A MULTIDIMENSIONAL ANALOGUE OF THE RADEMACHER-GAUSSIAN TAIL COMPARISON

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1 A MULTIDIMENSIONAL ANALOGUE OF THE RADEMACHER-GAUSSIAN TAIL COMPARISON PIOTR NAYAR AND TOMASZ TKOCZ Abstract We prove a menson-free tal comparson between the Euclean norms of sums of nepenent ranom vectors unformly strbute n centre Euclean spheres an properly rescale stanar Gaussan ranom vectors 00 Mathematcs Subject Classfcaton Prmary 60E5; Seconary 60G5, 60G50 Key wors probablty nequaltes, tal comparson, bouns for tal probabltes, Gaussan ranom vectors, unform strbutons n Euclean spheres Introucton Tal comparson bouns, such as Hoeffng s nequalty, have always playe a crucal role n probablty theory When specfe to concrete examples, very precse estmates for tal probabltes are usually known For nstance, f ɛ, ɛ, are nepenent ranom varables each takng values ± wth probablty an g, g, are nepenent stanar Gaussan ranom varables, then for every m, real numbers a,, a m an postve t, P P a ɛ + + a m ɛ m > t c P a g + + a m g m > t for some absolute constant c Ths nequalty was frst prove by Pnels n [5] wth c 446 Talagran n [6] treate the case of nepenent but not necessarly entcally strbute boune ranom varables by means of the Laplace transform establshng smlar Gaussan tal bouns Bobkov, Götze an Houré obtane a bgger constant c 0 n P, but ther nuctve argument was much smpler see [] Only very recently the best constant equal approxmately 38 has been foun see [] Oleszkewcz conjecture the followng multmensonal generalsaton of Pnels Raemacher-Gaussan tal comparson P: fx, let ξ, ξ, be nepenent ranom vectors unformly strbute n the Euclean unt sphere S R an let G, G, be nepenent stanar Gaussan ranom vectors n R wth mean zero an entty covarance matrx; there exsts a unversal constant C such that PN supporte n part by NCN grant DEC-0/05/B/ST/004

2 for every m, real numbers a,, a m an t > 0 we have m m G KO P a ξ > t C P a > t = Here an throughout, enotes the stanar Euclean norm n R Note that the normalsaton s chosen so that the vectors ξ an G / have the same covarance matrx Planly, when =, KO reuces to P For general, t s possble to euce KO wth C = O from Theorem n [4] The goal of ths note s to postvely resolve Oleszkewcz s conjecture We shall show the followng two theorems whch are our man results The latter wll easly follow from the former Theorem For every, nequalty KO hols wth C = C 0 = 397 Theorem Let X, X, be nepenent rotatonally nvarant ranom vectors havng values n the unt Euclean ball n R Let G, G, be nepenent stanar Gaussan ranom vectors n R wth mean zero an entty covarance matrx Then for every m, real numbers a,, a m an t > 0 we have m m G P a X > t C 0 P a > t, = where stans for the stanar Euclean norm n R an C 0 = 397 Remark Ths wll no longer hol f we only assume the bouneness of the X For example, conser nepenent X takng only two values ±, 0,, 0 each wth probablty Then for, say a = = a m = / m, t =, the rght-han se of goes to zero when goes to nfnty, whereas the left-han se oes not epen on Acknowlegements The authors woul lke to thank Krzysztof Oleszkewcz for ntroucng them nto the subject They are really grateful to Rafa l Lata la for a scusson concernng Theorem Proofs Our proof of Theorem s nuctve, nspre by the nuctve approach to the one mensonal case from [] = = In the nuctve step, usng the sphercal symmetry of our problem, we arrve at an nequalty comparng the Gaussan volume between centre an shfte balls Lemma 3 below Ths nequalty can be vewe as a multmensonal generalsaton of the two pont-nequalty erve n

