Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Are you dealing with discrete distributions? In general, probability distribution need not have a p.d.f. – Siméon Apr 7 '13 at 0:22

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 $$ f(n) = \frac{1}{2\pi}\int_0^{2\pi}M(i\theta)e^{-in\theta}\,d\theta $$

If you prefer to stay in the real realm, there is an interesting formula due to Post.

A related formula would be: $$ f(n) = \left.\frac{d^n}{dt^n}M(\log t)\right|_{t=0} $$

share|cite|improve this answer
We haven't learned anything about Fourier transforms. – user54609 Apr 7 '13 at 1:49
@user54609 If you were in elementary probability theory, there was no way to do it. You had to recognise a given mgf as an mgf of some random variable – BCLC Feb 6 at 7:15
Siméon, is Post's the same as this one? – BCLC Feb 6 at 7:16

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 $\mathbb{C}$.

share|cite|improve this answer
We haven't learned anything about Mellin transforms. :P Or how to work with complex numbers in that way. – user54609 Apr 7 '13 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 '13 at 9:16
This should work for appropriate discrete functions $f$ as well since piecewise continuity is enough – Cocopuffs Apr 7 '13 at 9:17

Yes, you can. First, convert your mgf into a characteristic function (i.e. replace $t \rightarrow it$). Next, invert the characteristic function to yield the pdf using an inverse Fourier transform.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.