5 we get easily the following lemma 26 n t has

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Unformatted text preview: te the distribution of N (s). Lemma 2.2. N (s) has a Poisson distribution with mean s. Proof. Now N (s) = n if and only if Tn s < Tn+1 ; i.e., the nth customer arrives before time s but the (n + 1)th after s. Breaking things down according to the value of Tn = t and noting that for Tn+1 > s, we must have ⌧n+1 > s t, and ⌧n+1 is independent of Tn , it follows that Zs P (N (s) = n) = fTn (t)P (tn+1 > s t) dt 0 Plugging in (2.12) now, the last expression is Zs ( t)n 1 · e (s t) dt = et (n 1)! 0 Zs n s = e tn 1 dt = e (n 1)! 0 s ( s)n n! which proves the desired result. Since this is our first mention of the Poisson distribution, we pause to derive some of its properties. Theorem 2.3. For any k 1 EX (X 1) · · · (X k + 1) = (2.13) k and hence var (X ) = Proof. X (X 1) · · · (X EX (X k + 1) = 0 if X k 1) · · · (X 1 X k + 1) = j e j! j =k = k 1 so 1 X j (j 1) · · · (j jk e (j j =k k )! = k + 1) k since the sum gives the total mass of the Poisson distribution. Using var (X ) = E (X (X 1)) + EX (EX )2 we conclude var (X ) = 2 + ( )2...
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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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