If p x 1 p p x 0 1 p and y poissonp then x p

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Unformatted text preview: arately, we have (i) k /k ! does not depend on n. (ii) There are k terms on the top and k terms on the bottom, so we can write this fraction as nn1 n k+1 · ··· n n n For any j we have (n j )/n ! 1 as n ! 1, so the second term converges to 1 as n ! 1. (iii) Skipping to the last term in (2.15), /n ! 0, so 1 k is fixed so ✓ ◆k 1 !1 k=1 n /n ! 1. The power (iv) We broke o↵ the last piece to make it easier to invoke one of the famous facts of calculus: (1 /n)n ! e as n ! 1. If you haven’t seen this before, recall that log(1 x) = x + x2 /2 + . . . 84 CHAPTER 2. POISSON PROCESSES so we have n log(1 /n) = + 2 /n + . . . ! as n ! 1. Combining (i)–(iv), we see that (2.15) converges to k k! ·1·e ·1 which is the Poisson distribution with mean . By extending the last argument we can also see why the number of individuals that arrive in two disjoint time intervals should be independent. Using the multinomial instead of the binomial, we see that the probability j people will go between 12:17 and 12:18 and k people wil...
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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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