Let $X$ and $Y$ be independent random variables with the same distribution over the reals. When does $X+Y\sim2X$ hold?
Edit: By "$\sim$" I intend to express that the left and right hand sides have the same distributions. If there's a better notation I'm open to other options.
Partial Progress
- This amounts to solving the functional equation $$\int_{-\infty}^\infty f(x-y)f(y)dy=\frac12f\left(\frac x2\right),$$ where $f$ would then be a probability density function corresponding to both $X$ and $Y$. In theory, restating the problem this way reduces a probability question of unknown difficulty to a simple calculus question, but nothing in my toolbox has been able to make heads or tails of it.
- Any normal distribution has this property of course, so there are plenty of solutions.
- Whether or not they have the same distribution, $X+Y$ and $2X$ have identical finite moments.
- The equation certainly doesn't always hold. Between uniform, beta, exponential, and gamma distributions nothing satisfies $X+Y\sim2X$.
- Informally, considering the fact that repeated summation would tend toward a normal distribution and the fact that multiplying by a constant does not push a distribution toward normality, I'm inclined to believe that the only solutions to this are those which are already normal.
- Update: Inspired by Kavi Rama Murthy in the comments, I took a look into infinite divisibility. If $X+Y\sim2X$, with a few integration tricks we can get $\sum^nX_i\sim nX$ if the $X_i$ are i.i.d with the same distribution as $X$. It isn't too much extra work to find that $X\sim\sum^n\left(\frac{X_i}n\right)$, hence $X$ must be infinitely divisible.
- Update: However, infinite divisibility is definitely not the only requirement. As mentioned before, the gamma distribution readily fails to satisfy $X+Y\sim2X$. Additionally, the students t distribution fails, and so would any stable but not strictly stable distribution.
- Update: Not all is lost. Every strictly stable distribution with a finite mean works, and the Cauchy distribution works despite its lack of moments. I'm not sure yet, but it might be the case that all strictly stable distributions qualify.