# How is the kurtosis of a distribution related to the geometry of the density function?

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?

• 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 '14 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 '14 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 '14 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. Feb 3 '14 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 '14 at 13:31

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. 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.

• "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). Feb 2 '14 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 '14 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. Feb 2 '14 at 23:20
• @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 '14 at 23:22

[NB this was written in response to another question on site; the answers were merged to the present question. This is why this answer seems to respond to a differently worded question. However much of the post should be relevant here.]

Kurtosis doesn't really measure the shape of distributions. Within some distribution families perhaps, you can say it describes the shape, but more generally kurtosis doesn't tell you terribly much about the actual shape. Shape is impacted by many things, including things unrelated to kurtosis.

If one does image searches for kurtosis, quite a few images like this one show up: which instead seem to be showing changing variance, rather than increasing kurtosis. For comparison, here's three normal densities I just drew (using R) with different standard deviations: As you can see, it looks almost identical to the previous picture. These all have exactly the same kurtosis. By contrast, here's an example that is probably nearer to what the diagram was aiming for The green curve is both more peaked and heavier tailed (though this display isn't well suited to seeing how much heavier the tail actually is). The blue curve is less peaked and has very light tails (indeed it has no tails at all beyond $\sqrt{6}$ standard deviations from the mean).

This is usually what people mean when they talk about kurtosis indicating the shape of the density. However, kurtosis can be subtle -- it doesn't have to work like that.

For example, at a given variance higher kurtosis can actually occur with a lower peak.

One must also beware the temptation (and in quite a few books it's openly stated) that zero excess kurtosis implies normality. There are distributions with excess kurtosis 0 that are nothing like normal. Here's an example: Indeed, that also illustrates the previous point. I could readily construct a similar-looking distribution with higher kurtosis than the normal but which is still zero at the center - a complete absence of peak.

There are a number of posts on site that describe kurtosis further. One example is here.

• But I did not say it? The book says it? Mar 26 '15 at 12:33
• I know that. I never said you said it. How would you suggest I respond to blatantly incorrect statements that you ask about? Just pretend they're not wrong? Mar 26 '15 at 12:35
• @Glen_b The pictures are not from the book. The book does not give illustrations. I used the goolge picture search for these illustrations. Mar 26 '15 at 12:44
• Some authors write of kurtosis as peakedness and some write of it as tail weight, but the skeptical interpretation that kurtosis is whatever kurtosis measures is the only totally safe story. Numerical examples given by Irving Kaplansky (1945) alone suffice to show that kurtosis bears neither interpretation unequivocally. (Kaplansky’s paper is one of a few that he wrote in the mid-1940s on probability and statistics. He is much better known as a distinguished algebraist.) Full reference and more within stata-journal.com/sjpdf.html?articlenum=st0204 Mar 26 '15 at 12:49
• There are books and papers claiming that kurtosis is peakedness, so my first clause remains correct as well as supportable as a statement on what is in the literature. What’s more crucial is how one regards Kaplansky’s examples and arguments. Nov 23 '18 at 23:57

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).

In a nutshell, 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.

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

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

• Everywhere you write "bimodal" do you perhaps mean "unimodal"?
– whuber
Feb 2 '14 at 23:06
• 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 '14 at 19:09
• – whuber
Feb 3 '14 at 19:23
• Kurtosis does not indicate bimodality, except in the extreme case where it is near its minimum, where it indicates something similar to the two-point equiprobable distribution. You can have bimodal distributions with every possible value of kurtosis. See ncbi.nlm.nih.gov/pmc/articles/PMC4321753 for examples. Oct 29 '18 at 12:20
• Yes; see the paper I linked. The first sentence of DeCarlo's abstract is absolutely wrong. If you don't want to read my paper, here is the math: take any symmetric bimodal distribution and mix it with a much wider symmetric distribution having the same median as the bimodal. The mixture is symmetric and bimodal for small $p$. By tweaking the wider distribution and the mixing $p$, you can make the kurtosis range to infinity. And you can get kurtosis as small as you want by using .5N(-1,v) + .5N(1,v), letting $v \rightarrow 0$. Symmetric and bimodal pdfs are easy to construct for all kurtosis. Oct 29 '18 at 14:07

A different kind of answer: We can illustrate kurtosis geometrically, using ideas from http://www.quantdec.com/envstats/notes/class_06/properties.htm: graphical moments.

