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\begin{document}

\title{On the tails of the distribution of the maximum of a smooth stationary
Gaussian process}
\author{Jean-Marc Bardet $^{1}$, Mario Wschebor $^{2}$ ~ \\
%EndAName
~\\
$^{1}$ Laboratoire de Statistique et de Probabilit\'es, URA CNRS 745, \\
Universit\'e Paul Sabatier, 118 route de Narbonne, \\
31062 Toulouse Cedex, France.\\
$^{2}$ Centro de Matem\'atica, Facultad de Ciencias, Universidad de la
Rep\'ublica,\\
Calle Igua $4225$, $11400$ Montevideo, Uruguay. \\
}
\maketitle

\begin{abstract}
This paper deals with the asymptotic behavior when the level tends to $%
+\infty $, of the tail of the distribution of the maximum of a stationary
Gaussian process on a fixed interval of the line. For processes satisfying
certain regularity conditions, we give a second order term for this
asymptotics.
\end{abstract}

\vspace{1cm} \noindent \newpage

\emph{Mathematics Subject Classification (1991):~}60Gxx, 60E05, 60G15, 65U05.

\emph{Key words:} Tail of Distribution of the Maximum, Stationary Gaussian
processes

\emph{Short Title:} Distribution of the Maximum.

\section{Introduction}

$X=\{X(t),t\in \lbrack 0,T]\}$, $T>0$ is a real-valued centered stationary
Gaussian process with continuous paths and $M_{T}=\underset{t\in \left[ 0,T%
\right] }{\max }X(t)$. We denote $F(u)=P(M_{T}\leq u)$ the distribution
function of the random variable $M_{T}$, $r(t)=E\left\{ X(s)X(s+t)\right\} $
the covariance function and $\lambda _{k}~(k=0,1,2,...)$ the spectral
moments of the process, whenever they are defined. With no loss of
generality we will assume that $\lambda _{0}=r(0)=1.$

Under certain regularity conditions, Piterbarg (1981, Theorem 2.2.) proved
that for each $T>0$ and any $u\in R$: 
\begin{equation}
\left | 1-\Phi (u)+\sqrt{\frac{\lambda _{2}}{2\pi }}T\phi
(u)-P(M_{T}>u)\right | \leq B~\exp \left( -\frac{u^{2}}{1+\rho }\right)
\label{piterbarg 1}
\end{equation}
for some constants $B>0$ and $\rho <1.$ $\Phi $ (respectively $\phi $)
denotes the standard normal distribution (respectively density).

The aim of this paper is to improve the description of the asymptotic
behavior of $P(M_{T}>u)$ as $u\rightarrow +\infty $ that follows from (\ref
{piterbarg 1}) replacing the bound for the error by an equivalent as $%
u\rightarrow +\infty $.\ More precisely, under the regularity conditions
required in Theorem \ref{jmbmw}, we will prove that: 
\begin{equation}
P(M_{T}\hspace{-1mm}>\hspace{-1mm}u) \hspace{-1mm} =\hspace{-1mm}\hspace{-1mm%
}1-\hspace{-1mm}\Phi (u)+\sqrt{\frac{\lambda _{2}}{2\pi }}T\phi (u)\hspace{%
-1mm}-\hspace{-1mm}\frac{T}{2\pi}(\lambda _{4}\hspace{-1mm}-\hspace{-1mm}%
\lambda _2^2)\sqrt{\frac 3 {\hspace{-1mm}\lambda_2 \lambda_4}}\hspace{-1mm} %
\left[ 1\hspace{-1mm}-\hspace{-1mm}\Phi \left( \sqrt{\frac{\lambda _{4}}{%
\lambda _{4}\hspace{-1mm}-\hspace{-1mm}\lambda _{2}^{2}}}u\right) \right] %
\left[ 1+o(1)\right]  \label{barmw}
\end{equation}

This contradicts Theorem 3.1. in Piterbarg's paper in which a different
equivalent is given in case $T$ is small enough (see also Aza\"is {t al.}
(1999)).

We will assume further that $X$ has $\mathcal{C}^{4}$ sample paths (this
implies $\lambda _{8}<\infty $) and that for every $n\geq 1$ and pairwise
different values $t_{1},..,t_{n}$ in $[0,T]$, the distribution of the set of 
$5n$ random variables ($X^{(j)}(t_{1}),..,X^{(j)}(t_{n}),j=0,1,2,3,4)$ is
non-degenerate. A sufficient condition for this to hold is the spectral
measure of the process not to be purely atomic or, if it is purely atomic,
that the set of atoms have an accumulation point in the real line (A proof
of this facts can be done in the same way as in Chap. 10 of Cram\'{e}r and
Leadbetter, 1967).

If $\xi $ is a random vector with values in \textsl{R}$^{n}$ whose
distribution has a density with respect to Lebesgue measure, we denote by \ $%
p_{\xi }(x)$ the density of $\xi $ at the point $x\in $\textsl{R}$^{n}$. 
{\large 1}$_{C}$ denotes the indicator function of the set $C$.

If $Y=\left\{ Y(t):t\in R\right\} $ is a process in $L^{2}$ we put $\Gamma
^{Y}(s,t)$ for its covariance function and $\displaystyle{\Gamma
_{ij}^{Y}(s,t)=\frac{\partial ^{i+j}\Gamma ^{Y}}{\partial s^{i}\partial t^{j}%
}(s,t)}$ for the partial derivatives, whenever they exist.

The proof \ of (\ref{barmw}) will consist in computing the density of the
distribution of the random variable $M_{T}$ and studying its asymptotic
behavior as $u\rightarrow +\infty $. Our main tool is the\ following
proposition which is a special case of the differentiation Lemma 3.3 in
Aza\"{i}s and Wschebor (1999):

\begin{prop}
\label{jmamw}Let $Y$ be a Gaussian process with $\mathcal{C}^{2}$ paths and
such that for every $n\geq 1$ and pairwise different values $t_{1},..,t_{n}$
in $[0,T]$ the distribution of the set of $3n$ random variables ($%
Y^{(j)}(t_{1}),..,Y^{(j)}(t_{n}),~j=0,1,2)$ is non-degenerate. Assume also
that $E\left\{ Y(t)\right\} =0,~\Gamma ^{Y}(t,t)=E\left\{ Y^{2}(t)\right\}
=1 $.

