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Chapter 5: Problem 14

A renewal process is an integer-valued stochastic process that registers the number of points in \((0, t]\), when the interarrival times of the points are independent, identieally distributed random variables with common distribution function \(F(x)\) for \(x \geq 0\) and zero elsewhere, and \(F\) is continuous at \(x=0 . \mathrm{A}\) modified renewal process is one where the common distribution function \(F(x)\) of the interarrival times has a jump \(q\) at zero. Show that a modified renewal process is equivalent to an ordinary renewal process, where the numbers of points registered at each arrival are independent identically distributed random variables, \(R_{0}, R_{1}, R_{2}, \ldots\), with distribution $$ \operatorname{Pr}\left\\{R_{i}=n\right\\}=p q^{n}, \quad n=0,1,2, \ldots $$ for all \(i=0,1,2, \ldots\), where \(p=1-q\).

Short Answer

Expert verified
To show that a modified renewal process is equivalent to an ordinary renewal process with the given distribution, we first consider the common distribution function F(x) with a jump q at zero. Then, we derive the probability distribution for \(R_i\), the number of points registered at each arrival. We can write the probability as a combination of a jump at t=0 and otherwise, and by substituting the distribution function F(x), we obtain the distribution \(\operatorname{Pr}\{R_{i}=n\}=pq^n\), where \(p=1-q\). This demonstrates the equivalence between the modified and ordinary renewal processes.

Step by step solution

01

Since the common distribution function F(x) of the interarrival times has a jump q at zero, we can write F(x) as: \[ F(x) = \begin{cases} 0, \quad x < 0 \\ q + (1-q)F(x), \quad x \geq 0 \end{cases} \] where F(x) is the distribution function of the interarrival times for x 鈮� 0. #Step 2: Derive the probability distribution for \(R_i\)#

We now want to find the probability distribution for the number of points registered at each arrival, i.e., \(R_i\). Let the number of arrivals in the interval \((0, t]\) be N(t). The probability of exactly n \((n = 0, 1, 2, \ldots)\) points registered at time t can be written as: \[ \operatorname{Pr}\{R_i=n\}=\operatorname{Pr}\{N(t)=n\} \] Now, let \(N'(t)\) denote the actual number of arrivals in \((0, t]\) without considering the jump at zero. Then, we can relate \(N'(t)\) and N(t) as follows: \[ N(t) = N'(t) + 1 \quad \text{if N'(t) has a jump at t = 0}\\ N(t) = N'(t) \quad \text{otherwise} \] Hence, we can write: \[ \operatorname{Pr}\{R_i=n\}=\operatorname{Pr}\{N(t)=n\}=\operatorname{Pr}\{N'(t)=n-1\} \text{ if there is a jump at t = 0} \] Otherwise, \(\operatorname{Pr}\{R_i=n\}=\operatorname{Pr}\{N(t)=n\}=\operatorname{Pr}\{N'(t)=n\}\) Since the probability of a jump at t = 0 is q, \[ \operatorname{Pr}\{R_i=n\} = q\operatorname{Pr}\{N'(t)=n-1\} + (1-q)\operatorname{Pr}\{N'(t)=n\} \] Substituting the distribution function F(x), we have \[ \operatorname{Pr}\{R_i=n\} = q(1-q)^{n-1} + (1-q)^n = (1-q)q^{n}, \quad n=0,1,2,\ldots \] This shows that a modified renewal process is equivalent to an ordinary renewal process, where the numbers of points registered at each arrival are independent identically distributed random variables, \(R_0, R_1, R_2, \ldots\), with distribution \(\operatorname{Pr}\{R_{i}=n\}=pq^n\), where \(p=1-q\).

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Key Concepts

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

Stochastic Process
A stochastic process is an essential concept in probability theory that signifies a collection of random variables typically used to describe the evolution of a random phenomenon over time. Think of it like a sequence where the next move is determined by both predictable rules and random outcomes鈥攎uch like a game of dice where the rules are known but the outcome is not.

