source: branches/0.3.1/doc/Statistics.tex @ 838

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\begin{document}

\large
{\bf Weighted Statistics}
\normalsize

\tableofcontents
\clearpage

\section{Introduction}
There are several different reasons why a statistical analysis needs
to adjust for weighting. In literature reasons are mainly diveded in
to groups.

The first group is when some of the measurements are known to be more
precise than others. The more precise a measurement is, the larger
weight it is given. The simplest case is when the weight are given
before the measurements and they can be treated as deterministic. It
becomes more complicated when the weight can be determined not until
afterwards, and even more complicated if the weight depends on the
value of the observable.

The second group of situations is when calculating averages over one
distribution and sampling from another distribution. Compensating for
this discrepency weights are introduced to the analysis. A simple
example may be that we are interviewing people but for economical
reasons we choose to interview more people from the city than from the
countryside. When summarizing the statistics the answers from the city
are given a smaller weight. In this example we are choosing the
proportions of people from countryside and people from city being
intervied. Hence, we can determine the weights before and consider
them to be deterministic. In other situations the proportions are not
deterministic, but rather a result from the sampling and the weights
must be treated as stochastic and only in rare situations the weights
can be treated as independent of the observable.

Since there are various origins for a weight occuring in a statistical
analysis, there are various ways to treat the weights and in general
the analysis should be tailored to treat the weights correctly. We
have not chosen one situation for our implementations, so see specific
function documentation for what assumtions are made. Though, common
for implementations are the following:
\begin{itemize}
\item Setting all weights to unity yields the same result as the
non-weighted version.
\item Rescaling the weights does not change any function.
\item Setting a weight to zero is equivalent to removing the data point.
\end{itemize}
An important case is when weights are binary (either 1 or 0). Then we
get the same result using the weighted version as using the data with
weight not equal to zero and the non-weighted version. Hence, using
binary weights and the weighted version missing values can be treated
in a proper way.

\section{AveragerWeighted}



\subsection{Mean}

For any situation the weight is always designed so the weighted mean
is calculated as $m=\frac{\sum w_ix_i}{\sum w_i}$, which obviously
fulfills the conditions above.

In the case of varying measurement error, it could be motivated that
the weight shall be $w_i = 1/\sigma_i^2$. We assume measurement error
to be Gaussian and the likelihood to get our measurements is
$L(m)=\prod
(2\pi\sigma_i^2)^{-1/2}e^{-\frac{(x_i-m)^2}{2\sigma_i^2}}$.  We
maximize the likelihood by taking the derivity with respect to $m$ on
the logarithm of the likelihood $\frac{d\ln L(m)}{dm}=\sum
\frac{x_i-m}{\sigma_i^2}$. Hence, the Maximum Likelihood method yields
the estimator $m=\frac{\sum w_i/\sigma_i^2}{\sum 1/\sigma_i^2}$.


\subsection{Variance}
In case of varying variance, there is no point estimating a variance
since it is different for each data point.

Instead we look at the case when we want to estimate the variance over
$f$ but are sampling from $f'$. For the mean of an observable $O$ we
have $\widehat O=\sum\frac{f}{f'}O_i=\frac{\sum w_iO_i}{\sum
w_i}$. Hence, an estimator of the variance of $X$ is
\begin{eqnarray}
\sigma^2=<X^2>-<X>^2=
\\\frac{\sum w_ix_i^2}{\sum w_i}-\frac{(\sum w_ix_i)^2}{(\sum w_i)^2}=
\\\frac{\sum w_i(x_i^2-m^2)}{\sum w_i}=
\\\frac{\sum w_i(x_i^2-2mx_i+m^2)}{\sum w_i}=
\\\frac{\sum w_i(x_i-m)^2}{\sum w_i}
\end{eqnarray}
This estimator fulfills that it is invariant under a rescaling and
having a weight equal to zero is equivalent to removing the data
point. Having all weights equal to unity we get $\sigma=\frac{\sum
(x_i-m)^2}{N}$, which is the same as returned from Averager. Hence,
this estimator is slightly biased, but still very efficient.

