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\begin{document}

\title{Stat 992: Lecture 03\\
Gaussian kernel smoothing.}
\author{Moo K. Chung \tt{mchung@stat.wisc.edu}}
\maketitle \thispagestyle{empty}

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\begin{enumerate}

\item  {\em Anisotropic kernel.} Let $t=(t_1,\cdots,t_n)' \in
\mathbb{R}^n.$ The $n$-dimensional isotropic Gaussian kernel is
given by the products of $1$-dimensional Gaussian kernel:
$$K(t)=\prod_{i=1}^n \frac{1}{\sqrt{2\pi}}e^{-t_i^2/2}.$$
The isotropic kernel under linear transform $t \to Ht$ changes the
shape of the kernel to anisotropic kernel
$$K_H(t) = K(H^{-1}t)/\det(H).$$
$\det H^{-1}$ is the Jacobian of the transformation that normalize
the density. It is the distribution of $n$-dimensional
multivariate normal with the covariance matrix $HH'$, i.e.
$N(0,HH')$. $H$ is also called the {\em bandwidth matrix} in the
context of kernel smoothing and it measures the amount of
smoothing. It can be shown that $K_{H}$ satisfy the definition of
the Dirac-delta function as the eigenvalues $\lambda_i$ of $H$ go
to zero, i.e.
$$\lim_{\lambda_i \to 0} K_{H}(t) = \delta(t).$$
The limit of the sequence of the Gaussian kernels gives the
Dirac-delta function and this is how we implement the Dirac-delta
function in computer programs. From now on we let $H \to 0$ if all
$\lambda_i \to 0$ and $H \to \infty$ if all $\lambda_i \to
\infty$.

\item {\em Kernel smoothing estimator.} Suppose we have a spatial
model \bq  Y(t) = \mu(t) + \epsilon(t) \label{eq:model1}\eq where
$e$ is a mean zero random field and $\mu$ is an unknown signal. In
image analysis, observations are so dense that we can take them to
be continues functional data. Then the kernel smoothing estimator
is given by \bq \hat \mu(t) = K_{H}*Y(t)= \int K_{H}(t-s) Y(s)
\;ds\label{eq:model2}\eq  As $H$ goes to zero, we are smoothing
less. To see this note that
$$\lim_{\sigma \to 0} \hat \mu(t) = \int \delta(t-s) Y(s) \; ds = Y(t)$$
 From equations (\ref{eq:model1}) and
(\ref{eq:model2}),
$$\mathbb{E} \hat \mu(t) = K_{H}*\mu(t) \to \mu(t) \mbox { as } \sigma \to 0$$
From the property of the Dirac-delta function as $H \to 0$, the
right hand side converges to the true but unknown signal $\mu$. So
we can see that our kernel estimator becomes more unbiased as $H
\to 0$.

{\bf Problem 4.} Investigate the property of the estimator when $H
\to \infty$. There are a couple of ways to prove what happens when
$H \to \infty$. It will be shown later that the estimator is a
solution to a heat equation and the condition $H \to \infty$ is
equivalent to the steady state when we diffuse heat for infinite
time.

Other properties of kernel smoothing estimator can be derived.
Assuming $|\mu| \leq \infty$, \bqn \mathbb{E} \hat \mu (t) \leq
\int K_H(t) \sup_{} \mu(t) \;dt \leq \sup_{} \mu(t).\eqn

Similarly we can bound from below so that \bq \inf_{} \mu(t) \leq
\mathbb{E} \hat \mu(t) \leq \sup_{} \mu(t) .
\label{eq:lecture3-inf}\eq

{\bf Problem 5.} Inequality (\ref{eq:lecture3-inf}) implies that a
smoothed signal will be smaller than the maximum and larger than
the minimum of the signal. Can you come up with shaper bounds than
(\ref{eq:lecture3-inf})? Hint: assume $\mu$ has a finite support
and work on the neighborhood of $t$.

Other interesting property is \bqn \int K_H*Y(t) \;
dt &=& \int_t \int_s K_H(t-s)Y(s) \; ds\; dt\\
&=& \int Y(s) \; ds.\eqn

So $\mathbb{E} \int \hat \mu(t) \; dt = \int \mu(t) \;dt$. If
$\mu(t)$ is a probability such as the transition probability in
Brownian motion, it implies that the total probability is
conserved even after smoothing.

\item {\em Covariance of the kernel estimator.} Let $R_Y$ be the
covariance function of field $Y$. It is trivially $R_Y(t,s) =
R_{\epsilon}(t,s) =\mathbb{E} [\epsilon(t)\epsilon(s)]$, the
covariance function of $\epsilon$. The covariance function of the
kernel smoother can be shown to be \bqn R(t,s) =\int \int
K_{H}(t-t')K_{H}(t-s')R_{\epsilon}(t',s') \;dt'
 \;ds'.\eqn

{\bf Problem 6.}  Is the variance of signal smaller after
smoothing? Come up with interesting inequality involving
variances. Note that ${\bf Var} \hat \mu(t) = R(t,t)$.
\end{enumerate}

\end{document}