Let $r\_1,r\_2,r\_3,\dotsc$ be an IID sequence of Rademacher random variables, so that $\mathbb P(r\_n=\pm1)=1/2$, and $a\_1,a\_2,\dotsc$ be a real sequence with $\sum\_na\_n^2=1$. For $S=\sum\_na\_nr\_n$, does the following inequality always hold?
$$
\mathbb P\left(\lvert S\rvert\ge1/\sqrt7\right)\ge1/2.\tag{\*}\label{star}
$$
Is this a known result, or conjecture?


Stated in another way, for each finite sequence $a\_1,a\_2,\dotsc,a\_N$ with $\sum\_na\_n^2=1$, if we look at all $2^N$ sums $\pm a\_1\pm a\_2\pm a\_3\pm\dotsb\pm a\_N$, do we always have at least as many with absolute value at least $1/\sqrt7$ as with absolute value strictly less than $1/\sqrt7$?


**Background**


The statement is very similar to a conjecture of Tomaszewski, that at least as many of the $2^N$ sums have absolute value less than or equal to 1 as have value greater than one. This conjecture is equivalent to the concentration inequality $\mathbb P(\lvert S\rvert\le1)\ge1/2$, and is still unproven, other than possibly in a (not currently peer reviewed, as far as I know) paper [Keller and Klein - Proof of Tomaszewski's conjecture on randomly signed sums](https://arxiv.org/abs/2006.16834) posted on the arXiv recently.


We can also consider more general anti-concentration inequalities, which bound $\mathbb P(\lvert S\rvert \ge x)$ from below. For $x=0$ the lower bound is trivially 1 and, for $x > 1$, it is $0$. For $0 < x < 1$, then, as in a [previous answer of mine](https://mathoverflow.net/questions/53669/anti-concentration-of-bernoulli-sums/53683#53683), the [Paley–Zygmund](https://en.wikipedia.org/wiki/Paley%E2%80%93Zygmund_inequality) inequality can be used to obtain
$$
\mathbb P(\lvert S\rvert \ge x) \ge (1-x^2)^2/3,
$$
but this is far from optimal. Also, for $x=1$, there does exist a strictly positive anti-concentration bound, as shown in the answers to another question of mine, [An $L^0$ Khintchine inequality](https://mathoverflow.net/questions/53855/an-l0-khintchine-inequality). It was shown by Oleszkiewicz, in 1996, that a (non-optimal) lower bound of $1/10$ applies when $x=1$ (in [On the Stein property of Rademacher sequences](https://www.math.uni.wroc.pl/%7Epms/publicationsArticle.php?nr=16.1&nrA=9&ppB=%20127&ppE=%20130)) and also conjectured that the optimal bound is $7/32$, but this is still open as far as I am aware.


Running a Monte Carlo simulation to randomly pick the values of $a$ and compute the probability suggests that the optimal lower bound for $\mathbb P(\lvert S\rvert\ge x)$ is piecewise constant in $x$, so that there is only a small set of $x$ values at which the bound changes (specifically, $x=0,1/\sqrt7,1/\sqrt5,1/\sqrt3,2/\sqrt6,1$). As mentioned, for $x=1$ it is an open conjecture, so my question here is regarding the smallest non-trivial value for $x$.


Note that \eqref{star} is best possible, in the sense that it does not hold if either of the inequalities are strict and, consequently, does not hold if either the $1/\sqrt7$ inside the probability or the $1/2$ outside is increased. Considering $a\_n=1/\sqrt7$ for $n\le7$, we obtain $\mathbb P(\lvert S\rvert > 1/\sqrt7)=29/64$, and considering $a\_n=1/\sqrt2$ for $n\le 2$ gives
$$
\mathbb P(\lvert S\rvert\ge1/\sqrt7)=\mathbb P(\lvert S\rvert > 0)=1/2.
$$
This example also shows that for any $x \gt 0$, the anti-concentration bound can never be better than 1/2, so \eqref{star} is also optimal in this sense, and it would immediately follow that we have optimal inequalities
$$
\mathbb P(\lvert S\rvert\ge x)\ge1/2 \tag{\*\*}\label{starstar}
$$
for all $0 \lt x \le 1/\sqrt7$. In fact, it is not difficult to prove \eqref{starstar} for $x\le0.693/\sqrt7$ as follows: Let $\lVert a\rVert\_\infty=\max\_n\lvert a\_n\rvert$. We split into two cases,


1. $\lVert a\rVert\_\infty\ge x$. In this case, choose $n$ such that $\lvert a\_n\rvert\ge x$. Flipping the sign of $r\_n$ does not affect the distribution of $S$, but if $\lvert S\rvert < x$ then it changes its value such that $\lvert S\rvert\ge x$, so that $\mathbb P(\lvert S\rvert\ge x)\ge\mathbb P(\lvert S\rvert \lt x)$, giving the result.
2. $\lVert a\rVert\_\infty < x$. In this case, if we let $\Phi(x)$ be the standard normal cumulatiive probability function, the [Berry–Esseen theorem](https://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem) gives
$$
\mathbb P(\lvert S\rvert\ge x)\ge 2\Phi(-x)-2C\lVert a\rVert\_\infty \gt 2\Phi(-x)-2Cx
$$
for a global constant $C$. We can use $C=0.56$ (as stated in the Wikipedia page, this was proven by Shevstova in 2010, in [An improvement of convergence rate estimates in the Lyapunov theorem](https://doi.org/10.1134/S1064562410060062)). Evaluating the right hand side with $x=0.693/\sqrt7$ gives a value greater than 1/2. *QED*


Finally, I mention that I have confirmed \eqref{star} numerically on a dense grid of values for $a$ and, by bounding the interpolation errors, should in principle lead to a (rather unsatisfactory) proof.



A Rademacher ‘root 7’ anti-concentration inequality
Addressed in Theorem 1.3 in [Dvořák and Klein - Probability mass of Rademacher sums beyond one standard deviation](https://arxiv.org/abs/2104.10005) (not yet peer reviewed). It describes a computer program that verifies $\Pr[\lvert S\rvert \geq 1/\sqrt{7} - \epsilon] \geq 1/2$, with concrete $\epsilon > 0$. Giving it more time (polynomial in $1/\epsilon$), it empirically seems we may take $\epsilon\to 0$.


Apologizes for the unsatisfactory answer :)


