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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 $c$.
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  • $\begingroup$ 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. $\endgroup$
    – Matt F.
    Jul 25 at 11:14
  • $\begingroup$ Thanks. One reference is Bishop's Pattern Recognition and Machine Learning, equations 2.113 - 2.115. $\endgroup$
    – ngiann
    Jul 25 at 11:20
  • $\begingroup$ Also here is the link to the book: microsoft.com/en-us/research/uploads/prod/2006/01/… $\endgroup$
    – ngiann
    Jul 25 at 11:25
  • $\begingroup$ The first equation in the post might need an analog of the $A\Lambda^{-1}A^T$ in Bishop’s 2.115. $\endgroup$
    – Matt F.
    Jul 25 at 12:46
  • $\begingroup$ My guess is not, as it is the b that is integrated out and not the x $\endgroup$
    – ngiann
    Jul 25 at 13:24

1 Answer 1

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

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    $\begingroup$ Thank you. This is indeed what I have been using, but it was based more on intuition which I tend to be wary of $\endgroup$
    – ngiann
    Jul 25 at 14:57

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