I have been reading up on characteristic functions for random variables and came across the following question:
"Let $X$ and $Y$ be independent and identically distributed random
variables. Show that $X$ and $Y$ cannot be chosen so that $X-Y$ has
the uniform distribution on the interval $(-1, 1)$."
I have approached the problem as follows:
Since both $X$ and $Y$ have the same distribution, they must also have the same characteristic function, call it $\phi(t)$.
Thusly, $X-Y$ has characteristic function $-\phi(t)^2$.
I know that the characteristic function associated with the uniform distribution is given by $$\phi_{U(-1, 1)} = \cases{\frac{e^{it} – e^{-it}}{2it}&if $t\neq$0\\\\1 & if $t = 0$ }$$
This is equivalent to
$$\phi_{U(-1, 1)} = \cases{\frac{\sin(t)}{t}&if $t\neq$0\\\\1 & if $t = 0$ }$$
But from here, I am stuck trying to show that $-\phi(t)^2$ cannot equal $\phi_{U(-1, 1)}$.
How can I go about showing this?
Best Answer
Your characteristic function computation isn't quite right: if $X$ and $Y$ are independent with characteristic function $\phi$, then $$\phi_{X-Y}(t)=\mathbb{E}[e^{it(X-Y)}]=\mathbb{E}[e^{itX}]\mathbb{E}[e^{-itY}]=\phi(t)\phi(-t)=|\phi(t)|^2$$ since $\phi(-t)=\overline{\phi(t)}$.
In particular, $\phi_{X-Y}(t)$ is always non-negative, whereas the characteristic function of the uniform distribution takes on both positive and negative values.