Given that $a$~$\mathcal{N}(\mu, \Sigma)$ and $\mu$~$\mathcal{N}(0, \sigma^2 I)$, Is $p(a|\Sigma, \sigma)=\int p(a|\mu, \Sigma)p(\mu|\sigma)d\mu$ again a Normal distribution? And how to estimate the mean and variance respectively? Thanks a lot!

  • $\begingroup$ Is this a self-study question? stats.stackexchange.com/tags/self-study/info $\endgroup$
    – Lucas
    Commented Mar 6, 2017 at 20:12
  • 1
    $\begingroup$ @Lucas It is not a homework but a question I have thought about when I read PRML. There is very detailed discussion of Normal distribution in that book. But I can not find the discussion about this one. $\endgroup$ Commented Mar 6, 2017 at 20:21

1 Answer 1


Yes, the marginal distribution is still a normal distribution with mean $0$ and variance $\Sigma+\sigma^2I$. Because $\int {{p_{a|\mu ,\Sigma }}(a){p_{\mu |\Sigma }}(\mu )du = } \int {{p_{a|0,\Sigma }}(a - \mu ){p_u}(\mu )du} $, this marginal distribution is the convolution of two gaussian distribution (see the definition of convolution). Then by convolution, we could get distribution of this marginal distribution.

  • $\begingroup$ Shouldn't it be mean $0$? : ) $\endgroup$ Commented Mar 6, 2017 at 21:05
  • $\begingroup$ Thanks for your answer! I also wonder what if the prior $p(y)\sim\mathcal{N}(0, I)$ and conditional distribution $p(x|y)\sim\mathcal{N}(Ay, \Sigma)$ are both normal distribution? Is the joint distribution $p(x, y)$ also normal distribution? I found all the notes haven't discussed this yet. $\endgroup$ Commented Mar 26, 2017 at 17:22

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