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?