mean and variance of reciprocal normal distribution

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If $X$ is a normal distributed with mean $\mu$ and variance $\sigma^2$. What would be the mean and variance of $Y = \dfrac{1}{X}$

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1 Answer

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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

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.

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