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\begin{document}

\title{Stat 992: Lecture 01\\
 Gaussian Random Fields.}
\author{Moo K. Chung \tt{mchung@stat.wisc.edu}}
\maketitle \thispagestyle{empty}

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\begin{enumerate}
\item {\em Spatiotemporal model.} Suppose we can measure
temperature $Y$ at position $x$ and time $t$ in a classroom $M \in
\mathbb{R}^3$. Since every measurement will be error-prone, we
model the temperature as
$$Y(x,t) = \mu(x,t) + \epsilon(x,t)$$
where $\mu$ is the real unknown signal and $\epsilon$ is
measurement error. The measurement error can be modelled as a
random variable. So at each point $(x,t) \in M \otimes
\mathbb{R}^+$, measurement error $\epsilon(x,t)$ is a random
variable. The collection of random variables $$\{\epsilon(x,t):
(x,t) \in M \otimes \mathbb{R}^+\}$$ is called a {\em stochastic
process}. For any stochastic process that contains a spatial
variable, it is called a {\em random field}. A formal measure
theoretic definition can be found in Geometry of Random Fields by
Adler (1980) and Introduction to the Theory of Random Processes by
Gikhman and Skorokhod (1969). Functional magnetic resonance images
(fMRI) is one example spatiotemporal modeling is necessary.

\item {\em Gaussian random fields.} A random vector
$X=(X_1,\cdots,X_m)$ is multivariate normal if $\sum_{i} c_i X_i$
is Gaussian for every possible choice of $c_i$.

A random fields $\epsilon(x) \in \mathbb{R}^N$ is a {\em Gaussian
random field} if $\epsilon(x_1),\cdots, \epsilon(x_m)$ is
multivariate normal for any $x_i \in M \subset \mathbb{R}^N$.
$\epsilon$ is a mean zero Gaussian field if $\mathbb{E}
\epsilon(x) = 0$ for all $x$. The covariance function of mean zero
field is defined as
$$R(x,y) = \mathbb{E} \epsilon(x)\epsilon(y).$$
The variance of field $\epsilon$ at fixed point $x$ is $R(x,x)$.
Mean zero Gaussian field is completely characterized by the
covariance function. A Gaussian random vector field is defined
similarly. $e(x)=(e_1(x),\cdots,e_n(x))'$ is a Gaussian vector
field if $e_i$ are Gaussian fields. We may generalize it further
to matrix fields and tensor fields. Diffusion tensor images (DTI)
can be modelled as a tensor field. Unfortunately, it will not be
Gaussian tensor fields although it is possible to make images more
Gaussian via kernel smoothing.


\item {\em Independence.} Two fields $e_1$ and $e_2$ are
independent if $e_1(x)$ and $e_2(y)$ are independent for every $x$
and $y$. For mean zero Gaussian fields, $e_1$ and $e_2$ are
independent if and only if the cross-covariance function
$$R(x,y)=\mathbb{E} e_1(x)e_2(y)=0$$
for all $x$ and $y$.

{\bf Problem 1.} For given two arbitrary mean zero Gaussian
fields, is there mapping that makes them independent?

\item {\em Infinite-dimensional vector space.} Let $\mathcal{L}$
be a collection of random fields. Suppose $e_1, e_2 \in
\mathcal{L}$. $\mathcal{L}$ forms a {\em vector space} if $c_1e_1
+ c_2e_2 \in \mathcal{L}$ for any $c_1$ and $c_2$. Obviously the
collection of Gaussian fields form a vector space. Since there is
no way to represent every element in $\mathcal{L}$ with finite
basis fields, it is an infinite-dimensional vector space. However,
there exists a finite vector space $L_p$ that is the closest to
$\mathcal{L}$ in the least-squares sense (Hilbert space theory).
Real and Complex Analysis by Rudin (1986) gives a very nice
undergraduate level introduction of Hilbert space. Functional
Analysis by Rudin (1991) and A Course in Functional Analysis by
Conway (1997) gives concise graduate level treatment of the
subject matter.

For any linear operator $f$, $f(\mathcal{L}) \subset \mathcal{L}$.
It can be shown that differentiation of Gaussian fields is again
Gaussian in the mean-square sense.
\end{enumerate}
\begin{figure}
\centering
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\includegraphics[scale=0.25]{ima01-1.eps}
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\caption{The first image is the realization of white noise which
would be unrealistic for modelling room temperature due to
discontinuity. The second image is the realization of a Gaussian
field which show realistic continuous measurements.}
\end{figure}

\end{document}