9
$\begingroup$

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?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
10
$\begingroup$

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.

Improve this answer
$\endgroup$
6
  • $\begingroup$ 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, 2015 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, 2015 at 0:08
  • 1
    $\begingroup$ 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? $\endgroup$
    – BigBendRegion
    Nov 11, 2017 at 22:53
  • $\begingroup$ is there a trade off in this case? $\endgroup$
    – Xiaoxiong Lin
    Apr 2, 2019 at 12:19
  • 1
    $\begingroup$ @filippo modelingwithdata.org/about_the_book.html $\endgroup$
    – Mark L. Stone
    Sep 30, 2019 at 11:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.