Then, if $\beta $ is a $\mathcal{C}^{2}$-function on $[0,T]$, 
\begin{equation*}
\frac{d}{du}P(Y(t)\leq u\beta (t),\forall t\in \lbrack
0,T])=L_{1}^{Y}(u,\beta )+L_{2}^{Y}(u,\beta )+L_{3}^{Y}(u,\beta ),~ %
\mbox{with}
\end{equation*}
\vspace{-0.5cm} 
\begin{equation}
L_{1}^{Y}(u,\beta )\hspace{-0.3cm}~=~\hspace{-0.3cm}\beta (0)P(Y^{\vdash
}(s)\leq u\beta ^{\vdash }(s),\forall s\in \lbrack 0,T]).p_{Y(0)}(\beta (0)u)
\label{L1}
\end{equation}
\vspace{-0.5cm} 
\begin{equation}
L_{2}^{Y}(u,\beta )\hspace{-0.3cm}~=\hspace{-0.3cm}~\beta (T)P(Y^{\dashv
}(s)\leq u\beta ^{\dashv }(s),\forall s\in \lbrack 0,T]).p_{Y(T)}(\beta (T)u)
\label{L2}
\end{equation}
\vspace{-0.5cm} 
\begin{equation}
\hspace{-3mm} L_{3}^{Y}(u,\beta )\hspace{-0.1cm}=\hspace{-0.1cm}-\hspace{%
-0.2cm}\int_{0}^{T}\hspace{-0.3cm}\beta (t)E\left\{ (Y^{t}(t) \hspace{-0.1cm}%
-\hspace{-0.1cm}\beta ^{t}(t)u)\mbox{\large 1}_{\{Y^{t}(s)\leq u\beta
^{t}(s),\forall s\in \lbrack 0,T]\}}\right\} p_{(Y(t),Y^{\prime }(t))}(\beta
(t)u,\beta ^{\prime }(t)u)dt.  \label{L3}
\end{equation}
Here the functions $\beta ^{\vdash },~\beta ^{\dashv },~\hspace{-0.1cm}\beta
^{t}$ and the (random) functions $Y^{\vdash },~Y^{\dashv },~Y^{t}$ are the
continuous extensions to $\left[ 0,T\right] $ of: 
\begin{equation}
\beta ^{\vdash }(s)=\frac{1}{s}\left( \beta (s)-\Gamma ^{Y}(s,0)\beta
(0)\right) ,~Y^{\vdash }(s)=\frac{1}{s}\left( Y(s)-\Gamma
^{Y}(s,0)Y(0)\right) ~ \mbox{for~} 0<s\leq T,  \label{gauche}
\end{equation}
\vspace{-0.5cm} 
\begin{equation}
\beta ^{\dashv }(s)\hspace{-0.1cm}=\hspace{-0.1cm}\frac{1}{T\hspace{-0.15cm}-%
\hspace{-0.15cm}s}\hspace{-0.1cm}\left( \beta (s)\hspace{-0.1cm}-\hspace{%
-0.1cm}\Gamma ^{Y}(s,T)\beta (T)\right) ,~ \hspace{-0.05cm}Y^{\dashv }(s)%
\hspace{-0.1cm}=\hspace{-0.1cm}\frac{1}{T\hspace{-0.15cm}-\hspace{-0.15cm}s}%
\hspace{-0.1cm}\left( Y(s)\hspace{-0.1cm}-\hspace{-0.1cm}\Gamma
^{Y}(s,T)Y(T)\right) ~ \hspace{-0.05cm} \mbox{for~} 0\hspace{-0.1cm}\leq 
\hspace{-0.05cm}s \hspace{-0.05cm}\hspace{-0.1cm}<T,  \label{droite}
\end{equation}
\vspace{-0.5cm} 
\begin{equation}
\beta ^{t}(s)=\frac{2}{(s-t)^{2}}\left[ \beta (s)-\Gamma ^{Y}(s,t)\beta (t)-%
\frac{\Gamma _{10}^{Y}(t,s)}{\Gamma _{11}^{Y}(t,t)}\beta ^{\prime }(t)\right]
\quad 0\leq s\leq T,~s\neq t,  \label{betat}
\end{equation}
\vspace{-0.5cm} 
\begin{equation}
Y^{t}(s)=\frac{2}{(s-t)^{2}}\left[ Y(s)-\Gamma ^{Y}(s,t)Y(t)-\frac{\Gamma
_{10}^{Y}(t,s)}{\Gamma _{11}^{Y}(t,t)}Y^{^{\prime }}(t)\right] \quad 0\leq
s\leq T,~s\neq t.  \label{Yt}
\end{equation}
\end{prop}

\bigskip We will repeatedly use the following Lemma. Its proof is elementary
and we omit it.

\begin{lem}
\label{laplace} Let $f$ and $g$ be real-valued functions of class $C^{2}$
defined on the interval $\left[ 0,T\right] $ of the real line verifying the
conditions:

1) $f$ has a unique minimum on $\left[ 0,T\right] $ at the point $t=t^{\ast
} $, and $f^{\prime }(t^{\ast })=0,$ $f"(t^{\ast })>0.$

2) Let $k=\inf \left\{ j:g^{(j)}(t^{\ast })\neq 0\right\} $ and suppose $%
k=0~,1~or~2$.

