Iosif Pinelis 1. Introduction and summary

Transcription

1 ESAIM: PS August 007, Vol., p DOI: 0.05/ps:00707 ESAIM: Probability and Statistics TOWARD THE BEST CONSTANT FACTOR FOR THE RADEMACHER-GAUSSIAN TAIL COMPARISON Iosif Pinelis Abstract. It is proved that the best constant factor in the Rademacher-Gaussian tail comparison is between two explicitly defined absolute constants c and c such that c.0 c. A discussion of relative merits of this result versus limit theorems is given. Mathematics Subject Classification. 60E5, 6G0, 6G5, 60G50, 6G5. Received June 7, 006. Revised November, 006. Introduction and summary Let ε,...,ε n be independent Rademacher random variables r.v. s), so that Pε i =)=Pε i = ) = for all i. Let S n := a ε + + a n ε n, where a,...,a n are any real numbers such that a + + a n =. The best upper exponential bound on the tail probability PZ x) for a standard normal random variable Z and a nonnegative number x is inf t 0 e tx E e tz =e x /. Thus, a factor of the order of magnitude of x is missing in this bound, compared with the asymptotics PZ x) x ϕx) asx,whereϕx) / / π is the :=e x density function of Z. Now it should be clear that any exponential upper bound on the tail probabilities for sums of independent random variables must be missing the x factor. Eaton [6] obtained an upper bound on PS n x), which is asymptotic to c PZ x) asx,where c := e , and he conjectured that PS n x) c x ϕx) forx>. The stronger form of this conjecture, PS n x) c PZ x) ) Keywords and phrases. Probability inequalities, Rademacher random variables, sums of independent random variables, Student s test, self-normalized sums. Department of Mathematical Sciences, Michigan Technological University, Houghton, Michigan 499 USA; ipinelis@mtu.edu c EDP Sciences, SMAI 007 Article published by EDP Sciences and available at or

2 TOWARD THE BEST CONSTANT FACTOR FOR THE RADEMACHER-GAUSSIAN TAIL COMPARISON 4 for all x R with c = c was proved by Pinelis [], along with a multidimensional extension. Edelman [7] proposed inequality PS n x) P Z x.5/x) for all x>0, but his proof appears to have a gap. A more precise upper bound, with ln c = in place of.5, was recently shown [0] to be a rather easy corollary of the mentioned result of []. Various generalizations and improvements of inequality ) as well as related results were given by Pinelis [,, 5 7, 9, 0] and Bentkus [ ]. Bobkov, Götze and Houdré BGH) [4] gave a simple proof of ) with a constant factor c.0. Their method was based on the Chapman-Kolmogorov identity for the Markov chain S n ). Such an identity was used, e.g., in [4] to disprove a conjecture by Graversen and Peškir [9] on max k n S k. In this paper, we shall show that a modification of the BGH method can be used to obtain inequality ) with a constant factor c.0 c,wherec is the best that is, the smallest) possible constant factor c in ). Let Φandr denote the tail function of Z and the inverse Mills ratio: Φx) :=PZ x) = x ϕu)du and r := ϕ Φ Theorem Main). For the least possible absolute constant factor c in inequality ) one has c := 4Φ ) c [c,c ] [.8,.], where ) and c := c + 50 ) +r ) c.0. Here we shall present just one application of Theorem, to self-normalized sums V n := X + + X n, X + + Xn where, following Efron [8], we assume that the X i s satisfy the orthant symmetry condition: the joint distribution of δ X,...,δ n X n is the same for any choice of signs δ,...,δ n {, }, so that, in particular, each X i is symmetrically distributed. It suffices that the X i s be independent and symmetrically but not necessarily identically) distributed. In particular, V n = S n if X i = a i ε i i. It was noted by Efron that i) Student s n statistic T n is a monotonic function of the self-normalized sum: T n = n V n/ Vn /n and ii) the orthant symmetry implies in general that the distribution of V n is a mixture of the distributions of normalized Rademacher sums S n. Thus, one obtains Corollary. Theorem holds with V n in place of S n. Various limit theorems for sums and self-normalized sums are available. In particular, the central limit theorem approximation for PS n x) andpv n x) issimplypz x), without any extra factor c. However, i) such asymptotic relations, without an upper bound on the rate of convergence, are impossible to use in statistical practice when one needs to be certain that the tail probability does not exceed a prescribed level; ii) when an upper bound say of the Berry-Esseen type) on the rate of convergence is available, usually it is relatively too large to be useful in statistics, especially in the tail area; iii) usually, large deviation asymptotics are valid at best in the zone x = on), and this zone is defined only qualitatively; iv) the summands X,...,X n are usually required to be identically, or nearly identically, distributed. If these conditions fail to hold then as Theorem, Corollary, and the discussion below in the beginning of Section show the asymptotic approximations may be inadequate. Also, it was pointed out in Theorem.8 of [], that, since the normal tail decreases fast, inequality ) even with c 4.46 implies that relevant quantiles of S n and V n may exceed the corresponding standard normal quantiles only by a relatively small amount; thus, one can use Corollary rather efficiently to test symmetry even for non-i.i.d. observations.

