The arcsine law of a simple. random walk

Transcription

1 The arcsine law of a simple random walk

2 The coin-tossing game and the simple random walk (RW) As you start playing your fortune is zero. Every second you toss a fair coin, each time either collecting 1 Euro from your opponent or paying it out. By the Law of Large Numbers, as the number of games played increases, the fraction of games with positive outcome approaches 1=2. In the course of the game, your total fortune is likely to uctuate, changing sign from positive to negative and back, so the question which may be asked is this: After a large number n of games have been played, what is the fraction of time when your total wealth remained in the winning zone (in `black')?

3 By the obvious symmetry, in the time between two consequitive `draws' (when your total wealth becomes zero) you are equally likely to stay in `black' or in `red'. As more games are played, more draws will occur, which seem to imply that the fraction of time spent in `black' goes to 1=2. However, the arcsine law explains that this intuition is completely wrong! The fraction is more likely to be close to extremities 0 and 1 than to be close to 1=2.

4 Zeroes of the RW Each possible path of the RW, (S j j =0 1 ::: n) (where S 0 = 0), has probability 2;n. Note that S n =0is only possible for n even. A piece of path between zeroes (draws) is called excursion (from zero). Let u 2n = P(S 2n =0) f 2n = P(S 2n =0 but S j 6=0for j =1 ::: n; 1) be the chances of a return to 0 and of the rst return, at time 2n. For a return to zero the RW goes some n steps up and some n steps down, hence u 2n = 2n 2 ;2n : n Lemma 1 f 2n = u 2n;2 2n :

5 First proof of Lemma 1: Introduce N n (a b) = n c =(n + b ; a)=2 c the number of paths from (wealth) a to b in n steps, N 6=0 n (a b) = the number of paths that do not visit 0 (except, perhaps, the endpoints) Nn(a 0 b) = the number of paths that visit 0 (endpoints not counted). Then by the rststep decomposition and symmetry N 6=0 (0 0) = N6=0 2n 2n;1 (1 0) + N6=0 2n;1 (;1 0) = 2N 6=0 2n;1 (1 0) = 2N6=0 2n;2 (1 1): By the reexion principle which yields N 6=0 N 0 2n;2(1 1) = N 2n;2(;1 1) 2n;2 (1 1) = N 2n;2(1 1) ; N2n;2(1 0 1) = N 2n;2(1 1) ; N 2n;2(;1 1) = 1 2n ; 2 n n ; 1 and nally N 6=0 2n (0 0) = 2N6=0 2n;2 (1 1) = 2 n 2n ; 2 : n ; 1

6 Second proof of Lemma 1. Let u 0 =1. The rst-zero decomposition of the RW yields u 2n = n X k=1 f 2k u 2n;2k : (1) Indeed, the probability to return to 0 at step 2n and visit 0 for the rst time at step 2k is equal to f 2k u 2n;2k. In the language of generating functions F (z) := 1 X n=1 f 2n z 2n the recursion translates as U(z) =1+U(z)F (z) U(z) := 1 X n=0 u 2n z 2n () F (z) =1; 1=U(z) (multiply (1) by z 2n and add up over n). Recalling the binomial series (1 + x) = 1 X with b:c: and checking n=0 n x n n := ( ; 1) ( ; n +1) n! 2n n =(;4) n ;1=2 n

7 we derive U(z) = 1 q1 ; z 2: Thus F (z) =1; q 1 ; z 2 whose binomial-series expansion yields f 2n =(;1) n+1 1=2 1 2n ; 2 = 2 ;2n+1 n n n ; 1 which is the same as f 2n = u 2n;2 2n :

