# 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.

• This notation by which the same symbol is used as the density function of different random variables is obnoxious. If $p$ is the density function of some random variable, then $p(3)$ should be the value of that function at $3.$ But which random variable? Jan 3, 2020 at 5:14

## 1 Answer

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