### first question of ex 3

parent a6b54745
 ... ... @@ -161,5 +161,29 @@ \text{s.t.} & \left\| A^\trans \nu \right\|_{\infty} \leqslant 1 \end{array} $} \end{center} \section*{Exercise 3 (Data Separation)} \paragraph{1.} First, we can remark that (Sep. 2) is feasible by simply taking the point$(0, \mathbf{1})$and bounded because the objective function is non-negative for$z \succeq 0$. \\ Let$(\omega, z)$a feasible point. Suppose that there exists$i$such that$z_i > \mathcal{L}(\omega, x_i, y_i)$. Then we let$z'$such that$z'_j = z_j$for$j \neq i$and$z'_i = \mathcal{L}(\omega, x_i, y_i)$. We have in particular : \vspace{-2mm} $$z'_i \geqslant 0 \qquad z'_i \geqslant 1 - y_i (\omega^\trans x_i)$$ And the other constraints are also verified because$z$is feasible. Because the factor of$z_i$is strictly positive in the objective function and because$z'_i < z_i$, the value associated to the point$(\omega, z')$is strictly inferior to the value of$(\omega, z)$. \\ Then for$(\omega^*, z^*)$optimal we have$z_i^* \leqslant \mathcal{L}(\omega^*, x_i, y_i)$for every$i$. Furthermore, because of constraints, this inequality is in fact an equality. So : \vspace{-2mm} $$z_i^* = \mathcal{L}(\omega^*, x_i, y_i)$$ In this case the optimal value is : $$p_2^* = \dfrac{1}{n \tau} \sum_{i = 1}^n \mathcal{L}(\omega^*, x_i, y_i) + \dfrac{1}{2} \| \omega^* \|_2^2 = \dfrac{1}{\tau} p_1(\omega^*)$$ Where$p_1(\omega)$is the value of the point$\omega$in the problem (Sep. 1). \\ Finally we can remark that for every$\omega$, the point$\left( \omega, \left( \mathcal{L}(\omega, x_i, y_i) \right)_{i = 1}^n \right)$is feasible for the problem (Sep. 2). We showed before that the optimal value can be expressed in this form then : $$p_2^* = \inf_\omega \dfrac{1}{\tau} p_1(\omega) = \dfrac{1}{\tau} p_1^* = \dfrac{1}{\tau} p_1(\omega^*)$$ We can now conclude : \begin{center} \fbox{\parbox{0.93\linewidth}{\centering The problem (Sep. 2) as a solution$(\omega^*, z^*)$of value$p_2^*$such that$\omega^*$is optimal for (Sep. 1),$z^*$is the vector of losses and$p_1^* = \tau p_2^*\$. }} \end{center} \paragraph{2.} \end{document} \ No newline at end of file
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