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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.

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.

I read this article and it said that it is impossible. But I could not understand its proof.

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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.

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How about this...

Uniform distribution on $[-1,1]$ has characteristic function

$$ \varphi(t) = \frac{\sin t}{t} $$

Suppose $X,Y$ are IID and $Z:=X+Y$ is uniformly distributed on $[-1,1]$. Of course $X,Y$ are bounded: $$ \mathbb P[X>1]^2 = \mathbb P[X>1, Y>1] \le \mathbb P[Z>2] = 0 $$ so $\mathbb P[X>2]=0$. Similarly $\mathbb P[X<-2]=0$. The characteristic function of a bounded random variable is an entire function. But the characteristic function $\psi(t)$ of $X$ and $Y$ satisfies $$ \psi(t)^2 = \varphi(t) $$ so it cannot be differentiable at $t=\pi$, where $\varphi(t)$ changes sign.

More... $\varphi(\pi)=0$ so $\psi(\pi)=0$. But if $\psi'(\pi)$ exists, we get $-1/\pi = \varphi'(\pi) = 2\psi(\pi)\psi'(\pi) = 0$, contradiction.

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  • $\begingroup$ can't one relax this assumption that X and Y have identical distribution? (excluding a delta-function distribution, I would think that the independence of X and Y is sufficient) $\endgroup$ – Carlo Beenakker Jan 9 '16 at 16:27
  • $\begingroup$ For non-identical distribution, see the link in Carlo's comment. $\endgroup$ – Gerald Edgar Jan 9 '16 at 16:28

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