# integration of Gaussian with prior mean [duplicate]

I want to calculate the following Integration $$\int \mathcal{N}\left(\mathbf{x} \mid \boldsymbol{\mu}, \boldsymbol{\Lambda}^{-1}\right) \cdot \mathcal{N}\left(\boldsymbol{\mu} \mid \mathbf{m},\left(\beta \boldsymbol{\Lambda}\right)^{-1}\right) d \boldsymbol{\mu}$$ and the answer is $$\mathcal{N}\left(\mathbf{x} \mid \boldsymbol{m}, (1 + \beta^{-1})\boldsymbol{\Lambda}^{-1}\right)$$ How to get this answer？

• What have you tried doing so far? The main trick to completing the integration is combining the exponentials into one and then completing the square within the exponential. Commented Jul 1, 2020 at 7:14
• The question is more about the marginal distribution in a Normal - Normal model that about integration per se. The solution is available from standard Bayesian textbooks, if need be. Commented Jul 1, 2020 at 7:49
• The technique is called completing the square.
– whuber
Commented Jul 1, 2020 at 13:31

## 1 Answer

There is a conclusion in Session 2.3.3 of PRML

Given a marginal Gaussian distribution for x and a conditional Gaussian distribution for y given x in the form \begin{aligned} p(\mathbf{x}) &=\mathcal{N}\left(\mathbf{x} \mid \boldsymbol{\mu}, \mathbf{\Lambda}^{-1}\right) \\ p(\mathbf{y} \mid \mathbf{x}) &=\mathcal{N}\left(\mathbf{y} \mid \mathbf{A} \mathbf{x}+\mathbf{b}, \mathbf{L}^{-1}\right) \end{aligned} the marginal distribution of y and the conditional distribution of x given y are given by $$p(\mathbf{y})=\mathcal{N}\left(\mathbf{y} \mid \mathbf{A} \boldsymbol{\mu}+\mathbf{b}, \mathbf{L}^{-1}+\mathbf{A} \mathbf{\Lambda}^{-1} \mathbf{A}^{\mathrm{T}}\right)$$

Here, we have $$p(\boldsymbol{\mu}) = \mathcal{N} (\boldsymbol{\mu}|\boldsymbol{m}, (\beta \mathbf{\Lambda})^{-1})$$ $$p(\boldsymbol{x}|\boldsymbol{\mu}) = \mathcal{N}(\boldsymbol{x}|\boldsymbol{\mu}, \mathbf{\Lambda}^{-1}) = \mathcal{N}(\boldsymbol{x}|\boldsymbol{I}\boldsymbol{\mu}+0, \mathbf{\Lambda}^{-1})$$ then, we have the marginal distribution $$p(\boldsymbol{x}) = \mathcal{N}(\boldsymbol{I}\boldsymbol{m}+0, \boldsymbol{\Lambda}^{-1}+\boldsymbol{I}(\beta \boldsymbol{\Lambda})^{-1}\boldsymbol{I})=\mathcal{N}(\boldsymbol{x}|\boldsymbol{m}, (1+\beta)\boldsymbol{\Lambda}^{-1})$$

Hence, $$\int p(\boldsymbol{x}|\boldsymbol{\mu})p(\boldsymbol{\mu})d\boldsymbol{\mu}=p(\boldsymbol{x})=\mathcal{N}(\boldsymbol{x}|\boldsymbol{m}, (1+\beta)\boldsymbol{\Lambda}^{-1})$$ Just as required.