# The volume-of-tube method for Gaussian random fields with inhomogeneous
variance

Satoshi Kuriki  The Institute of Statistical Mathematics, 10-3 Midoricho,
Tachikawa, Tokyo 190-8562, Japan<EMAIL_ADDRESS>,  Akimichi Takemura  The
Center for Data Science Education and Research, Shiga University, 1-1-1 Banba,
Hikone, Shiga 522-8522, Japan<EMAIL_ADDRESS>and  Jonathan
E. Taylor  Department of Statistics, Sequoia Hall, 390 Jane Stanford Way,
Stanford University, Stanford, CA 94305-4020, USA
<EMAIL_ADDRESS>
###### Abstract.

The tube method or the volume-of-tube method approximates the tail probability
of the maximum of a smooth Gaussian random field with zero mean and unit
variance. This method evaluates the volume of a spherical tube about the index
set, and then transforms it to the tail probability. In this study, we
generalize the tube method to a case in which the variance is not constant. We
provide the volume formula for a spherical tube with a non-constant radius in
terms of curvature tensors, and the tail probability formula of the maximum of
a Gaussian random field with inhomogeneous variance, as well as its Laplace
approximation. In particular, the critical radius of the tube is generalized
for evaluation of the asymptotic approximation error. As an example, we
discuss the approximation of the largest eigenvalue distribution of the
Wishart matrix with a non-identity matrix parameter. The Bonferroni method is
the tube method when the index set is a finite set. We provide the formula for
the asymptotic approximation error for the Bonferroni method when the variance
is not constant.

###### Key words and phrases:

Bonferroni method, Euler characteristic, Kac-Rice test, tail probability,
Weyl’s tube formula, Wishart matrix.

##  1\. Introduction: Tube with non-constant radius

Let $M\subset\mathbb{S}^{n-1}$ be a $d$-dimensional $C^{2}$-closed submanifold
of the ($n-1$)-dimensional unit sphere
$\mathbb{S}^{n-1}=S(\mathbb{R}^{n})=\\{u\in\mathbb{R}^{n}\mid\|u\|=1\\}$. We
assume that the linear-hull of $M$ is $\mathbb{R}^{n}$, and $M$ is a proper
subset of $\mathbb{S}^{n-1}$ (that is, $d<n-1$). Define a Gaussian random
field on $M$ by

(1.1) $X(u)=\sigma(u)\langle u,\xi\rangle,\quad u\in
M\subset\mathbb{S}^{n-1},$

where $\xi\in\mathbb{R}^{n}$ is a random vector distributed as the standard
Gaussian distribution $\mathcal{N}_{n}(0,I_{n})$, $\langle\cdot,\cdot\rangle$
denotes the standard inner product of $\mathbb{R}^{n}$, and $\sigma(\cdot)$ is
a positive $C^{2}$-function on $M$. Then, $X(\cdot)$ is a mean-zero Gaussian
random field on $M$ with variance $\operatorname{Var}(X(u))=\sigma(u)^{2}$ and
the correlation function $\operatorname{Corr}(X(u),X(v))=\langle u,v\rangle$.
(1.1) is the canonical representation of a Gaussian random field with a finite
Karhunen-Loève expansion and a smooth sample path.

The purpose of this study is to provide a formula for approximating the tail
probability of the maximum

(1.2) $\mathbb{P}\biggl{(}\max_{u\in M}X(u)>c\biggr{)}\quad\mbox{when $c$ is
large}.$

When the variance is constant, say, $\sigma(u)\equiv 1$, the recipes to
approximate the tail probability (1.2) are well established and known as the
volume-of-tube method, or simply, the tube method
([KS89],[JJ90],[Sun93],[KT01, KT09],[TK02, TK03]). The tail probability of
(1.2) corresponds to the $p$-value of a max-type statistical test and appears
in various statistical scenarios, such as construction of simultaneous
confidence bands, likelihood ratio test in singular models, change point
problems, order-restricted statistical inference, projection pursuit/testing
Gaussianity, and so on (e.g., [Nai86, Nai90],[Sun91],[KT08b],[KK13],[LK17]).
The tube method provides sufficiently accurate $p$-value formulas for such
problems.

In this study, we extend the tube formula to Gaussian random fields with
inhomogeneous variance. For later purposes, we define a standardized random
field on $M$ by

$Y(u)=\sigma(u)\bigl{\langle}u,\xi/\|\xi\|\bigr{\rangle},\ \ u\in M,\ \
\xi\sim\mathcal{N}_{n}(0,I_{n}).$

We sometimes use the abbreviations

$X_{\mathrm{max}}=\max_{u\in M}X(u),\quad Y_{\mathrm{max}}=\max_{u\in M}Y(u).$

Note that the support of $Y_{\mathrm{max}}$ is
$|Y_{\mathrm{max}}|\leq\max_{u\in M}\sigma(u)$. Because $\|\xi\|$ and
$\xi/\|\xi\|$ are independently distributed, we have

(1.3)
$\mathbb{P}\bigl{(}X_{\mathrm{max}}>c\bigr{)}=\mathbb{E}\biggl{[}\mathbb{P}\biggl{(}Y_{\mathrm{max}}>\frac{c}{\|\xi\|}\mid\|\xi\|\biggr{)}\biggr{]}.$

Moreover, because $\xi/\|\xi\|\sim\operatorname{Unif}(\mathbb{S}^{n-1})$, the
uniform distribution on $\mathbb{S}^{n-1}$,

(1.4) $\displaystyle\mathbb{P}\bigl{(}Y_{\mathrm{max}}>b\bigr{)}$
$\displaystyle=\mathbb{P}\biggl{(}\max_{u\in M}\sigma(u)\langle
u,\xi/\|\xi\|\rangle>b\biggr{)}$ $\displaystyle=\mathbb{P}\Bigl{(}\exists u\in
M,\,\langle u,\xi/\|\xi\|\rangle>b/\sigma(u)\Bigr{)}$
$\displaystyle=\frac{1}{\Omega_{n}}\operatorname{Vol}_{n-1}\bigl{(}(M)_{b/\sigma(\cdot)}\bigr{)},$

where

(1.5) $\displaystyle(M)_{f}=$
$\displaystyle\bigl{\\{}w\in\mathbb{S}^{n-1}\mid\exists u\in M,\,\langle
u,w\rangle>f(u)\bigr{\\}},$

$\operatorname{Vol}_{n-1}(\cdot)$ is the ($n-1$)-dimensional volume, and

(1.6)
$\Omega_{n}=\operatorname{Vol}_{n-1}(\mathbb{S}^{n-1})=\frac{2\pi^{n/2}}{\Gamma(n/2)}$

is the volume of ($n-1$)-dimensional unit sphere. Further, we observe that the
distribution of $X_{\mathrm{max}}$ is derived from that of $Y_{\mathrm{max}}$
in (1.4), which is proportional to the volume of the spherical set
$(M)_{b/\sigma(\cdot)}$ in (1.5). (It is shown that the relation between
$\mathbb{P}\bigl{(}Y_{\mathrm{max}}>\cdot\bigr{)}$ and
$\mathbb{P}\bigl{(}X_{\mathrm{max}}>\cdot\bigr{)}$, i.e., (1.3), is
essentially the Laplace/inverse Laplace transforms, and hence one-to-one.)

|  |  |  |
---|---|---|---|---
Figure 1. Tubes $(M)_{f}$.
(left: homogeneous variance, center and right: inhomogeneous variance)

Figure 1 depicts examples of $(M)_{f}$ with $f(\cdot)=b/\sigma(\cdot)$.
$(M)_{f}$ is a tube-like area about the set $M$. When $\sigma(u)$ is constant,
the width of $(M)_{f}$ is constant, and its volume can be evaluated using the
tube-volume formula by H. Hotelling [Hot39] and H. Weyl [Wey39]. The original
tube method for constant variance is used to evaluate the volume of (1.5) via
the tube-volume formula and thereafter transforms it into the distribution of
$X_{\mathrm{max}}$ by (1.4) (see [Sun93],[KT01]). As in the right panel of
Figure 1, the radius of tube may vanish. We call this phenomenon the
degeneracy of tube. We will see later the condition that such degeneration
occurs. In this study, we will extend Weyl’s tube-volume formula and derive
the distribution of $X_{\mathrm{max}}$ in a similar manner.

The tube treated here is spherical on the unit sphere. The volume of a tube
with non-constant radius when the ambient space is $\mathbb{R}^{n}$ is
discussed under different motivations (e.g., [Roc13],[Ros15]).

To approximate the tail probabilities of the maxima of random fields, R. Adler
and K. Worsley founded a methodology based on the Euler-Poincaré
characteristic of the excursion set ([AH76, Adl81],[Wor95]). Their method is
referred to as the expected Euler characteristic heuristic or the Euler
characteristic method. Later, this method has been developed and generalized
in theoretical and application aspects (e.g., [TA03, TA09], [Tay06], [TW08,
TW13]). For comprehensive treatments of this method, we refer to [AT07] and
[AT11]. [TK02] showed that the tube method and the Euler characteristic method
provide the same formula for the tail probability in (1.2) with
$\sigma(u)\equiv 1$.

As a related problem, [TLT16] formulated a test statistic for a global null
hypothesis in a regularized regression as the maximum of a Gaussian random
field with inhomogeneous variance:

(1.7) $\max_{\eta\in
K}\langle\eta,\xi\rangle,\quad\xi\sim\mathcal{N}_{n}(0,I_{n}),$

where $K$ is a closed, convex, stratified submanifold of $\mathbb{R}^{n}$. In
our notation, (1.7) is $\max_{u\in M}X(u)$ with $X(u)=\sigma(u)\langle
u,\xi\rangle$, $M=\\{\eta/\|\eta\|\mid\eta\in K\\}$ and $\sigma(u)=\|\eta\|$
for $u=\eta/\|\eta\|$. They derived the conditional distribution of the
maximum when $\eta^{*}=\operatorname{argmax}_{\eta\in
K}\langle\eta,\xi\rangle$ is given, and constructed a so-called Kac-Rice test
based on the conditional distribution. Differently from [TLT16], we treat the
unconditional distribution of the maximum. Another difference is the
assumption of convexity. In this paper, $M$ is a closed submanifold and not
(geodesically) convex. We will examine the effects of the non-convexity of the
set

(1.8) ${\sigma}M=\bigl{\\{}\sigma(u)u\in\mathbb{R}^{n}\mid u\in M\bigr{\\}}.$

It will be shown that only the supporting points of ${\sigma}M$ make
contributions to the volume of tube formula, and the degeneracy of the tube
(the right panel of Figure 1) occurs at the non-supporting points of
${\sigma}M$.

The outline of the paper is as follows. The main results are summarized in
Section 2. We present the approximate formulas for the tail probability of
$Y_{\mathrm{max}}$ and $X_{\mathrm{max}}$. To evaluate their approximation
error, we generalize the notion of the critical radius of a tube.
Consequently, based on the resulting formula, we provide its Laplace
approximation under the assumption that the set

(1.9) $M_{0}=\Bigl{\\{}u\in M\mid\sigma(u)=\max_{u\in M}\sigma(u)\Bigr{\\}}$

forms a closed $C^{2}$-manifold. In addition, we discuss the Bonferroni
method, which is the tube method when the index set is finite. We provide the
formula for the asymptotic approximation error for the Bonferroni method when
the variance is not constant. Examples are provided in Section 3. One of them
is the tail probability of the largest eigenvalue of the Wishart matrix when
the parameter matrix is not an identity matrix, which will be of independent
interest. All proofs for the main results are provided in Section 4.

##  2\. Main Results

###  2.1. Preliminary and notations

The index set $M$ is treated as a $C^{2}$-submanifold endowed with the metric
induced by the ambient space $\mathbb{R}^{n}$. Suppose that each point $u\in
M$ is represented by the local coordinates $t=(t^{i})_{1\leq i\leq d}$ as
$u=\varphi(t)$. We use conventions $\partial/\partial t^{i}=\partial_{i}$,
$\varphi_{i}=\partial_{i}\varphi$,
$\varphi_{ij}=\partial_{i}\partial_{j}\varphi$, etc. The tangent space of $M$
at $u\in M$ is

$T_{u}(M)=\operatorname{span}\\{\varphi_{i}\mid 1\leq i\leq d\\}.$

The metric tensor is $g_{ij}=\langle\varphi_{i},\varphi_{j}\rangle$, and the
affine connection is
$\Gamma_{ij,k}=\langle\varphi_{ij},\varphi_{k}\rangle=(1/2)(\partial_{i}g_{jk}+\partial_{j}g_{ik}-\partial_{k}g_{ij})$.
The volume element of $M$ at $u\in M$ is
$\det(G)^{\frac{1}{2}}\prod_{i}\mathrm{d}t^{i}|_{u}$, where $G=(g_{ij})_{1\leq
i,j\leq d}$ is the metric tensor in $d\times d$ matrix form. When needed, we
write $G=G_{u}$ to indicate the point $u\in M$ where the metric is defined.
The covariant derivative in the direction of $\partial/\partial t^{i}$ is
denoted by $\nabla_{i}$. The inverse of the metric matrix $G$ is denoted by
$G^{-1}=(g^{ij})_{1\leq i,j\leq d}$. Let

(2.1)
$R_{ij;kl}=\partial_{i}\Gamma_{jk,l}-\partial_{j}\Gamma_{ik,l}+\sum_{\alpha,\beta=1}^{d}(\Gamma_{ik,\alpha}\Gamma_{jl,\beta}-\Gamma_{il,\alpha}\Gamma_{jk,\beta})g^{\alpha\beta}$

be the curvature tensor.

Define the normal space of $M$ at $u$ by

(2.2) $N_{u}=T_{u}(M)^{\perp}\cap T_{u}(\mathbb{S}^{n-1}).$

Note that

$\mathbb{R}^{n}=T_{u}(\mathbb{R}^{n})=\operatorname{span}\\{u\\}\oplus
T_{u}(M)\oplus N_{u},$

where $\oplus$ is the orthogonal direct sum.

Let

(2.3) $\ell=\ell(u)=\log\sigma(u),$

and let

(2.4) $\displaystyle C=C_{u}=(c_{ij})_{1\leq i,j\leq d},$ $\displaystyle
c_{ij}=-\nabla_{i}\nabla_{j}\ell+\nabla_{i}\ell\nabla_{j}\ell=-\biggl{(}\partial_{i}\partial_{j}\ell-\sum_{k,l=1}^{d}\Gamma_{ij,k}\partial_{l}\ell
g^{kl}\biggr{)}+(\partial_{i}\ell)(\partial_{j}\ell).$

The gradient of $\ell(u)$ in (2.3) in the tangent space $T_{u}(M)$ is

$\nabla\ell=\nabla\ell(u)=\nabla|_{T_{u}(M)}\ell(u)=\sum_{i,j=1}^{d}\nabla_{i}\ell\varphi_{j}g^{ij}.$

The squared length of the gradient is
$\|\nabla\ell\|^{2}=\|\nabla\ell(u)\|^{2}=\sum_{i,j=1}^{d}(\nabla_{i}\ell)(\nabla_{j}\ell)g^{ij}$.

The $k$th elementary symmetric function of the eigenvalues of a matrix is
denoted by $\operatorname{tr}_{k}(\cdot)$ ([Mui82]) or
$\operatorname{detr}_{k}(\cdot)$ ([Wor95]). Note that
$\operatorname{tr}_{1}(\cdot)=\operatorname{tr}(\cdot)$, and
$\operatorname{tr}_{d}(A)=\det(A)$ if $A$ is $d\times d$, and let
$\operatorname{tr}_{0}(\cdot)=1$. We use the relation

$\det(I_{d}+A)=\sum_{k=0}^{d}\operatorname{tr}_{k}(A).$

For a finite set $I=\\{i_{1},\ldots,i_{n}\\}$ ($i_{1}<\cdots<i_{n}$), let

(2.5)
$\Sigma(I)=\\{(\pi_{1},\ldots,\pi_{n})\mid\\{\pi_{1},\ldots,\pi_{n}\\}=I\\}$

be the set of all permutations of elements of $I$. For $I$ such that $n=|I|$
is even, let

(2.6)
$\Pi(I)=\bigl{\\{}(\pi_{1},\ldots,\pi_{n})\in\Sigma(I)\mid\pi_{1}<\pi_{2},\,\pi_{3}<\pi_{4},\,\ldots,\pi_{n-1}<\pi_{n}\bigr{\\}}$

be the set of ordered parings of $I$. Let
$\Sigma_{n}=\Sigma(\\{1,\ldots,n\\})$ and $\Pi_{n}=\Pi(\\{1,\ldots,n\\})$.

