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Chapter 5: Problem 40

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
In summary, we've shown that if \(N_1(t)\) and \(N_2(t)\) are independent Poisson processes with rates \(\lambda_{1}\) and \(\lambda_{2}\), then their sum \(N(t) = N_{1}(t) + N_{2}(t)\) is also a Poisson process with rate \(\lambda_{1}+\lambda_{2}\), by verifying that \(N(t)\) satisfies the independent and stationary increments property and its counts follow a Poisson distribution.

Step by step solution

01

Verify Independent and Stationary Increments

For independent and stationary increments, we want to show that for any choice of time intervals \((t_1, t_2], (t_3, t_4], \dots, (t_{2n}, t_{2n+1}]\), where \(0 \leq t_1 < t_2 < t_3 < \dots < t_{2n+1}\), the increment counts \(N(t_2) - N(t_1), N(t_4) - N(t_3), \dots, N(t_{2n+1}) - N(t_{2n})\) are independent. Each of these increments can be written as the sum of the corresponding increments in \(N_1(t)\) and \(N_2(t)\): \[ N(t_{2i})-N(t_{2i-1}) = (N_1(t_{2i})-N_1(t_{2i-1})) + (N_2(t_{2i})-N_2(t_{2i-1})) \\ \] By independence of \(N_1(t)\) and \(N_2(t)\), \((N_1(t_{2i})-N_1(t_{2i-1}))\) and \((N_2(t_{2i})-N_2(t_{2i-1}))\) are independent. Since each increment in \(N_1(t)\) and \(N_2(t)\) are independent, their sum \(N(t)\) also has independent increments.
02

Verify Poisson Distribution

To prove that \(N(t)\) follows a Poisson distribution, we can use the fact that the sum of independent Poisson random variables is Poisson. Let \(X_1 \sim\) Poisson\((\lambda_1t)\) and \(X_2 \sim\) Poisson\((\lambda_2t)\) be the random variables representing the counts in the processes \(N_1(t)\) and \(N_2(t)\). Then, \(N(t) = X_1 + X_2 \sim\) Poisson\((\lambda t)\), where \(\lambda = \lambda_1+\lambda_2\). Now, we've verified that \(N(t)\) has both independent and stationary increments, and the counts follow a Poisson distribution with rate \(\lambda_{1}+\lambda_{2}\). Therefore, we have proven that \(N(t) = N_{1}(t) + N_{2}(t)\) is a Poisson process with rate \(\lambda_{1}+\lambda_{2}\).

