Regular Banach Spaces and Large Deviations of Random Sums

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1 Regular Banach Spaces and Large Devatons of Random Sums Arkad Nemrovsk Workng paper, verson of December 3, 004 Overvew A typcal result on large devatons of sums wth random terms states that f ξ t are ndependent scalar random varables wth zero means and such that ξ t has as lght tals as a Gaussan N (0, 4σ t ) random varable, specfcally, E expξ t /σ t } } O(), () and then S N = N ξ t, } Prob S N > t σ σ N O() exp O()t } () (from now on, all O() s are approprate postve absolute constants). Our goal s to get smlar results for the case when ξ t are ndependent random vectors wth zero means n a fnte-dmensonal vector space E equpped wth norm, S N = N ξ t and the lght tal condton () s stated as Note that a straghtforward guess E exp ξ t /σ t } } exp}. (3) Eξ t } = 0 t & (3) & ξ t } are ndependent Prob S N > t } σ σ N O() exp O()t } (4) s not true, as t s shown by the followng example: E = R n, x = x j x j, ɛt, j = t(mod n) (ξ t ) j =, where ɛ 0, otherwse, ɛ,... are ndependent random varables takng values ± wth probablty /, σ t =,. We have S n n = n σ σ n, Faculty of Industral Engneerng and Management, Technon Israel Insttute of Technology, Technon Cty, Hafa 3000, Israel, nemrovs@e.technon.ac.l

2 so that all we can hope for s a relaton of the type Eξ t } = 0 t & (3) & ξ t } are ndependent Prob S N > tθ } σ σ N O() exp O()t } (5) wth dependng on (E, ) factor Θ whch, as our example clearly shows, can be as large as dm E. Our major goal s to demonstrate that (5) ndeed holds true, provded that Θ s an upper bound on the constant of regularty κ(e, ) of (E, ), wth the latter noton defned as follows: Defnton. Let (E, ) be a Banach space, and let κ. () Space (E, ) s called κ-smooth, f the functon p(x) = x s contnuously dfferentable and x, y E : p(x + y) p(x) + Dp(x)[y] + κp(y). (6) () Space (E, ) (and the norm on E) s called κ-regular, f there exsts κ + [, κ] and a norm + on E such that (E, + ) s κ + -smooth and + s κ/κ + -compatble wth, that s, x E : x x + κ κ + x. (7) The constant κ(e, ) of regularty of E, s the nfnum of those κ for whch (E, ) s κ-regular. In the sequel we. Provde a number of nterestng examples of normed spaces wth nearly dmenson-ndependent constants of regularty. Specfcally, we demonstrate n Secton. that f p, then the norm X p on the space of m n matrces X: X p = σ(x) p [σ(x) s the vector of sngular values of X] s κ-regular wth κ = mn[p +, ( ln(mn[m, n]) + ) exp}].. Demonstrate (Secton 3) that (5) ndeed s true, provded that Θ κ(e, ). In partcular, (!!) If ξ t are ndependent random m n matrces wth zero mean such that E exp ξ t σt } } exp}, where X = σ(x) s the usual matrx norm of X, then N Prob ξ t t } ln(mn[m +, n + ]) σ σ N O() exp O()t }. Note also that every norm on R n s n-compatble wth Eucldean norm, and the latter s, of course, -smooth. Thus, every norm on R n s n-regular; n other words, the (R n, )-example, where the factor Θ = n n (5) s a must, s exactly the worst case. Major part of the results to follow were announced n [3]. Regular Banach spaces. Basc examples Example. Let p. The space (R n, p ) wth n 3 s κ p (n)-regular wth κ p (n) = mn (ρ )n ρ p mn[p, ln(n)] (8) ρ p ρ<

3 Proof. Let ρ <. We clam that n ths case the space (R n, ρ ) s (ρ )-smooth. Indeed, the functon p(x) = ρ s convex, contnuously dfferentable everywhere and twce contnuously dfferentable outsde of the orgn; for such a functon, (6) holds true f and only f D p(x)[h, h] κ + p(h) (x, h E, x 0); (9) snce p( ) s homogeneous of degree, the valdty of (9) for all x, h s equvalent to the valdty of the relaton for all h and all x normalzed by the requrement p(x) =. Gven such an x and h and assumng ρ >, we have ( ) Dp(x)[h] = x ρ ρ x ρ sgn(x )h D p(x)[h, h] = ( ) ( ) ρ x ρ ρ ( ) x ρ sgn(x )h }} 0 + ( x ρ ) ρ (ρ ) x ρ h (ρ ) }} = ( (ρ ) ( x ρ ) ρ ρ ) ρ ρ = (ρ ) h ρ = (ρ )p(h) ( ( h ) ρ ) ρ x ρ h as requred n (9) when κ + = ρ. In the case of ρ = relaton (9) wth κ + = ρ = s evdent. Now, when ρ [, p] and x R n, one has x ρ/ x p [, n ρ p ], so that (R n, p ) s κ-regular wth κ = (ρ )n ρ p, and (8) follows. Example. Let p. The norm X p = σ(x) p on the space R m n of m n real matrces, where σ(x) s the vector of sngular values of X, s κ p (m, n)-regular, wth κ p (m, n) = mn max[, ρ ](mn(m, n)) ρ p mn [max[, p ], ( ln(mn[m, n] + ) ) exp}]. ρ< ρ p (0) Proof. 0. We start wth the followng Lemma Let ρ. Then the space S n of symmetrc n n matrces wth the norm X ρ s κ-smooth wth κ = max[, ρ ]. () Proof. The statement s evdent when ρ = ; thus, from now on we assume that ρ >. A. We start wth the followng fact whch s mportant by ts own rght (for the proof, see Appendx): Proposton. Let be an open nterval on the axs, and f be a C functon on such that for certan θ ±, µ ± R one has (a < b, a, b ) : θ f (a) + f (b) + µ f (b) f (a) b a θ + f (a) + f (b) + µ + () Let, further, X n ( ) be the set of all n n symmetrc matrces wth egenvalues belongng to. The functon F (X) = Tr(f(X)) : X n ( ) R s C, and for every X X n ( ) and every H S n one has θ Tr(Hf (X)H) + µ Tr(H ) D F (X)[H, H] θ + Tr(Hf (X)H) + µ + Tr(H ). (3) 3

