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Chapter 7: Problem 6

Consider a renewal process \(\\{N(t), t \geqslant 0\\}\) having a gamma \((r, \lambda)\) interarrival distribution. That is, the interarrival density is $$ f(x)=\frac{\lambda e^{-\lambda x}(\lambda x)^{r-1}}{(r-1) !}, \quad x>0 $$ (a) Show that $$ P[N(t) \geqslant n]=\sum_{i=n r}^{\infty} \frac{e^{-\lambda t}(\lambda t)^{i}}{i !} $$ (b) Show that $$ m(t)=\sum_{i=r}^{\infty}\left[\frac{i}{r}\right] \frac{e^{-\lambda t}(\lambda t)^{i}}{i !} $$ where \([i / r]\) is the largest integer less than or equal to \(i / r\). Hint: Use the relationship between the gamma \((r, \lambda)\) distribution and the sum of \(r\) independent exponentials with rate \(\lambda\) to define \(N(t)\) in terms of a Poisson process with rate \(\lambda\).

Short Answer

Expert verified
In this exercise, we derived expressions for \(P[N(t) \geqslant n]\) and \(m(t)\) for a renewal process with gamma \((r, \lambda)\) interarrival distribution. Using the relationship between gamma distribution and sum of independent exponentials, and defining \(N(t)\) in terms of a Poisson process with rate \(\lambda\), we have: (a) \(P[N(t) \geqslant n] = \sum_{i = nr}^{\infty} \frac{e^{-\lambda t} (\lambda t)^i}{i!}\) (b) \(m(t) = \sum_{i=r}^{\infty}\left[\frac{i}{r}\right] \frac{e^{-\lambda t}(\lambda t)^{i}}{i!}\) These expressions give us the probability of having at least \(n\) arrivals in the time interval \((0, t]\) and the average number of renewals in that interval, respectively.

Step by step solution

01

Define the given interarrival distribution

Given that the interarrival density of the renewal process is a gamma distribution with parameters \(r\) and \(\lambda\), we have: \[f(x) = \frac{\lambda e^{-\lambda x} (\lambda x)^{r-1}}{(r-1)!}, \quad x > 0\]
02

Relationship between gamma distribution and sum of independent exponentials

Recall that a gamma distribution with shape parameter \(r\) and rate parameter \(\lambda\) can be represented as the sum of \(r\) independent exponential random variables with rate \(\lambda\). So, considering the given hint, we can define \(N(t)\) in terms of a Poisson process with rate \(\lambda\).
03

Find the probability \(P[N(t) \geqslant n]\)

Now we need to find the probability \(P[N(t) \geqslant n]\) using the given expression: \[P[N(t) \geqslant n] = \sum_{i = nr}^{\infty} \frac{e^{-\lambda t} (\lambda t)^i}{i!}\] Notice that since we have a Poisson process with rate \(\lambda\), the probability of having \(i\) arrivals in the time interval \((0, t]\) is given by the Poisson probability mass function: \[P[N(t) = i] = \frac{e^{-\lambda t} (\lambda t)^i}{i!}\] Therefore, the probability that the number of arrivals within the time interval \((0, t]\) is at least \(n\) is the sum of probabilities starting from \(nr\) to infinity, which is exactly the given expression.
04

Find the function \(m(t)\)

We also need to find the function \(m(t)\) using the given expression: \[m(t) = \sum_{i=r}^{\infty}\left[\frac{i}{r}\right] \frac{e^{-\lambda t}(\lambda t)^{i}}{i!}\] where \([\frac{i}{r}]\) is the largest integer less than or equal to \(\frac{i}{r}\). To derive this expression, we have to consider the renewal process and its embedded Poisson process with rate \(\lambda\). Since the interarrival times have a gamma distribution, the average number of renewals in the time interval \((0, t]\) is the sum of the average number of Poisson process events in each interarrival time period. So, we weight the Poisson probability mass function by the corresponding interarrival time period index \(\frac{i}{r}\), sum starting from \(i = r\) to infinity. Therefore, we have successfully derived the expressions for \(P[N(t) \geqslant n]\) and \(m(t)\) using the relationship between the gamma distribution and Poisson process as hinted.

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Key Concepts

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

Gamma Distribution
The gamma distribution is a continuous probability distribution that is widely used to model waiting times, such as the time until the occurrence of the r-th event in a Poisson process. In this context, the gamma distribution is parameterized by two parameters: shape parameter \( r \) and rate parameter \( \lambda \). This makes it suitable for describing the interarrival distribution in a renewal process. The probability density function of the gamma distribution is represented as:\[f(x) = \frac{\lambda e^{-\lambda x} (\lambda x)^{r-1}}{(r-1)!}, \quad x > 0\]Where:
  • \( \lambda \) is the rate parameter, controlling the scale of the distribution. Higher \( \lambda \) values result in a distribution that is more concentrated around zero.
  • \( r \) is the shape parameter, dictating the number of events or trials until a specific time period. The gamma distribution becomes the exponential distribution when \( r = 1 \).
By understanding the relationship between the parameters and the nature of the gamma distribution, one can effectively model systems where events occur continuously and independently over time.
Poisson Process
A Poisson process is a stochastic process that models a series of events occurring randomly in time or space, where these events happen with a constant known average rate, and the time between the events follows an exponential distribution. It is defined by a single parameter \( \lambda \), representing the average rate at which events occur per unit of time. A few essential characteristics of the Poisson process are:
  • The events are independent 鈥� knowing about previous events does not affect the probability of future events.
  • The probability of more than one event occurring in a very small time interval is negligible.
  • The number of events in any given time interval follows a Poisson distribution.
In the context of renewal processes, the Poisson process plays a crucial role in defining the probability distributions for the times between successive renewals. The process helps in expressing related probabilities and functions, such as in equation derivations where the interarrival times have a gamma distribution by leveraging this relationship.
Interarrival Distribution
The interarrival distribution is a critical component in renewal processes, representing the time between consecutive events or arrivals. In a typical renewal process, the interarrival times are often assumed to follow a certain probability distribution, which can be crucial in calculating the probabilities of interest, such as the number of events within a given time.In our problem, the interarrival distribution is specified as a gamma distribution, which means the time between renewals is modeled with a gamma probability density function. This adds layers of complexity compared to simpler models like the exponential distribution but more accurately models scenarios where multiple stages or events occur.Understanding the nature of the interarrival distribution helps in accurately deriving expressions for other functions in the renewal process, such as the probability \( P[N(t) \geq n] \), by interpreting it as a sum of independent exponential random variables. This interpretation is particularly useful in deriving solutions in renewal theory and related fields.
Exponential Random Variables
Exponential random variables are fundamental within probability theory when analyzing events occurring continuously over time. They are used to model the time between events in a Poisson process, thanks to their memoryless property, which means the probability of an event occurring in the future is independent of how much time has already passed.The probability density function of an exponential random variable is given by:\[f(x) = \lambda e^{-\lambda x}, \quad x \geq 0\]Exponential random variables are particularly useful in simplifying complex problems involving time-dependent events. When considering a gamma distribution characterized by shape parameter \( r \) and rate \( \lambda \), this can be viewed as the sum of \( r \) exponential random variables, each with the same rate \( \lambda \).
  • This relationship makes it easier to derive expressions for probabilities in renewal processes by using known properties of exponential distributions, such as their mean \( 1/\lambda \) and variance \( 1/\lambda^2 \).
  • It also allows leveraging the Poisson process framework to solve problems involving gamma distributed interarrival times, offering a structured way to approach such problems.

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Most popular questions from this chapter

If the mean-value function of the renewal process \(\\{N(t), t \geqslant 0\\}\) is given by \(m(t)=\) \(t / 2, t \geqslant 0\), what is \(P[N(5)=0\\} ?\)

Three marksmen take turns shooting at a target. Marksman 1 shoots until he misses, then marksman 2 begins shooting until he misses, then marksman 3 until he misses, and then back to marksman 1, and so on. Each time marksman \(i\) fires he hits the target, independently of the past, with probability \(P_{i}, i=1,2,3 .\) Determine the proportion of time, in the long run, that each marksman shoots.

In 1984 the country of Morocco in an attempt to determine the average amount of time that tourists spend in that country on a visit tried two different sampling procedures. In one, they questioned randomly chosen tourists as they were leaving the country; in the other, they questioned randomly chosen guests at hotels. (Each tourist stayed at a hotel.) The average visiting time of the 3000 tourists chosen from hotels was \(17.8\), whereas the average visiting time of the 12,321 tourists questioned at departure was \(9.0 .\) Can you explain this discrepancy? Does it necessarily imply a mistake?

Wald's equation can also be proved by using renewal reward processes. Let \(N\) be a stopping time for the sequence of independent and identically distributed random variables \(X_{i}, i \geqslant 1\) (a) Let \(N_{1}=N\). Argue that the sequence of random variables \(X_{N_{1}+1}, X_{N_{1}+2}, \ldots\) is independent of \(X_{1}, \ldots, X_{N}\) and has the same distribution as the original sequence \(X_{i}, i \geqslant 1\) Now treat \(X_{N_{1}+1}, X_{N_{1}+2}, \ldots\) as a new sequence, and define a stopping time \(\mathrm{N}_{2}\) for this sequence that is defined exactly as \(\mathrm{N}_{1}\) is on the original sequence. (For instance, if \(N_{1}=\min \left(n: X_{n}>0\right\\}\), then \(\left.N_{2}=\min \left[n: X_{N_{1}+n}>0\right\\} .\right)\) Similarly, define a stopping time \(N_{3}\) on the sequence \(X_{N_{1}+N_{2}+1}, X_{N_{1}+N_{2}+2}, \ldots\) that is identically defined on this sequence as \(N_{1}\) is on the original sequence, and so on. (b) Is the reward process in which \(X_{i}\) is the reward earned during period \(i\) a renewal Ireward process? If so, what is the length of the successive cycles? (c) Derive an expression for the average reward per unit time. (d) Use the strong law of large numbers to derive a second expression for the average reward per unit time. (e) Conclude Wald's equation.

For a renewal reward process consider $$ W_{n}=\frac{R_{1}+R_{2}+\cdots+R_{n}}{X_{1}+X_{2}+\cdots+X_{n}} $$ where \(W_{n}\) represents the average reward earned during the first \(n\) cycles. Show that \(W_{n} \rightarrow E[R] / E[X]\) as \(n \rightarrow \infty\)
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