source: trunk/doc/Statistics.tex @ 492

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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 measuremtns 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 origin for a weight occuring in a statistical analysis, there are various way 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 implementationare 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 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 weight 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{eqnarray}
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{eqnarray}
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 $\sigma_i^2=\frac{\sum w_i^2}{\left(\sum w_i\right)^2}\sum w_i(x-m)^2$ and we get the following estimator of the variance of the mean
 $\sigma_i^2=\frac{\sum w_i^2}{\left(\sum w_i\right)^3}\sum w_i(x-m)^2$. 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 geomtrical avergae 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

\beq
\frac{\sum_{\{i,j\}:x^-_i<x^+_j}1}{\sum_{i,j}1}.
\eeq

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 

\beq
\frac{\sum_{\{i,j\}:x^-_i<x^+_j}w^-_iw^+_j}{\sum_{i,j}w^-_iw^+_j}
\eeq

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 kenrel (usual scalar product). For weights we define the linear kernel to be $<x,y>=\frac{\sum w_xw_yxy}{\sum w_xw_y}$ 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

\beq 
y_i=\alpha+\beta (x-m_x)+\epsilon_i, 
\eeq 

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.

\beq
Q_0 = \sum \epsilon_i^2
\eeq

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

\beq
\frac{\partial Q_0}{\partial \alpha} = -2 \sum w_i(y_i - \alpha - \beta (x_i-m_x)=0
\eeq

and

\beq
\frac{\partial Q_0}{\partial \beta} = -2 \sum w_i(x_i-m_x)(y_i-\alpha-\beta(x_i-m_x)=0
\eeq

or equivalently

\beq
\alpha = \frac{\sum w_iy_i}{\sum w_i}=m_y
\eeq

and

\beq
\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)}
\eeq

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

\beq 
\textrm{Var}(\alpha )=\frac{w_i^2\frac{\sigma^2}{w_i}}{(\sum w_i)^2}=
\frac{\sigma^2}{\sum w_i}
\eeq

and
\beq
\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}
\eeq

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

\beq
s^2=\frac{\sum (y_i-\alpha-\beta (x_i-m_x))^2}{n-2}
\eeq

we suggest the following estimator

\beq
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}}
\eeq

\section{Outlook}
\subsection{Hierarchical clustering}
\label{hc}
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

\beq
d(x,y)=\frac{\sum w_i^xw_j^y(x_i-y_i)^2}{\sum w_i^xw_j^y} \eeq \\

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

\beq
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}
\eeq

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}