# Can you bound the third moment from the second moment?

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 $$\langle X^3\rangle$$?

• 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.
– whuber
Commented Mar 2, 2020 at 15:05
• That actually answers my question, thanks @whuber
– a06e
Commented Mar 2, 2020 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 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.

• +1, whuber. I changed the linked post's title and hence this minor edit to your post. Commented Nov 3, 2023 at 3:08