
\section{Introduction}

In Chapter~\ref{chap:challenges}, some challenges to the problem of estimating extreme value distributions from limited amounts of data were discussed. In the current chapter, this problem will be approached by exploiting the concept of the mean upcrossing rate. It will be shown that this opens the door to very robust and  reasonably accurate approximations to the extreme value distributions of stochastic processes, provided some reasonable conditions are satisfied. In the majority of books on extreme value statistics, which generally focus on asymptotic results for sequences of data, this approach is usually discussed under a heading that typically goes like the title of the current chapter. By stopping short of the asymptotic limits, this approach offers a uniquely applicable methodology for approximate extreme value analysis of a host of engineering problems. 

In all sections of this chapter, except the last, it is assumed that the stochastic process model is stationary. This is typically referred to as a short-term condition, which is highlighted because the environmental processes causing loads and motions of structures of interest to us, are changing their characteristics with time. E.g. in offshore engineering, the practical time window for sea states to be considered stationary, is typically chosen to be three hours.  Hence, in order to properly handle the estimation of extreme values over the design life of a structure, it is necessary to derive methods that allow us to obtain extreme value distributions for the long-term condition. This problem is discussed in Section~\ref{sec:ltev}.


\section{Average Rate of Level Crossings}\label{sec:arlc} \index{average rate of level crossings}


As an example, let us assume that the forces at a given location in a structure due to wind loads can be modeled as a stationary stochastic process $X(t)$ with smooth realizations, cf. Chapter~\ref{chap:mc}. In fact, in practice, it may be assumed that the variance spectrum, see Chapter~\ref{chap:mc}, of the process has compact support, implying that the realizations are infinitely smooth. It is now 
desirable to calculate how often an arbitrary realization of $X(t)$ can
be expected to exceed a given force level $a$. An equivalent
formulation of the same problem is the following: what is the
average number of $a$-upcrossings per unit time by $X(t)$? An
$a$-upcrossing \index{upcrossing} means that the level $a$ is exceeded with
positive slope. Figure~\ref{fig:6.1} shows part of a
realization of a narrow banded process, where there are three
upcrossings of the indicated level $a$. Each upcrossing is marked
with a small circle in Figure~\ref{fig:6.1}.


\begin{figure}[!hbt]
\begin{center}
%\setlength{\unitlength}{0.012500in}%
\dofig{8cm}{Figures/fig6_1.pdf}
\end{center}
\caption{A realization of a narrow banded process with three
upcrossings of the level $a$ during time $T$.} \label{fig:6.1}
\end{figure}

Let $N^+(a, \Delta t)$ denote the random number of times that
$X(t)$ upcrosses the level $a$ during the time interval $(t, t+
\Delta t)$. It has been assumed that $X(t)$ has smooth realizations, which means that they are differentiable and that the differentiated realizations are continuous. This means that
$x(t+\delta) \approx x(t) + \dot{x}(t) \, \delta$ for $0 \le
\delta \le \Delta t$ ($\Delta t$ small) for any realization $x(t)$
of $X(t)$. In other words, any realization $x(t)$ can be
approximated by a straight line in the interval $(t, t + \Delta
t)$. This implies that $x(t)$ crosses the level $a$ at most once
in this interval, see Figure~\ref{fig:6.2}. The conditions for one
upcrossing of the level $a$ in the interval $(t, t + \Delta t)$
then become
\begin{equation}\label{eq:6.1}
x(t) \le a
\end{equation}
and
\begin{equation}\label{eq:6.2}
x(t+\Delta t) \approx x(t) + \dot{x}(t) \, \Delta t > a.
\end{equation}

\begin{figure}[!hbt]
\begin{center}
%\setlength{\unitlength}{0.012500in}%
\dofig{8cm}{Figures/fig6_2.pdf}
\end{center}
\caption{Local approximation by a straight line at an upcrossing.}
\label{fig:6.2}
\end{figure}



To satisfy Eqs.~(\ref{eq:6.1}) and (\ref{eq:6.2}),
it is clearly necessary that $\dot{x}(t) > 0$. The conditions for one upcrossing can therefore
be written in the following way:
\begin{equation}\label{eq:6.3}
a - \dot{x}(t) \, \Delta t < x(t) \le a
\end{equation}
and
\begin{equation}\label{eq:6.4}
\dot{x}(t) > 0\,.
\end{equation}
It is seen that to calculate the probability
for an upcrossing, the joint density $f_{X(t) \dot{X}(t)}(x,
\dot{x})$ of $X(t)$ and  $\dot{X}(t)$ is needed. The way
stationarity has been defined, $f_{X(t) \dot{X}(t)}(x, \dot{x})$ is
automatically independent of $t$. Hence, we simply write $f_{X
\dot{X}}(x, \dot{x})$ to indicate this independence. From
Eqs.~(\ref{eq:6.3}) and (\ref{eq:6.4}), it is obtained that, for sufficiently small $\Delta t$,
\begin{equation}\label{eq:6.5}
\Prob\{ N^+(a, \Delta t) = 1 \} = \int_0^\infty \int_{a - \dot{x}
\Delta t}^a f_{X \dot{X}}(x, \dot{x}) \, d x \, d \dot{x}.
\end{equation}
Hence, for small $\Delta t$,
\begin{equation}\label{eq:6.6}
\int_{a - \dot{x} \Delta t}^a f_{X \dot{X}}(x, \dot{x}) \, d x =
\dot{x} \, \Delta t \, f_{X \dot{X}}(a, \dot{x}),
\end{equation}
provided that $f_{X \dot{X}}(x, \dot{x})$ is continuous. This
implies that
\begin{equation}\label{eq:6.7}
\Prob\{ N^+(a, \Delta t) = 1 \} = \Delta t \, \int_0^\infty
\dot{x} \, f_{X \dot{X}}(a, \dot{x}) \, d \dot{x}.
\end{equation}


Since it is assumed that the realizations can be approximated
locally by a straight line, then $p_n = \Prob\{ N^+(a, \Delta t) =
n \} $ is negligible for $n = 2, 3, \ldots$ compared to $p_1$ for
sufficiently small $\Delta t$. It follows that
\begin{align}\label{eq:6.8}
\EE[N^+(a, \Delta t)] &= \sum_{n=0}^\infty n \, p_n \approx 0 \cdot
p_0 + 1 \cdot p_1 + 2 \cdot 0 + 3 \cdot 0 + \ldots \nonumber \\
&= p_1 = \Delta t \, \int_0^\infty \dot{x} \, f_{X \dot{X}}(a,
\dot{x}) \, d \dot{x}.
\end{align}

The expected (or average) number of $a$-upcrossings per unit of time, which is
denoted by $\nu_X^+(a)$, is then given by the following expression:
\begin{align}\label{eq:6.9}
\nu_X^+(a ) = \lim_{\Delta t \to 0} \frac{1}{\Delta t} \,
\EE[N^+(a, \Delta t)] =  \int_0^\infty \dot{x} \, f_{X \dot{X}}(a,
\dot{x}) \, d \dot{x}.
\end{align}
$\nu_X^+(a)$ is referred to by several names. In this book we use mostly average (or mean) ($a$-)upcrossing rate and \index{average upcrossing rate} \index{mean upcrossing rate} average (or mean) ($a$-)upcrossing frequency.\index{average upcrossing frequency} \index{mean upcrossing frequency}  It is seen that $\nu_X^+(a )$ depends only on the level $a$. Because
$\nu_X^+(a )$ is independent of $t$, $\EE[N^+(a, T)] = \nu_X^+(a )
\, T$ for any value of $T$, cf. Eq.~(\ref{eq:6.8}), which was derived
under the assumption that $\Delta t$ is small.

Equation~(\ref{eq:6.9}) is a useful formula. It is often
referred to as the Rice formula \index{Rice formula} after its creator S. O. \citet{rice54}.
To get a feeling for the physical content of the Rice formula, one
may note that the right hand side of Eq.~(\ref{eq:6.9}) expresses a
sort of expectation value of the slope at upcrossing of the level
$a$ coupled with the probability of being at that level. The
greater the average slope at a given level, the more often an
arbitrary realization will upcross that level. In other words,
large average positive slope of the time histories implies shorter
cycles and thereby more frequent level crossings. At the same time
one must expect that the number of upcrossings of a given level is
coupled to the probability of reaching that level. The
Rice formula therefore appears to have a fairly plausible form
when it is subjected to closer scrutiny.

In the same way that an $a$-upcrossing was defined, an $a$-down\-crossing can be defined in a similar way. An $a$-downcrossing \index{downcrossing} implies that the level
$a$ is passed with negative slope. The expected number of
$a$-downcrossings per unit of time of a stationary process $X(t)$
is denoted by $\nu_X^-(a)$. To derive the formula for
$\nu_X^-(a)$, the following observation is made: it is obvious
that an $a$-downcrossing for the process $X(t)$ is equivalent with
a $(-a)$-upcrossing for the process $Y(t) = - X(t)$. Hence
$\nu_X^-(a) = \nu_Y^+(-a)$. From the relation $Y(t) = - X(t)$
follows that $f_{Y \dot{Y}}(y, \dot{y}) = f_{X \dot{X}}(-y, -
\dot{y})$. This, together with Eq.~(\ref{eq:6.9}), gives
\begin{align}\label{eq:6.10}
\nu_X^-(a ) &= \nu_Y^+( - a ) = \int_0^\infty \dot{y} \, f_{Y
\dot{Y}}(-a, \dot{y}) \, d \dot{y} = \int_0^\infty \dot{y} \, f_{X
\dot{X}}(a, - \dot{y}) \, d \dot{y} \nonumber \\
&= \int_0^{- \infty} \dot{x} \, f_{X \dot{X}}(a, \dot{x}) \, d
\dot{x} = - \int_{- \infty}^0 \dot{x} \, f_{X \dot{X}}(a, \dot{x})
\, d \dot{x}.
\end{align}
A common way of rewriting Eq.~(\ref{eq:6.10}) is
\begin{align}\label{eq:6.11}
\nu_X^-(a ) =  \int_{- \infty}^0 |\dot{x}| \, f_{X \dot{X}}(a,
\dot{x}) \, d \dot{x}.
\end{align}

An $a$-crossing is either an $a$-upcrossing or an
$a$-downcrossing. The expected number of $a$-crossings per unit of
time, denoted by $\nu_X(a)$, must satisfy the equation $\nu_X(a) =
\nu_X^+(a) + \nu_X^-(a)$. From Eqs.~(\ref{eq:6.9}) and
(\ref{eq:6.11}), it is then obtained that
\begin{align}\label{eq:6.12}
\nu_X(a ) =  \int_{- \infty}^\infty |\dot{x}| \, f_{X \dot{X}}(a,
\dot{x}) \, d \dot{x}.
\end{align}

For a stationary process, any $a$-upcrossing must by necessity be
followed by an $a$-downcrossing, and conversely. If that was not
the case, the realizations would have a mean drift in the positive
or negative direction; that is, the mean value would not be
constant. For a stationary process, the following relations apply:
\begin{align}\label{eq:6.13}
\nu_X^-(a) = \nu_X^+(a) = \frac{1}{2} \, \nu_X(a).
\end{align}