Define 
\begin{equation*}
h(u)=\int_{0}^{T}~g(t)~\exp \left[ -\frac{1}{2}u^{2}f(t)\right] ~dt.
\end{equation*}

Then, as $u\rightarrow \infty $: 
\begin{equation*}
h(u)\thickapprox \frac{g^{(k)}(t^{\ast })}{k!}\frac{1}{u^{k+1}}\exp \left[ -%
\frac{1}{2}u^{2}f(t^{\ast })\right] ~\int_{J}x^{k}~\exp \left[ -\frac{1}{4}%
f"(t^{\ast })x^{2}\right] ~dx,
\end{equation*}

where $J=\left[ 0,+\infty \right) ,~J=\left( -\infty ,0\right] ~or~J=\left]
-\infty ,+\infty \right[ $ according as $t^{\ast }=0,~t^{\ast
}=T~or~0<t^{\ast }<T$ respectively.
\end{lem}

We now turn to our result.

\begin{theo}
\label{jmbmw}Let $X=\left\{ X(t):t\in \left[ 0,T\right] \right\} $ be a
Gaussian centered stationary process with $C^{4}$-paths, covariance $r(.)$, $%
\lambda _{0}=1$, and such that for every $n\geq 1$ and pairwise different $%
t_{1},..,t_{n}$ in $[0,T]$, the distribution of the set of $\ 5n$ \ random
variables ($X^{(j)}(t_{1}),..,X^{(j)}(t_{n}),j=0,1,2,3,4)$ is
non-degenerate. We shall also assume the additional hypothesis that $%
r^{\prime }<0$ in a set dense in $[ 0,T] $.

Then (\ref{barmw}) holds true.
\end{theo}

\textsl{Proof}. We divide the proof into several steps.\newline

\textsl{Step 1}. Proposition\ref{jmamw} applied to the process $Y=X$ and the
function $\beta (t)=1$ for all $t\in \left[ 0,T\right] $ enables to write
the density $p_{M_{T}}$ of the distribution of the maximum $M_{T}$ as: 
\begin{equation}
p_{M_{T}}(u)=\left[ A_{1}(u)+A_{2}(u)+A_{3}(u)\right] .\phi (u),~~\mbox{with}
\label{primeraderivada}
\end{equation}
\vspace{-0.5cm} 
\begin{equation*}
A_{1}(u)=P(X^{\vdash }(s)\leq u\beta ^{\vdash }(s),\forall s\in \lbrack
0,T]),
\end{equation*}
\vspace{-0.5cm} 
\begin{equation*}
A_{2}(u)=P(X^{\dashv }(s)\leq u\beta ^{\dashv }(s),\forall s\in \lbrack
0,T]),
\end{equation*}
\vspace{-0.5cm} 
\begin{equation*}
A_{3}(u)=-\frac{1}{\sqrt{2\pi \lambda _{2}}}\int_{0}^{T}\hspace{-0.2cm}%
E\left\{ (X^{t}(t)\hspace{-0.1cm}-\hspace{-0.1cm}\beta ^{t}(t)u) %
\mbox{\large 1}_{\{X^{t}(s)\leq u\beta ^{t}(s),\forall s\in \lbrack
0,T]\}}\right\} dt.
\end{equation*}
Since $X$ is a stationary process and $\beta (t)\equiv 1$, it follows that
the processes $X$ and $\widetilde{X}$ - defined as $\widetilde{X}(t)=X(T-t)$
- have the same law, so that $P(X(s)\leq u$ for all $s\in \left[ 0,T\right]
~|X(0)=u)=P(X(s)\leq u$ for all $s\in \left[ 0,T\right] ~|X(T)=u)$. Hence, $%
A_{1}(u)=A_{2}(u)$.\newline

\textsl{Step 2}. We now consider $A_{1}(u)$ and write it in the form: 
\begin{equation*}
A_{1}(u)=P(Y(s)\leq u\gamma (s),\forall s\in \lbrack 0,T])
\end{equation*}
where $Y$ is the continuous extension to $\left[ 0,T\right] $ of: 
\begin{equation*}
Y(s)=\frac{X^{\vdash }(s)}{(E\left\{ \left[ X^{\vdash }(s)\right]
^{2}\right\} )^{1/2}}=\frac{s.X^{\vdash }(s)}{\left[ 1-r^{2}(s)\right] ^{%
\frac{1}{2}}}\quad s\in ]0,T]
\end{equation*}
and 
\begin{equation*}
\gamma (s)=\frac{\beta ^{\vdash }(s)}{(E\left\{ \left[ X^{\vdash }(s)\right]
^{2}\right\} )^{1/2}}=\sqrt{\frac{1-r(s)}{1+r(s)}}\quad s\in \lbrack 0,T].
\end{equation*}

Thus, Proposition \ref{jmamw} can be applied and: 
\begin{equation}
\frac{d}{du}A_{1}(u)=L_{1}^{Y}(u,\gamma )+L_{2}^{Y}(u,\gamma
)+L_{3}^{Y}(u,\gamma ).  \label{derivadaA1}
\end{equation}
$\bullet ~L_{1}^{Y}(u,\gamma )=0$ because $\gamma (0)=0$. \newline
\newline
$\bullet ~L_{2}^{Y}(u,\gamma )=\gamma (T)P(Y^{\dashv }(s)\leq u\gamma
^{\dashv }(s),\forall s\in \lbrack 0,T])\phi (\gamma (T)u)$ , with 
\begin{equation*}
\ \gamma ^{\dashv }(s)=\frac{1+r(T)-r(s)-r(T-s)}{(T-s)(1+r(T))\sqrt{%
1-r^{2}(s)}}\quad s\in ]0,T[\quad
\end{equation*}
\begin{equation*}
\gamma ^{\dashv }(0)=\frac{r^{\prime }(T)}{T(1+r(T))\sqrt{\lambda _{2}}}\leq
0\quad \text{and\quad }\gamma ^{\dashv }(T)=\frac{r^{\prime }(T)}{(1+r(T))%
\sqrt{1-r^{2}(T)}}\leq 0.
\end{equation*}
On the other hand: 
\begin{equation*}
P\left( Y^{\dashv }(s)\leq u\gamma ^{\dashv }(s),\forall s\in \lbrack
0,T]\right) \leq P\left( Y^{\dashv }(T)\leq u\gamma ^{\dashv }(T)\right) .
\end{equation*}
Check that 
\begin{equation*}
E(Y^{\dashv }(T))^{2})=E(Y^{\prime }(T))^{2}=\frac{\lambda
_{2}(1-r^{2}(T))-(r^{\prime }(T))^{2})}{(1-r^{2}(T))^{2}},
\end{equation*}
so that, since the non-degeneracy hypothesis implies that for each $T>0$, $%
E(Y^{\prime }(T))^{2})$ is non-zero, it follows that the numerator in the
right-hand member is stricly positive for $T>0$.\newline
Hence, 
\begin{equation}
P\left( Y^{\dashv }(s)\leq u\gamma ^{\dashv }(s),\forall s\in \lbrack
0,T]\right) \phi (\gamma (T)u)\leq C(T)\exp \left( -\frac{u^{2}}{2}%
F(T)\right) ,  \label{cota2}
\end{equation}
\bigskip $~~$with $~C(T)>0$, where $F(t)$ is the function 
\begin{equation*}
F(t)=\frac{\lambda _{2}(1-r(t))^{2}}{\lambda _{2}(1-r^{2}(t))-(r^{\prime
}(t))^{2}}~~~\mbox{for}~~t\in \lbrack -T,T],~t\neq 0.
\end{equation*}
which is well defined since the denominator does not vanish because of the
previous remark.

