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

TL;DR

This paper improves the lower bound on the proportion of signed sums of real numbers with unit squared sum that are within 1 in absolute value, surpassing the previous 3/8 barrier using a novel method.

Contribution

The authors introduce a new approach that improves the known bound from 3/8 to 13/32 for the fraction of signed sums within a certain bound.

Findings

Improved the lower bound from 3/8 to 13/32.

Established a new method surpassing previous limitations.

Demonstrated the effectiveness of the novel approach.

Abstract

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.

Equations

$$\sum _{i=1}^n{v}_i^2⩽\frac{2}{7}(1+\left|x\right|{)}^2.$$

$$\mathrm{Pr}[\left|x+Y\right|⩽1]⩾\frac{37}{98}.$$

$$\operatorname*{\mathbb{E}}(Y^{4})=3\Bigl{(}\sum_{i=1}^{n}v_{i}^{2}\Bigr{)}^{2}-2\sum_{i=1}^{n}v_{i}^{4}\leqslant 3\Bigl{(}\sum_{i=1}^{n}v_{i}^{2}\Bigr{)}^{2}\leqslant\frac{12}{49}(1+x)^{4}.$$

$$\mathrm{Pr}[\left|Y\right|⩾1+x]⩽\frac{E(Y^4)}{(1+x{)}^4}⩽\frac{12}{49}.$$

$$\mathrm{Pr}[\left|Y\right|<1+x]⩾\frac{37}{49}.$$

$$\mathrm{Pr}[-1-x<Y⩽0]⩾\frac{1}{2}\mathrm{Pr}[\left|Y\right|<1+x]⩾\frac{37}{98}.$$

$$\mathrm{Pr}[\left|x+Y\right|⩽1]⩾\mathrm{Pr}[-1-x<Y⩽0]⩾\frac{37}{98}.$$

$$\mathrm{Pr}[\left|S\right|⩽1]⩾\frac{37}{98}.$$

$$\frac{\left|v_1\right|+\left|v_2\right|+\left|v_{n-1}\right|+\left|v_n\right|}{4}⩽\sqrt{\frac{v_1^2+v_2^2+v_{n-1}^2+v_n^2}{4}}⩽\sqrt{\frac{1}{4}}=\frac{1}{2}.$$

$$\left|v_1\right|+\left|v_2\right|⩽\frac{\left|v_1\right|+\left|v_n\right|}{2}+\frac{\left|v_2\right|+\left|v_{n-1}\right|}{2}⩽1.$$

$$\left|X_T\right|⩽\left|X_{T-1}\right|+\left|v_T\right|⩽1-\left|v_T\right|+\left|v_T\right|=1.$$

$$\mathrm{Pr}[\left|S\right|⩽1\mid T,X_T]⩾\frac{37}{98}.$$

$$\mathrm{Pr}[\left|S\right|⩽1\mid T,X_T]⩾\mathrm{Pr}[Y_T⩽0\mid T,X_T]⩾\frac{1}{2}.$$

$$\left|Y_T\right|⩽\left|v_{n-1}\right|+\left|v_n\right|⩽2-\left|v_1\right|⩽2-\left|v_{n-1}\right|.$$

$$\mathrm{Pr}[\left|S\right|⩽1\mid T,X_T]⩾\mathrm{Pr}[Y_T⩽0\mid T,X_T]⩾\frac{1}{2}.$$

$$\sum _{i=T+1}^n{v}_i^2⩽1-\sum _{i=1}^T{v}_i^2⩽1-v_1^2-v_2^2⩽1-2v_{T+1}^2<1-2(1-X_T{)}^2.$$

$$1-2(1-X_T{)}^2=\frac{2}{7}(1+X_T{)}^2-\frac{1}{7}(4X_T-3{)}^2⩽\frac{2}{7}(1+X_T{)}^2.$$

$$\mathrm{Pr}[\left|S\right|⩽1\mid T,X_T]=\mathrm{Pr}[\left|X_T+Y_T\right|⩽1\mid T,X_T]⩾\frac{37}{98}.$$

