Are there any non-identical distributions which happen to have the same moment-generating function?
Moment-Generating Functions – Understanding Their Identity and Applications
distributionsmoment-generating-functionmoments
Related Solutions
Let me answer in reverse order:
2. Yes. If their MGFs exist, they'll be the same*.
Indeed it follows from the result you give in the post this comes from; if the MGF uniquely** determines the distribution, and two distributions have MGFs and they have the same distribution, they must have the same MGF (otherwise you'd have a counterexample to 'MGFs uniquely determine distributions').
* for certain values of 'same', due to that phrase 'almost everywhere'
** 'almost everywhere'
- No - since counterexamples exist.
Kendall and Stuart list a continuous distribution family (possibly originally due to Stieltjes or someone of that vintage, but my recollection is unclear, it's been a few decades) that have identical moment sequences and yet are different.
The book by Romano and Siegel (Counterexamples in Probability and Statistics) lists counterexamples in section 3.14 and 3.15 (pages 48-49). (Actually, looking at them, I think both of those were in Kendall and Stuart.)
Romano, J. P. and Siegel, A. F. (1986),
Counterexamples in Probability and Statistics.
Boca Raton: Chapman and Hall/CRC.
For 3.15 they credit Feller, 1971, p227
That second example involves the family of densities
$$f(x;\alpha) = \frac{1}{24}\exp(-x^{1/4})[1-\alpha \sin(x^{1/4})], \quad x>0;\,0<\alpha<1$$
The densities differ as $\alpha$ changes, but the moment sequences are the same.
That the moment sequences are the same involves splitting $f$ into the parts
$\frac{1}{24}\exp(-x^{1/4}) -\alpha \frac{1}{24}\exp(-x^{1/4})\sin(x^{1/4})$
and then showing that the second part contributes 0 to each moment, so they are all the same as the moments of the first part.
Here's what two of the densities look like. The blue is the case at the left limit ($\alpha=0$), the green is the case with $\alpha=0.5$. The right-side graph is the same but with log-log scales on the axes.
Better still, perhaps, to have taken a much bigger range and used a fourth-root scale on the x-axis, making the blue curve straight, and the green one move like a sin curve above and below it, something like so:
The wiggles above and below the blue curve - whether of larger or smaller magnitude - turn out to leave all positive integer moments unaltered.
Note that this also means we can get a distribution all of whose odd moments are zero, but which is asymmetric, by choosing $X_1,X_2$ with different $\alpha$ and taking a 50-50 mix of $X_1$, and $-X_2$. The result must have all odd moments cancel, but the two halves aren't the same.
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 $e^{tx}$.
Best Answer
Yes.
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,):
(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.
The blue curve corresponds to $\lambda=0$, a lognormal distribution. For the red curve, $\lambda = -1/4$ and for the gold curve, $\lambda = 1/2$.