the Matlab cross-entropy loss has this form:
loss = -sum(W*(T.*log(Y)))/N;
I would like to "clamp" it so that the log function output is bounded, for example it cannot be less than 100.
Can we do it?

 Accepted Answer

David Goodmanson
David Goodmanson on 3 Sep 2020
Edited: David Goodmanson on 6 Sep 2020
Hi Matt,
z = log(Y);
z(z<100) = 100;
loss = -sum(W*(T.*z))/N;
In the link you provided, they talk about a limit of -100 rather than +100. The former appears to make more sense. Lots of possibilities for a smooth differentiable cutoff, here is one, assuming Y>=0
Ylimit = -100;
loss = -sum(W*(T.*log(Y+exp(Ylimit)))/N;

3 Comments

That does seem reasonable but we have to remember that this is machine learning loss function and it has to be differentiable. So I am not sure if that answer will work.
Hi Matt,
see amended answer.
probably a smart approach.

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