###  2.2. Volume of tube formula

To state main theorems, we introduce the notion of self-overlap of tube.
Figure 1 depicts tube $(M)_{b/\sigma(\cdot)}$. Suppose we let $b$ be small.
Subsequently, the width of the tube becomes large, and therefore causing self-
overlap. More formally, $(M)_{b/\sigma(\cdot)}$ is said to have a self-overlap
if there exists a point $w\in(M)_{b/\sigma(\cdot)}$ such that the function
$u\mapsto\sigma(u)\langle u,w\rangle$ has two local maxima at the points
$u^{*},u^{**}\in M$ and $\sigma(u^{*})\langle
u^{*},w\rangle=\sigma(u^{**})\langle u^{**},w\rangle$. The threshold
$b=b_{\mathrm{cri}}$ causing such overlap is defined by

(2.7) $b_{\mathrm{cri}}=\sup\bigl{\\{}b\geq 0\mid(M)_{b/\sigma(\cdot)}\mbox{
has a self-overlap}\bigr{\\}}.$

For $x\geq 0$, let

$\bar{B}_{a,b}(x)=\begin{cases}\displaystyle\frac{\Gamma(a+b)}{\Gamma(a)\,\Gamma(b)}\int_{x}^{1}t^{a-1}(1-t)^{b-1}\,\mathrm{d}t&(0\leq
x\leq 1),\\\ 0&(x>1),\end{cases}$

be the upper probability of the beta distribution with parameter $(a,b)$. Now,
we state the following main results.

######  Theorem 2.1 (Volume of tube).

Let $\mathrm{d}u$ be the volume element of $M$ at $u\in M$. Let
$b_{\mathrm{cri}}$ be the threshold defined in (2.7). For $b\geq
b_{\mathrm{cri}}$,

(2.8) $\displaystyle\mathbb{P}\bigl{(}Y_{\mathrm{max}}>b\bigr{)}=$
$\displaystyle\frac{1}{\Omega_{n}}\operatorname{Vol}_{n-1}\bigl{(}(M)_{b/\sigma(\cdot)}\bigr{)}$
$\displaystyle=$
$\displaystyle\sum_{e=0}^{d}\frac{1}{(2\pi)^{e/2}\Omega_{d-e+1}}\int_{M}\frac{\det(I_{d}+C_{u}G_{u}^{-1})}{(1+\|\nabla\ell(u)\|^{2})^{\frac{1}{2}(d-e+1)}}$
$\displaystyle\qquad\qquad\times\bar{B}_{\frac{1}{2}(d-e+1),\frac{1}{2}(n-d+e-1)}\biggl{(}\frac{1+\|\nabla\ell(u)\|^{2}}{\sigma(u)^{2}}b^{2}\biggr{)}\,\zeta_{e}(u)\,\mathrm{d}u,$

where $\Omega_{k}$ is defined in (1.6), $G_{u}$ is the metric matrix at $u$,
$C_{u}$ is defined in (2.4),

(2.9)
$\zeta_{e}(u)=\begin{cases}\displaystyle\sum_{I\subset\\{1,\ldots,d\\},\,|I|=e}\frac{1}{(e/2)!}\sum_{\pi,\tau\in\Pi(I)}\operatorname{sgn}(\pi,\tau)\,\widetilde{R}_{\pi_{1}\pi_{2}}^{\tau_{1}\tau_{2}}\cdots\widetilde{R}_{\pi_{e-1}\pi_{e}}^{\tau_{e-1}\tau_{e}}\,&(\mbox{if
$e$ is even}),\\\ 0&(\mbox{if $e$ is odd}),\end{cases}$

with

(2.10)
$\widetilde{R}_{ij}^{kl}=\widetilde{R}_{ij}^{kl}(u)=\sum_{\alpha,\beta=1}^{d}\Bigl{(}R_{ij;\alpha\beta}-\bigl{(}g_{i\alpha}g_{j\beta}-g_{i\beta}g_{j\alpha}\bigr{)}\Bigr{)}\widetilde{g}^{\alpha
k}\widetilde{g}^{\beta l},$

$\widetilde{g}^{ij}$ the $(i,j)$th element of the inverse matrix
$\widetilde{G}_{u}^{-1}$ of $\widetilde{G}_{u}=G_{u}+C_{u}$,

$\operatorname{sgn}(\pi,\tau)=\operatorname{sgn}\begin{pmatrix}\pi_{1}&\cdots&\pi_{e}\\\
\tau_{1}&\cdots&\tau_{e}\end{pmatrix}$

the sign of permutation. In particular, $\zeta_{0}(u)=1$ and
$\zeta_{2}(u)=\frac{1}{2}\sum_{i,j=1}^{d}\widetilde{R}_{ij}^{ij}(u)$.

Moreover, (2.8) holds even if the range $M$ of the integral is replaced with

(2.11) $M_{\mathrm{cri}}=\biggl{\\{}u\in
M\mid\frac{1+\|\nabla\ell(u)\|^{2}}{\sigma(u)^{2}}b_{\mathrm{cri}}^{2}<1\biggr{\\}}.$

######  Remark 2.2 (Weyl’s tube formula).

When $\sigma(u)\equiv 1$, (2.8) becomes

$\displaystyle\mathbb{P}\bigl{(}Y_{\mathrm{max}}>b\bigr{)}=$
$\displaystyle\frac{1}{\Omega_{n}}\operatorname{Vol}_{n-1}\bigl{(}(M)_{b}\bigr{)}$
$\displaystyle=$
$\displaystyle\sum_{e=0}^{d}\frac{1}{(2\pi)^{e/2}\Omega_{d-e+1}}\int_{M}\,\zeta_{e}(u)\,\mathrm{d}u\times\bar{B}_{\frac{1}{2}(d-e+1),\frac{1}{2}(n-d+e-1)}\bigl{(}b^{2}\bigr{)},$

where $\zeta_{e}(u)$ is defined by (2.9) with

$\widetilde{R}_{ij}^{kl}=\sum_{\alpha,\beta=1}^{d}R_{ij;\alpha\beta}g^{\alpha
k}g^{\beta
l}-\bigl{(}\delta_{i}^{k}\delta_{j}^{l}-\delta_{i}^{l}\delta_{j}^{k}\bigr{)},\qquad\delta_{i}^{j}=\begin{cases}1&(i=j),\\\
0&(i\neq j).\end{cases}$

This is original Weyl’s tube formula [Wey39].

The theorem below characterizes the threshold $b_{\mathrm{cri}}$. This is also
useful for numerical calculation (Section 3).

######  Theorem 2.3.

The threshold $b_{\mathrm{cri}}$ appearing in Theorem 2.1 is

(2.12) $b_{\mathrm{cri}}^{2}=\sup_{u\in
M}b_{\mathrm{cri}}^{\prime}(u)^{2},\qquad
b_{\mathrm{cri}}^{\prime}(u)^{2}=\sup_{w\in
M\setminus\\{u\\}}\frac{\sigma(u)^{2}}{1+h(u,w)},$

where

(2.13)
$h(u,w)=\|\nabla\ell(u)\|^{2}+\biggl{(}\frac{\\{\sigma(u)/\sigma(w)-\langle
w,u-\nabla\ell(u)\rangle\\}_{+}}{\|P_{u}^{\perp}w\|}\biggr{)}^{2}.$

Here $x_{+}=\max\\{x,0\\}$, and $P_{u}^{\perp}$ is the orthogonal projection
in $\mathbb{R}^{n}$ onto the subspace $N_{u}$ defined in (2.2). That is,

$\|P_{u}^{\perp}w\|^{2}=1-\langle
u,w\rangle^{2}-\sum_{i,j=1}^{d}\langle\varphi_{i},w\rangle\langle\varphi_{j},w\rangle
g^{ij}.$

######  Remark 2.4 (Critical radius).

When $\sigma(u)\equiv 1$, (2.12) becomes

$b_{\mathrm{cri}}=\cos\theta_{\mathrm{cri}}=\sup_{u\in
M}\cos\theta_{\mathrm{cri}}^{\prime}(u),\quad\mbox{where}\ \
\tan\theta_{\mathrm{cri}}^{\prime}(u)=\inf_{w\in
M\setminus\\{u\\}}\frac{1-\langle w,u\rangle}{\|P_{u}^{\perp}w\|}.$

$\theta_{\mathrm{cri}}$ and $\theta_{\mathrm{cri}}^{\prime}(u)$ are referred
to as the global and local critical radius of the tube about $M$,
respectively. They are positive when $M$ is a $C^{2}$-closed manifold (e.g.,
[KT01, Remark 3.1]). We generalize the local critical radius
$\theta_{\mathrm{cri}}^{\prime}(u)$ to the inhomogeneous variance case by

$\tan^{2}\theta_{\mathrm{cri}}^{\prime}(u)=\inf_{w\in
M\setminus\\{u\\}}h(u,w).$

Then,

$b_{\mathrm{cri}}^{2}=\sup_{u\in
M}\bigl{[}\sigma(u)^{2}\cos^{2}\theta_{\mathrm{cri}}^{\prime}(u)\bigr{]}.$

In the formula (2.8), if $b$ is larger than $\sigma_{0}=\max_{u\in
M}\sigma(u)$, then the formula holds trivially as $0=0$. Therefore, Theorem
2.1 is meaningless unless $b_{\mathrm{cri}}<\sigma_{0}$ holds as follows.

######  Proposition 2.5.

The threshold $b_{\mathrm{cri}}$ in Theorem 2.1 satisfies

$b_{\mathrm{cri}}<\sigma_{0}=\max_{u\in M}\sigma(u).$

From Proposition 2.5, we immediately have $M_{0}\subset M_{\mathrm{cri}}$,
where $M_{0}$ and $M_{\mathrm{cri}}$ are defined in (1.9) and (2.11),
respectively.

###  2.3. Supporting points of ${\sigma}M$

In this subsection, we focus on the shape of ${\sigma}M$ in (1.8). Define a
half space and its boundary in $\mathbb{R}^{n}$ by

$\displaystyle L_{u}$ $\displaystyle=\\{z\in\mathbb{R}^{n}\mid\langle
z,u-\nabla\ell(u)\rangle\leq\sigma(u)\\},$ $\displaystyle\partial L_{u}$
$\displaystyle=\\{z\in\mathbb{R}^{n}\mid\langle
z,u-\nabla\ell(u)\rangle=\sigma(u)\\}.$

It is easy to see that

$\displaystyle\langle\sigma(u)u,u-\nabla\ell(u)\rangle=\sigma(u),\quad\biggl{\langle}\frac{\partial}{\partial
t^{i}}\bigl{[}\sigma(u)u\bigr{]},u-\nabla\ell(u)\biggr{\rangle}=0,$

that is, the hyperplane $\partial L_{u}$ is tangent to ${\sigma}M$ at
$\sigma(u)u$. Let

$\displaystyle M_{\mathrm{supp}}=\bigl{\\{}u\in M\mid\partial L_{u}\mbox{ is a
supporting hyperplane of }{\sigma}M\mbox{ with
}L_{u}\supset{\sigma}M\bigr{\\}}$

be the set of supporting points.

######  Proposition 2.6.

(i) $u\in M_{\mathrm{supp}}\iff\sigma(u)/\sigma(w)-\langle
w,u-\nabla\ell(u)\rangle\geq 0,\,\forall w\in M$.

(ii) $M_{\mathrm{supp}}\supset M_{\mathrm{cri}}$.

(iii) When $u\notin M_{\mathrm{supp}}$, the tube $(M)_{b/\sigma(\cdot)}$,
$b\geq b_{\mathrm{cri}}^{\prime}(u)$, is degenerated at $u$.

(i) means that, when $u\in M_{\mathrm{supp}}$, the argument of the truncation
function $\\{\cdot\\}_{+}$ in (2.13) is always nonnegative. (ii) implies that
the non-supporting points do not make contribution to the tube-volume formula
(2.8). The inclusion in (ii) is strict in general.

###  2.4. Tail probability of the maximum

Let

$\bar{G}_{\nu}(x)=\frac{1}{\Gamma(\nu/2)}\int_{x}^{\infty}t^{\nu/2-1}e^{-t/2}\,\mathrm{d}t$

be the upper probability of the chi-square distribution $\chi^{2}_{\nu}$ with
$\nu$ degrees of freedom. Let

$\bar{\Phi}(x)=\frac{1}{(2\pi)^{1/2}}\int_{x}^{\infty}e^{-t^{2}/2}\,\mathrm{d}t$

be the upper probability of the standard normal distribution
$\mathcal{N}(0,1)$.

######  Theorem 2.7 (Tube method).

Let $G$, $C$, $\mathrm{d}u$, and $b_{\mathrm{cri}}$ be defined as in Theorem
2.1. Then,

(2.14)
$\mathbb{P}\bigl{(}X_{\mathrm{max}}>c\bigr{)}=\mathbb{P}_{\mathrm{tube}}\bigl{(}X_{\mathrm{max}}>c\bigr{)}+O(\bar{G}_{n}(c^{2}/b_{\mathrm{cri}}^{2})),\quad
c\to\infty,$

where

(2.15)
$\displaystyle\mathbb{P}_{\mathrm{tube}}\bigl{(}X_{\mathrm{max}}>c\bigr{)}=\sum_{e=0}^{d}\frac{1}{(2\pi)^{e/2}\Omega_{d-e+1}}\int_{M}\frac{\det(I_{d}+C_{u}G_{u}^{-1})}{(1+\|\nabla\ell(u)\|^{2})^{\frac{1}{2}(d-e+1)}}$
$\displaystyle\times\bar{G}_{d-e+1}\biggl{(}\frac{1+\|\nabla\ell(u)\|^{2}}{\sigma(u)^{2}}c^{2}\biggr{)}\,\zeta_{e}(u)\,\mathrm{d}u.$

The range $M$ of the integral in (2.15) can be replaced with
$M_{\mathrm{cri}}$ in (2.11).

Each term of the tube formula (2.15) is of the order of
$\bar{G}_{\nu}(c^{2}/\sigma_{0}^{2})\asymp
c^{\nu-2}e^{-c^{2}/2\sigma_{0}^{2}}$, $\sigma_{0}=\max_{u\in M}\sigma(u)$,
whereas the order of the remainder term is
$\bar{G}_{n}(c^{2}/b_{\mathrm{cri}}^{2})\asymp
c^{n-2}e^{-c^{2}/2b_{\mathrm{cri}}^{2}}$. Here $f(c)\asymp g(c)$ denotes that
both $|f(c)/g(c)|$ and $|f(c)/g(c)|$ are bounded when $c\to\infty$. We use the
notation $f(c)\sim g(c)$ if $f(c)/g(c)\to 1$ as $c\to\infty$.

Proposition 2.5 affirms that the remainder term of the approximation (2.14) is
exponentially smaller than that of the main part. That is,

(2.16)
$\lim_{c\to\infty}\frac{1}{c^{2}}\log\biggl{|}\frac{\mathbb{P}_{\mathrm{tube}}\bigl{(}X_{\mathrm{max}}>c\bigr{)}}{\mathbb{P}\bigl{(}X_{\mathrm{max}}>c\bigr{)}}-1\biggr{|}=-\frac{1}{2}\biggl{(}\frac{1}{b_{\mathrm{cri}}^{2}}-\frac{1}{\sigma_{0}^{2}}\biggr{)}<0.$

######  Remark 2.8.

[TTA05, Theorem 4.3] proved that (2.16) holds when $\sigma(u)\equiv 1$ and
$X(u)$ does not necessarily have a finite Karhunen-Loève expansion, i.e., $n$
may be infinite. In other words, the remainder term
$O(\bar{G}_{n}(c^{2}/b_{\mathrm{cri}}^{2}))$ in (2.14) does not depend on $n$.
For this purpose, they described the threshold $b_{\mathrm{cri}}$ in terms of
$X(\cdot)$ irrespective of the dimension $n$ of the ambient space.

In the inhomegeneous variance case, by introducing a unit-variance Gaussian
random field

$\check{X}(u)=\frac{1}{\sigma(u)}X(u),\quad u\in M,$

$h(u,w)$ in (2.13) and hence $b_{\mathrm{cri}}$ in (2.12) can be written in
terms of $X(\cdot)$ by substituting

$\displaystyle\langle w,u\rangle=$
$\displaystyle\operatorname{Cov}\bigl{(}\check{X}(w),\check{X}(u)\bigr{)},$
$\displaystyle\langle w,\nabla\ell(u)\rangle=$
$\displaystyle\sum_{i,j=1}^{d}\operatorname{Cov}\bigl{(}\check{X}(w),\partial_{i}\check{X}(u)\bigr{)}\partial_{j}\ell(u)g^{ij}(u),\quad\partial_{i}=\partial/\partial
t_{i},$ $\displaystyle\|P_{u}^{\perp}w\|^{2}=$
$\displaystyle\operatorname{Var}\Bigl{(}\check{X}(w){\,|\,}\check{X}(u),\bigl{(}\partial_{i}\check{X}(u)\bigr{)}_{1\leq
i\leq d}\Bigr{)}$

into (2.13), where $(t^{i})_{1\leq i\leq d}$ is the local coordinates of $M$
around $u$. We conjecture that (2.16) holds even when $X(u)$ does not
necessarily have a finite Karhunen-Loève expansion.

In (2.15), the term of $e=0$ is the leading term when $c\to\infty$. In the
case where $\sigma(u)$ is not constant, this leading term can again be
approximated using Laplace’s method ([Erd56]).

