On the length of the longest consecutive switches

Transcription

1 On the length of the longest consecutive switches arxiv:8.0454v [math.pr] 2 Nov 208 Chen-Xu Hao, Ze-Chun Hu and Ting Ma College of Mathematics, Sichuan University, China November 3, 208 Abstract An unbiased coin is tossed n times independently and sequentially. In this paper, we will study the length of the longest consecutive switches, and prove that the limit behaviors are similar to that of the length of the longest head-run. Keywords: longest consecutive switches, longest head-run. Mathematics Subect Classification F5 Introduction An unbiased coin with two sides named by head and tail respectively, is tossed n times independently and sequentially. We use 0 to denote tail and to denote head. For simplicity, we assume that all the random variables in the following are defined in a probability space Ω,F,P. Let {X i,i } be a sequence of independent and identically distributed random variables with P{X 0} P{X } 2. Let S 0 0,S n X + +X n, n,2,..., and IN,K max 0 n N K S n+k S n, N K, N,K N.. Denote by Z N the largest integer for which IN,Z N Z N. Then Z N is the length of the longest head-run of pure heads in N Bernoulli trials. The statistic Z N has been long studied because it has extensive applications in reliability theory, biology, quality control, pattern recognition, finance, etc. Erdös and Rényi 970 proved the following result. Corresponding author: College of Mathematics, Sichuan University, Chengdu 60065, China address: matingting2008@yeah.net

2 Theorem.. Let 0 < C < < C 2 <. Then for almost all ω Ω, there exists a finite N 0 N 0 ω,c,c 2 such that [C logn] Z N [C 2 logn] if N N 0. Hereafter, we denote by log the logarithm with base 2, and by [x] the largest integer which is no more than x. Theorem. was extended by Komlós and Tusnády 975. Erdös and Révész 976 presented several sharper bounds of Z N including the following four theorems among other things. Theorem.2. Let ε be any positive number. Then for almost all ω Ω, there exists a finite N 0 N 0 ω,ε such that if N N 0, then Z N [logn logloglogn +logloge 2 ε]. Theorem.3. Let ε be any positive number. Then for almost all ω Ω, there exists an infinite sequence N i N i ω,ε i,2,... of integers such that Z Ni < [logn i logloglogn i +logloge +ε]. Theorem.4. Let {γ n } be a sequence of positive numbers for which n 2 γn. Then for almost all ω Ω, there exists an infinite sequence N i N i ω,{γ n } i,2,... of integers such that Z Ni γ Ni. Theorem.5. Let {δ n } be a sequence of positive numbers for which n 2 δn <. Then for almost all ω Ω, there exists a positive integer N 0 N 0 ω,{δ n } such that Z N < δ N if N N 0. These limit theorems have been extended by many authors. We refer to Guibas and Odlyzko 980, Samarova 98, Kusolitsch and Nemetz 982, Nemetz and Kusolitsch 982, Grill 987 and Vaggelatou The distribution function of Z N and some related problems have been studied by Goncharov 943, Földes 979, Arratia et al. 989, Novak 989, 99, 992, Schilling 990, Binswanger and Embrechts 994, Muselli 2000, Vaggelatou 2003, Túri 2009, Novak 207. Mao et al. 205 studied the large deviation behavior for the length of the longest head run. Anush 202 posed the definition of switch, and considered the bounds for the number of coin tossing switches. Li 203 considered the number of switches in unbiased coin-tossing, and established the central limit theorem and the large deviation principle for the total number of switches. According to Li 203, a head switch is the tail followed by a head and a tail switch is the head followed by a tail. Motivated by the study of the longest head-run and the work of Li 203, we will study the length of the longest consecutive switches in this paper. At first, we introduce some notations. For m,n N, define n+m S n m H X i X i, im+ n+m S n m T X i X i. im+ 2

