On a conjecture concerning the sum of independent Rademacher random variables

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1 On a conjecture concerning the sum of independent Rademacher rom variables Martien C.A. van Zuijlen arxiv: v1 [math.pr] 1 Dec 011 IMAPP, MATHEMATICS RADBOUD UNIVERSITY NIJMEGEN Heyendaalseweg AJ Nijmegen The Netherls author: M.vanZuijlen@science.ru.nl AMS 000 subject classifications: Primary 60E15, 60G50; secondary 6E15, 6N0. Keywords phrases: Sums of independent Rademacher rom variables, tail probabilities, upper bounds, concentration inequalities, rom walk, finite samples. Abstract For a fixed unit vector a = a 1,a,...,a n ) S n 1, we consider the n sign vectors ǫ = ǫ 1,ǫ,...,ǫ n ) { 1,1} n the corresponding scalar products a.ǫ = n i=1 a iǫ i. In [] the following old conjecture has been reformulated, stating that of the n sums of the form ±a i it is impossible that there are more with n i=1 ±a i > 1 than there are with n i=1 ±a i 1. The result is of interest in itself, but has also an appealing reformulation in probability theory in geometry. In this paper we will solve 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 rom variables, we will show that P n := P{ n S n n} 1/. Hence, there is already 50% of the probability mass between minus one one stard deviation. This lower bound is sharp, it is much better than for instance the bound that can be obtained from application of the Chebishev inequality our bound will have nice applications in rom walk theory. The method of proof is of interest in itself useful for obtaining exact expressions very precise estimations for the tail probabilities P n for finite samples. For instance, it turns out that already for n > 7 the tail probabilities are above 4719/ , whereas according to the central limit theorem the limit for these probabilities is The problem studied in this paper has been communicated to the author by Vidmantas Bentkus in the beginning of the year 010 this paper is devoted to his memory. 1

2 1 Introduction result For a positive integer n, let ǫ 1,ǫ,...,ǫ n be i.i.d. Rademacher rom variables let a = a 1,a,...,a n ) be a vector of positive real numbers with n i=1 a i = 1. The following unsolved problem has been presented in [3] is attributed to B. Tomaszewski. In [], Conjecture 1.1, this old conjecture has been reformulated: P a 1 ǫ 1 +a ǫ +...+a n ǫ n 1) 1/. This problem is at least 5 years old even in the uniform case where, a 1 = a =... = a n = n 1/, to our best knowledge the problem seems still to be unsolved. In the uniform case the maximum possible value of Sn n is n, where the conjecture states that S n := ǫ 1 +ǫ +...+ǫ n 1) P n := P{ S n n} = P{ n ǫ i n} 1/. ) The present paper solves the conjecture in this special case. It means that at least 50% of the probability mass is between minus one one stard deviation from the mean, which is quite remarkable. We note that i) the problem can be easily reformulated in terms of Binomialn,1/) rom variables since η i := ǫ i +1)/,i = 1,,...,n, are independent Bernoulli1/) rom variables hence P n := P{ S n n} = P{ n i=1 i=1 ǫ i n} = P{ n n T n n+ n }, 3) where T n = η 1 +η +...+η n is a Binomialn,1/) rom variable. Note that T n = S n +n. ii) easy calculations show that the sequence P n ) is not monotone in n; iii) the method we use is also suitable to obtain sharp lower bounds upper bounds for probabilities concerning a stard deviations a > 0): The result is as follows. P{ S n a n} = P{ n ǫ i a n}. i=1

