# Tag Info

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It means that $z$ has a (multivariate) normal distribution with 0 mean and identity covariance matrix. This essentially means each individual element of the vector $z$ has a standard normal distribution.

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In variational inference, the original objective is to minimize the Kullback-Leibler divergence between the variational distribution, $q(z \mid x)$, and the posterior, $p(z \mid x) = \frac{p(x, z)}{\int_z p(x, z)}$, given that the posterior can be difficult to directly infer with the Bayes rule, due to the denominator term, which can contain an intractable ...

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This is the analytical form of the KL divergence between two multivariate Gaussian densities with diagonal covariance matrices (i.e. we assume independence). More precisely, it's the KL divergence between the variational distribution q_{\boldsymbol{\phi}}(\mathbf{z}) = \mathcal{N}\left(\mathbf{z} ; \boldsymbol{\mu}, \mathbf{\Sigma} = \boldsymbol{\sigma}^... 4 The code is correct. Since OP asked for a proof, one follows. The usage in the code is straightforward if you observe that the authors are using the symbols unconventionally: sigma is the natural logarithm of the variance, where usually a normal distribution is characterized in terms of a mean \mu and variance. Some of the functions in OP's link even have ... 4 Let's assume the probability distributions are Gaussian (or normal) distributions. In other words, in the Bayes' rule \begin{align} p(z|x)=\frac{p(x|z)p(z)}{p(x)} \tag{1}\label{1} \end{align} The posterior p(z|x), the likelihood p(x|z), the prior p(z) and the evidence (or marginal) p(x) are Gaussian distributions. You can assume this because ... 3 The number of dimensions is a hyperparameter of your model, and you should do a hyperparameter search, like with any other parameters. There's also a tradeoff between dimension and training speed, so it should be small enough to be trainable in a reasonable time. 3 With Generative Adversarial Networks, all the generator cares about is fooling the discriminator. There's no requirement to be clever, or exhaustive, or make efficient use of the input space. As long as the discriminator returns "real" (vs. "fake") the generator "wins". The hope is that as the generator and discriminator are trained simultaneously, each ... 3 From this document, as you found here, X is an observed variable and Z is a hidden variable; p(X) is the density function of X. The posterior distribution of the hidden variables can then be written as follows using the Bayes’ Theorem:p(Z|X) = \frac{p(X|Z)p(Z)}{p(X)} = \frac{p(X|Z)p(Z)}{\int_Zp(X,Z)}$$Now base on what you post, if we denote ... 3 You can know it better, if you know the concept of entropy: Information entropy is the average rate at which information is produced by a stochastic source of data. The information content (also called the surprisal) of an event {\displaystyle E} is an increasing function of the reciprocal of the {\displaystyle p(E)} of the event, precisely {\... 2 The use of KL provides a more intuitive way of what the ELBO is attempting to maximize. Basically, we want to find a posterior approximation such that p(z\mid x) \approx q(z)\in\mathcal{Q}$$KL(q(z)\parallel p(z\mid x)) \rightarrow \min_{q(z)\in\mathcal{Q}}$$As a result of this, while finding this optimal posterior approximation, we maximize the ... 2 The only disadvantage and difference between these generative models and the method you describe, is the input. You describe inputting tags, where as for a GAN, or VAE, the generation segment of the model takes in some representation of a probability distribution. For a GAN, it's mostly random noise, and for a VAE it is some latent space (see nbros answer). ... 2 In the source code, the author defines sd by sd = 0.5 * tf.layers.dense(x, units=n_latent) which means that \operatorname{sd}\in \mathbb{R}^n. In particular, the support over sd includes negative numbers, which is something we want to avoid. Since standard deviations are always nonnegative, we can exponentiate to get us in the correct domain. ... 2 On page 5 of the VAE paper, it's clearly stated We let p_{\boldsymbol{\theta}}(\mathbf{x} \mid \mathbf{z}) be a multivariate Gaussian (in case of real-valued data) or Bernoulli (in case of binary data) whose distribution parameters are computed from \mathbf{z} with a MLP (a fully-connected neural network with a single hidden layer, see appendix \mathrm{... 2 Whilst you're right that for any continuous distribution P(X = x) = 0 \;; \forall x \in \mathcal{X} where \mathcal{X} is there support of the distribution, they are not referring to probabilities here, rather they are referring to density functions (though this should really be denoted with a lower case p to avoid confusion such as this). p(x|z) is a ... 2 To prove that the KL divergence does not satisfy the triangle inequality, you just need a counterexample. Definitions KL divergence Let's first recapitulate the definition of KL divergence for discrete probability distributions p and q (for simplicity).$$ D_{\text{KL}}(p\parallel q) = \sum_{x\in {\mathcal {X}}} p(x)\log \left( \frac {p(x)}{q(x)} \...