4 B. [ Let us ] apply Proposton. to = R, f(t) = t ρ wth θ = µ = 0, µ + = 0 and θ + = max ρ, (ths choce, as t s mmedately seen, satsfes ()). By Proposton, the functon F (X) = X ρ ρ on S n s twce contnuously dfferentable, and [ ] X, H : 0 D F (X)[H, H] θ + Tr(f (x)h ), θ + = max ρ,. (4) It follows that the functon p(x) = X ρ = (F (X)) ρ s contnuously dfferentable everywhere and twce contnuously dfferentable outsde of the orgn. For X 0 we have Dp(X)[H] = ρ (F (X)) ρ DF (X)[H], whence [ ] X 0 D p(x)[h, H] = ρ ρ (F (X)) ρ (DF (X)[H]) + ρ (F (X)) ρ D F (X)[H, H] }} (5) <0 ρ (F (X)) ρ θ + Tr(f (x)h ). Settng Z = ρ(ρ ) (F (X)) ρ f (X), p = ρ ρ, t s mmedately seen that Z p =. From (5) we have D p(x)[h, H] Θ + (ρ )Tr(ZH ) θ + (ρ ) Z p H p p = θ +(ρ ) H ρ = θ +(ρ ) H ρ. (6) Now, f X, Y S n are such that the segment [X; X + Y ] does not contan the orgn, then γ (0, ) : p(x + Y ) p(x) + Dp(X)[Y ] + D p(x + γy )[Y, Y ], and (6) mples that for the outlned X, Y one has p(x + Y ) p(x) + Dp(X)[Y ] + θ + (ρ )p(y ). Snce p s C, the resultng nequalty, by contnuty, s vald for all X, Y. 0. Now we can complete the justfcaton [ of Example ].. W.l.o.g. we may assume that m n. X Gven an m n matrx X, let S(X) = S m+n. One clearly has X T σ(x) ρ = X ρ = /ρ S(X) ρ, whence, by Lemma and due to the fact that the mappng X S(X) : R m n S m+n s lnear, the norm ρ, treated as a norm on R m n, s max[, ρ ]-smooth whenever ρ. Snce σ(x) R m for X R m n, for every ρ [, ) such that ρ p one has Thus, the space (R m n, p ) s κ-regular wth and we arrve at (0). κ =. Calculus of regular spaces We start wth the followng well-known fact: X p X ρ m ρ p X p. mn max[, ρ ]m ρ p, ρ< ρ p Proposton. Let E be fnte-dmensonal, be a norm on E, E be the space dual to E. be the norm on E dual to, ξ, x be the value of a lnear form x E on a vector x E. Let also f(x) = x : E R and f (ξ) = ξ : E R. The followng 6 propertes are equvalent to each other: 4

5 () (E, ) s κ-smooth; () f(x) = f (x)} s a sngleton for every x, and f (x) f (y), x y κ x y x, y E; (7) () f s contnuously dfferentable, and f ( ) s Lpschtz contnuous wth constant κ: f (x) f (y) κ x y x, y E; (8) (v) One has (v) One has (ξ, η E, x f (ξ), y f (η)) : ξ η, x y κ ξ η ; (ξ, η E, x f (ξ), y f (η)) : x y κ ξ η ; (v) One has (ξ, η E, x f (ξ)) : f (ξ + η) f (ξ) + η, x + κ η. Proof. () (): We are n the stuaton when f s contnuously dfferentable. Convolvng f( ) wth smooth nonnegatve kernels δ k ( ) wth unt ntegral and support shrnkng to orgn as k, we get a sequence f k ( ) of smooth functons convergng to f( ), along wth frst order dervatves, unformly on compact sets. We have f k (x + y) = f(x z + y)δ(z)dz [f(x z) + f (x z), y + κf(y)]δ(z)dz = f k (x) + f k (x), y + κf(y) From the resultng nequalty combned wth smoothness and convexty of f k t follows that 0 D f k (x)[h, h] κ h x, h E. Thus, f h = d =, then 4D f k (x)[h, d] = D f k (x)[h+d, h+d] D f k (x)[h d, h d] κ h+d 4κ, whence D f k (x)[h, d] κ whenever h = d =, or, whch s the same by homogenety, Consequently, f k(y) f k(x), h = 0 D f k (x)[h, d] κ h d x, h, d. D f k (x + t(y x))[y x, h]dt 0 κ y x h dt κ y x h, whence, takng maxmum over h wth h =, f k (y) f k (x) κ y x. As k, f k (x) converge to f (x), and we conclude that f ( ) possesses the requred Lpschtz contnuty, Q.E.D. () (): evdent () (): A convex functon on R n wth a sngleton dfferental at every pont clearly s contnuously dfferentable, so that n the case of () f s contnuously dfferentable. Besdes ths, n the case of () we have f(x + y) = f(x) + f (x), y + f (x + ty) f (x), y dt f(x) + f (x), y + 0 κt y dt = f(x) + f (x), y + κf(y), 0 whch mmedately mples (6) (recall that = f( )), Q.E.D. () (v): The functons f( ), f ( ) are the Legendre transforms of each other, so that x f (ξ) f and only f ξ f(x). Now let () be the case, and let ξ, η E and x f (ξ), y f (η). Then ξ = f (x), 5

6 η = f (y) and therefore, due to (), ξ η κ x y, so that (v) takes place. Vce versa, let (v) take place, and let x, y E, ξ f(x), η f(y). Then x f (ξ), y f (y), and therefore (v) says that ξ η κ x y. We conclude that f x = y, then ξ = η, that s, f(x) always s a sngleton, meanng that f s contnuously dfferentable, and that the nequalty n () takes place, that s, () holds true, Q.E.D. (v) (): Let (v) take place. If there exsts x E such that f(x) s not a sngleton, then, choosng ξ, η f(x) wth ξ η, we would have x f (ξ), x f (η), whence by (v) we should have ξ η, x x κ ξ η, whch s mpossble. Thus, f(x) s a sngleton for every x, so that f s contnuously dfferentable. Besdes ths, wth x, y E and ξ = f (x), η = f (y) we have x f (ξ), y f (η), whence, by (v), ξ η, x y κ ξ η. Snce ξ η, x y ξ η x y, we get ξ η x y κ ξ η, whence ξ η = f (x) f (y) κ x y, and thus () takes place. Now let () take place, and let us prove that (v) takes place as well, or, whch s the same n the case of (), that f (x) f (y), x y κ f (x) f (y). Settng g(u) = f(u) f (y), u y, we get a contnuously dfferentable convex functon on E such that g (x) g (y) κ x y and g (y) = 0. Due to these relatons, g(y + h) g(y) + κ h for all h. Now let e E be such that g (x), e = g (x) and e =. Due to g (u) g (v) κ u v, we have g(x g (x) κ e) g(x) g (x), g (x) κ e + κ g (x) κ e = g(x) g (x) κ + g (x) κ = g(x) g (x) κ. On the other hand, g attans ts global mnmum at y, so that g(x) g (x) κ whence g(x g (x) κ g(y) + κ h g(y + h) g(x) + g (x), y + h x g(y) + g (x) κ g (x), x y g (x) κ + g (x), h κ h. e) g(y). We now have + g (x), y + h x, Ths nequalty s vald for all h; settng h = g (x) κ e, the rght hand sde becomes g (x) κ. Thus, f (x) f (y), x y = g (x), x y g (x) κ = f (x) f (y) κ, Q.E.D. (v) (v): Let (v) take place, let ξ, η E and x f (ξ). Settng ξ t = ξ+tη, φ(t) = f (ξ t ), 0 t, we get an absolutely contnuous functon on [0, ] wth the dervatve whch s almost everywhere gven by φ (t) = η, x t, wth x t f (ξ t ). We have f (ξ + η) = φ() = φ(0) + 0 φ (t)dt = φ(0) + 0 η, x t dt = φ(0) + 0 [ η, x + η, x t x ]dt = φ(0) + η, x + 0 t (ξ + tη) ξ, x t x dt φ(0) + η, x + 0 t κ [ξ + tη] ξ dt = φ(0) + η, x + κ η = f (ξ) + η, x + κ η, where the nequalty s gven by (v). We end up wth the nequalty requred n (v), Q.E.D. (v) (): Let (v) be the case, let x E and ξ f(x), so that x f (ξ). We have [ ] f(x + y) = max η E [ ξ + η, x + y f (ξ + η)] max η E [ ] ξ + η, x + y f (ξ) η, x κ η = max η E ξ, x + y + η, y [ f (ξ) κ η ] = ξ, x f (ξ) + ξ, y + max η η, y }} κ η = f(x) + ξ, y + κ y. f(x) Ths relaton along wth the relaton f(x + y) f(x) + ξ, y mples that ξ s the Frechet dervatve of f at x, whence f s convex and dfferentable, and thus contnuously dfferentable functon on E whch satsfes the nequalty f(x + y) f(x) + f (x), y + κ y, Q.E.D. We have proved that () () () (v) (v) and (v) (v) (), meanng that all 6 propertes n queston are equvalent to each other. 6

