What Distribution Summed is Uniform?

Question

The sum of n independent uniform random variables forms a new random variable having an Irwin–Hall distribution with parameter n. (Approximately normal when n is large)

$$X_i \sim \cal{U},\quad \sum_{i=1}^n \mathit{X}_i \sim \mathcal{IH}(n)$$

But an IRV of what distribution will produce a uniformly distributed RV, when summed?

When I have, $$Z = \sum_{i=1}^{n}{Y_i} \sim \cal{U}$$ What is the distribution of $Y?$

I am particularly interested in the discrete case, and I suspect the particular solution will look like an upside-down bell curve.

Answer 1

It does not exist.

If such a distribution existed, the distribution of $Y$ would have compact support, therefore its characteristic function would be entire. Moreover the characteristic function of the uniform distribution would be the $n$'th power of the characteristic function of the distribution of the $Y_i$. But the characteristic function of the uniform distribution on $[a,b]$ is $$ \phi(t) = \frac{e^{ibt} - e^{iat}}{it(b-a)}$$ which has simple zeros at $2 \pi n/(b-a)$ for nonzero integers $n$, and therefore does not have an $n$'th root that is analytic at those points.

If by "discrete case" you mean a discrete uniform distribution, a similar thing happens. The discrete uniform distribution on $0, 1, \ldots, m$, say, has characteristic function $(m+1)^{-1} (1 + e^{it} + e^{2it} + \ldots + e^{mit})$, which has simple zeros at $t = 2 \pi i k/m$ where $k$ is not a multiple of $m$, and thus this can't be the sum of $n$ iid random variables.

Answer 2

I didn't see Robert Israel's answer when answering (or was writing when he answered), but here is a more elementary proof in the discrete case, I think.

Suppose you had such $Z$ uniform over $[0,M]$ and corresponding $Y_i$.

Then (in the discrete case): $$Pr(Z=M) = \prod_{i=1}^nPr(Y_i=M/n) = Pr(Y_i=M/n)^n.\\ Pr(Z=0) = \prod_{i=1}^nPr(Y_i=0) = Pr(Y_i=0)^n.$$ ($M/n$ is necessarily the maximum value with non-zero probability for $Y_i$)

Thus, $$Pr(Y_i=M/n) = Pr(Y_i=0).$$

But now you have $$Pr(Z=M/n) \geq Pr(Y_1=M/n)\prod_{i=2}^nPr(Y_i=0) + Pr(Y_n=M/n)\prod_{i=1}^{n-1}Pr(Y_i=0)\\ = 2Pr(Y_i=0)^n = 2Pr(Z=0).$$ And $Z$ is not uniform.