arXiv:2006.16834v4 [math.CO] 3 Aug 2021 upre yteIre cec onain(rn o 1612/ no. 2014290). (grant no. Foundation (grant Science dation Israel the by supported otdb h lr coasi Programme. Scholarship Clore the by ported usin1.1. Question Tomaszewski: Boguslav to attributed question, open sions qiaetformulations: equivalent value X oza n limn[9 rvdtelwrbudPr[ bound lower the proved [29] Kleitman and Holzman ntesqe,w s h rbblsi notation probabilistic the use we sequel, the In nteArl18 su of issue 1986 April the In Background 1.1 Introduction 1 ro fTmsesisCnetr nRnol indSums Signed Randomly on Conjecture Tomaszewski’s of Proof ∗ † = • • • eateto ahmtc,BrIa nvriy aa Gan, Ramat University, Ilan Bar Mathematics, of Department eateto ahmtc,BrIa nvriy aa Gan, Ramat University, Ilan Bar Mathematics, of Department ntefis ae tdigtepolm oza n Kleitman and Holzman problem, the studying paper first the In 2 1 {− afo h attosof partitions the of half hbse-yeinequality: Chebyshev-type u partitions: Sum hvsieult ilsalwrbudo o hsprobabil this for 0 of bound lower a yields inequality shev’s aladacube: a and ball A where tlathl h etcso h uelebten(ro)tet the on) (or between suppor lie parallel cube of the pair of any vertices for the that half true least it at Is it. containing nqaiyfrRdmce sums. Rademacher for inequality | ≤ ǫ ( 1 x 1 a 1 1 epoetefloigcnetr,det oazwk 18) L (1986): Tomaszewski to due conjecture, following the prove We u annvltosaelclcnetainieulte n nimpro an and inequalities concentration local are tools novel main Our ? 1 , + 1 + P } x . . . 2 n needn.I ttu htPr[ that true it Is independent. and i a + i 2 Consider + n each and 1 = x ǫ 3 n + a Let n x | with , 4 ,adwr o bet rv togrlwrbudfrPr[ for bound lower stronger a prove to able not were and ), h mrcnMteaia Monthly Mathematical American The n osdran Consider a 1 elnumbers real a , . . . , P ahnKeller Nathan x ǫ i i a sauiomyrno in hnPr[ Then sign. random uniformly a is = Let i n notosm,tesm ie ya ot1? most at by differ sums the sums, two into ± era ubr with numbers real be 1 X , uut4 2021 4, August n 1 dmninlbl n smallest a and ball -dimensional = ≤ a Abstract 1 a , i P ∗ ≤ 2 1 i n hdKlein Ohad and a , . . . , a n X i a hr emr ihvalue with more be there can , x | i = X 7 n yteBntoa SIre cec Foun- Science US-Israel Binational the by and 17) where , ≤ | n P uhthat such a p Israel. i Israel. | x P X Var( i omlzn uhthat such normalizing , { | i x a < i ihr u 2]peetdan presented [24] Guy Richard , i 2 } [email protected] X [email protected] † .I ttu hti tleast at in that true it Is 1. = ity. P 1] r nfrl itiue in distributed uniformly are ighprlnso h ball, the of hyperplanes ting )] ohyperplanes? wo | i X ≥ ≥ a ≤ | i 2 1 et 3 1 = / 2]peetdseveral presented [29] / e Berry-Esseen ved ?Nt htCheby- that Note 2? 1] ,wihi ih for tight is which 8, X n fthe Of . ≥ dmninlcube -dimensional = 1 / P 2. > eerhsup- Research . i n =1 1 2 P a hnwith than Research . n i | X x a i expres- , ≤ | i 2 1. = 1]. They conjectured that Pr[ X 1] 1/2, which would be tight, e.g., for X = a x + a x with | |≤ ≥ 1 1 2 2 any a , a < 1. This conjecture became known as Tomaszewski’s conjecture or Tomaszewski’s | 1| | 2| problem. As is suggested by its equivalent formulations, Tomaszewski’s problem and its variants naturally appear in diverse fields, including probability theory [2, 38], geometric analysis [32], optimization and operation research [1, 42], statistics [37], and theoretical computer science [11, 17, 43]. It became well-known, and was mentioned in lists of open problems in various fields (e.g., [22, 27]). A number of works obtained partial results toward Tomaszewski’s conjecture. Ben-Tal et al. [1] (who were not aware of the previous work on the conjecture and arrived to it indepen- dently, from applications to optimization) proved that Pr[ X 1] 1/3. Shnurnikov [41] | | ≤ ≥ improved their lower bound to 0.36. Boppana and Holzman [10] were the first to cross the 3/8 barrier, proving a lower bound of 0.406, and Boppana et al. [9] further improved the lower bound to 0.428. Very recently, Dvoˇr´ak et al. [13] proved a lower bound of 0.46, which is the best currently known bound. Several other authors proved the conjecture in special cases: Bentkus and Dzinzalieta [2] proved it in the case max a 0.16, Hendriks and van Zuijlen [26] proved it for n 9 (a | i| ≤ ≤ proof-sketch in that case, using different methods, was presented earlier by von Heymann [47]), van Zuijlen [45] proved it when all ai’s are equal, and Toufar [44] extended his result to the case where all ai’s but one are equal. In another direction, De et al. [11] presented an algorithm that allows approximating min Pr[ X 1] X = a x up to an additive error of ǫ. Unfortunately, { | |≤ | i i} the complexity of the algorithm is exp(O(1/ǫ6)). P 1.2 Our results In this paper we prove Tomaszewski’s conjecture.
Theorem 1.2. Let X = n a x , where n a2 = 1 and x are independent and uniformly i=1 i i i=1 i { i} distributed in 1, 1 . Then {− } P P Pr[ X 1] 1/2. (1) | |≤ ≥ Our result may be interpreted within the general context of tail bounds for Rademacher sums. A Rademacher sum is a random variable X = aixi, where the xi’s are i.i.d. Rademacher random variables (i.e., are uniformly distributed in 1, 1 ). Estimates on Rademacher sums P{− } were studied in numerous papers, both for their own sake (e.g., [8, 46]) and for the sake of ap- plications to statistics [19] and to optimization [1]. A main direction in this study is obtaining upper and lower bounds on the tail probability Pr[ X >t], aiming at showing that the tail of | | any such X behaves ‘similarly’ to the tail of a Gaussian random variable with the same variance (see, e.g., [2, 18, 38] for upper bounds and [28, 35, 36] for lower bounds). One of the upper bounds on the tail, proved by Dzindzalieta [15], asserts that
1 1 2 √ Pr[X>t] + (1 2 2 ), 1 Pr[X>t] 1/4, t 1, (2) ≤ ∀ ≥ which is tight for all t< √2, due to X = 1 x + 1 x . √2 1 √2 2 2 Furthermore, as was observed by Dzindzalieta [15], Theorem 1.2 implies the more general: Pr[ X 3 A refinement of Chebyshev’s inequality. We make repeated use of the following, rather standard, refinement of Chebyshev’s inequality: Proposition 1.4. Let X be a symmetric (around 0) random variable with Var(X) = 1, and let c ,...,c , d ,...,d R be such that 0 n 1 m ∈ 0= c c . . . c =1= d d . . . d . 0 ≤ 1 ≤ ≤ n 0 ≤ 1 ≤ ≤ m Then n 1 m − 2 2 2 (1 ci )Pr[X (ci, ci+1]] (di di 1)Pr[X di]. (5) − ∈ ≥ − − ≥ Xi=0 Xi=1 Proof outline. In the proof of Theorem 1.2, we consider several cases, according to the sizes of the ai’s, and prove the assertion in each of them separately using the tools described above. In particular, the case a + a 1 is covered by the aforementioned semi-inductive argument, 1 2 ≥ the case max a 0.31 is proved by the refined Berry-Esseen inequality, and the cases ‘in | i| ≤ the middle’ are proved via various combinations of the segment comparison argument and the refined Chebyshev inequality, sometimes using also the refined Berry-Esseen bound. Unfortunately, this part of the proof requires somewhat grueling computations, including two cases in which a light computer-aided check is applied. For the sake of readability, we divide the proofs into their ‘essential’ part and their ‘calculation’ part, and relay the calculations to the appendices. Organization of the paper. In Section 2 we introduce notation, a basic lemma, and a more detailed outline of the proof. The segment comparison argument (i.e., Theorem 1.3) is presented in Section 3. In Section 4 we present the refined Berry-Esseen bounds. The semi- inductive argument for the case a + a 1 is given in Section 5. The rest of the cases are 1 2 ≥ presented in Sections 6–8 and Appendices C–E. We conclude with several open problems in Section 9. 2 Preliminaries and Structure of the Proof This section presents notation, a basic lemma, and the structure of the proof in more detail. 2.1 Notation Standard notation. For n N, [n] denotes 1, 2,...,n . ∈ { } For A,B,C R, A + [B,C] denotes the segment [A + C,B + C]. ∈ The shorthand A , A ,...,A = x ,x ,...,x denotes the m equalities A = x , i [m]. 1 2 m 1 2 m i i ∈ The shorthands LHS and RHS denote the left hand side and the right hand side of an equation (or an inequality). Setting. Throughout the paper, X denotes a normalized Rademacher sum, that is, X = n a x , where n a2 = 1 and x are independent and uniformly distributed in 1, 1 . i=1 i i i=1 i { i} {− } Without loss of generality, we always assume P P a a . . . a > 0. 1 ≥ 2 ≥ ≥ n Note that sometimes, we pass from X to an auxiliary random variable, which we denote by X = m b x . In such cases, nothing is assumed on the b , unless stated otherwise explicitly. ′ i=1 i i { i} P 4 Precision. The paper contains explicit real numbers, which we present in decimal expansion. When we write these, we mean the exact value we write; we never write a rounded value and mean ‘a close’ number. We make such roundings by writing, e.g., π = 3.1416 10 5. ± − Notation for segments. We use the somewhat non-standard notation Pr[X [a, b]] Pr[X = a]/2 Pr[X = b]/2, a < b Pr [X a, b ]= ∈ − − . (6) ∈ h i (0, Otherwise Note that we have A B C = Pr [X A, C ]=Pr[X A, B ]+Pr[X B,C ] . (7) ≤ ≤ ⇒ ∈ h i ∈ h i ∈ h i Similarly, for a b we denote Pr[X a, b]] = Pr[X [a, b]] Pr[X = a]/2, and Pr[X ≤ ∈ h ∈ − ∈ [a, b ] = Pr[X [a, b]] Pr[X = b]/2. i ∈ − 2.2 A basic lemma – elimination of variables The following lemma allows eliminating several xi’s, by taking into account each possible value of these xi’s separately. The lemma exchanges Tomaszewski’s conjecture (1) by inequalities which are occasionally easier to approach. n Lemma 2.1. Let X = i=1 aixi be a Rademacher sum with Var(X) = 1 and let m Pm n 2 ai σ = 1 a and X′ = a′ x , with a′ = . (8) v − i i i i σ u i=1 i=m+1 u X X t Tomaszewski’s assertion (1) is equivalent to 2m 1 − m 2 Pr[X′ > T ] 2 − , (9) j ≤ Xj=0 where T = (1 a a )/σ ranges over all 2m options. { j} ± 1 ±···± m Proof. As X is symmetric, (1) is equivalent to Pr[X > 1] 1/4. By the law of total probability, m ≤ 2 Pr[X > 1] = j Pr[X′ > Tj], implying the lemma. Lemma 2.1 isP simple yet useful. For example, applying the lemma with m = 1, one can see that (1) is equivalent to the inequality Pr[X′ 0,t]] Pr[X′ > 1/t], (10) ∈ h ≥ where n 2 ai 1 a1 σ = 1 a , X′ = x , and t = − . (11) − 1 σ i 1+ a i=2 r 1 q X This readily implies the aforementioned observation of Dzindzalieta [15] that Tomaszewski’s conjecture is equivalent to the more general inequality (3). 5 2.3 Hard cases for the proof There are several classes of Rademacher sums which are ‘hard to handle’ with our tools. These classes motivate the partition into cases used in the proof. One of the obstacles we have to overcome, is the difficulty in distinguishing between the probabilities Pr[ X 1] and Pr[ X < 1]. This obstacle appeared in previous works as well, | | ≤ | | and is probably the reason for which the ‘barrier’ of 3/8 (which is the tight lower bound for Pr[ X < 1]) was not beaten for almost 25 years, until the work of Boppana and Holzman [10]. | | As a result, ‘hard’ cases for our proof are not only tightness examples for the conjecture, but also X’s for which Pr[ X < 1] is small. We list three such examples of Rademacher sums: | | 1. X = (x1 + x2)/√2, and more generally, X = aixi with Var(X)=1 and P n a + min a x > 1. (12) 1 n i i x 1,1 [ ]\{1} ∈{− } i=2 X 1 4 2. X = 2 i=1 xi. 1 P9 3. X = 3 i=1 xi. RademacherP sums X that belong to the first class are tightness examples for Theorem 1.2, in the sense that Pr[ X 1] = 1/2. These are the only tightness examples we are aware of. The | |≤ two latter classes are not tightness cases of the conjecture, but rather satisfy Pr[ X < 1] < 1/2, | | and inevitably complicate the proof. For example, in the third case, Pr[ X < 1] = 63 < 0.493, | | 128 which demonstrates that the improved Berry-Esseen bound we prove in Section 4 is almost optimal. Indeed, while the bound implies that Pr[ X < 1] 1/2 holds whenever max a | | ≥ i | i| ≤ 0.31, the example shows that this assertion fails when max a = 1/3. | i| 2.4 Structure of the proof The proof of Theorem 1.2 is split into seven cases, which we shortly overview: Case 1: a 0.31. This case is covered by the improved Berry-Esseen bound for Rademacher 1 ≤ sums presented in Section 4. Case 2: a + a 1. This case is covered by the semi-inductive argument (which uses the 1 2 ≥ ‘stopping time’ method), and is presented in Section 5. Case 3: a 0.55 and a +a 1. The proof in this case combines the ‘segment comparison’ 1 ≥ 1 2 ≤ argument (i.e., Theorem 1.3) with Chebyshev’s inequality. The proof in this case is a simple example of the proof strategy in some the following cases, and so we slightly detail about it. By Lemma 2.1, applied with m = 1, in order to verify Pr[ X 1] 1/2, it suffices to show | |≤ ≥ Pr[ X′ t] > Pr[ X′ > 1/t], (13) | |≤ | | where X ,t are as defined in (11). Since a – the largest weight of X – satisfies a t (which ′ 2′ ′ 2′ ≤ follows from the condition a + a 1), Theorem 1.3 implies that Pr[ X t] = Ω(t). As 1 2 ≤ | ′| ≤ Chebyshev’s inequality yields Pr[ X > 1/t] t2, (13) follows if t is sufficiently small, that is, | ′| ≤ if a is sufficiently large. This argument applies when a 0.55, and is presented in Section 6. 1 1 ≥ 6 Case 4: a [0.5, 0.55] and a + a 1. The proof in this case splits according to whether 1 ∈ 1 2 ≤ a2 is ‘large’ or ‘small’. If a2 is small, then the method of the previous case is sufficient. If a2 is large, we eliminate two variables (by applying Lemma 2.1 with m = 2) and prove the resulting inequality using Theorem 1.3 and the refined Chebyshev inequality (i.e., Proposition 1.4). This case is presented in Section 7. Case 5: a [0.31, 0.5] and a + a + a 1. The proof in this case splits according to the 1 ∈ 1 2 3 ≤ sizes of a2 and a3. If either of a2 or a3 is sufficiently small, then after elimination of one or two variables (respectively), the problem is reduced to a probabilistic inequality concerning a Rademacher sum with ‘sufficiently small’ weights, and follows from the improved Berry-Esseen inequality. Otherwise, we eliminate three variables and prove the assertion using Theorem 1.3 and Proposition 1.4. This case is demonstrated in Section 8 and treated in detail in Appendix C. Case 6: a [0.387, 0.5] and a + a + a 1. The proof method in this case is superficially 1 ∈ 1 2 3 ≥ similar to that of the previous case. However, a subtle difference in the details makes this case simpler to handle. The proof in this case is demonstrated in Section 8 and treated in detail in Appendix D. Case 7: a [0.31, 0.387] and a + a + a 1. The proof in this case splits according to 1 ∈ 1 2 3 ≥ whether there exists some k 4 with a ‘medium-sized’ a . ≥ k If there is no such ak, then the weights are partitioned into ‘large’ ones and ‘small’ ones. If there are at most four large weights, then by eliminating the four variables with largest weight, we reduce the problem to a probabilistic inequality concerning a Rademacher sum with ‘sufficiently small’ weights that can be handled easily. Otherwise, we eliminate five variables and show that the assertion follows from Proposition 1.4 and a ‘light’ semi-inductive argument. The latter part is however somewhat cumbersome, as after eliminating five variables, we have to deal with 25 = 32 summands simultaneously. In the case where there exists a medium-sized ak, we use an explicit bijection to prove a special segment comparison lemma that holds for it, which in turn allows deducing the assertion using Theorem 1.3 and Proposition 1.4. The proof in this case is demonstrated in Section 8 and treated in detail in Appendix E. Combination of the seven cases with induction over n (where the inductive assumption is used in Cases 2 and 7), completes the proof of Theorem 1.2. 3 Local Concentration Inequalities for Rademacher Sums In this section we present two concentration inequalities, that allow comparing the probabilities Pr[X I] and Pr[X J], where X is a Rademacher sum and I, J are segments or rays. Both ∈ ∈ results are used extensively throughout the paper. The first is a local concentration inequality: Theorem 3.1 (segment comparison). Let X = n a x , and write M = max a . For all i=1 i i i | i| A,B,C,D R such that ∈ P A min(B,C), 2M C A, and D C + min(2M, D B) B A, (14) | |≤ ≤ − − − ≤ − one has Pr[X C, D ] Pr[X A, B ]. (15) ∈ h i ≤ ∈ h i 7 The assertion of Theorem 3.1 holds for other types of segments as well; see Appendix A.6. We note that inequalities of the same type were obtained by the authors in [31], where they were used to obtain an alternative proof of another local tail inequality due to Devroye and Lugosi [12] and to study analytic properties of linear threshold functions. While the inequalities in [31] are qualitative (i.e., of the form Pr[X C, D ] c Pr[X A, B ] for some non-optimal ∈ h i ≤ ∈ h i constant c), for our purposes here a more exact inequality is required. Such an inequality is given in Theorem 3.1, which strictly supersedes [31, Lemma 3.1]. The usefulness of ‘segment comparison’ in the proof of Tomaszewski’s conjecture is apparent, as the conjecture itself can be rephrased as a segment comparison inequality: Pr [X 1, ] Pr [X 0, 1 ] . (16) ∈ h ∞i ≤ ∈ h i However, Theorem 3.1 alone is not sufficient for our needs since it allows deducing Pr[X I] ∈ ≤ Pr[X J] only for segments I, J that satisfy, in particular, I J . ∈ | | ≤ | | The second result we present is a simple-yet-powerful generalization of the classical Cheby- shev’s inequality, which allows handling cases where I > J . For example, it enables to deduce | | | | Pr [X 0, 1 ] Pr X √2, , which reminds of (16) but is of course much weaker. ∈ h i ≥ ∈ ∞ Lemma 3.2. Let X be a symmetric (around 0) random variable with Var(X) = 1, and let 0= c c . . . c =1= d d . . . d d = . 