Given the random LP: $K(x,\epsilon) = min_{a=(a_1,a_2)}\ a_1(w) + a_2(w)$ such that
$$\ a_1(w) - a_2(w) = x-\epsilon$$ and $$a_1(w), a_2(w), x\geq 0,$$ where $\epsilon\sim U(0,1)$ and $w$ is the outcome of random variable $\epsilon$. Assume $a_1^{*}(w)$ and $a_2^{*}(w)$ are optimal solution given some $x$ and some outcome $w$.
(a) Express $a_1^{*}(w)$, $a_2^{*}(w)$ and $K(x,\epsilon)$ as a function of $x$ and $\epsilon$.
(b) Find $F_{Q}(\cdot)$, the cumulative distribution of $K(x,\epsilon)$.
(c) Use part (b) to find $x^{*} = \arg\min E[K(x,\epsilon)]$.
My attempt:
(a) The dual problem is: $max_{\pi\geq 1} \ \pi(x-\epsilon)$. By strong duality theorem, we have: $a_1^{*}(w) + a_2^{*}(w) = max_{\pi\geq 1} \pi(x-\epsilon)$. In addition, since $a_1^{*}(w)$ and $a_2^{*}(w)$ are optimal solutions to the primal problem, $a_1^{*}(w) - a_2^{*}(w) = x - \epsilon$.
Thus, solving the system of two linear equations above, we obtain $$a_1^{*}(w) = \frac{x-\epsilon + max_{\pi\geq 1} \pi(x-\epsilon)}{2}$$
and $$a_2^{*}(w) = \frac{-x+\epsilon + max_{\pi\geq 1} \pi(x-\epsilon)}{2}$$
Thus, $K(x,\epsilon) = max_{\pi\geq 1} \pi(x-\epsilon)$.
(b) By definition, $F_{K}(\cdot) = P(K(x,\epsilon)\leq x) = P(\pi(x-\epsilon)\leq x) = P(x-\frac{x}{\pi}\leq \epsilon)$ for all $\pi\geq 1$.
Since $\epsilon\sim U(0,1)$, we obtain $F_{K}(\cdot) = P(x-\frac{x}{\pi}\leq \epsilon) = 1 - x + \frac{x}{\pi}$.
(c) Note that $E[K(x,\epsilon)] = E_{\epsilon}[K(x,\epsilon)] = \int_{0}^{1} [1-(1-x-\frac{x}{\pi})]\ d\epsilon = x+\frac{x}{\pi}$.
Since $x\geq 0$, the minimum value of $x+\frac{x}{\pi}$ obviously occurs at $x=0$, so $\fbox{$x^{*} = 0$}$
My question: Could someone please help review my solutions above and let me know if there are any mistakes in it, especially parts (a) + (c)? Any inputs would really be appreciated.
