Skewness of lognormal distribution

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

$$\left(\exp(\sigma^2)+2\right)\sqrt{\exp(\sigma^2)-1}$$

• 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. – chl Nov 27 '20 at 11:18
• 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 – Christoph Hanck Nov 27 '20 at 11:37
• 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 – Nick Cox Nov 27 '20 at 11:56
• +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. – 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

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

and

\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

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

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,

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