\section{Distribution of Peaks of a Narrow-banded Process}\label{sec:sdpnbp} \index{peak} \index{distribution of peaks}

Assume that $X(t)$ is a stationary process with zero mean value,
which is also narrow banded. What characterizes a realization of a
narrow banded process is that the amplitude and
length (period) of subsequent cycles vary slowly, as illustrated in
Figure~\ref{fig:6.1}. This implies that almost invariably, there
is only one maximum or peak value between an upcrossing and a
subsequent downcrossing of any level $a$, see
Figure~\ref{fig:6.4}. Because a zero mean value was assumed, the
mean number of peaks per unit of time will therefore be
approximately equal to the mean rate of zero-upcrossings, \index{zero-upcrossing} that is,
$\nu_X^+(0)$.

\begin{figure}[!hbt]
\begin{center}
%\setlength{\unitlength}{0.012500in}%
\dofig{8cm}{Figures/fig6_4.pdf}
\end{center}
\caption{(a) Peak between an upcrossing and a subsequent
downcrossing of the level $a$. (b) Two peaks between an upcrossing
and a subsequent downcrossing.} \label{fig:6.4}
\end{figure}


Let $X_p$ denote the size or height of an arbitrary peak of
$X(t)$. $X_p$ becomes a random variable. The probability
distribution of $X_p$ for a narrow banded process $X(t)$ with zero
mean value is now defined as
\begin{align}\label{eq:6.14}
\Prob\{ X_p > a \} = \frac{\nu_X^+(a)}{\nu_X^+(0)} \, , \quad (a
\ge 0).
\end{align}

The distribution $F_{X_p}(a)$ is then given as
\begin{equation}\label{eq:6.15}
F_{X_p}(a) = 1 - \frac{\nu_X^+(a)}{\nu_X^+(0)} \, , \quad a \ge 0,
\end{equation}
while $F_{X_p}(a) = 0$ for $a < 0$.


If $m_X \ne 0$, the following definition applies: $F_{X_p}(a) =  1 -
{\nu_X^+(a)}/{\nu_X^+(m_X)}$ for $a \ge m_X$ and $F_{X_p}(a) = 0$
for $a < m_X$. Because $\nu_X^+(a)$ is assumed to equal the mean
number of peaks per unit of time above the level $a$, then clearly
$\nu_X^+(a)$ will decrease with increasing $a$.  For all
processes of interest to us, it may be assumed that $\nu_X^+(a) \to 0$
when $a \to \infty$. This implies that $F_{X_p}(a)$ gets the
properties that a distribution must have; that is, $F_{X_p}(a)$ is a
nondecreasing function for increasing $a$, $F_{X_p}(a) \to 0$
when $a \to - \infty$ and $F_{X_p}(a) \to 1$ when $a \to \infty$.
It is tacitly assumed that $\nu_X^+(a)$ gets its largest
value when $a = m_X$, which is usually the case. As observed,
it is true for a Gaussian process. Note that it must also apply to
processes that are characterized by having only one maximum
between an upcrossing and a subsequent downcrossing of the mean
value level, that is, for infinitely narrow banded processes. The
assumption made is therefore quite reasonable.

If $\nu_X^+(a)$ can be differentiated with respect to $a$,
the density for peaks is obtained, assuming that $m_X = 0$,
\begin{equation}\label{eq:6.16}
f_{X_p}(a) = - \frac{1}{\nu_X^+(0)} \, \frac{d \nu_X^+(a)}{d a},
\quad a \ge 0,
\end{equation}
while $f_{X_p}(a) = 0$ for $a < 0$. Equations~(\ref{eq:6.15}) and
(\ref{eq:6.16}) are sometimes called the ''peak formulas''. It is
emphasized that they only apply to narrow banded processes.

For the sake of completeness, it should be mentioned that the peak
distributions can also be defined in the general case. However,
this would require that the exact mean number of peaks per unit of
time were used in definitions. The expressions then obtained
would be of limited practical use because they are difficult,
if not impossible, to calculate. The simplifications that are
sometimes introduced to make the expressions amenable to
calculations are often, in fact, ill defined.




\section{Average Upcrossing Rate and Distribution of Peaks of a Gaussian Process}\label{sec:uf} \index{distribution of peaks!Gaussian process} \index{average upcrossing rate!Gaussian process}


A random variable $X$ is normally distributed if the density of $X$ is
given as,
\begin{equation}\label{eq:6.17}
f_{X}(x) = \frac{1}{\sqrt{2 \pi} \sigma_X} \, \exp \Big\{ -
\frac{1}{2} \Big(\frac{x - m_X}{\sigma_X} \Big)^2 \Big\}
\end{equation}
for $\sigma_X > 0$. If $\sigma_X = 0$, then $X = m_X$. In this
case, $X$ may be considered as a degenerate normal variable.

Two random variables $X$ and $Y$ are called {\sl jointly normally
distributed} if the joint density of $X$ and $Y$ is given by the
equation,
\begin{align}\label{eq:6.18}
f_{XY}(x,y) &= \frac{1}{2 \pi \, \sigma_X \, \sigma_Y \sqrt{1 -
\rho_{XY}^2} } \, \exp \Big\{ - \frac{1}{2 \, (1 - \rho_{XY}^2)}
\Big[  \Big(\frac{x - m_X}{\sigma_X} \Big)^2 \nonumber \\
&- 2 \rho_{XY}\, \Big(\frac{x - m_X}{\sigma_X} \Big)\Big(\frac{y -
m_Y}{\sigma_Y} \Big) + \Big(\frac{y - m_Y}{\sigma_Y} \Big)^2 \Big]
\Big\}
\end{align}
where $\rho_{XY} = \EE [(X-m_X) (Y-m_Y) ] / \sigma_X \sigma_Y$ is the correlation coefficient for $X$ and $Y$. Invariably, $\mid \rho_{XY} \mid \le 1$, but to be precise, in Eq.~(\ref{eq:6.18}) it is assumed that $|\rho_{XY}| < 1$ and that $\sigma_X > 0$, $\sigma_Y > 0$.

If $X$ and $Y$ are uncorrelated, $\rho_{XY} = 0$ by definition.
For this case, Eq.~(\ref{eq:6.18}) assumes the form,
\begin{align}\label{eq:6.19}
&f_{XY}(x,y) = \frac{1}{2 \pi \, \sigma_X \, \sigma_Y } \, \exp
\Big\{ - \frac{1}{2} \Big[  \Big(\frac{x - m_X}{\sigma_X} \Big)^2
+ \Big(\frac{y - m_Y}{\sigma_Y} \Big)^2 \Big] \Big\} \nonumber \\
&= \frac{1}{\sqrt{2 \pi} \, \sigma_X } \, \exp \Big\{ -
\frac{1}{2} \Big(\frac{x - m_X}{\sigma_X} \Big)^2 \Big\} \cdot
\frac{1}{\sqrt{2 \pi} \, \sigma_Y } \, \exp \Big\{ - \frac{1}{2}
\Big(\frac{y - m_Y}{\sigma_Y} \Big)^2 \Big\} \nonumber \\
&= f_X(x) \cdot f_Y(y)
\end{align}

According to Eq.~(\ref{eq:6.19}), $X$ and $Y$ are
(statistically) independent variables. We have therefore shown
that two uncorrelated normally distributed (real) variables are
automatically independent. However, one should make a note of the
fact that this does not apply to other types of random variables.

A stochastic process $X(t)$ is called {\sl Gaussian} or {\sl normally
distributed} \index{stochastic process!Gaussian} if the random variable $Z = \sum_{j=1}^n c_j \,
X(t_j)$ is normally distributed for any (arbitrary) choice of $n \:
(= 1, 2, \ldots)$, constants $c_1, \ldots, c_n$, and times $t_1,
\ldots, t_n$. If $Y(t)$ is the response of a
linear, time-invariant system where the input process $F(t)$ is
Gaussian, $F(t)$ and $Y(t)$ are jointly normally distributed
variables for any time $t$, and $Y(t)$ also becomes a Gaussian
process.

If $X(t)$ is a stationary and differentiable Gaussian process, 
$X(t)$ and $\dot{X}(t)$ are jointly normally distributed for
any $t$. It can be shown that $\rho_{X(t) \,
\dot{X}(t)} = 0$, cf. \cite{woha85,namo13}; hence, $X(t)$ and $\dot{X}(t)$ are independent
variables, and it follows that,
\begin{align}\label{eq:6.20}
f_{X(t) \dot{X}(t)}(x, \dot{x}) &= f_{X(t)}(x) \cdot
f_{\dot{X}(t)}(\dot{x}) \nonumber \\
&= \frac{1}{2 \pi \, \sigma_X \, \sigma_{\dot{X}}} \, \exp \Big\{
- \frac{1}{2} \Big[ \Big(\frac{x - m_X}{\sigma_X} \Big)^2 +
\Big(\frac{\dot{x}}{\sigma_{\dot{X}}} \Big)^2 \Big] \Big\}
\end{align}
where the fact that $m_{\dot{X}} = 0$ for the
derivative $\dot{X}(t)$ of any stationary and differentiable
process $X(t)$ has been used. This can be seen as follows:
\begin{align}\label{eq:6.21}
\EE[\dot{X}(t)] = \lim_{N \to \infty} \frac{1}{N} \, \sum_{j=1}^N
\dot{x}_j(t) = \frac{d}{d t} \lim_{N \to \infty} \frac{1}{N} \,
\sum_{j=1}^N x_j(t) = \frac{d}{d t} \, \EE[{X}(t)]
\end{align}
Because $m_X = \EE[{X}(t)]$ = constant for a stationary process, it
follows immediately that $m_{\dot{X}} = 0$. Note also that
$f_{X(t) \dot{X}(t)}(x, \dot{x})$ is independent of $t$.

