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\begin{document}

\title{Stat 992: Lecture 12\\
Karhunen-Lo$\grave{e}$ve Expansion II.}
\author{Moo K. Chung \tt{mchung@stat.wisc.edu}}
\maketitle \thispagestyle{empty}

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\begin{enumerate}

\item {\em Global inference.} As one application of
Karhunen-Lo$\grave{e}$ve expansion, we show how we do a simple
inference on random fields. Assuming simple additive model
$$Y(x) = \mu(x) + \epsilon(x), x \in \Omega$$
where $\epsilon$ is true unknown signal while $\epsilon$ is a zero
mean Gaussian field with isotropic covariance function $R$. We are
interested in seeing if $$H_0: \mu(x) = 0 \mbox{ for all } x \in
\Omega.$$ The alternate hypothesis is
$$H_1: \mu(x) \neq 0 \mbox{ for some } x \in \Omega.$$
Assuming the continuity of signal $\mu$, the null hypothesis is
equivalent to the following hypothesis:
$$H_0':\int_{\Omega} \mu^2(x) \; dx=0.$$
So under $H_0$, our simple test statistic is
$$\int_{\Omega} Y^2(x) \; dx = \int_{\Omega} \epsilon^2(x)\; dx.$$
We can come up with different test statisticss but their
distribution might be very hard to compute analytically.


From Karhunen-Lo$\grave{e}$ve expansion, we have
$$\epsilon(x) = \sum_{j=0}^{\infty} \epsilon_j \phi_j(x)$$
with independent $\epsilon_j \sim N(0, \sigma_j^2)$ and $\sigma_j^2$
and $\phi_j$ are $i$-th eigensystem for covariance function $R$
with respect to domain $\Omega$. Then \bqn \int_{\Omega}
\epsilon^2(x)\; dx &=& \int_{\Omega} \big[\sum_j \epsilon_j \phi_j(x)\big]^2 \; dx\\
&=&\int_{\Omega} \sum_{i,j} \epsilon_i\epsilon_j\phi_i(x)\phi_j(x)\; dx\\
&=& \sum_{i,j} \epsilon_i \epsilon_j \int_{\Omega}\phi_i, \phi_j \; dx\\
&=& \sum_{j=0}^{\infty} \epsilon_j^2
 \eqn
This is the infinite sum of i.i.d. scaled $\chi^2$ random variable with different degrees
of freedom. There is no close form of the distribution so we will
approximate the distribution with a single scaled $\chi^2$ distribution
by matching the moments. See F.E. Satterthwaite, An approximate
distribution of estimates of variance components, Biometrics
2:110-114, 1946 for details. Approximately
$$ \sum_{j=0}^{\infty} \epsilon_j^2 \stackrel{D}{\sim} c\chi_{\nu}.$$

Note that
$$\mathbb{E} \sum_{j=0}^{\infty} \epsilon_j^2 = \sum_{j=0}^{\infty}
\sigma_j^2 < \infty$$ and
$$\mathbb{\bf Var} \sum_{j=0}^{\infty} \epsilon_j^2 = \sum_{j=0}^{\infty}
{\bf Var} \epsilon_j^2 = 2\sum_{j=0}^{\infty}\sigma_j^4.$$ To see
this we compute the higher order moment of Gaussian random
variable $Z_i \sim N(0,\sigma^2)$ via decomposition \bqn
\mathbb{E} [Z_1Z_2Z_3Z_4] &=& \mathbb{E}
[Z_1Z_2]\mathbb{E}[Z_3Z_4] + \mathbb{E}
[Z_1Z_3]\mathbb{E}[Z_2Z_4]\\
& +& \mathbb{E} [Z_1Z_4]\mathbb{E}[Z_2Z_3]. \eqn Now let $Z_1 =
Z_2 = Z_3 = Z_4$. Then $\mathbb{E} Z_j^4 = 3\sigma^4$. The
corresponding mean and variance of $c \chi_{\nu}$ are $c\nu$ and
$2c^2\nu$. Matching the mean and variance, we have $c\nu = \sum
\sigma_j^2$ and $c^2\nu = \sum \sigma_j^4$. Solving equations, $c
= \sum \sigma_j^4 / \sum \sigma_j^2$ and $\nu = (\sum \sigma_j^2)^2/ \sum
\sigma_j^4$. Based on this approximate
distribution, the inference can be performed\footnote{As pointed out by Prof. Tsui, the higher order moment of zero mean Gaussian random variable can be performed by differentiating the moment generating function $e^{-t^2/2}$ which yields the Hermite polynomials.}


 {\bf Problem 23.} Suppose we have a model for a vector field
$Y(x) = \mu(x) + \epsilon(x), x \in \Omega$ where $\epsilon$ is a
$n$-dimensional zero mean Gaussian vector field. Perform
hypothesis testing on null hypothesis $H_0: \mu(x) = 0$ for all $x
\in \Omega$.


\end{enumerate}

{\bf Homework 3.} Solve Problems 13-23 by March 1 11:00am. Solve
as many problems as you can.
\end{document}