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Let $x \in [-1/2,1/2]$ and $X_1,\ldots,X_n$ be drawn iid from the uniform distribution on $[-1,1]$.

Question. Given $\varepsilon \ge 0$ an integer $k \in [1,n]$, what is a good lower-bound on the probability $p_{n,k,\varepsilon}$ that: for every $x \in [-1/2,1/2]$, there exists $S \subseteq \{1,2,\ldots,n\}$ with $1 \le card(S) \le k$ such that $|\sum_{i \in S} X_i-x| \le \varepsilon$ ?

I'm particularly interested in the case $k=n$.

Notes.

  • If one could get a good lower-bound for the probability of approximating a fixed $x \in [-1/2,1/2]$ as a sum $\sum_{i \in S}X_i$, then one could use a covering argument on the interval $[-1/2,1/2]$ to get a uniform bound.

One could generalize the above problem as follows. Let $0 \le r < 1$ and $B_r^m$ be the origin-centered ball of radius $r$ in $\mathbb R^m$, and let $X_1,\ldots,X_n$ be drawn iid uniformly from the unit ball $B_1^m$.

Question. What is a good lower-lower bound for the probability $p_{m,r,n,k,\varepsilon}$ that: for every $x \in B_r^m$, there exists $S \subseteq \{1,2,\ldots,n\}$ with $1 \le card(S) \le k$ such that $\|\sum_{i \in S} X_i-x\| \le \varepsilon$ ?

As before, I'm particularly interested in the case $k=n$.

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  • $\begingroup$ What sort of dependence does $\varepsilon$ have on $n$? [If $n \varepsilon$ is bounded, this is going to be much different than the case where say $\varepsilon n > 10 \log(n)$.] $\endgroup$
    – Pat Devlin
    Commented Jun 16, 2020 at 19:22
  • $\begingroup$ You may assume $\varepsilon \ge e^{-Cn}$, for some fixed constant $C>0$, independent of $n$. $\endgroup$
    – dohmatob
    Commented Jun 16, 2020 at 19:25
  • $\begingroup$ Naturally, we'd need $\varepsilon > 2^{-n}$ or there won't be enough different sums to land in the $1/\varepsilon$ intervals needed. Can we assume $\varepsilon > e^{-n/100}$ if we feel like? Otherwise, if we have no control over $C$, then $\varepsilon$ might be much too small for this to be possible. $\endgroup$
    – Pat Devlin
    Commented Jun 16, 2020 at 20:11
  • $\begingroup$ Take $C= 1/100$ if you like. $\endgroup$
    – dohmatob
    Commented Jun 16, 2020 at 21:35
  • $\begingroup$ Even still, you're cutting it kind of close! :-) What kind of lower bound would you be happy with? Anything that converges to 1? $\endgroup$
    – Pat Devlin
    Commented Jun 16, 2020 at 21:43

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