Let $X$ be a **positive** real-valued random variable. Let $Y$ be an independent copy of $X$ and assume that the equality $X+Y=2X$ holds in distribution. Does this imply that $X$ is constant?



Distributional equation X+Y=2X
Yes, it does. As the random variable is positive, consider the Laplace transform $f(\lambda)=E \exp(-\lambda X)$ (instead of the Fourier one). 
Then, one has $f^2(\lambda)= f(2\lambda)$. 


Taking the logarithm, one gets $2\log f(\lambda)=\log f(2\lambda)$, and for the function $g(\lambda)=\frac{\log f(\lambda)}{\lambda}$ this implies $g(2\lambda)=g(\lambda)$. 


Finally, pass to the limit as $\lambda\to\infty$. Then $g(\lambda)$ tends to the essential infimum of the law of $X$. On the other hand, $g(e^s)$ is $\log 2$-periodic; as it has a limit as $s\to\infty$, this is a constant function. Hence, $g$ is a constant function, and thus $X$ is constant almost surely.


