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I have a continuous symmetric random variable $X$ for which $E\left(X^{2k}\right)=b^k.k!$. Since it's symmetric, all of its odd order moments are zero. And in my case, $b=2/n$ where $n$ is a large integer (if it's needed). Does $X$ have a well-known distribution? What's the distribution of $X$?

MohammadJavad Vaez
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    For a centered Laplace distribution you have $E[X^k]=k!b^k$ for all $k$ even – Snoop Aug 19 '23 at 19:23
  • @Snoop This is correct. Thanks, but it's not my answer. Since I'm looking for an X such that $E(X^k)=b^{\frac{k}{2}}(k/2)!$ for all $k$ even – MohammadJavad Vaez Aug 19 '23 at 22:11
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    of course - that's just a suggestion for a possible direction – Snoop Aug 19 '23 at 22:15
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    You might look for a distribution with the same moment generating function which is $\frac{1}{2} \sqrt{\pi } \sqrt{b} t e^{\frac{b t^2}{4}} \text{erf}\left(\frac{\sqrt{b} t}{2}\right)+1$. This, of course, simplifies for $b=4 but I don't know the name of such a distribution. – JimB Aug 20 '23 at 01:05

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@Snoop put you on the right path. Consider

$$Y=\text{sgn}(X)\sqrt{|X|}$$

where $X\sim \text{Laplace}(0,b)$.

I am too lazy to write down the steps so I'll use Mathematica:

dist = TransformedDistribution[Sign[x] Sqrt[Abs[x]], x \[Distributed] LaplaceDistribution[0, b]]

FullSimplify[Moment[dist, 2 k], k ∈ PositiveIntegers]
(* b^k k! *)


FullSimplify[Moment[dist, 2 k - 1], k ∈ PositiveIntegers]
(* 0 *)
JimB
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