3 44 I. PINELIS Κ x x Figure. Ratio of the Rademacher and Gaussian tails for n = 00 and a = = a 00 = 0.. Proof of Theorem Theorem follows immediately from Lemma, Theorem, and Lemma, stated in Section. below. In particular, Lemma implies that the upper bound h x) onps n x) provided by Theorem is somewhat better than the upper bound c PZ x), implied by Theorem. While S n represents a simplest case of the sum of independent non-identically distributed r.v. s, it is still very difficult to control in a precise manner. Figure shows the graph of the ratio κx) :=PS n x)/ PZ x) for n = 00 and a = = a n. One can see that even for such a fairly large value of n and equal coefficients a,...,a n,ratioκx) oscillates rather wildly. In view of this, the existence of a high-precision inductive argument in the general setting with possibly unequal a i s may seem very unlikely. However, such an argument will be presented in this paper. The key idea in the proof of Theorem is the construction of the upper bound h and, in particular, the function g defined by ) and ), which allows an inductive argument to prove Theorem, a refined version of Theorem. The proof of Theorem is based on a number of lemmas. The proofs of all lemmas are deferred to Section.... Statements of lemmas and the proof of Theorem Lemma. One has c c. For a [0, ) and x R, introduce gx) :=c hx) :=c +rx) ) ) Φx) = c 50 5 Φx)+ϕx) ) ; ) + 50 ) + 50 ) +r ) Φx) =c Φx);

4 TOWARD THE BEST CONSTANT FACTOR FOR THE RADEMACHER-GAUSSIAN TAIL COMPARISON 45 if x 0, if 0 <x, h x) := x if x<, gx) if x, hx) if x ; Ka, x) :=h u)+h v) h x), where 4) u := ua, x) := x a and 5) a v := va, x) := ) x + a a. 6) Theorem Refined). One has PS n x) h x) 7) for all x R. Lemma. One has g h on, ] and g h on [, ). Lemma. One has h h on R. Lemma 4. One has Ka, x) 0 for all a, x) [0, ) [, ]. Lemma 5. One has Ka, x) 0 for all a, x) [0, ) [, ). Now we can present Proof of Theorem. Theorem will be proved by induction in n. Itisobviousforn =. Letnown {,,...} and assume that Theorem holds with n inplaceofn. Note that for x 0 inequality 7) is trivial. For x 0, ), it follows by the symmetry of S n and Chebyshev s inequality. Therefore, assume without loss of generality that x and0 a n <. By the Chapman-Kolmogorov identity and induction, PS n x) = PS n x a n )+ PS n x + a n ) h uan,x) ) + h van,x) ) = h x)+ Ka n,x) for all x R. It remains to refer to Lemmas 4 and 5. Lemmas,, and 5 are much easier to prove than Lemma 4. The least elementary of Lemmas,, and 5 is Lemma, whose proof uses the following l Hospital-type rule for monotonicity. Proposition. [8] Let a<b. Letf and g be real-valued differentiable functions defined on the interval a, b) such that fb ) =gb ) =0. It is assumed that g and g do not take on the zero value on a, b). Suppose finally that f g on a, b); that is, f g switches from increase) to decrease) on a, b). Then f g or on a, b). This proposition follows immediately from [8, Proposition 4. and Remark 5.]. A significant difficulty in the proof of Lemma 4 is that the profiles of the function K given by the crosssections of the graph of K on rectangle R := {a, x) R :0<a<, <x< } or on its closure R := {a, x) R :0 a, x } )

5 46 I. PINELIS Figure. Partition of R =0, ), ). with a 0, ) or x, ) fixed are very complicated; in part, this is caused by the fact that the function h is defined in ) by three different expressions over the interval [, ). To overcome this difficulty, an idea is to partition R into pieces so that on each piece there is a direction in which the profiles of K are easier to deal with. This idea comes naturally from the following considerations. Remark. Observe that u = ua, x), ) and v = va, x), ) for all a, x) R. Therefore and in view of definitions 4) and ), the form of expression of K on R depends on whether u< and on whether v<. The curves u = andv = which are ellipses) partition R into 5 connected pieces as illustrated by Fig. ), one of which may be naturally cut further into two pieces by the line x = x, where x := = ) may be also defined by the following condition: a, x) R & u = & v = ) = x = x.

