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Sep 25, 2018 at 1:33 comment added gph yes, in the simple example it is a scalar, then it is generalized to vector version.
Sep 21, 2018 at 15:20 comment added user12075 you are right. I think the author means $t^{(t)}$ to be a scalar. I was thinking it as a one hot encoding, but it's not here.
Sep 21, 2018 at 7:34 comment added gph Thank you for help! I was too impatient to ask the question when I hadn't read the chapters before RNN because I had thought it simple. I read the underlying chapter again and wrote an answer below to prove the equation (10.18). By the way, I think $y^{(t)}$ here is a scalar at certain time t.
Sep 20, 2018 at 3:37 comment added user12075 $L$ is the cross-entropy loss defined in (10.14). For the indicator function, you are right that the condition is "scalar $i$"=="vector $y^{t}$", in this case the indicator function returns a vector whose k'th elements are indicator of whether that element in $i$ equals the k'th element of $y^{t}$. Similar like you compare a 1-D numpy array with a scalar in python.
Sep 20, 2018 at 3:16 comment added gph And the expression of $L$? I have thought it was based on assumption that :$p(y|x;\theta)\sim \mathcal{N}(y,\mu=\hat{y},\sigma=1)$, i.e Gaussian distribution. So $L=\sum_t\frac{1}{2}\Vert\hat{\boldsymbol{y}}^{(t)}-\boldsymbol{y}^{(t)}\Vert^2+Constant$ But it seems that here the authors use the cross entropy loss like $L=\sum_t \boldsymbol{y}^{(t)}t\log{\hat{\boldsymbol{y}}^{(t)}}$? So do you know what is the expression of loss $L$ here?Thank you!
Sep 20, 2018 at 2:59 comment added gph Thank you for your help! I have a further question that shouldn't $y^{(t)}$ be a vector $\boldsymbol{y}^{t}$ with the same length as $\hat{\boldsymbol{y}}^{(t)}$? Then how could it be equal to a scalar $i$?
Sep 20, 2018 at 2:54 vote accept gph
Sep 19, 2018 at 15:23 history answered user12075 CC BY-SA 4.0