Solved – Prove that sum of uniform distribution (-1,1) is also uniform (-n,n)?

If $d_i in U(-1,1)$ (uniform distribution between -1 and 1 – not sure what the canonical notation is for this), then it seems intuitive that $sum_{i=1}^n d_i in U(-n,n)$ and thus $$Pbig(sum_{i=1}^n d_i > 0big) = frac{1}{2}$$.

If this is true, it also seems intuitive that $Pbig(sum_{i=1}^n p_id_ibig) in U(-n,n)$ where $p_i in {-1,1}$. That is, we can apply the same ideas to a combination of additions/subtractions, not just a summation.

I'm not classically trained in statistics and don't know if this is a standard theorem or if I am way off base here. Is this true / false? If it is true, is it provable using classical statistical techniques?

Its false. The central limit theorem: (and variants) say that sums of independent bounded random variables converge to a normal distribution, essentially independently of how the original variables were distributed.

For a bit of intuition for why the sum of two uniform random variables over $[-1,1]$ is not uniform over $[-2,2]$, consider this: there is only one way to obtain a sum of $2$ — both random variables need to equal $1$. But there are many ways to obtain a sum of $0$ — we just need $d_1 = -d_2$ for any value of $d_1$. So some sums are “more likely'' than others.

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