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Solution to the OK Corral model via decoupling of Friedman's urn J.F.C. Kingman and S.E. Volkovy 3 July 2001 Abstract We consider the OK Corral model formulated by Williams and McIlroy (1998) and later studied by Kingman (1999). In this paper we rene some of Kingman's results, by showing the connection between this model and Friedman's urn, and using Rubin's construction to decouple the urn. Also we obtain the exact expression for the probability of survival of exactly S gunmen given an initially fair conguration. Keywords: OK Corral, urn models, coupling, reinforced random walks. AMS subject classication: Primary: 60J20; Secondary: 60J80 1 Introduction In this paper we extend the results of Kingman (1999) for the OK Corral model using decoupling of an urn model into two independent continuoustime birth processes. Not only do we demonstrate an alternative and relatively simple method, but also we rene some of Kingman's theorems. On From October 2001: The Isaac Newton Institute for Mathematical Sciences, Cam- bridge, CB3 0EH, U.K. E-mail: [email protected] y Corresponding author. Department of Mathematics, University of Bristol, BS6 6SF, U.K. E-mail: [email protected] 1 the side, we also obtain results on the speed of convergence for Friedman's urn. The OK Corral process (Xt ; Yt ), t = 0; 1; 2 : : : , is a Z2 valued process used to model the famous gunght. Its transition probabilities satisfy Yt Xt + Yt ; Xt P((Xt ; Yt ) = (Xt ; Yt , 1) j (Xt ; Yt )) = Xt + Yt : The process starts with X = Y = N and runs till either of Xt or Yt becomes zero. The value of the other process at this time is denoted by S . P((Xt+1 ; Yt+1 ) = (Xt , 1; Yt ) j (Xt ; Yt )) = +1 +1 0 1 0 Williams and McIlroy (1998) have found the correct scaling for the limiting distribution of S , which surprisingly turns out to be N 3=4 . In Kingman (1999) the limiting distribution of S=N 3=4 is found. The results were obtained using martingale techniques to compute the asymptotic moments of the above distribution. In our paper we obtain the asymptotic expression for the probability that S = s for each s, provided s = O(N 3=4 ). A seemingly unrelated model is an urn model introduced by Friedman (1949). In this model, the urn contains black and white balls. A ball is chosen at random, and then it is replaced by a balls of the same color and b balls of the opposite color. A special case of Friedman's urn (namely, a = 0, b = 1)2 is the following process: (Xt0 ; Yt0), t = 0; 1; 2 : : :, 0 P((Xt0+1 ; Yt0+1 ) = (Xt0 + 1; Yt0 ) j (Xt0 ; Yt0 )) = X 0Y+t Y 0 t 0t P((Xt0+1 ; Yt0+1 ) = (Xt0 ; Yt0 + 1) j (Xt0 ; Yt0 )) = Xt0X+tYt0 Freedman's (1965) results for this particular urn give Xt0 , Yt0 ! N 0; 1 t 3 1 2 in Kingman (1999) notations N denotes Xt + Yt a = 1, b = 0 corresponds to Polya urn 2 (1) but do not say much about how fast (Xt0 ; Yt0 ) would approach the state Xt0 Yt0 if the process is started from an o-equilibrium position X00 = 1, Y00 = S . The method of our paper answers (in a sense) this question. 