Tomaszewski's Problem on Randomly Signed Sums: Breaking the 3/8 Barrier

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 $S = \sum \pm v_i$. Holzman and Kleitman (1992) proved that at least 3/8 of these sums satisfy $|S| \le 1$. This 3/8 bound seems to be the best their method can achieve. Using a different method, we improve the bound to 13/32, thus breaking the 3/8 barrier.


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? Most examples with n = 2 and v 2 1 + v 2 2 = 1 show that 1 2 can't be replaced with a bigger number.
Holzman and Kleitman [7] proved that at least 3 8 of the sums satisfy |S| 1. This result was an immediate consequence of their main result: at least 3 8 of the sums satisfy the strict inequality |S| < 1, provided that each |v i | is strictly less than 1. This 3 8 bound for |S| < 1 is best possible: consider the example with n = 4 and v 1 = v 2 = v 3 = v 4 = 1 2 . So 3 8 seems to be a natural barrier to their method of proof.
Using a different method, we prove that more than 13 32 of the sums satisfy |S| 1. In other words, we break the 3 8 barrier. Our method, roughly speaking, goes like this. We will let the first few ± signs be arbitrary. But once the partial sum becomes near 1 in absolute value, we will show that the final sum still has a decent chance of remaining at most 1 in absolute value.
We can actually improve the 13 32 bound a tiny bit, to 13 32 + 9 × 10 −6 . Combining our method with other ideas, which could handle the tight cases for our analysis, may lead to further improvements of the bound. Still, the conjectured lower bound of 1 2 currently appears to be out of reach.
Ten years after Holzman and Kleitman [7] but independently, Ben-Tal, Nemirovski, and Roos [1] proved that at least 1 3 of the sums satisfy |S| 1; they say that the proof is mainly due to P. van der Wal. Shnurnikov [9] refined the argument of [1] to prove a 36% bound. Even though these two bounds are weaker than that of Holzman and Kleitman, the methods used to prove them are noteworthy. In particular, we will use the conditioning argument of [1] and the fourth moment method of [9].
Let Tomaszewski's constant be the largest constant c such that the fraction of sums that satisfy |S| 1 is always at least c. We now know that Tomaszewski's constant is between 13 32 and 1 2 . Both [7] and [1] conjecture that Tomaszewski's constant is 1 2 . De, Diakonikolas, and Servedio [2] developed an algorithm to approximate Tomaszewski's constant. Specifically, given an ǫ > 0, their algorithm will output a number that is within ǫ of Tomaszewski's constant. The running time of their algorithm is exponential in 1/ǫ 3 , so it's not clear that we can run their algorithm in a reasonable amount of time to improve the known bounds on Tomaszewski's constant.
The conjectured lower bound of 1 2 has been confirmed in some special cases. For example, von Heymann [6] and Hendriks and van Zuijlen [5] proved the conjecture when n 9. Also, van Zuijlen [10] and von Heymann [6] proved the conjecture when all of the |v i | are equal.
We will use the language of probability. Let Pr[A] be the probability of an event A. Let E(X) be the expected value of a random variable X. A random sign is a random variable whose probability distribution is the uniform distribution on the set {−1, +1}. With this language, we can restate our main result.
Main Theorem. Let v 1 , v 2 , . . . , v n be real numbers such that n i=1 v 2 i is at most 1. 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] > 13 32 . In Section 2 of this paper, we will provide a short proof of a bound better than 3 8 . In Section 3, we will refine the analysis to improve the bound to 13 32 and slightly beyond.

