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Consider a Poisson process of parameter \(\lambda\). Given that \(n\) events happen in time \(t\), find the density function of the time of the occurrence of the \(r\) th event \((r

Short Answer

Expert verified
The probability density function of the time of occurrence of the r-th event (r<n), given that n events happen in time t in a Poisson process with parameter λ, is a uniform distribution over the interval [0, t], i.e., \[f_{T_r | T_n = t}(t_r) = \begin{cases} \frac{1}{t}, & 0 \leq t_r \leq t \\ 0, &\text{otherwise} \end{cases} \]

Step by step solution

01

Understand the Poisson Arrival Times and Process

A Poisson process is a counting process that counts the number of events in non-overlapped intervals. The time of events in a Poisson process follows an exponential distribution, with arrival rate λ. Let X1, X2, … Xn denote the time intervals between consecutive events, and T_r denote the time of occurrence of the r-th event. Then, T_r can be expressed as the sum of r time intervals: \[T_r = X_1 + X_2 + ... + X_r\]
02

Use Joint pdf of Exponential Distributions

The joint pdf of n independent exponential random variables X1, X2, … Xn with rate parameter λ is given by: \[f_{X_1, X_2, ... X_n}(x_1, x_2, ... x_n) = \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i}\]
03

Obtain Conditional pdf of T_r

Since we want the density function of T_r given n events occurring in time t, we need to compute the conditional pdf f(T_r | T_n = t), where T_n is the time of the occurrence of the nth event. The conditional pdf can be obtained using the joint pdf and marginal pdf as follows: \[f_{T_r | T_n = t}(t_r) = \frac{f_{T_r, T_n}(t_r, t)}{f_{T_n}(t)}\]
04

Compute Joint and Marginal pdfs

To obtain the joint pdf of T_r and T_n, first compute the following Jacobian: \[J = \frac{\partial (x_1 + ... +x_r, x_1 + ... + x_n)}{\partial (t_r, t)} = \frac{\partial (t_r, t)}{\partial (t_r,t)}= 1\] Now, the joint pdf of T_r and T_n is given by: \[f_{T_r, T_n}(t_r, t) = J_{(T_r, T_n)} \times f_{X_1, X_2, ... X_n}(x_1, x_2, ... x_n) = \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i}\] Next, find the marginal pdf of T_n: \[f_{T_n}(t) = \int_{0}^{t} f_{T_r, T_n}(t_r, t) dt_r\]
05

Compute the Conditional pdf of T_r

Now we can substitute the expressions for the joint and marginal pdfs into the conditional pdf equation: \[f_{T_r | T_n = t}(t_r) =\frac{\lambda^n e^{-\lambda \sum_{i=1}^{n} x_i}}{\int_{0}^{t} \lambda^n e^{-\lambda \sum_{i=1}^{n} x_i} dt_r}\] At this point, notice that the term \(\lambda^n e^{-\lambda \sum_{i=1}^{n} x_i}\) is independent of t_r and can be treated as a constant. Therefore, \[f_{T_r | T_n = t}(t_r) = \frac{1}{\int_{0}^{t} dt_r} \propto 1\] This indicates that the conditional pdf of T_r is uniform over the interval [0, t]. So, we have: \[f_{T_r | T_n = t}(t_r) = \begin{cases} \frac{1}{t}, & 0 \leq t_r \leq t \\ 0, &\text{otherwise} \end{cases} \] This is the probability density function of the time of occurrence of the r-th event, given that n events happen in time t in a Poisson process with parameter λ.

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Key Concepts

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

Understanding the Probability Density Function (PDF)
A probability density function (PDF) is a vital concept in statistics and probability theory, instrumental for dealing with continuous random variables. It describes the likelihood of a random variable taking on a particular value. In other words, the PDF gives a relative sense of how 'dense' the probabilities are around a certain value for a continuous variable.

For a given random variable, the PDF is non-negative across its domain and its integral over the entire space is 1, meaning that the probability of the random variable falling within the entire range is certain – a total probability of 1.

