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Let $X$ be a random variable with continuous density $\rho(x)$. Assume that $X$ is symmetric and $\vert X\vert<L$. Since it has a bounded support, all moments of $X$ are well-defined. Let $m_i$ denote the moment $i$ of $X$, i.e. $$ m_i = \int_{-L}^{L} x^i \rho(x) dx. $$

Is there anyone knows if the following statement is true or not? $$ \frac{m_2}{2!}\times \frac{m_{4k}}{4k!}\geq \frac{m_{4k+2}}{(4k+2)!}. $$ for $k\geq 1$. Note that one may rewrite the above equation as $$ {4k+2\choose 2} m_2m_{4k}\geq m_{4k+2}. $$ The Above recursion is true for some common distributions such as uniform distribution and Gaussian distribution (even though it does not have a bounded support) but can we say in general if it is true?

If not, what are the necessary conditions to make it true? For example, if $m_2>L^2/15$ then it is true. But is there any other condition available with less restriction?

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up vote 0 down vote accepted

If $f(x)=\frac12a|x|^{a-1}$ for $|x|\leqslant1$, then $m_{2k}=\frac{a}{a+2k}$ for every integer $k$ hence $m_{2k}\sim\frac{a}{2k}$ when $a\to0$ and $$ {4k+2\choose 2}\frac{m_2\,m_{4k}}{m_{4k+2}}\sim c_ka, $$ for some $c_k\gt0$. In particular, when $a$ is small enough, the ratio is $\lt1$.

If the condition that the density is defined everywhere and continuous is important, consider the truncation of $f$ at level $h$, renormalized. Then, by continuity, the ratio is $\lt1$ for $h$ large enough. Likewise, modify $f$ at $x=\pm1$ to make it continuous everywhere.

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