Random Walks in the Plane: An Energy-Based Approach. Three prizes offered by SCM

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1 Société de Calcul Mathématique S Mathematical Modellig Compay, Corp Tools for decisio help sice 995 Radom Wals i the Plae: Eergy-Based pproach Three prizes offered by SCM Berard Beauzamy ugust 6 I. Itroductio We cosider a simple radom wal i the plae: a sequece of radom variables values, probability / i each case. Let S X X with be the sum of the first variables ; the iitial value (at time ) is. We are iterested i quatitative estimates of the behavior of S whe becomes large. well-ow result is Khitchie's law of the iterated logarith (94): almost surely, whe : S limsup ad Log Log S limif Log Log We mae two commets:. Such estimates are ot quatitative at all first attempt to obtai quatitative estimates, usig the origial probabilistic proof, was made by the author i : Siège social et bureaux :, Faubourg Sait Hooré, 758 Paris. Tel : Fax : Société oyme au capital de 56 Euros. RCS : Paris B SIRET : PE : 79Z

2 . Their probabilistic appearace is misleadig Looig at such a statemet, everyoe has the impressio that, for a give player, there are some uow forces which will, sooer or later, brig his fortue close to Khitchi's curves (a Khitchi curve is of the form y x Log Log x time, the game is oly govered by the rule, with equal probability. ). This is completely wrog; at ay What Khitchi's laws say, ad, more geerally, what ay result about radom wals says, is that there are more paths with some properties tha paths with other properties. They are ot idividual results about each path; they are results about the umber of paths with a give characteristic. Such results are i fact of combiatorial ature. For istace, at time, the proportio of paths which ever touched the curve y teds to whe. I order to prove such results, we tae here a completely differet approach, which is ot probabilistic aymore, but relies upo a cocept derived from "eergy absorptio". Our aim is also to obtai quatitative estimates, of the form: Give a curve the istat? y x II. Basic settigs, what is the proportio of paths which ever touched the curve before We cosider that, at time, a uit of eergy is put at the origi. This uit will the divide itself i two halves, at time, oe at the poit, ad oe at the poit,. More geerally, every time a divisio poit is met, the available eergy divides equally ito the two possible paths. So, for istace, at the time, 3 poits will receive some eergy, amely, receives /4,, receives /,, the sum is always. s it is well-ow, the values of that, obviously, at ay time, we have follows, for simplicity, we cosider oly the eve values): Lemma. - Let, paths from to, is: receives /4. t ay step, i this cofiguratio, S are eve if is eve, ad are odd if is odd. We observe S be the poit of coordiates N,. The followig Lemma is well-ow (i what,, with,...,. The umber of SCM Radom Wals Prizes, 6/8

3 Proof of Lemma If we wat to reach this poit i steps, we eed x times the value ad y times the value -, with x y ad x y, which gives x, y. So there are paths, which proves the result. We write N for the total umber of paths, startig at, fiishig at. possible x Sice there is a total of eergy equal to: possible paths at time, each poit, receives a amout of e, The repartitio of eergy is give by a biomial law: there is more eergy at the cetral poits ad very little eergy at the extreme poits (, ad, ). I this prelimiary approach, the total amout of eergy remais the same. Now, we itroduce a curve, y x, located i the upper half-plae (the same holds for the lower half-plae, of course), ad we wat to ivestigate the probability that the radom wal, up to time, S for all,...,. remais costatly below this curve, which meas that Our represetatio, i order to ivestigate this pheomeo, will be the fact that the curve absorbs the eergy. This meas that, for ay path which touches the curve, the correspodig eergy disappears. I this example, the poit seds its eergy to both B ad C, but B is o the curve we wat to ivestigate, so this part of the eergy disappears, ad we are left with ec e. SCM Radom Wals Prizes, 6/8 3

4 The curve we wat to ivestigate will be called the critical curve. It may be cosidered as a "blac frotier" (i the sese of a blac hole), meaig that it absorbs all eergy it receives, ad seds bac othig. We have: Propositio. - Let x y be ay critical curve, i the upper half-plae. The total eergy left, at time, is equal to the total probability to reach ay of the poits, below the curve, that is satisfyig, without ever touchig the curve at ay time before ( ). Proof of Propositio This is a mere rephrasig of the disparitio of eergy. y time a path touches the curve, it is aihilated, so what remais is the paths which ever touched the curve. If a time is fixed, ad a curve is fixed, we will call admissible a path with ever touches it (at ay time ). For ay poit i the plae, let N be the umber of admissible paths which reach, Propositio states that: ad ad Nad pad the probability to reach by a admissible path. e, pad, III. The case of a horizotal lie We ow compute the umber of admissible paths whe the critical curve is a horizotal lie segmet: Lemma 3. - Let y ( ) be a horizotal lie segmet. Let,, with coordiates,, be ay poit that the radom wal may reach, with. The umber of paths, startig at, fiishig at,, N, where, is the symmetric of,, which touch the horizotal segmet at a time before is with respect to the lie segmet. Proof of Lemma 3 This property is well-ow, uder the ame of "reflexio priciple" : SCM Radom Wals Prizes, 6/8 4

5 Let B be the first time a path touches the segmet (there may be several). There are as may paths from B to tha from B to. Sice the coordiates of, are,4, the symmetric of, is. So the,4 umber of paths which touch the segmet y at ay time before is, by Lemma : N,4 Therefore, the umber of paths which reach, without ever touchig the segmet y is: N ad, Propositio 4. - ssume that our critical curve is the lie segmet y. The distributio of eergy at time is: e, Ideed, this follows immediately from the previous Lemma. We observe that, due to the absorptio, this distributio of eergy is ot symmetric aymore. Startig at ad movig dowwards, it first icreases, reaches its maximum for some ad the decreases. Propositio 5. - ssume that our critical curve is the lie segmet y. The eergy left at time is: SCM Radom Wals Prizes, 6/8 5

