COINCIDENCES IN POISSON PATTERNS 1009 



ability exp ( — X5)) and the interval (0, y) must contain no coincidences 

 (probability Poiy)). One obtains finally 



.L-8 



PiiL) = e-'' 

 and similarly 



1 + t^e-'' f e^'P.Xy) dy 

 Jo 



(1-2) 



l + Xe-^^jT^ ^'Pr(y)dyj. (1-3) 



The solutions Pi(L) and P^iL) are determined uniquely by (1-1), 

 (1-2) and (1-3). For (1-1) determines them for ^ L ^ 6 and the in- 

 tegrations indicated in (1-2) and (1-3) will provide the solutions in ^ 

 L ^ (h + 1)6 when they are known in ^ L ^ 728. Pi{L) and P^iL) 

 are piecewise analytic; the analytic form of the solution changes 

 each time L passes an integer multiple of 5. These analytic expressions 

 soon become complicated and are less useful than the bounds and ap- 

 proximations given later on. 



To compute F{L), consider the last place before L at which either Pat- 

 tern No. 1 or No. 2 has a point. The probability that this last point lies 

 between x and x + dx and belongs to Pattern No. 1 is exp [—(X + n) 

 {L — x)]\ dx (Fig. 2). This term multiplied by Pi(.r) and integrated from 

 to L gives the probability of no coincidences if the last point is a circle. 

 A similar integral gives the probability if the last point is a cross. Finally 

 there is probability exp [— (X + ^)L] that neither pattern has a last 

 point [i.e., (0, L) empty]. Then 



F{L) = e 



-(X+M)i' 



1 + I' e^'^^^^(XP:(.T) -f f.P,ix)) dx~^ . (1-4) 



1.2 Solution by Laplace Transforms 

 For i = 1 or 2, let 



pXs) = f Pi(L)e-*^ dL. (1-5) 



Jo 



Replacing Pi(L) in (1-5) by (1-1) for ^ L ^ 5, by (1-2) for 5 ^ L, 



NO COINCIDENCES EMPTY 

 ! I 



'I ' CT) t+i i^^K 1 01 



X x + dx 



Fig. 2 — Patterns without coincidence. 



