Let $(Y_k)$ be i.i.d denote $\mu =\mathbb{E}[Y_1],\ \sigma ^2 = Var[Y_1]$ and there exists $c>0$ s.t $P(Y_1\geq c)=1$ $$ S_n=Y_1+...+Y_n,\ \ \ N_t=\sup\{n\geq 1 | S_n \leq t\}$$ for $t>0$
a) prove $\frac{N_t}{t} \overset{t\rightarrow \infty}{\rightarrow} 1/\mu$ a.s
b)prove weak convergence $$\frac{N_t-\frac{t}{\mu}}{\sigma \sqrt{t/\mu^3}}\rightarrow N(0,1)$$
This is a question from a test in probability. I tried denoting $m(x)=\mathbb{E}[N_t|Y_1]$, something we saw in the past. But since I dont know the density function I didnt know how to calculate it and continue. I came to the conclusion (without proving) that $$ m(x)=\mathbb{E}[P(Y_1>x)+P(Y_1<x)(1+N(x-Y_1)]$$
How do I continue?
Thank you,
The process from the question is called a $\textit{renewal process}$. The results from the problem are fairly common, but very important imo!
a) We'll show this with a squeeze. By definition of $N_t$, we necessarily have $S_{N_t} \leq t \leq S_{N_t+1}$. Therefore, $$ \frac{S_{N_t}}{N_t} \leq \frac{t}{N_t} \leq \frac{S_{N_{t}+1}}{N_t} $$ We wish to show both left and right terms converge to $\mu$ as $t \rightarrow \infty$. First, notice that $N_t \rightarrow \infty$ a.s. as $t \rightarrow \infty$. Now, the left term is easy, by strong law of large numbers, we have $$ \frac{S_{N_t}}{N_t} = \frac{1}{N_t} \sum_{i = 1}^{N_t}Y_i \rightarrow \mathbb{E}[Y_1] = \mu. $$ a.s. as $t \rightarrow \infty$. Now, assuming that $\mu < \infty$, we can show $$ \frac{S_{N_t + 1}}{N_t} = \frac{S_{N_t + 1}}{N_t+1}\cdot \frac{N_t + 1}{N_t} = \mu \cdot 1$$ a.s. as $t \rightarrow \infty$. Therefore, $N_t/t \rightarrow 1/\mu$ a.s..
b) We'd like to show that $$ W_t = \frac{N_t - t/\mu}{\sigma\sqrt{t/\mu^3}} \Rightarrow \mathcal{N}(0, 1)$$ The idea here is to start with known simple standard normal, and work forwards and backwards to show that $W_t$ approaches it as $t \rightarrow \infty$.
We know by simple properties that $S_n$ has mean $n\mu$ and variance $n\sigma^2$, thus, by CLT we have $$ T_n = \frac{S_n-n\mu}{\sigma\sqrt{n}} \Rightarrow\mathcal{N}(0, 1). $$ How should we relate $W_t$ and $T_n$? Notice that $\mathbb{P}(W_t \leq w) = \mathbb{P}(N_t \leq v) = \mathbb{P}(S_{v} \geq t)$ where $v = \lfloor t/\mu + \sigma w \sqrt{t/\mu^3}\rfloor$. This is because taking at most $v$ 'renewals' to reach $t$ implies that $S_{v}$ reaches at least $t$ because $Y_i > 0$ (with probability 1). Finally, we return to normalized $T_n$ as $$ S_{v} \geq t \Rightarrow T_{v} \geq \frac{t-v\mu}{\sigma \sqrt{v}} = \cdots = -\frac{w}{\sqrt{1 + w \sigma/\sqrt{t/\mu}}}. $$ Thankfully, as $t \rightarrow \infty$, we have that $v \rightarrow \infty$ and $w/\left(\sqrt{1+O(t^{-1/2})}\right) \rightarrow w$. If we let $W = \underset{t \rightarrow \infty}{\lim} W_t$ and $T = \underset{v \rightarrow \infty \text{ or } t \rightarrow \infty}{\lim} T_v = \mathcal{N}(0, 1)$, we can conclude as follows $$ \mathbb{P}(W \leq w) = \mathbb{P}(T \geq -w) = \mathbb{P}(T \leq w)$$
by symmetry of $\mathcal{N}$, and we are done.