Random Walks on Discrete and Continuous Circles. by Jeffrey S. Rosenthal School of Mathematics, University of Minnesota, Minneapolis, MN, U.S.A.

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1 Radom Walks o Discrete ad Cotiuous Circles by Jeffrey S. Rosethal School of Mathematics, Uiversity of Miesota, Mieapolis, MN, U.S.A (Appeared i Joural of Applied Probability 30 (1993), ) Summary. We cosider a large class of radom walks o the discrete circle Z/(), defied i terms of a piecewise Lipschitz fuctio, ad motivated by the geeratio gap process of Diacois. For such walks, we show that the time util covergece to statioarity is bouded idepedetly of. Our techiques ivolve Fourier aalysis ad a compariso of the radom walks o Z/() with a radom walk o the cotiuous circle. Keywords. Radom walk, Fourier aalysis, Time to statioarity, Geeratio gap. 1

2 1. Itroductio. This paper cosiders certai radom walks o the group Z/() of itegers mod. Our radom walks will coverge i total variatio distace to the uiform distributio U o Z/(). We shall be cocered with the rate of this covergece, as a fuctio of, the size of the group. It is kow that for simple radom walk o Z/(), where we move left or right oe space, or remai do t move, each with (say) probability 1 3, it takes O(2 ) steps to approach the uiform distributio i total variatio distace (see Diacois [1], Chapter 3C, Theorem 2). Ideed, it is easily verified that for ay radom walk o Z/() i which the size of a sigle step is bouded idepedetly of, at least O( 2 ) steps are required to approach uiformity. Various faster covergece results have bee obtaied whe the step distributio itself is chose radomly; see Hildebrad [4] ad Dou [3]. I this paper, we shall cosider radom walks whose step distributio grows liearly with. This study was motivated by the followig geeratio gap algorithm o Z/() suggested by Diacois [2]. Cosider the radom walk o Z/() which begis at the idetity, ad at each step moves to aother poit with probability proportioal to the distace (o Z/()) betwee the two poits. I other words, the further aroud the circle a poit is, the more likely the process is to jump there o its ext tur. If we cotiue to jump aroud o Z/() i this maer, how log (as a fuctio of ) will it take util our distributio is roughly uiform? The geeratio gap algorithm is a special case of the followig set-up. We let f be a o-egative, real-valued Lipschitz fuctio o the uit circle, ad we embed Z/() i i the obvious way. The, for each, we cosider the probability distributio P o Z/() iduced by usig the restrictio of f to Z/() to defie the step distributio. We shall allow to get large, but keep the fuctio f fixed. We show that uder such coditios, the total variatio distace of the radom walk o Z/() after m steps to the uiform distributio U ca be bouded by a quatity which is idepedet of, ad which goes to zero expoetially quickly as a fuctio of m. I this sese, we shall say that such radom walks coverge to uiform o Z/() i a costat umber of steps. This paper is orgaized as follows. Sectio 2 gives a precise statemet of our mai 2

3 result. Sectio 3 gives two examples of the use of this result. Fially, Sectio 4 proves the result, usig Fourier Aalysis o abelia groups. The key idea is to relate the radom walks (ad their Fourier coefficiets) o Z/() to the correspodig oes o the cotiuous circle. The result will the follow from stadard results about Fourier Aalysis. 2. Defiitios ad Mai Result. We begi with some stadard defiitios. Give two probability distributios P ad Q o Z/(), we defie their variatio distace by P Q = 1 2 We defie their covolutio P Q by P Q(s) = j Z/() j Z/() P (j) Q(j). P (j)q(s j) ; P Q is thus a ew probability distributio o Z/(), which represets the distributio after startig at the idetity, ad takig oe step accordig to P, the a secod step accordig to Q. Give a distributio P o Z/(), it iduces a radom walk o Z/() which starts at the idetity ad has step distributio give by P. Thus, its distributio after m steps is give by P m, the m-fold covolutio product of P with itself. We let U be the uiform distributio o Z/(), ad cosider the variatio distace P m U as a fuctio of m ad. (This assumes we have bee give a distributio P o Z/() for each.) The usual questio is, as a fuctio of, how large must m be to make the above variatio distace small? It is easily see usig Fourier aalysis that if P has bouded support (i.e. P is ozero oly o a eighbourhood of 0 Z/(), ad the size of this eighbourhood is bouded as a fuctio of ), the m must be of size O( 2 ) (for large ) to make the variatio distace small. I this paper, we cosider families of distributios P defied differetly, ad show that m eed oly be of size O(1) to make the variatio distace small. 3

