Can you bound the third moment from the second moment?

Question

Suppose $X$ is a random real variable with zero mean and finite second moment $\langle X^2\rangle$. Under what conditions can we give a bound (upper/lower) for the third moment $\langle X^3\rangle$?

Answer 1

There is no such universal bound.

Here is a cute fact to put this question in perspective:

Given real numbers $k^\prime \gt k \ge 0,$ there exist random variables $X$ that have a finite $k^\text{th}$ moment but infinite $k^{\prime\,\text{th}}$ moment.

It doesn't matter whether we consider these raw moments or central moments: see How to prove that the $r^\text{th}$ raw moment exists if and only if the $r^\text{th}$ central moment exists?.

To prove the fact, let $X$ be supported on $[1,\infty)$ with a density proportional to $(1 + x^2)^{-(k^\prime+1) /2}.$ Since $k^\prime \gt 0,$ this density integrates to a finite number, showing such distributions exist.

Because $1+x^2 \gt x^2,$

$$\mu_{k}(X) \propto \int_1^\infty \frac{x^k\,\mathrm{d}x}{(1 + x^2)^{(k^\prime+1)/2}} \le \int_1^\infty x^{k - k^\prime - 1}\,\mathrm{d}x = \frac{1}{k^\prime - k} \le \infty$$

but since $1+x^2 \le 2x^2,$

$$\mu_{k^\prime}(X) \propto \int_1^\infty \frac{x^{k^\prime}\,\mathrm{d}x}{(1 + x^2)^{(k^\prime+1)/2}} \gt \int_1^\infty \frac{x^{k^\prime}\,\mathrm{d}x}{(2 x^2)^{(k^\prime+1)/2}} \propto \int_1^\infty \frac{\mathrm{d}x}{x} = \lim_{x\to\infty}\log(x)$$

diverges, QED.

The question concerns the case $k^\prime=3$ and $k=2.$ To apply the foregoing analysis, let $\sigma^k$ be the given $k^{\text{th}}$ moment and let $X$ be the random variable described above. Then the $k^\text{th}$ moment of $$ \frac{\sigma\,X}{\mu_k(X)^{1/k}} $$ equals $\sigma^k$ but its $k^{\prime\,\text{th}}$ moment is infinite.