TITLE: How is it possible to take the inner product of states which belong to two different Hilbert spaces?
QUESTION [2 upvotes]: Question 1 In case of spontaneous breakdown of a continuous symmetry e.g. the ${\rm U(1)}$ symmetry, two different vacua can be labelled as $|\theta\rangle$ and $|\theta^\prime\rangle$, and they belong to different Hilbert spaces. I have often encountered inner products of the form $$\langle\theta|\theta^\prime\rangle=\delta(\theta-\theta^\prime)$$ for example, in the question here by @innisfree. Is it mathematically meaningful to take the inner product of two states which belong to two different Hilbert spaces?
Question 2 One must also look at the Wikipedia article here, which gives an expression of the form $\langle 0|J_0(0)|\theta\rangle\neq0$ where $J_0$ is the zeroth component of the current $J_\mu$ corresponding to the spontaneously broken symmetry.
$\bullet$ Which Hilbert space does $J_0$ act on? Does it act on the Hilbert space containing the vacuum $|0\rangle$ or that containing the vacuum $|\theta\rangle$?
$\bullet$ Which Hilbert space does the state $J_0|0\rangle$ or $J_0|\theta\rangle$ live? Does it belong to the Hilbert space containing the vacuum $|0\rangle$ or that containing the vacuum $|\theta\rangle$?
$\bullet$ In which Hilbert space is this scalar product $\langle 0|J_0(0)|\theta\rangle$ defined?

REPLY [1 votes]: Regarding question 1:
Two Hilbert spaces ${\cal H}_A$ and ${\cal H}_B$ can always be regarded as mutually orthogonal subspaces of a single Hilbert space ${\cal H}$. The algebra of observables might not contain any operators that connect these two subspaces, but I don't think there's any problem considering relationships like $\langle a|b\rangle=0$ with $|a\rangle\in{\cal H}_A$ and $|b\rangle\in{\cal H}_B$.
However, since the question refers to a spontaneously broken continuous symmetry, with the family ground states parameterized by a continuous parameter $\theta$, there is a problem with the equation $\langle \theta|\theta'\rangle=\delta(\theta-\theta')$. The problem is that this equation would imply the existence of an uncountably infinite number of mutually orthogonal state-vectors, which would imply that the Hilbert space is not separable. Quantum (field) theory is normally based on a separable Hilbert space. So in this case, I think the answer is that these different vacuum states cannot all belong to the same (separable) Hilbert space, so equations involving their inner products are dubious.
Granted, even introductory QM texts often write things like $\langle x|x'\rangle=\delta(x-x')$ in reference to the "eigenstates of the position operator", but that's also ill-defined for the same reason. The position operator does not have eigenstates, and it's not defined on the whole Hilbert space anyway because it's unbounded.
Regarding question 2:
I don't see how a local operator can connect two different SSB ground states. However, we can have a relationship of the form $\langle 0|J_0(x)|\pi\rangle\neq 0$, where $|0\rangle$ is a ground state and $|\pi\rangle$ is a state containing a single Goldstone boson, where that single boson lives in the world whose ground state is $|0\rangle$, not in a world whose ground state is $|\theta\rangle$ with $\theta\neq 0$. This relationship is shown, for example, on page 332 in this paper: "Spontaneous Breakdown of Symmetries and Zero-Mass States", https://projecteuclid.org/euclid.cmp/1103840121, accessed on 2018-10-26. The notation in the Wikipedia article is not very clear about this, but I suspect that this is what it meant.
To corroborate this guess about what the Wikipedia article might have meant, the same relationship (a local current connecting a vacuum state to a one-boson state, not another vacuum state) is shown in Weinberg, The Quantum Theory of Fields, Volume II, equation (19.2.34), in precisely the same context of a spontaneously broken global symmetry.


text:
\begin{document}

\title{Ruin problem for Brownian motion risk model with interest rate and tax payment}

\thispagestyle{empty}

{\author{Long Bai}
\address{Long Bai, Department of Actuarial Science, University of Lausanne, UNIL-Dorigny 1015 Lausanne, Switzerland}
\email{long.bai@unil.ch}

\author{Peng Liu}
\address{Peng Liu, Department of Actuarial Science, University of Lausanne,\\
	UNIL-Dorigny, 1015 Lausanne, Switzerland
}
\email{peng.liu@unil.ch}}

\bigskip

\date{\today}
 \maketitle

\bigskip
{\bf Abstract:} Let $\{B(t), t\ge 0\}$ be a  Brownian motion. Consider the Brownian motion risk model with interest rate collection and tax payment defined by
\BQN\label{Rudef}
\widetilde{U}_\Ga^\delta(t)=\widetilde{X}^\delta(t)-\Ga\sup_{s\in[0,t]}
\left(\widetilde{X}^\delta(s)e^{\delta(t-s)}-ue^{\delta(t-s)}\right),t\ge0,\EQN
with $$\widetilde{X}^\delta(t)=ue^{\delta t}+c \int_0^t e^{\delta (t-v)}dv-\sigma \int_0^t e^{\delta(t-v)}dB(v),$$
where $c>0,\gamma \in [0,1)$ and $\delta \in \R$ are three given constants. When $\delta=0$ and $\gamma \in (0,1)$ this is the risk model introduced from Albrecher and Hipp in \cite{AH2007} where the ruin probability in the infinite time horizon has been explicitly calculated. In the presence of interest rate $\delta\neq0$, the calculation of ruin probability for this risk process for both finite and infinite time horizon seems impossible. In the paper, based on asymptotic theory we propose an approximation for ruin probability and ruin time when the initial capital $u$ tends to infinity. Our results are of interest given the fact that this can be used as benchmark model in various calculations.\\
{\bf Key Words:} Brownian motion; force of interest; tax payment; ruin probability; ruin time.\\
{\bf AMS Classification:} Primary 60G15; secondary 60G70

\section{Introduction}
The risk reserve process of an insurance company without interest can be modelled by a stochastic process $\{ X(t),t\ge0\}$ given as
$$X(t)=u+ct-\sigma B(t),\ t\ge0,$$
see \cite{MR1458613,rolski2009stochastic}, where $u\ge0$ is the initial reserve, $c>0$ is the rate of premium received by the insurance company, and $\sigma B(t)$ is frequently referred as the loss rate of the insurance company.
If we add the effect of tax into the model, the new {\it claim surplus process} is
$$U_\gamma(t)=X(t)-\Ga\sup_{s\in[0,t]}\left(X(s)-u\right),\ t\ge0,$$
where $\gamma\in[0,1]$ is referred to the rate of tax. One of the most important characteristics in risk theory is the ruin probability and \cite{AH2007} shows that
$$\pk{\inf_{t\in[0,T]}U_\gamma(t)<0}=
1-\LT(1-e^{-\frac{cu}{\sigma^2}}\RT)^{\frac{1}{1-\gamma}}.$$

Due to the nature of the financial market, we shall consider a more general surplus process including interest rate, see \cite{rolski2009stochastic}, called a risk reserve process with constant force of interest and tax,  i.e., $\widetilde{U}_\Ga^\delta(t),\ t\geq 0$, in \eqref{Rudef}.

For $T\in(0,\IF]$, this contribution is concerned with the exact extreme of $$\psi^\delta_{\Ga,T}(u):=\pk{\inf_{t\in[0,T]}\widetilde{U}_\Ga^\delta(t)<0},$$
as $u\to\IF$. See \cite{rolski2009stochastic, DHJ13a,HX2007} for more studies on risk models with force of interest. Figure \ref{fig1} in Appendix depicts the ruin scenario.
 \\
When $\gamma=0$ and $\delta>0$, i.e. the risk reserve process with positive constant force of interest but without tax, $\psi^\delta_{0,T}(u)$ with $T\in(0,\IF)$ is investigated in \cite{ParisianBrownianfinite2017} and $\psi^\delta_{0,\IF}(u)$ is derived in \cite{Threshold2016,ParisianInfinite2018}.\\
Complementary, we investigate the asymptotic properties of the first passage time (ruin time) of $\widetilde{U}_\Ga^\delta(t)$ on the time interval $[0,T]$, given the ruin has ever  happened during  $[0,T]$. For any $u\geq 0$, and any $T\in(0,\IF]$, define the ruin time of the risk process $\widetilde{U}_\Ga^\delta(t)$ by
\BQN\label{tu}
\tau(u)=\inf\{t\geq 0:\widetilde{U}_\Ga^\delta(t)<0\}.
\EQN
We are interested in the approximate distribution of $\tau(u)|\tau(u)\leq T$, as $u\rw\IF$.\\
Brief organization of the rest of the paper: In Section 2 we first present our main results on the asymptotics of $\psi^\delta_{\Ga,T}(u)$ as $u\rw\IF$ and then
we display the approximation of the ruin time. All the  proofs are relegated to Section 3.

\section{Main Results}
Note that
\BQNY
\psi^\delta_{\Ga,T}(u)=\pk{\inf_{t\in[0,T]}U_\Ga^\delta(t)<0}
=\pk{\sup_{t\in[0,T]}(u-U_\Ga^\delta(t))>u},
\EQNY
where
\BQNY
U_\Ga^\delta(t)=\widetilde{U}_\Ga^\delta(t)e^{-\delta t},t\ge0.
\EQNY
Thus in the analysis of our main results, we consider $\pk{\sup_{t\in[0,T]}(u-U_\Ga^\delta(t))>u}$.\\
In the following theorem, $ \Psi(u), u\in\R$ denotes the survival function
of the standard normal distribution $N(0,1)$. Throughout this paper we write $ f(u)=h(u)(1+o(1))$ or $f(u) \sim h(u)$ if $ \lim_{u \to \infty} \frac{f(u)}{h(u)} = 1 $
and $ f(u) = o(h(u)) $ if $ \lim_{u \to \infty} \frac{f(u)}{h(u)} = 0 $.\\
We prefer to state our new results first, i.e., $\psi_{\gamma,T}^\delta(u)$ as $u\rightarrow\IF$. Cases $T=\IF$
and $T\in(0,\IF)$ are very different and will therefore be dealt with separately. We shall analyse first the case $T\in(0,\IF)$.
\BT\label{TheomOT}
 We have for $\gamma\in[0,1), \delta\in(-\IF,0)\cup(0,\IF)$ and $T\in(0,\IF)$
\BQN\label{result11}
\psi_{\gamma,T}^\delta(u) \sim \frac{2(1+e^{-2\delta T})}{1-\gamma+e^{-2\delta T}}\Psi\left(\frac{u+\frac{c}{\delta}(1-e^{-\delta T})}{a}\right),\ u\rw\IF,
\EQN
where $a^2=\frac{\sigma^2}{2\delta}(1-e^{-2\delta T}).$
\ET
\begin{remarks}
i) When $\gamma=0$, the result of \textbf{\netheo{TheomOT}} reduce to asymptotic ruin probability of the risk model without tax, i.e.
\BQNY
\psi_{\gamma,T}^\delta(u)\sim 2\Psi\left(\frac{u+\frac{c}{\delta}(1-e^{-\delta T})}{a}\right),
\EQNY
which corresponds to the result in \cite{ParisianBrownianfinite2017}.\\
ii) In \eqref{result11}, note that $\gamma$ is just related to the denominator of the constant part. The asymptotic result is increase as $\gamma$.  In fact, when $\gamma$ is bigger, it means that a company need to pay more tax before the ruin happens, thus the ruin probability should be increasing. Table \ref{tab1} is the simulated asymptotic results of $\psi_{\gamma,T}^\delta(u) $ in \netheo{TheomOT}, which also shows the increasing about $\gamma$.
\end{remarks}
\begin{table}[H]
\centering
\caption{The simulated asymptotic results of $\psi_{\gamma,T}^\delta$}\label{tab1}
\begin{tabular}{p{1cm}p{1cm}p{1cm}p{1.5cm}p{1cm}p{1cm}p{4cm}}
\hline
$u$   & $c$   & $\sigma$ & $\delta$ & $T$  & $\gamma$ & asymptotic results \\
\hline
5   & 0.1 & 1      & 0.05   & 20   &0.1    & 0.0363 \\
\hline
5   & 0.1 & 1      & 0.05   & 20  &0.2    & 0.0402 \\
\hline
5   & 0.1 & 1      &-0.05   & 20  &0.2    & 0.0455 \\
\hline
5   & 0.1 & 1      & 0.07   & 20   &0.1    & 0.0210 \\
\hline
5   & 0.1 & 1      & 0.07   & 30   &0.2    & 0.0229 \\
\hline
5   & 0.1 & 1      & -0.07   & 30   &0.2    & 0.0349 \\
\hline
5   & 0.1 & 1      & 0.1   & 20    &0.1   & 0.0090 \\
\hline
5   & 0.1 & 1      & 0.1   & 30    &0.2   & 0.0096 \\
\hline
5   & 0.1 & 1      & -0.1   & 30    &0.2   & 0.0136 \\
\hline
4   & 0.1 & 1      & 0.1   & 20    &0.1   & 0.0312 \\
\hline
4   & 0.1 & 1      & 0.1   & 30    &0.2   & 0.0333 \\
\hline
4   & 0.1 & 1      & -0.1   & 30    &0.2   & 0.0453 \\
\hline
\end{tabular}

