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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 !

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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.

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

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