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Log-Series Distribution

The terms in the series expansion of $\ln(1-\theta)$ about $\theta=0$ are proportional to this distribution.

$\displaystyle P(n)$ $\textstyle =$ $\displaystyle -{\theta^n\over n\ln(1-\theta)}$ (1)
$\displaystyle D(n)$ $\textstyle \equiv$ $\displaystyle \sum_{i=1}^n P(i) = {\theta^{1+n}\Phi(\theta,1,1+n)+\ln(1-\theta)\over\ln(1-\theta)},$  
      (2)

where $\Phi$ is the Lerch Transcendent. The Mean, Variance, Skewness, and Kurtosis


$\displaystyle \mu$ $\textstyle =$ $\displaystyle {\theta\over(\theta-1)\ln(1-\theta)}$ (3)
$\displaystyle \sigma^2$ $\textstyle =$ $\displaystyle -{\theta[\theta+\ln(1-\theta)]\over (\theta-1)^2[\ln(1-\theta)]^2}$ (4)
$\displaystyle \gamma_1$ $\textstyle =$ $\displaystyle {2\theta^2+3\theta\ln(1-\theta)+(1+\theta)\ln^2(1-\theta)\over \l...
...heta)[\theta+\ln(1-\theta)]
\sqrt{-\theta[\theta+\ln(1-\theta)]}} \ln(1-\theta)$  
      (5)
$\displaystyle \gamma_2$ $\textstyle =$ $\displaystyle {6\theta^3+12\theta^2\ln(1-\theta)+\theta (7+4\theta)\ln^2(1-\the...
...]^2}+{(1+4\theta+\theta^2)\ln^3(1-\theta)\over \theta[\theta+\ln(1-\theta)]^2}.$ (6)




© 1996-9 Eric W. Weisstein
1999-05-25