Why is there an \(i\) here (in \( \left(x_{n}\right)_{\boldsymbol{i}} \) )
\( \left(x_{n}\right)_{\boldsymbol{i}} \) denotes: sample \(n\), element \(i\). Note that in this case, \(w\) and \(x\) have the same number of elements. It might have been clearer if \(x\) was in bold font like this:
See the left hand side of the equation: the gradient of the loss \(\mathcal{L}\) is taken with respect to a single parameter \(w_i\) from the parameter vector \(\mathbf{w}\). This single parameter is obtained by indexing \(\boldsymbol{w}\) with \(i\).
MAE subgradient index i
Why is there an \(i\) here (in \( \left(x_{n}\right)_{\boldsymbol{i}} \) ) and there is no sigma with \(i\)?
$$\frac{\partial \mathcal{L}(\boldsymbol{w})}{\partial w_{i}}=\frac{1}{N} \sum_{n=1}^{N}-\left(x_{n}\right)_{i} \operatorname{sign}\left(y_{n}-\boldsymbol{x}_{n}^{\top} \boldsymbol{w}\right)$$
source: https://github.com/epfml/ML_course/blob/master/labs/ex02/solution-partial/solutions-theory-questions.pdf
Where can I find a more detailed solution?
\( \left(x_{n}\right)_{\boldsymbol{i}} \) denotes: sample \(n\), element \(i\). Note that in this case, \(w\) and \(x\) have the same number of elements. It might have been clearer if \(x\) was in bold font like this:
$$\frac{\partial \mathcal{L}(\boldsymbol{w})}{\partial w_{i}}=\frac{1}{N} \sum_{n=1}^{N}-\left(\boldsymbol{x}_{n}\right)_{i} \operatorname{sign}\left(y_{n}-\boldsymbol{x}_{n}^{\top} \boldsymbol{w}\right)$$
See the left hand side of the equation: the gradient of the loss \(\mathcal{L}\) is taken with respect to a single parameter \(w_i\) from the parameter vector \(\mathbf{w}\). This single parameter is obtained by indexing \(\boldsymbol{w}\) with \(i\).
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