13
$\begingroup$

A variational autoencoder (VAE) provides a way of learning the probability distribution $p(x,z)$ relating an input $x$ to its latent representation $z$. In particular, the encoder $e$ maps an input $x$ to a distribution on $z$. A typical encoder will output parameters $(\mu,\sigma)=e(x)$, representing the Gaussian distribution $\mathcal{N}(\mu,\sigma)$; this distribution is used as our approximation for $p(z|x)$.

Has anyone considered a VAE where the output is a Gaussian mixture model, rather than a Gaussian? Is this useful? Are there tasks where this is significantly more effective than a simple Gaussian distribution? Or does it provide little benefit?

$\endgroup$
3
  • 2
    $\begingroup$ arxiv.org/abs/1611.02648 $\endgroup$
    – shimao
    Commented Jun 12, 2018 at 3:08
  • 1
    $\begingroup$ @shimao, thank you! I've written an answer summarizing that, in case it's helpful to anyone else in the future. Thanks again. $\endgroup$
    – D.W.
    Commented Jun 12, 2018 at 5:15
  • $\begingroup$ @D.W. sorry for the late reply. I am just confused about something. Isn't VAE representing an infinite mixture of gaussians? $\endgroup$
    – floyd
    Commented Sep 11, 2019 at 15:41

1 Answer 1

10
$\begingroup$

Yes, it has been done. The following paper implements something of that form:

Deep Unsupervised Clustering with Gaussian Mixture Variational Autoencoders. Nat Dilokthanakul, Pedro A.M. Mediano, Marta Garnelo, Matthew C.H. Lee, Hugh Salimbeni, Kai Arulkumaran, Murray Shanahan.

They experiment with using this approach for clustering. Each Gaussian in the Gaussian mixture corresponds to a different cluster. Because the Gaussian mixture is in the latent space ($z$), and there is a neural network connecting $z$ to $x$, this allows non-trivial clusters in the input space ($x$).

That paper also mentions the following blog post, which experiments with a different variation on that architecture: http://ruishu.io/2016/12/25/gmvae/

Thanks to shimao for pointing this out.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.