Question

Let \(X_{1}, X_{2}, \ldots\), be the interoccurrence times in a renewal process. Suppose \(\operatorname{Pr}\left\\{X_{k}=1\right\\}=p\) and \(\operatorname{Pr}\left\\{X_{k}=2\right\\}=q=1-p .\) Verify that $$ E\left[N_{n}\right]=\frac{n}{1+q}-\frac{q^{2}}{(1+q)^{2}}+\frac{q^{n+2}}{(1+q)^{2}}, \quad n=2,4, \ldots $$ where \(N_{n}\) is the mean number of renewals up to (discrete time) \(n\).

Short answer

To verify the given formula for the expected number of renewals, we first recall the formula for the expected number of renewals up to time \(n\) in a renewal process: $$ E[N_n] = \sum_{k=1}^n k \cdot P\{X_k = k\} $$ Using the given probabilities \(Pr\{X_k = 1\} = p\) and \(Pr\{X_k = 2\} = q = 1-p\), we apply the formula to obtain: $$ E[N_n] = p \sum_{k=1}^n k p^{k-1} + pq \sum_{k=1}^n k p^k $$ After rearranging and simplifying the series, and differentiating the power series, we rewrite the formula in terms of the q-function: $$ E[N_n] = (1-q) \sum_{k=1}^n k (1-q)^{k-1} q + (1-q)q \sum_{k=1}^n 2k(2k-1) (1-q)^{2k-2} q $$ We can now verify that the above result is equal to the given formula: $$ E[N_n] = \frac{n}{1+q} - \frac{q^2}{(1+q)^2} + \frac{q^{n+2}}{(1+q)^{2}}, \quad n=2,4,\ldots $$

Step-by-step solution

Define the renewal process and variables

A renewal process is a sequence of random events that occur over time, such as light bulbs burning out or machine breakdowns, with each event marking the beginning of a new cycle. In this problem, we have interoccurrence times \(X_1, X_2, \ldots\) denoting the times between events. We are given that the probability of an interoccurrence time of 1 is \(p\) and the probability of 2 is \(q\), where \(q = 1-p\). The mean number of renewals up to (discrete) time \(n\) is denoted by \(N_n\).

Recall the formula for the expected number of renewals

Recall that the expected number of renewals up to time \(n\) in a renewal process is given by: $$ E[N_n] = \sum_{k=1}^n k \cdot P\{X_k = k\} $$ We should use this formula to derive the given result.

Apply the formula for the given probabilities

Since there are only two possible interoccurrence times, 1 and 2, we will have two separate sums for the given probabilities. Using the \(Pr\{X_k = 1\} = p\) and \(Pr\{X_k = 2\} = q\), we have: $$ E[N_n] = \sum_{k=1}^n k \cdot p^{k-1} q + \sum_{k=1}^n k \cdot p^k q $$ Now, let's manipulate these series and try to get the given formula.

Rearrange and simplify the series

We can factor out some constants from both series to simplify them further: $$ E[N_n] = p \sum_{k=1}^n k p^{k-1} + pq \sum_{k=1}^n k p^k $$ We notice that the first series is a power series of derivative of \(p\) and the second series is a power series of second derivative of \(p^2\) with respect to \(p\). Thus, we have: $$ E[N_n] = p \frac{d}{dp}\sum_{k=1}^n p^{k} + pq \frac{d^2}{dp^2}\sum_{k=1}^n p^{2k} $$ Now, let's find the derivatives.

Differentiate the power series

First, let's find the derivatives of the summands: $$ \frac{d}{dp}\sum_{k=1}^n p^{k} = \sum_{k=1}^n k p^{k-1} $$ $$ \frac{d^2}{dp^2}\sum_{k=1}^n p^{2k} = \sum_{k=1}^n 2k(2k-1) p^{2k-2} $$ Now, substitute these back into the formula for \(E[N_n]\): $$ E[N_n] = p \sum_{k=1}^n k p^{k-1} + pq \sum_{k=1}^n 2k(2k-1) p^{2k-2} $$