3 the nuctve step n [] Its proof leas us to somewhat subtle estmates for the Laplace transform of the frst coornate of ξ Lemma below We shall nee four lemmas We start wth a stanar one whch wll be use to prove numercal values of our constants We nclue ts proof for completeness see, eg Lemma n [3] Lemma Let an G be a stanar Gaussan ranom vector n R Then P G > /33 an P G > + /397 Proof Let G = g,, g, where g,, g are nepenent stanar Gaussan ranom varables Let S = = g We want to estmate P G > = P S > 0 Snce 0 = ES = ES {S>0} + ES {S 0}, usng the Cauchy-Schwarz nequalty we get E S = ES {S>0} ES {S 0} = ES {S>0} ES P S > 0 Moreover, by Höler s nequalty ES = E S 4/3 S /3 E S 4 /3 E S /3 Combnng these bouns an usng that Eg =, Eg 4 = 60 we obtan P S > 0 E S ES = 4 ES 4 ES = For the secon part, observe that P G > + P g 3, g + + g = P g 3 P g + + g P g > 3 33 > 397 The next lemma gves tght estmates for the Laplace transform of the frst coornate of a ranom vector unformly strbute n the unt sphere We hope these estmates are of nepenent nterest, n aton to playng a major role n our proof 3

4 Lemma For an b 0 let us enote Then for every we have J = J b = x / e bx x a b J + = + J + + J 3, b J 3 J + + b, 4 + c J J b, J + J 3 J + + Proof a Integratng by parts twce we get b J + = x + e bx x, whch, after computng the secon ervatve n the above expresson, easly leas to the esre relaton For the proof of b an c let us frst observe that ue to a these two assertons hol true for b = 0 We then show that for b > 0, b an c are equvalent Inee, part a yels J 3 J = Thus, b s equvalent to b J + + J + b J +, + J b = + After cancellng common factors on both ses ths becomes c + b + + b 4 + Let us fx b > 0 We shall show b by backwars nucton on We can use for +, that s the equalty to rewrte c n the form J +3 b J + + J = + J b J +3 J +, b = b + + b 4 +, whch becomes 3 J + J b + Frst notce that b an equvalently 3 hol for all 0 b for some large enough 0 b whch epens only on b To see ths observe that the left-han se 4

5 of 3 s strctly greater than, whereas the rght-han se for large s of orer 3/ + o/ Now suppose b hols for + 4, that s J J b an we want to show b nucton step By the above an the fact that 3 an b are equvalent, t s enough to show that b b + Ths follows from an the estmate 4 + b Clearly mmeately follows from b an c 4 + b + Remark Part mproves on Höler s nequalty whch gves J J +J 3 Remark Let us efne for an b 0 the normalse ntegrals J b = J b/j 0 so that they are the Laplace transforms of the probablty enstes: f an ξ s a ranom vector n R unformly strbute n the Euclean unt sphere S, we check that by rotatonal nvarance J 3 b = Ee v,ξ, for any vector v R of length b Part a for b = 0 gves J 3 0/J 0 = / Ths allows to smplfy b,c, rewrtten n terms of J to get for b a J + + = J + J 3, + + b, 4 + b J 3 J c J J + J + J 3 J b +, The followng lemma les at the heart of our nuctve argument It compares the stanar Gaussan measure of centre an shfte Euclean balls Lemma 3 Let an G be a stanar Gaussan ranom vector n R For every a 0, R + an a vector x R of length a we have P G R P G x R + a 5

6 Proof Snce for a = 0 we have equalty, t s enough to show that the rght-han se, ha, R = P G a e R + a s nonecreasng wth respect to a by rotatonal nvarance, for concreteness we can choose x = ae, where e =, 0,, 0 Usng Fubn s theorem we can wrte ha, R = S R +a { a + R +a r } r e r / φtt r, π 0 a R +a r where φt = π e t / The ervatve wth respect to a equals a ha, R = S R +a { r e r / π 0 [φ a R + a r ar + R + a r φ a + R + a r ar R + a r ]} r After changng the varables r = R + a x we see that ths s nonnegatve f an only f 0 x 3 [ e xa R +a x + ar + a e xa R +a x ar ] x 0 + a Ths conton can be further smplfe by ntegraton by parts usng x = x x 3 We obtan an equvalent nequalty x 3 + a x cosh ar + a x x 0 0 Let b = a R + a Then Observe that 0 + a = b R x coshbxx = x e bx x = J b Thus, the nequalty we want to show becomes J 3 b J b b R 6

