I need in particular to generate random numbers from a Four parameters Dagum distribution: http://www.mathwave.com/help/easyfit/html/analyses/distributions/dagum.html

And from the Generalized Extreme Value distribution http://www.mathwave.com/help/easyfit/html/analyses/distributions/gen_extreme.html

I can't find the formula/algorithm anywhere. I can't imagine that this is not solved quite often. I find it quite hard to do the inverse CDF of the generalized extreme value distribution... but maybe I'm overthinking it?


1 Answer 1


It's not that hard, but you need to be careful with the details and go in small steps, as usual.

For the Dagum distribution, we replace $F(x)$ with $U$, a Uniform$(0,1)$ variate, and work through the algebra, two (or so) steps at a time:

$$U = \left (1 + \left (\frac{x-\gamma}{\beta} \right )^{-\alpha} \right)^{-k}$$ $$U^{-1/k}-1 = \left(\frac{x-\gamma}{\beta}\right)^{-\alpha}$$ $$\beta \left(U^{-1/k}-1\right)^{-1/\alpha} = x - \gamma$$ $$x = \gamma + \beta \left(U^{-1/k}-1\right)^{-1/\alpha}$$

For the Generalized Extreme Value distribution, we have, for $k > 0$,:

$$U = \exp\left\{-(1+kz)^{-1/k}\right\}$$ $$(-\ln U)^{-k} = 1+kz$$ $$\frac{(-\ln U)^{-k}-1}{k} = z = \frac{x-\mu}{\sigma}$$ $$x = \sigma\frac{(-\ln U)^{-k}-1}{k}+\mu$$

When $k = 0$, we have, jumping ahead a bit,:

$$\ln (-\ln U) = -z$$ $$-\sigma(\ln (-\ln U)) + \mu = x$$


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