Does kurtosis measure center?

Question

Some have stated that kurtosis is the "movement of probability mass from the “shoulders” of a distribution into its center and tails" where "center" is defined as the range between $\mu \pm \sigma$. I was trying to find a proof for this but was unsuccessful. Can anyone point me to a proof of whether (i) larger kurtosis implies greater probability in the $\mu \pm \sigma$ range, or (ii) greater probability in the $\mu \pm \sigma$ range implies larger kurtosis? Does anyone have a proof of either of these conjectures?

Answer 1

Can anyone point me to a proof of whether (i) larger kurtosis implies greater probability in the μ±σ range, or (ii) greater probability in the μ±σ range implies larger kurtosis?

These are easily disproved by simple counterexamples. Let $P^*$ denote the "probability in the μ±σ range", i.e. $P^* =P[\mu-\sigma\le X\le\mu+\sigma]$, and let $\kappa$ denote the kurtosis.

Consider the following three symmetric distributions, each having $\mu=0$ and $P^* = P\left[X\in \{-1,1\}\right]$ (computations done in Sage): $$\begin{align}A:\quad P&=[.25,.25,.25,.25] \text{ on } X=[−2,−1,1,2]\implies\ P^*= 0.5;\ \kappa=1.36\\ B:\quad P&=[.30, .20, .20, .30] \text{ on } X=[−9,−1,1,9]\implies\ P^*= 0.4;\ \kappa=1.64\\ C:\quad P&=[.20,.30,.30,.20] \text{ on } X=[−2,−1,1,2]\implies\ P^*= 0.6;\ \kappa=1.45. \end{align}$$

Comparing A and B shows that it's neither the case that "(i) larger kurtosis implies greater probability in the μ±σ range", nor that "(ii) greater probability in the μ±σ range implies larger kurtosis". On the other hand, comparing A and C shows that it's sometimes the case that one increases or decreases when the other does likewise.

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Your question seems to be referring to an interpretation that the Wikipedia article attributes to J.J.A. Moors [1986], The Meaning of Kurtosis: Darlington Reexamined. I haven't read that paper, but virtually the same interpretation appeared almost forty years earlier in L. Guttman [1948], An Inequality for Kurtosis:

In this paper is developed a general inequality which describes the piling up of frequency around these two points for the case where the fourth moment exceeds the square of the variance. In a sense, it is shown how "U-shaped" a distribution must be according to its second and fourth moments.

[...]

In general, the smaller $\kappa-1$ is, the greater the probability that $\frac{X-\mu}{\sigma}$ is in a small interval around $+1$ or $-1$.

where I've substituted now-common symbols for Guttman's.

That might seem to suggest a general inverse relation between kurtosis and concentration near the two "shoulder points"; however, the statement is actually proved to hold only when the kurtosis ($\kappa$) is already sufficiently near (but not equal) to its minimum value of $1$. The proof goes as follows ...

If $X$ is a random variable with finite fourth moment and nonzero variance, say with mean $\mu$ and variance $\sigma^2$, then the kurtosis ($\kappa$) of the distribution can be expressed in terms of the standardized random variable $Z=\frac{X-\mu}{\sigma}$ as follows: $$\kappa=E[Z^4] = V[Z^2] + (E[Z^2])^2 = V[Z^2] + 1,\tag{1} $$ where we've used the fact that $Z$ must have mean $0$ and variance $1$ by virtue of the standardization; i.e., $$E[Z^2]=V[Z]+(E[Z])^2=V\left[\frac{X-\mu}{\sigma}\right]+\left(E\left[\frac{X-\mu}{\sigma}\right]\right)^2 = 1 + 0^2.$$

Note that $1\le \kappa\lt \infty,$ since $V[Z^2]\ge 0.$

By Chebyshev's inequality, for any $\epsilon>0$, $$P\left(\mid Z^2 - 1\mid \le \epsilon\right)>1- \frac{V[Z^2]}{\epsilon^2}=1-\frac{\kappa-1}{\epsilon^2}\tag{2} $$ so if $\kappa\to 1$, then for all $\epsilon>0$, $P\left(\mid Z^2 - 1\mid \le \epsilon\right)\to 1$.

Now, for any $0<\epsilon<1$, we have $$P\big{(}\mid Z^2 - 1\mid \le \epsilon\big{)}=P\left(\sqrt{1-\epsilon}\le |Z|\le \sqrt{1+\epsilon}\,\right)\\ =P\left(\{-\sqrt{1+\epsilon}\le Z\le -\sqrt{1-\epsilon}\,\} \cup \{\sqrt{1-\epsilon}\le Z\le \sqrt{1+\epsilon}\,\}\right)\\ =P\left(\underbrace{\{\mu-\sigma\sqrt{1+\epsilon} \le X\le \mu-\sigma\sqrt{1-\epsilon}\,\}}_{\text{an }\epsilon\text{-neighborhood of }\mu-\sigma} \cup \underbrace{\{\mu+\sigma\sqrt{1-\epsilon} \le X\le \mu+\sigma\sqrt{1+\epsilon}\,\}}_{\text{an }\epsilon\text{-neighborhood of }\mu+\sigma} \right).$$

Therefore, (2) implies that when $\kappa-1<\epsilon^2<1$ (i.e. when the kurtosis is already sufficiently small), it is the case that as $\kappa$ decreases, the distribution of $Z$ increasingly concentrates in small intervals around $\pm1,$ and the distribution of $X$ increasingly concentrates in small intervals around $\mu\pm\sigma$ (implying also that when $\kappa=1$, $Z\stackrel{\text{a.s.}}{=}\pm1$ and $X\stackrel{\text{a.s.}}{=}\mu\pm\sigma.$)

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NB: A "very large" kurtosis (in the sense of $\kappa\gg 1$) is driven by the occurrence of $|Z|>1$ (i.e., $|X-\mu|>\sigma$, which is to say, "$X$ outside the central $\mu\pm\sigma$ range"):

If $\kappa\gg 1 $, then (because $z^4\le 1$ when $|z|\le 1$ and $z^4\gt 1$ when $|z|\gt 1$) $$\begin{align}\kappa\ \ &=\ \ E[Z^4]\ \ =\ \ \underbrace{E[Z^4\,\big{|}\, |Z|>1]}_{\gg 1}\cdot \underbrace{P[|Z|>1]}_{<1}\ \ +\ \ \underbrace{E[Z^4\,\big{|}\, |Z|\le 1]}_{<1}\cdot \underbrace{P[|Z|\le 1]}_{<1}\\ \\ \ \ &\approx\ \ E[Z^4\,\big{|}\, |Z|>1]\cdot P[|Z|>1]. \end{align}$$ The only way to have $\kappa\gg 1$ is to have $E[Z^4\,\big{|}\, |Z|>1]\gg 1$, and vice versa, as all the other terms are less than $1$.