0
$\begingroup$

For a renewal process where $f(t)$ is the number of arrivals in time $t$ and $S_k$ is the $k^{th}$ time of arrival, how can we show:

$$f(\alpha S_k)/k \xrightarrow{\text{a.s.}}\alpha $$

as $k \to \infty$, where $0<\alpha<1 $, and $f(t)$ and $S_k$ belong to two separate renewal process of the same type with the same parameters?

We know $S_k/k\xrightarrow{\text{a.s.}} \overline{X}$ and $f(t)/t \xrightarrow{\text{a.s.}} 1/\overline{X}$, where $\overline{X}$ is the mean between two occurrence. How can I merge these two to obtain the above?

$\endgroup$
0
$\begingroup$

When combined with what you already know, doesn't $$ {f(\alpha S_k)\over k} = {f(\alpha S_k)\over\alpha S_k}\cdot{S_k\over k}\cdot\alpha $$ do the trick?

$\endgroup$
  • $\begingroup$ Can you provide me next steps? $\endgroup$ – Susan_Math123 Jul 9 '16 at 21:53
  • $\begingroup$ You've already stated the limit of $S_k/k$ as $k\to\infty$. What is the limit of $f(\alpha S_k)/\alpha S_k$ as $k\to\infty$? $\endgroup$ – John Dawkins Jul 9 '16 at 21:55
  • $\begingroup$ We have to prove two things to use the above's equation. First, $A_n \xrightarrow{a.s}A$ and $B_n \xrightarrow{a.s}B$ , then $A_n B_n\xrightarrow{a.s}A B$ . (This is a property, but I don't know where I can find it in the books). Second, we have to prove $\lim_{k \to \infty} \frac{f(\alpha S_k)}{\alpha k} = \lim_{t \to \infty} \frac{f(t)}{t}$. Why is this true? $\endgroup$ – Susan_Math123 Jul 9 '16 at 22:02
  • $\begingroup$ @Su20200 Please note that you wish to use that $\lim_{k \to \infty} = \frac{f(\alpha S_k)}{\alpha S_k} = \lim_{t \to \infty} \frac{f(t)}{t}$, rather than what you wrote. $\endgroup$ – Ritz Jul 11 '16 at 7:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.