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The kurtosis is to measure the peakedness and flatness of a distribution. The density function of the distribution, if it exists, can be viewed as a curve, and has geometric features (such as curvature, convexity, ...) related to its shape.

So I wonder whether the kurtosis of a distribution is related to some geometric features of the density function, which can explain the geometric meaning of kurtosis?

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I am asking for some relation in formula to some geometric quantity of the density curve, not just the vague meaning I pointed out in my post. Or it is fine to just have some explanation of why kurtosis has the geometric meaning –  Tim Feb 2 at 13:37
    
@Peter That is far from the truth. One can modify the geometry of the graph of the PDF almost arbitrarily without changing any specified (finite number of its) moments. –  whuber Feb 2 at 19:20
    
The closely related question at stats.stackexchange.com/questions/25010/… suggests what the right answer to this question should be. –  whuber Feb 2 at 23:07
    
@whuber while I agree and thank you for that example, I also wonder whether it doesn't says more about the remarkable property of that particular family of pdf than it does about kurtosis in general. –  user603 Feb 3 at 0:09
    
@user603 That's a good thing to wonder. However, the statement is not about this particular family: it just happens that for the lognormal distribution one can produce an explicit representation of a class of alternative PDFs with the same moments. It is special that all of the moments are the same, but perturbing most distributions in a way that fixes a finite number of their moments is not hard. (It's hard for certain discrete distributions, such as the Bernoulli, but they don't have PDFs.) –  whuber Feb 3 at 13:31
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2 Answers

The moments of a continuous distribution, and functions of them like the kurtosis, tell you extremely little about the graph of its density function.

Consider, for instance, the following graphs.

enter image description here

Each of these is the graph of a non-negative function integrating to $1$: they are all PDFs. Moreover, they all have exactly the same moments--every last infinite number of them. Thus they share a common kurtosis (which happens to equal $-3+3 e^2+2 e^3+e^4$.)

The formulas for these functions are

$$f_{k,s}(x) = \frac{1}{\sqrt{2\pi}x} \exp\left(-\frac{1}{2}(\log(x))^2\right)\left(1 + s\sin(2 k \pi \log(x)\right)$$

for $x \gt 0,$ $-1\le s\le 1,$ and $k\in\mathbb{Z}.$

The figure displays values of $s$ at the left and values of $k$ across the top. The left-hand column shows the PDF for the standard lognormal distribution.

Exercise 6.21 in Kendall's Advanced Theory of Statistics (Stuart & Ord, 5th edition) asks the reader to show that these all have the same moments.

One can similarly modify any pdf to create another pdf of radically different shape but with the same second and fourth central moments (say), which therefore would have the same kurtosis. From this example alone it should be abundantly clear that kurtosis is not an easily interpretable or intuitive measure of symmetry, unimodality, bimodality, convexity, or any other familiar geometric characterization of a curve.

Functions of moments, therefore (and kurtosis as a special case) do not describe geometric properties of the graph of the pdf. This intuitively makes sense: because a pdf represents probability by means of area, we can almost freely shift probability density around from one location to another, radically changing the appearance of the pdf, while fixing any finite number of pre-specified moments.

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"From this example alone it should be abundantly clear... any other familiar geometric characterization of a curve." I understand what you mean, but there is ground for reasonable divergence in the interpretation here. Another interpretation is that of Darlington, who shows how starting from a symmetric distribution, moving some mass at specific points increases/decrease the kurtosis (again, not a contradiction of your example, just a more 'positive' understanding). –  user603 Feb 2 at 23:11
    
@user603 I don't disagree, but I think that the "positive" approach overlooks the very special assumptions that are implicitly made in order for it to work at all. One could also begin with the graph of an extremely asymmetric PDF whose skewness is zero (they are not hard to construct). Thus that positive approach merely describes what happens to certain very special PDFs when the mass is moved around. Although that can be quite useful for intuition, it seems to have no logical bearing on the present question. –  whuber Feb 2 at 23:18
    
I agree for the skewness (and for your answer in general). But the kurtosis, as a function, has a minimum. That makes things slightly more interesting. –  user603 Feb 2 at 23:20
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@user603 Thank you; that's an insightful distinction. I don't think it changes any of the present conclusions in important ways but it certainly helps the intuition and points to an important difference between even and odd moments. –  whuber Feb 2 at 23:22
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For symmetric distributions (that is those for which the even centred moments are meaningful) kurtosis measures a geometric feature of the underlying pdf. It is not true that kurtosis measures (or is in general related) to the peakedness of a distribution. Rather, kurtosis measure how far the underlying distribution is from being symmetric and bimodal (algebraically, a perfectly symmetric and bimodal distribution will have a kurtosis of 1, which is the smallest possible value the kurtosis can have)[0].

In a nutshell[1], if you define:

$$k=E(x-\mu)^4/\sigma^4$$

with $E(X)=\mu,V(X)=\sigma^2$, then

$$k=V(Z^2)+1\ge1$$

for $Z=(X-\mu)/\sigma$.

This implies that $k$ can be seen as a measure of dispersion of $Z^2$ around its expectation 1. In other words, if you have a geometrical interpretation of the variance and the expectation, than that of the kurtosis follows.

[0] R. B. Darlington (1970). Is Kurtosis Really "Peakedness?". The American Statistician , Vol. 24, No. 2.

[1] J. J. A. Moors (1986).The Meaning of Kurtosis: Darlington Reexamined. The American Statistician, Volume 40, Issue 4.

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Everywhere you write "bimodal" do you perhaps mean "unimodal"? –  whuber Feb 2 at 23:06
    
@whuber: no. See [0] "the smaller the khurtosis, the greater the bimodality". A bimodal distribution where the modes are at $\mu\pm\sigma$ will have kurtosis of 1. –  user603 Feb 2 at 23:18
    
I guess I don't follow, because kurtosis has no essential relationship to bimodality or modes in general. Indeed, for any bimodal PDF of given kurtosis there are other PDFs with the same kurtosis having arbitrary numbers of modes, so kurtosis cannot possibly be interpreted as saying anything about modes. –  whuber Feb 2 at 23:20
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Yes, these examples work for symmetric distributions. An explicit one could be constructed from the pseudo-lognormal families: take one of those (infinitely modal) pdfs $f$ with a mean of $\mu$ and define a new pdf as $g(x) = (f(x)+f(2\mu-x))/2.$ By mixing in a small amount of $g$ with a minimum-kurtosis distribution you find that there are distributions with infinitely many modes whose kurtosis is arbitrarily close to the minimum value of $1$. Thus, at least, kurtosis says nothing whatsoever about bimodality. Since it does not, precisely what geometric property of the pdf is it describing? –  whuber Feb 3 at 19:09
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let us continue this discussion in chat –  whuber Feb 3 at 19:23
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