Raikov's theorem

Raikov’s theorem, named for Russian mathematician Dmitrii Abramovich Raikov, is a result in probability theory. It is well known that if each of two independent random variables ξ1 and ξ2 has a Poisson distribution, then their sum ξ=ξ1+ξ2 has a Poisson distribution as well. It turns out that the converse is also valid.

Statement of the theorem
Suppose that a random variable ξ has Poisson's distribution and admits a decomposition as a sum ξ=ξ1+ξ2 of two independent random variables. Then the distribution of each summand is a shifted Poisson's distribution.

Comment
Raikov's theorem is similar to Cramér’s decomposition theorem. The latter result claims that if a sum of two independent random variables has normal distribution, then each summand is normally distributed as well. It was also proved by Yu.V.Linnik that a convolution of normal distribution and Poisson's distribution possesses a similar property (Linnik's theorem on convolution of normal distribution and Poisson's distribution).

An extension to locally compact Abelian groups
Let $$X$$ be a locally compact Abelian group. Denote by $$M^1(X)$$ the convolution semigroup of probability distributions on $$X$$, and by $$E_x$$the degenerate distribution concentrated at $$x\in X$$. Let $$x_0\in X, \lambda>0$$.

The Poisson distribution generated by the measure $$\lambda E_{x_0}$$ is defined as a shifted distribution of the form

$$\mu=e(\lambda E_{x_0})=e^{-\lambda}(E_0+\lambda E_{x_0}+\lambda^2 E_{2x_0}/2!+\ldots+\lambda^n E_{nx_0}/n!+\ldots).$$

One has the following

Raikov's theorem on locally compact Abelian groups
Let $$\mu$$ be the Poisson distribution generated by the measure $$\lambda E_{x_0}$$. Suppose that $$\mu=\mu_1*\mu_2$$, with $$\mu_j\in M^1(X)$$. If $$x_0$$ is either an infinite order element, or has order 2, then $$\mu_j$$ is also a Poisson's distribution. In the case of $$x_0$$ being an element of finite order $$n\ne 2$$, $$\mu_j$$ can fail to be a Poisson's distribution.