Can the sum of two i.i.d. random variables be uniformly distributed? Let $U$ be a uniform random variable on the interval $[0,1]$.
It is exceedingly unlikely that $U$ can be written as a sum 
$U = X + Y$ where $X$ and $Y$ are independent identically distributed random variables.
Consider, for example, the discrete analogue where it is very clear that no such
decomposition is possible.
Furthermore, the moment generating function (mgf) of $U$ is given by $M(x):=\dfrac{e^x -1}{x}$. So that the mgf for $X$ and $Y$ would have to be the curious looking $\sqrt{M(x)}$.
Questions:


*

*Is there a simple proof that $U$ does not equal $X+Y$ as above?

*A necessary condition for the existence of $X$ and $Y$ is that the power 
series for $\sqrt{M(x)}$ possess exclusively positive coefficients.  Does it?
Thanks.
 A: Given
$$
(a_0 + a_1 x + a_2 x^2 + \ldots)^2 = 1 + \frac{x}{2} + \frac{x^2}{3!} + \frac{x^3}{4!} + \ldots
$$
we can solve (recursively) for the $a_i$.  We know
$$
a_0^2 + (2a_0 a_1)x + (2a_0 a_2 + a_1^2)x^2 + (2a_0 a_3 + 2a_1 a_2) x^3 + (2a_0 a_4+ 2a_1 a_3 + a_2^2)x^4
$$
$$
 = 1 + \frac{1}{2}x + \frac{1}{6}x^2 + \frac{1}{24}x^3 + \frac{1}{120}x^4 +\ldots,
$$
so by equating coefficients, we can start by solving for $a_0$, then use that to get $a_1$, then use that to get $a_2$, etc. We actually have a choice at $a_0 = \pm 1$, but we want positive coefficients so we take $a_0 = 1$. The first several values are then (starting at $a_0 = 1$):
$$
1, \frac{1}{4}, \frac{5}{96}, \frac{1}{128}, \frac{79}{92160}, \ldots
$$
At $a_{13}$ my computer gives a negative value:
$$
a_{13} = \frac{-20287103}{43878270659198976000},
$$
so I believe this answers Question 2.
I should add that I don't know enough about moments to be certain that the OP's condition (all positive coefficients) is indeed necessary.  However, I am certain that the condition is false.
A: Here's what I get...
The characteristic function of random variable $X$, is defined by
$$c_t=E\left[e^{i t X}\right].$$
Thus, since $$U = X + Y$$ then
$$c_{X+Y}= -\frac{\left(e^{i t}-1\right)^2}{t^2}$$.
Now, we need to invert the above. Using the "Inversion Theorem", the characteristic function $c_{X+Y}$ uniquely determines the pdf by
$$f x=\frac{1}{{2 \pi }}\int_{-\infty }^{\infty } c_{X+Y} e^{-i t x} \, dt$$.
Inverting the above, we get
$$f(x)_{X+Y}=\frac{1}{2} (\left| x-2\right| -2 \left| x-1\right| +y).$$
And, clearly, this is not uniformly distributed (i.e., $U \neq X + Y$).
I am unclear about what you mean RE part (b)
