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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)

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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 $x<q$ and $f(x) = x$ for all $x\geq q$. Your inequality can be written as: \begin{equation} E\left(\frac{\sum f(X_i)}{\sum X_i}\right)\leq \frac{E f(X)}{E(X)} \end{equation} 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.

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  • $\begingroup$ Your last inequality sign should be the other way around, I guess. $\endgroup$ – user74045 Oct 27 '19 at 18:39
  • $\begingroup$ @user74045 Thanks for the remark, hopefully that did not mess up the conclusion, this error is corrected. $\endgroup$ – Yan Oct 28 '19 at 21:57

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