0 ≤ 1 ≤ ≤ n 0 ≤ 1 ≤ ≤ m ≤ m+1 ∞ Then n 1 m − 2 2 2 (1 ci )Pr[X ci, ci+1 ] (di di 1)Pr[X di], (17) − ∈ h i ≥ − − ≥ Xi=0 Xi=1 and similarly, n 1 m − (1 c2)Pr[X c , c ] (d2 1) Pr [X d , d ] . (18) − i ∈ h i i+1i ≥ i − ∈ h i i+1i Xi=0 Xi=1 While we did not find this result in the literature, it is presumably known or even folklore. We note that a variant of Lemma 3.2 was used in the recent work of Dvoˇr´ak et al. [13]. Neither Theorem 3.1 nor Lemma 3.2 is sufficient for tackling Tomaszewski’s conjecture. However, a combination of these tools allows proving the conjecture in several significant cases. Organization. The proof of Theorem 3.1 uses an explicit injection, which maps the event on the left hand side of (15) to the event corresponding to the right hand side. We present several auxiliary bijections that will be used to construct our injection in Section 3.1, and then we prove Theorem 3.1 in Section 3.2. The simple and standard proof of Lemma 3.2 is presented in Appendix A.5. 3.1 Auxiliary bijections In this subsection we present three bijections on the discrete cube 1, 1 n (i.e., bijections from {− } the discrete cube to itself) that satisfy certain desired properties. The first two bijections were already introduced in [31] and are presented here for the sake of completeness. 8 3.1.1 Prefix flip Lemma 3.3 (Prefix flip). Let a = (a1, . . . , an) be a sequence of positive real numbers, let M = max a and let Q 0. There exists a bijection PF : 1, 1 n 1, 1 n such that for i i ≥ a,Q {− } → {− } any v 1, 1 n with ∈ {− } n X(v) := a v Q/2, (19) i i ≥ Xi=1 the image w = PFa,Q(v) satisfies n X(w)= a w (X(v) Q 2M, X(v) Q]. (20) i i ∈ − − − Xi=1 Construction of PF . Let v 1, 1 n satisfy (19). Let k > 0 be minimal so that a,Q ∈ {− } k a v Q/2. Define w = PF (v) by w = v for i k, and w = v for i > k. j=1 j j ≥ a,Q i − i ≤ i i WeP prove that PFa,Q satisfies the requirements of the lemma in Appendix A.1. We call the function PFa,Q a prefix flip. 3.1.2 Single coordinate flip Lemma 3.4 (Single coordinate flip). Let a = (a1, . . . , an) be a sequence of positive real numbers and let M = max a and m = min a . There exists a bijection SF : 1, 1 n 1, 1 n such i i i i a {− } → {− } that for any v 1, 1 n with ∈ {− } n X(v) := aivi > 0, (21) Xi=1 the image w = SF (v) is obtained from v by flipping a single coordinate from 1 to ( 1), and in a − particular, satisfies n X(w)= a w [X(v) 2M, X(v) 2m]. i i ∈ − − Xi=1 n Construction of SFa. Let v 1, 1 satisfy (21). Further assume that a1 . . . an > 0. ∈ {− } k ≥ ≥ Let k > 0 be the minimal value that maximizes the quantity j=1 vj. Define w = SFa(v) by w = v for all i = k, and w = v . i i 6 k − k P We prove that SFa satisfies the requirements of the lemma in Appendix A.2. We call the function SFa a single coordinate flip. 3.1.3 Recursive flip Lemma 3.5 (Recursive flip). Let a = (a1, . . . , an) be a sequence of positive real numbers and let M = max a and m = min a . There exists a bijection RF : 1, 1 n 1, 1 n such that i i i i a {− } → {− } any v 1, 1 n and its image w = RF (v) satisfy: ∈ {− } a 1) If X(v) > 0 then X(w) [X(v) 2M, X(v) 2m]; ∈ − − 2) If X(v) 0, then either w = v (and X(w)= X(v)) or X(w) [X(v) 2M, 0), ≤ − − ∈ − where X(u) := n a u for u 1, 1 n. i=1 i i ∈ {− } P 9 Construction of RF . Define the auxiliary injection F : v 1, 1 n X(v) > 0 1, 1 n a { ∈ {− } | }→{− } by F (v)= SF (v), − a and note that F is not defined for v X(v) 0 . Define RF : 1, 1 n 1, 1 n by { | ≤ } a {− } → {− } SFa(v), X(v) > 0 RFa(v)= , (F 1)k(v), X(v) 0 (− − ≤ where k 0 is minimal such that (F 1)k+1(v) does not exist, i.e., (F 1)k(v) / Image(F ). ≥ − − ∈ Informally, after defining RF (v) for all v X(v) > 0 using a single coordinate flip, we a { | } would like to define RF (v) for any other v as a simple negation: RF (v)= v. However, this a a − may breach the injectivity if there exists some w such that X(w) > 0 and SFa(w) = v. In 1 − such a case, we define v′ = SFa− ( v) and check whether v′ is ‘vacant’ (i.e., does not collide − − 1 with any SF (z)). If it is vacant, we set RF (v) = v ; otherwise, we continue applying SF− a a − ′ a and negating until we reach a vacant value. The auxiliary function F combines application of 1 SFa with negation, and hence, in each step we apply F − , as stated in the definition. We note that this construction is reminiscent of the way of constructing a bijection from two injections, used in the classical proof of the Cantor-Schr¨oder-Bernstein theorem by J. K¨onig. We call the function RFa a recursive flip. A concrete example that demonstrates the way RFa works is presented in Appendix A.4, and the proof that RFa satisfies the assertion of the lemma is presented in Appendix A.3. 3.2 Proof of Theorem 3.1 The proof of Theorem 3.1 is split into two different arguments, corresponding to how the condition D C +min(D B, 2M) B A in (14) is realized – either as D C +2M B A, − − ≤ − − ≤ − or as D C + (D B) B A. We formulate these two cases as two separate lemmas. − − ≤ − n Lemma 3.6. Let X = i=1 aixi be a Rademacher sum, and write M = maxi ai. For any A,B,C,D R such that ∈ P min( A , B ) C and D C + 2M B A, (22) | | | | ≤ − ≤ − we have Pr[X C, D ] Pr[X A, B ]. (23) ∈ h i ≤ ∈ h i n Lemma 3.7. Let X = i=1 aixi be a Rademacher sum, and write M = maxi ai. For any A,B,C,D R such that ∈ P A C and 2 max(M, D B) C A, (24) | |≤ − ≤ − we have Pr[X C, D ] Pr[X A, B ]. (25) ∈ h i ≤ ∈ h i The combination of Lemmas 3.6 and 3.7 immediately implies Theorem 3.1. Notice that the assumptions in both lemmas are slightly weaker than the assumptions in Theorem 3.1; this weakening of the assumptions will be needed in the sequel. Remark 3.8. The assertions of Lemmas 3.6 and 3.7 hold for other types of segments as well; see Appendix A.6. 10 Notation. Given a Rademacher sum X = n a x and a quadruple A,B,C,D R that i=1 i i ∈ satisfies either (22) or (24) (so that either Lemma 3.6 or Lemma 3.7 can be applied), or D C, P ≤ we write C, D A, B . Using this notation, the lemmas can be rewritten as the deduction h i≺X h i C, D A, B = Pr [X C, D ] Pr [X A, B ] . h i≺X h i ⇒ ∈ h i ≤ ∈ h i 3.2.1 Proof of Lemma 3.6 Proof of Lemma 3.6. We first show that we may assume A B , so that (22) is upgraded to | | ≤ | | A C and D C + 2M B A. (26) | |≤ − ≤ − Upgrading to (26). If A>B, Lemma 3.6 is vacant. Otherwise, if A B we are done. In the | | ≤ | | remaining case, we exchange A,B,C,D by B, A, C, D, so that (26) (and hence (22)) are sat- − − isfied by the new quadruple, and conclude (22) by noting Pr [X A, B ]=Pr[X B, A ]. ∈ h i ∈ h− − i Proving (23). We assume A,B,C,D satisfy (26) and prove (23). We may also assume Q := D B > 0, as otherwise D B, and [C, D] [A, B] implies (23). − ≤ ⊆ Consider the prefix flip map PF . For any v 1, 1 n with X(v) [C, D], we have a,Q ∈ {− } ∈ X(v) Q/2, since by (26), ≥ Q = D B D B + 2M C A 2C. − ≤ − ≤ − ≤ Hence, by Lemma 3.3, for any such v, the image w = PFa,Q(v) satisfies X(w) (X(v) Q 2M, X(v) Q]. ∈ − − − As C X(v) D and Q = D B, this implies X(w) (A, B]. Finally, since X(w)= B may ≤ ≤ − ∈ occur only if X(v)= D, the injectivity of PFa,Q implies Pr[X [C, D ] Pr[X (A, B ], (27) ∈ i ≤ ∈ i which is even slightly stronger than the assertion (23). This completes the proof. 3.2.2 Proof of Lemma 3.7 The proof of the lemma uses an explicit injection which we hereby describe. n Notation. Let X(x)= i=1 aixi be a Rademacher sum, and A,B,C,D,M be real numbers that satisfy the conditions of Lemma 3.7. Note that we may assume D>B, as otherwise, the P assertion of the lemma holds trivially. We partition the coefficients a into ‘large’ and ‘small’ ones. Let { i} L = i a (D B)/2 , S = [n] L. { | i ≥ − } \ Note that this partition depends only on the fixed parameters a and A,B,C,D. { i} In addition, for a subset I [n] and for x 1, 1 n, we write x := x and a := a , ⊆ ∈ {− } I I I I and correspondingly, aI xI = i I aixi. · ∈ Definition. Let v 1, 1 n be such that X(v) [C, D]. We define u = f( v) 1, 1 n in ∈ {− } P ∈ ∈ {− } two steps. First, we denote w = g(v) = (RF (v ), v ) and Q = (D B) (X(v) X(w)). aL L S v − − − Then, we set w, Q 0, u = f(v)= v ≤ ((wL, PFaS ,Qv (wS )), Qv > 0. 11 Motivation. We prove Lemma 3.7 by showing that f : v u injectively maps any v with 7→ X(v) [C, D] to some u with X(u) [A, B]. The single-coordinate-flip map of the large ∈ ∈ coordinates, SF , seemingly has the same property, that is, maps any v with X(v ) [C, D] aL ′ ′ ∈ to some w with X(w ) [A, B]. ′ ′ ∈ However, SF might fail to flip a single coordinate when a v 0. In such a case, we have aL L · L′ ≤ aS vS′ C (D B)/2, and a prefix-flip map PFaS ,D B : v′ u′ does satisfy X(u′) [A, B]. · ≥ ≥ − − 7→ ∈ This reasoning, of mapping v′ into either w′ or u′, proves that Pr [X C, D ] 2 Pr [X A, B ] , ∈ h i ≤ · ∈ h i as it results in a 2-to-1 map (each of v w and v u is injective). ′ 7→ ′ ′ 7→ ′ We wish to show the stronger inequality Pr [X C, D ] Pr [X A, B ]. For this, we ∈ h i ≤ ∈ h i construct f in two steps. At the first step we apply on the large coordinates the recursive-flip RFaL (which is closely related to SFaL ) to obtain w. Then, at the second step, conditioned on data available in wL, we choose whether to apply a prefix flip on the small coordinates or not. This results in a single 1-to-1 map. The usage of RFaL instead of SFaL is important in order for w to have guaranteed properties, even when a v 0. L L · L′ ≤ n Proof of Lemma 3.7. Let X = i=1 aixi and A,B,C,D,M be as in the statement of the lemma, and define the function f on the set v X(v) [C, D] as described above. P { | ∈ } Why is f injective? To compute the inverse map u v, first consider uL and apply the 1 7→ inverse recursive flip RFa−L to recover vL. Then, compute Qv (which depends only on vL and not on vS). Now, it is clear from the definition of f that we have 1 (RF−aL (uL), uS), Qv 0, v = 1 1 ≤ ((RF−aL (uL), PFa−S ,Qv (uS)), Qv > 0. Why is f into [A, B]? Let v satisfy X(v) [C, D]. We want to show that u = f(v) satisfies ∈ X(u) [A, B]. We consider two cases. ∈ Case 1: a v > 0. In this case, since for all i L we have (D B)/2 a M, the first L · L ∈ − ≤ i ≤ property of the recursive flip (presented in Lemma 3.5) implies X(w) [X(v) 2M, X(v) (D B)]. (28) ∈ − − − In particular, we have Q = (D B) (X(v) X(w)) [D B 2M, 0], and thus, by the v − − − ∈ − − definition of f, we set u = f(v)= w. By (28), we have X(u)= X(w) [X(v) 2M, X(v) (D B)] [C 2M, D (D B)] [A, B], ∈ − − − ⊆ − − − ⊆ where the last inclusion uses the assumption 2M C A. ≤ − Case 2: a v 0. In this case, the second property of RF (in Lemma 3.5) implies L · L ≤ aL a w [a v 2M, a v ], (29) L · L ∈ L · L − − L · L and hence, Q = (D B) (X(v) X(w) (D B) [2a v , 2M] = [D B 2M, D B 2a v ]. v − − − ∈ − − L · L − − − − L · L We further subdivide this case into two sub-cases. 12 Case 2a: Q 0. In this case, we have Q [D B 2M, 0], and hence, v ≤ v ∈ − − X(v) X(w) = (D B) Q [D B, 2M]. − − − v ∈ − As X(v) [C, D], this implies ∈ X(w) [C 2M, D (D B)] [A, B], ∈ − − − ⊆ where the last inclusion follows from the assumption 2M C A. ≤ − By the definition of the function f, we have u = w, and so, X(u)= X(w) [A, B]. ∈ Case 2b: Qv > 0. In this case, by the definition of f, we have u = f(v) = (wL, PFaS,Qv (wS)), and hence, we would like to apply Lemma 3.3 to the function PF : w u . To this end, aS ,Qv S 7→ S we have to prove a w Q /2. (30) S · S ≥ v To prove (30), note that the assumptions 2(D B) C A and A C imply − ≤ − | |≤ 2C C A 2(D B) D B. (31) ≥ − ≥ − ≥ − In addition, by (29) we have X(v) X(w)= a v a w 2a v . (32) − L · L − L · L ≥ L · L Since a w = a v and a v + a v = X(v) [C, D], (31) and (32) imply S · S S · S S · S L · L ∈ a w = X(v) a v C a v (D B (X(v) X(w)))/2= Q /2, S · S − L · L ≥ − L · L ≥ − − − v proving (30). As stated above, (30) allows us to apply Lemma 3.3 to the function PF : w u . aS ,Qv S 7→ S Since M ′ = maxi S ai (D B)/2, the lemma implies ∈ ≤ − X(u) X(w)= a u a w ( 2M ′ Q , Q ] − S · S − S · S ∈ − − v − v ((X(v) X(w)) 2(D B), (X(v) X(w)) (D B)] ⊆ − − − − − − = (X(v) X(w)) + ( 2(D B), (D B)]. − − − − − Therefore, X(u)= X(v) + (X(w) X(v)) + (X(u) X(w)) X(v) + ( 2(D B), (D B)]. − − ∈ − − − − Since X(v) [C, D], we have ∈ X(u) (C 2(D B), D (D B)] (A, B], ∈ − − − − ⊆ where the last inclusion follows from the assumption 2(D B) C A. − ≤ − What happens to the endpoints? Notice that in all cases, X(u) = A may hold only if X(v)= C, and X(u)= B may hold only if X(v)= D. Hence, the assertion of the lemma: Pr[X C, D ] Pr[X A, B ]. ∈ h i ≤ ∈ h i follows from the injectivity of f : v u. 7→ 13 4 Improved Berry-Esseen Type Inequalities for Rademacher Sums: Proving Theorem 1.2 for a 0.31 1 ≤ A natural approach toward proving Tomaszewski’s conjecture in the case where all the coeffi- cients ai are small, is using the classical Berry-Esseen theorem ([5, 20]), which allows approx- imating a sum of independent random variables by a Gaussian (i.e., a normally distributed random variable). Theorem 4.1 (Berry-Esseen). Let X1, X2,...,Xn be independent random variables, such that i : E[ X 3] < . Let X = n X . Then for all x, ∀ | i| ∞ i=1 i PX n E X 3 Pr x Pr[Z x] C i=1 | i| , (33) " Var(X) ≤ # − ≤ ≤ · Var(X)3/2 P where Z N(0, 1) is a standardp Gaussian and C is an absolute constant. ∼ For a Rademacher sum X = aixi with Var(X) = 1, the theorem yields P 3 2 Pr[X x] Pr[Z x] C ai C max ai ai = C max ai , | ≤ − ≤ |≤ · ≤ · i | | · i | | Xi Xi and consequently, Pr[ X 1] Pr[ Z 1] 2C max ai . (34) | | |≤ − | |≤ |≤ · i | | The best currently known upper bound on the constant C in (33) is C 0.56, obtained in [40]. ≤ Plugging it into (34) and noting that Pr[ Z 1] 0.682, (34) implies Tomaszewski’s conjecture | |≤ ≥ in the range max a 0.162, as was noted by Bentkus and Dzindzalieta [2]. i | i|≤ Using merely the general form of the Berry-Esseen theorem, this result cannot be improved much. Indeed, it was shown by Esseen [21] that the constant C in (33) satisfies C > 0.409, and hence, the best one can hope for by plugging an improved C into (34) is extending the range to max a 0.182/0.818 0.223. i | i|≤ ≤ In this section we show that refined Berry-Esseen type bounds can be obtained in the special case where X is a Rademacher sum. Our starting point is a smoothing inequality of Prawitz [39] which allows obtaining bounds on the cumulative distribution function of a random variable X (i.e., Pr[X x]), given partial itX ≤ knowledge of its characteristic function ϕX (t)= E[e ]. Prawitz’ inequality has many applica- tions (see Section 4.1). In particular, Prawitz himself suggested using his inequality to bound Pr[X Proposition 4.2. Let X = a x be a Rademacher sum with a a . . . a > 0 and i i i 1 ≥ 2 ≥ ≥ n Var(X) = 1. Then for any T > 0, q [0, 1], and x R we have P ∈ ∈ q 1 Pr[Z 14 (1 u) sin(πu Tux) sin(Tux) where k(u, x, T )= − − , sin(πu) − π 2 2 exp( v /2), a1v θ 2 1/a1 π exp( v /2) cos(a1v) , a1v 2 − 1/a2 ≤ g(v)= − − ≤ , h(v)= ( cos(a1v)) 1 , θ a1v π , (exp( v2/2) + 1, otherwise − ≤ ≤ − 1, otherwise 4 Z N(0, 1) is a standard Gaussian and θ = 1.778 10− is the unique root of the function ∼ ± y exp( y2/2) + cos(y) in the interval [0,π]. 7→ − While in this paper we bound ourselves to proving Proposition 4.2, it appears that one can obtain more general estimates about Rademacher sums using similar strategies; see Section 9. Our first application of Proposition 4.2 validates Tomaszewski’s conjecture in the range max a 0.31 (compared to 0.16 that can be obtained by the general Berry-Esseen bound): i i ≤ Proposition 4.3. Let X = a x be a Rademacher sum with max a 0.31 and Var(X) = 1. i i i i i ≤ Then P Pr[X < 1] Pr[Z < 1] 0.09115, (36) ≥ − and consequently, Pr[X < 1] 0.7501, Pr[ X < 1] 0.5002. (37) ≥ | | ≥ Proposition 4.3 follows from Proposition 4.2 directly, by substitution of suitable parameters. We note that the condition max a 0.31 cannot be relaxed significantly. Indeed, this is i i ≤ demonstrated by X = 1 9 x having max a = 1/3 and Pr[ X < 1] = 63 < 0.493. 3 i=1 i i i | | 128 Our second applicationP of Proposition 4.2 is a concrete estimate which we shall use in the proof of Tomaszewski’s conjecture in the range max a (0.31, 0.5). i i ∈ Proposition 4.4. Let X = a x be a Rademacher sum with max a 0.22 and Var(X) = 1. i i i i i ≤ Then, for any x 0, we have ≥ P Pr[X x] Pr[Z x] max(0.084, Pr[ Z a ]/2), (38) ≤ ≥ ≤ − | |≤ 1 where Z N(0, 1) is a standard Gaussian variable. ∼ Organization. In Section 4.1 we describe the inequality of Prawitz [39] and apply it to Rademacher sums, proving Proposition 4.2. In Section 4.2 we prove Propositions 4.3 and 4.4. In Appendix B we show how to practically evaluate the bound (35) to a required precision. 4.1 The smoothing inequality of Prawitz applied to Rademacher sums 4.1.1 Prawitz’ inequality In [39], H˚akan Prawitz proposed a way for bounding the cumulative distribution function of a random variable X, in terms of partial information on its characteristic function ϕX (t) = E[eitX ]. The main result of [39] (specifically, [39, (1b)]) reads: Theorem 4.5. Let X be a real-valued random variable, and assume that the characteristic function ϕ (t)= E[exp(itX)] is given for t T . Then for any x R, X | |≤ ∈ T 1 ixu 1 Pr[X 4.1.2 A refined inequality for Rademacher sums n For a Rademacher sum X = i=1 aixi, the characteristic function ϕX has the convenient form n P itX ϕX (t)= E[e ]= cos(ait). (41) Yi=1 Since (41) is defined for any t R, for any T > 0 we can substitute in (39) u/T in place of u: ∈ 1 1 ixuT Pr[X 2 log cos(aiv) log cos(aiv) ϕX (v) = exp log cos(aiv) = exp ai | | exp max | | , | | | | a2 ≤ i a2 i ! i i ! i X X (46) 2 where the ultimate inequality holds since ai = 1. The right hand side is maximized at i = 1. Indeed, P • For i with a v θ, by the previous case we have log cos(a v) /a2 v2/2. i ≤ | i | i ≤− • For i with a v [θ,π], as the function ψ : a log cos(a) /a2 increases in the range i ∈ 7→ | | a [θ,π], we get ψ(a v) ψ(a v). ∈ i ≤ 1 Since a = max a satisfies a v (θ,π], a combination of the two cases gives 1 i i 1 ∈ i : log cos(a v) /a2 max( v2/2, log cos(a v) /a2) = log cos(a v) /a2, ∀ | i | i ≤ − | 1 | 1 | 1 | 1 2 yielding ϕ (v) cos(a v) 1/a1 , as we claimed. X ≤ | 1 | Bounding ϕ (uT ) ϕ (uT ) . We claim that | X − Z | 2 2 1/a1 π exp( v /2) cos(a1v) , 0 a1v 2 ϕX (v) ϕZ (v) − − ≤ ≤ . (47) | − |≤ exp( v2/2) + 1, Otherwise ( − To see this, recall that ϕ (v) = exp( v2/2), and hence, for any v R, Z − ∈ ϕ (v) ϕ (v) exp( v2/2) + 1. | X − Z |≤ − In the case 0 a v π/2, on the one hand, similarly to (45), ≤ 1 ≤ ϕ (v)= cos(a v) exp( a2v2/2) = exp( v2/2) = ϕ (v). X i ≤ − i − Z Yi Xi 17 On the other hand, by virtue of (46) and cos(a v) 0, i ≥ 2 cos(aiv) exp(min(log(cos(aiv))/ai )). ≥ i Yi As the function a log(cos(a))/a2 decreases in the range a (0, π/2), one has that for all i, 7→ ∈ log(cos(a v))/a2 log(cos(a v))/a2, and thus, i i ≥ 1 1 2 2 1/a1 ϕZ (v) ϕX (v)= cos(aiv) exp(min(log(cos(aiv))/ai )) = cos(a1v) , ≥ ≥ i Yi yielding (47). Simplifying 2Re eiTuxK(u) . We claim that for all u (0, 1), ∈ (1 u) sin(πu T ux) sin(T ux) k(u, x, T ) := 2Re eiTuxK(u) = − − . (48) sin(πu) − π 1 u i sgn(u) To verify (48), notice that by definition, K(u) = −| | + (1 u ) cot(πu)+ and 2 2 − | | π exp(iT ux) = cos(T ux)+ i sin(T ux), and thus, for u (0, 1) we have ∈ 2 cos(T ux) 2 sin(T ux) k(u, x, T )= (1 u) ((1 u) cot(πu) + 1/π). 