\end{table}

\BT\label{TheomIF} We have for $\gamma\in[0,1), \delta >0$ and $T=\IF$
\BQNY
\psi_{\gamma,\IF}^\delta (u)\sim
\frac{1}{1-\gamma}\widehat{\mathcal{P}}^{\frac{c^2}{\sigma^2\delta}}[0,\IF)
\Psi\LT(\frac{\sqrt{2}}{\sigma}\sqrt{\delta u^2+2cu}\RT),\ u\rw\IF,
\EQNY
where for $-\IF\leq S_1<S_2\leq \IF$
$$
\widehat{\mathcal{P}}^{\frac{c^2}{\sigma^2\delta}}[S_1,S_2]=
\E{\sup_{t\in[S_1,S_2]}e^{\sqrt{2}B(t)-t-\LT(\sqrt{t}-\frac{c}{\sigma\sqrt{\delta}}\RT)^2
}}\in(0,\IF).
$$
\ET
\begin{remarks}
i) We have that
\begin{align*}
\frac{1}{1-\gamma}\widehat{\mathcal{P}}^{\frac{c^2}{\sigma^2\delta}}[0,\IF)
\Psi\LT(\frac{\sqrt{2}}{\sigma}\sqrt{\delta u^2+2cu}\RT)&\sim
\frac{1}{1-\gamma}\widetilde{\mathcal{P}}^{\frac{c^2}{\sigma^2\delta}}[0,\IF)
\Psi\LT(\frac{\sqrt{2}(\delta u+c)}{\sigma\sqrt{\delta}}\RT),
\end{align*}
where
\BQNY
\widetilde{\mathcal{P}}^{\frac{c^2}{\sigma^2\delta}}[0,\IF)=\E{\sup_{t\in [0, \IF)}
e^{\sqrt{2}B(t)-2t+\frac{2c}{\sigma\sqrt{\delta}}\sqrt{t}}}.
\EQNY
  The asymptotic result decreases when $\delta$ increases. Since $\widetilde{\mathcal{P}}^{\frac{c^2}{\sigma^2\delta}}[0,\IF)$ is a decreasing function of $\delta>0$  and $\Psi\LT(\frac{\sqrt{2}(\delta u+c)}{\sigma\sqrt{\delta}}\RT)$ is decreasing when $\delta$ increases and  $u\geq \frac{c}{\sqrt{2}\delta}$, the asymptotic of $\psi_{\gamma,\IF}^\delta(u)$ is also a decreasing function of $\delta$. The effect of $\delta$ is not an intuitionistic result from the original risk model.\\
Furthermore, comparing this result with it in \cite{Threshold2016,ParisianInfinite2018}, the scenario with tax is just $\frac{1}{1-\gamma}$ multiple of that without tax. Table \ref{tab2} is the simulated asymptotic results of $\psi_{\gamma,\IF}^\delta(u) $ in \netheo{TheomIF}.\\
ii) $\widehat{\mathcal{P}}^{\frac{c^2}{\sigma^2\delta}}[0,\IF)$ and  $\widetilde{\mathcal{P}}^{\frac{c^2}{\sigma^2\delta}}[0,\IF)$ can be considered as the generalised Piterbarg constants, see e.g., \cite{Pit96,MR1993262,debicki2011extremes,GeneralPit16}for various properties including the finiteness of related constants.\\
iii) We here interpret that the analysis of $\psi_{\gamma,\IF}^\delta(u) $ for the case $\delta<0$ is meaningless. We have
\BQNY
\sup_{t\in[0,\IF)}(u-U_\Ga^\delta(t))\geq \sup_{t\in[0,\IF)} \LT(\sigma\int_{0}^{t}e^{-\delta v}d B(v)-c\int_{0}^t e^{-\delta v}d v\RT)=\IF,\quad a.s.,
\EQNY
where $Var\LT(\sigma\int_{0}^{t}e^{-\delta v}d B(v)\RT)=\frac{\sigma^2}{2\delta}(1-e^{-2\delta t})$.
\end{remarks}

\begin{table}[H]
\centering
\caption{The simulated asymptotic results of $\psi_{\gamma,\IF}^\delta$}\label{tab2}
\begin{tabular}{p{1cm}p{1cm}p{1cm}p{1.5cm}p{1cm}p{4cm}}
\hline
$u$   & $c$   & $\sigma$ & $\delta$   & $\gamma$ & asymptotic results \\
\hline
5   & 0.1 & 1      & 0.05   &0.1    & 0.0467 \\
\hline
5   & 0.1 & 1      & 0.05   &0.2    & 0.0526 \\
\hline
5   & 0.1 & 1      & 0.07   &0.1    & 0.0256 \\
\hline
5   & 0.1 & 1      & 0.07   &0.2    & 0.0288 \\
\hline
5   & 0.1 & 1      & 0.1    &0.1   & 0.0113 \\
\hline
5   & 0.1 & 1      & 0.1    &0.2   & 0.0128 \\
\hline
4   & 0.1 & 1      & 0.1    &0.1   & 0.0378 \\
\hline
4   & 0.1 & 1      & 0.1    &0.2   & 0.0425 \\
\hline
\end{tabular}

