2

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:

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

  2. 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.

mlc
  • 5,608
user2052
  • 2,467

2 Answers2

0

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.

mathmandan
  • 2,114
  • 1
  • 19
  • 19
0

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)

PiE
  • 1,062