Originally Posted by
tukeywilliams Use the moment generating function of the normal distribution: $\displaystyle M_{X}(t) = e^{\mu t + \sigma^{2}t^{2}/2} $ to get the moments of the log normal. In particular, suppose that $\displaystyle X $ has a log normal distribution with parameters $\displaystyle \mu $ and $\displaystyle \sigma $. Show that $\displaystyle E(X^n) = e^{n \mu + \frac{1}{2}n^{2}\sigma^{2}} $. It then follows that $\displaystyle E(X) = e^{\mu + \frac{1}{2} \sigma^2} $ and $\displaystyle \text{Var}(X) = e^{2(\mu+\sigma^{2})}- e^{2 \mu + \sigma^{2}} $.
Also $\displaystyle X = e^{Y} $ where $\displaystyle Y $ is normally distributed with parameters $\displaystyle \mu $ and $\displaystyle \sigma $. You use the transformation theorem to get the pdf of the lognormal. Interestingly, the log normal distribution does not have a moment generating function.