######  Theorem 2.9 (Laplace approximation).

Assume that $\sigma(u)$ takes its maximum $\sigma_{0}$ on the
$d_{0}$-dimensional closed manifold in (1.9). Let $\mathrm{d}u_{0}(u)$ be the
volume element of $M_{0}$ at $u$. Let $G_{u}$ and $C_{u}$ be defined in
Theorem 2.1. Then,

(2.17)
$\displaystyle\mathbb{P}\bigl{(}X_{\mathrm{max}}>c\bigr{)}\sim\frac{1}{\Omega_{d_{0}+1}}\int_{M_{0}}\frac{\det(I_{d}+C_{u}G_{u}^{-1})^{\frac{1}{2}}}{\operatorname{tr}_{d-d_{0}}(C_{u}G_{u}^{-1})^{\frac{1}{2}}}\mathrm{d}u_{0}(u)\times\bar{G}_{d_{0}+1}(c^{2}/\sigma_{0}^{2}),\quad
c\to\infty.$

######  Corollary 2.10.

Assume that $\sigma(u)$ takes the unique maximum $\sigma_{0}$ at $u_{0}$. Let
$G_{0}=G_{u_{0}}$, $C_{0}=C_{u_{0}}$. Then,

$\mathbb{P}\bigl{(}X_{\mathrm{max}}>c\bigr{)}\sim\frac{\det(I_{d}+C_{0}G_{0}^{-1})^{\frac{1}{2}}}{\det(C_{0}G_{0}^{-1})^{\frac{1}{2}}}\,\bar{\Phi}(c/\sigma_{0}),\quad
c\to\infty.$

Corollary 2.10 is a special case of [Pit96, Theorem 8.2].

###  2.5. Bonferroni method as the tube method

So far we assumed that the index set $M$ is a manifold. In this subsection, we
consider the case where $M$ is a finite set.

Let $(X_{1},\ldots,X_{K})$ be a Gaussian random vector with moment structure
$\mathbb{E}[X_{i}]=0$, $\operatorname{Var}(X_{i})=\sigma_{i}^{2}$,
$\operatorname{Corr}(X_{i},X_{j})=\rho_{ij}$, $|\rho_{ij}|<1$. We consider the
distribution of the maximum

$X_{\mathrm{max}}=\max\\{X_{1},\ldots,X_{K}\\}.$

Such $X_{i}$’s are realized by

$X_{i}=\sigma_{i}\langle\varphi^{(i)},\xi\rangle,\quad\xi\sim\mathcal{N}_{n}(0,I_{n}),$

where $\varphi^{(i)}$’s are unit vectors such that
$\rho_{ij}=\langle\varphi^{(i)},\varphi^{(j)}\rangle$. $n$ is at most $K$.
$X_{\mathrm{max}}$ is the maximum of the random field $X(u)$, $u\in M$, in
(1.1) with

$M=\bigl{\\{}\varphi^{(i)}\in\mathbb{S}^{n-1}\mid 1\leq i\leq K\bigr{\\}}.$

Define

$Y_{i}=\sigma_{i}\langle\varphi^{(i)},\xi/\|\xi\|\rangle,\quad
Y_{\mathrm{max}}=\max\\{Y_{1},\ldots,Y_{K}\\}.$

The tube formula as well as its remainder order are given as a byproduct of
the previous sections.

######  Theorem 2.11.

(2.18)
$\mathbb{P}\bigl{(}Y_{\mathrm{max}}>b\bigr{)}=\sum_{i=1}^{K}\frac{1}{2}\bar{B}_{\frac{1}{2},\frac{1}{2}(n-1)}(b^{2}/\sigma_{i}^{2}),\quad
b\geq b_{\mathrm{cri}},$

and

(2.19)
$\mathbb{P}\bigl{(}X_{\mathrm{max}}>c\bigr{)}=\sum_{i=1}^{K}\bar{\Phi}(c/\sigma_{i})+\bar{G}_{n}(c^{2}/b_{\mathrm{cri}}^{2}),\quad
c\to\infty,$

where

(2.20) $b_{\mathrm{cri}}^{2}=\max_{i\neq
j}\frac{\sigma_{i}^{2}}{1+h(i,j)},\quad
h(i,j)=\frac{(\sigma_{i}/\sigma_{j}-\rho_{ij})_{+}^{2}}{1-\rho_{ij}^{2}}.$

(2.18) and (2.19) hold even if the summation $\sum_{i=1}^{K}$ is replaced with
$\sum_{i:\sigma_{i}>b_{\mathrm{cri}}}$.

The tube formula when the index set $M$ is finite is the Bonferroni method.
The next proposition corresponds to Proposition 2.5. Because
$\bar{\Phi}(c/\sigma_{i})\asymp c^{-1}e^{-c^{2}/2\sigma_{i}^{2}}$ and
$\bar{G}_{n}(c^{2}/b_{\mathrm{cri}}^{2})\asymp
c^{n-2}e^{-c^{2}/2b_{\mathrm{cri}}^{2}}$, Proposition 2.12 affirms that the
approximation error in (2.19) is asymptotically smaller than the leading term.

######  Proposition 2.12.

The threshold $b_{\mathrm{cri}}$ in Theorem 2.11 satisfies

(2.21) $b_{\mathrm{cri}}\leq\max_{1\leq i<j\leq
K}\sqrt{\frac{1+\rho_{ij}}{2}}\,\sigma_{0}<\sigma_{0},\ \
\sigma_{0}=\max_{1\leq i\leq K}\sigma_{i}.$

The factor $\max_{i<j}\sqrt{(1+\rho_{ij})/2}$ in (2.21) is
$b_{\mathrm{cri}}=\cos\theta_{\mathrm{cri}}$ when $\sigma_{i}\equiv 1$ (see
Remark 2.4), and is strictly less than 1.

######  Example 2.13.

Suppose that $K=3$, $(\sigma_{1},\sigma_{2},\sigma_{3})=(2,1,3)$,
$\rho_{12}=\rho_{23}=1/\sqrt{2}$, $\rho_{13}=1/2$. We have
$b_{\mathrm{cri}}=3\sqrt{3/7}=1.964$, which is less than

$\max_{i<j}\sqrt{\frac{1+\rho_{ij}}{2}}\max_{i}\sigma_{i}=\sqrt{\frac{1+1/\sqrt{2}}{2}}\times
3=2.772$

as proved in Proposition 2.12. When $\sigma_{i}\leq b_{\mathrm{cri}}$,
$\mathbb{P}(Y_{i}>b)\equiv 0$ for $b\geq b_{\mathrm{cri}}$, which makes no
contribution to $\mathbb{P}\bigl{(}Y_{\mathrm{max}}>b\bigr{)}$ in (2.18). In
this example, $\sigma_{2}\leq b_{\mathrm{cri}}$.

##  3\. Examples

###  3.1. Maximum of a Gaussian random process on a circle

We identify $\operatorname{Sym}(2)$, the set of $2\times 2$ real-symmetric
matrices, with $\mathbb{R}^{3}$ by the correspondence

$z=(z_{1},z_{2},z_{3})\ \leftrightarrow\
Z=\begin{pmatrix}z_{1}&z_{2}/\sqrt{2}\\\
z_{2}/\sqrt{2}&z_{3}\end{pmatrix},\qquad\|z\|_{\mathbb{R}^{3}}^{2}=\|Z\|_{\operatorname{Sym}(2)}^{2}=\operatorname{tr}\bigl{(}Z^{2}\bigr{)}.$

Let

$M=\\{\varphi(t)=h(t)h(t)^{\top}\mid
t\in[0,\pi)\\}\subset\operatorname{Sym}(2),\quad h(t)=(\cos t,\sin t)^{\top}.$

Because
$\|\varphi(t)\|_{\operatorname{Sym}(2)}^{2}=\operatorname{tr}\bigl{(}(h(t)h(t)^{\top})^{2}\bigr{)}=\|h(t)\|^{4}=1$
and $\varphi(0)=\varphi(\pi-)$, $M$ is a closed curve in the unit sphere of
$\operatorname{Sym}(2)$. Indeed, $M$ is a circle with radius $1/\sqrt{2}$. We
define a Gaussian random process on $[0,\pi)$ by

$X(t)=\sigma(t)\operatorname{tr}\bigl{[}h(t)h(t)^{\top}\Xi\bigr{]}=\sigma(t)h(t)^{\top}\Xi
h(t),\quad\Xi=\begin{pmatrix}\xi_{1}&\xi_{2}/\sqrt{2}\\\
\xi_{2}/\sqrt{2}&\xi_{3}\end{pmatrix},$

where $\xi_{1},\xi_{2},\xi_{3}$ are i.i.d. random variables from
$\mathcal{N}(0,1)$, and

$\sigma(t)=e^{\ell(t)}=e^{-m\sin^{2}t},\quad m\geq 0.$

Because
$\dot{\varphi}=\mathrm{d}\varphi/\mathrm{d}t=\dot{h}h^{\top}+h\dot{h}^{\top}$,
$\dot{h}=\mathrm{d}h/\mathrm{d}t=(-\sin t,\cos t)^{\top}$, the metric is
$g=\|\dot{\varphi}\|^{2}=2$, and the volume element of $M$ is
$\sqrt{2}\,\mathrm{d}t$. The first and second derivatives of
$\ell=-m\sin^{2}t$ are $\dot{\ell}=\mathrm{d}\ell/\mathrm{d}t=-m\sin 2t$ and
$\ddot{\ell}=\mathrm{d}^{2}\ell/\mathrm{d}t^{2}=-2m\cos 2t$, respectively. The
maximum of the variance is $1$ attained at $t=0$.

The tail probability of the maximum $X_{\mathrm{max}}=\max_{t\in[0,\pi)}X(t)$
is approximated by the tube method (2.15) as

$\mathbb{P}\bigl{(}X_{\mathrm{max}}>c\bigr{)}=\int_{0}^{\pi}\frac{1+(-\ddot{\ell}+(\dot{\ell})^{2})/2}{2\pi\,(1+(\dot{\ell})^{2}/2)}\,\bar{G}_{2}\biggl{(}\frac{1+(\dot{\ell})^{2}/2}{\sigma^{2}}c^{2}\biggr{)}\,\sqrt{2}\,\mathrm{d}t+O(\bar{G}_{3}\bigl{(}c^{2}/b_{\mathrm{cri}}^{2})\bigr{)},\
\ c\to\infty,$

and its Laplace approximation is

$\mathbb{P}\bigl{(}X_{\mathrm{max}}>c\bigr{)}\sim\sqrt{\frac{1-\ddot{\ell}(0)/2}{-\ddot{\ell}(0)/2}}\,\bar{\Phi}(c)=\sqrt{\frac{1+m}{m}}\,\bar{\Phi}(c),\quad
c\to\infty.$

The threshold $b_{\mathrm{cri}}$ is calculated as follows. The bases of the
tangent space and the normal space are given by
$\dot{\varphi}=\dot{h}h^{\top}+h\dot{h}^{\top}$ and $v=\dot{h}\dot{h}^{\top}$,
respectively. A simple calculation yields

$\psi(t)=\varphi(t)-\frac{1}{2}\dot{\ell}(t)\dot{\varphi}(t),$
$\displaystyle\frac{\sigma(t)}{\sigma(s)}-\langle s,\psi(t)\rangle=$
$\displaystyle
e^{\ell(t)-\ell(s)}-(h(t)^{\top}h(s))^{2}+\dot{\ell}(t)\dot{h}(t)^{\top}h(s)h(t)^{\top}h(s)$
$\displaystyle=$ $\displaystyle
e^{-m(\sin^{2}t-\sin^{2}s)}-\cos^{2}(s-t)-m\sin(2t)\sin(s-t)\cos(s-t),$
$\|P_{t}^{\perp}s\|=|v(t)^{\top}\varphi(s)|=(\dot{h}(t)^{\top}h(s))^{2}=\sin^{2}(s-t),$
$\displaystyle h(t,s)=$ $\displaystyle\frac{1}{2}m^{2}\sin^{2}(2t)$
$\displaystyle+\biggl{(}\frac{\\{e^{-m(\sin^{2}t-\sin^{2}s)}-\cos^{2}(s-t)-m\sin(2t)\sin(s-t)\cos(s-t)\\}_{+}}{\sin^{2}(s-t)}\biggr{)}^{2}\biggr{\\}}.$

Using these quantities, the threshold $b_{\mathrm{cri}}$ is obtained as

(3.1) $b_{\mathrm{cri}}^{2}=\sup_{t\neq s}\frac{\sigma(t)^{2}}{1+h(t,s)}.$

The objective function to be maximized in (3.1) contains the truncation
function $\\{\cdot\\}_{+}$. The set of $t$ where the truncation is active is

$\\{t\in[0,\pi)\mid\sigma(t)/\sigma(s)-\langle s,\psi(t)\rangle<0,\ \exists
s\in[0,\pi)\\},$

which is $\emptyset$ ($0\leq m<1$), $\\{0.5\pi\\}$ ($m=1$),
$[0.431\pi,0.568\pi]$ ($m=1.1$), $[0.366\pi,0.633\pi]$ ($m=1.5$),
$[0.266\pi,0.716\pi]$ ($m=10$). Therefore, we should include the truncation
when $m\geq 1$. Taking this into observation, we conducted the maximization in
(3.1). Numerical calculation suggests that the supremum in (3.1) is attained
when $t,s\to 0$. That is, it is conjectured that

(3.2) $b_{\mathrm{cri}}^{2}=\limsup_{t,s\to 0,\,t\neq
s}\frac{\sigma(t)^{2}}{1+h(t,s)}=\frac{1}{1+(1+m)^{2}}.$

This is strictly less than $\max_{t\in M}\sigma(t)^{2}=\sigma(0)^{2}=1$.

Figure 2 shows approximations to the tail probability of $X_{\mathrm{max}}$
using Monte Carlo simulations (10,000 replications), the tube method, and the
Laplace approximation. The tube method provides accurate approximations in all
configurations, but looks more accurate when $m$ is large. This observation is
consistent with (3.2). The Laplace approximation works well when $m$ is large.

|
---|---
$m=0$ | $m=1/16$
|
$m=1/4$ | $m=3/2$
Figure 2. Upper-tail probability of $\max_{t}X(t)$.
—: Simulation, $\bm{\cdots}$: Tube method, $\bm{--}$: Laplace approximation.

###  3.2. Largest eigenvalue of a Wishart matrix: Laplace approximation

The distribution of a $p\times p$ Wishart matrix with $\nu$ degrees of freedom
and matrix parameter $\Lambda$ is denoted by $\mathcal{W}_{p}(\nu,\Lambda)$.
The largest eigenvalue of the Wishart matrix $\mathcal{W}_{p}(\nu,\Lambda)$
can be formulated as the maximum of a Gaussian random field. Hence, its tail
probability can be approximated by the tube method. When the matrix parameter
$\Lambda$ is the identity matrix, the tube formula and its approximation error
are studied in [KT01, KT08a]. Here we derive the Laplace approximation formula
for the tail probability when $\Lambda$ is not proportional to the identity
matrix.

Without loss of generality, we assume

$\Lambda=\operatorname{diag}(\lambda_{1},\ldots,\lambda_{p}),\quad\lambda_{1}=\cdots=\lambda_{q}>\lambda_{q+1}\geq\cdots\geq\lambda_{p}>0.$

The largest eigenvalue $\ell_{1}$ of the Wishart matrix
$\mathcal{W}_{p}(\nu,\Lambda)$ is the square of the largest singular value
$s_{1}$ of $\Lambda^{\frac{1}{2}}\Xi$, where $\Xi=(\xi_{ij})$ is a
$p\times\nu$ random matrix consisting of i.i.d. standard Gaussian random
variables. From the definition of the largest singular value, we obtain

(3.3) $s_{1}=\max_{\|v\|=\|w\|=1}v^{\top}\Lambda^{\frac{1}{2}}\Xi w=\max_{u\in
M}\sigma(u)\langle u,\mathrm{vec}(\Xi)\rangle,$

where

(3.4) $\displaystyle\sigma(u)=\sqrt{v^{\top}\Lambda v},\qquad
u=\varphi(v,w)=\frac{(\Lambda^{\frac{1}{2}}\otimes I_{\nu})(v\otimes
w)}{\sqrt{v^{\top}\Lambda v}},$ $\displaystyle
M=\\{\varphi(v,w)\in\mathbb{R}^{p\nu}\mid\|v\|=\|w\|=1\\},$
$\displaystyle\mathrm{vec}(\Xi)=(\xi_{11},\xi_{21},\ldots,\xi_{p,1},\xi_{12},\ldots,\xi_{p,\nu})^{\top}.$

The maximizer of $\sigma(u)$ is

$M_{0}=\biggl{\\{}\varphi\biggl{(}\begin{pmatrix}v_{1}\\\
0\end{pmatrix},w\biggr{)}\in\mathbb{R}^{p\nu}\mid
v_{1}\in\mathbb{R}^{q},\,w\in\mathbb{R}^{\nu},\,\|v_{1}\|=\|w\|=1\biggr{\\}}.$

The dimensions of $M$ and $M_{0}$ are

(3.5) $d=\dim M=p+\nu-2,\quad d_{0}=\dim M_{0}=q+\nu-2.$

Fix a point $u_{0}=\varphi(v_{0},w_{0})\in M_{0}$,
$v_{0}=(1,0,\ldots,0)^{\top}$, $w_{0}=(1,0,\ldots,0)^{\top}$, and take local
coordinates of $M$ by

$\textstyle
v=\Bigl{(}\sqrt{1-\sum_{i=1}^{p-1}(t^{i})^{2}},t^{1},\ldots,t^{p-1}\Bigr{)}^{\top},\quad
w=\Bigl{(}\sqrt{1-\sum_{i=p}^{p+\nu-2}(t^{i})^{2}},t^{p},\ldots,t^{p+\nu-2}\Bigr{)}^{\top}.$

Then, $u_{0}=\varphi(v,w)|_{t=0}$.

