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

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 $$Y=1/X$$?

If $$X$$ is skew normal, then $$Y = 1/X$$ does not possess a first moment. Thus mean and variance are not defined.
• It is certainly possible to say something about the median of $Y$. You should write a new post with your questions. Dec 10, 2020 at 2:50
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 $$\Bbb D(x,a)=\exp(-x^2)\int_0^x\exp(t^2)\mathcal D(at)\,\mathrm dt.$$