# upper bound of the expectation of upper quantile ratio

We have a collection $$\boldsymbol{S}$$ of $$n$$ discrete random variables $$X_1$$, $$X_2$$, $$\dots$$, $$X_n$$ $$\overset{\small \text{i.i.d.}}{\small \sim}$$ $$\mathcal{D}$$, where $$\mathcal{D}$$ is a distribution over $$\{1, 2, \ldots, U\} \subset \mathbb{N}$$ with cumulative distribution function $$F_\mathcal{D}$$.

We define the subcollection that includes only the values in $$\boldsymbol{S}$$ that are above $$Q(p)$$, where $$Q$$ is the quantile function. That is:

$$\boldsymbol{S}_{\geq p} \overset{\small \text{def}}{=} \left\{X : X \in \boldsymbol{S} \text{ and } p\leq F_{\mathcal{D}}(X)\right\}$$

(in words: $$X \in \boldsymbol{S}_{\geq p}$$ if and only if it $$p$$ proportion of the population are smaller or equal to it)

(below we mark $$\pmb{\sum}\boldsymbol{C}$$ as the sum of all elements in collection $$\boldsymbol{C}$$)

Is the following true?

$$\forall n,\; \mathbb{E}\left[\frac{\pmb{\sum}\boldsymbol{S}_{\geq p}}{\pmb{\sum}\boldsymbol{S}}\right] \le \frac{\mathbb{E}\pmb{\sum}\boldsymbol{S}_{\geq p}}{\mathbb{E}\pmb{\sum}\boldsymbol{S}}$$

note this is an extracted step of a different question expectation of upper quantile proportion. This question is simpler and more focused. (+ if the statement above is true the previous question is also solved)

Your statement is true. This inequality is equivalent to the fact that the correlation of $$\frac{\sum S_{\geq p}}{\sum S}$$ and $$\sum S$$ is positive correlated. More precisely:
For a given $$q$$, let $$f(x) = 0$$ for all $$x and $$f(x) = x$$ for all $$x\geq q$$. Your inequality can be written as: $$$$E\left(\frac{\sum f(X_i)}{\sum X_i}\right)\leq \frac{E f(X)}{E(X)}$$$$ The left side equals to $$nE\left(\frac{f(X_1)}{X_1+\sum_{i\geq 1} X_i}\right)$$ since for $$i=1,2,...$$ the random variable $$\frac{f(X_i)}{\sum X_i}$$ has the same law. Since $$x\mapsto f(x)/(x+y)$$ and $$x\mapsto x+y$$ are increasing functions w.r.t $$x$$, the random variable $$f(X_1)/(X_1+Y)$$ and $$X_1+Y$$ (with $$Y=\sum_{i\geq 2} X_i$$, independent from $$X_1$$) is positive correlated. Then
$$E\left(\frac{f(X_1)}{X_1+Y}(X_1+Y)\right) - E\left(\frac{f(X_1)}{X_1+Y}\right) E(X_1+Y)\geq 0.$$ Rearranging the terms gets your inequality.