3 Then S m n H resp. S m n T denotes the number of head switches resp. tail switches in the trials {X m,x m+,...,x m+n }. Set S m n S n m H+Sm n T..2 Then S m n denotes the total number of switches in the sequence {X m,x m+,...,x m+n }. For i,n N, define H N n,i {S n m i m i+n n+ n }, n,...,n. Then ω H N n,i implies that there exists at least one sequence of consecutive switches of length n in the sequence {X i ω,x i+ ω,...,x i+n ω}. Define M i N max n N { n H N n,i },.3 which stands for the number of switches of the longest consecutive switches in the sequence {X i,x i+,...,x i+n }. When i, we denote M N instead of M N. We use A + A 2 + +A n instead of A A 2 A n when the sets A i,i,...,n are disoint. The rest of this paper is organized as follows. In Section 2, we give some illustration about the difference between M N and Z N. In Section 3, we present main results and some remarks. The proofs will be given in Section 4. 2 The difference of M N and Z N In this section, we want to give some illustration about the difference between M N and Z N. Suppose that N N,N 2. Then by the definitions of M N and Z N, we know that For k {,...,N}, define M N {0,,...,N }, Z N {0,,...,N}. A k N {M N k }, B k N {Z N k}. Consider a sequence {X i,...,x i+k } of length k. Without loss of generality, we assume that k is an even number. If it has pure heads, then we have only one case X i X i+k. If it has consecutive switches of length k, then we have the following two cases: X i,x i+ 0,,X i+k 2,X i+k 0, or X i 0,X i+,,x i+k 2 0,X i+k. 3

4 Based on the above rough analysis, we might conecture that PA k N 2PBk N. However, it is not true. In the following, we use some concrete examples to illustrate that the relation between M N and Z N is complex. Case : N 2 Table : M 2 0 Events {00} {} {0} {0} Number of Events 2 2 Table 2: Z Events {00} {0} {0} {} Number of Events 2 Case 2: N 3 Table 3: M Events {000} {} {00} {00} {0} {0} {0} {00} Number of Events Table 4: Z Events {000} {00} {00} {00} {0} {0} {0} {} Number of Events 4 2 4

5 Case 3: N 4 Table 5: M Events {0000} {} {000} {000} {0} {0} {00} {00} {00} {00} {000} {0} {0} {000} Number of Events {00} {00} Table 6: Z Events {0000} {000} {000} {0} {00} {0} {000} {000} {00} {0} {0} {00} {00} {00} {00} {} Number of Events Case 4: N 5 Table 7: M Events {00000} {} {0000} {0000} {0} {0} {00} {000} {000} {00} {000} {00} {00} {000} {00} {000} {0000} {0} {0000} {0} {00} {0000} {00} {000} {000} {00} {00} {000} {000} {00} Number of Events {00} {000} 5

6 Table 8: Z Events {00000} {0000} {0000} {000} {000} {00} {0} {} {0000} {0000} {000} {000} {00} {0} {0000} {000} {00} {00} {00} {000} {000} {00} {00} {0} {000} {000} {0} {00} {0} {000} {00} {00} Number of Events From the above concrete examples, we might say that the distributions of M N and Z N are very different. In the rest of this paper, we will show that as N, their limit behaviors are similar. 3 Main results and some remarks In this section, we will present several limit results on M N. At first, we present the following result, which is similar to the one about the length of the longest head run obtained by Rényi 970. Theorem 3.. We have lim N M N logn a.s. 3. Corresponding to Theorems.2-.5, we have the following four theorems. Theorem 3.2. Let ε be any positive number. Then for almost all ω Ω, there exists a finite N 0 N 0 ω,ε such that if N N 0. M N [logn logloglogn +logloge ε] : α N 3.2 Theorem 3.3. Let ε be any positive number. Then for almost all ω Ω, there exists an infinite sequence N i N i ω,εi,2,... of integers such that M Ni < [logn i logloglogn i +logloge+ε] : α 2 N i. 3.3 Theorem 3.4. Let {γ n } be a sequence of positive numbers for which n n 2 γn. Then for almost all ω Ω, there exists an infinite sequence N i N i ω,{γ n } i,2,... of integers such that M Ni γ Ni. Theorem 3.5. Let {δ n } be a sequence of positive numbers for which n n 2 δn <. Then for almost all ω Ω, there exists a positive integer N 0 N 0 ω,{δ n } such that M N < δ N if N N 0. 6

