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Is there any closed form solution for $\sigma$ in a bimodal Weibull distribution function written in the following form:

$$ P(\sigma) = 1- exp\Bigg(-\alpha\Big(\frac{\sigma}{\sigma_1}\Big)^{m1} -\alpha\Big(\frac{\sigma}{\sigma_2}\Big)^{m2}\Bigg) $$

where $P(\sigma)$ is the probability distribution function, $\alpha$ is constant and $\sigma_i$ and $m_i$ are the scale and the shape parameter respectively?

Morover it is possible to obtain a closed form solution for a multimodal Weibull distribution of order $n$ as:

$$ P(\sigma) = 1- exp\Bigg(\sum\limits_{i=1}^n-\alpha\Big(\frac{\sigma}{\sigma_i}\Big)^{mi}\Bigg) $$

Thanks in advance.

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    $\begingroup$ what does "solution" mean? moments? $\endgroup$
    – Carlo Beenakker
    Commented Mar 22, 2017 at 18:34
  • $\begingroup$ express $\sigma$ as a function of all the other terms. $\endgroup$
    – F.Danzi
    Commented Mar 22, 2017 at 23:16
  • $\begingroup$ just take the log of $1-P(\sigma)$ and you have a polynomial equation for $\sigma$, which you can feed to Wolfram alpha. $\endgroup$
    – Carlo Beenakker
    Commented Mar 23, 2017 at 7:02
  • $\begingroup$ I tried but it doesn't work. For the unimodal distribution it works and the solution is: $$\sigma = \sigma_i \Bigg(-\alpha log\Big (1-P(\sigma)\Big)\Bigg)^{1/m_i}$$ But not for the bimodal. $\endgroup$
    – F.Danzi
    Commented Mar 23, 2017 at 8:43

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There is no such closed form, except for very special cases.

Informal idea (implied by the comment of Carlo Beenakker): you can reduce the problem to that of solving

$$ ax^{\beta} + bx^{\delta} + c $$

which, even for positive integer $\beta$ and $\delta$ rarely has nice closed-forms (although there are some neat solutions in terms of special functions); this of course just gets much worse when the number of modes increases.

Your only hope is when the shape parameters are related, at which time there are some families of solutions. But, depending on what you're trying to use these solutions for, they might be much worse than using the problem itself as 'the answer'.

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