Show that, almost surely, $\lim_{t\to\infty} \frac{N_t}{t}=\frac{1}{E[T_1]}.$

Assume that $$T_i$$ for $$i=1,2,3,\dots$$ are i.i.d. random variables with such that $$E[T_i]<\infty$$ and $$0 with probability $$1$$. Let $$S_n=T_1+T_2+\dots+T_n$$. Define $$N_t=\sum_{n=1}^\infty I[S_n\le t]$$ Show that, almost surely, $$\lim_{t\to\infty} \frac{N_t}{t}=\frac{1}{E[T_1]}.$$

Since $$T_i$$ are iid with finite moment, then by Strong law of large number $$\frac{S_n}{n}\to E[T_1]$$ a.s.

I am stuck here... I am not sure how to connect with $$N_t$$? I only know that $$E[N_t]=E[\sum_{n\ge 1} I[S_n\le t]=\sum_{n\ge 1}P(S_n\le t)$$

By Markov's inequality, $$P(S_n\le t)\le t^{-1}E[S_n]=(n/t)E[T_1]$$

• See section "Asymptotic properties" in Renewal theory.
– Amir
Apr 20 at 6:26

Firstly, we know that $$N_t \rightarrow \infty$$ as $$t\rightarrow\infty$$. (I give a proof in the end.)

By SLLN: $$S_n/n\rightarrow \mathbb{E}[T_1]$$, we have $$S_{N_t}/N_t\rightarrow \mathbb{E}[T_1]$$.

By definition, we observed that $$S_{N_t}\leq t \leq S_{N_t+1}$$,

then we have, $$\frac{N_t}{t}\leq \frac{N_t}{S_{N_t}}$$,$$\frac{N_t+1}{t} \geq \frac{N_t+1}{S_{N_t+1}}$$

Taking $$t\rightarrow\infty$$, we have $$\lim_{t\rightarrow\infty}\frac{N_t}{t}=\lim_{t\rightarrow\infty}\frac{N_t}{S_{N_t}}=\frac{1}{\mathbb{E}[T_1]}$$

For the proof of $$N_t \rightarrow \infty$$ as $$t\rightarrow\infty$$, we only need to observe that {$$N_t\geq n$$}={$$S_n\leq t$$}.

Due to this and monotonicty of $$N_t$$, we find $$\mathbb{P}[\lim_t N_t\geq n]\geq\mathbb{P}[S_n\leq t]$$. Then let t goes to infinity, we have $$\lim_t N_t \geq n$$ almost surely for any $$n$$.