the diag problem

[ponykid]

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

[ykwon0407]

Hi Feng,

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. "?