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In Example \(7.6\), suppose that potential customers arrive in accordance with a renewal process having interarrival distribution \(F\). Would the number of events by time \(t\) constitute a (possibly delayed) renewal process if an event corresponds to a customer (a) entering the bank? (b) leaving the bank? What if \(F\) were exponential?

Short Answer

Expert verified
In summary, if an event corresponds to a customer entering the bank, the arrival process is a renewal process, regardless of whether the interarrival distribution F is exponential or not. However, if an event corresponds to a customer leaving the bank, the process cannot be considered a renewal process, even when the arrival distribution F is exponential, because the departure times depend on service times and queue waiting times, leading to potential dependencies between events.

Step by step solution

01

Scenario (a): Customers Entering the Bank

: Consider an event as a customer entering the bank. In this case, the arrival of each customer entering the bank will happen based on the interarrival distribution F. As the arrival of customers follows a renewal process, the interarrival times will be independent and identically distributed with the distribution F. Therefore, it can be considered a renewal process.
02

Scenario (b): Customers Leaving the Bank

: Now, let's consider an event as a customer leaving the bank. In this case, the time it takes for a customer to leave the bank depends on the service time provided by the bank to each customer. The service times can be different for different customers, and there might be dependencies between the departure times of the customers due to the variability in service times and queue waiting times. Thus, the interarrival times between customers leaving the bank will not follow the same distribution F and can be dependent, violating the conditions for a renewal process.
03

Exponential Interarrival Distribution F

: When the interarrival distribution F is exponential, it represents a Poisson process. In this case, the memoryless property of the exponential distribution applies, which means that the remaining time to the next arrival is independent of the time elapsed since the previous arrival. (a) For customers entering the bank, since the interarrival times remain independent and have the same exponential distribution, the process will still be considered a renewal process (in fact, a Poisson process in this case). (b) For customers leaving the bank, having an exponential interarrival distribution does not change the fact that the departure times depend on the service times and queue waiting times, resulting in potential dependencies between the departure events. Thus, the process for customers leaving the bank still cannot be considered a renewal process, even when the arrival distribution F is exponential.

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Key Concepts

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

Interarrival Distribution
Understanding the interarrival distribution is key to grasping the dynamics of customer arrivals in events such as entering or exiting a bank. The interarrival distribution, denoted as 'F' in our example, is the time between consecutive independent events in a stochastic process.

In a renewal process, these interarrival times are independent and identically distributed random variables, which means that each customer's arrival does not affect the next one's arrival time. The distribution 'F' can follow any pattern, but commonly it may be uniform, exponential, Gaussian, etc., depending on the context.
When dealing with scenario (a), customers entering the bank, with interarrival times being independent and identically distributed according to 'F', we find ourselves in the classic renewal process setup. Each new arrival effectively 'resets the clock', and therefore we can analyze these arrivals without considering the preceding ones.
Poisson Process
The Poisson process is a particular type of renewal process that is often used to model random events occurring over time, such as customers entering a bank. It is characterized by a constant rate, meaning that events happen independently at a steady average rate.

A crucial aspect of the Poisson process is its interarrival times follow an exponential distribution. This ties directly to the memoryless property: the waiting time for the next event does not depend on how much time has already passed. Therefore, whenever we have a renewal process with an exponential interarrival distribution, we're looking at a Poisson process in disguise.
For scenario (a), as customers enter the bank, if 'F' is exponential, it gives birth to a Poisson process representing the arrival of customers. It's the textbook case of a renewal process, thanks to this property.
Memoryless Property
The memoryless property is a fascinating feature of the exponential distribution that plays a central role in certain stochastic processes, like the Poisson process. It states that the probability of an event occurring in the future is the same, regardless of how much time has already elapsed, provided no event has occurred up to the present time.

In the context of our bank example, for scenario (a), this property implies that if 'F' is exponential, the time until the next customer enters is always the same on average, regardless of when the last customer arrived. But for scenario (b), concerning customers leaving the bank, the memoryless property of 'F' doesn't remedy the fact that there might be service time and queue waiting time dependencies.
Therefore, when the underlying interarrival distribution is exponential, it does grant the memoryless property to customer arrivals but does not solve dependency issues for customer departures.

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Most popular questions from this chapter

Consider a system that can be in either state 1 or 2 or \(3 .\) Each time the system enters state \(i\) it remains there for a random amount of time having mean \(\mu_{i}\) and then makes a transition into state \(j\) with probability \(P_{i j} .\) Suppose $$ P_{12}=1, \quad P_{21}=P_{23}=\frac{1}{2}, \quad P_{31}=1 $$ (a) What proportion of transitions takes the system into state \(1 ?\) (b) If \(\mu_{1}=1, \mu_{2}=2, \mu_{3}=3\), then what proportion of time does the system spend in each state?

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Consider a renewal process with mean interarrival time \(\mu .\) Suppose that each event of this process is independently "counted" with probability \(p\). Let \(N_{C}(t)\) denote the number of counted events by time \(t, t>0\). (a) Is \(N_{C}(t), t \geqslant 0\) a renewal process? (b) What is \(\lim _{t \rightarrow \infty} N_{C}(t) / t ?\)

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