# Limit of sum of (continuous) uniform distributions

In my stats courses at university, I've been working on transformations of distributions etcetera. However, one particular case has intrigued me for a while: the sum of continuous uniform distributions (on the interval $[0,1]$). I've already found out the general formula for the sum of $n$ uniform distributions on mathworld and other places, and I think I understand it.

However, what if you take the limit of $n$ going to infinity? For high $n$, the sum of $n$ uniform distributions looks like it's almost $0$ everywhere from $0$ to $n$, except for a small peak around $n/2$. This peak also shrinks if $n$ increases. Therefore, the limit of $n \rightarrow \infty$ should give a distribution that is zero everywhere, yet the integral over it should still give 1.

This made me think a bit of a Dirac Delta, except that here we don't have the "infinitely high peak" anywhere. So how does this work? Is the infinity somehow stuffed into the fact that the flat line extends to infinity too?

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The "Therefore" sentence is just fantasy but the intuition behind it can be made rigorous thanks to two arch-classical results, the law of large numbers and the central limit theorem. Calling $S_n$ the sum of $n$ i.i.d. random variables uniform on $(0,1)$, the first result states that for every positive $u$, $$\mathbb P(n/2-un\leqslant S_n\leqslant n/2+un)\to1,$$ and the second result states that, for every $x\leqslant y$, $$\mathbb P\left(n/2+x\sqrt{n/12}\leqslant S_n\leqslant n/2+y\sqrt{n/12}\right)\to\int_x^y\frac1{\sqrt{2\pi}}\mathrm e^{-t^2/2}\mathrm dt.$$ The explanation for the centering and normalization is that $\mathbb E(S_n)=n/2$ and $\mathrm{var}(S_n)=n/12$.