D.
We will make use of the following theorem of Chan and Lai (2006).
Theorem 7.2.2 [Theorem 2.1 in Chan and Lai (2006)] Suppose the Gaussian random field
{Y (t) : t ∈ RN } satisfies condition (7.2.5), in which rt (·) : SN −1 → R+ is a continuous
173

¯
function such that the convergence (7.2.3) is uniformly in D and supt∈D,v∈SN −1 rt (v) < ∞.
¯
Then as u → ∞,
P sup Y (t) ≥ u ∼ u2N/α Ψ(u)
t∈D

D

r
Hα (t)dt.

Lemma 7.2.3 Let {Wt (s) : t ∈ RN , s ∈ [0, ∞)N } be a Gaussian random field satisfying
(7.2.2) with
∀v ∈ SN −1 ,

rt (v) = Mt v α ,

where Mt are non-degenerate N × N matrices. Then for each t ∈ RN ,

r
Hα (t) = |detMt |Hα .

Proof

Let Wt (s) = Wt (Mt−1 s), ∀s ∈ [0, ∞)N . Then under the above conditions, Wt

satisfies
EWt (s) = − s α ,
Cov(Wt (s), Wt (v)) = s α + v α − s − v α .
Let BK = [0, K]N and define Mt BK = {s ∈ RN : ∃v ∈ BK such that s = Mt v}. Note that
Vol(Mt BK ) = |detMt |Vol(BK ) and sups∈B Wt (s) = sups∈Mt B Wt (s), it follows from
K
K
(7.2.4) that
∞
1
eu P sup Wt (s) ≥ u du
Vol(BK ) 0
K→∞
s∈BK
∞
Vol(Mt BK )
1
= lim
eu P
sup Wt (s) ≥ u du
K→∞ Vol(BK ) Vol(Mt BK ) 0
s∈Mt B

r
Hα (t) = lim

K

1
K→∞ Vol(Mt BK ) 0

∞

= |detMt | lim

174

eu P

sup
s∈Mt BK

Wt (s) ≥ u du.

By modifying the proofs in Qualls and Watanabe (1973), we can check that
∞
1
eu P
sup Wt (s) ≥ u du,
K→∞ Vol(Mt BK ) 0
s∈Mt BK

Hα = lim

completing the proof.
Now we can prove our main result.
Theorem 7.2.4 Let {X(x) : x ∈ SN } be a centered Gaussian random field satisfying condition (7.2.1) and let T ⊂ SN be an N -dimensional Jordan measurable set. Then as u → ∞,

P sup X(x) ≥ u ∼ cN/α Area(T )Hα u2N/α Ψ(u),
x∈T

where Area(T ) denotes the spherical area of T .

Proof

Let Mθ = c1/α diag(1, sin θ1 , . . . ,

N −1
i=1 sin θi ).

If N = 1, we set Mθ = c1/α . By

Lemma 7.2.1, condition (7.2.1) becomes

C(θ, θ + ξ) = 1 − ξ α rθ (ξ/ ξ )(1 + o(1)) as ξ → 0,

where rθ (τ ) = Mθ τ α , ∀τ ∈ SN −1 . Denote by D the domain of T under spherical coordinates. Then by Theorem 7.2.2, as u → ∞,

P sup X(x) ≥ u = P sup X(θ) ≥ u ∼ u2N/α Ψ(u)
x∈T

θ∈D

D

r
Hα (θ)dθ.

It follows from Lemma 7.2.3 that for any θ such that Mθ is non-degenerate( i.e.,

175

(7.2.6)

N −1
i=1 sin θi

=

0),
N −1

sinN −i θi Hα .

r
Hα (θ) = cN/α
i=1

Note that ( N −1 sinN −i θi )dθ is the spherical area element and Mθ are non-degenerate for
i=1
θ ∈ D almost everywhere, we obtain

D

r
Hα (θ)dθ = cN/α Area(T )Hα .

Plugging this into (7.2.6), we finish the proof.

7.2.2

Standardized Spherical Fractional Brownian Motion

Theorem 7.2.4 is an application of Lemma 7.2.1 and Theorem 7.2.2, and it provides a nice
formula since (7.2.1) has a simple form. More generally, the local behavior of covariance
function may be more complicated, but we can still apply Lemma 7.2.1 to find the corresponding local behavior of covariance function under spherical coordinates and then apply
Theorem 7.2.2 to obtain the asymptotics for the excursion probability. Here, we present an
example about spherical fractional Brownian motion on sphere.
Let o be a fixed point on SN . The Spherical Fractional Brownian Motion Bβ (x) is defined
as a centered real-valued Gaussian random field such that

B(o) = 0
E(B(x) − B(y))2 = d2β (x, y) ∀x, y ∈ SN ,

176

where β ∈ (0, 1/2]. It follows immediately that

Cov(B(x), B(y)) =

1 2β
d (x, o) + d2β (y, o) − d2β (x, y) .
2

Without loss of generality, we take o = (1, 0, . . . , 0) ∈ RN +1 , whose corresponding spherical
coordinate is (0, . . . , 0) ∈ RN . Define
B(x)
X(x) = β
,
d (x, o)

∀x ∈ SN \{o}.

Then the covariance is
d2β (x, o) + d2β (y, o) − d2β (x, y)
.
2dβ (x, o)dβ (y, o)

C(x, y) = Cov(X(x), X(y)) =

Note that under the spherical coordinates, d(x, o) = θ1 and d(y, o) = ϕ1 , together with
Lemma 7.2.1, we obtain that as d(x, y) → 0,

C(θ, ϕ) = Cov(X(θ), X(ϕ)) = 1 − (1 + o(1))

1
2β
2θ1

(ϕ1 − θ1 )2 + (sin2 θ1 )(ϕ2 − θ2 )2

N −1

sin2 θi

+ ··· +

(ϕN − θN

)2

i=1

Let

Mθ =

1
21/(2β) θ1

N −1

diag 1, sin θ1 , . . . ,

rθ (τ ) = Mθ τ 2β ,

sin θi ,
i=1

∀τ ∈ SN −1 .

177

β

.

Then as ξ = ϕ − θ → 0,

C(θ, θ + ξ) = 1 − ξ 2β rθ (ξ/ ξ )(1 + o(1)).

Let T ⊂ SN be an N -dimensional Jordan measurable set such that o ∈ T , and denote its
/ ¯
domain under spherical coordinates by D. Then by Theorem 7.2.2, as u → ∞,

P sup X(x) ≥ u = P sup X(θ) ≥ u ∼ uN/β Ψ(u)
x∈T

θ∈D

D

r
H2β (θ)dθ.

It follows from Lemma 7.2.3 that for any θ such that Mθ is non-degenerate( i.e.,

N −1
i=1 sin θi

=

0),
r
H2β (θ)

=

N −1

1
N
2N/(2β) θ1

sinN −i θi H2β .
i=1

Therefore,
N −1

P sup X(x) ≥ u ∼
x∈T

7.3

uN/β Ψ(u)2−N/(2β) H2β

D

−N
θ1

sinN −i θi dθ.
i=1

Smooth Isotropic Gaussian Fields on Sphere

In this section we consider the excursion probabilities for smooth isotropic Gaussian fields
on sphere.

178

7.3.1

Preliminaries

λ
Given λ > 0 and an integer n ≥ 0, the function Pn (t) is defined by the expansion
∞

(1 − 2rt + r2 )−λ

λ
rn Pn (t),

=

t ∈ [−1, 1],

n=0
λ
and Pn (t) is called the ultraspherical polynomial (or Gegenbauer polynomial ) of degree n.
0
If λ = 0, we follow Schoenberg (1942) and set Pn (t) = cos(n arccos t) = Tn (t), where Tn ,

n ≥ 0, are Chebyshev polynomials of the first kind defined by the expansion

1 − rt
=
1 − 2rt + r2

∞

rn Tn (t),

t ∈ [−1, 1].

n=0

λ
For reference later on, we need the following formulae on Pn .
0
o
(i). For all n ≥ 0, Pn (1) = 1, and if λ > 0 [cf. Szeg¨ (1975, p.80)],

λ
Pn (1) =

n + 2λ − 1
.
n

(7.3.1)

(ii). For all n ≥ 0,
d 0
1
P (t) = nPn−1 (t),
dt n

(7.3.2)

and if λ > 0 [cf. Szeg¨ (1975, p.81)],
o

d λ
λ+1
P (t) = 2λPn−1 (t).
dt n

(7.3.3)

The following theorem by Schoenberg (1942) characterizes the covariance function of an
isotropic Gaussian field on sphere [see also Gneiting (2012)].

179

Theorem 7.3.1 Let N ≥ 1, then a continuous function C(·, ·) : SN × SN → R is the
covariance of an isotropic Gaussian field on SN if and only if it has the form
∞
λ
an Pn ( x, y ),

C(x, y) =

x, y ∈ SN ,

n=0

where λ = (N − 1)/2, an ≥ 0,

∞
λ
n=0 an Pn (1)

< ∞.

∞
0
n=0 an Pn (1)

Remark 7.3.2 Note that for N = 1, λ = 0 and
∞
n=0 an

< ∞; while for N ≥ 2, λ = (N − 1)/2 and by (7.3.1),

equivalent to

∞
N −2 a
n
n=0 n

< ∞ is equivalent to
∞
λ
n=0 an Pn (1)

< ∞ is

< ∞.

λ
When N = 2, λ = 1/2 and Pn become Legendre polynomials. For more results on

isotropic Gaussian fields on S2 , we refer to a recent monograph by Marinucci and Peccati
(2011).
The following statement (S) is a smoothness condition for Gaussian fields on sphere. In
Lemma 7.3.3 below, we show that it implies C(·, ·) ∈ C 4 (SN × SN ).
(S). The covariance C(·, ·) of {X(x) : x ∈ SN } satisfies
∞
λ
an Pn ( x, y ),

C(x, y) =

x, y ∈ SN ,

n=0

where λ = (N − 1)/2, an ≥ 0, and

∞
8
n=1 n an

< ∞ if N = 1,

∞
N +6 a
n
n=1 n

< ∞ if

N ≥ 2.
Lemma 7.3.3 Let {X(x) : x ∈ SN } be an isotropic Gaussian field such that (S) is fulfilled.
Then the covariance C(·, ·) ∈ C 4 (SN × SN ) and hence X(·) ∈ C 2 (SN ) a.s.

180

Proof

λ
We first consider N ≥ 2. By Theorem 7.3.1, each Pn ( t, s ) is the covariance of an

isotropic Gaussian field on SN and hence

∀x, y ∈ SN .

λ
λ
λ
|Pn ( x, y )| ≤ Pn ( x, x ) = Pn (1),

(7.3.4)

λ
Combining (S) with (7.3.1), (7.3.3) and (7.3.4), together with the fact P0 (t) ≡ 1, we obtain

that there exist positive constants M1 and M2 such that
∞

sup

d4 λ
P (t)
dt4 n

an

t∈[−1,1] n=0

∞

∞
λ+4
an Pn−4 (1)

≤ M1

nN +6 an < ∞.

≤ M2

n=4

n=1

This gives C 4 (SN × SN ). The proof for N = 1 is similar once we apply both (7.3.2) and
(7.3.3).
By Schoenberg (1942) or Gneiting (2012), C(·, ·) is a covariance function on SN for every
N ≥ 1 if and only if it has the form
∞

bn x, y n ,

C(x, y) =

x, y ∈ SN ,

n=0

where bn ≥ 0 and

∞
n=0 bn

< ∞. Then we may state (S ) below as another form of

smoothness condition for Gaussian fields on sphere.
(S ). The covariance C(·, ·) of {X(x) : x ∈ SN } satisfies
∞

bn x, y n ,

C(x, y) =
n=0

where bn ≥ 0 and

∞
4
n=0 n bn

< ∞.

181

x, y ∈ SN ,

We obtain an analogue of Lemma 7.3.3 below. Since the proof is similar, it is omitted.
Lemma 7.3.4 Let {X(x) : x ∈ SN } be an isotropic Gaussian field such that (S ) is fulfilled.
Then the covariance C(·, ·) ∈ C 4 (SN × SN ) and hence X(·) ∈ C 2 (SN ) a.s.

7.3.2

Excursion Probability

Let χ(Au (X, SN )) be the Euler characteristic of excursion set Au (X, SN ) = {x ∈ SN :
X(x) ≥ u}. Denote by Hj (x) the Hermite polynomial, i.e.,

Hj (x) =

2
(−1)j ex /2

dj −x2 /2
e
.
dxj

Let ωj := Area(Sj ), where Sj is the j-dimensional unit sphere.
Lemma 7.3.5 Let {X(x) : x ∈ SN } be a centered, unit-variance, isotropic Gaussian field
satisfying (S). Suppose also that the joint distribution of (X(x), X(x), 2 X(x)) is nondegenerate for each x ∈ SN . Then
N

E{χ(Au

(X, SN ))}

(C )j/2 Lj (SN )ρj (u),

=

j=0

where

C =



 (N − 1)



n+N −1
∞
n=1
N

∞
2
n=1 n an

an

if N ≥ 2,
(7.3.5)
if N = 1,

2
ρ0 (u) = Ψ(u), ρj (u) = (2π)−(j+1)/2 Hj−1 (u)e−u /2 with Hermite polynomials Hj−1 for

182

j ≥ 1 and, for j = 0, . . . , N ,

 N ωN

 2
j ωN −j
N) =
Lj (S

 0


if N − j is even
(7.3.6)
otherwise

are the Lipschitz-Killing curvatures of SN .

Remark 7.3.6 In Lemma 7.3.5, if condition (S) is replaced by (S ), then it can be seen
from the proof below that C would be changed to a much simpler form
∞

C =

nbn .

(7.3.7)

n=1

Proof

Due to Theorem 12.4.1 in Adler and Taylor (2007), we only need to show that the

Lipschitz-Killing curvatures induced by X on SN is Lj (X, SN ) = (C )j/2 Lj (SN ).
The Riemannian structure induced by X on SN is defined as [cf. Adler and Taylor (2007,
p.305)]

X,SN

gx0

(ξx0 , σx0 ) := E{(ξx0 X) · (σx0 X)} = ξx0 σx0 C(x, y)|x=y=x0 ,

∀x0 ∈ SN ,

where ξx0 , σx0 ∈ Tx0 SN , the tangent space of SN at x0 . We may choose two smooth curves
on SN , say γ(t), τ (s), t, s ∈ [0, 1], such that γ(0) = τ (0) = x0 and γ (0) = ξx0 , τ (0) = σx0 .

183

We first consider N ≥ 2, then

ξx0 σx0 C(x, y)|x=y=x0
∂ ∂
∂ ∂
C(γ(t), τ (s))|t=s=0 =
=
∂t ∂s
∂t ∂s
=

∂
∂t

∞
λ
an Pn ( γ(t), τ (s) )|t=s=0
n=0

∞
λ+1
an (N − 1)Pn−1 ( γ(t), x0 ) γ(t), σx0 |t=0
n=1

∞
λ+2
an (N − 1)(N + 2)Pn−2 ( x0 , x0 ) ξx0 , x0 x0 , σx0

=

n=2
∞
λ+1
an (N − 1)Pn−1 ( x0 , x0 ) ξx0 , σx0

+
n=1
∞

λ+1
an (N − 1)Pn−1 (1)

=

ξx0 , σx0 = C ξx0 , σx0 ,

n=1

where the third and fourth equalities are from (7.3.3), the fifth equality comes from x0 , x0 =
1 and ξx0 , x0 = σx0 , x0 = 0, since the vector x0 is always orthogonal to its tangent space.
The case of N = 1 can be proved similarly once we apply (7.3.2) instead of (7.3.3).
Hence the induced metric is

X,SN

gx 0

(ξx0 , σx0 ) = C ξx0 , σx0 ,

∀x0 ∈ SN .

By the definition of Lipschitz-Killing curvatures, one has

Lj (X, SN ) = (C )j/2 Lj (SN ),

where Lj (SN ) are the original Lipschitz-Killing curvatures of SN given by (7.3.6). We finish
the proof.

