Commit cffb4be8 authored by nanored's avatar nanored
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first question of ex 3

parent a6b54745
......@@ -161,5 +161,29 @@
\text{s.t.} & \left\| A^\trans \nu \right\|_{\infty} \leqslant 1
\end{array} $}
\section*{Exercise 3 (Data Separation)}
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 :
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 $p_1^* = \tau p_2^*$.
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