Prove that counting process $(N_t)$ is such that $N_t/t\to1/\mathbb{E}\left[X\right]$

I am attempting exercise 1.8.1 from Brémaud's Markov Chains: Gibbs Fields, Monte Carlo Simulation, and Queues.

Let $S_1, S_2, \ldots$ be a sequence of i.i.d. random variables, with $\mathbb{P}(0 < S_1 < \infty) = 1$ and $\mathbb{E}\left[S_1\right] < \infty$. For $t \geq 0$, let $N_t = \sum_{n=1}^\infty {1_{\left(0, t\right]}\left(T_n\right)}$ where $T_n = S_1 + \cdots + S_n$. Prove that, almost surely, $$\lim_{t \to \infty} {\frac{N_t}{t}} = \frac{1}{\mathbb{E}\left[S_1\right]}$$

I am not sure how to get started. Apparently I should be applying the Strong Law of Large Numbers but I don't see immediately how it applies.

• Hint: $N_{S_n}/S_n = n/S_n \to 1/\mathbb{E}[S_1]$ a.s. Oct 6 '16 at 5:49
• @zhoraster Did you mean $N_{T_n} / T_n = n / T_n$? Oct 6 '16 at 20:49
Since $S_{N_t}\leqslant t\leqslant S_{N_t+1}$, we have $$\frac{S_{N_t}}{N_t}\leqslant \frac t{N_t} \leqslant \frac{S_{N_t+1}}{N_t}.$$ Now, $N_t\stackrel{t\to\infty}\longrightarrow\infty$ almost surely, so by the strong law of large numbers, $$\mathbb P\left(\lim_{t\to\infty} \frac{S_{N_t}}{N_t}= \mu \right) = 1.$$ Similarly, $$\frac{S_{N_t+1}}{N_t} = \frac{S_{N_t+1}}{N_t+1}\frac{N_t+1}{N_t}\stackrel{t\to\infty}\longrightarrow \mu$$ with probability $1$, so we conclude that $$\mathbb P\left(\lim_{t\to\infty} \frac t{N_t} =\mu\right)=1,$$ or equivalently $$\mathbb P\left(\lim_{t\to\infty} \frac{N_t}t =\frac1\mu\right)=1.$$
• Sorry, I don't entirely follow. Did you mean $T_{N_t} \leq t \leq T_{N_t + 1}$? Oct 6 '16 at 20:48
• @Arman Your notation is a bit nonstandard; generally $S_n$ denotes the time of the $n^{\mathrm{th}}$ renewal (where here you've used $T_n$). It is a bit more intuitive, as $S_n$ denotes a Sum whereas a capital $T$ is often used to denote a fixed Time. Oct 6 '16 at 21:09