\end{table}

We present below the approximation of the conditional passage time $\tau(u)|\tau(u)\leq T$ with $\tau(u)$ defined 	in \eqref{tu}.
\BT\label{ruintime}
For $T\in(0,\IF), \delta\in(-\IF,0)\cup(0,\IF)$ and $x>0$,  we have as $u\rw\IF$
\BQNY
\pk{u^2(T-\tau(u))>x|(\tau(u)\leq T)} \sim
\exp\LT(-\frac{\sigma^2e^{-2\delta T}x}{2a^2}\RT).
\EQNY
For $T=\IF, \delta>0$ and $x\in(-\frac{c}{\sigma\sqrt{\delta}},\IF)$, we have
\BQNY
\pk{u^{2}\LT(e^{-2\delta\tau(u)}-\LT(\frac{c}{\delta u+c}\RT)^2\RT)\leq x\big| \tau(u)<\IF}\sim \frac{\widehat{\mathcal{P}}^{\frac{c^2}{\sigma^2\delta}}[0,x+\frac{c}{\sigma\sqrt{\delta}}]}
{\widehat{\mathcal{P}}^{\frac{c^2}{\sigma^2\delta}}[0,\IF)},
\EQNY
as $u\rightarrow\IF$.
\ET
\begin{remark}
When $\gamma=0$, the result of the scenario $T=\IF$ corresponds to that in \cite{ParisianInfinite2018}.
\end{remark}
\section{Proofs}
Before giving the proofs of our main theorems, we need to introduce some notation which play a pivotal role in the proofs, starting with
\BQN\label{PS}
\mathcal{P}^a[0, S]=\E{ \sup_{t\in[0,S]}e^{\sqrt{2}B(t)-(1+a)\abs{t}}}\in (0,\IF),
\EQN
where $S, a$ are positive constants and
\BQNY
\mathcal{P}^a[0,\IF):=\underset{S\rw\IF}\lim\mathcal{P}^a[0,S]=1+\frac{1}{a}
\EQNY
where is known, see e.g., \cite{Pit96} or \cite{MR1993262}. \\
Moreover, Pickands constant defined by
$$\mathcal{H}[0,S]=\E{\sup_{t\in[0,S]} e^{\sqrt{2}B(t)-t}}\in(0,\IF).$$
It is known that $\lim_{T\rightarrow\infty} \frac{1}{T}\mathcal{H}[0,T]=1$, see  \cite{Pic1969, Berman82, Pit96,DebickiRol02, Mandjes2007,DMandjes2011,DiekerY} for various properties of $\mathcal{P}^a[0, S], \mathcal{H}[0,S]$ and its generalizations.\\
\prooftheo{TheomOT}
We define for any $\gamma \in (0,1)$ the random process $Z$ by
\BQN\label{ZZ1}
Z(s,t):=\frac{\sigma \int_0^t e^{-\delta v}dB(v)-c \int_0^t e^{-\delta v}dv-\Ga\left(\sigma \int_0^s e^{-\delta v}dB(v)-c \int_0^s e^{-\delta v}dv\right)}{1+\gamma(e^{-\delta s}-1)}, \quad s,t\ge 0,
\EQN
which is crucial for our analysis, then for any $u$ positive
\BQNY
\psi_{\gamma,T}^\delta(u)=\pk{\sup_{0\leq s\leq t\leq T}Z(s,t)>u}.
\EQNY
Define next the mean function of $Z(s,t)$
\BQNY
m(s,t):=\E{Z(s,t)}&=&\frac{1}{1+\gamma(e^{-\delta s}-1)}\left(-c\int_0^t e^{-\delta v}dv+\gamma c\int_0^s e^{-\delta v}dv\right)\\
&=&\frac{1}{1+\gamma(e^{-\delta s}-1)}\left(\frac{\gamma c}{\delta}\LT(1- e^{-\delta s}\RT)-\frac{c}{\delta}\LT(1- e^{-\delta t}\RT)\right)
\EQNY
and its variance function
\begin{align}
V^2_Z(s,t):&= \E{Z(s,t)-\E{Z(s,t)}}^2\label{VZ11}\\
&=\frac{1}{(1-\gamma+\gamma e^{-\delta s})^2}\E{\Bigl(\sigma\int_0^t e^{-\delta v}dB(v)-\gamma \sigma\int_0^s e^{-\delta v}dB(v)\Bigr)^2}\nonumber\\
&=\frac{\sigma^2}{2 \delta(1-\gamma+\gamma e^{-\delta s})^2}\LT(\LT(1-e^{-2\delta t}\RT)-\gamma(2-\gamma)\LT(1-e^{-2\delta s}\RT)\RT).\nonumber
\end{align}
By \nelem{VZ}, $V_Z^2(s,t)$ attains the unique maximum at the point$(0,T)$. We give the asymptotic expansion of the standard deviation function $V^2_Z(s,t)$ at this point. It follows that
\BQN\label{eq:VZ}
V_Z(s,t)=
a\LT(1-\frac{\Ga\sigma^2}{2a^2}(1-\gamma+e^{-2\delta T})s-\frac{\sigma^2e^{-2\delta T}}{2a^2}(T-t)\RT)+o((T-t)+s)
\EQN
as $(s,t)\rw(0,T)$; hence there exists a positive constant $\theta>0$ such that
\BQN\label{eq:det1}
  \abs{t-T-\gamma e^{\delta T}s}\le \mathbb{C} (\xx{V_Z}(0,T)-\xx{V_Z}(s,t))
\EQN
uniformly in $B_\theta:=\{(s,t): (s,t)\in[0,\theta]\times[T-\theta,T]\}$.
Besides, $m(0,T)=\frac{c}{\delta}(e^{-\delta T}-1)$.
Next, we study the asymptotic of the supremum of the Gaussian random  field defined on $B_\theta$.
Set below
\BQNY
\nu_u(s,t)=\frac{u-m(s,t)}{V_Z(s,t)},\ W(s,t)=Z(s,t)-m(s,t).
\EQNY
Setting $V^2_W(s,t)=Var(W(s,t))$ and $\overline{W}(s,t)=\frac{W(s,t)}{V_W(s,t)}$, it is clear that $V_W=V_Z$, for any $u>0$
\begin{align}\label{piZ}
\Pi(u)\le\pk{\sup_{(s,t)\in B}Z(s,t)>u}
\le \Pi(u)+\Pi_1(u),
\end{align}
where
\BQNY
&&\Pi(u)=\pk{\sup_{(s,t)\in B_\theta}\overline{W}(s,t)\frac{\nu_u(0,T)}{\nu_u(s,t)}>\nu_u(0,T)},\\
&&\Pi_1(u)=\pk{\sup_{(s,t)\in B\setminus B_\theta}\overline{W}(s,t)\frac{\nu_u(0,T)}{\nu_u(s,t)}>\nu_u(0,T)}.
\EQNY
Since
\begin{align}\label{bibi}
\frac{\nu_u(0,T)}{\nu_u(s,t)}&=1-\frac{V_W(0,T)-V_W(s,t)}{V_W(0,T)}+\frac{[m(s,t)-m(0,T)]V_W(s,t)}{(u-m(s,t))V_W(0,T)}\nonumber\\
&=1-\frac{V_W(0,T)-V_W(s,t)}{V_W(0,T)}+\frac{ce^{-\delta T}[(T-t)+\gamma e^{\delta T} s]V_W(s,t)+o((T-t)+s)}{(u-m(s,t))V_W(0,T)},
\end{align}
as $(s,t)\rightarrow(0,T)$, we have, in view of \eqref{eq:det1}, for any $\vp\in(0,1)$,  and sufficiently large $u$
\BQNY
1-\frac{V_W(0,T)-V_W(s,t)}{V_W(0,T)}\le \frac{\nu_u(0,T)}{\nu_u(s,t)}\le 1-(1-\vp)\frac{V_W(0,T)-V_W(s,t)}{V_W(0,T)}
\EQNY
uniformly in $(s,t)\in B_\theta$.
Consequently
\BQN
\pk{\sup_{(s,t)\in B_\theta}W_0(s,t)>\nu_u(0,T)}\le \Pi(u)\le \pk{\sup_{(s,t)\in B_\theta}W_\vp(s,t)>\nu_u(0,T)},
\EQN
where the random field $\{W_\vp(s,t), s,t\ge0\}$ is defined as
\BQNY
W_\vp(s,t):=\overline{W}(s,t)\left(1-(1-\vp)\frac{V_W(0,T)-V_W(s,t)}{V_W(0,T)}\right),\ \  \vp\in[0,1).
\EQNY
Direct calculations show that the standard deviation function $\sigma_{W_{\vp}}(s,t):=\sqrt{\E{(W_\vp(s,t))^2}}$ attains its unique maximum over $B_\theta$ at $(0,T)$ with $\sigma_{W_{\vp}}(0,T)=1$.
Thus, in the light of \eqref{eq:VZ}, we have
\BQN\label{eq:SWV}
\sigma_{W_\varepsilon}(s,t)=1-(1-\vp)\left(\frac{\Ga\sigma^2(1-\Ga+e^{-2\delta T})}{2a^2}s+\frac{\sigma^2e^{-2\delta T}}{2a^2}(T-t)\right)(1+o(1)),
\EQN
as $(s,t)\rw(0,T)$. Furthermore, it follows that
\BQN\label{cove1}
1-Cov(W_\vp(s,t),W_\vp(s',t'))=\frac{\sigma^2}{2a^2}\left(e^{-2\delta T}\mid t-t'\mid +\Ga^{2}\mid s-s'\mid\right)(1+o(1))
\EQN
as $(s,t), (s',t')\rw(0,T)$. In addition, we obtain
\BQNY
\E{(W_\vp(s,t)-W_\vp(s',t'))^2}\le \mathbb{C}(2e^{-2\delta T}|t-t'|+2\Ga^2|s-s'|)
\EQNY
for $(s,t), (s',t')\in B_\theta$. Consequently, by Theorem 8.2 of \cite{Pit96}
\BQN
\pk{\sup_{(s,t)\in B_\theta}W_\vp(s,t)>\nu_u(0,T)}\sim
\frac{2\LT(1+e^{-2\delta T}\RT)}{1-\gamma+e^{-2\delta T}}\Psi\left(\frac{u+\frac{c}{\delta}(1-e^{-\delta T})}{a}\right)\label{result}
\EQN
 as $u\rw\infty,\ \vp\rw 0$. Thus we obtain the asymptotic upper bound for $\Pi(u)$ on the set $B_{\theta}$. The asymptotic lower bound can be derived using the same arguments. In order to complete the proof we need to show further that
\BQN
\Pi_1(u)=o(\Pi(u))\ \ \text{as}\ u\rw\IF.
\label{WB}
\EQN
In the light of \eqref{bibi} for all $u$ sufficiently large
\BQNY
\sup_{(s,t)\in B\setminus B_\theta} Var\left(\overline{W}(s,t)\frac{\nu_u(0,T)}{\nu_u(s,t)}\right)\le (\rho(\theta))^2<1,
\EQNY
where $\rho(\theta)$ is a positive function in $\theta$ which exists due to the continuity of $V_W(s,t)$ in $B$.
Therefore, a direct application of Borell-TIS inequality as in \cite{AdlerTaylor}  implies
\BQNY
\pk{\sup_{(s,t)\in B \setminus B_\theta}\overline{W}(s,t)\frac{\nu_u(0,T)}{\nu_u(s,t)}>\nu_u(0,T)}\le 2\Psi\LT(\frac{\nu_u(0,T)-b}{\rho(\theta)}\RT)=o(\Pi(u)), \quad u\rw\IF,
\EQNY
where $b=\sup_{(s,t)\in B \setminus B_\theta}\E{\overline{W}(s,t)\frac{\nu_u(0,T)}{\nu_u(s,t)}}<\IF.$\\
Consequently, Eq. (\ref{WB}) is established.
\QED
\noindent
\prooftheo{TheomIF}
We have
\BQNY
\psi_{\gamma,\IF}^\delta(u)=\pk{\sup_{0<t\leq s\leq 1}\frac{Z(s,t)}{G_u(s,t)}>1},
\EQNY
 where
\begin{align}
Z(s,t)&=\sigma\int_0^{-\frac{1}{2\delta}lnt} e^{-\delta v}dB(v)-\Ga\sigma\int_0^{-\frac{1}{2\delta}lns} e^{-\delta v}dB(v),\label{ZZ2}\\
G_u(s,t)&=u+\Ga(s^{\frac{1}{2}}-1)u+c\int_0^{-\frac{1}{2\delta}lnt} e^{-\delta v}dv-c\Ga\int_0^{-\frac{1}{2\delta}lns} e^{-\delta v}dv\nonumber\\
&=u-\Ga \LT(u+\frac{c}{\delta}\RT)(1-s^{\frac{1}{2}})+\frac{c}{\delta}\LT(1-t^{\frac{1}{2}}\RT).\label{GU2}
\end{align}
The variance function of $Z(s,t)$ is given by
\begin{align}
V_Z^2(s,t)&=Var\left(\sigma\int_0^{-\frac{1}{2\delta}lnt} e^{-\delta v}dB(v)-\Ga\sigma\int_0^{-\frac{1}{2\delta}lns} e^{-\delta v}dB(v)\right)\label{VZ22}\\
&=\frac{\sigma^2}{2\delta}((1-t)-\Ga(2-\Ga)(1-s)).\nonumber
\end{align}
Let $M_u(s,t)=\frac{G_u(s,t)}{V_Z(s,t)}$, then for $t_u$ in \nelem{VIF}
\BQNY
M_u:=M_u(1,t_u)=\frac{\sqrt{2}}{\sigma}\sqrt{\delta u^2+2cu}=\frac{\sqrt{2\delta}}{\sigma}u(1+o(1)),
\EQNY
as $u\rw\IF$.
\BQNY
\frac{M_u(s,t)}{M_u}-1=\frac{[G_u(s,t)V_Z(1,t_u)]^2-[G_u(1,t_u)V_Z(s,t)]^2}{V_Z(s,t)G_u(1,t_u)[G_u(s,t)V_Z(1,t_u)+V_Z(s,t)G_u(1,t_u)]}
\EQNY
Since
\begin{align*}
&[G_u(s,t)V_Z(1,t_u)]^2-[G_u(1,t_u)V_Z(s,t)]^2\\
=&\{(u+\frac{c}{\delta})[1-\Ga(1-\sqrt{s})]-\frac{c}{\delta}\sqrt{t}\}^2\frac{\sigma^2}{2\delta}(1-t_u)\\
&-\{(u+\frac{c}{\delta})-\frac{c}{\delta}\sqrt{t_u}\}^2\frac{\sigma^2}{2\delta}[1-t-\Ga(2-\Ga)(1-s)]\\
\sim&\frac{\sigma^2}{2\delta}u^2[(\sqrt{t}-\sqrt{t_u})^2+\Ga(1-\Ga)(1-s)],
\end{align*}
then
\BQN\label{MM1}
\frac{M_u(s,t)}{M_u}-1\sim \frac{1}{2}[(\sqrt{t}-\sqrt{t_u})^2+\Ga(1-\Ga)(1-s)].
\EQN
Now we rewrite
\BQNY
\psi_{\gamma,\IF}^\delta(u)=\pk{\sup_{0<t\leq s\leq 1}\frac{Z(s,t)}{G_u(s,t)}M_u>M_u}.
\EQNY
The correlation function of $Z(s,t)$ is
\BQNY
r(s,s',t,t')

