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\begin{document}

\title{Stat 992: Lecture 14\\
Kernel Smoothing III.}
\author{Moo K. Chung \tt{mchung@stat.wisc.edu}}
\maketitle \thispagestyle{empty}

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{\bf Theorem.} If $Z \sim N(0,1)$, $$\mathbb{E} Z^{2m} = 1\cdot 3
\cdot 5 \cdots (2m-1).$$

{\em Proof.} (Prof. Tsui) For density $f(x) =
\frac{1}{\sqrt{2\pi}} e^{-x^2/2}$, $f'(x) = -xf(x)$. From
integration by parts \bqn\int x^{2m} f(x) \;dx&=&
-\int_{-\infty}^{\infty} x^{2m-1}f'(x) \;dx\\
&=& (2m-1)\int x^{2(m-1)}f(x) \;dx.\eqn Hence we have recurrence
$$\mathbb{E} Z^{2m} = (2m-1)\mathbb{E} Z^{2(m-1)}.$$
Solve this with initial condition $\mathbb{E} Z^0 = 1$. $\Box$


{\bf Problem 6.}  Is the variance of signal smaller after
smoothing? Come up with interesting inequality involving
variances. Note that ${\bf Var} \hat \mu(t) = R(t,t)$.\\ {\em
Partial solution.} For stochastic signal $Y(x) = \mu(x) +
\epsilon(x)$. We showed that the covariance function of
$K_{\sigma}*Y$ is \bqn R(x,y) &=& \int \int
K_{\sigma}(x-x')K_{\sigma}(y-y')R_Y(x',y')\;dx'dy'\\
&=&K_{\sigma}*_xK_{\sigma}*_yR_Y(x,y) \eqn where $*_x$ and $*_y$
indicate convolution with respect to argument $x$ and $y$
respectively. Let $x=y$ and we have
$${\bf Var} [K_{\sigma}*Y(x) ] = K_{\sqrt{2}\sigma}*{\bf Var}
Y(x).$$
Consider heat equation
$$\frac{1}{\sigma}\frac{\partial f}{\partial \sigma} = \Delta f$$
for nonnegative function $f \leq 0$. Suppose $x_0$ is the point
that gives local minimum of $f(x,\sigma)$ at fixed $\sigma$. Then
it is trivial to show that $\Delta f(x_0,\sigma) < 0$. Hence
$\frac{\partial f}{\partial \sigma} < 0$. Hence solution to the
heat equation is decreasing at $x_0$ with increasing $\sigma$.
Since the solution of heat equation is also a Gaussian kernel
smoother, $$K_{\sqrt{2}\sigma}*{\bf Var} Y(x_0) < {\bf Var}
Y(x_0)$$ where $x_0$ is now a point that gives local maximum of
${\bf Var}Y(x)$.

{\bf Problem 8.} Using the Fourier transform and its inverse
transform, implement 2D Gaussian kernel smoothing in computer. In
$\tt{MATLAB}$, you can perform finite fourier transform.\\ {\em
Solution.} (Partial solution given by Shijie Tang) We define
Fourier integral transform $\mathcal{F}$ such that
$$\mathcal{F} f(w) = \int_{\mathbb{R}^n} f(x)e^{-2\pi i \langle w,x\rangle} \;dx$$
and its inverse $$\mathcal{F}^{-1}f(x) = \int_{\mathbb{R}^n}
f(w)e^{2\pi i \langle x,w\rangle} \;dw.$$ See Folland's Real
Analysis for details. For $f,g \in L(\mathbb{R})$,
$\mathcal{F}^{-1}(\mathcal{F} f) = \mathcal{F}(\mathcal{F}^{-1} f)
= f$ almost everywhere. Then it can be shown that $$\mathcal{F}
(f*g) = \mathcal {F} f \cdot \mathcal {F}g.$$ Hence \bq
K_{\sigma}* Y = \mathcal{F}^{-1}[
\mathcal{F}K_{\sigma}\cdot\mathcal{F}Y]. \label{eq:ima14-ft}\eq
Gaussian kernel of the form $K(x) = e^{-\pi \|x\|^2}$ is invariant
under the Fourier transform. Then $K_{\sigma}(x)
=\frac{1}{(2\pi)^{n/2}\sigma^n}
K\big(\frac{x}{\sqrt{2\pi}\sigma}\big)$. It can be shown that for
an invertible linear transform $T$
$$\mathcal{F}(f \circ T) = |\det T|^{-1} \mathcal{F}f \circ T^{-1}.$$
So \bq \mathcal{F} K_{\sigma} (w) &=& K(\sqrt{2\pi}\sigma
w)\label{eq:eq:ima14-ft2}\eq In {\tt MATLAB}, we implement
Gaussian kernel smoothing using discrete fourier transform {\tt
FFT2} and its inverse {\tt IFFT2} which are implemented as the
Riemann sum of the integral operators. There is a way to speed
things up using the fast fourier transform.





\end{document}

{\bf Problem 11.} Come up with other method for proving iterated
kernel smoothing. One approach is to compute the convolution
directly. {\em Solution by Tulaya Limpiti.} She proved based on
two methods. By completing squares and using Fourier transform.
Note that
$$K_{\sigma} * K_{\sigma}(x) = \mathcal{F}^{-1}[(\mathcal{F}
K_{\sigma})^2](x) = \mathcal{F}^{-1}K(2\sqrt{2\pi} \sigma x)$$