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I will try to answer your questions directly (but I guess I won't be able to), otherwise, this can become quite confusing, given the inconsistencies that can be found across different sources. In $logp_{\theta}(x^1,...,x^N)=D_{KL}(q_{\theta}(z|x^i)||p_{\phi}(z|x^i))+\mathbb{L}(\phi,\theta;x^i)$ why is $\theta$ and param for $p$ and $q$? In a few words, ...

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Your three dimensional latent representation consists of two images of mean pixels and covariance pixels as shown in Fig. 3. Which represents a Gaussian distribution with the mean and covariance for each pixel in the latent representation. Each pixel value is a random variable. Now, have a close look at KL-loss Eq. 3 and it's corresponding description in the ...

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GANs generally produce better photo-realistic images but can be difficult to work with. Conversely, VAEs are easier to train but don’t usually give the best results. I recommend picking VAEs if you don’t have a lot of time to experiment with GANs and photorealism isn’t paramount. There are exceptions such as Google’s VQ-VAE 2 which can compete with GANs for ...

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The contradictory loss is the same loss function that the discriminator would normally use, except with deliberately incorrect labels. That is, when you train the generator, the output of the generator is fed to the discriminator, but instead of the correct label (typically $0$ for a false image), the opposite label is applied (e.g. $1$ for a real image). ...

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I don't want to think about the correctness of your supposed ELBO equation now. Nevertheless, it's true that the ELBO can be rewritten in different ways (e.g. if you expand the KL divergence below, by applying its definition, you will end up with a different but equivalent version of the ELBO). I will use the most common (and definitely most intuitive, at ...

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There are many variations of the original VAE (proposed in the 2013 paper Auto-Encoding Variational Bayes), with different purposes (such as the generation of discrete data or graphs). Of course, I cannot enumerate all of them, so here I will list only the ones I am currently aware of. Conditional variational auto-encoder (CVAE) (2015, NeurIPS) Adversarial ...

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One of the core contributions presented in the paper consists of understanding at a deeper level the objective function used in Beta VAE and improving it More specifically, the authors started from Beta VAE OF \frac{1}{N} \sum_{i=1}^{N}\left[\mathbb{E}_{q\left(z | x^{(i)}\right)}\left[\log p\left(x^{(i)} | z\right)\right]-\beta K L\left(q\left(z | x^{(i)}...

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VAE's try and model the distribution of your data. So it's not going to learn " images composed of random noise at each pixel" per se (though, if overfitting, it could remember them). But it would be very capable of learning the simple noise distribution from which you sampled your random pixels

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The VAE uses the ELBO loss, which is composed of the KL term and the likelihood term. The ELBO loss is a lower bound on the evidence of your data, so if you maximize the ELBO you also maximize the evidence of the given data, which is what you indirectly want to do, i.e. you want the probability of your given data (i.e. the data in your dataset) to be high (...

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If you are mathematically inclined, here is an article that discusses the reasoning. What I get as a take away is that the VAE forces the learned latent space to be Gaussian due to the KL divergence term in the loss function. So, now we have a known distribution to sample from to create input vectors to feed to the decoder, to produce say images of dogs, ...

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Kind of neither, although leaning towards the first definition of the Mean as a simple average of values. It's a distribution parameter of the Gaussian, so it's the expected average of samples as the number of samples approaches infinity. The distrinction is that you could draw three samples, -2, 0 and -1, from a Standard Normal - the mean of samples would ...

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I will only focus on the VAE because I am more familiar with it, but the explanations may also apply to the GAN and other generative models. In the case of the VAE, you train a neural network not only to generate images but to represent them compactly in a so-called latent space, so you train the VAE to do dimensionality reduction. More precisely, the VAE ...

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[Answering my own question after 5 months of studying VAE models] The point of the MMD-VAE or InfoVAE is not exactly to emphasise on the visual quality of generated samples. It is to preserve greater amount of information through the encoding process. The MMD formulation stems from introducing a mutual coefficient factor into the Evidence Lower BOund (ELBO) ...

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Whether a discrete or continuous class, you can model it the same. Denote the encoder $q$ and the decoder $p$. Recall the variational autoencoder's goal is to minimize the $KL$ divergence between $q$ and $p$'s posterior. i.e. $\min_{\theta, \phi} \ KL(q(z|x;\theta) || p(z|x; \phi))$ where $\theta$ and $\phi$ parameterize the encoder and decoder respectively. ...

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The thing is, the decoder samples from a latent mu and sigma, so you cant sample from a raw image directly. But if you’re trying to put a random image into the encoder of a trained VAE to match it to some sample image (via reconstruction loss), then your random input image will converge to the target sample. This will work when the following VAE ...

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