/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 80 Let \(T_{1}, T_{2}, \ldots\) den... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(T_{1}, T_{2}, \ldots\) denote the interarrival times of events of a nonhomogeneous Poisson process having intensity function \(\lambda(t)\). (a) Are the \(T_{i}\) independent? (b) Are the \(T_{i}\) identically distributed? (c) Find the distribution of \(T_{1}\).

Short Answer

Expert verified
The interarrival times (Ti) in a nonhomogeneous Poisson process are independent but not identically distributed. The distribution of T1 is given by the probability density function f(t) = \(\lambda(t)\exp(-\int_{0}^{t}\lambda(u)du)\).

Step by step solution

01

Answer (a) - Interarrival Time Independence

A nonhomogeneous Poisson process can be described as a counting process where the probability of an event happening in a small time interval [t, t+dt) is proportional to λ(t)dt. The number of events happening in non-overlapping intervals are independent. Therefore, the interarrival times (Ti) of events are also independent.
02

Answer (b) - Identical Distribution of Interarrival Times

In a nonhomogeneous Poisson process, the intensity function λ(t) can vary with time. Since the distribution of interarrival times depends on the intensity function, they are not identically distributed when λ(t) is not a constant function.
03

Answer (c) - Distribution of T1

To find the distribution of T1, we need to calculate the probability density function (pdf) of T1, which is denoted as f(t). The cumulative distribution function (CDF) of T1, denoted as F(t), is the probability that T1 < t, i.e., the probability that the process has at least one event before time t. Therefore, F(t) = 1 - Probability that there are no events before time t. Let N(t) be the number of events in the interval [0, t]. Then, P[N(t) = 0] represents the probability that there are no events before time t. For a nonhomogeneous Poisson process, the probability of N(t) = 0 can be given as: P[N(t) = 0] = \(\exp(-\int_{0}^{t}\lambda(u)du)\) So, F(t) = 1 - P[N(t) = 0] = 1 - \(\exp(-\int_{0}^{t}\lambda(u)du)\) Now, to find the pdf f(t), we need to differentiate F(t) with respect to t: f(t) = \(\frac{dF(t)}{dt}\) = \(\lambda(t)\exp(-\int_{0}^{t}\lambda(u)du)\) Thus, the distribution of T1 is given by the pdf: f(t) = \(\lambda(t)\exp(-\int_{0}^{t}\lambda(u)du)\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Interarrival Times
Understanding the concept of 'interarrival times' is crucial when dealing with process models such as the nonhomogeneous Poisson process. Interarrival times, represented as \(T_1, T_2, \ldots\), are the durations between consecutive events in a given stochastic process. Imagine you are waiting for buses at a stop, and the interarrival times would be the waiting periods between the arrival of each bus.

For the Poisson process, these interarrival times have significant properties. One key attribute is their independence, as mentioned in the exercise solution. This means the time between events does not influence the time between subsequent events; each waiting period is determined by its own set of circumstances, regardless of others.

However, in a nonhomogeneous scenario, these times are not identically distributed. This is because the intensity function \(\lambda(t)\), which governs the event occurrence rate, is time-dependent. Therefore, the likelihood of event spacing varies throughout the process. This time-dependency introduces variability in interarrival distributions, meaning your waiting time for buses changes depending on the time of day.
Intensity Function
The 'intensity function' of a Poisson process, denoted as \(\lambda(t)\), is a key concept that defines the instantaneous rate of event occurrences at any given time \(t\). Think of it as a dial that adjusts how frequently something might happen; it is the heartbeat of the nonhomogeneous Poisson process.

The intensity function in a homogeneous Poisson process is constant, meaning events occur with a steady rhythm. However, in a nonhomogeneous scenario, this function can vary with time, much like how the rhythm of your heart might quicken or slow down. This variability impacts the rate at which events are expected to occur and consequently affects the distribution of interarrival times.

Understanding the intensity function is fundamental when calculating the probability of events over time. It is integral to finding the cumulative distribution function (CDF) and probability density function (pdf) for the process, as it quantifies the probability of an event occurring within an infinitesimal time frame.
Probability Density Function
Finally, the 'probability density function' (pdf) is a mathematical expression that outlines how the density of probability is distributed across values of a continuous random variable; it describes the likelihood of a random variable taking on a specific value. In terms of a nonhomogeneous Poisson process, the pdf is used to determine the distribution of interarrival times.