Let us calculate the mean upcrossing rate $\nu_X^+(a)$.
Substituting from Eq.~(\ref{eq:6.20}) into Eq.~(\ref{eq:6.9}) gives
\begin{align}\label{eq:6.22}
\nu_X^+(a) &= \int_0^\infty  \frac{\dot{x}}{2 \pi \, \sigma_X \,
\sigma_{\dot{X}}} \, \exp \Big\{ - \frac{1}{2} \Big[ \Big(\frac{a
- m_X}{\sigma_X} \Big)^2 + \Big(\frac{\dot{x}}{\sigma_{\dot{X}}}
\Big)^2 \Big] \Big\} \, d \dot{x} \nonumber \\
&= \frac{1}{2 \pi} \, \frac{\sigma_{\dot{X}}}{\sigma_X } \, \exp
\Big\{ - \frac{1}{2} \Big(\frac{a - m_X}{\sigma_X} \Big)^2 \Big\}
\int_0^\infty  \frac{\dot{x}}{\sigma_{\dot{X}}} \, \exp \Big\{ -
\frac{1}{2} \Big(\frac{\dot{x}}{\sigma_{\dot{X}}} \Big)^2 \Big\}
\, d \Big(\frac{\dot{x}}{\sigma_{\dot{X}}} \Big) \nonumber \\
&= \frac{1}{2 \pi} \, \frac{\sigma_{\dot{X}}}{\sigma_X } \, \exp
\Big\{ - \frac{1}{2} \Big(\frac{a - m_X}{\sigma_X} \Big)^2 \Big\}.
\end{align}

It is seen that $\nu_X^+(a)$ decreases rapidly (with $\sigma_X$ as
reference scale) at each side of the mean value, where it assumes
its largest value
\begin{align}\label{eq:6.23}
\nu_X^+(m_X) = \frac{1}{2 \pi} \,
\frac{\sigma_{\dot{X}}}{\sigma_X}.
\end{align}

In many situations one would prefer to define the origin so that
$m_X = 0$. The expression on the rhs of
Eq.~(\ref{eq:6.23}) is therefore often referred to as the
mean zero upcrossing rate, under the tacit assumption that $m_X =
0$. Another corresponding parameter that is often met in the
literature, is the mean zero-crossing period $T_z$, which is
defined by,
\begin{align}\label{eq:6.24}
T_z = (\nu_X^+(0))^{-1} = 2 \pi \,
\frac{\sigma_X}{\sigma_{\dot{X}}}.
\end{align}

For a stationary Gaussian process $X(t)$ with mean value zero,
$\nu_X^+(a)$ is completely determined by the two standard
deviations $\sigma_X$ and $\sigma_{\dot{X}}$.  If the variance
spectrum $S_X(\omega)$ of $X(t)$ is known, $\sigma_X$ and
$\sigma_{\dot{X}}$ can be calculated by using the formulas,
\begin{align}\label{eq:6.25}
\sigma_X^2 = \int_{- \infty}^\infty S_X(\omega) \, d \omega
\end{align}
and
\begin{align}\label{eq:6.26}
\sigma_{\dot{X}}^2 = \int_{- \infty}^\infty \omega^2 \,
S_X(\omega) \, d \omega.
\end{align}

If $X(t)$ is also assumed to be narrow banded,
the density $f_{X_p}(a)$ of the peaks of $X(t)$ may be calculated. From Eqs.~(\ref{eq:6.16})
and (\ref{eq:6.22}), it is found that ($m_X = 0$),
\begin{equation}\label{eq:6.27}
f_{X_p}(a) = \left\{ \begin{array}{l} \frac{a}{\sigma_X^2} \, \exp
\Big\{ - \frac{a^2}{2 \sigma_X^2} \Big\}\, \:\: a \ge 0 \; \\
0\, , \quad a < 0.
\end{array} \right.
\end{equation}
A density of this type is called a {\sl Rayleigh density}, \index{Rayleigh-density} and $X_p$ becomes a Rayleigh distributed variable. An example of $f_{X_p}(a)$ is
shown in Figure~\ref{fig:6.5}


\begin{figure}[!hbt]
\begin{center}
%\setlength{\unitlength}{0.012500in}%
\dofig{8cm}{Figures/rayleigh.pdf}
\end{center}
\caption{The PDF $f_{X_p}(a)$ with $\sigma_X = 1.0$ of Rayleigh
distributed peaks.} \label{fig:6.5}
\end{figure}


Gaussian processes have great practical significance. This is
primarily due to the following two reasons. In many cases, important physical
phenomena that give rise to loads on structures can 
be modeled as Gaussian processes with a reasonable degree of
accuracy. Moreover, weakly damped structures usually make the
response more Gaussian than the load. In addition comes the fact
that a Gaussian process is particularly amenable to analytical
treatment.

From Eq.~\eqref{eq:6.22}, it is seen that $\nu_X^+(a)$ is proportional to $f_{X}(a)$, which is a consequence of the fact that $X(t)$ and $\dot{X}(t)$ are independent random
variables for any $t$. Of course, in general this is not the case. Returning to Eq.~\eqref{eq:6.9}, it may be rewritten as,
\begin{align}\label{eq:6.27a}
\nu_X^+(a ) = \int_0^\infty \dot{x} \, f_{X \dot{X}}(a,
\dot{x}) \, d \dot{x} = \int_0^\infty \dot{x} \, f_{\dot{X} \mid X}(\dot{x} \mid a) \,f_{X}(a) \, d \dot{x}
= \EE[\dot{X}^+ \mid X=a] \: f_{X}(a).
\end{align}
Here $\EE[\dot{X}^+ \mid X=a]$ denotes the average positive slope of $X(t)$ at the level $X(t) = a$, which in general depends on the level $a$. However, as pointed out by \cite{naga08jem}, for a wide range of response processes, this dependence appears to be surprisingly weak.


\section{Extreme Value Distributions by the Upcrossing Rate Method} \index{extreme value distribution!upcrossing rate method}
 
The starting point is that a response quantity has been modeled as
a stationary stochastic process. The goal now is to calculate the
probability distribution of the largest value of the response
process $X(t)$ during a specified time period. It is also a goal to
determine the probability distribution of the time to the first
exceedance of a given response level. These problems are very difficult to solve exactly, but by simplifying somewhat, one may often find reasonably accurate approximate solutions.

Let us denote the largest value that $X(t)$ assumes during the
time $T$ by $M(T)$. That is, $M(T) = \max \{ X(t) ; 0 \le t \le T
\}$. Also, let $\Theta(a)$ denote the time to the first exceedance \index{first exceedance}
of the level $a$. $M(T)$ and $\Theta(a)$ are random variables. If
it is convenient to emphasize that $M(T)$ and $\Theta(a)$ refer to
the process $X(t)$, the notation $M_X(T)$ and $\Theta_X(a)$ is used. 
Clearly,
\begin{align}\label{eq:6.28}
\Prob\{ M(T) \le a \} = \Prob\{ \Theta(a)> T \}
\end{align}
because both events $\{ M(T) \le a\}$ and $\{ \Theta(a) > T \}$
express the same, namely, that there are no exceedances of the
level $a$ during time $T$. Let us call this event $\mathcal{E}$.
Then $\mathcal{E} = \{ X(t) \le a \:\: \mbox{for all} \:\: t \in
(0, T) \}$, but this event can also be expressed as $\mathcal{E} =
\{ X(0) \le a \:\: \mbox{and} \:\: N^+(a,T) = 0 \} $. This is so
because if $X(0) \le a$ and there are no subsequent upcrossings of
$a$, there can be no exceedances. Hence, $\Prob\{
\mathcal{E}\} = \Prob\{ X(0) \le a \:\: \mbox{and} \:\: N^+(a,T) =
0 \} \to \Prob\{ N^+(a,T) = 0 \}$ when $a \to \infty$ because of
the law of marginal probability. When it is written that
$a \to \infty$ here, it means that $a$ assumes values that
are large compared to the typical values for the process considered;
it should not be strictly interpreted as meaning that $a$ grows
beyond all limits. In the chapter on asymptotic extreme value distributions, this will be different. 
Because the extreme values in most cases are much
larger than the typical values, the approximation
$\Prob\{ \mathcal{E} \} = \Prob\{ N^+(a,T) = 0 \}$ is introduced.

To determine $\Prob\{ N^+(a,T) = 0 \}$, the
following simplifying assumption is introduced: upcrossings of high levels are
statistically independent events. If the process $X(t)$ is not too
narrow banded, that is, neighbouring peaks tend to be of similar size, this is a reasonable approximation. This
simplification implies that the random number of upcrossings in an
arbitrary time interval of length $T$ is Poisson distributed with
parameter $\EE[N^+(a,T)] = \nu_X^+(a) \, T$. In particular, this
leads to the result
\begin{align}\label{eq:6.29}
\Prob\{  N^+(a,T) = 0 \} = \exp \{ - \nu_X^+(a) \, T \}.
\end{align}
A derivation of Eq.~(\ref{eq:6.29}) is given in the appendix at the end of this chapter.

From Eq.~(\ref{eq:6.29}), it is then obtained that
\begin{align}\label{eq:6.30}
F_{M(T)}(a) = \Prob\{ M(T) \le a \} = \exp \{ - \nu_X^+(a) \, T \}
\,, \quad (a \to \infty),
\end{align}
and
\begin{align}\label{eq:6.31}
F_{\Theta(a)}(\theta) = 1 - \Prob\{ \Theta(a) > \theta \} = 1 -
\exp \{ - \nu_X^+(a) \, \theta \} \,, \quad (a \to \infty),
\end{align}

Often, $\nu_X^+(a) \, T \ll 1$ for a relevant level $a$ and time
interval $(0, T)$, such that the probability of exceedance of $a$
during the time $T$ can be approximated as
\begin{align}\label{eq:6.32}
\Prob\{ \mbox{Exceedance} \} = F_{\Theta(a)}(T) \approx \nu_X^+(a)
\, T
\end{align}
because $e^x \approx 1 + x$ for $|x| \ll 1$.

The derivations above represent a successful attempt in deriving an approximate expression for the
distribution of the extreme value $M(T)$ \index{extreme value distribution} and for the time to first passage \index{time to first passage} $\Theta(a)$ for large values of $a$, and it was seen that these
distributions are determined by the mean level-upcrossing rate
$\nu_X^+(a)$. The only significant simplification that has been
adopted is the assumption that upcrossings of high levels are
independent. Regarding the response of a lightly damped
structure, the response maxima will have a tendency to occur in
clumps. In particular, large peaks will tend to occur
simultaneously as illustrated in Figure~\ref{fig:6.1}. The
assumption about independent upcrossings will then tend to be less
valid. In such cases, Eqs.~(\ref{eq:6.30}) and
(\ref{eq:6.31}) may give significant deviations from the correct
values, but always on the safe side in the sense that
Eq.~(\ref{eq:6.30}), for instance, leads to larger extreme value
estimates than the correct ones. It will be seen in the next chapter that the ACER method provides a practical and elegant solution to this specific problem. 

When the upcrossing rate is estimated from time series of limited length, there will be uncertainty due to sample variability. Hence, uncertainty quantification is an important issue in extreme value analysis, which is sensitive to small model changes. Since the upcrossing rate is, in fact, largely equivalent to one of the ACER functions, this problem is discussed in the next chapter.

In some situations, the relevant extreme values will be connected
to the smallest or minimum values of a process. Such a case can,
however, be easily recast to a study of maximum values by
observing that $\min \{ X(t); 0 \le t \le T \} = - \max \{ - X(t);
0 \le t \le T \}$.





\section{Extreme Values of Gaussian Processes}\label{sec:evgp} \index{extreme value!Gaussian process}

The particular case of a Gaussian process warrants special attention.
Hence, let $X(t)$ be a stationary Gaussian process with a mean level
upcrossing rate given by Eq.~(\ref{eq:6.22}). For simplicity, it is
assumed that $m_X = 0$. In any case, since changing the mean value
is equivalent to a constant shift of all realizations, nothing
is lost by this assumption.  