2 − − 2 − Substituting cot(πu) = cos(πu)/ sin(πu), we get sin(T ux) cos(πu) k(u, x, T ) = (1 u) cos(T ux) sin(T ux)/π. − − sin(πu) − Using the identity sin(α β) = sin(α) cos(β) sin(β) cos(α), we derive (48). − − Combining the bounds. Substituting the bounds (44) and (47) and the simplification (48) into (43), we obtain (35), namely, the assertion of Proposition 4.2. 4.2 Applications of the refined Berry-Esseen type inequalities 4.2.1 Tomaszewski’s conjecture for a 0.31 1 ≤ We prove Proposition 4.3, which implies Tomaszewski’s conjecture in the range a 0.31. 1 ≤ Proof of Proposition 4.3. Consider first the case a1 = 0.31. Applying Proposition 4.2 with a1 = 0.31, x = 1, T = 10 and q = 0.4, we obtain 5 Pr[Z < 1] Pr[X < 1] 0.09114 10− 0.09115. (49) − ≤ ± ≤ Consequently, Pr[X < 1] Pr[Z < 1] 0.09115 > 0.7501, and since X is a symmetric random ≥ − variable, Pr[ X < 1] = 2Pr[X < 1] 1 0.5002, | | − ≥ as asserted. To handle the case a1 < 0.31, note that the bound in the right hand side of (35) is increasing in a1, and thus, an application of Proposition 4.2 with a1 < 0.31 and the same values x,T,q as above, leads to a stronger lower bound on Pr[ X < 1]. This completes the proof. | | 18 4.2.2 A Berry-Esseen type inequality for Rademacher sums with a 0.22 1 ≤ To prove Proposition 4.4, we use the following lemma, whose proof is given in Appendix B.1. Lemma 4.6. Let X = a x be a Rademacher sum with max a 0.22 and Var(X) = 1. i i i i i ≤ Then for every x 0.35, we have ≥ P Pr[X x] Pr[Z x] 0.084, (50) ≤ ≥ ≤ − where Z N(0, 1) is a standard Gaussian variable. ∼ The proof of Lemma 4.6 proceeds by applying Proposition 4.2 to X, with suitably chosen parameters T,q, and obtaining a slightly stronger version of (50) for a finite set of x’s. Choosing this set of x’s fine enough, (50) follows for all x 0.35, by the monotonicity of x Pr[X x]. ≥ 7→ ≤ Proposition 4.4 follows from Lemma 4.6 and Lemma 3.7. Proof of Proposition 4.4. The proof is split according to the value of x. For x [0, a ) we have ∈ 1 Pr[X x] 1/2 Pr[Z x] Pr[ Z a ]/2, ≤ ≥ ≥ ≤ − | |≤ 1 implying (38). For x [0, 0.2], we have Pr[Z x] 0.58, and thus, Pr[X x] 1/2 Pr[Z ∈ ≤ ≤ ≤ ≥ ≥ ≤ x] 0.084, as asserted. For x 0.35, the assertion (38) follows directly from Lemma 4.6. − ≥ Hence, it is left to prove the assertion for x [max(a , 0.2), 0.35]. We show that in this ∈ 1 range, Pr[X x] Pr[Z x] 0.084. ≤ ≥ ≤ − Indeed, applying Lemma 3.7 to X, with the parameters A,B,C,D,M = x,x,x + ǫ, 2x + − ǫ/2, a and letting ǫ 0+, we obtain Pr[ X x] Pr[X (x, 2x]], and thus, 1 → | |≤ ≥ ∈ 1 Pr[X 0,x]] Pr[X 0, 2x]]. ∈ h ≥ 3 ∈ h (Note that the parameters A,B,C,D,M satisfy the assumptions of Lemma 3.7 since x a .) ≥ 1 Applying Lemma 4.6 with the parameter 2x (which can be done, as by assumption, 2x 0.4 > ≥ 0.35), and using the symmetry of X, we get 1 1 1 Pr[X x] 1/2+ Pr[X 0, 2x]] + (Pr[Z 2x] 0.584). ≤ ≥ 3 ∈ h ≥ 2 3 ≤ − Therefore, in order to complete the proof it is sufficient to show that 1 1 + (Pr[Z 2x] 0.584) Pr[Z x] 0.084, (51) 2 3 ≤ − ≥ ≤ − The inequality (51) indeed holds for all x [0.2, 0.35]. To see this, note that the function ∈ x Pr[Z 2x]/3 Pr[Z x] is decreasing in [0.2, 0.35], and thus, it suffices to verify (51) for 7→ ≤ − ≤ x = 0.35. At x = 0.35, the inequality holds, completing the proof. 5 Theorem 1.2 for a + a 1, via a Semi Inductive Argument 1 2 ≥ In this section we prove the following result. Proposition 5.1. Let 3 n N. Assume that for any Rademacher sum Z = m a x with ≤ ∈ i=1 i i m 19 There is no restriction in assuming n 3, as the assertion of Theorem 1.2 (namely, Pr[ X ≥ | |≤ 1] 1/2) holds trivially for Rademacher sums with n 2. Proposition 5.1 is only ‘semi- ≥ ≤ inductive’ in the sense that it assumes that Theorem 1.2 holds for all m Pr X′ [L ,L ] Pr X′ > R + Pr X′ > R , (52) ∈ 1 2 ≥ 1 2 where a + a 1 1 a + a L ,L = 1 2 − , − 1 2 , 1 2 σ σ and 1+ a a 1+ a + a R ,R = 1 − 2 , 1 2 . 1 2 σ σ Proof. The assertion follows immediately from Lemma 2.1 with m = 2, since Pr[X [L ,L ]] = ′ ∈ 1 2 1 Pr[X > L ] Pr[X >L ]. − ′ − 1 − ′ 2 Note that if a a and a + a 1 then the parameters L ,L ,R ,R satisfy 1 ≥ 2 1 2 ≥ 1 2 1 2 0 L 20 • Case 1: Z 0. We show that in this case, (54) holds as its right hand side is 0. ′ ≡ • Case 2: Z 0. In this case, we show that (54) follows from the inequality (52) applied to ′ 6≡ the Rademacher sum Z′/ Var(Z′). To show that the latter inequality holds, we note that when k is fixed, Z is a Rademacher sum on the n k variables x ,...,x . Hence, by ′ p − k+1 n applying Lemma 5.2 in the inverse direction, we may infer (52) for Z′ from the assertion Pr[ Z 1] 1/2 for an appropriate Rademacher sum Z on n k + 2 variables, which | | ≤ ≥ − holds due to the inductive hypothesis, since k 3. ≥ For the proof of (54), we observe the following relation between Y and Var(Z Y ). ′ ′| ′ Claim 5.3. Let X ,Y ,Z be as defined above, assume Y L and let s = (1 k (a )2)1/2. ′ ′ ′ ′ ≥ 1 − i=3 i′ We have 2 P Y ′(Y ′ L ) 1 s . (55) − 1 ≤ − Proof. By the definition of k, we have Y ′ L1 ak′ ak 1 . . . a3′ . Thus, − ≤ ≤ − ≤ ≤ k k k 2 2 1 s = (ai′ ) min ai′ ai′ (Y ′ L1) ai′ xi = (Y ′ L1)Y ′, − ≥ 3 i k ≥ − − i=3 ≤ ≤ i=3 i=3 X X X as asserted. The case Z 0. We observe that in this ‘singular’ case, the right hand side of (54) is zero, ′ ≡ and hence the inequality trivially holds. Indeed, Z 0 occurs in one of two cases: ′ ≡ • There does not exist k such that Y = k a x L . In this case, Y +Z = n a x < ′ i=3 i′ i ≥ 1 ′ ′ i=3 i′ i L , and thus, the right hand side of (54) is clearly equal to zero by (53). 1 P P • The minimal k such that Y := k a x L is k = n. By (53), in order to show ′ i=3 i′ i ≥ 1 that the r.h.s. of (54) is equal to zero, it is sufficient to prove that Y = Y + Z < R . P ′ ′ ′ 1 To see this, observe that by Claim 5.3, Y (Y L ) 1, and hence Y < R follows ′ ′ − 1 ≤ ′ 1 from R (R L ) > 1 (recall 0 L R by (53)). This latter inequality reads as 1 1 − 1 ≤ 1 ≤ 1 2(1 + a a )(1 a )/σ2 > 1, where σ2 = 1 a2 a2 and a a , and follows by 1 − 2 − 2 − 1 − 2 1 ≥ 2 2(1 + a a )(1 a )= σ2 + (a a )(2 + a a ) + (2a 1)2/2 + 1/2 >σ2. 1 − 2 − 2 1 − 2 1 − 2 2 − The case Z 0. Denote s = (Var(Z ))1/2 = ( n (a )2)1/2, so that 1 Z is a Rademacher ′ 6≡ ′ i=k+1 i′ s ′ sum with variance 1 on n k variables. We have to prove (54) which reads as − P Z L Y L Y Z R Y Z R Y Pr ′ 1 − ′ , 2 − ′ Pr ′ > 1 − ′ +Pr ′ > 2 − ′ . (56) s ∈ s s ≥ s s s s (Note that we assume k,Y ′ are fixed, and hence, the probabilities in (56) depend only on Z′.) We would like to deduce (56) from the assertion Pr[ Z 1] 1/2 for an auxiliary Rademacher | |≤ ≥ sum Z on n k + 2 variables, which holds due to the inductive hypothesis (since n k + 2 21 2 2 where σ′ = 1 b1 b2. (Concrete values of the possibly negative b1, b2 are given below.) We − − ′ define a Rademacher sum Z = b z + b z + σ Z , let p 1 1 2 2 s ′ 1+ b1 b2 1+ b1 + b2 R1′ ,R2′ = − , , σ′ σ′ and show that R1′ and R2′ satisfy R1 Y ′ R2 Y ′ R′ − and R′ − . (58) 1 ≤ s 2 ≤ s As Z is a Rademacher sum with variance 1 on n k + 2 Z′ Z′ Z′ Pr [L′ ,L′ ] Pr > R′ +Pr > R′ , s ∈ 1 2 ≥ s 1 s 2 which implies (56) via (58). So, it is only left to show that Z is well defined and that (58) holds. Why is Z well-defined? To show that there exist b , b R that satisfy (57), let 1 2 ∈ 2 2 2 + 2L1′ L2′ (L2′ ) (L1′ ) b1 = 2 2 and b2 = −2 2 . 2 + (L1′ ) + (L2′ ) 2 + (L1′ ) + (L2′ ) A direct computation shows that 2 2 2(L2′ L1′ ) σ′ = 1 b1 b2 = 2− 2 , (59) − − 2 + (L′ ) + (L′ ) q 1 2 2 2 and in particular, b1 + b2 < 1 (via (53) and the definition of L1′ ,L2′ in (57)). A further direct computation shows that the right equality in (57) holds as well. Proving (58). To verify the two inequalities in (58), notice that by adding L2′ to both sides of the first, and L to both sides of the second, and recalling the definitions of L ,L ,R ,R − 1′ 1 2 1 2 and L1′ ,L2′ ,R1′ ,R2′ , these inequalities are respectively equivalent to: s 1 s 1 + Y ′ and . σ′ ≤ σ σ′ ≤ σ As Y L 0 by (53), the former inequality clearly implies the latter; hence, we focus only ′ ≥ 1 ≥ on it. Note that similarly to (59), we have 2(L2 L1) σ = 2− 2 . 2+ L1 + L2 In addition, by (57), we have L L = (L L )/s. Using this and substituting the values of 2′ − 1′ 2 − 1 σ, σ , the inequality we seek to prove s + Y 1 , reads as ′ σ′ ′ ≤ σ 2 2 2 2 2 s (2 + L1′ + L2′ ) (2 + L1 + L2) + Y ′ . 2(L L ) ≤ 2(L L ) 2 − 1 2 − 1 Multiplying by 2(L L ) and substituting the values of L ,L from (57), we reduce to showing 2 − 1 1′ 2′ 2 2 2 2 2 2s + (L Y ′) + (L Y ′) + 2Y ′(L L ) 2+ L + L , 1 − 2 − 2 − 1 ≤ 1 2 Since s2 1 Y (Y L ) by Claim 5.3, it is sufficient to prove that ≤ − ′ ′ − 1 2 2 2 2 2 2Y ′(Y ′ L ) + (L Y ′) + (L Y ′) + 2Y ′(L L ) 2+ L + L . − − 1 1 − 2 − 2 − 1 ≤ 1 2 Simplifying this inequality, one sees it is equivalent to the inequality 2Y L 0, which indeed ′ · 1 ≥ holds since by (53), we have 0 L Y . This completes the proof of Proposition 5.1. ≤ 1 ≤ ′ 22 6 Theorem 1.2 for a 0.55, a + a < 1 1 ≥ 1 2 In the previous sections we handled Theorem 1.2 in the cases a 0.31 (Section 4.2), and 1 ≤ a + a 1 (Section 5, semi-inductively). In this section we handle the case (a 0.55) (a + 1 2 ≥ 1 ≥ ∧ 1 a2 < 1). Elimination step. Let X = n a x be a Rademacher sum with Var(X) = 1, a (1 + i=1 i i 1 ≥ √8)/7, and a + a < 1. (Note that 1 + √8/7 0.55.) We want to prove Pr[ X 1] 1/2. 1 2 P ≤ | |≤ ≥ By the case m = 1 of Lemma 2.1, it is sufficient to prove that Pr[X′ 0,t]] Pr[X′ > 1/t], (60) ∈ h ≥ where n 2 ai 1 a1 1 a1 σ = 1 a , a′ = , X′ = a′ x , and t = − = − − 1 i σ i i 1+ a σ i=2 r 1 q X (see (10) and (11) above). As we assume a1 + a2 < 1, we have i: a′ < t. (61) ∀ i Segment comparison step. Claim 6.1. Let X′,t be as defined above. Then: (a) Pr [X t, 2t ] 2Pr[X 0,t ] , and ′ ∈ h i ≤ ′ ∈ h i (b) k 2: Pr[X kt, (k + 1)t ] 4Pr[X 0,t ]. ∀ ≥ ′ ∈ h i ≤ ′ ∈ h i Proof. (a) Applying Lemma 3.7 to X , with the parameters A,B,C,D,M = t,t,t, 2t, a , we get ′ − 2′ Pr X′ t, 2t Pr X′ t,t . ∈ h i ≤ ∈ h− i (Note that 2M C A follows from (61)). By the symmetry of X , this implies (a). ≤ − ′ (b) Applying Lemma 3.6 to X , with the parameters A,B,C,D,M = t, 2t,kt, (k + 1)t, a , we ′ − 2′ get Pr [X kt, (k + 1)t ] Pr [X t, 2t ]. Using (a) as ′ ∈ h i ≤ ′ ∈ h− i Pr X′ t, 2t = Pr X′ t,t +Pr X′ t, 2t 4Pr X′ 0,t , ∈ h− i ∈ h− i ∈ h i ≤ ∈ h i the assertion (b) follows. Chebyshev-type inequality step. Applying the inequality (17) to X′, with c0, c1, c2, c3,...,c 1/t = 0, t, 2t, 3t,..., 1 and d0, d1 = 1, 1/t, ⌈ ⌉ we obtain 1/t 1 ⌈ ⌉− 2 1 (1 (kt) )Pr X′ kt, (k + 1)t 1 Pr[X′ 1/t]. − ∈ h i ≥ t2 − ≥ k=0 X By Claim 6.1, this implies 1/t 1 ⌈ ⌉− 2 2 1 1 + 2(1 t ) + 4 1 (kt) Pr X′ 0,t 1 Pr[X′ 1/t]. − − ∈ h i ≥ t2 − ≥ k=2 X 23 Dividing both sides by 1/t2 1, we obtain an inequality of the form − C Pr X′ 0,t Pr[X′ 1/t]. t · ∈ h i ≥ ≥ This inequality implies (60), provided C 1. Hence, it is left to verify: t ≤ 1/t 1 ? 1 + 2(1 t2) + 4 ⌈ ⌉− 1 (kt)2 k=2 1/2 Ct 1, with Ct = − 2 − , 0 Pr X′ t, 3t/2 Pr X′ 0,t and Pr X′ 3t/2, 1 Pr X′ 0,t . (63) ∈ h i ≤ ∈ h i ∈ h i ≤ ∈ h i The first inequality follows from Lemma 3.7, applied to X′ with the parameters A,B,C,D,M = 0,t,t, 3t/2, a2/σ, and the second inequality follows from Lemma 3.6, applied to X′ with the parameters A,B,C,D,M = 0, t, 3t/2, 1, a2/σ. To show that Lemmas 3.7 and 3.6 indeed can be applied (i.e., that the assumptions of the lemmas are satisfied), it is sufficient to verify: 2a ? 2a ? 2 t, (1 3t/2) + 2 t σ ≤ − σ ≤ with: a [0.5, 0.55], σ = 1 a2, (64) 1 ∈ − 1 1 a a 3+ 25q + 10a 63a2 t = − 1 , a 1 − 1 − 1 . 1+ a 2 ≤ 8 r 1 p These inequalities are proved in Appendix F.2. Chebyshev-type inequality step. Note that as a 1/2, we have t = (1 a )/(1 + a ) 1 ≥ − 1 1 ≤ 1/√3 < 2/3. Applying the inequality (17) to X , with ′ p c0, c1, c2, c3 = 0, t, 3t/2, 1 and d0, d1 = 1, 1/t, and using (63), we obtain 2 2 1 (1 + (1 t ) + (1 (3t/2) ))Pr X′ 0,t 1 Pr[X′ 1/t]. − − ∈ h i ≥ t2 − ≥ 24 Hence, the task of deducing (60) boils down to verifying: ? 2 2 1 + (1 t ) + (1 (3t/2) ) 2 C′ 1, with: C′ = − − , t (0, 1/3]. (65) t ≤ t (1/t2) 1 ∈ − This verification is done in Appendix F.3. 7.2 The case a (a 3+ 25+10a 63a2)/8 2 ≥ 1 − 1 − 1 Let X = aixi be a Rademacherp sum with Var(X) = 1, a1 [0.5, 0.55], a1 + a2 < 1, and ∈ a (a 3+ 25 + 10a 63a2)/8. The proof of Pr[ X 1] 1/2 is similar to the above 2 ≥ 1 −P 1 − 1 | | ≤ ≥ strategy, but this time, two variables are eliminated. p Elimination step. By Lemma 5.2, it it sufficient to show that Pr X′ 0, L + Pr X′ 0,L Pr[X′ > R ] + Pr[X′ > R ], (66) ∈ h − 1i ∈ h 2i ≥ 1 2 where X′,σ,L1,L 2,R1,R2 are as defined in Lemma 5.2. (But this time, since a1 + a2 < 1 we have L < 0). Note that unlike the previous subsection, X depends on n 2 variables. 1 ′ − Auxiliary estimates. We use several auxiliary estimates on L2,R1,R2: ? 2 ? ? L ,R max( 3 L2, √2),R max( 5 2L2, √3), 2 ≥ 3 1 ≥ − 2 2 ≥ − 2 1 a1 +qa2 1+ a1 a2 q 1+ a1 + a2 with: L2 = − ,R1 = − ,R2 = , σ σ σ (67) σ = 1 a2 a2, a [0.5, 0.55], a + a < 1, − 1 − 2 1 ∈ 1 2 q a a 3+ 25 + 10a 63a2 /8. 2 ≥ 1 − 1 − 1 q These inequalities are proved in Appendix F.4. The sub-case L 1. Applying the Chebyshev-type inequality (17) to X , with c , c = 0, 1 2 ≥ ′ 0 1 and d1, d2, d3 = 1, √2, √3, we obtain Pr X′ 0, 1 Pr[X′ √2] + Pr[X′ √3], ∈ h i ≥ ≥ ≥ which implies (66) via (67). The sub-case L2 < 1. Applying Lemma 3.7 to X′, with the parameters A,B,C,D,M = 0,L2,L2, 1, a3/σ, we obtain Pr X′ L , 1 Pr X′ 0,L . (68) ∈ h 2 i ≤ ∈ h 2i Notice that the assumptions of Lemma 3.7 are satisfied, as 2(D B) = 2(1 L ) L = C A − − 2 ≤ 2 − by (67), and 2a 1 a + a 2M 2 < − 1 2 = L = C A, ≤ σ σ 2 − by the assumption a1 + a2 < 1. Applying the Chebyshev-type inequality (17) to X′, with c0, c1, c2 = 0,L2, 1 and d1, d2, d3 = 1, 3 L2, 5 2L2, we obtain − 2 − 2 2 pPr X′ p 0,L + (1 L )Pr X′ L , 1 ∈ h 2i − 2 ∈ h 2 i ≥ (69) 2 2 2 2 (2 L )Pr X′ 3 L + (2 L )Pr X′ 5 2L . ≥ − 2 ≥ − 2 − 2 ≥ − 2 q q 25 By (68), this implies 2 2 Pr X′ 0,L Pr X′ 3 L + Pr X′ 5 2L , ∈ h 2i ≥ ≥ − 2 ≥ − 2 q q which, in turn, implies (66) via (67). This completes the proof. 8 The remaining case: a (0.31, 0.5) 1 ∈ Our proof is most involved in this range, although there are no remarkable tightness examples with a (0.31, 0.5) (except for X = 1 9 x , being the ‘lightest’ appearing in Section 2.3). 1 ∈ 3 i=1 i The full proof in this range is deferred to appendices C, D, and E, corresponding to three P main subcases. The general structure of the proof is to condition on the values of a few largest weights (say, a1, a2, a3), usually through elimination (Lemma 2.1), and to prove Tomaszewski’s assertion (1) regardless of the values we condition on. While the careful proof is somewhat cumbersome, it was plotted by considering several specific fixings of the large weights (say, a1, a2, a3 = 0.4, 0.3, 0.2) and proving the assertion (1) under these fixings. Then, the proof was generalized to capture any such fixing. We demonstrate the proof by presenting several (not fully) representative such fixings, and proving (1) under them. 8.1 Subcase a + a + a 1 1 2 3 ≤ The following example concisely demonstrates all the steps in the proof of this case (given in Appendix C). Assume a1, a2, a3 = 0.36, 0.2, 0.15, so that 3-elimination (Lemma 2.1) reduces us to proving the following inequality for all Rademacher sums X with Var(X ) = 1, whose largest weight is 0.15/(1 0.362 0.22 0.152)1/2 < 0.17: ′ ′ ≤ − − − Pr X′ 0, 0.32 + Pr X′ 0, 0.65 +Pr X′ 0, 0.76 + Pr X′ 0, 1.1 ∈ h i ∈ h i ∈ h i ∈ h i (70) ≥ Pr[X′ > 1.12] + Pr[X′ > 1.45] + Pr[X′ > 1.56] + Pr[X′ > 1.9] Using Chebyshev’s inequality (17), we lower bound an expression similar to the LHS of (70): Pr X′ 0, 0.32 + 0.9Pr X′ 0.32, 0.65 + ∈ h i ∈ h i 0.6Pr X′ 0.65, 0.76 + 0.5Pr X′ 0.76, 1 ∈ h i ∈ h i (71) ≥ (0.25 Pr[X′ > 1.12] + 0.25 Pr[X′ > 1.23]) + 0.5Pr[X′ > 1.42]+ (0.4Pr[X′ > 1.56] + 0.1Pr[X′ > 1.6]) + 0.5Pr[X′ > 1.75]. Can we deduce (70) from (71)? Not immediately. If we multiply (71) by 2 and see what is ‘missing’ in order to deduce (70), we get an inequality weaker than: Pr X′ 0, 0.32 + Pr X′ 0, 0.65 0.5Pr X′ 1.12, 1.23 + 0.2Pr X′ 1.56, 1.6 . ∈ h i ∈ h i ≥ ∈ h i ∈ h i Recall that the largest weight of X′ is at most 0.17, so using segment comparison (specifically, Lemma 3.6), we can prove this inequality by showing Pr [X 1.12, 1.23 ] Pr [X 0, 0.65 ] ′ ∈ h i ≤ ′ ∈ h i and Pr [X 1.56, 1.6 ] Pr [X 0, 0.65 ]. ′ ∈ h i ≤ ′ ∈ h i 26 8.2 Subcase a + a + a 1 and a (0.387, 0.5) 1 2 3 ≥ 1 ∈ The proof in this range is similar to that of Section 8.1, but the details are much simpler. We again give a demonstration that captures the essence of the proof (given in Appendix D). Assume a1, a2, a3 = 0.4, 0.35, 0.3, so that 3-elimination (Lemma 2.1) reduces us to prove the following inequality for all Rademacher sums X with Var(X ) = 1, whose largest weight is 0.3/(1 0.42 0.352 0.32)1/2 < 0.38: ′ ′ ≤ − − − Pr X′ 0.07, 0.69 + Pr X′ 0, 0.82 + Pr X′ 0, 0.94 ∈ h i ∈ h i ∈ h i (72) ≥ Pr[X′ > 1.57] + Pr[X′ > 1.7] + Pr[X′ > 1.83] + Pr[X′ > 2.58] We prove (72) even without the Pr [X 0.07, 0.69 ] term. Using Chebyshev’s inequality (17): ′ ∈ h i Pr X′ 0, 0.82 + 0.33Pr X′ 0.82, 0.94 + 0.12Pr X′ 0.94, 1 ∈ h i ∈ h i ∈ h i (73) ≥ 0.7Pr[X′ > 1.31] + 0.7Pr[X′ > 1.55] + 0.7Pr[X′ > 1.77] + 0.7Pr[X′ > 1.95]. Can we deduce (72) from (73)? Not immediately. If try to deduce 0.7 (72) from (73) we see · that what is ‘missing’, is weaker than: 0.4Pr X′ 0, 0.82 0.12Pr X′ 0.94, 1 0. ∈ h i − ∈ h i ≥ Using segment comparison, and specifically Lemma 3.6, barely applicable since (1 0.94) + 2 − · 0.38 = 0.82, we conclude Pr [X 0.94, 1 ] Pr [X 0, 0.82 ]. Also, we are equally satisfied ′ ∈ h i ≤ ′ ∈ h i with Pr X′ 0.94, 1 Pr X′ 0.82, 0.82 2Pr X′ 0, 0.82 , ∈ h i ≤ ∈ h− i ≤ ∈ h i which liberally follows from Lemma 3.6 (and whose analog holds true in the entire range). 