7 Remark 3.6. From Theorems , we can say that M N and Z N have similar limiting behaviors. Remark 3.7. The closely related result with respect to Theorems 3.4 and 3.5 is Guibas and Odlyzko 980, Theorem. Although their language is different from ours, we think that their results and ours are equivalent by using a transformation. 4 Proofs 4. Proof of Theorem 3. Step. We prove lim inf N M N logn a.s. 4. For any 0 < ε <, N N and N 2, we introduce the following notations: t [ εlogn]+, N [N/t], U k S tk+ t, k 0,,..., N, where S tk+ t is defined by.2. Then the sequence {U k,0 k N} of random variables are independent and identically distributed with It follows that U k t and P{U k t } 2 2 t 2 t. P{U 0 < t,u < t,,u N < t } By a simple calculation, we get that N Then by the Borel-Cantelli lemma, we get that N+ <. 2 t lim inf N By the arbitrariness of ε, we obtain 4.. Step 2. We prove lim sup N M N logn ε. M N logn 7 N+. 2 t a.s. 4.2

8 For any ε > 0 and N N, we introduce the following notations: u [+εlogn]+, A N N u+ k {S k u u }. We have P{S u k u } N and thus PA 2 u N. For any T N with Tε >, 2 [+εlogn] k N, it holds that PA k T 2 k k k T k 2 T+ε k <, ktε which together with the Borel-Cantelli lemma implies that k T u+ { PA k T i.o. P lim sup S i u u } 0. k It follows that lim sup k Let k T < n < k + T. By 4.3, we have i0 M k T a.s. 4.3 logkt M n M k+ T +εlogk + T +2εlogk T +2εlogn with probability for all but finitely many n. Hence 4.2 holds. 4.2 Proofs for Theorems The basic idea comes from [6]. For the reader s convenience, we spell out the details. At first, as in [6, Theorem 5], we give an estimate for the length of consecutive switches, which is very useful in our proofs for Theorems Theorem 4.. Let N,K N and let M N be defined in.3 with i. Then if N 2K. K +2 [ N K ] P M 2 K N < K K +2 [ 2 [N K ]] K To prove Theorem 4., we need the following lemma. Lemma 4.2. Let N,m N and let M m N P be defined in.3. Then M m 2N N N +2 2 N

9 Proof. Since {M i 2N,i N} are identically distributed, we only consider the case that i in the following. Let Then we have and Hence A {M 2N N }, A k {M k+ N N }, k 0,,...,N. A A 0 +Ā0A +Ā0ĀA 2 + +Ā0Ā A N A N, 4.6 PA 0 2 N, PĀ0Ā A k A k 2N, k,...,n. PA PA 0 +PĀ0A +PĀ0ĀA 2 + +PĀ0Ā A N A N 2 + N N 2 N +2 N 2. N Proof of Theorem 4.. Let N,K N with N 2K. Denote B {M + K K }, 0,,...,N K, l+k [ ] N 2K C l B, l 0,,...,. K lk Thenforanyl 0,,...,[ N 2K],wehaveC K l {M lk+ 2K K },andforanyl 0,,...,[ N 2K] K 2, we have C l C l+2. By Lemma 4.2, we know that for any l 0,,...,[ N 2K], K PC l K +2 2 K. Let D 0 C 0 +C C 2[ ]], 2 [N 2K K 4.7 D C +C C 2[ ] ]+. 2 [N 2K K By the independence of {C 0,C 2,...,C 2[ 2 [N 2K K ]]}, we have P D 0 P C 0 P C 2...PC 2[ ]] K +2 [ 2 [N 2K K ]] [N 2K K 2 K Similarly we have P D K +2 [ 2 [N 2K K ] ] K By the obvious fact that D 0 {M N K }, we get that PM N < K P D 0 9 K +2 [ 2 [N 2K K ]] K

10 In the following we prove that PD D 0 PD PD 0. To this end, by 4.7, it is enough to prove that for any i 2l, l 0, [ 2 [N 2K K ]], PD C i PD PC i. Below we give the proof for l 0 and the proofs for l,,[ 2 [N 2K ]] are similar. We omit K them. For i,...,k +, denote Then we have and F i {X i,,x i+k is the first section of consecutive switches of length K in the sequence X,,X 2K }, F i F, i ; C 0 K+ PF P{X X K has consecutive switches} 2 K, PF i P{X,,X +K has consecutive switches,x X } 2K, i 2,...,K. By the independence of {X,,2...,N}, we have i F i, PD F PD PF, 4. PD F 2 P D {X 2,,X K+ has consecutive switches, X X 2 } P D {X 2,,X K+ has consecutive switches, X X 2,X K+ } +P D {X 2,,X K+ has consecutive switches, X X 2,X K+ 0}} 2P D {X 2,,X K+ has consecutive switches, X X 2,X K+ } 2P D {X K+ } P {X 2,,X K has consecutive switches, X K 0, X X 2 } 2 KPD PD PF 2, 4.2 0

