Point estimate of $\sigma _g^2$

We get a point estimate of $\sigma _g^2$ by finding the maximum a posterior (MAP) over the marginal posterior distribution $p(\sigma_g^2,\vec{\tau_K}\vert Y)$. If we marginalise out $\beta _g$ then the marginal posterior is:

$$p(\sigma_g^2,\vec{\tau_K}\vert Y) = \vert U^{-1}\vert^{1/2}\vert X_{G}^TU^{-1}X_{G}\vert^{-1/2} \exp\... ...tilde{\beta}_g^TX_{G}^T U^{-1}X_{G}\tilde{\beta}_g \right)\right\} 1/\sigma_g^2$$ (43)

where

$$\tilde{\beta}_g = (X_{G}^T U^{-1} X_{G})^{-1}X_{G}^TU^{-1}\vec{\mu_{\beta_K}}$$ (44)

We then assume $\tau_k=1$ and look to find the MAP for $\sigma _g^2$. However, there is a question of parameterisation. The mode we get will depend on the parametrisation we use. For example, we could look to maximise with respect to $\sigma _g^2$, or $\sigma_g$, or $\log(\sigma_g^2)$, or $\phi_g = 1/\sigma_g^2$ etc., all of which will give us different MAPs. Note that as we reparameterise, the reference prior might change but the reference posterior always stays the same, see (2). Hence, a natural way to reparameterise such that the parameter we use gives us a uniform reference prior.

The parameterisation which gives us a uniform reference prior is $\theta=\log(\sigma_g^2)$. Hence, we need to solve:

$$\widehat{\theta} = \arg \max_{\theta} p(\sigma_g^2\vert Y,\vec{\tau_K=1})$$ (45)

where $p(\sigma_g^2\vert Y,\vec{\tau_K=1})$ is the marginal in equation 43 with $\vec{\tau_K=1}$. To solve for $\widehat{\theta}$ using this equation we use Brent's algorithm (3). We can then easily convert from $\widehat{\theta}$ to $\widehat{\sigma_g^2}$.