-1

Consider the weighted sum $S = \sum\limits_{i=1}^N p_i f(x_i), $ where $p_i \in [0, 1]$ and $f(x_i) \geq 0$. Also $f(x_i)$ is concave in $x_i$. I am wondering if $\frac{\sum\limits_{i=1}^N p_i}{N} \sum\limits_{i=1}^N f(x_i)$ is a upper bound on $S$.

one user
  • 534
  • That's not an upper bound, but Jensen's inequality gives another upper bound $,\frac{S}{\sum p_i} \le f\left(\frac{\sum p_ix_i}{\sum p_i}\right),$. – dxiv Jun 22 '22 at 18:44

1 Answers1

1

A simple counter-example suffices to disprove the claim. Let $f(x) = \sqrt{x}$ and $x_1 = 1, x_2 = 4$. Let $p_1 = 0, p_2 = 1$.

Then $S = f(4) = 2$ and the proposed upper bound is $\frac{1}{2}(f(1) + f(4)) = \frac{3}{2}$

egglog
  • 1,674
  • More generally, for any concave strictly increasing function, choosing $p_N = 1$ and the rest $p_i = 0$, assuming (which we can wlog) that the $x_i$ are ordered serves as counter example, as $f(x_N)$ will surely be greater than $\frac{1}{N}(f(x_1) + ... + f(x_N))$ – egglog Jun 22 '22 at 18:18