$$F(c)=\frac{1}{2}(1-3c^2).$$

$$\sum _{i=1}^n{v}_i^2⩽c(1+\left|x\right|{)}^2.$$

$$\mathrm{Pr}[\left|x+Y\right|⩽1]⩾F(c).$$

$$\operatorname*{\mathbb{E}}(Y^{4})\leqslant 3\Bigl{(}\sum_{i=1}^{n}v_{i}^{2}\Bigr{)}^{2}\leqslant 3c^{2}(1+x)^{4}.$$

$$\mathrm{Pr}[\left|Y\right|⩾1+x]⩽\frac{E(Y^4)}{(1+x{)}^4}⩽3c^2.$$

$$\mathrm{Pr}[\left|x+Y\right|⩽1]⩾\frac{1}{2}(1-3c^2)=F(c).$$

$$\mathrm{Pr}[\left|S\right|⩽1]>\frac{13}{32}.$$

$$v_n⩾v_1⩾v_{n-1}⩾v_2⩾v_3⩾\cdots ⩾v_{n-2}.$$

$$\mathrm{Pr}[\left|S\right|⩽1\mid T]⩾\{\begin{array}{ll}F\left(\frac{(K+1{)}^2-T}{(2K+1{)}^2}\right)& \text{if }T⩽\frac{3K+2}{2};\\ F\left(\frac{K}{4K+2}\right)& \text{if }T⩾\frac{3K+2}{2}.\end{array}$$

$$\sum_{i=1}^{K+1}v_{i}^{2}\geqslant\frac{1}{K+1}\Bigl{(}\sum_{i=1}^{K+1}v_{i}\Bigr{)}^{2}=\frac{1}{K+1}M_{K+1}^{2}>\frac{1}{K+1}\,.$$

$$\sum _{i=1}^T{v}_i^2>\frac{1}{K+1}+(T-K-1)v_{T+1}^2⩾\frac{1}{K+1}+(T-K-1)(1-X_T{)}^2.$$

$$\sum _{i=1}^T{v}_i^2⩾T{v}_{T+1}^2>T(1-X_T{)}^2.$$

Full text

Keywords: combinatorial probability; probabilistic inequalities

1 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 $S=\pm v_{1}\pm v_{2}\pm\dots\pm v_{n}$. In 1986, B. Tomaszewski (see Guy [3]) asked the following question: is it always true that at least $\frac{1}{2}$ of these sums satisfy $\left\lvert S\right\rvert\leqslant 1$? Most examples with $n=2$ and $v_{1}^{2}+v_{2}^{2}=1$ show that $\frac{1}{2}$ can’t be replaced with a bigger number.

Holzman and Kleitman [7] proved that at least $\frac{3}{8}$ of the sums satisfy $\left\lvert S\right\rvert\leqslant 1$. This result was an immediate consequence of their main result: at least $\frac{3}{8}$ of the sums satisfy the strict inequality $\left\lvert S\right\rvert<1$, provided that each $\left\lvert v_{i}\right\rvert$ is strictly less than 1. This $\frac{3}{8}$ bound for $\left\lvert S\right\rvert<1$ is best possible: consider the example with $n=4$ and $v_{1}=v_{2}=v_{3}=v_{4}=\frac{1}{2}$. So $\frac{3}{8}$ seems to be a natural barrier to their method of proof.

Using a different method, we prove that more than $\frac{13}{32}$ of the sums satisfy $\left\lvert S\right\rvert\leqslant 1$. In other words, we break the $\frac{3}{8}$ barrier. Our method, roughly speaking, goes like this. We will let the first few $\pm$ 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 $\frac{13}{32}$ bound a tiny bit, to $\frac{13}{32}+9\times 10^{-6}$. 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 $\frac{1}{2}$ 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 $\frac{1}{3}$ of the sums satisfy $\left\lvert S\right\rvert\leqslant 1$; 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 $\left\lvert S\right\rvert\leqslant 1$ is always at least $c$. We now know that Tomaszewski’s constant is between $\frac{13}{32}$ and $\frac{1}{2}$. Both [7] and [1] conjecture that Tomaszewski’s constant is $\frac{1}{2}$. De, Diakonikolas, and Servedio [2] developed an algorithm to approximate Tomaszewski’s constant. Specifically, given an $\epsilon>0$, their algorithm will output a number that is within $\epsilon$ of Tomaszewski’s constant. The running time of their algorithm is exponential in $1/\epsilon^{3}$, 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 $\frac{1}{2}$ has been confirmed in some special cases. For example, von Heymann [6] and Hendriks and van Zuijlen [5] proved the conjecture when $n\leqslant 9$. Also, van Zuijlen [10] and von Heymann [6] proved the conjecture when all of the $\left\lvert v_{i}\right\rvert$ are equal.

