91Ó°ÊÓ

The simple Poisson process of Section \(3.6\) is characterized by a constant rate \(\alpha\) 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 \([t, t+\Delta t]\) is \(\alpha(t) \cdot \Delta t+o(\Delta t)\). It can then be shown that the number of events occurring during an interval \(\left[t_{1}, t_{2}\right]\) has a Poisson distribution with parameter $$ \lambda=\int_{t_{1}}^{t_{2}} \alpha(t) d t $$ 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 $$ \alpha(t)=e^{a+b t} $$ 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 \([0, .9907]\) ?

Step by step solution

01

Understanding the Problem

We are given a nonhomogeneous Poisson process with intensity function \(\alpha(t) = e^{a + bt}\), where \(a = 2\) and \(b = 0.6\). Our goal is to first determine the expected number of events in specified time intervals.

02

Calculate Expected Number of Events [0,4]

To find the expected number of events in \([0, 4]\), we calculate the integral \(\lambda = \int_0^4 e^{2 + 0.6t} \, dt\). This requires the integration of the exponential function over the interval from 0 to 4.

03

Integration for Interval [0,4]

The integral transforms to \(\int_0^4 e^{2 + 0.6t} \, dt = e^2 \int_0^4 e^{0.6t} \, dt\). Solving, we have \(e^2 \cdot \left[ \frac{e^{0.6t}}{0.6} \right]_0^4 = \frac{e^2}{0.6} \left( e^{2.4} - 1 \right)\). Compute this value to find \(\lambda\).

04

Calculate Expected Number of Events [2,6]

Similarly, for the interval \([2, 6]\), we calculate \(\lambda = \int_2^6 e^{2 + 0.6t} \, dt\). The definite integral can be expressed as \(e^2 \cdot \left[ \frac{e^{0.6t}}{0.6} \right]_2^6 = \frac{e^2}{0.6} \left( e^{3.6} - e^{1.2} \right)\). Evaluate this expression for \(\lambda\).

05

Solve Integral for [0.9907] using Expectation

Using \(\alpha(t) = e^{2 + 0.6t}\), the expected number of events for the interval \([0, 0.9907]\) is given by \(\lambda = \int_0^{0.9907} e^{2 + 0.6t} \, dt\). This simplifies to \(\frac{e^2}{0.6} \left( e^{0.59442} - 1 \right)\). Evaluate this integral to get the parameter \(\lambda\) for the Poisson distribution.

06

Probability of At Most 15 Events [0, .9907]

Using the calculated \(\lambda\) in Step 5, determine the probability that at most 15 events occur. Use the cumulative Poisson probability: \(P(X \leq 15) = \sum_{k=0}^{15} \frac{\lambda^k e^{-\lambda}}{k!} \). Compute this sum using \(\lambda\) from Step 5.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Poisson distribution

The Poisson distribution is a fundamental probability distribution that describes the number of events occurring within a fixed interval of time or space.

• It is particularly useful for modeling situations where events are independent.
• The parameter \( \lambda \) represents the expected number of events, or the average rate, happening in the given interval.
• The key characteristic of a Poisson distribution is that it deals with rare events and large populations.

The Poisson distribution assumes that these events occur with a constant mean rate, which leads us into the concept of the intensity function, which can adjust this to account for varying rates over time or space.

Intensity function

In a nonhomogeneous Poisson process, the intensity function \( \alpha(t) \) is a vital concept. It defines how the rate of events varies over time.

• The intensity function \( \alpha(t) = e^{a + bt} \) suggests an exponential growth of rate, with constants \( a \) and \( b \) defining the specific characteristics.
• For example, in modeling HIV transmissions via blood transfusions, such an intensity function can capture increasing transmission rates over time.
• The function is integral to computing the expected number of events over the desired time interval \([t_1, t_2]\).

In our problem, we used the intensity function to calculate \( \lambda \) for different intervals, understanding how event occurrence varies as time progresses.

Expected number of events

The expected number of events is represented by the parameter \( \lambda \) in a Poisson distribution.

• To calculate \( \lambda \), integrate the intensity function \( \alpha(t) \) over the desired time interval \([t_1, t_2]\).
• This technique gives us the average number of events you can expect in that period.
• For example, \[ \lambda = \int_0^4 e^{2 + 0.6t} \, dt \]allows us to estimate the average events over 4 years.

The integration process might require techniques like substitution to solve complex exponential functions, but it ultimately leads us to the expected number of occurrences.

Cumulative probability

Cumulative probability in the context of Poisson distributions gives us the probability of observing up to a certain number of events.

• It is expressed as \( P(X \leq k) \), where \( k \) is the maximum number of events considered.
• This involves summing the probabilities of the event counts \( 0 \) to \( k \).
• The relevant formula for the Poisson distribution is:\[ P(X \leq k) = \sum_{i=0}^{k} \frac{\lambda^i e^{-\lambda}}{i!}\]

In practice, this means calculating each term in the series using a computed \( \lambda \) and adding them up to find the probability. For most practical problems, software or probability tables assist in these computations, especially for larger \( \lambda \) values.