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Let \(X_{1}, X_{2}, \ldots\) be independent positive continuous random variables with a common density function \(f\), and suppose this sequence is independent of \(N, a\) Poisson random variable with mean \(\lambda\). Define $$ N(t)=\text { number of } i \leqslant N: X_{i} \leqslant t $$ Show that \(\\{N(t), t \geqslant 0\\}\) is a nonhomogeneous Poisson process with intensity function \(\lambda(t)=\lambda f(t)\).

Short Answer

Expert verified
We have verified all properties required for a nonhomogeneous Poisson process: \(N(0) = 0\), the process has independent increments, and the distribution of \(N(t + \delta) - N(t)\) is a Poisson distribution with parameter \(\int_{t}^{t + \delta}\lambda(s) ds\). Thus, we have shown that \(\\{N(t), t \geqslant 0\\}\) is a nonhomogeneous Poisson process with intensity function \(\lambda(t) = \lambda f(t)\).

Step by step solution

01

Verify N(0) = 0

Since \(N(t)\) is the number of \(i \leqslant N\) such that \(X_i \leqslant t\), when \(t = 0\), all the \(X_i\) will be greater than 0 because they are positive continuous random variables. Therefore, there will be no \(X_i\) such that \(X_i \leqslant t=0\), which results in \(N(0) = 0\).
02

Verify Independent Increments

Consider two disjoint time intervals \((t_1, t_2]\) and \((t_3, t_4]\) with \(0 \leq t_1 < t_2 < t_3 < t_4\). Now we have to show that \((N(t_2) - N(t_1))\) is independent of \((N(t_4) - N(t_3))\). Since \(X_i\) and \(N\) are independent, any arrival that occurs in the time interval \((t_1, t_2]\) will not depend on an arrival that occurs in the time interval \((t_3, t_4]\). Therefore, the number of arrivals in each time interval will be independent.
03

Verify Poisson Distribution for N(t + δ) - N(t)

Let \(\delta > 0\). We need to show that \(N(t+\delta) - N(t) \sim \text{Poisson}\left(\int_{t}^{t + \delta}\lambda(s) ds\right)\). The conditional distribution of \(N(t + \delta) - N(t)\), given N = n, is a binomial distribution with parameters n and \(p_{\delta}(t) = P(t < X_i \leq t+\delta | N=n)\). Conditional probability: \begin{align*} p_{\delta}(t) &= P(t < X_i \leq t+\delta | N=n) \\ &= \frac{P(t < X_i \leq t+\delta, N=n)}{P(N=n)} \\ &= \frac{\int_t^{t+\delta} f(x) \,\text{d}x}{1} \end{align*} The distribution of \(N(t+\delta) - N(t)\), given \(N=n\), is a binomial distribution with parameters n and \(p_{\delta}(t)\). Now, we will find the distribution of \(N(t+\delta) - N(t)\). Using Binomial-Poisson convergence, the distribution of \(N(t+\delta) - N(t)\) is Poisson with parameter: \begin{align*} \int_{t}^{t + \delta}\lambda(s) ds &= \lambda \int_{t}^{t +\delta}f(s) ds \\ &= \lambda \cdot \left(\int_{t}^{t + \delta} f(x) \,\text{d}x\right) \end{align*} Thus, we have verified that \(N(t+\delta) - N(t)\) is Poisson distributed with the given parameter. We have shown that the given process has N(0) = 0, independent increments, and has a Poisson distribution for N(t + δ) - N(t) with parameter \(\lambda(t)=\lambda f(t)\). Therefore, the process \(\\{N(t), t \geqslant 0\\}\) is a nonhomogeneous Poisson process with intensity function \(\lambda(t) = \lambda f(t)\).

