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Show that if \(\left\\{N_{i}(t), t \geqslant 0\right\\}\) are independent Poisson processes with rate \(\lambda_{i}, i=1,2\), then \(\\{N(t), t \geqslant 0\\}\) is a Poisson process with rate \(\lambda_{1}+\lambda_{2}\) where \(N(t)=N_{1}(t)+N_{2}(t)\)

Short Answer

Expert verified
The process \(\\{N(t), t \geqslant 0\\}\) is a Poisson process with rate \(\lambda_{1}+\lambda_{2}\) where \(N(t)=N_{1}(t)+N_{2}(t)\), since it satisfies the following properties: \(N(0)=0\), the events in non-overlapping intervals are independent, and the distribution of the number of events in an interval of length \(t\) is Poisson with parameter \((\lambda_{1}+\lambda_{2})t\).

Step by step solution

01

Initial Condition

Since we are given that \(\\{N_{i}(t), t \geqslant 0\\}\) are Poisson processes with rates \(\lambda_{i}, i=1,2\), we have: \[N_{1}(0) = 0\] \[N_{2}(0) = 0\] Thus, their sum is: \[N(0) = N_{1}(0) + N_{2}(0) = 0\]
02

Independence of Non-overlapping Intervals

Let \(0 \leq s < t\) and consider the number of events in the intervals \([0, s)\) and \([s, t)\). We have to show that the events in these intervals are independent. Since \(N_{1}(t)\) and \(N_{2}(t)\) are both Poisson processes, we know that the number of events in these intervals for each process is independent. Now, for \(N(t)\): \begin{align*} N(t) &= N_{1}(t) + N_{2}(t) \\ N(s) &= N_{1}(s) + N_{2}(s) \end{align*} Given that the increments for \(N_1(t)\) and \(N_2(t)\) are independent, we can say that the increments for the sum of these processes are also independent.
03

Poisson Distribution with Parameter \((\lambda_{1}+\lambda_{2})t\)

Finally, we must show that the distribution of the number of events in an interval of length \(t\) is Poisson with parameter \((\lambda_{1}+\lambda_{2})t\). Let \(X_{1}(t)\) and \(X_{2}(t)\) represent the number of events in the interval \([0, t)\) for processes \(N_{1}(t)\) and \(N_{2}(t)\), respectively. Then, \(X(t)= N(t) = X_{1}(t) + X_{2}(t)\). It is given that \(X_{i}(t)\) are Poisson distributed with parameters \(\lambda_{i}t\), for \(i=1,2\). The sum of independent Poisson distributed variables is also Poisson distributed with the sum of their parameters. So, \[X(t) \sim Poisson(\lambda_{1}t + \lambda_{2}t)\] Therefore, the process \(\\{N(t), t \geqslant 0\\}\) is a Poisson process with rate \(\lambda_{1}+\lambda_{2}\) where \(N(t)=N_{1}(t)+N_{2}(t)\).

