Definition of sample excess kurtosis?

Question

Wikipedia http://en.wikipedia.org/wiki/Kurtosis#Sample_kurtosis calculates the sample excess kurtosis as

length(x) * sum((x - mean(x))^4)/(sum((x - mean(x))^2)^2) - 3

On the other hand, I learned from https://stackoverflow.com/a/21484052/156458 that, the function kurtosis form R package fBasics (actually from timeDate) calculate the sample excess kurtosis as:

sum((x - mean(x))^4)/as.numeric(var(x))^2)/length(x) - 3

where var() is the unbiased, so it is equivalent to

(length(x)-1)^2/length(x) * sum((x - mean(x))^4)/(sum((x - mean(x))^2)^2) - 3

So I wonder if the function kurtosis in the packages fBasics and timeDate is wrong? Thanks!

Answer 1

The Wikipedia equation uses the biased, maximum likelihood estimator for the sample variance (divide by $n$), while, as you say the function from the R-packages uses the bias-corrected formula (divide by $n-1$).

The difference in the two formulas is therefore that the maximum likelihood formula of Wikipedia multiplies the moment quotient by $n$, while the R-function multiplies by

$$ \frac {(n-1)^2}{n} = \frac {n^2 -2n +1}{n} = n -2 + \frac 1n$$

The paper Joanes, D. N., & Gill, C. A. (1998). Comparing measures of sample skewness and kurtosis. Journal of the Royal Statistical Society: Series D (The Statistician), 47(1), 183-189.

discusses in some detail different formulas for sample kurtosis (and skewness). They write that the Wikipedia formula you mention - call it "$g_2$" is typical of "older times" (citing Cramer 1946 textbook). Also they report that (at least when they wrote the paper), MINITAB and BMDP were using the same formula as R/timeDate -call it "$b_2$", while SAS, SPSS and Excel were using a third formula for kurtosis (which you can find in the Wikipedia article under "estimation of population kurtosis"), call it "$G_2$".

They perform various tests and they find: 1) In small samples, the differences may be even a reversion in sign (i.e. one formula to give negative kurtosis the other positive...). 2) For normal populations we have in mean-squared error, $$MSE(g_2) < MSE (b_2) < MSE (G_2) $$ For sample size $100$ the MSE-differences are negligible (+1%) to small (+9%).

3) They also compare the three in MSE terms, for a chi-square for various degrees of freedom and various sample sizes. Here the main results are:
a) For sample size $100+$ the three equations are very close, irrespective of degrees of freedom.
b) For smaller sample sizes, the ranking changes, with $G_2$ being better for d.f=$1$, and $b_2$ being better when d.f. are $50$ or $100$.

Overall, no clear "winner" emerges from this paper.

Answer 2

Neither version is "wrong" any more than the fact that there is more than one sample standard deviation makes one of them wrong.

e.g. sample standard deviations include:

• the ML version ($s_n$),

• the variance-unbiased version ($s_{n-1}$),

• there are also the MMSE-for-variance, the unbiased-for-standard-deviation version, the approximately-unbiased-for-standard-deviation version (all usually taken at the normal, giving divisors of $n+1$ and $n-1.5$ for the first and last respectively and a complicated expression for the middle one, which can be worked out from the expression for $c_4$ at the link), and some others.

None of them are 'wrong' as definitions of sample standard deviation. They're all sample estimates, and they all may legitimately be called 'standard deviation'. One of them is more common, but that doesn't make it more correct.

The same applies to other sample moments and quantities based on them, including the kurtosis, the skewness, sample covariances, measures of sample correlation, and so on.