Question

Consider a Poisson process with parameter \(\lambda .\) Let \(T\) be the time required to observe the first event, and let \(N(T / \kappa)\) be the number of events in the next \(T / \kappa\) units of time. Find the first two moments of \(N(T / \kappa) T\).

Short answer

In conclusion, for a Poisson process with parameter \(\lambda\), we find the first two moments of the product \(N(T/\kappa)T\), where \(T\) is the time required to observe the first event, and \(N(T/\kappa)\) is the number of events in the next \(T/\kappa\) units of time. The first moment (expectation) of the product is: \[E\left[N(T/\kappa) T\right] = \dfrac{T}{\kappa}\] and the second moment is: \[E\left[\left(N(T/\kappa)T\right)^2\right] = \dfrac{T}{\kappa}\left(\dfrac{1}{\lambda^2}+\dfrac{T}{\kappa}\right)\]

Step-by-step solution

Understand the Poisson Process and Moments

A Poisson process is a stochastic process (sequence of random variables) that describes the number of events that occur in fixed intervals of time or space. The parameter \(\lambda\) represents the average number of events per interval. In this case, the intervals are in units of time. The first moment (expectation) of a random variable is its mean value, and the second moment is the variance plus the square of the mean. We will compute these two moments for the product \(N(T/\kappa)T\).

Compute the Expectation

We can use the properties of expectation to simplify the problem. We need to find the expectation of the product: \[E\left[N(T/\kappa) T\right] = E\left[N(T/\kappa)\right] E[T]\] For a Poisson process with parameter \(\lambda\), the time until the first event (inter-arrival time) follows an exponential distribution with parameter \(\lambda\). Thus, the expectation of \(T\) is \(E[T] = \dfrac{1}{\lambda}\). Now, let's compute the expectation of the number of events \(N(T/\kappa)\) in the next \(T/\kappa\) units of time. Since it's a Poisson process, the number of events is distributed as a Poisson distribution with parameter \(\lambda' = \dfrac{\lambda}{\kappa}T\). The expectation of a Poisson distribution with parameter \(\mu\) is equal to \(\mu\). Therefore, the expectation of \(N(T/\kappa)\) is: \[E\left[N(T/\kappa)\right] = \dfrac{\lambda}{\kappa}T\] Finally, we can compute the expectation of the product: \[E\left[N(T/\kappa) T\right] = \left(\dfrac{\lambda}{\kappa}T\right) \left(\dfrac{1}{\lambda}\right) = \dfrac{T}{\kappa}\]

Compute the Variance

For the second moment of the product, we need to compute the variance: \[Var\left[N(T/\kappa)T\right]\] We will first compute the variance for \(N(T/\kappa)\). The variance of a Poisson distribution with parameter \(\mu\) is equal to \(\mu\). In this case, we have: \[Var\left[N(T/\kappa)\right] = \dfrac{\lambda}{\kappa}T\] Now, let's compute the covariance between \(N(T/\kappa)\) and \(T\). Since \(N(T/\kappa)\) and \(T\) are independent, their covariance is zero: \[Cov\left(N(T/\kappa), T\right) = 0\] Using the properties of variance for independent random variables, we can compute the variance of the product: \[Var\left[N(T/\kappa)T\right] = E[T]^2 Var\left[N(T/\kappa)\right] \] Substituting the computed values, we get: \[Var\left[N(T/\kappa)T\right] = \left(\dfrac{1}{\lambda}\right)^2 \left(\dfrac{\lambda}{\kappa}T\right) = \dfrac{T}{\lambda^2\kappa}\]

Compute the Second Moment

Now that we have the expectation and variance of the product, we can compute the second moment, which is the sum of the variance and the square of the expectation: \[E\left[\left(N(T/\kappa)T\right)^2\right] = Var\left[N(T/\kappa)T\right] + E\left[N(T/\kappa)T\right]^2\] Substituting the values, we get: \[E\left[\left(N(T/\kappa)T\right)^2\right] = \dfrac{T}{\lambda^2\kappa} + \left(\dfrac{T}{\kappa}\right)^2 = \dfrac{T}{\kappa}\left(\dfrac{1}{\lambda^2}+\dfrac{T}{\kappa}\right)\] In conclusion, the first moment (expectation) of the product \(N(T/\kappa)T\) is: \[E\left[N(T/\kappa) T\right] = \dfrac{T}{\kappa}\] and the second moment is: \[E\left[\left(N(T/\kappa)T\right)^2\right] = \dfrac{T}{\kappa}\left(\dfrac{1}{\lambda^2}+\dfrac{T}{\kappa}\right)\]

Key concepts

Stochastic Processes

When we talk about stochastic processes, we are referring to mathematical models that describe systems or phenomena that evolve over time in a way that is inherently random. These processes are critical in fields such as physics, engineering, finance, and of course, statistics. A fundamental example of a stochastic process is the Poisson process, which you have encountered in your exercise.

A Poisson process is used to model the occurrence of events that happen independently of each other within a continuous time frame. These events are also known as 'arrivals'. The distinguishing feature of the Poisson process is that it is characterized by a rate, often denoted by the symbol \(\lambda\), which represents the average number of events occurring in a unit of time. For instance, \(\lambda\) could be the average number of cars passing through an intersection per minute or the average number of phone calls received by an operator per hour.

One of the powerful aspects of the Poisson process is its memoryless property, which means that the probability of an event occurring is not influenced by the occurrence of previous events. This property allows us to analyze the process in a relatively straightforward manner, using the inter-arrival times between consecutive events.

Expectation and Variance

The concepts of expectation and variance are fundamental to statistics and play a central role in understanding stochastic processes. The expectation of a random variable is the long-term average value it will take after many realizations, commonly referred to as its mean. In the case of the Poisson process, for any time interval, the expected number of events (mean) is directly proportional to the length of the interval and the rate \(\lambda\).

The variance, on the other hand, measures the spread or variability around the mean. It tells us how much we can expect the number of events to fluctuate. In Poisson processes, it's particularly interesting to note that the variance is equal to the mean. This equivalence is unique to the Poisson distribution and is a critical concept when dealing with such processes. Hence, understanding these two moments is invaluable when you're trying to find the properties of random variables associated with a stochastic process, such as the one presented in your exercise.

Exponential Distribution

Let's delve into the exponential distribution, a key topic relevant to the exercise. The exponential distribution is intimately connected with the Poisson process; it describes the time between consecutive events in a Poisson process. In other words, if you're measuring the time until the next phone call at a call center, and calls come randomly but with a constant average rate, the time you'll be waiting follows an exponential distribution.

One of the most important features of the exponential distribution is that it is memoryless, paralleling the Poisson process's property we spoke about earlier. This essentially means that the probability of the event occurring in the next instant doesn't depend on how much time has already elapsed. Mathematically, if \(T\) represents the waiting time for the first event, and \(T\) follows an exponential distribution with rate \(\lambda\), then \(E[T]\) is the expectation you would compute, which is \(\frac{1}{\lambda}\).

In your exercise, as the occurrences of events are described by a Poisson process, the fact that these events' timing follows an exponential distribution is used to calculate the expected time until the first event and is pivotal for solving for the moments of \(N(T/\kappa)T\).