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Key Concepts

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

Independent Poisson processes
A Poisson process is a statistical model used to describe events that occur randomly over time. When dealing with independent Poisson processes, it's essential to understand what independence means in this context. Two Poisson processes, say \(\{ N_1(t) \} \) and \(\{ N_2(t) \} \), are considered independent if the occurrence of events in one does not affect the occurrence of events in the other.
This independence is particularly helpful when analyzing systems where multiple types of events happen simultaneously but don't influence each other. For instance, think of the number of phone calls and emails a customer service center receives. If these are modeled as independent Poisson processes, the phone calls (\( N_1(t) \)) happen independently of the emails (\( N_2(t) \)).
This independence is a foundational concept because it allows combining these processes effectively, as we shall see in the next sections.
Poisson distribution
The Poisson distribution is vital in probability and statistics when discussing the number of events in fixed intervals of time or space. It’s grounded on the assumption that these events happen independently, and the probability of two or more events occurring at precisely the same instant is zero.
The distribution is parameterized by \(\lambda\), which represents the average rate or expected number of events happening in a given interval. Mathematically, if \( X \) denotes the number of events, then \( X \) follows a Poisson distribution with parameter \( \lambda t \), expressed as \( X \sim Poisson(\lambda t) \). Here, \( t \) is the length of the interval being observed.
The beauty of the Poisson distribution lies in its simplicity and its suitability for modeling rare events. It transforms complex real-world phenomena into manageable probabilistic concepts, useful in fields ranging from telecommunications to biology.
Rate of a Poisson process
The rate of a Poisson process, denoted as \( \lambda \), is the average number of events expected in a unit time interval. It’s a crucial parameter that provides meaningful insight into the behavior of the process.
The rate determines how frequently events are likely to occur, representing the process's intensity. In any practical scenario, understanding this rate helps in forecasting and planning. For instance, if a bank knows it receives customers at a rate of 5 per hour, it can allocate resources more efficiently.
In combining Poisson processes, as seen in merging two independent processes \(\{ N_1(t) \} \) and \(\{ N_2(t) \}\) with different rates \(\lambda_1\) and \(\lambda_2\), the resulting process \(\{ N(t) \}\) will have a rate \(\lambda_1 + \lambda_2\). This additive property simplifies the analysis of such combined processes and is a significant aspect of studying Poisson processes.
Sum of Poisson processes
A fascinating property of Poisson processes is that the sum of independent Poisson processes is itself another Poisson process. When dealing with \(\{ N_1(t) \} \) and \(\{ N_2(t) \} \) that are independent Poisson processes with rates \(\lambda_1\) and \(\lambda_2\), the process \(\{ N(t) = N_1(t) + N_2(t) \} \) is also a Poisson process.
This combined process, \(\{ N(t) \} \), is characterized by the rate \(\lambda = \lambda_1 + \lambda_2\). This property is both intuitive and mathematically efficient, allowing us to treat complex problems as simple linear combinations.
Practically, this can be likened to combining traffic from two highways merging into one. Each highway carries traffic independently, and when they merge, the total traffic—not only the original individual ones but also the sum—flows through the new combined route. Understanding this concept is key in fields like network traffic management and resource optimization.

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Most popular questions from this chapter

Customers arrive at the automatic teller machine in accordance with a Poisson process with rate 12 per hour. The amount of money withdrawn on each transaction is a random variable with mean \(\$ 30\) and standard deviation \(\$ 50\). (A negative withdrawal means that money was deposited.) The machine is in use for 15 hours daily. Approximate the probability that the total daily withdrawal is less than \(\$ 6000\).

Let \(X_{1}, X_{2}, \ldots\) be independent and identically distributed nonnegative continuous random variables having density function \(f(x)\). We say that a record occurs at time \(n\) if \(X_{n}\) is larger than each of the previous values \(X_{1}, \ldots, X_{n-1}\). (A record automatically occurs at time 1.) If a record occurs at time \(n\), then \(X_{n}\) is called a record value. In other words, a record occurs whenever a new high is reached, and that new high is called the record value. Let \(N(t)\) denote the number of record values that are less than or equal to \(t\). Characterize the process \(\\{N(t), t \geqslant 0\\}\) when (a) \(f\) is an arbitrary continuous density function. (b) \(f(x)=\lambda e^{-\lambda x}\) Hint: Finish the following sentence: There will be a record whose value is between \(t\) and \(t+d t\) if the first \(X_{i}\) that is greater than \(t\) lies between \(\ldots\)

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

Consider \(n\) components with independent lifetimes which are such that component \(i\) functions for an exponential time with rate \(\lambda_{i} .\) Suppose that all components are initially in use and remain so until they fail. (a) Find the probability that component 1 is the second component to fail. (b) Find the expected time of the second failure. Hint: Do not make use of part (a).

If \(X_{1}\) and \(X_{2}\) are independent nonnegative continuous random variables, show that $$ P\left\\{X_{1}

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