Product of two multivariate normal distribution Let 
$$p(a\mid b)=  \mathcal{N}(a\mid Ab, S)$$
$$p(b) = \mathcal{N}(b\mid\mu, \Sigma)$$
I want to show that $p(a\mid b)p(b)$ is the pdf of a multivariate normal distribution? I tried explicitly writing down the pdf and computing product but I'm stuck.
 A: I'm going to construe the question as follows.
$X$ is a random vector taking values is $\mathbb R^m.$
$Y$ is a random variable taking values in $\mathbb R^n.$
$A\in \mathbb R^{m\times n}.$
$S\in \mathbb R^{m\times m}$ is symmetric and nonnegative-definite.
$\Sigma\in \mathbb R^{n\times n}$ is symmetric and nonnegative-definite.
$$
X\mid (Y= b) \sim \operatorname N(Ab,S).
$$
$$
Y\sim \operatorname N(\mu,\Sigma).
$$
The question is: What is the marginal probability distribution of $X$?
Notice that
$$
(X-Ab)\mid (Y=b) \sim \operatorname N(0,S).
$$
This is true of ALL values of $b.$ So we can say
$$
(X-AY)\mid Y \sim \operatorname N(0,S).
$$
In this expression $\text{“}\operatorname N(0,S)\text{''},$ one does not see $\text{“}Y\text{''}.$ From this one draws two conclusions:


*

*The random variable $X-AY$ is independent of $Y.$

*The marginal distribution of $X-AY$ is the same as this conditional distribution.


Therefore we have


*

*$X-AY \sim\operatorname N(0,S).$

*$Y\sim\operatorname N(\mu,\Sigma).$

*$Y$ and $X-AY$ are independent, and therefore $AY$ and $X-AY$ are independent.


And
$$
AY \sim \operatorname N(A\mu, A\Sigma A^\top).
$$
Finally we get
$$
X = (X-AY) + AY \sim \operatorname N(A\mu, S + A\Sigma A^\top).
$$