The distribution of $M(T)$ ($\ge 0$) for large values of $a$ is then,
according to Eqs.~(\ref{eq:6.22}) and (\ref{eq:6.30}), given by the
expression
\begin{align}\label{eq:6.33}
F_{M(T)}(a) = \exp \Big\{ - \nu_X^+(0) \, T \, \exp \Big( -
\frac{a^2}{2 \sigma_X^2} \Big) \Big\} \,, \quad (a \to \infty),
\end{align}

where the mean zero upcrossing rate enters, that is,
\begin{align}\label{eq:6.34}
\nu_X^+(0) = \frac{1}{2 \pi} \, \frac{\sigma_{\dot{X}}}{\sigma_X}.
\end{align}

The density of $M(T)$, $f_{M(T)}(a)$, can be calculated from
Eq.~(\ref{eq:6.33}) by $f_{M(T)}(a)$ $= dF_{M(T)}(a)/da$, and
it is given as follows (for large values of $a$):
\begin{align}\label{eq:6.33b}
f_{M(T)}(a) &= \frac{a}{\sigma_X^2}\, \nu_X^+(0) \, T \, \exp \Big(
- \frac{a^2}{2 \sigma_X^2} \Big) \exp \Big\{ - \nu_X^+(0) \, T \,
\exp \Big( - \frac{a^2}{2 \sigma_X^2} \Big) \Big\} \nonumber \\
&= \frac{a}{\sigma_X^2}\, \nu_X^+(0) \, T \, \exp \Big(
- \frac{a^2}{2 \sigma_X^2} \Big) \, F_{M(T)}(a).
\end{align}
Assuming that Eq.~(\ref{eq:6.33b}) is valid for all values of
$a$, Figure~\ref{fig:6.6} shows the density of $M(T)$ for various
values of $\nu_X^+(0) \, T$.


\begin{figure}[!hbt]
\begin{center}
%\setlength{\unitlength}{0.012500in}%
\dofig{12cm}{Figures/fig6_6b.pdf}
\end{center}
\caption{Various densities for a stationary Gaussian process $X(t)$. A:
Density of $X(t)$. B: Density of the peaks of $X(t)$ (narrow banded case).
C: Density of $M(T)$ for various values of $N = \nu_X^+(0) \, T$. }
\label{fig:6.6}
\end{figure}



A quantity of particular interest in connection with design of
structures is the level $a = \xi_p = \xi_p(T)$ \index{extreme value!quantile}, which with
probability $p$ is not exceeded during the time $T$, that is,
\begin{align}\label{eq:6.35}
F_{M(T)}(\xi_p) = p.
\end{align}
Because $p = \exp(\ln p)$, it follows from Eq.~(\ref{eq:6.33}) that
\begin{align}\label{eq:6.36}
\exp \Big( - \frac{\xi_p^2}{2 \sigma_X^2} \Big) = - \frac{\ln
p}{\nu_X^+(0) \, T}.
\end{align}
This leads to the formula
\begin{align}\label{eq:6.37}
\xi_p(T) = \sigma_X \, \sqrt{2 \ln \Big( \frac{\nu_X^+(0) \,
T}{\ln(1/p)}\Big)} .
\end{align}
This formula can then be used to find the response level that has
a probability of 1\%, say, of being exceeded ($p = 0.99$) during time
$T$.


The most probable extreme value,\index{extreme value!most probable}  denoted by $\hat{\xi} =
\hat{\xi}(T)$, is given to good approximation by the formula
\begin{equation}\label{eq:6.37a}
\hat{\xi} = \sigma_X\, \sqrt{2 \, \ln \big(\nu_X^+(0)\,T \big) }
\end{equation}
$\hat{\xi}$ is the value where the PDF of the extreme value $M(T)$
attains its maximum. It follows that $\hat{\xi} \approx
\xi_{0.37}$ because $\ln(1/0.37) \approx 1.0$ ($e^{-1} \approx 0.37$).

Specific values of the quantiles $\xi_p/\sigma_X$ for various
numbers of zero upcrossings are listed in
Table~\ref{tab:ev1}. Apart from providing a set of useful reference values, 
Table~\ref{tab:ev1} also clearly illustrates the slow increase in typical extreme values with increasing time for a stationary Gaussian process. 

\begin{table}[!h]
\begin{center}
\caption{\label{tab:ev1} Table of $\xi_p/\sigma_X$-values}
\vspace{2mm}
\begin{tabular}{c | c c c c}
\hline 
$\nu_X^+(0)\,T$   &    $10$   & $100$ & $1000$ & $10000$\\
\hline 
$p=0.37$ & $2.1460$ & $3.0349$ & $3.7169$ & $4.2919$\\
$p=0.50$ & $2.3105$ & $3.1533$ & $3.8143$ & $4.3765$\\
$p=0.90$ & $3.0176$ & $3.7028$ & $4.2797$ & $4.7876$\\
$p=0.95$ & $3.2474$ & $3.8924$ & $4.4448$ & $4.9357$ \\
$p=0.99$ & $3.7156$ & $4.2908$ & $4.7975$ & $5.2556$ \\
\hline
\end{tabular}
\end{center}
\end{table}


Another quantity that is often used as a measure of extreme
values is the expected largest value \index{extreme value!expected largest} during a given time $T$,
that is, $\EE[M(T)]$. This expected value can be calculated as,
\begin{align}\label{eq:6.38}
\EE[M(T)] = \int_0^\infty a \, \frac{d F_{M(T)}(a)}{d a} \, d a.
\end{align}
It is now convenient to introduce a new integration variable
$\eta$ defined by the equation $F_{M(T)}(a) = e^{- \eta}$, which implicitly defines $a$ as a function of $\eta$. Note that $a=0 \: (\infty)$ corresponds to $\eta = \infty \: (0)$. It
is obtained that,
\begin{align}\label{eq:6.39}
\frac{d F_{M(T)}(a)}{d a} \, d a = \frac{d F_{M(T)}(a)}{d \eta}
\,\frac{d \eta}{d a} \, d a = \frac{d F_{M(T)}(a)}{d \eta} \, d
\eta = - e^{- \eta} \, d \eta \,.
\end{align}
Substituted into Eq.~(\ref{eq:6.38}), this leads to the equation,
\begin{align}\label{eq:6.40}
\EE[M(T)] = \int_0^\infty a(\eta) \, e^{- \eta} \, d \eta \,,
\end{align}
where $a = a(\eta)$ is a function of $\eta$. The way
$\eta$ is defined, $\eta = \nu_X^+(0) \, T \, \exp \{ - a^2 / (2
\sigma_X^2)\}$. Our focus is on large $a$-values, that is, small
$\eta$-values, and it is obtained by solving with respect to $a$:
\begin{align}\label{eq:6.41}
a &= \sigma_X \, \sqrt{2 \ln(\nu_X^+(0) \, T) - 2 \ln \eta}
\nonumber \\
&= \sigma_X \, \sqrt{2 \ln(\nu_X^+(0) \, T)} \, \Big\{ 1 -
\frac{\ln \eta}{\ln(\nu_X^+(0) \, T)} \Big\}^{1/2} \nonumber
\\
&= \sigma_X \, \sqrt{2 \ln(\nu_X^+(0) \, T)} \, \Big\{ 1 -
\frac{\ln \eta}{2 \ln(\nu_X^+(0) \, T)} - \frac{(\ln \eta)^2}{8
(\ln(\nu_X^+(0) \, T))^2} + \ldots \Big\}\,.
\end{align}


It can be shown that the main contribution to the integral in
Eq.~(\ref{eq:6.40}) comes from small $\eta$-values, and it is found that,
\begin{align}\label{eq:6.42}
\EE[M(T)] &\approx  \sigma_X \, \sqrt{2 \ln(\nu_X^+(0) \, T)} \,
\Big\{ 1 - \frac{ \int_0^\infty \ln \eta \, e^{- \eta} \, d
\eta}{2 \ln(\nu_X^+(0) \, T)} - \frac{\int_0^\infty (\ln \eta)^2
\, e^{- \eta} \, d \eta}{8 (\ln(\nu_X^+(0) \, T))^2}
+ \ldots \Big\} \nonumber \\
&= \sigma_X \, \sqrt{2 \ln(\nu_X^+(0) \, T)} \, \Big\{ 1 +
\frac{\lambda_E}{2 \ln(\nu_X^+(0) \, T)} - \frac{ \frac{\pi^2}{6} +
\lambda_E^2 }{8 (\ln(\nu_X^+(0) \, T))^2} + \ldots \Big\}
\end{align}
because $\int_0^\infty \ln \eta \, e^{- \eta} \, d \eta = - \lambda_E$
and $\int_0^\infty (\ln \eta)^2 \, e^{- \eta} \, d \eta = \pi^2/6
+ \lambda_E^2$, where $\lambda_E = 0.5772...$ denotes Euler's constant.
Usually, $\ln(\nu_X^+(0) \, T)$ is sufficiently large to warrant
the following approximation
\begin{align}\label{eq:6.43}
\EE[M(T)] \approx  \sigma_X \, \Big\{ \sqrt{2 \ln(\nu_X^+(0) \, T)}
+ \frac{\lambda_E}{\sqrt{2 \ln(\nu_X^+(0) \, T)}} \Big\}
\end{align}

Similarly, it is found that,
\begin{align}\label{eq:6.44}
\EE[M(T)^2] = \int_0^\infty a(\eta)^2\, e^{- \eta} \, d
\eta \approx 2 \sigma_X^2 \,
\big\{ \ln(\nu_X^+(0) \, T) + \lambda_E \big\}
\end{align}

From Eqs.~(\ref{eq:6.44}) and (\ref{eq:6.43}), the
expression for the variance of $M(T)$ may be derived. It is obtained that,
\begin{align}\label{eq:6.45}
\sigma_{M(T)}^2 = \EE[M(T)^2] - \EE[M(T)]^2 \approx \frac{\pi^2
\sigma_X^2}{12 \ln(\nu_X^+(0) \, T)}.
\end{align}
One may note that $\EE[M(T)] \to \infty$, while $\sigma_{M(T)} \to
0$ when $T \to \infty$. In Figure~\ref{fig:6.6} it is clearly seen
how the mean value of $M(T)$ increases, while the standard
deviation decreases with increasing values of $T$.

