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

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    $\begingroup$ 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 $1/X$ ~ $InverseGamma(a,b)$. $\endgroup$ – wolfies Oct 17 '13 at 6:06
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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.

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  • $\begingroup$ But $v\sim\text{Inverse-Gamma}$? I have $$v^{-p/}=v^{-p/2+1-1}=v^{-(p/2-1)}-1$$ $\endgroup$ – user30490 Oct 17 '13 at 3:39
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Your scale parameter seems to be problematic. Here is the relationship between Gamma and Inv-Gamma distributions:

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.

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