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Variational Autoencoders Explained

(Source: anotherdatum.com)

Latent space¶

You know every image of a digit should contain, well, a single digit. An input in $\mathbb{R}^{28×28}$ doesn’t explicitly contain that information. But it must reside somewhere… That somewhere is the latent space.

Photo by Samuel Zeller on Unsplash

You can think of the latent space as $\mathbb{R}^{k}$ where every vector contains $k$ pieces of essential information needed to draw an image. Let’s say the first dimension contains the number represented by the digit. The second dimension can be the width. The third – the angle. And so on.

We can think of the process that generated the images as a two steps process. First the person decides – consciously or not – all the attributes of the digit he’s going to draw. Next, these decisions transform into brushstrokes.

VAE tries to model this process: given an image $x$, we want to find at least one latent vector which is able to describe it; one vector that contains the instructions to generate $x$. Formulating it using the law of total probability, we get $P(x) = \int P(x|z)P(z)dz$.

Let’s pour some intuition into the equation:

  • The integral means we should search over the entire latent space for candidates.
  • For every candidate $z$, we ask ourselves: can $x$ be generated using the instructions of $z$? Is $P(x|z)$ big enough? If, for instance, $z$ encodes the information that the digit is 7, then an image of 8 is impossible. An image of 1, however, might be possible, since 1 and 7 look similar.
  • We found a good $z$? Good! But wait a second… Is this $z$ even likely? Is $P(z)$ big enough? Let’s consider a given image of an upside down 7. A latent vector describing a similar looking 7 where the angle dimension is set to 180 degrees will be a perfect match. However, that $z$ is not likely, since usually digits are not drawn in 180 degrees angle.

The VAE training objective is to maximize $P(x)$. We’ll model $P(x|z)$ using a multivariate Gaussian $\mathcal{N}(f(z), \sigma^2 \cdot I)$.

$f(z)$ will be modeled using a neural network. $\sigma$ is a hyperparameter that multiplies the identity matrix $I$.

You should keep in mind that $f$ is what we’ll be using when generating new images using a trained model. Imposing a Gaussian distribution serves for training purposes only. If we’d use a Dirac delta function (i.e. $x = f(z)$ deterministically), we wouldn’t be able to train the model using gradient descent!

More Info: anotherdatum.com

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