# HW6_Sol - ECEN 646 Homework 6 Solutions 1 C ¸inlar...

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ECEN 646 Homework 6 Solutions 1 C ¸inlar Exercise (8.4), Chapter 4 We are interested in every 13 th arrival in a Poisson process. (a) The interarrival times in a Poisson process are exponentially dis- tributed. The lucky arrivals are separated by 12 unlucky ones and so the time between 2 lucky arrivals is the sum of 13 exponential random variables i.e. an Erlang(13) random variable. f T ( t ) = λe - λt ( λt ) 12 12! t 0 (b) Let N t be the number of arrivals in the interval [0 , t ]. The number of gifts M t given would then be equal to the number of integer multiples of 13 contained in N t . The event { M t = k } is the same as the event { 13 k N t 13 k + 12 } P ( M t = k ) = P (13 k N t 13 k + 12) = 12 summationdisplay i =0 P ( N t = 13 k + i ) = 12 summationdisplay i =0 e - λt ( λt ) 13 k + i (13 k + i )! No further simplification is possible. 2 C ¸inlar Exercise (8.10), Chapter 4 We have a family of Poisson processes out of which we choose one based on the outcome of a random experiment. The outcome of the random ex- periment takes value i with probability π ( i ). (a) Conditioned on the event { X = i } , the process N is a Poisson process with rate λ ( i ). So, the unconditional distribution of N t can be found by conditioning on the value of X . 1 This study resource was shared via CourseHero.com

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P { N t = k } = summationdisplay i E P ( N t = k | X = i ) P ( X = i ) = summationdisplay i E π ( i ) e - λ ( i ) t ( λ ( i ) t ) k k ! k = 0 , 1 , ... (b) For a process to have independent increments, the events { N t + s N t } and { N t } should be independent. In this case, it suffices to show that P ( N t + s N t = k 1 | N t = k 2 ) negationslash = P ( N t + s N t = k 1 ) P ( N t + s N t = k 1 | N t = k 2 ) = P ( N t + s N t = k 1 and N t = k 2 ) P ( N t = k 2 ) Since the process in question is no longer a simple Poisson process, we can’t assume that the two events in the numerator are independent. However, conditioned on the the event { X = i } , the process is Poisson. Hence, we can use the law of total probability to evaluate the numerator just as we did in part ( a ). P ( N t + s N t = k 1 and N t = k 2 ) = summationdisplay i E P ( N t + s N t = k 1 and N t = k 2 | X = i ) P ( X = i ) = summationdisplay i E π ( i ) P ( N t + s ( i ) N t ( i ) = k 1 ) P ( N t ( i ) = k 2 ) = summationdisplay i E π ( i ) e - λ ( i ) s ( λ ( i ) s ) k 1 k 1 ! e - λ ( i ) t ( λ ( i ) t ) k 2 k 2 ! The denominator has already been evaluated in part ( a ). P ( N t = k 2 ) = summationdisplay i E π ( i ) e - λ ( i ) t ( λ ( i ) t ) k 2 k 2 !
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