8.3 Subcase a + a + a 1 and a (1/3, 0.387) 1 2 3 ≥ 1 ∈ The proof in this case is a bit more complicated than in the other cases, and the demonstration we give does not capture the entire set of arguments we use. However it represents perhaps the most exotic argument, and features a semi-inductive argument along with a 5-elimination. Assume a1 = a2 = a3 = a4 = a5 = 0.34, so that 5-elimination (Lemma 2.1) reduces us to proving the following inequality for all Rademacher sums X with Var(X ) = 1, whose largest weight is 0.34/(1 5 0.342)1/2 < 0.6: ′ ′ ≤ − · 4Pr X′ 0, 1.01 +5Pr X′ 0.04, 1.01 +Pr X′ 1.08, 2.06 ∈ h i ∈ h i ∈ h i (74) ≥ Pr[X′ > 1.01] + 9 Pr[X′ > 2.06] + 5 Pr[X′ > 3.1] + Pr[X′ > 4.15]. We prove (74) even with only the first term on the LHS. Using Chebyshev’s inequality (17): 3Pr X′ 0, 1 9Pr[X′ > 2] + 15 Pr[X′ > 3]. (75) ∈ h i ≥ 27 In order to deduce (74) from (75), we have to show Pr X′ 0, 1.01 Pr[X′ > 1.01]. (76) ∈ h i ≥ This inequality is actually not easy to prove. However, it is implied by Tomaszewski’s asser- tion (1) for the variable X which depends on n 5 weights (where n is the number of weights ′ − in the original Rademacher sum X we discuss)! We comment that in general we should use (3) instead of (1) to prove (76), which is possible as (3) for X′ is implied from Tomaszewski’s assertion (1) on n 4 weights. − 9 Open Problems As was mentioned in the introduction, we believe the methods developed in this paper can be applied to obtain further results on the distribution of Rademacher sums. We conclude the paper with several related open questions. Tail bounds for Rademacher sums. Consider the following general problem. Problem 9.1. Let be the class of all Rademacher sums with variance 1. Characterize the X following function, defined for all x R: ∈ F (x)= sup Pr[X>x]. X ∈X While different sub-cases of this problem were studied in many papers (see, e.g., [16, 38] and the numerous references therein), only a few exact results on it are known (e.g., [2, 38]). Theorem 1.2 continues the series of exact results, showing that F (x) = 1/4 for all x [1, √2). ∈ A well-known conjecture, due to Hitczenko and Kwapie´n[28], concerns F (x) for x = 1. − Conjecture 9.2 ([28]). Let X = aixi be a Rademacher sum with Var(X) = 1. Then 7 P Pr[X 1] . ≥ ≥ 64 Conjecture 9.2 is an evident counterpart of Tomaszewski’s conjecture – while the latter states that Pr[X > 1] must be somewhat small, the former declares that Pr[X 1] must be ≥ somewhat large. The best currently known result toward Conjecture 9.2 is Pr[X > 1] 1/20 (whenever ≥ X x ), proved by Oleszkiewicz [36] more than 20 years ago. Using our methods (specifically, 6≡ 1 Proposition 4.2 and Lemma 2.1 with m 3) and additional tools, Dvoˇr´ak and the second ≤ author [14] proved the stronger bound Pr[X 1] 6/64, along with the sharp bound Pr[X > ≥ ≥ 1] 1 . ≥ 16 Improved Berry-Esseen type bounds for Rademacher sums. Proposition 4.4 shows that for a Rademacher sum X = a x with Var(X)=1 and 0 a a = 0.22, and for any i i ≤ i ≤ 1 x 0, ≥ P Pr[X x] Pr[Z x] Pr[Z 0, a ]], ≤ ≥ ≤ − ∈ h 1 where Z is a standard Gaussian. The following conjecture is a natural extension: Conjecture 9.3. Let X = a x be a Rademacher sum with Var(X) = 1 and i: 0 < a a . i i ∀ i ≤ 1 Let Z N(0, 1) be a standard Gaussian, and x R. Then ∼ P ∈ 1 Pr[X x] Pr[Z x] Pr[Z (0, a1)] < a1. (77) | ≤ − ≤ |≤ ∈ √2π 28 n Note that first inequality in (77) is tight, as is demonstrated by X = i=1 xi/√n for n odd, and x = 1/√n ǫ.We note that it follows from [34, Theorem 1.3 and Remark 1.4(c)] that for n − P X = i=1 xi/√n, this inequality holds for any n,x. 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[45] M. C. A. van Zuijlen, On a conjecture concerning the sum of independent Rademacher random variables, 2011. Available at arXiv:1112.4988v1. [46] M. Veraar, A note on optimal probability lower bounds for centered random variables, Colloq. Math., 113 (2008), pp. 231–240. [47] F. von Heymann, Ideas for an old analytic enigma about the sphere that fail in intriguing ways, manuscript, 2012. Available at http://www.mi.uni-koeln.de/opt/wp-content/uploads/2017/02/Cube sphere.pdf. 31 Appendix A Proofs for Section 3 A.1 Proof of Lemma 3.3 Proof. We define PF (v) for all v 1, 1 n that satisfy X(v) Q/2, and show that the a,Q ∈ {− } ≥ map is injective, into 1, 1 n, and satisfies (20). Then, PF can be arbitrarily completed to {− } a,Q a bijection on 1, 1 n that satisfies the assertion of the lemma. {− } Given v, consider the partial sums s (v)= k a v , and let k [n] be minimal such that k j=1 j j ∈ s (v) Q/2. Note that k is well-defined, since s (v)= X(v) Q/2. Define w = PF (v) by k ≥ Pn ≥ a,Q vi, i k wi = − ≤ . ( vi, i>k It turns out that k, and hence v, can be recovered from w (which implies that PFa,Q is injective): i wi, i k k = min i [n] ajwj Q/2 , and vi = − ≤ . ∈ ≤− w , i>k j=1 ( i X Hence, to complete the proof we have to show that for all v and w = PF (v), we have a,Q X(w) (X(v) Q 2M, X(v) Q]. ∈ − − − To see this, note that X(v) X(w) = 2s (v). As k is minimal with the property s (v) Q/2, − k k ≥ and as M = max a , we have s (v) [Q/2, Q/2+ M), and hence, i i k ∈ X(w)= X(v) 2s (v) (X(v) Q 2M, X(v) Q], − k ∈ − − − as asserted. A.2 Proof of Lemma 3.4 Proof. We define SF (v) for all v 1, 1 n that satisfy X(v) > 0, and show that the map a ∈ {− } is injective, into 1, 1 n, and that v and w = SF (v) differ in exactly one coordinate. Then, {− } a SF can be arbitrarily completed to a bijection on 1, 1 n that satisfies the assertion of the a {− } lemma. Without loss of generality, assume a1 a2 . . . an > 0. ≥ i≥ ≥ Given v, consider its partial sums Si(v) = j=1 vj, where S0(v) = 0. (Note that these partial sums differ from the sums s (v) considered in Lemma 3.3). Let k [n] be minimal such i P ∈ that S (v) S (v), for all j [n] (i.e., the lowest amongst the indices in which the maximal k ≥ j ∈ partial sum is attained). Then, define w = SFa(v) by vi, i = k wi = − . ( vi, i = k 6 Let us temporarily suppose that Sk(v) > 0; we prove this assertion at the end of the proof. First, we observe that SFa(v) is obtained from v by flipping a ‘positive’ coordinate (i.e., vk = 1). To see this, notice that if k > 1, then vk = 1, as otherwise Sk 1(v) > Sk(v), − contradicting the definition of k. If k = 1, one has v1 = S1(v) > 0, yielding, once again, vk = 1. Second, we observe that k, and hence v, can be recovered from w, via the formula wi, i = k k = 1+max 0 i 32 • If k > 1, then since vk = 1, we have Sk 1(w)= Sk 1(v)= Sk(v) 1. − − − Moreover, for all i k one has Si(w)= Si(v) 2 Hence, SFa(v) is an injection, as asserted. To conclude the proof, we show that for the chosen k, we have Sk(v) > 0. By the definition of k, this assertion is equivalent to maxi [n] Si(v) > 0. The latter follows from Abel’s summation formula using the assumptions X(v)=∈ n a v > 0 and a a 0, with a = 0: i=1 i i i − i+1 ≥ n+1 n P n 0 < X(v)= aivi = (ai ai+1)Si(v) a1 max Si(v). − ≤ i [n] Xi=1 Xi=1 ∈ This completes the proof. A.3 Proof of Lemma 3.5 Proof of Lemma 3.5. Let RFa and F be defined as in Section 3.1. Why is RFa well defined? To show RFa is well-defined, we have to prove that for any v with X(v) 0, there exists k 0 such that F k(v) / Image(F ). ≤ ≥ − ∈ Indeed, if there was no such k, we would enter a loop (as 1, 1 n is finite), that is, 1 k 1 k′ {− } (F − ) (v) = (F − ) (v) for some k < k′. Applying F on both sides, k times, we would ′′ get v = (F 1)k (v), with k > 0, contradicting the assumption v / Domain(F ). − ′′ ∈ Why is RFa a bijection? It is sufficient to show RFa is an injection. Suppose RFa(v) = RFa(u). We consider three cases. • If X(v) > 0 and X(u) > 0, then v = u because SFa is injective, due to Lemma 3.4. 1 k • If X(v) 0 and X(u) > 0, then we have (F − ) (v)=RFa(v)=RFa(u)= F (u), with ≤ 1 k − 1 k+1 − k as above. In particular, (F − ) (v) = F (u), and so (F − ) (v) exists, contradicting the definition of k. ′ • If X(v) 0 and X(u) 0, then we have (F 1)k(v) = (F 1)k (u). Without loss of ≤ ≤ − − generality, assume k′ k, and thus, after applying F to both sides, k times, we obtain ′′ ≥ v = (F 1)k (u), with k 0. If k = 0 then v = u and we are done. Otherwise, we get a − ′′ ≥ ′′ contradiction, since v / Domain(F ). ∈ Why does RFa satisfy the asserted properties? The first property of RFa holds directly by Lemma 3.4. We show that the second property holds as well. Let v be such that X(v) 0, ≤ and so, w := RF (v)= (F 1)k(v) for some k 0. We consider three cases. a − − ≥ • If k = 0, then w = v and X(w)= X(v). − − 33 • If k = 1, then w = F 1(v), and so X( w) > 0 and − − − RF ( w)=SF ( w)= F ( w)= v. a − a − − − − Thus, from the first property of RFa (i.e., Lemma 3.4), we have X(v)= X( v)= X(RF ( w)) [X( w) 2M, X( w) 2m] . − − a − ∈ − − − − This implies X(v) [X(w) + 2m, X(w) + 2M], and hence, ∈ X(w) [X(v) 2M, X(v) 2m]. ∈ − − • If k > 1, let u = (F 1)k 1(v), so that w = F 1(u). Clearly, u Domain(F ) and so − − − − ∈ X(u) > 0. By the first property of RFa (i.e., Lemma 3.4), we have 1 1 0 > X(u)= X( u) X(RF− ( u)) 2M = X(F − (u)) 2M = X( w) 2M. − − ≥ a − − − − − Moreover, ( w) Domain(F ) implies X( w) > 0. Finally, since X(w) = X( w), we − ∈ − − − have X(w) ( 2M, 0). This completes the proof. ∈ − A.4 An example of the bijection Recursive Flip Example A.1. Let X(v) = a v + a v + a v , where a a a > 0 and a > a + a . 1 1 2 2 3 3 1 ≥ 2 ≥ 3 1 2 3 Clearly, M = max a = a and m = min a = a . The set v : X(v) > 0 consists of i i 1 i i 3 { } v1 = (1, 1, 1), v2 = (1, 1, 1), v3 = (1, 1, 1), v4 = (1, 1, 1). − − − − For these vectors, we define RF (v)=SF (v) and F (v)= SF (v), and so, we have a a − a RF (v1) = (1, 1, 1), RF (v2) = (1, 1, 1), a − a − − RF (v3) = ( 1, 1, 1), RF (v4) = ( 1, 1, 1), a − − a − − − and F (v1) = ( 1, 1, 1), F (v2) = ( 1, 1, 1), F (v3) = (1, 1, 1), F (v4) = (1, 1, 1). − − − − Now, we consider the four remaining vectors: w1 = ( 1, 1, 1), w2 = ( 1, 1, 1), w3 = ( 1, 1, 1), w4 = ( 1, 1, 1). − − − − − − − − We have w1, w3 / Image(F ), and hence, we set ∈ RF (w1)= w1 = (1, 1, 1), and RF (w3)= w3 = (1, 1, 1). a − a − − For w2, we have F 1(w2)= v1 = (1, 1, 1) Image(F ), and thus, we move forward to F 2(w2)= − ∈ − F 1(v1)= v4 = (1, 1, 1) / Image(F ). Hence, we define − − − ∈ 2 2 2 RF (w )= F − (w ) = ( 1, 1, 1). a − − Similarly, for w4 we have F 2(w4) = (1, 1, 1) / Image(F ), and thus, we define − − ∈ 4 2 4 RF (w )= F − (w ) = ( 1, 1, 1). a − − − It is easy to see that RFa is indeed a bijection that satisfies the assertion of Lemma 3.5. 34 A.5 Proof of Lemma 3.2 Let X be a symmetric (around 0) random variable with Var(X) = 1. Observe that: 2 2 ½ E ½ E (1 X ) X <1 = (X 1) X >1 . (79) − {| | } − {| | } The equality (79) holds, since by linearity of expectation, 2 2 2 2 ½ ½ E (1 X )½ X <1 E (X 1) X >1 = E (1 X ) X <1 X >1 = E (1 X ) = 0, − {| | } − − {| | } − {| | }∪{| | } − where the ultimate equality is satisfied as E[X2] = Var(X)+ E[X]2 =1+0 2 = 1. Notice that (79) immediately implies Chebyshev’s inequality. Indeed, by upper bounding the left hand side of (79) with Pr[ X < 1] and lower bounding the right hand side of (79) by | | (t2 1) Pr[ X t] for any t 1, one obtains Pr[ X < 1] (t2 1) Pr[ X t], or equivalently, − | |≥ ≥ | | ≥ − | |≥ (t2 1) Pr[ X t] + Pr[ X 1] 1, − | |≥ | |≥ ≤ yielding the Chebyshev bound Pr[ X t] 1/t2. | |≥ ≤ It is however clear that there is extra freedom in the derivation of this inequality. Specifically, for any set of real numbers 0= c c . . . c =1= d d . . . d d = , 0 ≤ 1 ≤ ≤ n 0 ≤ 1 ≤ ≤ m ≤ m+1 ∞ we can upper bound the left hand side of (79) as n 1 2 − 2 E ½ (1 X ) X <1 (1 ci )Pr[ X [ci, ci+1)], (80) − {| | } ≤ − | |∈ i=0 X and lower bound the right hand side as m m 2 2 2 2 E ½ (X 1) X >1 (di 1) Pr[ X [di, di+1)] = (di di 1)Pr[ X di]. (81) − {| | } ≥ − | |∈ − − | |≥ i=1 i=1 X X Using the symmetry of X, we may deduce n 1 m − (1 c2)Pr[X c , c ] (d2 1) Pr [X d , d ] , (82) − i ∈ h i i+1i ≥ i − ∈ h i i+1i Xi=0 Xi=1 and similarly, n 1 m − 2 2 2 (1 ci )Pr[X ci, ci+1 ] (di di 1)Pr[X di], (83) − ∈ h i ≥ − − ≥ Xi=0 Xi=1 proving Lemma 3.2. A.6 Comparison lemmas for other types of segments One may consider 9 types of segments T (a, b), characterized by having each of its ends a, b open, closed, or semi-open (the latter meaning that the probability at that end counts as 1/2, which we denote by or ). h i 35 • It turns out that if C, D A, B (that is, if A,B,C,D satisfy the assumptions of h i ≺X h i either Lemma 3.6 or Lemma 3.7), then for any segment type T we have Pr[X T (C, D)] Pr[X T (A, B)]. ∈ ≤ ∈ To see this, one can track the proofs of Lemmas 3.6 and 3.7 and notice that all injections we use, map v to w in such a way that if X(v) (C, D) then X(w) (A, B), if X(v)= C ∈ ∈ then X(w) [A, B), and if X(v) = D then X(w) (A, B]. The only exception is in ∈ ∈ Lemma 3.6 in the case A Pr[X S(C, D)] Pr[X T (A, B)] ∈ ≤ ∈ for different segment types S, T , as is demonstrated in (27). Appendix B Proofs for Section 4 B.1 Proof of Lemma 4.6 Proof of Lemma 4.6. Since in the proof we use the value of a1 only through substitution in Proposition 4.2, we may assume a = 0.22 rather than a 0.22 (as otherwise, the lower 1 1 ≤ bound only gets better). Consider first the range x 1.65. Applying Proposition 4.2 with a = 0.22, x = 1.65, ≥ 1 T = 14.5, and q = 0.4, we obtain Pr[Z < 1.65] Pr[X < 1.65] < 0.0314, (84) − and consequently, Pr[X < 1.65] > 0.919. In particular, for all x 1.65 we have ≥ Pr[X x] > 1 0.084 Pr[Z x] 0.084, ≤ − ≥ ≤ − as claimed in the lemma. Now, let x [0.35, 1.65]. Instead of delicately analysing the bound (35) as x varies, it is ∈ sufficient to show that for some finite sequence x m with 0.35 = x < x <... Pr[Z 36 This would readily imply Pr[Z x 93 = (0.35, 0.358, 0.366, 0.374, 0.38, 0.386, 0.39, 0.395, 0.399, 0.403, 0.406, 0.409, 0.412, { i}i=0 0.415, 0.417, 0.419, 0.421, 0.423, 0.425, 0.427, 0.428, 0.429, 0.43, 0.431, 0.432, 0.433, 0.434, 0.435, 0.436, 0.437, 0.438, 0.439, 0.44, 0.441, 0.442, 0.443, 0.444, 0.445, 0.446, 0.447, 0.448, 0.449, 0.45, 0.451, 0.452, 0.453, 0.454, 0.455, 0.456, 0.457, 0.458, 0.459, 0.46, 0.461, 0.462, 0.463, 0.464, 0.466, 0.468, 0.47, 0.472, 0.474, 0.476, 0.478, 0.481, 0.484, 0.487, 0.49, 0.494, 0.499, 0.504, 0.51, 0.517, 0.526, 0.537, 0.55, 0.567, 0.589, 0.61, 0.63, 0.65, 0.67, 0.69, 0.71, 0.73, 0.76, 0.8, 0.85, 0.91, 0.98, 1.07, 1.2, 1.38, 1.65). (87) This completes the proof. B.2 Numeric integration in our proofs In the proofs of Proposition 4.3 and of Lemma 4.6, we evaluate multiple times the right hand side of (35), with various parameters. The proof requires precision of 10 5 in the results of ± − the evaluations. The Python3 program we used for the evaluation (provided in https://github.com/IamPoosha/tomaszewski-problem/blob/master/formal_verification.py) uses the open-source procedure scipy.integrate.quad, wrapping standard adaptive integra- tors from the QUADPACK library, which estimates the total integration error as being below 10 10. In addition, there are implicit numerical errors of order 2 53 10 16, caused by the − − ≈ − finite precision of double-precision floating point numbers. These estimates are well below the allowed 10 5 error, which we obtain by requiring all inequalities to hold with a ‘safety margin’ ± − of 2 10 5. As the functions involved in the numerical integration do not oscillate excessively · − in the integration range, relying on numeric integration in our scenario is standard. Nevertheless, since the numeric integration is not a 100% rigorous proof, the accompanied program makes also a slower, more straightforward evaluation of the integrals via Riemann sums. In this subsection we analyze the error of this evaluation and show it is bounded, as 5 required, by 10− . Given a piecewise-differentiable function f(u) defined on a finite domain (a, b), together with a bound on its derivative f (u) B for all u (a, b), it is easy to check that for any | ′ | ≤ ∈ N N, ∈ N b b a 2k 1 B(b a)2 f(u)du − f a + − (b a) − . (88) a − N · 2N − ≤ 4N Z k=1 X b By choosing N large enough (as a function of B), one may evaluate a f(u)du using (88), with any required precision rate. We henceforth compute a bound B associated with the R integrals involved in (35), and subsequently, re-compute the evaluations of (35), using (88) with a sufficiently large N. 1A computer program verifying (86) for the 93 elements of the se- quence {xi}, as well as the inequalities (49) and (84), is provided in https://github.com/IamPoosha/tomaszewski-problem/blob/master/formal_verification.py. All in- equalities hold with a spare of at least 2 · 10−5, allowing tolerating precision errors of up to ±10−5. 37 For the sake of convenience, let us recall (35). We have to evaluate the right hand side of the inequality q Pr[Z In the proof of Proposition 4.3, we substitute into (89) the parameters a1 = 0.31, x = 1, T = 10, and q = 0.4. In the proof of Lemma 4.6, the parameters are a1 = 0.22, T = 14.5, q = 0.4, and various values x [0, 1.65]. In order to estimate the integrals S ,S ,S ,S by the ∈ 1′ 2′ 3′ 4′ method described above, we upper bound the absolute value of the derivatives of the functions appearing in (89), for our choices of the parameters. The derivative bound corresponding to Si′ is called Bi. Regarding B One can easily check that the derivative d exp( u2/2)/√2π is negative for 4 du − u 0, and is minimized at u = 1. This gives the bound ≥ B exp( 1/2)/√2π < 1/4. 