4 To defie our radom walk, let f be a o-egative real-valued fuctio o the cotiuous circle, which satisfies the followig piecewise Lipschitz ad positivity coditios: (A1) The circle ca be decomposed ito J itervals I 1, I 2,..., I J such that for some positive costats L ad α, ad for 1 j J, f(x) f(y) L x y α for all x, y It(I j ). (A2) f(x) > 0 for some x It(I j ), for some j; Idetify with the iterval [0, 2π) i the obvious way. For each, defie a measure P o Z/() by P (j) = f ( ) 2πj f ( ), j Z/(). 2πs 1 s=0 (Note that the hypotheses o f imply that 1 s=0 f ( ) 2πs is positive for all but fiitely may.) The measure P is thus obtaied by regardig Z/() as sittig iside ad usig the values of f o Z/() as weights for P. Let P m be the m-fold covolutio product of P with itself, ad let U be the uiform distributio o Z/(). The mai result of this paper is the followig. Theorem 1. Uder the above assumptios, there are positive costats A ad B (depedig o f but ot o or m) such that the radom walk o Z/() satisfies P m U A e Bm, for all m ad for all but fiitely may. We shall prove the above Theorem usig Fourier Aalysis ad the Upper Boud Lemma of Diacois ad Shashahai (see Diacois [1]). The proof is preseted i Sectio 4. The key idea of the proof is to relate the radom walks o the discrete circles Z/() to a sigle radom walk o the cotiuous circle, iduced by the same fuctio f. For large, the radom walk o Z/() will be similar to the radom walk o, ad thus the rates of covergece will be related to the sigle rate of covergece for the radom walk o. 4

5 I Sectio 3 below, we cosider two examples of the use of Theorem 1. We coclude this Sectio with a remark about the ecessity of the restrictios o f. Remark. The requiremet that f satisfy a Lipschitz coditio (except at a fiite umber of poits), or some similar coditio, is ideed ecessary. It is ot sufficiet that f be merely, say, a bouded L 1 fuctio o. Ideed, let p 1 < p 2 < p 3 <... be a icreasig sequece of prime umbers, ad defie the fuctio f o by 0 if x = 2πj for some i ad some iteger j 0, ±1 f(x) = p i 1 otherwise. The f is a bouded L 1 fuctio which does ot satisfy (A1) above. O the other had, o Z/(p i ), the iduced radom walk is simple radom walk (it moves distace oe to the right or left, or does ot move, each with probability 1/3). Thus, for = p i, O( 2 ) steps are required to approach uiform. Hece, the coclusio of Theorem 1 is ot satisfied. 3. Examples. I this Sectio we preset two simple examples of the use of Theorem 1. Example 1. The Geeratio Gap process. Diacois [2] has proposed the followig process o Z/(). At each step, move from a poit x to a poit y with probability proportioal to the distace from x to y aroud the circle Z/(). Ituitively, each step of this algorithm attempts to move as far as possible from the previous positio (aalogous to childre attemptig to be as differet as possible from their parets). I terms of Theorem 1, we ca formulate this process by defiig a fuctio f o [0, 2π) by f(x) = mi(x, 2π x). The measures P iduced by this fuctio f are precisely those that geerate the Geeratio Gap process. Thus, Theorem 1 shows that a costat umber of steps suffices to approach the uiform distributio o Z/(). I other words, the Geeratio Gap process mixes up very quickly. 5

6 Example 2. Radom walk with large step size. Cosider the radom walk o Z/() which at each step moves to oe of the d earest eighbours (d 1) with equal probability. I the cotext of Theorem 1, this is the radom walk o Z/() iduced from the fuctio f(x) = { 1, x πd or x 2π πd 0, πd < x < 2π πd Thus, agai oly a costat umber of steps are required to approach the uiform distributio o Z/(). This is i cotrast to the O( 2 ) steps that are required whe the step size does ot grow with. Results of Dou [3] imply that for most choices of probability measures supported o d poits of Z/(), a costat umber of steps suffices to approach the uiform distributio. There are obviously may other examples of uses of Theorem 1; ew examples ca be obtaied simply by varyig the fuctio f. 4. Proof of Theorem 1. I this Sectio we prove Theorem 1. We begi with a review of the relevat facts from Fourier aalysis. Give a probability measure P o Z/(), we defie its Fourier coefficiets by 1 (1) a,k = e 2πikj/ P (j), 0 k 1. Similarly, give a probability measure P o, we defie its Fourier coefficiets by (2) a k = e ikx P (dx), k Z. We let U be the uiform distributio o, ad let U be the uiform distributio o Z/(). The Upper Boud Lemma of Diacois ad Shashahai (see Diacois [1], Chapter 3C) states that for the radom walk o Z/() with step distributio give by P, its variatio distace to uiform distributio U after m steps satisfies (3) P m U a Z/(),k 2m. 4 6 k=1

