Mean and variance of the reciprocal of a random variable My random variable $X$ is normally distributed with mean $b$ and variance $p$. I defined a new random variable, $Y$, such that $Y=\frac{1}{X}$. Does anyone know how to find the mean of $Y$ and $Y^2$?
 A: This is equivalent to asking for
$\int_{- \infty}^\infty \frac{1}{x} P(X=x) \, dx$ and $\int_{- \infty}^\infty \frac{1}{x^2} P(X=x) \, dx$.
Considering the second, let's choose some arbitrary $a > 0$:
$$\eqalign {
\int_{- \infty}^\infty \frac{1}{x^2} P(X=x) \, dx
&= \int_{- \infty}^{-a} \frac{1}{x^2} P(X=x) \, dx\\
 &+ \int_{-a}^a \frac{1}{x^2} P(X=x) \, dx
 + \int_a^\infty \frac{1}{x^2} P(X=x) \, dx
\\
&\ge \int_{-a}^a \frac{1}{x^2} P(X=x) \, dx
\\
&\ge \int_{-a}^a \frac{1}{x^2} \min\left(P(X=-a), P(X=a)\right) \, dx
\\
&= \min\left(P(X=-a), P(X=a)\right) \int_{-a}^a \frac{1}{x^2} \, dx
\to \infty
}$$
where first we split up the integral and note that the tails are nonnegative,
and then use the fact that the normal pdf is quasiconcave. So $\mathbb{E}\left[\frac{1}{X^2}\right]$ doesn't exist.
We can do exactly the same thing to show that $\mathbb{E}\left[\frac{1}{\lvert X \rvert}\right]$ doesn't exist (since $\int_{0}^a \frac{1}{X} dx$ also diverges).
But $\frac{1}{X}$ is a little trickier, since one of the tail integrals is negative-valued.
Still, $\mathbb{E}[\frac{1}{X} \mid \frac{1}{X} < a]$ doesn't exist, since the integral diverges.
I think that, because $P(\frac{1}{X} < a) > 0$, using basically the same argument as the law of total expectation, this then implies that the overall expectation cannot exist. But I'm not 100% on that.
A: Actually, I wrote a paper in 1991 addressing this issue, available on Research Gate. In short, the mean and variance do not exist.
More precisely, if one defines a generalised inverse normal distribution by its density ($\alpha>1,\sigma>0$)
$$f(z|\mu,\sigma,\alpha) = \dfrac{K(\alpha,\mu,\sigma)}{|z|^\alpha}\exp\left\{-(z^{-1}-\mu)^2/2\sigma^2\right\}$$
then
$$K(\alpha,\mu,\sigma)^{-1}=\sigma^{\alpha-1}e^{-\mu^2/2\sigma^2}2^{(\alpha-1)/2}\Gamma((\alpha-1)/2){}_1F_1\left(\frac{1}{2}(\alpha-1);\frac{1}{2};\frac{\mu^2}{2\sigma^2}\right)$$
where ${}_1F_1$ is a confluent hypergeometric function. From this derivation of the normalising constant, one deduces that the mean only exists for $\alpha>2$ (while the inverse normal distribution corresponds to $\alpha=2$) and the variance only exists for $\alpha>3$. The mean is given by
$$\mathbb{E}[X]=\frac{\mu}{\sigma^2}\,\frac{{}_1F_1\left(\frac{1}{2}(\alpha-3);\frac{3}{2};\frac{\mu^2}{2\sigma^2}\right)}{{}_1F_1\left(\frac{1}{2}(\alpha-1);\frac{1}{2};\frac{\mu^2}{2\sigma^2}\right)}$$
A much simpler argument as to why the mean does not exist is that, if $f$ is the normal $\text{N}(b,p)$ density, 
$$x^{-1} f(x)\equiv \frac{e^{-b^2/2p}}{\sqrt{2\pi p}}x^{-1}$$
at zero, which is not integrable.