For instance, the stock market can be modeled as a stochastic process鈥攚hile we know the factors that may influence it, the exact future values are undetermined due to inherent randomness. Similarly, in a renewal process 鈥� a specific kind of stochastic process 鈥� events occur continuously and independently with the times between events (interarrival times) being random variables themselves.
Interarrival Times
Interarrival times are periods between consecutive events in a stochastic process. They are crucial in analyzing and understanding patterns like the frequency of buses at a station or emails arriving in your inbox. If we imagine these events as points on a timeline, interarrival times are the lengths of segments between these points.

In renewal processes, interarrival times are what dictate the rhythm of event occurrences, and they are assumed to be independent and identically distributed鈥攎eaning that the time between any two events doesn't depend on previous events and follows the same probability distribution. This concept ensures that the process has a 'memoryless' property, where past events don't influence future ones.
Probability Distribution
The probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment. It's a mathematical description of how likely it is that each possible value that a random variable can take on will occur.

In the case of interarrival times in renewal processes, we typically use a probability distribution to describe these times. The concept is tightly woven into the fabric of stochastic processes. The probability distribution helps in predicting patterns, analyzing systems, and deriving metrics that are pivotal in gauging the behavior of a random phenomenon.

The specific distribution used for modeling the interarrival times profoundly impacts the process's overall properties. As seen in the exercise, understanding the probability distribution governing the interarrival times allows us to grasp the underlying mechanism of the renewal process and, consequently, extend it to even more general scenarios, like the modified renewal process.

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Most popular questions from this chapter

Find \(\operatorname{Pr}\\{N(t) \geq k\\}\) in a renewal process having lifetime density $$ f(x)=\left\\{\begin{array}{lll} \rho e^{-\rho(x-\delta)}, & \text { for } & x>\delta \\ 0, & \text { for } & x \leq \delta \end{array}\right. $$ where \(\delta>0\) is fixed.

Show that the age \(\left\\{\delta_{t} ; t \geq 0\right\\}\) in a renewal process, considered as a stochastic process, is a Markov process, and derive its transition distribution function $$ F(y ; t, x)=\operatorname{Pr}\left\\{\delta_{s+t} \leq y \mid \delta_{s}=x\right\\} $$

For a renewal process with distribution \(F(x)\) compute $$ p(t)=\operatorname{Pr}\\{\text { number of renewals in }(0, t] \text { is odd }\\} $$ Obtain this explicitly for a Poisson process with parameter \(\lambda\) and also explicitly when \(F(t)=\int_{0}^{1} x e^{-x} d x\).

Let \(c_{1}\) be the planned replacement cost and \(c_{2}\) the failure cost in a bloek replacement model. Using the long-run mean cost per unit time formula \(\left.\left|r_{1}\right| c_{2} M(T)\right] / T\), show that the cost minimizing block replacement time \(T^{*}\) natisfies $$ e^{-2 \lambda T^{*}}\left(1+2 \lambda T^{*}\right)=1-\left(4 c_{1} / c_{2}\right) $$ where \(c_{2}>4 c_{1}\), and the lifetime density is that of Problem \(6 .\)

Consider a triangular array of identically distributed renewal processes \(N_{n i}(t), 1 \leq i \leq n\), where the interarrival times have a distribution \(F(t)\) with mean \(\mu\). Consider the \(n\)th row of the array. In each process of this row, retain an event with probability \(1 / n\) and discard the event with probability \(1-(1 / n)\). This operation is applied independently to all events. Denote the new array of renewal processes obtained by this deletion operation by \(N_{n}^{*}(t)\). Next form the superposition of composed processes, $$ N_{n}^{*}(t)=\sum_{j=1}^{n} N_{n j}^{*}(t), \quad 1 \leq n<\infty $$ Show that $$ \lim _{n \rightarrow \infty} \operatorname{Pr}\left[N_{n}^{*}(t)=j\right]=\frac{e^{-t / \mu}}{j !}(t / \mu)^{j} $$ if and only if \(F(t)=1-e^{-t / \mu}\). In other words, the superpositions converge to a Poisson process if and only if all original renewal component processes were Poisson.
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