Are there any non-identical distributions which happen to have the same moment-generating function?

  • 4
    $\begingroup$ See Wikipedia under Moment-generating function#Important properties $\endgroup$
    – onestop
    Commented Mar 21, 2012 at 15:17
  • $\begingroup$ @onestop that answers it! if you want to put that as an answer I'll accep tit. $\endgroup$
    – user9437
    Commented Mar 21, 2012 at 15:52

1 Answer 1



In an exercise, Stuart & Ord (Kendall's Advanced Theory of Statistics, 5th Ed., Ex. 3.12) quote a 1918 result of TJ Stieltjes (which apparently appears in his Oeuvres Completes,):

If $f$ is an odd function of period $\frac{1}{2}$, show that

$$\int_0^\infty x^r x^{-\log x} f(\log x) dx = 0$$

for all integral values of $r$. Hence show that the distributions

$$dF = x^{-\log x}(1 - \lambda \sin(4\pi \log x))\ dx, \quad 0 \le x \lt \infty;\quad 0 \le |\lambda| \le 1,$$

have the same moments whatever the value of $\lambda$.

(In the original, $|\lambda|$ appears only as $\lambda$; the restriction on the size of $\lambda$ arises from the requirement to keep all values of the density function $dF$ non-negative.) The exercise is easy to solve via the substitution $x = \exp(y)$ and completing the square. The case $\lambda=0$ is the well-known lognormal distribution.

enter image description here

The blue curve corresponds to $\lambda=0$, a lognormal distribution. For the red curve, $\lambda = -1/4$ and for the gold curve, $\lambda = 1/2$.

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    $\begingroup$ But the lognormal distribution does not have a moment-generating function. $\endgroup$
    – onestop
    Commented Mar 21, 2012 at 16:44
  • 7
    $\begingroup$ That's an excellent point, onestop, and I have to agree with it. I took the question in the sense of "having the same set of moments" and I should have pointed out that change of interpretation. When the mgf exists as a function (and not just as a formal power series) then it can be inverted to produce a unique density to which it corresponds. $\endgroup$
    – whuber
    Commented Mar 21, 2012 at 16:55
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    $\begingroup$ For the record, both @onestop and I were implicitly discussing the existence of an mgf in a neighborhood of $0.$ This sense is often assumed because one of the most basic uses of an mgf is to expand its MacLaurin series (Taylor series around $0$) to compute or analyze the moments and this requires the function to be defined in a neighborhood, not just at $0.$ $\endgroup$
    – whuber
    Commented Feb 1, 2019 at 20:27
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    $\begingroup$ @whuber: That's OK, but it seems to be understood implicitely so often that one forgets that mgf's can be useful otherwise too. See also (the links in) stats.stackexchange.com/questions/389846/… $\endgroup$ Commented Feb 2, 2019 at 17:59
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    $\begingroup$ OK, that makes sense. So if the mgf exists, it must be unique. However, because it's only real-analytic and not complex-analytic, we can have two different mgf's with the same derivatives at the origin. For instance, we could have two piecewise-analytic mgfs which agree on a neighborhood of 0 but then diverge beyond that, and which would thus have the same moments. If I get what you are saying correctly, that is. $\endgroup$ Commented Jun 10, 2022 at 8:38

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