4 ≤ − Regarding B3 In order to give a bound B3 on the derivative, we use (fg)′ = f ′g + fg′, implying (fg) f g + f g . We clearly have exp( (T u)2/2) 1, and by the argument | ′| ≤ | ′|| | | || ′| | − |≤ of the previous case (applied with T u in place of u), d exp( (T u)2/2) exp( 1/2)T 2T/3. du − ≤ − ≤ To bound k(u, x, T ), recall the simplification (48) and notice that as sin(v) v for all v R, | | ≤ | | ∈ and πu(1 u) sin(πu) when u [0, 1], we have − ≤ ∈ π Tx T ux k(u, x, T ) | − | + 1 + 2T x/π. | |≤ π π ≤ To bound d k(u, x, T ) , we write s = π Tx, and obtain | du | − d Tx (1 u)s cos(su) (1 u)π cos(πu) sin(su) sin(su) k(u, x, T )+ cos(T ux)= − − = du π sin(πu) − sin(πu)2 − sin(πu) u(1 u) su cos(su) sin(su) = − s2 − + sin(πu) · (su)2 sin(su) (1 2u) sin(πu) πu(1 u) cos(πu) + s − − − . · su · u(1 u) sin(πu) − (90) 38 It can be verified that for any u [0, 1] we have ∈ (1 2u) sin(πu) πu(1 u) cos(πu) u(1 u) sin(πu) and πu(1 u) sin(πu), | − − − |≤ − − ≤ and for all v R, we have ∈ v cos(v) sin(v) v2/2 and sin(v) v . | − |≤ | | ≤ | | Plugging these inequalities into (90), we obtain the bound d 1 1 k(u, x, T ) ( s 2/2+ s + Tx) ((Tx)2/2+ π2) = (Tx)2/(2π)+ π. du ≤ π | | | | ≤ π We conclude d 2 d 2 B3 max k(u, x, T ) exp( (T u) /2) + k(u, x, T ) exp( (T u) /2) ≤ u [0,q] | | du − du − ≤ ∈ (1 + 2Tx/π) 2T/3 + ((Tx)2/(2π)+ π ) 1 . ≤ · · Regarding B Using the previous bounds, together with the bounds h(T u) 1 and 2 | |≤ 2 v exp( v /2), a1v<θ d − − 1/(a2 ) 1 d h(v)= ( cos(a v)) 1 sin(a v)/a , θ < a v<π = h(T u) T, du − 1 − 1 1 1 ⇒ du ≤ 0, π < a1v we conclude B (1 + 2Tx/π) T + ((Tx)2/(2π)+ π) 1. 2 ≤ · · 2 The inequality cos(u) 1/(a1 ) 1 sin(u) 1 we use, holds for any u R and a (0, 1). This can | | − ≤ ∈ 1 ∈ be checked easily for a 0.22, 0.31 , which are the only cases we need. 1 ∈ { } Regarding B1 Using the previous bounds, together with the bounds g(T u) 1 + exp( (π/(2a ))2/2) < 1.1 | |≤ − 1 and 1/(a2 ) 1 2 d cos(a v) 1 sin(a v)/a v exp( v /2), a v <π/2 d g(v)= 1 − 1 1 − − 1 = g(T u) T, du 2 ⇒ du ≤ ( v exp( v /2), π/2 < a1v − − we conclude B (1 + 2Tx/π) T + ((Tx)2/(2π)+ π) 1.1. 1 ≤ · · The aforementioned Python3 program verifies (49), (84), and the inequality (86) for the 93 elements of the sequence x , via (88) with the bounds B ,B ,B ,B given above. In all { i} 1 2 3 4 cases, the inequalities hold with a safety margin (i.e., difference between the two sides) of at 5 5 least 2 10− . This allows the program to estimate the integrals to within 10− (total) additive · 5 error, while leaving a spare error bound of 10− for any other numeric inaccuracies occurring outside of the numeric integration procedure integrate implementing (88). The running time of this verification is less than 10 minutes on a modern machine. 39 Appendix C Theorem 1.2 for a (0.31, 0.5), a + a + a 1 1 ∈ 1 2 3 ≤ The proof of Theorem 1.2 in this range splits into three cases. If a2 is sufficiently small, elimination of one variable and the improved Berry-Esseen bound of Section 4 are sufficient for proving the assertion. If a3 is sufficiently small, elimination of two variables and the improved Berry-Esseen bound do the work. The hard case is when both a2 and a3 are not very small. To handle this case, we eliminate three variables and combine segment comparison with a Chebyshev-type inequality. Unfortunately, the proof is quite long and rather cumbersome. C.1 The case a 0.19 2 ≤ We begin with eliminating one variable (i.e., applying Lemma 2.1 with m = 1). The lemma implies that in order to derive Pr[ X 1] 1/2, it is sufficient to prove | |≤ ≥ Pr[X′ >t] + Pr[X′ > 1/t] 1/2, (91) ≤ where X′,t are defined as in (60). The maximal weight of X is a , which satisfies a 0.19/ 3/4 < 0.2195. Hence, applying ′ 2′ 2′ ≤ Proposition ?? to X gives Pr[X > x] Pr[Z>x] + 0.088 for any x 0. Thus, (91) follows ′ ′ ≤ p ≥ from ? Pr[Z>t]+Pr[Z > 1/t] 1/2 2 0.088, ≤ − · (92) 1 a with: a (0.31, 0.5), t = − 1 ,Z N(0, 1). 1 ∈ 1+ a ∼ r 1 This inequality is verified in Appendix F.5. Note that the proof in this case does not use the assumption a + a + a 1. Thus, the assertion holds in the case (a (0.31, 0.5)) (a 1 2 3 ≤ 1 ∈ ∧ 2 ≤ 0.19) (a + a + a > 1) as well. ∧ 1 2 3 C.2 The case a 0.19, a 0.15 2 ≥ 3 ≤ We begin with eliminating two variables. By Lemma 2.1, applied with m = 2, in order to prove Pr[ X 1] 1/2 for X = n a x , it is sufficient to verify | |≤ ≥ i=1 i i P 4 Pr[X′ > T ] 1, (93) k ≤ Xk=1 with X = n a x , σ = 1 a2 a2, a = a /σ and T = (1 a a )/σ as in Lemma 2.1. ′ i=3 i′ i − 1 − 2 i′ i k ± 1 ± 2 The maximal weight of X is a , which satisfies a 0.15/ 1/2 = 0.15√2. Hence, an applica- P ′p 3′ 3′ ≤ tion of Proposition 4.4 to X gives ′ p Pr[X′ x] Pr[Z x] max(0.084, Pr[ Z a′ ]/2) ≤ ≥ ≤ − | |≤ 3 Pr[Z x] max(0.084, Pr[ Z 0.15√2]/2) ≥ ≤ − | |≤ = Pr[Z x] 0.084, ≤ − and thus, Pr[X > x] Pr[Z>x] + 0.084, for all x 0. As T 0 for all k (since a + a ′ ≤ ≥ k ≥ 1 2 ≤ 2a 1), (93) follows from the inequality 1 ≤ 4 ? Pr[Z > T ] 1 4 0.084, k ≤ − · k=1 (94) X 1 a a with: Z N(0, 1), 0.19 a a 1/2, σ = 1 a2 a2, T = ± 1 ± 2 . ∼ ≤ 2 ≤ 1 ≤ − 1 − 2 k σ q 40 This inequality is verified in Appendix F.6. Note that like in the previous case, the proof does not use the assumption a + a + a 1. Thus, the assertion holds in the case (a 1 2 3 ≤ 1 ∈ (0.31, 0.5)) (a 0.19) (a 0.15) (a + a + a > 1) as well. ∧ 2 ≥ ∧ 3 ≤ ∧ 1 2 3 C.3 The case a 0.19, a 0.15 2 ≥ 3 ≥ Elimination step. We begin with a variant of the ‘variable elimination lemma’, eliminating three variables. This lemma will be used several more times in the sequel. Note that in our case, as a1 < 1/2, we have n> 4, and thus, we may eliminate three variables and have σ> 0. Lemma C.1. Let n 4. Let X = n a x be a Rademacher sum with Var(X) = 1, and ≥ i=1 i i write X = a x + a x + a x + σX , such that Var(X ) = 1 (hence, σ = 1 a2 a2 a2). 1 1 2 2 3 3 ′ P ′ − 1 − 2 − 3 The assertion p Pr[ X 1] 1/2 | |≤ ≥ is equivalent to the following inequality involving X′: Pr[X′ [ L ,L ]] + Pr[X′ [ L ,L ]] ∈ − 1 2 ∈ − 3 4 (95) ≥ Pr X′ > R1 +Pr X′ > R2 + Pr X′ > R3 + Pr X′ > R4 , where 1 a a a 1 a a + a 1 a + a a 1 a a a L ,L ,L ,L = − 1 − 2 − 3 , − 1 − 2 3 , − 1 2 − 3 , − | 1 − 2 − 3| 1 2 3 4 σ σ σ σ and 1+ a a a 1+ a a + a 1+ a + a a 1+ a + a + a R ,R ,R ,R = | 1 − 2 − 3|, 1 − 2 3 , 1 2 − 3 , 1 2 3 . 1 2 3 4 σ σ σ σ Proof. The assertion follows from Lemma 2.1, applied to X with the parameter m = 3, and the equality 4 Pr[X′ [ L ,L ]] + Pr[X′ [ L ,L ]] = 2 Pr[X′ >L ]. ∈ − 1 2 ∈ − 3 4 − j Xj=1 As we assume a + a + a 1, we have L ,L ,L ,L 0, and thus, by symmetry of X , 1 2 3 ≤ 1 2 3 4 ≥ ′ the inequality (95) we have to prove follows from 4 4 4 Pr X′ 0,L Pr X′ R , = i Pr X′ R ,R , (96) ∈ h ii ≥ ∈ h i ∞i · ∈ h i i+1i i=1 i=1 i=1 X X X with R = . 5 ∞ 41 Auxiliary estimates. To proceed, we use the following auxiliary estimates on Li,Ri. ? ? ? L 0, L 1/2, i 1, 2, 3, 4 : R (1 + i)/2, 3 ≥ 4 ≥ ∀ ∈ { } i ≥ with: a [0.31, 0.5], a [0.19, a ], a [0.15, a ], a + a + a 1, 1 ∈ p 2 ∈ 1 3 ∈ 2 p 1 2 3 ≤ σ = 1 a2 a2 a2, − 1 − 2 − 3 (97) 1 a + a a q1 a a a 1+ a a a L = − 1 2 − 3 , L = − | 1 − 2 − 3|,R = | 1 − 2 − 3|, 3 σ 4 σ 1 σ 1+ a a + a 1+ a + a a 1+ a + a + a R = 1 − 2 3 R = 1 2 − 3 ,R = 1 2 3 , 2 σ 3 σ 4 σ These inequalities are proved in Appendix F.7. Chebyshev-type inequality step. We reduce (96) into a more convenient inequality, by applying a Chebyshev-type inequality. Denote L¯3 = min(L3, 1) and L¯4 = min(L4, 1). Let c0, c1, c2, c3, c4 = 0, L¯3, max(L¯3, 1/2), L¯4, 1 d0, d1, d2, d3, d4 = 1,R1, max(R1, 3/2)p,R2, max(R2, 4/2) (98) d , d , d , d , d = R , max(R , 5p/2),R , max(R , 6p/2), . 5 6 7 8 9 3 3 4 4 ∞ p p Note that by the definition of L¯3, L¯4 and (97), we have 0= c c c c c =1= d d . . . d d = . 0 ≤ 1 ≤ 2 ≤ 3 ≤ 4 0 ≤ 1 ≤ ≤ 8 ≤ 9 ∞ Hence, we may apply the Chebyshev-type inequality (18) to X′, with the parameters c0,...,c4 and d0,...,d9, to obtain 2 1 Pr X′ 0, L¯ + (1 L¯ )Pr X′ L¯ , max(L¯ , 1/2) + · ∈ 3 − 3 ∈ 3 3 2 h D p2 Ei +(1 max(L¯ , 1/2) )Pr X′ max(L¯ , 1/2), L¯ + (1 L¯ )Pr X′ L¯ , 1 − 3 ∈ 3 4 − 4 ∈ 4 p h D p Ei ≥ (99) 4 2 (R 1)Pr X′ R , max(R , (i + 2)/2) + i − ∈ i i i=1 X h D p Ei 2 + max R , (i + 2)/2 1 Pr X′ max(R , (i + 2)/2),R , i − ∈ i i+1 p h D p Ei with R = . We now claim that the following inequality, coupled with (99), implies (96). 5 ∞ Pr X′ 0,L + Pr X′ 0,L + ∈ h 1i ∈ h 2i 2 2 (2L 1)Pr X′ L , 1/2 + (2L 2)Pr X ′ L , √1 + 3 − ∈ 3 4 − ∈ 4 h D p Ei h D Ei ≥ (100) 2 2 (3 2R )Pr X′ R , 3/2 + (4 2R )Pr X′ R , √2 + − 1 ∈ 1 − 2 ∈ 2 2 h D p Ei 2 h D Ei (5 2R )Pr X′ R , 5/2 + (6 2R )Pr X′ R , √3 . − 3 ∈ 3 − 4 ∈ 4 h D p Ei h D Ei To see this, note that if 0 L 1/2 L 1 R 3/2 R 4/2 R 5/2 R 6/2, (101) ≤ 3 ≤ ≤ 4 ≤ ≤ 1 ≤ ≤ 2 ≤ ≤ 3 ≤ ≤ 4 ≤ p p p p p 42 then (99) reads 2 1 Pr X′ 0,L + (1 L )Pr X′ L , 1/2 + · ∈ h 3i − 3 ∈ 3 1 h 2 D p Ei Pr X′ 1/2,L + (1 L )Pr X′ L , 1 2 ∈ 4 − 4 ∈ h 4 i h Dp Ei ≥ 2 1 (R 1)Pr X′ R , 3/2 + Pr X′ 3/2,R + 1 − ∈ 1 2 ∈ 2 2 h D p Ei h Dp Ei (R 1)Pr X′ R , √2 +1Pr X′ √2,R + 2 − ∈ 2 ∈ 3 2 h D Ei 3 h D Ei (R 1)Pr X′ R , 5/2 + Pr X′ 5/2,R + 3 − ∈ 3 2 ∈ 4 2 h D p Ei h Dp Ei (R 1)Pr X′ R , √3 +2Pr X′ √3, , 4 − ∈ 4 ∈ ∞ h D Ei h D Ei and thus, we have 2(99) + (100) = (96). If (101) is not satisfied, then 2(99) + (100) is even stronger than (96), and implies it. To see this, observe that (101) can fail in three possible ways: • R > (i + 2)/2 for some 1 i 4. In this case, the contribution of the region R ,R i ≤ ≤ h i i+1i to the right hand side of 2(99) + (100) becomes p 2 2 2(R 1)Pr X′ R ,R 2(( (i + 2)/2) 1)Pr X′ R ,R i − ∈ h i i+1i ≥ − ∈ h i i+1i = i Pr X′ R ,R , · p ∈ h i i+1i compared to i Pr [X R ,R ] in the right hand side of (96), while the contribution · ′ ∈ h i i+1i of the other regions remains unchanged. Hence, in this case the inequality 2(99) + (100) implies (96). • L3 > 1 or L4 > 1. If L3 > 1, then the left hand side of 2(99) + (100) becomes Pr X′ 0,L + Pr X′ 0,L +2Pr X′ 0, 1 , ∈ h 1i ∈ h 2i ∈ h i which is not larger than the left hand side of (96), being 4 Pr [X 0,L ]. As the i=1 ′ ∈ h ii right hand side is unchanged, (96) follows in this case as well. P Similarly, if L 1 Therefore, in order to prove (96), it is sufficient to show (100), which we rewrite in the form 6 1 Pr X′ L ,L + Pr X′ 0,L c Pr X′ d , e , (102) 2 ∈ h− 1 1i ∈ h 2i ≥ i ∈ h i ii i=1 X with d , d , d , d , d , d = L ,L ,R ,R ,R ,R , c = i 2d2, e = i/2. 1 2 3 4 5 6 3 4 1 2 3 4 i − i i Notice we might have e d , in which case Pr [X d , e ] = 0. p i ≤ i ′ ∈ h i ii Segment comparison step. In order to prove (102), we use the following lemma, which is proved in Appendix G: Lemma C.2. Let X′,Li,Ri, ci, di, ei be defined as above. We have: 1. i: d , e ′ 0,L , and thus Theorem 3.1 implies: Pr [X d , e ] Pr [X 0,L ]. ∀ h i ii≺X h 2i ′ ∈ h i ii ≤ ′ ∈ h 2i 2. 6 max(c , 0) 3/2. i=1 i ≤ 3. Pi B(X′) max(ci, 0) 1, where B(X′)= i [6] di, ei X′ L1,L1 . ∈ ≤ { ∈ | h i 6≺ h− i} P Observe that (102) is trivially implied by combination of the following two inequalities: c Pr X′ d , e C Pr X′ 0,L , (103) i ∈ h i ii ≤ · ∈ h 2i i B(X′) ∈X with C = i B(X′) max(ci, 0), and ∈ P 1 c Pr X′ d , e (1 C)Pr X′ 0,L + Pr X′ L ,L . (104) i ∈ h i ii ≤ − ∈ h 2i 2 ∈ h− 1 1i i B(X′) 6∈X We deduce both inequalities from Lemma C.2. Inequality (103) follows immediately from the first item of Lemma C.2. We now reason about (104). Let p = min(Pr [X 0,L ] , Pr [X L ,L ]). Notice that the first item of Lemma C.2, ′ ∈ h 2i ′ ∈ h− 1 1i together with the definition of B(X′), implies that i / B(X′): Pr X′ d , e p. ∀ ∈ ∈ h i ii ≤ Hence, we may deduce (104), as (a) 1 c Pr X′ d , e max(c , 0) p (1 C)p + p i ∈ h i ii ≤ i · ≤ − 2 ≤ i B(X′) i B(X′) 6∈X 6∈X (105) (b) 1 (1 C)Pr X′ 0,L + Pr X′ L ,L , ≤ − ∈ h 2i 2 ∈ h− 1 1i where inequality (a) uses the second item of Lemma C.2 and the definition of C, and inequal- ity (b) follows from the definition of p, via 1 C 0, which is a rephrasing of the third item − ≥ of Lemma C.2. This completes the proof. 44 Appendix D Theorem 1.2 for a [0.387, 0.5], a + a + a 1 1 ∈ 1 2 3 ≥ The proof of Theorem 1.2 in this range is similar to – but much easier than – the proof in the range a (0.31, 0.5), a + a + a 1, presented in Appendix C. If a is sufficiently small, 1 ∈ 1 2 3 ≤ 3 then the Berry-Esseen argument of Appendix C yields the assertion. If a3 is not very small, the proof goes by elimination of three variables and combination of Chebyshev-type inequalities and segment comparison. The difference from Appendix C is that the slightly different assumptions on the parameters allow for a simple Chebyshev-type argument to work. A similar argument fails in the range of Appendix C (specifically, the inequality (107) below, does not hold there), and thus, an exhausting detour is needed. D.1 The case a 0.15 3 ≤ By assumption, we have a + a + a 1 and a a a 0.5, and thus, a 0.25. Hence, 1 2 3 ≥ 3 ≤ 2 ≤ 1 ≤ 2 ≥ if a 0.15 then the assertion Pr[ X 1] 1/2 follows from the argument of Appendix C.2, 3 ≤ | |≤ ≥ which applies whenever (a (0.31, 0.5)) (a 0.19) (a 0.15), as was shown at the end 1 ∈ ∧ 2 ≥ ∧ 3 ≤ of Appendix C.2. D.2 The case a 0.15 3 ≥ Elimination step. By Lemma C.1, it is sufficient to prove Pr X′ L ,L + Pr X′ 0,L + Pr X′ 0,L ∈ h− 1 2i ∈ h 3i ∈ h 4i (106) ≥ Pr X′ R , + Pr X′ R , + Pr X′ R , + Pr X′ R , , ∈ h 1 ∞i ∈ h 2 ∞i ∈ h 3 ∞i ∈ h 4 ∞i where σ, X′,L1,L2,L3,L4,R1,R2,R3,R4 are as defined in Lemma C.1. Auxiliary estimate. To proceed, we use the following auxiliary inequality in Li,Ri. ? R2 1 R2 1 R2 1 R2 1 max 1 L2, 3/2 L2, 1/2 min 1 − , 2 − , 3 − , 4 − , − 3 − 4 ≤ 1 2 3 4 with: a [0.387, 0.5], 0.15 a a a , a + a + a 1, 1 ∈ ≤ 3 ≤ 2 ≤ 1 1 2 3 ≥ σ = 1 a2 a2 a2, (107) − 1 − 2 − 3 1 a + a a q1+ a a a 1 a + a + a L = − 1 2 − 3 , L = 1 − 2 − 3 ,R = − 1 2 3 , 3 σ 4 σ 1 σ 1+ a a + a 1+ a + a a 1+ a + a + a R = 1 − 2 3 R = 1 2 − 3 ,R = 1 2 3 . 2 σ 3 σ 4 σ The proof of (107) is presented in Appendix F.8. Chebyshev-type step. Let c = 1/ max 1 L2, 3/2 L2, 1/2 , and − 3 − 4 (108) d = 1/ min R2 1, (R2 1)/2, (R2 1)/3, (R2 1)/4 . 1 − 2 − 3 − 4 − We show that c 2 LHS of (106) E[(1 X′ )½ X′ < 1 ], (109) ≥ 2 · − {| | } and d 2 RHS of (106) E[(X′ 1)½ X′ > 1 ]. (110) ≤ 2 · − {| | } 45 As by (107) we have c d, the assertion (106) follows from (109) and (110) by the Chebyshev- ≥ type equality (79). Proving (109). We consider three sub-cases. Sub-case 1: L L 1. In this case, it is clear that 4 ≥ 3 ≥ 2 LHS of (106) 2Pr X′ 0, 1 E[(1 X′ )½ X′ < 1 ], ≥ ∈ h i ≥ − {| | } implying (109) (by noting c/2 = 1). Sub-case 2: L 1 > L . In this case, the Chebyshev-type inequality (80) with c , c , c = 4 ≥ 3 0 1 2 0,L3, 1 yields 2 2 2Pr X′ 0,L + 2(1 L )Pr X′ L , 1 E[(1 X′ )½ X′ < 1 ]. ∈ h 3i − 3 ∈ h 3 i ≥ − {| | } The left hand side can be bounded from above by 2 max(1, 2(1 L )) Pr X′ 0,L + Pr X′ 0,L , − 3 · ∈ h 3i ∈ h 4i and thus (109) follows (by noting c/2 = 1/ max(1, 2(1 L2))). − 3 Sub-case 3: L4 < 1. In this case, (80) with c0, c1, c2, c3 = 0,L3,L4, 1 yields 2 2 2Pr X′ 0,L + 2(1 L )Pr X′ L ,L + 2(1 L )Pr X′ L , 1 ∈ h 3i − 3 ∈ h 3 4i − 4 ∈ h 4 i 2 (111) E[(1 X′ )½ X′ < 1 ]. ≥ − {| | } We claim that Pr [X L , 1 ] Pr [X L ,L ] = 2Pr[X 0,L ], and consequently, ′ ∈ h 4 i ≤ ′ ∈ h− 3 3i ′ ∈ h 3i the LHS of (111) is upper bounded by 2 2 max 1 + 2(1 L ), 2(1 L ) Pr X′ 0,L + Pr X′ 0,L , − 4 − 3 · ∈ h 3i ∈ h 4i thus implying (109) (by noting c/2 = 1/ max(3 2L2, 2 2L2)). − 4 − 3 The claim Pr [X L , 1 ] Pr [X L ,L ] follows from a segment comparison argument ′ ∈ h 4 i ≤ ′ ∈ h− 3 3i – namely, Lemma 3.6, applied to X with the parameters A,B,C,D,M = L ,L ,L , 1, a /σ. ′ − 3 3 4 4 To see that the lemma can be applied, we have to check that D C + 2M B A. Using − ≤ − M a /σ, this follows from ≤ 3 1 L + 2a /σ 2L . − 4 3 ≤ 3 Rearranging, we have to prove σ 3 a + a 5a . This indeed holds, since ≤ − 1 2 − 3 σ 1 3a2 5/2 4a 3 a + a 5a , ≤ − 3 ≤ − 3 ≤ − 1 2 − 3 where the ultimate inequalityq holds as a + a 1/2+ a , and the penultimate inequality 1 3 ≤ 2 follows from (1 2a )(21 38a ) (5/2 4a )2 (1 3a2)= − 3 − 3 0. − 3 − − 3 4 ≥ Proving (110). The Chebyshev-type inequality (81) with the parameters d0, d1, d2, d3, d4 = 1, 1 + 1/d, 1 + 2/d, 1 + 3/d, 1 + 4/d, where d is as defined above, yields p p p p 4 2 1 E[(X′ 1)½ X′ > 1 ] Pr[ X′ 1+ i/d]. − {| | } ≥ d | |≥ i=1 X p This implies (110), provided that R 1+ i/d holds for all i [4]. The latter reads 1/d i ≥ ∈ ≤ (R2 1)/i, which indeed follows from the definition of d. i − p 46 Appendix E Theorem 1.2 for a [0.31, 0.387], a + a + a 1 1 ∈ 1 2 3 ≥ The proof of Theorem 1.2 in this region is a bit more intricate than the proof in the other regions. A reason for the extra difficulty is that after eliminating the variables x1,x2,x3 by applying Lemma C.1, we have to show the LHS of (95) is the RHS, where the LHS includes ≥ the term Pr[X [ L ,L ]], with L < 0. It is harder for us to exploit this quantity, as ′ ∈ − 1 2 1 the segment [ L ,L ] does not contain 0, which is essential for an approach based on the − 1 2 Chebyshev-type inequality (79). In Appendix D we overcome this difficulty by just neglecting the term Pr[X [ L ,L ]]. However, this is not possible in the case a 1/3, where this term ∈ − 1 2 1 ≈ is inherently needed. In the proof, we consider two cases, which we further subdivide into two sub-cases each. 1. There are intermediate-sized weights. The two sub-cases of this case are: • There exists k 4 with a [a + a + a 1, 1 a a ]. ≥ k ∈ 1 2 3 − − 1 − 2 • There exist k>j> 1 with a + a + a 1 and a , a 1 2a . 