We will use the language of probability. Let $\Pr[A]$ be the probability of an event $A$. Let $\operatorname*{\mathbb{E}}(X)$ 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 $\sum_{i=1}^{n}v_{i}^{2}$ is at most $1$. Let $a_{1}$, $a_{2}$, …, $a_{n}$ be independent random signs. Let $S$ be $\sum_{i=1}^{n}a_{i}v_{i}$. Then $\Pr[\left\lvert S\right\rvert\leqslant 1]>\frac{13}{32}.$

In Section 2 of this paper, we will provide a short proof of a bound better than $\frac{3}{8}$. In Section 3, we will refine the analysis to improve the bound to $\frac{13}{32}$ and slightly beyond.

2 Beating the 3/8 bound

In this section, we will give the simplest proof we can of a bound better than $\frac{3}{8}$. Namely, we will prove a bound of $\frac{37}{98}$, 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 $\left\lvert x\right\rvert\leqslant 1$. Let $v_{1}$, $v_{2}$, …, $v_{n}$ be real numbers such that

[TABLE]

Let $a_{1}$, $a_{2}$, …, $a_{n}$ be independent random signs. Let $Y$ be $\sum_{i=1}^{n}a_{i}v_{i}$. Then

[TABLE]

Proof.

By symmetry, we may assume that $x\geqslant 0$. The fourth moment of $Y$ is

[TABLE]

So, by the fourth moment version of Chebyshev’s inequality111Shnurnikov [9] used the fourth moment in a similar situation.,

[TABLE]

Looking at the complement,

[TABLE]

Because $Y$ has a symmetric distribution,

[TABLE]

Recall that $x\leqslant 1$. Hence if $-1-x<Y\leqslant 0$, then $\left\lvert x+Y\right\rvert\leqslant 1$. Therefore

[TABLE]

∎

Next we will use Lemma 1 to go beyond the $\frac{3}{8}$ bound.

Theorem 2**.**

Let $v_{1}$, $v_{2}$, …, $v_{n}$ be real numbers such that $\sum_{i=1}^{n}v_{i}^{2}$ is at most $1$. Let $a_{1}$, $a_{2}$, …, $a_{n}$ be independent random signs. Let $S$ be $\sum_{i=1}^{n}a_{i}v_{i}$. Then

[TABLE]

Proof.

By inserting [math]’s, we may assume that $n\geqslant 4$. By permuting, we may assume that the four largest $\left\lvert v_{i}\right\rvert$ are $\left\lvert v_{n}\right\rvert\geqslant\left\lvert v_{1}\right\rvert\geqslant\left\lvert v_{n-1}\right\rvert\geqslant\left\lvert v_{2}\right\rvert$. By the quadratic mean inequality,

[TABLE]

So $\left\lvert v_{1}\right\rvert+\left\lvert v_{2}\right\rvert+\left\lvert v_{n-1}\right\rvert+\left\lvert v_{n}\right\rvert\leqslant 2$. Because of our ordering,

[TABLE]

Given an integer $t$ from [math] to $n$, let $X_{t}$ be the partial sum $\sum_{i=1}^{t}a_{i}v_{i}$ and let $Y_{t}$ be the remaining sum $\sum_{i=t+1}^{n}a_{i}v_{i}$. Let $T$ be the smallest nonnegative integer $t$ such that $t=n-1$ or $\left\lvert X_{t}\right\rvert>1-\left\lvert v_{t+1}\right\rvert$. In a stochastic process such as ours, $T$ is called a stopping time, defined by the stopping rule in the previous sentence222A similar stopping rule was implicitly used by Ben-Tal et al. [1] and refined by Shnurnikov [9]. In addition, [9] pointed out the value of having $\left\lvert v_{1}\right\rvert+\left\lvert v_{2}\right\rvert\leqslant 1$.. Note that $T\geqslant 2$, since $\left\lvert v_{1}\right\rvert+\left\lvert v_{2}\right\rvert\leqslant 1$. By the stopping rule, $\left\lvert X_{T-1}\right\rvert\leqslant 1-\left\lvert v_{T}\right\rvert$. Hence by the triangle inequality,

