aa r X i v : . [ m a t h . P R ] S e p The diminishing segment process
Gergely Ambrus Alfr´ed R´enyi Institute of Mathematics, Hungarian Academy of Sciences,PO Box 127, 1364 Budapest, Hungary; e-mail: [email protected]
P´eter Kevei Centro de Investigaci´on en Matem´aticas, Jalisco S/N, Valenciana,Guanajuato, GTO 36240, Mexico; e-mail: [email protected]
Viktor V´ıgh Department of Mathematics and Statistics, University of Calgary2500 University Drive NW Calgary, AB, Canada T2N 1N4 e-mail: [email protected]
Abstract
Let Ξ = [ − , n recursively in thefollowing manner: for every n = 0 , , . . . , let Ξ n +1 = Ξ n ∩ [ a n +1 − , a n +1 + 1], where the point a n +1 is chosen randomly on the segmentΞ n with uniform distribution. For the radius ρ n of Ξ n we prove that n ( ρ n − /
2) converges in distribution to an exponential law, and weshow that the centre of the limiting unit interval has arcsine distribu-tion.
Keywords:
Arcsine law; Continuous state space Markov chain;Poisson–Dirichlet law; Intersection of convex discs.
We consider the following stochastic process. Let Ξ = [ − , n recursively in the following manner: for every n = 0 , , . . . ,let Ξ n +1 = Ξ n ∩ [ a n +1 − , a n +1 + 1] , where the point a n +1 is chosen randomly on the segment Ξ n with uniformdistribution. After n steps, one obtains the segmentΞ n = [ Z n − ρ n , Z n + ρ n ] . The (centre, radius) process ( Z n , ρ n ) is a continuous state space Markovchain. The radius sequence ( ρ n ) is monotonically decreasing, and it is easy Supported by the OTKA grants 75016 and 76099. Supported by the Analysis and Stochastics Research Group of the Hungarian Academyof Sciences. Supported by the OTKA grant 75016, and by NSERC of Canada.
1o see that with probability 1, ρ n → /
2. Moreover, ( Z n ) is a convergentsequence, assuming values on [ − / , / n →∞ Z n by Z .We are interested in the most straightforward questions: (1) What is the distribution of Z n and ρ n for a given n ?(2) What is the asymptotic behaviour of the radius?(3) What is the limit distribution of the centre? Our work was motivated by the following problem formulated by B´alintT´oth (T´oth, 2010) some 20 years ago with K being the unit disc of theplane. Let K = K be a convex body in the R d that contains the origin, anddefine the process ( K n , p n ), n >
1, similarly to the above construction: let p n +1 be a uniform random point in K n , and set K n +1 = K n ∩ ( p n +1 + K ).Clearly, ( K n ) is a nested sequence of convex bodies which converge to a non-empty limit object, again a convex body in R d . What can we say about thedistribution of this limit body? When K is the unit disc the limit objectis almost surely a convex disc of constant width 1. The present note dealswith the 1-dimensional analogue of this problem. Intriguingly, apart fromalmost trivial results, nothing is known about the related questions even inthe plane.Another direction of generalising the problem treated here is to choosethe subsequent centres according to a previously fixed distribution instead ofthe uniform one. Research on this version is currently in progress.The paper is organised as follows. In Section 2 we give a recursion for thedensity function of ρ n , which allows us to explicitly calculate the expectationfor small n ’s. In Section 3 we show that n ( ρ n − /
2) converges in distributionto an exponential law, which actually shows the rapidness of the process.Section 4 contains the limit distribution of the centre, while in the last sectionwe derive a somewhat unexpected connection between the process and thePoisson–Dirichlet distribution.