3 Theorem 1.1. Let ǫ 1,ǫ,...,ǫ n be independent Rademacher rom variables, so that P{ǫ 1 = 1} = P{ǫ 1 = 1} = 1/ let S n P n be defined as in 1) ). Moreover, let where Then, for k = 1,,..., we have N := {0,1,,3,...} = C k, k=1 C k = {n : k n+1 < k +1) } = = {k 1,k,k +1,...,k +1) 3,k +1) }. 4) Q k := P k+1) = min n C k P n, 5) Q + k := P k = max n C k P n. 6) Moreover, the sequence Q k ) is monotone increasing, the sequence Q+ k ) is monotone decreasing lim k Q k = lim k Q + k = Φ1) 0.68, 7) where Φ is the stard normal distribution function. It follows that = 1, for n = 1, = 1, for n =, 0.55, for n = 3,4,5,6,7, P n so that certainly for n N P n so that certainly for n , for n = 8,9,...,14, 0.60, for n 15. P n 1/, = 1, for n = 1, = 1, for n =, 0.88, for n = 3,4,5,6,7,. 0.8, for n = 8,9,...,14, , for n , for n 4. P n 7/8. Note that trivially P 1 = 1, P = 1/ by definition P 0 = 1. 3

4 Remark 1.. It will turn out that our method of proof also gives a way to compute P n, for n = 1,,..., in a recursive way. More precisely, let n C kn for some k n n = k n 1+i for some i {0,1,...,k n }, then + k n 1 k= { P{S k = k}+ +P{S k n = k n }+ k j=1,j=odd i j=1,j=odd Proof of the Theorem P n = 1/+ P{S k +j = k +1} P{S k +j = k n +1} k j=1,j=even i j=1,j=even Note that because of the symmetry in the distribution of S n we have P{S k +j = k} P{S k +j = k n }. P n := P{ n S n n} = P{1 n S n 1 1+ n}. We will compare the probability P n with P n 1 := P{ n 1 S n 1 n 1}. Notice that S n takes on values with positive probability only in the set D n, where D n = {0,,,4, 4,...,n, n} for n is even D n = {1, 1,3, 3,...,n, n} for n is odd. Also note that the different values of D n have at least a distance, so that there can be at most one element of D n in a half open interval of lenght. In fact P n = P n 1 P{S n 1 B n }+P{S n 1 Ãn} = P n 1 P{S n 1 B n }+P{S n 1 A n }, 8) } + where B n = [ n 1,1 n), Ã n = n 1,1+ n] Since A n = [ 1 n, n 1). 0 < n n 1 < 1 the difference between P n P n 1 will be rather small only some integers have to be taken into account. In fact, it follows that B n = 1 n n 1) 0,1), A n = 1+ n n 1) 1,) 4

5 A n B n = [ 1 n,1 n) =, where C is the length of the interval C. It is clear that A n B n contains exactly one even integer one odd integer the set B n contains at most one negative integer the set A n contains at most two negative integers. If we restrict ourselves to the even integers, then exactly one of the sets A n B n contains one even integer. The other set contains no even integer. The same statement holds if we restrict ourselves to the odd integers. The main problem is that we have to show that P n 1/, for n {,3,4,...}. We will solve the problem separately for blocks of integers related with two adjacent squares. Note that N = {0,1,,3,...} = C k, where C k,1 := {k,k +,...,k+1) 3} = C k, := {k 1,k +1,...,k +1) } = k=1 C k = C k,1 C k,, 9) { {n C k : n is even}, if k is even, {n C k : n is odd}, if k is odd, { {n C k : n is even}, if k is odd, {n C k : n is odd}, if k is even. Note that C k contains k+1 non-negative integers, C k,1 contains k non-negative integers, C k, contains k +1 non-negative integers. In case k is even odd), then C k,1 consists of all even odd) integers in C k C k, consists of all odd even) integers in C k, since Furthermore, we have [k = even] [k 1 = odd]. { k,1 k} A n B n = [ 1 n,1 n), for n = k 1 { 1 k, k} A n B n = [ 1 n,1 n), for n C k \{k 1}. Moreover, for n C k we have P{S n 1 A n B n } = P{S n 1 [ 1 n,1 n)} = { P{S n 1 = 1 k}, for n C k,1, P{S n 1 = k}, for n C k,. For each n C k by using 8) we can compute P n from P n 1 by subtracting P{S n 1 B n } or adding P{S n 1 A n }. It will turn out that for the integers n C k, \{k 1} we have 5

