Download Solution to the OK corral model via a decoupling of Friedman`s urn

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.
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