A express fy in terms of fx b sketch fy in the case x

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Unformatted text preview: t E2 is the same as E1 , so P (E2 ) = 0.00091. Using the pmf of the Poisson distribution with mean 7 yields: P (E3 ) = P {N (3.5) ≤ 2} = 2 P {N (3.5) = 0} + P {N (3.5) = 1} + P {N (3.5) = 2} = e−7 (1 + 7 + 72 ) = 0.0296. Event E4 is the same as event E3 , so P (E4 ) = 0.0296. Event E5 is the same as the event that the number of calls that arrive by time t is less than or equal to t2 two, so P (E5 ) = P {N (t) ≤ 2} = P {N (t) = 0} + P {N (t) = 1} + P {N (t) = 2} = e−2t (1 + 2t + (22) ). Event E6 is just the complement of E5 , so: P (E6 ) = 1 − P (E5 ) = 1 − e−2t (1 + 2t + (2t)2 2 ). 3.5. POISSON PROCESSES 85 (b) As a function of t, P (E6 ) is the CDF for the time of the arrival time of the third call. To get the pdf, differentiate it to get f (t) = e−2t 2 1 + 2t + (2t)2 2 = e−2t (2t)2 = e−2t − 2 − 4t 23 t2 . 2 This is the gamma density with parameters r = 3 and λ = 2. (c) The expected time between arrivals is 1/λ, so the expected time until the tenth arrival is 10/λ = 5. Example 3.5.5 Consider a Poisson process...
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