7 Proposton.3 () Let p [, ], and let (E, ) be fnte-dmensonal κ-smooth spaces, =,..., m >. The space E = E... E m equpped wth the norm ( n ) /p (x,..., x m ) = x p (the rght hand sde s max x when p = ) s κ + -regular wth = κ + = mn ρ p [κ + ρ ]m ρ p mn[κ + p, [κ + ln(m) ] exp}]. (9) () Let be κ-smooth norms on E. Then the norm x = m x = s mκ-regular on E. () Let (E, ) be κ-regular, α > 0, and let L be a lnear subspace n E. Then (L, α ) s κ-regular. Proof. (): To prove (), let p (x ) = x. A. Let ρ [, ) be such that ρ p, and let r = ρ/. Our local goal s to prove Lemma The norm on E = E... E m defned as (x,..., x m ) = ( x,..., x m m ) ρ s κ + -smooth, wth κ + = κ + ρ (0) Proof. We have p(x,..., x m ) ( x,..., x m m ) ρ = (p (x ),..., p m (x m )) r. From ths observaton t mmedately follows that p( ) s contnuously dfferentable. Indeed, ρ, whence r, so that the functon y r s contnuously dfferentable everywhere on R m + except for the orgn; the functons p (x ) are contnuously dfferentable by assumpton. Consequently, p(x) s contnuously dfferentable everywhere on E = E... E m, except, perhaps, the orgn; the fact that p s contnuous at the orgn s evdent. Invokng Proposton., n order to prove Lemma t suffces to verfy that p (x) p (y) κ + x y () for all x, y. Snce p s contnuous, t suffces to prove ths relaton for a dense n E E set of pars x, y, for example, those for whch all blocks x E n x are nonzero. Wth such x, the segment [x, y] contans fntely many ponts u such that at least one of the blocks u s zero; these ponts splt [x, y] nto fntely many consecutve segments, and t suffces to prove that p (x ) p (y ) κ + x y when x, y are endponts of such a segment. Snce p s contnuous, to prove the latter statement s the same as to prove smlar statement for the case when x, y are nteror ponts of the segment. The bottom lne s as follows: n order to prove () for all pars x, y, t suffces to prove the same statement for those pars x, y for whch every segment [x, y ] does not pass through the orgn of the correspondng E. Let x, y be such that [x, y ] does not pass through the orgn of E, =,..., m. Same as n the tem () () of the proof of Proposton., for every there exsts a sequence of C convex functons p t ( ) > 0} on E convergng to p ( ) along wth frst order dervatves unformly on compact sets and such that D p t (u )[h, h ] κ h (u, h E ). () Functons p t (u) = (p t (u ),..., p t m(u m )) r clearly are convex, C (recall that p t ( ) > 0) and converge to p( ), along wth ther frst order dervatves, unformly on compact sets. It follows that p (y) p (x), h = lm t 0 D p t (x + t(y x))[y x, h]dt. (3) 7

8 Settng F (y,..., y m ) = y r ym, r y 0, we have p t (u) = F r (p t (u ),..., p t m(u m )). Now let u [x, y], and let v E. We have ( ) Dp t (u)[v] = r F r (p t (u ),..., p t m(u m )) r(p t (u )) r Dp t (u )[v ] ( ) ( ) D p t (u)[v, v] = r r F r (p t (u ),..., p t m(u m )) r(p t (u )) r Dp t (u )[v ] }} 0 +F r (p t (u ),..., p t m(u m )) [ (r )(p t (u )) r (Dp t (u )[v ]) + (p t (u )) r D p t (u )[v, v ] ] F r (p t (u ),..., p t m(u m )) [ (r )(p t (u )) r (Dp t (u )[v ]) + κ(p t (u )) r p (v ) ] whence 0 D p t (u)[v, v] F r (p t (u ),..., p t m(u m )) [ (r )(p t (u )) r (Dp t (u )[v ]) + κ(p t (u )) r p (v ) ]. (4) Takng nto account that p ( ) are bounded away from zero on [x, y] and that p t ( ) converge, along wth frst order dervatves, to p ( ) unformly on compact sets as t, the rght hand sde n bound (4) converges, as t, unformly n u [x, y] and v, v, to ( Ψ(u, v) = u ρ ) ρ [ (r ) u ρ 4 By evdent reasons, Dp (u )[v ] u v, whence ] (Dp (u )[v ]) + κ u ρ v. Ψ(u, v) When ρ >, we have ( ) u ρ ρ [ ( = [ρ + κ 4] u ρ }} κ + 4(r ) u ρ ) ρ ] v + κ u ρ v u ρ v (5) u ρ v = ( ( u ρ ) ρ ρ ( u ρ ) ρ ρ ) ρ ρ ( v ρ ( ( v ) ρ ) ρ, ) ρ and (5) mples that Ψ(u, v) κ + v. Ths nequalty clearly s vald for ρ = as well. Recallng the orgn of Ψ(, ), we conclude that for every ɛ > 0 there exsts t ɛ such that t t ɛ, u [x, y], v 0 D p t (u)[v, v] κ + v + ɛ. The resultng nequalty va the same reasonng as n the proof of tem () () of Proposton. mples that t t ɛ, u [x, y] D p t (u)[v, w] (κ + + ɛ) v w v, w. In vew of ths bound and (3), we conclude that p (y) p (x), h (κ + + ɛ) y x h for all h, whence p (y) p (x) (κ + + ɛ) y x. Snce ɛ > 0 s arbtrary, we arrve at (). 8

9 B. When ρ p, we have ( x,..., x m m ) p ( x,..., x m m ) ρ m ρ p ( x,..., x m m ) p, whch combnes wth Lemma to mply that the norm n () s κ-regular wth κ = [ρ + κ ]m ρ p, for every ρ [, p], and () follows. (): To prove (), consder the norm (x,..., x m ) = m / x xm m on E E... E. As t s mmedately seen, ths norm s κ-smooth. If, further, (x,..., x m ) = x, then x x m x x E... E, whence s mκ-regular. The norm n () s nothng but the restrcton of on the mage of E under the embeddng x (x,..., x) of E nto E... E, and t remans to use (). (): Evdent. To proceed, we need the followng fact: Lemma 3 Let (E, ) be a fnte-dmensonal κ-regular space. Then there exsts κ-smooth norm + on E such that (x E) : x x + x. (6) Proof. By defnton, there exsts κ + [, κ] and a norm π( ) on E whch s κ + -smooth and such that (x E) : x π (x) µ x, µ = κ/κ +, or, whch s the same, ξ E : π (ξ) ξ µ π (ξ), (7) where E s the space dual to E and π, are the norms on E conjugate to π,, respectvely. In the case of µ, let us take + π( ), thus gettng a κ + -smooth (and thus κ-smooth as well) norm on E satsfyng (6). Now let µ >, so that γ = /(µ ) (0, ). Let us set q (ξ) = γπ (ξ) + ( γ) ξ, so that q ( ) s a norm on E. We have Further, by Proposton. we have ξ E : q (ξ) ξ γµ + γ q (ξ) = q (ξ). (8) (ξ, η E, x π (ξ)) : π (ξ + η) π (ξ) + η, x + κ + π (η), whence, due to ξ + η ξ + η, y for all ξ, η and every y from the subdfferental D(ξ) of at the pont ξ, (ξ, η E, x π (ξ), y D(ξ)) : q (ξ +η) q (ξ)+ η, x+y + γ κ + π (η) q (ξ)+ η, x+y + γ κ + q (η) (note that π ( ) q ( ) by (7)). Snce π (ξ) + D(ξ) = q (ξ) and γ κ + = (µ )κ + κ, we get (ξ, η E, z q (ξ)) : q (ξ + η) q (ξ) + η, z + κ q (η). By the same Proposton., t follows that the norm + q( ) on E such that q ( ) s the conjugate of q( ) s κ-smooth. At the same tme, (8) mples (6). Lemma 3 allows to prove the followng modfcaton of Proposton.3.(,): 9

