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Actually, the cross-entropy loss function would be appropriate here, since it measures the "distance" between a distribution $q$ and the "true" distribution $p$. You are right, though, that using a loss function called "cross_entropy" in many APIs would be a mistake. This is because these functions, as you said, assume a one-hot ...


6

The probability density is used to 'measure how good' the parameters are because it is a natural way of quantifying if these parameters are good for the observed data. Also, as the notation often causes some confusion, $L(\theta | x)$ denotes the probability of all of your observed data, not just one value. Also the "$|$" may cause confusion as it ...


4

It seems your question is concerned with how an empirical mean works. It is indeed true that, if all $x^{(i)}$ are independent identically distributed realisations of a random variable $X$, then $\lim_{n \rightarrow \infty} \frac{1}{n}\sum_{i=1}^n f(x^{(i)}) = \mathbb{E}[f(X)]$. This is a standard result in statistics known as the law of large numbers.


3

I think that the normalisation factor is assumed to be non-zero. So, in practice, I guess, you must eventually check that $P(z \mid b, a)$ is non-zero (even though, I guess, it will likely never be zero because of round-off errors in computers). The formula to calculate $b'(s')$ comes from its definition, which is based on Bayes' theorem, where the ...


3

Also, in general, in the conditional expectation, which distribution do we compute the expectation with respect to? From what I have seen, in $\mathbb{E}[X|Y]$, we always calculate the expected value over distribution $X$. No, for $\mathbb{E}[X|Y]$ we take expectation of $X$ with respect to the conditional distribution $X|Y$, i.e. $$\mathbb{E}[X|Y] = \...


3

Yes. In Machine Learning we consider that the samples in your training set are sampled from an underlying distribution called the data generating distribution. Generative models classify the samples by trying to learn the distribution of the data. In most cases, either the model is incapable of doing so, or the training samples aren't enough to properly ...


3

Well, there are some questions here... Does it (Deep Learning) try to learn a continuous distribution based on the training-set and its corresponding mappings, and map unseen examples from this learned distribution? Yes. Talking about Deep Artificial Neural Networks, they try to learn continuous distribution using continuous activation functions in each ...


3

Random variables You do not necessarily need to understand the concept of a random variable (r.v.) to understand the concept of a probability distribution, but the concept of a random variable is strictly connected to the concept of a probability distribution (given that each random variable has an associated probability distribution), so, before proceeding, ...


2

A probability distribution in ML is the same as a probability distribution elsewhere. A probability distribution (or probability function, or probability mass function, or probability density function) is any function that accepts as input elements of some specific set $x \in X$, and produces as output, real-valued numbers between 0 and 1 (inclusive), such ...


2

Introduction: MAP finds a point estimate! As opposed to your apparently current belief, in maximum a posteriori (MAP) estimation, you are looking for a point estimate (a number or vector) rather than a full probability distribution. The MAP estimation can be seen as a Bayesian version of the maximum likelihood estimation (MLE). Therefore, I will first ...


2

Lets start with question 1) how does JS-divergence handles zeros? by definition: \begin{align} D_{JS}(p||q) &= \frac{1}{2}[D_{KL}(p||\frac{p+q}{2}) + D_{KL}(q||\frac{p+q}{2})] \\ &= \frac{1}{2}\sum_{x\in\Omega} [p(x)log(\frac{2 p(x)}{p(x)+q(x)}) + q(x)log(\frac{2 q(x)}{p(x)+q(x)})] \end{align} Where $\Omega$ is the union of the domains of $p$ and ...


2

The language used here is confusing me, because it is discussing a "distribution", as in a "probability distribution", but then refers to inputs, which are data gathered from outside of any probability distribution. Based on the limited information my studying of machine learning has taught me so far, my understanding is that the machine learning algorithm (...


2

This is the definition of conditional probability + Total probability decomposition formula: $p(y|x) = \frac{p(y,x}{p(x)} = \frac{p(x,y)}{\sum_{y'}p(x,y')}$. The idea is to use some unsupervised learning algorithm to learn the distribution $p(x,y)$ for every possible value of $y$, and by using the previous formula you can find $p(y|x)$.


2

The probabilistic models that represent distributions implicitly are, for example, the GANs. (Goodfellow is one of the authors of the original GAN model). In the paper Variational Inference using Implicit Distributions (2017), the authors write Implicit distributions are probability models whose probability density function may be intractable, but there is ...


