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I have a random variable $X$ for which it is known a priori that $P(X > x) = \exp(-ax^b)$, i.e. the CDF is given by $F_X(x) = 1-\exp(-ax^b)$.

I would like to determine the values of $a$ and $b$ but I am currently struggling to find an appropriate transformation of this "exponential-power" distribution that would allow me using the fitting functions with standard distributions.

Would you have any idea of transformation (or any alternative) that would help this purpose?

Edit: As indicated by the valuable comments, I was indeed looking for a Weibull distribution. I tried fo fit this model using fitdist in R. The results are shown on the figure. Weibull fitting

At first glance, it looks pretty satisfying (except for the tail on the Q-Q plot). However, the Kolmogorov-Smirnov goodness of fit test dramatically fails (p-value < 1e-14). Should I reject the model based on the KS test results?

G. Vaurs
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  • Would you please post a link to the data? – James Phillips Aug 07 '19 at 16:42
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    @user158565 - It's a Weibull distribution. $F_X(0) = 0$. – jbowman Aug 07 '19 at 18:19
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    What language are you working with? It makes a difference, as "fitting functions with standard distributions" is a language-specific reference, and your distribution is one of the standard ones (the Weibull.) – jbowman Aug 07 '19 at 18:23
  • @user158565 - $\exp\{-(ax)^n\} = \exp\{-bx^n\}$, where $b = a^n$. $1 - \exp\{-a0^n\} = 1-\exp\{-0\} = 0$. – jbowman Aug 07 '19 at 18:35
  • @jbowman Sorry. Something is wrong with me. basically I switched $x^n$ and $n^x$. – user158565 Aug 07 '19 at 18:40
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    @user158565 - Ha ha! It's happened to me before too, no worries :). Funny how our minds fail us sometimes... – jbowman Aug 07 '19 at 19:01
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    @jbowman I thought I was the only one who did that. – James Phillips Aug 07 '19 at 20:01
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    Many of the ideas in https://stats.stackexchange.com/questions/111010/interpreting-qqplot-is-there-any-rule-of-thumb-to-decide-for-non-normality carry over to this thread. Tests are of very limited use here, especially with large sample sizes. The conclusion "this distribution isn't perfectly fit by a Weibull" is neither surprising nor constructive. Reject Weibull if and only if you can find a better fitting distribution. – Nick Cox Aug 08 '19 at 09:16

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