0
$\begingroup$

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).

$\endgroup$
2
  • 2
    $\begingroup$ 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. $\endgroup$ Jul 29, 2011 at 13:48
  • $\begingroup$ 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) $\endgroup$ Jul 29, 2011 at 14:21

1 Answer 1

5
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ I have just notice the same. But you probably mean f(x) = c e^{-\sqrt{t}|x|} i.e. symmetrized exponential distribution. $\endgroup$ Jul 29, 2011 at 13:36
  • $\begingroup$ 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?) $\endgroup$
    – Did
    Aug 8, 2011 at 13:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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