6 e Proof of Propositio 5 Let us loo at the picture below: The critical lie segmet y has two effects : No poit, above this segmet receives ay eergy at all; there is a drop of total eergy equal to the probability to reach this poit; For every poit strictly below this segmet, there is a drop of eergy equal to the probability to reach its symmetric. Sice both terms are equal, the total drop of eergy (that is the total eergy "swallowed" by the segmet), istead of reachig the poits, ot o the segmet, is. Now, there is the sigle poit, which is its ow symmetric ad should be couted oly oce; this proves Propositio 5. Corollary 6. - Let be a give threshold. The probability that the radom wal ever reaches this threshold at ay time t is : It teds to whe. p ad Ideed, whe, SCM Radom Wals Prizes, 6/8 ad, sice. 6

7 IV. The critical curve y x We ow ivestigate the critical curve y x.. Iitial step t the begiig, we put a eergy equal to at the origi. The oly admissible move (that is, a path which does ot touch the critical curve) is X, which leads us to the poit, ; this poit receives a eergy equal to / (the other half is lost). Now, we compute the eergy which arrives o the vertical V 4. Oly the poits with eve coordiates may be reached, amely,,,,, 4 (see picture above). We have : The symmetric 4,4. Therefore : N,,, of, with respect to the lie segmet y has coordiates N,, The eergy set by, to ay of the poits, is, taig ito accout the atteuatio: SCM Radom Wals Prizes, 6/8 7

8 e 3 3 4, 4 We fid: e. 6 e,, e,,, 4 B. Geeral step We ow ivestigate the geeral step of the iductio. We discretize the x axis usig the poits of coordiate Let be the poit of coordiates,,, with ad the y axis usig the poits.,..., (so there are poits). We wat to compute the eergy received by ay of the poits eergy received by all the B,,. I order to simplify our otatio, we set, such owig the,. The poits ad B are below the curve if, respectively, ad. Let e be the eergy received by each poit at the ow) ; we wat to compute eb, the eergy available at the ad st iductio step (supposed to be th iductio step. The umber of paths from to B is : Let B be the symmetric of N B ( ) B with respect to the curve y B,. The umber of paths from to B is: N B 4 ; the coordiates of B are: We observe, i these formulas, that ad caot have the same parity: the sum the differece alteratively odd ad eve. must be odd, otherwise there is o path. Ideed, the squares Therefore, for ad, the amout of eergy set by to B is: or are SCM Radom Wals Prizes, 6/8 8

9 e e B 4 The total amout of eergy available at each poit B is obtaied, summig upo : e eb 4 d the total amout of eergy available at time is the sum of these quatities, summig upo : e ev 4 C. Complete formula at step Let us set, for we have, for every : w ;, 4, ;, ;, ;, e w w w ad: ;, ;, ;, e V w w w Here, V deotes the vertical at time with. We wat to prove that, below the curve, that is the set of all poits ev whe., D. alysis of a sigle term Let us loo at the fial sum i the above expressio, that is : SCM Radom Wals Prizes, 6/8 9

10 D ; 4 Simple computatios show that it may be writte: D ; 3 ssume that is eve, m, so is odd,. The umber of terms i the above sum is m ; this umber is miimal for (we have term). It is maximal (amely ) if we have ad D ;. this case, 3 We also observe that the sum exteds o both sides of the media term., that is. I s we already sait, let V deote the vertical at ; The quatity e V e D ;,, that is the set of poits, D may ve viewed as the proportio of eergy set by. We have:, to the whole vertical V. It is a "atteuatio coefficiet", which is. We observe that the total quatity of eergy at step, ev satisfies: ; e V e D e e V,, so it is decreasig at each step (this was obvious, sice some eergy disappears ad o eergy is created). Moreover, the maximum value of the eergy is decreasig at each step: max max e e,, We also observe that the itroductio of the critical curve leads to the fact that some eergy disappears, because some paths do ot exist aymore. It does ot create ay ew path. Therefore, all "blocs" of terms which had their eergy tedig to zero i the ormal scheme will have the same property i the ew scheme. SCM Radom Wals Prizes, 6/8

11 V. Prizes offered fter this presetatio of the subect, ad of some basic properties, we would lie to offer three prizes.. Prize : 5 Euros I the case of the critical curve y vertical, after atteuatio due to the curve, teds to. x (see above), prove that the eergy received by the th Recall that, i the above otatio, V is the set of poits o the curve. So we wat to show that ev, whe. th vertical which lie below the Of course, sice this quatity is decreasig, it is eough to do it for a subsequece, but we wat quatitative estimates, amely, for ay, we wat to ow explicitly the value of such that for all, ev. B. Prize : Euros Fix ay ad cosider Khitchi's critical curve y xlog Log x. Prove that ev, whe ; same as above : we wat explicit quatitative estimates. C. Prize 3 : Euros Fix ay lim ev ad cosider Khitchi's critical curve y xlog Log x ; same as above : we wat explicit quatitative estimates.. Prove that D. Geeral rules for participatio Each prize will be give oly oce, to the best cotributio. cotributio will be cosidered oly if it is well writte ad complete. ll details should be give. Please sed the cotributios (i Eglish or i Frech) to cotact@scmsa.com o later tha Jue 3 th, 7. Everyoe may participate (idividuals, istitutios, ad so o). SCM Radom Wals Prizes, 6/8