A simple calculation yields the metric tensor $G$ and the Hesse matrix $C$ of
$\ell=\log\sigma$ evaluated at $u_{0}$ as

(3.6) $G=\begin{pmatrix}I_{q-1}&0&0\\\ 0&\frac{1}{\lambda_{1}}\Lambda_{2}&0\\\
0&0&I_{\nu-1}\end{pmatrix},\quad C=\begin{pmatrix}0&0&0\\\
0&I_{p-q}-\frac{1}{\lambda_{1}}\Lambda_{2}&0\\\
0&0&0\end{pmatrix},\quad\Lambda_{2}=\operatorname{diag}(\lambda_{q+1},\ldots,\lambda_{p}).$

We substitute (3.5) and (3.6) into (2.17). Because the integrand of
$\int_{M_{0}}\mathrm{d}u_{0}$ is constant owing to symmetry, the integration
is replaced by multiplication with the constant

$\operatorname{Vol}_{d_{0}}(M_{0})=\frac{1}{2}\operatorname{Vol}_{d_{0}}\bigl{(}\mathbb{S}^{q-1}\times\mathbb{S}^{\nu-1}\bigr{)}=\frac{\Omega_{q}\Omega_{\nu}}{\Omega_{1}}.$

Hence,

$\mathbb{P}(s_{1}>c)\sim\frac{1}{\Omega_{q+\nu-1}}\frac{\Omega_{q}\Omega_{\nu}}{\Omega_{1}}\frac{\det(I_{p-q})^{\frac{1}{2}}}{\det\bigl{(}I_{p-q}-\frac{1}{\lambda_{1}}\Lambda_{2}\bigr{)}^{\frac{1}{2}}}\,\bar{G}_{(q+\nu-2)+1}(c^{2}/\lambda_{1}).$

######  Theorem 3.1.

The tail probability of the largest eigenvalue $\ell_{1}$ of the Wishart
matrix $\mathcal{W}_{p}(\nu,\Lambda)$, where $\Lambda$ has the eigenvalues
$\lambda_{1}=\cdots=\lambda_{q}>\lambda_{q+1}\geq\cdots\geq\lambda_{p}>0$, is

$\mathbb{P}(\ell_{1}>c)\sim\frac{1}{\prod_{i=q+1}^{p}(1-\lambda_{i}/\lambda_{1})^{1/2}}\frac{\Omega_{q}\Omega_{\nu}}{\Omega_{q+\nu-1}\Omega_{1}}\,\bar{G}_{q+\nu-1}(c/\lambda_{1}),\quad
c\to\infty.$

######  Remark 3.2.

When $\nu=1$, Theorem 3.1 reduces to the formula for the weighted sum of the
chi-square random variables:

$\mathbb{P}\Biggl{(}\sum_{i=1}^{p}\lambda_{i}\xi_{i}^{2}>c\Biggr{)}\sim\frac{1}{\prod_{i=q+1}^{p}(1-\lambda_{i}/\lambda_{1})^{1/2}}\,\bar{G}_{q}(c/\lambda_{1}),\quad
c\to\infty,$

where $\xi_{i}^{2}\sim\chi^{2}_{1}$ i.i.d. ([Ber75, Theorem 2]). For exact
calculation, see [KT16] and references therein.

###  3.3. Largest eigenvalue of a $2\times 2$ Wishart matrix

When the size of Wishart matrix is small, we can derive the tube method
approximation without using Laplace’s method. We consider the Wishart
distribution $\mathcal{W}_{2}(\nu,\Lambda)$,
$\Lambda=\operatorname{diag}(\lambda_{1},\lambda_{2})$
($\lambda_{1}\geq\lambda_{2}$). Define $\sigma(u)$, $\varphi(v,w)$ and $M$ as
in (3.4) with $p=2$, and consider the distribution $s_{1}$ in (3.3).

Fix a point $u_{0}=\varphi(v_{0},w_{0})\in M$,
$v_{0}=\bigl{(}\cos\theta,\sin\theta\bigr{)}^{\top}$,
$w_{0}=(1,0,\ldots,0)^{\top}$, and take local coordinates of $M$ about $u$ by

$\textstyle v=\bigl{(}\cos t_{1},\sin t_{1}\bigr{)}^{\top},\quad
w=\Bigl{(}\sqrt{1-\sum_{i=2}^{\nu}t_{i}^{2}},t_{2},\ldots,t_{\nu}\Bigr{)}^{\top}.$

Then, $u_{0}=\varphi(v,w)|_{t_{1}=\theta,t_{2}=\cdots=t_{\nu}=0}$.

A simple calculation yields

(3.7) $\sigma^{2}=\lambda_{1}\cos^{2}\theta+\lambda_{2}\sin^{2}\theta,\qquad
q=\frac{1+\|\nabla\ell\|^{2}}{\sigma^{2}}=\lambda_{1}^{-1}\cos^{2}\theta+\lambda_{2}^{-1}\sin^{2}\theta,$

and

$G=(g_{ij})=\operatorname{diag}\bigl{(}g_{11},\underbrace{1,\ldots,1}_{\nu-1}\bigr{)},\qquad
g_{11}=\frac{\lambda_{1}\lambda_{2}}{\sigma^{4}},\qquad\widetilde{G}=(\widetilde{g}_{ij})=G+C=I_{\nu}.$

The volume element of $M$ is

(3.8)
$\sqrt{g_{11}}\mathrm{d}\theta\,\mathrm{d}w=\frac{\sqrt{\lambda_{1}\lambda_{2}}}{\sigma^{2}}\mathrm{d}\theta\,\mathrm{d}w,$

where $\mathrm{d}w$ is the volume element of $\mathbb{S}^{\nu-1}$, and

(3.9)
$\det(I_{\nu}+CG^{-1})=\det(G+C)\det(G)^{-1}=g_{11}^{-1}=\frac{\sigma^{4}}{\lambda_{1}\lambda_{2}}.$

According to (2.1), the curvature tensor is

$R_{ij;kl}=\begin{cases}\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk}&(\mbox{if
$i,j,k,l\geq 2$}),\\\ 0&(\mbox{otherwise}),\end{cases}$

where $\delta_{ij}$ is the Kronecker delta. Suppose that $i<j$ and $k<l$. If
$i,k\geq 2$, then both $R_{ij;kl}$ and $g_{ik}g_{jl}-g_{il}g_{jk}$ are equal
to $\delta_{ik}\delta_{jl}-\delta_{il}\delta_{jk}$, and hence
$R_{ij;kl}-(g_{ik}g_{jl}-g_{il}g_{jk})$ vanishes. If $i=1<k$ or $k=1<i$, then
$R_{ij;kl}=0$ and $g_{ik}g_{jl}-g_{il}g_{jk}=0$. Remaining case is

$R_{ij;kl}-(g_{ik}g_{jl}-g_{il}g_{jk})=-g_{11}\quad\mbox{when }i=k=1<j=l.$

Therefore, the function $\zeta_{e}$ in Theorem 2.1 is

(3.10) $\zeta_{e}=\begin{cases}1&(e=0),\\\ -(\nu-1)g_{11}&(e=2),\\\
0&(\mbox{otherwise}).\end{cases}$

Substituting (3.7), (3.8), (3.9), and (3.10) into (2.15), we have the
approximation formula for the largest eigenvalue $\ell_{1}$ as

(3.11) $\displaystyle\mathbb{P}_{\mathrm{tube}}(\ell_{1}>c)=$
$\displaystyle\mathbb{P}_{\mathrm{tube}}(s_{1}>\sqrt{c})$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int_{0}^{2\pi}\int_{\mathbb{S}^{\nu-1}}\frac{\sqrt{\lambda_{1}\lambda_{2}}}{\sigma^{2}}\mathrm{d}\theta\,\mathrm{d}w\frac{1}{\Omega_{\nu+1}}\frac{\sigma^{4}}{\lambda_{1}\lambda_{2}}(\sigma^{2}q)^{-\frac{1}{2}(\nu+1)}\bar{G}_{\nu+1}(qc)$
$\displaystyle+\frac{1}{2}\int_{0}^{2\pi}\int_{\mathbb{S}^{\nu-1}}\frac{\sqrt{\lambda_{1}\lambda_{2}}}{\sigma^{2}}\mathrm{d}\theta\,\mathrm{d}w\frac{1}{2\pi\Omega_{\nu-1}}\frac{\sigma^{4}}{\lambda_{1}\lambda_{2}}(\sigma^{2}q)^{-\frac{1}{2}(\nu-1)}\bar{G}_{\nu-1}(qc)$
$\displaystyle\qquad\qquad\times\biggl{(}-(\nu-1)\frac{\lambda_{1}\lambda_{2}}{\sigma^{4}}\biggr{)}$
$\displaystyle=$
$\displaystyle\frac{\Omega_{\nu}}{2\Omega_{\nu+1}}\int_{0}^{2\pi}(\sigma^{2}q)^{-\frac{1}{2}(\nu+1)}\biggl{[}\frac{\sigma^{2}}{\sqrt{\lambda_{1}\lambda_{2}}}\bar{G}_{\nu+1}(qc)-\sqrt{\lambda_{1}\lambda_{2}}q\bar{G}_{\nu-1}(qc)\biggr{]}\mathrm{d}\theta.$

Next, we obtain the threshold $b_{\mathrm{cri}}$. We prepare a lemma for this
purpose.

######  Lemma 3.3.

For $a>0$,

$\inf_{|y|<1}\frac{(a-by)_{+}^{2}}{1-y^{2}}=(a^{2}-b^{2})_{+}.$

###### Proof.

When $|b|\leq a$, $a-by\geq 0$ for all $|y|<1$, and the left-hand side is

$\inf_{|y|<1}\frac{(a-by)^{2}}{1-y^{2}}=\frac{(a-by)^{2}}{1-y^{2}}\Big{|}_{y=b/a}=a^{2}-b^{2}.$

When $|b|>a$, $a-by=0$ at $-1<y=a/b<1$, and the left-hand side is 0. ∎

As in (3.4), let $u=\varphi(v,w)$ and
$\widetilde{u}=\varphi(\widetilde{v},\widetilde{w})$, where

$\varphi(v,w)=\frac{(\Lambda^{\frac{1}{2}}\otimes I_{\nu})(v\otimes
w)}{\sigma(u)},\quad\sigma(u)=\sqrt{v^{\top}\Lambda v}.$

We will evaluate

$1/b_{\mathrm{cri}}^{2}=\inf_{u\neq\widetilde{u}}\frac{1}{\sigma(u)^{2}}\biggl{\\{}1+\|\nabla\ell(u)\|^{2}+\biggl{(}\frac{\\{\sigma(u)/\sigma(\widetilde{u})-\langle\widetilde{u},u-\nabla\ell(u)\rangle\\}_{+}}{\|P_{u}^{\perp}\widetilde{u}\|}\biggr{)}^{2}\biggr{\\}}.$

Let $v=(\cos t,\sin t)^{\top}$, $\widetilde{v}=(\cos s,\sin s)^{\top}$. Then,
simple calculations yield

$\sigma(u)=\sqrt{\lambda_{1}\cos^{2}t+\lambda_{2}\sin^{2}t},\quad\sigma(\widetilde{u})=\sqrt{\lambda_{1}\cos^{2}s+\lambda_{2}\sin^{2}s},$
$\displaystyle\langle\widetilde{u},u\rangle=\frac{\lambda_{1}\cos t\cos
s+\lambda_{2}\sin t\sin
s}{\sigma(u)\sigma(\widetilde{u})}(\widetilde{w}^{\top}w),$
$\displaystyle\langle\widetilde{u},\nabla\ell(u)\rangle=\frac{(\lambda_{1}-\lambda_{2})\sin(2t)\sin(t-s)}{2\sigma(u)\sigma(\widetilde{u})}(\widetilde{w}^{\top}w),$
$\displaystyle\|P^{\perp}_{u}\widetilde{u}\|=\frac{\sqrt{\lambda_{1}\lambda_{2}}|\sin(t-s)|}{\sigma(u)\sigma(\widetilde{u})}\sqrt{1-(\widetilde{w}^{\top}w)^{2}},\qquad\|\nabla\ell(u)\|^{2}=\frac{(\lambda_{1}-\lambda_{2})^{2}\sin^{2}(2t)}{4\lambda_{1}\lambda_{2}},$

and hence,

$\displaystyle\frac{\\{\sigma(u)/\sigma(\widetilde{u})-\langle\widetilde{u},\psi(u)\rangle\\}_{+}^{2}}{\|P_{u}^{\perp}\widetilde{u}\|^{2}}$
$\displaystyle=\frac{\\{\sigma(u)^{2}-[(\lambda_{1}\cos t\cos
s+\lambda_{2}\sin t\sin
s)-\frac{1}{2}(\lambda_{1}-\lambda_{2})\sin(2t)\sin(t-s)](\widetilde{w}^{\top}w)\\}_{+}^{2}}{\lambda_{1}\lambda_{2}\sin^{2}(t-s)(1-(\widetilde{w}^{\top}w)^{2})}.$

By means of Lemma 3.3,

$\displaystyle\inf_{w\neq\widetilde{w}}\frac{\\{\sigma(u)/\sigma(\widetilde{u})-\langle\widetilde{u},\psi(u)\rangle\\}_{+}^{2}}{\|P_{u}^{\perp}\widetilde{u}\|^{2}}$
$\displaystyle=\frac{\\{\sigma(u)^{4}-[(\lambda_{1}\cos t\cos
s+\lambda_{2}\sin t\sin
s)-\frac{1}{2}(\lambda_{1}-\lambda_{2})\sin(2t)\sin(t-s)]^{2}\\}_{+}}{\lambda_{1}\lambda_{2}\sin^{2}(t-s)}$
$\displaystyle=\frac{\\{\sigma(u)^{4}\sin^{2}(t-s)\\}_{+}}{\lambda_{1}\lambda_{2}\sin^{2}(t-s)}=\frac{\sigma(u)^{4}}{\lambda_{1}\lambda_{2}},$

which is independent of $\widetilde{v}$. Therefore, we have the threshold
$b_{\mathrm{cri}}$ as follows.

$1/b_{\mathrm{cri}}^{2}=\inf_{u}\frac{1}{\sigma(u)^{2}}\biggl{\\{}1+\frac{(\lambda_{1}-\lambda_{2})^{2}\sin^{2}(2t)}{4\lambda_{1}\lambda_{2}}+\frac{\sigma(u)^{4}}{\lambda_{1}\lambda_{2}}\biggr{\\}}=\inf_{u}\biggl{(}\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}\biggr{)}=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}.$

The results of this subsection are summarized as follows.

######  Theorem 3.4.

The upper probability of the largest eigenvalue $\ell_{1}$ of the $2\times 2$
Wishart matrix $\mathcal{W}_{2}(\nu,\Lambda)$,
$\Lambda=\operatorname{diag}(\lambda_{1},\lambda_{2})$, is

$\mathbb{P}(\ell_{1}>c)=\mathbb{P}_{\mathrm{tube}}(\ell_{1}>c)+O(\bar{G}_{2\nu}(c/b_{\mathrm{cri}}^{2})),\
\ c\to\infty,$

where $\mathbb{P}_{\mathrm{tube}}(\ell_{1}>c)$ is given in (3.11), and

$b_{\mathrm{cri}}=\sqrt{\frac{\lambda_{1}\lambda_{2}}{\lambda_{1}+\lambda_{2}}}.$

Figure 3 shows the tail probability of the largest eigenvalue of the $2\times
2$ Wishart matrix $\mathcal{W}_{2}(4,\Lambda)$ for various $\Lambda$’s. The
simulation is based on 10,000 replications. The tube formula shows a good
performance for estimating the tail probability of the regions used in testing
hypotheses. The Laplace approximation also works well when largest eigenvalue
is dominating. This is consistent with the fact that when $\lambda_{1}=1$,
$b_{\mathrm{cri}}=\sqrt{1+1/\lambda_{2}}$ is a decreasing function in
$\lambda_{2}$.

|
---|---
$\Lambda=\operatorname{diag}(1,1)$ | $\Lambda=\operatorname{diag}(1,7/8)$
|
$\Lambda=\operatorname{diag}(1,3/4)$ | $\Lambda=\operatorname{diag}(1,1/4)$
Figure 3. Upper-tail probability of the largest eigenvalue of
$\mathcal{W}_{2}(4,\Lambda)$.
—: Simulation, $\bm{\cdots}$: Tube method, $\bm{--}$: Laplace approximation.

##  4\. Proofs

In this section we prove theorems and propositions of Section 2. We represent
elements of $\mathbb{R}^{n}$ as column vectors. The inner product of
$u,w\in\mathbb{R}^{n}$ is denoted by $\langle u,w\rangle$ or $u^{\top}w$
interchangeably.

