Inverse laplace transform of $e^{-(xh)^k}$.

Question

In the paper, MIXED POISSON DISTRIBUTIONS ASSOCIATED WITH HAZARD FUNCTIONS OF EXPONENTIAL MIXTURES by Wakoli and Otteno, the authors mention that:

Hesselager et. al. (1998) have stated the following theorem: "A distribution with a completely monotone hazard function is a mixed exponential distribution".

Here, by hazard function, they mean $\lambda(t)=\frac{f(t)}{S(t)}$ where $f(t)$ and $S(t)$ are the PDF and survival functions respectively.

Now, the Weibull distribution can have either monotonically increasing or monotonically decreasing hazard rates. So, it too must be a mixed exponential distribution.

I then wonder what the mixing density might be. In remark-1 of the paper referenced above, the authors state:

The survival function of an exponential mixture is the Laplace transform of the mixing distribution.

The survival function of the Weibull distribution is given by:

$$S(x) = e^{-(x h)^k}$$

So, the mixing distribution, $g(\lambda; k, h)$ must be the inverse Laplace transform of the survival function above. So, how do I find it.

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My attempt:

We want $g(\lambda; k,h)$ such that:

$$\int\limits_0^\infty e^{-\lambda x} g(\lambda; k,h)d \lambda = e^{-(x h)^k}$$

Integrating by parts we get:

$$g(0; k,h)+\int\limits_0^\infty g'(\lambda;k,h)e^{-\lambda x}d \lambda = xe^{-(xh)^k}$$

Not sure how to make further progress.

Answer 1

Berstein's theorem asserts that if $f$ is completely monotone it is the Laplace transform of a positive measure. This gives a numerical way of testing if $f$ is completely monotone: If $f$ is completely monotone, all the matrices $M$ (where $M=(M_{i,j}) =(f(x_i+x_j))$ for given $x_i\in\mathbb R$) must be psd. If any of them has a negative eigenvalue, the function $f$cannot be a Laplace transform, and hence cannot be completely monotone.

I used this method to convince myself that neither the hazard function nor the survival function of the Weibull distribution is completely monotone, at least for generic parameters $h$ and $k$. This seems to doom the OP's quest, at least in its most general form.

Of course these conclusions are based on trusting my use of the numerical analysis software for calculating eigenvalues, so this is not a formal proof.

Added note 12 Nov 2019. When I first posted I had confused Bernstein's theorem with Widder's "Necessary and sufficient conditions for the representation of a function by a doubly infinite Laplace integral", Bull. Amer. Math. Soc.40 (1934), 321-326. It is elementary that $f$ completely monotone implies all the $M$ matrices are psd. The converse, that a continuous $f$ for which all $M$ are psd must be the 2-sided Laplace tranform of a positive measure, is the meat of Widder's paper, is much harder, and is not used in this answer.