6 TOWARD THE BEST CONSTANT FACTOR FOR THE RADEMACHER-GAUSSIAN TAIL COMPARISON 47 Thus, one comes to the following definitions of the pieces : LLe := {a, x) R: u<,v }; LG := {a, x) R: u<,v > }; GL := {a, x) R: u>,v <,x x }; GL := {a, x) R: u>,v <,x>x }; GG := {a, x) R: u>,v >,a< }; GG := {a, x) R: u>,v >,a }; GE := {a, x) R: u>,v = }; ELe := {a, x) R: u =,v }; EG := {a, x) R: u =,v >,a< }; EG := {a, x) R: u =,v >,a }; A := {a, x) R: a =0}; A := {a, x) R: a =}; X, := {a, x) R: x =, 0 <a< }; X, := {a, x) R: x =, a< }; X, := {a, x) R: x =, a<}; X := {a, x) R: x = }, where u and v are defined by 5) and 6). Here, for example, the L in the first position in symbol LLe refers to less than in inequality u<, while the ligature Le in the second position refers to less than or equal to in inequality v. Similarly, G and E in this notation refer to greater than and equal to, respectively. Symbol A refers here to a fixed value of a, and X to a fixed value of x. Observe that the set R is the union of the pieces LLe,...,X. Indeed, let {C} denote, for brevity, the set {a, x) R: C}, wherec stands for a condition. Then, basically following the just explained meaning of the notation for the pieces, one has LLe }{{ LG } GL GL }{{ } {u< } {u>,v< } GG GG }{{} {u>,v> } GE }{{} {u> } ELe EG EG }{{} {u=,v> } }{{} {u=} = R. 9) In fact, the sets on the left-hand side of 9) form a partition of R.) It is also clear that the union A A X, X, X, X equals the boundary of R. This verifies the entire union-observation, which is illustrated in Figure, where only the labels of the two-dimensional members of the partition of R are shown. It will be understood that the function K defined by 4)) is extended to A by continuity, so that Ka, x) := gx) a, x) A. 0)

7 48 I. PINELIS Another difficulty to overcome in the proof of Lemma 4 is that the expressions for h and hence for K contain the transcendental function Φ. Yet, fixing an arbitrary value of u, v, or x, one will thereby fix the value of one of the three terms in expression 4) h u), h v), or h x), respectively, so that the partial) derivative of K in say) a with the value of u or v or x fixed will be of the form k := A e A + A e A4,whereA,...,A 4 are algebraic expressions in a, x) or, equivalently, in a, u) or a, v). Assuming that sign A is constant and nonzero on a given piece say P ) of rectangle R, the sign of k on P is the same as or opposite to that of k := A e A A4 /A +. Therefore, the sign on P ) of the derivative say k )of k in say) a is the same as that of a certain algebraic expression. Since the equations of the boundaries of the pieces are algebraic as well, the sign pattern of k on P can be determined in a completely algorithmic manner, according to a result by Tarski [5, 0, ]. The implementation of this scheme will require a great amount of symbolic and numerical computation. We have done that with the help of Mathematica TM 5., which is rather effective and allows complete and easy control over the accuracy. The Tarski algorithm is implemented in Mathematica 5. via Reduce and related commands. In particular, command Reduce[cond && cond &&..., {var,var,...,}, Reals] returns a simplified form of the system of algebraic conditions equations or inequalities) cond, cond,... over real variables var, var,... However, the execution of such a command may take a long time if the algebraic system is more than a little complicated; in such cases, Mathematica can use some human help. On each of the pieces ELe,...,X we shall consider its own coordinate system, the one that will be most convenient for us. In particular, we shall use the coordinate pair a, v) oneachofthe pieces LLe and LG; the pair a, x) on each of the pieces GL,GL,andGG ;andthepaira, u) on the piece GG. The choice of these coordinate systems was motivated by a consideration of contour plots of the function K on R and on particular pieces. For instance, it will be shown in the proofs of Lemmas 6 and 7 below that the monotonicity pattern of the function K in a with v fixed over each of the pieces LLe and LG is decreasing), increasing), or switching from decrease to increase as a increases). Similarly over the other two-dimensional pieces: it will be shown that K is i) or in a with x fixed over each of the pieces GL and GL ; ii) in a with u fixed over GG ;and iii) in a with x fixed over GG. The monotonicity patterns of K over the one-dimensional pieces of R are somewhat easier to establish. Remark. Since the function h defined by ) is upper-semicontinuous on R and continuous on [, ), the function K is upper-semicontinuous on the compact rectangle R and therefore attains its maximum on R. We conclude that K attains its maximum over R on at least one of the sets LLe,...,X. Inviewofthisremark,themainlemma Lemma4 willbeeasilyobtainedfromthefollowingseriesof lemmas. Lemma 6 LLe). The function K does not attain a maximum on LLe. Lemma 7 LG). The function K does not attain a maximum on LG. Lemma 8 GL ). The function K does not attain a maximum on GL. Lemma 9 GL ). The function K does not attain a maximum on GL. Lemma 0 GG ). The function K does not attain a maximum on GG. Lemma GG ). The function K does not attain a maximum on GG. Lemma GE). One has K 0 on GE. Lemma ELe). One has K 0 on ELe.