###  4.1. Notation

Recall that the point of $M$ is written as $u=\varphi(t)$, $t=(t^{i})_{1\leq
i\leq d}$. Write $\partial/\partial t^{i}=\partial_{i}$, and
$\varphi_{i}=\partial_{i}\varphi$,
$\varphi_{ij}=\partial_{i}\partial_{j}\varphi$,
$\sigma_{i}=\partial_{i}\sigma$, etc. The metric tensor and the connection
coefficient are $g_{ij}=\langle\varphi_{i},\varphi_{j}\rangle$ and

$\Gamma_{ij,k}=\langle\varphi_{ij},\varphi_{k}\rangle=\frac{1}{2}(\partial_{i}g_{jk}+\partial_{j}g_{ik}-\partial_{k}g_{ij}),$

respectively. Note that $\partial_{k}g_{ij}=\Gamma_{ik,j}+\Gamma_{jk,i}$.

The tangent space of $\mathbb{S}^{n-1}=S(\mathbb{R}^{n})$ and $M$ at
$u=\varphi$ are written as

$T_{u}(\mathbb{S}^{n-1})=\\{w\in\mathbb{R}^{n}\mid u^{\top}w=0\\},\qquad
T_{u}(M)=\operatorname{span}\\{\varphi_{i}\mid 1\leq i\leq d\\},$

respectively.

An orthonormal basis of the normal space $N_{u}$ in (2.2) is denoted by
$n_{a}=n_{a}(u)$, $1\leq a\leq n-d-1$. That is,

$N_{u}=T_{u}(M)^{\perp}\cap
T_{u}(\mathbb{S}^{n-1})=\operatorname{span}\\{n_{a}\mid 1\leq a\leq n-d-1\\}.$

Note that

(4.1) $\mathbb{R}^{n}=T_{u}(\mathbb{R}^{n})=\operatorname{span}\\{u\\}\oplus
T_{u}(M)\oplus N_{u}.$

Throughout this section, we use Einstein’s convention that $a_{i}b^{i}$ means
$\sum_{i}a_{i}b^{i}$ unless it causes confusion.

###  4.2. Jacobian

Let $z\in\mathbb{R}^{n}$ be fixed. The maximal point $u\in M$ of the function
$u\mapsto\sigma(u)\langle u,z\rangle$ satisfies the “normal equation”:

(4.2) $\displaystyle
0=\partial_{i}(\sigma\langle\varphi,z\rangle)=\langle\sigma_{i}\varphi+\sigma\varphi_{i},z\rangle,\quad
1\leq i\leq d.$

Because of (4.1), any $z\in\mathbb{R}^{n}$ is represented by
$z=r\varphi+r^{i}\varphi_{i}+s^{a}n_{a}$. Then, (4.2) becomes
$0=\sigma_{i}r+\sigma r^{j}g_{ij}$, and hence
$r^{j}=-\sigma^{-1}\sigma_{i}rg^{ij}=-\ell_{i}rg^{ij}$, where
$\ell_{i}=\partial_{i}\ell=\partial_{i}\log\sigma$ and $g^{ij}$ is the
$(i,j)$th element of $(g_{ij})^{-1}$. Therefore, at point $u$ such that
$\max_{u}X(u)=X(u)$,

(4.3) $z=r\varphi-\ell_{i}rg^{ij}\varphi_{j}+s^{a}n_{a}=r\psi+s^{a}n_{a},$

where

$\psi=\varphi-\ell_{i}g^{ij}\varphi_{j}=u-\nabla\ell,\quad\nabla\ell=\ell_{i}g^{ij}\varphi_{j}.$

$\nabla\ell$ is the gradient of $\ell$ in the tangent space $T_{u}(M)$. This
defines a local one-to-one transformation between
$z\leftrightarrow(r,t^{i},s^{a})$. We derive the Jacobian of this
transformation as follows.

Taking the derivatives of (4.3), we obtain

$\mathrm{d}z=\mathrm{d}r\psi+r\mathrm{d}\psi+\mathrm{d}s^{a}n_{a}+s^{a}\mathrm{d}n_{a},$

where

$\displaystyle\mathrm{d}\psi=$
$\displaystyle\mathrm{d}\varphi-\mathrm{d}[\ell_{j}g^{ji}\varphi_{i}]=\varphi_{k}\mathrm{d}t^{k}-\partial_{k}[\ell_{j}g^{ji}\varphi_{i}]\mathrm{d}t^{k}$
$\displaystyle=$
$\displaystyle[\varphi_{k}-\ell_{kj}g^{ji}\varphi_{i}-\ell_{j}(\partial_{k}g^{ji})\varphi_{i}-\ell_{j}g^{ji}\varphi_{ki}]\mathrm{d}t^{k},$
$\displaystyle\mathrm{d}n_{a}=$
$\displaystyle(\partial_{k}n_{a})\mathrm{d}t^{k}.$

We will evaluate $\varphi^{\top}\mathrm{d}z$, $\varphi_{l}^{\top}\mathrm{d}z$,
$n_{b}^{\top}\mathrm{d}z$, in turn.

(i) To obtain $\varphi^{\top}\mathrm{d}z$, we have

$\displaystyle\varphi^{\top}\psi=$ $\displaystyle 1,$
$\displaystyle\varphi^{\top}\mathrm{d}\psi=$
$\displaystyle-\ell_{j}g^{ji}\varphi^{\top}\varphi_{ki}\mathrm{d}t^{k}=\ell_{j}g^{ji}g_{ki}\mathrm{d}t^{k}=\ell_{k}\mathrm{d}t^{k},$
$\displaystyle\varphi^{\top}\mathrm{d}n_{a}=$
$\displaystyle(\varphi^{\top}\partial_{k}n_{a})\mathrm{d}t^{k}=-(\varphi_{k}^{\top}n_{a})\mathrm{d}t^{k}=0.$

Summarizing the above,

(4.4) $\varphi^{\top}\mathrm{d}z=\mathrm{d}r+r\ell_{k}\mathrm{d}t^{k}.$

(ii) To obtain $\varphi_{l}^{\top}\mathrm{d}z$, we have

$\displaystyle\varphi_{l}^{\top}\psi=$
$\displaystyle-\ell_{j}g_{li}g^{ji}=-\ell_{l},$
$\displaystyle\varphi_{l}^{\top}\mathrm{d}\psi=$
$\displaystyle[g_{lk}-\ell_{kj}g^{ji}g_{il}-\ell_{j}(\partial_{k}g^{ji})g_{li}-\ell_{j}g^{ji}\varphi_{l}^{\top}\varphi_{ik}]\mathrm{d}t^{k}$
$\displaystyle=$
$\displaystyle[g_{lk}-\ell_{kl}+\ell_{j}g^{ji}(\partial_{k}g_{li})-\ell_{j}g^{ji}\varphi_{l}^{\top}\varphi_{ik}]\mathrm{d}t^{k}$
$\displaystyle=$
$\displaystyle[g_{lk}-\ell_{kl}+\ell_{j}g^{ji}(\Gamma_{lk,i}+\Gamma_{ik,l})-\ell_{j}g^{ji}\Gamma_{ik,l}]\mathrm{d}t^{k}$
$\displaystyle=$
$\displaystyle[g_{lk}-(\ell_{kl}-\ell_{j}g^{ji}\Gamma_{lk,i})]\mathrm{d}t^{k}=[g_{lk}-\nabla_{k}\nabla_{l}\ell]\mathrm{d}t^{k},$

where

$\nabla_{l}\ell=\partial_{l}\ell,\quad\nabla_{k}\nabla_{l}\ell=\partial_{k}\partial_{l}\ell-\Gamma_{kl,i}g^{ij}\ell_{j}$

are the covariant differentials. Moreover,

$\displaystyle\varphi_{l}^{\top}\mathrm{d}n_{a}=$
$\displaystyle\varphi_{l}^{\top}\partial_{k}n_{a}\mathrm{d}t^{k}=-\varphi_{lk}^{\top}n_{a}\mathrm{d}t^{k}=h_{kl,a}\mathrm{d}t^{k},$

where we let $h_{kl,a}=-\varphi_{lk}^{\top}n_{a}$. Summarizing the above,

(4.5)
$\varphi_{l}^{\top}\mathrm{d}z=-\ell_{l}\mathrm{d}r+r[g_{lk}-\nabla_{l}\nabla_{k}\ell]\mathrm{d}t^{k}+s^{a}h_{kl,a}\mathrm{d}t^{k}.$

(iii) To obtain $n_{b}^{\top}dz$, we have $n_{b}^{\top}\psi=0$, and
$n_{b}^{\top}\mathrm{d}\psi$ and $n_{b}^{\top}\mathrm{d}n_{a}$ consist of
terms containing $\mathrm{d}t^{k}$. Therefore,

(4.6) $n_{b}^{\top}\mathrm{d}z=(\mbox{terms containing
}\mathrm{d}t^{k})+\delta_{ab}\mathrm{d}s^{a},$

where $\delta_{ab}$ is the Kronecker delta.

Combining (4.4), (4.5), and (4.6), we obtain

(4.7) $\begin{pmatrix}\varphi^{\top}\\\ \varphi_{l}^{\top}\\\
n_{b}^{\top}\end{pmatrix}\mathrm{d}z=\begin{pmatrix}1&r\nabla_{k}\ell&0\\\
-\nabla_{l}\ell&r[g_{lk}-\nabla_{k}\nabla_{l}\ell]+s^{a}h_{kl,a}&0\\\
0&*&\delta_{ab}\end{pmatrix}\begin{pmatrix}\mathrm{d}r\\\ \mathrm{d}t^{k}\\\
\mathrm{d}s^{a}\end{pmatrix},$

where entries marked by $*$ do not affect the following calculations. The
determinant of the matrix on the right-hand side of (4.7) is

$\det(r[g_{lk}-\nabla_{k}\nabla_{l}\ell+\nabla_{k}\ell\nabla_{l}\ell]+s^{a}h_{kl,a})=\det(r(G+C)+H(s)),$

where $c_{kl}=-\nabla_{k}\nabla_{l}\ell+\nabla_{k}\ell\nabla_{l}\ell$,
$C=(c_{kl})$, $G=(g_{kl})$, and $H(s)=(s^{a}h_{kl,a})$ as before. $H(s)$ is
the second fundamental form in the normal direction $s^{a}n_{a}\in N_{u}$ in
(2.2). When needed, we write $C=C_{u}$, $G=G_{u}$, and $H(s)=H_{u}(s)$ to
indicate the point $u$ where these quantities are defined.

On the other hand,

$\det(\varphi,\varphi_{l},n_{b})=\det(G)^{\frac{1}{2}}.$

Taking the wedge product of all rows in (4.7),

$\det(G)^{\frac{1}{2}}\bigwedge_{i=1}^{n}\mathrm{d}z_{i}=r^{d}\mathrm{d}r\,\det(G+C+r^{-1}H(s))\,\bigwedge_{i=1}^{d}\mathrm{d}t^{i}\,\bigwedge_{a=1}^{n-d-1}\mathrm{d}s^{a}.$

Moreover, let

$\textstyle s=\sqrt{\sum_{a=1}^{n-d-1}(s^{a})^{2}},\qquad
v=\Bigl{(}\sum_{a=1}^{n-d-1}s^{a}n_{a}\Bigr{)}\big{/}s\in S(N_{u}),$

where $S(N_{u})$ is the unit sphere of the linear space $N_{u}$. We conclude
that the local one-to-one correspondence

$z=r\psi+sv,\quad\psi=u-\nabla\ell$

has the Jacobian

$\mathrm{d}z=\bigl{|}\det(I_{d}+CG^{-1})\det(I_{d}+(s/r)H(v)(G+C)^{-1})\bigr{|}\,r^{d}\mathrm{d}r\,\mathrm{d}u\,s^{n-d-2}\mathrm{d}s\,\mathrm{d}v,$

where $\mathrm{d}z$ is the $n$-dimensional Lebesgue measure at $z$,
$\mathrm{d}u=\det(G)^{\frac{1}{2}}\prod_{i=1}^{d}\mathrm{d}t^{i}$ is the
volume element of $M$ at $u$, and $\mathrm{d}v$ is the volume element of
$S(N_{u})$. Note that

(4.8) $\|z\|^{2}=r^{2}(1+\|\nabla\ell\|^{2})+s^{2},\qquad\sigma(u)\langle
u,z/\|z\|\rangle=\frac{\sigma(u)}{\sqrt{1+\|\nabla\ell\|^{2}+(s/r)^{2}}}.$

###  4.3. $b_{\mathrm{cri}}$ and critical radius $\theta_{\mathrm{cri}}$

In this subsection, we prove Theorem 2.3 and Propositions 2.5 and 2.6.

In the previous subsection, we proved that

$u\in M$ annihilates the gradient of map $u\mapsto\sigma(u)\langle u,z\rangle$
$\displaystyle\iff$ $\displaystyle z\in\\{r\psi(u)+sv\mid(r,s,v)\in Q_{u}\\},$

where

$Q_{u}=\\{(r,s,v)\mid r>0,\ s\geq 0,\ v\in S(N_{u})\\}.$

Here, we will find additional conditions on $z$ (the set $V_{u}$) such that

$u\in M$ is the unique maximal point of the map $u\mapsto\sigma(u)\langle
u,z\rangle$ $\displaystyle\iff$ $\displaystyle
z\in\\{r\psi(u)+sv\mid(r,s,v)\in Q_{u}\\}\cap V_{u}.$

Once the set $V_{u}$ is identified, and the event that multiple $u$’s have the
same maximal values has a probability of zero, then the upper probability of
$Y_{\mathrm{max}}$ is written as

(4.9)
$\mathbb{P}\bigl{(}Y_{\mathrm{max}}>b\bigr{)}=\int\mathds{1}_{\bigl{\\{}\sigma(u)\langle
u,z/\|z\|\rangle>b,\,z\in
V_{u}\bigr{\\}}}|p(u,r,s,v)|\,\mathrm{d}u\,\mathrm{d}r\,\mathrm{d}s\,\mathrm{d}v,$

where

(4.10) $\displaystyle
p(u,r,s,v)=\frac{1}{(2\pi)^{n/2}}e^{-\frac{1}{2}\\{r^{2}(1+\|\nabla\ell(u)\|^{2})+s^{2}\\}}\det(I_{d}+C_{u}G_{u}^{-1})$
$\displaystyle\times
r^{d}s^{n-d-2}\det(I_{d}+(s/r)H_{u}(v)(G_{u}+C_{u})^{-1}).$

###### Proof of Theorem 2.3.

From the definition, we write

$V_{u}=\bigl{\\{}z\in\mathbb{R}^{n}\mid\sigma(w)\langle
w,z\rangle<\sigma(u)\langle u,z\rangle,\ z=r\psi(u)+sv,\ \forall w\in
M\setminus\\{u\\}\bigr{\\}}.$

Then, $V_{u}\cap V_{u^{\prime}}=\emptyset$, $u\neq u^{\prime}$, and the
complement $\mathbb{R}^{n}\setminus\bigcup_{u\in M}V_{u}$ has the Lebesgue
measure zero by Sard’s theorem.

The restriction $z\in V_{u}$ is to prevent the overlap of the integration
(4.9). If $b$ is sufficiently large, we do not need to impose this
restriction. For fixed $u$, we find a threshold of $b$, say
$b_{\mathrm{cri}}^{\prime}(u)$, as the infimum of $b$ such that

$\displaystyle\\{(r,s,v)\in Q_{u}\mid\sigma(u)\langle
u,z/\|z\|\rangle>b,\,z=r\psi(u)+sv\in V_{u}\\}$ $\displaystyle=\\{(r,s,v)\in
Q_{u}\mid\sigma(u)\langle u,z/\|z\|\rangle>b,\,z=r\psi(u)+sv\\}\ \
\mbox{a.s.},$

or equivalently

(4.11) $A_{u}(b)\subset B_{u}\ \ \mbox{a.s.},$

where

$\displaystyle A_{u}(b)=$ $\displaystyle\,\\{(r,s,v)\in
Q_{u}\mid\sigma(u)\langle u,z/\|z\|\rangle>b,\,z=r\psi(u)+sv\\},$
$\displaystyle B_{u}=$ $\displaystyle\,\\{(r,s,v)\in Q_{u}\mid
z=r\psi(u)+sv\in V_{u}\\}.$

The threshold $b_{\mathrm{cri}}$ in Theorem 2.3 is obtained by
$b_{\mathrm{cri}}=\sup_{u\in M}b_{\mathrm{cri}}^{\prime}(u)$.

