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 $z \succeq0$. \\

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}

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}

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 :

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 $p_1^*=\tau p_2^*$.