Let us conclude this discussion of the extreme value distribution of a stationary Gaussian process by showing that it approaches one of the three asymptotic extreme value distributions discussed in Chapter~\ref{chap:aevt} , viz. the Gumbel distribution.\index{Gumbel distribution} The expression for $F_{M(T)}(a)$ given by Eq.~\eqref{eq:6.33} can be written as,
\begin{align}\label{eq:6.45a}
F_{M(T)}(a) = \exp \Big\{ -  \exp \Big( - h(a)
 \Big) \Big\} \,, \quad (a \to \infty),
\end{align}
where $h(a) = \frac{a^2}{2 \sigma_X^2} + \ln(\nu_X^+(0) \, T)$. Let $a_0$ denote the solution of the equation $h(a) = 0$. This gives $a_0 = \sigma_X \sqrt{2 \ln(\nu_X^+(0) \, T)}$. It can be verified that the range of values of $a$ where most of the extreme value distribution ''lives'', cf. Figure~\ref{fig:6.6}, satisfies $|a - a_0 | << a_0$ for $a_0 \to \infty$. Then $h(a) = h(a) - h(a_0) \approx h'(a_0)\, (a - a_0)$ for large values of $a_0$. Hence, it follows that asymptotically,  
\begin{align}\label{eq:6.45b}
F_{M(T)}(a) &\approx \exp \big\{ -  \exp \big( - h'(a_0) \,(a - a_0) \big) \big\} \nonumber \\ 
&= \exp \left\{ -  \exp \left( - \frac{a - a_0}{\sigma_0}
 \right) \right\} \,, \quad (a_0 \to \infty) \,,
\end{align}
where $\sigma_0 = \sigma_X / \big(\sqrt{2 \ln(\nu_X^+(0) \, T)} \big)$. This is clearly an extreme value distribution of the Gumbel type. It may be noted that the mean value of this Gumbel distribution is $a_0 + 0.5772 \sigma_0$, cf. Section~\ref{sec:alevd}, which agrees with the mean value given in Eq.~\eqref{eq:6.43}. Similarly, the variance of this Gumbel distribution is $\pi^2 \sigma_0^2 / 6$, which coincides with the variance derived in Eq.~\eqref{eq:6.45}.

The Gumbel distribution limit will follow for a large class of extreme value models of the form given in Eq.~\eqref{eq:6.45a} with a differentiable function $h$ with the properties that $h(a) \to \infty$ when $a \to \infty$, and $h'(a) > 0$ for $a > a_l$ for some value $a_l$. 




\section{The Crossing Rate of Transformed Processes}\label{ex:cr} \index{average upcrossing rate!transformed process}


Assume that two stationary and differentiable processes $X(t)$ and
$Y(t)$ satisfy the equation
\begin{align}\label{eq:tp1}
Y(t) = h\big[X(t)\big],  
\end{align}
where $h(\cdot)$ is a given differentiable function. The following useful result
can then be shown: the upcrossing rate $\nu_Y^+(b)$ of $Y(t)$ is
determined by the upcrossing rate $\nu_X^+(a)$ of $X(t)$ by the
relation \citep{naes83oe,grig84},
\begin{align}\label{eq:tp2}
\nu_Y^+(b) = \sum_{j=1}^n \nu_X^+(a_j),  
\end{align}
where $a_1, \ldots, a_n$ denote all possible $x$-solutions of the
equation $b = h(x)$.

Let us apply this result on the example $Y(t) = X(t)^2$, where
$X(t)$ is a stationary Gaussian process with mean value zero. It
can be shown that $Y(t)$ is also stationary. In this particular
case, $h(x) = x^2$ so that the equation $b = h(x)$ has the
solutions $a_1 = \sqrt{b}$ and $a_2 = - \sqrt{b}$ ($b \ge 0$).
According to Eqs.~\eqref{eq:tp2} and (\ref{eq:6.22}), it is obtained that,
\begin{align}\label{eq:tp3}
\nu_Y^+(b) = \frac{1}{ \pi} \, \frac{\sigma_{\dot{X}}}{\sigma_X }
\, \exp \Big\{ - \frac{b}{ 2 \sigma_X^2} \Big\}, \quad b \ge 0.  
\end{align}

The extreme value distribution for the $Y(t)$ process then becomes
\begin{align}\label{eq:tp4}
F_{M_Y(T)}(b) = \exp \{ - \nu_Y^+(b) \, T \} = \exp \Big\{ -
\nu_Y^+(0) \, T \, \exp \Big( - \frac{b}{2 \sigma_X^2} \Big)
\Big\}, 
\end{align}
where
\begin{align}\label{eq:tp5}
\nu_Y^+(0) = \frac{1}{ \pi} \, \frac{\sigma_{\dot{X}}}{\sigma_X }
\: ( = 2  \, \nu_X^+(0)). 
\end{align}
Note that $F_{M_Y(T)}(b)$ is in fact a Gumbel distribution. Analogously to the Gaussian case, one will find that, for example,
\begin{align}\label{eq:tp6}
\EE[M_Y(T)] =  2 \sigma_X^2 \, \big\{ \ln(\nu_Y^+(0) \, T) + \lambda_E
\big\}\,. 
\end{align}

For the similar case of $Z(t) = X(t)\, | X(t)|$, then $\nu_Z^+(b) =
\nu_Y^+(|b|)/2$ because in this case the equation $b = h(x) = x |\,x|$
has only one solution, viz. $a = \sign(b) \sqrt{|b|}$.



\section{Hermite Moment Models}\label{sec:hmm} \index{Hermite moment model}


When only statistical moments of a response process are available, it has been proposed to use Hermite moments to capture non-Gaussian behaviour and its effect on extreme response statistics \citep{wint85,wint88}. Assuming that a stationary response process $X(t)$ can be related to a stationary standard Gaussian process $U(t)$ by a strictly increasing function $g$ by $X = g(U)$, an approximation to $g$ is sought in terms of Hermite polynomials. Specifically,

\begin{align}\label{eq:hmm1}
\frac{X - \mu_X}{\sigma_X} &= X_0  = g(U) \approx \kappa \Big( U + \Sigma_{n=3}^N \: g_n \, He_{n-1} (U)\Big) \nonumber \\
&= \kappa \Big( U + g_3 \,(U^2 - 1) + g_4 \, (U^3 - 3 U) + \ldots \Big).
\end{align} 

The expansion coefficients $g_n$ control the shape of the standardized distribution, while the $\kappa$ parameter is a scaling factor ensuring that $X_0(t)$ has unit variance. For $N = 4$, the $g_n$ can be expressed in terms of the central moments $\alpha_n = \EE(X_0^n)$, assuming that $\alpha_4 > 3$ \citep{wint88},
\begin{align}\label{eq:hmm2}
g_3 = \frac{\alpha_3}{4 + 2 \sqrt{1 + 1.5(\alpha_4 - 3)}},
\end{align} 
\begin{align}\label{eq:hmm3}
g_4 = \frac{\sqrt{1 + 1.5(\alpha_4 - 3)} - 1}{18},
\end{align}
and
\begin{align}\label{eq:hmm4}
\kappa = \big(1 + 2 \, g_3^2 + 6 \, g_4^2 \big)^{- 1/2}.
\end{align}
The condition $\alpha_4 > 3$ corresponds to a 'softening' response, signifying that the tails of the distribution is wider than the Gaussian. The opposite case, $\alpha_4 < 3$, is discussed by \cite{wint88}. For some recent work on this method, cf. \cite{zhzhlu19}. Having obtained an approximation of the function $g$ in terms of statistical moments estimated from the recorded or simulated response time series, the result from the previous section can now be applied to determine the crossing rate of the response process to estimate extreme value statistics by the point process method.  

Previously, a common model adopted in ocean engineering was that the ocean surface elevation could be regarded as a Gaussian random field. However, in recent years there has been a development toward implementing also non-Gaussian, second order wave field models. This topic is briefly discussed in Section~\ref{sec:ngf} on extremes of non-Gaussian random fields. With a Gaussian model of the random wave field, the dynamic response of marine structures to the ocean waves would then appear to be represented as a transformation of a Gaussian input process. There will be several examples of such modelling in this book, and it will be clear that, in general, a simple, marginal transformation between the input and output processes will not be adequate. However, it is a rather interesting observation that the upcrossing rate for a wide range of nonlinear dynamical models is largely determined by the probability density \citep{naga08jem}. This indicates that in such cases a marginal transformation may give reasonably good results also for the upcrossing rate.  


\section{Return Period}\label {sec:rp} \index{return period}

Let $Z$ be a random variable, and let
\begin{align}\label{eq:6.46}
p = \Prob\{ Z > z \} = 1 - F_Z(z).
\end{align}

Assume that a series of independent observations of
$Z$ can be made. The mean number of observations to the first time the
observed (measured) value of $Z$ exceeds $z$, is called the {\sl return period} for exceedance of $z$, and it is denoted by $\tilde{R}(z)$. It can be shown that,
\begin{align}\label{eq:6.47}
\tilde{R}(z) = \frac{1}{p} = \frac{1}{ 1 - F_Z(z)}.
\end{align}
This equation can be explained by recognizing that, on average,
$1/p$ trials must be conducted before an event of probability $p$
occurs.

Note that $\tilde{R}(z)$ refers to the number of observations and
that these are assumed to be statistically independent. To express
the return period in terms of time, knowledge about the time
interval between the observations is needed. If the observation interval is
$\Delta t$, the return period specified in terms of time, which is
denoted by $R(z)$, will be given as,
\begin{align}\label{eq:6.48}
R(z) = \Delta t \, \tilde{R}(z).
\end{align}
The observation interval $\Delta t$ must be chosen sufficiently
long such that the individual observations become approximately
independent. Note that $R(\xi_p(T)) = T / (1-p)$, where $\xi_p(T)$
is given by Eq.~(\ref{eq:6.37}) and $T$ is the ''observation'' interval.

A design load with a probability of $10^{-2}$ of being exceeded
during one year is often used in connection with the design of
offshore structures. If $X(t)$ denotes a relevant load
process considered for such a design provision, and $\xi$
denotes the corresponding load level, then $\Prob \{ Z > \xi \} =
0.01$, where $Z = \max(X(t); 0 \le t \le 1\, \mbox{year})$. The
return period for exceedance of $\xi$ then becomes,
\begin{align}\label{eq:6.49}
\tilde{R}(\xi) = \frac{1}{\Prob \{ Z > \xi \}} = \frac{1}{ 0.01} =
100.
\end{align}
The reference period in this case is one year; therefore, $R(\xi) =
100$ years.


It should be mentioned that time varying loads caused by, for
example, ocean waves cannot generally be considered as stationary
over an extended period of time. This implies that quantities such as
yearly maxima must be calculated by using so-called long-term
statistics. This is discussed in the next section.






\section{Long Term Extreme Value Distributions}\label{sec:ltev}\index{extreme value distribution!long-term}

Clearly, the estimation of the extreme loads or load effects on e.g a marine structure subjected to the ocean environment over the design life of the structure must take into account the changing
weather conditions. This is done in a consistent manner by
invoking an appropriate long-term statistical method.

There are basically three different approaches to estimating
characteristic long-term extreme values. These methods are based
on (1) all peak values\index{peak value}, (2) all short-term
extremes, or (3) the long-term extreme value. A more detailed
description follows, where $X(t)$ denotes a zero-mean stochastic
process, for example the wave elevation or a corresponding load
effect, that reflects the changing environmental conditions.
Therefore, $X(t)$ is a nonstationary process\index{nonstationary
process}. Let $T$ denote the long-term time duration, e.g. 1 year,
or a service life of, e.g., 30 years, and let $\tilde{T}$ denote the duration of each
short-term weather condition, assuming that $T= K\, \tilde{T}$,
where $K$ is a large integer. The long-term situation is
considered to be a sequence of $K$ short-term conditions, where
each short-term condition \index{short-term condition} is assumed
to be stationary. Significantly, around the world there are different kinds of weather conditions, also at sea. A coarse characterization of sea states:
\begin{itemize}
	\item[-] extratropical, with slowly varying wave conditions
	\item[-] tropical, with rare hurricanes that represent very rapidly changing weather conditions
\end{itemize}
In this section, the discussions and derivations are limited to extratropical conditions.