Beating the 3/8 bound
In this section, we will give the simplest proof we can of a bound better than 3 8 . Namely, we will prove a bound of 37 98 , which is a little more than 37.75%. In Section 3, we will improve the bound further.
We begin with a lemma. Roughly speaking, this lemma can be used to show that if a partial sum is a little less than 1, then the final sum has a decent chance of remaining less than 1 in absolute value. Lemma 1. 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. Let Y be n i=1 a i v i . Then Proof. By symmetry, we may assume that x 0. The fourth moment of Y is So, by the fourth moment version of Chebyshev's inequality 1 , Looking at the complement, Because Y has a symmetric distribution, Recall that x 1.
Next we will use Lemma 1 to go beyond the 3 8 bound.
Proof. By inserting 0's, we may assume that n 4. By permuting, we may assume that the four largest |v i | are |v n | |v 1 | |v n−1 | |v 2 |. By the quadratic mean inequality, 1 Shnurnikov [9] used the fourth moment in a similar situation.
So |v 1 | + |v 2 | + |v n−1 | + |v n | 2. Because of our ordering, Given an integer t from 0 to n, let X t be the partial sum In a stochastic process such as ours, T is called a stopping time, defined by the stopping rule in the previous sentence 2 . Note that T 2, since |v 1 | + |v 2 | 1. By the stopping rule, |X T −1 | 1 − |v T |. Hence by the triangle inequality, We will condition on T and X T . We claim that By averaging over T and X T , this claim implies the theorem. To prove the claim, we may assume by symmetry that X T 0. We will divide the proof of the claim into three cases, depending on T .
Case 2: T = n − 2. In this case, Case 3: T n − 3. In this case, by the stopping rule, We can bound the final expression as follows: Hence the hypotheses of Lemma 1 are satisfied with x = X T and Y = Y T . By Lemma 1, we conclude that 98 .

Further improvement
In this section, we will improve the lower bound to 13 32 , which is 40.625%. At the end, we will sketch how to improve the bound further, to 13 32 + 9 × 10 −6 . Let us examine where the proof of Theorem 2 is potentially tight. Looking at its Case 3, we see that the proof is potentially tight when T = 2 and |v 1 | = |v 2 | = |v 3 | = 1 4 . But that scenario is impossible: if T = 2, then by the stopping rule, |v 1 | + |v 2 | > 1 − |v 3 |. This suggests that we can sharpen the bound on n i=T +1 v 2 i in terms of T and X T . Another idea is that our final bound on Pr[|S| 1], instead of being the worst-case conditional bound, may be taken to be a weighted average of the conditional bounds, with weights corresponding to the distribution of T .
First, we state the following generalization of Lemma 1. Given a number c, define F (c) by Lemma 3. Let c be a nonnegative 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. Let Y be n i=1 a i v i . Then Pr[|x + Y | 1] F (c).
Proof. By symmetry, we may assume that x 0. As in the proof of Lemma 1, the fourth moment of Y satisfies

