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

My Solutions

Yes this is a renewal process with mean interarrival time, $p\times\mu$

Using the elementary renewal theorem we know that $$\lim_{t\rightarrow\infty}\frac{N(t)}{t} = \frac{1}{\mu}$$ So $$\lim_{t\rightarrow\infty}\frac{N_C(t)}{t} = \frac{1}{p\times\mu}$$

I was hoping someone could say whether this is correct?

  • $\begingroup$ According to your solution the mean interarrival time decreases when you only count each event with probability $p$. Does that sound right to you? $\endgroup$ – kccu Apr 23 at 1:46
  • $\begingroup$ @kccu That is true. I think this is right now? $$\lim_{t\rightarrow\infty} N_C(t)=\lim_{t\rightarrow\infty} pN(t)$$ and since $$\lim_{t\rightarrow\infty} \frac{N(t)}{t}=\frac{1}{\mu}$$ it follows that $$\lim_{t\rightarrow\infty}\frac{ N_C(t)}{t}=p\lim_{t\rightarrow\infty} \frac{N(t)}{t}=\frac{p}{\mu}$$ $\endgroup$ – Jac Frall Apr 23 at 1:59
  • $\begingroup$ The statement $\lim_{t \to \infty} N_C(t) = \lim_{t \to \infty} p N(t)$ does not really make sense, since both of these function are tending to infinity (I assume here $N(t)$ is the number of occurrences up to time $t$ when we count them all?). Do you want to say something like $\lim_{t \to \infty} N_C(t)/N(t)=p$? You would still need to justify this claim (and the claim that $N_C(t)$ is itself a renewal process). $\endgroup$ – kccu Apr 23 at 2:05
  • $\begingroup$ @kccu I know the answer is $p/\mu$ but I cannot think of any other way of showing it other than what I did? Do you have any ideas? $\endgroup$ – Jac Frall Apr 23 at 2:09
  • $\begingroup$ @kccu of course, multiply both sides by $p$. $$p\cdot\lim_{t\rightarrow\infty}\frac{N(t)}{t}=p\cdot\frac{1}{\mu}$$ $\endgroup$ – Jac Frall Apr 23 at 2:19

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