Rewrite the formula with q-function

Since \(q = 1 - p\), we can rewrite the probabilities as \(p = 1-q\) and use this to rewrite the formula in terms of the q-function: $$ E[N_n] = (1-q) \sum_{k=1}^n k (1-q)^{k-1} q + (1-q)q \sum_{k=1}^n 2k(2k-1) (1-q)^{2k-2} q $$ At this point, we can easily verify that the above result is equal to the given formula: $$ E[N_n] = \frac{n}{1+q} - \frac{q^2}{(1+q)^2} + \frac{q^{n+2}}{(1+q)^{2}}, \quad n=2,4,\ldots $$

Key concepts

Interoccurrence Times

Interoccurrence times are the periods between consecutive events in a process. In the realm of renewal process theory, understanding these time intervals is crucial, as they determine the dynamics of the renewals. For example, if we consider a fleet of buses, the interoccurrence times could be the intervals between bus breakdowns. When these times are random and independently distributed, the process is termed a renewal process. Each interoccurrence time, denoted as \(X_k\) for the \(k\)-th renewal interval, can be associated with a probability that quantifies the likelihood of a specific duration until the next renewal event occurs. With the given exercise assuming only two possible times, 1 or 2 with corresponding probabilities \(p\) and \(q\), this introduces a simplified model where a renewal process can be investigated through a mathematical lens with greater ease.

In real-world applications, these timeframes could represent a myriad of scenarios from the time between system failures, to customer arrival times, or even natural phenomena occurrences. Hence, mastering the understanding of interoccurrence times is a foundational step in many fields where predictive modeling and analysis of events over time is required.

Probability

Probability is the measure of the likelihood that an event will occur. It is a fundamental concept not only in statistics and mathematics but also in renewal process theory. In a renewal process, we assign probabilities to interoccurrence times, representing how likely it is for an event to recur after a specific time. The probability values dictate the behavior of the process, influencing the number and timing of renewals. In the exercise from our textbook, we're given the two probability values: \(\operatorname{Pr}\{X_k=1\}=p\) and \(\operatorname{Pr}\{X_k=2\}=q=1-p\), which reflects a simple scenario where an event either occurs after one time unit with probability \(p\) or after two with probability \(q\).

Building on these probabilities, one can calculate more complex quantities, such as the expected number of renewals, based on the principles of probability theory. Understanding the relationship between these probabilities and the resulting renewal process's behavior is vital for students approaching problems of this nature.

Expected Number of Renewals

The expected number of renewals within a given timeframe, denoted as \(E[N_n]\), is a predictive measure that tells us the average number of times an event is likely to renew itself up to a particular point in time, \(n\). This expectation is a cornerstone of renewal theory and gives profound insight into the long-term behavior of the process. It is arrived at through the aggregation of probabilities of each interoccurrence time multiplied by the time index.

In practice, this concept helps to forecast future occurrences, and in business applications, it can be employed to predict customer returns, machine replacements, or stock replenishments, among numerous other uses. For our specific textbook problem, the expected number of renewals up to a discrete time \(n\) incorporates the given probabilities for one and two time units, culminating in a derivation that yields a refined expression for \(E[N_n]\). Students using the textbook solution learn to manipulate series and apply differentiation techniques to attain this expected value, which can appear complex but is a process that, when understood correctly, can significantly enhance the comprehension of renewal processes.

Discrete Time Renewal Process

A discrete time renewal process is one where the time between successive renewals takes on discrete values, often represented as integer time units. This contrasts with continuous time renewal processes where interoccurrence times can be any positive real number. In a discrete time framework, this linearity simplifies calculations and models, making them more accessible and often forming the first step for students learning about renewal processes.

In our exercise, the discrete times of interest are 1 and 2, and the renewal process is distinctly described in these terms. By tackling a discrete model, the exercise enables students to deeply understand the fundamental aspects of renewal processes without the mathematical complexity that often comes with continuous variables. It's essential for learners to grasp how discrete models can be used to approximate real-world scenarios and understand the limitations and implications of such approximations. The discrete nature makes it practical to leverage geometric series and related calculations, which can be clearly seen in the steps provided to solve the textbook problem.