Question

Suppose that the times between successive arrivals of customers at a single- server station are independent random variables having a common distribution \(F .\) Suppose that when a customer arrives, he or she either immediately enters service if the server is free or else joins the end of the waiting line if the server is busy with another customer. When the server completes work on a customer, that customer leaves the system and the next waiting customer, if there are any, enters service. Let \(X_{n}\) denote the number of customers in the system immediately before the \(n\) th arrival, and let \(Y_{n}\) denote the number of customers that remain in the system when the \(n\) th customer departs. The successive service times of customers are independent random variables (which are also independent of the interarrival times) having a common distribution \(G\). (a) If \(F\) is the exponential distribution with rate \(\lambda\), which, if any, of the processes \(\left\\{X_{n}\right\\},\left[Y_{n}\right\\}\) is a Markov chain? (b) If \(G\) is the exponential distribution with rate \(\mu\), which, if any, of the processes \(\left\\{X_{n}\right\\},\left\\{Y_{n}\right\\}\) is a Markov chain? (c) Give the transition probabilities of any Markov chains in parts (a) and (b).

Short answer

In this problem, we determined whether processes {X_n} and {Y_n} are Markov chains under two specific conditions: (a) For interarrival times following an exponential distribution, process {X_n} is a Markov chain, but process {Y_n} is not. (b) For service times following an exponential distribution, both processes {X_n} and {Y_n} are Markov chains. We also found the transition probabilities for {X_n} and {Y_n} when they form Markov chains: For {X_n}: 1. P(X_{n+1} = i+1 | X_n = i) = λ / (λ + μ) when i > 0 2. P(X_{n+1} = i-1 | X_n = i) = μ / (λ + μ) when i > 1 For {Y_n}: 1. P(Y_{n+1} = i | Y_n = i) = λ(λ + μ(i - 1)) / ((λ + μi)(λ + μ(i - 1))) 2. P(Y_{n+1} = i-1 | Y_n = i) = μ / (λ + μi) when i > 1

Step-by-step solution

Analyze the process {X_n} when F is exponential

Consider the process {X_n}, the number of customers in the system immediately before the nth arrival. Since the interarrival times are exponentially distributed, the arrivals follow a Poisson process. A Poisson process has independent increments, which means that the number of customers arriving in non-overlapping time intervals are independent, and the process {X_n} is a Markov chain.

Analyze the process {Y_n} when F is exponential

The process {Y_n} represents the number of customers remaining in the system when the nth customer departs. Since this process depends on both the interarrival times and the service times at the server, and we are given that the interarrival times are exponential, the server becomes free in random instances that might not correspond to the arrivals. This means that the process {Y_n} does not possess the Markov property when only considering the interarrival times and is not a Markov chain. The answer to part (a) is that the process {X_n} is a Markov chain when interarrival times follow an exponential distribution. (b) Determine if the processes {X_n} or {Y_n} is a Markov chain when service times follow an exponential distribution.

Analyze the process {X_n} when G is exponential

Again, {X_n} represents the number of customers in the system immediately before the nth arrival. The arrival process is not specified in the second part and thus does not influence whether or not {X_n} is a Markov chain. It solely depends on the process from part (a). Therefore, {X_n} is still a Markov chain in this case.

Analyze the process {Y_n} when G is exponential

In part (b), we are given that the service times are exponentially distributed, and the service times are independent with a common distribution G. The exponential service times have a memoryless property, which means that the remaining service time after waiting for any time is still exponentially distributed. This property makes {Y_n} a Markov chain when service times follow an exponential distribution. The answer to part (b) is that both processes {X_n} and {Y_n} are Markov chains when the service times follow an exponential distribution. (c) Find the transition probabilities for the Markov chains found in parts (a) and (b).

Transition probabilities for {X_n}

For the {X_n} process, we have a Markov chain when the interarrival times follow an exponential distribution. This is a birth-death process because the arrival and departure rates are constant. To find the transition probabilities, we must consider the following cases: 1. An arrival occurs in the system without someone leaving: Transition probability P(X_{n+1} = i+1 | X_n = i) = λ / (λ + μ) when i > 0 2. A departure occurs in the system without someone arriving: Transition probability P(X_{n+1} = i-1 | X_n = i) = μ / (λ + μ) when i > 1

Transition probabilities for {Y_n}

For the {Y_n} process, we have a Markov chain when the service times follow an exponential distribution. The transition probabilities for {Y_n} are as follows: 1. A new customer enters the system and immediate service is available: Transition probability P(Y_{n+1} = i | Y_n = i) = λ(λ + μ(i - 1)) / ((λ + μi)(λ + μ(i - 1))) 2. A customer leaves the system after his service is complete and the system has more customers: Transition probability P(Y_{n+1} = i-1 | Y_n = i) = μ / (λ + μi) when i > 1 Thus, we have answered all parts of the problem and derived the transition probabilities for the Markov chains for the given processes.

Key concepts

Exponential Distribution

The exponential distribution is crucial to understanding many processes where we're interested in the time until an event occurs. For instance, it can represent the time between arrivals of customers at a service station as seen in the exercise.

What sets the exponential distribution apart is its memoryless property. This means that no matter how much time has already passed, the probability of the event occurring in the next moment remains unchanged. Mathematically, this property is represented as:
\[ P(T > t + s | T > s) = P(T > t) \].

In the context of the exercise, this property implies that as customers arrive, the time until the next arrival is always exponentially distributed, regardless of when the last arrival happened. This characteristic directly contributes to one of the processes, \({X_n}\), being a Markov chain. Because the future state (number of customers before the nth arrival) only depends on the current state, past states, which are in effect, the 'memory', do not influence the process—the hallmark attribute of a Markov chain.

Poisson Process

A Poisson process is a mathematical model that describes a series of events occurring independently and at a constant average rate. It's a type of counting process where we're typically interested in the number of events happening in a fixed period of time.

Linked with the exponential distribution through interarrival times (the time between consecutive events), a Poisson process assumes that these interarrival times are exponentially distributed with a rate parameter \(\text{λ}\). The number of events in non-overlapping intervals are independent, a property known as independent increments.

In the textbook problem, the independence of events (customer arrivals) and the memoryless property of the exponential distribution mean the number of arrivals in a non-overlapping time interval doesn't affect arrivals in another, which is why the process \({X_n}\) is considered a Poisson process and subsequently a Markov chain.

Transition Probabilities

Transition probabilities are the backbone of Markov chains, determining the likelihood of moving from one state to another. They're crucial for understanding the behavior of the system over time. For a process to be a Markov chain, the transition probability from state i to state j should only depend on the current state, not the sequence of events that led there.

In the exercise, after determining that both \({X_n}\) and \({Y_n}\) can be considered Markov chains under different conditions, we calculate the transition probabilities. These calculations are based on whether a new customer arrives causing an increase (a 'birth') or a customer receives service and leaves (a 'death').

For \({X_n}\), transition probabilities involve scenarios of arrival-only or departure-only events. For \({Y_n}\), probabilities adapt to include the memoryless property of the service times. Both sets of probabilities require careful attention to detail, as they consider the rates at which customers arrive and receive service (\(\text{λ}\) and \(\text{μ}\) respectively), and involve different cases depending on the number of customers present.