Skip to content
Discussion options

You must be logged in to vote

Hello @EthanMarx 👋

I don't think there is an easy way to find an efficient inverse for an autoregressive flow.

However, it should be possible to train one! You could take your trained MAF $z = f(x)$ and train an IAF $x = g(z)$ to solve $x = g(f(x))$ or $z = f(g(z))$. Both training objectives have advantages. The former requires access to data points $x$, but no back-propagation through $f$. The latter requires back-propagation through $f$, but only needs latent (Gaussian) samples $z$.

Replies: 1 comment 3 replies

Comment options

You must be logged in to vote
3 replies
@EthanMarx
Comment options

@francois-rozet
Comment options

@EthanMarx
Comment options

Answer selected by EthanMarx
Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment
Category
Q&A
Labels
None yet
2 participants