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So I know if we have a normal likelihood $P(\mathbf{y|b}) = \mathcal{N}(\mathbf{y}|\mathbf{Gb}, \mathbf{\Sigma}_y)$ and a normal prior $P(\mathbf{b}|\mathbf{\theta}) = \mathcal{N}(\mathbf{b} | \mu_p, \mathbf{\Sigma}_p)$ specified by some hyperparameters $\theta$, then the marginal likelihood is given by $$ P(\mathbf{y}|\theta) = \int P(\mathbf{y|b}) P(\mathbf{b}|\theta) \, \mathrm{d}\mathbf{b} = \mathcal{N}(\mathbf{y}|\mathbf{G}\mu_p, \mathbf{G} \mathbf{\Sigma}_p \mathbf{G}^\top + \mathbf{\Sigma}_y) $$

However, if we additionally know that $b_i \ge 0 \; \forall \; i$, we may instead want to use a half-normal prior where $$ P(\mathbf{b}|\mathbf{\theta}) = \begin{cases} 2^{m}\mathcal{N}(\mathbf{b} | \mu_p, \mathbf{\Sigma}_p) & \quad \text{if } b_i \ge 0 \; \forall \; i\\ 0 & \quad \text{otherwise}\\ \end{cases} $$ where $m$ is the length of $\mathbf{b}$. I want to be able to calculate the marginal likelihood in the case of a half-normal prior like this. We can write down an expression for it just by changing the domain of the integral to $[0 , \infty]$ such that $$ P(\mathbf{y}|\theta) = \int_{0}^{\infty} P(\mathbf{y|b}) P(\mathbf{b}|\theta) \, \mathrm{d}\mathbf{b} = \; ? $$ Is there any analytical solution for the above? If not, does anyone have suggestions for how I might go about evaluating it?

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  • $\begingroup$ that isn't the correct density for a half-normal prior. you have to double the height. $\endgroup$ – Taylor Aug 6 '17 at 22:10
  • $\begingroup$ @Taylor quite right! infact, wouldn't we need to multiply by 2 to the power of the number of elements in $b$? $\endgroup$ – CBowman Aug 7 '17 at 20:53
  • $\begingroup$ right. I was thinking univariate $\endgroup$ – Taylor Aug 7 '17 at 21:22

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