# Unbiased estimators of skewness and kurtosis

The skewness and kurtosis are defined as: $$\zeta_3 = \frac{E[(X-\mu)^3]}{E[(X-\mu)^2]^{3/2}} = \frac{\mu_3}{\sigma^3}$$ $$\zeta_4 = \frac{E[(X-\mu)^4]}{E[(X-\mu)^2]^2} = \frac{\mu_4}{\sigma^4}$$

The following formulae are used to calculate sample skewness and kurtosis: $$z_3 = \frac{\frac{1}{n}\sum_{i=1}^{n} [(x_i-\bar x)^3]}{(\frac{1}{n}\sum_{i=1}^{n}[(x_i-\bar x)^2])^{3/2}}$$ $$z_4 = \frac{\frac{1}{n}\sum_{i=1}^{n} [(x_i-\bar x)^4]}{(\frac{1}{n}\sum_{i=1}^{n}[(x_i-\bar x)^2])^2}$$

My question is: are these estimators unbiased? I don't know whether I should use unbiased standard deviation or the biased one in the denominator.

In general, if we have a function $f$ whose variables are unbiased estimators, then can we say $f$ is an unbiased estimator as well?

Note that what you are probably calling the unbiased standard deviation is a biased estimator of standard deviation Why is sample standard deviation a biased estimator of $\sigma$? , although before taking the square root it is an unbiased estimator of variance.
• Thanks! It seems like the formulae of "good" estimators are very lengthy. If I use other simpler ones, does that really cause any serious problems? Btw, I've always misconceived that sample std is an UNBIASED estimator of $\sigma$, this also answers my second question. Jun 21 '15 at 4:17