Stochastic

# The mean is 1 variance 1 2 the density function is ft

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Unformatted text preview: iewing the red and green Poisson processes as being constructed by starting with one rate + µ Poisson process and ﬂipping coins with probability p = /( + µ) to decide the color, we see that the probability of interest is 9 X ✓9◆ k=6 If we suppose for simplicity that k pk (1 p)9 k = µ so p = 1/2, this expression becomes 9✓◆ 1 X9 1 + 9 + (9 · 8)/2 + (9 · 8 · 7)/3! 140 · = = = 0.273 512 k 512 512 k=6 2.4.3 Conditioning Let T1 , T2 , T3 , . . . be the arrival times of a Poisson process with rate , let U1 , U2 , . . . Un be independent and uniformly distributed on [0, t], and let V1 < . . . Vn be the Ui rearranged into increasing order . This section is devoted to the proof of the following remarkable fact. Theorem 2.14. If we condition on N (t) = n, then the vector (T1 , T2 , . . . Tn ) has the same distribution as (V1 , V2 , . . . Vn ) and hence the set of arrival times {T1 , T2 , . . . , Tn } has the same distribution as {U1 , U2 , . . . , Un }. Why is this true? We begin by ﬁn...
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## This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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