1 Feb, 2016
Let $\phi$ be the precision parameter. The formulations of g-prior is
\[\beta | \phi \sim N(0, \frac{g}{\phi} \paren{X'X}^{-1}), \pi(\phi) \propto 1/\phi\]for $n \approx 100-1000, p \approx 16-30$
$\pi(y | M_\gamma) = \int~like*prior~d\beta d\phi$
For $M_{null}$, $R_\gamma^2 = 0, p_\gamma=0$. So, bayes factor = $\frac{\pi(y|M_\gamma)}{\pi(y|M_{null})}= \frac{(1+g)^{(n-1-p_\gamma)/2}}{[1+g(1-R_\gamma^2)]^{(n-1)/2}}$
As $g \rightarrow \infty$, BF($M_\gamma: M_{null}$) $\rightarrow 0$. This mean that non-informative prior on $\beta$ will help reject the model even if it is correct. (Bartlett Paradox).
As $R_\gamma^2 \rightarrow 1$, BF $\rightarrow (1+g)^{-p_\gamma/2}$
Since $M_\gamma$ approaches to a model with a perfect fit when $R_\gamma^2 \rightarrow 1$, BF should go to $\infty$. (Information Paradox)
Both paradoxes found in Langer et. al 2008.
Since fixing $g$ leads to paradoxes, we need to put a prior on $g$.
$y = \mu_0(x) + \epsilon$, $\mu_0$ is an unknown function, continuously differentiable up to order $s$. $\mu_0^{(\nu)}$ is Lipschitz cont of order $\nu-[\nu]$, $[x]$ is the floor of $x$. Stronger than continuous.
When using GP, you need Lipschitz continuity.
$y = \mu(x) + \epsilon$, $\mu \sim GP(0,\kappa(.,.; \phi))$. e.g. exponential correlation function $\kappa(x_i,x_j;\phi) = e^{-\norm{x_i-x_j}\phi}$, matern class of correlation function.