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Unformatted text preview: this section again) that by time s there
arrivals T1 , T2 , T3 , T4 that occurred at times t1 , t2 , t3 , t4 . We
waiting time for the ﬁfth arrival must have ⌧5 > s t4 , but
memory property of the exponential distribution (2.6)
P (⌧5 > s t4 + t⌧5 > s t4 ) = P (⌧5 > t) = e can use Figure
have been four
know that the
by the lack of
t This shows that the distribution of the ﬁrst arrival after s is exponential( )
and independent of T1 , T2 , T3 , T4 . It is clear that ⌧6 , ⌧7 , . . . are independent
of T1 , T2 , T3 , T4 , and ⌧5 . This shows that the interarrival times after s are
independent exponential( ), and hence that N (t + s) N (s), t 0 is a Poisson
process.
From Lemma 2.5 we get easily the following:
Lemma 2.6. N (t) has independent increments: if t0 < t1 < . . . < tn , then
N (t1 ) N (t0 ), N (t2 ) N (t1 ), . . . N (tn ) N (tn 1) are independent Why is this true? Lemma 2.5 implies that N (tn ) N (tn 1 ) is independent of
N (r), r tn 1 and hence of N (tn 1 ) N (tn 2 ), . . . N (t1...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell University (Engineering School).
 Spring '10
 DURRETT
 The Land

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