[TABLE]

Also by the stopping rule, if $T<n-1$, then $\left\lvert X_{T}\right\rvert>1-\left\lvert v_{T+1}\right\rvert$.

We will condition on $T$ and $X_{T}$. We claim that

[TABLE]

By averaging over $T$ and $X_{T}$, this claim implies the theorem. To prove the claim, we may assume by symmetry that $X_{T}\geqslant 0$. We will divide the proof of the claim into three cases, depending on $T$.

Case 1: $T=n-1$. In this case, $\left\lvert Y_{T}\right\rvert=\left\lvert v_{n}\right\rvert\leqslant 1$. Recall that $0\leqslant X_{T}\leqslant 1$. Hence if $Y_{T}\leqslant 0$, then $\left\lvert S\right\rvert=\left\lvert X_{T}+Y_{T}\right\rvert\leqslant 1$. Therefore by symmetry,

[TABLE]

Case 2: $T=n-2$. In this case,

[TABLE]

Recall that $1-\left\lvert v_{n-1}\right\rvert<X_{T}\leqslant 1$. Hence if $Y_{T}\leqslant 0$, then $\left\lvert S\right\rvert=\left\lvert X_{T}+Y_{T}\right\rvert\leqslant 1$. Therefore by symmetry,

[TABLE]

Case 3: $T\leqslant n-3$. In this case, by the stopping rule,

[TABLE]

We can bound the final expression as follows:

[TABLE]

Hence the hypotheses of Lemma 1 are satisfied with $x=X_{T}$ and $Y=Y_{T}$. By Lemma 1, we conclude that

[TABLE]

∎

3 Further improvement

In this section, we will improve the lower bound to $\frac{13}{32}$, which is $40.625\%$. At the end, we will sketch how to improve the bound further, to $\frac{13}{32}+9\times 10^{-6}$.

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 $\left\lvert v_{1}\right\rvert=\left\lvert v_{2}\right\rvert=\left\lvert v_{3}\right\rvert=\frac{1}{4}$. But that scenario is impossible: if $T=2$, then by the stopping rule, $\left\lvert v_{1}\right\rvert+\left\lvert v_{2}\right\rvert>1-\left\lvert v_{3}\right\rvert$. This suggests that we can sharpen the bound on $\sum_{i=T+1}^{n}v_{i}^{2}$ in terms of $T$ and $X_{T}$.

Another idea is that our final bound on $\Pr[\left\lvert S\right\rvert\leqslant 1]$, 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

[TABLE]

Lemma 3**.**

Let $c$ be a nonnegative number. Let $x$ be a real number such that $\left\lvert x\right\rvert\leqslant 1$. Let $v_{1}$, $v_{2}$, …, $v_{n}$ be real numbers such that

[TABLE]

Let $a_{1}$, $a_{2}$, …, $a_{n}$ be independent random signs. Let $Y$ be $\sum_{i=1}^{n}a_{i}v_{i}$. Then

[TABLE]

Proof.

By symmetry, we may assume that $x\geqslant 0$. As in the proof of Lemma 1, the fourth moment of $Y$ satisfies

[TABLE]

So, by the fourth moment version of Chebyshev’s inequality,

[TABLE]

Following the proof of Lemma 1, by taking the complement and then using the symmetry of $Y$, we have

[TABLE]

∎

Now we will use Lemma 3 to prove our $\frac{13}{32}$ lower bound.

Theorem 4**.**

Let $v_{1}$, $v_{2}$, …, $v_{n}$ be real numbers such that $\sum_{i=1}^{n}v_{i}^{2}$ is at most $1$. Let $a_{1}$, $a_{2}$, …, $a_{n}$ be independent random signs. Let $S$ be $\sum_{i=1}^{n}a_{i}v_{i}$. Then

[TABLE]

Proof.