10 Proposton.4 () Let p [, ], and let (E, ) be fnte-dmensonal κ-regular spaces, =,..., m >. The space E = E... E m equpped wth the norm ( n (x,..., x m ) = x p = ) /p (the rght hand sde s max x when p = ) s κ ++ -regular wth κ ++ = mn ρ p [κ + ρ ]m ρ p mn[κ + p, [κ + ln(m) ] exp}]. (9) () Let be κ-regular norms on a fnte-dmensonal space E. Then the norm s mκ-regular on E. x = m x = Proof s readly gven by Lemma 3 combned wth the correspondng tems of Proposton.3. E.g., to prove (), note that by Lemma 3 we can fnd κ-smooth norms q ( ) on E such that q (x ) x q (x ) for every and all x E. Applyng Proposton.3.() to the spaces (E, q ( )), we get that the ( /p norm q(x,..., x m ) = q p )) (x on E... E m s κ + -regular wth κ + gven by (9). Takng nto = account the evdent relaton q (x,..., x m ) (x,..., x m ) q (x,..., x m ) and recallng the defnton of regularty, we conclude that s κ ++ -regular, as requred. 3 Sums of random vectors n regular spaces In ths Secton, we consder the stuaton as follows. We are gven a fnte-dmensonal κ-regular space (E, ), a Polsh space Ω wth Borel probablty measure µ and a sequence F 0 =, Ω} F F... of σ-sub-algebras of the Borel σ-algebra of Ω. We denote by E t, t =,,... the condtonal expectaton w.r.t. F t, and by E E 0 the expectaton w.r.t. µ. The queston we are nterested n as follows: Gven a martngale-dfference ξ t } wth values n E, so that ξ t s a F t -measurable random vector n E such that E t ξ t } 0, t =,,..., what can we say about typcal norms of the assocated sums S n = ξ t. In the sequel, we denote by + a κ + -smooth norm on E whch s (κ/κ + )-compatble wth : x x + (κ/κ + ) x x E, (30) and set p(x) = x +. 0

11 3. Bounds on second moments of S n Our frst observaton s nearly tautologcal: Proposton 3. Assume that E-valued martngale-dfference ξ = ξ t } s square summable: Then Proof. Snce + s κ + -smooth, we have E ξ t } σ t <. E S n } κ σt. (3) p(s t+ ) p(s t ) + Dp(S t )[ξ t+ ] + κ + p(ξ t+ ) whence, takng expectatons and makng use of the fact that ξ s martngale-dfference, E p(s t+ )} E p(s t )} + κ + E p(ξ t+ )} E p(s t )} + κe ξ t+ } (we have used the rght nequalty n (30)). From ths recurrent nequalty we get E S n } n + κ E ξ t } κ σt. The left hand sde n ths nequalty, by (30), s E S n }, and (3) follows. 3. Large devatons for S n, I Theorem 3. Let E-valued martngale-dfference ξ = ξ t } and reals σ t > 0 be such that Then for all n and Ω 0 one has n Prob ξ t > 5Ω κ n E t exp ξt σ t } } exp}, t =,,... (3) σ t 3 exp Ω }. (33) Proof. 0. Let us set σ t = κ / σ t κ / + and η t = ξ t σ t, so that η t + (κκ + ) η t = (κκ + ) ξ t (κ κ + σt ) = ξ t σt, whence ξ t = σ t η t, E t ηt +} exp}, Et η t } = 0, S n = Observe that by the moment nequalty σ t η t. (34) 0 τ E t expτ ηt +} } expτ}. (35) Further, snce expx} xl l! for all x 0, t follows from (35) that E t ηt l + } el!, l = 0,,... (36) 0. Let ( n ) / ω n = σ t. (37)

12 Let us prove by nducton n n that f ɛ 3 + κ +, (38) then (P n ) : 0 τ ωn E expɛ τp(s n )} } expωnτ}, Base n = : evdent n vew of (35). Step n n + : Assume that (P n ) holds true. Denotng by s the norm on E conjugate to + and by φ, x the value of a lnear functonal φ E at a vector x E, we have: E expɛ τp(s n+ )} } E expɛ τ [ ] } p(s n ) + σ n+ p (S n ), η n+ + κ + σ n+ η n+ + } E expɛ τp(s n )} expɛ τ [ σ ] } n+ p (S n ), η n+ + κ + σ n+ η n+ + } = E expɛ τp(s n )}E n expɛ τ [ σ (39) ] }} n+ p (S n ), η n+ + κ + σ n+ η n+ + } Now let 0 τ ωn+. We have E n expɛ τ [ σ ] } n+ p (S n ), η n+ + κ + σ n+ η n+ + } } = + ɛ τe n σn+ p (S n), η n+ + κ + σ n+ η n+ + + l= } [ɛ + ɛ τe n κ+ σ n+ η n+ + + l! E n τ [ σ ]] } n+ p (S n ), η n+ + κ + σ n+ η n+ l + l= } [ + ɛ τe n κ+ σ n+ η n+ + + l! E n (ɛ ( τ) l p (S n ) l σl n+ η n+ l + + κ + σ n+ η n+ + l= } (ɛ ) } = + ɛ τe n κ+ σ n+ η n+ + + l! En τκ + σ n+ η n+ l + + l= l= E n expκ+ ɛ τ σ n+ η n+ }} } + + l! (ɛ τ) l E n (4p(S n )) l σ n+ l η n+ l + l= Further, expκ + ɛ τ σ n+ } + l= expκ + ɛ τ σ n+ } + l= } l! (ɛ τ) l E n (4p(S n )) l σ n+ l η n+ l + ( ) l l! (ɛ τ σ n+ ) l (4p(S n )) (el!) [ [ }} ] ( 7ɛ τ σ n+ p(sn) expκ + ɛ τ σ n+ } + exp } 3 0. We need the followng l= θ n+ expκ + ɛ τ σ n+} + exp } l! l= ] l ( expκ + ɛ τ σ n+} + exp } l= ( expκ + ɛ τ σ n+} + l= [ɛ l! En τ [ σ ]] } n+ p (S n), η n+ + κ + σ n+ η n+ l + ) 3 l ( (θn+p(s n)) l l! (θ n+ p(s n )) l l! (θ n+ p(s n )) l l! ) ) l ]} } l! (ɛ τ) l E n p (S n) l σl n+ η n+ l + [snce p (u) 4p(u)] [we have used (35) combned wth κ + ɛ τ σ n+ ] ) 3 l ) ( l= expκ + ɛ τ σ n+} + (expθ n+ p(s n )} θ n+ p(s n )) Lemma 4 Let 0 τ and θ = (7ɛ τ σ n+ ). Then 3 l ) [we have used (36)] expκ + ɛ τ σ n+} + (expθp(s n )} θp(s n )) exp κ + ɛ τ σ n+ + 75ɛ 4 τ σ n+p(s n ) } (4) (40) (4)

