Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a very simple experience $E$ that takes a time $T$ to complete, $T$ is uniformly distributed in $[1;2]$.

I consider doing a sequence of such experiences $E_i$ ($i\le n$). By Central Limit Theorem, the total time $T_n=\sum_{i\le n} T_i$ is normally distributed when $n\rightarrow\infty$.

Now, I consider that I fix the time $\tau$, and repeat the experience $E$, until I reach $\tau$, and I count the number $N_\tau$ of finished experiences $E$. When $\tau\rightarrow\infty$, what will be the kind of distribution of $N_\tau$ ?

Thank you !

share|cite|improve this question
up vote 1 down vote accepted

The central limit theorem you allude to states that $S_n=\sum\limits_{k=1}^nT_k$ is such that $S_n=n\,\mu+\sqrt{n}\,\sigma\,Z_n$ where $\mu$ and $\sigma^2$ are the mean and variance of every $T_k$, and $Z_n$ converges in distribution to a standard normal random variable when $n\to\infty$. The process $(N_t)_{t\geqslant0}$ is characterized by the identities $[N_t=n]=[S_n\leqslant t\lt S_{n+1}]$ for every $n\geqslant0$, where $S_0=0$.

Roughly speaking, when $n$ and/or $t$ is large, $S_n\approx n\mu$ hence $N_t=n$ solves $S_n\approx t$, that is, $n\mu\approx t$. Indeed, a rigorous result is that $N_t/t\to1/\mu$ almost surely, when $t\to\infty$.

Likewise, roughly speaking, when $n$ and/or $t$ is large, $S_n\approx n\mu+\sqrt{n}\,\sigma\,Z$ where $Z$ is a standard normal random variable hence $S_n\approx t$ when $n\mu\approx t-\sqrt{n}\,\sigma\,Z\approx t-\sqrt{t/\mu}\,\sigma\,Z$. Indeed, a rigorous result is that $(N_t-t/\mu)/\sqrt{t}$ converges in distribution to a centered normal random variable with variance $\sigma^2/\mu^3$.

The process $(N_t)_{t\geqslant0}$ is the counting process associated to the arrival process $(S_n)_{n\geqslant0}$ and the results quoted above are usually called renewal limit theorems.

share|cite|improve this answer
Thank you very much for your answer and the link to the renewal limit theorems ! – Xoff Jul 21 '12 at 15:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.