According to a paper the equation with integral $\int_{-\infty}^{\infty}dx \rho_0(\lambda)e^{-b\lambda}=1/N$ (#1) where $\rho_0(\lambda)$ is a distribution function, $N$ is a natural number, can be written in other way :

$\sum_{n=1}^{\infty} \frac {(-b)^n} {n!} <\lambda^n>+lnN=0$ (#2)

Here $<\lambda^n>$ is the nth cumulant of the distribution function.

My question is how to transform the initial eq step by step?

The paper notes that 'by the fact that the Fourier transform of $rho_0(\lambda)$ is the exponential of the cumulant generating function [10], we finnd that Eq. implies' #2. However it's not so clear how to get #2.


This is the definition. With $X$ a random variable with pdf $\rho$ let $f(t) = \Bbb{E}[e^{-X t}]= \int_{-\infty}^\infty e^{-xt}\rho(x)dx$ if $\rho$ decays faster than $e^{r |x|}$ then $f$ is analytic around $t=0$ since $f(0)=1$ then $g(t) = \ln f(-t)$ is analytic around $t=0$ that is $g(t) = \sum_{n=1}^\infty \frac{t^n}{n!}g^{(n)}(0)$.

It is useful because if $X,Y$ are independent then $e^{-Xt},e^{-Yt}$ are independent so $\Bbb{E}[e^{-Xt}e^{-Yt}] =\Bbb{E}[e^{-Xt}]\Bbb{E}[e^{-Yt}]$ and $f_{X+Y}(t) = f_X(t)f_Y(t), g_{X+Y}(t) = g_X(t)+g_Y(t)$.

$g^{(n)}(0)$ is called the $n$-th cumulant. Here you are told one value $f(b) = 1/N$ so that $-\ln N =g(-b) = \sum_{n=1}^\infty \frac{(-b)^n}{n!}g^{(n)}(0)$

  • $\begingroup$ Thank you, it's clear except how to get 1st and 2nd cumulants : $<\lambda>$ and $\sigma$ like in normal distribution? $\endgroup$ – Mikhail Gaichenkov Apr 25 at 7:44
  • $\begingroup$ If you know $f(0),f'(0)$ then you know $g(0)= \ln f(0),g'(0) = -f'(0)/f(0)$ $\endgroup$ – reuns Apr 25 at 19:24
  • $\begingroup$ Thank you, it's clear about $g(0)$. However, what about the case $\rho\frac{1}{\sqrt{2\pi\sigma^2} } e^{ -\frac{(x-\mu)^2}{2\sigma^2} }$ Does it mean we have to calculate the integral and get $f(t)$? $\endgroup$ – Mikhail Gaichenkov Apr 27 at 19:54
  • $\begingroup$ $f^{(n)}(t) = \Bbb{E}[(-X)^n e^{-X t}]$ so $f(0) = 1, f'(0) = -\mu,f''(0) = \sigma^2$ and hence $g(0)=0, g'(0) = \mu,g''(0) = ..$ $\endgroup$ – reuns Apr 27 at 20:57

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