7 For a fxe b, the rght-han se as a functon of R s clearly ecreasng, so gven our assumpton R + t s enough to conser R = +, whch follows from Lemma b Remark The statement for = remans true an was prove n [], where t playe a key role n the nuctve proof of Pnels nequalty P The last lemma wll help us use the sphercal symmetry of our problem Lemma 4 Let X be a rotatonally nvarant ranom vector n R Let x R an t > 0 be such that t > x Then P X + x > t = P X > θ x + t + θ x x, where θ s the frst coornate of an nepenent of X ranom vector unformly strbute n the unt sphere S n R Proof Let ξ be an nepenent of X ranom vector unformly strbute n the unt sphere S n R By rotatonal nvarance X has the same strbuton as Rξ, where R = X We have P X + x > t = P R + R ξ, x + x > t an by the rotatonal nvarance of ξ, ξ, x has the same strbuton as θ x wth θ beng the frst coornate of ξ The nequalty R +Rθ x + x > t s equvalent to R > θ x + t + θ x x or R < θ x t + θ x x, but the secon case oes not hol as the rght-han se s negatve, for we assume that t > x Proof of Theorem We fx an procee by nucton on m For m = we have to check that for 0 < t < a we have G C 0 P a > t Ths follows because P G > /33 by Lemma Suppose the asserton s true for m We shall show t for m+ We can assume that the a are nonzero By homogenety we can also assume that m+ = a = If t a + + we trvally boun the rght-han se as follows: m+ G a P a > t = P + G > t P G > +, = 7

8 an by Lemma we get C 0 P m+ = a G > t Now suppose t > a + + Notce that n partcular t > a Conser v = m+ = a ξ By nepenence an rotatonal nvarance, m+ P a ξ > t = P a e + v > t = Lemma 4 apple to X = v an x = a e yels m+ P a ξ > t = E θ P v v > θ a + = t + θ a a As a consequence, by the nepenence of θ an v, an the nuctve hypothess, m+ m+ G P a ξ > t C 0 E θ P G m+ a > θ a + t = + θ a a = = The vector m+ m+ = a G = has the same strbuton as a G = G Therefore, applyng agan Lemma 4 yels m+ P a ξ > t C 0 P G + a e > t = To fnsh the nuctve step t suffces to show that m+ G a P G + a e > t P a > t = P + G > t = Ths follows from Lemma 3 apple to a = a an R = t a + > +, whch completes the proof Proof of Theorem Let ξ, ξ, be nepenent ranom vectors unformly strbute n the unt Euclean sphere S R, nepenent of the sequence X, X, Snce X s rotatonal nvarant, t has the same strbuton as R ξ, where R = X Note that almost surely 0 R Applyng KO contonally on the R we get m m P a X > t = E R m P = ξ m a = R ξ > t = = C 0 E R m = P G m = m = a R G > t To fnsh the proof notce that for any fxe numbers R [0, ] we have m m = P a R = G > t P a G > t 8

9 References [] Bobkov, S, Götze, F, Houré, Ch, On Gaussan an Bernoull covarance representatons Bernoull 7 00, no 3, [] Bentkus, V K, Dznzaleta, D, A tght Gaussan boun for weghte sums of Raemacher ranom varables Bernoull 05, no, 3 37 [3] Lata la, R, Oleszkewcz, K, Small ball probablty estmates n terms of wths Stua Math , no 3, [4] Leoux, M, Oleszkewcz, K, On measure concentraton of vector-value maps Bull Pol Aca Sc Math , no 3, 6 78 [5] Pnels, I, Extremal probablstc problems an Hotellng s T test uner a symmetry conton Ann Statst 994, no, [6] Talagran, M, The mssng factor n Hoeffng s nequaltes Ann Inst H Poncaré Probab Statst 3 995, no 4, Potr Nayar, nayar@mmuweupl Tomasz Tkocz, ttkocz@prncetoneu Unversty of Pennsylvana Prnceton Unversty Wharton Statstcs Department Mathematcs Department 3730 Walnut St Fne Hall Phlaelpha, PA 904 Prnceton, NJ Unte States Unte States 9