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?
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$\begingroup$ From this equation you can conclude that the variance is zero and hence that $X$ is (almost surely) constant. $\endgroup$– user35593Commented May 24, 2016 at 7:23
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$\begingroup$ And what if there is no variance ;-)? $\endgroup$– GagarCommented May 24, 2016 at 7:36
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3$\begingroup$ I don't see how to prove the variance is finite. The issue is whether functions of the form $e^{ict}$ are the only characteristic functions satisfying $\phi(t)^2=\phi(2t)$ for all $t$. It seems likely. $\endgroup$– Brendan McKayCommented May 24, 2016 at 7:48
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$\begingroup$ The fact that the variable is positive should play a crucial role, otherwise one may take a Cauchy random variable (which even does not have an expectation), and there exists solutions which even are not stable random variables (see Feller, An Introduction to Probability Theory and Its Applications Volume 2, chapter XVII.3 (f)) $\endgroup$– GagarCommented May 24, 2016 at 7:57
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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.
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