11 PD F 3 suppose that K 3 P D {X 3,,X K+2 has consecutive switches, X 2 X 3 } P D {X 3,,X K+2 has consecutive switches, X 2 X 3, X K+,X K+2 0,} +P D {X 3,,X K+2 has consecutive switches, X 2 X 3, X K+,X K+2,0} 2P D {X 3,,X K+2 has consecutive switches, X 2 X 3, X K+,X K+2 0,} 2P {X 3,,X K has consecutive switches, X 2 X 3,X K } P D {X K+,X K+2 0,} 2 K 2P D {X K+,X K+2 0,} 4P D {X K+,X K+2 0,} PF 3. By the definition of D, we know that P D {X K+,X K+2 0,} P D {X K+,X K+2 0,0}, P D {X K+,X K+2 0,} P D {X K+,X K+2,}, P D {X K+,X K+2 0,} P D {X K+,X K+2,0}, which together with 4.3 implies that 4.3 PD F 3 PD PF 3 {P {X K+,X K+2 0,} D P {XK+,X K+2 0,0} D +P } {X K+,X K+2 0,} D P {XK+,X K+2,} D PF Similarly, if K 4, we have that PD F i PD PF i, i 4,...,K. 4.5 Finally, by the definitions of D and F K+, we know that D F K+ F K+. Hence we have PD F K+ PF K+ PD PF K By 4., 4.2, 4.4, 4.5 and 4.6, we obtain PD C 0 It is easy to check that K+ i PD F i K+ i PF i PD PD PC 0. PD D 0 PD PD 0 P D D0 P D P D 0.

12 Then by 4.8 and 4.9, we get P D 0 D K +2 [ 2 K As to the right-hand side of 4.7, we have 2 [N 2K K ]]++[ 2 [N 2K K ] ] i when [ N 2K ] is even, the exponential part on the right-hand side is equal to K [ N 2K ] + [ N 2K ] [ N 2K ] +2 +; 2 K 2 K K ii when [ N 2K ] is odd, the exponential part on the right-hand is equal to K [ N 2K ] [ [ N 2K ] 2 K 2 + ] [ N 2K ] +2 2 K 2 K Hence PM N < K P D 0 D By 4.0 and 4.8, we complete the proof. To prove Theorem 3.2, we need the following lemma. +. K +2 [ N 2K K ] K Lemma 4.3. Let {α, } be a sequence of positive numbers. Suppose that lim α a > 0. Then we have and Proof. α log If a > 2, then p loga > and thus + logα α α log α log log α log α 2 < + if a > 2, + if a 2. + loga logα If a 2, then p loga and thus + logα log α 2 loga + + loga logα loga + logα. p logα <. loga p logα. loga 2

13 Proof of Theorem 3.2. Let N be the smallest integer with α N +. Since lim /2, there exists M N such that M > 2 and Then by Theorem 4., we have P { M N < α N } where e : /2 2, > M. 2 M : β + β + β +2 β [ 2 [N ]] 2 +2 [ 2 [N ]] + 2 M+ M+ M+ M M+ [ 2 [N ]] N N 2 2 e N 2 +2, +2. By lim + x x x e, we have Without loss of generality, we assume that e 2 for any > M. By α N, we have +2 [ 2 [N ]] 2 lim e e. 4.9 logn logloglogn +logloge ε, and thus N 2+ε loglogn, log < loglogn 2. loge 3

14 Thus M+ e N 2 +2 For any ε > 0, by 4.9, we have M+ M+ M+ ε e e 2ε loge e 2ε loge 2 N loglogn +2 log. M+ lim e 2 loge e 2 ε loge e ln2 2 ε 2 2ε > 2. e 2 2+ε loglogn loge Then by Lemma 4.3 and the Borel-Cantelli lemma, we complete the proof. To prove Theorem 3.3, we need the following version of Borel-Cantelli lemma. Lemma 4.4. [2] Let A,A 2,... be arbitrary events, fulfilling the conditions n PA n and Then P lim supa n. n lim inf n k<l n PA ka l k<l n PA kpa l Proof of Theorem 3.3. Let δ > 0. Let N N δ be the smallest integer for which α 2 N [ +δ ] with α 2 N given by 3.3. Let A {M N < α 2 N },. 4.2 By Theorem 4., we have PA : [ ] N [+δ ]+3 [ +δ ]+ 2 [+δ ]+ [+δ ]+3 2 [+δ ]+ f N +δ ]+3 2 [+δ]+ [ [ +δ ]+ 2 [+δ ]+ [ +δ ]+3 [+δ ]+3 N 2 ]+ [+δ [ +δ ]+,

