# Tag Info

5

$\|x\| = |x|$ denotes the absolute value norm, which is a special case of the $L_1$ norm defined on the 1-D vector spaces formed by real or complex numbers. $\|\textbf{x}\|_1 = \sum_{i=1}^n|x_i|$ denotes the Taxicab / Manhattan norm, relating to how a Taxi would drive along a rectangular grid of roads to reach a point $(x, y)$ from $(0,0)$. $\|\textbf{x}\|... 5 When lambda = 0 as in TD(0), how does the method learn? As it appears, with lambda = 0, there will never be a change in weight and hence no learning. I think the detail that you're missing is that one of the terms in the sum (the final "iteration" of the sum, the case where$k = t$) has$\lambda$raised to the power$0$, and anything raised to the power$0$... 4 Here is a paper with the mathematical definition of each term: Let Nt,n,σ,L be all target functions that can be implemented using a neural network of depth t, size n, activation function σ, and when we restrict the input weights of each neuron to be |w|1 + |b| ≤ L. 3 It doesn't seem that it is a "proper" symbol. I guess that$\sup$simply refers to the supremum, that is, you want to select actions that maximize the quantity that comes to the right of$\sup$, while$\text{dist}$is simply a proxy for any possible distance between distributions. For example, you can replace$\text{dist}$with the Kullback-Leibler ... 3 The square brackets$[]$in$[\tau_{ij}]^\alpha$and$[\eta_{ij}]^\beta$may be just a way of emphasing that the elements$\tau_{ij} \in \mathbb{R}$and$\eta_{ij} \in \mathbb{R}$of respectively the matrices$\mathbf{\tau} \in \mathbb{R}^{n \times n}$and$\mathbf{\eta} \in \mathbb{R}^{n \times n}$(where$n$is the number of nodes in the graph) are ... 3 This is standard backpropagation. The gradient term you see is in fact a vector of partial derivatives where each element is the partial derivative of the log-likelihood with respect to each element of the parameter vector$\theta$. Therefore, it has the same dimensionality as$\theta$. Each element of the parameter vector is then updated with the respective ... 3 The first part of this answer is a little background that might bolster your intuition for what's going on. The second part is the more practical and direct answer to your question. The gradient is just the generalization of the derivative to multivariable functions. The gradient of a function at a certain point is a vector that points in the direction of ... 2$1-\sum_i(e_i-a_i)^2\sum$- there just means sum. It is greek letter for S. You can rewrite the above formula as$1 -[(e_1 - a_1)^2+(e_2-a_2)^2+(e_3-a_3)^2+\ldots ]\sum$just helps us avoid writing dozens of$+$signs. Read more here. What they are doing here is subtracting taking difference of expected value$e_1$and the actual value$a_1$... 2 To understand this equation first you need to understand the context in which it is first introduced. We have two neural networks (i.e.$D$and$G$) that are playing a minimax game. This means that they have competing goals. Let's look at each one separately: Generator Before we start, you should note that throughout the whole paper the notion of the data-... 2 At page 130 of the same book, the author states that$\hat{p}_\text{data}$is an empirical distribution defined by the training data. Similarly, at page 129, he states that$p_\text{data}$is the true distribution that generates the set$\mathbb{X} = \{ \boldsymbol{x}^{(1)}, \dots, \boldsymbol{x}^{(m)} \}$. What is the difference between$\hat{p}_\text{data}...

2

This equation and more information of it can be found in Expectation Maximization Wikipedia site and the explanation there was as follows (formula there in two parts): Some more explanation from same page: In statistics, an expectation–maximization (EM) algorithm is an iterative method to find maximum likelihood or maximum a posteriori (MAP) estimates of ...

2

So, what is the purpose of the new index for $V$ in Chapter 7, and why is it more important at this particular chapter? My guess would be that your intuition is correct, and that it's mostly introduced just to clarify exactly which "version" of our value function approximator is going to be used in any particular equation. In previous chapters, which ...

2

In full: The limit, as standard deviation $\sigma$ tends towards zero, of the gradient with respect to vector $\mathbf{x}$, of the expectation - where perturbation $\epsilon$ follows the normal distribution with mean 0 and variance $\sigma^2$ times identity vector $[1,1,1,1...]$ * - of any function $f$ of $\mathbf{x}$ plus $\epsilon$ is equal to the ...

2

The dot ($.$) at the end of $T(s,a,.)$ shows all possible states that we can go from state $S$ by doing action $a$. As you know there are some probabilities here for choosing those states, that the sum of these probabilities is equal to 1. Hence, $T(s,a,.)$ is a probability distribution.

2

I assume this is an iterative function. It means the current $V(S_t)$ is the sum of the previous plus some adjustment. The arrow is like an assignment. In code, you would do vst = vst + alpha * (gt - vst) So vst will be overwritten.

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

Your first option is correct: $$r(s,a) = \mathbb{E}\left[R_t|S_{t-1}=s,A_{t-1}=a\right]=\sum_{r\in \mathcal{R}}\left[r\sum_{s'\in \mathcal{S}}p(s',r|s,a)\right]$$ It's partly a matter of taste, but I prefer not moving the $r$ into the double sum, because its value does not change in the "inner loop". There is a small amount of intuition to be had that way ...

1

In reinforcement learning, you can distinguish algorithms based on the functions they use to ultimately find the policy (which is the goal in RL anyway!). algorithms that attempt to find an optimal value function (an example is Q-learning, which attempts to find a state-action value function), then derive the policy from the value function algorithms that ...

1

I agree that this notation is unclear. I would interpret it as follows: Given that the expression is supposed to denote the average norm $|p_i|$ is likely the cardinality of the set $\{p_i\}$. In that case the expression would just be the sum over all norms divided by the number of norms, resulting in the average norm. The authors likely use this ...

1

This is a commonly used notation in theoretical computer science. $[m]$ is not the variable $m$, but is instead the set of integers from $1$ to $m$ inclusive. The empirical error equation thus reads in English: The cardinality of a set consisting of the elements $i$ of the set of integers $[m]$ such that the hypothesis given input $x_i$ disagrees with ...

1

Yes, since $\tilde{\theta}$ is a vector, to define its distribution one needs a covariance matrix. Here $I$ is the identity matrix, which means that the noise has a zero-mean normal distribution with standard deviation $\sigma$, and different components of this noise are uncorrelated.

1

The change in nomenclature from what you expected is Lisa Meeden's choice, for unknown reasons. Those with whom she published in the past used $\epsilon$ to represent error, the result of a loss function. Why she did not use $\alpha$ may be because the letter $a$ was used elsewhere in the formula, but, if that was the reasons, that wasn't a great one. ...

Only top voted, non community-wiki answers of a minimum length are eligible