# Changeset 1153

Ignore:
Timestamp:
Feb 26, 2008, 3:01:02 AM (16 years ago)
Message:

Removing text not being true anymore

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 r1125 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$\exp\left(-\frac{\sum w_xw_y(x-y)^2}{\sum w_xw_y}\right)\f$, which fulfills the conditions listed in the introduction. Is this kernel a proper kernel? Yes, following the proof of the non-weighted 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$\exp\left(-N\frac{\sum w_xw_y(x-y)^2}{\sum w_xw_y}\right)\f$. \section Regression