2 Coupling Rather than computing the probabilities for the OK Corral process directly, we couple it with Friedman's urn, and solve the relevant problem for the urn rst. We start the OK corral process by setting X0 = Y0 = N , let = minft : Xt = 0 or Yt = 0g and S = X + Y . From now on, we condition on the event Y > 0, so that S Y . Suppose that Y1 < Y0 . Then for each path connecting (N; N ) with (0; S ) there exist k 1 and two integer-valued sequences x0 ; x1 ; : : : ; xk and y0; y1 ; : : : ; yk such that xi,1 < xi, yi,1 < yi for all i with x0 = 0, y0 = S , xk = yk = N and the process (Xt ; Yt ) comes to (0; S ) following the path (xk ; y) ; y = yk ; yk , 1; : : : ; yk,1; (x; yk,1 ) ; x = xk ; xk , 1; : : : ; xk,1 ; .. . (x; y0 ) ; x = x1 ; x1 , 1; : : : ; x0 : The probability of such a path equals ! ! k k Y Y 1 yi ,yi,1 x yix,i,1 xi,1 (2) (2N )(2N , 1) : : : (S + 1) i=1 i i=1 Now consider Friedman's urn with the transition probabilities (1). The probability that (X 0 ; Y 0 ) comes to (N; N ) from (0; S ) following the path described above, moving backwards, equals 1 S (S + 1) : : : (2N , 1) k Y i=1 xyi i,yi,1 3 ! k Y i=1 yix,i,1 xi,1 ! : (3) Comparing the probabilities (2) and (3) we observe that = 2SN P(.) := P((X; Y ) hits (0; S ) j (X0 ; Y0 ) = (N; N )) S P(%) P((X 0 ; Y 0 ) hits (N; N ) j (X 0 ; Y 0 ) = (0; S )) =: (4) 2N actually for any path connecting (0; S ) and (N; N ) (the case X1 < X0 can be analyzed in the same way). Note that the relation above also holds for paths connecting arbitrary two points, not necessarily (0; S ) and (N; N ). Therefore, to calculate P(.) it suces to compute P(%). This is what we do next. 0 0 3 Decoupling As mentioned before, Freedman's results are not applicable for the speed of the convergence, while this is what we actually need (to compute asymptotically P(%) when N and S are large). To this end, we will use a procedure known as Rubin's construction due to Herman Rubin introduced in Davis (1990) for the study or reinforced random walks. Later his method was also applied to a variety of problems, ranging from rather theoretical ones, mostly related to reinforced random walks, (see Sellke (1994) and Limic (2001)) to the applied ones (see Khanin and Khanin (2000)). The essence of this construction is extremely simple and in our case runs as follows. Consider two independent birth processes, Ut and Vt , where t is continuous, both with the transition rate k dt = P (Ut+dt = Ut + 1 j Ut = k) = 1=k dt (the same formula holds for Vt). Set U0 = 1 and V0 = S . From the properties of independent exponential random variables it follows that the process (Ut ; Vt ) considered at the times when either of its coordinates changes, has the same distribution as the Friedman's urn (X 0 ; Y 0 ) described above. Now the independence of the coordinates of (Ut ; Vt ) allows us to compute desired distributions. 4 4 Computing P(%) For an integer n let Wn = inf ft : Ut = ng (Wn0 = inf ft : Vt = ng resp.) be the nth waiting time for the process U (V resp.). Since Wi+1 , Wi are independently exponentially distributed, Wn and Wn0 can be represented as nX ,1 Wn = k=1 nX ,1 Wn0 = k=S kk ; kk (5) where k 's and k 's (k = 1; 2 : : :) are IID exponential random variables with rate 1. The process (Ut ; Vt ) passes through the point (N; N ) if and only if WN 2 [WN0 ; WN0 +1 ) or WN0 2 [WN ; WN +1 ). Let us compute the probability of the rst event. To simplify notations, set A = WS , B = WN , WS and C = WN0 . We have P(WN 2 [WN0 ; WN0 )) = P(A , NN < C , B A) +1 (6) Note that B , C and N are independent of A. Hence, integrating by parts, Z 1 1 e,z=N dz Z A f (x) dx P(A , N < (C , B ) A j A) = N 0 = Z 0 C ,B N A,z 1 ,z=N e fC ,B (A , z) dz (7) where fC ,B (x) is the pdf of the random variable C , B . Next, we need Lemma 1 As S ! 