I've found two different bias-corrected estimators for the kurtosis.

The first one is used in various software packages, such as MATLAB, and is called bias-corrected in the respective documentations:

$k_0 = \frac{n-1}{(n-2) (n-3)}\left((n+1) \frac{\frac{1}{n} \sum\limits_{i=1}^n (x_i-\overline{x})^4}{\left(\frac{1}{n} \sum\limits_{i=1}^n (x_i-\overline{x})^2\right)^2}-3 (n-1)\right)+3.$

The second one can be found in Harald Cramér's book Mathematical Methods of Statistics on page 352 and is also the result of using the HStatistic function from the mathStatica package for Mathematica. It is said to be unbiased:

This image has been taken from another answer by wolfies: https://stats.stackexchange.com/a/307576/178794,

where $m_i$ is the sample central moment: $m_r = \frac{1}{n} \sum\limits_{i=1}^n (x_i - \overline{x})^r.$

Simulation Results

With both of these estimators being bias-corrected, the average of each of the kurtosis estimates from multiple independent trials should reflect the population kurtosis quite well. I've tried to test this assumption in Mathematica. For the first test, I've used the standard normal distribution and averaged the kurtosis estimates of 100,000 independent trials:

Mathematica simulation for calculating the average of multiple bias-corrected kurtosis estimates using samples from a standard normal distribution.

As expected, the average of the kurtosis estimates reflect the population kurtosis quite well. However, when performing the same test with another distribution (in this case the chi-squared distribution with one degree of freedom) the estimates are extremely off:

Mathematica simulation for calculating the average of multiple bias-corrected kurtosis estimates using samples from a chi-squared distribution.


Can someone provide an explanation why the averages of the bias-corrected kurtosis estimates ($k_0$ and $h_4$) are that off in my simulation? I'm especially surprised by the error of $h_4$, an estimate that is said to be unbiased.

  • $\begingroup$ As you have noted in your answer below, $h_4$ is an unbiased estimator of the 4th central moment ... not of population kurtosis. General unbiased estimators of ratios of moments do not usually exist ... but can be found for products of moments. This is because products of moments are symmetric functions, while ratios are not. $\endgroup$
    – wolfies
    Jan 24, 2019 at 18:16

1 Answer 1


I figured it out by myself.

Unfortunately, I made a mistake during my experiments: I overlooked the fact that the kurtosis is defined as

$\operatorname{Kurt}[X] = \frac{\operatorname{E}[(X-\mu)^4]}{(\operatorname{E}[(X-\mu)^2])^2},$

i.e., the fourth central moment divided by the square of the second central moment (notation taken from Wikipedia). In my original question, I erroneously mistook the unbiased estimator for the fourth central moment, $\frac{m_2^2 (9-6 n) n^2+m_4 \left(n^3-2 n^2+3 n\right) n}{(n-3) (n-2) (n-1) n}$, as an unbiased estimator for the kurtosis (and completely neglected the division by the squared second central moment).

Furthermore, there doesn't seem to be a general unbiased sample kurtosis estimator. The bias depends on the shape of the distribution. Here is a nice overview of different sample kurtosis estimators that also compares their relative performance.


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