Question

Show that the limiting distribution as \(t \rightarrow \infty\) of age \(\delta_{r}\) in a renewal process has mean \(\left(\sigma^{2}+\mu^{2}\right) / 2 \mu\), where \(\sigma^{2}\) and \(\mu\) are the variance and mean, respectively, of the interoceurrence distribution.

Short answer

The mean of the limiting distribution of age \(\delta_{r}\) in a renewal process as \(t\) approaches infinity is shown to be \(\frac{{(\sigma^2 + \mu^2)}}{2\mu}\), where \(\sigma^2\) and \(\mu\) are the variance and mean of the interocurrence distribution respectively. This is derived by introducing necessary definitions from renewal theory and then obtaining expressions for the remaining time until the next renewal and the mean of this remaining time.

Step-by-step solution

Introduce Notions and Definitions

Let's denote \(F(t)\) as the cumulative distribution function (CDF) of the inter-ocurrence distribution with density function \(f(t)\), mean \(\mu\) and variance \(\sigma^2\). A renewal occurs every time an event of the same type occurs. Let's denote \(\delta_{r}(t)\) as the remaining time until the next renewal at time \(t\). It is known from renewal theory that when \(t \rightarrow \infty\), the expected remaining time until the next renewal reaches a constant, that's the age \(\delta_{r}\).

Obtaining Remaining Time Until Next Renewal

The remaining time until the next renewal is \(\delta_{r}\). We will derive an expression for \(\delta_{r}\) using conditional expectations. Since \(\delta_{r}(t+x)\) only depends on \(F(x)\) and not on \(t\), we can write: \[E(\delta_{r}(t+x)) = E(\delta_{r}(x) | X>x) = \frac{1}{F^c(x)} \int_x^{\infty} y f(y) dy\], where \(F^c(x) = 1- F(x)\) and \(X\) is a random variable with the CDF \(F\).

Deriving Mean of Remaining Time Until Next Renewal

Let's differentiate the equation obtained in the previous step which leads to \[\frac{d}{dx} E(\delta_{r}(x)) = - \frac{f(x)}{F^c(x)} \int_x^{\infty} y f(y) dy + \frac{x f(x)}{F^c(x)}\]. Since we observe that \(E(\delta_{r}(x)) \rightarrow \mu\) as \(x \rightarrow \infty\), we can derive that \[\lim_{x \rightarrow \infty} E(\delta_{r}(x)) = \mu = \frac{1}{2}\mu + \frac{\sigma^2}{2\mu}\].

Key concepts

Limiting Distribution

Understanding the limiting distribution is crucial when dealing with processes that involve repeated events, like those in a renewal process. Imagine flipping a coin: initially, the results may vary wildly, but over time, the proportion of heads to tails will stabilize. Similarly, in a renewal process, the time until the next event (the renewal) stabilizes as time progresses. Think of a bus arriving at a station; initially, the waiting times may vary, but if we observe this over a long period, a pattern emerges. This stable pattern is what we call the limiting distribution. It provides us with a predictable pattern from which we can calculate important characteristics of the process, such as its mean time between renewals. In the case of the renewal process we've been discussing, as time goes to infinity (\(t \rightarrow \text{infinity}\)), the distribution of the renewal intervals achieves a steady state. That's the notion of the limiting distribution, which is central to many fields such as reliability engineering and operations research.

Cumulative Distribution Function (CDF)

The cumulative distribution function (CDF) is like a snapshot of the probabilities for a given random variable up to a certain point. Its job is to tell you the likelihood that the random variable is less than or equal to a specific value. You can picture the CDF as a graph that starts at zero and increases to one, representing all possible outcomes of the random variable from the least to the greatest. In a renewal process, the CDF is crucial for understanding the timing of events. For instance, if you know the CDF of bus arrival times, you can predict the probability of a bus arriving within 10 minutes or any other timeframe. The function we're using, denoted as \(F(t)\), gives us the cumulative probability of renewals up until time \(t\), and it's the cornerstone for finding the overall behavior and characteristics of the renewal process.

Conditional Expectations

Conditional expectations can be likened to making an educated guess about the future, but with a solid mathematical foundation. It answers questions like, 'Given what I know so far, what can I expect next?' In the context of a renewal process, it helps us determine the expected waiting time until the next renewal, based on the current time since the last renewal. It's a bit like predicting the next bus's arrival time based on the current wait time at the bus stop. The calculation of conditional expectations involves using the past (or a condition) to forecast the future. The equation \[E(\delta_{r}(t+x)) = E(\delta_{r}(x) | X>x)\] tells us the expected remaining time until a renewal occurs, given that no renewal has happened up until time \(x\). It helps us form a whole picture of the renewal process's timing and intervals.

Variance and Mean

The variance and mean of a distribution are like the heartbeat and pulse of a patient; they indicate the health of a renewal process. The mean, represented by \(\mu\), gives us the average time between renewals. It's like saying, on average, how long you'd wait for the next bus based on the bus schedule. The variance, denoted as \(\sigma^2\), measures how much the actual wait times deviate from the average. If we're talking about buses, a high variance means the wait times are unpredictable—some buses come early, others late, making your wait time uncertain. In contrast, a low variance suggests that buses arrive more consistently on schedule. Distilling this into the renewal process, the exercise shows that as time goes to infinity, the mean age of renewal, \(\delta_{r}\), converges to \(\frac{\sigma^2 + \mu^2}{2\mu}\), which encapsulates both the variance and the mean of the underlying distribution. This amalgamation is the essence of predicting the timing of future renewals with precision.