/*! 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 10 Consider a renewal process with ... [FREE SOLUTION] | 91影视

91影视

Consider a renewal process with mean interarrival time \(\mu\). Suppose that each event of this process is independently "counted" with probability \(p .\) Let \(N_{C}(t)\) denote the number of counted events by time \(t, t>0\). (a) Is \(N_{C}(t), t \geqslant 0\) a renewal process? (b) What is \(\lim _{t \rightarrow \infty} N_{C}(t) / t ?\)

Short Answer

Expert verified
(a) Yes, \(N_{C}(t)\) is a renewal process for \(t \geq 0\). (b) \(\lim_{t \rightarrow \infty} \frac{N_{C}(t)}{t} = \frac{p}{\mu}\)

Step by step solution

01

Determining if \(N_{C}(t)\) is a renewal process

A renewal process is a stochastic process that describes the arrival times of certain events, in which the interarrival intervals are statistically independent and have a common distribution. In other words, a renewal process must satisfy the following conditions: 1. The events occur one at a time. 2. The time between consecutive events (interarrival time) has a common distribution. 3. The interarrival times are independent of each other. In our given problem, each event of this process is independently counted with probability \(p\). Since events are individually counted and each event is counted independently of the others, the resulting counting process, \(N_{C}(t)\), will also have independent interarrival times. Consequently, \(N_{C}(t)\) is a renewal process for \(t \geq 0\). So, the answer to part (a) is "yes".
02

Calculating the limiting value of \(N_{C}(t)/t\)

We need to find the limit of \(N_{C}(t)\) divided by \(t\) as \(t\) goes to infinity. First, let's find the expected value of \(N_{C}(t)\). From the definition of a renewal process, we know that the mean interarrival time is 饾渿. Since each event is counted with probability \(p\), the expected number of counted events in a period of length \(t\) is: \[E\left[N_{C}(t)\right] = \frac{t}{\mu} \cdot p\] Now, we can calculate the limiting value of \(N_{C}(t)/t\) as \(t\) goes to infinity: \[\lim_{t \rightarrow \infty} \frac{N_{C}(t)}{t} = \lim_{t \rightarrow \infty} \frac{E\left[N_{C}(t)\right]}{t} = \lim_{t \rightarrow \infty} \frac{\frac{t}{\mu} \cdot p}{t}\] \[= \lim_{t \rightarrow \infty} (p \cdot \frac{1}{\mu}) = \frac{p}{\mu}\] So, the answer to part (b) is \(\frac{p}{\mu}\).

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 91影视!

Key Concepts

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

Stochastic Process
A stochastic process is a mathematical object usually defined as a collection of random variables, which can represent the evolution of some system of random values over time. It is a concept used in probability theory to describe systems that evolve in a random manner.

The idea is that a stochastic process is influenced by some forms of random events, and its development over time is not deterministic but instead follows a probabilistic model. In the realm of renewal processes, the stochastic process is utilized to model and predict events happening at various times.
  • Think of it like observing a queue, where customers arrive at random intervals. Each point in this queue鈥檚 timeline is influenced by stochasticity.
  • Processes are identified by the statistical properties of their path, or history, as they may accomplish different form or be used for different kind of data interpretations, like Gaussian processes for normal distributions.

In our exercise, the renewal process itself is a stochastic process. Specifically, it describes the counting of events over time governed by stochastic variables, like whether or not each event happens and gets "counted".
Interarrival Time
Interarrival time refers to the time between two consecutive events in a stochastic system. It is an essential concept in renewal processes because it helps to determine the frequency and regularity of occurrences in a system.

In topics like queuing theory or renewal theory, the interarrival time might be modeled with different probability distributions to reflect the real-world system in which these events occur. Here are a few key points about interarrival time:
  • The interarrival time can be constant, but often it follows a probability distribution which can be exponential, Poisson, or any other relevant form depending on the system鈥檚 nature.
  • For a true renewal process, the interarrival times are assumed to be statistically independent and identically distributed (i.i.d).

In the scenario described in our exercise, the interarrival time has a mean value denoted by \(\mu\). The independence of these intervals ensures that even when applying a probability \(p\) to "count" the occurrences, each counting process maintains the integrity of a renewal process.
Probability Theory
Probability theory is the branch of mathematics concerned with analyzing random phenomena. It serves as the foundation upon which renewal processes, and indeed our entire exercise, are built.

This theory allows us to quantify the likelihood of various outcomes and to understand the underlying patterns within seemingly random events. Some important aspects include:
  • Probability provides the framework for determining how likely it is that a particular event will occur, whether it's part of a renewal process or any stochastic model.
  • In our exercise, probability theory is crucial for determining the "counting" mechanism. Each event is "counted" with a probability \(p\), which ultimately affects the resulting stochastic process.

In renewal processes, concepts from probability theory are employed to predict how the process changes over time, using mathematical tools to determine properties like expected value and to make predictions about future behavior. Specifically, the expectation that each event is independently "counted" examines the likelihood of each occurrence, while the result \(\frac{p}{\mu}\) is a consequence of understanding these probabilities over a long time horizon.

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 semi-Markov process in which the amount of time that the process spends in each state before making a transition into a different state is exponentially distributed. What kind of process is this?

Suppose that the interarrival distribution for a renewal process is Poisson distributed with mean \(\mu .\) That is, suppose $$ P\left\\{X_{n}=k\right\\}=e^{-\mu} \frac{\mu^{k}}{k !}, \quad k=0,1, \ldots $$ (a) Find the distribution of \(S_{n}\). (b) Calculate \(P\\{N(t)=n\\}\).

An airport shuttle bus picks up all passengers waiting at a bus stop and drops them off at the airport terminal; it then returns to the stop and repeats the process. The times between returns to the stop are independent random variables with distribution \(F\), mean \(\mu\), and variance \(\sigma^{2}\). Passengers arrive at the bus stop in accordance with a Poisson process with rate \(\lambda\). Suppose the bus has just left the stop, and let \(X\) denote the number of passengers it picks up when it returns. (a) Find \(E[X]\). (b) Find \(\operatorname{Var}(X)\). (c) At what rate does the shuttle bus arrive at the terminal without any passengers? Suppose that each passenger that has to wait at the bus stop more than \(c\) time units writes an angry letter to the shuttle bus manager. (d) What proportion of passengers write angry letters? (e) How does your answer in part (d) relate to \(F_{e}(x)\) ?

For an interarrival distribution \(F\) having mean \(\mu\), we defined the equilibrium distribution of \(F\), denoted \(F_{e}\), by $$F_{e}(x)=\frac{1}{\mu} \int_{0}^{x}[1-F(y)] d y$$ (a) Show that if \(F\) is an exponential distribution, then \(F=F_{e}\). (b) If for some constant \(c\), $$ F(x)=\left\\{\begin{array}{ll} 0, & x

Satellites are launched according to a Poisson process with rate \(\lambda\). Each satellite will, independently, orbit the earth for a random time having distribution \(F\). Let \(X(t)\) denote the number of satellites orbiting at time \(t\). (a) Determine \(P\\{X(t)=k\\}\). Hint: Relate this to the \(M / G / \infty\) queue. (b) If at least one satellite is orbiting, then messages can be transmitted and we say that the system is functional. If the first satellite is orbited at time \(t=0\), determine the expected time that the system remains functional. Hint: Make use of part (a) when \(k=0\).

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.