Skewness of lognormal distribution Does anyone know how to derive the equation of the skewness of a lognormal distribution?
$$\left(\exp(\sigma^2)+2\right)\sqrt{\exp(\sigma^2)-1}$$
 A: 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

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


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