184

Theorem 7.3.7 Suppose the conditions in Lemma 7.3.5 hold. Then, under the notations
therein, there exists α > 0 such that as u → ∞,
N

P

2 2
(C )j/2 Lj (SN )ρj (u) + o(e−αu −u /2 ).

sup X(x) ≥ u =
x∈SN

(7.3.8)

j=0

Remark 7.3.8 Under the conditions in Theorem 7.3.7, the covariance function C satisfies
(7.2.1) with α = 2. Also note that when α = 2, Pickands’ constant H2 = π −N/2 . Then
one can check that the approximation in Theorem 7.2.4 only provides the leading term of
the approximation in Theorem 7.3.7. This also affects the errors in two approximations: the
2

error in the former one is only o(1), while the error in the latter one is o(e−αu ).
Proof

The result is an immediate consequence of Lemma 7.3.5 and Theorem 14.3.3 in

Adler and Taylor (2007).
If the set SN is replaced by a more general subset T ⊂ SN , by simply revising Lemma
7.3.5 and applying Theorem 14.3.3 in Adler and Taylor (2007) again, we obtain the following
corollary.
Corollary 7.3.9 Suppose the conditions in Lemma 7.3.5 hold. Let T ⊂ SN be a k-dimensional,
locally convex, regular stratified manifold [cf. Adler and Taylor (2007)], then there exists
α > 0 such that as u → ∞,
k

P sup X(x) ≥ u =
x∈T

2 2
(C )j/2 Lj (T )ρj (u) + o(e−αu −u /2 ),

j=0

where Lj (T ) are the Lipschitz-Killing curvatures of SN [cf. Adler and Taylor (2007)], C
and ρj (u) are as in Lemma 7.3.5.

185

Example 7.3.10 Canonical field on SN , whose covariance structure is given by C(x, y) =
x, y . Since C(x, y) = cos d(x, y), it satisfies

1
C(x, y) = 1 − d2 (x, y)(1 + o(1)),
2

as d(x, y) → 0.

Applying Theorem 7.2.4, one can get an approximation to the excursion probability. However, by applying Theorem 7.3.7, we will get a more precise approximation for N ≥ 2.
Example 7.3.11 Consider the Hamiltonian of the pure p-spin model on SN −1

HN,p (x) =

1

N

Ji ,...,ip xi1 · · · xip ,
N (p−1)/2 i ,...,i =1 1
p
1

∀x = (x1 , · · · , xN ) ∈ SN −1 ,

where Ji1 ,...,ip are independent standard Gaussian random variables. Then

E{HN,p (x)HN,p (y)} =

1
N p−1

x, y p .

Let
∞

X(x) =

bp HN,p (x),
p=2

where (bp )p≥2 is a sequence of positive numbers such that

∞
p
p=2 2 bp

< ∞, and HN,p and

HN,p are independent for any p = p . Then X is a smooth Gaussian random field on sphere
with covariance
∞

C(x, y) =
p=2

b2
p
N p−1

186

x, y p .

Example 7.3.12 We show how to apply Corollary 7.3.9. If T is the semisphere of dimension
one, then L0 (T ) = 1 and L1 (T ) = π. If T is the semisphere of dimension two, then
L0 (T ) = 1, L1 (T ) = π and L2 (T ) = 2π. Basically, L0 (T ) is the Euler characteristic, Lk (T )
is the volume and Lk−1 (T ) is half of the surface area. Usually, one may use Steiner’s formula
[Adlar and Taylor (2007, p.142)] to compute the Lipschitz-Killing curvatures exactly.
2
Example 7.3.13 Consider the covariance structure C(x, y) = 1 − π d(x, y), which can be

verified to be a valid covariance on sphere. Since d(x, y) = arccos x, y , we can write
2
C(x, y) = 1 − arccos x, y =
π

∞
n=0

(2n)!
4n (n!)2 (2n + 1)

x, y 2n+1

∞

bn x, y n ,

:=
n=0

it is easy to check that

∞
n
n=0 nb

= ∞, hence Theorem 7.3.7 is not applicable. In fact,

C(x, y) is not smooth neither. Instead, we may use Theorem 7.2.2 to get the approximation
to excursion probability.

187

Chapter 8
Excursion Probability of Anisotropic
Gaussian and Asymptotically
Gaussian Random Fields

8.1

Preliminaries

For vectors u, v ∈ Rd , the relation u ≤ v means ui ≤ vi for all i and u < v means ui < vi for
all i, also we let uv := (u1 v1 , · · · , ud vd ). For t = (t1 , · · · , td ) ∈ Rd and ζ = (ζ1 , · · · , ζd ) > 0,
define
d

It,ζ =

[ti , ti + ζi ).
i=1

Let

·

be the Euclidean norm of a vector, · be the greatest integer function, µ(·) be the

volume of set. For bounded and Jordan measurable set D ⊂ Rd and δ > 0, define

[D]δ = {t + u : t ∈ D, u ≤ δ}.

188

Let 0 < α ≤ 1, p = (p1 , · · · , pd ) with 0 < pi ≤ 2 for all 1 ≤ i ≤ d and let {Wt (u) : u ∈
[0, ∞)d } be a Gaussian random field such that Wt (0) = 0 and
d

E(Wt (u)) = −

Cov(Wt (u), Wt (v)) =

+

−

p
ui i

p

p

α

udd

u11

pi , · · · ,
pi
d
d
i=1 ui
i=1 ui
i=1
p
p
d
α
udd
u11
pi
ui
rt
pi , · · · ,
pi
d
d
i=1 ui
i=1 ui
i=1
p
p
d
α
vd d
v1 1
pi
vi
rt
p ,··· ,
pi
d
d
vi i
i=1
i=1 vi
i=1
d
α
|u1 − v1 |p1
|ui − vi |pi rt
,···
d
|ui − vi |pi
i=1
i=1

rt

,

,

|ud − vd |pd
d
i=1 |ui

− vi |pi

,

(8.1.1)

where rt : S = {v ∈ [0, ∞)d :

d
i=1 vi

= 1} → R+ is a continuous function satisfying

sup |rt (v) − rs (v)| → 0,

as t − s → 0.

(8.1.2)

v∈S

In particular, we define
∞

HK (t) =

0

ey P

sup
0≤ui ≤K,∀i

Wt (u) > y dy,
(8.1.3)

H(t) = lim K −d HK (t).
K→∞

Let L be a slowly varying function, define

∆c,i = min{x > 0 : xαpi L(x) = c−2 },

189

∀1 ≤ i ≤ d,

(8.1.4)

2

2

−
−
and ∆c = (∆c,1 , · · · , ∆c,d ). For example, if L(x) ≡ 1, then ∆c = (c αp1 , · · · , c αpd ).

Our goal is to investigate the asymptotic property of centered Gaussian fields satisfying the
following condition:

E(X(t)X(t + u))
d

|ui |pi

= 1 − (1 + o(1))

d

α

|ui

L

i=1

|pi

|u1 |p1

rt

d
p
i=1 |ui | i

i=1

,··· ,

|ud |pd
d
p
i=1 |ui | i

(8.1.5)
,

as u → 0, uniformly over t ∈ [D]δ .
Theorem 8.1.1 Suppose Gaussian random field X satisfies condition (8.1.5), where 0 <
α ≤ 1, p = (p1 , · · · , pd ) with 0 < pi ≤ 2 for all 1 ≤ i ≤ d, and rt : S → R+ is a continuous
function such that the convergence in (8.1.2) is uniform in t ∈ [D]δ and supt∈[D] ,v∈S rt (v) <
δ
∞. Then as c → ∞,
d

P sup X(t) > c
t∈D

8.2

∆−1
c,i

∼ Ψ(c)
i=1

H(t)dt.
D

Asymptotically Gaussian Random Fields

2
For c > 0, let Xc be random fields such that EXc (t) = 0, EXc (t) = 0 for all c and t. Define

ρc (t, u) = E(Xc (t)Xc (u)). We impose the following conditions for Xc .
(C). There exist 0 < α ≤ 1, p = (p1 , · · · , pd ) with 0 < pi ≤ 2 for all 1 ≤ i ≤ d and a slowly

190

varying function L such that as u → 0,

ρ(t, t + u)
d

= 1 − (1 + o(1))

|ui

|pi

d

α

|ui |pi rt

L

i=1

i=1

|u1 |p1
d
p
i=1 |ui | i

,··· ,

|ud |pd
d
p
i=1 |ui | i

uniformly over t ∈ [D]δ and c > 0.
(B1). As c → ∞,
P{Xc (t) > c − y/c} ∼ Ψ(c − y/c)
uniformly over t ∈ [D]δ and positive, bounded values of y.
(B2). The convergence in (8.1.2) is uniform over t ∈ [D]δ , with supt∈[D] ,v∈S rt (v) < ∞.
δ
Moreover, for any a > 0, a = {a1/p1 , · · · , a1/pd } and positive integers mi , as c → ∞,

{c[Xc (t + ak∆c ) − Xc (t)] : 0 ≤ ki < mi }|Xc (t) = c − y/c
⇒ {Wt (ak) : 0 ≤ ki < mi }
uniformly over positive, bounded values of y.
(B3). There exists a positive function h such that limy→∞ h(y) = 0 and

P{Xc (t + u∆c ) > c − γ/c, Xc (t) ≤ c − y/c} ≤ h(y)Ψ(c),

for all u ≥ 0 (u is a vector) and γ > 0.
(B4). Let pi0 = min{pi , 1 ≤ i ≤ d}, a = (a1/p1 , · · · , a1/pd ). There exist nonincreasing func∞
tions Na on R+ and positive constants γa such that γa → 0 and Na (γa )+ 1 ω s Na (γa +

191

ω)dω = o(a

d/pi
0)

as a → 0, and

sup

P

Xc (v) > c, Xc (t) ≤ c − y/c

≤ Na (γ)Ψ(c),

v∈It,a∆c

for all γa ≤ γ ≤ c and s > 0.
q

(B5). There exists a nonincreasing function f : [0, ∞) → R+ such that f (y) = O(e−y ) for
some q > 0 and for all γ ≥ 0 and c sufficiently large,
d

|ui |pi

P{Xc (t) > c − γ/c, Xc (t + u∆c ) > c − γ/c} ≤ Ψ(c − γ/c)f
i=1

uniformly in t and t + u∆c belonging to [D]δ .
For K > 0 and a > 0, let At = (At (K, a, c))

{t + ak∆c : 0 ≤ ki < mi , k ∈ Zd },

where mi = K/a1/pi and a = (a1/p1 , · · · , a1/pd ). As a discrete set, At will be used to
approximate It,K∆c .
Lemma 8.2.1 Under (C) and (B1)-(B3),
∞

HK,a (t)

0

ey P

sup

Wt (ak) > y dy

0≤ki <mi ,∀i

is uniformly continuous in t ∈ [D]δ and supt∈[D] HK,a (t) < ∞. Moreover, for γ ≥ 0, as
δ
c → ∞,
P

sup Xc (u) > c − γ/c
u∈At

∼ Ψ(c − γ/c)(1 + HK,a (t))

uniformly in t ∈ [D]δ .

192

(8.2.1)

Proof

Let ε > 0. By (B3), there exists y ∗ > γ such that h(y ∗ ) < ε/( d mi ) and
i=1

0≤P

sup Xc (u) > c − γ/c − P{Xc (t) > c − γ/c}
u∈At

sup Xc (u) > c − γ/c, c − y ∗ /c < Xc (t) ≤ c − γ/c

−P

u∈At

=P

(8.2.2)

sup Xc (u) > c − γ/c, Xc (t) ≤

c − y ∗ /c

u∈At
d

mi h(y ∗ )Ψ(c) < εΨ(c),

≤
i=1

since card(At ) =

d
i=1 mi .

By (B1), there exists ξc → 0 such that

2
|P{Xc (t) > c − y/c}/Ψ(c − y/c) − 1| = O(ξc )

(8.2.3)

−1
2
uniformly for γ ≤ y ≤ y ∗ ; we can also assume that ξc (y ∗ −γ) ∈ Z. Since eξc = 1+ξc +O(ξc )

and Ψ(c − y/c) ∼ ey Ψ(c), (8.2.3) implies

P{c − (y + ξc )/c < Xc (t) ≤ c − y/c}
2
2
= (1 + O(ξc ))ey+ξc Ψ(c) − (1 + O(ξc ))ey Ψ(c) ∼ ξc ey Ψ(c).

By (B2), uniformly for t ∈ [D]δ and γ ≤ y ≤ y ∗ ,

P

sup Xc (u) > c − γ/c, c − (y + ξc )/c < Xc (t) ≤ c − y/c
u∈At

∼P

sup

Wt (ak) > y − γ P{c − (y + ξc )/c < Xc (t) ≤ c − y/c}

0≤ki <mi ,∀i

∼P

sup

Wt (ak) > y − γ ξc ey Ψ(c).

0≤ki <mi ,∀i

193

(8.2.4)

−1
Summing (8.2.4) over y = jξc + γ for j = 0, 1, · · · , ξc (y ∗ − γ) − 1, we obtain

P

sup Xc (u) > c − γ/c, c − y ∗ /c < Xc (t) ≤ c − γ/c
u∈At
y ∗ −1

∼ Ψ(c)
γ

ey P

Wt (ak) > y − γ dy

sup
0≤ki <mi ,∀i

y ∗ −γ−1

∼ Ψ(c − γ/c)
0

ey P

sup

Wt (ak) > y dy.

0≤ki <mi ,∀i

Plugging this in (8.2.2) and letting y ∗ → ∞, we get (8.2.1).
∞
Since 0 ey P{Wt (ak) > y}dy < ∞ for all k and At is a finite set, HK,a (t) is finite and

its uniform continuity follows from (8.1.1) and (8.1.2), with the fact described in (B2) that
the convergence in (8.1.2) is uniform over t ∈ [D]δ . Racall that supt∈[D] ,v∈S rt (v) < ∞,
δ
yielding the finiteness of supt∈[D] HK,a (t).
δ

Theorem 8.2.2 Let K > 0. Assume (C) and (B1)-(B4). Then as c → ∞,

P

sup

Xc (u) > c

u∈It,K∆c

∼ Ψ(c)(1 + HK (t))

uniform over t ∈ [D]δ , where HK (t) is defined in (8.1.3) and is finite and uniformly continuous in t ∈ [D]δ .

194

Proof

Let a > 0. By (B4) and (8.2.1), we have for all large c,

0≤ P

Xc (u) > c − P

sup
u∈It,K∆c

sup Xc (u) > c

Ψ(c)

u∈At

≤ P c − γa /c < sup Xc (u) ≤ c
u∈At

+

P

Xc (v) > c, Xc (u) ≤ c − γa /c

sup

Ψ(c)

(8.2.5)

v∈Iu,a∆c

u∈At

d

(K/a1/pi )Na (γa )

≤ 2(Ψ(c − γa /c) − Ψ(c))(1 + HK,a (t))/Ψ(c) +

i=1
d

≤ 3(eγa − 1)(1 + HK,a (t)) + (K d /a i=1 1/pi )Na (γa ).
d

By (B4), for any ε > 0, we can choose a∗ small enough such that Na (γa )/a i=1 1/pi < ε/K d
and 3(eγa − 1) < ε for all 0 < a ≤ a∗ . Therefore, by (8.2.1) and (8.2.5),

(1 − ε)(1 + HK,a (t)) ≤ P

sup

Xc (u) > c

u∈It,K∆c

Ψ(c) ≤ (1 + 2ε)(1 + HK,a (t)) + ε,

for all large c and all t ∈ [D]δ and 0 < a ≤ a∗ . By uniform continuity of Wt (u), HK,a (t) →
HK (t) as a ↓ 0. Therefore,

1 + HK,a (t) ≤ 1 + HK (t) ≤ (1 + ε)(1 + HK,a∗ (t)),

(8.2.6)

for all t ∈ [D]δ and 0 < a ≤ a∗ .
First note that M

1 + supt∈[D] HK (t) < ∞ in view of (8.2.6) and Lemma 8.2.1,
δ

therefore
|HK (t) − HK,a∗ (t)| ≤ M ε

195

(8.2.7)

for all t ∈ [D]δ , by (8.2.6) with a = a∗ . Because HK,a∗ is uniformly continuous by Lemma
8.2.1,
∀ t − u < δ ∗ , t, u ∈ [D]δ ,

|HK,a∗ (t) − HK,a∗ (u)| ≤ ε,

(8.2.8)

for some δ ∗ > 0. It follows from (8.2.7) and (8.2.8) that, if t − u < δ ∗ ,

|HK (t) − HK (u)|
≤ |HK (t) − HK,a∗ (t)| + |HK (u) − HK,a∗ (u)| + |HK,a∗ (t) − HK,a∗ (u)|
≤ 2M ε + ε.
Hence HK (t) is uniformly continuous in t ∈ [D]δ .
Combining (8.2.7) and the definition of M yields that for all large c and t ∈ [D]δ ,

−εM − ε(1 − ε)M ≤ P

sup

Xc (u) > c

u∈It,K∆c

Ψ(c) − (1 + HK (t))

≤ 2εM + ε(1 + 2ε)M.
Since ε is arbitrary, this proves the theorem.

