# 13 k and hence var x proof x x 1 x ex x k

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Unformatted text preview: iven by ( t)n 1 fTn (t) = e t · for t 0 (2.12) (n 1)! and 0 otherwise. 80 CHAPTER 2. POISSON PROCESSES Proof. The proof is by induction on n. When n = 1, T1 has an exponential( ) distribution. Recalling that the 0th power of any positive number is 1, and by convention we set 0!=1, the formula reduces to fT1 (t) = e t and we have shown that our formula is correct for n = 1. To do the induction step, suppose that the formula is true for n. The sum Tn+1 = Tn + ⌧n+1 , so breaking things down according to the value of Tn , and using the independence of Tn and tn+1 , we have Zt fTn+1 (t) = fTn (s)ftn+1 (t s) ds 0 Plugging the formula from (2.12) in for the ﬁrst term and the exponential density in for the second and using the fact that ea eb = ea+b with a = s and b= (t s) gives Zt Zt n 1 n1 s s ( s) ( t s) tn e ·e ds = e ds (n 1)! (n 1)! 0 0 nn t =et n! which completes the proof. 2.2 Deﬁning the Poisson Process In this section we will give two deﬁnitions of the Poisson process with rate . The ﬁrst...
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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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