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I have a centered random variables $Y_1,Y_2$ which first 10 moments are given respectively by

$$ 0, 1, 0, 6, 0, 90, 0, 2520, 0, 113400 $$

$$0, 1, 0, 32/3, 0, 36847/100, 0, 436879364/15435, 0, 71230188570971/17781120$$

May be someone is able to conjecture what is the law of $Y_1$ and/or $Y_2$. I.e. may be some "well-known" law has the same moments.

The above values are limit cases of parametrised family of laws in (www.mimuw.edu.pl/~pmilos/moments.pdf) which I am in fact interested about. In principle, I could calculate higher momements using some recursive formula but its gets too complicated (even for mathematica).

flag	asked Jul 29 2011 at 12:24 Piotr Miłoś 746●3●9

2

How you got the moments of $Y_2$? Because in OEIS is no sequence with the given numerators: oeis.org/search?q=1%2C32%2C36847. This reduces the chance that $Y_2$ is "well-known". Maybe some more information is helpful. – André Schlichting Jul 29 2011 at 13:48

Oh, I did not know about this website. Thanks. And answering to your question: it is limit case (as p->+\infty) of a sequence which is in pdf file mentioned above) – Piotr Miłoś Jul 29 2011 at 14:21

Andrew · Answer 1 · 2011-07-29 13:31:08Z UTC

5

The first sequence can be obtained with the pdf $f(x)=e^{-\sqrt2|x|}/\sqrt2\;$, the moments of order $2n$ being $(2n)!/2^n$.

answered Jul 29 2011 at 13:31

Andrew
948●1●4●9

	I have just notice the same. But you probably mean f(x) = c e^{-\sqrt{t}\|x\|} i.e. symmetrized exponential distribution. – Piotr Miłoś Jul 29 2011 at 13:36
	Piotr, you could read again Andrew's solution, you will see it is correct. (And while we are at it, what about thanking the guy?) – Didier Piau Aug 8 2011 at 13:52

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