Question

Consider a renewal process with mean interarrival time \(\mu .\) Suppose that each event of this process is independently "counted" with probability \(p\). Let \(N_{C}(t)\) denote the number of counted events by time \(t, t>0\). (a) Is \(N_{C}(t), t \geqslant 0\) a renewal process? (b) What is \(\lim _{t \rightarrow \infty} N_{C}(t) / t ?\)

Short answer

(a) No, \(N_{C}(t), t \geqslant 0\) is not a renewal process since the interarrival times of counted events are not independent and identically distributed. (b) The limit \(\lim_{t \rightarrow \infty} N_{C}(t) / t\) is equal to the rate of counted events, which is \(p / \mu\).

Step-by-step solution

Define the renewal process

A renewal process is a point process that has the property of independent and identically distributed (IID) interarrival times. Specifically, denote \(T_i\) as the interarrival time between the \(i\)-th and the \((i-1)\)-th event. Then, if \((T_1, T_2, \ldots)\) is a sequence of IID random variables and \(T_i \geq 0\), we have a renewal process.

Check if \(N_{C}(t), t \geqslant 0\) is a renewal process

Now we need to check if \(N_{C}(t), t\geq 0\) is a renewal process. Let \(T_i^c\) be the interarrival time between the \(i\)-th and the \((i-1)\)-th counted events. Since each event is independently counted with probability \(p\), the distribution of \(T_i^c\) is the sum of \(N\) IID random variables with distribution \(T_i\), where \(N\) is a geometric random variable with success probability \(p\). Thus, the interarrival times \(T_i^c\) are Not independent and identically distributed, which means that \(N_{C}(t), t\geq 0\) is not a renewal process. Answer (a): No, \(N_{C}(t), t \geqslant 0\) is not a renewal process.

Calculate the rate of counted events

Since each event of the original renewal process happens with mean inter-event time \(\mu\), the rate (number of events per unit time) of the process is \(1/\mu\). The rate for the counted events is \(p / \mu\) because the original events of the renewal process are independently counted with probability \(p\).

Find the limit \(\lim _{t \rightarrow \infty} N_{C}(t) / t\)

The limit \(\lim _{t \rightarrow \infty} N_{C}(t) / t\) represents the long-term average number of counted events per unit time when the event-counting is applied to the renewal process. This limit is equal to the rate of counted events, which is calculated in Step 3. Answer (b): The limit of \(\lim_{t \rightarrow \infty} N_{C}(t) / t\) is equal to the rate of counted events, which is \(p / \mu\).