13 Proof. Observe, frst, that for x 0 one has and, second, that for x, y 0 one has Therefore expx} x x expx}, (43) expx}y expx + y}. (44) expκ + ɛ τ σ n+} + (expθp(s n )} θp(s n )) [ ] expκ + ɛ τ σ n+} + (θp(sn)) expθp(s n )} ( ) = expκ + ɛ τ σ n+} θp(sn ) + exp θp(s } n) }}}} x y expκ + ɛ τ σ n+} + exp θp(s n) + θp(sn) } [by (43)] [by (44)] expκ + ɛ τ σ n+} + expκ + ɛ τ σ n+} [ exp 3 θp(s n) } ] = exp κ + ɛ τ σ n+ + 3 θp(s n) } 4 0. Combnng (39), (40), (4) and (4), we arrve at the relaton 0 τ ωn+ E expɛ τp(s n+ )} } E exp [ ɛ τ + 75ɛ 4 τ σ }} n+] p(sn ) + κ + ɛ τ σ n+ (45). Now let 0 τ ω n+ and let µ = τ + 75ɛ τ σ n+. Then, settng ρ = σ n+ ω, n µ + 75ɛ σ n+ ωn+ ωn+ 4 +ρ(+75ɛ ) +ρ+ρ < ωn = ω n (snce 75ɛ ). Consequently, by (P n ) one has = + 75ɛ σ n+ = ωn + σ (ω n+ n + σ n+ ) ωn [ ] +ρ + 75ɛ ρ (+ρ) E exp[ɛ τ + 75ɛ 4 τ σ n+]p(s n )} } = E expɛ µp(s n )} } expµω n}, and (45) mples that 0 τ E expɛ τp(s ωn+ n+ )} } exp } κ + ɛ τ σ n+ + µωn = exp τ [ κ + ɛ σ ]} n+ + [ + 75ɛ τ σ n+]ω n = expτωn+χ}, (46) where χ = κ +ɛ ρ + ρ ɛ (τωn+) + ρ ρ +ρ Takng nto account that κ + ɛ /, τω n+ and 75ɛ /, we get and (46) mples (P n+ ). χ ρ + ρ + + ρ + ρ,. 3

14 5 0. Now we are ready to prove (33). Let us set ɛ = 5 κ +, so that (38) holds true. Then for Ω > 0 one has Prob S n > 5Ω } κ Prob S n + > 5Ω κ } κ + /κ (we have used (P n ) wth τ = ω n ). σt σ t [snce + ] = Prob } S n + > Ωɛ ω n < E expɛ ωn S n +} } exp Ω } exp Ω } Refnements n the case of bounded ξ t. In ths case, constants n Theorem 3. can be mproved. Theorem 3. Let ξ = ξ t } be a E-valued martngale-dfference and σ t > 0 be reals such that ξ t σ t. Then Prob S n > Ω κ n σt Ω exp }. (47) 4.4 The proof s completely smlar to the one of Theorem 3.. Note that a result smlar to the one of Theorem 3. can be easly derved from Talagrand Inequalty. Here s ths nequalty (n slghtly extended form presented n []): Theorem 3.3 Let (E t, Et ) be fnte-dmensonal normed spaces, t =,..., n, F be the drect product of E,..., E n equpped wth the norm (x,..., x n ) F = x t E t, µ t be Borel probablty dstrbutons on the unt balls of E t and µ be the product of these dstrbutons. Gven a closed convex set A F, let dst(x, A) = mn y A x y F. Then Talagrand Inequalty mmedately mples the followng result: E µ exp } 4 dst (x, A)} µ(a). (48) Theorem 3.4 Let (E, ) be κ-regular space, and let ξ,..., ξ n be ndependent random vectors n E wth zero means and σ t be reals such that ξ t σ t, t =,..., n. Then Ω κ Prob S n > Ω n σt exp Ω }. (49) 3 Proof. Let F be the drect product of n copes of E equpped wth the norm (x,..., x n ) F = x t, let ζ t = (σ t ) ξ t, and let Q be the set of all x = (x,..., x n ) F such that S(x) n σ t x t E satsfes S(x). Note that Q s a closed convex set n F, and that S(x) r f and only f x rq. Let Θ = σt. Our frst observaton s that Q contans F -ball of the radus ρ = (Θ) centered at the orgn. Indeed, f (x,..., x n ) F ρ, then S(x) n σ t x t Θ x t Θρ =. 4

15 Next observe that f ζ = (ζ,..., ζ n ), then for every γ > 0 one has Indeed, we have S(ζ) = n Prob S(ζ) > γ} Prob ζ γq} κθ γ. (50) σ t ζ t = n ξ t ; by Proposton 3., we have E S(ζ) } κθ, and (50) follows from the Tschebyshev nequalty. Let us fx γ > κθ and set A = γq; note that A s closed convex set n F symmetrc w.r.t. the orgn and contanng the centered at the orgn F -ball of radus γρ; besdes ths, by (50). Observe that Prob ζ A} κθ γ > 0 (5) s >, x F \(sa) dst(x, A) > (s )γρ. (5) Indeed, for s, x from the premse of ths mplcaton, the set B = x + (s )A does not ntersect A; snce A contans the F -ball of radus γρ centered at the orgn, B contans F -ball of the radus (s )γρ centered at x. Snce B A =, the concluson n (5) follows. Applyng (48) to the dstrbuton of ζ, we get E exp 4 dst (ζ, A)} } (we have used (5)). In vew of (5), ths bound mples Probζ A} κθ γ ρ (s ) s > Prob ζ sa = sγq} κθ γ exp γ } (53) 4 Snce ζ αq f and only f n ξ t > α, we arrve at (s >, γ > κθ) : Prob } (s ) ξ t > γs κθ γ exp γ 8Θ } (we have substtuted the value of ρ). Gven Ω κ and settng γ = κθ s = Ω/ κ, we arrve at (49). 3.3 Large devatons for S n, II Theorem 3.5 Let α (0, ], and let E-valued martngale-dfference ξ = ξ t } and reals σ t > 0 be such that E t exp ξt α σt α } exp}, t =,,... (54) Then for all n and Ω 0 one has n Prob ξ t > Ω κ n σ t C α exp C α Ω α }, (55) where C α depends solely on α (0, ] and s contnuous n α > 0. In partcular, for approprately chosen c α > 0 dependng solely on α (0, ] and contnuous n α, one has for all n : n ξ t α E exp n (c α κ σ t ) } exp}. (56) α 5

