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Following are similar to but different from previous posts here and here

  1. Given two distributions which admit moments of all orders, if all the moments of two distributions are the same, then are they identical distributions a.e.?
  2. Given two distributions which admit moment generating functions, if they have the same moments, are their moment generating functions the same?

Thanks!

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In accordance with question #2, I believe in general, if two functions have the same MGF (if it exists in an open neighborhood of 0) then they follow the same distribution. Unfortunately, I do not know of the proof, as it is quite complex. Hope that helps just a little bit. –  nicefella Feb 3 at 2:31
    
@nicefella The proof is relatively easy: evaluating the MGF at imaginary values gives the characteristic function which can be inverted to produce the distribution. The inversion works provided the MGF is analytic in a neighborhood of the origin. –  whuber Feb 3 at 15:23
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1 Answer

up vote 6 down vote accepted

Let me answer in reverse order:

2. Yes. If their MGFs exist, they'll be the same*.

see here and here for example

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'

  1. 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.

example of same moments, different densities

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:

enter image description here

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.

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Thanks! In your reply to my second question, what does "for certain values of 'same'" mean? Can you give counterexamples to my first question? –  Tim Feb 3 at 2:39
    
It's simply a reference to the necessary qualification caused by the 'almost everywhere' that is in the previous question. So counterexamples could look at density functions that were the same almost everywhere but differed at a countable subset of points - I already gave an example to you previously. –  Glen_b Feb 3 at 2:51
    
For my first question, (according to your answer yes to my second question and to my question in my previous post), do all the counterexamples belong to the case when not both distributions admit moment generating functions? –  Tim Feb 3 at 3:00
    
That it must be so is a consequence of the statement "If the mgf is finite in an open interval containing zero, then the associated distribution is characterized by its moments" in cardinal's answer I believe I linked to. If an mgf is not finite in that sense, that's the only way for the distribution not to be characterized by its moments. –  Glen_b Feb 3 at 4:22
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The first question was answered at stats.stackexchange.com/questions/25010/… and in the OP's recent question at stats.stackexchange.com/questions/84158/…. Feller's example is attributed to Stieltjes (way before Feller's time) in Stuart & Ord. –  whuber Feb 3 at 15:20
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