The following properties of the function $F$ are elementary and will be
useful in our calculations.

(a) $F$ has a continuous extension at $t=0$.

(b) $\displaystyle{F(t)>F(0)=\frac{\lambda _{2}^{2}}{\lambda _{4}-\lambda
_{2}^{2}}}$ for $t\neq 0$ because:\newline
\vspace{10mm} 
\begin{tabular}{ll}
\vspace{4mm}~~~~~~ & * $\displaystyle{F^{\prime}(t)=\frac
{2\lambda_2(1-r(t))r^{\prime}(t)((r^{\prime}(t))^2-(\lambda_2-r^{\prime%
\prime}(t))(1-r(t))} {(\lambda_2(1-r^2(t))-(r^{\prime}(t))^2)^2} }$ and \\ 
\vspace{4mm} & * $r^{\prime}(t) <0$ for $t\in A \subset [0,T]$ with $A$
dense in $[0,T]$, and \\ 
\vspace{4mm} & * For $t \neq 0$, \\ 
& $\displaystyle{(r^{\prime}(t))^2-(\lambda_2-r^{\prime\prime}(t))(1-r(t))=}$
\\ 
\vspace{4mm} & \hspace{1cm} $\displaystyle{\left ( E
(X^{\prime}(t)-X^{\prime}(0))(X(t)-X(0)) \right ) ^2- \left (E
(X^{\prime}(t)-X(0))^2\right ) \left ( E(X(t)-X(0))^2\right ) <0 }$, \\ 
\vspace{4mm} & from Cauchy-Schwartz inequality (and non-degeneracy
hypothesis).
\end{tabular}

(c) $F^{\prime }(0)=0.$

(d) $\displaystyle{F^{\prime \prime }(0)=\frac{\lambda _{2}(\lambda
_{2}\lambda _{6}-\lambda _{4}^{2})}{9(\lambda _{4}-\lambda _{2}^{2})^{2}}}$. 
\newline
\newline
From (\ref{cota2}) and (b) it follows that 
\begin{equation}
L_{2}^{Y}(u,\gamma )\leq D(T)\exp \left( -\frac{u^{2}}{2}\left( \delta (T)+%
\frac{\lambda _{2}^{2}}{\lambda _{4}-\lambda _{2}^{2}}\right) \right) ,
\label{l2}
\end{equation}
with $D(T)>0$ and $\delta (T)>0$ for all $T>0$. 
\begin{equation*}
\bullet ~L_{3}^{Y}(u,\gamma )=-\int_{0}^{T}\hspace{-0.2cm}\gamma (t)E\left\{
(Y^{t}(t)\hspace{-0.1cm}-\hspace{-0.1cm}\gamma ^{t}(t)u)1_{\{Y^{t}(s)\leq
u\gamma ^{t}(s),\forall s\in \lbrack 0,T]\}}\right\} p_{(Y(t),Y^{\prime
}(t))}(u\gamma (t),u\gamma ^{\prime }(t))dt,
\end{equation*}
with 
\begin{eqnarray*}
p_{(Y(t),Y^{\prime }(t))}(u\gamma (t),u\gamma ^{\prime }(t)) &=&\frac{1}{%
2\pi (E\left\{ (Y^{\prime }(t))^{2}\right\} )^{1/2}}\exp \left( -\frac{u^{2}%
}{2}\left( \gamma ^{2}(t)+\frac{(\gamma ^{\prime }(t))^{2}}{E\left\{
(Y^{\prime }(t))^{2}\right\} }\right) \right)  \\
&=&\frac{1}{2\pi (E\left\{ (Y^{\prime }(t))^{2}\right\} )^{1/2}}\exp \left( -%
\frac{u^{2}}{2}F(t)\right) .
\end{eqnarray*}
We know that \ \ $\displaystyle{\min_{t\in \lbrack 0,T]}F(t)=F(0)=\frac{%
\lambda _{2}^{2}}{\lambda _{4}-\lambda _{2}^{2}}}$.

Also check that: 
\begin{equation*}
\gamma (0)=0,\gamma ^{\prime }(0)=\frac{1}{2}\sqrt{\lambda _{2}}>0;\quad
\gamma ^{0}(0)=0;\quad \lim_{t\rightarrow 0}\frac{d\gamma ^{t}(t)}{dt}=\frac{%
1}{12}\frac{\lambda _{2}\lambda _{6}-\lambda _{4}^{2}}{\lambda _{4}-\lambda
_{2}^{2}}>0
\end{equation*}
and $E\left\{ (Y^{0}(0))^{2}\right\} >0$ from the non-degeneracy condition.

