Sharp upper bounds for the deviations from the mean of the sum of independent Rademacher random variables
SSharp upper bounds for the deviations from the meanof the sum of independent Rademacher randomvariables
Harrie Hendriks and Martien C.A. van Zuijlen
IMAPP, MATHEMATICSRADBOUD UNIVERSITY NIJMEGENHeyendaalseweg 1356525 AJ NijmegenThe Netherlandse-mail authors: [email protected], [email protected]
AMS 2000 subject classifications:
Primary 60E15, 60G50; secondary 62E15, 62N02.
Keywords and phrases:
Sums of independent Rademacher random variables, tail prob-abilities, upper bounds, concentration inequalities, random walk, finite samples.
Abstract
For a fixed unit vector a = ( a , a , ..., a n ) ∈ S n − , i.e. (cid:80) ni =1 a i = 1, we consider the2 n sign vectors (cid:15) = ( (cid:15) , (cid:15) , ..., (cid:15) n ) ∈ {− , } n and the corresponding scalar products a.(cid:15) = (cid:80) ni =1 a i (cid:15) i . In [1] the following old conjecture has been reformulated. It statesthat among the 2 n sums of the form (cid:80) ± a i there are not more with | (cid:80) ni =1 ± a i | > | (cid:80) ni =1 ± a i | ≤ . The result is of interest in itself, but has also anappealing reformulation in probability theory and in geometry. In this paper we willsolve an extension of this problem in the uniform case where all the a ’ s are equal.More precisely, for S n being a sum of n independent Rademacher random variables,we will give, for several values of ξ, precise lower bounds for the probabilities P n := P {− ξ √ n ≤ S n ≤ ξ √ n } or equivalently for Q n := P {− ξ ≤ T n ≤ ξ } , where T n is a standardized Binomial random variable with parameters n and p = 1 / . These lower bounds are sharp and much better than for instance the bound that canbe obtained from application of the Chebishev inequality. In case ξ = 1 Van Zuijlensolved this problem in [3]. We remark that our bound will have nice applications inprobability theory and especially in random walk theory. a r X i v : . [ m a t h . P R ] O c t Introduction and result
Let (cid:15) , (cid:15) , ..., be a sequence of i.i.d. Rademacher random variables and for positive integers n let a n = ( a n , a n , ..., a nn ) be a unit-vectors in R n , so that (cid:80) ni =1 a in = 1. The followingproblem has been presented in [2] and is attributed to B. Tomaszewski. In [1], Conjecture1.1, this old problem has been reformulated as follows: P ( | a n (cid:15) + a n (cid:15) + ... + a n (cid:15) n | ≤ ≥ , for n = 1 , , ... This conjecture is at least 25 years old and seems still to be unsolved. In the uniform casewhere, a n = a n = ... = a nn = n − / , the maximum possible value of S n √ n is √ n, where S n := (cid:15) + (cid:15) + ... + (cid:15) n (1)and the conjecture, stating that for integers n ≥ , P {| S n | ≤ √ n } = P {| n (cid:88) i =1 (cid:15) i | ≤ √ n } ≥ / , has been solved recently by M.C.A. van Zuijlen. See [3]. It means that at least 50% of theprobability mass is between minus one and one standard deviation from the mean, whichis quite remarkable. We note thati) S n can be easily expressed in terms of sums of independent Bernoulli(1/2) randomvariables since ( (cid:15) i + 1) / S n is distributed as 2 B n − n , where B n is a binomial random variable with parameters n and 1 / . It follows that S n / √ n is distributed as T n , where T n is a binomial randomvariable with parameters n and p = 1 / . ii) easy calculations show that the sequence ( P n ) is not monotone in n .In this paper we shall generalize Van Zuijlen’s result and derive sharp lower bound forprobabilities concerning ξ standard deviations: P n := P {| S n | ≤ ξ √ n } = P {| n (cid:88) i =1 (cid:15) i | ≤ ξ √ n } , (2)where ξ ∈ (0 , . Note that trivially P = (cid:40) , for ξ = 1;0 , for ξ < . Throughout the paper n and k will denote nonnegative integers. Our result is as follows.2 heorem 1. Let (cid:15) , (cid:15) , ..., (cid:15) n be independent Rademacher random variables, so that P { (cid:15) = 1 } = P { (cid:15) = − } = 1 / and let S n and P n be defined as in (1) and (2), where ξ ∈ (0 , . Define n k = 2 (cid:38) k ξ + k (cid:39) − k − , C k = { n : n k ≤ n < n k +1 } , and Q − k := P n k +1 − . Then, with Φ indicating the standard normal distribution function, we have for k ≥ a. P n = P {| S n | ≤ ξ √ n } = P {| S n | ≤ k } , for n ∈ C k ,b. Q − k = min n ∈ C k P n , c. the sequence ( Q − k ) is strictly monotone increasing in k ,Moreover,d. lim k →∞ Q − k = 1 − ξ ) ,e. Q − = P n − ≤ P n , for all n ≥ n . A consequence of Theorem 1 is the following result.