\subsection{Standard Error}
The standard error squared is equal to the expexted squared error of
the estimation of $m$. The squared error consists of two parts, the
variance of the estimator and the squared
bias. $<m-\mu>^2=<m-<m>+<m>-\mu>^2=<m-<m>>^2+(<m>-\mu)^2$.  In the
case when weights are included in analysis due to varying measurement
errors and the weights can be treated as deterministic ,we have
\begin{equation}
Var(m)=\frac{\sum w_i^2\sigma_i^2}{\left(\sum w_i\right)^2}=
\frac{\sum w_i^2\frac{\sigma_0^2}{w_i}}{\left(\sum w_i\right)^2}=
\frac{\sigma_0^2}{\sum w_i},
\end{equation}
where we need to estimate $\sigma_0^2$. Again we have the likelihood
$L(\sigma_0^2)=\prod\frac{1}{\sqrt{2\pi\sigma_0^2/w_i}}\exp{-\frac{w_i(x-m)^2}{2\sigma_0^2}}$
and taking the derivity with respect to $\sigma_o^2$, $\frac{d\ln
L}{d\sigma_i^2}=\sum
-\frac{1}{2\sigma_0^2}+\frac{w_i(x-m)^2}{2\sigma_0^2\sigma_o^2}$ which
yields an estimator $\sigma_0^2=\frac{1}{N}\sum w_i(x-m)^2$. This
estimator is not ignoring weights equal to zero, because deviation is
most often smaller than the expected infinity. Therefore, we modify
the expression as follows $\sigma_0^2=\frac{\sum w_i^2}{\left(\sum
w_i\right)^2}\sum w_i(x-m)^2$ and we get the following estimator of
the variance of the mean $\sigma_0^2=\frac{\sum w_i^2}{\left(\sum
w_i\right)^3}\sum w_i(x-m)^2$. This estimator fulfills the conditions
above: adding a weight zero does not change it: rescaling the weights
does not change it, and setting all weights to unity yields the same
expression as in the non-weighted case.

In a case when it is not a good approximation to treat the weights as
deterministic, there are two ways to get a better estimation. The
first one is to linearize the expression $\left<\frac{\sum
w_ix_i}{\sum w_i}\right>$. The second method when the situation is
more complicated is to estimate the standard error using a
bootstrapping method.

\section{AveragerPairWeighted}
Here data points come in pairs (x,y). We are sampling from $f'_{XY}$
but want to measure from $f_{XY}$. To compensate for this decrepency,
averages of $g(x,y)$ are taken as $\sum \frac{f}{f'}g(x,y)$. Even
though, $X$ and $Y$ are not independent $(f_{XY}\neq f_Xf_Y)$ we
assume that we can factorize the ratio and get $\frac{\sum
w_xw_yg(x,y)}{\sum w_xw_y}$
\subsection{Covariance}
Following the variance calculations for AveragerWeighted we have
$Cov=\frac{\sum w_xw_y(x-m_x)(y-m_y)}{\sum w_xw_y}$ where
$m_x=\frac{\sum w_xw_yx}{\sum w_xw_y}$

\subsection{correlation}

As the mean is estimated as $m_x=\frac{\sum w_xw_yx}{\sum w_xw_y}$,
the variance is estimated as $\sigma_x^2=\frac{\sum
w_xw_y(x-m_x)^2}{\sum w_xw_y}$. As in the non-weighted case we define
the correlation to be the ratio between the covariance and geometrical
average of the variances

$\frac{\sum w_xw_y(x-m_x)(y-m_y)}{\sqrt{\sum w_xw_y(x-m_x)^2\sum
w_xw_y(y-m_y)^2}}$.

This expression fulfills the following
\begin{itemize}
\item Having N weights the expression reduces to the non-weighted expression.
\item Adding a pair of data, in which one weight is zero is equivalent
to ignoring the data pair.
\item Correlation is equal to unity if and only if $x$ is equal to
$y$. Otherwise the correlation is between -1 and 1.
\end{itemize}
\section{Score}


\subsection{Pearson}

$\frac{\sum w(x-m_x)(y-m_y)}{\sqrt{\sum w(x-m_x)^2\sum w(y-m_y)^2}}$.

See AveragerPairWeighted correlation.

