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I have a probability distribution that is defined through it's Laplace transform by :

$$L(t) = \mathbb E(e^{-tX}) = e^{1 - \frac{1+t}{t}\ln(1+t)}$$

Using R and the invLT package, i have a numerical inversion that is quite convincing. Then, i tried through fitdistrplus to estimate the resulting density by MLE on weibull, gamma, pareto and burr models, which produced the following plots:

Graphs of the fitdistr results

It look like the density that i am looking for lives on the positive real axis, has only one mode, and behave quite nicely.

My goal is now to recover an analytic expression for the coresponding CDF (at least). I tried doing the Broomwitch integration as this wikipedia page suggests, but i am not skilled enough in complex integration to make it through...

Do you think that an analytic expression could be recovered for this CDF ? How should i proceed ?

Thanks for your input.

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$\require{\AMScd}$$\require{\mathabx}$$\newcommand\al{\alpha}\newcommand\be\beta$ $\newcommand{\Conv}{\mathop{\ast}}$Mathematica cannot invert this transform:

enter image description here


However, for real $t>0$ we can write \begin{align} L(t)&=\exp\Big\{1-\frac1t\,\int_0^t(1+\ln(1+u))\,du\Big\} \\ &=\exp\Big\{-\frac1t\,\int_0^t\ln(1+u)\,du\Big\} \\ &=\exp\Big\{-\int_0^1\ln(1+zt)\,dz\Big\} \\ &=\prod_{j=1}^n\exp\Big\{-\int_{(j-1)/n}^{j/n}\ln(1+zt)\,dz\Big\} \\ &=\prod_{j=1}^n\exp\Big\{-\frac1n\,\ln\Big(1+\frac jn\,t\Big)+O\Big(\frac{t/n^2}{1+t(j-1)/n}\Big)\Big\} \\ &=\exp\{O(t/n)\}\prod_{j=1}^n\Big(1+\frac jn\,t\Big)^{-1/n}. \end{align} Note that $[0,\infty)\ni t\mapsto(1+\beta t)^{-\alpha}$ is the Laplace transform of the density (say $f_{\al,\be}$) of the gamma distribution with parameters $\al,\be$. Thus, the inverse Laplace transform in question is the density function $$f:=\lim_{n\to\infty}{\DeclareMathOperator{\st}{\ast}}\Conv_{j=1}^n f_{1/n,\,j/n},$$ an infinite infinitesimal convolution of gamma densities. In particular, $f$ is indeed supported on the interval $(0,\infty)$.

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  • $\begingroup$ Yeah, that is where it came from, this is the 'continuous' convolution of all exponentials with rates between 0 and 1. The Laplace transform converges to this nice expression, so i was hoping we could obtain something for the CDF / the density. $\endgroup$
    – lrnv
    Oct 9, 2020 at 13:04
  • $\begingroup$ An explicit expression seems very unlikely, taking in particular into account that Mathematica cannot produce it. $\endgroup$ Oct 9, 2020 at 13:15
  • $\begingroup$ I finaly used a numerical approach to simulate the distribution (wich is what i wanted), and it 'seems' to be OK. However, a full expression would have been better. Sorry for not telling you where it came from and wasting your time, and thanks anyway for your oppinon ! Although the result is negative, i'll still accept your ansewr ;) $\endgroup$
    – lrnv
    Oct 9, 2020 at 15:24
  • $\begingroup$ All right. I am curious, how did you simulate the distribution? $\endgroup$ Oct 9, 2020 at 16:34
  • $\begingroup$ Hahaha. As i said on the upper post, I used the invLT::iv.OpC function in R. It does a numerical inverse laplace transform to recover observations of the density; Then i cumsumed to obtain a discretisation of the CDF. But this is not optimal if you only want to generate from the distribution, then you should take a look at this paper : link.springer.com/article/10.1007/s11222-008-9103-x The coresonding code is there : kent.ac.uk/smsas/personal/msr/rlaptrans.html . $\endgroup$
    – lrnv
    Oct 9, 2020 at 17:02

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