Is there a bound on the tail distribution of a finite sum of i.i.d Weibull random variables?

  • $\begingroup$ You'll need to be more precise about "nice", I think. $\endgroup$
    – jonsca
    Dec 13, 2012 at 5:42
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    $\begingroup$ There are some references and comments that might be of some help in this post $\endgroup$
    – Glen_b
    Dec 13, 2012 at 8:22
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    $\begingroup$ Glen_b's reference sounds promising, but the Hölder inequality lets you relate the sum of Weibull random variables with shape $k \gt 1$ to the sum of exponential random variables, for which you have exponential bounds stats.stackexchange.com/questions/4816/…. For $k \lt 1$, you can't get exponential bounds because the tail of each term decays too slowly. $\endgroup$ Dec 13, 2012 at 9:37
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    $\begingroup$ By the way, there is a vote to close this as an exact duplicate of a past question, which essentially asks for a tail bound for $k=1/2$. However, I don't agree that this is an exact duplicate because $k=1/2$ was not specified here, and seems counter-indicated by the request for exponential bounds which are impossible for $k=1/2$. $\endgroup$ Dec 14, 2012 at 1:11
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    $\begingroup$ When the shape is $<1$, the distribution can be shown to be subexponential as defined in the book 'Modelling Extremal Events for Insurance and Finance' by Embrechts , Klüppelberg and Mikosch (chap 1). This means that the survival $S(x)$ of the convolution is equivalent to the survival of the max of the same number of i.i.d. variables for large $x$. Not a bound yet. $\endgroup$
    – Yves
    Oct 30, 2014 at 14:50

2 Answers 2


The convolution form is not closed, so it is impossible to do in the continuos way. You can use a Taylor Series Sum instead to approach the convolution. If you're not looking for a nice solution, and want to be more practical, try doing simple simulation and choose a nice fit.


For the asymptotic tail, search for "sub-Weibull" distributions. They develop concentration properties in a similar manner we cope with sub-exponentials or sub-gaussians.


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