Let $W$ denote the vector of parameters that describes the
short-term environmental condition. $W$ can be considered as a
random vector variable. For simplicity, let us assume that $W=(H_s,
T_s)$, where $H_s$ is the significant wave height
\index{significant wave height} and $T_s$ a suitable spectral
period \index{spectral period} (generic notation). For example, $T_s$ may represent the spectral peak period $T_p$ or the mean zero-crossing period $T_z$. In principle, the analysis is
entirely similar if $W$ contains more parameters, e.g. dominant
wave direction, wind speed, etc.



\subsection{All peak values}\label{ssec:apv}\index{long-term extreme
	value distribution!all peak values}

A peak value of $X(t)$\index{peak value}, denoted generically by
$X_p$, is defined here as the maximum value of $X(t)$ between two
consecutive zero upcrossings. For each short-term condition, let
$F_{X_p|H_sT_s}(\xi|h_s, t_s)$ denote the conditional distribution of the
peak value. \cite{batt70} showed that the long-term distribution
$F_{X_p}(\xi)$ of the peak value $X_p$ is given as follows:
\begin{equation}\label{apv1}
F_{X_p}(\xi) = \frac{1}{\overline{\nu_X^+(0)}} \int_{h_s}
\int_{t_s} \nu_X^+(0|h_s, t_s)\, F_{X_p|H_sT_s}(\xi|h_s,
t_s)f_{H_sT_s}(h_s, t_s) \, dh_s dt_s \,,
\end{equation}
where $\overline{\nu_X^+(0)}$ denotes the long-term average
zero-upcrossing rate given by
\begin{equation}\label{apv2}
\overline{\nu_X^+(0)} = \int_{h_s} \int_{t_s} \nu_X^+(0|h_s,
t_s)\, f_{H_sT_s}(h_s, t_s) \, dh_s dt_s \,.
\end{equation}
Here, $\nu_X^+(0|h_s, t_s)$ denotes the average zero-upcrossing
rate for the short-term stationary condition characterized by
$H_s=h_s$ and $T_s=t_s$.

In practical applications, a commonly adopted statistical
distribution for the peak values in a short-term condition is the
Rayleigh distribution, that is,
\begin{equation}\label{apv4}
F_{X_p|H_sT_s}(\xi|h_s, t_s) = 1 - \exp\Big( - \frac{\xi^2}{2\, \sigma_X(h_s,
	t_s)^2}\Big) \,.
\end{equation}
In the modelling of ocean waves, it is sometimes appropriate to use a more accurate distribution of the peak values, or wave crest heights. A distribution that is frequently used, is one proposed by \cite{forr00}. 

Under the assumption that all peak values can be considered as
statistically independent, which may not always be very accurate,
the peak value $\xi_q$ with a probability $q$ of being exceeded
per year is found by solving the following equation:
\begin{equation}\label{apv3}
F_{X_p}(\xi_q) = 1 - \frac{q}{S^{(1y)} \cdot
	\overline{\nu_X^+(0)}} \,,
\end{equation}
where $S^{(1y)} = 365 \cdot 24 \cdot 3600$ denotes the number of seconds in a year. The short-term duration $\tilde{T}$ does not enter into this analysis. In the long run, the relative frequency of the
various sea states is reflected in the joint density $f_{H_sT_s}(h_s, t_s)$,
which can be approximated by using an appropriate scatter
diagram\index{scatter diagram}, if that is available. An example of
such a scatter diagram is shown in Table~\ref{tab:10.2}.

Let the scatter diagram be divided into $m$ intervals for the
$h_s$-values and $n$ intervals for the $t_s$-values. It may often
be an acceptable approximation to assume that $\nu_X^+(0|h_s,
t_s)=T_z^{-1} \approx c \, T_s^{-1}$ for a fixed constant $c$, cf. Eq.~\eqref{eq:6.24}.
Equation~(\ref{apv1}) with the Rayleigh approximation\index{Rayleigh approximation} for
$F_{X_p|H_sT_s}(\xi|h_s, t_s)$ can then be approximately expressed 
in the following form:
\begin{equation}\label{apv5}
F_{X_p}(\xi) \approx \frac{1}{\overline{t_s^{-1}}} \sum_{i=1}^m
\sum_{j=1}^n \Big\{1 - \exp\Big( - \frac{\xi^2}{2\, \sigma_X(h_i,
	t_j)^2}\Big) \Big \} \frac{K_{ij}}{t_j \, K} \,,
\end{equation}
where
\begin{equation}\label{apv6}
\overline{t_s^{-1}} = \sum_{i=1}^m \sum_{j=1}^n
\,\frac{K_{ij}}{t_j \, K} \,.
\end{equation}
Here, $K_{ij}$ equals the number of observations in condition $(i,j)$, that is,
in the $h_s$-interval $(h_i - \Delta h/2, h_i + \Delta h/2)$ and the 
$t_s$-interval $(t_j - \Delta t/2, t_j + \Delta t/2)$; $i= 1,
\ldots, m$, $j= 1, \ldots, n$. $K = \sum_{i=1}^m \sum_{j=1}^n K_{ij}$ is the total number of
observations, or sea states. Also note that the values for $h_s$ and $t_s$ included on the scatter diagram in Table~\ref{tab:10.2} are upper-class limits, that is, $h_i + \Delta h/2$ and $t_j + \Delta t/2$.



\begin{center}
	\begin{table}
		\caption{Scatter diagram northern North Sea, 1973 -- 2001. Values
			given for $h_{{\rm s}}$ and $t_{{\rm p}}$ are upper-class limits.
			\label{tab:10.2}} \scalebox{0.75}{
			\begin{tabular}
				{|p{35pt}|p{17pt}|p{17pt}|p{17pt}|p{17pt}|p{25pt}|p{25pt}|p{25pt}|p{25pt}|p{25pt}|p{25pt}|p{25pt}|p{17pt}|p{17pt}|p{17pt}|p{17pt}|p{17pt}|p{15pt}|p{15pt}|p{20pt}|}
				\cline{1-20} {\textbf{h}$_{\rm s}~$\textbf{(m)}}&
				\multicolumn{19}{|p{437pt}|} {\hspace{7cm}{\textbf{t}$_{\rm p}$\textbf{(s)}}}  \\
				\cline{2-20} & \textbf{3}& \textbf{4}& \textbf{5}& \textbf{6}&
				\textbf{7}& \textbf{8}& \textbf{9}& \textbf{10}& \textbf{11}&
				\textbf{12}& \textbf{13}& \textbf{14}& \textbf{15}& \textbf{16}&
				\textbf{17}& \textbf{18}& \textbf{19}& \textbf{20}&
				\textbf{$>$ 20} \\
				\cline{1-20} \textbf{0.5}& 18& 15& 123& 113& 110& 390& 260& 91&
				38& 42& 32& 3&
				19& 13& 9& 1& 3& 2& 7 \\
				\textbf{1.0}& 16& 49& 675& 433& 589& 1442& 1802& 959& 273& 344&
				125& 33& 64& 29& 13& 1& 7& 1& 6 \\
				\textbf{1.5}& 5& 32& 417& 893& 1107& 1486& 2757& 1786& 636& 731&
				299& 121& 92& 43& 18& 10& 5& 2& 13 \\
				\textbf{2.0}& 1& 0& 102& 741& 1290& 1496& 2575& 1968& 780& 868&
				492& 200& 116& 51& 31& 8& 4& 4& 8 \\
				\textbf{2.5}& 0& 0& 9& 256& 969& 1303& 2045& 1892& 803& 941& 484&
				181& 157& 58& 23& 19& 5& 1& 8 \\
				\textbf{3.0}& 0& 0& 1& 45& 438& 1029& 1702& 1898& 705& 957& 560&
				218& 196& 92& 40& 11& 4& 2& 5 \\
				\textbf{3.5}& 0& 0& 1& 4& 124& 650& 1169& 1701& 647& 865& 456&
				237& 162& 100& 36& 12& 6& 1& 5 \\
				\textbf{4.0}& 0& 0& 2& 0& 33& 270& 780& 1369& 573& 868& 427& 193&
				157& 91& 51& 13& 3& 0& 1 \\
				\textbf{4.5}& 0& 0& 0& 0& 3& 90& 459& 1017& 466& 761& 380& 127&
				137& 86& 31& 23& 6& 5& 0 \\
				\textbf{5.0}& 0& 0& 0& 0& 0& 15& 228& 647& 408& 737& 354& 119&
				96& 50& 32& 18& 2& 4& 1 \\
				\textbf{5.5}& 0& 0& 0& 0& 0& 2& 68& 337& 363& 580& 283& 94& 92&
				31& 24& 10& 6& 2& 0 \\
				\textbf{6.0}& 0& 0& 0& 0& 0& 1& 20& 166& 221& 418& 307& 63& 76&
				24& 13& 9& 4& 0& 0 \\
				\textbf{6.5}& 0& 0& 0& 0& 0& 0& 5& 50& 140& 260& 257& 59& 49& 20&
				12& 4& 2& 2& 2 \\
				\textbf{7.0}& 0& 0& 0& 0& 0& 0& 0& 23& 90& 180& 193& 41& 53& 20&
				5& 3& 3& 0& 0 \\
				\textbf{7.5}& 0& 0& 0& 0& 0& 0& 0& 6& 25& 93& 121& 45& 46& 17& 5&
				5& 0& 1& 0 \\
				\textbf{8.0}& 0& 0& 0& 0& 0& 0& 0& 3& 14& 50& 84& 26& 47& 11& 6&
				0& 1& 0& 0 \\
				\textbf{8.5}& 0& 0& 0& 0& 0& 0& 0& 0& 7& 25& 45& 23& 25& 20& 8&
				0& 0& 0& 0 \\
				\textbf{9.0}& 0& 0& 0& 0& 0& 0& 0& 1& 2& 12& 30& 22& 20& 19& 0&
				0& 0& 0& 0 \\
				\textbf{9.5}& 0& 0& 0& 0& 0& 0& 0& 0& 1& 2& 20& 21& 14& 7& 1& 1&
				0& 1& 0 \\
				\textbf{10.0}& 0& 0& 0& 0& 0& 0& 0& 0& 0& 2& 5& 4& 21& 6& 2& 0&
				0& 0& 0 \\
				\textbf{10.5}& 0& 0& 0& 0& 0& 0& 0& 0& 0& 3& 4& 8& 9& 12& 2& 0&
				0& 0& 0 \\
				\textbf{11.0}& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 2& 0& 4& 3& 1& 0& 1&
				0& 0 \\
				\textbf{11.5}& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 2& 1& 2& 3& 0& 0& 0&
				0& 0 \\
				\textbf{12.0}& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 2& 1& 0& 0&
				0& 0 \\
				\textbf{12.5}& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0& 0& 0&
				0& 0 \\
				\textbf{13.0}& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 0& 1& 0&
				0& 0 \\
				\cline{1-20}
			\end{tabular}
		}
	\end{table}
\end{center}





\subsection{All short-term extremes} \index{long-term extreme
	value distribution!all short-term extremes}

The conditional distribution $F_{\tilde{X}|H_s T_s}(\xi|h_s, t_s)$ of the
largest peak value, $\tilde{X}$, during a short-term condition is
clearly given by the expression
\begin{equation}\label{ase1}
F_{\tilde{X}|H_s T_s}(\xi|h_s, t_s) = \big(\,
F_{X_p|H_s T_s}(\xi|h_s, t_s) \, \big)^{k^{(st)}} \,,
\end{equation}
where $k^{(st)}= \nu_X^+(0|h_s, t_s) \, \tilde{T}$ is the number
of peak values during the short-term condition specified by
$H_s=h_s$ and $T_s=t_s$. The validity of Eq.~(\ref{ase1}) is
again based on the assumption that all peak values are
independent.