Lemma 8.2.3 Under (C) and (B1)-(B4), supt∈[D] ,K≥1 K −d HK (t) < ∞ and {K −d HK :
δ
K ≥ 1} is uniformly equicontinuous on [D]δ , that is,

sup
K≥1,t,s∈[D]δ , t−s ≤ε

|K −d HK (t) − K −d HK (s)| → 0,

196

as ε → 0.

Proof

Let
a = (a1/p1 , · · · , a1/pd ),
d

N (K, a) = (

mi = K/a1/pi ,

d

a−1/pi ) .

mi )/(
i=1

i=1

Note that the integrand of HK,a (t) involves the set {ak : 0 ≤ ki < mi }, which can be
partitioned into N (K, a) + 1 disjoint subsets Lj such that card(Lj ) =
1 ≤ j ≤ N (K, a) and card(LN (K,a)+1 ) =

d
i=1 mi

− N (K, a)

d
i=1

d
i=1

a−1/pi for

a−1/pi . It is possible

that card(LN (K,a)+1 ) = 0, in this case LN (K,a)+1 is regarded as an empty set. We can
therefore use the arguments at the end of the proof of Lemma 8.2.1 to bound
N (K,a)+1

K −d

P
j=1

sup Wt (ak) > y
k∈Lj

−P

sup Ws (ak) > y
k∈Lj

and thereby establish the uniform equicontinuity and boundedness of {K −d HK,a : K ≥ 1}
on [D]δ . Moreover, by partitioning the cube [0, K)d similarly into K d cubes, it can be shown
that supK≥1,t∈[D] |K −d HK (t) − K −d HK,a (t)| → 0, as a → 0. Hence we can proceed as in
δ
(8.2.7) and (8.2.8) but with HK,a and HK replaced by K −d HK,a and K −d HK respectively
to prove the uniform equicontinuity and boundedness of {K −d HK : K ≥ 1}.

Lemma 8.2.4 Under (C) and (B1)-(B5), there exist constants sK → 0 as K → ∞ such
that
P

sup
u∈It,K∆c

Xc (u) > c,

sup

Xc (v) > c

≤ sK K d Ψ(c)

v∈B\It,K∆c

for c large enough, uniformly over t ∈ [D]δ and over subsets B of [D]δ .

197

(8.2.9)

Proof

Let a > 0 and 0 < q < q . Then by the property of f described in (B5),
d

Ga

d

|wi |pi

exp

|wi |pi

f

i=1

w∈aZd

< ∞.

i=1

Let n be positive integers that are large enough such that
d

d

|wi |pi

exp
d |w |pi ≥np1 a
i=1 i

w∈aZd ,

i=1

|wi |pi

f

d

< εa i=1 1/pi ;

i=1

and K > 0 be large enough such that
d

d

(1 − 2n/mi ) Ga < εa i=1 1/pi ,

1−

i=1

where mi = K/a1/pi . Let F1,t = {t + ak∆c : n ≤ ki < mi − n, k ∈ Zd }, F2,t = At \F1,t ,
Bt = {t + ak∆c ∈ B\It,K∆c , k ∈ Zd }, guv = min{c − γa , (

d
i=1 |(ui

− vi )/∆c,i |pi )q }. Then

by (B5),

P Xc (u) > c − (γa + guv )/c, Xc (v) > c − (γa + guv )/c
d

|(ui − vi )/∆c,i |pi

≤ Ψ(c − (γa + guv )/c)f

(8.2.10)

i=1
d

≤

2eguv Ψ(c − γa /c)f

|(ui − vi )/∆c,i |pi ,
i=1

for all large c. For u ∈ F1,t and v ∈ Bt ,
(

d
i=1 |(ui
d
i=1 (mi

d
i=1 |(ui

− vi )/∆c,i |pi )q . Noting that card(F1,t ) ≤

− 2n) =

d
i=1 mi [1 −

d
i=1 (1 − 2n/mi )],

198

− vi )/∆c,i |pi ≥ np1 a and guv ≤
d
i=1 mi ,

and that

card(F2,t ) ≤

u∈At

=

u∈F1,t +

d
i=1 mi

−

u∈F2,t ,

we obtain from (8.2.10) that for all large c,

P Xc (u) > c − (γa + guv )/c, Xc (v) > c − (γa + guv )/c
u∈At v∈Bt
d

≤ 2Ψ(c − γa /c)

d

mi

|wi

exp

i=1

w∈aZd ,

d |w |pi ≥np1 a
i=1 i

d

|pi

|wi |pi

f

i=1

i=1

d

+ 1−

(1 − 2n/mi ) Ga
i=1

≤ 4εK d Ψ(c − γa /c).
(8.2.11)
Define λw = minu∈At guw if w ∈ Bt , and λw = 0 if w ∈ At . Then

sup

P

Xc (u) > c,

u∈It,K∆c

≤

sup

Xc (v) > c

v∈B\It,K∆c

P Xc (u) > c − (γa + guv )/c, Xc (v) > c − (γa + guv )/c
u∈At v∈Bt

+

P
w∈At ∪Bt

sup

Xc (z) > c, Xc (v) ≤ c − (γa + λw )/c .

z∈Iw,a∆c

199

(8.2.12)

On the right-hand side of (8.2.12), the first sum can be bounded by (8.2.11) and the second
sum by

Xc (z) > c, Xc (u) ≤ c − γa /c

sup

P

z∈Iu,a∆c

u∈At

+

Xc (z) > c, Xc (v) ≤ c − (γa + λv )/c

sup

P

v∈Iv,a∆c

v∈Bt

(8.2.13)

d

≤

mi Na (γa )Ψ(c) +
i=1

≤ K d a−

Na (γa + λv )Ψ(c)
v∈Bt

d 1/p
i Na (γa )Ψ(c) +
i=1

Na (γa + λv )Ψ(c),
v∈Bt

in view of (B4) and that card(At ) =

d
i=1 mi .

To bound the last sum

v∈Bt

in (8.2.13),

first consider the case d = 1. Since λv ≥ min{c−γa , (ak p )q } if ak p ≤ inf u∈At (|v −u|/∆c )p <
a(k + 1)p , and since Na is nonincreasing, it follows that

Na (γa + λv )
v∈Bt
∞

Na (γa + (ak p )q ) + Na (c)µ(B)/(a∆c )

≤2
k=1
∞

(8.2.14)
Na (γa + (a1/p k)pq ) + Na (c)µ(B)/(a∆c )

=2
k=1

≤2

∞

a−1/p

Na (γa + y pq )dy + µ(B)Na (c)/(a∆c ) .

0
∞
Making change of variable by y pq replaced by ω, and noting that Na (γa ) + 1 ω s Na (γa +
∞
ω)dω = o(a) for all s ≥ 0 as described in (B4), yield that a−1/p 0 Na (γa + y q )dy = o(1)
c−γ

a
as a → 0. Moreover, in view of (8.1.4), Na (c)/∆c = O( c/2−γ ω s Na (γa + ω)dω) = o(a1/p )
a

as a → 0 and c → ∞, for s > 2/(αp). Therefore,

200

v∈Bt

Na (γa + λv ) ≤ ε for all large c and

small a.
d
p q
i=1 j i ) }

In general, for d > 1, note that λv ≥ min{c − γa , (a
d
i=1 |(ui

d
i=1 (j

− vi )/∆c,i |pi < a

if a

d
p
i=1 j i

≤

+ 1)pi . Let U = {σ ⊂ {1, · · · , d} : card(σ) = d − 1},

since Na is nonincreasing, it follows that

Na (γa + λv )
v∈Bt
∞

≤2

j=1

+
≤2

≤2
+

j pi

+ 2j) Na γa + a

q

i=1

σ∈U l∈σ

Na (c)µ(B)
d
1/pi ∆ )
c,i
i=1 (a
∞
2d
(a−1/pl K
j=1

+

d

(a−1/pl K

2d

+ 2j) Na (γa + dq (a

1/pi
p q
0 j) i0 )

(8.2.15)

σ∈U l∈σ

Na (c)µ(B)
d
1/pi ∆ )
c,i
i=1 (a
∞
−1/pi
0
2d a
(a−1/pl K
0
σ∈U l∈σ

+ 2a

−1/pi
0 y)

p q
Na (γa + dq y i0 )

Na (c)µ(B)
d
1/pi ∆ )
c,i
i=1 (a

≤ εK d−1 ,
for all large c and small a, as can be shown by arguments similar to those in the case d = 1.
Combining (8.2.11)-(8.2.15) yields the desired conclusion.

Theorem 8.2.5 Assume (C) and (B1)-(B5). Let c → ∞ and c = o(∆c,i ) for all i and
hence c ∆c = o(1) as c → ∞. Then

P

sup

Xc (u) > c

u∈It, c ∆c

201

∼ H(t)Ψ(c) d ,
c

(8.2.16)

sup

P

Xc (u) > c,

u∈It, c ∆c

sup

= o(Ψ(c) d ),
c

Xc (v) > c

v∈B\It, ∆
c c

(8.2.17)

as c → ∞, uniformly over t ∈ [D]δ and over subsets B of [D]δ , where H(t) is defined in
(8.1.3) and is uniformly continuous and bounded below on D.

Proof

Let ε > 0. There exists K ∗ such that sK ≤ ε/3 for all K ≥ K ∗ . For fixed t ∈ D

and K ≥ K ∗ , define

Λ = {u ∈ K∆c Zd : Iu,K∆c ⊂ It, c ∆c },
Λ = {u ∈ K∆c

Zd

(8.2.18)

: Iu,K∆c ∩ It, c ∆c = ∅},

Ju = Iu,K∆c .

Covering It, c ∆c by rectangles with edges length K∆c,i , 1 ≤ i ≤ d. and letting B be a subset
of [D]δ containing It, c ∆c , we have

u∈Λ

sup Xc (v) > c − P

sup Xc (v) > c,

v∈Ju

P

v∈Ju

sup

Xc (w) > c

w∈B\Ju

(8.2.19)
≤P

sup

Xc (u) > c

u∈It, c ∆c

≤

P
u∈Λ

sup Xc (v) > c .
v∈Ju

By Theorem 8.2.2 and Lemma 8.2.4, as c → ∞,

(1 + HK (u) − sK K d )

(1+o(1))Ψ(c)
u∈Λ

≤P

sup

(8.2.20)
Xc (u) > c

≤ (1 + o(1))Ψ(c)

u∈It, ∆
c c

(1 + HK (u)),
u∈Λ

uniformly over t ∈ [D]δ . In view of c ∆c → 0 and the uniform equicontinuity in Lemma
8.2.3, we can choose c∗ large enough so that |K −d HK (u) − K −d HK (t)| ≤ ε/3 for all c ≥ c∗ ,
√
c

≥ K ≥ K ∗ , t ∈ D and u ∈ Λ. Putting this and the bound sK ≤ ε/3 in (8.2.20) and
202

dividing (8.2.20) by Ψ(c) d , we obtain that for all c ≥ c∗ ,
c
(1 − ε)(K −d HK (t) − 2ε/3) ≤ P

sup

√
c

≥ K ≥ K ∗ and t ∈ D,

Xc (u) > c

u∈It, c ∆c

(Ψ(c) d )
c
(8.2.21)

≤ (1 + ε)(K −d HK (t) + ε/3),
since card(Λ) ∼ card(Λ) ∼ K −d d . By Lemma 8.2.3, M
c

supt∈[D] ,K≥1 K −d HK (t) < ∞.
δ

Therefore, it follows from (8.2.21) that

sup

sup P

Xc (u) > c

u∈It, c ∆c

t∈D

for all c ≥ c∗ and

√
c

(Ψ(c) d ) − K −d HK (t) ≤ εM + 2ε/3,
c

(8.2.22)

≥ K ≥ K ∗ . Letting c → ∞ in (8.2.22) yields

˜
sup |K −d HK (t) − K −d HK (t)| ≤ 2εM + 4ε/3,
˜

t∈D

˜
if K, K ≥ K ∗ , establishing that {K −d HK } is uniformly Cauchy. Hence K −d HK (t) converges
uniformly in t ∈ D to H(t), which is also bounded by M . We can therefore proceed as in the
second paragraph of the proof of Theorem 8.2.2 to show that H(t) is uniformly continuous
in t ∈ D. Moreover, taking K large enough such that supt∈D |K −d HK (t) − H(t)| ≤ ε/3, it
follows from (8.2.22) that

sup P
t∈D

sup

Ψ(c) d − H(t) ≤ ε(M + 1)
c

Xc (u) > c

u∈It, c ∆c

for all c ≥ c∗ , proving (8.2.16).
We next show that inf t∈D H(t) > 0. For the function f in (B5), we can choose a > 0 large
enough so that

k=0 f

a

d
p
i=1 |ki | i

≤ 1/2. Let K large enough and mi = K/a1/pi ,
203

1 ≤ i ≤ d, and At = At (K, a, c) as defined before so that card(At ) =

d
i=1 mi .