In the provided solution, the distribution of the first interarrival time \(T_1\) is described by a pdf denoted \(f(t)\). This function is not fixed but is tied intricately to the intensity function \(\lambda(t)\). The pdf illustrates how likely it is that the first event in our Poisson process occurs at any specific time \(t\).

The process to find the pdf from the intensity function involves creating the cumulative distribution function (CDF) first. The CDF is the integral of the intensity function over time, which represents the accumulation of probabilities up until a certain time. Differentiating this CDF with respect to time gives us the pdf, \(f(t) = \lambda(t)\exp(-\int_{0}^{t}\lambda(u)du)\), revealing the probability density of \(T_1\) at any given moment. This function is indispensable for predictive modeling and understanding the dynamic nature of event occurrences in a nonhomogeneous Poisson process.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider a two-server system in which a customer is served first by server 1, then by server 2, and then departs. The service times at server \(i\) are exponential random variables with rates \(\mu_{i}, i=1,2 .\) When you arrive, you find server 1 free and two customers at server 2 -customer \(\mathrm{A}\) in service and customer \(\mathrm{B}\) waiting in line. (a) Find \(P_{A}\), the probability that \(A\) is still in service when you move over to server 2 . (b) Find \(P_{B}\), the probability that \(B\) is still in the system when you move over to server 2 . (c) Find \(E[T]\), where \(T\) is the time that you spend in the system. Hint: Write $$ T=S_{1}+S_{2}+W_{A}+W_{B} $$ where \(S_{i}\) is your service time at server \(i, W_{A}\) is the amount of time you wait in queue while \(A\) is being served, and \(W_{B}\) is the amount of time you wait in queue while \(B\) is being served.

A two-dimensional Poisson process is a process of randomly occurring events in the plane such that (i) for any region of area \(A\) the number of events in that region has a Poisson distribution with mean \(\lambda A\), and (ii) the number of events in nonoverlapping regions are independent. For such a process, consider an arbitrary point in the plane and let \(X\) denote its distance from its nearest event (where distance is measured in the usual Euclidean manner). Show that (a) \(P[X>t\\}=e^{-\lambda \pi t^{2}}\), (b) \(E[X]=\frac{1}{2 \sqrt{2}}\).

Consider an infinite server queuing system in which customers arrive in accordance with a Poisson process with rate \(\lambda\), and where the service distribution is exponential with rate \(\mu\). Let \(X(t)\) denote the number of customers in the system at time \(t\). Find (a) \(E[X(t+s) \mid X(s)=n] ;\) (b) \(\operatorname{Var}[X(t+s) \mid X(s)=n]\). Hint: Divide the customers in the system at time \(t+s\) into two groups, one consisting of "old" customers and the other of "new" customers. (c) Consider an infinite server queuing system in which customers arrive according to a Poisson process with rate \(\lambda\), and where the service times are all exponential random variables with rate \(\mu .\) If there is currently a single customer in the system, find the probability that the system becomes empty when that customer departs.

If \(X_{i}, i=1,2,3\), are independent exponential random variables with rates \(\lambda_{i}\), \(i=1,2,3\), find (a) \(\left.P \mid X_{1}

Let \(S(t)\) denote the price of a security at time \(t .\) A popular model for the process \(\\{S(t), t \geqslant 0\\}\) supposes that the price remains unchanged until a "shock" occurs, at which time the price is multiplied by a random factor. If we let \(N(t)\) denote the number of shocks by time \(t\), and let \(X_{i}\) denote the \(i\) th multiplicative factor, then this model supposes that $$ S(t)=S(0) \prod_{i=1}^{N(t)} X_{i} $$ where \(\prod_{i=1}^{N(t)} X_{i}\) is equal to 1 when \(N(t)=0 .\) Suppose that the \(X_{i}\) are independent exponential random variables with rate \(\mu ;\) that \(\\{N(t), t \geqslant 0\\}\) is a Poisson process with rate \(\lambda ;\) that \(\\{N(t), t \geqslant 0\\}\) is independent of the \(X_{i} ;\) and that \(S(0)=s\). (a) Find \(E[S(t)]\). (b) Find \(E\left[S^{2}(t)\right]\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.