2

@The Pointer the $2^n$ came from the question: How many function do we need to have if each of the $n$ inputs can be missing? example: $f_1(\text{missing}, x_2, x_3, \dots, x_n)$ for $x_1$ missing $f_2(x_1, x_2, \text{missing}, x_4, \text{missing}, \dots, x_n)$ for $x_3$ and $x_5$ missing. So this problem is a combinatorial one and the event for each $x_i$ ...


2

This question is very general in the sense that the reason may differ depending on the area of ML you are considering. Below are two different areas of ML where the KL-divergence is a natural consequence: Classification: maximizing the log-likelihood (or minimizing the negative log-likelihood) is equivalent to minimizing KL divergence as typical used in DL-...


2

Some lines above the author says By the law of large numbers we can also assume that the total input to the node has a Gaussian distribution hence we can assume $X \sim \mathcal{N}(0,1)$ with the $X$ domain being continuous Then he says the input vector is assumed to be binary which changes the domain from continuous to discrete so we can discretize ...


2

The idea behind this kind of reasoning is that there is a "true" distribution (unknown to us, mere mortals) and that the data is generated following this distribution. But what we don't really know the shape of the distribution, all we know is the distribution of the data that we have. This is called the empirical distribution. Let's see a simple ...


1

In ML we always deal with unknown probability distributions from which the data comes. The most common way to calculate the distance between real and model distribution is $KL$ divergence. Why Kullback–Leibler divergence? Although there are other loss functions (e.g. MSE, MAE), $KL$ divergence is natural when we are dealing with probability distributions. It ...


1

Consider the case of binary classification, i.e. you want to classify each input $x$ into one of two classes: $y_1$ or $y_2$. For example, in the context of object classification, $y_1$ could be "cat" and $y_2$ could be "dog", and $x$ is an image that contains one main object. In certain cases, $x$ cannot be easily classified. For example,...


1

To add to nbro's answer, I'd say also that much of the time the distance measure isn't simply a design decision, rather it comes up naturally from the model of the problem. For instance, minimizing the KL divergence between your policy and the softmax of the Q values at a given state is equivalent to policy optimization where the optimality at a given state ...


1

I did not read those two specified linked/cited papers and I am not currently familiar with the total variation distance, but I think I can answer some of your questions, given that I am reasonably familiar with the KL divergence. When you compute the $D_{KL}$ between two polices, what does that tell you about them The KL divergence is a measure of "...


1

If you're using a discount factor less than 1, you should be able to compute a maximum return (likewise, a minimum return) based on the max (min) reward you can earn at each timestep. However, this issue you bring up is usually cited as a difficulty with C51. I think people tend to simply use fixed values for the min/max return (or just make rough estimates)....


1

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


1

Yes you can, provided you know about $f$ and $g$. Expression $X3 = f(X1, g(X1))$can be written as $X3 = h(X1)$ where $h$ takes into account both $f$ and $g$. After this finding the PDF is simple by differentiating the CDF: $$ F_{X3} (x3) = P(X3 \leq x3) = P(h(X1) \leq x3) = P(X1 \leq h^{-1}(x3))$$ $$ \frac {d F_{X3} (x3)}{dx3} = \frac {d P(X1 \leq h^{-1}(...


1

Extracting a joint PDF just means that you create a model that models the behavior of several variables combined instead of in isolation. If these variables aren't independent and your loss functions is influenced by all of them, you obviously have to learn this joint PDF to minimize your loss. So I don't see this statement as particularly mysterious.


1

Dirichlet is the Multi Variate version of the Beta distribution. In general, these distributions can be thought to model the probability of modelling a probability distribution. The support Dirichlet distribution is defined as follows: $$ S_K = \{ x:0 \leq x_k \leq 1, \sum_{k=1}^K x_k=1 \} $$ and the PDF is defined as: $$Dir(x|\alpha) = \frac{1}{B(\...


1

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, ...


1

Intuitively, this is similar to the case when you are making predictions but you don't have all the necessary information to make the most accurate prediction or maybe there isn't a single accurate prediction, so you have a set of possible predictions (rather than a single prediction). For example, if you hadn't seen the last Liverpool game (in the ...


1

When we say that we have $N$ points that were "drawn from the probability distribution or probability density", this means that every point $x_n$ had the correct probability $p(x_n)$ of being sampled from the distribution when we were sampling our $n^{th}$ point. For example, suppose that we wish to compute/estimate the expected value of a distribution ...


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