\documentclass[11pt,twocolumn]{article}
\usepackage{times,amsmath,amsfonts}
\usepackage[dvips]{graphicx,graphics}


\setlength{\textheight}{10.2in} \setlength{\textwidth}{7.3in}
\setlength{\columnsep}{0.3in} \setlength{\topmargin}{0in}
\setlength{\headheight}{-0.7in} \setlength{\headsep}{0in}
\setlength{\oddsidemargin}{-0.6in}  % Centers text.


% Add the period after section numbers.  Adjust spacing.
%\newcommand{\Section}[1]{\vspace{-5pt}\section{\hskip -1em.~~#1}\vspace{-3pt}}
%\newcommand{\SubSection}[1]{\vspace{-2pt}\subsection{\hskip -1em.~~#1}
%        \vspace{-3pt}}
\newcommand{\bqn}{\begin{eqnarray*}}
\newcommand{\eqn}{\end{eqnarray*}}
\newcommand{\bq}{\begin{eqnarray}}
\newcommand{\eq}{\end{eqnarray}}

\begin{document}

\title{Stat 992: Lecture 02\\
Linear operators on fields.}
\author{Moo K. Chung \tt{mchung@stat.wisc.edu}}
\maketitle \thispagestyle{empty}

%\begin{figure}
%\centering
%\renewcommand{\baselinestretch}{1}
%\includegraphics[scale=0.8]{homework05-1.eps}
%\end{figure}

\begin{enumerate}

\item {\em Differentiation.} The continuity and differentiability
of a random fields boil down to the convergence of sequences of
random fields. The sequence of random fields $X_h(t)$ converges to
$X(t)$ as $h \to 0$ in the mean-square sense if
$$\lim_{h \to 0} \mathbb{E} |X_h(t) - X(t)|^2=0.$$
We denote it by $\lim_{h \to 0} X_h(t) = X(t)$ if there is no
ambiguity. The convergence in $L_2$ normal implies the convergence in $L_1$ norm
as we can see in the following arguemnt due to Prof. Tsui.
$$\mathbb{E}|X_h(t) -X(t)|^2 = -{\bf Var} \big[X_h(t) - X(t)\big]^2- \big[\mathbb{E}X_h(t) - X(t)\big]^2.$$
Now let $X_h \to X$ in mean squre. The right hand side is negative and both terms should
converge to zero.

The mean-square derivative is defined as
$$\frac{\partial X(t)}{\partial t} = \lim_{h \to 0} \frac{X(t+h)-X(t)}{h}.$$
Partial derivatives are defined similarly. Since differentiation
is a linear operator, differentiation of a Gaussian field is again
Gaussian with a different covariance function. For zero mean
Gaussian fields, the covariance function is $R(t,s) = \mathbb{E}
X(t)X(s)$. So the covariance function of the derivative field $X'$
is
$$\mathbb{E} \Big[\frac{d X(t)}{d t}
\frac{d X(s)}{d s} \Big]=\frac{\partial ^2 R(t,s)}{\partial t
\partial s}.$$
{\bf Problem 2.} Given zero mean Gaussian field $X(t), t
=(t_1,\cdots,t_n) \in \mathbb{R}^n$, compute the expectation of
the determinant of the Hessian of $X(t)$. Hessian matrix field
$H(t)$ is given as $H(t) = \Big(\frac{\partial^2 X(t)}{\partial
t_i \partial t_j}\Big).$ Do it for $n=2$. 12 bonus points for
doing it in general $n$. Hint: for general $n$, you need to know
how to decompose the expectation of the products as the sum of
pairwise expectations in Gaussian random variables.

\item {\em Integration.} Integral is defined as the limit of
Riemann sum. Let $\cup_{i=1}^n\Omega_i$ be a partition of $\Omega
\subset \mathbb{R}^n$, i.e. $\Omega = \cup_{i=1}^n \Omega_i$ and
$\Omega_i \cap \Omega_j = \emptyset$ if $i \neq j$. Let $t_i \in
\Omega_i$ and $\mu(\Omega_i)$ be the volume of $\Omega_i$. Then
$$\int_{\Omega} X(t) \; dt = \lim_{\mbox{all }\mu(\Omega_j) \to 0} \sum_{i=1}^n X(t_i)\mu(\Omega_i).$$
Multiple integration is defined similarly. When we integrate a
Gaussian field, it is the limit of a linear combination of
Gaussian random variables so it is again Gaussian.

{\bf Problem 3.} Let $\Omega = [0,1]^n \subset \mathbb{R}^n$. Let
$X(t)$ be a zero mean Gaussian fields in $\Omega$ with covariance
function $R$. Find the distribution of $\int_{\Omega} X(t) \; dt$.


%\begin{figure}
%\centering
%\renewcommand{\baselinestretch}{1}
%\includegraphics[scale=0.25]{ima01-1.eps}
%\includegraphics[scale=0.25]{ima01-2.eps}
%\caption{}
%\end{figure}

\item {\em Linear filter.} We will investigate the linear system
of the form
$$Y(t) = \int K(t,s)X(s) \; ds.$$
where $K$ is called the kernel of the integral. This is called the
{\em convolution} of $K$ and $X$ and denoted by \bq Y(t) =
K*X(t).\label{eq:lecture2-con}\eq In fMRI analysis, for given
input $X \in \mathbb{R}^4$ and observation $Y \in \mathbb{R}^4$,
estimating arbitrary $K$ in linear system (\ref{eq:lecture2-con})
is an important problem. Formulate deconvolution analysis via the
finite element method (FEM). Data will be provided if you want to
take this as a project. James Ramsay worked on this problem a few
years ago. \item {\em Dirac-delta.} Formulation
(\ref{eq:lecture2-con}) is also called kernel smoothing. See
Functional Data Analysis by Ramsay and Silverman (1997) for a
brief introduction to smoothing in the context of nonparametric
regression. We assume the kernel to be isotropic, i.e. $K(t,s) =
K(t-s)$ and $\int K(t) \; dt =1.$ Further we may assume $K$ to be
unimodal with some parameter $\sigma$ such that
$$ \lim_{\sigma \to 0} K(t;\sigma) \to \delta(t)$$
the Dirac-delta function. Read Quantum Mechanics by Dirac, P.A.M.
(1958) firsthand where he introduced the Dirac-delta for the first
time. It is a special case of {\em generalized functions} or {\em
distributions}. For mathematical treatment of distribution theory,
read Generalized functions by Gelfand and Shilov (1964) and
Green's Functions and Boundary Value Problem by Stakgold (1997).
It is defined as
$$\int \delta(t-s)f(s) \;ds = f(t)$$
for any function $f$. An alternate definition is  $\delta(t)=0$ if
$ t\neq 0$ and $\int \delta(t) \;dt =1$. This is why it is usually
called the {\em impulse function} in engineering literature.

{\bf HOMEWORK 01.} Submit solutions to problems 1 to 3 by Feb 2
11:00am. If you don't like these problems, there is no need to
submit it. To get full credit, you only need to solve two third of
total homework problems. Some problems are intentionally vague.
\end{enumerate}


\end{document}