Because of (4.8), the set $A_{u}(b)$ is

(4.12) $\displaystyle A_{u}(b)=$ $\displaystyle\,\biggl{\\{}(r,s,v)\in
Q_{u}\mid\frac{\sigma(u)}{\sqrt{1+\|\nabla\ell(u)\|^{2}+(s/r)^{2}}}>b\biggr{\\}}$
$\displaystyle=$ $\displaystyle\,\bigl{\\{}(r,s,v)\in
Q_{u}\mid(s/r)^{2}<\sigma(u)^{2}/b^{2}-(1+\|\nabla\ell(u)\|^{2})\bigr{\\}}.$

Next, we characterize the set $B_{u}$. Note first that

$\displaystyle r\psi(u)+sv\in V_{u}$ $\displaystyle\iff$
$\displaystyle\sigma(w)\langle w,r\psi(u)+sv\rangle<\sigma(u)\langle
u,r\psi(u)+sv\rangle,\ \ \forall w\in M\setminus\\{u\\}$ $\displaystyle\iff$
$\displaystyle\frac{s}{r}\langle w,v\rangle<\sigma(u)/\sigma(w)-\langle
w,\psi(u)\rangle,\ \ \forall w\in M\setminus\\{u\\}.$

$M\setminus\\{u\\}$ is partitioned as $M_{u}^{+}\sqcup M_{u}^{-}$, where

$\displaystyle M_{u}^{+}=$ $\displaystyle\\{w\in
M\mid\sigma(u)/\sigma(w)-\langle w,\psi(u)\rangle>0\\},$ $\displaystyle
M_{u}^{-}=$ $\displaystyle\\{w\in M\mid\sigma(u)/\sigma(w)-\langle
w,\psi(u)\rangle\leq 0\\}\setminus\\{u\\}.$

We consider two cases $M_{u}^{-}=\emptyset$ and $M_{u}^{-}\neq\emptyset$,
separately.

(i) When $M_{u}^{-}=\emptyset$,

(4.14) $\displaystyle B_{u}=$ $\displaystyle\,\biggl{\\{}(r,s,v)\in
Q_{u}\mid\mbox{``if }\langle w,v\rangle>0,\mbox{ then }\
\frac{s}{r}<\frac{\sigma(u)/\sigma(w)-\langle w,\psi(u)\rangle}{\langle
w,v\rangle}\mbox{''},\ \ \forall w\in M\setminus\\{u\\}\biggr{\\}}$
$\displaystyle=$ $\displaystyle\,\biggl{\\{}(r,s,v)\in
Q_{u}\mid\frac{s}{r}<\inf_{w\in M\setminus\\{u\\}:\langle
w,v\rangle>0}\frac{\sigma(u)/\sigma(w)-\langle w,\psi(u)\rangle}{\langle
w,v\rangle}\biggr{\\}}.$

Therefore, $A_{u}(b)\subset B_{u}$ iff

$\displaystyle\sqrt{\frac{\sigma(u)^{2}}{b^{2}}-(1+\|\nabla\ell(u)\|^{2})}<$
$\displaystyle\inf_{v\in S(N_{u})}\inf_{w\in M\setminus\\{u\\}:\langle
w,v\rangle>0}\frac{\sigma(u)/\sigma(w)-\langle w,\psi(u)\rangle}{\langle
w,v\rangle}$ $\displaystyle=$ $\displaystyle\inf_{w\in
M\setminus\\{u\\}}\inf_{v\in S(N_{u}):\langle
w,v\rangle>0}\frac{\sigma(u)/\sigma(w)-\langle w,\psi(u)\rangle}{\langle
w,v\rangle}$ $\displaystyle=$ $\displaystyle\inf_{w\in
M\setminus\\{u\\}}\frac{\sigma(u)/\sigma(w)-\langle
w,\psi(u)\rangle}{\|P^{\perp}_{u}w\|},$

where $P^{\perp}_{u}$ is defined in the statement of Theorem 2.3. Here we used

$\sup_{v\in S(N_{u}):\langle w,v\rangle>0}\langle
w,v\rangle=\|P^{\perp}_{u}w\|.$

(ii) When $M_{u}^{-}\neq\emptyset$,

(4.15) $\displaystyle B_{u}=$ $\displaystyle\,\biggl{\\{}(r,s,v)\in
Q_{u}\mid\mbox{``}\langle w,v\rangle<0\ \mbox{and}\
\frac{s}{r}>\frac{-\sigma(u)/\sigma(w)+\langle w,\psi(u)\rangle}{-\langle
w,v\rangle}\mbox{''},\ \ \forall w\in M\setminus\\{u\\}\biggr{\\}}$
$\displaystyle=$ $\displaystyle\,\biggl{\\{}(r,s,v)\in Q_{u}\mid\sup_{w\in
M\setminus\\{u\\}}\langle w,v\rangle<0,\ \frac{s}{r}>\sup_{w\in
M\setminus\\{u\\}}\frac{-\sigma(u)/\sigma(w)+\langle
w,\psi(u)\rangle}{-\langle w,v\rangle}\biggr{\\}}.$

If $A_{u}(b)$ is not empty, then there exist $(r,s,v),(r,s,-v)\in A_{u}(b)$,
but this contradicts the definition of $B_{u}$. Therefore, $A_{u}(b)\subset
B_{u}$ holds iff $A_{u}(b)=\emptyset$. That is,

$\frac{\sigma(u)^{2}}{b^{2}}-(1+\|\nabla\ell(u)\|^{2})\leq 0.$

Combining (i) and (ii), $b_{\mathrm{cri}}^{\prime}(u)$ is given by

$\sqrt{\frac{\sigma(u)^{2}}{b_{\mathrm{cri}}^{\prime}(u)^{2}}-(1+\|\nabla\ell(u)\|^{2})}=\inf_{w\in
M\setminus\\{u\\}}\frac{\\{\sigma(u)/\sigma(w)-\langle
w,\psi(u)\rangle\\}_{+}}{\|P^{\perp}_{u}w\|},$

or equivalently,

$b_{\mathrm{cri}}^{\prime}(u)^{2}=\frac{\sigma(u)^{2}}{1+\inf_{w\in
M\setminus\\{u\\}}h(u,w)},$

where $h(u,w)$ is defined in (2.13). The threshold
$b_{\mathrm{cri}}=\sup_{u\in M}b_{\mathrm{cri}}^{\prime}(u)$ is derived as
(2.12). ∎

###### Proof of Proposition 2.6.

(i) is just a restatement.

(ii). Let $\underline{h}(u)=\inf_{w:w\neq u}h(u,w)$. Suppose that $u\notin
M_{\mathrm{supp}}$. Then, $h(u,w)=\|\nabla\ell(u)\|^{2}$ for some $\exists
w\,(\neq u$), $\underline{h}(u)=\|\nabla\ell(u)\|^{2}$,

$\frac{1+\|\nabla\ell(u)\|^{2}}{\sigma(u)^{2}}b_{\mathrm{cri}}^{2}=\frac{1+\underline{h}(u)}{\sigma(u)^{2}}b_{\mathrm{cri}}^{2}=\frac{1+\underline{h}(u)}{\sigma(u)^{2}}\sup_{u^{\prime}}\frac{\sigma(u^{\prime})^{2}}{1+\underline{h}(u^{\prime})}\geq
1,$

which means that $u\notin M_{\mathrm{cri}}$. Therefore,
$M_{\mathrm{supp}}\supset M_{\mathrm{cri}}$.

(iii). We follow the proof of Theorem 2.3. Recall that $u\notin
M_{\mathrm{supp}}\iff M_{u}^{-}\neq\emptyset$. When $M_{u}^{-}\neq\emptyset$,

$\\{(r,s,v)\in Q_{u}\mid\sigma(u)\langle u,z/\|z\|\rangle>b,\,z=r\psi(u)+sv\in
V_{u}\\}=A_{u}(b)\cap B_{u}=\emptyset$

for $b\geq b_{\mathrm{cri}}^{\prime}(u)$, meaning that there is no point
$z/\|z\|$ such that the maximum of $\sigma(u)\langle z/\|z\|,u\rangle$ is
attained at $u$. In other words, the radius of tube $(M)_{b/\sigma(\cdot)}$
vanishes at $u$ for any $b\geq b_{\mathrm{cri}}^{\prime}(u)$. ∎

######  Lemma 4.1.

For $z\in V_{u}$, the matrix $G_{u}+C_{u}+(s/r)H_{u}(v)$ is non-negative
definite.

###### Proof.

From (4.3),

(4.16) $\liminf_{w\to u}\frac{\sigma(u)/\sigma(w)-\langle
w,\psi(u)\rangle-(s/r)\langle w,v\rangle}{\|w-u\|^{2}}\geq 0.$

To evaluate the left-hand side, we introduce a local coordinate system
$t=(t^{i})$ so that $u=u(0)$, $w=u(t)$. Because

$\displaystyle w=$
$\displaystyle\;u+\partial_{i}ut^{i}+\frac{1}{2}\partial_{i}\partial_{j}ut^{i}t^{j}+o(\|t\|^{2}),$
$\displaystyle\ell(w)=$
$\displaystyle\;\ell(u)+\partial_{i}\ell(u)t^{i}+\frac{1}{2}\partial_{i}\partial_{j}\ell(u)t^{i}t^{j}+o(\|t\|^{2}),$

we obtain

$\|w-u\|^{2}=2(1-\langle
w,u\rangle)=g_{ij}(u)t^{i}t^{j}+o(\|t\|^{2}),\quad\langle
w,v\rangle=-\frac{1}{2}h_{ij}(u;v)t^{i}t^{j}+o(\|t\|^{2}),$

where $g_{ij}(u)$ and $h_{ij}(u;v)$ are the $(i,j)$th elements of $G_{u}$ and
$H_{u}(v)$, respectively, and

$\displaystyle\frac{\sigma(u)}{\sigma(w)}-\langle w,\psi(u)\rangle=$
$\displaystyle e^{-\ell(w)+\ell(u)}-\langle w,u-\nabla\ell(u)\rangle$
$\displaystyle=$ $\displaystyle
1-\partial_{i}\ell(u)t^{i}-\frac{1}{2}\partial_{i}\partial_{j}\ell(u)t^{i}t^{j}+\frac{1}{2}\partial_{i}\ell(u)\partial_{j}\ell(u)t^{i}t^{j}$
$\displaystyle-\langle
u+\partial_{i}ut^{i}+\frac{1}{2}\partial_{i}\partial_{j}ut^{i}t^{j},u-g^{kl}(u)\partial_{k}\ell(u)\partial_{l}u\rangle+o(\|t\|^{2})$
$\displaystyle=$ $\displaystyle
1-\partial_{i}\ell(u)t^{i}-\frac{1}{2}\partial_{i}\partial_{j}\ell(u)t^{i}t^{j}+\frac{1}{2}\partial_{i}\ell(u)\partial_{j}\ell(u)t^{i}t^{j}$
$\displaystyle-1+\partial_{i}\ell(u)t^{i}-\frac{1}{2}\langle\partial_{i}\partial_{j}u,u\rangle
t^{i}t^{j}+\frac{1}{2}\langle\partial_{i}\partial_{j}u,\partial_{l}u\rangle
g^{kl}(u)\partial_{k}\ell(u)t^{i}t^{j}$ $\displaystyle+o(\|t\|^{2})$
$\displaystyle=$
$\displaystyle\frac{1}{2}\bigl{(}g_{ij}(u)-\nabla_{i}\nabla_{j}\ell(u)+\nabla_{i}\ell(u)\nabla_{j}\ell(u)\bigr{)}t^{i}t^{j}+o(\|t\|^{2}).$

(4.16), by letting $\|t\|\to 0$, means that the matrix
$(G_{u}+C_{u}+(s/r)H_{u}(v))G_{u}^{-1}$ with $C_{u}=(c_{ij}(u))$,
$c_{ij}(u)=-\nabla_{i}\nabla_{j}\ell(u)+\nabla_{i}\ell(u)\nabla_{j}\ell(u)$,
is non-negative definite. ∎

###### Proof of Proposition 2.5.

We prove the existence of a constant $c<1$ such that

(4.17) $\frac{\sigma(u)^{2}}{1+h(u,w)}\leq
c\,\sigma_{0}^{2}\quad\forall(u,w)\in M\times M,\,u\neq w.$

We divide $M\times M\setminus\\{(u,u)\mid u\in M\\}$ into three parts.

(i) Recall that $M_{0}=\\{u\in M\mid\sigma(u)=\sigma_{0}\\}$ is the maximizer
of $\sigma(u)$. Suppose that $u\in M$ is in the neighborhood of $M_{0}$, and
$w$ is sufficiently close to $u$. By introducing the same local coordinates
$u=u(0)$, $w=u(t)$ as in the proof of Lemma 4.1, we see

(4.18) $\frac{\sigma(u)}{\sigma(w)}-1+\langle
w,\nabla\ell(u)\rangle=\frac{1}{2}\bigl{(}g_{ij}(u)-\nabla_{i}\nabla_{j}\ell(u)+\nabla_{i}\ell(u)\nabla_{j}\ell(u)\bigr{)}t^{i}t^{j}+o(\|t\|^{2}),$

which is positive because at $u\in M_{0}$, the gradient is
$\nabla_{i}\ell(u)=0$, and the Hesse matrix $(\nabla_{i}\nabla_{j}\ell(u))$ is
non-positive definite. Let $M_{1}\supset M_{0}$ be an open subset of $M$ and
$\varepsilon>0$ such that (4.18) is positive for all $u\in M_{1}$ and $\langle
w,u\rangle>1-\varepsilon$.

Because

$\frac{\sigma(u)}{\sigma(w)}-\langle w,\psi(u)\rangle-(1-\langle
w,u\rangle)=\frac{\sigma(u)}{\sigma(w)}-1+\langle w,\nabla\ell(u)\rangle>0,$

we have

$h(u,w)\geq\frac{\\{\sigma(u)/\sigma(w)-\langle
w,\psi(u)\rangle\\}_{+}}{\|P_{u}^{\perp}w\|}>\frac{1-\langle
w,u\rangle}{\|P_{u}^{\perp}w\|}.$

Moreover,

$\frac{1-\langle w,u\rangle}{\|P_{u}^{\perp}w\|}\geq\inf_{u,w\in M,u\neq
w}\frac{1-\langle
w,u\rangle}{\|P_{u}^{\perp}w\|}=\tan^{2}\theta_{\mathrm{cri}},$

where $\theta_{\mathrm{cri}}$ is the global critical radius of $M$ when
$\sigma(u)\equiv 1$, and is positive (Remark 2.4).

In summary, on the region

(4.19) $\\{(u,w)\in M_{1}\times M\mid\langle w,u\rangle>1-\varepsilon,\,u\neq
w\\},$
$\frac{\sigma(u)^{2}}{1+h(u,w)}\leq\sigma_{0}^{2}\cos^{2}\theta_{\mathrm{cri}}.$

(ii) Let $\varepsilon>0$ be the number chosen in (i). We can choose an open
subset $M_{1}^{\prime}$ of $M$, $M_{0}\subset M_{1}^{\prime}\subset M_{1}$
such that $\sigma(u)>\sigma_{0}(1-\varepsilon^{2})$ and
$\|\nabla\ell(u)\|<\varepsilon^{2}$ for $u\in M_{1}^{\prime}$. Then, for $u\in
M_{1}^{\prime}$ and $w$ such that $\langle w,u\rangle\leq 1-\varepsilon$,

$h(u,w)\geq\frac{\sigma(u)}{\sigma(w)}-\langle
w,u-\nabla\ell(u)\rangle\geq\frac{\sigma(u)}{\sigma(w)}-\langle
w,u\rangle-\|\nabla\ell(u)\|\geq\varepsilon-2\varepsilon^{2}=\varepsilon^{\prime},$

which means that on the region

(4.20) $\\{(u,w)\in M_{1}^{\prime}\times M\mid\langle w,u\rangle\leq
1-\varepsilon\\},$
$\frac{\sigma(u)^{2}}{1+h(u,w)}\leq\sigma_{0}^{2}/(1+\varepsilon^{\prime}).$

(iii) $M_{1}^{\prime}$ contains $M_{0}$, and outside $M_{1}^{\prime}$ the
$\sigma(u)$ is strictly less that $\sigma_{0}=\max_{u\in M}\sigma(u)$. Let
$\sigma_{1}=\max_{u\in M\setminus M_{1}^{\prime}}\sigma(u)<\sigma_{0}$. On the
region

(4.21) $\\{(u,w)\in(M\setminus M_{1}^{\prime})\times M\mid u\neq w\\},$

$h(u,w)\geq 0$ and

$\frac{\sigma(u)^{2}}{1+h(u,w)}\leq\sigma_{1}^{2}.$

The union of (4.19), (4.20) and (4.21) covers $M\times M\setminus\\{(u,u)\mid
u\in M\\}$. Therefore, we have (4.17) with the constant

$c=\max\bigr{\\{}\cos^{2}\theta_{\mathrm{cri}},1/(1+\varepsilon^{\prime}),\sigma_{1}^{2}/\sigma_{0}^{2}\bigl{\\}}<1.$

∎

Lemma 4.1 affirms that $p(u,r,s,v)$ in (4.10) is non-negative. We have the
refinement of (4.9) as follows.