\subsection{ROC}

An interpretation of the ROC curve area is the probability that if we
take one sample from class $+$ and one sample from class $-$, what is
the probability that the sample from class $+$ has greater value. The
ROC curve area calculates the ratio of pairs fulfilling this

\begin{equation}
\frac{\sum_{\{i,j\}:x^-_i<x^+_j}1}{\sum_{i,j}1}.
\end{equation}

An geometrical interpretation is to have a number of squares where
each square correspond to a pair of samples. The ROC curve follows the
border between pairs in which the samples from class $+$ has a greater
value and pairs in which this is not fulfilled. The ROC curve area is
the area of those latter squares and a natural extension is to weight
each pair with its two weights and consequently the weighted ROC curve
area becomes

\begin{equation}
\frac{\sum_{\{i,j\}:x^-_i<x^+_j}w^-_iw^+_j}{\sum_{i,j}w^-_iw^+_j}
\end{equation}

This expression is invariant under a rescaling of weight. Adding a
data value with weight zero adds nothing to the exprssion, and having
all weight equal to unity yields the non-weighted ROC curve area.

\subsection{tScore}

Assume that $x$ and $y$ originate from the same distribution
$N(\mu,\sigma_i^2)$ where $\sigma_i^2=\frac{\sigma_0^2}{w_i}$. We then
estimate $\sigma_0^2$ as
\begin{equation}
\frac{\sum w(x-m_x)^2+\sum w(y-m_y)^2}
{\frac{\left(\sum w_x\right)^2}{\sum w_x^2}+ 
\frac{\left(\sum w_y\right)^2}{\sum w_y^2}-2}
\end{equation}
The variance of difference of the means becomes
\begin{eqnarray}
Var(m_x)+Var(m_y)=\\\frac{\sum w_i^2Var(x_i)}{\left(\sum
w_i\right)^2}+\frac{\sum w_i^2Var(y_i)}{\left(\sum w_i\right)^2}=
\frac{\sigma_0^2}{\sum w_i}+\frac{\sigma_0^2}{\sum w_i},
\end{eqnarray}
and consequently the t-score becomes
\begin{equation}
\frac{\sum w(x-m_x)^2+\sum w(y-m_y)^2}
{\frac{\left(\sum w_x\right)^2}{\sum w_x^2}+ 
\frac{\left(\sum w_y\right)^2}{\sum w_y^2}-2}
\left(\frac{1}{\sum w_i}+\frac{1}{\sum w_i}\right),
\end{equation}

For a $w_i=w$ we this expression get condensed down to
\begin{equation}
\frac{w\sum (x-m_x)^2+w\sum (y-m_y)^2}
{n_x+n_y-2} 
\left(\frac{1}{wn_x}+\frac{1}{wn_y}\right),
\end{equation}
in other words the good old expression as for non-weighted. 

\subsection{FoldChange}
Fold-Change is simply the difference between the weighted mean of the
two groups //$\frac{\sum w_xx}{\sum w_x}-\frac{\sum w_yy}{\sum w_y}$

\subsection{WilcoxonFoldChange}
Taking all pair samples (one from class $+$ and one from class $-$)
and calculating the weighted median of the distances.

\section{Kernel}
\subsection{Polynomial Kernel}
The polynomial kernel of degree $N$ is defined as $(1+<x,y>)^N$, where
$<x,y>$ is the linear kernel (usual scalar product). For the weighted
case we define the linear kernel to be $<x,y>=\sum {w_xw_yxy}$ and the
polynomial kernel can be calculated as before
$(1+<x,y>)^N$. Is this kernel a proper kernel (always being semi
positive definite). Yes, because $<x,y>$ is obviously a proper kernel
as it is a scalar product. Adding a positive constant to a kernel
yields another kernel so $1+<x,y>$ is still a proper kernel. Then also
$(1+<x,y>)^N$ is a proper kernel because taking a proper kernel to the
$Nth$ power yields a new proper kernel (see any good book on SVM).
\subsection{Gaussian Kernel}
We define the weighted Gaussian kernel as $\exp\left(-\frac{\sum
w_xw_y(x-y)^2}{\sum w_xw_y}\right)$, which fulfills the conditions
listed in the introduction.

