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
$$v^{-1}\sim\text{Gamma}\left(\frac{p}{2}-1,\frac{s}{2}\right)$$
Best Answer
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.