# Deducing a probability distribution from its moment-generating function

It's pretty trivial to get a moment-generating function from a p.d.f. (provided $\sum e^{tx}f(x)$ isn't too difficult to evaluate), but since moment-generating functions uniquely determine a probability distribution function, is there a way to "back-generate" the p.d.f from the m.g.f.?

Edit: I'm talking about a discrete distribution.

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Are you dealing with discrete distributions? In general, probability distribution need not have a p.d.f. –  Siméon Apr 7 at 0:22

Let $\mathcal{M}(g)(s) = \int_0^{\infty} x^{s-1} g(x) dx$ be the Mellin transform; then the moment-generating function of a smooth enough p.d.f $f$ is given by $\mathcal{M}(f(-\log(x))(-s)$;

so given a nice enough moment-generating function $h(s) = E[e^{sX}] = \int_{-\infty}^{\infty} e^{sx} f(x) dx$, we recover $f$ as $$f(x) = \mathcal{M}^{-1}(h(-s))(-e^x)$$ where $\mathcal{M}^{-1}$ is given by the Mellin inversion theorem: $$\mathcal{M}^{-1}h(x) = \frac{1}{2\pi i}\int_{c - i\infty}^{c + i \infty} x^{-s} h(s) ds$$ for an appropriate real number $c$, where the integral is understood to be along a line in $\mathbb{C}$.

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We haven't learned anything about Mellin transforms. :P Or how to work with complex numbers in that way. –  user54609 Apr 7 at 1:50
@EricDong I'm not sure you'll be able to do it in a significantly different way. Probably the best thing to do is to use a table. –  Cocopuffs Apr 7 at 9:16
This should work for appropriate discrete functions $f$ as well since piecewise continuity is enough –  Cocopuffs Apr 7 at 9:17
The question is to inverse a Laplace/Fourier transform. I take the example of a discrete distribution $f(n)$ on the natural numbers with moment-generating-function $$M(t) = \sum_{n=0}^\infty f(n)e^{nt}$$ with radius of convergence $R \geq 1$. Fourier inversion here is $$f(n) = \frac{1}{2\pi}\int_0^{2\pi}M(i\theta)e^{-in\theta}\,d\theta$$
A related formula would be: $$f(n) = \left.\frac{d^n}{dt^n}M(\log t)\right|_{t=0}$$