Then by

(B1) and (B5), as c → ∞,

sup Xc (u) > c

P

u∈At

≥

P(Xc (u) > c) −
u∈At

≥

P(Xc (u) > c, Xc (v) > c)
v∈At ,v=u

(8.2.23)

(1 + o(1))Ψ(c)/2
u∈At
d

mi Ψ(c)/2,

= (1 + o(1))
i=1

uniformly in t ∈ D and mi ≥ 2. Combining (8.2.23) with Theorem 8.2.2 yields

1 + HK (t) = lim P

sup

c→∞

Xc (u) > c

Ψ(c)

u∈It,K∆c
d

≥ lim sup P
c→∞

sup Xc (u) > c
u∈At

Ψ(c) ≥

mi /2
i=1

uniformly in t ∈ D and mi ≥ 2. Since limK→∞ K −d HK (t) = H(t), it then follows that
H(t) ≥ ( d a−1/pi )/2 for all t ∈ D.
i=1
Finally, to prove (8.2.17), apply Lemma 8.2.4 to obtain that for all t ∈ D and large c,

sup

P

Xc (u) > c,

u∈It, c ∆c

≤

P
u∈Λ

sup

Xc (v) > c

v∈B\It, c ∆c

sup Xc (v) > c, sup Xc (v) > c
v∈Ju

v∈B\Ju

≤ card(Λ)sK K d Ψ(c).
Since sK → 0 as K → ∞ and card(Λ) ∼ K −d d as c /K → ∞, (8.2.17) follows.
c

204

Theorem 8.2.6 Assume (C) and (B1)-(B5). Then as c → ∞,
d

P sup Xc (t) > c

∆−1
c,i

∼ Ψ(c)

t∈D

H(t)dt.
D

i=1

Proof

A basic idea of the proof is to cover the set D by rectangles with edges length

c ∆c,i ,

1 ≤ i ≤ d, and also c → ∞ and c ∆c,i → 0 for all i and hence c ∆c → 0 as

c → ∞. Define Λ, Λ and Ju as in (8.2.18) but with K∆c Zd replaced by c ∆c Zd , Iu,K∆c by
Iu, c ∆c , and It, c ∆c by D. Then (8.2.19) still holds with these new definitions of Λ, Λ and Ju
and also with B replaced by [D]δ . Labeling it as (8.2.19 ), the upper and lower bounds
in (8.2.19 ) are both asymptotically equivalent to ( d
c

d
−1 d
i=1 ∆c,i ) ( c )Ψ(c) D H(t)dt

=

Ψ(c)( d ∆−1 ) D H(t)dt, since c ∆c → 0 and H(t) is continuous. This finishes the proof.
i=1 c,i

8.3

Proof of Theorem 8.1.1

In view of Theorem 8.2.6, we only need to show that (B1)-(B5) holds for such Gaussian
fields. (B1) is obvious. To show (B2), it follows from (8.1.5) that as c → ∞,

E{c[X(t + u∆c ) − X(t)]|X(t) = c − y/c}
= −c[1 − ρ(t, t + u∆c )](c − y/c)
d

|ui |pi

→−
i=1

α

rt

|u1 |p1
d
p
i=1 |ui | i

205

(8.3.1)
,··· ,

|ud |pd
d
p
i=1 |ui | i

,

Cov{c[X(t + u∆c ) − X(t)], c[X(t + v∆c ) − X(t)]|X(t) = c − y/c}
= c2 [ρ(t + u∆c , t + v∆c ) − ρ(t, t + u∆c )ρ(t, t + v∆c )]
d

→

+

−

p

p

α

p
ui i

udd

u11

p ,··· ,
pi
d
d
ui i
i=1
i=1 ui
i=1
p
p
d
α
vd d
v1 1
pi
vi
rt
pi , · · · ,
pi
d
d
i=1 vi
i=1 vi
i=1
d
α
|u1 − v1 |p1
pi
,···
|ui − vi |
rt
d
p
i=1 |ui − vi | i
i=1

rt

(8.3.2)

,

|ud − vd |pd
d
i=1 |ui

− vi |pi

.

Since {c[X(t + ak∆c ) − X(t)] : 0 ≤ ki < mi } is multivariate normal, (B2) then follows.
Let γ > 0, φ be the density of standard normal. Since Ψ(c − z/c) ∼ ez Ψ(c) for all z ≥ 0
and there exist constants B > 0, B > 0 such that P{Wt (u) > z − γ} ≤ B exp(−B z 2 ), it
follows from (8.3.1) and (8.3.2) that as c → ∞,

P{X(t + u∆c ) > c − γ/c, X(t) ≤ c − y/c}
c−y/c

= P{X(t) ≤ c − y/c}
= P{X(t) ≤ c − y/c}
∞

≤ (1 + o(1))Ψ(c)
y

−∞
∞
y

P{X(t + u∆c ) > c − γ/c|X(t) = x}φ(x)dx
P{X(t + u∆c ) > c − γ/c|X(t) = c − z/c}c−1 φ(c − z/c)dz

ez P{Wt (u) > z − γ}dz

≤ h(y)Ψ(c),
where h(y) → 0 as y → ∞, establishing (B3).

206

To show (B5) holds, note that

P{X(t) > c, X(t + u∆c ) > c}
≤ P{X(t) + X(t + u∆c ) > 2c}
1/2
2c2
1 + ρ(t, t + u∆c )

∼Ψ

c2
1 + ρ(t, t + u∆c ) 1/2
c2
+
exp −
2
1 + ρ(t, t + u∆c )
2

= Ψ(c)

c2 1 − ρ(t, t + u∆c )
≤ Ψ(c) exp −
2
2

.

By (8.1.5), there exists η > 0 such that c2 [1 − ρ(t, t + u∆c )] ≥ η(

d
p α
i=1 |ui | i ) L(

d
p
i=1 |ui | i )

for all t, t + u∆c ∈ [D]δ . Hence (B5) holds with f (y) = Bλ exp(−y λ ) with 0 < λ < α, for
some Bλ > 0.
p ζ/2
Finally we turn to (B4). Let a > 0, a = {a1/p1 , · · · , a1/pd }, 0 < ζ < α, 1 < ξ < 2 i0 ,

κ=

∞
−r
r=0 ξ

and wr = ξ −r /(2κ). Define

Br = {t + 2−r ak∆c : 0 ≤ ki < 2r , k ∈ Zd },
F =

sup

X(u) > c ,

u∈It,a∆c

E−1 = {X(t) ≤ c − γ/c},
Er =

sup X(v) ≤ c − γ(1 − w0 − · · · − wr )/c)

for r ≥ 0,

v∈Br

recalling that

∞
r=0 wr

= 1/2. Note that Br ⊂ Br+1 ⊂ It,a∆c and that by the continuity of

207

∞
r=0 P(Er−1

X, P(F ∩ E−1 ) ≤

c
∩ Er ). Moreover,

c
P(Er−1 ∩ Er )

≤ 2r+d

P{X(v) ≤ c − γ(1 − w0 − · · · − wr−1 )/c),

sup
v∈It,a∆c ,ε∈{0,1}d \{0}

(8.3.3)

X(v + ε2−r a∆c ) > c − γ(1 − w0 − · · · − wr )/c}.
Given X(v) = c − y/c, the conditional distribution of c[X(v + ε2−r a∆c ) − X(v)] is normal
with mean −c(c − y/c)[1 − ρ(v, v + ε2−r a∆c )] < 0 and variance c2 [1 − ρ2 (v, v + ε2−r a∆c )],
d
−rpi )ζ
i=1 2

which is bounded by B(a

sup

P

for some B > 0, in view of (8.1.5). Hence

c[X(v + ε2−r a∆c ) − X(v)] > wr y|X(v) = c − y/c

ε∈{0,1}d
d

≤

2d exp

− C(wr

y)2

ζ

2−rpi

a

(8.3.4)

i=1
−rpi
0

≤ 2d exp − C(wr y)2

ζ

da2

for some C > 0.
Let η

p ζ
2 i0 /ξ 2 > 1. Combining (8.3.3) and (8.3.4) with fact P{X(v) ∈ c − y/c} ∼

Ψ(c)ey dy then yields

P(F ∩ E−1 )
∞

2−r

≤ (1 + o(1))Ψ(c)
r=0

(8.3.5)

∞

exp[y

− Cη r y 2 /(4aζ dζ κ2 ) + C

r]dy

γ/2

for some C > 0. Let γa = aζ/3 . Then for large c and γa ≤ γ ≤ c, (8.3.5) is bounded above

208

by Ψ(c)Na (γ), where
∞

∞

2−r

Na (γ) = 2

exp[y − Cη r y 2 /(4aζ dζ κ2 ) + C r]dy

γ/2

r=0

∞
satisfies Na (γa ) + 1 ω s Na (γa + ω)dω = o(al ) for all s > 0 and l > 0.

8.4

Example: Standardized Fractional Brownian Sheet

For a given vector H = (H1 , · · · , Hd ) ∈ (0, 1)d , a d-fractional Brownian sheet B H = {B H (t) :
t ∈ Rd } with Hurst index H is a real-valued, centered Gaussian field with covariance function
given by
d

E(B H (t)B H (s))

=
i=1

1
|ti |2Hi + |si |2Hi − |ti − si |2Hi ,
2

t, s ∈ Rd .

¯
Let D ⊂ Rd such that D having no intersection with any coordinate, define the standardized
field
B H (t)

X(t) =

Var(B H (t))

,

t ∈ D.

It follows that
d

E(X(t)X(s)) =

|ti |2Hi + |si |2Hi − |ti − si |2Hi
,
2|ti si |Hi
i=1

and hence

1
E(X(t)X(t + u)) = 1 − (1 + o(1))
2

209

d
i=1

ui 2Hi
,
ti

(8.4.1)

as u → 0, uniformly over t ∈ [D]δ . Hence (8.1.5) is satisfied with pi = 2Hi for 1 ≤ i ≤ d,
α = 1, L(

d
2H
i=1 |ui | i )

≡ 1/2 and

|u1 |2H1

rt

d
2H
i=1 |ui | i

,··· ,

d

|ud |2Hd

|ui

=

d
2H
i=1 |ui | i

−1 d

|2Hi

i=1

i=1

ui 2Hi
.
ti

In other word,
d

rt (v) =

vi

|t |2Hi
i=1 i

t ∈ D, v ∈ S.

,

Applying Theorem 8.1.1, we obtain

P sup X(t) > c
t∈D

c2
∼ Ψ(c)
2

1
d
i=1 2Hi

H(t)dt,

(8.4.2)

D

where
∞

H(t) = lim K −d
K→∞

0

ey P

sup
0≤ui ≤K,∀i

Wt (u) > y dy,

and {Wt (u) : u ∈ [0, ∞)d } is a Gaussian random field such that Wt (0) = 0 and
d

E(Wt (u)) = −

Cov(Wt (u), Wt (v)) =

ui 2Hi
,
ti

i=1
2H
d
ui i

2H
+ ui i − |ui − vi |2Hi

|ti |2Hi

i=1

.

By similar discussions in Lemma 7.2.3, we obtain further that

P sup X(t) > c
t∈D

1
d
i=1 2Hi

c2
∼ Ψ(c)
2

d

H
D

210

1

|t |2Hi
i=1 i

dt,

(8.4.3)

where H is the Pickands’ constant defined by
∞

H = lim K −d
K→∞

0

ey P

sup

W (u) > y dy,

0≤ui ≤K,∀i

and {W (u) : u ∈ [0, ∞)d } is a Gaussian random field such that W (0) = 0 and
d

2H
ui i ,

E(W (u)) = −
i=1
d

2H
2H
ui i + ui i − |ui − vi |2Hi .

Cov(W (u), W (v)) =
i=1

Especially, when H = (1/2, · · · , 1/2), then H = 1 and thus
d

P sup X(t) > c

∼

Ψ(c)2−d c2d
D

t∈D

−1

|ti |

dt.

(8.4.4)

i=1

This result is very similar to Example 2.2 in Chan and Lai (2006).
It is worth mentioning here that we may also apply Piterbarg’s result to get the approximation. Due to the covariance structure (8.4.1), applying Theorem 7.1 on Page 108 in
Piterbarg (1996a), we obtain
d

P sup X(t) > c

d

1

D

t∈D

1

2 2Hi |ti |2Hi

∼ Ψ(c)c i=1 Hi H

−1

dt,

i=1

which is the same as (8.4.3).
Remark 8.4.1 In certain sense, Theorem 8.1.1 generalize Theorem 7.1 on Page 108 in

211

Piterbarg (1996a), since the latter one is the case that

t ∈ D, v ∈ S,

rt (v) = Ct v,

where Ct is some nondegenerate d × d matrix.

8.5

Example: Standardized Random String Processes

We study an anisotropic Gaussian field which is the solution to a stochastic partial differential
equation in Mueller and Tribe (2002). We write the original process {Ut (x) : t ≥ 0, x ∈ R}
in Mueller and Tribe (2002) as {U (t) : t1 ≥ 0, t2 ∈ R}. Then it is a centered Gaussian
field with stationary increments and U (0) = 0. It has the following covariance structure: for
t2 , s2 ∈ R, t1 = s1 ≥ 0,
E{(U (t) − U (s))2 )} = |t2 − s2 |,
and for t2 , s2 ∈ R, t1 > s1 ≥ 0,

E{(U (t) − U (s))2 )} =

√

|t − s2 |
t1 − s1 F √2
,
t1 − s1

where
F (a) = (2π)−1/2 +

1
G(a − z)G(a − z )(|z| + |z | − |z − z |)dzdz
2 R R

= −(2π)−1/2 +

G(a − z)|z|dz
R
a

= −(2π)−1/2 + 4G(a) + 2a

G(z)dz,
0

212

2

x
and G(x) = √1 e− 4 . Now we define the standardized field
4π

X(t) =

U (t)
Var(U (t))

t ∈ R+ × R.

,

(8.5.1)

Note that

Var(U (t)) =

=

√

√

|t |
t1 F √2
t1
t2

t1

1
2 − 2
− √ + √ e 4t1
π
2π

|t
√2 |

+ |t2 |

t1

0

1 − z2
√ e 4 dz,
π

and

E(X(t)X(s)) =

Var(U (t)) + Var(U (s)) − E{(U (t) − U (s))2 )}
2 Var(U (t))Var(U (s))

.

Thus we obtain that as u → 0,

E(X(t)X(t + u)) = 1 − (1 + o(1)) rt,1

|u2 |
|u1 |

|u1 | + rt,2

|u2 |
|u1 |

where

rt,1

rt,2

|u2 |
|u1 |
|u2 |
|u1 |

1
=
2Var(U (t))

u2

1
2 − 2
− √ + √ e 4|u1 | ,
π
2π
|u2 |

√
2
1
1
|u1 | √ − z4
=
e
dz.
2Var(U (t)) 0
π

213

|u2 | ,

|u1 | + |u2 |) ≡ 1 and

Hence (8.1.5) is satisfied with p1 = 1/2, p2 = 1, α = 1, L(
|u1 |

rt

|u1 | + |u2 |
=

,

1
|u1 | + |u2 |

|u2 |
|u1 | + |u2 |
rt,1

|u2 |
|u1 |

|u1 | + rt,2

|u2 |
|u1 |

|u2 |

In other words,
rt (v) = rt,1

v2
v
v1 + rt,2 2 v2 ,
v1
v1

t ∈ D, v ∈ S.

Hence we can apply Theorem 8.1.1 to get the approximation to the excursion probability.
However, Piterbarg’s result is not applicable for such case.

214

Chapter 9
Vector-valued Smooth Gaussian
Random Fields

9.1

Joint Excursion Probability

Let {(X(t), Y (s)) : t ∈ T, s ∈ S} be an R2 -valued, centered, unit-variance Gaussian random
field, where T and S are rectangles in RN . Let

ρ(t, s) = E{X(t)Y (s)},

ρ(T, S) =

sup

E{X(t)Y (s)}.

t∈T,s∈S

We will make use of the following conditions.
(C1). X, Y ∈ C 2 (RN ) almost surely and their second derivatives satisfy the uniform meansquare H¨lder condition: there exist constants L, η > 0 such that
o

E(Xij (t) − Xij (t ))2 ≤ L t − t 2η ,

∀t, t ∈ T, i, j = 1, . . . , N,

E(Yij (s) − Yij (s ))2 ≤ L s − s 2η ,

∀s, s ∈ S, i, j = 1, . . . , N.

(C2). For every (t, t , s) ∈ T 2 × S with t = t , the Gaussian vector

(X(t), X(t), Xij (t), X(t ), X(t ), Xij (t ), Y (s), Y (s), Yij (s), 1 ≤ i ≤ j ≤ N )
215

is non-degenerate; and for every (s, s , t) ∈ S 2 × T with s = s , the Gaussian vector

(Y (s), Y (s), Yij (s), Y (s ), Y (s ), Yij (s ), X(t), X(t), Xij (t), 1 ≤ i ≤ j ≤ N )

is non-degenerate.
(C3). For all (t, s) ∈ T × S such that ρ(t, s) = ρ(T, S),

(E{Xij (t)Y (s)})i,j∈ζ(t,s) ,

(E{X(t)Yi j (s)})i ,j ∈ζ (t,s)

are both negative semidefinite, where

ζ(t, s) = {n : E{Xn (t)Y (s)} = 0, 1 ≤ n ≤ N },
ζ (t, s) = {n : E{X(t)Yn (s)} = 0, 1 ≤ n ≤ N }.