Corollary 2.
For n ≥ we have P n ≥ / , for ξ = 1;3 / , for ξ ∈ [ (cid:112) / , / , for ξ ∈ [ (cid:112) / , (cid:112) / . More generally, if < ξ ≤ and n ≥ , n odd, then we have √ n + 1 ≤ ξ < √ n − , n = 2 (cid:24) n − (cid:25) and for all n ≥ n P n ≥ P n − = (cid:18) n − n − / (cid:19) − ( n − . It is worthwhile to clarify in a plot the structure of the probabilities P n ( (cid:96) ) = P {| S n | = (cid:96) } ,where n and (cid:96) are nonnegative integers such that n + (cid:96) is even. See Figure 1.3 l ll ll lll lll llll llll lllll lllll llllll llllll lllllll lllllll llllllll llllllll lllllllll lllllllll llllllllll llllllllll lllllllllll . . . . . . n p r obab ili t y n=2 n=7 n=14 l l l l l l l l l l ll l l l l l l l l ll l l l l l l l l ll l l l l l l l ll l l l l l l l ll l l l l l l ll l l l l l l ll l l l l l ll l l l l l ll l l l l ll l l l l ll l l l ll l l l ll l l ll l l ll l ll l ll ll lll l l l l l l l l l l l l l l l l l l l l Figure 1: Graph of probabilities P n ( (cid:96) ), n + (cid:96) even. (Dotted lines connect points with constant (cid:96) = 0 , , , . . . , upwards in graph. The square symbols indicate the points ( n, P n ) for ξ = 1.The vertical lines separate the regions C k , k = 1 , , . . . )4 Preliminaries
Be given independent Rademacher random variables ε i , i = 1 , , , . . . , as defined in Theo-rem 1, and let S n = (cid:80) ni =1 ε i such that S = 0. Define P n ( k ) = P ( | S n | ≤ k ) . Since n + S n is even it follows that P n ( k ) = P n ( k −
1) if n + k is odd.A basic property is the symmetry of the distribution of S n : P { S n = k } = P { S n = − k } . Moreover, ε n being independent of S n − , P { ( S n − = k + 1 & ε n = −
1) or ( S n − = − k − ε n = +1) } = 2 P { S n − = k + 1 & ε n = − } = P { S n − = k + 1 } and, replacing ε n by the equally distributed − ε n , P { ( S n − = k + 1 & ε n = +1) or ( S n − = − k − ε n = − } = P { S n − = k + 1 } . This leads to the following properties for P n ( k ). Remark 3.
Suppose n + k is even, n ≥ , then P n ( k ) = P n − ( k −
1) + P { ( S n − = k + 1 & ε n = −
1) or ( S n − = − k − ε n = +1) } = P n − ( k −
1) + P { S n − = k + 1) ,P n ( k ) = P n − ( k + 1) − P { ( S n − = k + 1 & ε n = +1) or ( S n − = − k − ε n = − } = P n − ( k + 1) − P { S n − = k + 1) . Suppose n + k is even, n ≥ , then P { S n − = k − } = (cid:18) n − n + k − (cid:19) − ( n − = n + kn (cid:18) n n + k (cid:19) − n = n + kn P { S n = k } , P { S n − = k + 1 } = P { S n − = − k − } = n − kn P { S n = − k } = n − kn P { S n = k } . Suppose n + k is even, n ≥ , k − ≥ . It follows that P n − ( k − − P n +1 ( k −
1) = P n ( k −
2) + P { S n − = k − } − P n ( k − − P { S n = k } = P { S n − = k − } − P { S n = k } = kn P { S n = k } . Furthermore, for n ≥ k ≥ , P { S n = k } P { S n +2 = k } = P { S n = k } P { S n +1 = k + 1 } × P { S n +1 = k + 1 } P { S n +2 = k } = n + 1 + k + 1 n + 1 × n + 2 − kn + 2= ( n + 2) − k ( n + 2) − ( n + 2) . In particular, if k ≤ n +2 then P { S n = k } ≥ P { S n +2 = k } , with equality only if k = n +2 . orollary 4. Suppose n + k is even, n > k (i.e. n ≥ k + 2 ), then P n ( k ) = P n ( k + 1)
Suppose k ≥ , n ≥ k and n + k is even. If n + 2 ≥ k and (cid:96) ≥ , such that n + 1 + 2 (cid:96) < ( k +1) k ( n + 2) , then P n ( k − < P n +1+2 (cid:96) ( k − .Proof. For (cid:96) = 0 we remark that P n +1 ( k − − P n ( k −
