Unbiased estimators of skewness and kurtosis

Question

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?

Answer 1

See pp. 8-9 of these notes from Klemens, B. (2008). Modeling with data: tools and techniques for scientific computing. Also look at Doane, D. P., & Seward, L. E. (2011). Measuring skewness: a forgotten statistic?. Journal of statistics education, 19(2). 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 if the function is convex or concave.

Answer 2

are these estimators unbiased?

Quite evidently, they are not. In fact, you can easily choose an underlying distribution $F$ of $X$ such that $E[z_3] \neq \zeta_3$ and/or $E[z_4] \neq \zeta_4$ when the sample size $n$ is finite.

For example, fix $n = 2$, let $X$ be a discrete random variable following the probability mass function $P(X = -1) = 0.4$, $P(X = 1/2) = 0.4$, $P(X = 1) = 0.2$. Let $X_1, X_2 \text{ i.i.d.} \sim X$, it then follows that \begin{align*} z_3 = 0, \quad z_4 = 1. \end{align*} On the other hand, since \begin{align*} & \mu = E[X] = -0.4 + 0.2 + 0.2 = 0, \\ & \sigma^2 = E[X^2] = 0.4 + 0.1 + 0.2 = 0.7, \\ & \mu_3 = E[X^3] = -0.4 + 0.05 + 0.2 = -0.15, \\ & \mu_4 = E[X^4] = 0.4 + 0.025 + 0.2 = 0.625, \end{align*} we have \begin{align*} & \zeta_3 = \frac{\mu_3}{\sigma^{3/2}} = \frac{-0.15}{0.586} = -0.256, \\ & \zeta_4 = \frac{\mu_4}{\sigma^{2}} = \frac{0.625}{0.49} = 1.276, \\ \end{align*}

Therefore, $E[z_3] \neq \zeta_3$, $E[z_4] \neq \zeta_4$.

By this example, it can be seen that replacing the denominator in the definition of $z_3$ and $z_4$ with the bias-corrected standard deviation wouldn't help either.

One thing to note though, for any $F$, provided that $X_1, \ldots, X_n$ are i.i.d. draws from $F$, it can be shown that $z_3$ and $z_4$ converges to $\zeta_3$ and $\zeta_4$ with probability 1 as $n \to \infty$. This is just an easy application of SLLN: for example, by SLLN, \begin{align*} & \frac{1}{n}\sum_{i = 1}^n(X_i - \bar{X})^3 \\ =& \frac{1}{n}\sum_{i = 1}^n (X_i - \mu)^3 - \frac{3}{n}\sum_{i = 1}^n(X_i - \mu)^2(\mu - \bar{X}) \\ & + \frac{3}{n}\sum_{i = 1}^n(X_i - \mu)(\mu - \bar{X})^2 + (\mu - \bar{X})^3 \\ \to & E[(X - \mu)^3] \quad\text{ with probability 1}, \end{align*} whence $z_3 \to \zeta_3$ with probability $1$. Similarly, one can show that $z_4 \to \zeta_4$ with probability $1$. One caveat is, this result is not equivalent to $E[z_k] \to \zeta_k$, $k = 3, 4$, because almost surely convergence does not guarantee convergence by moments. That is, we still cannot claim $z_k$ are asymptotically unbiased unless we impose more conditions such as uniform integrability.

In summary, we can say that $z_k$ are consistent estimators of $\zeta_k$, but in general they are not unbiased, or even asymptotically unbiased estimators of $\zeta_k$.