The Lognormal Distribution

If $Y$ is a log-normal random variable then $\ln Y$ is normally distributed.
$Y = e^{X}$ where $X \sim N(\mu , \sigma^2)$
We write $Y \sim \ln N(\mu, \sigma^2)$

$$\begin{align} P(Y < y) = P(X < \ln y ) = \int_{-\infty}^{\ln y} p(x) dx \end{align}$$
Recall
$$\begin{align} \frac{d}{dy} \int^{f(y)}_{g(y)} p(x) dx = p(f(y))f^{‘}(y) – p(g(y))g^{‘}(y) \end{align}$$
so that
$$\begin{align} \frac{d}{dy} P(Y < y) = p(\ln y) \frac{1}{y} = \tilde{p} (y) \end{align}$$

where $\tilde{p} (y)$ is the probability density function for $Y$

Mean
$$\begin{align} \mathbb{E}[e^X] & =\frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} e^{x} e^{-\frac{(x-\mu)^2}{2\sigma^2}} dx \\ & = \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} e^{-\frac{\left(x^2 – 2\mu x + \mu^2 – 2\sigma^2 x\right)}{2\sigma^2}} dx \\ &= \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} e^{-\frac{\left( x-(\mu + \sigma^2)\right)^2}{2 \sigma^2}}e^{\frac{\sigma^4+2\mu \sigma^2}{2 \sigma^2}} dx\\ &= e^{\frac{\sigma^4+2\mu \sigma^2}{2 \sigma^2}}\\ &= e^{\frac{\sigma^2+2\mu}{2}} \end{align}$$
Variance:
$$\begin{align} \mathbb{E}[Y^2] & = \mathbb{E}[e^{2X}] \\ & = \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} e^{2x} e^{-\frac{\left(x-\mu\right)^2}{2\sigma^2}} dx \\ & = \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} e^{-\frac{\left(x^2-2(\mu+2\sigma^2)x+\mu^2\right)}{2\sigma^2}} dx \\ & = \frac{1}{\sqrt{2\pi \sigma^2}} \int_{-\infty}^{\infty} e^{-\frac{\left(x-(\mu+2\sigma^2)\right)^2}{2\sigma^2}} e^{\frac{4\sigma^4+4\sigma^2\mu}{2\sigma^2}}dx \\ = e^{2\sigma^2+2\mu} \end{align}$$

Alternatively we know $2X \sim N(2\mu,4\sigma^2)$ so from earlier $\mathbb{E}[e^{2X}] = e^{2\mu +2\sigma^2}$
The variance is therefore
$$\begin{align} \mathbb{E}[Y^2]-\mathbb{E}[Y]^2 &= e^{2\sigma^2+2\mu} – e^{\sigma^2+2\mu}\\ &=e^{2\mu}\left(e^{2\sigma^2}-e^{\sigma^2}\right) \end{align}$$