16 Proof. 0. Let ρ (/α) /α. Let us set Observe that whence, due to E t ξ t } = 0, also 0. We have η t = χ ξt >σ tρ}ξ t, ζ t = η t E t η t } }} δ t, ω t = ξ t ζ t. E t ζ t } = 0, (57) E t ω t } = 0. (58) δ t E t χ ξt >σ tρ} ξ t } = E t exp ξt α σt α [ }σ t exp ξt α σt α ]} ] }[ ξ t /σ t ]χ ξt >σ tρ} σ t [max [z z ρ exp zα }] E t exp ξt α σt α } } σ t exp} max [z z ρ exp zα }] = σ t exp}ρ exp ρ α } [due to ρ α /α ] (59) Consequently, Settng ω t = ξ t [η t δ t ] ξ t η t + δ t σ t exp}ρ exp ρ α } + ξ t χ ξt σ t ρ}. σ t = σ t ρ α, we have from (60): }} E t exp ωt σ t E t exp 0.5 [ ] }} σ t exp}ρ exp ρ α } + ξ t χ ξt σ t ρ} σ t ρ α [ E t exp 0.5 exp} exp ρ α }ρ α + [ ξ t σt ]}} [ ]ρ α χ ξt σ t ρ} E t exp 0.5 exp} exp ρ α }ρ α + [ ξ t α σt α ]}} [ ]χ ξt σ t ρ} E t max exp exp} exp ρ α }ρ α }, exp [ ξ t α σt α ] }]} exp exp} exp ρ α }ρ α } + exp} exp0.5 exp}} + exp} exp}. (60) It follows that wth one has Thus, σ t = ρ α σt (6) E t exp ωt σ t } } exp}. (6) 3 0. We have E t η t } E t ηt }, whence E t ζt } E t 4 ηt } } = 4E t ξt χ ξt >σ t ρ} = 4σt E t exp ξt α σt α } [ ξ t σt exp ξ t α σ α 4σt E t exp ξ t α σ α [ } max z exp z α } ]} = 4σ t exp}ρ exp ρ α } t z ρ E t ζt σt } 4 exp}ρ exp ρ α }. t }χ ξt >σ t ρ} ]} [snce ρ (/α) /α ] Thus, ζ t s F t -measurable random vector (by ts orgn) such that E t ζ t } = 0, E t ζt σt } 4 exp}ρ exp ρ α }, (63) 6

17 (see (57)). Besdes ths, ω t s F t -measurable random vector (by ts orgn) such that E t ω t } = 0, E t exp ωt σ t } } exp}, (64) (see (58), (6)) Applyng Theorem 3. to random vectors ω,..., ω n and takng nto account (64), we get n Prob ω t ρ α/ κ n σ t C exp C ρ α } (65) (from now on, C are approprate postve absolute constants), whence, recallng (6), n Prob ω t ρ κ n σt C 3 exp C 4 ρ α }. (66) Further, by (63) and Proposton 3. as appled to random vectors ζ,..., ζ n, we have } ( n ) E ζ t C 5 κ ρ exp ρ α }, whence by Tchebyshev nequalty n Prob ζ t ρ κ n σ t σ t C 6 exp C 7 ρ α }. Combnng ths nequalty wth (65) and takng nto account that ξ t = ω t + ζ t, we conclude that n Prob ξ t [ + ]ρ κ n σt C 7 exp C 8 ρ α } whenever ρ (/α) /α, and (55) follows. (56) s an mmedate corollary of (55). Indeed, let us fx n, and let D = κ n σ t. For c > 0, consder the random varable θ c = n ξ t α /(cd) α. By (55), for t > 0 we have ψ(t) Probθ c > t} = Prob n ξ t > ct /α D} C α exp Cα c α t}; settng c = (C α ) /α, we get ψ(t) C α exp t}, whence Eexpθ c }} = Thus, E exp } n ξt α C α D } α (C α ln( + C α )) /α. 0 expt}dψ(t) = + 4 Refnements n Gaussan case 0 expt}ψ(t)dt + C α 0 exp t}dt = + C α. + C α, whence, by Moment Inequalty, (56) holds true wth c α = We are about to refne the above results for the case when ξ = ξ t } s a sequence of ndependent Gaussan random vectors wth zero mean n a fnte-dmensonal normed space (E, ). 7

18 4. The basc fact We start wth the followng fact whch seems to be mportant by ts own rght. Let Φ(t) = exp s φ(r) /}ds, φ(r) : exp s /}ds = r. π π t Proposton 4. Let η N (0, I k ), and let B be a closed convex set n R k such that Then Probη B} θ >. (67) } 0 < α < Probαη B} exp φ (θ) α. (68) Equvalently: for a closed and convex set B and ζ N (0, Σ) one has Prob ζ B} δ < Prob ζ γb} ( δ)γ exp φ } γ >. (69) Proof s based on the followng fact []: (!) For every γ > 0, ɛ 0 and every closed set X R k such that Probη X} γ one has where dst(a, X) = mn x X a x. Prob dst(η, X) > ɛ} Φ(φ(γ) + ɛ) Now let η, ζ be ndependent N (0, I k ) random vectors, and let The vector αη + α ζ s N (0, I k ), so that p(α) = Probαη B}. Probdst(αη + α ζ, B) > t} Φ(φ(θ) + t) (70) by (!). On the other hand, let αη B, and let e = e(η) be a unt vector such that e T [αη] > max x B et x. If ζ s such that α e T ζ > t, then dst(αη + α ζ, B) > t, whence αη B Prob ζ : dst(αη + } α ζ, B) > t Φ(t/ α ), whence for all t 0 such that δ(t) φ(θ) + t t/ α 0 one has p(α)φ(t/ α ) Probdst(αη + α ζ, B) > t} Φ(φ(θ) + t) p(α) = Φ(φ(θ)+t) Φ(t/ α ) = t/ exp (s+δ(t)) /}ds α exp s /}ds t/ α exp s / sδ(t) δ (t)/}ds t/ α exp s /}ds t/ α Settng n the resultng nequalty t = φ(θ)( α ) α, we get p(α) exp φ (θ) α }. exp tδ(t)/ α δ (t)/}. 8

19 4. Gaussan verson of Theorem 3. Theorem 4. Let (E, ) be κ-regular and let ξ, ξ,... be ndependent Gaussan random vectors n E wth zero means. Settng δ t = E ξ t }, one has n Ω 3 Prob ξ t > Ω κ n Proof. Let U be the unt ball of. By Proposton 3., we have E S n } κ δ t δ t }} ω n whence by Tschebyshev nequalty for every β > one has, Prob ω n S n βu } β <. Ω exp }. (7). Applyng (69) (whch s legtmate snce S n s Gaussan wth zero mean), we arrve at Prob ωn S n γβu } β < ( β )γ exp φ } γ >, or, whch s the same, whenever Ω >, one has Prob S n > Ω κ n δt nf exp φ ( β )Ω <β<ω β } (7) Maxmzng the rato φ ( β )/β n β, we get from (7) Ω 3 Prob S n > Ω κ n δ t exp Ω. }. Dscusson. Invokng Proposton 4., t easy to demonstrate that f η s a Gaussan vector wth zero mean n E, δ = E η } and σ = O()δ wth properly chosen absolute constant O(), then E exp η σ } } exp}. On the other hand, f σ s such that the latter nequalty holds true, then, by Jensen s nequalty, expe η σ }} exp}, that s, σ δ. Thus, n the case when ξ t are ndependent Gaussan vectors wth zero means, the probabltes of large devatons Prob S n Ω } κ can be bounded by both Theorem 3. and Theorem 4.. Both bounds are δt of the same type O() exp O()Ω }; however, the absolute constants n the second bound are better than n the frst one. Indeed, snce the quanttes σ t arsng n Theorem 3. should be δ t, bound from Theorem 3. s not better than Prob S n > Ω } κ δt 3 exp Ω 5 }, whle the rght hand sde n the bound gven by Theorem 4. s exp Ω. }. 9

