Let $Y\sim N(\mu,\sigma^2)$ and $X=e^Y$. That is, $X$ is log-normally distributed. I am trying to figure out what is the expected value of $X$ based on the moment-generating function of $Y$.
The moment-generating function of the normally distributed function $Y$ is $\exp\left(t\mu+\frac{1}{2}\sigma^2t^2\right)$. Now
$$EX=Ee^Y$$
and I don't know where to go from here.
Best Answer
Note that $E[e^Y] = M_Y(1)$ where $M_Y$ denotes the moment generating function of $Y$. Thus $E[e^Y] = e^{\mu+ \frac{1}{2}\sigma^2}$