8 The discrete arcsine distribution Let p 2k 2n be the probability that 2k out of 2n seconds the RW spends in the winning zone, 0 k n. Lemma 2. p 2k 2n = u 2k u 2n;2k = 2k k 2n ; 2k n ; k 2 ;n : (2) Proof. Check that f 2n = u 2n;1 ; u 2n, from which the `no draws' probability is P(S 1 6=0 ::: S 2n 6=0)=1;f 2 ;f 4 ;:::;f 2n = 1;(1;u 2 );(u 2 ;u 4 );:::;(u 2n;2;u 2n )=u 2n and (2) follows for k =0and k = n ; 1. Proceed by induction. If the RW spends 2k 6= 0 2n seconds in the winning zone, then there is a return to 0. Let 2r be the return time. Then either the rst excursion is positive (so 2r 2k) and (S 2r+1 ::: S 2n ) stays 2k ; 2r seconds in the winning zone, or the rst excursion is negative (so 2(n ; r) 2k) and

9 (S 2r+1 ::: S 2n ) spends 2k seconds in the winning zone. Thence kx n;k X p 2k 2n = 1 f 2 2r p 1 2k;2r 2n;2r+ r=1 2 r=1 which by induction and (1) is equal to f 2r p 2k 2n;2r 1 2 u 2n;2ku 2k u 2ku 2n;2k = u 2k u 2n;2k: Example: for 20 tosses with probability about 0 35 one of the players will never be in the winning zone! And the chance of parity 10: 10 is only The probability law p 2k 2n = 2k 2n ; 2k 2 ;2n k =0 ::: n k n ; k is `the discrete arcsine' distribution for the time spent in the winning zone.

10 Large n approximation Recall Stirling's formula: n! p 2n(n=e) n (for n!1), from which p 2k 2n 1 k!1 n ; k!1: qk(n ; k) Thus the chance that the fraction of time in the winning zone assumes a value between 1=2 and x (1=2 <x<1) is X k:1=2<k=n<x p 2k 2n 1 n which is a Riemann sum for 1 Z x 1=2 k n (1 ; k n ) ;1=2 dt qt(1 ; t) = 2 arcsin p t ; 1 2 : By symmetry, the probability for k=n 1=2 goes to 1=2.

11 Theorem (the arcsine law) The probability that in n games the fraction of time spent in the winning zone is at most x (0 < x < 1) approaches as n!1. 1 Z x The U-shaped function 1 0 dt q t(1 ; t) 1 q t(1 ; t) is at in the middle of [0 1] and goes to 1 at the endpoints, thus it is much more likely that the fraction belongs to [0 ][[1; 1] than that it is within from 1=2.

12 The number of draws The reason why the fraction of time in the winning zone does not go to 1=2 is that it takes too long to get back to 0. In particular, the expected rst return time is innite: 1X k=1 2kf 2k = 1 X k=1 u 2k;2 = 1 because u 2k 1= p k. If this expectation were nite, the Law of Large Numbers would force the number of draws to grow approximately linearly with n. In fact, this number if of the order of p n. Let q 2n r be the probability that the RW returns to 0 exactly r times during the rst 2n seconds. We know already that q 2n 0 = u 2n, and from (1) also q 2n 1 = u 2n Theorem For r =0 1 ::: q 2n r = 1 2n ; r 2 2n;r n :

13 From this (with some calculus) one can deduce that q 2n r 1 p n e ;r2 =n for n!1and r = o(n 2=3. This entails: Corollary The probability that the number of draws in n games does not exceed x p n converges to s 2 Z x 2 =2 0 e;t dt (this is a half of the `bell-shaped curve', the law of the modulus of a normal variable).

14 The last draw Thus as n grows we are likely to observe longer and longer excursions, when one of the players remains in the winning zone. It is easy to derive the law of meander, i.e. the last, incomplete, excursion before given time. Theorem. The time from the last return to zero to 2n follows the discrete arcsine law. Proof. P(S 2k=0 S 2k+1 6=0 ::: S 2n 6=0)= u 2k P(S 1 6=0 ::: S 2n;2k 6= 0)=u 2k u 2n;2k: We see that the number of games after the last draw before a given time n is typically of the order of n.