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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.

Hints please; thank you!

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    $\begingroup$ Hint: $N_{S_n}/S_n = n/S_n \to 1/\mathbb{E}[S_1]$ a.s. $\endgroup$
    – zhoraster
    Oct 6, 2016 at 5:49
  • $\begingroup$ @zhoraster Did you mean $N_{T_n} / T_n = n / T_n$? $\endgroup$ Oct 6, 2016 at 20:49

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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 $$\mathbb P\left(\lim_{t\to\infty} \frac{N_t}t =\frac1\mu\right)=1. $$

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  • $\begingroup$ Sorry, I don't entirely follow. Did you mean $T_{N_t} \leq t \leq T_{N_t + 1}$? $\endgroup$ Oct 6, 2016 at 20:48
  • $\begingroup$ @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. $\endgroup$
    – Math1000
    Oct 6, 2016 at 21:09
  • $\begingroup$ I thought as much, but this was the notation chosen by Brémaud in the book. $\endgroup$ Oct 6, 2016 at 21:11
  • $\begingroup$ Fair enough. It's worth mentioning that the notation I used in my post is that from the course in which I studied renewal theory three years ago, as well as the (entirely different) text that I was reviewing the other day. $\endgroup$
    – Math1000
    Oct 6, 2016 at 21:13

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