8 TOWARD THE BEST CONSTANT FACTOR FOR THE RADEMACHER-GAUSSIAN TAIL COMPARISON 49 Lemma 4 EG ). One has K 0 on EG. Lemma 5 EG ). One has K 0 on EG. Lemma 6 A ). One has K =0on A. Lemma 7 A ). One has K 0 on A. Lemma 8 X, ). One has K 0 on X,. Lemma 9 X, ). One has K 0 on X,. Lemma 0 X, ). One has K 0 on X,. Lemma X ). One has K 0 on X... Proofs of the lemmas Proof of Lemma. Let n =anda = a =.ThenPS n )= 4 = c PZ ). Proof of Lemma. This follows from the well-known and easy-to-prove fact that the inverse Mills ratio r is increasing. Proof of Lemma. On interval, 0], one has h =<.6 <h0) h, sinceh is decreasing. On interval 0, ], one similarly has h = < 0.5 <h) h. On interval [, ], one has h = g h, by Lemma. On interval [, ), one has h = h. It remains to consider the interval, ). For x, ), one has h x) = px) := x. One has p ) =h ) = 0 and, for some constant C>0, h x)/p x) =Cx ϕx), so that h p on 0, ). Hence, in view of the l Hospital-type rule for monotonicity provided by Proposition, one has h p or on 0, ), and so, inf h, ) h =min h [, ] p =min {, } h p.0 >, whence h <hon, ). Proof of Lemma 4. This lemma follows from Remark and Lemmas 6, proved below. Indeed, by the conclusion of Remark, the function K attains its maximum over R on at least one of the sets LLe,...,X. By Lemmas 6, none of the sets LLe,...,GG canbeasetonwhichk attains its maximum over R. Therefore, K attains its maximum over R on at least one of the sets GE,...,X. Finally, Lemmas imply that this maximum is no greater than 0. Proof of Lemma 5. Let 0 a<andx. Then u andv, where u and v are defined by 5) and 6), as before. Therefore and in view of Lemma and ), one has h u) hu), h v) =hv), and h x) =hx), so that Ka, x) c Φu)+ Φv) Φx)) 0, as shown in the mentioned proof in [4]. Proof of Lemma 6 LLe). Expressing x and u in view of 5) and 6) in terms of a and v as xa, v) := a v a and ũa, v) :=v, respectively, one has a a Ka, x) =ka, v) :=k LLe a, v) := since u>andv> onr. Fora, x) R, let + gv) gxa, v)) a, x) LLe, ) ũa, v) D a k)a, v) :=4Φ ) xa, v) a ) k a, v); ) D a,a k)a, x) := 5 a ) D a k) ) a, va, x). x a) ϕx) )

