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So the Inverted Gamma probability density function is: $\displaystyle{f(x; \alpha, \beta) = \frac{\beta^\alpha}{\Gamma(\alpha)} x^{-\alpha - 1}\exp\left(-\frac{\beta}{x}\right)}$

The equation I'm dealing with is:

$\displaystyle{f(\sigma; ?, ?) = \frac{2}{\Gamma\left(\frac{v}{2}\right)}\left(\frac{v\hat{\sigma}^2}{2}\right)^{\frac{v}{2}} \frac{1}{\sigma^{v+1}}\exp\left[\frac{-v\hat{\sigma}^2}{2\sigma^2}\right]}$

with parameters $\displaystyle{v}$ and $\displaystyle{\hat{\sigma}}$

Clearly $\displaystyle{\alpha = \frac{v}{2}}$, but what's $\displaystyle{\beta}$?

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  • $\begingroup$ OP writes: So the Inverted Gamma probability density function is ... blah. ......... There are many different competing names and functional forms for distributions. The thing above that you refer to as an Inverted Gamma ... I would call an Inverse Gamma, the latter describing the pdf of $1/X$, when $X$~Gamma$(a,b)$, where $a$ = your $\alpha$, and $b$ = 1/(your $\beta$). $\endgroup$ – wolfies Jul 17 '13 at 15:38
  • $\begingroup$ Ah yes, I assume you mean this pdf: $g(x;\alpha,\beta) = \beta^{\alpha}\frac{1}{\Gamma(\alpha)} x^{\alpha-1} e^{-\beta x} \quad \text{ for } x \geq 0 \text{ and } \alpha, \beta > 0$? If so, I still can't seem to figure out what $\alpha$ and $\beta$ are since $x = \sigma$ but there is a $\sigma^2$... and $\alpha$ and $\beta$ can't be functions of $\sigma$ $\endgroup$ – Trts Jul 17 '13 at 17:03
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Look at Arnold Zellner's appendix on distributions in his bayesian textbook. You have an inverted gamma, often the posterior distribution of a standard deviation. The inverse gamma is different, it is the often the posterior distribution of a variance, sigma squared. The distribution of a precision would be a gamma.

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