sampling from the product of two conditional distributions Assume x,y,z are dependent variables.
I want to sample z from p(z|x,y) but direct sampling is impossible.
Instead, I can sample z from p(z|x) and p(z|y) independently. But I would like to sample z using all information (x,y) I have.
If I sample z from:
$z \sim \dfrac{p(z|x)p(z|y)}{\int_{z'} p(z'|x)p(z'|y)dz'}$
would this sampling be a reasonable approximation compared to $z \sim p(z|x,y)$ ?
If it is practical usage, could you give me an example?
Or is there better way to sample z from p(z|x,y) given above assumptions?
EDIT
I also considered $$z \sim \frac{1}{2}[p(z|x)+p(z|y)]$$, when sampling (and evaluation) from $z \sim p(z|x,y)$  is not available.
So I'm trying to show that $z \sim \frac{1}{2}[p(z|x)+p(z|y)]$ is better than $z \sim p(z|x)$ or $z \sim p(z|y)$ by showing KL divergence is smaller:
$$ KL\left(p(z|x,y) \middle\| \frac{1}{2}[p(z|x)+p(z|y)]\right) < KL\left(p(z|x,y) \middle\| p(z|x)\right) $$
$$\int{p(z|x,y)\log{\frac{p(z|x,y)}{\frac{1}{2}[p(z|x)+p(z|y)]}}}dz < \int{p(z|x,y)\log{\frac{p(z|x,y)}{p(z|x)}}}dz$$
-->
$$-\int{p(z|x,y)\log{p(z|x)}}dz + \int{p(z|x,y)\log\left({\frac{1}{2}[p(z|x)+p(z|y)]}\right)}dz > 0$$
From above inequality, I've noticed that if the support of p(z|x) cannot cover p(z|x,y), the first term goes to infinity, while second term uses p(z|y) to make the support wider and possibly cover p(z|x,y), resulting inequality is validated.
In the case where p(z|x) cannot cover p(z|x,y), $\frac{1}{2}[p(z|x)+p(z|y)]$ which uses x,y information looks better approach.
Could this reasoning be sufficient to use $\frac{1}{2}[p(z|x)+p(z|y)]$ other than p(z|x) ?

This is not correct:
$$p(z|x,y)\ne p(z|x)p(z|y)\Big/ \int p(z|x)p(z|y)\,\text dz$$
in general. The fundamental reason is that the marginal conditionals $p(z|x)$ and $p(z|y)$ do not incorporate any information about the joint distribution of $(x,y)$.
To wit,
$$p(z|x,y)=\frac{p(x,y,z)}{p(x,y)}=\frac{p(x,z)p(y|x,z)}{p(x,y)}$$
while
$$p(z|x)p(z|y) = \dfrac{p(x,z)p(y,z)}{p(x)p(y)}$$
For instance, it can happen that $$\int p(x,z)p(y,z)\,\text d z=+\infty$$
Another illustration of the difference is that, when $p(z|x)=p(z|y)=p(z|x,y)$ (as when $x$ and $y$ are in bijection),$$p(z|x)p(z|y)=p(z|x,y)^2$$
which is not proportional to $p(z|x,y)$ and usually more concentrated / less variable.
