Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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