So, by the fourth moment version of Chebyshev's inequality,
Following the proof of Lemma 1, by taking the complement and then using the symmetry of Y , we have Now we will use Lemma 3 to prove our 13 32 lower bound. Theorem 4. Let v 1 , v 2 , . . . , v n be real numbers such that n i=1 v 2 i is at most 1. 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] > 13 32 .
Proof. By inserting 0's, we may assume that n 4. By symmetry, we may assume that each v i is nonnegative. By permuting, we may assume that the v i are ordered as follows: Except for the oddballs v n and v n−1 , the order is decreasing. As in Theorem 2, we have v 1 + v 2 + v n−1 + v n 2 and v 1 + v 2 1. Given an integer t from 0 to n, let M t be the sum t i=1 v i . Let K be the smallest nonnegative integer t such that t = n − 1 or M t > 1 − v t+1 . The parameter K measures how spread out the v i are. Note that K 2, since v 1 + v 2 1. By the definition of K, Given an integer t from 0 to n, define the sums X t and Y t as in Theorem 2. Note that |X t | M t . Following Theorem 2, let T be the smallest nonnegative integer t such that We will bound from below the conditional probability Pr[|S| 1 | T ]. Namely, we will prove the two-piece lower bound We will actually prove the same lower bound on the refined conditional probability Pr[|S| 1 | T, X T ]. To prove this claim, we may assume by symmetry that X T 0. We will divide the proof of the claim into five cases, depending on T . and T n − 3. By the quadratic mean inequality, Hence, by splitting our sum into two parts, we get As a simpler bound, Multiplying the second-to-last inequality by 2T −K−1 2K+1 and the last inequality by 3K+2−2T 2K+1 , both multipliers being nonnegative by the case assumption, we get Therefore, looking at the complementary sum, we get We can bound the bracketed expression as follows: Plugging this inequality back into the previous one, we get Hence the hypotheses of Lemma 3 are satisfied with c = (K+1) 2 −T (2K+1) 2 , x = X T , and Y = Y T . By Lemma 3, we conclude that Case 4: 3K+2 2 T n − 3. As in Case 3, we can bound T i=1 v 2 i as follows: Compare this bound with the combined bound from Case 3: Note that our bound on T i=1 v 2 i is the same as this bound from Case 3 when T = 3K+2 2 . So we can repeat the remainder of Case 3 to get the same lower bound on Pr[|S| 1 | T, X T ] when T = 3K+2 2 . The bound on Pr[|S| 1 | T, X T ] in Case 3 was When T = 3K+2 2 , this bound becomes So we get the same bound in our current case.
Case 5: T = K n − 3. By the quadratic mean inequality, We can bound the final expression as follows: Plugging this inequality back into the previous one, we get This is the same inequality we derived at the beginning of Case 3. So we can repeat the remainder of Case 3 to get the same lower bound: In summary, we have proved our claim on conditional probability: Next, we will use this conditional bound to derive a lower bound on the unconditional probability Pr[|S| 1].
As mentioned above, we always have T K. In fact, assuming that K n − 4, we have T = K if the signs a 1 , . . . , a K are all equal, and otherwise T K + 2. This follows from observing that if a 1 , . . . , a K are not all equal, then |X K | Here we have used our conditional bounds, the fact that they are nondecreasing in T , and the inequality K + 2 3K+2 2 . Note that this lower bound on Pr[|S| 1] remains valid without assuming that K n − 4. Indeed, if K = n − 3 it is still true that Pr[T = K] = 1 2 K−1 , and while T = K + 1 = n − 2 may occur in this case, it yields a conditional bound of 1 2 as shown in Case 2 above, which is even better than our stated lower bound. The values n − 2 and n − 1 for K are of course covered by the conditional bound of 1 2 in Cases 1 and 2 above. Thus, to conclude our proof it suffices to show that holds for all K 2. Substituting the relevant expressions into the formula for F and performing routine manipulations, the latter is shown to be equivalent to 64(K 2 + K) < 2 K−1 (40K 2 + 40K − 15), which indeed holds for K 2.
Can we improve this 13 32 lower bound? Yes, a little. The idea is to replace the fourth moment with the more flexible pth moment, where p is a parameter to be optimized. To do so, we will need Khintchine's inequality. This inequality was first proved by Khintchine [8] in a weaker form and later proved by Haagerup [4] with the optimal constants. Namely, given p 2, let B p be the constant where Γ is the gamma function. For example, B 2 = 1, B 3 = 2 2/π, and B 4 = 3.
Theorem 5 (Khintchine's inequality). Let p be a real number such that p 2. Let v 1 , v 2 , . . . , v n be real numbers. Let a 1 , a 2 , . . . , a n be independent random signs. Let S be For the improved lower bound, choose with foresight p = 3.95937. In Lemma 3, replace the fourth moment with the pth moment and apply Khintchine's inequality (with S = Y ), which allows us to replace the function F with the function G defined by G(c) = 1 2 (1 − B p c p/2 ). Use this revised lemma in Theorem 4. The resulting lower bound is G( 1 4 ), which is bigger than 13 32 + 9 × 10 −6 . We omit the details.