norm.tex

聚类算法全集以及内附数据集

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\def\IR{\mbox{I\kern-0.22em R}}

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\subsection*{Normalization of the
             generalized multidimensional Gaussian distribution}

Jacobi formula for the surface of the $n$-dimensional (hyper)sphere
\cite[0.1.6, p.~15]{Bronstein et al 1996}:
\[ A_n(r) = \frac{2\pi^{\frac{n}{2}} r^{n-1}}{\Gamma(\frac{n}{2})},
   \qquad n \ge 2. \]
Integral formula \cite[0.9.6, No.~1, p.~184]{Bronstein et al 1996}:
\[ \int_0^\infty z^\beta e^{-\alpha z} \mbox{d}z
   = \frac{\Gamma(\beta+1)}{\alpha^{\beta+1}}, \qquad
   \alpha, \beta \in \IR, \quad \alpha > 0, \quad \beta > -1. \]
In order to determine the normalization factor, we have to compute
the following integral
\[ \int_0^\infty
   \underbrace{\frac{2\pi^{\frac{n}{2}} x^{n-1}}
                    {\Gamma\!\left(\frac{n}{2}\right)}}
   _{\mbox{\parbox{32mm}{surface area of the $n$-di\-men\-sional
                         (hyper) sphere with radius $x$}}}
   \quad
   \underbrace{e^{-\frac{1}{2}(bx^a+c)}
   \hbox to0pt{$\phantom{\displaystyle
               \frac{\frac{n}{2}}{\left(\frac{n}{2}\right)}}$\hss}}
   _{\mbox{\parbox{18mm}{generalized \\ distribution \\ function}}}
   \quad \mbox{d}x. \]
Note that this formula works also for $n = 1$, although the formula
for the surface of the $n$-dimensional (hyper)sphere is valid only
for $n \ge 2$. For $n = 1$ the fraction evaluates to 2, which takes
care of the fact that the integral has to be doubled in order to take
the area from $-\infty$ to 0 into account.
\begin{eqnarray*}
\int_0^\infty \frac{2\pi^{\frac{n}{2}} x^{n-1}}
                   {\Gamma\!\left(\frac{n}{2}\right)}
              \;e^{-\frac{1}{2}(bx^a+c)} \;\mbox{d}x
& = & \frac{2\pi^{\frac{n}{2}} e^{-\frac{c}{2}}}
           {\Gamma\!\left(\frac{n}{2}\right)}
      \int_0^\infty x^{n-1} e^{-\frac{1}{2}bx^a} \;\mbox{d}x
\end{eqnarray*}
In order to reduce this integral to the known formula stated above,
we substitute $z$ for $x^a$.
\begin{eqnarray*}
z = x^a
& \Rightarrow & x = \sqrt[a]{z}, \\
& \Rightarrow & \frac{\mbox{d}z}{\mbox{d}x} = a x^{a-1}
~~\Rightarrow~~ \mbox{d}z = a (\sqrt[a]{z})^{a-1} \mbox{d}x
                          = a z^{\frac{a-1}{a}} \mbox{d}x
~~\Rightarrow~~ \mbox{d}x = \frac{\mbox{d}z}{a\,z^{\frac{a-1}{a}}}
                          = \frac{1}{a} z^{\frac{1-a}{a}} \mbox{d}z, \\
& \Rightarrow & x^{n-1}   = (\sqrt[a]{z})^{n-1} = z^{\frac{n-1}{a}}.
\end{eqnarray*}
Note that the integration bounds---from 0 to $\infty$---do not change.
By inserting the above equations into the integral formula we arrive at
\begin{eqnarray*}
\int_0^\infty \frac{2\pi^{\frac{n}{2}} x^{n-1}}
                   {\Gamma\!\left(\frac{n}{2}\right)}
              \;e^{-\frac{1}{2}(bx^a+c)} \;\mbox{d}x
& = & \frac{2\pi^{\frac{n}{2}} e^{-\frac{c}{2}}}
           {\Gamma\!\left(\frac{n}{2}\right)}
      \int_0^\infty x^{n-1} e^{-\frac{1}{2}bx^a} \;\mbox{d}x \\
& = & \frac{2\pi^{\frac{n}{2}} e^{-\frac{c}{2}}}
           {\Gamma\!\left(\frac{n}{2}\right)}
      \int_0^\infty z^{\frac{n-1}{a}} e^{-\frac{1}{2}bz}
      \frac{1}{a} z^{\frac{1-a}{a}} \mbox{d}z \\
& = & \frac{2\pi^{\frac{n}{2}} e^{-\frac{c}{2}}}
           {\Gamma\!\left(\frac{n}{2}\right) a}
      \int_0^\infty z^{\frac{n}{a}-1} e^{-\frac{1}{2}bz} \mbox{d}z \\
& = & \frac{2\pi^{\frac{n}{2}} e^{-\frac{c}{2}}}
           {\Gamma\!\left(\frac{n}{2}\right) a}
      \frac{\Gamma\!\left(\frac{n}{a}\right)}
           {\left(\frac{b}{2}\right)^{\frac{n}{a}}}.
\end{eqnarray*}
Therefore the desired normalization factor is
\[ \gamma = \frac{\Gamma\!\left(\frac{n}{2}\right) a
                  \left(\frac{b}{2}\right)^{\frac{n}{a}}}
                 {2\pi^{\frac{n}{2}} e^{-\frac{c}{2}}
                  \Gamma\!\left(\frac{n}{a}\right)}. \]