Start with the definition of kurtosis: $$\DeclareMathOperator{\E}{\mathbb{E}} k = \E \left( \frac{X-\mu}{\sigma} \right)^4 =\int \left(\frac{x-\mu}{\sigma}\right)^4 f(x) \; dx$$ where $$f$$ is the density of $$X$$, $$\mu, \sigma^2$$ respectively expectation and variance. The nonnegative function under the integral sign integrates to the kurtosis, and gives contribution to kurtosis from around $$x$$. We can call it the kurtosis density, and plotting it shows the kurtosis graphically. (Note that in this post we are not using the excess kurtosis $$k_e=k-3$$ at all).

In the following I will show a plot of graphical kurtosis for some symmetric distributions, all centered at zero and scaled to have variance 1. Note the virtual absence of contribution to the kurtosis from the center, showing that kurtosis does not have much to do with "peakedness".

• Exactly right. The areas under the curves you show from -1 to +1 are between 0 and 1 for all distributions, and are between 0 and 0.5 for all continuous distributions for which the density of $Z^2$ is decreasing on the [0,1] range. Those two theorems are proven in my paper ncbi.nlm.nih.gov/pmc/articles/PMC4321753 . A third theorem proven therein is that, for any sequence of distributions for which kurtosis tends to infinity, the ratio of the area under the range from $-b$ to $+b$ to kurtosis tends to zero, for every fixed $b$. Oct 29 '18 at 12:13

Kurtosis is not related to the geometry of the distribution at all, at least not in the central portion of the distribution. In the central portion of the distribution (within the $$\mu \pm \sigma$$ range) the geometry can show an infinite peak, a flat peak, or bimodal peaks, both in cases where the kurtosis is infinite, and in cases where the kurtosis is less than that of the normal distribution. Kurtosis measures tail behavior (outliers) only. See https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4321753/

Edit 11/23/2018: Since writing this post, I have developed some geometrical perspectives on kurtosis. One is that excess kurtosis can indeed be visualized geometrically in terms of deviations from the expected 45 degree line in the tails of the normal quantile-quantile plot; see Does this Q-Q plot indicates leptokurtic or platykurtic distribution?

Another (perhaps more physical than geometrical) interpretation of kurtosis is that kurtosis can be visualized as the point of balance of the distribution $$p_V(v)$$, where $$V = \{(X - \mu)/\sigma \}^4$$. Note that (non-excess) kurtosis of $$X$$ is equal to $$E(V)$$. Thus, the distribution of $$V$$ balances at the kurtosis of $$X$$.

Another result that shows that geometry in the $$\mu \pm \sigma$$ range is nearly irrelevant to kurtosis is given as follows. Consider the pdf of any RV $$X$$ having finite fourth moment. (Thus the result applies to all empirical distributions.) Replace the mass (or geometry) within the $$\mu \pm \sigma$$ range arbitrarily to get a new distribution, but keep the mean and standard deviation of the resulting distribution equal to $$\mu$$ and $$\sigma$$ of the original $$X$$. Then the maximum difference in kurtosis for all such replacements is $$\le 0.25$$. On the other hand, if you replace the mass outside the $$\mu \pm \sigma$$ range, keeping the center mass as well as $$\mu$$, $$\sigma$$ fixed, the difference in kurtosis is unbounded for all such replacements.

• Rather than just continuing to refer people to a paper in most of your posts, would you mind summarizing the arguments here? See the help here under "always provide context for links", in particular where it says "always quote the important part". It's not necessarily to literally quote it where the argument is extensive, but at least a summary of the argument is needed. You just make a couple of sweeping statements and then link to a paper. The statement that kurtosis measures tail behavior is (absent context) misleading (demonstrably so) Apr 2 '17 at 0:20
• ... but it's impossible to disagree with arguments you don't present here, and perhaps arrive at a more nuanced conclusion. Apr 2 '17 at 0:39
• My arguments are clearly laid out here: en.wikipedia.org/wiki/… Comments welcome! BTW, kurtosis IS a measure of tail weight, just not the same as others that have been considered. It measures tail weight via E(Z^4), which is a measure of tail weight since the values |Z|<1 contribute so little to it. By the same logic, E(Z^n), for higher even powers n, are also tail weight measures. Oct 30 '17 at 22:37
• Hi Peter, Please visit stats.stackexchange.com/help/merging-accounts to merge your accounts so that you can modify your old posts.
– whuber
Nov 23 '18 at 21:31