2 j k ≤ j k ≥ − 1 2. There are no intermediate-sized weights. The two sub-cases of this case are: • There are at most 4 ‘large’ weights of size 1 a a . ≥ − 1 − 2 • There are at least 5 ‘large’ weights. The proof in the first three sub-cases goes through elimination of three variables, similarly to the proofs in Appendices C and D. The only significant difference is in the segment comparison step, which is somewhat more complex and differs between the three sub-cases. The strategy in the fourth sub-case is different. It goes through elimination of five variables and a combination of a Chebyshev-type inequality with a semi-inductive argument. E.1 Case 1: There are intermediate-sized weights E.1.1 Sub-case 1: There exists k 4 with a [a + a + a 1, 1 a a ] ≥ k ∈ 1 2 3 − − 1 − 2 Elimination step. By Lemma C.1, it is sufficient to prove Pr[X′ [ L ,L ]] + Pr[X′ [ L ,L ]] ∈ − 1 2 ∈ − 3 4 (112) ≥ Pr X′ > R1 +Pr X′ > R2 + Pr X′ > R3 + Pr X′ > R4 , where σ, X′,L1,L2,L 3,L4,R1,R2,R3,R4 are as defined in Lemma C.1. Segment comparison step. The following segment comparison lemma plays an important role in the proof, allowing us to take into account the segment [ L ,L ] in a Chebyshev-type − 1 2 approach. Lemma E.1. Let X = n a x be a Rademacher sum, such that Var(X) = 1, a +a +a 1, i=1 i i 1 2 3 ≥ 0 an . . . a2 a1 0.387. Suppose there exists k 4 with ak [a1+a2+a3 1, 1 a1 a2]. ≤ ≤ ≤ ≤ P≤ n ≥ ∈ − − − Let σ = 1 a2 a2 a2, a = a /σ, X = a x and − 1 − 2 − 3 i′ i ′ i=4 i′ i p 1 a1 a2 a3 1 a1 a2 + a3 1+ a1 a2 a3 L = − − − , L = −P − , L = − − 1 σ 2 σ 4 σ be as in Lemma C.2. Then Pr[X′ (L , 1)] 2Pr[X′ ( L ,L ]]. (113) ∈ 4 ≤ ∈ − 1 2 47 Proof. As Pr[X′ (L4, 1)] = b 1,1 Pr[X′ (L4, 1) xk = b], it is clearly sufficient to show ∈ ∈{− } ∈ ∧ P b 1, 1 : Pr[X′ (L , 1) x = b] Pr[X′ ( L ,L ]]. (114) ∀ ∈ {− } ∈ 4 ∧ k ≤ ∈ − 1 2 We handle these two cases (corresponding to the value of b) separately. Proving (114) for b = 1. Applying Lemma 3.7 to the Rademacher sum X = X a x , − ′′ ′ − k′ k with the parameters A,B,C,D = L + a ,L + a ,L + a , 1+ a and M a , we get 1 k′ 2 k′ 4 k′ k′ ≤ 3′ Pr[X′ (L , 1) x = 1] Pr[X′ (L ,L ) x = 1]. (115) ∈ 4 ∧ k − ≤ ∈ 1 2 ∧ k − Note that we apply the lemma with open segments instead of half-open segments; the lemma indeed holds in this setting, as is shown in Appendix A.6. To verify that the assumptions of the lemma are satisfied, note that the assumption 0 A C holds since a a + a + a 1, ≤ ≤ k ≥ 1 2 3 − the assumption 2M C A holds since a a , and the assumption D C + D B B A ≤ − 1 ≥ 3 − − ≤ − is equivalent to a a 1 σ, and follows from 2 − 3 ≤ − a a a (1 2a ) 3 0.387 1 < 1 2/3 1 σ, 2 − 3 ≤ 1 − − 1 ≤ · − − ≤ − where the first and last inequalities are implied by a + a + a p1 (the latter, via the Cauchy- 1 2 3 ≥ Schwarz inequality). By (115), the assertion (114) for b = 1 will follow once we show − Pr[X′ (L ,L ] x = 1] Pr[X′ ( L ,L ]]. (116) ∈ 1 2 ∧ k − ≤ ∈ − 1 2 We prove this by constructing an explicit injective map. Let n 3 Ω= z = (z ,...,z ) 1, 1 − X′(z) (L ,L ] z = 1 , 4 n ∈ {− } ∈ 1 2 ∧ k − and n 3 Ω′ = z = (z ,...,z ) 1, 1 − X′(z) ( L ,L ] . 4 n ∈ {− } ∈ − 1 2 We define f :Ω Ω by setting (f(z)) = z for all i = k, and → ′ i i 6 zk, X′(z) ( L1,L2] (f(z))k = ∈ − . z , X (z) ( L ,L ] (− k ′ 6∈ − 1 2 That is, we flip the k’th coordinate of z iff X (z) ( L ,L ], and leave the other coordinates ′ 6∈ − 1 2 unchanged. It is clear that f is injective. To see that Range(f) Ω , notice that ⊂ ′ 2L 2a′ L ( L ), (117) − 1 ≤ k ≤ 2 − − 1 where the first inequality holds since a a + a + a 1 and the second inequality holds k ≥ 1 2 3 − since a 1 a a . As when X (z) ( L ,L ] we have X (f(z)) = X (z) + 2a , the k ≤ − 1 − 2 ′ 6∈ − 1 2 ′ ′ k′ inequality (117) implies X′(z) (L , L ]= X′(f(z)) ( L ,L ]. ∈ 1 − 1 ⇒ ∈ − 1 2 Since when X (z) ( L ,L ] we have f(z)= z, the assertion (116) follows. ′ ∈ − 1 2 48 Proving (114) for b = 1. Notice that Pr[X′ (L , 1) x = 1] = Pr[X′ (L 2a′ , 1 2a′ ) x = 1]. ∈ 4 ∧ k ∈ 4 − k − k ∧ k − We proceed by proving Pr[X (L 2a , 1 2a ) x = 1] Pr[X ( L ,L ]] by a slight ′ ∈ 4 − k′ − k′ ∧ k − ≤ ′ ∈ − 1 2 variation of the proof of (114) for b = 1 presented above. By (116), it suffices to prove − Pr[X′ (L 2a′ , 1 2a′ ) x = 1] Pr[X′ (L ,L ] x = 1]. (118) ∈ 4 − k − k ∧ k − ≤ ∈ 1 2 ∧ k − This inequality is equivalent to Pr[X (L a , 1 a )] Pr[X (L + a ,L + a ]], where ′′ ∈ 4 − k′ − k′ ≤ ′′ ∈ 1 k′ 2 k′ X = X a x , as defined above. By subtracting Pr[X (L a ,L + a ]] from both sides, ′′ ′ − k′ k ′′ ∈ 4 − k′ 2 k′ the latter is equivalent to Pr[X′′ (L + a′ , 1 a′ )] Pr[X′′ (L + a′ ,L a′ ]]. (119) ∈ 2 k − k ≤ ∈ 1 k 4 − k To prove (119), we apply Lemma 3.7 to X , with the parameters A,B,C,D = L + a ,L ′′ 1 k′ 4 − a ,L +a , 1 a and M a , and open segments instead of half-open segments (Appendix A.6). k′ 2 k′ − k′ ≤ 3′ To verify that the assumptions of the lemma are indeed satisfied, note that the assumption 0 A C holds since a a + a + a 1, the assumption 2M C A = 2a holds trivially, ≤ ≤ k ≥ 1 2 3 − ≤ − 3′ and the assumption D C + D B B A is equivalent to σ 1+ a a , being clear. This − − ≤ − ≤ 1 − 2 completes the proof. Auxiliary estimate. To proceed, we use the following auxiliary inequality in Li,Ri. 3 2L2 ? R2 1 R2 1 R2 1 max 1/2, − 4 min R2 1, 2 − , 3 − , 4 − , 3 ≤ 1 − 2 3 4 with: a [0.31, 0.387], a a a , a + a + a 1, 1 ∈ 3 ≤ 2 ≤ 1 1 2 3 ≥ 1+ a a a 1 a + a + a (120) σ = 1 a2 a2 a2, L = 1 − 2 − 3 ,R = − 1 2 3 , − 1 − 2 − 3 4 σ 1 σ q1+ a a + a 1+ a + a a 1+ a + a + a R = 1 − 2 3 R = 1 2 − 3 ,R = 1 2 3 , 2 σ 3 σ 4 σ The proof of (120) is presented in Appendix F.9. Chebyshev-type step. Let c = 1/ max (3 2L2)/3, 1/2 , and − 4 (121) d = 1/ min R2 1, (R2 1)/2, (R2 1)/3, (R2 1)/4 . 1 − 2 − 3 − 4 − We show that c 2 LHS of (112) E[(1 X′ )½ X′ < 1 ], (122) ≥ 2 · − {| | } and d 2 RHS of (112) E[(X′ 1)½ X′ > 1 ]. (123) ≤ 2 · − {| | } As by (120) we have c d, the assertion (112) follows from (122) and (123) by the Chebyshev- ≥ type equality (79). 49 Proving (122). First, we claim that L 1/√2. 3 ≥ To see this, note that the assumption a + a + a 1 implies, via the Cauchy-Schwarz 1 2 3 ≥ inequality, a2 + a2 + a2 1/3, and thus σ = 1 a2 a2 a2 2/3. Using this, along 1 2 3 ≥ − 1 − 2 − 3 ≤ with the assumptions a 0.387 and a a , we obtain 1 ≤ 2 ≥ 3 p p 1 a1 + a2 a3 1 a1 1 0.387 1 L3 = − − − − > . σ ≥ 2/3 ≥ 2/3 √2 As L 1/√2, we have p p 3 ≥ 2 Pr[X′ [ L ,L ]] 2Pr[X′ 0,L ]]+2max 0, 1 L Pr[X′ (L ,L ]]. (124) ∈ − 3 4 ≥ ∈ h 3 { − 3} · ∈ 3 4 We consider two sub-cases. Sub-case 1: L 1. Let L¯ = min L , 1 . The Chebyshev-type inequality (80), applied 4 ≥ 3 { 3 } with the parameters c0, c1, c2 = 0, L¯3, 1, yields ¯ ¯2 ¯ 2 2Pr[X′ 0, L ] + 2(1 L )Pr[X′ (L , 1)] E[(1 X′ )½ X′ < 1 ], ∈ 3 − 3 ∈ 3 ≥ − {| | } which together with (124) and the definition of c implies (122). Sub-case 2: L < 1. Since L 1/2, the Chebyshev-type inequality (80), applied with 4 3 ≥ the parameters c , c , c , c = 0,L ,L , 1, implies 0 1 2 3 3 4p 2 c 2 c Pr[X′ 0,L ]] + Pr[X′ (L ,L ]] + c(1 L )Pr[X′ (L , 1)] E[(1 X′ )½ X′ < 1 ]. ∈ h 3 ∈ 3 4 − 4 ∈ 4 ≥ 2 · − {| | } Thus, in order to deduce (122) it suffices to show 2 2Pr[X′ 0,L ]] + Pr[X′ ( L ,L ]] c Pr[X′ 0,L ]] + c(1 L )Pr[X′ (L , 1)]. (125) ∈ h 3 ∈ − 1 2 ≥ ∈ h 3 − 4 ∈ 4 To prove this, note that Lemma E.1 implies Pr[X′ (L , 1)] 2Pr[X′ ( L ,L ]] 2Pr[X′ 0,L ]]. ∈ 4 ≤ ∈ − 1 2 ≤ ∈ h 3 If 2c(1 L2) 1, then (125) follows as c 2 by definition. Otherwise, it suffices to check − 4 ≤ ≤ 2 (2 (2c(1 L ) 1)) Pr[X′ 0,L ]] c Pr[X′ 0,L ]]. − − 4 − ∈ h 3 ≥ ∈ h 3 This inequality follows from (2 (2c(1 L2) 1)) c, which holds by the definition of c. This − − 4 − ≥ completes the proof. Proving (123). Inequality (123) is the same as (110) and is proved in the same way. (Note that the slightly different assumptions on X′ here do not affect the proof.) E.1.2 Sub-case 2: There exist k>j> 1 with a + a + a 1 and a , a 1 2a 2 j k ≤ j k ≥ − 1 Elimination step. By Lemma C.1, it is sufficient to prove 4 Pr X′ L ,L + Pr X′ 0,L + Pr X′ 0,L Pr[X′ > R ], (126) ∈ h− 1 2i ∈ h 3i ∈ h 4i ≥ i i=1 X 50 with ai 2 2 2 X′ = x , σ = 1 a a a , σ i − 1 − j − k i [n] 1,j,k ∈ X\{ } q and 1 a a a 1 a a + a 1 a + a a 1+ a a a L ,L ,L ,L = − 1 − j − k , − 1 − j k , − 1 j − k , 1 − j − k 1 2 3 4 σ σ σ σ 1 a + a + a 1+ a a + a 1+ a + a a 1+ a + a + a R ,R ,R ,R = − 1 j k , 1 − j k , 1 j − k , 1 j k . 1 2 3 4 σ σ σ σ Auxiliary inequalities. To proceed, we use the following auxiliary inequalities in Li,Ri. ? ? L L 1/2, i 1, 2, 3, 4 : R 1 + (i/2) 4 ≥ 3 ≥ ∀ ∈ { } i ≥ with: 1 2a p a a a 0.387, σ = p1 a2 a2 a2, − 1 ≤ k ≤ j ≤ 1 ≤ − 1 − j − k 1 a + a a 1+ a a a q 1 a + a + a (127) L = − 1 j − k , L = 1 − j − k ,R = − 1 j k , 3 σ 4 σ 1 σ 1+ a a + a 1+ a + a a 1+ a + a + a R = 1 − j k ,R = 1 j − k ,R = 1 j k . 2 σ 3 σ 4 σ The proof of (127) is presented in Appendix F.10. Segment comparison step. By Lemma 3.7, applied to the Rademacher sum X′ with the parameters A,B,C,D = L ,L ,L , 1 and M a /σ, we have − 1 2 4 ≤ 2 Pr X′ L , 1 Pr X′ L ,L . (128) ∈ h 4 i ≤ ∈ h− 1 2i To verify that the assumptions of the lemma are satisfied, note that the assumption A C, ≤ A C holds since a +a < 1, the assumption 2M C A (being 2a /σ (2 2a 2a )/σ) − ≤ j k ≤ − 2 ≤ − j − k holds since a + a + a 1, and the assumption D C + D B B A is equivalent to 2 j k ≤ − − ≤ − σ 2 a 2a , which holds as ≤ − 1 − j (2 a 2a )2 σ2 = (a2 (1 2a )2) + (a a )(8 9a 5a ) + 5(0.4 a )(2 3a ) > 0. − 1 − j − k − − 1 1 − j − 1 − j − 1 − 1 Chebyshev-type step. Denote L¯ = min L , 1 and L¯ = min L , 1 . The Chebyshev-type 3 { 3 } 4 { 4 } inequality (17), applied with the parameters c0, c1, c2, c3 = 0, L¯3, L¯4, 1 and d0, d1, d2, d3, d4 = 1, 3/2, 4/2, 5/2, 6/2, gives 2 2 p 2PrpX′ p0, L¯ p+ 2(1 L¯ )Pr X′ L¯ , L¯ + 2(1 L¯ )Pr X′ L¯ , 1 ∈ 3 − 3 ∈ 3 4 − 4 ∈ 4 4 1 (129) 2 Pr[X′ 1 + (i/2)]. ≥ · 2 ≥ i=1 X p By (127) and (128), 2 Pr X′ L ,L Pr X′ L , 1 max 0, 2(1 L ) Pr X′ L , 1 . ∈ h− 1 2i ≥ ∈ h 4 i ≥ { − 4 } ∈ h 4 i Using in addition the inequality L 1/2 that holds by (127), we obtain 3 ≥ 2 LHS of (126) 2Pr X′ 0p,L + max 0, 2(1 L ) Pr X′ L ,L ≥ ∈ h 3i { − 3 } ∈ h 3 4i 2 + max 0, 2(1 L ) Pr X′ L , 1 { − 4 } ∈ h 4 i LHS of (129). ≥ 51 On the other hand, as i : R 1 + (i/2) by (127), we have ∀ i ≥ 4 p 4 RHS of (129) = Pr[X′ 1 + (i/2)] Pr[X′ R ] RHS of (126). ≥ ≥ ≥ i ≥ i=1 i=1 X p X Therefore, (126) follows from (129). E.2 Case 2: There are no intermediate-sized weights In Appendix E.1.1 we covered the case where there exists a weight a with a [a + a + a i i ∈ 1 2 3 − 1, 1 a a ]. We hence assume the inexistence of such weights. That is, we may partition the − 1 − 2 weights into ‘big’ and ‘small’ ones, B and S: • B = i [n] a > 1 a a . (Notice that 1, 2, 3 B, as a + a + a 1.) { ∈ | i − 1 − 2} { } ⊂ 1 2 3 ≥ • S = i [n] a < a + a + a 1 . (Notice that S = [n] B.) { ∈ | i 1 2 3 − } \ We divide this case into two sub-cases, according to the size of B. E.2.1 Sub-case 1: B 4 | |≤ The proof in this case is very similar to the proof in Appendix E.1.2, except for a slightly more complicated segment comparison step. Elimination step. By Lemma C.1, it is sufficient to prove Pr X′ L ,L + Pr X′ 0,L + Pr X′ 0,L ∈ h− 1 2i ∈ h 3i ∈ h 4i (130) ≥ Pr X′ R , + Pr X′ R , + Pr X′ R , + Pr X′ R , , ∈ h 1 ∞i ∈ h 2 ∞i ∈ h 3 ∞i ∈ h 4 ∞i where σ, X′,L1,L2,L3,L4,R1,R2,R3,R4 are as defined in Lemma C.1. Note that by (127), we have L L 1/2 and i: R 1+ i/2. (131) 4 ≥ 3 ≥ ∀ i ≥ p p (The assertion (127) applies whenever the weights a1, aj, ak of the three eliminated variables satisfy 1 2a a a . This holds for a , a , a , since a + a + a 1 by assumption.) − 1 ≤ k ≤ j 1 2 3 1 2 3 ≥ Segment comparison step. Like in Appendix E.1.2, we claim that Pr X′ L , 1 Pr X′ L ,L . (132) ∈ h 4 i ≤ ∈ h− 1 2i We would like to deduce this inequality from Lemma 3.6, but for using the lemma we need a good upper bound on the maximal weight of the Rademacher sum it is applied to. In order to obtain such a bound, we eliminate also the variable x4 and use the fact that due to the definition of B,S and the assumption B 4, we have | |≤ i 5: a a + a + a 1. (133) ∀ ≥ i ≤ 1 2 3 − The argument goes as follows. To prove (132), it suffices to check b 1, 1 : Pr[X′ L , 1 x = b] Pr[X′ L ,L x = b]. ∀ ∈ {− } ∈ h 4 i∧ 4 ≤ ∈ h− 1 2i∧ 4 52 Defining ai′ = ai/σ and X′′ = X′ a4′ x4 = i 5 ai′ xi, this boils down to showing − ≥ b 1, 1 : Pr X′′ L ba′ , 1P ba′ Pr X′′ L ba′ ,L ba′ . ∀ ∈ {− } ∈ 4 − 4 − 4 ≤ ∈ − 1 − 4 2 − 4 This follows from Lemma 3.6, applied to the Rademacher sum X ′′ with the parameters A,B,C,D,M = L ba′ ,L ba′ ,L ba′ , 1 ba′ , a′ . − 1 − 4 2 − 4 4 − 4 − 4 5 Let us verify that the assumptions of the lemma are satisfied. By (133), all weights of X′′ are indeed bounded by M = a (a +a +a 1)/σ. To verify the assumption C min A , B , 5′ ≤ 1 2 3 − ≥ {| | | |} it is sufficient to check 0 B C. B C is equivalent to L L which holds as a a , ≤ ≤ ≤ 2 ≤ 4 1 ≥ 3 and B > 0 holds since a + a a + a < 1. Finally, the assumption D C + 2M B A is 1 2 − 3 4 − ≤ − equivalent to σ 5 3a 5a 3a , which holds since ≤ − 1 − 2 − 3 (5 3a 5a 3a )2 σ2 = (a a )(30 18a 40a 10a )+ − 1 − 2 − 3 − 2 − 3 − 1 − 2 − 3 (a a )(80 114a 66a ) + (2 4a )(12 31a ) > 0, 1 − 2 − 1 − 2 − 1 − 1 where all expressions here are nonnegative as a a a 0.387. 3 ≤ 2 ≤ 1 ≤ Chebyshev-type step. The Chebyshev-type step is almost identical to the corresponding step in Appendix E.1.2, and thus we describe it very briefly. Denoting L¯ = min L , 1 and L¯ = min L , 1 , the Chebyshev-type inequality (17), applied 3 { 3 } 4 { 4 } with c0, c1, c2, c3 = 0, L¯3, L¯4, 1 and d0, d1, d2, d3, d4 = 1, 3/2, 4/2, 5/2, 6/2, gives 2 2 2Pr X′ 0, L¯ + 2(1 L¯ )Pr X′ L¯ , L¯ p+ 2(1 pL¯ )PrpX′ pL¯ , 1 ∈ 3 − 3 ∈ 3 4 − 4 ∈ 4 4 1 (134) 2 Pr[X′ 1 + (i/2)]. ≥ · 2 ≥ i=1 X p The LHS of 130 is the LHS of (134) due to (131) and (132), while the RHS of (130) is the ≥ ≤ RHS of (134) due to (131). Therefore, (130) follows from (134). E.2.2 Sub-case 2: B 5 | |≥ Elimination step. We begin with eliminating 5 variables. Let 5 n 2 ai σ = 1 a and X′ = x . v − i σ i u i=1 i=6 u X X By Lemma 2.1, applied with mt= 5, it is sufficient to prove 31 Pr[X′ > T ] 8, (135) k ≤ Xi=0 where T 31 range over all options { k}k=0 1 a a a a a T ,...,T = ± 1 ± 2 ± 3 ± 4 ± 5 . 0 31 σ We order the Tk’s according to the bit-representation of k, that is: 5−i 1 5 ( 1) k/2 a T = − i=1 − ⌊ ⌋ i , k σ P so that, for example 1+ a a + a + a a T = 1 − 2 3 4 − 5 . 22 σ 53 Auxiliary inequalities. To proceed, we use the following auxiliary inequalities in the Tk’s. ? T max(T , T ) 1, 12 · 10 17 ≥ ? T , T , T 1, 18 20 24 ≥ ? T , T , T 1 + (3/3), 11 13 14 ≥ ? T , T , T , T , T ,p T 1 + (9/3), 19 21 22 25 26 28 ≥ (136) ? T , T , T , T , T , T p1 + (15/3), 15 23 27 29 30 31 ≥ i 1 5 ( 1) k/2 a p 5 with: T = − i=1 − ⌊ ⌋ i , σ = 1 a2, k σ v − i u i=1 P u X 1 a a a a a a a t0.387. − 2 − 4 ≤ 5 ≤ 4 ≤ 3 ≤ 2 ≤ 1 ≤ The proof of (136) is presented in Appendix F.11. Note that in light of Appendix E.1.2, we may assume a + a + a 1 for any distinct j,k > 1 2 j k ≥ in B. In particular, as B 5, we may assume that a +a +a 1, and thus, the assumptions | |≥ 2 4 5 ≥ of (136) hold true in our region. Reduction step. Note that if a + b 0, then Pr[X′ > a]+Pr[X′ > b] 1. Indeed, denoting a b ≥ ≤ X′′ = X′ + −2 xn+1, where xn+1 is a Rademacher random variable independent of x1,...,xn, we have a + b Pr[X′ > a]+Pr[X′ > b]=2Pr X′′ > 2Pr[X′′ > 0] 1. (137) 2 ≤ ≤ Since in our range, i, j : a + a < 1, we have ∀ i j T0 + T7 > 0, T1 + T6 > 0, T2 + T9 > 0, T4 + T10 > 0, T4 + T17 > 0, T8 + T3 > 0, T16 + T5 > 0. Thus, by (137), (i, j) P : Pr[X′ > T ]+Pr[X′ > T ] 1, Pr[X′ > T ]+Pr[X′ > min(T , T )] 1, ∀ ∈ i j ≤ 4 10 17 ≤ where P = (0, 7), (1, 6), (2, 9), (3, 8), (5, 16) . Hence, in order to prove (135), it is sufficient to { } prove Pr[X′ > max(T , T )] + Pr[X′ > T ] 2. (138) 10 17 i ≤ i=0,1,2,3,4,5,6,7,8,9,10,16,17 6 X Semi-inductive step. We further claim that Pr[X′ > max(T , T )] + Pr[X′ > T ] 1/2. (139) 10 17 12 ≤ To see this, rewrite this inequality as Pr[X′ 0, T ]] Pr[X′ > max(T , T )]. (140) ∈ h 12 ≥ 10 17 We deduce (140) from the assertion of Theorem 1.2 for a Rademacher sums on n 4 variables, − whose correctness we may assume by induction. 54 Note that by Lemma 2.1, applied with m = 1, for any Rademacher sum Z′ with Var[Z′] = 1 and any 0 Pr[Z′ 0,t]] Pr[Z′ > 1/t] (141) ∈ h ≥ follows from Tomaszewski’s assertion Pr[ Z 1] 1/2 for the Rademacher sum Z = b x + | | ≤ ≥ 1 1 σ Z , where b = (1 t2)/(1+t2) and σ = 1 b2 (see (10) and (11))). Furthermore, the range ′ ′ 1 − − 1 0 Pr[X′ 0, T ]] Pr[X′ > 1/T ] (142) ∈ h 12 ≥ 12 from Tomaszewski’s assertion for Rademacher sums on n 5+1= n 4 variables, which holds − − by the induction hypothesis. As max T , T 1/T by (136), the assertion (140) follows from (142). { 10 17}≥ 12 Chebyshev-type inequality step. By combining (138) with (139) and replacing Pr[X′ >t′] with 1/2 Pr[X 0,t ]] for t = T , T , T , we are left with proving − ′ ∈ h ′ ′ 18 20 24 Pr[X′ > T ] Pr[X′ 0, T ]]. (143) i ≤ ∈ h j i 11,13,14,15,19,21,22,23,25,26,27,28,29,30,31 j 18,20,24 ∈{ X } ∈{ X } Since T , T , T 1 by (136), it is sufficient to prove 18 20 24 ≥ 1 Pr[X′ 0, 1]] Pr[X′ > T ]. (144) ∈ h ≥ 3 i i 11,13,14,15,19,21,22,23,25,26,27,28,29,30,31 ∈{ X } The Chebyshev-type inequality (17), applied to the Rademacher sum X′ with c0, c1 = 0, 1 and the sequence di = 1 + (i/3), i = 1, 2, 3,..., yields p 1 ∞ Pr[X′ 0, 1]] Pr[X′ > 1 + (i/3)]. (145) ∈ h ≥ 3 i=1 X p The assertion (144) follows from (145) instantly, via the lower bounds on the Ti’s proved in (136). This completes the proof of Theorem 1.2. Appendix F Proofs of inequalities In this appendix we prove a series of inequalities that are used at various stages of the proof of Theorem 1.2. Polynomial inequalities. We usually choose to prove inequalities through Positivstellensatz, i.e., representation as a combination of terms that are transparently positive. For example, in order to prove that 2a2 ab + 1/16 0 holds for any a b 0, we just write − ≥ ≥ ≥ 2a2 ab + 1/16 = a(a b + 1/2) + (a 1/4)2 0. − − − ≥ 55 F.1 Proof of Inequality (62) Recall we have to prove the following, in the range 0 Case 1: 1/2 t 1 1/2. In this case we have to prove 1 + 2(1 t2) 1/(t2) 1, ≤ ≤ − − ≤ − which is equivalent to q p 2t4 4t2 + 1 0. − ≥ This is a simple quadratic inequality in t2, which holds in particular when t2 1 1/2, as ≤ − required. p Case 2: 1/3 t< 1/2. In this case the inequality states 1 + 2(1 t2) + 4(1 4t2) 1/t2 1, ≤ − − ≤ − which is equivalent to 18t4 8t2 + 1 0. − ≥ Applying the inequality 2ab a2 + b2, we see that 8t2 √72t4 1 + 18t4, as required. ≤ ≤ ≤ Case 3: 0 1/t 1 ⌈ ⌉− 1/t 1 1 1 1 (kt)2 (1 (xt)2)dx = 1 t2 . − ≤ − t − − 3 t − k=2 Z1 X Thus, it is sufficient to prove 1 + 2(1 t2) + 4(1/t 1 (1/t t2)/3) 1/t2 1. Rearranging, − − − − ≤ − this is equivalent to 2t4 8t + 3 0. Since t< 1/3, one trivially has 3 8t 0, as required. − ≥ − ≥ F.2 Proof of Inequality (64) Recall we have to prove 2a2 t and (1 3t/2) + 2a2 t, with σ ≤ − σ ≤ 2 2 1 a1 a1 3+ 25 + 10a1 63a1 a1 [0.5, 0.55], σ = 1 a1, t = − , a2 − − . ∈ − 1+ a1 ≤ 8 q r p Note the first inequality follows from the second one, as t 1/3 < 2/3 in our parameters ≤ range. Rewriting the second inequality to depend only on a (by writing σ, t in terms of a ), 1 p 1 we are required to prove: 4 1 a2 + 25 + 10a 63a2 13 11a . − 1 1 − 1 ≤ − 1 q q Notice 1 a2 1 a2/2, and so after rearranging, it is sufficient to show − 1 ≤ − 1 p 25 + 10a 63a2 (2a2 11a + 9)2, 1 − 1 ≤ 1 − 1 or equivalently, a4 11a3 + 55a2 52a + 14 0. We present it as a sum of squares: 1 − 1 1 − 1 ≥ 11 2 75 38 2 14 a4 11a3 + 55a2 52a +14 = a2 a + 3 + a + > 0. 