9 40 I. PINELIS The coefficients of the partial derivatives in ) and ) are chosen in order to make D a,a k)a, x) equal to an algebraic expression and even a polynomial) in a and x, and at that signd a,a k)a, x) =sign Dak) ) a, va, x). With Mathematica 5., one can therefore use the command Reduce[daakx<=0 && 0<a< && Sqrt[]<x<Sqrt[]] where daakx stands for D a,a k)a, x)), which outputs False, meaning that D a,a k>0onr. By ), this implies Dak) ) a, va, x) > 0fora, x) R, sothatda k)a, v) isincreasingina for every fixed value of v; more exactly, D a k)a, v) isincreasingina a v),a v) ) for every fixed value of v, ], where a and a are certain functions, such that ) a, xa, v) LLe v, ] & a a v),a v) )). Thus, for every fixed value of v, ] the sign pattern of D a k)a, v) ina a v),a v) ) is or + or +; that is, D a k)a, v) may change sign only from to + as a increases. By ), k a, v) has the same sign pattern. Hence, for every fixed value of v, ] one has ka, v) or or in a a v),a v) ).Now Lemma6follows. Proof of Lemma 7 LG). This proof is almost identical to that of Lemma 6, except that the term gv) in ) is replaced here by hv), and the interval, ] is replaced by, 5 ). However, both terms gv) and hv) are constant for any fixed value of v. Proof of Lemma 8 GL ). One has a, x) GL <x x &0<a<a x)), 4) where a x) := x 6 x.since <u<v< ongl where again u and v are defined by 5) and 6)), one has Ka, x) =ka, x) :=k GL a, x) :=gu)+gv) gx) a, x) GL. 5) For a, x) R, let D a k)a, x) := k 5 Φ ) ) a / a, x) + ax)5 + v)ϕv) ; 6) D a,a k)s, x) := D ak) a, x) a ) + ax) 5 + v) e ax a, 7) where s is substituted for a in the right-hand side of 7). Then D a,a k)s, x) is a polynomial in s and x. Using again the Mathematica command Reduce, namely Reduce[daak[s,x]>0 && Sqrt[]<x<Sqrt[] && 0<s<], 8) where daak[s,x] stands for D a,a k)s, x), one sees that for every s, x) R D a,a k)s, x) > 0 <x<x & s, x) <s<s, x)), 9) where x is a certain number between and, and s, and s, are certain functions. In fact, x is a root of a certain polynomial of degree and, for each x,x ), the values s, x) ands, x) are two of the roots s of the polynomial D a,a k)s, x). ) Next, Reduce[daak[95/00,x]<=0 && Sqrt[]<x<=xx]

10 TOWARD THE BEST CONSTANT FACTOR FOR THE RADEMACHER-GAUSSIAN TAIL COMPARISON 4 produces False; here xx stands for x recall the definition of GL and 8); that is, D a,a k) 95 00,x) > 0 x,x ]. Hence and in view of 9), s, x) < <s,x) x,x ]. 0) On the other hand, setting s x) := a x) ) with a as in 4), one has s > on,x ]. Hence, by 0), s >s, on,x ]. So, in view of 9), the sign pattern of D a,a k)s, x) ins s x), ) is or +, depending on whether s x) s, x) or not, for each x,x ]. Now 7) and ) imply that the sign pattern of Dak) a, x) ina 0,a x)) is or +, so that D a k)a, x) or in a 0,a x)), for each x,x ]. Also, D a k)0,x) = 0 for all x R. So,the sign pattern of D a k)a, x) ina 0,a x)) is or +, for each x,x ]; in view of 6), k a, x) hasthe same sign pattern. Thus, ka, x) or in a 0,a x)), for each x,x ]. Recalling 4), we complete the proof of Lemma 8. Proof of Lemma 9 GL ). This proof is similar to that of Lemma 8. Here one has a, x) GL x <x< ) &0<a<a x), ) where a x) := x x. Relation 5) holds here, and we retain definitions 6) and 7). Definition ) is replaced here by s x) := a x). ) Letting s := s x )= , 9 one can see that s x) s x x, ). 4) On the other hand, using instead of 8) the Mathematica command Reduce[daak[s,x]>0 && xx<x<sqrt[] && ss<s<, Quartics->True], where xx stands for x and ss stands for s, one sees that D a,a k)s, x) > 0&x <x< ) &s <s< ) x <x<x & s <s<s, x), 5) where x is a root of a certain polynomial of degree 0 and s, x) isthesamerootins of the polynomial D a,a k)s, x) ass, x) in 9). Hence, in view of 5) and 4), the sign pattern of D a,a k)s, x) in s s x), ) is or +, foreachx x, ). D Now 7) and ) imply that the sign pattern of ak) a, x) in a 0,a x)) is or +, so that D a k)a, x) or in a 0,a x)), for each x x, ). Also, D a k)0,x) = 0 for all x R. So, the sign pattern of D a k)a, x) ina 0,a x)) is or +, for each x x, ); in view of 6), k a, x) has the same sign pattern. Thus, ka, x) or in a 0,a x)), for each x x, ). Recalling ), we complete the proof of Lemma 9. Proof of Lemma 0 GG ). Expressing x and v in view of 5) and 6) in terms of a and u as xa, u) := a u + a and ṽa, u) :=u +, respectively, one has a a Ka, x) =ka, u) :=k GG a, u) :=gu)+hṽa, u)) g xa, u)) a, x) GG ; 6)

