Question

Events occur according to a Poisson process with rate \(\lambda\). Any event that occurs within a time \(d\) of the event that immediately preceded it is called a \(d\) -event. For instance, if \(d=1\) and events occur at times \(2,2.8,4,6,6.6, \ldots\), then the events at times \(2.8\) and \(6.6\) would be \(d\) -events. (a) At what rate do \(d\) -events occur? (b) What proportion of all events are \(d\) -events?

Short answer

In summary: (a) The rate at which \(d\)-events occur is \(\lambda_d = \lambda (1 - e^{-\lambda d})\). (b) The proportion of all events that are \(d\)-events is \(1 - e^{-\lambda d}\).

Step-by-step solution

Memoryless property and exponential distribution

We know that events occur according to a Poisson process with rate \(\lambda\). The time between events follows an exponential distribution with parameter \(\lambda\). The exponential distribution has the memoryless property, which means that the time until the next event is independent of how long it has been since the last event.

Find the probability of having a d-event

To find the rate at which \(d\)-events occur, we need to determine the probability of an event being a \(d\)-event. Using the exponential distribution, the probability density function \(f(t)\) of the time \(T\) between events is given by: \[f(t) = \lambda e^{-\lambda t}\] We can find the probability of having a \(d\)-event by integrating this probability density function over the interval \([0, d]\): \[P(T \leq d) = \int_0^d \lambda e^{-\lambda t} dt\]

Integrate the probability density function

Integrating the probability density function, we get: \[P(T \leq d) = -e^{-\lambda t} \Big|_0^d = 1 - e^{-\lambda d}\]

Calculate the rate of d-events

The rate at which \(d\)-events occur is the probability of having a \(d\)-event multiplied by the overall rate \(\lambda\) of events occurring: \[\lambda_d = \lambda \times P(T \leq d) = \lambda (1 - e^{-\lambda d})\] So, the rate at which \(d\)-events occur is \(\lambda_d = \lambda (1 - e^{-\lambda d})\).

Calculate the proportion of d-events

To calculate the proportion of \(d\)-events among all events, divide the rate of \(d\)-events by the total rate of events occurring: \[Proportion \ of \ d-events = \frac{\lambda_d}{\lambda} = \frac{\lambda (1 - e^{-\lambda d})}{\lambda} = 1 - e^{-\lambda d}\] So, the proportion of all events that are \(d\)-events is \(1 - e^{-\lambda d}\). To summarize: (a) The rate at which \(d\)-events occur is \(\lambda_d = \lambda (1 - e^{-\lambda d})\). (b) The proportion of all events that are \(d\)-events is \(1 - e^{-\lambda d}\).

Key concepts

Exponential Distribution

The exponential distribution is crucial in the study of continuous time events, especially when it comes to Poisson processes. Imagine waiting for buses that come randomly but with a consistent average rate. How long you'll have to wait for the next bus follows what we call 'exponential distribution'. The key parameter here is the rate \(\lambda\), which tells us the average number of events happening in a unit time interval.

The formula for the probability density function (PDF) of the exponential distribution is: \[ f(t) = \lambda e^{-\lambda t} \] where \(t\) represents time. This function gives us the likelihood of an event occurring at a specific time after the previous event. The exponential distribution is unique because it is memoryless, which brings us neatly to our next important concept.

Memoryless Property

This property is both intriguing and counterintuitive. It means that the probability of an event occurring in the future is independent of how much time has already elapsed. In other words, no matter how long you've been waiting, your expected wait time remains the same. Applying this to our bus stop analogy, even if you've already waited 10 minutes, your expected wait time for the next bus is still the same as when you first arrived.

Mathematically, the memoryless property can be demonstrated like this: For any non-negative numbers \(s\) and \(t\), the probability that the wait will exceed \(s + t\) given that it has exceeded \(s\) is the same as the probability that it will exceed \(t\):\[ P(T > s + t | T > s) = P(T > t) \] This property is exclusive to the exponential (for continuous variables) and geometric (for discrete variables) distributions. It simplifies a lot of calculations concerning the timing of future events in a Poisson process.

Probability Density Function

To truly comprehend the exponential distribution, we need to understand the probability density function (PDF). The PDF is a function that describes the relative likelihood for a continuous random variable to take on a given value. In the context of the exponential distribution, the PDF tells us how likely it is that an event will occur at any given instant within a continuous interval.

For the exponential distribution, the PDF is defined by \[ f(t) = \lambda e^{-\lambda t} \], where \(e\) is the base of the natural logarithm, and \(\lambda\) is the rate of events per time unit. To find the probability of an event occurring within a specific interval, we would integrate this function over that interval. For example, in the exercise provided, the integration of the PDF from \(0\) to \(d\) gives us the probability of a \(d\)-event, as shown by: \[ P(T \leq d) = \int_0^d \lambda e^{-\lambda t} dt \] This way, the PDF not only gives us a snapshot of an event's instant likelihood but, with integration, it also allows us to calculate the chance of an event occurring within a broader time frame.