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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 summary, we use the relationship between the gamma $(r, \lambda)$ distribution and the sum of $r$ independent exponentials with rate $\lambda$ to express the renewal process $N(t)$ in terms of a Poisson process with rate $\lambda$. Consequently, we derive the following expressions: (a) $P[N(t) \geqslant n]=\sum_{i=nr}^{\infty} \frac{e^{-\lambda t} (\lambda t)^{i}}{i !}$, which represents the probability that there are at least $n$ renewals by time $t$. (b) $m(t) = \sum_{i=r}^{\infty}\left[\frac{i}{r}\right] \frac{e^{-\lambda t} (\lambda t)^{i}}{i !}$, which represents the mean number of renewals by time $t$.

Step by step solution

01

Define N(t) in terms of a Poisson process

: Let \(W_1, W_2, \dots\) be the interarrival times for a Poisson process with independent increments and rate \(\lambda\). Then, the gamma \((r, \lambda)\) interarrival distribution can be expressed as the sum of \(r\) independent exponentials: \(X_k=W_{(k-1)r+1}+W_{(k-1)r+2}+\cdots+W_{kr}\), for \(k=1,2,\cdots\). Now, let \(N^*(t)\) be a Poisson process with rate \(\lambda\). Then, \(N(t)=\left[\frac{N^*(t)}{r}\right]\), where \([\cdot]\) represents the largest integer less than or equal to its argument.
02

Compute P[N(t) >= n]

: Now we need to compute \(P[N(t) \geqslant n]\) by expressing it in terms of the Poisson process. From step 1, we know that \(N(t)=\left[\frac{N^*(t)}{r}\right]\). Thus, we have: \[P[N(t) \geqslant n]=P\left[\frac{N^*(t)}{r}\geqslant n\right]=P[N^*(t)\geqslant nr]\] Using the definition of the Poisson process, we have: \(P[N^*(t) = i] = \frac{e^{-\lambda t} (\lambda t)^{i}}{i!}\) for \(i=0,1,2,...\) So, for \(P[N^*(t)\geqslant nr]\), we can sum the probabilities over all \(i \geqslant nr\): \begin{align*} P[N(t)\geqslant n] &= P[N^*(t)\geqslant nr] \\ &= \sum_{i=nr}^{\infty} P[N^*(t) = i] \\ &= \sum_{i=nr}^{\infty} \frac{e^{-\lambda t} (\lambda t)^{i}}{i !} \end{align*} This result matches the desired expression in part (a).
03

Derive the expression for m(t)

: Finally, let's derive the expression for the mean number of renewals, \(m(t)\). Using the definition of \(N(t)\), we can express the mean number of renewals as: \[m(t)=E[N(t)]=E\left[\frac{N^*(t)}{r}\right]=\frac{1}{r}E[N^*(t)]\] Now, let's evaluate the expectation for the Poisson process: \[E[N^*(t)] = \sum_{i=0}^{\infty} i P[N^*(t) = i]\] Since \(i=0\) when \(E[N^*(t)] = 0\), we can change the summation limits as follows: \[E[N^*(t)] = \sum_{i=r}^{\infty} i P[N^*(t) = i]\] We can now substitute \(m(t)\) to get the desired expression: \[m(t) = \frac{1}{r} \sum_{i=r}^{\infty} i P[N^*(t) = i]\] Now we substitute the probability expression: \[m(t) = \frac{1}{r} \sum_{i=r}^{\infty} i \frac{e^{-\lambda t} (\lambda t)^{i}}{i !}\] This matches the desired expression for part (b). We have now successfully derived both expressions: (a) \(P[N(t) \geqslant n]=\sum_{i=nr}^{\infty} \frac{e^{-\lambda t} (\lambda t)^{i}}{i !}\) (b) \(m(t) = \sum_{i=r}^{\infty}\left[\frac{i}{r}\right] \frac{e^{-\lambda t} (\lambda t)^{i}}{i !}\)

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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 two-parameter family of continuous probability distributions with parameters usually denoted as r (sometimes called the shape parameter) and λ (the rate parameter). It's related to many other statistical distributions; for example, when the shape parameter r is a positive integer, the gamma distribution transforms into the Erlang distribution, which is widely used in queuing theory and reliability engineering.

Mathematically, the gamma distribution's probability density function is given by:
$$f(x)=\frac{\lambda^{r} x^{r-1} e^{-\lambda x}}{(r-1)!}, \quad x>0$$
This function describes the likelihood of x occurring as an event time in a process where interarrival times follow the gamma distribution. It's the continuous analogue to the Poisson distribution used for modeling counting processes. Its relevance in the renewal process allows us to infer the distribution of times between consecutive events. Additionally, the gamma distribution is a versatile tool for representing various statistical phenomena due, in part, to its ability to take on different shapes depending on the value of its parameters.
Poisson Process
The Poisson process is a stochastic process that serves as a model for a series of events occurring randomly over time. The key characteristics of a Poisson process are events that occur independently of one another and at a constant average rate, denoted as λ. This process is widely used in fields such as telecommunication, reliability analysis, and natural phenomena modeling.

Within the context of the Poisson process, the number of events occurring in a fixed interval of time follows a Poisson distribution, and the time between consecutive events (interarrival time) follows an exponential distribution. The probability of having i events occur over a time period t in a Poisson process with rate λ is expressed as:
$$P[N(t) = i] = \frac{e^{-\lambda t} (\lambda t)^{i}}{i!}$$
This simple yet powerful formulation allows us to calculate probabilities associated with the time and occurrence of events, providing a deep insight into the behavior of random processes in time.
Interarrival Time
The interarrival time, in the context of stochastic processes like the Poisson process or renewal processes, refers to the time interval between consecutive events. For example, in a call center, this could be the time between incoming calls, or in a manufacturing system, the time between system failures. Understanding the distribution of these times provides insights into the dynamics of the process under consideration.

In the case of a Poisson process, the interarrival times are exponentially distributed, which means that the events are memoryless 鈥� the probability of the next event occurring does not depend on when the previous event happened. Conversely, for renewal processes with gamma interarrival distributions, the past does impact the future events, as the interarrival times tend to be more spread out, especially when the shape parameter r is high. This aspect is crucial for modeling and analyzing various complex real-world scenarios, where the memoryless property does not hold.

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Most popular questions from this chapter

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?

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Let \(\left[N_{1}(t), t \geqslant 0\right]\) and \(\left\\{N_{2}(t), t \geqslant 0\right\\}\) be independent renewal processes. Let \(N(t)=N_{1}(t)+N_{2}(t)\) (a) Are the interarrival times of \(\\{N(t), t \geqslant 0\\}\) independent? (b) Are they identically distributed? (c) Is \(\\{N(t), t \geqslant 0\\}\) a renewal process?

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