20 5 Extensons to sem-scalar case In ths secton we extend the results for Gaussan case to the stuaton of random sums of the form S n = ζ t f t wth determnstc vectors f t E and ndependent random scalars ξ t whch are symmetrcally dstrbuted on the axs wth lght tal of the dstrbutons. Snce multplyng ξ t by determnstc postve reals and dvdng f t by the same reals does not affect the stuaton, n the sequel we normalze ξ t by the condton 5. Basc results ξ t ξ t, E exp4ξ t } } exp}. (73) We start wth results whch can be vewed as modfcatons of Proposton 4.. Proposton 5. Let ξ t be ndependent and symmetrcally dstrbuted random reals such that Eξ t } σ > 0, t =,..., n (74) and ether () ξ t /, t =,..., n, or () E exp4ξ t } } exp}, t =,..., n, and let ξ = (ξ,..., ξ t ). Let, further, A be a closed convex symmetrc w.r.t. the orgn set n E such that Then for every ϑ > one has n the case of (): ProbS n A} µ > ν Prob ξ ϑa} exp (ϑ ) σ } 8 σ 4 σ 4 +. (75) (76) n the case of (): σ (ϑ ) Probξ ϑa} O() exp O() }, (77) σ(ϑ ) + ln n wth properly chosen postve absolute constants O(). Proof. A. We start wth the followng Lemma 5 Under the premse of () (and thus under the premse of () as well), the set A contans the centered at the orgn -ball of the radus ρ = σ. (78) Proof. Assume, on the contrary to what should be proved, that there exsts a A wth a = ρ. Then there exsts a vector p, p =, such that p T x < p T a ρ for all x A; snce A s symmetrc w.r.t. the orgn, we have max x A pt x < ρ. (79) Consder the random varable ζ = p T ξ, and let θ = p Eξ }, so that θ σ. We have Eζ 4 } = 6 <j p p jeξ }Eξ j } + p 4 Eξ 4 } γ(θ) 3θ + 4 0

21 (snce p 4 Eξ4 } p 8 [Eexp4ξ }} ] p 4 due to t 4 8 [exp4t } ]). Recallng that Probζ ρ} Probζ A} > ν, we get σ4 σ 4 + θ = Eζ } ρ Prob0 ζ ρ} + Eζ 4 } Probζ > ρ} < ρ + ( ν)γ(θ), whence ρ > θ ( ν)γ(θ) φ(θ). (80) Observe that φ (θ) 0 provded that 3θ ν +/4 3θ, whch defntely s the case when ν > 3, as guaranteed by the orgn of ν (note that σ due to Eexp4ζt }} exp}). Consequently, (80) combnes wth θ σ to mply that ρ > φ(σ ) = σ ν 3σ 4 + /4, whence, by (75), ρ > σ /, whch s a contradcton. B. Recall that by Talagrand Inequalty, for a sequence of ndependent random real varables ξ, =,..., n, takng values n [ /, /] and a closed set A n R n one has E } 4 expdst (ξ, Conv(A))} Probξ A}. (8) C. W.l.o.g., let A be a compact set; by Lemma 5, A contans -ball of radus ρ centered at the orgn. Let ϑ > and x ϑa. Consder the norm n whch A s the unt ball; snce x ϑa, the -ball B of the radus ϑ centered at x does not ntersect A. Snce the -ball of the radus ρ(ϑ ), centered at x, s contaned n B, ths ball does not ntersect A as well. Thus, x ϑa dst (x, A) dst (x, Conv(A)) > ρ(ϑ ). (8) D. Assume that () s the case. Combnng (8) wth (8), we arrve at Prob ξ ϑa} µ exp ρ (ϑ ) /4} wth ρ gven by (78), as requred n (76). E. Now assume that () s the case. Let L > 0, let Ξ be the event ξ : ξ L/} and p be the probablty of the event ξ Ξ}. We have p = Prob ξ > L/} E exp4ξ L } } n exp L }. (83) Applyng (8) to the condtonal, by the condton ξ Ξ, dstrbuton of ξ (whch s agan a dstrbuton wth ndependent coordnates), we get E exp dst (ξ,a) 4L } } Ξ } p Prob ξ A µ p Ξ } E exp dst (ξ,a) 4L }χ ξ Ξ ( p) µ p } Probξ Ξ & ξ ϑa} Prob ξ Ξ & dst (ξ,a) 4L ρ (ϑ ) 4L } exp ρ (ϑ ) 4L }E exp dst (ξ,a) 4L }χ ξ Ξ exp ρ (ϑ ) 4L } ( p) Probξ ϑa} p + exp σ (ϑ ) 8L µ p = exp σ (ϑ ) 8L } ( p) µ p } ( p) µ p Assumng σ (ϑ ) 4, let L = ln n + σ(ϑ ). Wth ths L, the resultng bound combnes wth (83) to mply (77).

22 Proposton 5. descrbes a famly of probablty dstrbutons P on R n wth the followng common property: for every closed convex set A centered at the orgn, the probablty mass P (γa) of γ- enlargement of A rapdly approaches as γ grows, provded that P (A) s not small. A shortcomng of the representaton of ths phenomenon as gven by Proposton 5. s that that what s and what s not small depends on the parameter σ; for small value of the parameter, not small actually means close to. We are about to present a slghtly modfed verson of Proposton 5. n whch every fxed postve value of P (A) s not small. Proposton 5. Let ξ t be ndependent and symmetrcally dstrbuted random reals, and let the dstrbutons of ξ t possess denstes p t ( ) such that p t ( ) 3 3σ (84) and ether () ξ t /, t =,..., n, or () E exp4ξ t } } exp}, t =,..., n, and let ξ = (ξ,..., ξ t ). Let, further, A be a closed convex symmetrc w.r.t. the orgn set n E, and let µ ProbS n A} > 0. (85) Then for every ϑ > one has n the case of (): Prob ξ ϑa} µ exp µ6 σ 6 (ϑ ) } 56 (86) n the case of (): Probξ ϑa} µ µ 6 σ 6 (ϑ ) exp O() µ 3 σ 3 (ϑ ) + ln(n/µ) } (87) wth properly chosen postve absolute constant O(). Proof. A. We start wth the followng Lemma 6 Let a R n, a =, and let ζ = a T ξ. Then ρ (0, /] Prob ζ ρ} ρ /3 σ. (88) Proof. Observe, frst, that σ t Eξ t } σ. (89) Indeed, for every δ 3 3σ, we have Prob ξ t > δ} δ 3, whence 3σ σ t δ ( δ 3 ). Maxmzng 3σ over δ, we arrve at (89). Snce p t ( ) are even and E exp4ξt } } exp}, the generaton functons φ t (y) = Eexpxy}} are real-valued, even and C ; besdes ths, φ t (0) = σt σ, φ (4) t (y) E } ξt 4 8 E exp4ξt } } e 8. Consequently, for y 5σ the remander n the thrd order Taylor expanson of φ t(y), taken at the orgn, does not exceed e 4 8 y4 σ y /4, whence y 5σ φ t (σ) σ t y / + σ y /4 σ y /4 exp σ y /4}. (90) Now let α = max a t and φ(y) be the generatng functon of ζ: φ(y) = n φ t (a t y). By (90), we have t y 5σα φ(y) n exp a t σ y /4} exp σ a y /4}. (9)