7 Similarly, for the radom walk o with step distributio give by P, its variatio distace to the uiform distributio U o satisfies (4) P m U <k< k 0 a k 2m. The task at had is to show that for the problem uder cosideratio, the sum i (3) ca be bouded by A e Bm for some positive costats A ad B idepedet of. To that ed, we let f be a o-egative fuctio o which satisfies (A1) ad (A2) above. This implies that M = sup f(x) f(y) x,y is fiite. Without loss of geerality, we take α 1 (which must be true if f is ot piecewise costat). For coveiece, we assume that f(x)dx = 1; if ot, we ca divide f by its L 1 orm, ad modify the costats L ad M appropriately. We let P be the probability measure o defied by dp = f(x)dx, ad (for sufficietly large ) let P be the probability measure o Z/() defied by P (j) = f ( ) 2πj f ( ). 2πs 1 s=0 We defie the Fourier coefficiets a,k ad a k by (1) ad (2) above. The pla will be to show that for large, a,k will be close to a k. This will allow us to boud the sum i (3) idepedetly of. Remark. The remaider of this sectio essetially amouts to obtaiig various bouds o a,k ad a k ad o the relatioship betwee them. There is of course a log history of bouds o Fourier coefficiets, ad we do ot claim ay great ovelty i our methods or results. I particular, the bouds of Lemma 2 (a) ad Propostio 4 (b) follow easily from stadard techiques such as those i [5] (see Theorem therei). However, for the discrete Fourier coefficiets a,k ad the coectio betwee a,k ad a k, our bouds such 7

8 as Lemma 2 (b) ad Propositio 4 (c) do ot appear to follow immediately from stadard results. We proceed as follows. fuctios o by the equatio For each > 0, we defie the operator T o the set of (T g)(x) = g ( ) [x/2π] /2π where [y] deotes the greatest iteger ot exceedig y. Thus, (T g) is a slight modificatio [ ] of the fuctio g, which is costat o itervals of the form 2πj, 2π(j+1). The beefit of the operator T comes from otig that e ikx (T k f)(x)dx = 0. Furthermore, (T g) provides a lik betwee the fuctio g o, ad the restrictio of the fuctio g to Z/(), as the followig Lemma shows. Lemma 2. For ay fuctio g o with g(x) 1 ad g(x) g(y) x y for all x ad y, if 3, (a) (b) Proof. at 2πj. (T (gf))(x) g(x)f(x) dx (T (gf))(x)dx 4MJπ 1 ( ) 2πj g P (j) 4MJπ, ( π ) α + 2π(L + M). ( π ) α + 2π(L + M). For (a), we break up ito itervals, each of legth 2π, with midpoits O J of these itervals, f will have a discotiuity, ad we ca oly boud (T (gf))(x) g(x)f(x) by 2M. O the other pieces, the fuctio gf is easily see to satisfy a Lipschitz coditio with L replaced by L+M, so (T (gf))(x) g(x)f(x) (L+M) ( π α. ) We coclude that (T (gf))(x) g(x)f(x) dx ( π (2M)(J) + ) = 4MJπ ( π ) α (L + M) dx ( ) α 2π + 2π(L + M). 8

9 For (b), we first ote that 1 g( 2πj )P (j) = = = 1 2π ( g( 2πj 2πj )f( ) 1 ( f( 2πj ) (T (gf))(x)dx (T f)(x)dx ) 2π (T (gf))(x)dx (T f)(x)dx ). Hece, Now, 1 1 g( 2πj )P (j) (T (gf))(x)dx 1 = 1 (T f)(x)dx g( 2πj )P (j) 1 (T f)(x)dx. (T f)(x)dx = (f (T f)(x)) dx f (T f)(x) dx, so the result follows from part (a) by settig g(x) = 1. The Lemma ad the triagle iequality immediately imply Corollary 3. g(x)f(x)dx 1 ( ) 2πj g P (j) 8MJπ 9 ( π ) α + 4π(L + M).