######  Proposition 4.2.

Let $\mathrm{d}u$ and $\mathrm{d}v$ be the volume elements of $M$ and
$S(N_{u})$, respectively. Let $b_{\mathrm{cri}}$ be the threshold defined in
(2.7). Then, for all $b$,

$\mathbb{P}\bigl{(}Y_{\mathrm{max}}>b\bigr{)}=\int_{M}\mathrm{d}u\int_{0}^{\infty}\mathrm{d}r\int_{0}^{\infty}\mathrm{d}s\,\mathds{1}_{\Bigl{\\{}\frac{\sigma(u)r}{\sqrt{r^{2}(1+\|\nabla\ell\|^{2})+s^{2}}}>b\Bigr{\\}}\cap
B_{u}}\int_{S(N_{u})}\mathrm{d}v\,p(u,r,s,v),$

where $B_{u}=(\ref{Bu1})$ if $M_{u}^{-}=\emptyset$, $(\ref{Bu2})$ if
$M_{u}^{-}\neq\emptyset$. For $b\geq b_{\mathrm{cri}}$,

(4.22)
$\mathbb{P}\bigl{(}Y_{\mathrm{max}}>b\bigr{)}=\int_{M}\mathrm{d}u\int_{0}^{\infty}\mathrm{d}r\int_{0}^{\infty}\mathrm{d}s\,\mathds{1}_{\Bigl{\\{}\frac{\sigma(u)r}{\sqrt{r^{2}(1+\|\nabla\ell\|^{2})+s^{2}}}>b\Bigr{\\}}}\int_{S(N_{u})}\mathrm{d}v\,p(u,r,s,v).$

###  4.4. Volume of tube formula: Intrinsic representation

In this subsection, we prove Theorem 2.1.

We evaluate the integral (4.22). The following lemma is useful for managing
the determinant factor in $p(u,r,s,v)$ of (4.10).

For a finite set $I=\\{i_{1},\ldots,i_{n}\\}$, the notations $\Sigma(I)$ and
$\Pi(I)$ are defined in (2.5) and (2.6), respectively. Recall that
$\Sigma_{d}=\Sigma(\\{1,\ldots,d\\})$ and $\Pi_{d}=\Pi(\\{1,\ldots,d\\})$.

######  Lemma 4.3.

Let $H=(h_{ij})$ be a $d\times d$ square Gaussian random matrix of mean zero.
Let

$s_{ij;kl}=\mathbb{E}[h_{ik}h_{jl}-h_{il}h_{jk}]=\mathbb{E}\left[\det\begin{pmatrix}h_{ik}&h_{il}\\\
h_{jk}&h_{jl}\end{pmatrix}\right].$

Then,

(4.23)
$\mathbb{E}[\det(H)]=\begin{cases}\displaystyle\frac{1}{(d/2)!}\sum_{\pi,\tau\in\Pi_{d}}\operatorname{sgn}(\pi)\operatorname{sgn}(\tau)s_{\pi_{1}\pi_{2};\tau_{1}\tau_{2}}s_{\pi_{3}\pi_{4};\tau_{3}\tau_{4}}\cdots
s_{\pi_{d-1}\pi_{d};\tau_{d-1}\tau_{d}}&(\mbox{if $d$ is even}),\\\
0&(\mbox{if $d$ is odd}).\end{cases}$

###### Proof.

The case $d$ is odd is easy. Suppose that $d$ is even.

$\displaystyle\mathbb{E}[\det(H)]=$
$\displaystyle\sum_{\tau\in\Sigma_{d}}\operatorname{sgn}(\tau)\mathbb{E}[h_{1\tau_{1}}h_{2\tau_{2}}\cdots
h_{d,\tau_{d}}]$ $\displaystyle=$
$\displaystyle\sum_{\tau\in\Sigma_{d}}\operatorname{sgn}(\tau)\frac{1}{(d/2)!}\sum_{\pi\in\Pi_{d}}\mathbb{E}\bigl{[}h_{\pi_{1}\tau_{\pi_{1}}}h_{\pi_{2}\tau_{\pi_{2}}}\bigr{]}\mathbb{E}\bigl{[}h_{\pi_{3}\tau_{\pi_{3}}}h_{\pi_{4}\tau_{\pi_{4}}}\bigr{]}\cdots\mathbb{E}\bigl{[}h_{\pi_{d-1},\tau_{\pi_{d-1}}}h_{\pi_{d},\tau_{\pi_{d}}}\bigr{]}$
$\displaystyle=$
$\displaystyle\frac{1}{(d/2)!}\sum_{\tau\in\Sigma_{d}}\sum_{\pi\in\Pi_{d}}\operatorname{sgn}(\tau)\operatorname{sgn}(\pi)\mathbb{E}[h_{\pi_{1}\tau_{1}}h_{\pi_{2}\tau_{2}}]\cdots\mathbb{E}[h_{\pi_{d-1},\tau_{d-1}}h_{\pi_{d},\tau_{d}}]$
$\displaystyle=$
$\displaystyle\frac{1}{(d/2)!}\sum_{\tau\in\Pi_{d}}\sum_{\pi\in\Pi_{d}}\operatorname{sgn}(\tau)\operatorname{sgn}(\pi)\Biggl{(}\sum_{\sigma\in\Sigma(\\{1,2\\})}\operatorname{sgn}(\sigma)\mathbb{E}\bigl{[}h_{\pi_{1}\tau_{\sigma_{1}}}h_{\pi_{2}\tau_{\sigma_{2}}}\bigr{]}\Biggr{)}\times\cdots$
$\displaystyle\qquad\times\Biggl{(}\sum_{\sigma\in\Sigma(\\{d-1,d\\})}\operatorname{sgn}(\sigma)\mathbb{E}\bigl{[}h_{\pi_{d-1},\tau_{\sigma_{1}}}h_{\pi_{d},\tau_{\sigma_{2}}}\bigr{]}\Biggr{)}$
$\displaystyle=$
$\displaystyle\frac{1}{(d/2)!}\sum_{\tau\in\Pi_{d}}\sum_{\pi\in\Pi_{d}}\operatorname{sgn}(\tau)\operatorname{sgn}(\pi)\bigl{(}\mathbb{E}[h_{\pi_{1}\tau_{1}}h_{\pi_{2}\tau_{2}}]-\mathbb{E}[h_{\pi_{1}\tau_{2}}h_{\pi_{2}\tau_{1}}]\bigr{)}\times\cdots$
$\displaystyle\qquad\times\bigl{(}\mathbb{E}[h_{\pi_{d-1},\tau_{d-1}}h_{\pi_{d},\tau_{d}}]-\mathbb{E}[h_{\pi_{d-1},\tau_{d}}h_{\pi_{d},\tau_{d-1}}]\bigr{)},$

which is the right-hand side of (4.23). The second equality is due to the
moment formula for a product of Gaussian random variables. ∎

######  Remark 4.4.

Lemma 4.3 can be stated in terms of the commutative algebra of double forms.
See [Fed59],[TA03],[KT09].

###### Proof of Theorem 2.1.

We apply Lemma 4.3 in the evaluation of the integral (4.22) with respect to
$\mathrm{d}v$. Recall that $\mathrm{d}v$ is the volume element of the
($n-d-2$)-dimensional unit sphere $S(N_{u})$, and hence,
$\mathrm{d}v/\Omega_{n-d-1}$ is the uniform distribution on the sphere.

Suppose that $\eta$ is the standard Gaussian distribution of space
$N_{u}=T_{u}(M)^{\perp}\cap T_{u}(\mathbb{S}^{n-1})$. Then,
$\eta/\|\eta\|\sim\mathrm{d}v/\Omega_{n-d-1}$ and
$\|\eta\|^{2}\sim\chi^{2}_{n-d-1}$ are independently distributed. Let
$\widetilde{H}(v)=(\widetilde{h}_{ij}(v))=H(v)(G+C)^{-1}$ for short. The
determinant in $p(u,r,s,v)$ in (4.10) is expanded as

(4.24) $\displaystyle\det\bigl{(}I_{d}+(s/r)\widetilde{H}(v)\bigr{)}=$
$\displaystyle\sum_{e=0}^{d}(s/r)^{e}\operatorname{tr}_{e}\bigl{(}\widetilde{H}(v)\bigr{)}$
$\displaystyle=$
$\displaystyle\sum_{e=0}^{d}(s/r)^{e}\\!\\!\\!\\!\sum_{I:\,I\subset\\{1,\ldots,d\\},\,|I|=e}\det\bigl{(}\widetilde{H}_{I}(v)\bigr{)},\qquad\widetilde{H}_{I}(v)=\bigl{(}\widetilde{h}_{ij}(v)\bigr{)}_{i,j\in
I}.$

Hence, the integral with respect to $\mathrm{d}v$ involved in (4.22) is

(4.25) $\displaystyle\int\det\bigl{(}\widetilde{H}_{I}(v)\bigr{)}\mathrm{d}v=$
$\displaystyle\Omega_{n-d-1}\mathbb{E}\bigl{[}\det\widetilde{H}_{I}(\eta/\|\eta\|)\bigr{]}$
$\displaystyle=$
$\displaystyle\frac{\Omega_{n-d-1}}{\mathbb{E}\bigl{[}\|\eta\|^{e}\bigr{]}}\mathbb{E}\bigl{[}\det\widetilde{H}_{I}(\eta)\bigr{]}=\frac{\Omega_{n-d-1}}{\mathbb{E}\bigl{[}(\chi^{2}_{n-d-1})^{e/2}\bigr{]}}\mathbb{E}\bigl{[}\det\widetilde{H}_{I}(\eta)\bigr{]}$
$\displaystyle=$
$\displaystyle\frac{\Omega_{n-d+e-1}}{(2\pi)^{e/2}}\mathbb{E}\bigl{[}\det\widetilde{H}_{I}(\eta)\bigr{]},$

and $\mathbb{E}\bigl{[}\det\widetilde{H}_{I}(\eta)\bigr{]}$ is expressed by
(4.23) with $s_{ij,kl}=\widetilde{R}_{ij}^{kl}$ in (2.10), because the
curvature tensor is characterized as

(4.26)
$R_{ij;kl}=\bigl{(}g_{ik}g_{jl}-g_{il}g_{jk}\bigr{)}+\mathbb{E}[h_{ik}(\eta)h_{jl}(\eta)-h_{il}(\eta)h_{jk}(\eta)],$

where $H(v)=(h_{ij}(v))$ is the second fundamental form, and $\eta$ is a
standard Gaussian vector in $N_{u}$ ([Wey39]). (4.26) is the Gauss equation
for a submanifold of the unit sphere.

On the other hand, the integral with respect to $\mathrm{d}r$ and
$\mathrm{d}s$ in (4.22) is a linear combination of

(4.27)
$\displaystyle\frac{1}{(2\pi)^{n/2}}\int_{0}^{\infty}\int_{0}^{\infty}\mathds{1}_{\bigl{\\{}\frac{r^{2}(1+\|\nabla\ell\|^{2})}{r^{2}(1+\|\nabla\ell\|^{2})+s^{2}}>\frac{1+\|\nabla\ell\|^{2}}{\sigma^{2}}b^{2}\bigr{\\}}}r^{d}s^{n-d-2}(s/r)^{e}e^{-\frac{1}{2}\\{r^{2}(1+\|\nabla\ell(u)\|^{2})+s^{2}\\}}\mathrm{d}r\,\mathrm{d}s$
$\displaystyle=(1+\|\nabla\ell\|^{2})^{-\frac{1}{2}(d-e+1)}\frac{1}{(2\pi)^{n/2}}\int_{0}^{\infty}\int_{0}^{\infty}\mathds{1}_{\bigl{\\{}\frac{r^{2}}{r^{2}+s^{2}}>\frac{1+\|\nabla\ell\|^{2}}{\sigma^{2}}b^{2}\bigr{\\}}}r^{d-e}s^{n-d+e-2}e^{-\frac{1}{2}(r^{2}+s^{2})}\mathrm{d}r\,\mathrm{d}s$
$\displaystyle=(1+\|\nabla\ell\|^{2})^{-\frac{1}{2}(d-e+1)}\frac{1}{2(2\pi)^{n/2}}\int_{\frac{1+\|\nabla\ell\|^{2}}{\sigma^{2}}b^{2}}^{1}t^{(d-e-1)/2}(1-t)^{(n-d+e-3)/2}\mathrm{d}t\int_{0}^{\infty}e^{-\frac{1}{2}R^{2}}R^{n-1}\mathrm{d}R$
$\displaystyle=(1+\|\nabla\ell\|^{2})^{-\frac{1}{2}(d-e+1)}\frac{1}{\Omega_{d-e+1}\Omega_{n-d+e-1}}\bar{B}_{\frac{1}{2}(d-e+1),\frac{1}{2}(n-d+e-1)}\biggl{(}\frac{1+\|\nabla\ell\|^{2}}{\sigma^{2}}b^{2}\biggr{)}.$

Combining (4.24), (4.25), and (4.27), we have (2.8).

Moreover, the point $u$ such that the argument of
$\bar{B}_{\frac{1}{2}(d-e+1),\frac{1}{2}(n-d+e-1)}(\cdot)$ is always greater
than or equal to 1 makes no contribution to (2.8). Hence, the range of
integration $M$ in (2.8) can be replaced with $M_{\mathrm{cri}}$. The proof of
Theorem 2.1 is completed. ∎

###  4.5. Tail probability of the maximum

In this subsection, we prove Theorem 2.7.

###### Proof of Theorem 2.7.

The right-hand side of (2.8) is denoted by $F(b)$. Because
$\mathbb{P}\bigl{(}Y_{\mathrm{max}}>~{}\cdot\bigr{)}$ is monotonically
decreasing, Theorem 2.1 implies that

$\mathds{1}(b\geq
b_{\mathrm{cri}})F(b)\leq\mathbb{P}\bigl{(}Y_{\mathrm{max}}>b\bigr{)}\leq\mathds{1}(b\geq
b_{\mathrm{cri}})F(b)+\mathds{1}(b<b_{\mathrm{cri}}),$

or

$-\mathds{1}(b<b_{\mathrm{cri}})F(b)\leq\mathbb{P}\bigl{(}Y_{\mathrm{max}}>b\bigr{)}-F(b)\leq-\mathds{1}(b<b_{\mathrm{cri}})F(b)+\mathds{1}(b<b_{\mathrm{cri}}).$

We use (1.3). By letting $b:=c/\|\xi\|^{2}$, and taking the expectation with
respect to $\|\xi\|^{2}\sim\chi^{2}_{n}$,

$|\mathbb{P}\bigl{(}X_{\mathrm{max}}>c\bigr{)}-\mathbb{E}[F(c/\|\xi\|^{2})]|\leq(C+1)\mathbb{P}(c/\|\xi\|^{2}<b_{\mathrm{cri}})=(C+1)\bar{G}_{n}(c/b_{\mathrm{cri}}),$

where $C=\max_{0\leq b\leq\sigma_{0}}|F(b)|<\infty$, $\sigma_{0}=\max_{u\in
M}\sigma(u)$. Noting that the relationship

$\mathbb{E}\Bigl{[}\bar{B}_{\frac{1}{2}k,\frac{1}{2}(n-k)}(c^{2}/\|\xi\|^{2})\Bigr{]}=\bar{G}_{k}(c^{2}),\quad\|\xi\|^{2}\sim\chi^{2}_{n},$

we see that $\mathbb{E}[F(c/\|\xi\|^{2})]$ is
$\mathbb{P}_{\mathrm{tube}}\bigl{(}X_{\mathrm{max}}>c\bigr{)}$ in (2.15). ∎

###  4.6. Laplace approximation

In this subsection, we prove Theorem 2.9 and Corollary 2.10.

###### Proof of Theorem 2.9 and Corollary 2.10.

The leading term of the tube formula (2.15) is

$\displaystyle\mathbb{P}_{\mathrm{tube}}\bigl{(}X_{\mathrm{max}}>c\bigr{)}\sim$
$\displaystyle\frac{1}{\Omega_{d+1}}\int_{M}\mathrm{d}u\frac{\det(I_{d}+CG^{-1})}{(1+\|\nabla\ell\|^{2})^{\frac{1}{2}(d+1)}}\,\bar{G}_{d+1}\biggl{(}\frac{1+\|\nabla\ell\|^{2}}{\sigma^{2}}c^{2}\biggr{)}$
$\displaystyle=$
$\displaystyle\frac{1}{\Omega_{d+1}}\int_{M}\mathrm{d}u\frac{\det(I_{d}+CG^{-1})}{(1+\|\nabla\ell\|^{2})^{\frac{1}{2}(d+1)}}\int_{t>\frac{1+\|\ell\|^{2}}{\sigma^{2}}c^{2}}\mathrm{d}t\,t^{\frac{1}{2}(d+1)-1}e^{-t/2}$
$\displaystyle=$
$\displaystyle\frac{1}{(2\pi)^{\frac{1}{2}(d+1)}}\int_{M}\mathrm{d}u\det(I_{d}+CG^{-1})\int_{r>c}\frac{\mathrm{d}r\,r^{d}}{\sigma^{d+1}}e^{-\frac{1}{2}\bigl{(}\frac{r}{\sigma}\bigr{)}^{2}(1+\|\nabla\ell\|^{2})}.$

Here, we treat the two typical cases: (i) The maximum of the variance is
attained at only $M_{0}=\\{u_{0}\\}\subset M$. (ii) The minimizer of variance
$M_{0}$ forms a $d_{0}$-dimensional closed manifold.

The values evaluated at $u=u_{0}$ are denoted by the subscript 0, i.e.,
$\sigma_{0}=\sigma(u_{0})$, $G_{0}=G_{u_{0}}$, and $C_{0}=C_{u_{0}}$.

