$$p(a\mid b)= \mathcal{N}(a\mid Ab, S)$$

$$p(b) = \mathcal{N}(b\mid\mu, \Sigma)$$

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.

  • $\begingroup$ 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? $\endgroup$ Jan 3, 2020 at 5:14

1 Answer 1


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


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