As a consequence:

\begin{equation*}
\left| \int_{0}^{T}\hspace{-0.2cm}\gamma (t)E\left\{ Y^{t}(t) \mbox{\large 1}%
_{\{Y^{t}(s)\leq u\gamma ^{t}(s),\forall s\in \lbrack 0,T]\}}\right\}
p_{(Y(t),Y^{\prime }(t))}(u\gamma (t),u\gamma ^{\prime }(t))dt\right| {\leq }
\end{equation*}
\begin{equation*}
\leq \int_{0}^{T}\hspace{-0.2cm}\gamma (t)\frac{(E\left\{
(Y^{t}(t))^{2}\right\} )^{1/2}}{2\pi (E\left\{ (Y^{\prime }(t))^{2}\right\}
)^{1/2}}\exp \left( -\frac{u^{2}}{2}F(t)\right) dt\leq C\int_{0}^{T}\hspace{%
-0.2cm}\gamma (t)\exp \left( -\frac{u^{2}}{2}F(t)\right) dt,
\end{equation*}
where $C$ is a positive constant. Also, from Lemma \ref{laplace} and the
properties of the functions $F$ and $\gamma $, one gets: 
\begin{equation*}
\int_{0}^{T}\hspace{-0.2cm}\gamma (t)\exp \left( -\frac{u^{2}}{2}F(t)\right)
dt\leq C_{1}\frac{1}{u^{2}}\exp \left( -\frac{u^{2}}{2}\frac{\lambda _{2}^{2}%
}{\lambda _{4}-\lambda _{2}^{2}}\right) .
\end{equation*}
$C_{1}$ a positive constant. Hence, %\begin{mathletters}
\begin{equation*}
\int_{0}^{T}\hspace{-0.2cm}\gamma (t)E\left\{ Y^{t}(t)\mbox{\large 1}%
_{\{Y^{t}(s)\leq u\gamma ^{t}(s),\forall s\in \lbrack 0,T]\}}\right\}
p_{(Y(t),Y^{\prime }(t))}(u\gamma (t),u\gamma ^{\prime }(t))dt=
\end{equation*}
%\end{mathletters}
\vspace{-0.5cm} 
\begin{equation}
=\mathcal{O}\left( \frac{1}{u^{2}}\exp \left( -\frac{u^{2}}{2}\frac{\lambda
_{2}^{2}}{\lambda _{4}-\lambda _{2}^{2}}\right) \right) .  \label{term1}
\end{equation}

On the other hand, 
\begin{equation*}
\left| \int_{0}^{T}\hspace{-0.2cm}\gamma (t)E\left\{ \gamma ^{t}(t)%
\mbox{\large 1}_{\{Y^{t}(s)\leq u\gamma ^{t}(s),\forall s\in \lbrack
0,T]\}}\right\} p_{(Y(t),Y^{\prime }(t))}(u\gamma (t),u\gamma ^{\prime
}(t))dt\right| \leq
\end{equation*}
\vspace{-0.5cm} 
\begin{equation*}
\leq A_{2}\int_{0}^{T}\hspace{-0.2cm}\gamma (t)\gamma ^{t}(t)\exp \left( -%
\frac{u^{2}}{2}F(t)\right) dt,
\end{equation*}
where $A_{2}$ a positive constant. Since the function $g(t)=\gamma (t)\gamma
^{t}(t)$ verifies $g(0)=g^{\prime }(0)=0$, and $g^{\prime \prime }(0)\neq 0$%
, Lemma \ref{laplace} implies: 
\begin{equation*}
\int_{0}^{T}\hspace{-0.2cm}\gamma (t)E\left\{ \gamma ^{t}(t)\mbox{\large 1}%
_{\{Y^{t}(s)\leq u\gamma ^{t}(s),\forall s\in \lbrack 0,T]\}}\right\}
p_{(Y(t),Y^{\prime }(t))}(u\gamma (t),u\gamma ^{\prime }(t))dt=
\end{equation*}
\begin{equation}
=\mathcal{O}\left( \frac{1}{u^{3}}\exp \left( -\frac{u^{2}}{2}\frac{\lambda
_{2}^{2}}{\lambda _{4}-\lambda _{2}^{2}}\right) \right).  \label{term2}
\end{equation}

\bigskip From (\ref{term1}) and (\ref{term2}) follows: 
\begin{equation}
L_{3}^{Y}(u,\gamma )=\mathcal{O}\left( \frac{1}{u^{2}}\exp \left( -\frac{%
u^{2}}{2}\frac{\lambda _{2}^{2}}{\lambda _{4}-\lambda _{2}^{2}}\right)
\right) ,  \label{l3}
\end{equation}
and from (\ref{l2}) and (\ref{l3}): 
\begin{equation}
~~\frac{d}{du}A_{1}(u)=\mathcal{O}\left( \frac{1}{u^{2}}\exp \left( -\frac{%
u^{2}}{2}\frac{\lambda _{2}^{2}}{\lambda _{4}-\lambda _{2}^{2}}\right)
\right) .  \label{deriva1}
\end{equation}
Further, observe that since $Y$ is continuous and $\gamma (s)>0$ for $s\in %
\left] 0,T\right] $, $\gamma (0)=0$, if $Y(0)>0$ the event $\left\{ Y(s)\leq
u\gamma (s),\forall s\in \lbrack 0,T]\right\} $ does not occur for positive $%
u$, and if $Y(0)<0$, the same event occurs if $u$ is large enough. This
implies that 
\begin{equation*}
A_{1}(u)=P(Y(s)\leq u\gamma (s),\forall s\in \lbrack 0,T])\rightarrow
P(Y(0)<0)=\frac{1}{2}\quad \text{ as }u\rightarrow +\infty
\end{equation*}
and so, 
\begin{equation}
~A_{1}(u)-\frac{1}{2}=-\int_{u}^{\infty }\frac{d}{dv}A_{1}(v)dv=\mathcal{O}%
\left( \frac{1}{u^{3}}\exp \left( -\frac{u^{2}}{2}\frac{\lambda _{2}^{2}}{%
\lambda _{4}-\lambda _{2}^{2}}\right) \right)  \label{a1}
\end{equation}
on applying (\ref{deriva1}).