1 − 1 1 − 1 1 − 2 1 4 1 − 75 75 56 F.3 Proof of Inequality (65) Recall we have to prove 1 + (1 t2) + (1 (3t/2)2) 1/t2 1, − − ≤ − in the range t2 (0, 1/3]. This inequality is equivalent to 13t4 16t2 + 4 0. This latter ∈ − ≥ inequality is quadratic in t2, and holds whenever t2 (8 √12)/13. In particular, it holds for ≤ − t2 1/3, as asserted. ≤ F.4 Proof of Inequality (67) Recall we are required to prove ? 2 ? ? L ,R max( 3 L2, √2),R max( 5 2L2, √3) 2 ≥ 3 1 ≥ − 2 2 ≥ − 2 1 a + a q1+ a a 1+ a +qa with: L = − 1 2 ,R = 1 − 2 ,R = 1 2 , σ = 1 a2 a2, 2 σ 1 σ 2 σ − 1 − 2 q where a [0.5, 0.55], a + a < 1, and a a 3+ 25 + 10a 63a2 /8. 1 ∈ 1 2 2 ≥ 1 − 1 − 1 p F.4.1 Proving R max( 3 L2, √2) 1 ≥ − 2 In the inequality R 3 Lp2, both sides are positive, and squaring shows equivalence to the 1 ≥ − 2 inequality p 5a2 4a a + (5a2 1) 0. (146) 2 − 1 2 1 − ≥ Considering the left hand side as a quadratic function of a2, we find it has no zeros, as ∆ = ( 4a )2 4 5 (5a2 1) = 20 84a2 < 0, − 1 − · · 1 − − 1 where the ultimate inequality holds since a 1/2. Thus, (146) holds for any value of a . 1 ≥ 2 ? In a similar way, in the inequality R √2, squaring shows equivalence to the inequality 1 ≥ ? 3a2 2(1 + a )a + 3a2 + 2a 1 0. (147) 2 − 1 2 1 1 − ≥ Considering the left hand side as a quadratic function of a2, we have ∆ = 4(1 + a )2 4 3 (3a2 + 2a 1) = 16 16a 32a2 0, 1 − · · 1 1 − − 1 − 1 ≤ where the ultimate inequality holds since a 1/2. Thus, (147) holds for any value of a . 1 ≥ 2 F.4.2 Proving R max( 5 2L2, √3) 2 ≥ − 2 The inequality R √3 holdsp since the assumption a 1/2 implies 2 ≥ 1 ≥ 1+ a1 + a2 1 + 1/2 R2 = = √3. 1 a2 a2 ≥ 1 (1/2)2 − 1 − 2 − p p 57 ? Regarding the inequality R 5 2L2, squaring shows equivalence to 8a2 + (6 2a )a + 2 ≥ − 2 2 − 1 2 (8a2 2a 2) 0. Solving this quadratic inequality in a shows that it holds whenever 1 − 1 − ≥ p 2 2 2 (2a1 6) + (6 2a1) 32(8a1 2a1 2) a2 − − − − − , ≥ p 16 or equivalently, a a 3+ 25 + 10a 63a2 /8, which is precisely the assumption we 2 ≥ 1 − 1 − 1 made on a2. p F.4.3 Proving L 2/3 2 ≥ By squaring and rearranging, one sees the inequality is equivalent to 13a2 + (18 18a )a + (13a2 18a + 5) 0. 2 − 1 2 1 − 1 ≥ We already observed (in the proof of R 5 2L2) that 8a2 +(6 2a )a +(8a2 2a 2) 0 2 ≥ − 2 2 − 1 2 1 − 1 − ≥ holds in our range of parameters. Thus, it is sufficient to prove p 3 13a2 + (18 18a )a + (13a2 18a + 5) (8a2 + (6 2a )a + (8a2 2a 2)) 0, 2 − 1 2 1 − 1 − 2 2 − 1 2 1 − 1 − ≥ or equivalently, a2 + (9 15a )a + (a2 15a + 8) 0. This indeed holds in our range, since 2 − 1 2 1 − 1 ≥ a2 0 holds trivially, 9 15a 0 holds as a 0.55, and a2 15a + 8 0 holds whenever 2 ≥ − 1 ≥ 1 ≤ 1 − 1 ≥ a 0.553, whereas we assume a 0.55. This completes the proof. 1 ≤ 1 ≤ F.5 Proof of Inequality (92) Recall we have to prove ? Pr[Z>t]+Pr[Z > 1/t] 0.324, (148) ≤ 1 a1 for a1 (0.31, 0.5), t = − , and a standard Gaussian Z N(0, 1). We claim that the ∈ 1+a1 ∼ function f : t Pr[Z >tq] + Pr[Z > 1/t] is decreasing in (0, 1), and thus, it is sufficient to 7→ verify (148) only at the maximal t = 1/3, achieved at a1 = 1/2. For this value of t, we indeed have Pr[Z > 1/3] + Pr[Z > √3] 0.324, as required. ≤ p To see that fpis decreasing in (0, 1), note that 1/(2t2 ) 1 e− t2/2 f ′(t)= 2 e− . √2π t − ! Letting s = 1/t2 so that s 1, we see that the assertion f (t) 0 is equivalent to log(s) ≥ ′ ≤ ≤ (s 1 )/2. This indeed holds for all s 1, as at s = 1 the two sides are equal, and the derivative − s ≥ of the l.h.s. (namely, 1/s) is never larger than the derivative of the r.h.s. (namely, (1+1/s2)/2), by the arithmetic vs. geometric means inequality. This completes the proof. F.6 Proof of Inequality (94) Recall we have to prove 4 ? Pr[Z > T ] 0.664, k ≤ k=1 (149) X 1 a a with: Z N(0, 1), 0.19 a a 1/2, σ = 1 a2 a2, T = ± 1 ± 2 . ∼ ≤ 2 ≤ 1 ≤ − 1 − 2 k σ q 58 Let f(a , a ) = 4 Pr[Z > T ], so that we wish to prove f(a , a ) 0.664 in the region 1 2 k=1 k 1 2 ≤ = [0.19, 1/2]2. Experimental evidence strongly suggests that in this region, f is maximized D P at (a , a ) = (1/2, 1/2), where f(1/2, 1/2) = 0.6596 10 4 < 0.664. However, we could not 1 2 ± − find a short analytic proof of this, and so we (rigorously) prove (149) numerically. Using a bound on the gradient of f, we verify f(a , a ) 0.664 for all (a , a ) , by 1 2 ≤ 1 2 ∈ D sampling a sufficiently fine net. Specifically, by Lagrange’s mean value theorem, if for some function g : R that is continuous in and differentiable in the interior , the gradient D → D D◦ g satisfies g(¯a) C for anya ¯ = (a , a ) , then for anya, ¯ a¯ we have ∇ k∇ k2 ≤ 1 2 ∈ D◦ ′ ∈ D g(¯a) g(¯a′)+ C a¯ a¯′ . (150) ≤ − 2 Bounding the gradient of f. Recall f = 4 Pr[Z > T ] for a Gaussian Z N(0, 1). k=1 k ∼ Hence, for i = 1, 2 we have P 4 ∂f 1 ∂Tk 2 = − exp( Tk /2). ∂ai √2π ∂ai − Xk=1 For T = (1+ a y + a y )/σ with y ,y 1, 1 , we have k 1 1 2 2 1 2 ∈ {− } ∂Tk aiTk yi = 2 + . ∂ai σ σ Hence, using 0 a , a 1/2, σ 1/√2 and k : T 0, we obtain ≤ 1 2 ≤ ≥ ∀ k ≥ ∂f 4 2 max(a1, a2) 1 max exp( Tk /2) Tk + ∂a ≤ √ k − σ2 σ i 2π 4 2 max(exp( Tk /2)(Tk + √2)). ≤ √2π k − A simple analysis shows that the function T exp( T 2/2)(T + √2) has a global maximum k 7→ − k k at T = (√6 √2)/2, where it attains a value < 1.69. Thus, for i = 1, 2 we have ∂f < 2.7, k − ∂ai and therefore, f √2 2.7 < 4. k∇ k2 ≤ · Sampling f. Consider the set of points A2, where A = 0.185 + 0.0015ℓ ℓ Z, 0 ℓ 210 . { | ∈ ≤ ≤ } Any pointa ¯ [0.19, 0.5]2 has somea ¯ A2 with a¯ a¯ 0.0015/√2. Hence, applying (150) ∈ ′ ∈ k − ′k2 ≤ (with the bound f < 4), we get k∇ k2 0.0015 f(¯a) 2 a¯′ A : f(¯a′) 0.6597. (151) ∀ ∈ ≤ 211 A computer program which verifies the 2 inequalities included in (151) is provided in https://github.com/IamPoosha/tomaszewski-problem/blob/master/formal_verification.py. 59 F.7 Proof of Inequality (97) Recall we have to prove ? ? ? L 0, L 1/2, i 1, 2, 3, 4 : R (1 + i)/2, 3 ≥ 4 ≥ ∀ ∈ { } i ≥ with: a [0.31, 0.5], a [0.19, a ], a [0.15, a ], a + a + a 1, 1 ∈ p 2 ∈ 1 3 ∈ 2 p 1 2 3 ≤ σ = 1 a2 a2 a2, − 1 − 2 − 3 1 a + a a q1 a a a 1+ a a a L = − 1 2 − 3 , L = − | 1 − 2 − 3|,R = | 1 − 2 − 3|, 3 σ 4 σ 1 σ 1+ a a + a 1+ a + a a 1+ a + a + a R = 1 − 2 3 R = 1 2 − 3 ,R = 1 2 3 , 2 σ 3 σ 4 σ Proving L 0 3 ≥ Follows from a + a + a 1 and 0 < a . 1 2 3 ≤ 2 Proving L 1/2 4 ≥ We wish to provep 1 a a a σ/√2. Notice we have − | 1 − 2 − 3|≥ a a a 1/3, | 1 − 2 − 3|≤ as a a a 0.5 0.19 0.15 0.16 and 1 − 2 − 3 ≤ − − ≤ (a a a )= a (a a ) a (a + a + a )/3 1/3. − 1 − 2 − 3 3 − 1 − 2 ≤ 3 ≤ 1 2 3 ≤ Notice also σ2 1 0.312 0.192 0.152 0.8453. Hence, we conclude with ≤ − − − ≤ 1 a a a 2/3 0.8453/2 σ/√2. − | 1 − 2 − 3|≥ ≥ ≥ p Proving R 1 1 ≥ We verify the stronger inequality 1 a + a + a σ. Notice that as we assume a 0.19 and − 1 2 3 ≥ 2 ≥ a 0.15, we have 3 ≥ 1 a + a + a 1.34 a and σ2 0.9414 a2. − 1 2 3 ≥ − 1 ≤ − 1 Thus, it is sufficient to check (1.34 a )2 0.9414 a2. This indeed holds, as − 1 ≥ − 1 (1.34 a )2 (0.9414 a2) = (1 2a )(0.84 a ) + 0.0142 > 0. − 1 − − 1 − 1 − 1 Proving R 3/2 2 ≥ Recall σ √1 p0.312 0.192 0.152, hence R (1 + a )/σ 1.15/√0.8453 > 3/2. ≤ − − − 2 ≥ 3 ≥ p Proving R √2 3 ≥ Recall σ √1 0.312 0.192 0.152, hence R (1 + a )/σ 1.31/√0.8453 > √2. ≤ − − − 3 ≥ 1 ≥ Proving R 5/2 4 ≥ Simply, R 1 +p 0.31 + 0.19 + 0.15 > 5/2. 4 ≥ p 60 F.8 Proof of Inequality (107) Recall we have to prove ? R2 1 R2 1 R2 1 R2 1 max 1 L2, 3/2 L2, 1/2 min 1 − , 2 − , 3 − , 4 − , − 3 − 4 ≤ 1 2 3 4 with: a [0.387, 0.5], 0.15 a a a , a + a + a 1, 1 ∈ ≤ 3 ≤ 2 ≤ 1 1 2 3 ≥ σ = 1 a2 a2 a2, (152) − 1 − 2 − 3 1 a + a a q1+ a a a 1 a + a + a L = − 1 2 − 3 , L = 1 − 2 − 3 ,R = − 1 2 3 , 3 σ 4 σ 1 σ 1+ a a + a 1+ a + a a 1+ a + a + a R = 1 − 2 3 R = 1 2 − 3 ,R = 1 2 3 , 2 σ 3 σ 4 σ In principle, a proof that max A min B for two sets A, B consists of A B comparisons. ≤ | | · | | In our case, A = 3 and B = 4. We effectively reduce B to 3 by showing that (R2 1)/3 | | | | | | 3 − ≤ (R2 1)/4, and then we verify the 3 3 = 9 remaining inequalities. 4 − · Proving (R2 1)/3 (R2 1)/4 3 − ≤ 4 − Recall 0.15 a a a 1/2. Hence, ≤ 3 ≤ 2 ≤ 1 ≤ (3(R2 1) 4(R2 1))σ2 = σ2 + (1 a a )(0.9+ a + a )+ 4 − − 3 − − 1 − 2 1 2 + (a 0.15)(13.85 + 14a + 14a a ) + 0.1775 > 0. 3 − 1 2 − 3 Proving 1 L2 R2 1 − 3 ≤ 1 − This inequality reads as 2 L2 +R2, or equivalently, σ2 (1 a +a )2 +a2. As a +a +a 1, ≤ 3 1 ≤ − 1 2 3 1 2 3 ≥ it is sufficient to prove the inequality 1 a2 a2 (1 a + a )2 + 2(1 a a )2. − 1 − 2 ≤ − 1 2 − 1 − 2 Since 0.25 a a 1/2, we have ≤ 2 ≤ 1 ≤ (1 a + a )2 + 2(1 a a )2 (1 a2 a2) = (1 2a )(2 a 2a )+ a (4a 1) 0. − 1 2 − 1 − 2 − − 1 − 2 − 1 − 2 − 1 2 2 − ≥ Proving 1 L2 (R2 1)/2 − 3 ≤ 2 − This inequality reads as (1+a a +a )2+2(1 a +a a )2 3σ2 0. By the Cauchy-Schwarz 1− 2 3 − 1 2− 3 − ≥ inequality, we have a2 + a2 + a2 (a + a + a )2/3 1/3, 1 2 3 ≥ 1 2 3 ≥ and hence, σ2 2/3. Thus, it suffices to show (1 + a a + a )2 + (1 a + a a )2 2 0, ≤ 1 − 2 3 − 1 2 − 3 − ≥ which clearly holds as (1 + A)2 + (1 A)2 2 for any A. − ≥ Proving 1 L2 (R2 1)/3 − 3 ≤ 3 − This inequality reads as 4σ2 3(1 a + a a )2 +(1+ a + a a )2. Since σ2 2/3 and ≤ − 1 2 − 3 1 2 − 3 ≤ a a , it suffices to prove 3(1 a )2 +(1+ a )2 8/3. This indeed holds, as 3 ≤ 2 − 1 1 ≥ 3(1 a )2 +(1+ a )2 = 4(a 1/2)2 + 3 > 8/3. − 1 1 1 − 61 Proving 3/2 L2 (R2 1)/2, which implies 3/2 L2 min(R2 1, (R2 1)/2) − 4 ≤ 1 − − 4 ≤ 1 − 2 − This inequality reads as 2σ2 (1 A)2 +(1+ A)2/2, with A = a + a + a . Since σ2 2/3, ≤ − − 1 2 3 ≤ it suffices to show (1 A)2 +(1+ A)2/2 4/3. − ≥ This indeed holds, as (1 A)2 +(1+ A)2/2 = 4/3 + (3A 1)2/6. − − Proving 3/2 L2 (R2 1)/3 − 4 ≤ 3 − This inequality reads as 11σ2/2 3(1 + a a a )2 +(1+ a + a a )2. Set α = 141/365, ≤ 1 − 2 − 3 1 2 − 3 then 6(1 + a a a )2 + 2(1 + a + a a )2 11σ2 1 − 2 − 3 1 2 − 3 − = (a α)(16 + 19α + 19a 8a 16a ) + 4(a a )(1 + α a + a )+ 1 − 1 − 2 − 3 2 − 3 − 2 3 +23(a 6(1 + α)/23)2 + 23(a 6(1 + α)/23)2 > 0, 2 − 3 − where the last inequality follows from the assumptions a a a 1/2 and a 0.387 > α. 3 ≤ 2 ≤ 1 ≤ 1 ≥ Proving 1/2 R2 1 ≤ 1 − This inequality reads as 3 σ2 (1 a + a + a )2. Recall σ2 2/3, so we conclude with 2 · ≤ − 1 2 3 ≤ 3 (1 a + a + a )2 = (1 + (a + a + a 1) + (1 2a ))2 1 σ2. − 1 2 3 1 2 3 − − 1 ≥ ≥ 2 · Proving 1/2 (R2 1)/2 ≤ 2 − This inequality reads as (1 + a a + a )2 2σ2 0. This indeed holds, since 1 − 2 3 − ≥ (1 + a a + a )2 2σ2 = (a a )(2 a 3a + 2a ) + ((2a )2 (1 a )2) + 4a2 0, 1 − 2 3 − 1 − 2 − 1 − 2 3 1 − − 3 3 ≥ where the last inequality follows from the assumptions 0 a a a 1/2 and 2a ≤ 3 ≤ 2 ≤ 1 ≤ 1 ≥ a + a 1 a . 1 2 ≥ − 3 Proving 1/2 (R2 1)/3 ≤ 3 − This inequality reads as 5 σ2 (1 + a + a a )2. Recall σ2 2/3. Using the inequality 2 · ≤ 1 2 − 3 ≤ a + a a a 0.387, we get 1 2 − 3 ≥ 1 ≥ 5 5 (1 + a + a a )2 1.3872 > σ2. 1 2 − 3 ≥ 3 ≥ 2 · F.9 Proof of Inequality (120) We prove a slight strengthening of the inequality: 3 2L2 ? 10 ? R2 1 R2 1 R2 1 max 1/2, − 4 min R2 1, 2 − , 3 − , 4 − , 3 ≤ 27σ2 ≤ 1 − 2 3 4 with: a [0.31, 0.387], a a a , a + a + a 1, 1 ∈ 3 ≤ 2 ≤ 1 1 2 3 ≥ 1+ a a a 1 a + a + a σ = 1 a2 a2 a2, L = 1 − 2 − 3 ,R = − 1 2 3 , − 1 − 2 − 3 4 σ 1 σ q1+ a a + a 1+ a + a a 1+ a + a + a R = 1 − 2 3 R = 1 2 − 3 ,R = 1 2 3 , 2 σ 3 σ 4 σ Proving this involves checking 6 inequalities, and so we proceed. 62 Proving 1/2 10/(27σ2) ≤ This inequality is equivalent to a2 +a2 +a2 7/27. This indeed holds, as we have a2 +a2 +a2 1 2 3 ≥ 1 2 3 ≥ (a + a + a )2/3 1/3 by the Cauchy-Schwarz inequality. 1 2 3 ≥ Proving (3 2L2)/3 10/(27σ2) − 4 ≤ This inequality is equivalent to 27σ2 18(1 + a a a )2 10. This indeed holds, as − 1 − 2 − 3 ≤ 10 27σ2 + 18(1 + a a a )2 = 9(a a )(4 4a a 5a )+ − 1 − 2 − 3 1 − 3 − 2 − 1 − 3 + 9(a a )(4 5a 5a ) + (3a 1)(33a 1) 0, 1 − 2 − 1 − 2 1 − 1 − ≥ where the ultimate inequality uses the assumptions a a a 0.4 and a 1/3. 3 ≤ 2 ≤ 1 ≤ 1 ≥ Proving 10/(27σ2) (R2 1)/2, which implies 10/(27σ2) min R2 1, (R2 1)/2 ≤ 1 − ≤ { 1 − 2 − } This inequality reads as 27(1 a + a + a )2 27σ2 20, and is satisfied, since − 1 2 3 − ≥ 27(1 a + a + a )2 27σ2 27 (1 + 0.2)2 27 2/3 > 20, − 1 2 3 − ≥ · − · where the first inequality uses a + a a = (a + a + a ) 2a 1 2a > 0.2 and 2 3 − 1 1 2 3 − 1 ≥ − 1 a2 + a2 + a2 1/3. 1 2 3 ≥ Proving 10/(27σ2) (R2 1)/3 ≤ 3 − This inequality is equivalent to 27(1 + a + a a )2 27σ2 30. This indeed holds, as 1 2 − 3 − ≥ 27(1 + a + a a )2 27σ2 27(1 + 1/3)2 27 2/3 = 30, 1 2 − 3 − ≥ − · where the inequality uses a 1/3, a a and a2 + a2 + a2 1/3. 1 ≥ 2 ≥ 3 1 2 3 ≥ Proving 10/(27σ2) (R2 1)/4 ≤ 4 − This inequality is equivalent to 27(1 + a + a + a )2 27σ2 40. This indeed holds, as 1 2 3 − ≥ 27(1 + a + a + a )2 27σ2 27 22 27 2/3 = 90, 1 2 3 − ≥ · − · where the inequality uses a + a + a 1 and a2 + a2 + a2 1/3. 1 2 3 ≥ 1 2 3 ≥ F.10 Proof of Inequality (127) Recall we are required to prove: ? ? ? L L 1/2, i 1, 2, 3, 4 : R 1 + (i/2), 4 ≥ 3 ≥ ∀ ∈ { } i ≥ with: 1 2a p a a a 0.387, σ = p1 a2 a2 a2, − 1 ≤ k ≤ j ≤ 1 ≤ − 1 − j − k 1 a + a a 1+ a a a q 1 a + a + a L = − 1 j − k , L = 1 − j − k ,R = − 1 j k , 3 σ 4 σ 1 σ 1+ a a + a 1+ a + a a 1+ a + a + a R = 1 − j k ,R = 1 j − k ,R = 1 j k . 2 σ 3 σ 4 σ 63 Proving L L 1/2 4 ≥ 3 ≥ The inequality L Lp holds trivially, being equivalent to a a . The inequality L 1/2 4 ≥ 3 j ≤ 1 3 ≥ reads p 2(1 a + a a )2 1 a2 a2 a2. − 1 j − k ≥ − 1 − j − k As a a 1 2a , it suffices to check 2(1 a )2 1 a2 2(1 2a )2, or equivalently, j ≥ k ≥ − 1 − 1 ≥ − 1 − − 1 11a2 12a + 3 0. 1 − 1 ≥ This indeed holds, as the two roots of the quadratic 11a2 12a + 3 are > 0.387 (just barely). 1 − 1 Proving R 3/2 1 ≥ This inequalityp is equivalent to 2(1 a + a + a )2 3σ2 0, which holds as − 1 j k − ≥ 2(1 a + a + a )2 3σ2 = (a (1 2a ))(9 14a + 5a + 4a )+ − 1 j k − j − − 1 − 1 j k + (a (1 2a ))(13 22a + 5a ) + 7(11a2 12a + 3) > 0. k − − 1 − 1 k 1 − 1 Proving R 4/2 2 ≥ This inequalityp is equivalent to (1 + a a + a )2 2σ2 0, which holds as 1 − j k − ≥ (1 + a a + a )2 2σ2 = (a (1 2a ))(5 4a 2a + 3a )+ 1 − j k − k − − 1 − 1 − j k + (a a )(4 5a 3a ) + (4a 2)2 > 0. 1 − j − 1 − j 1 − Proving R 5/2 3 ≥ This inequalityp is equivalent to 2(1 + a + a a )2 5σ2 0, which holds since 1 j − k − ≥ 2(1+a +a a )2 5σ2 = (a a )(4+4a 3a 7a )+10(a2 (1 2a )2)+(47a2 36a +7), 1 j − k − j − k 1 − j − k j − − 1 1 − 1 where the inequality 47a2 36a + 7 > 0 holds for all a as its discriminant is 362 4 7 47 < 0. 1 − 1 1 − · · Proving R 6/2 4 ≥ This inequalityp is equivalent to (1 + a + a + a )2 3σ2 0, which holds since a + a + a 1 j k − ≥ 1 j k ≥ a + 2(1 2a ) 2 3a > 0.8, and thus, 1 − 1 ≥ − 1 (1 + a + a + a )2 3σ2 1.82 3 > 0. 1 j k − ≥ − 64 F.11 Proof of Inequality (136) Recall we have to prove ? T max(T , T ) 1, 12 · 10 17 ≥ ? T , T , T 1, 18 20 24 ≥ ? T , T , T 1 + (3/3), 11 13 14 ≥ ? T , T , T , T , T ,p T 1 + (9/3), 19 21 22 25 26 28 ≥ ? T , T , T , T , T , T p1 + (15/3), 15 23 27 29 30 31 ≥ i 1 5 ( 1) k/2 a p 5 with: T = − i=1 − ⌊ ⌋ i , σ = 1 a2, k σ v − i u i=1 P u X 1 a a a a a a a t0.387. − 2 − 4 ≤ 5 ≤ 4 ≤ 3 ≤ 2 ≤ 1 ≤ Simplification. As 0 < a a a a a , it is easy to check that 5 ≤ 4 ≤ 3 ≤ 2 ≤ 1 T T T , 18 ≤ 20 ≤ 24 T T T , 11 ≤ 13 ≤ 14 T T min(T , T ) T T , 19 ≤ 21 ≤ 22 25 ≤ 26 ≤ 28 T T T T T T . 15 ≤ 23 ≤ 27 ≤ 29 ≤ 30 ≤ 31 Hence, it is sufficient to verify the five inequalities T max(T , T ) 1, T 1, T √2, 12 · 10 17 ≥ 18 ≥ 11 ≥ T √4, T √6. 19 ≥ 15 ≥ Proving T max(T , T ) 1 12 · 10 17 ≥ Note that the assumption a + a + a 1 implies a a (1 a )/2, and thus, a 1/3. 2 4 5 ≥ 2 ≥ 4 ≥ − 2 2 ≥ Furthermore, using the assumptions a a a , a + a + a 1 and a 0.4, we have 3 ≥ 4 ≥ 5 2 4 5 ≥ 2 ≤ 3 3 3 a + a + a (a + a ) (1 a ) 0.6 a . 3 4 5 ≥ 2 · 4 5 ≥ 2 · − 2 ≥ 2 · ≥ 2 Thus, we have a 1/3 and a + a + a a 0. (153) 2 ≥ 3 4 5 − 2 ≥ Now we can prove the required inequality. Since max(T , T ) (T + T )/2, it suffices to 10 17 ≥ 10 17 prove T (T + T )/2 1, which reads (1 a + a + a a a )(1 a ) σ2. This indeed 12 10 17 ≥ − 1 2 3 − 4 − 5 − 3 ≥ holds, as (1 a + a + a a a )(1 a ) σ2 = − 1 2 3 − 4 − 5 − 3 − (a + a + a 1)( a + a + a + a ) + 2(a a a a )+ 2 4 5 − − 2 3 4 5 2 3 − 4 5 +(a a )(3a 1) + (a a )(a + a + a 1) 0, 2 − 3 2 − 1 − 2 1 2 3 − ≥ where all terms in the left hand side of the ultimate inequality are nonnegative by the assump- tions a + a + a 1, a a a a a , and the inequality (153). 2 4 5 ≥ 1 ≥ 2 ≥ 3 ≥ 4 ≥ 5 65 Proving T 1 18 ≥ This inequality holds since (1 + a a a + a a )2 σ2 (1 a )2 (1 2a2 1/3) = 3(a 1/3)2 0. 1 − 2 − 3 4 − 5 − ≥ − 3 − − 3 − 3 − ≥ Here, the first inequality uses the assumptions a a , a a , a2 a2, and the inequality 1 ≥ 2 4 ≥ 5 1 ≥ 3 a2 + a2 + a2 1/3, which follows from the assumption a + a + a 1 via the Cauchy-Schwarz 2 4 5 ≥ 2 4 5 ≥ inequality. Proving T √2 11 ≥ This inequality holds since (1 a + a a + a + a )2 2σ2 (2 2 0.387)2 2 (1 1/3) > 0, − 1 2 − 3 4 5 − ≥ − · − · − where the first inequality uses a + a + a 1, a + a 2 0.387, and a2 + a2 + a2 1/3. 2 4 5 ≥ 1 3 ≤ · 2 4 5 ≥ Proving T √4 19 ≥ This inequality holds since (1 + a a a + a + a )2 4σ2 (2 a a )2 4(2/3 a2 a2)= 1 − 2 − 3 4 5 − ≥ − 2 − 3 − − 2 − 3 = 3(a + a 2/3)2 + 2(a a )2 0, 2 3 − 2 − 3 ≥ where the first inequality uses a + a + a 1 and a2 + a2 + a2 1/3. 1 4 5 ≥ 1 4 5 ≥ Proving T √6 15 ≥ Note that the assumption a + a + a 1 implies a (a + a )/2 (1 a )/2. As 2 4 5 ≥ 3 ≥ 4 5 ≥ − 2 a a < 0.4, this in turn implies 2 ≤ 1 a a a (1 a )/2 0.4 (1 0.4)/2 = 0.1. 1 − 3 ≤ 1 − − 2 ≤ − − Therefore, we have T √6, as 15 ≥ σ2(T 2 6) = (1 a +a +a +a +a )2 6σ2 (2 a +a )2 6(2/3 a2) 1.92 6 (5/9) > 0, 15 − − 1 2 3 4 5 − ≥ − 1 3 − − 1 ≥ − · where the first inequality uses a + a + a 1, a2 + a2 + a2 1/3, and a2 0, and the second 2 4 5 ≥ 2 4 5 ≥ 3 ≥ inequality uses a2 1/9 (which holds since 3a a + a + a 1) and a a 0.1. 1 ≥ 1 ≥ 1 2 3 ≥ 1 − 3 ≤ Appendix G Proof of Lemma C.2 n Let us recall the assertion we have to prove. Let X = i=1 aixi be a Rademacher sum with variance 1, where the weights ai satisfy a3 a2 a1 1/2, a1 + a2 + a3 1, a1 0.31, { } n ≤ ≤ P≤ ≤ ≥ a 0.19, a 0.15. Let X = a x , where i: a = a /σ and σ = 1 a2 a2 a2. 2 ≥ 3 ≥ ′ i=4 i′ i ∀ i′ i − 1 − 2 − 3 Denote P p 1 a a a 1 a a + a 1 a + a a 1 a a a L ,L ,L ,L = − 1 − 2 − 3 , − 1 − 2 3 , − 1 2 − 3 , − | 1 − 2 − 3| 1 2 3 4 σ σ σ σ 66 and 1+ a a a 1+ a a + a 1+ a + a a 1+ a + a + a R ,R ,R ,R = | 1 − 2 − 3|, 1 − 2 3 , 1 2 − 3 , 1 2 3 . 1 2 3 4 σ σ σ σ Furthermore, let (c , d , e ) = (1 2L2,L , 1/2) 1 1 1 − 3 3 (c , d , e ) = (2 2L2,L , p2/2) 2 2 2 − 4 4 (c , d , e ) = (3 2R2,R ,p3/2) 3 3 3 − 1 1 (c , d , e ) = (4 2R2,R , p4/2) 4 4 4 − 2 2 (c , d , e ) = (5 2R2,R , p5/2) 5 5 5 − 3 3 (c , d , e ) = (6 2R2,R , p6/2). 