# Posterior distribution of $\sigma^2$

In chapter 9 of Jim Albert's Bayesian computation with R it's mentioned that, in the context of Normal Linear Regression, the posterior joint density is:

$$g(\beta, \sigma^2 | y) =g(\beta|y, \sigma^2)g(\sigma^2|y)$$

And it's stated (without proof) that the marginal posterior distribution of $$\sigma^2$$ ( i.e. $$g(\sigma^2|y)$$ ) is $$Inv. Gamma((n-k)/2, S/2)$$ where n = # of observations, k = # of parameters, and $$S = (y-X\hat\beta)^T(y-X\hat\beta)$$.

I was wondering how one get's to this (he assumed uninformative prior of $$g(\beta, \sigma^2) \propto 1/\sigma^2$$)?

I tried the following but got stuck: $$g(\sigma^2|y) = \int_{\beta} g(\sigma^2, \beta|y)d\beta \propto \int_{\beta} g(y |\sigma^2, \beta)g(\sigma^2, \beta)d\beta \propto \int_{\beta} g(y |\sigma^2, \beta)\frac{1}{\sigma^2}d\beta \\ =\int_{\beta} (2\pi)^{-n/2}(\sigma^2)^{-n/2-1}e^{-\frac{S/2}{\sigma^2}}d\beta$$

So without doing the integral I can account for the $$S/2$$ (kind-of, if I estimate $$S$$ using $$\hat\beta$$), and $$n/2$$ in the inverse gamma. But how do I integrate, and will I then get $$(\sigma^2)^{k/2}$$ ?

• It's a not so short story :) See Gill, Bayesian Methods, pp. 79-80 – Sergio Aug 3 at 12:24
• @Sergio Unfortunately this book is not available to me from my university databases... – Maverick Meerkat Aug 4 at 8:12
• You can read pp. 79-80 in books.google.it/… – Sergio Aug 4 at 8:40
• nope, it only lets me go until p. 74 – Maverick Meerkat Aug 4 at 8:56
• I can read pp. 79-80. – Sergio Aug 4 at 9:32