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\begin{document}

\title{Stat 992: Lecture 23\\
Maxima of random fields.}
\author{Moo K. Chung \tt{mchung@stat.wisc.edu}}
\maketitle \thispagestyle{empty}

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\begin{enumerate}

\item {\em Suprema distributions.} Following previous lectures,
for zero mean random fields $Y$, type I-error $\alpha$ was given
by \bqn \alpha &=& P\Big(\bigcup_{x \in \Omega} \{Y(x) > h\}\Big)\\
&=& P\Big( \sup_{x \in \Omega} Y(x) > h\Big).\eqn So we need to
find the distribution of a random variable $\sup_{x \in \Omega}
Y(x)$. This is a difficult math problem. Let us compute some
simple suprema distributions for Gaussian processes.


{\bf Theorem.}({\em reflection principle}) Consider a Brownian
motion $B(x)$ in interval $\Omega=[0,u]$ with $P(B(0)=0)=1$. Then
 $$ P\Big(\sup_{0 \leq x \leq u}B(x) > h\Big) =\frac{2}
{\sqrt{2\pi}u}\int_h^{\infty} e^{x^2/2u} \;dx.$$ {\em Proof.} Note
this proof is not rigorous but intuitive. For a rigorous proof,
see Freedman's Brownian Motion and Diffusion. Let $M_u = \sup_{0
\leq x \leq u} B(x)$. Define hitting time $$T_h = \inf \{x: 0\leq
x, B(x) = h\}.$$ By treating $x$ to be time, $T_h$ is the first
time the Brownian motion hits $h$. Note that $$P(M_u \geq h) =
P(T_h \leq u).$$ So by knowing the distribution of $T_h$, we know
the distribution of $M_u$. Define a reflected process $\widetilde
B(x)$ to be
$$\widetilde B(x) = \Big[
\begin{array}{c}
  B(x) \mbox{ for } x < T_h,\\
  h - [B(x) -h ] \mbox { for } x \geq T_h.\\
\end{array}$$
Note that $B$ and $\widetilde B$ are equal in distribution from
symmetry. Then \bqn P(M_u \geq h) &=& P(M_u \geq h, B(u) \geq h) +
P(M_u \geq h, B(u) < h)\\
&=& P(B(u) \geq h) +  P(\widetilde
B(u) > h)\\
&=& 2P(B(u) > h)\eqn

\item {\em Smooth Gaussian process.} Read R.J. Adler's On
Excursion Sets, Tube Formulae, and Maxima of Random Fields (1999).
The suprima of Brownian motion is simple due to its independent
increment properties. Now consider 1-dimensional smooth isotropic
Gaussian random process $Y(x), x \in \Omega=[0,1] \subset
\mathbb{R}$. Let $N_h$ to be the number of times $Y$ crosses over
$h$ from below ({\em upcrossing}) in $[0,1]$. Then \bqn
P\Big(\sup_{x \in
[0,1]} Y(x) > h\Big) &=& P(N_h \geq 1 \mbox{ or } Y(0) > h)\\
&\leq& P(N_h \geq 1) + P(Y(0) >h)\\
&\leq& \mathbb{E} N_h + P(Y(0) >h).\eqn Let $\sigma^2 = \mathbb{E}
Y^2(x)$. It can be shown that from Rice formula (1945),
$$\mathbb{E} N_h = C_1\exp(h^2/2\sigma^2)$$
for some constant $C_1$ given in Adler (1999). Also note that
$P(Y(0) > h) =1-\Phi(\frac{h}{\sigma})$ where $\Phi$ is the
cumulative distribution function of the standard normal. Then

From the well known inequality (Feller, 1968, p.193), \bqn
\big(1-\frac{\sigma^2}{h^2}\big)\frac{\sigma}{\sqrt{2\pi}
h}e^{-h^2/2\sigma^2} \leq 1-\Phi \big(\frac{h}{\sigma}\big) \leq
\frac{\sigma}{\sqrt{2\pi} h}e^{-h^2/2\sigma^2} \eqn So \bqn
P\Big(\sup_{x \in [0,1]} Y(x) > h\Big) \leq \Big[C_1
+\frac{C_2}{h}\Big] e^{h^2/2\sigma^2}.\eqn In fact we can show
that \bqn P\Big(\sup_{x \in [0,1]} Y(x) > h\Big) = \Big[C_1
+\frac{C_2}{h} +O(h^{-2})\Big] e^{h^2/2\sigma^2}.\eqn

{\bf Problem 36.} Simulate 10,000 smooth zero mean isotropic
Gaussian fields in $[0,1]$ in {\tt MATLAB} and check the validity
Rice formula by the Monte-Carlo method. You need to plot
$\mathbb{E} N_h$ as a function of $h$. Also plot $P\Big(\sup_{x
\in [0,1]} Y(x) > h\Big)$ as a function of $h$ based on your
simulation and check the above asymptotic result is correct.


\end{enumerate}
\end{document}