$X \sim U[-1,1]$, is there $Y$ independent of $X$ s.t $X+Y$ and $\frac{Y}{2}$ have the same distribution?

Question

I have thought that if there exist such $Y$ then we can look at characteristic functions of $X + Y$ and $\frac{Y}{2}$ to get:

$$ \phi_{X+Y}(t)= \phi_X(t)\phi_Y(t) = \phi_{\frac{Y}{2}}(t) \\ \frac{\exp(it) - \exp(-it)}{2it} \phi_Y(t) = \phi_\frac{Y}{2}(t) \\ \frac{\sin(t)}{t}\phi_Y(t)=\phi_\frac{Y}{2}(t)$$

But I don't see how to proceed with this approach. Is there anything I could do with that?

Answer 1

Show by induction that $\phi_Y(2^{-k}\pi) = 0$ for $k \in \mathbb{N_+}$ and from it follows that $\phi$ is not continous at $t = 0$. Contradiction.

Answer 2

Hint: $\text{Var}(X+Y) = \text{Var}(Y/2)$. Simplify and use the fact that a variance must be nonnegative.