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}