tn given that there were n arrivals before time t is

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Unformatted text preview: Proof. Again we consider only the case k = 2 and check the second definition given in Theorem 2.7. It is clear that the sum has independent increments and N1 (0) + N2 (0) = 0. The fact that the increments have the right Poisson distribution follows from Theorem 2.4. We will see in the next chapter that the ideas of compounding and thinning are very useful in computer simulations of continuous time Markov chains. For the moment we will illustrate their use in computing the outcome of races between Poisson processes. 90 CHAPTER 2. POISSON PROCESSES Example 2.5. A Poisson race. Given a Poisson process of red arrivals with rate and an independent Poisson process of green arrivals with rate µ, what is the probability that we will get 6 red arrivals before a total of 4 green ones? Solution. The first step is to note that the event in question is equivalent to having at least 6 red arrivals in the first 9. If this happens, then we have at most 3 green arrivals before the 6th red one. On the other hand if there are 5 or fewer red arrivals in the first 9, then we have had at least 4 red arrivals and at most 5 green. V...
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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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