Let $p_i \in [0,1]$ and $\sum_{i} p_i = 1$, and furthermore let $a_i$ and $b_i$ be positive real numbers. Is the inequality $$ \sum_{i} p_i \frac{a_i}{b_i} \leq \frac{\sum_{i} p_i a_i}{\sum_{i} p_i b_i}$$ true?
$\begingroup$
$\endgroup$
1
-
2$\begingroup$ Your inequality is linear with respect to, say, $a_1$. And the coefficient of $a_1$ in LHS may be both more and less than in RHS. Thus, without additional restrictions there is no chance to be true. $\endgroup$– Fedor PetrovCommented Jun 15, 2023 at 10:28
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
7
No. This reads $${\mathbb E}[f(a,b)]\le f({\mathbb E}[a,b])$$ for an arbitrary discrete probability. This is true if and only if $f$ is a concave function (Jenssen Inequality). But $f(x,y)=\frac xy$ is not concave (nor convex).
You might prefer the convex function $g(x,y)=\frac{x^2}y$, for which we thus have $${\mathbb E}\left[\frac{a^2}b\right]\ge \frac{{\mathbb E}[a]^2}{{\mathbb E}[b]}.$$
answered Jun 15, 2023 at 7:24
-
$\begingroup$ By $f({\mathbb E}[a,b])$ you mean $f({\mathbb E}[a],{\mathbb E}[b])$? $\endgroup$ Commented Jun 15, 2023 at 7:44
-
-
2$\begingroup$ Nice answer. Without graphing them I never would have thought that $x^2/y$ is convex while $x/y$ isn't! $\endgroup$ Commented Jun 15, 2023 at 7:56
-
4$\begingroup$ @YaakovBaruch. The convexity of $x^2/y$ over ${\mathbb R}\times(0,+\infty)$ is well-known in mathematical fluid dynamics, because of the kinetic energy, written in terms of the linear momentum and the mass density. $\endgroup$ Commented Jun 15, 2023 at 7:59
-
$\begingroup$ For what it worth, restricting to the lines $x=ay+b$ makes the convexity question of these functions pretty clear. $\endgroup$ Commented Jun 15, 2023 at 10:25
$\begingroup$
$\endgroup$
Here is a concrete counterexample: $p_1 = \frac 1 2 = p_2, a_1 = 1, a_2 = 2, b_1 = 2, b_2 = 1$. Then $$p_1 \frac{a_1}{b_1} + p_2 \frac{a_2}{b_2} = \frac 1 2 \cdot \frac 1 2 + \frac 1 2 \cdot \frac 2 1 = \frac 1 4 + 1 = \frac{11}{4}.$$ On the other hand, $$\frac{p_1 a_1 + p_2 a_2}{p_1 b_1 + p_2 b_2} = 1 < \frac{11}{4}.$$
