Thank you for your sharing your code,
I have some questions.
In your paper, appendix A, For categorical variable y∗, Varp(y∗|x∗,ω)(y∗) = Ep(y∗|x∗,ω)(y∗⊗2)−Ep(y∗|x∗,ω)(y∗)⊗2 and it is Ep(y∗|x∗,ω){diag(y∗)}−Ep(y∗|x∗,ω)(y∗)⊗2 because y∗ is one-hot encoded.
However,
You known, the output of softmax is not one-hot encoded variable, each position is the probability of the classes , for example,
the output is [0.9,0.1,0.1](not one-hot), the ground trouth is 1,0,0, so do we really can use the "it is Ep(y∗|x∗,ω){diag(y∗)}−Ep(y∗|x∗,ω)(y∗)⊗2 because y∗ is one-hot encoded. "?
I do not think I understand your question. What do you mean by do we really can use the "it is Ep(y∗|x∗,ω){diag(y∗)}−Ep(y∗|x∗,ω)(y∗)⊗2 because y∗ is one-hot encoded. "?
Thank you for your sharing your code, I have some questions. In your paper, appendix A, For categorical variable y∗, Varp(y∗|x∗,ω)(y∗) = Ep(y∗|x∗,ω)(y∗⊗2)−Ep(y∗|x∗,ω)(y∗)⊗2 and it is Ep(y∗|x∗,ω){diag(y∗)}−Ep(y∗|x∗,ω)(y∗)⊗2 because y∗ is one-hot encoded.
However, You known, the output of softmax is not one-hot encoded variable, each position is the probability of the classes , for example, the output is [0.9,0.1,0.1](not one-hot), the ground trouth is 1,0,0, so do we really can use the "it is Ep(y∗|x∗,ω){diag(y∗)}−Ep(y∗|x∗,ω)(y∗)⊗2 because y∗ is one-hot encoded. "?
Feng