Monkeys and Walks. Muhammad Waliji. August 12, 2006

Transcription

1 Mokeys ad Walks Muhammad Waliji August 12, relimiaries We will be dealig with outcomes resultig from radom processes. A outcome of the process will sometimes be deoted ω. Note that we will idetify the set A with the evet that the radom outcome ω belogs to A. Hece, A will mea A occurs, ad A B meas either A or B occurs, whereas A B meas both A ad B occur. For a sequece of evets, A 1, A 2,..., ote that m1 meas that iftely may of the A occur, or that A occurs iftely ofte A i.o.. The evets A 1,..., A are said to be idepedet if A 1 A A 1 A. A ifte collectio of evets is said to be idepedet if every fte subcollectio is idepedet. A 2 The Ifte-Mokey Theorem Before we state the ifte-mokey theorem, we will prove a useful lemma. Cosider the evets A 1, A 2,.... Lemma 2.1 Borel-Catelli. If A <, the A i.o. 0. Furthermore, if the evets A are idepedet, the if A, we have A i.o. 1. roof. Suppose that A <

2 The, m1 A lim m lim m ad hece by 2.1, the limit i 2.2 must be 0. For the coverse, it is eough to show that 0 m1 ad so it is also eough to show that Note that sice 1 x e x, ad by idepedece, m+k m+k exp A A A m+k A. 2.4 But sice, takig k, the last sum i 2.4 diverges, this establsihes 2.3, thereby completig the proof. Loosely speakig, the ifte-mokey theorem states that if a mokey hits keys o a typewriter at radom for a ifte amout of time, he will almost surely produce the etire collected works of Shakespeare! Eve better, he will almost surely do this iftely ofte! Here is the precise statemet: Theorem 2.2 Ifte Mokey. Cosider a ifte-legth strig produced from a fte alphabet by pickig each letter idepedetly at radom, uiformly from the alphabet say the alphabet has letters. Fix a strig S of legth m from the same alphabet. Let E k be the evet the m-substrig startig at positio k is the strig S. The, iftely may of the E k occur with probability 1. roof. Note that the evets E mj+1 are idepedet for j 0, 1,.... Furthermore, E k 1 m. Hece, m 1 E mj+1, j1 so by Lemma 2.1, E mj+1 i.o. 1. j1 2

3 3 Radom Walks i Z d Cosider a Radom Walk a drukard s walk i the d dimesioal lattice Z d. Suppose the that drukard starts out at the origi, ad at each step he moves to ay adjacet poit with equal probability. The questio is, what is the chace that he will retur to the origi? Actually, the questio that we will cosider is, what is the chace that he will retur to the origi iftely ofte? Let us fix some otatio. Let X be the positio of the drukard after steps 0, 1,.... Let i deote the probability that a evet occurs for a radom walk startig a positio i. Let p ij deote the probability that after startig at positio i, the walker is at positio j after steps. Let f ij deote the probability that after startig at positio i, the walker reaches positio j for the first time after stepts. Let f ij deote the probability that the walker evetually reaches j after startig out at i. So, f ij f ij. Deftio 3.1. A positio i is recurret if f 1. It is trasiet if f < 1. Lemma 3.2. Suppose a radom walker starts out at positio i. The, i X i i.o. { 1 if i is recurret 0 if i is trasitive 3.1 roof. Let 1 1 < < k. Let A ij 1,..., k be the evet that X 1 j ad X t j for the other t < k. The, X k Now, i A ij 1,..., k f 1 ij f 2 1 f k k 1 A k : 1,..., k A ij 1,..., k is the evet that X t j at least k times. Also, i A k A ij 1,..., k 1 1< < k f 1 ij f 2 1 f k k f 1 ij f 1 ij k k f 2 1 f 2 k 1 k k 1 +1 f k f k k 1 f ij f k

4 Takig the limit as k i 3.2, we get i X j i.o. i A k k1 lim ia k k { fij if f 1 0 if f < Replacig j with i i 3.3, we obtai 3.1. Before we aswer the questio posed at the begg of the sectio, we would like to have a Borel-Catelli like lemma for the radom walk. However, the X are maifestly ot idepedet. Still, the lemma will hold. Lemma 3.3. The followig hold: i i X i i.o. 0 p <. i X i i.o. 1 p. roof. Note that uder 3.1, i ad are equivalet. We will prove i. Note that follows from Lemma 2.1. We will here prove. We ca coditio p ij o the first t st X t j: Hece, 1 p ij f l ij p l. l0 k1 p k k 1 f k l p l k1 l0 1 l0 l0 1 + p l p l f kl+1 l1 p l f k l f ad therefore, k1 p k 1 f f 3.4 4

5 Now, sice f < 1, 3.4 gives a upper boud for the partial sums of thereby completig the proof. The questio that we posed at the begg of the sectio was aswered by olya i Theorem 3.4 olya. If d 1, 2, the the probability that the walker returs to the origi iftly ofte is 1. For d 3 the probability is 0. This is proved by calculatig the sum ad applyig Lemma 3.3. We will give a proof for the case d 1, 2. First, recall Stirlig s formula:! 2π +1/2 e Therefore, Also, recall 2 p p π k0 k k roof whe d 1. Note that p m 00 0 if m is odd. If m 2, the every particular walk of legth 2 has probability 1/2 2. If of the steps are to the right, ad are to the left, the walker will have retured to the origi. There are 2 2 ways of doig this. Therefore, p /2 2. Summig over all, we have p /2 2 which diverges, ad so we are doe by Lemma π roof whe d 2. Now, the probability of ay particular walk of legth is 1/4. I order that we retur to the origi, there must be the same umber of moves left as there are right, ad there must be the same umber of moves dow as there are up. Agai, we have p m 00 0 for odd m. If m 2, there are 2 k k k k k0 5

6 ways of makig the same umber of moves left ad right ad the same umber of moves up ad dow. Hece, we have p k k k k k ! 4 k!k! k! k! k !!! 4!! k!k! k! k! k k k k0 2 1 π Summig over, we fid that p2 00 diverges, ad so we are doe by Lemma 3.3. For d 3, we will fid that p 2 00 O d/2 by some sort of iductive proof. 4 Further Study The followig problem was proposed to me by Dr Laszlo Babai. Cosider the radom walk i Z startig at the origi, where at each step the walker moves either to the right or to the left, each with probability p, or he stays put with probability 1 2p. It ca easily be show that the probabability of returig to the origi i steps is O 1/2, as above. Now, cosider a phased radom walk o Z d, where at step, we move i the directio of dimesio mod d, with the same probabilities as before. The, it is trivial to prove that the probability of retur to the origi i steps is O d/2. Is it possible to fid a alterate proof to olya s Theorem by approximatig the Radom Walk o Z d by the hased Radom Walk o Z d? 6