The long-term distribution\index{extreme value!long-term} of the
short-term extreme peak values is often approximated by the expression,
\begin{equation}\label{ase2}
F_{\tilde{X}}(\xi) = \int_{h_s} \int_{t_s} 
F_{\tilde{X}|H_sT_s}(\xi|h_s, t_s) \,
f_{H_sT_s}(h_s, t_s) \, dh_s dt_s \,.
\end{equation}


Although the error is usually not significant, the averaging done in Eq.~\eqref{ase2} is not quite correct in the sense that it is not a so-called ergodic average \citep{naes84baor}, which would be the correct approach. To achieve this, Eq.~\eqref{ase2} has to be modified to read \citep{krog85},
\begin{equation}\label{ase2b}
F_{\tilde{X}}(\xi) = \exp \left\{ \int_{h_s} \int_{t_s} \ln F_{\tilde{X}|H_sT_s}(\xi|h_s, t_s) \,
f_{H_sT_s}(h_s, t_s) \, dh_s dt_s \right\} \,.
\end{equation}

The root of this problem of averaging resides in the very notion of long-term statistics, and how it has to be interpreted. By its very definition, it is a notion built on evolution in time. And all information that is extracted about its properties is obtained by taking time averages along the observed time histories. That is precisely what is wrong with Eq.~\eqref{ase2}, because it applies a simple ensemble average. On the other hand, Eq.~\eqref{ase2b}, which is an ergodic average, results from taking the appropriate time averages.

Assuming for illustration that $\tilde{T}$ = 3~h, an estimate of
the value $\xi_q$, which has a probability $q$ of being exceeded
per year, is in this case determined by the equation
\begin{equation}\label{ase3}
F_{\tilde{X}}(\xi_q) = 1 - \frac{q}{365 \cdot 8}
\end{equation}

If Eq.~\eqref{ase2} is used, a relation analogous to Eq.~(\ref{apv5}) would be (with
$\tilde{T}$ = 3~h and $T_z^{-1} \approx c \, T_s^{-1}$)
\begin{equation}\label{ase4}
F_{\tilde{X}}(\xi) \approx  \sum_{i=1}^m \sum_{j=1}^n \Big\{1 -
\exp\Big( - \frac{\xi^2}{2\, \sigma_X(h_i,
	t_j)^2}\Big) \Big \}^{\frac{60^2 \times 3 \times c}{t_j}}\,\, \frac{K_{ij}}{K} \,.
\end{equation}





\subsection{The long-term extreme value} \index{long-term extreme
	value distribution!long-term formulation} \label{ssec:ltev}



The distribution of the extreme value
$\hat{X}=\hat{X}(T)$, that is, the global extreme value over a
long-term period $T$, can be expressed as follows \citep{naes84baor},
\begin{equation}\label{lev1}
F_{\hat{X}}(\xi) = \exp \Big\{ - T \, \int_{h_s} \int_{t_s}
\nu_X^+(\xi|h_s, t_s)\, f_{H_sT_s}(h_s, t_s) \, dh_s dt_s \Big\} \,,
\end{equation}
where $\nu_X^+(\xi|h_s, t_s)$ denotes the average $\xi$-upcrossing
rate for the short-term stationary situation characterized by
$H_s=h_s$ and $T_s=t_s$.

From Eqs.~(\ref{eq:6.33}) and (\ref{eq:6.34}), it follows that for the case of a
zero-mean Gaussian process, Eq.~(\ref{lev1}) would read,
\begin{align}\label{lev2}
&F_{\hat{X}}(\xi) = \nonumber\\ &\exp \Big\{ - T \, \int_{h_s}
\int_{t_s} \frac{\sigma_{\dot{X}}(h_s, t_s)}{2\, \pi \sigma_X(h_s,
	t_s) } \exp \Big( -  \frac{\xi^2}{2 \sigma_X(h_s, t_s)^2}
\Big)\,\, f_{H_sT_s}(h_s, t_s) \, dh_s dt_s \Big\} \,,
\end{align}
where the standard deviations $\sigma_X$ and $\sigma_{\dot{X}}$ in
the long-term situation become functions of the environmental
parameters $h_s$ and $t_s$, as indicated.

With $T$ = 1 year = $S^{(1y)}$ seconds, the value
$\xi_q$, which has a probability $q$ of being exceeded per year,
is now calculated from the equation,
\begin{equation}\label{lev3}
F_{\hat{X}}(\xi_q) = 1 - q \,.
\end{equation}

With reference to Table~\ref{tab:10.2}, Eq.~(\ref{lev2})
can then be expressed as a relation analogous to
Eq.~(\ref{apv5}) in the following way (with $T=1$~year and
$T_z^{-1} \approx c \, T_s^{-1}$):
\begin{equation}\label{lev4}
F_{\hat{X}}(\xi) \approx \exp\Big\{ - \sum_{i=1}^m \sum_{j=1}^n 
\frac{S^{(1y)} \times c }{t_j} \, \exp\Big( -
\frac{\xi^2}{2\, \sigma_X(h_i, t_j)^2}\Big) \,\, \frac{K_{ij}}{K}
\Big\} \,.
\end{equation}



For the purpose of estimating extreme load effects, the use of
scatter diagrams calls for a certain amount of caution. If the
scatter diagram is too coarse, leading to poor resolution in the
tail regions, the long-term extreme value estimates may become
inaccurate. In such cases, it is recommended to use a properly
adapted smooth joint density of the parameters characterizing the
short-term sea states. For our purposes, the joint density of
$W=(H_s, T_s)$ is needed. For North Sea applications, the spectral
period $T_s$ is often the spectral peak period $T_p$ due to
the fact that a commonly adopted spectral model is the JONSWAP
spectrum,\index{wave spectrum!JONSWAP} which is usually parameterized by the significant wave
height and the spectral peak period. The marginal distribution of $H_s$ is now often modeled as one of the following two alternatives:
\begin{itemize}
	\item[-] a three-parameter Weibull distribution 
	\item[-] a combination of a lognormal and a Weibull distribution 
\end{itemize}
The following probabilistic model given by \cite{have80} and \cite{hany86} has been frequently adopted as the latter model. Expressed in terms of probability densities, it assumes the form,
\begin{equation}\label{ew10}
f_{H_s}(h_s) = \frac{1}{\sqrt{2 \pi}\alpha h_s} \, \exp \Big\{ -
\frac{(\ln h_s - \theta)^2}{2\, \alpha^2} \Big\}, \qquad h_s \le \eta,
\end{equation}
and
\begin{equation}\label{ew11}
f_{H_s}(h_s) = \frac{\beta}{\rho} \Big( \frac{h_s}{\rho}
\Big)^{\beta -1} \, \exp \Big\{ - \big( \frac{h_s}{\rho}
\big)^\beta \Big\}, \qquad h_s > \eta,
\end{equation}
where the value of the transition parameter $\eta$ separating the
lognormal model for the smaller values of $H_s$ from the Weibull
model for the larger values, will depend on the geographic
location. A requirement is that $\lim_{h_s \uparrow \eta} f_{H_s}(h_s) = \lim_{h_s \downarrow \eta} f_{H_s}(h_s)$, that is, $f_{H_s}(h_s)$ is continuous at $\eta$.

This marginal density for the significant wave height is complemented
by the conditional density of the spectral peak period $T_p$ given the
value of $H_s$ using a lognormal model:\index{lognormal model}
\begin{equation}\label{ew12}
f_{T_p|H_s}(t_p|h_s) = \frac{1}{\sqrt{2 \pi}\sigma t_p} \, \exp
\Big\{ - \frac{(\ln t_p - \mu)^2}{2\, \sigma^2} \Big\} \,,
\end{equation}
where the parameters $\mu$ and $\sigma$ are assumed to depend on
the significant wave height $h_s$ in the following manner:
\begin{equation}\label{ew13}
\mu = a_1 + a_2 h_s^{a_3} \,,
\end{equation}
\begin{equation}\label{ew14}
\sigma^2 = b_1 + b_2 \exp(-b_3h_s) \,,
\end{equation}
for suitably chosen constants $a_i$ and $b_i$, $i=1,2,3$.

The joint density for the environmental parameters is then obtained by
multiplying the marginal density for the significant wave height with
the conditional density for the spectral peak period, that is, 
\begin{equation}\label{ew15}
f_W(w)= f_{H_s T_p}(h_s, t_p) = f_{H_s}(h_s) \, f_{T_p|H_s}(t_p|h_s)\,.
\end{equation}

The following set of parameter values was cited by
\cite{have02} for locations in the northern North Sea (Statfjord
area): $\alpha=0.6565$, $\theta=0.77$, $\eta=2.90$,
$\beta=2.691$, $\rho=1.503$, $a_1=1.134$, $a_2=0.892$,
$a_3=0.225$, $b_1=0.005$, $b_2=0.120$, $b_3=0.455$.





\subsection{Simplified methods} \index{simplified methods!long-term} \label{ssec:sm}

In this chapter three alternative long-term statistical approaches to estimate the extreme wave-induced response for ULS design checks at a given annual probability of exceedance (or return period), have been  outlined. It is emphasized that a long-term approach is the most accurate approach, if it can be achieved at all. If computer models can be used, a long-term analysis is possible with the computational power that is accessible today. However, such analyses may still become a challenge in terms of required computer time. Therefore, simplified design methods are still very popular, but such methods would always need to be validated against the full long-term approach. 

The applicability of simplified approaches depends on the character of the response, especially whether it can be considered quasistatic or dynamic, which response values are relevant and also which accuracy is required; that is, whether the analysis is carried out in pre-engineering or the detailed design phase. In this connection, the fact that nonlinear hydrodynamic effects might be present and cause sum- or difference-frequency excitation, respectively, should be considered. 

