# Bayes' Theorem and Factorization [closed]

When applying Bayes' Theorem to observed variable $$\mathbf{x} = (x_1, x_2, \ldots, x_n)$$ and latent variable $$\mathbf{z} = (z_1, z_2,\ldots, z_m)$$, assuming a prior $$p(\mathbf{z}) = \prod_{i=1}^m p(z_i)$$ which factorizes over dimensions, does this imply the posterior $$p(\mathbf{z} \vert \mathbf{x})$$ is also factorized?

$$p(\mathbf{z} \vert \mathbf{x}) = \frac{p(\mathbf{x} \vert \mathbf{z}) p(\mathbf{z})}{p(\mathbf{x})} = \frac{p(\mathbf{x} \vert \mathbf{z})}{p(\mathbf{x})} \prod_{i=1}^m p(z_i)$$

If this is the case, how does the probability mass/density of the factor $$\frac{p(\mathbf{x} \vert \mathbf{z})}{p(\mathbf{x})}$$ 'distribute' itself over the dimensions of $$p(\mathbf{z}) = \prod_{i=1}^m p(z_i)$$ - equally amongst all dimensions?

In other words, if the posterior can be written as:

$$p(\mathbf{z} \vert \mathbf{x}) = \prod_{i=1}^m p(z_i \vert \mathbf{x}) =\prod_{i=1}^m \alpha_i p(z_i)$$

What would the $$\alpha_i$$ be?

• By factorisation of $p(z|x)$, I'd assume something like $\prod p(z_i|x)$,but the first equation you wrote is always correct given that $p(z)$ is factorized. – gunes Oct 12 at 13:22
• Thanks, I edited the question to make it more clear, but I'm not sure if that directly translates to factorization of the posterior. – Eweler Oct 12 at 13:32

There is no reason for $$p(\mathbf x|\mathbf z)$$ to factorise in a product of functions of the $$z_i$$ (for a given $$\mathbf x$$). The OP seems to ignore the fact that $$p(\mathbf x|\mathbf z)$$ is a function of $$\mathbf z$$ as a whole and hence that$$p(\mathbf x|\mathbf z)\prod_{i=1}^m p(z_i)$$has no reason to turn into$$\prod_{i=1}^m \alpha_i p_i(z_i)$$unless the $$\alpha_i$$'s are themselves functions of $$\mathbf z$$, such that $$\prod_{i=1}^m \alpha_i(\mathbf z) \propto p(\mathbf x|\mathbf z)$$ which ruins the purpose of the decomposition.