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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 $$P\big(\sum_{i=1}^n d_i > 0\big) = \frac{1}{2}$$.

If this is true, it also seems intuitive that $P\big(\sum_{i=1}^n p_id_i\big) \in U(-n,n)$ 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?

drjrm3
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    The duplicate includes extensive, detailed analyses of the sums of uniform distributions. The difference between a uniform distribution on $[0,1]$ and one on $[-1,1]$ is of no consequence whatever, because it's just a change in the units of measurement. – whuber Jan 06 '16 at 21:07

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Its false. The central limit theorem: https://en.wikipedia.org/wiki/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.

Aaron
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  • That makes a lot of sense and I had heard that before but totally forgot about (I feel silly now). Does the claim about the probability of the sum being positive still hold, though? Wouldn't the sum converge to a normal distribution with a mean of zero, leaving the probability that the sum is positive as 1/2? – drjrm3 Jan 06 '16 at 20:47
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    +1. Here is how we could create a histogram for 10000 instances with $n=5$ in R: `hist(rowSums(matrix(runif(50000,-1,1),ncol=5)))` This shows clearly that the result is not uniform at all. Simulate first, think later. Or: computers are cheap, and thinking hurts (which I think is due to Uwe Ligges). – Stephan Kolassa Jan 06 '16 at 20:49
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    Yes -- everything in your scenario is symmetric: there is nothing to privilege ``positive'' over ``negative'' (i.e. we could rename them and nothing would change), so it must be that the probability that the sum is positive is exactly $1/2$. – Aaron Jan 06 '16 at 20:50
  • Thanks, is there a 'statistics' way of saying this? It could be easy to say $\sum d_i \in U(-n,n)$ so $P(\sum d_i > 0) = 1/2$ or $\sum d_i \in N(0,n)$ so it is 1/2, but how can I mathematically describe the distribution so that it easily follows that the probability is 1/2? – drjrm3 Jan 06 '16 at 20:53
  • Scratch that ... this leads me to a whole new question which I will ask in a another question ... – drjrm3 Jan 06 '16 at 20:55
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    It is easy to formalize the symmetry argument -- your uniform random variables $d_1$ are distributed identically to $-1 \cdot d_1$. So if $X = \sum_i d_i$ and $Y = -1\cdot\sum_i d_i$, then $X$ and $Y$ are identically distributed, but $X = -Y$. Hence $\Pr[X > 0] = \Pr[Y < 0] = \Pr[X < 0]$, and you are done. – Aaron Jan 06 '16 at 20:57