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



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, $\lambda = 1/2$.

  • 5
    $\begingroup$ But the lognormal distribution does not have a moment-generating function. $\endgroup$ – onestop Mar 21 '12 at 16:44
  • 4
    $\begingroup$ 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 Mar 21 '12 at 16:55
  • $\begingroup$ Its not true that the lognormal dont have mgf, its only that its not defined on an open interval containing zero $\endgroup$ – kjetil b halvorsen Apr 21 '17 at 7:10
  • 1
    $\begingroup$ 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 Feb 1 at 20:27
  • 1
    $\begingroup$ @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/… $\endgroup$ – kjetil b halvorsen Feb 2 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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