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Key Concepts

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

Independent Random Variables
Understanding independent random variables is crucial for grasping the problem at hand. In probability theory, two random variables are independent if the occurrence of one does not affect the probability of the occurrence of the other. In simple terms, knowing the outcome of one variable tells us nothing about the other.
In this exercise, the random variables \(X_1, X_2, \ldots\) are independent and have a common density function \(f\). This means that the behavior of each \(X_i\) doesn't influence any other \(X_j\).
Additionally, the sequence \(X_1, X_2, \ldots\) is independent of the Poisson random variable \(N\) with mean \(\lambda\).
  • Key takeaway: Independence ensures that the sum of these events over time can be simplified, making calculations manageable.
Poisson Distribution
The Poisson distribution is a statistical distribution that shows the likelihood of a given number of events happening in a fixed interval of time or space. It is characterized by the mean number of events \(\lambda\).
In the context of this problem, \(N\) is a Poisson random variable with mean \(\lambda\). The Poisson distribution helps quantify the probability of a number of events (arrivals, in this case) occurring within a specified time period.
  • Essential property: The process is memoryless, meaning past events do not affect future event probabilities.
It's used in this exercise to model the number of random arrivals up to time \(t\) that are within a certain bound \(X_i \leq t\).
Intensity Function
The intensity function \(\lambda(t)\) in a nonhomogeneous Poisson process shifts the regular Poisson process into one where the rate of occurrence can vary over time. It quantifies how the rate of events at a small time interval around \(t\) might differ from another time interval.
Here, \(\lambda(t) = \lambda f(t)\), meaning it's modulated by the underlying density function \(f(t)\). This structure ensures that as the density of \(X_i\) changes with time, so does the rate at which events pop up in the Poisson process.
  • Functionality: Adjusts the conventional Poisson rate \(\lambda\) based on the relevant density function \(f(t)\).
This function is the heart of defining the nonhomogeneous process in the exercise.
Binomial-Poisson Convergence
Binomial-Poisson convergence is a concept that outlines how a binomial distribution approaches a Poisson distribution under specific conditions. This convergence plays a pivotal role when the number of trials is large, and the probability of success is small.
In this exercise, the difference \(N(t+\delta) - N(t)\) conditional on \(N=n\) follows a binomial distribution, but the solution moves towards showing this is Poisson distributed.
  • Reason for convergence: If we have many arrivals but very few events in each small interval, it mirrors typical Poisson scenarios.
This is achieved as the calculation simplifies, iterating reasoning that transforms discrete (binomial) to continuous (Poisson) distribution modeling effectively.

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Most popular questions from this chapter

Teams 1 and 2 are playing a match. The teams score points according to independent Poisson processes with respective rates \(\lambda_{1}\) and \(\lambda_{2} .\) If the match ends when one of the teams has scored \(k\) more points than the other, find the probability that team 1 wins. Hint: Relate this to the gambler's ruin problem.

(a) Let \(\\{N(t), t \geqslant 0\\}\) be a nonhomogeneous Poisson process with mean value function \(m(t) .\) Given \(N(t)=n\), show that the unordered set of arrival times has the same distribution as \(n\) independent and identically distributed random variables having distribution function $$ F(x)=\left\\{\begin{array}{ll} \frac{m(x)}{m(t)}, & x \leqslant t \\ 1, & x \geqslant t \end{array}\right. $$ (b) Suppose that workmen incur accidents in accordance with a nonhomogeneous Poisson process with mean value function \(m(t) .\) Suppose further that each injured man is out of work for a random amount of time having distribution F. Let \(X(t)\) be the number of workers who are out of work at time \(t\). By using part (a), find \(E[X(t)]\)

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) \(E[X]=\frac{1}{2 \sqrt{2}}\).

The lifetime of a radio is exponentially distributed with a mean of ten years. If Jones buys a ten-year-old radio, what is the probability that it will be working after an additional ten years?

Shocks occur according to a Poisson process with rate \(\lambda\), and each shock independently causes a certain system to fail with probability \(p .\) Let \(T\) denote the time at which the system fails and let \(N\) denote the number of shocks that it takes. (a) Find the conditional distribution of \(T\) given that \(N=n\). (b) Calculate the conditional distribution of \(N\), given that \(T=t\), and notice that it is distributed as 1 plus a Poisson random variable with mean \(\lambda(1-p) t .\) (c) Explain how the result in part (b) could have been obtained without any calculations.

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