11 4 I. PINELIS note that For a, u) R, let a, x) GG = a, ua, x) ) 0, ), ). 7) da, u) := k a, u) 5 Φ ) a ) / /ϕṽa, u)); 8) d a a, u) := d a, u) ϕṽa, u)) a ) ϕ xa, u)) ; d u a, u) := d ϕṽa, u)) a, u) u ϕ xa, u)) a. The idea of the proof of Lemma 0 is to show that da, u) < 0 and hence k a, u) < 0 for all a, u) R 0 := [0, ] [, ]. This is done by considering separately the interior and the four sides of the rectangle R 0,at that using the fact that d a a, u) andd u a, u) arepolynomialsina, u, and a. Employing the command Reduce[da[a,u]==0 && du[a,u]==0 && 0<a</Sqrt[] &&Sqrt[]<u<Sqrt[], {a,u}, Reals], where da[a,u] and du[a,u] stand for d a a, u) andd u a, u), one sees that the system of equations d a a, u) = 0=d u a, u) has a unique solution a,u ) 0.98,.57770) in a, u) 0, ), ), and da,u ) 0.44 < 0. Let us consider next the values of d on the boundary of the rectangle 0, ), ). First, d0,u) is an increasing affine function of u, andd0, ) 0.4 < 0. Hence, d0,u) < 0 u, ). Second, 7 e 5+4 u+u )/6 d u,u)= u + 5 ) u + 5 ) u is decreasing in u and takes on value 9 75 < 0atu =, so that d u,u) < 0foru, ) and d,u) in u, ). Moreover, d, ) < 0. Thus, d,u) < 0 u, ). Third, a ) ϕṽa, )) ϕ xa, d )) a, ) = p a)+ a p a), where p a) andp a) are certain polynomials in a. Therefore, the roots a of d a, ) are among the roots of the polynomial p, a) :=p a) a )p a), which has exactly two roots a 0, ). Of the latter roots, one is not a root of d a, ). Also, d 0, ) = > 0and d, ) < 0. Hence, d a, ) has exactly one root, a 0.4, in a 0, ) and, moreover, da, ) in a 0,a ]andda, ) in a [a, ). Besides, da, ) < 0. Thus, da, ) < 0 a 0, ). Fourth very similar to third), a ) ϕṽa, )) ϕ xa, )) d a, )=p a)+ a p a), where p a)andp a) are certain polynomials in a, different from the polynomials p a)andp a) in the previous paragraph. Therefore, the roots a of d a, ) are among the roots of the polynomial p, a) :=p a) a )p a), which has exactly two roots a 0, ). Of the latter roots, one is not a root of d a, ). Also, d 0, ) = > 0and d, ) < 0. Hence, d a, ) has exactly one root, a , in a 0, )

12 TOWARD THE BEST CONSTANT FACTOR FOR THE RADEMACHER-GAUSSIAN TAIL COMPARISON 4 and, moreover, da, ) in a 0,a ]andda, ) in a [a, ). Besides, da, ) 0.58 < 0. Thus, da, )< 0 a 0, ). We conclude that da, u) < 0 a, u) [0, ] [, ]. By 8), the same holds for k a, u). It remains to recall 6) and 7). Proof of Lemma GG ). In view of Lemma, Ka, x) =g h)ua, x)) + hva, x)) gx) a, x) GG. 9) One has u > 0onGG ) and v c > 0onR. Sinceg = 50 5Φ+ϕ and h = c Φ are decreasing on [0, ), we conclude that Ka, x) is decreasing in a on GG. Proof of Lemma GE). One has a, x) GE x <x< ) &a = a x) := 4 x x 4, where, as before, x is defined by 8). Therefore, for all a, x) GE Ka, x) =kx) :=k GE x) :=Ka x),x)=g ua x),x) ) + g ) gx), and it suffices to show that k 0on[x, ]. For x [x, ], let k x) := k x) Φ ) ϕx)5 + x) ; k x) :=k x) 5 x + 5) 4 x ) / ρx) / 6 ϕ ua x),x) ), /ϕx) where ρx) :=x + x 4 x +. Then k x) =p x) + ρx) p x), where p x) andp x) aresome polynomials in x and 4 x. Hence, the roots of k x) are among the roots of p, x) :=p x) ρx)p x) = p,, x)+ 4 x p,, x), where p,, x) andp,, x) are some polynomials in x. Hence, the roots of k x) are among the roots of p,, x) 4 x )p,, x) 04 x ) 4, which is a polynomial of degree and has exactly one root in [x, ]. Also, k x ) > 0 and k ) < 0. Therefore, k and hence k have the sign pattern + on [x, ]. Next, k x ) < 0andk )=0,sothatk and hence k have the sign pattern + on[x, ]. It follows that k does not have a local maximum on x, ). At that, kx ) < 0andk ) = 0. Thus, k 0on[x, ]. Proof of Lemma ELe). This proof is very similar to that of Lemma GE). One has a, x) ELe <x x & a = a x) := x 6 x where, as before, x is defined by 8). Therefore, for all a, x) LLe Ka, x) =kx) :=k ELe x) :=Ka x),x)=g ) + g va x),x) ) gx), ), 0)

