# Normality of the sum of uniformly distributed random variables

As noted in the recent answer by Yuval Peres, the sum of independent uniformly distributed random variables (r.v.'s) cannot have a normal distribution.

The question is, what happens without the independence condition? More specifically:

Do there exist positive real numbers $$a_1,a_2,\dots$$ and r.v.'s $$U_1,U_2,\dots$$ such that for each natural $$i$$ the r.v. $$U_i$$ is uniformly distributed on the interval $$[-a_i,a_i]$$ and the sum $$U_1+U_2+\cdots$$ has the standard normal distribution?

Yes. Let the random variables $$V_i$$ be independent, with $$V_i$$ uniform on $$[-\frac 1i,\frac 1i]$$. Let $$N$$ be an independent standard normal. Inductively define $$Z_i\in \{\pm 1\}$$ by $$Z_i= \begin{cases} 1&\text{if \text{sgn} (N-\sum_{j Notice that $$\sum Z_iV_i=N$$ almost surely - since the $$|V_i|$$ approach 0, but are almost surely not summable, the partial sums overshoot successively in one direction then the other, converging to $$N$$.

An observation that we need is the following: the involution that negates all of the $$V$$’s and $$N$$ preserves probabilities; and the values of the $$Z$$ variables are unchanged. In particular for any $$a$$ and $$i$$, we have $$\mathbb P(V_i-a,\, Z_i=-1)$$.

Now we claim that $$U_i=Z_iV_i$$ has the same distribution as $$V_i$$. To see this, notice that $$\mathbb P(U_i-a,\, Z_i=-1)$$. But by the above observation, $$\mathbb P(V_i>-a,\,Z_i=-1)=\mathbb P(V_i, so that $$\mathbb P(U_i as required.

added comment: @Michael Hardy pointed out the use of conditional convergence in this answer. It turns out this is essential: there does not exist a collection of dependent uniform random variables where the sum is (non-trivially) normal and the convergence is absolute. To see this, note that if $$\sum a_i<\infty$$, then the sum is compactly supported, so cannot be normal. If $$\sum a_i=\infty$$, then $$\mathbb E\sum_{i=1}^M|U_i|=\sum_{i=1}^M a_i/2$$ by linearity of expectation. Since $$\sum_{i =1}^M |U_i|\le \sum_{i=1}^M a_i$$, we deduce $$\mathbb P(\sum_{i=1}^M |U_i|\ge \sum_{i=1}^M a_i/4)\ge 1/4$$. In particular, the limit superior of these sets has measure at least $$1/4$$, but on this set the sum of the absolute values diverges.

• Can you clarify — what is the pdf for the sequence $u_1, u_2, \ldots$? – Matt F. Oct 14 '19 at 9:05
• Very nice! However, it may seem unclear when you write "by symmetry (under replacing each random variable by its negation)" but actually (of course) do not negate the $Z_i$'s. Can you detail this symmetry consideration? – Iosif Pinelis Oct 14 '19 at 15:05
• @MattF. : Sorry. I have no idea what the joint pdf looks like. (I do know the marginal PDFs though!) – Anthony Quas Oct 14 '19 at 18:26
• @IosifPinelis : I tried to expand the explanation in the way that you suggested. – Anthony Quas Oct 14 '19 at 18:27
• @MichaelHardy: I haven't seen conditoinal convergence in probability before. On the other hand, what I'm doing here is a version of my favourite party trick: essentially using extra randomness (in this case the $N$ which is the target random variable) to build something doing what you want (Notice that the $N$ once built in to the $Z$'s is obtained from the $Z$'s rather than the other way around). I have generally done things a bit like this in dynamical systems, where I call the technique "Coupling and Splicing". The symmetry is a bonus making things work out nicely. – Anthony Quas Oct 15 '19 at 6:21