# Tag Info

6

It depends on your loss function, but you probably need to tweak it. If you are using an update rule like loss = -log(probabilities) * reward, then your loss is high when you unexpectedly got a large reward—the policy will update to make that action more likely to realize that gain. Conversely, if you get a negative reward with high probability, this will ...

4

I guess the issue is you lost track of where the samples came from and since you requested a math explanation I'll try to go step by step using my notation and without checking other material to avoid being biased by how other authors present it So we start from $$L(D,G) = E_{x \sim p_{r}(x)} \log(D(x)) + E_{x \sim p_{g}(x)}\log(1 - D(x))$$ then you apply ...

4

It's the same thing, first version is the special case of the more general one. In the first case you only have two classes, it's binary cross-entropy, and they also included iteration over batch of samples. In the second case you have multiple classes and in the current form it's only for a single sample. In the first case there is only one output, if you ...

3

The sentences coming from the same document, author, etc., are unlikely to be independent, that is, the occurrence of a sentence $s_i$ in a certain document $d$ is likely correlated with the occurrence of another sentence $s_j$. If they are not independent, they can also not be independent and identically distributed (which is a stronger condition). The same ...

2

The cross-entropy loss will always be positive because the probability is in the range $[0, 1]$, so $-ln(p)$ will always be positive.

2

In short: It is easy to quantify information, but it is not easy to quantify its usefulness I'm not sure how exactly you are looking to formalise your experiment, but it might be helpful to consider these points: There is no such thing as an absolute measure of information. The amount of information contained in some dataset is dependent on the underlying ...

2

Why wouldn't they work? Each neuron's output is equal to a function over the sum of all its weights multiplied by their corresponding neurons. If that function is the Sigmoid function, then the output is squashed from $[0,1]$. If the entire layer uses a SoftMax function, then the output of all neurons is squashed from $[0,1]$ and their sum equals 1. In other ...

1

To provide a good answer would fill several pages. To keep it very simple try many different loss functions on your model. Your goal is to have the highest performance based on some desired prediction metric (e.g., RMSE, MAE, MAPE, etc.). You almost always have plenty of time to try many loss functions so you don't need to have a full understanding, and ...

1

Alright. Consider an ordinary neural network, so, in the last layer, we have, $z^{[L]} = W^{[L]} a^{[L-1]} + b^{[L]}$, where $a^{[L]} = \sigma(z^{[L]})$, where $\sigma$ is the softmax activation: $$\sigma(\mathbf z)_{i} = \frac{e^{z_i}}{\sum_k e^{z_k}}$$ I think, one of the most effective ways of not to get confused about all these matrices with different ...

1

Yes, $E$ is the cross-entropy function and a direct generalization of the binary case. For the binary case, probability to belong to the class $1$ is given by a sigmoid function $\sigma(x)$ of the output $x$, and the probability to belong to the class $0$ is $1 - \sigma(x)$. Therefore the binary crossentropy will give:  -\sum_i (l_i \log \sigma(x) + (1 - ...

1

$\textbf{Remark.}$ I'd leave this as a comment if I could. Regarding notation (which I believe may be the cause of your issue here), the loss function is better written as \begin{align*} \operatorname{Loss} &= \frac{1}{m}\sum_{i=1}^m \left(\log D\big(x^{(i)}\big) + \log\Big(1-D\big(G\big(z^{(i)}\big)\big)\right)\\ &\approx \mathbb{E}_x[\log D(x)] + \...

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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 ...

1

Short answer: larger gradients That is not the derivative of the softmax function. $t - o$ is the combined derivative of the softmax function and cross entropy loss. Cross entropy loss is used to simplify the derivative of the softmax function. In the end, you do end up with a different gradients. It would be like if you ignored the sigmoid derivative when ...

1

You don't need to manage negative rewards separately, if you implemented the algorithm correctly it will work regardless if the rewards are negative or not. You seem to be using rewards for the loss but you should be using the return which is the sum of the rewards for some state action pair from that point until the end of trajectory. You also seem to be ...

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I want to make it so that if the correct label is 3, then it will penalize the model less heavily if it classifies a 4 than a 7 because 4 is closer numerically to 3 than 7 is. How do I do this? Really you should not, because the symbols used (Arabic numerals) do not have direct relation to quantity in the same way e.g. tally counts or dots do. They are good ...

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https://stats.stackexchange.com/questions/260505/machine-learning-should-i-use-a-categorical-cross-entropy-or-binary-cross-entro Is relevant. based on my reading when you have a NN and do Binary crossentropy on what you might call 'Linked category data' the accuracy can tend to be better than in a Categorical crossentropy model. The binary aspect implies ...

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Christopher Olah's blog post describes it better that I ever could. Basically, most data we come across can't be separated with a single line, but with some kind of curve. Non-linearities allow us to distort the input space in ways that make the data linearly separable, making classification more accuarate.

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