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Just wondering why in the literature Weibull distribution is always defined for positive shapes, whereas the extension in the negative direction is possible and has many useful properties.

Suppose $X \propto \mathrm{Weibull}(\theta, \lambda)$, i.e. is Weibull-distributed with shape $\theta > 0$ and scale $\lambda > 0$ with PDF $$ p_X(x) = \frac{\theta}{\lambda}\left(\frac{x}{\lambda}\right)^{\theta-1}e^{-\left(x/\lambda\right)^\theta}, \quad x > 0. $$

$Y=1/X$ then follows the inverse Weibull distribution with PDF $$ p_{Y}(y) = \left|\left(1/y\right)'_y\right| p_X\left(1/y\right) = \frac{\theta}{\lambda y^2} \left(\frac{1}{\lambda y} \right)^{\theta-1} e^{-\left(1/\lambda y\right)^\theta} = \frac{\theta}{\lambda^{-1}} \left(\frac{y}{\lambda^{-1}}\right)^{-\theta-1} e^{-\left(y/\lambda^{-1} \right)^{-\theta}}. $$ I.e. if negative shapes would be allowed, we could say $Y \propto \mathrm{Weibull}(-\theta,\lambda^{-1})$ (just $\theta$ in the front of $p_X(x)$ has to be replaced with $|\theta|$). CDF/mean/mode etc also require very minor adaptation for the negative shapes.

I guess the same trick is possible for the whole generalized Gamma family, and e.g. inverse Gamma would then become its member.

alyst
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  • http://en.wikipedia.org/wiki/Fr%C3%A9chet_distribution – whuber May 20 '14 at 16:30
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    @whuber ah, indeed! And both generalize to [GEV](https://en.wikipedia.org/wiki/Generalized_extreme_value_distribution) – alyst May 20 '14 at 17:33
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    They also both generalize to [Type-2 Gumbel Distribution](https://en.wikipedia.org/wiki/Type-2_Gumbel_distribution), although the parametrization is different. – alyst May 22 '14 at 11:24

1 Answers1

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There is no good reason not to do such a generalization, which would unite the Weibull and inverse Weibull distributions. So reasons must be historical or accidental.

Also see the comments thread.

kjetil b halvorsen
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