I have problems with the following question:
Let $X$ be the random variable such that $X\sim U[-1,1]$. Does there exist a random variable $Y$, independent to $X$, such that $X+Y\sim 2Y$?
I thought that uniqueness of characteristic function would help. So, we ask if there exists a variable Y with characteristic function $\phi(t)$, such that $\frac{\sin(t)}{t}\phi(t)=\phi(2t)$. (And, of course $\phi(0)=1)$
I would be thankful for help.
Edit:
Proof by contradiction.
I will show that zeros of $\phi$ have $0$ as their accumulation point:
Notice that if $t\in(-\pi,\pi)$ we have: if $\phi(2t)=0$ then $\phi(t)=0 (*) $.
We just have to know if there exists any zero of $\phi$ in $(-\pi,\pi)$. $\phi(\pi)=\frac{2}{\pi}\phi(\frac{\pi}{2})$ and $\phi(-\pi)=\frac{-2}{\pi}\phi(\frac{-\pi}{2})$. So if $\phi(\frac{\pi}{2})\neq0$ and $\phi(-\frac{\pi}{2})\neq0$, then either $\phi(\pi)$ and $\phi(-\pi)$ or $\phi(\frac{\pi}{2})$ and $\phi(-\frac{\pi}{2})$ have opposite signs, so there exists zero of $\phi$ in $(-\pi,\pi)$. We know that $\phi(0)=1$, hence from $(*)$ there is a sequence of zeros of $\phi$ which converges to 0. Then, because $\phi$ is continuous $\phi(0)=0$.
Is it correct?