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\begin{document}

\title{Stat 992: Lecture 19\\
Multiple comparisons II.}
\author{Moo K. Chung \tt{mchung@stat.wisc.edu}}
\maketitle \thispagestyle{empty}

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\begin{enumerate}


\item {\em Bonferroni correction.} One standard method for dealing
with multiple comparison is to use the Bonferroni correction. Note
that the probability measure is additive, i.e. \bq
P(\bigcup_{j=1}^m E_j) \leq \sum_{j=1}^m P(E_j).\eq This
inequality and its variations are called Bonferroni inequalities
and it has been used in the construction of simultaneous
confidence intervals and multiple comparisons when the number of
hypotheses are small. In the previous lecture, the type I error
was given by \bq \alpha &=& P\Big(\bigcup_{j=1}^m \{Y(x_j) >
h\}\Big|\mathbb{E} Y=0\Big)\\
&\leq& \sum_{j=1}^m P\big(Y(x_j)>h|\mathbb{E} Y
=0\big)\label{eq:lec19-1}\eq So by controlling each type I error
separately so that
$$P\big(Y(x_j)>h|\mathbb{E} Y =0\big) <\alpha/m,$$
we can construct level $\alpha$ test. For image of size $95 \times
68$, there are $m=95 \times 68 = 6460$ hypotheses to test. So
$\alpha/m=0.05/6460= 7.7 \times 10^{-6}$. For $Y(x_j) \sim
N(0,1)$, the corresponding threshold is computed by
\begin{verbatim}
>> norminv(1-7.7*10^(-6),0,1)
ans =
    4.3229
\end{verbatim}
By setting $h=4.33$, any image intensity larger than the threshold
is taken as statistically significant signal. The problem with the
Bonferroni correction is that in most cases it is considerably too
conservative since each pixel or voxel intensities are strongly
correlated and in turn it gives highly correlated statistics
across pixels. In some cases, the size of image tend to be $200
\times 200 \times 100$, which gives extremely high threshold of
$5.57$.
\begin{verbatim}
>> norminv(1-0.05/4000000,0,1)
ans =
    5.5733
\end{verbatim}
See Miller, R. G. Jr. Simultaneous Statistical Inference. New
York: Springer-Verlag, 1991.

{\bf Problem 29.} Based on slightly different version of
Bonferroni inequalities, it is possible to have a slightly better
inequality than (\ref{eq:lec19-1}) so that the threshold $h$ is
lower than the above Bonferroni correction. Come up with a better
less conservative method than the one presented in this lecture.

\item {\em Random fields approach.} Assume $\mathbb{E} Y=0$.
\bqn\alpha(h) &=& P\Big(\bigcup_{j=1}^m \{Y(x_j) > h\}\Big)\\
&=&1-  P\Big(\bigcap_{j=1}^m \{Y(x_j) \leq h  \}\Big)\\
&=& 1 - P(\max_{x_j \in \Omega} Y(x_j) \leq h)\\
&=& P(\max_{x_j \in \Omega} Y(x_j) > h)\eqn So in order to
construct $\alpha$-level test, we need to know the distribution of
the maximum of $m$ correlated Gaussian random variables (see HW
problem 28). For continuous mean zero random fields $Y$, following
the similar argument, we get
\bqn \alpha(h) &=& P\Big(\bigcup_{x \in \Omega} \{Y(x) > h\}\Big)\\
&=&  P(\sup_{x \in \Omega} Y(x) > h).\eqn Analytically computing
the exact distribution of the maxima of random fields is hard.
Multiple testing procedures based on the distribution of the
maxima of fields is usually called random field approach and the
$P$-value based on the maximum field is called {\em corrected}
$P$-value to distinguish it from uncorrected $P$-value.

Let $Z=\sup_{x \in \Omega} Y(x)$ and $F_Z$ to be the cumulative
distribution of $Z$. Then given $\alpha =0.05$, we compute
$h=1-F_Z^{-1}(\alpha)$. Then the region of statistically
significant signal is given by $\{x \in \Omega: Y(x) > h\}$.


\end{enumerate}

{\bf Homework 4.} Solve Problems 24-32 and any other previous
problems where complete solutions are not given. Due March 22
Monday 11:00am.
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