13 44 I. PINELIS and it suffices to show that k 0on[,x ]. For x [,x ], let k x) := 500 Φ ) k x) ϕx)5 + x) ; x + 5) k x) :=k x) x ) / ρx) / 9ϕ va x),x) ), /ϕx) where ρx) :=x +x x +. Then k x) =p x) + ρx) p x), where p x) andp x) aresome polynomials in x and x. Hence, the roots of k x) are among the roots of p, x) :=p x) ρx) p x) = p,, x)+ x p,, x), where p,, x) andp,, x) are some polynomials in x. Hence, the roots of k x) are among the roots of p,, x) x )p,, x) x ) 4, which is a polynomial of degree and has exactly one root in [,x ]. Also, k ) < 0and k x ) > 0. Therefore, k and hence k have the sign pattern + on[,x ]. Next, k ) = 0 and k x ) > 0, so that k and hence k have the sign pattern + on[,x ]. It follows that k does not have a local maximum on,x ). At that, k ) = 0 and kx ) < 0. Thus, k 0 on [,x ]. Proof of Lemma 4 EG ). This proof is similar to that of Lemma. One has a, x) EG x <x< &a = a x) := x 6 x ). ) Therefore, for all a, x) EG Ka, x) =kx) :=k EG x) :=Ka x),x)=g )+h va x),x) ) gx), and it suffices to show that k 0on[x, ]. For x [x, ], let k x) := 500 Φ ) k x) ϕx)5 + x) ; x + 5) k x) :=k x) x ) / ρx) 9/ 9 r )ϕ va x),x) ), /ϕx) where ρx) :=x +x x +. Then k x) =p x)+ x p x), where p x) andp x) are some polynomials in x. Hence, the roots of k x) are among the roots of p, x) := p x) x )p x) /r ) ) x ) 4, which is a polynomial of degree 4 and has exactly one root in [x, ], x #.60. Also, k x ) > 0andk ) < 0. Therefore, k and hence k have the sign pattern + on [x, ], so that max [x, ] k = k x # ) < 0. It follows that k < 0 and hence k on [x, ]. At that, kx ) < 0. Thus, k 0on[x, ].

14 TOWARD THE BEST CONSTANT FACTOR FOR THE RADEMACHER-GAUSSIAN TAIL COMPARISON 45 Proof of Lemma 5 EG ). One has a, x) EG <x< &a = a x) := x + ) 6 x. ) Therefore, for all a, x) EG Ka, x) =kx) :=k EG x) :=Ka x),x)=g ) + h va x),x) ) gx), and it suffices to show that k 0on, ). For the functions a defined in 0) and )) and a definedin)),andforx, ), one has a x) a x); also, hz) is decreasing in z and va, x) isincreasingina. Hence, h va x),x) ) h va x),x) ), so that k EG k EG on [x, ). Similarly, in view of Lemma one has h va x),x) ) g va x),x) ) g va x),x) ) x,x ], so that k EG k ELe on,x ]. Now Lemma 5 follows, because it was shown in the proofs of Lemmas and 4, respectively, that k ELe 0 on [,x ]andk EG 0on[x, ]. Proof of Lemma 6 A ). This is trivial. Proof of Lemma 7 A ). This is also trivial, in view of 0). Proof of Lemma 8 X, ). On X,, one has u< v. Also, by Lemma, g h on [, ). Therefore, for all a, x) X, Ka, x) ka) :=k X,a) := ua, ) + g va, ) ) g ), ) and it suffices to show that k 0on[0, ). For a [0, ), let k a) := 000 Φ ) k a), λa) := a λa) ) > 0; a k a) :=k a) a) 4 a ) 4 λa) / ϕ va, ) ) ). Then k a) = a p a) +p a), where p a) andp a) are some polynomials in a. Hence, the roots of k a) are among the roots of p, a) := a )p a) p a), which is a polynomial of degree and has exactly two roots in [0, ). Of those two roots, one is not a root of k a), so that k a) has at most one root in [0, ). Also, k 0) = 5 > 0andk )= 7 < 0. Therefore, k and hence k have the sign pattern + on [0, ], so that k on [0, ]. At that the values 6 of k at points 0, 0,and 7 0 are approximately 5 < 0, 48 > 0, and 44 < 0, respectively. Therefore, k and hence k have the sign pattern + on [0, ) and, moreover, the only local maximum of k on 0, ] occurs only between 6 0 and 7 0 ; in fact, it occurs at a and equals < 0. It remains to note that k0) < 0. Proof of Lemma 9 X, ). This proof is similar to that of Lemma. In place of 9), here one still has relation ) for all a, x) X,,sinceu< v on X, as well. Since ua, ) and va, ) in a [, ), one has k on [, ), so that the maximum of k on [, )equalsk ), which is negative, in view of Lemma 8 and the continuity of k.

