# Theorem 213 suppose n1 t nk t are independent poisson

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Unformatted text preview: ) = 0 by deﬁnition, it only remains to check that the components Xi = Ni (t + s) Ni (s) are independent and have the right Poisson distributions. To do this, we note that if X1 = j and X2 = k , then there must have been j + k arrivals between s and s + t, j of which were assigned 1’s and k of which were assigned 2’s, so P (X1 = j, X2 = k ) = e =e ( t)j +k (j + k )! j · p (1 p)k (j + k )! j !k ! j ( (1 p)t)k pt ( pt) e (1 p)t j! k! t (2.16) so X1 = Poisson( pt) and X2 = Poisson( (1 p)t). For the general case, we use the multinomial to conclude that if pj = P (Yi = j ) for 1 j m then P (X1 = k1 , . . . Xm = km ) =e t ( t)k1 +···km (k1 + · · · km )! k1 p · · · pk m m (k1 + · · · km )! k1 ! · · · km ! 1 = m Y j =1 e pj t ( pj )kj kj ! which proves the desired result. The thinning results generalizes easily to the nonhomogeneous case: Theorem 2.12. Suppose that in a Poisson process with rate , we keep a point that lands at s with probability p(s). Then the result is a nonhomogeneous Poisson process with rate p(s). 89 2.4. TRANSFORMATIONS For an appl...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).

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