Calculating Marginal Probability of the product of two Gaussian PDFs The problem I am interested in is as follows:
Let
$$
P(x_k|x_{k-1}) \sim\mathcal{N}(F x_{k-1}, Q)\\
P(x_{k-1}|z^{k-1}) \sim \mathcal{N}(\mu, R)
$$
I would like to calculate
$$
P(x_k|z^{k-1})=\int_{x_{k-1}}P(x_k|x_{k-1})P(x_{k-1}|z^{k-1}) dx_{k-1} 
$$
where $z^{k-1} = (z_1,z_2, ... , z_{k-1})$.
This is a product of two gaussian pdfs marginalized over $x_{k-1}$.
$$
P(x_k|z^{k-1}) = \int_{x_{k-1}} \mathcal{N}(F x_{k-1}, Q)\ \mathcal{N}(\mu, R) dx_{k-1} 
$$
The solution is:
$$
P(x_k|z^{k-1}) \sim \mathcal{N}(F \mu, FRF^T +Q)
$$
I would like to find this solution using the integral formulation above. This is an exercise I am doing related to the Forward Algorithm for Gaussian HMM's, Kalman Filters, Bayes Filters.
 A: The marginal probability we want to calculate is a convolution of two Gaussian PDFs, with respect to the term $Fx_{k-1}$.

Consider the following compact notation of Gaussian functions:
$$N(\mu, \Sigma) = \frac{1}{\sqrt{(2\pi)^k |\Sigma|}}e^{-\frac{1}{2}\mu^T\Sigma^{-1}\mu}$$
where $\mu \in \mathbb{R}^k, \Sigma \in \mathbb{R}^{k \times k}  $.
Note the equality:
$$N(\mu,\Sigma) = |A| N(A\mu,A\Sigma A^T)$$
from $\mu^T\Sigma^{-1}\mu = \mu^TA^T(A^T)^{-1}\Sigma^{-1}A^{-1}A\mu = (A\mu)^T(A\Sigma A^T)^{-1}(A\mu) $
and $|AB|=|A||B|, |A|=|A^T|$
It is known that the convolution of two Gaussian PDFs is also Gaussian.
(For a scalar derivation of the expression below refer to page 3 of this notes.)
$$\int N(x-\tau-\mu^F,\Sigma^F) \cdot N(\tau-\mu^G,\Sigma^G)d\tau = N(x-(\mu^F+\mu^G),\Sigma^F+\Sigma^G)$$
Now we have all the tools to adress the question.
Write the relevant PDFs in the compact form:
$$P(x_k|x_{k−1}) = N(x_k - Fx_{k−1},Q)$$
$$P(x_{k−1}|z^{k−1}) = N(x_{k−1}-\mu,R) = N(Fx_{k−1}-F\mu,FRF^T) $$
And apply the expression for the convolution of two Gaussians, using $x = x_k, \tau = Fx_{k−1}, d\tau = |F|dx_{k−1}, \mu^F=0, \Sigma^F=Q, \mu^G=F\mu$ and $\Sigma^G=FRF^T$,
to obtain
$$P(x_{k}|z^{k−1}) = N(x_k-F\mu, Q+FRF^T)$$
