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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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 Spring '08
 Zahrn
 The Land

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