\sim 1-\frac{1}{2}|t-t'|-\frac{1}{2}\Ga^2|s-s'|
\EQNY
for $s,s'\rw1$, $t-t_u, t'-t_u\rw0$.\\
In addition, we obtain for some $\theta_0,\mathbb{C},$
\BQNY
\mathbb{E}\left(Z(s,t)-Z(s',t')\right)^2\leq\mathbb{C}(|t-t'|+\Ga^2|s-s'|)
\EQNY
for $ s,t\in[1-\theta_0,1]\times[0,\theta_0]$.\\
\def\TPi{\widetilde{\Pi}}
Note that for any small $\theta_1,\theta_2\in(0,1)$, set $B=\{(s,t):0<t\leq s\leq 1\}$ and $\Delta_\theta=[1-\theta_1,1]\times[0,\theta_2]$,
\BQNY
\Pi(u):=\pk{\sup_{(s,t)\in\Delta_\theta}\frac{Z(s,t)}{G_u(s,t)}M_u>M_u}\leq \psi_{\gamma,\IF}^\delta(u)\leq\Pi(u) +\TPi(u)
\EQNY
with
\BQNY
\TPi(u):=\pk{\sup_{t\in B\setminus\Delta_\theta}\frac{Z(s,t)}{G_u(s,t)}M_u>M_u}
\EQNY
 In the following, we shall focus on the asymptotics of $\Pi(u)$ as $u\to\IF$, and finally we show that
 \BQN\label{PIPI}
 \TPi(u)=o(\Pi(u)),\ \ \ u\to\IF.
 \EQN
We set $\delta_1(u)=\LT(\frac{(lnu)^q}{u}\RT)^2$, $\delta_2(u)=2\sqrt{t_u}\frac{(lnu)^q}{u}+\LT(\frac{(lnu)^q}{u}\RT)^2$ for some $q>1$ and
$$D_u=[1-\delta_1(u),1]\times[0,t_u+\delta_2(u)]=\LT[0,\LT(\frac{(lnu)^q}{u}\RT)^2\RT]\times\LT[0,\LT(\sqrt{t_u}+\frac{(lnu)^q}{u}\RT)^2\RT],\ \ \Theta_u=\Delta_\theta\setminus D_u.$$
Clearly, for $u$ large enough,
\BQNY
\Pi_1(u):=\pk{\sup_{t\in D_u}\frac{Z(s,t)}{G_u(s,t)}M_u>M_u}\leq \Pi(u)\leq \Pi_1(u)+\Pi_2(u),
\EQNY
with
\BQNY
\Pi_2(u)=\pk{\sup_{t\in \Theta_u}\frac{Z(s,t)}{G_u(s,t)}M_u>M_u}
\EQNY
Next, we give a tight upper bound  for $\Pi_2(u)$ which will finally imply that, for some small $\theta_1, \theta_2\in(0,1)$
\BQNY
\Pi_2(u)=o(\Pi_1(u)),\ \ \ u\to\IF.
\EQNY
By (\ref{MM1}) we have that for any small $\vn_o>0$, there exists some small $\theta_1, \theta_2>0$ such that
\BQNY
\frac{M_u}{M_u(s,t)}\leq\frac{1}{1+\frac{1-\vn_0}{2}[(\sqrt{t}-\sqrt{t_u})^2+\Ga(1-\Ga)(1-s)]}
\EQNY
holds for all $(s,t)\in \Delta_\theta$. Furthermore, for any $t\in\Theta_u$
\BQNY
1+\frac{1-\vn_0}{2}[(\sqrt{t}-\sqrt{t_u})^2+\Ga(1-\Ga)(1-s)]\geq 1+\frac{1-\vn_0}{2}\min(1, \Ga(1-\Ga))\LT(\frac{(lnu)^q}{u}\RT)^2
\EQNY
implying $\left(set \overline{Z}(s,t)=\frac{Z(s,t)}{V_Z(s,t)}\right)$
\BQN
\Pi_2(u)
&=&\pk{\sup_{(s,t)\in\Theta_u} \overline{Z}(s,t)\frac{M_u}{M_u(s,t)} >M_u}\nonumber\\
&\leq& \mathbb{Q}_1M_u^4\Psi\left(M_u \LT(1+\frac{1-\vn_0}{2}\min(1, \Ga(1-\Ga))\LT(\frac{(lnu)^q}{u}\RT)^2\RT)\right)
\EQN
holds for all $u$ large, with some constant $\mathbb{Q}_1>0$.
For any small $\vn>0$, define below
$$ \mathcal{B}_u^{\vn\pm}(\Delta):= \Bigl\{ \sup_{(s,t)\in \Delta}  \frac{\xi^{\pm}(s,t)}{[1+(\frac{1}{2}\pm\vn)(\sqrt{t}-\sqrt{t_u})^2][1+(\frac{1}{2}\Ga(1-\Ga)\pm\vn)(1-s)]}>M_u\Bigr\},\ \Delta\subset \R^2,$$
where $\{\xi^{\pm}(s,t), s, t\geq 0\}$ is a zero-mean stationary Gaussian field with continuous sample paths and correlation function
\BQNY
 \Ga^{\pm}_\xi(s,t)=\exp\LT(-\LT(\frac{{\Ga}^2}{2}\mp\vn\RT)|s|-\LT(\frac{1}{2}\mp\vn\RT)|t|\RT).
\EQNY
Next we analyse $\Pi_1(u)$, as $u\rw\IF$. By (\ref{MM1}), when $u$ large enough, for any $t\in D_u$,
\BQNY
&&\frac{M_u}{M_u(s,t)}\geq\frac{1}{[1+(\frac{1}{2}+\vn)(\sqrt{t}-\sqrt{t_u})^2]
[1+(\frac{1}{2}\Ga(1-\Ga)+\vn)(1-s)]},\\
&& \frac{M_u}{M_u(s,t)}\leq
\frac{1}{[1+(\frac{1}{2}-\vn)(\sqrt{t}-\sqrt{t_u})^2][1+(\frac{1}{2}\Ga(1-\Ga)-\vn)(1-s)]}.
\EQNY
Then
\BQNY
\mathbb{P}\{\mathcal{B}_u^{\vn+}(D_u)\}\leq\Pi_1(u)\leq\mathbb{P}\{\mathcal{B}_u^{\vn-}(D_u)\},\ u\rw\IF.
\EQNY
Thus we just need establish the asymptotic behavior of $\pi^{\pm}(u):=\mathbb{P}\{\mathcal{B}_u^{\vn\pm}(D_u)\}$, then according to the continuous of the results which can be seen from the following calculation, setting $\vn\rw 0$, we will gain the precision estimates of $\Pi_1(u)$. Below we mainly show the calculation of $\pi^{+}(u)$.

$$D_u=[1-\delta_1(u),1]\times[0,t_u+\delta_2(u)]=\LT[0,\LT(\frac{(lnu)^q}{u}\RT)^2\RT]\times\LT[0,\LT(\sqrt{t_u}+\frac{(lnu)^q}{u}\RT)^2\RT]$$
For any positive constant $S_1,S_2$, define
\BQNY
&&\Delta^1_k=1-u^{-2}[(k+1)S_1,kS_1],\ \ k=0,1,2,3\ldots\\
&&\Delta^2_{-1}=[0,t_u],\ \ \Delta^2_k=[(\sqrt{t_u}+\sqrt{kS_2}u^{-1})^2,(\sqrt{t_u}+\sqrt{(k+1)S_2}u^{-1})^2],\ \ k=0,1,,3\ldots
\EQNY
and let further for $u>0$
\BQNY
h_1(u)=\lfloor S_1^{-1}(\ln u)^{2q}\rfloor+1,\ \
h_2(u)=\LT\lfloor \frac{\LT(\sqrt{t_u}+\frac{(\ln u)^q}{u}\RT)^2-t_u}{(\sqrt{t_u}+\sqrt{(k+1)S_2}u^{-1})^2
-(\sqrt{t_u}+\sqrt{kS_2}u^{-1})^2}\RT\rfloor+1,
\EQNY
where $h_1(u),h_2(u)\rw\IF$, as $u\rw\IF$.
By Bonferroni's inequality we have
\BQN\label{piz1}
\pi_1^+(u):=\pk{\mathcal{B}^{\vn+}_u(\Delta^1_0\times(\Delta^2_{-1}\cup\Delta^2_0))}
\leq\pi^{+}(u)\leq\pi_1^+(u)+\pi_2^+(u),
\EQN
where
\begin{align*}
\pi_2^+(u)=&\sum_{k_1=1}^{h_1(u)}\pk{\mathcal{B}^{\vn+}_u(\Delta^1_{k_1}\times(\Delta^2_{-1}\cup\Delta^2_0))}
+\sum_{k_2=1}^{h_2(u)}\pk{\mathcal{B}^{\vn+}_u(\Delta^1_0\times\Delta^2_{k_2})}\nonumber\\
&+\sum_{k_1=1}^{h_1(u)}\sum_{k_2=1}^{h_2(u)}\pk{\mathcal{B}^{\vn+}_u(\Delta^1_{k_1}\times\Delta^2_{k_2})}\\
=:& I_1+I_2+I_3.\nonumber
\end{align*}
By \textbf{\nelem{PST}}, we get as $u\rw\IF$
\BQN
\pi_1^+(u)\sim\mathcal{P}^{\frac{\Ga(1-\Ga)+2\vn}{\Ga^2-2\vn}}[0,(\Ga^2-2\vn)\frac{\delta}{\sigma^2}S_1]
\widehat{\mathcal{P}}^{\frac{1+2\vn}{1-2\vn},\frac{c^2}{\sigma^2\delta}(1-2\vn)}[0,(1-2\vn)\frac{\delta}{\sigma^2}(\frac{c}{\delta}+\sqrt{S_2})^2]\Psi(M_u).
\EQN
Next we calculate the required asymptotic bounds for $I_1(u)$, $I_2(u)$ and $I_3(u)$ and show that as $u\rw\IF,\ S_i\rw\IF,\ i=1,2 $
\BQN\label{II}
I_1(u)=o(\pi_1^+(u)),I_2(u)=o(\pi_1^+(u)),I_3(u)=o(\pi_1^+(u)).
\EQN
By \textbf{\nelem{PST}}, we drive that
\begin{align*}
I_1(u)=&\sum_{k_1=1}^{h_1(u)}\pk{\mathcal{B}^{\vn+}_u(\Delta^1_{k_1}\times(\Delta^2_{-1}\cup\Delta^2_0))}\\
\leq&\sum_{k_1=1}^{h_1(u)}\pk{\sup_{(s,t)\in(\Delta^1_{k_1}\times(\Delta^2_{-1}\cup\Delta^2_0))}
\frac{\xi^{+}(s,t)}{1+(\frac{1}{2}+\vn)(\sqrt{t}-\sqrt{t_u})^2}>M_u[1+(\frac{1}{2}\Ga(1-\Ga)+\vn)u^{-2}k_1S_1]}\\
\leq&\mathcal{H}[0,(\Ga^2-2\vn)\frac{\delta}{\sigma^2}S_1]
\widehat{\mathcal{P}}^{\frac{1+2\vn}{1-2\vn},\frac{c^2}{\sigma^2\delta}(1-2\vn)}[0,(1-2\vn)\frac{\delta}{\sigma^2}(\frac{c}{\delta}+\sqrt{S_2})^2]\frac{1}{\sqrt{2\pi}}\\
&\times\sum_{k_1=1}^{h_1(u)}\frac{1}{M_u[1+(\frac{1}{2}\Ga(1-\Ga)+\vn)u^{-2}k_1S_1]}
\exp\LT(-\frac{M_u^2[1+(\frac{1}{2}\Ga(1-\Ga)+\vn)u^{-2}k_1S_1]^2}{2}\RT)(1+o(1))\\
=&\mathcal{H}[0,(\Ga^2-2\vn)\frac{\delta}{\sigma^2}S_1]
\widehat{\mathcal{P}}^{\frac{1+2\vn}{1-2\vn},\frac{c^2}{\sigma^2\delta}(1-2\vn)}[0,(1-2\vn)\frac{\delta}{\sigma^2}(\frac{c}{\delta}+\sqrt{S_2})^2]\\
&\times\Psi(M_u)\sum_{k_1=1}^{h_1(u)}\exp\LT(-\frac{\delta}{\sigma^2}(\Ga(1-\Ga)+2\vn)k_1S_1\RT)(1+o(1))
\end{align*}
as $u\rw\IF$.
Similarly
\begin{align*}
I_2(u)\leq& \mathcal{P}^{\frac{\Ga(1-\Ga)+2\vn}{\Ga^2-2\vn}}[0,(\Ga^2-2\vn)\frac{\delta}{\sigma^2}S_1]
\widehat{\mathcal{P}}^{0,\frac{c^2}{\sigma^2\delta}(1-2\vn)}
[0,(1-2\vn)\frac{\delta}{\sigma^2}(\frac{c}{\delta}+\sqrt{S_2})^2]\\
&\times\Psi(M_u)\sum_{k_2=1}^{h_2(u)}\exp\LT(-\frac{\delta}{\sigma^2}(1+2\vn)k_2S_2\RT)(1+o(1)),
\end{align*}
as $u\rw\IF$. Moreover,
\begin{align*}
I_3(u)=&\sum_{k_1=1}^{h_1(u)}\sum_{k_2=1}^{h_2(u)}\pk{\mathcal{B}^{\vn+}_u(\Delta^1_{k_1}\times\Delta^2_{k_2})}\\
\leq&\sum_{k_1=1}^{h_1(u)}\sum_{k_2=1}^{h_2(u)}\pk{\sup_{(s,t)\in(\Delta^1_{k_1}\times(\Delta^2_{-1}\cup\Delta^2_0))}\xi^{+}(s,t)
>M_u[1+(\frac{1}{2}\Ga(1-\Ga)+\vn)u^{-2}k_1S_1+(\frac{1}{2}+\vn)u^{-2}k_2S_2]}\\
=&\mathcal{H}[0,(\Ga^2-2\vn)\frac{\delta}{\sigma^2}S_1]
\widehat{\mathcal{P}}^{0,\frac{c^2}{\sigma^2\delta}(1-2\vn)}[0,(1-2\vn)\frac{\delta}{\sigma^2}(\frac{c}{\delta}+\sqrt{S_2})^2]\\
&\times\Psi(M_u)\sum_{k_1=1}^{h_1(u)}\sum_{k_2=1}^{h_2(u)}\exp\LT(-\frac{\delta}{\sigma^2}[(\Ga(1-\Ga)+2\vn)k_1S_1+(1+2\vn)k_2S_2]\RT)
(1+o(1)),
\end{align*}
as $u\rw\IF$. By letting $S_1,S_2\rw\IF$, (\ref{II}) is proved.\\
Thus we finish the calculation of $\pi^+(u)$.\\
For $\pi^-(u)$, we just need notice that the equations (\ref{piz1}) are replaced by
\BQNY
\pi_1^-(u):=\pk{\mathcal{B}^{\vn-}_u(\Delta^1_0\times(\Delta^2_{-1}\cup\Delta^2_0))}\leq\pi^{-}(u)\leq\pi_1^-(u)+\pi_2^-(u)
\EQNY
and
\begin{align*}
\pi_2^-(u)=&\sum_{k_1=1}^{h_1(u)}\pk{\mathcal{B}^{\vn-}_u(\Delta^1_{k_1}\times(\Delta^2_{-1}\cup\Delta^2_0))}
+\sum_{k_2=1}^{h_2(u)}\pk{\mathcal{B}^{\vn-}_u(\Delta^1_0\times\Delta^2_{k_2})}\\
&+\sum_{k_1=1}^{h_1(u)}\sum_{k_2=1}^{h_2(u)}\pk{\mathcal{B}^{\vn-}_u(\Delta^1_{k_1}\times\Delta^2_{k_2})}\\
=&: J_1+J_2+J_3.
\end{align*}
Then by \textbf{\nelem{PST}}, we get
\BQNY
\pi_1^-(u)\sim\mathcal{P}^{\frac{\Ga(1-\Ga)-2\vn}{\Ga^2+2\vn}}[0,(\Ga^2+2\vn)\frac{\delta}{\sigma^2}S_1]
\widehat{\mathcal{P}}^{\frac{1-2\vn}{1+2\vn},\frac{c^2}{\sigma^2\delta}(1+2\vn)}
[0,(1+2\vn)\frac{\delta}{\sigma^2}(\frac{c}{\delta}+\sqrt{S_2})^2]\Psi(M_u),
\EQNY
and similarly
\BQNY
J_1(u)=o(\pi_1^-(u)),J_2(u)=o(\pi_1^-(u)),J_3(u)=o(\pi_1^-(u)),\ as\  u\rw\IF,\ S_i\rw\IF,\ i=1,2.
\EQNY
Thus,
$$\lim_{\vn\rw 0} \pi^+(u)=\lim_{\vn\rw 0} \pi^-(u)=\Pi_1(u).$$
Finally, by (\ref{MM1}) and \textbf{\nelem{VIF}}, we can choose some small $\theta_1, \theta_2>0$ so that for any $u$ sufficiently large
\BQN
\sup_{(s,t)\in B\setminus\Delta_\theta}Var\LT\{\frac{Z(s,t)}{G_u(s,t)}M_u\RT\}\leq\sup_{(s,t)\in B\setminus\Delta_\theta}\LT(\frac{M_u}{M_u(s,t)}\RT)^2\leq (\rho(\theta_1, \theta_2))^2< 1
\EQN
where $\rho(\theta_1, \theta_2)$ is a positive function in $\theta_1$ and $\theta_2$ which exists due to the continuity of $M_u(s,t)$ in $B$. Additionally, by the almost surely continuity of random field, we have, for some constant $b>0$

