Is a moment-generating function a Fourier transform of a probability density function?

In other words, is a moment generating function just the spectral resolution of a probability density distribution of a random variable, i.e. an equivalent way to characterize a function in terms of it's amplitude, phase and frequency instead of in terms of a parameter?

If so, can we give a physical interpretation to this beast?

I ask because in statistical physics a cumulant generating function, the logarithm of a moment generating function, is an additive quantity that characterizes a physical system. If you think of energy as a random variable, then it's cumulant generating function has a very intuitive interpretation as the spread of energy throughout a system. Is there a similar intuitive interpretation for the moment generating function?

I understand the mathematical utility of it, but it's not just a trick concept, surely there's meaning behind it conceptually?

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    $\begingroup$ I believe it's the characteristic function that more resembles the Fourier transform. The moment generating function is a Laplace transform. $\endgroup$
    – Placidia
    Jun 9, 2014 at 14:06
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    $\begingroup$ Interesting: "The Laplace transform is related to the Fourier transform, but whereas the Fourier transform resolves a function or signal into its modes of vibration, the Laplace transform resolves a function into its moments" princeton.edu/~achaney/tmve/wiki100k/docs/… Then I guess the question is - how, intuitively, does a Laplace transform decompose a function into it's moments, and is there a geometric interpretation of this? $\endgroup$
    – bolbteppa
    Jun 9, 2014 at 14:15
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    $\begingroup$ It does it by virtue of the Taylor series expansion of the exponential function. $\endgroup$
    – Placidia
    Jun 9, 2014 at 14:18
  • $\begingroup$ Now everything nearly makes sense! However, what exactly is a moment, intuitively? I know this: "Broadly speaking a moment can be considered how a sample diverges from the mean value of a signal - the first moment is actually the mean, the second is the variance etc... " dsp.stackexchange.com/a/11032 However, what does that mean intuitively? What is the sample when calculating the 1st/2nd/3rd/4th moment of say, x^2 (taking a Laplace transform of x^2)? Is there a geometric interpretation? $\endgroup$
    – bolbteppa
    Jun 9, 2014 at 14:26

1 Answer 1


The MGF is

$M_{X}(t)=E\left[ e^{tX} \right]$

for real values of $t$ where the expectation exists. In terms of a probability density function $f(x)$,

$M_{X}(t)=\int_{-\infty}^{\infty} e^{tx}f(x) dx.$

This is not a Fourier transform (which would have $e^{itx}$ rather than $e^{tx}$.

The moment generating function is almost a two-sided Laplace transform, but the two-sided Laplace transform has $e^{-tx}$ rather than $e^{tx}$.

  • 3
    $\begingroup$ +1 As an aside: the characteristic function is the one that's more closely related to the Fourier transform (in that case, again, there's the small issue of a minus sign) - the c.f. is $E(e^{itX})$, while - up to multiplicative constants - the usual Fourier transform would be $E(e^{-itX})$. These connections prove to be quite useful at times, such as finding lists of useful properties of Fourier or Laplace transforms that usually carry directly over, or being able to look up extensive tables of Fourier or Laplace transforms when finding MGFs or cfs. $\endgroup$
    – Glen_b
    Jun 10, 2014 at 0:53
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    $\begingroup$ And of course the most useful property is that the MGF of the sum of two independent random variables is the product of their moment generating functions. This is equivalent to the rule that the Fourier transform of the convolution of two functions is the product of their Fourier transforms. $\endgroup$ Jun 10, 2014 at 3:21

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