# Why do Variational Bayes methods assume that the likelihood $p(x|z)$ is tractable while the posterior is not?

I am trying to understand the motivation behind Variational Bayes. I get that the posterior $p(z|x)$ can be intractable, when we would have to compute the evidence with $p(x) = \int p(x|z)p(z) \text{d}z$.

However, all tutorials I have read on Variational Bayes so far simply assume that the likelihood $p(x|z)$ is easy to compute. Why? Why does the same assumption not hold for the posterior?

If we can define $p(x|z)$ analytically with our model, why can't we do the same for $p(z|x)$?

• I'm not an expert in VA, but in general likelihood is a function of data given parameters, i.e. a distribution assumed over data, where the distributions of priors come from the prior distributions, so you have all the stuff needed to evaluate it. Posterior is a distribution of parameters (unobserved!) given data. To obtain the posterior you need likelihood, priors and Bayes theorem.
– Tim
Commented Jun 15, 2018 at 9:26

$$p(z|x) = \frac{p(x|z) p(z)}{p(x)}.$$