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Unformatted text preview: = Theorem 2.4. If Xi are independent Poissson( i ) then
X1 + · · · + Xk = Poisson( 1 + ··· + n ). Proof. It su ces to prove the result for k = 2, for then the general result follows
P (X1 + X2 = n) =
m=0 P (X1 = m)P (X2 = n
e 1 ( m
1) m! ·e 2 m) ( 2 )n m
(n m)! 82 CHAPTER 2. POISSON PROCESSES Knowing the answer we want, we can rewrite the last expression as
e ( 1+ 2) ( 1 n ✓ ◆✓
+ 2 )n X n
1 + 2 ◆m ✓ ◆n 2
1 + 2 m The sum is 1, since it is the sum of all the probabilities for a binomial(n, p)
distribution with p = 1 /( 1 + 2 ). The term outside the sum is the desired
Poisson probability, so have proved the desired result.
The property of the Poisson process in Lemma 2.2 is the ﬁrst part of our
second deﬁnition. To start to develop the second part we prove a Markov
Lemma 2.5. N (t + s) N (s), t
of N (r), 0 r s. 0 is a rate Poisson process and independent Why is this true? Suppose for concreteness (and so that we
2.2 at the beginning of...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.
- Spring '10
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