At the ( n + 1)st step, we separate two cases according to the location of a n +1 . First, if a n +1 ∈ [ Z n + ρ n − , Z n − ρ n + 1], then no change occurs to thesegment, and thus Ξ n +1 = Ξ n . In the second case, when a n +1 is close to oneof the endpoints of Ξ n , the centre moves and the length decreases. Introduceyet two new random processes measuring the change of the location of thecentre by ε n X n = Z n − Z n − , n > , ε n = ± X n > Z n = Z n − , then of no consequence, let ε n = 1).Thus, for n > X n = ( − ρ n − ) /ρ n − uniform on [0 , ρ n − − /
2] w.prob. (2 ρ n − − /ρ n − , (1)moreover, P ( ε n = 1 | X n = 0) = 12 . By definition, Z n = n X i =1 ε i X i , (2)and it is easy to see that ρ n = 1 − n X i =1 X i =: 1 − S n . Thus, with probability 1, P ∞ X i = 1 /
2. By an inductive argument, itfollows that S n has a continuous distribution. Denote by f n ( x ) the probabilitydensity function of S n . Using the Markov property and (1) it is easy toexpress f n +1 in terms of f n : for 0 x / P ( S n +1 > x ) = Z / P ( S n +1 > x | S n = y ) f n ( y ) d y = P ( S n > x ) + Z x f n ( y ) 1 − x − y d y, whence differentiating f n +1 ( x ) = f n ( x ) x − x + 2 Z x f n ( y )1 − y d y. (3)The first few examples are, for 0 x / f ( x ) = 2 f ( x ) = 2 x − x − − x ) f ( x ) = 2 x (2 − x )(1 − x ) + 4(1 − x )1 − x ln(1 − x ) + 4(ln(1 − x )) . In order to calculate the expectation of S n (and thus E ρ n ), we considerthe Taylor series expansion of f n about 0: f n ( x ) = ∞ X k =0 c n,k x k . c n +1 ,k = k + 2 k k − X j =0 c n,j , (4)and hence E S n = Z / xf n ( x )d x = ∞ X k =0 c n,k − ( k +2) k + 2 = 14 ∞ X j =0 c n − ,j ∞ X k = j +1 − k k , (5)which may be useful for obtaining a recursive expression involving only theexpectations. These formulas also enable us to efficiently compute the ex-pectations for relatively small n ’s, see the figure below.
10 20 30 40 50 60 700.260.270.280.290.300.310.32
Figure 1: The first 70 values of n (1 / − E S n ) We determine the asymptotic behavior of ρ n , the radius of the n th segment. Theorem 1.
We have the distributional convergence n (cid:18) ρ n − (cid:19) D −→ Exp(4) . Accordingly, for any k > , lim n →∞ E n k (cid:18) ρ n − (cid:19) k = k !4 k . roof. Since ρ n = 1 − S n , it is equivalent to prove the corresponding limittheorems for 1 / − S n .It follows from (1) that S n +1 = ( S n w.prob. S n / (1 − S n )uniform on [ S n , /
2] w.prob. (1 − S n ) / (1 − S n ).Observe that given S n , S n +1 has the same distribution as max { S n , U } , where U is a uniform random variable on [ S n / , / S n . Let 0 α /
2, and U α , U α , . . . be independent, uniform random variables on [1 / − α/ , / M αn = max i n U αi . It is well-known in extreme valuetheory (Billingsley, 1995, p.192), and easy to check that n (cid:18) − M αn (cid:19) D −→ Exp(4 / (1 + 2 α )) . (6)We say that X stochastically dominates (or just dominates) Y if P ( X >x ) > P ( Y > x ) for all x ∈ R .To obtain a lower estimate, we use that S n /
2, and thus an easyinduction argument shows that M n dominates S n . From this and (6) weobtain P (cid:18) n (cid:18) − S n (cid:19) > x (cid:19) > P (cid:18) n (cid:18) − M n (cid:19) > x (cid:19) → e − x , i.e. lim inf n →∞ P (cid:18) n (cid:18) − S n (cid:19) > x (cid:19) > e − x , x > . The almost sure convergence S n → / / − S n / ≈ /
4, hence for n sufficiently large, S n behaves approximately like M n does. This heuristic idea is made preciseas follows. Fix the small positive numbers β > ε >
0. Since S n → / n sufficiently large, P ( S βn < / − ε ) ε . Moreover, if S βn > / − ε ,then S n is minored by M εn − βn . This can be shown again by induction. Thus,by (6), P (cid:18) n (cid:18) − S n (cid:19) > x (cid:19) = P (cid:18) S n < − xn (cid:19) P (cid:18) S n < − xn (cid:12)(cid:12)(cid:12) S βn > − ε (cid:19) + ε P (cid:18) M ε (1 − β ) n < − xn (cid:19) + ε → e − x (1 − β )1+2 ε + ε , x > n →∞ P (cid:18) n (cid:18) − S n (cid:19) > x (cid:19) e − x (1 − β )1+2 ε + ε. Since ε and β are arbitrary, this gives the distributional convergence.To prove the convergence of the moments, it is enough to show that for any k >
1, the sequence { n k (cid:0) − S n (cid:1) k } ∞ n =1 is uniformly integrable (Billingsley,1995, Theorem 25.12). The fact that S n dominates M / n readily implies P (cid:18) n (cid:18) − S n (cid:19) > x (cid:19) e − x , which shows uniform integrability.In particular, E ρ n = 1 / − / (4 n ) + o ( n − ) and Var( ρ n ) ∼ / (16 n ). Thequestion of obtaining an exact or asymptotic formula for E S n using only therecursive relations (4) and (5) remains open. In this section, we determine the limit behaviour of Z n . Let F ( x ) denote thecumulative distribution function of Z ; based on the definition of Z , it followsthat F ( x ) = 0 if x − / F ( x ) = 1 if x > / Theorem 2.
