Almost everybody is familiar with Cramer's theorem: a sum $X+Y$ of of independent random variables is normal if and only if both $X$ and $Y$ are normal. Are there any other classes of distributions which can replace normality in the above statement?
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3$\begingroup$ Another example is Raikov's theorem: en.wikipedia.org/wiki/Raikov%27s_theorem $\endgroup$– LuciaCommented Apr 29, 2014 at 21:28
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1$\begingroup$ See also infinitely divisible distributions $\endgroup$– Nate EldredgeCommented Apr 29, 2014 at 22:05
3 Answers
As has been already stated, the analogous statement (I believe) is known to hold for stable distributions (I can't find a precise reference for this, but results along these lines are discussed in sections 6 and 7 of this survey article of E. Lukacs).
It is also worth mentioning that there is an longstanding problem of Mark Kac which seeks to extend the spirit of these results to a more general setting. Note that we may reformulate Cramer's theorem as follows:
Theorem (Cramer) Let $\sigma_1$ and $\sigma_2$ be positive functions such that: $$\int_{-\infty}^{\infty} e(-\xi x) d\sigma_1 (x) \times \int_{-\infty}^{\infty} e(-\xi x) d\sigma_2(x) = e^{-\xi^2/4}.$$ Then $\sigma_1(x)$ and $\sigma_2$ are Gaussians.
Seeking to generalize this, Kac's problem is the following:
(Kac's Problem) Let $\phi(x,y) = \frac{1}{ \frac{1}{x} +\frac{1}{y} -1} $ and assume that $$ \phi \left( \int_{-\infty}^{\infty} e(-\xi x) d\sigma_1 (x), \int_{-\infty}^{\infty} e(-\xi x) d \sigma_2(x)\right) = \frac{1}{1+\xi^2}.$$ Is it true that $\sigma_1(x) = \alpha_1 e^{-\beta_1 |x|}$ and $\sigma_2(x) = \alpha_2 e^{-\beta_2 |x|}$?
This is problem 178 from the Scottish book, where it is dated September 11, 1938.
Poisson Point process in $\mathbb{R}^d$ is an example. If you have two homogeneous Poisson point processes $\Phi_X$ and $\Phi_Y$ with intensities $\lambda$ and $\mu$, their superposition is also a Poisson process with intensity $\lambda+\mu$. In other words, the sum of number of points of two Poisson point processes in a compact set $A$, follows Poisson law with sum of their intensity.
For the other direction, Raikov's theorem says that sum of if the sum of two independent RVs follows Poisson then each one of them should be Poisson.
The general concept you're looking for is that of a stable distribution. See (e.g.) here for background on discrete stable distributions.
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4$\begingroup$ Stable certainly gives the if. Does it give the only if as well? $\endgroup$ Commented Apr 29, 2014 at 22:23