6 negative contributions due to P{S n 1 B n }, whereas for the integers n C k,1 {k 1} we have positive contributions due to P{S n 1 A n }. For instance, for n = 3 we obtain P{S Ã3} = P{S A 3 } = P{ǫ 1 +ǫ [ 1 3, )} = P{ǫ 1 +ǫ = } = 1/ = 1/4, for n = 4 we obtain P{S 3 Ã4} = P{S 3 A 4 } = P{ǫ 1 +ǫ +ǫ 3 [ 3, 3)} = P{ǫ 1 +ǫ +ǫ 3 = 3} = 1/ 3 = 1/8, for n = 7 we obtain P{S 6 B 7 } = P{ǫ 1 +ǫ +...+ǫ 6 [ 6,1 7)} = P{ǫ 1 +ǫ +...+ǫ 6 = } = More generally, we obtain the following lemma: Lemma.1. For k =,3,... we have the following. i) For n = k 1 we have so that k A n = [ 1 k 1, k ), P{S n 1 A n } = P{S k = k} = k kk 1)/ 1 ; P{S k n 1 B n } = ii) For n C k,1 we have for n = k +i, with i = 0,1,...,k 1, so that 1 k A n = [ 1 k +i, k +i 1), P{S n 1 A n } = P{S k +i 1 = k +1} = k +i 1 kk 1)/+i 1 ; P{S k +i 1 n 1 B n } = 0. iii) For n C k, \{k 1} we have for n = k +1+i, with i = 0,1,...,k 1, so that k B n = [ k +i,1 k +1+i), P{S n 1 B n } = P{S k +i = k} = k +i kk 1)/+i ; P{S k +i n 1 A n } = 0. 6

7 n P{S n 1 A n } P{S n 1 B n } P n Table 1: Table for C 1 n P{S n 1 A n } P{S n 1 B n } P n 3 0) 3 = ) ) ) ) Table : Table for C For k =,3,4 also for general k {,3,...,} the results of the lemma above are illustrated in the Tables -5. The starting points where n = 1 n = are indicated in Table 1 separately. The proof of the lemma will be given in Section 3. 7

8 n P{S n 1 A n } P{S n 1 B n } P n 8 7 ) ) ) ) ) ) ) Table 3: Table for C 3 n P{S n 1 A n } P{S n 1 B n } P n ) ) ) ) ) ) ) ) ) 0.60 Table 4: Table for C 4 8

9 n A n B n k 1 [ 1 k 1, k ) k [ 1 k, k 1) k +1 [ k,1 k +1) k + [ 1 k +, k +1) k +3 [ k +,1 k +3) k +4 [ 1 k +4, k +3) k +1) 3 [ 1 k +1) 3, k +1) 4) k +1) [ k +1) 3,1 k +1) ) Table 5: Intervals for C k n integer P{S n 1 A n } P{S n 1 B n } k 1 k A n P{S k = k} = k k 1 k A n P{S k 1 = k +1} = k kk 1)/ 1) k kk 1)/ 1) k +1 k B n P{S k = k} = k k + 1 k A n P{S k +1 = k +1} = k +1 kk 1)/) k +3 k B n P{S k + = k} = k k +4 1 k A n P{S k +3 = k +1} = k +3 kk 1)/+1) k k 1 k +1 4 kk+1)/ ) k+1) 4 k +1) 3 1 k A n P{S k+1) 4 = k +1} = k+1) kk 1)/) k + kk 1)/+1) k + 3 kk+1)/ 1) k+1) 3 k +1) k B n P{S k+1) 3 = k} = k+1) Table 6: Probabilities for C k 9