Remark 9.1.1 Note that
∂ρ
(t, s) = E{Xi (t)Y (s)},
∂ti

∂ 2ρ
(t, s) = E{Xij (t)Y (s)},
∂ti ∂tj

∂ρ
(t, s) = E{X(t)Yi (s)},
∂si

∂ 2ρ
(t, s) = E{X(t)Yij (s)}.
∂si ∂sj

Therefore, similarly to Remark 3.1.2, in order to verify (C3), it suffices to consider those
points (t, s) ∈ T ×S such that t ∈ ∂k T with 0 ≤ k ≤ N −2 or s ∈ ∂k S with 0 ≤ k ≤ N −2.
We decompose T and S into several faces as
N

T =

N

∂k T =
k=0

N

J,
k=0 J∈∂k T

S=
l=0

216

N

∂l S =

L.
l=0 L∈∂l S

For each J ∈ ∂k T and L ∈ ∂l S, define the number of extended outward maxima above level
u as
E
Mu (X, J) := #{t ∈ J : X(t) ≥ u, X|J (t) = 0, index( 2 X|J (t)) = k,

/
ε∗ Xj (t) ≥ 0 for all j ∈ σ(J)},
j
E
Mu (Y, L) := #{s ∈ L : Y (s) ≥ u, Y|L (s) = 0, index( 2 Y|L (s)) = l,

ε∗ Yj (s) ≥ 0 for all j ∈ σ(L)};
/
j
and define the number of maxima above level u as
Mu (X, J) := #{t ∈ J : X(t) ≥ u, X|J (t) = 0, index( 2 X|J (t)) = k},
Mu (Y, L) := #{s ∈ L : Y (s) ≥ u, Y|L (s) = 0, index( 2 Y|L (s)) = l}.
Similarly to Lemma 2.3.1, we have the following result.
Lemma 9.1.2 Under (C1) and (C2), the following relation holds for each u > 0:
N
E
E
{Mu (X, J) ≥ 1, Mu (Y, L) ≥ 1} a.s.

sup X(t) ≥ u, sup Y (s) ≥ u =
t∈T

s∈S

k,l=0 J∈∂k T,L∈∂l S

It follows from Lemma 9.1.2 that
N
E
E
P{Mu (X, J) ≥ 1, Mu (Y, L) ≥ 1}

P sup X(t) ≥ u, sup Y (s) ≥ u ≤
t∈T

s∈S

k,l=0 J∈∂k T,L∈∂l S
N
E
E
E{Mu (X, J)Mu (Y, L)}.

≤
k,l=0 J∈∂k T,L∈∂l S

(9.1.1)

217

On the other hand, by the Bonferroni inequality, Lemma 9.1.2 implies
N
E
E
P{Mu (X, J) ≥ 1, Mu (Y, L) ≥ 1}

P sup X(t) ≥ u, sup Y (s) ≥ u ≥
t∈T

s∈S

k,l=0 J∈∂k T,L∈∂l S

N
E
E
E
P{Mu (X, J) ≥ 1, Mu (J , X) ≥ 1, Mu (Y, L) ≥ 1}

−
k,l=0 J,J ∈∂ T,J=J
k
L∈∂l S
N

E
E
E
P{Mu (X, J) ≥ 1, Mu (Y, L) ≥ 1, Mu (L , Y ) ≥ 1}

−
k,l=0

J∈∂k T
L,L ∈∂l S,L=L

N
E
E
E
E
P{Mu (X, J) ≥ 1, Mu (J , X) ≥ 1, Mu (Y, L) ≥ 1, Mu (L , Y ) ≥ 1}.

−
k,l=0 J,J ∈∂ T,J=J
k
L,L ∈∂l S,L=L

E
E
E
E
Let pij = P{Mu (X, J) = i, Mu (Y, L) = j}, then P{Mu (X, J) ≥ 1, Mu (Y, L) ≥ 1} =
∞
i,j=1 pij

E
E
and E{Mu (X, J)Mu (Y, L)} =

∞
i,j=1 ijpij ,

and hence

E
E
E
E
E{Mu (X, J)Mu (Y, L)} − P{Mu (X, J) ≥ 1, Mu (Y, L) ≥ 1}
∞

∞

=

(ij − 1)pij ≤
i,j=1

[i(i − 1)j + j(j − 1)i]pij
i,j=1

E
E
E
E
E
E
= E{Mu (X, J)[Mu (X, J) − 1]Mu (Y, L)} + E{Mu (Y, L)[Mu (Y, L) − 1]Mu (X, J)}.

218

We therefore obtain the following lower bound for the excursion probability,
N
E
E
E{Mu (X, J)Mu (Y, L)}

P sup X(t) ≥ u, sup Y (s) ≥ u ≥
t∈T

s∈S

k,l=0 J∈∂k T,L∈∂l S

E
E
E
E
E
E
− E{Mu (X, J)[Mu (X, J) − 1]Mu (Y, L)} − E{Mu (Y, L)[Mu (Y, L) − 1]Mu (X, J)}
N
E
E
E
E{Mu (X, J)Mu (X, J )Mu (Y, L)}

−
k,l=0 J,J ∈∂ T,J=J
k
L∈∂l S
N

E
E
E
E{Mu (X, J)Mu (Y, L)Mu (Y, L )}.

−2
k,l=0

J∈∂k T
L,L ∈∂l S,L=L

(9.1.2)

We will show that the upper bound in (9.1.1) makes the major contribution and the other
terms in the lower bound in (9.1.2) are super-exponentially small.
Lemma 9.1.3 Let Di be compact sets in RN , i = 1, 2, 3. Let

{(ξ1 (x1 ), ξ2 (x2 ), ξ3 (x3 )) : (x1 , x2 , x3 ) ∈ D1 × D2 × D3 }

be an R3 -valued, C 2 , centered, unit-variance, non-degenerate Gaussian random field and let

ρ12 (x1 , x2 ) = Eξ1 (x1 )ξ2 (x2 ),

ρ12 =

ρ13 (x1 , x3 ) = Eξ1 (x1 )ξ3 (x3 ),

ρ13 =

ρ23 (x2 , x3 ) = Eξ2 (x2 )ξ3 (x3 ),

ρ23 =

219

sup
x1 ∈D1 ,x2 ∈D2

sup
x1 ∈D1 ,x3 ∈D3

sup
x2 ∈D2 ,x3 ∈D3

ρ12 (x1 , x2 ),
ρ13 (x1 , x3 ),
ρ23 (x2 , x3 ).

If ρ12 ≥ ρ13 ∨ ρ23 , then there exists some constant α > 0 such that as u → ∞,

sup
x1 ∈D1 ,x2 ∈D2 ,x3 ∈D3

= o exp

E{|ξ1 (x1 )ξ2 (x2 )ξ3 (x3 )|m 1{ξ (x )≥u,ξ (x )≥u,ξ (x )≥u} }
1 1
2 2
3 3

u2
− αu2 −
1 + ρ12

(9.1.3)
,

where m is a fixed positive number.

Proof

Let ξ(x1 , x2 , x3 ) = [ξ1 (x1 ) + ξ2 (x2 ) + ξ3 (x3 )]/3, then there exists a positive number

m such that for all (x1 , x2 , x3 ) ∈ D1 × D2 × D3 and u large enough,
E{|ξ1 (x1 )ξ2 (x2 )ξ3 (x3 )|m 1{ξ (x )≥u,ξ (x )≥u,ξ (x )≥u} }
1 1
2 2
3 3
≤ E{(ξ1 (x1 ) + ξ2 (x2 ) + ξ3 (x3 ))m 1{ξ (x )≥u,ξ (x )≥u,ξ (x )≥u} }
1 1
2 2
3 3
≤ E{(ξ1 (x1 ) + ξ2 (x2 ) + ξ3 (x3

))m

(9.1.4)

1{[ξ1 (x1 )+ξ2 (x2 )+ξ3 (x3 )]/3≥u} }

= E{(3ξ(x1 , x2 , x3 ))m 1{ξ(x ,x ,x )≥u} }.
1 2 3
It follows from the assumption ρ12 ≥ ρ13 ∨ ρ23 that

sup
x1 ∈D1 ,x2 ∈D2 ,x3 ∈D3

Var(ξ(x1 , x2 , x3 ))

3 + 2[ρ12 (x1 , x2 )) + ρ13 (x1 , x3 ) + ρ23 (x2 , x3 )]
9
x1 ∈D1 ,x2 ∈D2 ,x3 ∈D3
3 + 6ρ12
1 + 2ρ12
≤
=
,
9
3
=

sup

and hence ρ12 ∈ (−1/2, 1). Combining this with (9.1.4), we see that for any ε > 0, as
2

3u
u → ∞, the first line in (9.1.3) is o(exp{εu2 − 2(1+2ρ ) }). Now the result follows by taking
12

220

α to be a positive number less than

3
1
1 − ρ12
−
=
.
2(1 + 2ρ12 ) 1 + ρ12
2(1 + ρ12 )(1 + 2ρ12 )

Lemma 9.1.4 Let D1 , . . . , Dn be compact sets in RN , where n ≥ 3, and let

(ξ1 (x1 ), ξ2 (x2 ), ξ3 (x3 ), . . . , ξn (xn ) : xi ∈ Di , i = 1 . . . , n)

be an Rn -valued, C 2 , centered, unit-variance, non-degenerate Gaussian random vector. Let
m be a fixed positive number and

ρ12 (x1 , x2 ) = E{ξ1 (x1 )ξ2 (x2 )},

ρ12 =

sup
x1 ∈D1 ,x2 ∈D2

ρ12 (x1 , x2 ).

Then
lim u−2 log E{|ξ1 (x1 )ξ2 (x2 )|m 1{ξ (x )≥u,ξ (x )≥u} |ξ3 (x3 ) = · · · = ξn (xn ) = 0}
1 1
2 2
1
.
≤−
1 + ρ12 (x1 , x2 )

u→∞

If

{(x1 , . . . , xn ) ∈ D1 × · · · × Dn :
(9.1.5)
ρ12 (x1 , x2 ) = ρ12 , E{(ξ1 (x1 ) + ξ2 (x2 ))ξi (xi )} = 0, ∀i = 3, . . . , n} = ∅,

221

then there exists some constant α > 0 such that as u → ∞,

sup
xi ∈Di ,i=1,...,n

= o exp

Proof

E{|ξ1 (x1 )ξ2 (x2 )|m 1{ξ (x )≥u,ξ (x )≥u} |ξ3 (x3 ) = · · · = ξn (xn ) = 0}
1 1
2 2
− αu2 −

u2
1 + ρ12

.

Let ξ(x1 , x2 ) = [ξ1 (x1 ) + ξ2 (x2 )]/2, then there exists a positive number m such

that for all xi ∈ Di , i = 1, . . . , n and u large enough,
E{|ξ1 (x1 )ξ2 (x2 )|m 1{ξ (x )≥u,ξ (x )≥u} |ξ3 (x3 ) = · · · = ξn (xn ) = 0}
1 1
2 2
≤ E{[(ξ1 (x1 ) + ξ2 (x2 ))/2]m 1{[ξ (x )+ξ (x )]/2≥u} |ξ3 (x3 ) = · · · = ξn (xn ) = 0}
1 1
2 2
= E{(ξ(x1 , x2 ))m 1{ξ(x ,x )≥u} |ξ3 (x3 ) = · · · = ξn (xn ) = 0}.
1 2
Note that

Var(ξ(x1 , x2 )|ξ3 (x3 ) = · · · = ξn (xn ) = 0) ≤ Var(ξ(x1 , x2 )) =

1 + ρ12 (x1 , x2 )
,
2

where the equality holds if and only if ξ(x1 , x2 ) is independent of (ξ3 (x3 ), . . . , ξn (xn )). Now
our result follows from the continuity of the conditional expectation and the compactness of
Di , i = 1, . . . , n.
The following result is similar to Lemma 3 in Piterbarg (1996b).
Lemma 9.1.5 Let (X, Y ) = {(X(t), Y (s)) : t ∈ K ⊂ RN , s ∈ D ⊂ RN } be an R2 -valued,
centered, unit-variance Gaussian random field satisfying (C1) and (C2). Then for any ε > 0,

222

there exists δ > 0 such that for K with diam(K) ≤ δ and u large enough,

E{Mu (X, K)[Mu (X, K) − 1]Mu (Y, D)} ≤ Vol(K) exp

−

u2
+ εu2 ,
2βX (K, D)

where

βX (K, D) =

sup

Var

t∈K,s∈D,e∈SN −1

X(t) + Y (s)
2

X(t)= Y (s)=0,
2 X(t)e=0

.

Similarly, for any ε > 0, there exists δ > 0 such that for D with diam(D) ≤ δ and u large
enough,

E{Mu (X, K)Mu (Y, D)[Mu (Y, D) − 1]} ≤ Vol(D) exp

−

u2
+ εu2 ,
2βY (K, D)

where

βY (K, D) =

Proof

sup

Var

t∈K,s∈D,e∈SN −1

X(t) + Y (s)
2

X(t)= Y (s)=0,
2 Y (s)e=0

.

The proof will be similar to the original proof of Lemma 3 in Piterbarg (1996b).

The only difference is that the integral here involves both X and Y exceeding u. But we
may apply the arguments for proving Lemma 9.1.3 and Lemma 9.1.4 to handle the double
integral, to make it bounded above by the integral of (X + Y )/2 exceeding u, and then the
desired result follows.

Lemma 9.1.6 Let (X, Y ) = {(X(t), Y (s)) : t ∈ J ⊂ RN , s ∈ L ⊂ RN } be an R2 -valued,
centered, unit-variance Gaussian random field satisfying (C1), (C2) and (C3). Then there
223

exists α > 0 such that as u → ∞,

E{Mu (X, J)[Mu (X, J) − 1]Mu (Y, L)} = o exp

−

u2
− αu2
1 + ρ(J, L)

,

E{Mu (X, J)Mu (Y, L)[Mu (Y, L) − 1]} = o exp

u2
−
− αu2
1 + ρ(J, L)

,

(9.1.6)

where ρ(J, L) = supt∈J,s∈L ρ(t, s).

Proof We only prove the first line in (9.1.6), since the proof for the second line is the same.
The set J may be covered by congruent cubes Ki with disjoint interiors, edges parallel to
coordinate axes and sizes so small that the conditions of Lemma 9.1.5 are satisfied for each
union of two neighboring cubes Ki and Kj . Then

E{Mu (X, J)[Mu (X, J) − 1]Mu (Y, L)}
≤E

[Mu (X, Kj ) − 1] Mu (Y, L)

Mu (X, Ki )
i

=E

j

Mu (X, Kj ) −

Mu (X, Ki )
i

j

i

E{Mu (X, Ki )2 Mu (Y, L)} +

=
i

Mu (X, Ki ) Mu (Y, L)

E{Mu (X, Ki )Mu (X, Kj )Mu (Y, L)}−
i=j

−

E{Mu (X, Ki )Mu (Y, L)}
i

E{Mu (X, Ki )[Mu (X, Ki ) − 1]Mu (Y, L)} +

=
i

E{Mu (X, Ki )Mu (X, Kj )Mu (Y, L)}.
i=j

(9.1.7)

224

Then by the Kac-Rice formula and Lemma 9.1.3, there exists α > 0 such that for u large
enough,

E{Mu (X, Ki )Mu (X, Kj )Mu (Y, L)} ≤ exp

−

|i−j|≥2

u2
− α u2 .
1 + ρ(J, L)

(9.1.8)

If Ki and Kj are neighboring, say j = i + 1, we have

E{Mu (X, Ki ∪ Ki+1 )[Mu (X, Ki ∪ Ki+1 ) − 1]Mu (Y, L)}
= E{[Mu (X, Ki ) + Mu (X, Ki+1 )][Mu (X, Ki ) + Mu (X, Ki+1 ) − 1]Mu (Y, L)}
= 2E{Mu (X, Ki )Mu (X, Ki+1 )Mu (Y, L)} + E{Mu (X, Ki )[Mu (X, Ki ) − 1]Mu (Y, L)}
+ E{Mu (X, Ki+1 )[Mu (X, Ki+1 ) − 1]Mu (Y, L)}.
Applying Lemma 9.1.5, we see that for u large enough,

E{Mu (X, Ki )[Mu (X, Ki ) − 1]Mu (Y, L)} +
i

E{Mu (X, Ki )Mu (X, Kj )Mu (Y, L)}
|i−j|=1

≤ exp

−

u2
+ εu2 .
2βX (J, L)
(9.1.9)

It is obvious that βX (J, L) ≤

1+ρ(J,L)
,
2

and we will show

βX (J, L) <

1 + ρ(J, L)
.
2

225

(9.1.10)

By the definition of βX (J, L) in Lemma 9.1.5, if βX (J, L) =

1+ρ(J,L)
,
2

then due to the

¯ ¯
continuity, there are some (t, s) ∈ J × L and e ∈ SN −1 such that

Var

X(t) + Y (s)
2

X(t)= Y (s)=0,
2 X(t)e=0

=

1 + ρ(J, L)
.
2

(9.1.11)

This implies

ρ(t, s) = ρ(J, L),

E{X(t) Y (s)} = E{Y (s) X(t)} = 0.

