/*! 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 12 Events occur according to a Pois... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Events occur according to a Poisson process with rate \(\lambda\). Any event that occurs within a time \(d\) of the event that immediately preceded it is called a \(d\) -event. For instance, if \(d=1\) and events occur at times \(2,2.8,4,6,6.6, \ldots\), then the events at times \(2.8\) and \(6.6\) would be \(d\) -events. (a) At what rate do \(d\) -events occur? (b) What proportion of all events are \(d\) -events?

Short Answer

Expert verified
In summary: (a) The rate at which \(d\)-events occur is \(\lambda_d = \lambda (1 - e^{-\lambda d})\). (b) The proportion of all events that are \(d\)-events is \(1 - e^{-\lambda d}\).

Step by step solution

01

Memoryless property and exponential distribution

We know that events occur according to a Poisson process with rate \(\lambda\). The time between events follows an exponential distribution with parameter \(\lambda\). The exponential distribution has the memoryless property, which means that the time until the next event is independent of how long it has been since the last event.
02

Find the probability of having a d-event

To find the rate at which \(d\)-events occur, we need to determine the probability of an event being a \(d\)-event. Using the exponential distribution, the probability density function \(f(t)\) of the time \(T\) between events is given by: \[f(t) = \lambda e^{-\lambda t}\] We can find the probability of having a \(d\)-event by integrating this probability density function over the interval \([0, d]\): \[P(T \leq d) = \int_0^d \lambda e^{-\lambda t} dt\]
03

Integrate the probability density function

Integrating the probability density function, we get: \[P(T \leq d) = -e^{-\lambda t} \Big|_0^d = 1 - e^{-\lambda d}\]
04

Calculate the rate of d-events

The rate at which \(d\)-events occur is the probability of having a \(d\)-event multiplied by the overall rate \(\lambda\) of events occurring: \[\lambda_d = \lambda \times P(T \leq d) = \lambda (1 - e^{-\lambda d})\] So, the rate at which \(d\)-events occur is \(\lambda_d = \lambda (1 - e^{-\lambda d})\).
05

Calculate the proportion of d-events

To calculate the proportion of \(d\)-events among all events, divide the rate of \(d\)-events by the total rate of events occurring: \[Proportion \ of \ d-events = \frac{\lambda_d}{\lambda} = \frac{\lambda (1 - e^{-\lambda d})}{\lambda} = 1 - e^{-\lambda d}\] So, the proportion of all events that are \(d\)-events is \(1 - e^{-\lambda d}\). To summarize: (a) The rate at which \(d\)-events occur is \(\lambda_d = \lambda (1 - e^{-\lambda d})\). (b) The proportion of all events that are \(d\)-events is \(1 - e^{-\lambda d}\).

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.

Exponential Distribution
The exponential distribution is crucial in the study of continuous time events, especially when it comes to Poisson processes. Imagine waiting for buses that come randomly but with a consistent average rate. How long you'll have to wait for the next bus follows what we call 'exponential distribution'. The key parameter here is the rate \(\lambda\), which tells us the average number of events happening in a unit time interval.

The formula for the probability density function (PDF) of the exponential distribution is: \[ f(t) = \lambda e^{-\lambda t} \] where \(t\) represents time. This function gives us the likelihood of an event occurring at a specific time after the previous event. The exponential distribution is unique because it is memoryless, which brings us neatly to our next important concept.
Memoryless Property
This property is both intriguing and counterintuitive. It means that the probability of an event occurring in the future is independent of how much time has already elapsed. In other words, no matter how long you've been waiting, your expected wait time remains the same. Applying this to our bus stop analogy, even if you've already waited 10 minutes, your expected wait time for the next bus is still the same as when you first arrived.

Mathematically, the memoryless property can be demonstrated like this: For any non-negative numbers \(s\) and \(t\), the probability that the wait will exceed \(s + t\) given that it has exceeded \(s\) is the same as the probability that it will exceed \(t\):\[ P(T > s + t | T > s) = P(T > t) \] This property is exclusive to the exponential (for continuous variables) and geometric (for discrete variables) distributions. It simplifies a lot of calculations concerning the timing of future events in a Poisson process.
Probability Density Function
To truly comprehend the exponential distribution, we need to understand the probability density function (PDF). The PDF is a function that describes the relative likelihood for a continuous random variable to take on a given value. In the context of the exponential distribution, the PDF tells us how likely it is that an event will occur at any given instant within a continuous interval.

For the exponential distribution, the PDF is defined by \[ f(t) = \lambda e^{-\lambda t} \], where \(e\) is the base of the natural logarithm, and \(\lambda\) is the rate of events per time unit. To find the probability of an event occurring within a specific interval, we would integrate this function over that interval. For example, in the exercise provided, the integration of the PDF from \(0\) to \(d\) gives us the probability of a \(d\)-event, as shown by: \[ P(T \leq d) = \int_0^d \lambda e^{-\lambda t} dt \] This way, the PDF not only gives us a snapshot of an event's instant likelihood but, with integration, it also allows us to calculate the chance of an event occurring within a broader time frame.

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 the gambler's ruin problem where on each bet the gambler either wins 1 with probability \(p\) or loses 1 with probability \(1-p\). The gambler will continue to play until his winnings are either \(N-i\) or \(-i\). (That is, starting with \(i\) the gambler will quit when his fortune reaches either \(N\) or \(0 .\) ) Let \(T\) denote the number of bets made before the gambler stops. Use Wald's equation, along with the known probability that the gambler's final winnings are \(N-i\), to find \(E[T]\). Hint: Let \(X_{j}\) be the gambler's winnings on bet \(j, j \geqslant 1\). What are the possible values of \(\sum_{j=1}^{T} X_{j} ?\) What is \(E\left[\sum_{j=1}^{T} X_{j}\right]\) ?

A system consists of two independent machines that each function for an exponential time with rate \(\lambda .\) There is a single repairperson. If the repairperson is idle when a machine fails, then repair immediately begins on that machine; if the repairperson is busy when a machine fails, then that machine must wait until the other machine has been repaired. All repair times are independent with distribution function \(G\) and, once repaired, a machine is as good as new. What proportion of time is the repairperson idle?

Mr. Smith works on a temporary basis. The mean length of each job he gets is three months. If the amount of time he spends between jobs is exponentially distributed with mean 2, then at what rate does \(\mathrm{Mr}\). Smith get new jobs?

Each time a certain machine breaks down it is replaced by a new one of the same type. In the long run, what percentage of time is the machine in use less than one year old if the life distribution of a machine is (a) uniformly distributed over \((0,2)\) ? (b) exponentially distributed with mean \(1 ?\)

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.