# 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 Mar 2 at 15:05
• That actually answers my question, thanks @whuber – 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 $$1+x^2 \gt x^2,$$

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

but

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

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.