0

I have a question about the solution of exercise 2.15 part (a) from Boyd & Vandenberghe's Convex Optimization. The exercise says let $x$ be a real-values random variable with $\operatorname{prob}(x = a_i) = p_i, i = 1, \ldots, n$, where $a_1 < \ldots < a_n$. Which of the following conditions are convex in $p$?

(a) $\alpha \leq \mathbb{E} f(x) \leq \beta$, where $\mathbb{E}f(x)$ is the exc+pected value of $f(x) $, i.e., $\mathbb{E} f(x) = \sum_i^n p_i f(a_i)$. (The function $f: \mathbb{R} \to \mathbb{R} $ is given.)

The answer in the solution manual does not satisfy me. It says: $\mathbb{E} f(x) = \sum_i^n p_i f(a_i)$, so the constraint is equivalent to two linear inequalities $$\alpha \leq \sum_i^n p_i f(a_i) \leq \beta.$$

Can anybody help me understand why $f(x)$ is not important in this argument? $f$ could be non linear or even non convex function and the solution did'nt say any thing about it.

SaraK
  • 21

1 Answers1

1

What the authors want you to show is that the condition $\alpha\le \mathbb Ef(x)\le \beta$ is convex in $p=(p_1,...,p_n)$. This means if $q=(q_1,...,q_n)$ is another probability vector and $\epsilon\in[0,1]$ then $$ \alpha\le \sum_{i=1}^np_i f(a_i)\le \beta\quad\text{ and }\quad \alpha\le \sum_{i=1}^nq_i f(a_i)\le \beta $$ implies $$ \alpha\le \sum_{i=1}^n\Big(\epsilon p_i+(1-\epsilon)q_i\Big) f(a_i)\le \beta\,. $$ To me this looks trivial.

Kurt G.
  • 14,198