\BQNY
\pk{\sup_{t\in B\setminus\Delta_\theta}\frac{Z(s,t)}{G_u(s,t)}M_u>b}\leq \frac{1}{2}.
\EQNY
Therefore, a direct application of the Borell inequality (e.g., \textbf{Theorem D.1} of \cite{Pit96})
 implies
 \BQNY
 \TPi(u)=\pk{\sup_{t\in B\setminus\Delta_\theta}\frac{Z(s,t)}{G_u(s,t)}M_u>M_u}\leq 2\Psi\LT(\frac{M_u-b}{\rho(\theta_1, \theta_2)}\RT)=o(\Pi(u))\
 \ as\ \ u\rw \IF.
 \EQNY
 Consequently, Eq. (\ref{PIPI}) is established, and thus the proof is complete.
 \QED
 \prooftheo{ruintime} For $T\in(0,\IF)$, first note that
\BQN\label{eq:PP}
\pk{u^2(T-\tau(u))>x|(\tau(u)\leq T)}=\frac{\pk{\sup_{0\leq s\leq t\leq T_x(u)}Z(s,t)>u}}{\pk{\sup_{0\leq s\leq t\leq T}Z(s,t)>u}}
\EQN
for any $x,u>0,$ where $T_x(u)=T-xu^{-2}$ and $Z(s,t)$ is the same as in \eqref{ZZ1}. Next for any $x,u>0$, we follow the similar argumentation as in the proof \netheo{TheomOT}
$$\pk{\sup_{0\leq s\leq t\leq T_x(u)}Z(s,t)>u}\sim \mathcal{P}^{\frac{1-\gamma+e^{-2\delta T}}{\gamma}}\lim_{S\rw\IF}\mathcal{P}^{1}\LT[\frac{\sigma^2e^{-2\delta T}}{2a^2}x,\frac{\sigma^2e^{-2\delta T}}{2a^2}S\RT]\Psi\LT(\frac{\sqrt{2}}{\sigma}\sqrt{\delta u^2+2cu}\RT),$$
as $u\rw\IF$.
By Remark 2.3 of \cite{Threshold2016},  we have
\begin{align*}
\lim_{S\rw\IF}\mathcal{P}^{1}\LT[\frac{\sigma^2e^{-2\delta T}}{2a^2}x,\frac{\sigma^2e^{-2\delta T}}{2a^2}S\RT]
&=\lim_{S\rw\IF}\E{ \sup_{t\in\LT[\frac{\sigma^2e^{-2\delta T}}{2a^2}x,\frac{\sigma^2e^{-2\delta T}}{2a^2}S\RT]}e^{\sqrt{2}B(t)-2t}}\\
&=\lim_{S\rw\IF}\E{ \sup_{t\in\LT[0,\frac{\sigma^2e^{-2\delta T}}{2a^2}(S-x)\RT]}e^{\sqrt{2}B(t)-2t-\frac{\sigma^2e^{-2\delta T}}{2a^2}x}}\\
&=e^{-\frac{\sigma^2e^{-2\delta T}}{2a^2}x}\lim_{S\rw\IF}\E{ \sup_{t\in\LT[0,\frac{\sigma^2e^{-2\delta T}}{2a^2}(S-x)\RT]}e^{\sqrt{2}B(t)-2t}}\\
&=e^{-\frac{\sigma^2e^{-2\delta T}}{2a^2}x}\mathcal{P}^{1}[0,\IF)\\
&=2e^{-\frac{\sigma^2e^{-2\delta T}}{2a^2}x}.
\end{align*}
Therefore, we conclude by an application of Theorem \ref{TheomOT} and equation(\ref{eq:PP}) that
$$\pk{u^2(T-\tau_1(u))>x|(\tau_1(u)\leq T)}\rightarrow \exp\LT(-\frac{\sigma^2e^{-2\delta T}x}{2a^2}\RT), $$
as $u\rightarrow \IF$, for any $x>0$.\\
For the case $T=\IF$, we have
\BQNY
\pk{u^{2}\LT(e^{-2\delta\tau_u}-\LT(\frac{c}{\delta u+c}\RT)^2\RT)\leq x\big| \tau_u<\IF}
=\frac{\pk{\underset{0<t\leq t_u+u^{-2}x}{\sup_{0<t\leq s\leq 1}}\frac{Z(s,t)}{G_u(s,t)}>1}}{\pk{\sup_{0<t\leq s\leq 1}\frac{Z(s,t)}{G_u(s,t)}>1}}
\EQNY
where $Z(s,t)$ and $G_u(s,t)$ are the same as in \eqref{ZZ2} and \eqref{GU2}.
We follow the similar argumentation as in the proof \netheo{TheomIF}
\BQNY
\pk{\underset{0<t\leq t_u+u^{-2}x}{\sup_{0<t\leq s\leq 1}}\frac{Z(s,t)}{G_u(s,t)}>1}\sim \frac{1}{1-\gamma}\widehat{\mathcal{P}}^{1,\frac{c^2}{\sigma^2\delta}}[0,x+\frac{c}{\sigma\sqrt{\delta}}]
\Psi\LT(\frac{\sqrt{2}}{\sigma}\sqrt{\delta u^2+2cu}\RT),
\EQNY
as $u\rw\IF$.
Thus we get the results by an application of Theorem \ref{TheomIF}.
\QED
\section{Appendix}
\noindent
Here  we give several Lemmas which are used in the proofs.

