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 $\langle X^3\rangle$?

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    $\begingroup$ What kinds of "conditions" do you have in mind? You need something, because the third moment can be undefined (consider the Student $t$ distribution with 3 df, for instance), it can be infinite, and it can be negatively infinite: there are no general bounds. $\endgroup$ – whuber Mar 2 at 15:05
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    $\begingroup$ That actually answers my question, thanks @whuber $\endgroup$ – becko Mar 2 at 23:50

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 Mathematical statistics (moments vs. central moments).

To prove the fact, let $X$ be supported on the positive real numbers 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 $1+x^2 \gt x^2,$

$$\mu_{k}(X) \propto \int_0^\infty \frac{x^k\,\mathrm{d}x}{(1 + x^2)^{(k^\prime+1)/2}} \le \int_0^\infty x^{k - k^\prime - 1}\,\mathrm{d}x = \frac{1}{k^\prime - k} \le \infty$$


$$\mu_{k^\prime}(X) \propto \int_0^\infty \frac{x^{k^\prime}\,\mathrm{d}x}{(1 + x^2)^{(k^\prime+1)/2}} \gt \int_1^\infty \frac{x^{k^\prime}\,\mathrm{d}x}{(2 x^2)^{(k^\prime+1)/2}} \propto \int_1^\infty \frac{\mathrm{d}x}{x} = \lim_{x\to\infty}\log(x)$$

diverges, QED. The inequality in the last line (a) dropped the area from $0$ to $1$ and (b) uses $2x^2 \ge 1 + x^2$ when $x \ge 1.$

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.

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