TITLE: Calculate the expected value of $Y=e^X$ where $X \sim N(\mu, \sigma^2)$
QUESTION [30 upvotes]: I got a problem of calculating $E[e^X]$, where X follows a normal distribution $N(\mu, \sigma^2)$ of mean $\mu$ and standard deviation $\sigma$.
I still got no clue how to solve it. Assume $Y=e^X$. Trying to calculate this value directly by substitution $f(x) = \frac{1}{\sqrt{2\pi\sigma^2}}\, e^{\frac{-(x-\mu)^2}{2\sigma^2}}$ then find $g(y)$ of $Y$ is a nightmare (and I don't know how to calculate this integral to be honest).
Another way is to find the inverse function. Assume $Y=\phi(X)$, if $\phi$ is differentiable, monotonic, and have inverse function $X=\psi(Y)$ then $g(y)$ (PDF of random variable $Y$) is as follows: $g(y)=f[\psi(y)]|\psi'(y)|$.
I think we don't need to find PDF of $Y$ explicitly to find $E[Y]$. This seems to be a classic problem. Anyone can help?

REPLY [3 votes]: Since mgf of $X$ is $\;M_x(t)=e^{\large\mu t + \sigma^2 t^2/2}$
$$E[Y]=E[e^X]=M_x(1)=e^{\large\mu + \sigma^2/2}$$