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\begin{document}

\title{Stat 992: Lecture 05\\
Diffusion equations.}
\author{Moo K. Chung \tt{mchung@stat.wisc.edu}}
\maketitle \thispagestyle{empty}

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\begin{enumerate}

\item {\em Laplacian of Gaussian kernel.} To implement iterated
kernel smoothing, we need to understand the relationship between
kernel smoothing and diffusion equations. Let $H=\sigma I_n$ and
represent isotropic kernel in $n$-dimensional spherical
coordinates:
$$K_{\sigma}(r) = \frac{1}{(2\pi)^{n/2}\sigma^n}\exp
\big[-\frac{r^2}{2\sigma^2}\big]$$ where $r= \sum_{i=1}^n x_i^2$.
Laplacian $\Delta$ in rectangular coordinates is given by $\Delta
= \sum_{i=1}^n \frac{\partial^2}{\partial t_i^2}.$ In spherical
coordinates it can be shown that
$$\Delta K_{\sigma} = \frac{\partial^2 K_{\sigma}}{\partial r^2} +
\frac{n-1}{r}\frac{\partial K_{\sigma}}{\partial r}.$$ Note that
$$ \frac{\partial K_{\sigma}}{\partial r}=
-\frac{r}{\sigma^2}K_{\sigma}, \frac{\partial^2
K_{\sigma}}{\partial r^2} = \big[\frac{r^2}{\sigma^4}-
\frac{1}{\sigma^2}\big]K_{\sigma}.$$ Thus \bq \Delta K_{\sigma} =
\big[\frac{r^2}{\sigma^4}- \frac{n}{\sigma^2}\big]K_{\sigma}.
\label{eq:lec5-2}\eq

On the other hand \bq \frac{\partial K_{\sigma}}{\partial \sigma}
= \big[\frac{r^2}{\sigma^3}-\frac{n}{\sigma}\big]K_{\sigma}
\label{eq:lec5-3}\eq


From (\ref{eq:lec5-2}) and (\ref{eq:lec5-3}), we have \bq
\frac{1}{\sigma} \frac{\partial K_{\sigma}}{\partial \sigma} =
\Delta K_{\sigma}. \label{eq:lec5-4}\eq

This equation is an isotropic diffusion equation but it doesn't
look like it. Let $2t=\sigma^2$. Then differential $dt = \sigma
d\sigma$ and equation (\ref{eq:lec5-4}) transforms to
$$\frac{\partial K_{\sigma}}{\partial t} = \Delta K_{\sigma}.$$

Now apply convolution on both sides and get
$$\frac{\partial K_{\sigma} * Y(x)}{\partial t} = \Delta \big[K_{\sigma}*Y(x)\big].$$
Hence $K_{\sigma}*Y(x)$ is a solution of an isotropic heat
equation \bq \frac{\partial f}{\partial t} = \Delta f
\label{eq:lec5-5}\eq with initial condition $f(x,0) = Y(x)$ after
time $t=\sigma^2/2$. Note that $\lim_{t \to 0} K_{\sigma}
* Y(x) = Y(x)$ so the solution satisfies the initial condition. So
if we diffuse observation $Y(x)$ for time duration of
$t=\sigma^2/2$. it should be equivalent to kernel smoothing with
bandwidth $\sigma$. This is a remarkable result that has many
important ramifications\footnote{{\bf Project 3.} investigate the
statistical properties of diffusion smoother in 1D and higher
dimensions. Need to do stochastic simulations.}.

{\bf Problem 9.} Prove the uniqueness of the solution. Look at
Greens functions and fundamental solutions. One reference would be
Stakgold's Greens Functions and Boundary Value Problems (1979).

{\bf Problem 10.} Laplacian of image is used to detect edges but
since images are discrete, partial derivatives are undefined. So
traditionally kernel smoothing is applied before estimating the
Laplacian. Implement Laplacian-based edge detector using above
kernel smoothing and apply it to {\tt sagittal.data}. Experiment
with different filter size $\sigma$ to see which one detect the
best. Can you somehow choose $\sigma$ automatically?

\item {\em Iterated Gaussian kernel smoothing.} We will call the
solution to (\ref{eq:lec5-5}) as {\em diffusion smoother}. It is
identical to Gaussian kernel smoother so it should inherit all the
statistical properties of kernel smoothing estimator and vice
versa. When we diffuse observation $Y$ for duration $\delta t =
\sigma^2/2$ we get $K_{\sigma} * Y$. Now we diffuse $K_{\sigma}
* Y$ for additional $\delta t = \sigma^2/2$ and get
$K_{\sigma}* K_{\sigma} * Y$. The total time it has been diffused
is $\sigma^2/2 + \sigma^2/2 = \sigma^2$. Hence it must be
equivalent to kernel smoothing with bandwidth $\sqrt 2 \sigma$,
i.e. $$K_{\sigma} * K_{\sigma} * Y = K_{\sqrt 2 \sigma} * Y.$$ Let
$K_{\sigma} * K_{\sigma} = K_{\sigma}^{(2*)}$. Then
$K_{\sigma}^{(m*)} = K_{\sqrt{m} \sigma}$.

An alternate proof is that noticing that $K_{\sigma}*K_{\sigma}$
is the probability density of the sum of two i.i.d. Gaussian
random variables with variance $\sigma^2$ which gives a Gaussian
with variance $2\sigma^2$.

 {\bf Problem 11.} Come up with other methods for proving this.
One approach is to compute the convolution directly.

\end{enumerate}


\end{document}