# Marginal Gaussian distribution

I have a confusion over an integral involving a multivariate and a univariate Gaussian. We know that in the case of two multivariate Gaussians the following is true: $$\int \mathcal{N}(\mathbf{y}|\mathbf{A}\mathbf{x} + \mathbf{b}, \mathbf{S}) \mathcal{N}(\mathbf{b}|\mathbf{m}, \mathbf{C}) \mathbf{db} = \mathcal{N}(\mathbf{y}|\mathbf{A}\mathbf{x} + \mathbf{m}, \mathbf{S} + \mathbf{C})$$

Could somebody please tell me what the integral evaluates to in the following case where $$b$$ is now a scalar: $$\int \mathcal{N}(\mathbf{y}|\mathbf{A}\mathbf{x} + b\mathbf{1}, \mathbf{S}) \mathcal{N}(b|m, c^2) db$$

where

• $$\mathbf{1}$$ is a column of ones of the same dimension as $$\mathbf{y}$$
• $$\mathcal{N}(b|m, c^2)$$ is the univariate Gaussian distribution with mean $$m$$ and standard devation of $$c$$.
• Can you give a reference for the formula you know? The notation in these integrals is not quite standard, so they are not clear to me immediately. Jul 25 at 11:14
• Thanks. One reference is Bishop's Pattern Recognition and Machine Learning, equations 2.113 - 2.115. Jul 25 at 11:20
• Also here is the link to the book: microsoft.com/en-us/research/uploads/prod/2006/01/… Jul 25 at 11:25
• The first equation in the post might need an analog of the $A\Lambda^{-1}A^T$ in Bishop’s 2.115. Jul 25 at 12:46
• My guess is not, as it is the b that is integrated out and not the x Jul 25 at 13:24

The distribution$$\mathcal{N}(\mathbf{b}|\mathbf{m}, \mathbf{C})$$ simplifies into the distribution$$\mathcal{N}({b}|{m}, {c^2})$$ over the diagonal when $$\mathbf{m}={m}\boldsymbol{1}\quad \text{and}\quad \mathbf{C}=c^2\boldsymbol{1}\boldsymbol{1}^\top$$