\BEL\label{VZ}
The variance function $V^2_Z(s,t)$ in \eqref{VZ11} attains its unique global maximum over set $B:=\{(s,t):0\leq s\leq t<T\}$ at $(s_0,t_0)$, with $s_0=0$ and
$t_0= T$. Further
$
a^2:=V_Z^2(0,T)=\frac{\sigma^2}{2\delta}(1-e^{-2\delta T}).
$
\EEL
\noindent
\prooflem{VZ}
It is obvious that $t_0=T$, then
\begin{align*}
\frac{\partial V_Z^2(s,T)}{\partial s}&=\frac{\gamma \sigma^2 e^{-\delta s}}{(1-\gamma+\gamma e^{-\delta s})^3}\left((1-e^{-2\delta T})-\gamma(2-\gamma)\LT(1-e^{-2\delta s}\RT)\right)-\frac{\gamma (2-\gamma) \sigma^2 e^{-2\delta s}}{(1-\gamma+\gamma e^{-\delta s})^2}\\

&=\frac{\gamma \sigma^2 e^{-\delta s}}{(1-\gamma+\gamma e^{-\delta s})^3}\left(
(1-e^{-2\delta T})-\gamma(2-\gamma)-(1-\gamma)(2-\gamma) e^{-\delta s}
\right),
\end{align*}
we have when $\delta<0$, $V_Z^2(s,T)$ attains the maximum only at $s=0$ and when $\delta>0$, $V_Z^2(s,T)$ attains the maximum only at $s=0$ and $s=T$, since
\BQNY
V_Z^2(0,T)=\frac{\sigma^2}{2\delta}(1-e^{-2\delta T}),\ \  \
V_Z^2(T,T)=\frac{\sigma^2(1-\gamma)^2}{2\delta(1-\gamma+\gamma e^{-\delta T})}(1-e^{-2\delta T}),
\EQNY
and $
\frac{(1-\gamma)^2}{(1-\gamma+\gamma e^{-\delta T})}<1,
$
hence the claim follows. \QED

\BEL\label{VIF}
Let $F_u(s,t)=\frac{V_Z(s,t)}{G_u(s,t)}$ with  $G_u(s,t)$ in \eqref{GU2} and $V_Z(s,t)$ in \eqref{VZ22}. Then for $u$ sufficiently large, the function $F_u(s,t), 0<t \leq s\leq 1 $ attains its maximum at the unique point $(s,t)=(1,t_u)$, where
$$t_u=\LT(\frac{c}{\delta u+c}\RT)^2.$$
\EEL
\noindent\prooflem{VIF} Note that
\BQNY
\frac{\partial V_Z^2(s,t)}{\partial t}=2V_Z(s,t)\frac{\partial V_Z(s,t)}{ \partial t}=-\frac{\sigma^2}{2\delta}, \ \ \frac{\partial V_Z^2(s,t)}{\partial s}=2V_Z(s,t)\frac{\partial V_Z(s,t)}{\partial s}=\frac{\sigma^2}{2\delta}\Ga(2-\Ga).
\EQNY
Then we have
\BQNY
\frac{\partial F_u(s,t)}{\partial t}&=&\frac{\partial V_Z(s,t)}{\partial t}\cdot\frac{1}{G_u(s,t)}-\frac{V_Z(s,t)}{G_u^2(s,t)}\LT(-\frac{c}{2\delta}t^{-\frac{1}{2}}\RT)\\
&=&\frac{1}{2G_u^2(s,t) V_z(s,t)}\left(\frac{\partial V_Z^2(s,t)}{\partial t}G_u(s,t)+V_Z^2(s,t)\frac{ct^{-\frac{1}{2}}}{\delta}\right)\\
&=&\frac{\sigma^2 t^{-1/2}}{4\delta G_u^2(s,t) V_Z(s,t)}\left\{[(1-\Ga)^2+(2-\Ga)\Ga s]\frac{c}{\delta}-(u+\frac{c}{\delta})(1-\Ga+\Ga s^{\frac{1}{2}})t^{\frac{1}{2}}\right\}.
\EQNY
So $t_u\rw 0$, as $u\rw\IF$.
\begin{align*}
\frac{\partial F_u(s,t)}{\partial s}&=\frac{\partial V_Z(s,t)}{\partial s}\cdot\frac{1}{G_u(s,t)}-\frac{V_Z(s,t)}{G_u^2(s,t)}(\frac{1}{2}\gamma u s^{-\frac{1}{2}}+\frac{c\Ga}{2\delta} s^{-\frac{1}{2}})\\
&=\frac{1}{2G_u^2(s,t) V_Z(s,t)}\left[\frac{\partial V_Z^2(s,t)}{\partial s}G_u(s,t)-V_Z^2(s,t)(\gamma u s^{-\frac{1}{2}}+\frac{c\Ga}{\delta} s^{-\frac{1}{2}})\right]\\
&=\frac{\Ga \sigma^2 s^{-\frac{1}{2}}}{4\delta G_u^2(s,t) V_Z(s,t)}\left((2-\gamma)[(1-\Ga)(u+\frac{c}{\delta})-\frac{c}{\delta}t^{\frac{1}{2}}]s^{\frac{1}{2}}-
[(1-\gamma)^2-t](u+\frac{c}{\delta})\right)
\end{align*}
so for u large enough, $F_u(\cdot,t_u)$ only can reach its maximum only at $s=1$ or $s=t_u$.
$$F_u(1,t)=\frac{\sqrt{\frac{\sigma^2}{2\delta}(1-t)}}{u+\frac{c}{\delta}(1-t^{\frac{1}{2}})},\ \ \ F_u(t,t)=\frac{\sqrt{\frac{\sigma^2}{2\delta}(1-t)}}{u+\frac{\Ga}{1-\Ga}ut^{\frac{1}{2}}+\frac{c}{\delta}(1-t^{\frac{1}{2}})}<F_u(1,t).$$
So $F_u(s,t)$ attains the maximum point only at $s=1$.
Let $\frac{\partial F_u(1,t)}{\partial t}=0$, we get $t_u=\LT(\frac{c}{\delta u+c}\RT)^2$.
\QED

\BEL\label{PST}
$\{\xi(s,t), s, t\geq 0\}$ is a zero-mean stationary Gaussian field with continuous sample paths and correlation function
$ \Ga_\xi(s,t)=\exp\LT(-(a_1|s|+a_2|t|)\RT)$ for some positive constants $a_i, i=1,2$. Further, $\lim_{u\rightarrow\IF}t_u u^2=c>0$ and $\lim_{u\rightarrow\IF}\frac{f(u)}{u}=d>0$. Then for some positive constants $S_i, i=1,2, b_i, i=1,2$,
\BQNY
&&\pk{\sup_{(s,t)\in[1-S_1u^{-2},1]\times[0,(\sqrt{S_2}u^{-1}+\sqrt{t_u})^2]}
\frac{\xi(s,t)}{1+b_1(1-s)+b_2(\sqrt{t}-\sqrt{t_u})^2}>f(u)}\\
&&\sim\mathcal{P}^{\frac{b_1}{a_1}}[0,a_1d^2S_1]\widehat{\mathcal{P}}^{\frac{b_2}{a_2},d^2a_2c}[0,a_2d^2(\sqrt{c}+\sqrt{S_2})^2]\Psi(f(u)),
\EQNY
where
$$
\widehat{\mathcal{P}}^{\frac{b_2}{a_2},d^2a_2c}[0,a_2d^2(\sqrt{c}+\sqrt{S_2})^2]:=
\E{\sup_{t\in[0,a_2d^2(\sqrt{c}+\sqrt{S_2})^2]}e^{
\sqrt{2}B(t)-t-\frac{b_2}{a_2}(\sqrt{t}-\sqrt{d^2a_2c})^2}}.
$$
\EEL
\noindent
The proof of this lemma follows along the same lines of \cite{Uniform2016}[Theorem 2.1].
\COM{\prooflem{PST}
For all $u>0$, we have
\BQNY
&&\pk{\sup_{(s,t)\in[1-S_1u^{-2},1]\times[0,(\sqrt{S_2}u^{-1}+\sqrt{t_u})^2]}\frac{\xi(s,t)}{1+b_1(1-s)+b_2(\sqrt{t}-\sqrt{t_u})^2}>f(u)}\\
&=&\pk{\sup_{(s,t)\in[0,S_1]\times[-u\sqrt{t_u},\sqrt{S_2}]}\frac{\xi(1-su^{-2},(\sqrt{t_u}+tu^{-1})^2)}{1+b_1u^{-2}s+b_2u^{-2}t^2}>f(u)}\\
&=&\frac{1}{\sqrt{2\pi}}\int_{-\IF}^{+\IF}e^{-v^2/2}\pk{\sup_{(s,t)\in[0,S_1]\times[-u\sqrt{t_u},\sqrt{S_2}]}\frac{\xi(1-su^{-2},(\sqrt{t_u}+tu^{-1})^2)}{1+b_1u^{-2}s+b_2u^{-2}t^2}>f(u)|\xi(1,t_u)=v}dv\\
&=&\frac{1}{\sqrt{2\pi}f(u)}e^{-f^2(u)/2}\int_{-\IF}^{+\IF}e^{w-\frac{w^2}{2f^2(u)}}\pk{{\sup_{(s,t)\in[0,S_1]\times[-u\sqrt{t_u},\sqrt{S_2}]}\mathcal{X}_u(s,t)>w|\xi(1,t_u)=f(u)-\frac{w}{f(u)}}}dw
\EQNY
where
$$\mathcal{X}_u(s,t)=f(u)\left(\frac{\xi(1-su^{-2},(\sqrt{t_u}+tu^{-1})^2)}{1+b_1u^{-2}s+b_2u^{-2}t^2}-f(u)\right)+w.$$
For the family of Gaussian distributions inside the integral
\BQNY
&&E(\mathcal{X}_u(s,t)|\xi(1,t_u)=f(u)-\frac{w}{f(u)})\\
&=&f(u)\left(E\left(\frac{\xi(1-su^{-2},(\sqrt{t_u}+tu^{-1})^2)}{1+b_1u^{-2}s+b_2u^{-2}t^2}|\xi(1,t_u)=f(u)-\frac{w}{f(u)}\right)-f(u)\right)+w\\
&=&-f^2(u)\left(1-\frac{r_\xi(su^{-2},2\sqrt{t_u}u^{-1}t+u^{-2}t^2)}{1+b_1u^{-2}s+b_2u^{-2}t^2}\right)
+w\left(1-\frac{r_\xi(su^{-2},2\sqrt{t_u}u^{-1}t+u^{-2}t^2)}{1+b_1u^{-2}s+b_2u^{-2}t^2}\right);
\EQNY
\BQNY
E\LT(\mathcal{X}_u(0,0)|\xi(1,t_u)=f(u)-\frac{w}{f(u)}\RT)=E\LT(\mathcal{X}^2_u(0,0)|\xi(1,t_u)=f(u)-\frac{w}{f(u)}\RT)=0;
\EQNY
\BQNY
&&Var(\mathcal{X}_u(s,t)-\mathcal{X}_u(s',t')|\xi(1,t_u)=f(u)-\frac{w}{f(u)})\\
&=&f^2(u)\left(Var\left[\frac{\xi(1-u^{-2}s,(\sqrt{t_u}+u^{-1}t)^2)}{1+b_1u^{-2}s+b_2u^{-2}t^2}-
\frac{\xi(1-u^{-2}s',(\sqrt{t_u}+u^{-1}t')^2)}{1+b_1u^{-2}s'+b_2u^{-2}t'^2}\right]\right.\\
&&-\left.\left[\frac{r_\xi(u^{-2}s,2\sqrt{t_u}u^{-1}t+u^{-2}t^2)}{1+b_1u^{-2}s+b_2u^{-2}t^2}
-\frac{r_\xi(u^{-2}s',2\sqrt{t_u}u^{-1}t'+u^{-2}t'^2)}{1+b_1u^{-2}s'+b_2u^{-2}t'^2}\right]^2\right).
\EQNY
Letting $u\rw\IF$, we obtain that
\BQN
E(\mathcal{X}_u(s,t)|\xi(1,t_u)=f(u)-\frac{w}{f(u)})=-d^2(b_1|s|+b_2t^2+a_1|s|+a_2|2\sqrt{c}t+t^2|)(1+o(1)),
\EQN
here o(1) depends linearly on $w$; also
\BQNY
Var(\mathcal{X}_u(s,t)-\mathcal{X}_u(s',t')|\xi(1,t_u)=f(u)-\frac{w}{f(u)})=d^2[2a_1|s-s'|+2a_2|(\sqrt{c}+t)^2-(\sqrt{c}+t')^2|](1+o(1)),
\EQNY
here $o(1)$ is uniform in $w$. Thus, there exists a constant $C>0$ such that for all $t, s$ and for all large enough $u$,
\BQN
Var(\mathcal{X}_u(s,t)-\mathcal{X}_u(s',t')|\xi(1,t_u)=f(u)-\frac{w}{f(u)})\leq C|s-s'|+C|t-t'|,
\EQN
Therefore for any $w$,
\BQNY
&&\lim_{u\rw\IF}\pk{{\sup_{(s,t)\in[0,S_1]\times[-u\sqrt{t_u},\sqrt{S_2}]}\mathcal{X}_u(s,t)>w|\xi(1,t_u)=f(u)-\frac{w}{f(u)}}}\\
&=&\pk{\sup_{(s,t)\in[0,S_1]\times[-\sqrt{c},\sqrt{S_2}]}d\sqrt{2a_1}\tilde{B}(s)+d\sqrt{2a_2}B((\sqrt{c}+t)^2)-d^2(b_1|s|+b_2t^2+a_1|s|+a_2|2\sqrt{c}t+t^2|)>w}
\EQNY
Hence the Borel theorem implies the dominated convergence theorem as $u\rw\IF$ in the integrand
$$\int_{-\IF}^{+\IF}e^{w-w^2/2f^2(u)}\pk{{\sup_{(s,t)\in[0,S_1]\times[-u\sqrt{t_u},\sqrt{S_2}]}\mathcal{X}_u(s,t)>w|\xi(1,t_u)=f(u)-\frac{w}{f(u)}}}dw,$$
and the limit is finite and equal to
\BQNY
&&\int_{-\IF}^{+\IF}e^w\pk{\sup_{(s,t)\in[0,S_1]\times[-\sqrt{c},\sqrt{S_2}]}d\sqrt{2a_1}\tilde{B}(s)+d\sqrt{2a_2}B((\sqrt{c}+t)^2)-d^2(b_1|s|+b_2t^2+a_1|s|+a_2|2\sqrt{c}t+t^2|)>w}dw\\
&=&\mathbb{E}\exp\LT(\sup_{(s,t)\in[0,S_1]\times[-\sqrt{c},\sqrt{S_2}]}d\sqrt{2a_1}\tilde{B}(s)+d\sqrt{2a_2}B((\sqrt{c}+t)^2)-d^2(b_1|s|+b_2t^2+a_1|s|+a_2|2\sqrt{c}t+t^2|)\RT)\\
&=&\mathbb{E}\exp\LT(\sup_{(s,t)\in[0,S_1]\times[0,(\sqrt{c}+\sqrt{S_2})^2]}d\sqrt{2a_1}\tilde{B}(s)+d\sqrt{2a_2}B(t)-d^2(b_1|s|+b_2(\sqrt{t}-\sqrt{c})^2+a_1|s|+a_2|t-c|)\RT)\\
&=&\mathcal{P}_1^{\frac{b_1}{a_1}}[0,a_1d^2S_1]\widehat{\mathcal{P}}_1^{\frac{b_2}{a_2},d^2a_2c}[0,a_2d^2(\sqrt{c}+\sqrt{S_2})^2]
\EQNY
\QED}

