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Cauchy Distribution

\begin{figure}\begin{center}\BoxedEPSF{Cauchy_Distribution.epsf}\end{center}\end{figure}

The Cauchy distribution, also called the Lorentzian Distribution, describes resonance behavior. It also describes the distribution of horizontal distances at which a Line Segment tilted at a random Angle cuts the x-Axis. Let $\theta$ represent the Angle that a line, with fixed point of rotation, makes with the vertical axis, as shown above. Then

$\displaystyle \tan\theta$ $\textstyle =$ $\displaystyle {x\over b}$ (1)
$\displaystyle \theta$ $\textstyle =$ $\displaystyle \tan^{-1}\left({x\over b}\right)$ (2)
$\displaystyle d\theta$ $\textstyle =$ $\displaystyle - {1\over 1+{x^2\over b^2}} {dx\over b} = - {b\,dx\over b^2+x^2},$ (3)

so the distribution of Angle $\theta$ is given by
\begin{displaymath}
{d\theta\over\pi} = -{1\over\pi} {b\,dx\over b^2+x^2}.
\end{displaymath} (4)

This is normalized over all angles, since
\begin{displaymath}
\int_{-\pi/2}^{\pi/2} {d\theta\over \pi} = 1
\end{displaymath} (5)

and
$\displaystyle -\int_{-\infty}^\infty {1\over \pi} {b\,dx\over b^2+x^2}$ $\textstyle =$ $\displaystyle {1\over \pi}\left[{\tan^{-1}\left({b\over x}\right)}\right]_{-\infty}^\infty$  
  $\textstyle =$ $\displaystyle {1\over \pi} [{\textstyle{1\over 2}}\pi-(-{\textstyle{1\over 2}}\pi)] =1.$ (6)


\begin{figure}\begin{center}\BoxedEPSF{CauchyDistribution.epsf scaled 650}\end{center}\end{figure}

The general Cauchy distribution and its cumulative distribution can be written as

$\displaystyle P(x)$ $\textstyle =$ $\displaystyle {1\over\pi} {{\textstyle{1\over 2}}\Gamma\over (x-\mu)^2+({\textstyle{1\over 2}}\Gamma)^2}$ (7)
$\displaystyle D(x)$ $\textstyle =$ $\displaystyle {1\over 2}+{1\over\pi}\tan^{-1}\left({x-\mu\over b}\right),$ (8)

where $\Gamma$ is the Full Width at Half Maximum ($\Gamma=2b$ in the above example) and $\mu$ is the Mean ($\mu=0$ in the above example). The Characteristic Function is
$\displaystyle \phi(t)$ $\textstyle =$ $\displaystyle {1\over \pi}\int_{-\infty}^\infty {e^{it(\Gamma x/2-\mu)}\over 1+x^2}\,dx$  
  $\textstyle =$ $\displaystyle {e^{-i\mu t}\over\pi}\int_{-\infty}^\infty {\cos(\Gamma tx/2)\over 1+(\Gamma x/2)^2}\,dx$  
  $\textstyle =$ $\displaystyle e^{-i\mu t-\Gamma\vert t\vert/2}.$ (9)

The Moments are given by
$\displaystyle \mu_2$ $\textstyle =$ $\displaystyle \sigma^2 = \infty$ (10)
$\displaystyle \mu_3$ $\textstyle =$ $\displaystyle \left\{\begin{array}{ll} 0 & \mbox{for $\mu=0$}\\  \infty & \mbox{for $\mu\not=0$}\end{array}\right.$ (11)
$\displaystyle \mu_4$ $\textstyle =$ $\displaystyle \infty,$ (12)

and the Standard Deviation, Skewness, and Kurtosis by
$\displaystyle \sigma^2$ $\textstyle =$ $\displaystyle \infty$ (13)
$\displaystyle \gamma_1$ $\textstyle =$ $\displaystyle \left\{\begin{array}{ll} 0 & \mbox{for $\mu=0$}\\  \infty & \mbox{for $\mu\not=0$}\end{array}\right.$ (14)
$\displaystyle \gamma_2$ $\textstyle =$ $\displaystyle \infty.$ (15)


If $X$ and $Y$ are variates with a Normal Distribution, then $Z\equiv X/Y$ has a Cauchy distribution with Mean $\mu=0$ and full width

\begin{displaymath}
\Gamma={2\sigma_y\over\sigma_x}.
\end{displaymath} (16)

See also Gaussian Distribution, Normal Distribution


References

Spiegel, M. R. Theory and Problems of Probability and Statistics. New York: McGraw-Hill, pp. 114-115, 1992.



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© 1996-9 Eric W. Weisstein
1999-05-26