# Joint densities and conditional independece

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?

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.

• very useful re-arranging, which brings up important details about the conditional probabilities in the assumed factorization model. – user1571823 Sep 26 '18 at 10:24