Index: trunk/doc/Statistics.doxygen
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 trunk/doc/Statistics.doxygen (revision 1152)
+++ trunk/doc/Statistics.doxygen (revision 1153)
@@ 302,34 +302,18 @@
and calculating the weighted median of the distances.
+\section Distance
+
\section Kernel
\subsection polynomial_kernel Polynomial Kernel
The polynomial kernel of degree \f$N\f$ is defined as \f$(1+)^N\f$, where
\f$\f$ is the linear kernel (usual scalar product). For the weighted
case we define the linear kernel to be \f$=\sum {w_xw_yxy}\f$ and the
+case we define the linear kernel to be
+\f$=\frac{\sum {w_xw_yxy}}{\sum{w_xw_y}\f$ and the
polynomial kernel can be calculated as before
\f$(1+)^N\f$. Is this kernel a proper kernel (always being semi
positive definite). Yes, because \f$\f$ is obviously a proper kernel
as it is a scalar product. Adding a positive constant to a kernel
yields another kernel so \f$1+\f$ is still a proper kernel. Then also
\f$(1+)^N\f$ is a proper kernel because taking a proper kernel to the
\f$Nth\f$ power yields a new proper kernel (see any good book on SVM).
+\f$(1+)^N\f$.
+
\subsection gaussian_kernel Gaussian Kernel
We define the weighted Gaussian kernel as \f$\exp\left(\frac{\sum
w_xw_y(xy)^2}{\sum w_xw_y}\right)\f$, which fulfills the conditions
listed in the introduction.

Is this kernel a proper kernel? Yes, following the proof of the
nonweighted kernel we see that \f$K=\exp\left(\frac{\sum
w_xw_yx^2}{\sum w_xw_y}\right)\exp\left(\frac{\sum w_xw_yy^2}{\sum
w_xw_y}\right)\exp\left(\frac{\sum w_xw_yxy}{\sum w_xw_y}\right)\f$,
which is a product of two proper kernels. \f$\exp\left(\frac{\sum
w_xw_yx^2}{\sum w_xw_y}\right)\exp\left(\frac{\sum w_xw_yy^2}{\sum
w_xw_y}\right)\f$ is a proper kernel, because it is a scalar product and
\f$\exp\left(\frac{\sum w_xw_yxy}{\sum w_xw_y}\right)\f$ 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
+We define the weighted Gaussian kernel as \f$\exp\left(N\frac{\sum
+w_xw_y(xy)^2}{\sum w_xw_y}\right)\f$.
\section Regression