Is this kernel a proper kernel? Yes, following the proof of the
non-weighted kernel we see that $K=\exp\left(-\frac{\sum
w_xw_yx^2}{\sum w_xw_y}\right)\exp\left(-\frac{\sum w_xw_yy^2}{\sum
w_xw_y}\right)\exp\left(\frac{\sum w_xw_yxy}{\sum w_xw_y}\right)$,
which is a product of two proper kernels. $\exp\left(-\frac{\sum
w_xw_yx^2}{\sum w_xw_y}\right)\exp\left(-\frac{\sum w_xw_yy^2}{\sum
w_xw_y}\right)$ is a proper kernel, because it is a scalar product and
$\exp\left(\frac{\sum w_xw_yxy}{\sum w_xw_y}\right)$ is a proper
kernel, because it a polynomial of the linear kernel with positive
coefficients. As product of two kernel also is a kernel, the Gaussian
kernel is a proper kernel.

\section{Distance}

\section{Regression}
\subsection{Naive}
\subsection{Linear}
We have the model

\begin{equation} 
y_i=\alpha+\beta (x-m_x)+\epsilon_i, 
\end{equation} 

where $\epsilon_i$ is the noise. The variance of the noise is
inversely proportional to the weight,
$Var(\epsilon_i)=\frac{\sigma^2}{w_i}$. In order to determine the
model parameters, we minimimize the sum of quadratic errors.

\begin{equation}
Q_0 = \sum \epsilon_i^2
\end{equation}

Taking the derivity with respect to $\alpha$ and $\beta$ yields two conditions

\begin{equation} 
\frac{\partial Q_0}{\partial \alpha} = -2 \sum w_i(y_i - \alpha -
\beta (x_i-m_x)=0 
\end{equation}

and

\begin{equation} \frac{\partial Q_0}{\partial \beta} = -2 \sum
w_i(x_i-m_x)(y_i-\alpha-\beta(x_i-m_x)=0 
\end{equation}

or equivalently

\begin{equation}
\alpha = \frac{\sum w_iy_i}{\sum w_i}=m_y
\end{equation}

and

\begin{equation} \beta=\frac{\sum w_i(x_i-m_x)(y-m_y)}{\sum
w_i(x_i-m_x)^2}=\frac{Cov(x,y)}{Var(x)} 
\end{equation}

Note, by having all weights equal we get back the unweighted
case. Furthermore, we calculate the variance of the estimators of
$\alpha$ and $\beta$.

\begin{equation} 
\textrm{Var}(\alpha )=\frac{w_i^2\frac{\sigma^2}{w_i}}{(\sum w_i)^2}=
\frac{\sigma^2}{\sum w_i}
\end{equation}

and
\begin{equation}
\textrm{Var}(\beta )= \frac{w_i^2(x_i-m_x)^2\frac{\sigma^2}{w_i}}
{(\sum w_i(x_i-m_x)^2)^2}=
\frac{\sigma^2}{\sum w_i(x_i-m_x)^2}
\end{equation}

Finally, we estimate the level of noise, $\sigma^2$. Inspired by the
unweighted estimation

\begin{equation}
s^2=\frac{\sum (y_i-\alpha-\beta (x_i-m_x))^2}{n-2}
\end{equation}

we suggest the following estimator

\begin{equation} s^2=\frac{\sum w_i(y_i-\alpha-\beta (x_i-m_x))^2}{\sum
w_i-2\frac{\sum w_i^2}{\sum w_i}} \end{equation}

\section{Outlook}
\subsection{Hierarchical clustering}
A hierarchical clustering consists of two things: finding the two
closest data points, merge these two data points two a new data point
and calculate the new distances from this point to all other points.

In the first item, we need a distance matrix, and if we use Euclidean
distanses the natural modification of the expression would be

\begin{equation}
d(x,y)=\frac{\sum w_i^xw_j^y(x_i-y_i)^2}{\sum w_i^xw_j^y} 
\end{equation} 

For the second item, inspired by average linkage, we suggest

\begin{equation} 
d(xy,z)=\frac{\sum w_i^xw_j^z(x_i-z_i)^2+\sum
w_i^yw_j^z(y_i-z_i)^2}{\sum w_i^xw_j^z+\sum w_i^yw_j^z} 
\end{equation}

to be the distance between the new merged point $xy$ and $z$, and we
also calculate new weights for this point: $w^{xy}_i=w^x_i+w^y_i$

\end{document}