TITLE: mean and variance of reciprocal normal distribution
QUESTION [11 upvotes]: If $X$ is a normal distributed with mean $\mu$ and variance $\sigma^2$. What would be the mean and variance of $Y = \dfrac{1}{X}$

REPLY [19 votes]: Mean and variance do not exist. For the mean to exist, the integral
$\int^\infty_{-\infty} e^{-\frac{(x-\mu)^2}{2\sigma^2}}\frac{1}{|x|} \text{d}x$
needs to be finite. This is clearly not the case.
Note it is necessary that mean exists for variance to exist.
See 
http://en.wikipedia.org/wiki/Inverse_distribution#Reciprocal_normal_distribution
Note inverse gaussian is something completely different. It is connected to brownian motion hitting a level. I changed the title. The thing you are referring to is a reciprocal normal.