Tomaszewski's problem on randomly signed sums, revisited

Let $v_1$, $v_2$, ..., $v_n$ be real numbers whose squares add up to 1. Consider the $2^n$ signed sums of the form $S = \sum \pm v_i$. Boppana and Holzman (2017) proved that at least 13/32 of these sums satisfy $|S| \le 1$. Here we improve their bound to $0.427685$.


Introduction
Let v 1 , v 2 , . . . , v n be real numbers such that the sum of their squares is at most 1. Consider the 2 n signed sums of the form S = ±v 1 ± v 2 ± · · ·± v n . In 1986, B. Tomaszewski (see Guy [3]) asked the following question: is it always true that at least 1 2 of these sums satisfy |S| 1?
Boppana and Holzman [2] proved that at least 13 32 = 0.40625 of the sums satisfy |S| 1. Actually, they proved a slightly better bound of 0.406259. See their paper for a discussion of earlier work on Tomaszewski's problem.
In this note, we will improve the bound to 0.427685. We will sharpen the Boppana-Holzman argument by using a Gaussian bound due to Bentkus and Dzindzalieta [1].
We will use the language of probability. Let Pr[A] be the probability of an event A. A random sign is a random variable whose probability distribution is the uniform distribution on the set {−1, +1}. With this language, we can state our main result.
Let a 1 , a 2 , . . . , a n be independent random signs. Let S be n i=1 a i v i . Then Pr[|S| 1] > 0.427685.

Proof of the improved bound
In this section, we will prove the bound of 0.427685. We will follow the approach of Boppana and Holzman [2], replacing their fourth-moment method with a Gaussian bound.
Let Q be the tail function of the standard normal (Gaussian) distribution: Note that Q is a decreasing, positive function. Bentkus and Dzindzalieta [1] proved the following Gaussian bound on randomly-signed sums. See their paper for a discussion of earlier work on such bounds.

Theorem 1 (Bentkus and Dzindzalieta
Let a 1 , a 2 , . . . , a n be independent random signs. Let Given a positive number c, define F (c) by Note that F is a decreasing function bounded above by 1 2 . A calculation shows that F ( 1 4 ) > 0.427685. We will need the following lemma, which quantitatively improves Lemma 3 of Boppana and Holzman [2].
Lemma 2. Let c be a positive number. Let x be a real number such that |x| 1. Let v 1 , v 2 , . . . , v n be real numbers such that Let a 1 , a 2 , . . . , a n be independent random signs.
Proof. By symmetry, we may assume that x 0. Let By the Bentkus-Dzindzalieta inequality (Theorem 1), we have Taking the complement, we obtain We will also need the following lemma, which says that F satisfies a certain weightedaverage inequality.
Lemma 3. Let K be an integer such that K 2. Then Proof. Let Since c 1 c 2 and F is a decreasing function, we see that for K 2 we have Therefore it is sufficient to show that the following inequality holds for 0 ξ 1/25: Once we show that F (x) is a concave function in the region 0 < x 1/4 + 3/100, we conclude that the left hand side of the inequality is also concave in ξ in the region 0 ξ 1/25 and we need only check the inequality for ξ = 0 and for ξ = 1/25. We will show that Q(1/ √ x ) is convex in x in the region 0 < x 1/3. Recall that Q satisfies the ordinary differential equation Q ′′ (x) = −xQ ′ (x) and that Q ′ (x) < 0 for all x. Thus, for which is positive if 1−3x > 0. It follows that Q(x −1/2 ) is convex in the region 0 < x 1/3. Therefore F (x) is concave in the region 0 < x 1/3. Inequality (1) holds trivially for ξ = 0, and one can check by calculation that it also holds for ξ = 1/25 (and even for ξ = 1/9).
Finally, we will use these two lemmas to prove our main theorem.
Proof of Main Theorem. We will follow the proof of Theorem 4 of Boppana and Holzman [2] nearly line for line. Their proof uses a different function F . Closely examining their proof, we see that they use four properties of F : it is bounded above by 1 2 , satisfies their Lemma 3 (our Lemma 2), is a nonincreasing function (on the set of positive numbers), and satisfies the weighted-average inequality of Lemma 3. Our function F has those same four properties. Hence we reach the same conclusion: Pr[|S| 1] F ( 1 4 ). A calculation shows that F ( 1 4 ) > 0.427685.