# Relationship between inverse gamma and gamma distribution

I have the following posterior distribution for $v$ $$f(v)\propto v^{-p/2}\exp\left(-\frac{1}{v}\frac{s}{2}\right)$$ and so clearly $$v\sim\text{Inverse-Gamma}\left(\frac{p}{2}-1,\frac{s}{2}\right)$$

Now can I say that $$v^{-1}\sim\text{Gamma}\left(\frac{p}{2}-1,\frac{s}{2}\right)$$

• There are many competing ways of expressing Inverse distributions. Accordingly, if you fail to provide the functional forms you are using, there is nothing 'clear' about the above. The definition I use is that if $X$~$Gamma(a,b)$ with pdf $$f(x) = \frac{x^{a-1} e^{-\frac{x}{b}}}{b^a \Gamma (a)}$$ then $1/X$ ~ $InverseGamma(a,b)$. Oct 17, 2013 at 6:06
• The 'competing ways' comment is important. Some answers below assume standardization across authorities that simply doesn't exist, and folks shouldn't answer based on their favorite book. If you are just writing it down, I'd say write down the version of the inverse gamma for clarity. On the other hand if you are actually going to be sampling, then you need match up against the pdf implemented in software you will be using. In many cases you literally have to plug values in to figure it out. It is a nomenclature/documentation failure I wish the various packages would remedy. Dec 19, 2020 at 1:09

Yes, but I think the first parameter of the Gamma should be $1-p/2$ instead of $1+p/2$. $$v \sim \text{Gamma}(1-p/2, s/2)$$ I'm using the shape-rate parametrization, as in here.
• But $v\sim\text{Inverse-Gamma}$? I have $$v^{-p/}=v^{-p/2+1-1}=v^{-(p/2-1)}-1$$
A random variable X is said to have the inverse Gamma distribution with parameters $\alpha$ and $\theta$ if 1/X has the Gamma($\alpha$, $1/\theta$) distribution.