\COM{In the following theorem we present some results used in the proof of our main theorems; denote the Euler Gamma function by $\Gamma(\cdot)$.

\BT\label{ThmPiter}
Let $T_1,T_2$ be two positive constants, and \xH{let} $\{X(t_1,t_2),(t_1,t_2)\in\lbrack0,T_1]\times\lbrack0,T_2]\}$ be a zero-mean \cJ{Gaussian random field} with standard \cc{deviation} function $\sigma(\cdot,\cdot)$ and correlation function $r(\cdot,\cdot,\cdot,\cdot)$. Assume \xH{that} $\sigma(\cdot,\cdot)$ \cc{attains its unique maximum \xx{on} $\lbrack0,T_1]\times\lbrack0,T_2]$ at  $(t_1^\ast,t_2^\ast)$}, and further
\BQN\label{PF}
\sigma(t_1,t_2)=1-b_{1}|t_1-t_1^\ast|^{\beta_{1}}(1+o(1))-b_{2}|t_2-t_2^\ast|^{\beta_{2}}(1+o(1)),
\EQN
as $(t_1,t_2)\rightarrow(t_1^\ast,t_2^\ast)$ for some positive constants $b_i,\beta_i, i=1,2.$
Let, moreover,
\[
r(t_1,t_1^{\prime},t_2,t_2^{\prime})=1-(a_1|t_1-t_1^{\prime}|^{\alpha_{1}}
+a_2|t_2-t_2^{\prime}|^{\alpha_{2}})(1+o(1))
\]
as $(t_1,t_2),(t_1^{\prime
},t_2^{\prime})\rightarrow(t_1^\ast,t_2^\ast)$ for some positive constants \cc{$a_i, i=1,2$ and $\alpha_i\in(0,2], i=1,2.$}
In addition, there \cL{exist two positive constants $G, \mu$ with $\mu\in(0,2]$} such that
$$
\E{(X(t_1,t_2)-X(t_1',t_2'))^2}\le G(|t_1-t_1^{\prime}|^{\mu}
+|t_2-t_2^{\prime}|^{\mu})
$$
for any $(t_1,t_2),(t_1^{\prime},t_2^{\prime})\in\lbrack0,T_1]\times\lbrack0,T_2]$.
Then as $u\to \IF$
$$\pk{\sup_{(t_1,t_2)\in\lbrack0,T_1]\times\lbrack0,T_2]}X(t_1,t_2)>u}= \left(\prod_{i=1}^2 A_i\right)\Psi(u)(1+o(1)),$$

where for $i=1,2,$
\BQNY
&&A_i:=\left\{
\begin{array}{ll}
\mathcal{H}_{\alpha_i}a_i^{1/\alpha_i} b_i^{-1/{\beta_i}}\ \widehat{I}_i \  \Gamma\left(\frac{1}{\beta_i}+1\right)u^{\frac{2}{\alpha_i}-\frac{2}{\beta_i}},& \hbox{if} \ \ \alpha_i<\beta_i,\\
\widehat{\mathcal{P}_{\alpha_i}^{b_i/a_i}},&  \hbox{if} \ \ \alpha_i=\beta_i,\\
1,&  \hbox{if} \ \ \alpha_i>\beta_i,\\
\end{array}
 \right.\\
&&\widehat{\mathcal{P}_{\alpha_i}^{b_i/a_i}}:=\left\{
            \begin{array}{ll}
      \widetilde{\mathcal{P}}_{\alpha_i}^{b_i/a_i}      , & \hbox{if } t_i^\ast\in(0,T_i) ,\\
\mathcal{P}_{\alpha_i}^{b_i/a_i}, & \hbox{if } t_i^\ast=0\ \text{or}\ T_i,
              \end{array}
            \right.\ \
\widehat{I}_i:=\left\{
            \begin{array}{ll}
  2    , & \hbox{if } t_i^\ast\in(0,T_i) ,\\
1, & \hbox{if } t_i^\ast=0\ \text{or}\ T_i.
              \end{array}
            \right.
\EQNY
\ET
 \noindent\prooftheo{ThmPiter} In the context of Piterbarg \cite{Pit96} and Fatalov \cite{Fatalov1992},  condition \eqref{PF} is formulated as
\BQN\label{PF2}
\sigma(t_1,t_2)=1-(b_{1}|t_1-t_1^\ast|^{\beta_{1}}+b_{2}|t_2-t_2^\ast|^{\beta_{2}})(1+o(1)),\ \ (t_1,t_2)\rightarrow(t_1^\ast,t_2^\ast)
\EQN
In fact, conditions \eqref{PF} and \eqref{PF2} play the same roles in the proof, since only bounds of the form
\BQNY
(b_{1}|t_1-t_1^\ast|^{\beta_{1}}+b_{2}|t_2-t_2^\ast
|^{\beta_{2}})(1-\epsilon)\le 1-\sigma(t_1,t_2)\le (b_{1}|t_1-t_1^\ast|^{\beta_{1}}+b_{2}|t_2-t_2^\ast
|^{\beta_{2}})(1+\epsilon)\
\EQNY
for any $\epsilon>0,\ \ \text{as }(t_1,t_2)\rightarrow(t_1^\ast,t_2^\ast),$ are needed.
Therefore, the claims follow by  similar argumentations as in Piterbarg \cite{Pit96} and Fatalov \cite{Fatalov1992}. \QED\\}

\begin{figure}[H]
  \centering

  \includegraphics[width=15cm,height=8cm]{ruintime7.eps}\\
  \vspace{-0.7cm}
  \caption{Ruin times}\label{fig1}
\end{figure}
Figure \ref{fig1} shows the ruin time $\tau(u)$  of a surplus process $U_\Ga^\delta(t)$.

{\bf Acknowledgement}: Thanks to Enkelejd Hashorva for his suggestions. Thanks to  Swiss National Science Foundation Grant no. 200021-166274.

\bibliographystyle{plain}

 \bibliography{gamLA}

\end{document}

config: arxiv


TITLE: The difference between mathematics and statistics?
QUESTION [2 upvotes]: I am learning how to construct proofs. I always wondered why stats is considered "different" to math? I know that it is an open-ended question.

REPLY [1 votes]: There are two primary aspects of statistics. The first is the development of statistical theory and statistical tests. While not all properties of these statistical tests are always rigorously proven, this first aspect still fundamentally remains in the realm of mathematics.
The second aspect is the choice of which statistical tests are appropriate for achieving certain objectives. This is truly a science and cannot come from purely mathematical thinking. Methods are often employed because they empirically give good results, or because they align with certain intuition that the user intends to follow. Real-life systems are given mathematical models before statistics can be applied, models that are often not clear-cut.


TITLE: Uncertainty of a PT100 temperature calibration
QUESTION [2 upvotes]: Description of the Problem:
I have a PT100 that is calibrated by an external lab and used in our lab as a reference for calibrating thermocouples. In the end, I would like to have a complete uncertainty budget for temperatures measured with the calibrated thermocouples.
I would really appreciate any help in the calculation of the uncertainties. I'll describe my thought process below.
What I did so far:
The PT100 was sent to a lab to calibrate and I got back a table that contains three columns of data:

the true temperature $T_\mathrm{90}$ as defined in the ITS-90
the temperature $T$ measured with the PT100
the measurement uncertainty $u(T)$ from the temperature measurement with the PT100

From this data I learn that the temperature measured with the PT100 is too low, which gets worse with higher temperature. For example, at 300 °C, the PT100 measures 299,5 °C and at 500 °C, the PT100 measures 498 °C (values are rounded, in the sheet from the lab all values are given with a scale of two).
I would like to have a function that corrects for this error and did the following (with OriginLab 2021):

Plot $T$ versus $T_\mathrm{90}$ whereas I put $T$ on the x-axis (thought it would be easier to fit it with a quadratic function that way, see step 2)
Fit the data with a function in the form of $y = y_0 + aT + bT^2$, which gives me the coefficients $y_0$, $a$ and $b$ together with their uncertainties

Now in the function $y$ I can enter the temperature measured with the PT100 and get the corrected value.
For the calculation of the uncertainty $u(y)$ I did the following, using the laws of error propagation:

Calculate the derivatives $\partial_\mathrm{y_0}y=1$, $\partial_\mathrm{a}y=T$, $\partial_\mathrm{b}y=T^2$ and $\partial_\mathrm{T}y=a+2bT$ (with $\partial_\mathrm{x}y=\frac{\partial y}{\partial x}$)
Calculate the uncertainty $u(y)$ as $u(y)=\sqrt{u(y_0)^2+[T u(a)]^2+[T^2 u(b)]^2+[(a + 2bT) u(T)]^2}$

What I got:
Here is a plot of the temperature difference from the uncorrected temperature of the PT100 and the corrected temperature of the PT100 to the temperature $T_\mathrm{90}$, together with the uncertainties $u(T)$ and $u(y)$:

Now my final questions are:

Is this approach correct? Because later I would like to use $y$ and $u(y)$ instead of $T_\mathrm{90}$ to perform a similar correction for the thermocouples.
Is it correct that the error of $y$ rises with the temperature? (Why is it not constant?)

