# Tag Info

## Hot answers tagged loss-functions

9

I see no reason why decaying learning rates should create the kinds of jumps in losses that you are observing. It should "slow down" how quickly you "move", which in the case of a loss that otherwise consistently shrinks really should, at worst, just lead to a plateau in your losses (rather than those jumps). The first thing I observe in your code is that ...

7

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 ...

7

In general a cost function can be negative. The more negative, the better of course, because you are measuring a cost the objective is to minimise it. A standard Mean Squared Error function cannot be negative. The lowest possible value is $0$, when there is no output error from any example input. How can our cost function which is mean squared error have ...

7

The Focus of This Question "How can ... we process the data from the true distribution and the data from the generative model in the same iteration? Analyzing the Foundational Publication In the referenced page, Understanding Generative Adversarial Networks (2017), doctoral candidate Daniel Sieta correctly references Generative Adversarial Networks, ...

7

The MSE can be defined as $(\hat{y} - y)^2$, which should be equivalent to $(y - \hat{y})^2$ They are not just "equivalent". It is actually the exact same function, with two different ways to write it. $$(\hat{y} - y)^2 = (\hat{y} - y)(\hat{y} - y) = \hat{y}^2 -2\hat{y}y + y^2$$ $$(y - \hat{y})^2 = (y -\hat{y})(y - \hat{y}) = y^2 -2y\hat{y} + \hat{y}^2$$ ...

6

The derivative of $\mathcal{L_1}(y, x) = (\hat{y} - y)^2 = (f(x) - y)^2$ with respect to $\hat{y}$, where $f$ is the model and $\hat{y} = f(x)$ is the output of the model, is \begin{align} \frac{d}{d \hat{y}} \mathcal{L_1} &= \frac{d}{d \hat{y}} (\hat{y} - y)^2 \\ &= 2(\hat{y} - y) \frac{d}{d \hat{y}} (\hat{y} - y) \\ &= 2(\hat{y} - y) (1) \\ &...

5

The loss function used is the triplet loss function. Let me explain it part by part. Notation The $f^a_i$ means the anchor input image. The $f^p_i$ means the postive input image, which corresponds to the same people as the anchor image. The $f^n_i$ corresponds to the negative sample, which is a different person(input image) then the anchor image. The ...

5

As a rule of thumb, mean squared error (MSE) is more appropriate for regression problems, that is, problems where the output is a numerical value (i.e. a floating-point number or, in general, a real number). However, in principle, you can use the MSE for classification problems too (even though that may not be a good idea). MSE can be preceded by the ...

4

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 ...

4

I'll cover both L2 regularized loss, as well as Mean-Squared Error (MSE): MSE: L2 loss is continuously-differentiable across any domain, unlike L1 loss. This makes training more stable and allows for gradient-based optimization, as opposed to combinatorial optimization. Using L2 loss (without any regularization) corresponds to the Ordinary Least Squares ...

3

In general I agree with @nbro answer, nevertheless sticking strictly to this specific question I'd like to share some speculations: what the author of the question provides us with is the Loss Function Shape so I'll try to use the full information here to compare the 2 minima looking at the LF steepness we observe the Left LM is in a steeper region than ...

3

For the first question, RMSE and Euclidean distance have no difference, not that i know of. For the second question, you only need the common loss function for normal tasks. MSE is a common loss function used in linear regression tasks as well as loss function similar in nature like the RMSE. For classification tasks, Cross Entropy Loss is preferred. For ...

3

Depends on what does 1 represent in your task. If you are trying to predict household prices and 1 represents \$1, I think the average validation loss is good. If 1 represents \$10000 in this case, probably something is not right. But remember that there are 2 parts contributing to the overall loss. The mse loss and the l2 penalty loss. (Also remember that ...

3

The validation loss settles exactly at an error of one. Probably means there's something off with either the kind of data validation set has or with something in the training. An exact validation loss of one almost definitely means there's something off. I'd recommend before doing anything thoroughly go through your data or see if there's anything to debug ...

3

Welcome to AI Stack exchange! You're right, as the network is initialised randomly, the resultant function is essentially impossible to get your head around. This is because most of the time the network has >4 dimensions (4 can be graphed with some effort and a lot of color), and as such is literally beyond human comprehension via graphing. So what do we ...

