For a normal likelihood $$ P(\mathbf{y}|\mathbf{b}) = \mathcal{N}(\mathbf{Gb}, \mathbf{\Sigma}_y) $$ and a normal prior $$ P(\mathbf{b}) = \mathcal{N}(\mathbf{\mu}_p, \mathbf{\Sigma}_p) $$ I'm trying derive the evidence (or marginal likelihood) $P(\mathbf{y})$ where $$ P(\mathbf{y}) = \int P(\mathbf{y, b}) \, \mathrm{d}\mathbf{b} = \int P(\mathbf{y|b}) P(\mathbf{b}) \, \mathrm{d}\mathbf{b} = \mathcal{N}(\mu_{\mathrm{ML}}, \mathbf{\Sigma}_{\mathrm{ML}}) $$ Could anyone point me at a source for this derivation (or reproduce it)?

I tried doing it in an indirect way by first finding the posterior $P(\mathbf{b|y})$ and then via Bayes' theorem writing $$ \ln{P(\mathbf{y})} = \ln{P(\mathbf{y|b})} + \ln{P(\mathbf{b})} - \ln{P(\mathbf{b|y})}, $$ then by grouping terms quadratic in $\mathbf{y}$ and using the Woodbury identity I managed to show that the marginal likelihood covariance is $\mathbf{\Sigma}_{\mathrm{ML}} = \mathbf{G} \mathbf{\Sigma}_p \mathbf{G}^{\top} + \mathbf{\Sigma}_y$, but I haven't been able to get the mean $\mu_{\mathrm{ML}}$.

I'm sure the integral can be evaluated directly to get the answer, but I'm not sure how to go about it.



1 Answer 1


The key here is to rewrite $y$ as $$y = Gb + e$$ where $G$ is known, $b\sim N(\mu_p,\Sigma_p)$ and, independently, $e\sim N(0,\Sigma_y)$. Now $y$ is just the sum of two independent normals which is fully determined by its mean and covariance matrix. The mean is $$E[Y] = E[Gb] + E[e] = GE[b] + 0 = G\mu_p$$ and the variance is $$Var[y] = Var[Gb] + Var[e] = GVar[b]G^\top + Var[e] = G\Sigma_pG^\top +\Sigma_y$$. Since $b$ and $e$ are independent, the covariance is zero. Thus the marginal likelihood is $$y\sim N(G\mu_p,G\Sigma_pG^\top + \Sigma_y).$$


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