By inserting [math]’s, we may assume that $n\geqslant 4$. By symmetry, we may assume that each $v_{i}$ is nonnegative. By permuting, we may assume that the $v_{i}$ are ordered as follows:

[TABLE]

Except for the oddballs $v_{n}$ and $v_{n-1}$, the order is decreasing. As in Theorem 2, we have $v_{1}+v_{2}+v_{n-1}+v_{n}\leqslant 2$ and $v_{1}+v_{2}\leqslant 1$.

Given an integer $t$ from [math] to $n$, let $M_{t}$ be the sum $\sum_{i=1}^{t}v_{i}$. 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 $K\geqslant 2$, since $v_{1}+v_{2}\leqslant 1$. By the definition of $K$, observe that $M_{K-1}\leqslant 1-v_{K}$ and hence $M_{K}\leqslant 1$. Also, if $K<n-1$, then $M_{K}>1-v_{K+1}$ and hence $M_{K+1}>1$.

Given an integer $t$ from [math] to $n$, define the sums $X_{t}$ and $Y_{t}$ as in Theorem 2. Note that $\left\lvert X_{t}\right\rvert\leqslant M_{t}$. Following Theorem 2, let $T$ be the smallest nonnegative integer $t$ such that $t=n-1$ or $\left\lvert X_{t}\right\rvert>1-v_{t+1}$. Note that $T\geqslant K$. As before, we have $\left\lvert X_{T-1}\right\rvert\leqslant 1-v_{T}$ and $\left\lvert X_{T}\right\rvert\leqslant 1$. Also, if $T<n-1$, then $\left\lvert X_{T}\right\rvert>1-v_{T+1}$.

We will bound from below the conditional probability $\Pr[\left\lvert S\right\rvert\leqslant 1\mid T]$. Namely, we will prove the two-piece lower bound

[TABLE]

We will actually prove the same lower bound on the refined conditional probability $\Pr[\left\lvert S\right\rvert\leqslant 1\mid T,X_{T}]$. To prove this claim, we may assume by symmetry that $X_{T}\geqslant 0$. We will divide the proof of the claim into five cases, depending on $T$.

Case 1: $T=n-1$. The proof of this case is the same as Case 1 of Theorem 2, which yields the stronger bound $\Pr[\left\lvert S\right\rvert\leqslant 1\mid T,X_{T}]\geqslant\frac{1}{2}$.

Case 2: $T=n-2$. The proof of this case is the same as Case 2 of Theorem 2, which yields the stronger bound $\Pr[\left\lvert S\right\rvert\leqslant 1\mid T,X_{T}]\geqslant\frac{1}{2}$.

Case 3: $K+1\leqslant T\leqslant\frac{3K+2}{2}$ and $T\leqslant n-3$. By the quadratic mean inequality,

[TABLE]

Hence, by splitting our sum into two parts, we get

[TABLE]

As a simpler bound,

[TABLE]

Multiplying the second-to-last inequality by $\frac{2T-K-1}{2K+1}$ and the last inequality by $\frac{3K+2-2T}{2K+1}$, both multipliers being nonnegative by the case assumption, we get

[TABLE]

Therefore, looking at the complementary sum, we get

[TABLE]

We can bound the bracketed expression as follows:

[TABLE]

Plugging this inequality back into the previous one, we get

[TABLE]

Hence the hypotheses of Lemma 3 are satisfied with $c=\frac{(K+1)^{2}-T}{(2K+1)^{2}}$, $x=X_{T}$, and $Y=Y_{T}$. By Lemma 3, we conclude that

[TABLE]

Case 4: $\frac{3K+2}{2}\leqslant T\leqslant n-3$. As in Case 3, we can bound $\sum_{i=1}^{T}v_{i}^{2}$ as follows:

[TABLE]

Because $T\geqslant\frac{3K+2}{2}$, this inequality implies

[TABLE]

Compare this bound with the combined bound from Case 3:

[TABLE]

Note that our bound on $\sum_{i=1}^{T}v_{i}^{2}$ is the same as this bound from Case 3 when $T=\frac{3K+2}{2}$. So we can repeat the remainder of Case 3 to get the same lower bound on $\Pr[\left\lvert S\right\rvert\leqslant 1\mid T,X_{T}]$ when $T=\frac{3K+2}{2}$. The bound on $\Pr[\left\lvert S\right\rvert\leqslant 1\mid T,X_{T}]$ in Case 3 was

[TABLE]

When $T=\frac{3K+2}{2}$, this bound becomes

[TABLE]

So we get the same bound in our current case.

