How to prove $E[e^x]=e^{E[x]+Var[x]/2}$ when $x$ is a centered Gaussian random variable?

Question

$E[]$ is the expectation and $Var[]$ is the variance.

Answer 1

The formula for the expectation of $e^X$, with $X$ normal $(0 , \sigma^2)$, is: $$ \int_{\mathbb R} e^x \left(\frac{1}{\sqrt {2 \pi \sigma}} e^{-\frac{x^2}{2 \sigma^2}}\right) dx = \frac{1}{\sqrt{2\pi\sigma}}\int_{\mathbb R} e^{x - \frac{x^2}{2\sigma^2}} dx \\ = \frac{1}{\sqrt{2 \pi \sigma}} e^{\sigma^2 / 2} \int_{\mathbb R} e^{-(1 - \frac{x}{\sqrt{2 \sigma}})^2} dx = e^{\sigma^2/2} $$

To get the third step, I completed the square $x - \frac{x^2}{2\sigma^2}$ to a perfect square. For the fourth, I set $\frac{t}{\sqrt 2} = 1 - \frac{x}{\sqrt 2\sigma}$ and simplified using the usual Gaussian integral $\int_{\mathbb R} e^{-t^2 / 2}dt = \sqrt{2\pi}$ to conclude.

Of course, if $X$ is not centered, say $X = N(\mu , \sigma^2)$, then you can still carry out a similar thing and conclude that the complete formula is $e^{\mu + \sigma^2/2}$.