By (C3), E{Y (s) 2 X(t)} becomes negative semidefinite. But E{X(t) 2 X(t)} is always
negative definite due to the constant variance, so that

E{Y (s) 2 X(t)} + E{X(t) 2 X(t)} e = 0,

∀e ∈ SN −1 .

This contradicts (9.1.11) and hence (9.1.10) holds. Plugging (9.1.8) and (9.1.9) into (9.1.7),
we finish the proof.

Lemma 9.1.7 Let {(X(t), Y (s)) : t ∈ T, s ∈ S} be an R2 -valued, centered, unit-variance
Gaussian random field satisfying (C1), (C2) and (C3). Then there exists α > 0 such that
as u → ∞,

E
E
E
E{Mu (X, J)Mu (X, J )Mu (Y, L)} = o exp

−

u2
− αu2
1 + ρ(J, L)

,

E
E
E
E{Mu (X, J)Mu (Y, L)Mu (Y, L

u2
−
− αu2
1 + ρ(J, L)

,

)} = o exp

where J and J are different faces of T , L and L are different faces of S.

226

(9.1.12)

Proof

We only prove the first line in (9.1.12), since the proof for the second line is the

same. If the two faces J and J are not neighboring, by similar arguments in Lemma 2.3.6
and Lemma 9.1.3, it is straightforward to verify that the high moment in (9.1.12) is superexponentially small. Thus we turn to considering the case when J and J are neighboring,
¯ ¯
i.e., I := J ∩ J = ∅. Without loss of generality, assume

σ(J) = {1, . . . , m, m + 1, . . . , k},
σ(J ) = {1, . . . , m, k + 1, . . . , k + k − m},
σ(L) = {1, . . . , l},

where 0 ≤ m ≤ k ≤ k ≤ N and k ≥ 1. If k = 0, we consider σ(J) = ∅ by convention.
Under such assumption, J ∈ ∂k T , J ∈ ∂k T , dim(I) = m and L ∈ ∂l S. We assume also
that all elements in ε(J) and ε(J ) are 1.
We first consider the case k ≥ 1. By the Kac-Rice metatheorem,

E
E
E
E{Mu (X, J)Mu (X, J )Mu (Y, L)}

J
∞
0

J

u

L

u

u

∞

dzk+1 · · ·

0

dy

dx

dx

ds

dt

dt

∞

∞

∞

≤

∞

∞

dzk+k −m

0

dwm+1 · · ·

0

dwk

E{ det 2 X|J (t)||det 2 X|J (t )||det 2 Y|L (s)||X(t) = x, X(t ) = x , Y (s) = y,
X|J (t) = 0, Xk+1 (t) = zk+1 , . . . , Xk+k −m (t) = zk+k −m ,
X|J (t ) = 0, Xm+1 (t ) = wm+1 , . . . , Xk (t ) = wk , Y|L (s) = 0}
× pt,t ,s (x, x , y, 0, zk+1 , . . . , zk+k −m , 0, wm+1 , . . . , wk , 0)
:=

A(t, t , s) dtdt ds,
J×J ×L

227

(9.1.13)

where pt,t ,s (x, x , y, 0, zk+1 , . . . , zk+k −m , 0, wm+1 , . . . , wk , 0) is the density of

(X(t), X(t ), Y (s), X|J (t), Xk+1 (t), . . . , Xk+k −m (t),
X|J (s), Xm+1 (s), . . . , Xk (s), Y|L (s))
evaluated at (x, x , y, 0, zk+1 , . . . , zk+k −m , 0, wm+1 , . . . , wk , 0).
Similarly to Lemma 3.1.7, by Lemma 9.1.4 and continuity, if

¯
I0 := {(t, s) ∈ I × S :ρ(t, s) = ρ(T, S), E{Xi (t)Y (s)} = E{X(t)Yj (s)} = 0,
∀i = 1, . . . , k + k − m, j = 1, . . . , l} = ∅,
E
E
E
then E{Mu (X, J)Mu (X, J )Mu (Y, L)} is super-exponentially small. Therefore, similarly

to the proof in Lemma 3.1.8, we only need to consider the alternative case, which is I0 = ∅.
Define

B(I0 , δ) := {(t, t , s) ∈ J × J × S : d((t, s), I0 ) ∨ d((t , s), I0 ) < δ},

where δ is a small positive positive number to be specified. Then the difference between
J×J ×L A(t, t

, s)dtdt ds and B(I ,δ) A(t, t , s)dtdt ds is super-exponentially small. Hence
0

we turn to estimating B(I ,δ) A(t, t , s) dtdt ds.
0
Due to (C3), we may choose δ small enough such that for all (t, t , s) ∈ B(I0 , δ),

ΛJ∪J (t, s) = −(E{[X(t) + Y (s)] 2 X(t)})i,j=1,...,k+k −m

are positive definite.

228

Let {e1 , e2 , . . . , eN } be the standard orthonormal basis of RN . For t ∈ J and s ∈ J , let
et,t = (t − t)T / t − t and let αi (t, t , s) = ei , ΛJ∪J (t, s)et,t , then
N

N

ei , ΛJ∪J (t, s)et,t ei =

ΛJ∪J (t, s)et,t =

αi (t, t , s)ei .

(9.1.14)

i=1

i=1

There exists some α0 > 0 such that

et,t , ΛJ∪J (t, s)et,t ≥ α0

(9.1.15)

for all t and t . Since all elements in ε(J) and ε(J ) are 1, we have the following representation,

t = (t1 , . . . , tm , tm+1 , . . . , tk , bk+1 , . . . , bk+k −m , 0, . . . , 0),
t = (t1 , . . . , tm , bm+1 , . . . , bk , tk+1 , . . . , tk+k −m , 0, . . . , 0),
where ti ∈ (ai , bi ) for all i ∈ σ(J) and tj ∈ (aj , bj ) for all j ∈ σ(J ). Therefore,

ei , et,t ≥ 0,

∀ m + 1 ≤ i ≤ k,

ei , et,t ≤ 0,

∀ k + 1 ≤ i ≤ k + k − m,

ei , et,t = 0,

∀ k + k − m < i ≤ N.

(9.1.16)

Let

Di = {(t, t , s) ∈ B(I0 , δ) : αi (t, t , s) ≥ βi },
Di = {(t, t , s) ∈ B(I0 , δ) : αi (t, t , s) ≤ −βi },

if m + 1 ≤ i ≤ k,
if k + 1 ≤ i ≤ k + k − m,

m

D0 =

(t, t , s) ∈ B(I0 , δ) :

αi (t, t , s) ei , et,t ≥ β0 ,
i=1

229

(9.1.17)

where β0 , β1 , . . . , βk+k −m are positive constants such that β0 +

k+k −m
i=m+1 βi

< α0 . It follows

from (9.1.16) and (9.1.17) that, if (t, s) does not belong to any of D0 , Dm , . . . , Dk+k −m ,
then by (9.1.14),
k+k −m

N

αi (t, t , s) ei , et,t ≤ β0 +

ΛJ∪J (t, s)et,t , et,t =

βi < α0 ,
i=m+1

i=1

which contradicts (9.1.15). Thus D0 ∪ ∪k+k −m Di is a covering of B(I0 , δ), by (9.1.13),
i=m+1

E
E
E
E{Mu (X, J)Mu (X, J )Mu (Y, L)}
k+k −m

≤

A(t, t , s) dtdt ds.

A(t, t , s) dtdt ds +
D0

i=m+1

Di

We first show that D A(t, t , s) dtdt ds is super-exponentially small.
0

D0 A(t, t

, s) dtdt ds
∞

∞

≤

dx

dtdt ds
D0

u

u

dy p X (t), X (t ), Y (s) (0, 0, 0)
|J
|L
|J

× pX(t),Y (s) (x, y| X|J (t) = 0, X|J (t ) = 0, Y|L (s) = 0)
× E{|det 2 X|J (t)||det 2 X|J (t )||det 2 Y|L (s)||X(t) = x, Y (s) = y,
X|J (t) =

X|J (t ) =

Y|L (s) = 0}.

230

(9.1.18)

Note that

Var(X(t) + Y (s)| X|J (t), X|J (t ), Y|L (s))
≤ Var(X(t) + Y (s)|X1 (t), . . . , Xm (t), X1 (t ), . . . , Xm (t ))
= Var(X(t) + Y (s)|X1 (t), . . . , Xm (t), X1 (t) +
Xm (t) +

1
X1 (t), t − t + t − t 1+η Yt,t , . . . ,

m
Xm (t), t − t + t − t 1+η Yt,t )

= Var(X(t) + Y (s)|X1 (t), . . . , Xm (t),

1
X1 (t), et,t + t − t η Yt,t , . . . ,

m
Xm (t), et,t + t − t η Yt,t )

≤ Var(X(t) + Y (s)|

1
X1 (t), et,t + t − t η Yt,t , . . . ,

= Var(X(t) + Y (s)|

X1 (t), et,t , . . . ,

m
Xm (t), et,t + t − t η Yt,t )

Xm (t), et,t ) + o(1).
(9.1.19)

Hence there exist constants C2 > 0 and ε0 > 0 such that for t − t sufficiently small,

Var((X(t) + Y (s))/2| X|J (t), X|J (t ), Y|L (s))
ρ(T, S) + 1
− C2
≤
2

m
2
αi (t, t
i=1

(9.1.20)

ρ(T, S) + 1
− ε0 ,
, s) + o(1) <
2

where the last inequality is due to the fact that (t, t , s) ∈ D0 implies
m

m
2
αi (t, t

i=1

2
αi (t, t

, s) ≥
i=1

, s)| ei , et,t

|2

1
≥
m

m

2

αi (t, t , s) et,t , ei
i=1

β2
≥ 0.
m

Similarly, we can use the techniques in the proof of Theorem 2.3.8 to show that for
i = m + 1, . . . , k + k − m, D A(t, t , s) dtdt ds are super-exponentially small, .
i

231

Now, the approximation obtained still contains the absolute values of determinants, which
are hard to be computed. However, we will show that removing these absolute values only
causes exponentially small difference, and then we will get the approximation based on the
mean Euler characteristic of the excursion set.
Proposition 9.1.8 Let {(X(t), Y (s)) : t ∈ T, s ∈ S} be an R2 -valued, centered, unitvariance Gaussian random field satisfying (C1), (C2) and (C3). Then there exists α > 0
such that for any J ∈ ∂k T , L ∈ ∂l S, as u → ∞,
E
E
E{Mu (X, J)Mu (Y, L)}

= (−1)k+l
J

L

E{det 2 X(t)det 2 Y (s)1{X(t)≥u, ε∗ X (t)≥0 for all j ∈σ(J)}
/
j j

× 1{Y (s)≥u, ε∗ Y (s)≥0 for all j ∈σ(L)} | X|J (t) =
/
j j
× p X (t), Y (s) (0, 0)dtds + o exp
|J
|L

−

where ρ(T, S) = supt∈T,s∈S ρ(t, s).

232

Y|L (s) = 0}

u2
− αu2
1 + ρ(T, S)

,

(9.1.21)

Proof

To simplify the proof, let us consider the case when k = l = N , and the proof for

general cases is similar. By the Kac-Rice formula,

E
E
E{Mu (X, J)Mu (Y, L)}

E{|det 2 X(t)||det 2 Y (s)|1{X(t)≥u, index( 2 X(t))=N }
J L

=

× 1{Y (s)≥u, index( 2 Y (s))=N } | X(t) =

Y (s) = 0}p X(t), Y (s) (0, 0)dtds
∞

= (−1)N +N
J

L

p X(t), Y (s) (0, 0)dtds

u

∞
u

E{det 2 X(t)det 2 Y (s)

× 1{index( 2 X(t))=N } 1{index( 2 Y (s))=N } |X(t) = x, Y (s) = y, X(t) =
× pX(t),Y (s) (x, y| X(t) =

Y (s) = 0)dxdy
∞

:=
J

L

Y (s) = 0}

p X(t), Y (s) (0, 0)dtds

∞

A(x, y, t, s)dxdy.
u

u

Similarly to the proof in Lemma 3.1.5, define

¯ ¯
O(J, L) = {(t, s) ∈ J × L : ρ(t, s) = ρ(T, S), E{X(t) Y (s)} = E{Y (s) X(t)} = 0},
Uδ (J, L) = {(t, s) ∈ J × L : d((t, s), O(J, L)) < δ},
where δ is a small positive number to be specified. Then, similarly, we only need to estimate
∞
Uδ (J,L)

p X(t), Y (s) (0, 0)dtds

∞

A(x, y, t, s)dxdy.
u

u

Due to (C3), we may choose δ small enough such that E{[X(t) + Y (s)] 2 Y (s)} and
E{[X(t) + Y (s)] 2 X(t)} are negative definite for all (t, s) ∈ Uδ (J, L). Also note that as

233

δ → 0, both E{X(t) Y (s)} and E{Y (s) X(t)} tend to 0, therefore,

E{Xij (t)|X(t) = x, Y (s) = y, X(t) =

Y (s) = 0}


−1 
1

ρ(T, S) 

ρ(T, S)
1


1
 x − ρ(T, S)y 
=
(E{Xij (t)X(t)}, E{Xij (t)Y (s)}) 
,
1 − ρ(T, S)2
y − ρ(T, S)x



 x 
 
y


= (1 + o(1))(E{Xij (t)X(t)}, E{Xij (t)Y (s)}) 

and similarly,

E{Yij (s)|X(t) = x, Y (s) = y, X(t) =

=

Y (s) = 0}




1
 x − ρ(T, S)y 
(E{Yij (s)X(t)}, E{Yij (s)Y (s)}) 
.
1 − ρ(T, S)2
y − ρ(T, S)x

Note that E{X(t) 2 X(t)} and E{Y (s) 2 Y (s)} are both negative definite, E{X(t) 2 Y (s)}
and E{Y (s) 2 X(t)} both negative semidefinite. There exists ε0 ∈ (0, 1 − ρ(T, S)) such that
for δ small enough and all (x, y) ∈ [u, ∞)2 with (ε0 + ρ(T, S))y < x < (ε0 + ρ(T, S))−1 y,
Σ1 (x, y) := E{ 2 X(t)|X(t) = x, Y (s) = y, X(t) =

Y (s) = 0},

Σ2 (x, y) := E{ 2 Y (s)|X(t) = x, Y (s) = y, X(t) =

Y (s) = 0},

are both negative definite. Define

∆1 (x, y) =

2 X(t) − Σ

1 (x, y),

∆2 (x, y) =

234

2 Y (s) − Σ (x, y).
2

Then, conditioning on (X(t) = x, Y (s) = y, X(t) =

Y (s) = 0), ∆1 (x, y) and ∆2 (x, y)

are both centered Gaussian random matrices. We write
∞

∞

A(x, y, t, s)dxdy =
u

u
∞

dy
u

(ε0 +ρ(T,S))y
∞

∞

dy

+

(ε0 +ρ(T,S))−1 y

∞

u

(ε0 +ρ(T,S))−1 y

∞

dx

A(x, y, t, s)dx +
u

A(x, y, t, s)dx+

(ε0 +ρ(T,S))−1 x

A(x, y, t, s)dy.

