Can sum of two random variables be uniformly distributed

Question

Say $X$ and $Y$ are two random variables where $X\in [-\alpha,\alpha]$, $Y\in [-\alpha,\alpha]$ and $Z=X+Y$. Is it possible to find two independent random variables with certain pdf (not necessarily identically distributed) that force $Z$ to be uniformly distributed (i.e. $Z\sim \mathcal{U}[-2\alpha,2\alpha]$)?

As the sum of $N$ random variables with zero mean resembles Gaussian distribution with zero mean, I suspect it is not possible to find two such random variables. Do you know any counterexample?

Answer 1

Let $ X \sim {\mathcal U}[-\alpha, \alpha] $ and let $ Y $ be a discrete r.v. taking values $ \pm\alpha $ with equal probabilities. Then $ X+Y \sim {\mathcal U}[-2\alpha, 2\alpha] $.

I believe that the sum of $ n \geq 3 $ independent r.v.'s distributed on $ [-\alpha, \alpha] $ cannot be uniform on $ [-n\alpha, n\alpha] $, but I have no proof of this.

Answer 2

Here is another one.

Recall if $U_n$ are IID with Bernoulli distribution $$ \mathbb P[U_n=0]=\frac{1}{2},\qquad \mathbb P[U_n=1]=\frac{1}{2}, $$ then $$Z = \sum_{n=1}^\infty U_n 2^{-n} $$ is uniformly distributed on $[0,1]$. So let $$ X = \sum_{n\text{ even}} U_n 2^{-n},\qquad Y = \sum_{n\text{ odd}} U_n 2^{-n} $$ to get independent $X,Y$ with $X+Y=Z$.