# Integrating out parameters in a Gaussian distribution

I have seen in a text book the following,

$$\int \mathcal{N} \left(t;s,\beta^2\right) \mathcal{N} \left(t;\langle \mu, x \rangle + \frac{y \Sigma v}{w},\frac{\Sigma^2(1-w)}{w}\right)dt \propto \mathcal{N} \left(s;\langle \mu, x \rangle + \frac{y \Sigma v}{w},\frac{\Sigma^2(1-w)}{w}+\beta^2\right).$$

I am unsure as to how integrals are evaluated in normal distributions. Could someone please shed some light for me on this. Thanks

• I am beginning to think that this is not correct. I have tried various methods but never can get the same answer as above – George Sep 16 '16 at 22:50

The notation is tending to obscure what's going on. If you:

1. write the densities out as actual densities,

2. combine the exponents ($e^A.e^B=e^{A+B}$),

3. expand the terms in the exponent and collect like terms in the exponent (i.e. terms in $t^2$, $t$ and the constants)

4. complete-the-square in $t$ (i.e. write $at^2-bt$ as $a(t-\frac{b}{2a})^2+\text{<something>}$)

5. take the remaining terms not involving $t$ in the exponent outside the integral.

6. Then you can cross out the integral in $t$ (since being an unnormalized Gaussian density it now integrates to a constant, and the constant goes into the constant of proportionality).

7. What's left should be proportional to the RHS.

You can see many discussions of these calculations on site, some with fairly complete derivations (e.g. search on completing the square). Though they're not identical to your calculation, they all work in the same general way.