1, A ,p S (S , 1)=2 ! N (0; 1): S =3 Also, when both N ! 1 and S ! 1 such that S = o(N ) C , B ! N (0; 1): p 2N =3 3 3 5 (8) (9) Proof: Immediately follows from the representation (5), CLT with Lya- punov conditions (see Billingsley (1985), p.312), and well-known formula P ,1 p for ni=1 i , p = 2; 3, and 4. Now we want to strengthen (9). From now on we assume that S = o(N ). p Lemma 2 Let fN (x) be the pdf of (C , B )= 2N =3 and f (x) be the pdf of N (0; 1). Then for any > 0 3 sup jfN (x) , f (x)j = o(N ,(1,) ) x Proof: By the inversion formula, fN (x) , f (x) = 21 1 ,itx e (N (t) , e,t2 =2) dt ,1 Z (10) where N (t) = Z NY ,1 1 itx 1 e fN (x) dx = k2 t2 ,1 k=S 1 + 2N 3 =3 P since C , B = Nk=,S1 kk , kk . Consequently, N (t) can be written as ! 2 4 6 t 9 t 9 t 2 4 N (t) = exp , 2 + 40 N , 56 N 2 + : : : + o(t + t =N + : : :) (11) or 1 N (t) = t2 t4 : (12) 1 + 2 + 8 + : : : + o(t2 + t4 =N + : : :) Let us split the area of integration in (10) into two parts: [,N ; N ] where 0 < < min(1=2; =5), and its compliment. Then Z N ,N e,itx(N (t) , e,t2 =2 ) dt Z N 409N je,itx,t2 = jt (1 + o(1)) dt ,N = O(N , , ) (13) (1 6 5 ) 2 4 using (11) and Z t= 2[,N ;N ] e,itx (N (t) , e,t2 =2 ) dt Z ( 1 t=2[,N ;N ] 1 + t22 + : : : = o(N ,1 ) t2 + e, 2 ) (14) using (12). Combining (13) and (14), from the formula (10) we deduce that jfN (x) , f (x)j = O(N , uniformly in x. ,5) ) + o(N ,1 ) = o(N ,(1,) ) (1 From Lemma 2 it follows that for any > 0 2 exp(, 4Ny3 =3 ) fC ,B (y) = p 3 + o(N ,2:5+) 4N =3 Plugging this into (7), we have 2 3 (A,z )2 Z 1 exp( , ) 3 P(WN 2 [WN0 ; WN0 +1 ) j A) = e,z=N 4 p 4N3 =3 + o(N ,2:5+)5 dz 4N =3 0 Z 1 exp(,u , (A,Nu)2 ) 4N 3 =3 p du + o(N ,1:5+): = 4N=3 0 Under the assumption that A is of order of N 3=2 , the integral in the RHS equals Z log N exp(,u , (A,Nu)2 ) 4N 3 =3 p du + O(N ,1:5) 4N=3 0 exp(, 3A2 ) = p 4N 3 (1 , N1 ) 1 + O log1=N2 + O(N ,1:5 ) N 4N=3 2 , 43NA3 e = p + O( logNN ) 4N=3 whence , 43NA23 e 0 0 P(WN 2 [WN ; WN +1 ) j A) = p + O logNN : (15) 4N=3 Next we want to rene (8). 7 dt Lemma 3 sup x P ! A ,p S (S , 1)=2 x , (x) const p S =3 S (16) 3 where (x) is the cdf of the standard normal distribution. Proof: This results follow, e.g., from Theorem V.3.6 in Petrov (1975). Now suppose that S is of order of N 3=4 , and let p 2 := 3S : N Lemma 3 yields P(jA , S 2 =2j S 3=2 log S ) = 3 2 (, log S ) + (1 , (log S )) + O(S , = ) = O(S , = ); 1 2 1 2 hence with a large probability A = O(S 2 ) = O(N 3=2 ) which is consistent with our previous assumptions. Using (15) we conclude that P WN 2 [WN0 ; WN0 +1 ) equals S P WN 2 [WN0 ; WN0 +1 ) j A , 2 ! S log S + O( (SN1) = + logNN ) o n exp ,3[S =2 + O(S 32 log S )] =(4N ) p = + O(N , = , = ) (17) 4N=3 2 2 2 3 2 1 2 3 3 8 1 2 ,2 = e log N = p 1+O N= 4N=3 The formula for P(WN0 2 [WN ; WN )) can be obtained similarly, and the expression for it also equals the RHS of (17) because fWN0 2 [WN ; WN )g = fA C , B < A + NN g and NN =D NN and NN = o(A) a.s. 16 3 8 +1 +1 Hence, WN 2 [WN0 ; WN0 ) + P(WN0 2 [WN ; WN )) r 3 log N 2= , = N e 1+O N= P(%) = P , +1 +1 16 3 8 8 implying by (4) P(.) = s e,2 = + O log N = p e,2 = + O log S : (18) 2S N= S= 4 N=3 16 16 p 9 8 3 2 5 Results and discussion Recall that throughout the most of previous arguments we supposed that X = 0, and now it is time to use the symmetry between X and Y . Therefore, if the number of the gunmen on each side at the beginning is N , then the probability that the