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Q: (Probability) distribution from full set of moments ( Answered ,   1 Comment )
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 Subject: (Probability) distribution from full set of moments Category: Science > Math Asked by: borel-ga List Price: \$15.00 Posted: 02 Apr 2006 21:08 PDT Expires: 02 May 2006 21:08 PDT Question ID: 714834
 ```I'm looking for detailed explanation (or even proof) of how to reconstruct a distribution from full set of moments. I've found this page so far, but it doesn't contain a proof and it's also unclear about exact conditions: http://www.plmsc.psu.edu/~www/matsc597/probability/moments/node3.html#SECTION00030000000000000000 This should be fairly classical mathematical result.```
 ```Hi borel The 'missing link' is SEARCH TERMS: The Characteristic Function of a Probability Distribution Its Taylor series gives you moments The Characteristic Function as a Moment-Generating Function it's Fourier tranfrom is the distribution http://www2.sjsu.edu/faculty/watkins/charact.htm see also http://en.wikipedia.org/wiki/Characteristic_function http://mathworld.wolfram.com/UniformDistribution.html cumulant http://mathworld.wolfram.com/UniformDistribution.html etc. Hope this explains the one-to-one correspondence Hedgie``` Request for Answer Clarification by borel-ga on 02 Apr 2006 23:47 PDT ```Thank you for these links, they seem to be pointing to the right directions. However, I was hoping for more detailed information: existence, uniqueness, more detailed conditions in general etc. Something that is more "formal" (in mathematical sense).``` Clarification of Answer by hedgie-ga on 19 Dec 2006 12:17 PST ```Thanks for the rating Borel. For the issues you mention existence .., I suggest to combine the corresponding theorems for the Taylor expansion and Fourier Transform.```
 borel-ga rated this answer: ```Although these links are not exactly what I was hoping for, they have pointed me to the right direction. The detailed answer I hoped for is obviously only in textbooks and not on the internet. Thank you Hedgie.```
 ```Well I'm guessing you will have a hard time finding the answer to this question in a formal proof at least. My final stats course book (in a stats major) gives the theorem that if the moment generating functions for distributions X and Y are equal (mx(t)=my(t)) for all values of t then X and Y have the same prob. dist. However it adds "the proof of this theorem is beyond the scope of this text" Now having all the moments to a function seems like a hard task in most cases for me, as many cases there are infinite moments. However having the m.g.f. in closed form allows you of course to generate any moment you want. Thusly if the mgfs are equal all moments must be equal, from this we know that E{Y^k)=E(X^k) for all k such that k is in Z+ Where to go from there is where I get lost, obviously intuition leads you to Y=X but intuition doesn't amount to much.```