Since (ε0 + ρ(T, S))−1 > 1, the last two integrals above are super-exponentially small.
Meanwhile,
(ε0 +ρ(T,S))−1 y

∞

dy
u

A(x, y, t, s)dx

(ε0 +ρ(T,S))y
(ε0 +ρ(T,S))−1 y

∞

=

dy
u

(ε0 +ρ(T,S))y

E{det(∆1 (x, y) + Σ1 (x, y))det(∆2 (x, y) + Σ2 (x, y))

× 1{index(∆ (x,y)+Σ (x,y))=N } 1{index(∆ (x,y)+Σ (x,y)))=N } |
1
1
2
2
|X(t) = x, Y (s) = y, X(t) =

Y (s) = 0}pX(t),Y (s) (x, y| X(t) =

Y (s) = 0)dx.

Using the same arguments in the proof of Lemma 2.3.2, we see that removing the two
indicator functions above only causes a super-exponentially small difference. Therefore,
there exists α > 0 such that for u large enough,

E
E
E{Mu (X, J)Mu (Y, L)}
∞

=
J

L

p X(t), Y (s) (0, 0)dtds

∞

u

u

pX(t),Y (s) (x, y| X(t) =

× E{det 2 X(t)det 2 Y (s)|X(t) = x, Y (s) = y, X(t) =
+ o exp

−

u2
− αu2
1 + ρ(T, S)

.

235

Y (s) = 0)

Y (s) = 0}dxdy

Define the excursion set

Au (X, T ) = {t ∈ T : X(t) ≥ u},
Au (Y, S) = {s ∈ S : Y (s) ≥ u},
Au (X, T ) × Au (Y, S) = {(t, s) ∈ T × S : X(t) ≥ u, Y (s) ≥ u}.
Let
X|J (t) = 0, index( 2 X|J (t)) = i,

µi (X, J) := #{t ∈ J : X(t) ≥ u,

ε∗ Xj (t) ≥ 0 for all j ∈ σ(J)},
/
j
µi (Y, L) := #{s ∈ L : Y (s) ≥ u, Y|L (s) = 0, index( 2 Y|L (s)) = i,
ε∗ Yj (s) ≥ 0 for all j ∈ σ(L)},
/
j
where ε∗ = 2εj − 1 and the index of a matrix is defined as the number of its negative
j
eigenvalues. Then, it follows from the Morse theorem that the Euler characteristic of the
excursion set can be represented as
N

k

(−1)k

ϕ(Au (X, T )) =

(−1)i µi (X, J),
i=0

k=0 J∈∂k T

l

N

(−1)l

ϕ(Au (Y, S)) =
l=0 L∈∂l S

(−1)i µi (Y, L).
i=0

Since for two sets D1 and D2 , ϕ(D1 × D2 ) = ϕ(D1 )ϕ(D2 ), we have

ϕ(Au (X, T ) × Au (Y, S)) = ϕ(Au (X, T )) × ϕ(Au (Y, S))
N

k

(−1)k+l

=
k,l=0 J∈∂k T,L∈∂l S

(9.1.22)

l

(−1)i µ
i=0

i (J)

(−1)j µ
j=0

236

j (L)

.

Now we can state our result as follows.
Theorem 9.1.9 Let {(X(t), Y (s)) : t ∈ T, s ∈ S} be an R2 -valued, centered, unit-variance
Gaussian random field satisfying (C1), (C2) and (C3). Then there exists α > 0 such that
as u → ∞,

P sup X(t) ≥ u, sup Y (s) ≥ u
t∈T
N

s∈S
E
E
E{Mu (X, J)Mu (Y, L)} + o exp

=
k,l=0 J∈∂k T,L∈∂l S

= E{ϕ(Au (X, T ) × Au (Y, S))} + o exp

−

−

u2
− αu2
1 + ρ(T, S)

u2
− αu2
1 + ρ(T, S)

,
(9.1.23)

where ρ(T, S) = supt∈T,s∈S ρ(t, s).

Proof The first equality in (9.1.23) follows from the combination of (9.1.1), (9.1.2), Lemma
9.1.6 and Lemma 9.1.7. The second equality in (9.1.23) follows from applying Proposition
9.1.8 and (9.1.22).

237

Example 9.1.10

Let T = S = [0, 1], then

P sup X(t) ≥ u, sup Y (s) ≥ u = P{X(0) ≥ u, Y (0) ≥ u, X (0) < 0, Y (0) < 0}
t∈T

s∈S

+ P{X(0) ≥ u, Y (1) ≥ u, X (0) < 0, Y (1) > 0}
+ P{X(1) ≥ u, Y (0) ≥ u, X (1) > 0, Y (0) < 0}
+ P{X(1) ≥ u, Y (1) ≥ u, X (1) > 0, Y (1) > 0}
∞

1

+ (−1)

pX (t) (0)dt

0

u

∞

0

pX(t),Y (0),Y (0) (x, y, z|X (t) = 0)

−∞

u

× E{X (t)|X(t) = x, Y (0) = y, Y (0) = z, X (t) = 0}dxdydz
∞

1

pX (t) (0)dt

+ (−1)
0

u

∞

∞

u

pX(t),Y (1),Y (1) (x, y, z|X (t) = 0)

0

× E{X (t)|X(t) = x, Y (1) = y, Y (1) = z, X (t) = 0}dxdydz
∞

1

pY (s) (0)ds

+ (−1)
0

u

∞

0
−∞

u

pX(0),Y (s),X (0) (x, y, z|Y (s) = 0)

× E{Y (s)|X(0) = x, Y (s) = y, X (0) = z, Y (s) = 0}dxdydz
∞

1

+ (−1)
0

pY (s) (0)ds

u

∞

∞

u

pX(1),Y (s),X (1) (x, y, z|Y (s) = 0)

0

× E{Y (s)|X(1) = x, Y (s) = y, X (1) = z, Y (s) = 0}dxdydz
1
0

∞

1

+
0

pX (t),Y (s) (0, 0)dtds

u

∞
u

pX(t),Y (s) (x, y|X (t) = Y (s) = 0)

× E{X (t)Y (s)|X(t) = x, Y (s) = y, X (t) = Y (s) = 0}dxdy
+ o exp

−

u2
− αu2
1 + ρ(T, S)

.

238

9.2

Vector-valued Gaussian Processes

Let {(X(t), Y (t)) : t ∈ T } be an R2 -valued, centered, unit-variance Gaussian process, where
T = [a, b] is a finite interval in R. We want to estimate the following probability

P{∃t ∈ T such that X(t) ≥ u, Y (t) ≥ u}.

Let
ρ(t) = E{X(t)Y (t)},

ρ(T ) = sup ρ(t).
t∈T

We will make use of the following conditions.
(D1). X, Y ∈ C 2 (RN ) almost surely, and for each pair (t, s) ∈ T 2 with t = s,

(X(t), X (t), X (t), Y (t), Y (t), Y (t), X(s), X (s), Y (s), Y (s))

is non-degenerate.
(D2). For all t ∈ T such that ρ(t) = ρ(T ) (hence E{X (t)Y (t)} + E{X (t)Y (t)} = 0),
E{X (t)Y (t)} = −E{X(t)Y (t)} = 0.

239

Define
◦

◦

◦

◦

◦

◦

µ(X, T ) = #{t ∈ T : Y (t) > X(t) ≥ u, X (t) = 0, X (t) < 0},
µ(Y, T ) = #{t ∈ T : X(t) > Y (t) ≥ u, Y (t) = 0, Y (t) < 0},
µ(X = Y, T ) = #{t ∈ T : X(t) = Y (t) ≥ u, X (t)Y (t) < 0},
µ(X, a) = 1{Y (a)>X(a)≥u,X (a)<0} ,

(9.2.1)

µ(Y, a) = 1{X(a)>Y (a)≥u,Y (a)<0} ,
µ(X, b) = 1{Y (b)>X(b)≥u,X (b)>0} ,
µ(Y, b) = 1{X(b)>Y (b)≥u,Y (b)>0} .
Lemma 9.2.1 Under (D1), for each u > 0, we have

{∃t ∈ T such that X(t) ≥ u, Y (t) ≥ u}
◦

◦

◦

= {µ(X, T ) ≥ 1} ∪ {µ(Y, T ) ≥ 1} ∪ {µ(X = Y, T ) ≥ 1}
∪ {µ(X, a) ≥ 1} ∪ {µ(Y, a) ≥ 1} ∪ {µ(X, b) ≥ 1} ∪ {µ(Y, b) ≥ 1} a.s.

Proof

Note that

{∃t ∈ T such that X(t) ≥ u, Y (t) ≥ u}
= {∃t ∈ T such that (X ∧ Y )(t) ≥ u}.
Then the result follows similarly to Lemma 2.3.1.

240

Lemma 9.2.2 Under (D1) and (D2), there exists some constant α > 0 such that as u → ∞,
◦

Eµ(X, T ) = o exp
◦

Eµ(Y, T ) = o exp

u2
− αu2
1+ρ
u2
−
− αu2
1+ρ
−

,
.

◦

◦

Proof We only show the proof for Eµ(X, T ), since the proof for Eµ(Y, T ) will be the same.
By the Kac-Rice formula,
∞

b

◦

Eµ(X, T ) =

a

pX (t) (0)

∞

dx
u

x

dy pX(t),Y (t) (x, y|X (t) = 0)

× E{|X (t)|1{X (t)<0} |X(t) = x, Y (t) = y, X (t) = 0}.
We only need to show that P{Y (t) > X(t) ≥ u|X (t) = 0} is super-exponentially small. But

P{Y (t) > X(t) ≥ u|X (t) = 0} ≤ P{(X(t) + Y (t))/2 ≥ u|X (t) = 0},

and due to (D2), for each t ∈ T such that ρ(t) = ρ(T ),

Var((X(t) + Y (t))/2|X (t) = 0) < Var((X(t) + Y (t))/2) = (1 + ρ(T ))/2.

By continuity, we obtain

sup Var((X(t) + Y (t))/2|X (t) = 0) < (1 + ρ(T ))/2,
t∈T

completing the proof.

241

Lemma 9.2.3 Under (D1) and (D2), there exists some constant α > 0 such that as u → ∞,

◦

◦

E{µ(X = Y, T )[µ(X = Y, T ) − 1]} = o exp

Proof

−

u2
− αu2
1 + ρ(T )

.

By the Kac-Rice formula,
◦

◦

E{µ(X = Y, T )[µ(X = Y, T ) − 1]}
b

b

=
a

a
∞

pX(t)−Y (t),X(s)−Y (s) (0, 0)dtds
∞

×
u

u

dxdy pX(t),X(s) (x, y|X(t) − Y (t) = 0, X(s) − Y (s) = 0)

× E{|X (t) − Y (t)||X (s) − Y (s)|1{X (t)Y (t)<0} 1{X (s)Y (s)<0} |
|X(t) = x, X(s) = y, X(t) − Y (t) = 0, X(s) − Y (s) = 0}.

Similarly to the proof of Lemma 3 in Piterbarg (1996b), we will write T as the union of
several small intervals, and then it suffices to prove that there exists α > 0 such that

Var(X(t)|X(t) − Y (t), X(s) − Y (s)) <

1 + ρ(T )
−α,
2

∀|t − s| < δ

(9.2.2)

and
lim u−2 log P{X(t) ≥ u, Y (s) ≥ u|X(t) − Y (t) = 0, X(s) − Y (s) = 0}

u→∞

1
<−
−α,
1 + ρ(T )

(9.2.3)
∀|t − s| ≥ δ,

where δ is a small positive number to be specified.

242

Note that for all t ∈ T ,

Var(X(t)|X(t) − Y (t)) = 1 −

1 + ρ(t)
1 + ρ(T )
(1 − ρ(t))2
=
≤
.
2(1 − ρ(t))
2
2

But for those t such that ρ(t) = ρ(T ), E{X(t)(X (t) − Y (t))} = −E{X(t)Y (t)} = 0 by
(D2), it then follows from continuity that

sup Var(X(t)|X(t) − Y (t), X (t) − Y (t)) <
t∈T

1 + ρ(T )
.
2

(9.2.4)

Therefore (9.2.2) follows by noting that as |t − s| → 0,

Var(X(t)|X(t) − Y (t), X(s) − Y (s))
= (1 + o(1))Var(X(t)|X(t) − Y (t), X (t) − Y (t)).

Now we turn to proving (9.2.3). By continuity, it suffices to show that there is not any
pair (t, s) with |t − s| ≥ δ such that

lim u−2 log P{X(t) ≥ u, Y (s) ≥ u|X(t) − Y (t) = 0, X(s) − Y (s) = 0}

u→∞

(9.2.5)

1
=−
.
1 + ρ(T )

243

By (9.2.4) and the following evident inequality

P{X(t) ≥ u, Y (s) ≥ u|X(t) − Y (t) = 0, X(s) − Y (s) = 0}
≤ min P{X(t) ≥ u|X(t) − Y (t) = 0, X(s) − Y (s) = 0},
P{Y (s) ≥ u|X(t) − Y (t) = 0, X(s) − Y (s) = 0},
P{[X(t) + Y (s)]/2 ≥ u|X(t) − Y (t) = 0, X(s) − Y (s) = 0} ,
if (9.2.5) holds, then we have

ρ(t) = ρ(s) = ρ(T ),
E{X(t)[X(s) − Y (s)]} = E{Y (t)[X(s) − Y (s)]} = 0,

(9.2.6)

E{X(s)[X(t) − Y (t)]} = E{Y (s)[X(t) − Y (t)]} = 0,
and

Var([X(t) + Y (s)]/2|X(t) − Y (t) = 0, X(s) − Y (s) = 0) =

1 + ρ(T )
.
2

(9.2.7)

But by the conditional formula for Gaussian variables, (9.2.6) implies

Var([X(t) + Y (s)]/2|X(t) − Y (t) = 0, X(s) − Y (s) = 0)
1 + E{X(t)X(s)}
2ρ(T ) + E{X(t)X(s)} − 1
− (1 − ρ(T )) =
2
2
1 + ρ(T )
<
,
2
=

which contradicts (9.2.7). Thus there is no pair (t, s) with |t − s| ≥ δ such that (9.2.5) holds,
and hence (9.2.3) is true, completing the proof.

244

Lemma 9.2.4 Under (D1) and (D2), there exists some constant α > 0 such that as u → ∞,
◦

◦

max E{µ(X = Y, T )µ(X, a)}, E{µ(X = Y, T )µ(Y, a)},
◦

◦

E{µ(X = Y, T )µ(X, b)}, E{µ(X = Y, T )µ(Y, b)}
= o exp

−

u2
− αu2
1 + ρ(T )

,

◦

Proof

We only show the proof for E{µ(X = Y, T )µ(X, b)}, since the other terms can be

proved similarly. By the Kac-Rice metatheorem,
b

◦

E{µ(X = Y, T )µ(X, b)} =

a

E{|X (t) − Y (t)|1{X(t)≥u,X (t)Y (t)<0}

× 1{Y (b)>X(b)≥u,X (b)>0} |X(t) − Y (t) = 0}dt.
We only need to show that there exists α > 0 such that for δ small enough,

lim u−2 log P{X(t) ≥ u, X(b) − Y (b) < 0, X (b) > 0|X(t) − Y (t) = 0}

u→∞

1
<−
−α,
1 + ρ(T )

(9.2.8)
∀|t − b| < δ,

and
lim u−2 log P{X(t) ≥ u, X(b) ≥ u, Y (b) ≥ u|X(t) − Y (t) = 0}

u→∞

1
−α,
<−
1 + ρ(T )

(9.2.9)
∀|t − b| ≥ δ.

