Stochastic orders of summands when sum has fixed distribution

Question

Suppose $X,X',Y,Y'$ are independent random variables. We know that $X+Y \overset{d}{=} X'+Y'\overset{d}=Uniform[0,1]$ and $X\prec X'$ in the sense that $P(X> x)\le P(X'>x)$ for any $x\in \mathbb{R}$. Is it true that $Y'\prec Y$?

Answer 1

This is an answer to the earlier version of the question where $X+Y$ was not required to have uniform distribution. This is false. If this implication is true then the following must also be true: $X,X',Y,Y'$ independent, $X+Y \overset {d} =X'+Y'$ and $X\overset {d}=X'$ implies $Y\overset {d}=Y'$. It is well known that there exist characteristic functions $\phi, \phi_1,\phi_2$ such that $\phi (t) \phi_1(t)=\phi (t) \phi_2(t)$ for all $t$ but $\phi_1 \neq \phi_2$. Such an example is available in Volume 2 of Feller's book. [ See Curiosities ii) in Section 2 of the chapter on Characteristic Functions]. Now just consider independent random variables $X,X',Y,Y'$ such that $X$ and $X'$ have characteristic function $\phi $, $Y$ has characteristic function $\phi_1$ and $Y'$ has characteristic function $\phi_2$ to get a contradiction.