15 46 I. PINELIS Proof of Lemma 0 X, ). This proof is similar to that of Lemma 9. In place of ), here one has Ka, x) ka) :=k X, a) :=g ua, ) ) + g va, ) ) g ), for all a, x) X,,sinceu andv onx,.sincek in a [, ), the maximum of k on [, ) equals k ) < 0. Proof of Lemma X ). This follows immediately from Lemma 5. References [] V. Bentkus, A remark on the inequalities of Bernstein, Prokhorov, Bennett, Hoeffding, and Talagrand. Lithuanian Math. J. 4 00) [] V. Bentkus, An inequality for tail probabilities of martingales with differences bounded from one side. J. Theoret. Probab. 6 00) 6 7 [] V. Bentkus, On Hoeffding s inequalities. Ann. Probab. 004) [4] S.G. Bobkov, F. Götze, C. Houdré, On Gaussian and Bernoulli covariance representations. Bernoulli 7 00) [5] G.E. Collins, Quantifier elimination for the elementary theory of real closed fields by cylindrical algebraic decomposition. Lect. Notes Comput. Sci. 975) 4 8. [6] M.L. Eaton, A probability inequality for linear combinations of bounded random variables. Ann. Statist. 974) [7] D. Edelman, An inequality of optimal order for the tail probabilities of the T statistic under symmetry. J. Amer. Statist. Assoc ) 0. [8] B. Efron, Student s t test under symmetry conditions. J. Amer. Statist. Assoc ) [9] S.E. Graversen, G. Peškir, Extremal problems in the maximal inequalities of Khintchine. Math. Proc. Cambridge Philos. Soc. 998) [0] S. Lojasiewicz, Sur les ensembles semi-analytiques. Actes du Congrès International des Mathématiciens Nice, 970). Tome, Gauthier-Villars, Paris 970) 7 4. [] I. Pinelis, Extremal probabilistic problems and Hotelling s T test under a symmetry condition. Ann. Statist. 994) [] I. Pinelis, Optimal tail comparison based on comparison of moments. High dimensional probability Oberwolfach, 996). Birkhäuser, Basel Progr. Probab ) [] I. Pinelis, Fractional sums and integrals of r-concave tails and applications to comparison probability inequalities Advances in stochastic inequalities Atlanta, GA, 997). Amer. Math. Soc., Providence, RI. 4 Contemp. Math.,. 999) [4] I. Pinelis, On exact maximal Khinchine inequalities. High dimensional probability, II Seattle, WA, 999). Birkhäuser Boston, Boston, MA Progr. Probab ) [5] I. Pinelis, Birkhäuser, Basel L Hospital type rules for monotonicity: applications to probability inequalities for sums of bounded random variables. J. Inequal. Pure Appl. Math. 00) Article 7, 9 pp. electronic). [6] I. Pinelis, Binomial upper bounds on generalized moments and tail probabilities of super)martingales with differences bounded from above. IMS Lecture Notes Monograph Series 5 006) -5. [7] I. Pinelis, On normal domination of super)martingales. Electronic Journal of Probality 006) [8] I. Pinelis, On l Hospital-type rules for monotonicity. J. Inequal. Pure Appl. Math ) art40 electronic). [9] I. Pinelis, Exact inequalities for sums of asymmetric random variables, with applications. Probability Theory and Related Fields 007) DOI 0.007/s [0] I. Pinelis, On inequalities for sums of bounded random variables. Preprint 006) [] A.A. Tarski, A Decision Method for Elementary Algebra and Geometry. RAND Corporation, Santa Monica, Calif. 948).