When working with a PDF, one can find the probability that a random variable falls within a certain interval by integrating the PDF over that interval. For example, if you want to calculate the probability of a random variable being between a and b, you would integrate the PDF from a to b.
Exponential Distribution and its Role in Poisson Processes
The exponential distribution plays a fundamental role in the context of Poisson processes. It is commonly used to model the time between events in a Poisson process, which is characterized by constant average rate \( \lambda \).

The PDF of an exponential distribution is defined as:

\[ f(x;\lambda) = \begin{cases} \lambda e^{-\lambda x}, & x \geq 0,\ 0, & x < 0.\end{cases} \]
The mean and standard deviation of an exponential distribution are both equal to \(1/\lambda\), indicating that the average time between events is \(1/\lambda\) and the variability of this time is also \(1/\lambda\).

In our case, the solution to the exercise relies on the memoryless property of the exponential distribution. This property states that the probability of an event occurring in the next interval is independent of how much time has already passed, a key characteristic that allows the independent events in a Poisson process to be modeled.
Conditional Probability in Complex Scenarios
Conditional probability is the probability of an event occurring given that another event has already occurred. It's an essential concept for understanding dependencies between events and is represented mathematically as \( P(A|B) \) – the probability of event A occurring given event B.

In the context of the Poisson process problem we are analyzing, conditional probability allows us to determine the likelihood of the r-th event happening at a certain time, assuming that n events occur within a certain timeframe t.

The conditional probability density function (PDF) is used to describe the probability of a continuous random variable given some condition. For example, finding \( f_{T_r | T_n = t}(t_r) \) involves computing how probable it is for the r-th event to occur at time \( t_r \) given n events in total happen by time t.

In broader statistics and probability studies, mastering conditional probability is crucial for understanding a wide range of problems, from simple scenarios to complex stochastic processes like the Poisson process.

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Most popular questions from this chapter

Consider a pure birth process having infinitesimal parameters \(\lambda_{n}=\lambda n^{2}\), where \(\lambda>0\) is fixed. Given that at time 0 there is a single particle, determine $$ P_{\infty}(t)=1-\sum_{k=1}^{\infty} P_{k}(t) $$

Consider a birth and death process with infinitesimal parameter \(\lambda_{n}, \mu_{n}\). Show that the expected length of time for reaching state \(r+1\) starting from state 0 is $$ \sum_{n=0}^{r} \frac{1}{\lambda_{n} \pi_{n}} \sum_{k=0}^{n} \pi_{k} $$ For the definition of \(\pi_{n}\) see Eq. (4.5).

Consider a Yule process with parameter \(\beta\) and initial state \(N=1\). Suppose the first individual is also subject to death, with the probability of death in the interval \(t\) to \(t+h\), given that the individual is living at time \(t\), being \(\mu h+o(h)\). Compute the distribution of the number of offspring due to a single individual and his descendants at the time of death of the original parent.

Consider an infinitely many-server queue with an exponential service time distribution with parameter \(\mu\). Suppose customers arrive in batches with the interarrival time following an exponential distribution with parameter \(\lambda\). The number of arrivals in each batch is assumed to follow the geometric distribution with parameter \(\rho(0<\rho<1)\), i.e., Pr \(\\{\) number of arrivals in a batch has size \(k\\}\) \(=\rho^{k-1}(1-\rho)(k=1,2, \ldots)\) Formulate this process as a continuous time Markov chain and determine explicitly the infinitesimal matrix of the process.

Let \(\\{X(t), t \geq 0\\}\) and \(\\{Y(t), t \geq 0\\}\) be two independent Poisson processes with parameters \(\lambda_{1}\) and \(\lambda_{2}\), respectively. Define $$ Z(t)=X(t)-Y(t), \quad t \geq 0 $$ This is a stochastic process whose state space consists of all the integers (positive, negative, and zero). Let $$ P_{n}(t)=\operatorname{Pr}\\{Z(t)=n\\}, \quad n=0, \pm 1, \pm 2, \ldots $$ Establish the formula $$ \sum_{n=-\infty}^{\infty} P_{n}(t) z^{n}=\exp \left(-\left(\lambda_{1}+\lambda_{2}\right) t\right) \exp \left(\lambda_{1} z t+\left(\lambda_{2} / z\right) t\right), \quad|z| \neq 0 $$ Compute \(E(Z(t))\) and \(E\left(Z(t)^{2}\right) .\)

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