Property of renewal point processes

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?

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?

• Can you provide me next steps? – Susan_Math123 Jul 9 '16 at 21:53
• 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 $k\to\infty$? – John Dawkins Jul 9 '16 at 21:55
• 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? – Susan_Math123 Jul 9 '16 at 22:02
• @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. – Ritz Jul 11 '16 at 7:16