What is the mean and variance of $1/X$ if $X$ is skew normal distributed?

Question

The density function of skew normally distributed random variable $X$ is given as

$$f(x)=\frac{2}{\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)\Phi\left(\alpha \left(\frac{x-\mu}{\sigma}\right)\right)$$ where $\phi$ is standard Gaussian density, $\Phi$ is standard Gaussian cumulative distribution function with location $\mu$, scale $\sigma$ and shape $\alpha$ parameters.

What is the mean and variance of the random variable $Y=1/X$?

Answer 1

If $X$ is skew normal, then $Y = 1/X$ does not possess a first moment. Thus mean and variance are not defined.

Answer 2

Moments of the form $\mathsf EX^{-n}$ for $n=1,2,\dots$ do not exist for the skew normal distribution because this distribution has positive density in a neighborhood of the origin. If, however, you define $\mathsf EX^{-1}$ in the sense of the Cauchy Principal Value $$ \mathsf EX^{-1}=\lim_{\epsilon\to 0^+}\int_{\Bbb R\setminus(-\epsilon,\epsilon)}\frac{f_X(x)}{x}\,\mathrm dx, $$ then an explicit expression is given by Peng (2008) in the form $$ \mathsf EX^{-1}=\frac{\sqrt 2}{\sigma}\mathcal D\left(\frac{\mu}{\sqrt 2\sigma}\right)-\frac{2\sqrt 2\alpha}{\sigma\sqrt\pi}\Bbb D\left(\frac{\mu}{\sqrt 2\sigma},\sqrt{1+\alpha^2}\right)+\sqrt{\frac{2}{\pi\sigma^2}}\exp\left(-\frac{\mu^2}{2\sigma^2}\right)\log(\alpha+\sqrt{1+\alpha^2}), $$ where $$ \mathcal D(z)=\exp(-z^2)\int_0^z\exp(t^2)\,\mathrm dt $$ is the Dawson function and $$ \Bbb D(x,a)=\exp(-x^2)\int_0^x\exp(t^2)\mathcal D(at)\,\mathrm dt. $$