Let us assume the joint density $p(x,y,z)$ is factorized as $p(y)p(z|y)p(x|z)$. Hence, $x \perp y|z$.

Now, the posterior distribution of z is: $p(z|x,y)=\frac{p(x,y,z)}{p(x,y)}$, where $p(x,y)=\int p(x,y,z)dz$.

Is it correct to rewrite $p(x,y)$ as $p(x)p(y)$, or the variables are only independet given $z$? So given that $z$ is integrated out, is there a density function where $x$ and $y$ exist together?


1 Answer 1


Some re-arranging of your initial assumption yields: $p(x,y,z)=p(y|z)p(x|z)p(z)$

which you might find a more enlightening way of looking at this. As you correctly stated, if you know z, x and y are independent, but if you don't, they are not. If you know x, this gives you information about z and thus in turn about y.

More mathematically, you stated that $p(x,y)=\int p(x,y,z)dz$

Also, $p(x,y)=p(x|y)p(y)$, writing the former in integral form and applying conditional probability once, I obtain:

$p(y)p(x|y)=p(y)\int p(z|y)p(x|z)dz $

and thus

$p(x|y) = \int p(z|y)p(x|z)dz \neq p(x)$

Hopefully this final expression is somewhat instructive to look at. If you want to know $p(x|y)$, y does contain information about x, because it gives you information about z, and z gives you information about x.

  • $\begingroup$ very useful re-arranging, which brings up important details about the conditional probabilities in the assumed factorization model. $\endgroup$ Sep 26, 2018 at 10:24

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