We show (9.2.8) first. Note that we only need to consider the case when ρ(b) = ρ(T ). Under this situation, by (D2), either E{X (b)Y (b)} < 0 or E{Y (b)X(b)} < 0. If E{X (b)Y (b)} <

245

0, then as |t − b| → 0,

E{X(t)X (b)|X(t) − Y (t) = 0}
= E{X(t)X (b)} −
= (1 + o(1))

(1 − ρ(t))E{X (b)[X(t) − Y (t)]}
2(1 − ρ(t))

E{X (b)Y (b)}
.
2

It then follows from Lemma 2.3.10 that
lim u−2 log P{X(t) ≥ u, X (b) > 0|X(t) − Y (t) = 0}

u→∞

1
<−
−α,
1 + ρ(T )

(9.2.10)
∀|t − s| < δ.

For the alternative case, E{Y (t)X(t)} < 0, consider Taylor’s expansion,

X(b) − Y (b) = X(t) − Y (t) + (b − t)(X (t) − Y (t)) + (b − t)1+η Zt,b ,

where η > 0 and Zt,b is a Gaussian random field with uniformly finite variance. Then as
|t − b| → 0,

E{X(t)[X(b) − Y (b)]|X(t) − Y (t) = 0}
= E{X(t)[X(b) − Y (b)]} −

(1 − ρ(t))E{[X(t) − Y (t)][X(b) − Y (b)]}
2(1 − ρ(t))

= −(1 + o(1))(b − t)E{X(b)Y (b)},

246

and also Var(X(b)−Y (b)|X(t)−Y (t)) ≤ C1 (t−b)2 for some positive constant C1 . Therefore,
E{X(t)[X(b) − Y (b)]|X(t) − Y (t) = 0}
Var(X(t)|X(t) − Y (t))Var(X(b) − Y (b)|X(t) − Y (t))
≥ −C2 E{X(b)Y (b)} > 0,
for some positive constant C2 . It then follows from Lemma 2.3.10 that
lim u−2 log P{X(t) ≥ u, X(b) − Y (b) < 0|X(t) − Y (t) = 0}

u→∞

1
−α,
<−
1 + ρ(T )

(9.2.11)
∀|t − s| < δ.

Now, (9.2.10) and (9.2.11) imply (9.2.8).
We turn to proving (9.2.9). Note that

P{X(t) ≥ u, X(b) ≥ u, Y (b) ≥ u|X(t) − Y (t) = 0}
≤ P{X(t) ≥ u, [X(b) + Y (b)]/2 ≥ u|X(t) − Y (t) = 0},
and
1 + ρ(T )
,
2
1 + ρ(T )
Var([X(b) + Y (b)]/2|X(t) − Y (t)) ≤
.
2
Var(X(t)|X(t) − Y (t)) ≤

Due to the regularity condition (D1), we obtain

lim u−2 log P{X(t) ≥ u, [X(b) + Y (b)]/2 ≥ u|X(t) − Y (t) = 0}

u→∞

<−

1
−α,
1 + ρ(T )

∀|t − b| ≥ δ,

and then (9.2.9) follows.

247

Define the excursion set Au (T, X ∧ Y ) = {t ∈ T : (X ∧ Y )(t) ≥ u}. Then Morse theorem
gives
◦

◦

◦

◦

ϕ(Au (T, X ∧ Y )) = µ(X, T ) − µ (X, T ) + µ(Y, T ) − µ (Y, T )
(9.2.12)

◦

+ µ(X = Y, T ) + µ(X, a) + µ(Y, a) + µ(X, b) + µ(Y, b),
where
◦

◦

◦

◦

µ (X, T ) = #{t ∈ T : Y (t) > X(t) ≥ u, X (t) = 0, X (t) > 0},
µ (Y, T ) = #{t ∈ T : X(t) > Y (t) ≥ u, Y (t) = 0, Y (t) > 0},

and the rest terms on the right hand side of (9.2.12) are defined in (9.2.1).
Now we obtain the following mean Euler characteristic approximation.
Theorem 9.2.5 Let {(X(t), Y (t)) : t ∈ R} be an R2 -valued, centered, unit-variance Gaussian process satisfying (D1) and (D2), and let T = [a, b] be a closed finite interval in R.
Then there exists some constant α > 0 such that as u → ∞,

P{∃t ∈ T such that X(t) ≥ u, Y (t) ≥ u}
◦

= E{µ(X = Y, T )} + E{µ(X, a)} + E{µ(Y, a)} + E{µ(X, b)} + E{µ(Y, b)}
+ o exp

−

(9.2.13)

u2
− αu2
1 + ρ(T )

= E{ϕ(Au (T, X ∧ Y ))} + o exp

−

u2
− αu2
1 + ρ(T )

,

where ρ(T ) = supt∈T ρ(t).

Proof By Lemma 9.2.1, we can find the upper bound and lower bound, similarly to (2.3.1)
and (2.3.2) respectively, for the excursion probability. Applying Lemma 9.2.2, Lemma 9.2.3
248

and Lemma 9.2.4 yields the first equality in (9.2.13). The last line of (9.2.13) follows from
(9.2.12).

Remark 9.2.6 Based on the proof of Lemma 9.2.3 and Lemma 9.2.4, the term E{µ(X =
◦

◦

Y, T )} in the approximation (9.2.13) can be replaced by a simpler one E{µ (X = Y, T )},
where

◦

◦

µ (X = Y, T ) = #{t ∈ T : X(t) = Y (t) ≥ u}.

It follows from the Kac-Rice metatheorem that

b

◦

E{µ (X = Y, T )} =

a

E{|X (t) − Y (t)|1{X(t)≥u} |X(t) − Y (t) = 0}dt.

249

BIBLIOGRAPHY

250

BIBLIOGRAPHY
[1] R. J. Adler (1981), The Geometry of Random Fields. Wiley, New York.
[2] R. J. Adler (2000), On excursion sets, tube formulas and maxima of random fields.
Ann. Appl. Probab. 10, 1–74.
[3] R. J. Adler and J. E. Taylor (2007), Random Fields and Geometry. Springer, New
York.
[4] R. J. Adler and J. E. Taylor (2011), Topological complexity of smooth random func´
´ e
tions. Lecture Notes in Mathematics, 2019. Ecole d’Et´ de Probabilit´s de Saint-Flour.
e
Springer, Heidelberg.
[5] R. J. Adler, J. E. Taylor and K. J. Worsley (2012), Applications of Random Fields
and Geometry: Foundations and Case Studies. In preparation.
[6] A. B. Anshin (2006), On the probability of simultaneous extremes of two Gaussian
nonstationary processes. Theory Probab. Appl. 50, 353–366.
[7] M. Arendarczyk and K. Debicki (2011), Asymptotics of supremum distribution of a
Gaussian process over a Weibullian time. Bernoulli 17, 194–210.
[8] M. Arendarczyk and K. Debicki (2012), Exact asymptotics of supremum of a stationary Gaussian process over a random interval. Stat. Prob. Lett 82, 645–652.
[9] A. Auffinger (2011), Random Matrices, Complexity of Spin Glasses and Heavy Tailed
Processes. Ph.D. Thesis, New York University.
[10] J.-M. Aza¨ J.-M. Bardet and M. Wschebor (2002), On the tails of the distribution
ıs,
of the maximum of a smooth stationary Gaussian process. ESAIM Probab. Statist. 6,
177–184.
[11] J.-M. Aza¨ and C. Delmas (2002), Asymptotic expansions for the distribution of the
ıs
maximum of Gaussian random fields. Extremes. 5, 181–212.
[12] J.-M. Aza¨ and M. Wschebor (2005), On the distribution of the maximum of a
ıs
Gaussian field with d parameters. Ann. Appl. Probab. 15, 254–278.
251

[13] J.-M. Aza¨ and M. Wschebor (2008), A general expression for the distribution of the
ıs
maximum of a Gaussian field and the approximation of the tail. Stoch. Process. Appl.
118 (2008), 1190–1218.
[14] J.-M. Aza¨ and M. Wschebor (2009), Level Sets and Extrema of Random Processes
ıs
and Fields. John Wiley & Sons, Hoboken, NJ.
[15] K. Bartz, S. C. Kou and R. J. Adler (2011), Estimating thresholding levels for random
fields via Euler characteristics. Preprint.
[16] C. Borell (1975), The Brunn-Minkowski inequality in Gauss space. Invent. Math. 30,
205–216.
[17] N. Chamandy, K. J. Worsley, J. E. Taylor and F. Gosselin (2008), Tilted Euler characteristic densities for central limit random fields, with application to “bubbles”. Ann.
Statist. 36, 2471–2507.
[18] H. P. Chan and T. L. Lai (2006), Maxima of asymptotically Gaussian random fields
and moderate deviation approximations to boundary crossing probabilities of sums
of random variables with multidimensional indices. Ann. Probab. 34, 80–121.
[19] D. Cheng and Y. Xiao (2012), The mean Euler characteristic and excursion probability
of Gaussian random fields with stationary increments. arXiv:1211.6693.
[20] K. Debicki (2002), Ruin probability for Gaussian integrated processes. Stoch. Process.
Appl. 98, 151–174.
[21] K. Debicki, K. M. Kosi´ski, M. Mandjes, and T. Rolski (2010), Extremes of multidin
mensional Gaussian processes. Stoch. Process. Appl. 120, 2289–2301.
[22] A. B. Dieker (2006), Extremes of Gaussian processes over an infinite horizon. Stoch.
Process. Appl. 115, 207–248.
[23] T. Gneiting. (2012), Strictly and non-strictly positive definite functions on spheres.
arXiv:1111.7077v4.
[24] J. Husler and V. I. Piterbarg (1999), Extremes of a certain class of Gaussian processes.
Stoch. Process. Appl. 83, 257–271.
[25] S. G. Kobelkov (2005), The ruin problem for the stationary Gaussian processes. Theory Probab. Appl. 49, 155–163.
252

[26] T. J. Kozubowski, M. M. Meerschaert, F. J. Molz and S. Lu (2004), Fractional Laplace
model for hydraulic conductivity. Geophysical Res. Lett. 31, L08501.
[27] A. Ladneva and V. Piterbarg (2000), On double extremes of Gaussian stationary
processes. website: http://ladneva.stormpages.com/article.pdf
[28] T. J. Kozubowski, M. M. Meerschaert and K. Podgorski (2006), Fractional Laplace
motion. Adv. Appl. Probab. 38, 451–464.
[29] H. Landau and L. A. Shepp (1970), On the supremum of a Gaussian process. Sankya
32, 369–378.
[30] J. Liu (2012), Tail approximations of integrals of Gaussian random fields. Ann.
Probab. 40, 1069–1104.
[31] M. B. Marcus and L. A. Shepp (1970), Sample behaviour of Gaussian processes. Proceedings of the 6th Berkeley Symposium on Mathematics, Statistics and Probability,
Vol. 2, University of California Press, Berkeley, CA, 423–442.
[32] D. Marinucci and G. Peccati (2011), Random Fields on the Sphere. Representation,
Limit Theorems and Cosmological Applications. Cambridge University Press.
[33] D. Marinucci and S. Vadlamani (2013), High-frequency asymptotics for LipschitzKilling curvatures of excursion sets on the sphere. arXiv:1303.2456.
[34] G. Matheron (1973), The intrinsic random functions and their applications. Adv.
Appl. Probab. 5, 439–468.
[35] C. Mueller and R. Tribe (2002), Hitting properties of a random string. Electronic J.
Probab. 7, 1–29.
[36] Y. Nardi (2006), Gaussian, and Asymptotically Gaussian, Random Fields and Their
Maxima. Ph.D. dissertation, Hebrew Univ. of Jerusalem.
[37] Y. Nardi, D. O. Siegmund and B. Yakir (2008), The distribution of maxima of approximately Gaussian random fields. Ann. Statist. 36, 1375–1403.
[38] J. Pickands III (1969a), Asyptotic properties of the maximum in a stationary Gaussian
processes. Trans. Amer. Math. Soc. 145, 75–86.

253

[39] J. Pickands III (1969b), Upcrossing probabilities for stationary Gaussian processes.
Trans. Amer. Math. Soc. 145, 51–73.
[40] V. I. Piterbarg (1996a), Asymptotic Methods in the Theory of Gaussian Processes and
Fields. Translations of Mathematical Monographs, Vol. 148, American Mathematical
Society, Providence, RI.
[41] V. I. Piterbarg (1996b), Rice’s method for large excursions of Gaussian random fields.
Technical Report NO. 478, Center for Stochastic Processes, Univ. North Carolina.
[42] V. I. Piterbarg and S. Stamatovic (2001), On maximum of Gaussian non-centered
fields indexed on smooth manifolds. Asymptotic methods in probability and statistics
with applications (St. Petersburg, 1998), 189–203, Stat. Ind. Technol., Birkh¨user
a
Boston, Boston, MA.
[43] V. I. Piterbarg and S. Stamatovic (2005), Crude asymptotics of the probability of
simultaneous high extrema of two Gaussian processes: the dual action function. Russ.
Math. Surv. 60, 167–168.
[44] J. Potthoff (2010), Sample properties of random fields III: differentiability. Commun.
Stoch. Anal. 4, 335–353.
[45] C. Qualls and H. Watanabe (1973), Asymptotic properties of Gaussian random Fields.
Trans. Amer. Math. Soc. 177, 155–171.
[46] I. J. Schoenberg (1942), Positive definite functions on spheres. Duke Math. J. 9,
96–108.
[47] K. Shafie, B. Sigal, D. O. Siegmund and K. J. Worsley (2003), Rotation space random
fields with an application to fMRI data. Ann. Statist. 31, 1732–1771.
[48] M. L. Stein (1999), Interpolation of Spatial Data: Some Theory for Kriging. Springer,
New York.
[49] M. L. Stein (2012), On a class of space-time intrinsic random functions. Bernoulli, to
appear.
[50] J. Sun (1993), Tail Probabilities of the Maxima of Gaussian Random Fields. Ann.
Probab. 21, 34–71.

254

[51] G. Szeg¨ (1975), Othogonal Polynomials. American Mathematical Society, Provio
dence, RI.
[52] A. Takemura and S. Kurki (2002), On the equivalence of the tube and Euler characteristic methods for the distribution of the maximum of Gaussian fields over piecewise
smooth domains. Ann. Appl. Probab. 12, 768-796.
[53] J. E. Taylor and R. J. Adler (2003), Euler characteristics for Gaussian fields on
manifolds. Ann. Probab. 31, 533–563.
[54] J. E. Taylor, A. Takemura, and R. J. Adler (2005), Validity of the expected Euler
characteristic heuristic. Ann. Probab. 33, 1362–1396.
[55] J. E. Taylor and K. J. Worsley (2007), Detecting sparse signals in random fields, with
an application to brain mapping. J. Amer. Statist. Assoc. 102, 913–928.
[56] J. E. Taylor and K. J. Worsley (2008), Random fields of multivariate test statistics,
with applications to shape analysis. Ann. Statist. 36, 1–27.
[57] B. S. Tsirelson, I. A. Ibragimov and V. N. Sudakov (1976), Norms of Gaussian sample functions. Preceedings of the 3rd Japan-USSR Symposium on Probability Theory
(Tashkent, 1975), Lecture Notes in Mathematics, Vol. 550, Springer-Verlag, Berlin,
20–41.
[58] R. Wong (2001), Asymptotic Approximations of Integrals. SIAM, Philadelphia, PA.
[59] K. J. Worsley (1994), Local maxima and the expected Euler characteristic of excursion
sets of χ2 , F , and t fields. Adv. Appl. Probab. 26, 13–42.
[60] K. J. Worsley (1996), The geometry of random images. Chance 9, 27–39.
[61] Y. Xiao (2009), Sample path properties of anisotropic Gaussian random fields. In:
A Minicourse on Stochastic Partial Differential Equations, (D. Khoshnevisan and F.
Rassoul-Agha, editors), Lecture Notes in Math. 1962, pp. 145–212. Springer, New
York.
[62] Y. Xue and Y. Xiao (2011), Fractal and smoothness properties of space-time Gaussian
models. Frontiers Math. China 6, 1217–1246.
[63] A. M. Yaglom (1957), Some classes of random fields in n-dimensional space, related
to stationary random processes. Th. Probab. Appl. 2, 273–320.
255