ebook img

On a property of the simple random walk on $\mathbb{Z}$ PDF

0.09 MB·
Save to my drive
Quick download
Download
Most books are stored in the elastic cloud where traffic is expensive. For this reason, we have a limit on daily download.

Preview On a property of the simple random walk on $\mathbb{Z}$

On a property of the simple random walk on Z Norio Konno,1 ∗ Hayato Saigo2 †and Hiroki Sako3 ‡ 1 DepartmentofAppliedMathematics,FacultyofEngineering,YokohamaNationalUniversity,, Hodogaya,Yokohama,240-8501,Japan 2 NagahamaInstituteofBio-SciencesandTechnology Tamura,Nagahama,526-0829,Japan 3 FacultyofEngineering,NiigataUniversity, 7 Nishi-ku,Niigata950-2181,Japan 1 0 2 n a Abstract Thesubject of thispaper is thesimple random walk on Z. Wegive avery simple answer tothefollowing J 7 problem: under the condition that a random walk has already spent α-percent of the traveling time on the positive 2 ] side Z≥0,what is theprobability that therandom walk is now on thepositive side? R P The symmetric random walks which step 2n-times can be decomposed in thefollowing two ways: (1) how many . h t timesthewalk stepsonthepositiveside,(2) whetherthelast stepisonthepositivesideoronthenegativeside. To a m answer the problem above, we clarify the number of the walks classified by (1) and (2). It has been already known [ 1 thatthedistributionofthenumberindicated by(1)makesthearcsine law. Combiningwiththedecomposition with v 6 respect to (2), we obtain a decomposition of thearcsine law into theMarchenko-Pastur law. 4 9 7 0 . 1 1 Introduction 0 7 1 : An invisible simple random walker on Z suddenly telephoned to you at the time t = 2n and informed you v i X that she has spent 70 persent of time on the positive side, from t=0 to t=2n. r a Question: What is the probability that she is on the positive side now (t=2n)? Answer: 70 persent. ...But WHY? Inthe presentpaper,we will explainWHY (andto show some remarkablecorollalryconnecting classical ∗[email protected] †h [email protected][email protected] Key words. Random walk, Arcsine law, Marchenko-Pastur law 1 probability theory and noncommutative probability). In Section 2 we briefly review basic definitons and knownresults onthe simple randomwalkonZ. We provethe maintheorem(Theorem5)andasacorollary we explain WHY (Theorem 6) in Section 3. In the last section we show that the theorem implies the new limit theorem connecting simple random walk with positive ends and the Marchenko-Pasturlaw. 2 Preliminaries on the simple random walk on Z notation: discrete time interval [m,n]:={m,m+1,...,n} for integers satisfying m≤n. Definition 1 (P ,P0) We denote the set of all possible paths of simple random walk X(t) on discrete time N N interval [0,N]={0,1,2,3,...,N} by P . The set of all possible paths of simple random walk X(t) on time N interval [0,N] such that X(N)=0 is denoted by P0. N These sets can be written as P = {X: [0,N]→Z | X(0)=0,|X(t)−X(t−1)|=1 (t∈[1,N])}, N P0 = {X ∈P | X(N)=0}. N N The number of the elements of P is 2N. The number of the elements of P0 can be written as follows: