Let $X$ and $Y$ be independent and identically distributed random variables.
Can $X+Y$ be a uniform distribution?
(Please prove.)


In other words, is a uniform distribution divisible?


The meaning of "divisible" is described at [Wikipedia](https://en.wikipedia.org/wiki/Indecomposable_distribution).



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> If, further, it can be expressed as the distribution of the sum of two or more independent identically distributed random variables, then it is divisible: Z = X1 + X2.
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I read [this article](http://godplaysdice.blogspot.jp/2008/04/uniform-distribution-as-sum.html) and it said that it is impossible.
But I could not understand its proof.



divisibility of uniform distribution
Here is a direct argument.


Suppose independent $X\_1,X\_2 \sim X$, and $X\_1+X\_2$ is uniform on $[0,1]$. 


$X$ is supported on $[0,1/2]$. 


For any $0 \lt \alpha \lt 1/4$,
$\alpha = P\left(X\_1+X\_2 \in [0,\alpha]\right) \le P(X \in [0,\alpha])^2$ so $P(X \in [0,\alpha]) \ge \sqrt{\alpha}$. Similarly, $P(X \in [1/2-\alpha,1/2]) \ge \sqrt{\alpha}$.


Consider the event that $X\_1+X\_2 \in [1/2-\alpha,1/2+\alpha]$, that the sum is within $\alpha$ of $1/2$. The probability is $2\alpha.$ Within this event are the disjoint events that $X\_1 \in [0,\alpha] \wedge X\_2 \in [1/2-\alpha,1/2]$ and that $X\_2 \in [0,\alpha] \wedge X\_1 \in [1/2-\alpha,1/2]$. (Disjointness is implied by $\alpha \lt 1/4$.) These each have probability at least $\alpha$ and they are disjoint, so in fact they have probability exactly $\alpha$, and $P(X \in [0,\alpha])=P(X \in [1/2-\alpha,1/2]) = \sqrt{\alpha}$.


This completely determines the distribution of $X$:


$$\textrm{CDF}\_X(x) = \begin{cases} \sqrt{x} & x \in [0,1/4] \newline 1-\sqrt{1/2-x} & x\in[1/4,1/2]\end{cases}.$$


However, this doesn't work. The convolution square is not uniform. You can see this without integrating because the event that $X\_1+X\_2 \in [0.49,0.51]$ must get its whole probability from $X\_1 \in [0,0.01] \wedge X\_2 \in [0.49,0.5]$ and $X\_2 \in [0,0.01] \wedge X\_1 \in [0.49,0.5]$. However, $X$ has positive density around $0.25$, which means that the probability that $X\_1+X\_2 \in [0.49,0.51]$ is greater than $0.02$. Thus, the convolution square of $X$ is not uniform.


