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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 !
( (1 p)t)k
pt ( pt)
e (1 p)t
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
(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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