# Is there an interpretation of the hyper skewness?

Let $X$ be a random variable. The standardized $n$th moment of $X$ is defined as $$\frac{E[(X-\mathbb{E}[X])^n]}{\mbox{Var}[X]^{n/2}}.$$ Special cases are the skewness ($k=3$) and the kurtosis $k=4$. The skewness is a measure for the asymmetry of a distribution while the kurtosis measures how peaked the distribution is. In my financial engineering project, I have to work with $k=5$ which is referred to as hyper skewness. As a benchmark, it is common to consider the normal distribution which has zero skewness and kurtosis equal to three. However, the hyper skewness of the normal distribution is also equal to zero, so at first sight it does not tell anything more about the distribution.

I was wondering if there is a useful interpretation of the hyper skewness? Does someone know any literature about this feature? If there is no any available literature then I can perhaps compute the hyper skewness for a variety of distributions and try to find an interpretation. However, some known literature would spare me some time.

• What might be more natural than the fifth central moment is the fifth cumulant: $\kappa_5 = \mu_5 - 10\mu_3 \mu_2$, where $\mu_k$ is the $k$th central moment. Like the fifth central moment, the fifth cumulant is fifth-degree homogeneous and translation-invariant, but it is also "cumulative", i.e. if $X_1,\ldots,X_n$ are independent random variables then $\kappa_5(X_1+\cdots+X_n) = \kappa_5(X_1) + \cdots + \kappa_5(X_n)$. $\qquad$ – Michael Hardy May 6 '16 at 15:39
• Thanks, I was aware of the cumulant function. How does that contribute to an actual interpretation of the hyper skewness? Ideally, I would obtain a similar interpretation as for the skewness/kurtosis. But perhaps that is not possible. – Cavents May 6 '16 at 17:32
• – Guilherme Thompson May 6 '16 at 18:00

Here's the method to construct an interpretation for anything. Think of relevant English words. Map Mathmatical ideas to English descriptions in a one-to-one fashion.

The expectation of a random variable is called $E(X)$.

English: $E(X)$ is the average value of $X$.

The variation from mean is called $(X-E(X))$.

English: $(X-E(X))$ is the difference between the random variable and its average value.

$X^n$ is power redistribution of the bariable.

English: $X^n$ compresses low values and stretches high values relative to $1$. More so, if $n$ is large. Less so if $n$ is small.

Put these translations together to get the "meaning" of $E((X-E(X))^n)$. The other part is just a constant.