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\begin{document}

\title{Stat 992: Lecture 13\\
Kernel Smoothing II.}
\author{Moo K. Chung \tt{mchung@stat.wisc.edu}}
\maketitle \thispagestyle{empty}

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 {\bf Problem 4.} Investigate the property of the kernel estimator
 $K_{H}*Y$ 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 amount of time.

{\em Solution.} We will show it for 1D isotropic case. General
case follows similarly. Note that $K_{\sigma} \leq
\frac{1}{\sqrt{2\pi} \sigma}$. So $$\big|K_{\sigma}*Y(x)\big| \leq
\frac{1}{\sqrt{2\pi}\sigma} \big|\int Y(x) \; dx \big|.$$ As
$\sigma \to \infty$, it converges to 0. So it seems the kernel
smoothing is completely useless when the bandwidth is extremely
large. However things change dramatically if we use truncated
kernel $\widetilde{K}_{\sigma}$. Assume $Y$ is defined in finite
domain $\Omega$. We truncate and normalize kernel such that
$$\widetilde{K}_{\sigma} =\frac{K_{\sigma}(x)\mathbf{1}_{\Omega}(x)}
{\int_{\Omega} K_{\sigma}(x) \; dx} =\frac{e^{-x^2/2\sigma^2}
\mathbf{1}_{\Omega}(x)}{\int_{\Omega} e^{-x^2/2\sigma^2} \;dx}$$
where $\mathbf{1}_{\Omega}(x)$ is an indicator function. Then
trivially $$\lim_{\sigma \to \infty} \widetilde{K}_{\sigma} =
\frac{\mathbf{1}_{\Omega}(x)}{\int_{\Omega} \; dx}.$$ Hence $$
\lim_{\sigma \to \infty} \widetilde{K}_{\sigma} *Y(x) =
\frac{\int_{\Omega} Y(x) \; dx}{ \int_{\Omega} \;dx}$$ which is
the average signal over $\Omega$.

The kernel smoothing corresponds to the solution of a heat
equation
$$\frac{\partial f}{\partial t} = \frac{\partial^2 f}{\partial x^2}$$
with initial condition $f(x,0) = Y(x)$ at time $t=2\sigma^2$. As
$\sigma \to \infty$, $t \to \infty$. Heat diffusion after infinite
amount of time corresponds to steady state $\frac{\partial
f}{\partial t}=0$ and the same conclusion can be drawn.


{\bf Problem 5.} \bq \inf_{} \mu(t) \leq \mathbb{E} \hat \mu(t)
\leq \sup_{} \mu(t) . \label{eq:lecture3-inf}\eq Inequality
(\ref{eq:lecture3-inf}) implies that the 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 $t$.

{\em Solution.} We will prove it for isotropic case but
anisotropic case is similar. We need to assume $|\mu| < \infty$
and $\mu$ to have compact support. Let us partition $\mathbb{R}^n
= \cup_{j=1}^m \Omega_j$ be a partition. Then \bqn \mathbb{E} \hat
\mu(t) &=& K_{\sigma}* \mu(t)
\\&=& \sum_{j=1}^m \int_{\Omega_j} K_{\sigma}(t-s)\mu(s) \;ds\\
&\leq& \sum_{j=1}\epsilon_j\sup_{s \in \Omega_j} \mu(s),\eqn where
$\epsilon_j=\int_{\Omega_j} K_{\sigma}(t-s) \;ds$ and
$\sum_{j=1}^m \epsilon_j =1$. Suppose $\arg\max_{1 \leq j \leq m}
\sup_{s \in \Omega_j} \mu(s) = j_0$. We can construct partition
such that only single partition $\Omega_{j_0}$ gives the maximum.
So the global maximum is obtained in partition $\Omega_{j_0}$,
i.e.
$$\sup_{s \in \mathbb{R}^n} \mu(s)= \sup_{s \in \Omega_{j_0}}
\mu(s).$$ Then the largest value of $\sum_{j=1}^m\epsilon_j\sup_{s
\in \Omega_j} \mu(s)$ is obtained if $\epsilon_{j_0}=1$ and other
$\epsilon_j$'s vanish which do not happen since we can always make
$0 <\epsilon_{j_0} < 1$. Hence
$$\sum_{j=1}^m\epsilon_j\sup_{s \in \Omega_j} \mu(s) < \sup_{s \in
\mathbb{R}^n} \mu(s).$$
Similarly we can bound by below so that
$$\inf_{s \in \mathbb{R}^n} \mu(s) < \sum_{j=1}^m\epsilon_j\sup_{s \in \Omega_j}
\mu(s).$$
 There is a degenerate situation this proof breaks down.
Consider $\mu(t) = c < \infty$. In that case, we should have a
single partition $\Omega_{j_0}=\mathbb{R}^n$ and everything
collapses to constant $\inf_{s \in \mathbb{R}^n} \mu(s) =
K_{\sigma}*\mu = \sup_{s \in \mathbb{R}^n} \mu(s) =c.$ {\bf
Problem 24.} Can you come up with a shaper bound than this? There
is. :) Hint: For fixed $t$ consider a partition $\Omega_{j_0}$
that contains $t$.
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