The distribution of Z is a translated arcsine law: for − / x / , F ( x ) = 2 π arcsin p x + 1 / . Proof.
By (2), we have to determine the limit distribution of P ε i X i . Clearly,this is not affected by the steps where X i = 0. We introduce the thinnedprocess ( e Z n , ˜ ρ n ) as follows: for n >
1, let ξ n be independent Bernoulli(1/2)random variables, and e X n be a uniform random variable on [0 , / − P n − e X i ].The centre of the segment after the n th step of the thinned process is givenby e Z n = P n ξ i e X i , and the radius is ˜ ρ n = 1 − P n e X i . Plainly, Z D = ∞ X i =1 ξ i e X i . Introduce r n = 2 ˜ ρ n −
1. If U , U , . . . are i.i.d. Uniform(0 ,
1) random vari-ables, then setting ( e Z , r ) = (0 , Z n +1 = e Z n + 12 ξ n +1 (1 − U n +1 ) r n ,r n +1 = U n +1 r n . (7)Notice that after choosing ( e Z , r ), the process ( e Z , r ) , ( e Z , r ) , . . . is a scaledand translated copy of the original one, which implies the distributional equa-tion Z D = r Z ′ + e Z , (8)where Z ′ is independent from ( e Z , r ), and has the same distribution as Z .Thus, for every x ∈ [ − / , / F ( x ) = Z / F (cid:18) x − y − y (cid:19) d y + Z / F (cid:18) x + y − y (cid:19) d y = (1 − x ) Z x −∞ F ( z )(1 − z ) d z + (1 + 2 x ) Z ∞ x F ( z )(1 + 2 z ) d z = (1 − x ) Z x − / F ( z )(1 − z ) d z + (1 + 2 x ) Z / x F ( z )(1 + 2 z ) d z + 1 + 2 x . This also shows that F is continuously differentiable, and by differentiatingwe arrive at F ′ ( x ) = − Z x − / F ( z )(1 − z ) d z + 2 Z / x F ( z )(1 + 2 z ) d z + 4 xF ( x )1 − x + 12 . Once again, we derive that F is twice differentiable (being the reason forstarting with the distribution function rather than the density function),whence F ′′ ( x ) = F ′ ( x ) 4 x − x . Taking into account that F ′ ( x ) is a density function on [ − / , /
2] yieldsthat the solution is F ′ ( x ) = 1 π p (1 / x )(1 / − x ) , the desired density function. Setting v i = U U . . . U i (1 − U i +1 ), (7) implies that e Z n = 1 / P n − v i ξ i +1 ,hence the limit Z has the infinite series representation Z = 12 ∞ X i =0 v i ξ i +1 . (9)7t is easy to check that P ∞ v i = 1 almost surely, thus v := ( v , v , . . . ) ∈ ∆,where ∆ = ( x = ( x , x , . . . ) : ∞ X i =0 x i = 1 , x i > , i = 0 , , . . . ) is the infinite dimensional simplex. The construction of the random vector v implies that it has the so-called GEM (Griffiths–Engen–McCloskey) distri-bution with parameter 1 (see the residual allocation model in Bertoin, 2006,p.89). This distribution appears in various contexts, such as prime factori-sation of a random integer (Hirth, 1997); and in particular, the decreasingreordering of the GEM distribution is the so-called Poisson–Dirichlet distri-bution, which is one of the most important distributions in fragmentationtheory, see Bertoin, 2006.Using this terminology and (9), Theorem 2 can be reformulated as: if v = ( v , v , . . . ) is a GEM distributed random vector, and ξ, ξ , ξ , . . . is aniid sequence of Bernoulli random variables such that P ( ξ = 1) = 1 / P ( ξ = − v , then P ∞ i =0 v i ξ i +1 has arcsine distribu-tion. This theorem was first proved by Donnelly and Tavar´e (1987), usingthe construction of the Poisson–Dirichlet distribution by means of an inho-mogeneous Poisson process. Later Hirth (1997) also gave a proof by usingthe method of moments. As far as we know, our proof, solving an integralequation based on the distributional equality (8), is new. Acknowledgements.
Our thanks are due to Imre B´ar´any for introducingthe problem to us and for illuminating discussions, and to Juan Carlos Pardofor bringing the connection with the Poisson–Dirichlet law to our attention.The authors are also grateful to the unknown referee for a number of com-ments and suggestions that improved the paper. While carrying out theresearch, the first named author held a pleasant visiting scholarship at theIsaac Newton Institute for Mathematical Sciences, University of Cambridge.
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Random Fragmentation and Coagulation Processes ,Cambridge University Press.[2] Billingsley, P. (1995)
Probability and Measure , Wiley–Interscience Publi-cation, New York.[3] Donnelly, P. and Tavar´e, S. (1987) The population genealogy of theinfinitely-many neutral alleles model.