Is a power of an inverse gamma random variable itself inverse gamma?

Question

If $X$ is an inverse gamma distributed random variable, then would $X^p$ also be distributed as inverse gamma?

I found someone asked about the square root of inverse gamma, but I didn't find a direct answer to that question.

Answer 1

The short answer is no, in general a power of an inverse gamma random variable is not itself distributed as inverse gamma.

This can be seen by directly computing the distribution of the transformed random variable.

In what follows, I'll assume $p>0$.

With the transformation $Y = X^p$ we have $f_Y(y)= f_X(y^{1/p}) \left|\frac{d y^{1/p}}{dy}\right|$.

If

$$f_X(x) \propto x^{-(\alpha+1)} \exp(-\beta/x); \quad x>0, \alpha,\beta>0\,,$$

then (if I didn't make a mistake somewhere)

$$f_Y(y) \propto y^{-(\alpha/p) -1} \exp(-\beta/y^{1/p}) ; \quad y>0, \alpha,\beta>0\,.$$

As you see, because of the power of $y$ in the $\exp$ term, the result isn't inverse-gamma. Indeed, at least for some values of $p$, it looks like it's a particular case of what Mathematica calls the Generalized Inverse Gamma (see here, scroll down 2/3 of the page), with shift-parameter $\mu=0$ and second shape parameter $\gamma$ related to $p$.