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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?

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See pp. 8-9 of http://modelingwithdata.org/pdfs/moments.pdf . Also look at http://www.amstat.org/publications/jse/v19n2/doane.pdf for some useful perspectives to get your thinking in the right frame of mind.

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.

A nonlinear function of an unbiased estimator is not necessarily going to be unbiased ("almost surely" won't be). The direction of the bias can be determined by Jensen's Inequality https://en.wikipedia.org/wiki/Jensen%27s_inequality if the function is convex or concave.

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  • $\begingroup$ 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. $\endgroup$ – SiXUlm Jun 21 '15 at 4:17
  • $\begingroup$ You have to decide if you want the best answer you can get, or not. if you want the best answer, then pay the price in complication if needed. $\endgroup$ – Mark L. Stone Jun 22 '15 at 0:08
  • $\begingroup$ Bias is not necessarily bad. You have to consider variance as well. The closeness of the estimator to the estimand can be measured using expected squared deviation from estimator to estimand, which is equal to variance of the estimator plus squared bias of the estimator. In many cases there is a "variance bias trade-off" where the increase in bias is more than offset by the reduction in variance. I would bet that this is true for the estimates of kurtosis and skewness. Someone want to post some research on this? $\endgroup$ – Peter Westfall Nov 11 '17 at 22:53
  • $\begingroup$ is there a trade off in this case? $\endgroup$ – Xiaoxiong Lin Apr 2 at 12:19
  • $\begingroup$ @Xiaoxiong Lin I believe there is a trade off in this case, but have not carefully studied that tradeoff specifically for skewness and kurtosis. $\endgroup$ – Mark L. Stone Apr 2 at 12:37

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