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It is usual to use second, third and fourth moments of a distribution to describe certain properties. Do partial moments or moments higher than fourth describe any useful properites of a distribtution.

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That depends on what you mean by "useful properties". For instance, all of the even moments are required to characterize sub-Gaussian random variables, which are those variables whose tail is majorized by that of a Gaussian random variable. A random variable X is sub-Gaussian if and only if there exists a non-negative number b such that $E[X^{2k}] \leq \frac{(2k)!b^{2k}}{2^k k!}$ for all $k \geq 1$.

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Not sure if this is what you had in mind, but a bounded random variable (i.e. one which almost surely takes values in some bounded subset of $\mathbb R$) is uniquely determined by its moments.

Off the top of my head, one way to see this - probably not the most efficient! - is that the distribution of a random variable $X$ is uniquely determined by the set of values ${\mathbb E}f(X)$ as $f$ runs over all bounded continuous functions $\mathbb R\to \mathbb R$; and now since $X$ is a.s. bounded, the Weierstrass approximation theorem implies that $X$ is uniquely determined by the set of values $\mathbb E p(X)$ as $p$ runs over all polynomials.

Thus in some settings, knowing the moments is equivalent to knowing the distribution. (This point of view is much loved by those working in "free probability", but that's another story altogether...)

I should also mention that the question of just when we can say that the moment sequence determines the distribution has also been studied: see here for instance.

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The typical example I remember being given at school is that of kurtosis, which uses fourth moment and is easy to understand.

Wiki says: "Higher kurtosis means more of the variance is the result of infrequent extreme deviations, as opposed to frequent modestly sized deviations."

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  • $\begingroup$ The question asks about moments higher than fourth though... $\endgroup$ – j.c. Sep 20 '10 at 20:01
  • $\begingroup$ Ah. Indeed it does. My mistake: read this in a hurry, clearly. Will leave answer up unless anyone has serious objections. $\endgroup$ – Spencer Sep 20 '10 at 20:27

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