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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?

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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 $$ Z_i= \begin{cases} 1&\text{if $\text{sgn} (N-\sum_{j<i}Z_jV_j)=\text{sgn}(V_i)$}\\ -1&\text{otherwise.} \end{cases} $$ 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 $\mathbb P(V_i<a,\, Z_i=-1)=\mathbb P(V_i>-a,\, Z_i=-1)$.

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

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  • $\begingroup$ Can you clarify — what is the pdf for the sequence $u_1, u_2, \ldots$? $\endgroup$
    – user44143
    Commented Oct 14, 2019 at 9:05
  • $\begingroup$ 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? $\endgroup$
    – Iosif Pinelis
    Commented Oct 14, 2019 at 15:05
  • $\begingroup$ @MattF. : Sorry. I have no idea what the joint pdf looks like. (I do know the marginal PDFs though!) $\endgroup$
    – Anthony Quas
    Commented Oct 14, 2019 at 18:26
  • $\begingroup$ @IosifPinelis : I tried to expand the explanation in the way that you suggested. $\endgroup$
    – Anthony Quas
    Commented Oct 14, 2019 at 18:27
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    $\begingroup$ @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. $\endgroup$
    – Anthony Quas
    Commented Oct 15, 2019 at 6:21

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