2 Independence For disjoint sets A 1 A 2 A n the random vari ables N A 1 N A 2

2 independence for disjoint sets a 1 a 2 a n the

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2. ( Independence ) For disjoint sets A 1 , A 2 , . . . , A n the random vari- ables N ( A 1 ) , N ( A 2 ) , . . . , N ( A n ) are independent. Quick exercise 12.3 Suppose that the locations of defects in a certain type of material follow the two-dimensional Poisson process model. For this material it is known that it contains on average five defects per square meter. What is the probability that a strip of length 2 meters and width 5 cm will be without defects? In Figure 7.4 the locations of the buildings the architect wanted to distribute over a 100-by-300-m terrain have been generated by a two-dimensional Poisson process. This has been done in the following way. One can again show that given the total number of points in a set, these points are uniformly distributed over the set. This leads to the following procedure: first one generates a value n from a Poisson distribution with the appropriate parameter ( λ times the area), then one generates n times a point uniformly distributed over the 100- by-300 rectangle. Actually one can generate a higher-dimensional Poisson process in a way that is very similar to the natural way this can be done for the one-dimensional process. Directly from the definition of the one-dimensional process we see that it can be obtained by consecutively generating points with exponentially distributed gaps. We will explain a similar procedure for dimension two. For s > 0, let M s = N ( C s ) , where C s is the circular region of radius s , centered at the origin. Since C s has area πs 2 , M s has a Poisson distribution with parameter λπs 2 . Let R i denote the distance of the i th closest point to the origin. This is illustrated in Figure 12.2. Note that R i is the analogue of the i th arrival time for the one-dimensional Poisson process: we have in fact that R i s if and only if M s i. In particular, with i = 1 and s = t , P ( R 2 1 t ) = P R 1 t = P ( M t > 0 ) = 1 e λπt . In other words: R 2 1 is Exp ( λπ ) distributed. For general i , we can similarly write P ( R 2 i t ) = P R i t = P ( M t i ) .
12.4 Higher-dimensional Poisson processes 175 . . R 1 . . R 2 × × × × × × × × × × × × × × × × × × × × × × × + + + + + + + + + + + + + + + + + + + + + + + . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ................ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ........................ . . . . . . . . . . . . . . . . . . . . . . . Fig. 12.2. The Poisson process in the plane, with the two circles of the two points closest to the origin. So P ( R 2 i t ) = 1 e λπt i 1 j =0 ( λπt ) j j ! , which means that R 2 i has a Gam ( i, λπ ) distribution—as we saw on page 157. Since gamma distributions arise as sums of independent exponential distribu- tions, we can also write R 2 i = R 2 i 1 + T i , where the T i are independent Exp ( λπ ) random variables (and where R 0 = 0). Note that this is quite similar to the one-dimensional case. To simulate the two-dimensional Poisson process from a sequence U 1 , U 2 , . . . of independent U (0 , 1) random variables, one can therefore proceed as follows (recall from Section 6.2 that (1 ) ln( U i ) has an Exp ( λ ) distribution): for i = 1 , 2 , . . .

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• Spring '08
• WEBER
• Normal Distribution, The Land, Probability distribution, Probability theory, probability density function, quick exercises