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Unformatted text preview: 1 e p 2p 2 Let Ym = Poisson(pm ) be independent. At this point we have shown kXi Yi k p2 . With a little work one can show i k(X1 + · · · + Xn ) k(X1 , · · · , Xn ) and the desired result follows. (Y1 + · · · + Yn )k n X (Y1 , · · · , Yn )k kX m m=1 Ym k Theorem 2.9 is useful because it gives a bound on the di↵erence between the distribution of Sn and the Poisson distribution with mean n = ESn . To bound the bound it is useful to note that ! n n X X 2 pm max pk pm m=1 k m=1 so the approximation is good if maxk pk is small. This is similar to the usual heuristic for the normal distribution: the sum is due to small contributions from a large number of variables. However, here small means that it is nonzero with small probability. When a contribution is made it is equal to 1. The last results handles problem (i). To address the problem of varying arrival rates mentioned in (ii), we generalize the deﬁnition. Nonhomogeneous Poisson processes. We say that {N (s), s Poisson process with rate (r) if (i) N (0) = 0, (ii) N (t) has indepe...
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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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