Can sum of two random variables be uniformly distributed

Solution 1:

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 $X+Y=Z$.

Solution 2:

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.