   Chapter 3, Problem 120SE

Chapter
Section
Textbook Problem

The simple Poisson process of Section 3.6 is characterized by a constant rate α at which events occur per unit time. A generalization of this is to suppose that the probability of exactly one event occurring in the interval [t, t + Δt] is α(t) · Δt + o(Δt). It can then be shown that the number of events occurring during an interval (t1, t2] has a Poisson distribution with parameter μ  =  ∫ t 2 t 1 α ( t ) d t The occurrence of events over time in this situation is called a nonhomogeneous Poisson process. The article ‘inference Based on Retrospective Ascertainment ” (J. Amer. Stat. Assoc., 1989: 360-372). considers the intensity function α ( t )  =  e a  +  b t as appropriate for events involving transmission of HIV (the AIDS virus) via blood transfusions. Suppose that a = 2 and b = .6 (close to values suggested in the paper), with time in years. a. What is the expected number of events in the interval [0, 4]? In [2, 6]? b. What is the probability that at most 15 events occur in the interval [0, .9907]?

a.

To determine

Find the expected number of events in the interval [0,4]and[2,6]

Explanation

Given info:

The number of events occurring during an interval [t1,t2]  has a Poisson distribution with parameter μ=t1t2α(t)dt. The intensity function α(t)=ea+bt are appropriate for events involving transmission of HIV via blood transfusion. a = 2, b = 0.6.

Calculation:

The expected number of events in the interval [0,4]and[2,6]:

for[0,4]:

μ=04e2+0.6tdt=e204e0.6tdt=e2[e0.6t0.6]04=e20

b.

To determine

Find the probability that at most 15 events occur in the interval [0,0.9907]

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