Case (i). Let

$q=\frac{1}{\sigma^{2}}(1+\|\nabla\ell\|^{2})=e^{-2\ell}(1+\ell_{k}\ell_{l}g^{kl}).$

The derivative with respect to $t^{i}$ is denoted by the subscript, for
example, $\ell_{i}=\partial\ell/\partial t^{i}$. Because
$\ell_{i}|_{u_{0}}=0$,

$\displaystyle\partial_{i}q=$
$\displaystyle-2\ell_{i}e^{-2\ell}(1+\ell_{k}\ell_{l}g^{kl})+e^{-2\ell}(2\ell_{ik}\ell_{l}g^{kl}+\ell_{k}\ell_{l}\partial_{i}g^{kl}),$
$\displaystyle\partial_{i}q|_{u_{0}}=$ $\displaystyle 0,$
$\displaystyle\partial_{i}\partial_{j}q|_{u_{0}}=$
$\displaystyle-2\ell_{ij}e^{-2\ell}+e^{-2\ell}(2\ell_{ik}\ell_{jl}g^{kl})=\frac{2}{\sigma^{2}}(-\ell_{ij}+\ell_{ik}g^{kl}\ell_{lj}).$

In matrix notation,

$\bigl{(}\partial_{i}\partial_{j}q|_{u_{0}}\bigr{)}_{1\leq i,j\leq
d}=C_{0}+C_{0}G_{0}^{-1}C_{0}$

and about the point $u_{0}=u(t)|_{t=0}$,

(4.28)
$q=\frac{1}{\sigma_{0}^{2}}\bigl{(}1+t^{\top}(C_{0}+C_{0}G_{0}^{-1}C_{0})t\bigr{)}+o(\|t\|^{2}),\
\ t=(t^{1},\ldots,t^{d})^{\top},$

where $u_{0}=\varphi(0)$. The volume element at $u_{0}$ is
$\mathrm{d}u=\det(G_{0})^{\frac{1}{2}}\prod_{i=1}^{d}\mathrm{d}t_{i}$.
Therefore, according to the standard arguments of Laplace’s method (e.g.,
[Erd56]),

$\displaystyle\mathbb{P}_{\mathrm{tube}}\bigl{(}X_{\mathrm{max}}>c\bigr{)}\sim$
$\displaystyle\frac{1}{(2\pi)^{\frac{1}{2}(d+1)}}\det(I_{d}+C_{0}G_{0}^{-1})\int_{r>c}\frac{\mathrm{d}r\,r^{d}}{\sigma_{0}^{d+1}}\,e^{-\frac{1}{2}\bigl{(}\frac{r}{\sigma_{0}}\bigr{)}^{2}}$
$\displaystyle\times(2\pi)^{d/2}\det(C_{0}+C_{0}G_{0}^{-1}C_{0})^{-\frac{1}{2}}\det(G_{0})^{\frac{1}{2}}\biggl{(}\frac{r}{\sigma_{0}}\biggr{)}^{-d}$
$\displaystyle\sim$
$\displaystyle\det(I_{d}+C_{0}G_{0}^{-1})^{\frac{1}{2}}\det(C_{0}G_{0}^{-1})^{-\frac{1}{2}}\bar{\Phi}\biggl{(}\frac{c}{\sigma_{0}}\biggr{)},$

where

$C_{0}=(-\nabla_{i}\nabla_{j}\ell(u_{0}))_{1\leq i,j\leq
d}=(-\nabla_{i}\nabla_{j}\log\sigma(u_{0}))_{1\leq i,j\leq d}.$

Case (ii). When $\dim M_{0}=d_{0}>0$, the matrix $C+CG^{-1}C$ evaluated at
$u\in M_{0}$ in (4.28) is of rank $d-d_{0}$. By redefining the local
coordinates $t=(t^{i})$, we can assume that $C$ at $u$ is of the form
$C=\operatorname{diag}(0,C_{22})$ where $C_{22}$ is
$(d-d_{0})\times(d-d_{0})$. Accordingly, we write

$G=\begin{pmatrix}G_{11}&G_{12}\\\ G_{21}&G_{22}\end{pmatrix},\ \ \mbox{and
let}\ \ G_{22\cdot 1}=G_{22}-G_{21}G_{11}^{-1}G_{12}.$

Then, for $u\approx u_{0}=u(t)|_{t=0}\in M_{0}$,

$q(u)=\frac{1}{\sigma_{0}^{2}}\bigl{(}1+\bar{t}^{\top}(C_{22}+C_{22}G_{22\cdot
1}^{-1}C_{22})|_{u_{0}}\bar{t}\bigr{)}+o(\|\bar{t}\|^{2}),\ \
\bar{t}=\bigl{(}t^{d_{0}+1},\ldots,t^{d}\bigr{)}^{\top},$
$\mathrm{d}u=\mathrm{d}u_{0}\prod_{i=d_{0}+1}^{d}\mathrm{d}t^{i}\det(G_{22\cdot
1})^{\frac{1}{2}},\quad\mathrm{d}u_{0}=\det(G_{11})^{\frac{1}{2}}\prod_{i=1}^{d_{0}}\mathrm{d}t^{i},$

and

$\displaystyle\mathbb{P}_{\mathrm{tube}}\bigl{(}X_{\mathrm{max}}>c\bigr{)}\sim$
$\displaystyle\frac{1}{(2\pi)^{\frac{1}{2}(d+1)}}\int_{M_{0}}\mathrm{d}u_{0}\det(I_{d-d_{0}}+C_{22}G_{22\cdot
1}^{-1})\int_{r>c}\frac{\mathrm{d}r\,r^{d}}{\sigma_{0}^{d+1}}e^{-\frac{1}{2}\bigl{(}\frac{r}{\sigma_{0}}\bigr{)}^{2}}$
$\displaystyle\times(2\pi)^{\frac{1}{2}(d-d_{0})}\det(C_{22}+C_{22}G_{22\cdot
1}^{-1}C_{22})^{-\frac{1}{2}}\det(G_{22\cdot
1})^{\frac{1}{2}}\biggl{(}\frac{r}{\sigma_{0}}\biggr{)}^{-(d-d_{0})}$
$\displaystyle\sim$
$\displaystyle\frac{1}{\Omega_{d_{0}+1}}\int_{M_{0}}\mathrm{d}u_{0}\frac{\det(I_{d-d_{0}}+C_{22}G_{22\cdot
1}^{-1})^{\frac{1}{2}}}{\det(C_{22}G_{22\cdot
1}^{-1})^{\frac{1}{2}}}\times\bar{G}_{d_{0}+1}\biggl{(}\frac{c^{2}}{\sigma_{0}^{2}}\biggr{)}$
$\displaystyle=$
$\displaystyle\frac{1}{\Omega_{d_{0}+1}}\int_{M_{0}}\mathrm{d}u_{0}\frac{\det(I_{d}+CG^{-1})^{\frac{1}{2}}}{\operatorname{tr}_{d-d_{0}}(CG^{-1})^{\frac{1}{2}}}\times\bar{G}_{d_{0}+1}\biggl{(}\frac{c^{2}}{\sigma_{0}^{2}}\biggr{)}.$

The asymptotic equivalence on the last line is due to the uniformity of the
Laplace approximation on the compact set $M_{0}$. ∎

###  4.7. Approximation error of the Bonferroni method

In this subsection, we prove Theorem 2.11 and Proposition 2.12. For this
purpose, we will provide analogues to Theorems 2.1 and 2.3 when $M$ is a
finite set.

###### Proof of Theorem 2.11.

Recall that $X_{i}=\sigma_{i}\langle\varphi^{(i)},\xi\rangle$,
$Y_{i}=\sigma_{i}\langle\varphi^{(i)},\xi/\|\xi\|\rangle$,
$\xi\sim\mathcal{N}_{n}(0,I_{n})$, and
$\rho_{ij}=\langle\varphi^{(i)},\varphi^{(j)}\rangle$. If $b$ is sufficiently
large, say $b\geq b_{\mathrm{cri}}$, the events $\\{Y_{i}>b\\}$ are mutually
exclusive, and

$\mathbb{P}\bigl{(}\max\\{Y_{1},\ldots,Y_{K}\\}>b\bigr{)}=\sum_{i=1}^{K}\mathbb{P}(Y_{i}>b),$

which is (2.18).

To identify the threshold $b_{\mathrm{cri}}$, we will follow the proof in
Section 4.3. Let $i$ be fixed. Let

$N_{i}=T_{\varphi^{(i)}}(\mathbb{S}^{n-1})$

be the space in $\mathbb{R}^{n}$ orthogonal to $\varphi^{(i)}$. Almost all
$z\in\mathbb{R}^{n}$ is written as

$z=r\varphi^{(i)}+sv,\ \ (r,s,v)\in Q_{i},$

where

$Q_{i}=\\{(r,s,v)\mid r>0,\ s\geq 0,\ v\in S(N_{i})\\}.$

Let

$V_{i}=\\{z\in\mathbb{R}^{n}\mid\sigma_{j}\langle\varphi^{(j)},z\rangle<\sigma_{i}\langle\varphi^{(i)},z\rangle,\
z=r\varphi^{(i)}+sv,\ \forall j\neq i\\}.$

Instead of (4.11), we will find the threshold $b_{\mathrm{cri}}^{\prime}(i)$
of $b$ satisfying $A_{i}(b)\subset B_{i}$ a.s., where

$\displaystyle A_{i}(b)=$ $\displaystyle\,\\{(r,s,v)\in
Q_{i}\mid\sigma_{i}\langle\varphi^{(i)},z/\|z\|\rangle>b,\,z=r\varphi^{(i)}+sv,\
\forall j\neq i\\},$ $\displaystyle B_{i}=$ $\displaystyle\,\\{(r,s,v)\in
Q_{i}\mid z=r\varphi^{(i)}+sv\in V_{i}\\}.$

The threshold $b_{\mathrm{cri}}$ in Theorem 2.11 is given by $\max_{i\in
I}b_{\mathrm{cri}}^{\prime}(i)$.

As in (4.12), the set $A_{i}(b)$ is

$A_{i}(b)=\bigl{\\{}(r,s,v)\in
Q_{i}\mid(s/r)^{2}<\sigma_{i}^{2}/b^{2}-1\bigr{\\}}.$

Let $I=\\{1,\ldots,K\\}$. We divide the index set $I\setminus\\{i\\}$ as
$I_{i}^{+}\sqcup I_{i}^{-}$, where

$I_{i}^{+}=\bigl{\\{}j\in
I\mid\sigma_{i}/\sigma_{j}-\rho_{ij}>0\bigr{\\}},\qquad
I_{i}^{-}=\bigl{\\{}j\in I\mid\sigma_{i}/\sigma_{j}-\rho_{ij}\leq
0\bigr{\\}}\setminus\\{i\\}.$

(i) When $I_{i}^{-}=\emptyset$,

$\displaystyle B_{i}=$ $\displaystyle\biggl{\\{}(r,s,v)\in
Q_{i}\mid\frac{s}{r}<\min_{j\in
I\setminus\\{i\\}:\langle\varphi^{(j)},v\rangle>0}\frac{\sigma_{i}/\sigma_{j}-\rho_{ij}}{\langle\varphi^{(j)},v\rangle}\biggr{\\}}.$

Therefore, $A_{i}(b)\subset B_{i}$ iff

$\displaystyle\sqrt{\frac{\sigma_{i}^{2}}{b^{2}}-1}<$ $\displaystyle\inf_{v\in
S(N_{i})}\min_{j\in
I\setminus\\{i\\}:\langle\varphi^{(j)},v\rangle>0}\frac{\sigma_{i}/\sigma_{j}-\rho_{ij}}{\langle\varphi^{(j)},v\rangle}$
$\displaystyle=$ $\displaystyle\min_{j\in I\setminus\\{i\\}}\inf_{v\in
S(N_{i}):\langle\varphi^{(j)},v\rangle>0}\frac{\sigma_{i}/\sigma_{j}-\rho_{ij}}{\langle\varphi^{(j)},v\rangle}=\min_{j:j\neq
i}\frac{\sigma_{i}/\sigma_{j}-\rho_{ij}}{\sqrt{1-\rho_{ij}^{2}}}.$

Here we used

$\sup_{v\in
S(N_{i}):\langle\varphi^{(j)},v\rangle>0}\langle\varphi^{(j)},v\rangle=\sqrt{1-\rho_{ij}^{2}}.$

(ii) When $I_{i}^{-}\neq\emptyset$,

$B_{i}=\biggl{\\{}(r,s,v)\in
Q_{i}\mid\mbox{``}\langle\varphi^{(j)},v\rangle<0\ \mbox{and}\
\frac{s}{r}>\frac{-\sigma_{i}/\sigma_{j}+\rho_{ij}}{-\langle\varphi^{(j)},v\rangle}\mbox{''},\
\ \forall j\in I\setminus\\{i\\}\biggr{\\}}.$

$A_{i}(b)\subset B_{i}$ holds iff $A_{i}(b)=\emptyset$. That is,

$\frac{\sigma_{i}^{2}}{b^{2}}-1\leq 0.$

Combining (i) and (ii), $b_{\mathrm{cri}}^{\prime}(i)$ is given by

$\sqrt{\frac{\sigma_{i}^{2}}{b_{\mathrm{cri}}^{\prime}(i)^{2}}-1}=\min_{j\in
I\setminus\\{i\\}}\frac{(\sigma_{i}/\sigma_{j}-\rho_{ij})_{+}}{\sqrt{1-\rho_{ij}^{2}}},$

or equivalently,

$b_{\mathrm{cri}}^{\prime}(i)^{2}=\frac{\sigma_{i}^{2}}{1+\min_{j\in
I\setminus\\{i\\}}h(i,j)},$

where $h(i,j)$ is given in (2.20). The threshold in Theorem 2.11 is
$b_{\mathrm{cri}}=\max_{i\in i}b_{\mathrm{cri}}^{\prime}(i)$. In (2.18), the
term
$\mathbb{P}(Y_{i}>b)=(1/2)\bar{B}_{\frac{1}{2},\frac{1}{2}(n-1)}(b^{2}/\sigma_{i}^{2})$
for $\sigma_{i}<b_{\mathrm{cri}}$ is always zero and can be omitted.

Finally, we apply the proof of Theorem 2.7 in Section 4.5 to (2.18) to prove
(2.19) of Theorem 2.11.

∎

###### Proof of Proposition 2.12.

We evaluate

(4.29)
$\frac{1+h(i,j)}{\sigma_{i}^{2}}=\frac{1}{\sigma_{i}^{2}}\biggl{(}1+\frac{(\sigma_{i}/\sigma_{j}-\rho_{ij})_{+}^{2}}{1-\rho_{ij}^{2}}\biggr{)}.$

(i) In the case $\rho_{ij}<1\leq\sigma_{i}/\sigma_{j}$,

$(\ref{(1+h)/s})\geq\frac{1}{\sigma_{i}^{2}}\biggl{(}1+\frac{(1-\rho_{ij})^{2}}{1-\rho_{ij}^{2}}\biggr{)}=\frac{1}{\sigma_{i}^{2}}\frac{2}{1+\rho_{ij}}.$

(ii) In the case $\rho_{ij}<\sigma_{i}/\sigma_{j}<1$,

$\displaystyle(\ref{(1+h)/s})=\frac{1}{\sigma_{i}^{2}}\biggl{(}1+\frac{(\sigma_{i}/\sigma_{j}-\rho_{ij})^{2}}{1-\rho_{ij}^{2}}\biggr{)}=$
$\displaystyle\frac{1}{\sigma_{j}^{2}}\biggl{(}1+\frac{(\sigma_{j}/\sigma_{i}-\rho_{ij})^{2}}{1-\rho_{ij}^{2}}\biggr{)}$
$\displaystyle>$
$\displaystyle\frac{1}{\sigma_{j}^{2}}\biggl{(}1+\frac{(1-\rho_{ij})^{2}}{1-\rho_{ij}^{2}}\biggr{)}=\frac{1}{\sigma_{j}^{2}}\frac{2}{1+\rho_{ij}}.$

(iii) In the case $\sigma_{i}/\sigma_{j}\leq\rho_{ij}<1$,

$(\ref{(1+h)/s})=\frac{1}{\sigma_{i}^{2}}\geq\frac{1}{\sigma_{j}^{2}\rho_{ij}^{2}}>\frac{1}{\sigma_{j}^{2}}\frac{2}{1+\rho_{ij}}.$

In each case,

$(\ref{(1+h)/s})\geq\min\biggl{\\{}\frac{1}{\sigma_{i}^{2}},\frac{1}{\sigma_{j}^{2}}\biggr{\\}}\frac{2}{1+\rho_{ij}},$

and hence

$\frac{1}{b_{\mathrm{cri}}^{2}}\geq\min_{i\neq
j}\frac{1}{\sigma_{i}^{2}}\frac{2}{1+\rho_{ij}}\geq\biggl{(}\min_{i}\frac{1}{\sigma_{i}^{2}}\biggr{)}\biggl{(}\min_{i<j}\frac{2}{1+\rho_{ij}}\biggr{)}.$

∎

### Acknowledgments

The authors are grateful to Robert Adler for bringing [Ros15] to their
attention.

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