10 To complete the proof one has to show that the other P n for n C k cannot be smaller than P k+1). This can be done by comparing a contribution due to B n with the lower but adjacent contribution due to A n : Lemma.. For i = 0,1,...,k 1, let δ i := P{S k +i 1 = k +1} P{S k +i = k} = Then we have The proof of this lemma will be given in Section 3. k 1+i k +i kk 1)/+i 1 k 1+i k +i kk 1)/+i ). 10) δ 0 δ 1... δ k 1 < 0. 11) The remaining line of proof of Theorem 1.1 is now as follows. Because of the monotony of the δ s we have k 1 [P{S k = k}+kδ 0 0] [P{S k = k}+ δ i 0] [P k P k+1) ] = so that we obtain indeed k kk 1)/ 1 k k i=0 P{S k = k}+kδ 0 k 1 kk 1)/ 1 P k P k+1). Next, using that the δ s are negative, we have for k =,3,..., since also Q k := P k+1) = min n C k P n, P k+1) P k+1) 4... P k 1 k k kk 1)) = 0, P k+1) P k+1) 3 P k+1) 5... P k. Since trivially P k 1 P k, it also follows that which completes the proof of the theorem. Q + k := P k = max n C k P n, 10

11 3 Proof of the Lemmas In this section we will deliver the proof of Lemma.1 the proof of Lemma.. Proof of Lemma.1: Suppose for the moment that n 1 is even so that the values of S n 1 can be only even integers with positive probability. Since A n B n = since the interal A n B n is half open, there will be exactly one even integer in A n B n, so that either P{S n 1 A n } = 0 or P{S n 1 B n } = 0. The same argument applies in case n 1 is odd. To prove the first part of the lemma, let n = k 1. Since we have 1 k < 1 k 1 k < k k A n = [ 1 k 1, k ). 1) Moreover, there is exactly one possible value of S n 1 in the interval A n B n = [ 1 k 1,1 k 1) since n 1 = k k are either both even or both odd, the only possible value of S n 1 in the interval A n B n is k A n, which implies that P{S n 1 B n } = 0 P{S n 1 A n } = P{S n 1 = k}. This last probability can easily be calculated by using 3). Next, let n C k,1, so that n = k +i, for some i {0,1,...,k 1}. Note that so that 1 k +i 1 k k < 1 k +i, { 1 k, k} A n B n = [ 1 k +i,1 k +i). Moreover, since n 1 = k + i 1, 1 + k are either both even or both odd, the only possible value of S n 1 in the interval A n B n is 1 k A n, which implies that P{S n 1 B n } = 0 P{S n 1 A n } = P{S n 1 = 1+k}. This last probability can easily be calculated again. Finally, for n C k, \{k 1}, a similar reasoning can be followed. Proof of Lemma.: We will use for i = 0,1,...,k 1 the identity n n+1 n n n 1 = n 1 j n+1 { }, j j +1 j j +1 with n = k 1+i j = kk 1)/+i 1, obtain k 1+i kk 1)/+i 1 k δ i = k +i kk 1)+i < 0. 11

12 Moreover, for i {0,1,...,k }, we find [ [δ i δ i+1 ] k 1+i kk 1)/+i 1 k kk 1)+i k +i k ) 1+i+ kk 1)/+i k +i+ k kk 1)+i+ ] 4 [ kk 1)+i k +i)k +i+1) {kk 1)/+i}{kk +1)/+i+1}{kk 1)+i+} ] [+3i+k 0]. This last inequality holds trivially, so that 11) holds. Acknowledgement. The author is indebted to Harrie Hendriks for careful reading the manuscript for giving useful comments, which have led to a simplification in the proof of the main theorem. References [1] P. Billingsley: Probability measure, Wiley, New York, [] R. Holzman D.J. Kleitman: On the product of sign vectors unit vectors, Combinatorica 1 3) 199) [3] R.K. Guy: Any answers anent these analytical enigmas?, Amer. Math. Monthly ),