10 Usig the Lemma ad the Corollary, we prove Propositio 4. The Fourier coefficiets a,k ad a k defied by equatios (1) ad (2) satisfy (a) For all k 0, a k < 1. (b) For all k 0, a k 4MJπ k (c) For ay 3 ad 0 k 1, a,k a k 8MJπ (d) For ay 3 ad 0 < k 1, a,k < 12MJπ k ( ) α π + 2π(L + M). k ( π α + 4π(L + M) ) ( π ) α + 6π(L + M). k (e) For each k > 0, there is a umber b k, 0 < b k < 1, such that a,k < b k for all Proof. sufficietly large. For (a), we recall that a k = e ikx f(x)dx, ad ote that by the assumptios o f, it is positive o some ope iterval, o which e ikx does ot have costat argumet. Thus, the iequality i the statemet a k < e ikx f(x) dx = f(x)dx = 1 is strict. so that For (b), we recall that a k = S 1 e ikx (T k f)(x)dx = 0, e ikx ( ) f(x) (T k f)(x) dx f(x) (T k f)(x) dx, 10

11 ad the boud ow follows from Lemma 2 (a), with g(x) = 1 for all x. For (c), recall that a,k a k = 1 e 2πikj/ P (j) e ikx P (dx), ad use Corollary 3 with g(x) = e ikx. Statemet (d) is immediate from statemets (b) ad (c), the triagle iequality, ad the observatio that k = k <. For (e), we ote that (a) ad (b) imply that a k < 1 ad a k 0. Thus, if we let a = max { a k, k > 0}, the a < 1. Hece, if we set b k = a k + 1 a 2, the 0 < b k < 1. Also, part (c) implies that a,k < b k provided is chose large eough that 8MJπ + 4π(L + M) ( ) α 2π < 1 a 2. Propositio 4 allows us to complete the proof of Theorem 1, but we first record a corollary about the radom walk o itself. Corollary 5. The radom walk o iduced by the measure P coverges to the uiform distributio U o expoetially quickly i total variatio distace. Proof. From equatio (4) above, the variatio distace P m U 2 is bouded by the S1 sum 1 a k 2m ; 4 k Z k 0 it suffices to boud this sum by a expressio of the form C 1 e C 2m. From part (b) of Propositio 4, there is a costat C 3 such that a k < C 3 k. The, writig α a k 2m = k Z k 0 0 k (2C 3 ) 1/α a k 2m + 11 k >(2C 3 ) 1/α a k 2m,

12 we see that the secod sum ca be easily be bouded by a itegral, ad show to be less tha a expoetially decayig fuctio of m. There are oly a fiite of terms i the first sum, so the first sum decays expoetially by part (a) of Propositio 4. We ow proceed to the proof of Theorem 1. From part (d) of Propositio 4, there is a costat C 4 such that a,k < C 4 k α usig part (e) of Propositio 4, the variatio distace P m 1 1 a,k 2m k=1 1 k=1 for 1 < k <. The usig equatio (3) above, ad ( mi( C ) 2m 4 k α, b k) 1 (b k ) 2m <k (2C 4 ) 1/α U is bouded by (2C 4 ) 1/α <k< ( ) 2m C4 where i this last expressio we sum over all itegers k, icludig those greater tha. This last expressio is clearly idepedet of. Furthermore, the expressio ca be bouded as i the proof of Corollary 5. Ideed, the first sum i the expressio cosists of a fiite umber of terms, ad clearly decays expoetially with m. The secod ca easily be bouded by a itegral ad show to also decay expoetially with m. k α Hece the sum decays expoetially quickly i m, uiformly i (for sufficietly large), provig Theorem 1. Ackowledgemets. I am very grateful to Persi Diacois for suggestig the Geeratio Gap process, ad for may helpful discussios. I thak the referee for useful commets. 12

13 REFERENCES [1] P. Diacois (1988), Group Represetatios i Probability ad Statistics, IMS Lecture Series volume 11, Istitute of Mathematical Statistics, Hayward, Califoria. [2] P. Diacois, persoal commuicatio. [3] C.C. Dou (1991), Eumeratio ad Radom Walk o Groups, preprit. [4] M.V. Hildebrad (1990), Rates of Covergece of Some Radom Processes o Fiite Groups, Ph.D. dissertatio, Mathematics Departmet, Harvard Uiversity. [5] T. Kawata (1972), Fourier Aalysis i Probability Theory. Academic Press, New York. 13

Integrable Functions. { f n } is called a determining sequence for f. If f is integrable with respect to, then f d does exist as a finite real number

Integrable Functions. { f n } is called a determining sequence for f. If f is integrable with respect to, then f d does exist as a finite real number MATH 532 Itegrable Fuctios Dr. Neal, WKU We ow shall defie what it meas for a measurable fuctio to be itegrable, show that all itegral properties of simple fuctios still hold, ad the give some coditios