3

I know that gradient descent allows you to find the local minimum of a function. What I don't know is what exactly that function IS. It's usually called the loss function (and, in general, objective function) and often denoted as $\mathcal{L}$ or $L$ (or something like that, i.e. it is not really important how you denote it). The specific function used as a ...

3

The book Deep Learning by Goodfellow, Bengio, and Courville says (Sec 8.3.3, p 292 in my copy) states that Unfortunately, in the stochastic gradient case, Nesterov momentum does not improve the rate of convergence. I'm not sure why this is, but the theoretical advantage depends on a convex problem, and from this, it sounds like the practical advantage ...

3

I've implemented this exact scenario before; your approach would most likely be successful, but I think it could be simplified. Therefore, when deciding on which action to pick, agent sets Q-values to 0 for all the illegal moves while normalizing the values of the rest. In DQN, the Q-values are used to find the best action. To determine the best action in ...

3

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

When you use the softmax activation function is usually as a last layer of your network and to get an output that is a vector. Now your confusion is about shapes, so let's review a bit of calculus. If you have a function $$f:\mathbb{R}\rightarrow\mathbb{R}$$ the derivative is a function on its own and you have $$f':\mathbb{R}\rightarrow\mathbb{R}.$$ If you ...

3

I will first explain briefly to you the difference between supervised learning and reinforcement learning to make sure that you don't have any misunderstandings. In supervised learning you are provided with some data $\{(\textbf{x}_i, y_i)\}_{i=1}^n$ where $\textbf{x}_i$ are the features for data point $i$ and $y_i$ is its true label. Now, the aim of ...

2

Let's start at the beginning. GANs are models that can learn to create data that is similar to the data that we give them. When training a generative model other than a GAN, the easiest loss function to come up with is probably the Mean Squared Error (MSE). Kindly allow me to give you an example (Trickot L 2017): Now suppose you want to generate cats ; you ...

2

It may come as a surprise, after learning all about dynamic programming, temporal difference learning, SARSA/Q-learning, to then discover that there is yet another whole dimension to reinforcement learning (on top of choices for on-policy/off-policy, model-based/model-free, bootstrap/monte carlo etc). That is value-based vs policy-based methods. Policy-based ...

2

Penalty (barrier function) is perfectly valid and simplest method for simplex type constraint (L1 norm is simplex constraint on absolute values). Any type of barrier function may work, logarithmic, reciprocal or quadratic. All of them supported by any major framework(pytorch, tensorflow), just add them to loss function. You would need some hyperparameter ...

2

Classic Question Not only can the reliability or accuracy needs be asymmetrically distributed between category boundaries, but the asymmetry might not be describable in terms of a first degree polynomial translation of the error surface. The question data set are images belonging to class $c \in \Big( A: \{A_1, \, ..., \, A_n\} \land B: \{B_1, \, ..., \, ... 2 You can find an implementation of the REINFORCE algorithm (as defined in your question) in PyTorch at the following URL: https://github.com/JamesChuanggg/pytorch-REINFORCE/. First of all, I would like to note that a policy can be represented or implemented as a neural network, where the input is the state (you are currently in) and the output is a "... 2 I don't think that the concept of generalization is (directly) related to the "shape" of the function close to the point where it attains a minimum. The concept of generalisation refers to when a trained model is able "perform well" on unseen data (that is, data not seen during the training phase). If a trained model does not generalise well, then it might ... 2 First of all you made a mistake, equation 8 in the paper is defined with$\frac{\partial L(\theta)}{\partial s_t}$not$\frac{\partial L(\theta)}{\partial\theta}$. Loss is defined as:$L(\theta) = - \mathbb{E}_{w^s \sim p_{\theta}}[r(w^s)]$If we use definition of expectation (for discrete case):$\mathbb{E}[X] = \sum\limits_{i} p_i(x_i)x_i$we get ... 2 But, if$t^i - o^i\$ is negative, doesn't the power of 2 eliminate any negative result? In the loss function, yes that is correct, and is what you want - a measurement that gets higher due to any difference between the predicted and correct results. Minimising the value of that measurement is a goal for the optimiser. How can then exist any negative ...

2

There are a few things you could do to improve this NN, but are probably worth covering in different questions. Your main problem though is that you forgot to reset the gradient after each training batch. You need to call optim.zero_grad() in order to do this, at the start of each training loop. Otherwise, using PyTorch, the gradient values keep ...

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