# Matching pdf with the Inverted Gamma Distribution

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 $\displaystyle{\beta}$?

• 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 $\beta$). – wolfies Jul 17 '13 at 15:38
• 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 $\sigma$ – Trts Jul 17 '13 at 17:03