23 Now let / ρ > 0. Consder the functon h(x) = ρ χ x ρ along wth the functon g = h h ( stands for convoluton). Functon g clearly s nonnegatve and g(x) when x ρ. Observe that the Fourer transform of g s the functon sn (ρy) ρy [0, ]. Denotng by p( ) the densty of ζ, we have Prob ζ ρ} p(x)g(x)dx = π φ(y) sn (ρy) ρy dy π φ(y) sn (ρy) ρy dy + π ρy dy π exp σ y 4 } sn (ρy) ρy dy + α 5πρσ y 5σα 5σα π exp σ z 4ρ } sn (z) z dz + α 5πρσ π exp σ z 4ρ }dz + α 5πρσ = 4ρ πσ + α 5σρ Besdes ths, the unform norm of the densty of ζ clearly does not exceed the mnmum, over t, of the unform norms of the denstes of a t ξ t, that s, t does not exceed 3. We conclude that 3σα In the case of α ρ /3, (9) yelds [ Prob ζ ρ} mn Now let α < ρ /3. Invokng (9), we have Prob ζ ρ} Prob ζ ρ /3 } σ ρ 3 3σα, 4ρ Prob ζ ρ} ρ/3 3 3σ. + α ] πσ 5σρ [ 4ρ /3 + α ] [ ρ/3 4 + ] ρ /3 σ. π 5ρ /3 σ π 5 Thus, n all cases Prob ζ ρ} ρ /3 σ, as clamed. B. We now clam that under the premse of Proposton 5., A contans the centered at the orgn -ball of radus ρ=(µσ ). Indeed, otherwse, same as n the proof of Lemma 5, we could fnd a vector p, 3 p = and ρ < ρ such that A s contaned n the strpe x : p T x < ρ}, that s, wth ζ = p T ξ one has Prob ζ < ρ } Probξ A} = µ. On the other hand, ρ < ρ /, where the latter nequalty follows from the fact that σ (ndeed, σ σt by (89), whle σt = Eξt } E 8 [exp4ξ t } ]} e 8 ). Applyng Lemma 6, we get Prob ζ < ρ } (ρ ) / σ < µ, whch s a contradcton. C. Now we can complete the proof n exactly the same way as n the case of Proposton 4.. Specfcally, same as n tem C of the latter proof, relaton (8) wth ρ gven by B holds true. In the case of () ths observaton combnes wth Talagrand Inequalty (8) to yeld the relaton Prob ξ ϑa} µ exp (ϑ ) ρ } = 4 µ exp (ϑ ) ρ } = σ 6 (ϑ ) 4 µ exp µ6 }, 56 as requred n (77). In the case of (), the same reasonng as n tem E of the proof of Proposton 4. results n (87). 5. Sem-scalar verson of Theorem 3. Theorem 5. Let f t be determnstc vectors from a normed fnte-dmensonal space (E, ) and ξ t be ndependent symmetrcally dstrbuted random scalars such that and ether () ξ t, t =,..., n, or () E exp4ξt } } exp}, t =,..., n. Assume that Eξ t } σ > 0, t =,..., n E (9) } ξ t f t < Θ. (93) 3

24 Then n the case of (): Prob } ξ t f t ΩΘ O() exp O()σ 6 Ω } (94) wth approprate postve absolute constants O(); n the case of (): } σ 6 Ω Prob ξ t f t ΩΘ O() exp O() ln n + σ 3 Ω } (95) Proof. Let ξ = (ξ,..., ξ n ), B = y E : y rθ} and A = s R n : s t f t B}. Then A s a convex closed symmetrc w.r.t. the orgn set such that Probξ A} r by (93). Settng r = σ4 + σ, we get Prob ξ A} > ν σ4 4 σ 4 +. Applyng Proposton 5., we get In the case of (): n the case of (): References Prob n ξ t f t > ΩΘ} = Prob ξ Ωr A } O() exp O()Ω r σ } Prob n O() exp O()σ 6 Ω }. ξ t f t > ΩΘ} = Prob ξ Ωr A } O() exp O() Ω r σ ln n+ωr σ } O() exp O() σ 6 Ω ln n+σ 3 Ω }. [] Borell, C., The Brunn-Mnkowsk nequalty n Gauss space Inventones Mathematcae 30 () (975), [] Johnson, W.B., Schechtman, G., Remarks on Talagrand s devaton nequalty for Rademacher functons, Banach Archve /6/90, Sprnger Lecture Notes 470 (99), pp [3] Nemrovsk, A., On tractable approxmatons of randomly perturbed convex constrants Proceedngs of the 4nd IEEE Conference on Decson and Control Mau, Hawa USA, December 003, Appendx: Proof of Proposton. Let f k (t)} be a sequence of polynomals convergng to f, along wth the frst and the second dervatves, unformly on every compact subset of. For a polynomal p(t) = N j=0 p jt j the functon P (X) = Tr( j p jx j ) s a polynomal on S n. Let now X, H S n, let λ s = λ s (X) be the egenvalues of X, X = UDagλ}U T be the egenvalue decomposton of X, and let Ĥ be such that H = UĤU T. We have P (X) = n s= p(λ s(x)) (a) DP (X)[H] = Tr( N N j= s=0 Xs HX N s = Tr(p (X)H) = n s= p (λ s (X))Ĥss (b) Further, let γ be a closed contour n the complex plane encrclng all the egenvalues of X. Then DP (X)[H] = Tr(p (X)H) = πı p (z)tr((zi X) H)dz γ D P (X)[H, H] = πı p (z)tr((zi X) H(zI X) H)dz = n Ĥ st p (z) πı s, (z λ s )(z λ t ) dz. γ γ (96) 4

25 Computng the resduals, we get D P (X)[H, H] = s,t Γ s,t [p]ĥ st, Γ s,t [p] = p (λ s ) p (λ t ) λ s λ t, λ s λ t p (λ s ), λ s = λ t (97) Substtutng p = f k nto (96.a, b) and (97), we see that the sequence of polynomals F k (X) = Tr(f k (X)) converges, along wth the frst and the second order dervatves, unformly on compact subsets of X n ( ); by (96.a), the lmtng functon s exactly F (X). We conclude that F (X) s C on X n ( ) and that the frst and the second dervatves of ths functon are lmts, as k, of the correspondng dervatves of F k (X), so that for X = UDagλ}U T X n ( ) (where U s orthogonal) and every H = UĤU T S n we have DF (X)[H] = s f (λ s )Ĥss = Tr(f (X)H) D F (X)[H, H] = s,t Γ (98) s,t[f]ĥ st So far, we dd not use (). Invokng the rght nequalty n (), we get D F (X)[H, H] [ ] f s,t θ (λ s )+f (λ t ) + + µ + Ĥ st = θ + s f (λ s ) t Ĥ st + µ + s,t Ĥ st = θ + Tr(Dagf (λ ),..., f (λ n )}Ĥ ) + µ + Tr(Ĥ ) = θ + Tr(f (X)H ) + µ + Tr(H ), whch s the rght nequalty n (3). The dervaton of the left nequalty n (3) s smlar. 5