Does variational auto-encoder output the variational distribution of the latent variable or the distribution of the input x?

Question

In the simple case of mixture of gaussians(with known variance), we have 2 latent variables $\mu$ and $z$. In the vaiational auto-encoder, we assume that the model is infinite mixture of gaussians. If we assume that the variance is known then we have 2 latent variables $\mu$ and $z$. the encoder the responsible for calculating the variational distribution of $z$ (its mean and variance). and the decoder is responsible for calculating the variational distribution of the $\mu$ (it's mean and variance)
So the variance $\sigma$ that the generative network is outputting Is NOT the variance of $x$
My questions is
Is my viewpoint correct?
I read many articles on the internet that say $P(x|z, w) = \mathcal{N}(\mu(z, w),\sigma(z, w))$ where $\mu(.)$ and $\sigma(.)$ are the generative network with parameters $w$. Isn't that wrong?

PS: I know that training and results are the same in both cases

Answer 1

In a GMM, $z$ is a latent variable, but the $\mu$'s are a model parameter.

You seem to be trying to draw an analogy between GMMs and VAEs -- it's true you can think of a VAE as an infinite mixture model, where each component has $\mu = f(z; \theta)$ where $f$ is the decoder. Here $\theta$, not $\mu$, are the model parameters.

and the decoder is responsible for calculating the variational distribution of the $\mu$

No -- $\mu$ isn't a latent variable, so there's no need for any variational approximation for it.

Yes, $x|z$ has distribution $\mathcal{N}(f(z;\theta))$ -- this is simply how the VAE model is defined.