\bigskip \textsl{Step 3}. We will now give an equivalent for $A_{3}(u).$
Introduce the following notations:

\begin{itemize}
\item  for $t\in \left] 0,T\right[ $, $Z_{t}(s)$ is the centered Gaussian
process 
\begin{equation*}
Z_{t}(s)=\frac{X^{t}(s)}{(E\left\{ (X^{t}(s))^{2}\right\} )^{1/2}},s\in
\lbrack 0,T].
\end{equation*}

\item  for $t\in \left[ 0,T\right] $, 
\begin{equation*}
\delta _{t}(s)\hspace{-1mm}=\hspace{-1mm}\frac{\beta ^{t}(s)}{(E\left\{
(X^{t}(s))^{2}\right\} )^{1/2}}\hspace{-1mm}=\hspace{-1mm}\frac{\sqrt{%
\lambda _{2}}(1-r(t-s))}{\sqrt{\lambda _{2}(1-r^{2}(t-s))-(r^{\prime
}(t-s))^{2}}}
\end{equation*}
\begin{equation*}
=\hspace{-1mm}\sqrt{F(t-s)}~\mbox{for}~s\in \lbrack 0,T],~s\neq t
\end{equation*}
and 
\begin{equation*}
\delta _{t}(t)=\frac{\lambda _{2}}{\sqrt{\lambda _{4}-\lambda _{2}^{2}}}.
\end{equation*}

Hence, 
\begin{equation*}
\delta _{t}^{\prime }(t)=\frac{F^{\prime }(0)}{2\sqrt{F(0)}}=0.
\end{equation*}

\item  $B_{3}(u;t)=P(Z_{t}(s)\leq u\delta _{t}(s),\forall s\in \lbrack
0,T]), $

$\widetilde{B}_{3}(u;t)=E\left\{ X^{t}(t)1_{\left\{ Z_{t}(s)\leq u\delta
_{t}(s),\forall s\in \lbrack 0,T]\right\} }\right\} $

so that 
\begin{equation}
A_{3}(u)=u\sqrt{\frac{\lambda _{2}}{2\pi }}\int_{0}^{T}B_{3}(u;t)~dt-\frac{1%
}{\sqrt{2\pi \lambda _{2}}}\int_{0}^{T}\widetilde{B}%
_{3}(u;t)~dt=S_{3}(u)-T_{3}(u).  \label{a3comosuma}
\end{equation}
We will consider in detail the behavior of the first term as $u\rightarrow
+\infty $.\newline
\end{itemize}

