# Are there normalized equivalents to Skewness and Kurtosis?

What would be the normalized equivalent to Skewness that would have the same unit as the data? Similarly, what would be the normalized equivalent to Kurtosis? Ideally, these functions should be linear with respect to the data, meaning that if all observations were to be multiplied by a factor n, the resulting normalized skewness and kurtosis would be multiplied by the same factor n. The benefit of having such normalized equivalents would be to be able to overlay them on top of a standard box-and-whisker plot.

• What a fun question! – Alexis Dec 7 '14 at 4:07
• I'm not sure how enlightening it would be to illustrate these on graphs. The reason we illustrate standard deviations is that they give a natural measure of the dispersion of the data (if it's normally distributed): 65% of the observations lie inside the interval. I don't think there are such natural visual interpretations for the third and fourth moments. – Ben Kuhn Dec 7 '14 at 4:17
• What are you trying to show about your data? If it's a certain qualitative behavior of the distribution, might a violin plot be preferable? But yes, anyway, it's a fun question. – Ben Kuhn Dec 7 '14 at 4:18
• One can get a sense of skewness and kurtosis by looking at a histogram showing the distribution of one's dataset, but it will give a very subjective perception of these measures. I would like to depict them on two linear scales, one for skewness parallel to the axis of the box-and-whisker plot, the other orthogonal to it. This could be depicted as a separate box overlaid on top of the primary box. The taller that box, the more skewed the data is. The wider, the more pointy (high kurtosis). – Ismael Ghalimi Dec 7 '14 at 4:30
• And thanks for the link to the violon plot. It's really clever. – Ismael Ghalimi Dec 7 '14 at 4:34

Skewness measures are deliberately unitless.

The usual moment-skewness is a standardized third moment, $E[(\frac{X-\mu}{\sigma})^3]$.

If you center but don't standardize, you have $\mu_3=E[(X-\mu)^3]$... which is plainly then in cubed units.

If you wanted something in the same units as $X$, you'd have to take the cube-root, in the same way that we take square root of variance and get something in the same units of the original data. (However - beware, because many packages won't take cube roots of negative numbers, you might have to compute it as: $\quad\text{sign}(X-\mu)\times |E(X-\mu)^3|^{1/3}\:$.)

I'm not sure how useful that will be.

For some other skewness measures, like the two Pearson skewness measures, you just multiply by $\sigma$.

For sample skewness measures where $\sigma$ and $\mu$ are generally not known, as with sample skewness, you'd typically replace them by their own sample estimates.

Kurtosis follows the same pattern - for moment kurtosis, you'd need to take fourth roots of the unstandardized fourth moment to get something that scaled with the data.

For some of the other measures of kurtosis, they would only need to be multiplied by $\sigma$.

Skewness and kurtosis are shape characteristics. So, if I tell you that the thing, a ball, is round it shouldn't matter what is a radius of the thing. It can be a small ball or a big ball. On the other hand, when I say small ball or a large cube I'm referring to the size of the object, not the shape.

In this regard, the standard deviation is the size of the distribution, that's why skewness and kurtosis are normalized by size. You could also say that standard deviation belongs to mechanics, and skewness and kurtosis to geometry. Therefore, no, we don't need to have them in units of measure of the variable. Size and shape are separate. A large and a small ball are equally round, i.e. size doesn't matter in this case :)

Denoting vectors which is distributed in region $$R$$, let's assume the zeroth and the first moment is already normalized. The second moment is calculated with $$M_2 = \int_R {x x^T} |dx|$$, so if we can find diagonalization $$M_2 = P\Lambda^2 P^T$$, then we will be able to define $$x'=\Lambda^{-1} P^Tx$$ so that $$M_2$$ is normalized:

$$M'_{2ij} = \int_R {(\Lambda^{-1}P^T x)(\Lambda^{-1} P^T x)^T} |dx|$$ $$= \Lambda^{-1} P^T (\int_R { xx^T |dx|}) P \Lambda^{-1}$$ $$= \Lambda^{-1} P^T P \Lambda^2 P^T P \Lambda^{-1} = I$$

Geometrical meaning of the second moment is "orientation", which is justified by the fact that diagonalization normalizes the second moment. When skewness is calculated under this normalization, it is called Mardia's skewness.