6 6 6 − 4 4 We have to prove three statements: p 1. i: d , e ′ 0,L , and thus by Theorem 3.1, Pr [X d , e ] Pr [X 0,L ]. ∀ h i ii≺X h 2i ′ ∈ h i ii ≤ ′ ∈ h 2i 2. max(c , 0) 3/2. i i ≤ 3. Pi B(X′) max(ci, 0) 1 where B(X′)= i [6] di, ei X′ L1,L1 . ∈ ≤ { ∈ | h i 6≺ h− i} Unfortunately,P the proof we present below is grueling and not enlightning. G.1 Proving d , e ′ 0, L h i ii ≺X h 2i First, note that if for some i, we have e < d , then clearly, Pr [X d , e ] Pr [X 0,L ]. i i ′ ∈ h i ii ≤ ′ ∈ h 2i Thus, we henceforth assume d e . i ≤ i In order to prove d , e ′ 0,L , we have to verify: h i ii≺X h 2i (a) 0 min(L , d ); ≤ 2 i (b) 2M d 0; ≤ i − (c) e d + min(2M, e L ) L 0, i − i i − 2 ≤ 2 − where M = max a a /σ. i{ i′ }≤ 3 The assertion (a) holds trivially, as the assumption a + a + a 1 implies L ,R 0 for all i. 1 2 3 ≤ i i ≥ The assertion (b) holds for all i, since 2a 2a 1 a a a 2M 3 3 + − 1 − 2 − 3 = L d , ≤ σ ≤ σ σ 2 ≤ i where the second inequality uses again the assumption a + a + a 1. 1 2 3 ≤ In the following paragraphs, we verify the assertion (c) for each i [6]. ∈ Proving (c) for i = 1: Checking 1/2 L + ( 1/2 L ) L − 3 − 2 ≤ 2 The inequality is equivalent to √2σp 3 3a pa + a . Both sides are positive, so after ≤ − 1 − 2 3 squaring we should prove Q := (3 3a a + a )2 2(1 a2 a2 a2) 0. − 1 − 2 3 − − 1 − 2 − 3 ≥ This indeed follows from the assumptions a + a 1, a 1/2, and 0 a a , as 1 2 ≤ 1 ≤ ≤ 3 ≤ 2 Q = (1 a a )(3 3a 3a + 6a ) + 4(1 a )(1 2a )+ a (4a + 3a ) 0. − 1 − 2 − 1 − 2 3 − 1 − 1 3 2 3 ≥ 67 Proving (c) for i = 2: Checking 1 L + min(2M, 1 L ) L − 4 − 2 ≤ 2 ? We split into two sub-cases according to whether a 1/5. 3 ≤ Sub-Case 1: a 1/5. We show that in this case, 1 L + (1 L ) L . This inequality 3 ≥ − 4 − 2 ≤ 2 reads 2 2L + L , or equivalently, ≤ 2 4 2σ 3 2a 2a + 2a a a a . ≤ − 1 − 2 3 − | 1 − 2 − 3| Notice a a a a 0.5 2 0.2 a , and thus, a a a a . Hence, it is − 3 ≤ 1 − 2 − 3 ≤ − · ≤ 3 | 1 − 2 − 3| ≤ 3 sufficient to prove 2σ 3 2a 2a + a , or equivalently (since both sides are positive), ≤ − 1 − 2 3 Q := (3 2a 2a + a )2 4(1 a2 a2 a2) 0. − 1 − 2 3 − − 1 − 2 − 3 ≥ This assertion indeed follows from the assumptions a + a + a 1, a 1/5, as 1 2 3 ≤ 3 ≥ Q = 6(1 a a a )2 + 16a (1 a a a ) + 2(a a )2 + (3a + 1)(5a 1) 0. − 1 − 2 − 3 3 − 1 − 2 − 3 1 − 2 3 3 − ≥ Sub-Case 2: a 1/5. We show that in this case, 3 ≤ 1 L + 2M 1 L + 2a /σ L . − 4 ≤ − 4 3 ≤ 2 Unfolding this, we have to prove σ + a + a + a + a a a 2. 1 2 3 | 1 − 2 − 3|≤ If a a + a , this inequality reads σ + 2a 2, which holds since a 1/2 and σ 1. 1 ≥ 2 3 1 ≤ 1 ≤ ≤ If a a + a , we should prove σ 2 2a 2a , or equivalently (as both sides are positive), 1 ≤ 2 3 ≤ − 2 − 3 Q := 4(1 a a )2 (1 a2 a2 a2) 0. − 2 − 3 − − 1 − 2 − 3 ≥ This indeed follows from the assumptions a + a + a 1, 0 a a a , and a 1/5, as 1 2 3 ≤ ≤ 3 ≤ 2 ≤ 1 3 ≤ Q = (a a )(a + a ) + (1 5a )(1 a a ) + (2 3a )(1 2a a ) 0. 1 − 2 1 2 − 3 − 2 − 3 − 2 − 2 − 3 ≥ Proving (c) for i = 3: Checking 3/2 R + 2M L − 1 ≤ 2 We prove the stronger assertion p 1 2a 3/2 + 3 L , − σ σ ≤ 2 p which is equivalent to the inequality 3/2σ 2 a a a . Denote t := a + a + a 1. ≤ − 1 − 2 − 3 1 2 3 ≤ By the Cauchy-Schwarz inequality, we have a2 + a2 + a2 t2/3. Hence, the above inequality p 1 2 3 ≥ follows from 3 (1 t2/3) 2 t. r2 · − ≤ − By squaring, the latter inequality is equivalent to 3t2 8t + 5 0, which indeed holds for any − ≥ t 1, as 3t2 8t + 5 = (1 t)(5 3t) 0. ≤ − − − ≥ Proving (c) for i = 4: Checking √2 R + 2M L − 2 ≤ 2 We prove the stronger assertion √2 R + 2a /σ L , or equivalently, σ √2(1 a ). By − 2 3 ≤ 2 ≤ − 2 squaring, we should prove Q := 2(1 a )2 (1 a2 a2 a2) 0. − 2 − − 1 − 2 − 3 ≥ This indeed holds, as Q = (1 2a )2 + (a2 a2)+ a2 0. − 2 1 − 2 3 ≥ 68 Proving (c) for i = 5: Checking 5/2 R + 2M L − 3 ≤ 2 We prove the stronger assertion 5/p2 R + 2a /σ L , or equivalently, − 3 3 ≤ 2 p 5/2 σ 2(1 a ). · ≤ − 3 As a a a , this inequality followsp from 3 ≤ 2 ≤ 1 5/2 1 3a2 2(1 a ). · − 3 ≤ − 3 p q The latter inequality is equivalent (via squaring) to the quadratic inequality 23a2 16a +3 0, 3 − 3 ≥ that holds for any a R, as its discriminant is ∆ = 20 < 0. 3 ∈ − Proving (c) for i = 6: Checking √3 R + 2M L − 4 ≤ 2 This inequality follows from the stronger inequality √3 R + 2a /σ L , that is equivalent − 4 3 ≤ 2 to √3σ 2. The latter inequality is clear, as σ< 1 < 2/√3. ≤ G.2 Proving max(c , 0) 3/2 i i ≤ Reminder of the assumptions and notation. Recall the region we consider is 0.15 P ≤ a a a 1/2, a + a + a 1, a 0.19, a 0.31. In particular, in our region we have 3 ≤ 2 ≤ 1 ≤ 1 2 3 ≤ 2 ≥ 1 ≥ σ2 1 0.312 0.192 0.152 = 0.8453. (154) ≤ − − − 2 Throughout the proof below, we write ci′ = σ ci. Proof strategy. We prove the following three statements, which together clearly imply the assertion. (a) max(c , 0) + max(c , 0) 0.58; 1 3 ≤ (b) max(c , 0) + max(c , 0) + max(c , 0) 0.92; 2 4 5 ≤ (c) c 0. 6 ≤ The assertion (c) is immediate. Indeed, as a 0.31, a 0.19, a 0.15, we have 1 ≥ 2 ≥ 3 ≥ 2 2 2 2 2 2 c′ = 6(1 a a a ) 2(1 + a + a + a ) < 6(1 0.31 ) 2(1 + 0.31 + 0.19 + 0.15) < 0, 6 − 1 − 2 − 3 − 1 2 3 − − and hence, c = c /σ2 0. It is thus left to prove (a) and (b). 6 6′ ≤ In the proof we use the following obvious claim. Claim G.1. Let T be a finite set, and let ci i T be a set of values, and B > 0. In order to { } ∈ T prove i T max(ci, 0) B, it is sufficient to verify 2| | 1 inequalities: ∈ ≤ − P S T, S = : c B. ∀ ⊆ 6 ∅ i ≤ i S X∈ G.2.1 Proving max(c , 0) + max(c , 0) 0.58 1 3 ≤ By Claim G.1, it is sufficient to verify c 0.58, c 0.58 and c + c 0.58. 1 ≤ 3 ≤ 1 3 ≤ 69 Checking c 0.58 1 ≤ This inequality is equivalent to 0.58σ2 c 0. We have − 1′ ≥ 2 2 2 2 0.58σ c′ = 2(1 a + a a ) 0.42σ 2(1 a ) 0.42 0.08 > 0, − 1 − 1 2 − 3 − ≥ − 1 − ≥ where we used the assumptions a a , a 1/2, and σ 1. 3 ≤ 2 1 ≤ ≤ Checking c 0.58 3 ≤ This inequality is equivalent to 0.58σ2 c 0. That is, we should prove − 3′ ≥ 2 2 2 0.58σ c′ = 2(1 + a a a ) 2.42σ 0. − 3 | 1 − 2 − 3| − ≥ If a 0.34, then 1 ≥ 2 2.42σ2 2 2.42(1 0.342 0.192 0.152) > 0, − ≥ − − − − and the assertion follows. If a 0.34 then a a a 0.34 a , and thus, it is sufficient to prove 1 ≤ | 1 − 2 − 3|≥ − 1 2(1.34 a )2 2.42(1 a2 0.192 0.152) > 0. − 1 − − 1 − − This quadratic inequality indeed holds for all a 0.34. 1 ≤ Checking c + c 0.58 1 3 ≤ This inequality is equivalent to 0.58σ2 c c 0. Since a 0.19, a 0.15, we have − 3′ − 1′ ≥ 2 ≥ 3 ≥ 2 2 2 2 0.58σ c′ c′ = 2( a a a + 1) + 2(1 a + a a ) 3.42σ − 3 − 1 | 1 − 2 − 3| − 1 2 − 3 − 2 + 2(1 a )2 3.42(1 a2 0.192 0.152) ≥ − 1 − − 1 − − = 5.42(a 100/271)2 + 2872913/67750000 > 0. 1 − G.2.2 Proving max(c , 0) + max(c , 0) + max(c , 0) 0.92 2 4 5 ≤ By Claim G.1, it is sufficient to verify the seven inequalities c 0.92, c 0.92, c 0.92, 2 ≤ 4 ≤ 5 ≤ c + c 0.92, c + c 0.92, c + c 0.92, c + c + c 0.92. 2 4 ≤ 2 5 ≤ 4 5 ≤ 2 4 5 ≤ Checking c 0.92 2 ≤ This inequality is equivalent to 0.92σ2 c 0, which is in turn equivalent to − 2′ ≥ 2(1 a a a )2 1.08σ2 0. − | 1 − 2 − 3| − ≥ In our region, a a a 0.16 a + 0.01 and a + a + a a , and thus, a a a 1 − 2 − 3 ≤ ≤ 3 − 1 2 3 ≤ 3 | 1 − 2 − 3|≤ a + 0.01. As σ2 1 0.312 0.192 a2, we have 3 ≤ − − − 3 2(1 a a a )2 1.08σ2 2(0.99 a )2 1.08(0.87 a2) > 0, − | 1 − 2 − 3| − ≥ − 3 − − 3 where the ultimate inequality holds since a (a + a + a )/3 1/3. 3 ≤ 1 2 3 ≤ 70 Checking c 0.92 4 ≤ This inequality is equivalent to 0.92σ2 c 0, which is in turn equivalent to − 4′ ≥ 2(1 + a a + a )2 3.08σ2 0. 1 − 2 3 − ≥ As σ2 0.8453 by (154) and 1 + a a + a 1+ a 1.15, we have ≤ 1 − 2 3 ≥ 3 ≥ 2(1 + a a + a )2 3.08σ2 2 1.152 3.08 0.8453 > 0. 1 − 2 3 − ≥ · − · Checking c 0.92 5 ≤ This inequality is equivalent to 0.92σ2 c 0, which is in turn equivalent to − 5′ ≥ 2(1 + a + a a )2 4.08σ2 0. (155) 1 2 − 3 − ≥ Write d = a a and recall d 0. We claim that 2 − 3 ≥ a 0.19 d . 3 ≥ | − | Indeed, on the one hand we have a = a d 0.19 d, since a 0.19. On the other hand, 3 2 − ≥ − 2 ≥ as a + a + a 1, we have a (1 a )/2, and thus, 1 2 3 ≤ 2 ≤ − 3 1 a a (d 0.19) = 2a a + 0.19 2a − 3 + 0.19 = 2.5a 0.31 > 0, 3 − − 3 − 2 ≥ 3 − 2 3 − where the ultimate inequality holds since a 0.15. As a 0.31, in order to prove (155) it is 3 ≥ 1 ≥ sufficient to show 2(1.31 + d)2 4.08(1 0.312 0.192 (0.19 d)2) 0. − − − − − ≥ This inequality, which reads 6.08d2 + 3.6896d + 0.038864 0, indeed holds for all d 0. ≥ ≥ Checking c + c 0.92 2 4 ≤ This inequality is equivalent to 0.92σ2 c c 0, which is in turn equivalent to the inequality − 2′ − 4′ ≥ 2(1 + a a + a )2 + 2(1 a a a )2 5.08σ2 0. 1 − 2 3 − | 1 − 2 − 3| − ≥ As in the proof of c 0.92 above, we may use the bound a a a 0.01 + a , and so, it 2 ≤ | 1 − 2 − 3|≤ 3 is sufficient to prove 2(1 + a a + a )2 + 2(0.99 a )2 5.08σ2 0. 1 − 2 3 − 3 − ≥ Differentiating the left hand side with respect to a , we get 4(0.01 + a a + 4.54a ) which 3 1 − 2 3 is clearly positive. Hence, it suffices to verify the inequality for the minimal possible a3 in our region, that is, for a3 = 0.15. Write d = a a . We clearly have 0 0.31 d a . Hence, it is sufficient to prove 1 − 2 ≤ − ≤ 2 2(1 + d + 0.15)2 + 2(0.99 0.15)2 5.08(1 0.312 (0.31 d)2 0.152) 0, − − − − − − ≥ or equivalently, 7.08d2 + 1.4504d + 0.066876 0, which indeed holds for all d 0. ≥ ≥ 71 Checking c + c 0.92 2 5 ≤ This inequality is equivalent to 0.92σ2 c c 0, which is in turn equivalent to − 2′ − 5′ ≥ 2(1 + a + a a )2 + 2(1 a a a )2 6.08σ2 0. (156) 1 2 − 3 − | 1 − 2 − 3| − ≥ ? The proof splits into two sub-cases according to whether a a + a . 1 ≥ 2 3 Sub-case 1: a a + a . In this case we have 1 + a + a a 1 + 2a 1.38 and 1 ≥ 2 3 1 2 − 3 ≥ 2 ≥ a a a 0.5 0.19 0.15 = 0.16. Recalling σ 0.8453, we deduce | 1 − 2 − 3|≤ − − ≤ 2 2 2 2 0.92σ c′ c′ = 2(1 + a + a a ) + 2(1 a a a ) 6.08σ − 2 − 5 1 2 − 3 − | 1 − 2 − 3| − ≥ 2 1.382 + 2(1 0.16)2 6.08 0.8453 > 0. ≥ · − − · Sub-case 2: a a + a . In this case, the inequality (156) reads 1 ≤ 2 3 2(1 + a + a a )2 + 2(1 + a a a )2 6.08σ2 0, 1 2 − 3 1 − 2 − 3 − ≥ or equivalently, 4(1+a a )2 +4a2 6.08σ2 0. Write d = a a , so that a a 0.31 d . 1 − 3 2 − ≥ 1 − 3 2 ≥ 3 ≥ | − | We have 2 2 2 2 0.92σ c′ c′ = 4(1 + a a ) + 4a 6.08σ − 2 − 5 1 − 3 2 − 4(1 + d)2 + 4(0.31 d)2 6.08(1 0.312 2(0.31 d)2) ≥ − − − − − = 20.16d2 2.0192d + 0.057264 > 0, − where the ultimate inequality holds since the quadratic polynomial in d has ∆ < 0.54 < 0. − Checking c + c 0.92 4 5 ≤ This inequality is equivalent to 0.92σ2 c c 0. We have − 4′ − 5′ ≥ 2 2 2 2 2 2 2 0.92σ c′ c′ = 2(1+a a +a ) +2(1+a +a a ) 8.08σ = 4(1+a ) +4(a a ) 8.08σ . − 4− 5 1− 2 3 1 2− 3 − 1 2− 3 − Using the inequalities 1 + a 1.31 and σ2 0.8453, we get 1 ≥ ≤ 2 2 2 2 2 0.92σ c′ c′ = 4(1 + a ) + 4(a a ) 8.08σ 4 1.31 8.08 0.8453 > 0. − 4 − 5 1 2 − 3 − ≥ · − · Checking c + c + c 0.92 2 4 5 ≤ This inequality is equivalent to 0.92σ2 c c c 0, which is in turn equivalent to − 2′ − 4′ − 5′ ≥ 4(1 + a )2 + 4(a a )2 + 2(1 a a a )2 10.08σ2 0. 1 2 − 3 − | 1 − 2 − 3| − ≥ ? The proof splits to two sub-cases, according to whether a a + a . 1 ≥ 2 3 Sub-case 1: a a + a . In this case, a a a 0.5 0.19 0.15 0.16, and hence, 1 ≥ 2 3 | 1 − 2 − 3|≤ − − ≤ it is sufficient to verify 4(1 + a )2 + 2 0.842 10.08σ2 0. 1 · − ≥ Since a a + a 0.34 and σ2 0.8453, we have 1 ≥ 2 3 ≥ ≤ 4(1 + a )2 + 2 0.842 10.08σ2 4 1.342 + 2 0.842 10.08 0.8453 > 0. 1 · − ≥ · · − · 72 Sub-case 2: a a + a . In this case, we should prove 1 ≤ 2 3 Q := 4(1 + a )2 + 4(a a )2 + 2(1 + a a a )2 10.08(1 a2 a2 a2) 0. 1 2 − 3 1 − 2 − 3 − − 1 − 2 − 3 ≥ This indeed holds, as Q = (a 0.3)(16.08a 4a 4a + 16.824) + 7.04(a + a 65/176)2+ 1 − 1 − 2 − 3 2 3 − + 9.04(a a )2 + 767/110000 > 0. 2 − 3 Summarizing. Combining the above bounds, we have 6 max(ci, 0) = (max(c1, 0) + max(c3, 0)) + (max(c2, 0) + max(c4, 0) + max(c5, 0)) + max(c6, 0) Xi=1 0.58 + 0.92 + 0 = 1.5, ≤ as asserted. G.3 Proving i B(X′) max(ci, 0) 1 ∈ ≤ The proof consistsP of two steps. First, we show that either B(X′) 1, 3, 5, 6 or B(X′) ⊆ { } ⊆ 2, 4, 5, 6 . Then, we verify the assertion in each of these cases separately. { } G.3.1 Proving that B(X ) 1, 3, 5, 6 or B(X ) 2, 4, 5, 6 ′ ⊆ { } ′ ⊆ { } To prove this, it is sufficient to show that for any (i, j) 1, 3 2, 4 we have ∈ { } × { } d , e ′ L ,L or d , e ′ L ,L . (157) h i ii≺X h− 1 1i h j ji≺X h− 1 1i By Lemma 3.6, in order to prove d , e ′ L ,L for some k,ℓ, it is sufficient to show h k ℓi≺X h− 1 1i a e d or e d + 2 3 2L . ℓ ≤ k ℓ − k σ ≤ 1 (Note that the other assumption of Lemma 3.6, namely, L d , holds in our case for all k, by 1 ≤ k the definition of the dk’s.) The latter condition can be rewritten as 1 a1 a2 2a3 e d 2L′, with L′ = − − − . (158) ℓ − k ≤ σ Hence, in order to verify (157) for some (i, j), we have to show min(e d , e d ) max(0, 2L′). (159) i − i j − j ≤ In the following paragraphs we show this for all (i, j) 1, 3 2, 4 . ∈ { } × { } Case 1: i = 1 and j = 2 We prove a stronger inequality: min(e d , e d ) 0, or equivalently, 1 − 1 2 − 2 ≤ max 2(1 a + a a )2, (1 a a a )2 σ2. − 1 2 − 3 − | 1 − 2 − 3| ≥ If a a + a , it is sufficient to show (1 a + a + a )2 σ2 0. This indeed holds, as 1 ≥ 2 3 − 1 2 3 − ≥ (1 a + a + a )2 σ2 = (a 0.15)(2.3 + 2a + 2a 2a ) + (a 0.19)(2.68 + 2a 2a ) − 1 2 3 − 3 − 3 2 − 1 2 − 2 − 1 + (1 2a )(0.84 a ) + 0.0142 > 0. − 1 − 1 73 If a a + a , it is sufficient to prove 1 ≤ 2 3 1 Q := (1 a + a a )2 + (1 + a a a )2 σ2 0, − 1 2 − 3 2 1 − 2 − 3 − ≥ as a maximum between two quantities is no smaller than their average. This indeed holds, as Q = 2(1 a )(a a ) + 2(a a )2 + (1 a a a )2/2 0. − 3 2 − 3 1 − 2 − 1 − 2 − 3 ≥ Case 2: i = 1 and j = 4 We prove a stronger inequality: e d + e d 0, or equivalently, 3 2 0. Since 1 − 1 4 − 4 ≤ √2 − σ ≤ σ 0.8453, we have ≤ 3 2 3 2 < 0. √2 − σ ≤ √2 − 0.8453 Case 3: i = 3 and j = 2 We prove a stronger inequality: e d + e d L /2, or equivalently, 3 − 3 2 − 2 ≤ ′ 3/2 + 1 2/σ (1 a a 2a )/(2σ). − ≤ − 1 − 2 − 3 This latter is further equivalentp to (√6 + 2)σ 5 a a 2a . By squaring and using the ≤ − 1 − 2 − 3 assumption a a , it is sufficient to prove 2 ≥ 3 f(a , a , a ) := (5 a 1.5(a + a ))2 (10 + 4√6)σ2 0. 1 2 3 − 1 − 2 3 − ≥ ∂f It is easy to verify that = (8√6+22)a1 +3a2 +3a3 10, which is positive in our region since ∂a1 − a 0.31 and a , a > 0. Hence, f(0.31, a , a ) f(a , a , a ). Furthermore, by the definition 1 ≥ 2 3 2 3 ≤ 1 2 3 of σ and the convexity of the function x x2, we have 7→ f(a , (a + a )/2, (a + a )/2) f(a , a , a ). 1 2 3 2 3 ≤ 1 2 3 Hence, writing t = (a2 + a3)/2, we conclude with 14.07 2 180.4126 68.61 √6 f(a , a , a ) f(0.31,t,t) = (29 + 8√6) t + − · > 0. 1 2 3 ≥ − √ 457 8 6 + 29 Case 4: i = 3 and j = 4 We prove a stronger inequality: e d +e d L , or equivalently, 3 2a a + a a a 3− 3 4− 4 ≤ ′ − 2 − 3 | 1− 2− 3|≥ ((2 + √3)/√2) σ. By squaring, this is equivalent to · (3 2a a + a a a )2 (√12 + 7/2)σ2 0. − 2 − 3 | 1 − 2 − 3| − ≥ If a a + a , we should prove 1 ≥ 2 3 f(a , a , a ) := (3+ a 3a 2a )2 (√12 + 7/2)σ2 0. 1 2 3 1 − 2 − 3 − ≥ It is clear that f(a , a , a ) f(a + a , a , a ), and thus, we may assume a = a + a , which 1 2 3 ≥ 2 3 2 3 1 2 3 is covered in the other case. 74 If a a + a , the inequality reads 1 ≤ 2 3 f(a , a , a ) := (3 a a )2 (√12 + 7/2)σ2 0. 1 2 3 − 1 − 2 − ≥ It is clear that f(a , a , 0.15) f(a , a , a ). Furthermore, by the convexity of the function 1 2 ≤ 1 2 3 x x2, we have f((a + a )/2, (a + a )/2, a ) f(a , a , a ). Hence, writing t = (a + a )/2, 7→ 1 2 1 2 3 ≤ 1 2 3 1 2 we conclude with 6 2 10.28 √3 + 89.99 f(a , a , a ) f(t,t, 0.15) = (11 + 4√3) t + · > 0. 1 2 3 ≥ − √ 584 4 3 + 11 G.3.2 Proving i B(X′) max(ci, 0) 1 in the case B(X′) 2, 4, 5, 6 ∈ ≤ ⊆ { } By the inequality P max(c , 0) 0.92 proved in Subsection G.2.2 and the easy inequality i 2,4,5 i ≤ c 0 proved above,∈{ we see} that B(X ) 2, 4, 5, 6 implies 6 ≤ P ′ ⊆ { } max(c , 0) max(c , 0) + max(c , 0) + max(c , 0) + max(c , 0) 0.92 < 1, i ≤ 2 4 5 6 ≤ i B(X′) ∈X as asserted. G.3.3 Proving i B(X′) max(ci, 0) 1 in the case B(X′) 1, 3, 5, 6 , a3 0.2 ∈ ≤ ⊆ { } ≤ We show that in thisP case, we actually have 5 / B(X ), and hence B(X ) 1, 3, 6 . Thus, the ∈ ′ ′ ⊆ { } inequality max(c , 0) + max(c , 0) 0.58 proved in Subsection G.2.1 and the inequality c 0 1 3 ≤ 6 ≤ imply i B(X′) max(ci, 0) 0.58 < 1, as required. ∈ ≤ ShowingP 5 B(X ). To show this, we have to verify d , e ′ L ,L . By (158), it is 6∈ ′ h 5 5i ≺X h− 1 1i sufficient to show e d max(0, 2L ). We show the stronger inequality e d 1.5L . The 5 − 5 ≤ ′ 5 − 5 ≤ ′ inequality reads 1+ a + a a 3(1 a a 2a ) 5/2 1 2 − 3 − 1 − 2 − 3 , − σ ≤ 2σ or equivalently, (5 a p a 8a )2 10σ2 0. This indeed holds, as the left hand side is − 1 − 2 − 3 − ≥ (0.2 a )(65.2 16a 16a 74a ) + 6(a + a 17/30)2 + 5(a a )2 + 1/30 > 0. − 3 − 1 − 2 − 3 1 2 − 1 − 2 G.3.4 Proving i B(X′) max(ci, 0) 1 in the case B(X′) 1, 3, 5 , a3 0.2 ∈ ≤ ⊆ { } ≥ As c 0, it followsP from Claim G.1 that in order to prove the assertion, it is sufficient to verify 6 ≤ the seven inequalities c 1, c 1, c 1, c +c 1, c +c 1, c +c 1, c +c +c 1. 1 ≤ 3 ≤ 5 ≤ 1 3 ≤ 1 5 ≤ 3 5 ≤ 1 3 5 ≤ Proving c 1, c 1, c 1, c + c 1 1 ≤ 3 ≤ 5 ≤ 1 3 ≤ This was already done in Subsections G.2.1 and G.2.2. Proving c + c 1 1 5 ≤ This inequality is equivalent to σ2 c c 0, which is in turn equivalent to − 1′ − 5′ ≥ 2(1 + a + a a )2 + 2(1 a + a a )2 5σ2 0. 1 2 − 3 − 1 2 − 3 − ≥ By rearranging, we have to show 4(1 + a a )2 + 4a2 5σ2 0. The left hand side is at least 2 − 3 1 − ≥ 4 + 4 0.312 5 0.8453 > 0.15 > 0, as required. · − · 75 Proving c + c 1 3 5 ≤ This inequality is equivalent to σ2 c c 0, which is in turn equivalent to − 3′ − 5′ ≥ 2(1 + a + a a )2 + 2(1 + a a a )2 7σ2 0. 1 2 − 3 | 1 − 2 − 3| − ≥ Thus, it is sufficient to show 2(1 + a )2 + 2(1 a + a + a )2 7σ2 0. Since a a 0.2, 1 − 1 2 3 − ≥ 2 ≥ 3 ≥ it is enough to verify 2(1 + a )2 + 2(1.4 a )2 7(0.92 a2) 0. 1 − 1 − − 1 ≥ This indeed holds for all a 0.31, as the left hand side is equal to (a 0.31)(11a + 1.81) + 1 ≥ 1 − 1 0.0411 > 0. Proving c + c + c 1 1 3 5 ≤ This inequality is equivalent to σ2 c c c 0, which is in turn equivalent to − 1′ − 3′ − 5′ ≥ 2(1 + a + a a )2 + 2(1 a + a a )2 + 2(1 + a a a )2 8σ2 0. 1 2 − 3 − 1 2 − 3 | 1 − 2 − 3| − ≥ As a a 0.2, it is sufficient to verify 2 ≥ 3 ≥ (1 + a )2 + (1 a )2 + (1.4 a )2 4(0.92 a2) 0. 1 − 1 − 1 − − 1 ≥ This holds for all a , as the left hand side is equal to 7a2 2.8a + 0.28 = 7(a 0.2)2 0. 1 1 − 1 1 − ≥ 76