2) = P { S n = k } >
0. For (cid:96) ≥ P n +1 ( k − − P n ( k − > P n +1 ( k − − P n +1+2 (cid:96) ( k − P n +1 ( k − − P n ( k −
2) = P { S n = k } and P n +1 ( k − − P n +1+2 (cid:96) ( k −
1) = (cid:96) (cid:88) i =1 kn + 2 i P { S n +2 i = k } , this inequality will follow from the claim P { S n = k } > (cid:80) (cid:96)i =1 kn +2 i P { S n +2 i = k } .If (cid:96) = 1, then P { S n = k } ≥ P { S n +2 = k } > > kk +2 ≥ kn +2 . If (cid:96) > n + 2 > k ,then P { S n = k } > P { S n +2 (cid:96) = k } >
0, so that it is sufficient to show that (cid:80) (cid:96)i =1 kn +2 i ≤ Corollary 6.
Let n k , k = 1 , , , . . . , be an increasing sequence of integers such that n ≥ , n k + k is odd, n k + 1 ≥ k and n k +1 − < ( k +1) k ( n k + 1) . Then for n k ≤ m < n k +1 − wehave P { S n − = 0 } = P n − (0) < · · · < P n k − ( k − < P n k +1 − ( k − < P m ( k ) , which for k = 1 reduces to P { S n − = 0 } = P n − (0) < P m (1) . Proof.
For k ≥ n ≥
2, apply Theorem 5 with n = n k − (cid:96) = ( n k +1 − n k − / k = 1 and n = 0 we have n = 3 and the claim in the corollary is trivial. Let ξ > {| S n | ≤ ξ √ n } . Let k be the integer such that n + k iseven and k ≤ ξ √ n < k + 2. Then {| S n | ≤ ξ √ n } = {| S n | ≤ k } . Notice that such k satisfiesthe inequalities n + k ≤ n + ξ √ n < n + k + 22 = n + k n + k = (cid:106) n + ξ √ n (cid:107) and hence k = κ ( n ) := 2 (cid:22) n + ξ √ n (cid:23) − n. It follows immediately that κ ( n + 2) ≥ κ ( n ), κ (0) = 0. Moreover κ ( n + 1) − κ ( n ) = 2 (cid:22) n + 1 + ξ √ n + 12 (cid:23) − n − − (cid:22) n + ξ √ n (cid:23) + n = 2 (cid:22) n + 1 + ξ √ n + 12 (cid:23) − (cid:22) n + ξ √ n (cid:23) − , so that κ ( n + 1) − κ ( n ) is odd and greater than or equal to − a and b are nonnegative integers we have ξ √ a < κ ( a ) + 2 and κ ( b ) ≤ ξ √ b, so that a < ( κ ( a ) + 2) κ ( b ) b. (3)From the inequality (cid:98) a (cid:99) − (cid:98) b (cid:99) < a − b + 1 one concludes κ ( n + 1) − κ ( n ) < n + 1 + ξ √ n + 1 − n − ξ √ n + 1= 2 + ξ √ n + 1 − ξ √ n = 2 + ξ √ n + 1 + √ n . It follows that for ξ ≤ κ ( n + 1) − κ ( n ) ≤
1, since then it is an odd number strictly lessthan 3. As a matter of fact, already for ξ < √ κ ( n + 1) − κ ( n ) ≤
1. In the sequelassume that ξ ≤
1. Then we have the basic properties κ ( n + 1) − κ ( n ) = ± , κ (0) = 0 , κ ( n ) ≤ ξ √ n, κ ( n + 2) ≥ κ ( n ) . For k ≥
1, define n k := min { n | κ ( n + 1) ≥ k } = 2 (cid:38) k ξ + k (cid:39) − k − . (4)It is clear that n k is strictly increasing in k . Moreover κ ( n k + 1) = k and κ ( n k ) = k − κ ( n k −
1) = k − k ≥ k = 1 and n ≥ n k + k is odd. In case ξ = 1it is easy to see that n k = k −
1. Since n k is decreasing in ξ , it follows for ξ ≤ n k ≥ k − m ≤ n k +1 we have κ ( m ) < k + 1, so that κ ( m ) ≤ k . On the other hand,if κ ( m ) ≤ k −
2, it follows for all n ≤ m that κ ( n ) ≤ k −
1, so that m < n k + 1 and7ince κ ( n k ) = k − m < n k . We conclude that for n k ≤ m ≤ n k +1 we have k − ≤ κ ( m ) ≤ k , so that P m ( k ) = P {| S m | ≤ ξ √ m } . (5)Since k ≤ ξ √ n k + 1 and 0 < ξ ≤ k ≤ n k + 1.From Inequality (3) we obtain n k +1 − < ( k + 1) k ( n k + 1) . Provided that n k − ≥ k , Theorem 5 leads to the inequality P n k − ( k − < P n k +1 − ( k − n k ≥ k ≥ k = 1 and ξ <
1. The main result, Theorem 1, in factfollows from Corollary 6. More specifically,
Corollary 7.