Case 5: $T=K\leqslant n-3$. By the quadratic mean inequality,

[TABLE]

We can bound the final expression as follows:

[TABLE]

Plugging this inequality back into the previous one, we get

[TABLE]

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:

[TABLE]

In summary, we have proved our claim on conditional probability:

[TABLE]

Next, we will use this conditional bound to derive a lower bound on the unconditional probability $\Pr[\left\lvert S\right\rvert\leqslant 1]$.

As mentioned above, we always have $T\geqslant K$. In fact, assuming that $K\leqslant n-4$, we have $T=K$ if the signs $a_{1}$, …, $a_{K}$ are all equal, and otherwise $T\geqslant K+2$. This follows from observing that if $a_{1}$, …, $a_{K}$ are not all equal, then $\left\lvert X_{K}\right\rvert\leqslant\sum_{i=1}^{K-1}v_{i}-v_{K}\leqslant 1-v_{K+1}$ and $\left\lvert X_{K+1}\right\rvert\leqslant\sum_{i=1}^{K-1}v_{i}-v_{K}+v_{K+1}\leqslant 1-v_{K+2}$, by the definition of $K$ and the ordering of the $v_{i}$.

This shows that for $K\leqslant n-4$ we have $\Pr[T=K]=\frac{1}{2^{K-1}}$ and $\Pr[T\geqslant K+2]=1-\frac{1}{2^{K-1}}$. Therefore

[TABLE]

Here we have used our conditional bounds, the fact that they are nondecreasing in $T$, and the inequality $K+2\leqslant\frac{3K+2}{2}$. Note that this lower bound on $\Pr[\left\lvert S\right\rvert\leqslant 1]$ remains valid without assuming that $K\leqslant n-4$. Indeed, if $K=n-3$ it is still true that $\Pr[T=K]=\frac{1}{2^{K-1}}$, and while $T=K+1=n-2$ may occur in this case, it yields a conditional bound of $\frac{1}{2}$ 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 $\frac{1}{2}$ in Cases 1 and 2 above.

Thus, to conclude our proof it suffices to show that

[TABLE]

holds for all $K\geqslant 2$. Substituting the relevant expressions into the formula for $F$ and performing routine manipulations, the latter is shown to be equivalent to

[TABLE]

which indeed holds for $K\geqslant 2$. ∎

Can we improve this $\frac{13}{32}$ lower bound? Yes, a little. The idea is to replace the fourth moment with the more flexible $p$th 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 $p\geqslant 2$, let $B_{p}$ be the constant

[TABLE]

where $\Gamma$ is the gamma function. For example, $B_{2}=1$, $B_{3}=2\sqrt{2/\pi}$, and $B_{4}=3$.

Theorem 5** (Khintchine’s inequality).**

Let $p$ be a real number such that $p\geqslant 2$. Let $v_{1}$, $v_{2}$, …, $v_{n}$ be real numbers. Let $a_{1}$, $a_{2}$, …, $a_{n}$ be independent random signs. Let $S$ be $\sum_{i=1}^{n}a_{i}v_{i}$. Then

[TABLE]

For the improved lower bound, choose with foresight $p=3.95937$. In Lemma 3, replace the fourth moment with the $p$th moment and apply Khintchine’s inequality (with $S=Y$), which allows us to replace the function $F$ with the function $G$ defined by $G(c)=\frac{1}{2}(1-B_{p}c^{p/2})$. Use this revised lemma in Theorem 4. The resulting lower bound is $G(\frac{1}{4})$, which is bigger than $\frac{13}{32}+9\times 10^{-6}$. We omit the details.