Question

Consider a renewal process \(N(t)\) with associated distribution function \(F(x)\). Define \(m_{k}(t)=E\left[N(t)^{h}\right]\). Show that \(m_{k}(t)\) satisfies the renewal equation $$ m_{k}(t)=z_{k}(t)+\int_{0}^{t} m_{k}(t-\tau) d F(\tau), \quad k=1,2, \ldots $$ where $$ z_{k}(t)=\int_{0}^{t} \sum_{j=0}^{k-1}\left(\begin{array}{l} k \\ j \end{array}\right) m_{j}(t-\tau) d F(\tau) $$ Ilint: Use the renewal argument.

Short answer

The exercise proves the given renewal equation through the application of the renewal argument to calculate \(m_k(t)\), subsequently expressing the expected value as an integral over the distribution function and determining \(z_k(t)\) by manipulation of this integral. Ultimately the renewal equation is established: \(m_{k}(t)=z_{k}(t)+\int_{0}^{t} m_{k}(t-\tau) d F(\tau)\).

Step-by-step solution

Applying the Renewal Argument to Calculate \(m_{k}(t)\)

The renewal argument states that \(N(t)\) is composed of \(N(t-\tau)\) for \(\tau < t\), plus a renewal that happens at time \(\tau\) that ages to time \(t\). Based on this, express \(N(t)\) as a sum of powers, \(N(t)^k = \sum_{h=0}^{k}\) \(\binom{k}{h}\) \(N(t-\tau)^h N(\tau)^{k-h}\), and then take the expected value, \(m_k(t)\), on both sides.

Expressing the Expected Value as an Integral

By use of the law of total expectation, express \(E[N(t)^k]\) as an integral over the distribution function \(F(x)\), resulting in \(m_{k}(t) = \int_0^t E(N(t-\tau)^k N(\tau)^{k-k}) dF(\tau)\) along with the term \(E[(N(\tau))^{k-k}]\) when \(\tau < t\).

Deriving \(z_k(t)\)

By manipulating the integral in step 2, derive \(z_k(t)\). Keep terms corresponding to \(j = 0, ..., k-1\) on one side and name them \(z_k(t)\), which by definition equals \(\int_0^t \sum_{j=0}^{k-1}\binom{k}{j}m_j(t-\tau) dF(\tau)\). The remaining term corresponds to \(m_k(t-\tau)dF(\tau)\). The integral for all \(\tau\) from 0 to t of this term gives the second term on the right hand side of the renewal equation.

Establishing the Renewal Equation

Combine the results of steps 2 and 3 to form the renewal equation: \(m_{k}(t)=z_{k}(t)+\int_{0}^{t} m_{k}(t-\tau) d F(\tau)\).

Key concepts

Renewal Equation

The renewal equation is a fundamental concept in the study of renewal processes, which are models for events that occur repeatedly over time. A key application of these processes is in reliability theory, where they're used to model systems that require repair or replacement at random intervals.

In the context of the given exercise, we consider a renewal process denoted by \(N(t)\), where \(t\) represents time. The renewal equation defines the expected number of renewals, or occurrences, of an event by time \(t\). This equation takes into account the entire history of the process by integrating over the possible times \(\tau\) that a renewal could have occurred, with each renewal contributing to the future renewals. Intuitively, it combines the initial renewals, represented by \(z_k(t)\), with the renewals occurring after some time \(\tau\).

To visualize this, imagine flipping a coin repeatedly and tracking the number of heads (the 'event'). The renewal equation would predict the expected number of heads at any given time, taking into account the history of flips. This is similar to predicting system replacements in a mechanical setting, such as light bulbs in a building where each renewal is the installation of a new bulb after one burns out.

Using the Renewal Argument

Applying the renewal argument, as seen in the exercise, helps to express \(m_{k}(t)\), the expected value of the number of renewals raised to the \(k\)-th power, as the sum of all possible scenarios where renewals have occurred at varying times in the past. This creates a recursive relationship where the expected value at time \(t\) depends on all the previous expected values, hence the integral involving \(m_{k}(t-\tau)\).

Expected Value

Expected value, often denoted as \(E(X)\), is a key concept in probability and statistics. It represents the average outcome one would expect from an event or process if it were repeated many times. In simple terms, it's like the 'long-term average' for a random event.

In our renewal process example, \(m_{k}(t)\) denotes the expected value, specifically the expected number of renewals raised to the power of \(k\) up to time \(t\). The expected value tells us what to expect on average from the renewal process over time. It's crucial in understanding the behavior of random processes and making predictions about future occurrences.

For instance, if \(N(t)\) tracked the number of people entering a store up to time \(t\), then \(m_{k}(t)\) could help the store owner estimate the total number of groups of \(k\) people or fewer entering the store by that time. This statistic is invaluable for resource planning and assessing service requirements. Calculating the expected value involves integrating over the probability or distribution function, adding another layer of complexity as it accounts for the variability over time.

Distribution Function

A distribution function, often referred to as the cumulative distribution function (CDF), is a fundamental concept in statistics which describes the probability that a real-valued random variable \(X\) is less than or equal to a certain value. More formally, for any number \(x\), the distribution function \(F(x)\) represents the probability that \(X\) is less than or equal to \(x\), that is, \(F(x) = P(X \leq x)\).

In the context of renewal processes, the distribution function \(F(x)\) provides the probabilities of the time intervals between consecutive renewals. When we integrate over this function, as seen in the renewal equation, we essentially 'sum up' (integrate) the contributions of all past renewals to the expected number up to the current time \(t\). This process accounts for the inherent randomness in the timing of each renewal event. Understanding the distribution function is critical for applying the renewal argument since it captures the stochastic nature of the time between renewals.

Imagine a bus arriving at a station. The distribution function would describe the probability of the bus's arrival time. In a renewal process, we'd use this distribution to predict the probability of the next bus arriving within a certain timeframe based on the historical pattern of arrivals.

Law of Total Expectation

The law of total expectation is a theorem in probability theory that relates the expected value of a random variable to the expected values within sub-populations of a probability space. It's essentially an 'averaging' of averages across different scenarios, or conditional expectations. The law states that the expected value of a random variable can be computed by averaging over the expected values conditioned on a second random variable.

This law can be expressed as \(E(X) = E[E(X|Y)]\), where \(E(X|Y)\) is the expected value of \(X\) given \(Y\), and we average over all values of \(Y\). In the renewal process, we use the law of total expectation to break down the calculation of the expected number of renewals into more manageable pieces, accounting for the number of renewals at different times \(\tau\) up to time \(t\).

For example, consider a classroom of students taking multiple-choice tests throughout a semester. The law of total expectation allows us to calculate the overall average score by first finding the average scores per test (conditional on each test), and then averaging these results across all tests. Similarly, in renewal processes, the law allows us to understand the contribution of each point in time to the overall expected renewals, which is key in deriving the renewal equation highlighted in the exercise.