2
$\begingroup$

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 $(\sigma^2)^{k/2}$ ?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.