We apply again Proposition \ref{jmamw} to compute the derivative of $%
B_{3}(u;t)$ with respect to $u$. For $t\in \left] 0,T\right[ $: 
\begin{equation*}
\frac{d}{du}B_{3}(u;t)=L_{1}^{Z_{t}}(u,\delta _{t})+L_{2}^{Z_{t}}(u,\delta
_{t})+L_{3}^{Z_{t}}(u,\delta _{t}),~\mbox{where}
\end{equation*}
$*~L_{1}^{Z_{t}}(u,\delta _{t})=\delta _{t}(0)P(Z_{t}^{\vdash }(s)\leq
u\delta _{t}^{\vdash }(s),\forall s\in \lbrack 0,T])\phi (\delta _{t}(0)u)$.
Then : 
\begin{equation*}
L_{1}^{Z_{t}}(u,\delta _{t})\leq \frac{1}{\sqrt{2\pi }}\sqrt{F(t)}\exp
\left( -\frac{u^{2}}{2}F(t)\right) ,
\end{equation*}
so that as $u\rightarrow +\infty :$ 
\begin{equation}
\int_{0}^{T}L_{1}^{Z_{t}}(u,\delta _{t})dt\leq C_{1}\frac{1}{u}\exp \left( -%
\frac{u^{2}}{2}F(0)\right)  \label{l1Z}
\end{equation}
for some constant $C_{1}$. \newline
\newline
* $L_{2}^{Z_{t}}(u,\delta _{t})=\delta _{t}(T)P(Z_{t}^{\dashv }(s)\leq
u\delta _{t}^{\dashv }(s),\forall s\in \lbrack 0,T])\phi (\delta _{t}(T)u)$.
In the same way: 
\begin{equation*}
L_{2}^{Z_{t}}(u,\delta _{t})\leq \frac{1}{\sqrt{2\pi }}\sqrt{F(T-t)}\exp
\left( -\frac{u^{2}}{2}F(T-t)\right) .
\end{equation*}
and: 
\begin{equation}
\int_{0}^{T}L_{2}^{Z_{t}}(u,\delta _{t})dt\leq C_{2}\frac{1}{u}\exp \left( -%
\frac{u^{2}}{2}F(0)\right) .  \label{l2Z}
\end{equation}
for some constant $C_{2}$. 
\begin{equation*}
*~L_{3}^{Z_{t}}(u,\delta _{t})=\hspace{-0.2cm}-\hspace{-0.2cm}\int_{0}^{T}%
\hspace{-0.2cm}\delta _{t}(x)E\left( (Z_{t}^{x}(x)\hspace{-0.1cm}-\hspace{%
-0.1cm}\delta _{t}^{x}(x)u)1_{\{Z_{t}^{x}(s)\leq u\delta _{t}^{x}(s),\forall
s\in \lbrack 0,T]\}}\right) p_{(Z_{t}(x),Z_{t}^{\prime }(x))}(u\delta
_{t}(x),u\delta _{t}^{\prime }(x))dx.
\end{equation*}
~\newline
We first consider the density in the integrand: 
\begin{equation*}
p_{(Z_{t}(x),Z_{t}^{\prime }(x))}(u\delta _{t}(x),u\delta_{t}^{\prime }(x))=
\end{equation*}
\vspace{-0.5cm} 
\begin{equation}
=\frac{1}{2\pi \sqrt{E\left\{ (Z_{t}^{\prime }(x))^{2}\right\} }}\exp \left(
-\frac{u^{2}}{2}\left( F(x-t)+\frac{(F^{\prime }(x-t))^{2}}{4F(x-t)E\left\{
(Z_{t}^{\prime }(x))^{2}\right\} }\right) \right) .  \label{pZ}
\end{equation}
Define 
\begin{equation*}
G_{t}(x)=F(x-t)+\frac{(F^{\prime }(x-t))^{2}}{4F(x-t)E\left\{ (Z_{t}^{\prime
}(x))^{2}\right\} }.
\end{equation*}
Check that 
\begin{equation*}
\min_{x\in \lbrack 0,T]}G_{t}(x)=G_{t}(t)=F(0),\quad ~G_{t}^{\prime
}(t)=F^{\prime }(0)=0.
\end{equation*}
Moreover, for $x\in \lbrack 0,T]$ one has: 
\begin{equation*}
G_{t}(x)=F(0)+\frac{(x-t)^{2}}{2}F^{\prime \prime }(0)+\frac{((x-t)F^{\prime
\prime }(0))^{2}}{4F(0)E\left\{ (Z_{t}^{\prime }(t))^{2}\right\} }+\mathcal{O%
}((x-t)^{3}),
\end{equation*}
and thus 
\begin{equation*}
G_{t}^{\prime \prime }(t)=F^{\prime \prime }(0)+\frac{(F^{\prime \prime
}(0))^{2}}{2F(0)E\left\{ (Z_{t}^{\prime }(t))^{2}\right\} }.
\end{equation*}
Also, 
\begin{eqnarray*}
X^{t}(x)&=&\frac{2}{(x-t)^{2}}\lambda _{2}(X(t)+(x-t)X^{\prime }(t)+\frac{%
(x-t)^{2}}{2}X^{\prime \prime }(t)+\frac{(x-t)^{3}}{6}X^{\prime \prime
\prime }(t))-.. \\
\hspace{1cm}&-&X(t)(1-\lambda _{2}(x-t)^{2})-X^{\prime }(t)(-\lambda
_{2}(x-t)+\lambda _{4}\frac{(x-t)^{3}}{6})+\mathcal{O}((x-t)^{4}),
\end{eqnarray*}
and\newline
\begin{equation*}
Z_{t}(x)=\frac{1}{\lambda _{2}\sqrt{\lambda _{4}-\lambda _{2}^{2}}+\mathcal{O%
}((x-t)^{2})}\lambda _{2}(X^{\prime \prime }(t)+\lambda _{2}X(t))+
\end{equation*}
\vspace{-0.5cm} 
\begin{equation*}
\hspace{3cm}+\frac{(x-t)}{3}(\lambda _{2}X^{\prime \prime \prime
}(t)+\lambda _{4}X^{\prime }(t))+\mathcal{O}((x-t)^{2}).
\end{equation*}
It follows that: 
\begin{equation*}
E\left\{ (Z_{t}^{\prime }(t))^{2}\right\} =\frac{\lambda _{2}\lambda
_{6}-\lambda _{4}^{2}}{9\lambda _{2}(\lambda _{4}-\lambda _{2}^{2})}=\frac{%
F^{\prime \prime }(0)}{F(0)},
\end{equation*}
and 
\begin{equation*}
G_{t}^{\prime \prime }(t)=\frac{3}{2}F^{\prime \prime }(0)=\frac{\lambda
_{2}(\lambda _{2}\lambda _{6}-\lambda _{4}^{2})}{6(\lambda _{4}-\lambda
_{2}^{2})^{2}}.
\end{equation*}
\ 

We also have: 
\begin{equation*}
\delta _{t}^{x}(s)=\frac{2}{(s-x)^{2}}\left( \delta _{t}(s)-\Gamma
^{Z_{t}}(x,s).\delta _{t}(x)-\frac{\Gamma _{10}^{Z_{t}}(x,s)}{\Gamma
_{11}^{Z_{t}}(x,x)}\delta _{t}^{\prime }(x)\right).
\end{equation*}
\vspace{-0.5cm} 
\begin{equation*}
\delta _{t}^{t}(s)=\frac{2}{(t-s)^{2}}\left( \delta _{t}(s)-\Gamma
^{Z_{t}}(t,s).\delta _{t}(t)-\frac{\Gamma _{10}^{Z_{t}}(t,s)}{\Gamma
_{11}^{Z_{t}}(t,t)}\delta _{t}^{\prime }(t)\right) =
\end{equation*}
\vspace{-0.5cm} 
\begin{equation*}
=\frac{2}{(t-s)^{2}}{\left( \sqrt{F(t-s)}-\sqrt{F(0)}.E\left\{
Z_{t}(t)Z_{t}(s)\right\} \right) }\text{ for }s\neq t,
\end{equation*}
\vspace{-0.5cm} 
\begin{equation*}
\delta _{t}^{t}(t)=\frac{1}{2}\frac{F^{\prime \prime }(0)}{\sqrt{F(0)}}+%
\sqrt{F(0)}E\left\{ \left( Z_{t}^{\prime }(t)\right) ^{2}\right\} =\frac{3}{2%
}\frac{F^{\prime \prime }(0)}{\sqrt{F(0)}}=\frac{(\lambda _{2}\lambda
_{6}-\lambda _{4}^{2})}{6(\lambda _{4}-\lambda _{2}^{2})^{3/2}}>0,
\end{equation*}
where the last inequality is a consequence of the non-degeneracy condition.