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Key Concepts

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

Independent Poisson Processes
Imagine you have two separate queues in a store, and you want to count how many customers arrive in each queue over a given period. If these queues operate independently, then the timing of customer arrivals in one line doesn't affect the other. This is exactly how independent Poisson processes work.
Poisson processes are often used to model random events that occur independently over time. Each process has a rate, called \( \lambda_i \), which represents the average number of occurrences per time unit.
  • For independent Poisson processes \( \{ N_1(t), t \geq 0 \} \) and \( \{ N_2(t), t \geq 0 \} \), their increments, or counts in different time intervals, are independent from each other.
  • This means knowing the result of one process gives you no information about the finished counts of the other at any given time.
Both processes run concurrently without interfering, ensuring true randomness in their counts.
Independent and Stationary Increments
Increments in a Poisson process refer to the number of events occurring in disjointed time intervals. For a Poisson process to have independent and stationary increments, any two non-overlapping time interval increments must be independent, and the distribution of each increment is only a function of its length.
For example, consider the increments of customer arrivals at the store
  • Independent increments mean if you look at one queue's customers from 1pm to 2pm and another queue from 3pm to 4pm, these counts are independent.
  • Stationary increments imply, whether you check between 1pm-2pm or 3pm-4pm, the expected number of arrivals is the same, provided each is one hour long.
Thus, for the sum process \( N(t) = N_1(t) + N_2(t) \), demonstrating both independence and sameness across any span of time is crucial to establishing a true Poisson Process.
Sum of Poisson Random Variables
The magic of Poisson processes extends to their additivity. When you add two independent Poisson processes, the result is another Poisson process. This is quite convenient and has a solid mathematical foundation.
Let's denote \( N_1(t) \) and \( N_2(t) \) as separate Poisson processes with rates \( \lambda_1 \) and \( \lambda_2 \) respectively.
  • The sum \( N(t) = N_1(t) + N_2(t) \) becomes a new Poisson process.
  • The combined rate of this new process is the sum of the individual rates: \( \lambda = \lambda_1 + \lambda_2 \).
This additivity property helps to simplify complex systems into a more manageable single Poisson process, especially when modeling real-world tasks that have multiple independent streams of events.
Poisson Distribution
The Poisson distribution is a probability distribution that counts how often an event can occur within a fixed period. It's characterized by the parameter \( \lambda \), which is the expected count of events.
The distribution is particularly useful for modeling rare events in large populations. The key properties include:
  • The mean of the distribution equals \( \lambda \).
  • The variance also equals \( \lambda \), indicating the spread of the distribution.
In a system with multiple independent Poisson processes, such as \( N_1(t) \) and \( N_2(t) \) with rates \( \lambda_1 \) and \( \lambda_2 \) respectively, their combined process \( N(t) = N_1(t) + N_2(t) \) follows a Poisson distribution with rate \( \lambda_1 + \lambda_2 \). This property makes it a vital tool in counting processes and stochastic modeling.
Rate of Poisson Process
The rate of a Poisson process, denoted by \( \lambda \), is a fundamental parameter that defines how frequently events occur on average within a given time frame. It's the backbone of understanding and calculating probabilities within the Poisson framework.
To illustrate, if \( \lambda = 5 \), on average, 5 events occur per unit time. This average is key for predicting future occurrences. For multiple independent processes like \( N_1(t) \) and \( N_2(t) \),
  • Their rates add up to give a new combined process: \( \lambda = \lambda_1 + \lambda_2 \).
  • This summed rate helps predict the behavior of the summed process \( N(t) = N_1(t) + N_2(t) \).
Understanding the rate \( \lambda \) is crucial for applications across various domains, including telecommunications, finance, and traffic engineering, where event timing and prediction are key concerns.

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Most popular questions from this chapter

A viral linear DNA molecule of length, say, 1 is often known to contain a certain "marked position," with the exact location of this mark being unknown. One approach to locating the marked position is to cut the molecule by agents that break it at points chosen according to a Poisson process with rate \(\lambda .\) It is then possible to determine the fragment that contains the marked position. For instance, letting \(m\) denote the location on the line of the marked position, then if \(L_{1}\) denotes the last Poisson event time before \(m\) (or 0 if there are no Poisson events in \([0, m])\), and \(R_{1}\) denotes the first Poisson event time after \(m\) (or 1 if there are no Poisson events in \([m, 1])\), then it would be learned that the marked position lies between \(L_{1}\) and \(R_{1} .\) Find (a) \(P\left[L_{1}=0\right\\}\), (b) \(P\left(L_{1}x\right\\}, m

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 \(E\left[S^{2}(t)\right]\)

In a certain system, a customer must first be served by server 1 and then by server \(2 .\) The service times at server \(i\) are exponential with rate \(\mu_{i}, i=1,2 .\) An arrival finding server 1 busy waits in line for that server. Upon completion of service at server 1 , a customer either enters service with server 2 if that server is free or else remains with server 1 (blocking any other customer from entering service) until server 2 is free. Customers depart the system after being served by server \(2 .\) Suppose that when you arrive there is one customer in the system and that customer is being served by server \(1 .\) What is the expected total time you spend in the system?

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.

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be independent and identically distributed exponential random variables. Show that the probability that the largest of them is greater than the sum of the others is \(n / 2^{n-1}\). That is, if $$ M=\max _{j} X_{j} $$ then show $$ P\left\\{M>\sum_{i=1}^{n} X_{i}-M\right\\}=\frac{n}{2^{n-1}} $$ Hint: What is \(P\left[X_{1}>\sum_{i=2}^{n} X_{i}\right\\} ?\)
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