Let ξ ≤ and n k , k = 1 , , , . . . be defined as (4). Then, for k ≥ and all m satisfying n k − ≤ m < n k , we have P { S n − = 0 } = P n − (0) ≤ P n k − ( k − < P m ( k −
1) = P {| S m | ≤ ξ √ m } . In particular, for m ≥ n we have P { S n − = 0 } ≤ P {| S m | ≤ ξ √ m } , with equality only for m = n − .Proof of Theorem 1. Claim a) has been dealt with in (5). Claims b), c) and e) followdirectly from the above Corollary 7. Finally, Claim d) follows from the Central LimitTheorem.It is the condition ξ ≤ n k + 1 ≥ k , needed in Corollary 6. For ξ > P n k +1 − ( k − > P n k − ( k −
2) as can be seen from the followingexamples. For ξ = √
2, we have n = 7 , n = 12 and P n − (3) = < = P n − (2). For ξ = 1 . k = 22: n = 399 = 20 − n = 438 and P n − (21) < . < P n − (20).For ξ = 1 .
01 and k = 202, n k = 39999 = 200 − n k +1 = 40398 and P n − (201) < . < P n − (200).Concerning Corollary 2 we note the following. It is straightforward to see that n = , for ξ = 1 , , for ξ ∈ [ (cid:112) / , , , for ξ ∈ [ (cid:112) / , (cid:112) / , so that P {− ≤ S n − ≤ } = / , for ξ = 1 , / , for ξ ∈ [ (cid:112) / , , / , for ξ ∈ [ (cid:112) / , (cid:112) / . n = 2 (cid:24) ξ +22 (cid:25) − − (cid:108) ξ (cid:109) −
1, which isequivalent to 2 √ n + 1 ≤ ξ < √ n − n + 38 < ξ + 12 ≤ n + 58 . Since n is odd, the open interval ( n +38 , n +58 ) does not contain an integer. Thus, for such ξ , n = 2 (cid:38) ξ + 12 (cid:39) − − (cid:24) n + 58 (cid:25) − (cid:24) n − (cid:25) and for all n ≥ n , we have from Theorem 1 P n ≥ P n − = P { S n − = 0 } = (cid:18) n − n − / (cid:19) − ( n − . In case ξ = (cid:112) / , we obtain for k ∈ { , , ... } n k = 2 (cid:38) k ξ + k (cid:39) − k − (cid:24) k + k (cid:25) − k − (cid:40) k − , for k = even , k , for k = odd . In this case n = 2 , n = 7 , n = 18 , n = 31 , so that C = [2 , , C = [7 , , C = [18 , C is P n − = P = P {− ξ √ n − ≤ S n − ≤ ξ √ n − } = P {− ≤ S ≤ } = P { S = 0 } == P { B = 3 } = 516 . Here the B n denote the binomial random variables as in the Introduction. Also, P n − = P = P {− ≤ S ≤ } = 2 P { S = 1 } = 2 P { B = 9 } = 1215532768 ≥ . In case ξ = (cid:112) / , hence we obtain for k ∈ { , , ... } n k = 2 (cid:38) k + k (cid:39) − k − (cid:24) k + 2 k (cid:25) − k − (cid:40) k − , for k = even , k + , for k = odd . . n = 2 , n = 5 , n = 14 , n = 23 , n = 38 , n = 53 with blocks C = [2 , , C =[5 , , C = [14 , , C = [23 , , C = [38 , . The minimal value in C is obtained for P n − = P {− ξ √ n − ≤ S n − ≤ ξ √ n − } = P {− ≤ S ≤ } = P { S = 0 } == P { B = 2 } = 38 . Also, P n − = P {− ≤ S ≤ } = 2 P { S = 1 } = 2 P { B = 7 } = 4291024 ≥ . In case ξ = 1 we obtain for k ∈ { , , ... } n k = 2 (cid:38) k ξ + k (cid:39) − k − k − . We obtain for integers k ≥ , C k = { k − , k , ..., ( k + 1) − } , with length m k = 2 k + 1 . Now n = 0 , n = 3 , n = 8 , n = 15 , so that C = [0 , , C = [3 , , C = [8 , . Theminimal value in C is obtained for P n − = P {− ξ √ n − ≤ S n − ≤ ξ √ n − } = P {− ≤ S ≤ } = P { S = 0 } == P { B = 1 } = 12 . The minimal value in C is obtained for n = n − P n − = P = P {− ≤ S ≤ } = 2 P { S = 1 } = 2 P { B = 4 } = 3564 ≥ . Appendix
In this section we state and prove the lemma needed in the proof of Theorem 5.