A bottom fixed structure with a natural period below 3 seconds, say, could be considered to have a quasi-static behaviour under steady wave loading and a design approach with appropriately chosen wave height and period would be relevant. For structures with natural periods above 3 seconds, the quasistatic approach might still be used in combination with the use of a dynamic amplification factor (DAF) determined by a stochastic analysis for relevant sea states, if the DAF is limited, say, to less than 1.5. The simplified methods would be relevant for early design phases while a stochastic dynamic approach should be used in the detailed design phase.



\subsubsection*{Method of equivalent storms} \index{simplified methods!method of equivalent storms}

Based on earlier work by \cite{jawh72} and \cite{hahe78},
\cite{trva95} proposed an alternative approach to the calculation
of the long-term extreme load or load effect. In their
approach, the focus is on storm events, similar to what is done in
a peaks-over-threshold analysis. This approach is particularly relevant for tropical areas with rare hurricanes. Hence, the long-term situation is considered as a sequence of storm events.
The method is based on the assumption that the distribution of the
storm extreme response value can be approximated by a Gumbel
extreme value distribution conditional on the most probable
extreme response for that storm. The distribution of the most
probable extreme value itself is assumed to follow a generalized
Pareto distribution, which is determined by fitting to data. By
invoking the rule of total probability, as exemplified by
Eq.~(\ref{ase2}), the long-term extreme response value
distribution can be calculated.


\subsubsection*{Contour line method} \index{simplified methods!method of contour lines}


In recent years, the environmental contour line\index{contour line}
approach \citep{wiudcobjha93,hawi08} has been advocated as a rational basis for choosing the
appropriate short-term design storms leading to load and response
extremes corresponding to a prescribed return period, e.g. 100 years, or
equivalently, a prescribed annual probability of exceedance, which
otherwise has to be obtained from a long-term analysis.

Environmental contour line plots are convenient tools for complicated structural dynamic systems where a full long-term response
analysis is extremely time consuming. Environmental contour lines make
it possible to obtain reasonable long-term extremes by
concentrating the short-term considerations to a few sea states in the scatter diagram.

The contour line approach can be applied for an offshore site if
the joint probability density for the significant wave height and the spectral
peak period is available in the form of a joint model as described
by Eqs.~(\ref{ew10})-(\ref{ew12}). This joint model must
be fitted to the available data given in the
form of a scatter diagram\index{scatter diagram} such as the one in
Table~\ref{tab:10.2}. As demonstrated by \cite{mogaay05}, prediction of extreme values is very sensitive to the amount of environmental data available to represent the long-term variability of the sea states. The fitting of appropriate analytical densities ensures a smoothing and can facilitate a reasonably accurate representation of the long-term extreme response.

Contour lines corresponding to a constant
annual exceedance probability can be obtained by transforming the
joint model to a space consisting of independent, standard
Gaussian variables and then using the inverse first order
reliability method (IFORM),\index{Inverse First Order Reliability Method}\index{IFORM} (see, e.g., \cite{wiudcobjha93}). In the standard Gaussian space, the contour line corresponding to an
annual exceedance probability of $q$ will be circles with radius
$r = \Phi^{-1}(1-q/2920)$, where $\Phi$ denotes the distribution of a
standard Gaussian variable, and 2920 = $365 \times 8$ is the
number of 3-hour sea states per year. Transforming these circles
back to the physical parameter space provides the $q$-probability
contour lines. Approximate contour lines can be obtained by
determining the probability density for the point defined by the
marginal $q$-probability significant wave height and the
conditional median spectral peak period, and then estimating the
$q$-probability contour line by the line of constant probability
density. Contour lines based on the joint model discussed in
Subsection~\ref{ssec:ltev} are plotted in Fig.~\ref{fig:std1}, cf.
\cite{have02}. Even for the most complicated systems, simple
methods may often be used to identify the most
critical range of the $q$-probability contour line regarding a
prediction of the $q$-probability response extreme. 

The advantage of this method is that analyses of only a few sea states are required. As the
most unfavorable sea state along the $q$-probability contour line
is identified, a proper estimate for the $q$-probability response
is taken as the $p$-fractile of the distribution of the 3-hour
extreme response value. It is important to note that the median
3-hour extreme value for this sea state, i.e. $p=0.50$, will not
represent a proper estimate for the $q$-probability extreme value
because this characteristic value will not account for the inherent
randomness of the 3-hour extreme value. The fractile level, $p$,
will depend on the aimed exceedance probability target, $q$, and
the degree of nonlinearity of the system. For most practical
systems, $p=0.90$ seems reasonable for $q=10^{-2}$, while $p=0.95$
may be more adequate for $q=10^{-4}$. As an alternative to using a $p$-fractile above 0.5, the desired load effect may be obtained by multiplying the median or expected maximum value with a factor of 1.2 - 1.3. Anyway, this simplified contour line method ideally needs to be validated by a full long-term analysis for the relevant type of environmental conditions and load effects.


\bigskip
\begin{figure}[!htb]
	\begin{center}
		\dofig{8cm}{Figures/pic12022.pdf} \caption{\label{fig:std1}
			Environmental contour line plot for the wave conditions in the Statfjord area }
	\end{center}
\end{figure}





\section{Appendix}

In this appendix the validity of
Eq.~(\ref{eq:6.29}) is demonstrated. The derivation of the formula for the
mean rate of upcrossings of a given level, as it is expressed in
Eq.~(\ref{eq:6.9}), was based on assumptions about the
probability of the number of upcrossings during short time
intervals. Using similar notation as in Section~\ref{sec:arlc}, that is, $p_n
= p_n(\Delta t) = \Prob \{ N^+(a, \Delta t) = n \}$, it was
assumed in particular that $p_n / p_1 \to 0$, when $\Delta t \to 0$
for $n \ge 2$. In other words, the probability of occurrence of
two or more upcrossings during a short time interval can be
neglected compared with the probability of one upcrossing during
the same time interval. To simplify the notation somewhat, $\lambda = \nu_X^+(a)$ is used. According to Eq.~(\ref{eq:6.9}), it can
therefore be assumed that for sufficiently small $\Delta t$,
\begin{equation}\label{eq:A2.1}
p_1(\Delta t) = \lambda \Delta t
\end{equation}
and
\begin{equation}\label{eq:A2.2}
p_n(\Delta t) = 0 \, , \quad n \ge 2.
\end{equation}
Because $\sum_{n=0}^\infty p_n = 1$, it follows that,
\begin{equation}\label{eq:A2.3}
p_0(\Delta t) = 1 - \lambda \Delta t.
\end{equation}

To proceed, it is now required to use the assumption that the upcrossings
of the level $a$ are independent events. This implies that the
number of upcrossings in a given time interval is statistically
independent of the number of upcrossings in another,
nonoverlapping time interval. This may be used to derive the
following equation:
\begin{align}\label{eq:A2.4}
p_0(t + \Delta t) &= \Prob\{ \mbox{No upcrossings in} \: (0,t+\Delta t)\} \nonumber \\
&= \Prob\{ [\mbox{No upcrossings in} \: (0,t)] \:\: \mbox{and} \:\: [\mbox{No upcrossings in} \: (t, t + \Delta t)]
\} \nonumber \\
&= p_0(t) \, p_0(\Delta t)
\end{align}
Equation~(\ref{eq:A2.4}) together with Eq.~(\ref{eq:A2.3}) gives,
\begin{align}\label{eq:A2.5}
\frac{p_0(t + \Delta t) - p_0(t)}{\Delta t} = - \lambda \,
p_0(t).
\end{align}
Strictly speaking, this equation is only approximately correct.
However, on the basis of the assumptions made, it is realized
that the approximation becomes better the smaller $\Delta t$
becomes. This leads to the differential equation,
\begin{align}\label{eq:A2.6}
\frac{d p_0(t)}{ d t} = - \lambda \,
p_0(t),
\end{align}
which has the solution $p_0(t) = C \, \exp ( - \lambda t)$, where
$C$ is a constant. Clearly $p_0(0) = 1$. This gives $C = 1$. The
solution is therefore,
\begin{equation}\label{eq:A2.7}
p_0(t) = e^{ - \lambda t},
\end{equation}
which corresponds to Eq.~(\ref{eq:6.29}).


While in the process, the expression for
$p_n(t)$ will also be derived. It is realized that $n \ge 1$ upcrossings in the
interval $(0, t+\Delta t)$ can occur as follows: \{ $n$
upcrossings in $(0,t)$ and 0 upcrossings in $(t,t+\Delta t)$ \} or
\{ $n-1$ upcrossings in $(0,t)$ and 1 upcrossing in $(t,t+\Delta
t)$ \}, etc. This can be expressed by the equation,
\begin{align}\label{eq:A2.8}
p_n(t + \Delta t) = \sum_{i=0}^n p_{n-i}(t) \, p_i(\Delta t) = p_n(t)\, p_0(\Delta t) + p_{n-1}(t) \, p_1(\Delta t).
\end{align}
The last equality follows from Eq.~(\ref{eq:A2.2}).

Similarly to the preceding derivation, this leads to the differential
equations,
\begin{align}\label{eq:A2.9}
\frac{d p_n(t)}{ d t} = - \lambda \, p_n(t) + \lambda \,
p_{n-1}(t) \, , \quad n = 1, 2, \ldots
\end{align}
with the initial conditions $p_n(0) = 0$ ($n \ge 1$). These
equations can be solved in several ways. One way is to introduce
the auxiliary functions $u_n(t)$, $n = 0, 1, \ldots$ defined by
$p_n(t) = e^{- \lambda t} \, u_n(t)$. When this is substituted
into Eq.~(\ref{eq:A2.9}), it leads to the equations,
\begin{align}\label{eq:A2.10}
\frac{d u_n(t)}{ d t} = \lambda \,
u_{n-1}(t) \, , \quad n = 1, 2, \ldots\,,
\end{align}
with initial conditions $u_n(0) = 0$, $n = 1, 2, \ldots$. In
particular,
\begin{align}\label{eq:A2.11}
\frac{d u_1(t)}{ d t} = \lambda \,
u_0(t) = \lambda,
\end{align}
which gives $u_1 (t) = \lambda \, t$, because $u_1 (0) = 0$.
Successive solution of Eq.~(\ref{eq:A2.10}) gives,
\begin{align}\label{eq:A2.12}
u_n(t) = \frac{(\lambda\, t)^n}{n !},
\end{align}
and thereby,
\begin{equation}\label{eq:A2.13}
p_n(t) = \frac{(\lambda\, t)^n}{n !} \, e^{ - \lambda t},
\end{equation}
which holds for $n = 0, 1, 2, \ldots$ ($0! = 1$).

The Poisson process\index{Poisson process} is a frequently used model for phenomena
characterized by events that occur approximately independent of
each other. Subject to certain conditions, the Poisson process can
be used to model the stream of telephone calls through a telephone
exchange or the stream of cars passing through a road crossing.

