Question

Individuals join a club in accordance with a Poisson process with rate \(\lambda\). Each new member must pass through \(k\) consecutive stages to become a full member of the club. The time it takes to pass through each stage is exponentially distributed with rate \(\mu\). Let \(N_{i}(t)\) denote the number of club members at time \(t\) who have passed through exactly \(i\) stages, \(i=1, \ldots, k-1 .\) Also, let \(\mathrm{N}(t)=\left(N_{1}(t), N_{2}(t), \ldots, N_{k-1}(t)\right)\) (a) Is \(\\{\mathbf{N}(t), t \geqslant 0\\}\) a continuous-time Markov chain? (b) If so, give the infinitesimal transition rates. That is, for any state \(\mathrm{n}=\) \(\left(n_{1}, \ldots, n_{k-1}\right)\) give the possible next states along with their infinitesimal rates.

Short answer

(a) Yes, \(\boldsymbol{N}(t)\) is a continuous-time Markov chain because both the Poisson process for individuals joining the club and the exponential distribution for passing through each stage are memoryless. (b) The infinitesimal transition rates are: - From state \(\boldsymbol{n}\) to \((n_1 + 1, n_2, \ldots, n_{k-1})\): \(\lambda\) - From state \(\boldsymbol{n}\) to \((n_1, \ldots, n_{i-1}, n_i + 1, \ldots, n_{k-1})\) for \(i = 2, \ldots, k-1\): \(n_{i-1}\mu\) - From state \(\boldsymbol{n}\) to \((n_1, \ldots, n_{i-1} - 1, n_i + 1, \ldots, n_{k-1})\) for \(i = 2, \ldots, k-1\): \(n_i\mu\)

Step-by-step solution

(a) Determine if N(t) is a continuous-time Markov chain

To check if N(t) is a continuous-time Markov chain, we have to ensure that the future evolution of the process depends only on the present state and not on the history of the system. Given that individuals join the club according to a Poisson process with rate \(\lambda\), the time between arrivals follows an exponential distribution. The time it takes to pass through each stage is also exponentially distributed with rate \(\mu\). Both distributions are memoryless, meaning the future depends only on the present state and not on how much time has passed. Therefore, \(\\{ \boldsymbol{N}(t), t\geq 0 \\}\) is a continuous-time Markov chain.

(b) Finding the infinitesimal transition rates

The next step is to find the infinitesimal transition rates for possible next states from any given state \(n = (n_1, n_2, \ldots, n_{k-1})\). Note that transitions can occur in two ways: 1. A new member joins the club and enters the first stage. 2. A member passes from one stage to the next stage. Consider the following possible next states and their transition rates: 1. State \(\boldsymbol{n}^\prime = (n_1+1, n_2, \ldots, n_{k-1})\): A new member enters the first stage; the infinitesimal transition rate is \(\lambda\). For \(i\) from \(2\) to \(k-1\): 2. State \(\boldsymbol{n}^\prime = (n_1, \ldots, n_{i-1}, n_i + 1, \ldots, n_{k-1})\): A member moves from stage \(i-1\) to stage \(i\); the infinitesimal transition rate is \(n_{i-1}\mu\). 3. State \(\boldsymbol{n}^\prime = (n_1, \ldots, n_{i-1}-1, n_i+1, \ldots, n_{k-1})\): A member passes from stage \(i\) to the final stage and becomes a full member; the infinitesimal transition rate is \(n_{k-1}\mu\). In summary, the infinitesimal transition rates are: - From state \(\boldsymbol{n}\) to \((n_1 + 1, n_2, \ldots, n_{k-1})\): \(\lambda\) - From state \(\boldsymbol{n}\) to \((n_1, \ldots, n_{i-1}, n_i + 1, \ldots, n_{k-1})\) for \(i = 2, \ldots, k-1\): \(n_{i-1}\mu\) - From state \(\boldsymbol{n}\) to \((n_1, \ldots, n_{i-1} - 1, n_i + 1, \ldots, n_{k-1})\) for \(i = 2, \ldots, k-1\): \(n_i\mu\)

Key concepts

Poisson Process

In the world of stochastic processes, the Poisson process is a prime example of a model that captures the essence of random events that occur independently and at a constant average rate. Imagine something akin to emails dropping into your inbox at an unpredictable yet steady pace—that's akin to a Poisson process.

The crucial aspect of the Poisson process is the arrival times of events are scattered randomly, making it a go-to model for various systems such as the number of customers arriving at a store, phone calls at a call center, or as in our exercise, individuals joining a club. Its defining characteristic is its rate, \( \lambda \), representing the average number of occurrences within a unit of time.

In educational terms, we can illustrate the Poisson process using the example of flipping a coin. Even though tosses are random, over a long period, we notice an average rate of heads or tails—that's the heart of the Poisson process.

Exponential Distribution

When time plays a role in your random process, and you're keeping an eye on the wait times between occurrences, the exponential distribution takes center stage. It's the continuous counterpart to the geometric distribution found in discrete settings.

Let's unwrap this with the analogy of waiting for a bus. If buses arrive randomly but on average every 15 minutes, how long would your wait be if you just missed one? The exponential distribution quantifies that wait time—your personal moment of patience. Mathematically, it's characterized by a rate, \( \mu \), which is the flip side of the average wait time. The fascinating tidbit here is that no matter when you start timing, your expected waiting time always resets, which ties into the memoryless property we'll discuss later on.

Going back to our club members, each stage they pass through mirrors waiting for that proverbial bus. They spend an exponentially distributed amount of time at each station, all thanks to the steady rate \( \mu \) guiding their journey.

Transition Rates

In the dance of stochastic processes, transition rates are the rhythm to the Markov chain's movements. It's the precise metric that describes the tempo at which our system hops from one state to another. In simpler terms, it's the likelihood of change happening in a tiny window of time.

If the process were a video game, transition rates would be the rules determining how quickly a player moves from one level to the next. In our case, the club members are 'players', and each stage is a 'level'. The rates \( \lambda \) and \( \mu \) lay down the law for their progress.

For any given state of the system, transition rates allow us to foresee how members move through stages or how new members join the club. It's these fundamental rules that make it possible to predict the evolution of the system over time and ultimately assist in understanding the complex dynamics at play within continuous-time Markov chains.

Memoryless Property

The memoryless property is a fascinating aspect of certain probability distributions, making them unique in the world of statistics. Picture a goldfish with its purportedly short-term memory—a while after swimming away, it circles back as if discovering its castle for the first time. Exponential distributions and the Poisson processes they're connected to are just like our fishy friend; they 'reset' with every new observation.

What this means is once a member of our club survives to a particular point in time, the likelihood of their advancing forward doesn't depend on how long they've already been in the process. For instance, no matter how long someone has been waiting to progress from one membership stage to the next, their chance of moving on in the next moment remains unchanged.

The memoryless property simplifies analysis and calculations. It's the reason that even though life and processes might seem complex, we're able to model them predictably—reframing the intricate dance into a step-by-step number where each move is fresh and unburdened by the past.