15 where f [+δ ]+3 2 [+δ ]+ 2 [+δ ]+ [ +δ ]+3. As in 4.9, we have Then there exists M N such that M > 2 and By 3.3 we have which implies that Then by 4.22 and 4.24, we get lim f e f 2, > M logn logloglogn +logloge+ε [ +δ ]+, N 2 [+δ ]+ loglogn 2 ε loge. PA M PA + M+ +δ ]+3 f [ ]+ loglogn [ +δ 2 ε loge Let 0 < ε 0 < satisfy [ +δ ] [logn logloglogn +logloge+ε] > ε 0 logn,,2,... Then logn < ε 0 +δ, which implies that loglogn < log ε 0 ++δlog Hence we have M+ +δ ]+3 f [ ]+ loglogn [ +δ 2 ε loge M+ M+ f f [ +δ loglogn ]+3 [+δ]+ 2 logloge ε [ +δ ]+3 log +δlog ε 0 [+δ]+ 2 logloge ε For any given positive number ε, by 4.23, we have +δ ]+3 [ +δ lim f [ +δ ]+ 2 ε loge 2 +δ 2 ε 2, 4.28 when δ is small enough. By Lemma 4.3, 4.25, 4.27 and 4.28, we get that n PA n. 5

16 Recall that {A, } is defined in 4.2. For i <, we define { {MNi < α B i, 2 N }, if N i α 2 N, Ω, otherwise, { } C i, M N i N N i < α 2 N. We claim that PA PB i, PC i, +o as In fact, by the definitions of A,B i, and C i,, we know that { } A B i, C i, M N i α 2 N + 2α 2 N < α 2 N By Lemma 4.2, we have P M N i α 2 N + 2α 2 N < α 2 N P 4.30 and 4.3 imply Similar to 4.29, we have By 4.29 and 4.32, we get which implies that M N i α 2 N + 2α 2 N < α 2 N α 2N +2 2 α 2N as. 4.3 PA i A PA i PC i, +o as PA i A PA ipa +o as, PB i, PA i A PA i PA +o PB i, +o as By the fact that n PA n and 4.33, we know that 4.20 holds. By Lemma 4.4, we complete the proof. Proof of Theorem 3.4. Let A n {M n γ n }. By Theorem 4. and Taylor s formula, we have PA n P{M n < γ n } γ n +2 [ 2 [ n γn ]] 2 [ γn n ]] γ n Define A {n N : n 2γ n < }. Then we have k Ak2 γk k A γ n +2 2 γn [ 2 k 2 k <. k/2 2k/2 6 k γn +2 +o. 2 γn

17 It follows that there exists a subsequence {n k,k } such that n k and n k 2 γn k. 2γ nk Then PA n n 2 PA nk k { γnk +2 [ 2 γn k 2 k γ nk +2 2 γn k k γ nk +2 2 γn k k k k [ nk ]] } +o γ nk nk 2γ nk 3 2 +o n k 2γ nk + n k 2 γn k + k γ nk o k 2 γn k γ nk +2 2 γn k 32 +o Then the first condition in Lemma 4.4 holds. By following the method in the proof of Theorem 3.3, we have which implies that PĀiĀ PĀiPĀ PA i A PA i PA +o as, +o as. Then we get that 4.20 holds and thus by Lemma 4.4 we complete the proof. Proof of Theorem 3.5. Let B n {M n δ n }. By Theorem 4. we have PB n P{M n < δ n } δ n +2 n δn δn By Taylor s formula, we have δ n +2 n δn n δn +2, 2 δn δ n 2 δn which together with 4.35 implies that n PB n 2 δn +2 2 n 2 δn < +. δn δ n n n By the Borel-Cantelli lemma, we get PB n i.o. 0 and thus obtain the result. Acknowledgments This work was supported by National Natural Science Foundation of China Grant No and 8784 and the Fundamental Research Funds for the Central Universities of China. 7 n

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