Edit after reading @Urb's answer:
Origin shows $R^2$ as 1.
To reduce the uncertainties I thought: What if I try to fit the data with $y = y_0 + bT^2$, leaving out the $aT$ term? Since this didnt really work I fitted the difference $T_\mathrm{90}-T$ vs. $T$.
The error is $u(y)=\sqrt{u(y_0)^2+[T^2 u(b)]^2+[2bT\cdot u(T)]^2}$.
The corrected temperature is then $T_c=T + y$ and the uncertainty is $u(T_c)=\sqrt{u(T)^2+u(y)^2}$.
If I plot the difference $T_\mathrm{90}-T_c$ vs. $T$ I see that the uncertainties are much smaller, even in the range of $u(T)$ (but also growing with temperature, which I understood now why). Does this calculation make any sense?
Additional question: In this scenario $T_\mathrm{90}$ had no uncertainty. But what do I do if it would have an uncertainty? How do I incorporate this in my calculations?
--> Saw that its the same as above with $a=1$... However, the uncertainty seems to be way lower...

REPLY [2 votes]: The approach looks valid to me, given that the purpose is to obtain a better estimation of the actual temperature, which the model clearly does. I'm guessing that you chose the quadratic function by looking at the plot. Although there are only six points, you could do some checks: is the coefficient of determination $R^2$ high? Are the coefficients of the regression statistically significant doing a $t$-test? Is the dependence really quadratic? (all this seems ok given that none of the residuals $T_{90}-y$ in the figure look large compared with the others). Assuming the fit was done by OLS regression, one could still check if more assumptions hold: normality of the residuals, homoscedasticity, etc. but failure to accomplish any of these won't bias the coefficients, so the estimation is still good; just make sure not to predict the temperature for new values outside the range of temperatures (I'm guessing $[\sim 20,500]\ ^\circ \rm C$) where the fit was done.
The error of $y$ rising with temperature seems correct. Since the functional form of $y(T)$ is quadratic in $T$, for a fixed error $u(T)$ (which is what it seems from the errors in the black curve) the error in $y$ gets inflated as $T$ grows. A simplistic analysis shows that the interval $\left[y(T-u(T)),\ y(T+u(T)))\right]$ grows linearly with $T$.


TITLE: When is the minimizer of the Lagrangian the solution to the primal problem?
QUESTION [1 upvotes]: This is a problem that came up when I studied the proof of the Newman-Pearson lemma/theorem, both in Van Trees' book "Detection Estimation, and Modulation theory" and Kay's "Fundamentals of statistical signal processing: detection theory".
The problem is (the functions are basically integrals of probability density functions, but that is not important here (I think))
\begin{aligned}
\min_x f(x)\\
\text{s.t. }  h(x) = 0
\end{aligned}
and I wish to find the solution to this problem, $x^*$. In the books, this is solved by forming the Lagrangian:
$$L(x,\lambda) = f(x) + \lambda h(x)$$
and minimizing this wrt $x$, giving the dual function
$$g(\lambda) = L(\hat{x}(\lambda), \lambda) = \min_x L(x,\lambda) .$$
We can then choose $\lambda^*$ in such a way that the constraint in the first problem is satisfied, i.e., $h(\hat{x}(\lambda^*) = 0.$ (We assume that there exists such a $\lambda^*$.)
My question: when does this procedure ensure that the obtained solution $x^*$ solves the original problem? In other words, what has to be assumed about $f(x)$, $h(x)$, $L(x,\lambda)$ and $g(\lambda)$ for this to hold?
My thoughts: It seems like this is some special case of basically solving the dual problem, instead of the primal problem, but stopping half way through. It seems reasonable to me that if there is a unique minimizer for the Lagrangian, this should be the one minimizing the original problem, but I base this on (probably) flawed intuition.
A related question, asking about (basically) the converse to my question is here.
Edit
I feel I need to clarify something. To me, the way this problem is solved seems to be kind of a "shortcut" to solving the dual problem. Instead of finding $\lambda^*$ by maximizing the dual function, it is found by looking at the primary constraint. Is this always the case? Or does this only hold if we have strong duality? Or is there another sufficient condition?

REPLY [1 votes]: You basically, If I got you write, ask when Strong Duality holds.
Well, there are countless of edge cases but the simplest one is Slater Condition.
Simply:

If the objective function is convex.
If the Equality Constraints are affine.
If the inequality constraints are convex.
If the domain all above are defined on is convex.

The above states a Convex Problem.
Slater Condition simply says:

The Domain isn't empty.
There is a point in the domain which holds all inequality constraints in strict form or all inequality constraints are affine.

If (5) and (6) hold then the problem is guaranteed to hold Strong Duality -> The dual solution is identical to the primal solution.


TITLE: Characterization of being a submersion
QUESTION [1 upvotes]: Let $\mathcal{M}$ be a smooth $m$-dimensional manifold and let $F\colon \mathcal{M} \longrightarrow \mathbb{R}^{n}$ be any smooth map, with m $\geq$ n . I want to prove that, if $p \in \mathcal{M}$:
$$ \text{$F$ is a submersion at $p$} \iff \lbrace d_{p}F^{1},\dots,d_{p}F^{n}\rbrace  \text{ are l.i. in $(T_{p}\mathcal{M})^{*}$} $$
Were $d_{p}f := \lambda^{-1}_{f(p)} \circ T_{p}f \colon T_{p}\mathcal{M} \longrightarrow \mathbb{R} \in (T_{p}\mathcal{M})^{*}$, $T_{p}f\colon T_{p}\mathcal{M}\longrightarrow T_{f(p)}\mathcal{M}$ is the tangent map at $p$ and $\lambda_{p}\colon V \longrightarrow T_{p}V$ is the canonical isomorphism between a finite dimensional vector space over $\mathbb{R}$ and its tangent space. We can extend the definition of differential of a smooth map $F\colon \mathcal{M} \longrightarrow V$ by putting $D_{p}F := \lambda^{-1}_{F(p)} \circ T_{p}F \colon T_{p}\mathcal{M} \longrightarrow V$, where $V$ is a finite dimensional vector space over $\mathbb{R}$.
This is trivial if you take $\mathcal{M} = \mathbb{R}^{m}$ since you can identify the tangent map with the differential from ordinary calculus, but I want to formalize the proof as much as possible in the general case using the definitions I gave.

REPLY [2 votes]: Linear algebra Fact: Let $A:V\rightarrow \mathbb{R}^n$ be a linear transformation of vector spaces, then $A$ is surjective iff $\pi_1A,\pi_2A,...,\pi_nA$ are linearly independent members of $V^*$. (Where $\pi_i:\mathbb{R}^n\rightarrow \mathbb{R}$ is ith projection).
Proof: Equip $V$with any inner product, then there exists $v_1,...,v_n\in V$ such that $\pi_iA(x)=x \circ  v_i$ for all $x\in V , i\in [n]$. $\dim(Im(A))=\dim(V)-dim(Ker(A))=dim(V)-dim(\{v_1,...,v_n\}^{\perp})=dim(span(\{v_1,...,v_n \}))$.
Thus $A$ is surjective iff $dim(Im A)=n$ iff $dim(span(\{v_1,...,v_n\}))=n$ iff $v_1,...,v_n$ are L.I. iff $\pi_1A,...,\pi_nA$ are L.I. $\square$
Identify the tangent space of any point of $\mathbb{R}^n$ with $\mathbb{R}^n$
Now let's go back to your setting, then for any $v\in T_pM$: $dF^i_p(v)=<v,F^i>=<v,\pi_i \circ f>=<Df|_pv,\pi_i>=\pi_i(Df|_pv)$.
So $dF^i_p=\pi_i\circ Df|_p$. Now apply the previous fact on the linear transformation $Df|_p:T_pM\rightarrow \mathbb{R}^n$ and you will be done.


TITLE: Proposition 2.9 of Hatcher's Algebraic K-theory and Vector Bundles
QUESTION [0 upvotes]: https://www.math.cornell.edu/~hatcher/VBKT/VB.pdf
Reading through Hatcher's proof of the the induced exact sequence of $\widetilde{K}$ groups, I've run into a few issues.

I'm unsure of how there is an induced projection $E/h \to X/A$, I tried forming some type of commutative diagram, but was unsuccessful.
He also says that "A trivialisation of $E$ over $A$ determines the sections $s_i : A \to E$ which form a basis in each finer over $A$." I'm not sure how such a section is determined, and why we are guaranteed that it forms a basis in each fiber.
The last thing I don't understand is "Via a local trivialisation, each section $s_i$ can be regarded as a map from $A \cap U_j$ to a single fiber".

REPLY [1 votes]: Regarding the induced projection.
If I am not mistaken, the equivalence relation is defined to be $h^{-1}(x,v)\sim h^{-1}(y,v)$, and we have maps $E\overset{p}{\to} X\overset{q}{\to} X/A$.
To obtain a map $E/h \to X/A$ we need to show that $q\left(p\left(h^{-1}\left(x,v\right)\right)\right)=q\left(p\left(h^{-1}\left(y,v\right)\right)\right)$, but $p\left(h^{-1}\left(x,v\right)\right)=x\in A$, and similarly for $y$, and indeed $q$ maps take them to the same point $*\in X/A$.
To simplify notations, let's denote $B=p^{-1}(A)\subseteq E$, and for $a\in A$ denote $F_a=p^{-1}(a)$ (the fiber over $a$).
A trivialization is a map $h:B \to A\times \mathbb{C}^n$, which satisfies some conditions.
For each $e_i\in \mathbb{C}^n$ define $s_i: A \to E$ by $s_i(a)=h^{-1}(a,e_i)$ (which is indeed a section of $p$).
One of the conditions $h$ must satisfy is that for each $a\in A$, $h\mid_{F_a}:F_a\to \{a\}\times \mathbb{C}^n\cong\mathbb{C}^n$ is an isomorphism (of vector spaces), so since the $e_i$-s are a basis for $\mathbb{C}^n$, it follows that the $s_i(a)$-s are basis for $F_a$.
The trivialization $h$ allows you to project everything to $A \times \mathbb{C}^n$, and by further projection, to any of its constituents.
So you can this to obtain from $s_i\mid_{A\cap U_j}: A\cap U_j \to B$, a new map $A\cap U_j \to \mathbb{C}$.
Now you can apply the Tietze extension theorem, to extend if to a map $U_j \to \mathbb{C}$.
Using the trivialization yet again, you get the extend map $s_{ij}: U_j \to E$.


TITLE: Let X be a Poisson random variables with $\lambda = 3.$ Find $E(X\mid X>2).$
QUESTION [4 upvotes]: So I write this out as,
$$E(X\mid X>2) = \frac 1 {P(X>2)}\sum_{k=2}^\infty k\frac{3^k e^{-3}}{k!}$$
For $P(X>2)$ I get .80085 and when I multiply by the sum I get 3.55947. My book, however, says that the answer is 4.16525.
PS, I computed $P(X>2)$ by subtracting $P(X=0) + P(X=1)$ from $1.$ I hope I didn't mess this up somehow so I assume I made a mistake in computing the summation.
Thank you in advance. Pardon any mistakes in my math. Hopefully it's not something too silly :P

REPLY [4 votes]: You are asked to require $X>2$, so $2$ itself is not included. Your computation correctly gives $E[X \mid X \geq 2]$.