Lemma 8.
Suppose k ≥ and n + k even. If n + 2 ≥ k and (cid:96) ≥ , such that n + 1 + 2 (cid:96) < ( k +1) k ( n + 2) , then (cid:80) (cid:96)i =1 kn +2 i ≤ .Proof. The goal is to prove the inequality (cid:96) (cid:88) i =1 kn + 2 i ≤ . It is easy to see that k/ ( n + 2 i ) + k/ ( n + 2 (cid:96) + 2 − i ) is decreasing in i for i ≤ (cid:96)/
2. Thereforeit is sufficient to prove k/ ( n + 2) + k/ ( n + 2 (cid:96) ) < /(cid:96) , or equivalently (cid:96)n + 2 + (cid:96)n + 2 (cid:96) ≤ k . (6)10ince the left hand is increasing in (cid:96) , it is sufficient for given n to consider the maximallyallowed (cid:96) . In the same way, given (cid:96) it is sufficient to prove the inequality for the minimallyallowed n .The condition n + 1 + 2 (cid:96) < ( k +1) k ( n + 2) is equivalent to 2 (cid:96) − < k +1 k ( n + 2). Thus forany n such that n + 2 ≥ k , (cid:96) = k is an allowed value for (cid:96) . The corresponding minimalvalue of n is n = k −
2. It follows that for (cid:96) ≤ k Inequality (6) holds:2 k − (cid:96)n + 2 − (cid:96)n + 2 (cid:96) ≥ k − kn + 2 − kn + 2 k ≥ k − kk − kk = 0 . Next consider the case (cid:96) ≥ k + 1. Then the condition 2 (cid:96) − < k +1 k ( n + 2) leads to n + 2 > (cid:96) − k + 1 k = k + ( (cid:96) − k − k −
12 ) + (cid:96) − k − k + 1) ≥ k + ( (cid:96) − k − k −
12 ) . (7)In case (cid:96) = k + 1 it means that n + 2 > k , and because n + k is even, n + 2 ≥ k + 2.Substituting (cid:96) = k + 1 and n = k we get Inequality (6) for (cid:96) = k + 1:2 k − (cid:96)n + 2 − (cid:96)n + 2 (cid:96) = 2( k + 2 k + 4) k ( n + 2)( n + 2 (cid:96) ) ≥ . For the case (cid:96) ≥ k +2 we conclude from Inequality (7) that n +2 ≥ k +( (cid:96) − k − k − )+ .Substituting (cid:96) = k + 2 + j and n = k + ( (cid:96) − k − k − ) − we get2 k − (cid:96)n + 2 − (cid:96)n + 2 (cid:96) = 2 j ( k −
2) + j (2 k − k ( n + 2)( n + 2 (cid:96) ) . Since the right hand side is nonnegative for j ≥ k ≥ k ≥ (cid:96) ≥ k + 2.If k = 1, n odd, then from n + 1 + 2 (cid:96) < ( k +1) k ( n + 2) it follows that 2 (cid:96) − < n + 2),which implies 2 (cid:96) − ≤ n + 2) −
2, so that the maximal (cid:96) is (cid:96) = (3 n + 5) /
2. Again2 k − (cid:96)n + 2 − (cid:96)n + 2 (cid:96) = ( n + 1)( n + 5)2( n + 2)( n + 2 (cid:96) ) ≥ . References [1] R. Holzman and D.J. Kleitman: On the product of sign vectors and unit vectors.,Combinatorica (3) (1992), 303-316.[2] R.K. Guy: Any answers anent these analytical enigmas?, Amer. Math. Monthly93