0
$\begingroup$

Assume that $T_i$ for $i=1,2,3,\dots$ are i.i.d. random variables with such that $E[T_i]<\infty$ and $0<T_i<\infty$ 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, $$ P(S_n\le t)\le t^{-1}E[S_n]=(n/t)E[T_1] $$

$\endgroup$
1
  • $\begingroup$ See section "Asymptotic properties" in Renewal theory. $\endgroup$
    – Amir
    Apr 20 at 6:26

1 Answer 1

2
$\begingroup$

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 $n$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .