Bayesian Inference Intractability

Question

When looking at Bayesian posteriors

$$ p(z \mid x) = \frac{p(x \mid z)p(z)}{\int p(x \mid z')p(z')dz'} $$

The denominator commonly intractable. I understand this is due to the possibility of high dimensionality, but I'm not sure why a Monte-Carlo estimate approach is not commonly used e.g.

$$ \int p(x \mid z')p(z')dz' = \mathbb{E}_z(p(x \mid z)) \approx \frac{1}{L}\sum_{i=1}^{L} p(x \mid z_i) $$

Particularly when we usually choose a prior $p(z)$ which we can easily sample. Any help would be appreciated.

Answer 1

In principle, you certainly could do that. In practice, it is often too inefficient. In high dimensions, the Monte-Carlo estimate often has very poor accuracy (the estimator has high variance), so the number $L$ of samples needed to get a reasonable estimate is infeasibly large.