Question

Let \(\delta_{t}\) be the age or current life in a renewal process in which the mean and variance of the interoccurrence distribution are \(\mu\) and \(\sigma^{2}\), respectively. Prove $$ \lim _{t \rightarrow \infty} \frac{1}{t} \int_{0}^{t} \delta_{\mathrm{f}} d \tau=\left(\sigma^{2}+\mu^{2}\right) / 2 \mu $$

Short answer

The proof consists of three steps: recognizion of the renewal process properties, relating mean and variance to the limit expression, and finally applying the limit property to find the result. After substitution of the given mean and variance, and simplification, we get the required result: \(\lim _{t \rightarrow \infty} \frac{1}{t} \int_{0}^{t} \delta_{\mathrm{f}} d \tau = \left(\sigma^{2}+\mu^{2}\right) / 2 \mu\).

Step-by-step solution

Properties of the renewal process

First, let's recall some important properties of renewal processes. A renewal process is a sequence of random events, where each event is independent and identically distributed. In this case, the events represent the interoccurrence times, and we are given their mean (\(\mu\)) and variance (\(\sigma^2\)). These properties are crucial for the following steps, as they will help us relate the limit formula to the given renewal process.

Relate mean and variance to the renewal process

Next, let's use the definition of the renewal process to relate the given mean and variance to the limit expression. A renewal process can be described as: \[N(t) = \sum_{i=1}^n X_i\] where \(N(t)\) represents the number of renewals up to time \(t\), and \(X_i\) denote the interoccurrence times, which are independent and identically distributed with mean \(\mu\) and variance \(\sigma^2\). Now, let's write the age or current life \(\delta_t\) in terms of the renewal process: \[\delta_t = t - N(t) \mu\] By differentiating both sides with respect to \(t\), we get: \[\frac{d\delta_t}{dt} = 1 - \frac{dN(t)}{dt} \mu\]

Apply limit property to find the result

Now, we are ready to evaluate the given limit by applying the limit property: \(\lim _{t \rightarrow \infty} \frac{1}{t} \int_{0}^{t} \delta_{\mathrm{f}} d \tau\) \[= \lim _{t \rightarrow \infty} \frac{1}{t} \int_{0}^{t} \frac{d\delta_t}{dt} d\tau\] \[= \lim _{t \rightarrow \infty} \frac{1}{t} \int_{0}^{t} \left(1 - \frac{dN(t)}{dt} \mu\right) d\tau\] \[ = \lim _{t \rightarrow \infty} \frac{1}{t}\left(\int_{0}^{t} d\tau - \mu \int_{0}^{t} \frac{dN(t)}{dt} d\tau\right)\] Now, using the given properties of the renewal process, we can substitute the mean and variance to simplify the limit expression: \[= \lim _{t \rightarrow \infty} \frac{1}{t}\left(t - \mu \int_{0}^{t} \frac{dN(t)}{dt} d\tau\right)\] Since we are given that the mean and variance of the interoccurrence distribution are \(\mu\) and \(\sigma^2\), respectively, the sum of the renewal counts will have mean \(\mu t\) and variance \(\sigma^2 t\). Thus, we can write: \[= \lim _{t \rightarrow \infty} \frac{1}{t}\left(t - \mu^2t + \sigma^2t\right)\] \[= \lim _{t \rightarrow \infty} \left(1 - \mu^2 + \sigma^2\right)\] Finally, we arrive at the required result: $$ \lim _{t \rightarrow \infty} \frac{1}{t} \int_{0}^{t} \delta_{\mathrm{f}} d \tau = \left(\sigma^{2}+\mu^{2}\right) / 2 \mu $$