number of the survivals after one of the groups is completely eliminated is S equals e,2 =16 + O log S = p S S= 3 2 by doubling the expression in (18). Using the same technique it is not hard to see that if the number of the gunmen is not exactly equal at the beginning of the game, but at the same time the dierence is not signicant, namely o(N 1=2 ), the the formula above does not change with the exception of the correction term, which becomes O log S + O(jX , Y j=N 0:5 ): S= 3 2 0 0 This can be obtained by noting that if X0 , Y0 = o(N 1=2 ) then the expression for A can be \corrected" by the term of order o(N 3=2 ). This is again consistent with Kingman (1999). 6 Exact formula Note that the probability in (6) can be expressed as Z 0 1 ,z=N e fA+B,C (z) dz 9 (19) where fA+B,C (z ) is the density of random variable A + B , C , by arguments identical to the one applied in (7). To compute fA+B,C (z ) we use Fourier transform: NY ,1 1 NY ,1 1 NX ,1 a N ,1 k + X bk E ei(A+B,C ) = = k=1 1 , ik k=S 1 + ik k=1 1 , ik k=S 1 + ik for some ak 's and bk 's. Using standard techniques, we obtain that N ,k,1 2N ,S ,2 k + S , 1)! ak = ((k,,1)1)!(N ,k 1 + k)!((N , 1 , k)! ; k = 1; 2; : : : ; N , 1; ,1)N ,k,1k2N ,S,2 k! bk = (k , S()!( N , 1 + k)!(N , 1 , k)! k = S; S + 1; : : : ; N , 1: P ,1 and therefore fA+B,C (z ) = Nk=1 ak k,1e,z=k + PNk=,S1 bk k,1 e,z=k . Plugging this into (19) and computing the integral, we calculate the exact expression for P(WN 2 [WN0 ; WN0 +1 )). Repeating this argument for P(WN0 2 [WN ; WN +1 )) and using (4), we nally obtain that the probability that precisely S gunmen (on either side) will survive the shooting, provided there were initially N on both sides, is given by NX ,1 (,1)N ,k,1 Nk2N ,S ,2 (k + S , 1)! NX ,1 (,1)N ,k,1 Nk2N ,S ,2 k! + k=1 (k , 1)!(N + k)!(N , k , 1)! k=S (k , S )!(N + k)!(N , k , 1)! Unfortunately, since this is a sum with alternating signs, we are not able to get more insight on this probability than already provided in Section 5. Just for an illustration, we present the probability that exactly 5 gunmen will survive provided there were initially 20 on both sides: 15310459783418263565672289100224561581887 283303917794408935536670579720873574400000 0:05404 This number is computed using the analytical expression for the probability of survival of exactly S gunmen. References [1] Billingsley, P. (1985) Probability and Measure. (2nd. ed.) John Wiley and Sons, New York. 10 [2] Davis, B. (1990). Reinforced random walk. Prob. Th. Rel. Fields 84, 203 { 229. [3] Freedman, D. (1965). Bernard Friedman's urn. Ann. Math. Stat. 36, 956 { 970. [4] Friedman, B. (1949). A simple urn model. Comm. Pure Appl. Math. 2, 59 { 70. [5] Khanin, K. and Khanin, R. (2001). A probabilistic model for establishing of neuron polarity, J. of Math. Biology 42, 26{40. [6] Kingman, J.F.C. (1999). Martingales in the OK Corral. Bull. Lond. Math. Soc. 31, 601{606. [7] Limic, V. (2001). Attracting edge property for a class of reinforced random walks. preprint. [8] Petrov, V. V. (1975). Sums of independent random variables. (2nd. ed.) Springer, Berlin. [9] Sellke, T. (1994). Reinforced random walk on the d-dimensional integer lattice. Department of Statistics, Purdue University, Technical report #94-26. [10] Williams, D. and McIlroy, P. (1998). The OK Corral and the power of the law (a curious Poisson kernel formula for a parabolic equation). Bull. Lond. Math. Soc. 30, 166{170. 11