Does anyone know how to derive the equation of the skewness of a lognormal distribution?


  • $\begingroup$ Welcome on this site. You can use LaTeX to format your post. Also, please note that if this is homework, you should probably add the self-study tag and review its wiki. $\endgroup$ – chl Nov 27 '20 at 11:18
  • $\begingroup$ That may depend on the principles from which you are asked to derive the result, but you can combine the definition of Pearson's skewness coefficient (en.wikipedia.org/wiki/Skewness) with the moments of the log-normal reported at, e.g., en.wikipedia.org/wiki/Log-normal_distribution $\endgroup$ – Christoph Hanck Nov 27 '20 at 11:37
  • 2
    $\begingroup$ Watch out: the skewness of a lognormal can be extraordinarily large in principle but on the other hand sample skewness is bounded as a function of sample size. As a result samples from a lognormal can deny their parentage. This possibly cryptic remark is made more concrete in stata-journal.com/article.html?article=st0204 $\endgroup$ – Nick Cox Nov 27 '20 at 11:56
  • $\begingroup$ +1 Although moments of lognormal distribution have been mentioned in many posts here on CV, I haven't located one that shows how the moments can be found. $\endgroup$ – whuber Nov 27 '20 at 16:56

When a variable $X$ has a Normal distribution with mean $\mu$ and standard deviation $\sigma \gt 0,$ we say that $Z=e^X$ has a Lognormal$(\mu,\sigma)$ distribution.

The laws of logarithms show that $\mu$ (an additive location parameter for the Normal family of distributions) determines the scale of $Z.$ Because the skewness of a variable does not depend on its scale, we may take $\mu$ to be any convenient value. Choosing $\mu=0,$ use the Normal density (which is proportional to the exponential of $-x^2/(2\sigma^2)$) to compute the (raw) $k^\text{th}$ moment of $Z$ via the substitution $y = x - k\sigma^2:$

$$\begin{aligned} \mu_k(\sigma) &=E\left[Z^k\right] = E\left[\exp(X)^k\right] = E\left[\exp(kX)\right]\\\ &= \frac{1}{\sigma\sqrt{2\pi}}\int_{\mathbb{R}} \exp\left(\frac{1}{2\sigma^2}x^2 + kx\right)\,\mathrm{d}x\\ &= \frac{1}{\sigma\sqrt{2\pi}}\exp\left(k^2\sigma^2/2\right)\int_{\mathbb{R}} \exp\left(\frac{1}{2\sigma^2}x^2 + kx - k^2\sigma^2/2\right)\,\mathrm{d}x\\ &= \frac{1}{\sigma\sqrt{2\pi}}\exp\left(k^2\sigma^2/2\right)\int_{\mathbb{R}} \exp\left(\frac{1}{2\sigma^2}\left[x - k\sigma^2\right]^2\right)\,\mathrm{d}x\\ &= \exp\left(k^2\sigma^2/2\right)\left[\frac{1}{\sigma\sqrt{2\pi}}\int_{\mathbb{R}} \exp\left(\frac{1}{2\sigma^2}y^2\right)\,\mathrm{d}y\right]\\ &= \exp\left(k^2\sigma^2/2\right). \end{aligned}\tag{*}$$

For $k=1$ this shows the mean is $\exp(\sigma^2/2)$ and from this we may compute the central moments from the Binomial Theorem as

$$\begin{aligned} \mu^\prime_k(\sigma) &= E\left[(Z - E[Z])^k\right] = E\left[\sum_{i=0}^k \binom{k}{i} Z^i E(Z)^{k-i}\right] \\ &= \sum_{i=0}^k \binom{k}{i}(-1)^{i-k} \mu_i(\sigma) \mu_1(\sigma)^{k-i} \end{aligned}.\tag{**}$$

Applying this to $k=2,3$ gives

$$\mu^\prime_2(\sigma) = \mu_0(\sigma)\mu_1(\sigma))^2 - 2\mu_1(\sigma)\mu_1(\sigma) + \mu_2(\sigma) = e^{\sigma^2}\left(e^{\sigma^2}-1\right)$$


$$\begin{aligned}\mu^\prime_3(\sigma) &= -\mu_0(\sigma)\mu_1(\sigma)^3 + 3\mu_1(\sigma)\mu_1(\sigma)^2 - 3\mu_2(\sigma)\mu_1(\sigma) + \mu_3(\sigma) \\ &= e^{3\sigma^2/2}\left(2 - 3 e^{\sigma^2} + e^{3\sigma^2}\right) \\ &= e^{3\sigma^2/2}\left(e^{\sigma^2}+2\right)\left(e^{\sigma^2}-1\right)^2. \end{aligned}$$

By definition, the skewness is

$$\operatorname{Skew}(Z) = \frac{\mu^\prime_3(\sigma)}{\mu^\prime_2(\sigma)^{3/2}} = \frac{e^{3\sigma^2/2}\left(e^{\sigma^2}+2\right)\left(e^{\sigma^2}-1\right)^2}{\left[e^{\sigma^2}\left(e^{\sigma^2}-1\right)\right]^{3/2}} = \left(e^{\sigma^2}+2\right)\sqrt{e^{\sigma^2}-1}.$$

Comments and Generalizations

Higher standardized central moments (e.g. the kurtosis) are readily computed in the same way: $(*)$ and $(**)$ reduce the problem to polynomial algebra (the variable is $\exp(\sigma^2/2)$).

Because $\mu$ is a scale parameter for the Lognormal family (corresponding to a scale factor of $e^\mu$), it can be introduced into the formulas $(*)$ directly, where its $k^\text{th}$ power $\left(e^\mu\right)^k = e^{k\mu}$ will multiply the result, giving the general formulas

$$\mu_k(\mu,\sigma) = E\left[Z^k\right] = \exp\left(k\mu + k^2\sigma^2\right)$$

and then, of course,

$$\mu^\prime_k(\mu,\sigma) = E\left[\right(Z - E[Z]\left)^k\right] = e^{k\mu} \mu^\prime_k(\sigma).$$


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