Let $X_1,\dots, X_n$ be independent non-negative random variables (with finite expectation and variance), and $0 < m < n$ be a fixed integer such that there exists a subset $S\subseteq [n]$ of size $\lvert S\rvert =m$ with
- $\sum_{k\in S} \mathbb{E}[X_k] \geq 100$; and
- $\sum_{k\notin S} \mathbb{E}[X_k] \leq 1$
and
- $\sum_{k\in S} \operatorname{Var}[X_k] \leq \frac{1}{100} \left(\sum_{k\in S} \mathbb{E}[X_k]\right)^2$; and
- $\sum_{k\notin S} \operatorname{Var}[X_k] \leq \frac{1}{100}$.
In other words, there is an unknown subset of elements that, altogether, are "heavy", and its complement is "light." If I define $T$ as the indices of the $2m$ top values among $X_1,\dots, X_n$, what is the probability that $\sum_{[n]\setminus T} \mathbb{E}[X_i] < 2$? (where the probability is taken over the realization of the $X_i$'s). In other terms, what is the probability that we removed (almost) all the weight of the"heavy set"?
If $m=1$, then this is clear that is happens with high probability by Chebyshev's inequality: if $S=\{i\}$, then the probability that $X_i$ is less than $50$ is at most $\frac{1}{25}$, and similarly the probability that the sum of all other elements (so a fortiori any of them) is greater than 2 is at most $\frac{1}{100}$, so that overall $X_i$ is the greatest element (and therefore among the two top elements) with probability at least $19/20$. Is there a way to generalize this argument to $m \geq 2$?
PS: as a disclaimer, this question was first asked on Math.SE 10 days ago; I am cross-posting it because, in spite of this delay and a bounty I put on it (and expired this morning), it has not received any answer. (And I'm hoping it may be more suited to this website.) I hope this is OK; however, if this appears unacceptable to the moderators, feel free to close this question...
