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I already search for this question but I can't find any convincing explanation so I want to ask it here. my problem is with softmax activation function and cross-entropy.why they can produce a better estimation for multi-classification problems? I search for it and people keep talking about the advantage of softmax like : Each value ranges between 0 and 1 or The sum of all values is always 1 and etc.

but this paper directly said: It is unclear why the log-softmax loss would perform better than other loss alternatives.

I want to know is there any proof for better performances of soft-max and cross-entropy function or just we conclude it by experimental.

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The below answer is copy and paste from arxiv.org/pdf/1711.07758.pdf

Supposing the dataset has input $X$ and label $Y$,
the task is to find a good prediction of $Y$ using $X$.

The prediction $\hat{Y}$ needs to maximize the conditional entropy $H(\hat{Y} \mid X)$ while preserving the same distribution with data $(X, Y)$.

This is formulated as:


$$ \begin{gathered} \min -H(\hat{Y} \mid X) \\ \text { s.t. } P(X, Y)=P(X, \hat{Y}), \sum_{\hat{Y}} P(\hat{Y} \mid X)=1 \end{gathered} $$


This optimization question can be solve by lagrangian multiplier method:


$$ \begin{aligned} L = & \sum_{X, \hat{Y}} P(X) P(\hat{Y} \mid X) \log (P(\hat{Y} \mid X))+\omega_0\left(1-\sum_{\hat{Y}} P(\hat{Y} \mid X)\right) \\ & +\sum_{X, Y} \omega_i(P(X, Y)-P(X) P(\hat{Y}=Y \mid X)) \end{aligned} $$


The above equation can be equivalently written with the original defined predicate function in (Berger et al.,
1996):


$$ \begin{aligned} L = & \sum_{X, \hat{Y}} P(X) P(\hat{Y} \mid X) \log (P(\hat{Y} \mid X))+\omega_0\left(1-\sum_{\hat{Y}} P(\hat{Y} \mid X)\right) \\ & +\sum_i \omega_i\left(\sum_{X, Y} P(X, Y) f_i(X, Y)-\sum_{X, \hat{Y}} P(X) P(\hat{Y} \mid X) f_i(X, \hat{Y})\right) \end{aligned} $$


where $f_i(X, Y)$ is predicate function,
which equalizes 1 when $(X, Y)$ satisfies a certain status:


$$ f_i(X, Y)=\left\{\begin{array}{lr} 1, & X=x_i, Y=y_i \\ 0, & \text { others } \end{array}\right. $$


The solution to the above problem is:


$$ \begin{gathered} P_\omega(\hat{Y}=y \mid X=x)=\frac{1}{Z_\omega(x)} \exp \left(\sum_i \omega_i f_i(x, y)\right) \\ Z_\omega(x)=\sum_y \exp \left(\sum_i \omega_i f_i(x, y)\right) \end{gathered} $$

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As far as I know, there is no proof published.

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  • $\begingroup$ This is being automatically flagged as low quality, probably because it is so short. At present it is more of a comment than an answer by our standards. Can you expand on it? We can also turn it into a comment. $\endgroup$ Commented Aug 11, 2018 at 16:18

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