N N Theorem 1 (Feller [1], Chap III Sec 4 Theorem1-(a)) 2n , N =2n, (n=0,1,2,3,...) #P0 = n N (cid:26) 0(cid:0), (cid:1) N is odd. Definition 2 (positive side) Let X(t) be a simple random walk on Z. For positive integer t, we say that “X(t) is on the positive side” if X(t)≥0 and X(t−1)≥0. In such a case, we also say that the path X is on the positive side at t. Definition 3 (P+) We denote by P+ the set of all the random walks X(t) defined for t∈[0,N] which are N N on the positive side at N. Definition 4 (soujourn time) Let X(t) be the simple random walk on Z. The soujourn time on the positive side T is defined by N T :=#{t∈[1,N] | X(t) is on the positive side.}. N 2 Definition 5 (P ,P0 ,P+ .) We define P by N,m N,m N,m N,m P ={p∈P | p is a path for which T =m is satisfied.}. N,m N N We also define P0 :=P ∩P0 and P+ :=P ∩P+. N,m N,m N N,m N,m N For #P and #P0 , the following general formula is well-known. 2n,2k 2n,2k Theorem 2 (Arcsine law for simple random walk, Feller [1], Chap III Sec 5 Theorem 1) 2k 2n−2k #P = . 2n,2k (cid:18)k(cid:19)(cid:18) n−k (cid:19) Theorem 3 (Uniform principle, Feller [1], Chap III Sec 2 Theorem 3) 1 2n #P0 = , for every k ∈[0,n]. 2n,2k n+1(cid:18)n(cid:19) Then how about #P+ ? In the case that k = n, it is obvious that #P+ = #P , so Theorem 2 2n,2k 2n,2n 2n,2n gives the following: Proposition 4 (Feller [1], Chap III Sec 4 Theorem 1-(c)) 2n #P+ = . 2n,2n (cid:18)n(cid:19) Theorem 5 in the next section gives a general formula for #P+ . As a collorary (Theorem 6), we get an 2n,2k answer to the question “WHY” in the introduction. 3 Main Theorem Lemma 1 1 2n−2l 2l−2 #P+ = · ·2 . 2n,2k n−l+1 (cid:18) n−l (cid:19) (cid:18)l−1(cid:19) l∈X[1,k] Proof. Since the proposition is trivial for the case of k=0,n, we focus on the case of k∈[1,n−1]. It is obvious that for X ∈P+ 2n,2k {t∈[1,2n−1] | X(t)=0}6=∅. Let 2τ := Max{t ∈ [1,2n−1] | x(t) = 0}. Then τ ∈ [n−k,n−1]. We can totally decompose the set P+ by the values of 2τ. It is easy to see that X(t) is on the positive side for 2k−(2n−2τ) times in 2n,2k [0,2τ] because X(t) is always (i.e. for 2n−2τ times) on the positive side; and more strongly, 3 • X(2τ +1)=1 • X(t)≥1 on [2τ +1,2n−1] • X(2n)=X(2n−1)+1 or X(2n−1)−1. Itiseasytoseethatthenumberofpathsdefinedon[2τ+1,2n]satisfyingX(2τ+1)=1andX(t)≥1for all t∈[2τ+1,2n−1] is nothing but 2·#P+ . Hence, fromthe aboveargumentand Theorem 2n−2τ−2,2n−2τ−2 3, we obtain #P+ = #P0 ·(2·#P+ ) 2n,2k 2τ,2k−(2n−2τ) 2n−2τ−2,2n−2τ−2 τ∈[nX−k,n−1] 1 2τ 2(n−τ)−2 = ·2 τ +1(cid:18)τ (cid:19) (cid:18)(n−τ)−1(cid:19) τ∈[nX−k,n−1] 1 2n−2l 2l−2 = ·2 , n−l+1(cid:18) n−l (cid:19) (cid:18)l−1(cid:19) l∈X[1,k] where l :=n−τ. ✷ Lemma 2 Let k,l,n be positive integers and 1≤k ≤n−1. Then the identity 1 2n−2l 2l−2 k 2k 2n−2k · ·2 = n−l+1 (cid:18) n−l (cid:19) (cid:18)l−1(cid:19) n(cid:18)k(cid:19)(cid:18) n−k (cid:19) l∈X[1,k] holds. Proof. We show the identity by induction with respect to k. (i) The case that k =1 is trivial. (ii) Suppose that the lemma holds for k =m<n−1. Then we have 1 2n−2l 2l−2 · ·2 n−l+1 (cid:18) n−l (cid:19) (cid:18)l−1(cid:19) l∈[1X,m+1] 1 2n−2l 2l−2 1 2n−2m−1 2m = · ·2 + · ·2 . n−l+1 (cid:18) n−l (cid:19) (cid:18)l−1(cid:19) n−m (cid:18) n−m−1 (cid:19) (cid:18)m(cid:19) l∈X[1,m] 4 By the induction hypothesis, the above quantity is m 2m 2n−2m 1 2n−2m−2 2m + · ·2 n(cid:18)m(cid:19)(cid:18) n−m (cid:19) n−m (cid:18) n−m−1 (cid:19) (cid:18)m(cid:19) m 2m (2n−2m)(2n−2m−1) 2n−2m−2 1 2n−2m−2 2m = + · ·2 n(cid:18)m(cid:19) (n−m)2 (cid:18) n−m−1 (cid:19) n−m (cid:18) n−m−1 (cid:19) (cid:18)m(cid:19) 2 2m 2n−2m−2 m = (2n−m−1)+1 n−m(cid:18)m(cid:19)(cid:18) n−m−1 (cid:19)nn o 2 2m+2 (m+1)2 2n−2m−2 (n−m)(2m+1) = n−m(cid:18)m+1(cid:19)(2m+2)(2m+1)(cid:18) n−m−1 (cid:19) n m+1 2m+2 2n−2m−2 = n (cid:18)m+1(cid:19)(cid:18) n−m−1 (cid:19) m+1 2(m+1) 2n−2(m+1) = . n (cid:18) m+1 (cid:19)(cid:18) n−(m+1) (cid:19) This is nothing but the identity for k=m+1. By (i) and (ii) the identity is proved for any k ∈[1,n−1]. ✷ The following is the main theorem of the present paper: Theorem 5 k 2k 2n−2k #P+ = . 2n,2k n(cid:18)k(cid:19)(cid:18) n−k (cid:19) Proof. For k =0, it is trivial. For k =n, it follows from Proposition 4. The equation for the case of 0<n<k directly follows from Lemma 1 and Lemma 2. ✷ As a corollary,we obtain an explanation for WHY: Theorem 6 For the simple random walk on Z, the identity below holds. k P(X(2n)isonthepositiveside| T =2k)= . 2n n Proof. By the definition of the conditional probability and from Theorem 2 and Theorem 5, we have #P+ k 2k 2n−2k k P(X(2n) is on the positive side | T =2k)= 2n,2k = n k n−k = . 2n #P (cid:0)2k (cid:1)(cid:0)2n−2k (cid:1) n 2n,2k k n−k (cid:0) (cid:1)(cid:0) (cid:1) ✷ This is why “the probability that we can now find on the positive side a simple random walker who has already spent 70 percent of time on the positive side is 70 persent.” 5 4 Marchenko-Pastur Law For the Brownian motion x(τ), it is well known that 1 r dx P(µ(τ ∈[0,t] | x(τ)≥0)≤rt)= , π Z x(1−x) 0 p where µ(dx) = dx denotes the Lebesgue measure of the real line. The fact above is called the Arcsine law [1]. Based on the theorems in the previous section and on the standardtransition from digital to analog,we easily obtain the arcsine law for the Brownian motion x(τ) with positive ends: Theorem 7 We have 2 r xdx 2 r x P(µ(τ ∈[0,t] | x(τ)≥0)≤rt)= = dx, π Z x(1−x) π Z r1−x 0 0 p and dually, 2 r (1−x)dx 2 r 1−x P(µ(τ ∈[0,t] | x(τ)≤0)≤rt)= = dx, π Z x(1−x) π Z r x 0 0 p where µ(dx)=dx denotes the Lebesgue measure of the real line. The probability law above is nothing but the Marchenko-Pastur law, which plays fundamental and universal roles in the theory of random matrices and free probability. It strongly suggests that there is some hidden relationship between the classical probability, combinatorics, random matrices and quantum probability. References [1] W.Feller, An Introduction to Probability Theory and its Applications 1 (2nd edn.), JohnWiley &Sons, New York, 1957. 6

See more

The list of books you might like

Most books are stored in the elastic cloud where traffic is expensive. For this reason, we have a limit on daily download.