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\newpage

\subsection*{Normalization of the multidimensional Cauchy distribution}

Integral formula \cite[0.9.6, No.~53, p.~189]{Bronstein et al 1996}:
\[ \int_0^\infty \frac{z^{\alpha-1}}{1 +z^\beta} \mbox{d}z
   = \frac{\pi}{\beta \sin \frac{\alpha \pi}{\beta}}, \qquad
   \alpha, \beta \in \IR, \quad 0 < \alpha < \beta. \]
In order to determine the normalization factor, we have to compute
the following integral
\[ \int_0^\infty
   \underbrace{\frac{2\pi^{\frac{n}{2}} x^{n-1}}
                    {\Gamma\!\left(\frac{n}{2}\right)}}
   _{\mbox{\parbox{32mm}{surface area of the $n$-di\-men\-sional
                         (hyper) sphere with radius $x$}}}
   \quad
   \underbrace{\frac{1}{bx^a+c}
   \hbox to0pt{$\phantom{\displaystyle
               \frac{\frac{n}{2}}{\left(\frac{n}{2}\right)}}$\hss}}
   _{\mbox{\parbox{18mm}{distribution \\ function}}}
   \quad \mbox{d}x. \]
As in the preceding section this formula also works for $n = 1$.
\begin{eqnarray*}
\int_0^\infty \frac{2\pi^{\frac{n}{2}} x^{n-1}}
                   {\Gamma\!\left(\frac{n}{2}\right)}
              \frac{1}{bx^a+c} \;\mbox{d}x
& = & \frac{2\pi^{\frac{n}{2}}}
           {\Gamma\!\left(\frac{n}{2}\right)c}
      \int_0^\infty \frac{x^{n-1}}{\frac{b}{c}x^a +1} \;\mbox{d}x
\end{eqnarray*}
In order to reduce this integral to the known formula stated above,
we substitute $z$ for $x \left(\frac{b}{c}\right)^{\frac{1}{a}}$.
\begin{eqnarray*}
z = x\left(\frac{b}{c}\right)^{\frac{1}{a}}
& \Rightarrow & x = z\left(\frac{c}{b}\right)^{\frac{1}{a}}, \\
& \Rightarrow & \frac{\mbox{d}z}{\mbox{d}x}
                = \left(\frac{b}{c}\right)^{\frac{1}{a}}
~~\Rightarrow~~ \mbox{d}z =
                \left(\frac{b}{c}\right)^{\frac{1}{a}}\mbox{d}x
~~\Rightarrow~~ \mbox{d}x =
                \left(\frac{c}{b}\right)^{\frac{1}{a}}\mbox{d}z, \\
& \Rightarrow & x^{n-1}
  = \left(z\left(\frac{c}{b}\right)^{\frac{1}{a}}\right)^{n-1}
  = z^{n-1}\left(\frac{c}{b}\right)^{\frac{n-1}{a}}.
\end{eqnarray*}
Note that the integration bounds---from 0 to $\infty$---do not change.
By inserting the above equations into the integral formula we arrive at
\begin{eqnarray*}
\int_0^\infty \frac{2\pi^{\frac{n}{2}} x^{n-1}}
                   {\Gamma\!\left(\frac{n}{2}\right)}
              \frac{1}{bx^a+c} \;\mbox{d}x
& = & \frac{2\pi^{\frac{n}{2}}}
           {\Gamma\!\left(\frac{n}{2}\right)c}
      \int_0^\infty \frac{x^{n-1}}{\frac{b}{c}x^a +1} \;\mbox{d}x \\
& = & \frac{2\pi^{\frac{n}{2}}}
           {\Gamma\!\left(\frac{n}{2}\right)c}
      \int_0^\infty
      \frac{z^{n-1}\left(\frac{c}{b}\right)^{\frac{n-1}{a}}}
           {z^a +1}
      \left(\frac{c}{b}\right)^{\frac{1}{a}} \mbox{d}z \\
& = & \frac{2\pi^{\frac{n}{2}}}
           {\Gamma\!\left(\frac{n}{2}\right)c}
           \left(\frac{c}{b}\right)^{\frac{n}{a}}
      \int_0^\infty \frac{z^{n-1}}{1 +z^a} \;\mbox{d}z \\
& = & \frac{2\pi^{\frac{n}{2}}\left(\frac{c}{b}\right)^{\frac{n}{a}}}
           {\Gamma\!\left(\frac{n}{2}\right)c}
      \frac{\pi}{a \,\sin\frac{n \pi}{a}}.
\end{eqnarray*}
Therefore the desired normalization factor is
\[ \gamma = \frac{\Gamma\!\left(\frac{n}{2}\right)
                  ac \,\sin\frac{n \pi}{a}}
                 {2\pi^{\frac{n}{2}+1}
                  \left(\frac{c}{b}\right)^{\frac{n}{a}}}. \]

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\def\etal{{\it et al.~}}

\begin{thebibliography}{99}
\bibitem[Bronstein \etal 1996]{Bronstein et al 1996}
  I.N.~Bronstein, K.A.~Semendjajew, G.~Grosche, V.~Ziegler,
  and D.~Ziegler.
  {\it Teubner Taschenbuch der Mathematik}, Part~1.
  Teubner, Leipzig, Germany 1996
\end{thebibliography}

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