\begin{itemize}
\item  Note that since $E\left\{ Z_{t}(t)Z_{t}(s)\right\} \leq 1$, 
\begin{equation*}
\delta _{t}^{t}(s)\geq \frac{2}{(t-s)^{2}}{\left( \sqrt{F(t-s)}-\sqrt{F(0)}%
\right) }\text{ for }s\neq t,
\end{equation*}
so that 
\begin{equation*}
\underset{s,t\in \left[ 0,T\right] }{\inf }\delta _{t}^{t}(s)>0.
\end{equation*}
On the other hand, it is easy to see that $\delta _{t}^{x}(s)$ is a
continuous function of the triplet $(x,t,s)$ and a uniform continuity
argument shows that one can find $\tau >0$ in such a way that if $\left|
x-t\right| \leq \tau $ then 
\begin{equation*}
\delta _{t}^{x}(s)\geq c>0~ \mbox{for all} ~s\in \lbrack 0,T].
\end{equation*}
Thus, for $\left| x-t\right| \leq \tau $, using the Landau-Shepp-Fernique
inequality (see Fernique, 1974): 
\begin{equation*}
\frac{\hspace{-0.2cm}\delta _{t}(x)}{\sqrt{E\left\{ (Z_{t}^{\prime
}(x))^{2}\right\} }}E\left( (Z_{t}^{x}(x)\hspace{-0.1cm}-\hspace{-0.1cm}%
\delta _{t}^{x}(x)u)1_{\{Z_{t}^{x}(s)\leq u\delta _{t}^{x}(s),\forall s\in
\lbrack 0,T]\}}\right) =-\frac{\hspace{-0.2cm}\delta _{t}(x)\delta
_{t}^{x}(x)u}{\sqrt{E\left\{(Z_{t}^{\prime }(x))^{2}\right\} }}(1+R)
\end{equation*}
where $R\leq \alpha _{1}.\exp (-\alpha _{2}.u^{2})$ and $\alpha _{1},\alpha
_{2}$ positive constants independent of $t,$ $x$ and $u$.
\end{itemize}

So, 
\begin{equation}
L_{3}^{Z_{t}}(u,\delta _{t})=\frac{u}{2\pi }\int_{0}^{T}\hspace{-0.2cm}\frac{%
\hspace{-0.2cm}\delta _{t}(x)\delta _{t}^{x}(x)}{\sqrt{E\left\{
(Z_{t}^{\prime }(x))^{2}\right\} }}(1+R)\exp \left[ -\frac{1}{2}u^{2}G_{t}(x)%
\right] ~dx .  \label{l3Z}
\end{equation}

Using the fact that $B_{3}(+\infty ;t)=1$ for every $t\in \left[ 0,T\right] $
we can write: 
\begin{equation*}
S_{3}(u)=u\sqrt{\frac{\lambda _{2}}{2\pi }}T-u\sqrt{\frac{\lambda _{2}}{2\pi 
}}\int_{0}^{T}dt\int_{u}^{+\infty }\frac{d}{dv}B_{3}(v;t)~dv.
\end{equation*}

The same method of \ Lemma \ref{laplace}, plus 
\begin{equation*}
\frac{\hspace{-0.2cm}\delta _{t}(t)\delta _{t}^{t}(t)}{\sqrt{E\left\{
(Z_{t}^{\prime }(t))^{2}\right\} }}=\frac{3}{2}\sqrt{F"(0)F(0)}
\end{equation*}
and (\ref{l1Z}),\ (\ref{l2Z}), (\ref{l3Z}), (\ref{pZ}) show that 
\begin{equation}
S_{3}(u)=u\sqrt{\frac{\lambda _{2}}{2\pi }}T-\left( 1+o(1)\right) \frac{T}{%
2\pi}\sqrt{\frac{3\left( \lambda _{4}-\lambda _{2}^{2}\right)}{\lambda _2} }
\exp \left( -\frac{u^{2}}{2}\frac{\lambda _{2}^{2}}{\lambda _{4}-\lambda
_{2}^{2}}\right).  \label{s3}
\end{equation}
The second term in (\ref{a3comosuma}) can be treated in a similar way, only
one should use the full statement of Lemma 3.3 in Aza\"{i}s and Wschebor
(1999) instead of Proposition \ref{jmamw}, thus obtaining: 
\begin{equation}
T_{3}(u)=\mathcal{O}(\frac{1}{u}). \exp \left( -\frac{u^{2}}{2}\frac{\lambda
_{2}^{2}}{\lambda _{4}-\lambda _{2}^{2}}\right).  \label{t3}
\end{equation}
Then, (\ref{s3}) together with (\ref{t3}) imply that as $u\rightarrow
+\infty $: 
\begin{equation}
A_{3}(u)=u\sqrt{\frac{\lambda _{2}}{2\pi }}T-\left( 1+o(1)\right) \frac{T}{%
2\pi}\sqrt{\frac{3\left( \lambda _{4}-\lambda _{2}^{2}\right)}{\lambda _2} }
\exp \left( -\frac{u^{2}}{2}\frac{\lambda _{2}^{2}}{\lambda _{4}-\lambda
_{2}^{2}}\right) .  \label{a3}
\end{equation}
Replacing (\ref{a1}), (\ref{a3}) into (\ref{primeraderivada}) and
integrating, one obtains (\ref{barmw}). ~\newline
~\newline
\textbf{Acknowledgment.} The authors thank Professors J-M. Aza\"{i}s, P.
Carmona and C. Delmas for useful talks on the subject of this paper.

\section{References}

Aza\"{i}s, J-M., Cierco-Ayrolles, C. and Croquette, A. (1999). Bounds and
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stationary Gaussian process. \textit{ESAIM Probab. Statist.}, 3, 107-129.%
\newline
\newline
Aza\"{i}s, J-M. and Wschebor, M. (1999). On the Regularity of the
Distribution of the Maximum of One-parameter Gaussian Processes. \textit{%
Submitted}.\newline
\newline
Cram\'{e}r, H. and Leadbetter, M.R. (1967). \textit{Stationary and Related
Stochastic Processes}.\ J. Wiley \& Sons,\ New-York.\newline
\newline
Fernique, X. (1974). \textit{R\'{e}gularit\'{e} des trajectoires des
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\newline
\newline
Piterbarg, V.I. (1981). Comparison of distribution functions of maxima of
Gaussian processes. \textit{Th. Prob. Appl.}, 26, 687-705.

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