It is well known that $\text{excess kurtosis} \geq \text{skew}^2 - 2$, at least in a population. However, what is the relationship between skew and excess kurtosis in a finite sample?

Define excess kurtosis as:

$\gamma_2= \mu_4/\sigma^4 - 3$, so that the excess kurtosis of a normal distribution = 0.



(Kudos to @Silverfish for originally raising this issue in a comment)

  • 3
    $\begingroup$ Your question relates to relations between sample quantities, so you should give sample definitions of the quantities you ask about; those are the ones for which multiple definitions exist, and those definitions will affect the answer. It may actually be (a bit) easier to work with ordinary kurtosis rather than excess kurtosis. $\endgroup$
    – Glen_b
    Commented Jan 13, 2015 at 17:23
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    $\begingroup$ The relationship for ordinary kurtosis is $\beta_2 \geq \beta_1+1$ (where $\beta_1=\gamma_1^2$ is the square of skewness). $\endgroup$
    – Glen_b
    Commented Jan 13, 2015 at 17:31

2 Answers 2


A discussion on the limits of the sample skewness and kurtosis is available here. The author gives proper references to the original proofs, and the cited results are: $$ |g_1| \le \frac{n-2}{\sqrt{n-1}} = \sqrt{n-1} - \frac{1}{\sqrt{n-1}} $$ $$ b_2 = g_2 + 3 \le \frac{n^2-3n+3}{n-1} = n -2 + \frac1{n-1} $$ So for $n=10$, you can't have skewness greater than 2.89, and excess kurtosis, greater than 5.11.


Some broad discussion on how to understand the problem in the absence of sample definitions.

Since the quoted relationship applies to distributions, if you treat the ecdf as the cdf of a distribution, and apply those population definitions you gave, the relationship must still hold. That is, if you use $n$ denominators on all the averages in the sample definitions (including the calculation of $\hat{\sigma}^2$), so that they're expected values on that distribution, the relationship should be what you stated.

So, by defining your central sample moments all as $m_k=\frac{1}{n}\sum_i (x_i-\bar{x})^k$, you must get the same result you quoted; no additional algebra is required.

If you subsequently want to use different definitions, by writing the new ones as functions of the old ones just mentioned (pulling out scaling terms for any non-$n$ denominators), you should be able to then derive the relationships you seek (which should still asymptotically go to the relationship you mention)

So, for example, if you use the sample definition here:

$g_2 = \frac{m_4}{m_2^2}-3\,$,

and the equivalent for skewness,

$g_1 = \frac{m_3}{m_2^{3/2}}\,$,

the proof that established the population relationship will still apply.

If instead you used the definition for sample skewness here (note that this would leave you with inconsistent definitions of the variance estimates!), then you can simply write

$b_1 = g_1 \frac{m_2^{3/2}}{s^3} = g_1 (\frac{n-1}{n})^{3/2}$

and then use the relationship you quoted to derive one between $g_2$ and $b_1$. And so on for other definitions (you might like to try it with $G_1$ mentioned in the wikipedia article on skewness, for example).


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