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I'm learning about VAEs, and need to go this deep to understand them. However, the question is for Bayes theorem with probability distributions. I learnt about Bayes theorem from this video. Excellent explanation, with a simple example. There, are 4 events, a person is librarian ($P(H)$), or not ($P(\lnot H)$), and a person matches the description ($P(E)$), or not ($P(\lnot E)$). We want to know the probability of a person is librarian, given it matches the description. And we're calculating it in a nice visual way. It helps a lot that the probabilities are single numbers.

It is a lot harder to understand this with probability distributions, and I'm having a hard time trying to make up a real world example.

Or is it right to think about $P(H)$, and $P(\lnot H)$ together as a discrete probability distribution where on the x axis 0 is $P(\lnot H)$, and 1 is $P(H)$, and same with $E$?

In VAEs $p(z)$, and $q(z|x)$ are both normal probability distributions, and I can't really imagine a real world example similar.

Say example $p(z)$ is the probability of outcomes if two fair dices are rolled as in the picture:

enter image description here

What could $p(x|z)$ could be as a probability distribution? Say $z$ (the value of outcome) is 6, and we're checking the probability of one of the cubes is $x$ (6 possibilities).

However, in the VAE example, we know $x$ (in fact, that's the only thing we know for sure, and $p(z)$. Say, following for the above example, what's the probability of one of the cubes has the value of 6 ($x$), given ($z$).

Are those correct examples? If not, what could be an understandable, real world example, representing a VAE kind of situation with probabilities?

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It is not necessary for $p(z)$ and $q(z|x)$ to be normal distributions in a VAE. We can have discrete VAEs too wherein $z$ are discrete. The distribution you've shown for $z$ is completely valid. It's a multinomial/categorical distribution. However, a mistake in your example is that you take $z$ to be observed (you know the value of the sum $S$ = $z$). In VAEs, $z$ is supposed to be latent/unobserved, and $x$ is observed. So I don't think your example is valid.

Here's my attempt at a real-world example using Discrete VAE. Let's say your data $X$ is the output of rolling a die, so $X = \{1, 3, 5, 6, 1, 4, 2, ...\}$. You want to figure out what the probabilities of the six faces are. The die is loaded, so it isn't just 1/6 for each face.

Now comes the modeling part. You assume that there is some hidden (latent) variable $z$ that determines how the die is loaded. If $z = 0$, then the die favors {1,2,3}. If $z = 1$, the die favors {4,5,6}. So the generative story is as follows

  1. Sample $z$ from $p(z)$ = Bernoulli($\theta = 0.5$) - like a coin toss
  2. Sample $x$ from $p(x|z)$ = Multinoulli($[p_{01}, p_{02}, p_{03}, p_{04}, p_{05}, p_{06}]$) if $z$ is 0 or Multinoulli($[p_11..p_16]$) if $z$ is 1 (just a vague model)

You'd expect that $[p_{01}, p_{02}, p_{03}]$ are higher as compared to $[p_{04}, p_{05}, p_{06}]$, and vice-versa for $z = 1$ case.

The parameters you learn with VAE-style training are 12 $p_{ij}$s for the decoder part. For encoder, you could approximate the posterior with $q_{\phi}(z | x) = Bern(\phi)$.

Effectively, you are gaining two things by modeling the die problem as above.

  1. Given this data $X$, can I tell if my die is {1,2,3}-ish loaded or {4,5,6}-ish loaded? This comes from "Encoder" q(z | x). I'm being vague about what $z$ means because that's what it is. It's unknown. We just hope that it means something (like {4,5,6}-ish).
  2. If I sample $z$ (0 or 1) - can I tell what $x$ I'd get? This is "decoding" -- $p(x|z)$. This gives me the power to "generate" data that looks almost similar to rolls of the die.

Another example - data $X$ = heights of students (continuous), and hidden $z$ could be the grades they belong to (discrete).

Check this answer for an understanding of continuous VAEs with a better example. Here, $x$ is an image (discrete pixel data of MNIST digits), and $z$ is continuous. The directions/vectors in z-space could be "italic", "bold" etc. Generative story for a variational auto encoder

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  • $\begingroup$ Excellent explanation, much appreciated. According to your explanation I just got confused about what does $X$ mean in a training step. When we input a single $X$ from a batch to training, do we look at it as $X_i$ from the possible $X$s, or it is $X$ itself containing the possible $X_i$s inside? To follow your answer, do we input ${1,3,5,6...}$ as $X$ from a batch (so there are multiple like this in the batch), or it is the batch itself, and we input 1,3,5,6 separately? $\endgroup$ Dec 27, 2020 at 13:26
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    $\begingroup$ Thank you :) $X$ is the entire dataset; $x_i$ would be a single training example ($x_i$ = 3, $x_i$ = 1), etc. You could batch it too. Note that categorical variables are often one-hot encoded - so when you batch it, your input will be of shape $(B, 6)$, as in, $3 = [0, 0, 1, 0, 0, 0]$. As I mentioned, this is a vague model and probably not of much use - since the decoder is strong enough to model $p(X)$. There is something called posterior collapse in VAEs, wherein the decoder is so strong that it doesn't even use $z$. That is bad - because we want to know the unseen hidden variable $z$ info $\endgroup$
    – stbv
    Dec 27, 2020 at 17:54
  • $\begingroup$ In the case of MNIST images a single image is $x_i$, and the whole dataset (all batches) is $X$ right? Otherwise it doesn't make too much sense to me, and will be the case when I think I understand the process, then realise that I don't $\endgroup$ Dec 27, 2020 at 19:14
  • $\begingroup$ Yes, a single image is $x_i$. The image is represented as a matrix though - of size (32, 32, 3) for example - the RGB pixels of a 32*32 image. It is later flattened into a 32*32*3 vector. Each pixel can take values from [0...255] assuming 8-bits. So it's as if each image is equivalent to 32*32*3 rolls of a 256-faced die 😅 enormous complexity! So don't confuse a die with an image. $\endgroup$
    – stbv
    Dec 27, 2020 at 21:52

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