Log-normal distribution

Example

Let $\mu \in \R$ and $\sigma > 0$, and let $Y$ be normal with mean $\mu$ and variance $\sigma^2$. The law of $X = e^Y$ is the log-normal distribution with parameters $\mu$ and $\sigma^2$; so $X > 0$ and $\ln X$ is normal. Its density is $$f(x) = \frac{1}{x \sigma \sqrt{2\pi}} \exp\left(-\frac{(\ln x - \mu)^2}{2 \sigma^2}\right) \qquad (x > 0),$$ and $f(x) = 0$ for $x \le 0$. Check: for $0 < a < b$, $P(a < X < b) = P(\ln a < Y < \ln b)$, and the substitution $y = \ln x$, $dy = dx / x$, turns the normal density of $Y$ into $f$. The moments are $E[X^n] = E[e^{nY}] = e^{n\mu + n^2 \sigma^2 / 2}$ for every real $n$, by the moment-generating function of the normal law. So $E[X] = e^{\mu + \sigma^2/2}$ and $\operatorname{Var}(X) = (e^{\sigma^2} - 1) e^{2\mu + \sigma^2}$. The median is $e^{\mu}$, which is smaller than the mean.