Raikov's theorem

Suppose that a random variable ${\displaystyle \xi }$ has a Poisson distribution and admits a decomposition as a sum ${\displaystyle \xi =\xi _{1}+\xi _{2}}$ of two independent random variables. Then the distribution of each summand is a shifted Poisson distribution.

Raikov's theorem is similar to Cramér’s decomposition theorem. The latter result claims that if a sum of two independent random variables has a 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 [ru]).

Let ${\displaystyle X}$ be a locally compact Abelian group. Denote by ${\displaystyle M^{1}(X)}$ the convolution semigroup of probability distributions on ${\displaystyle X}$, and by ${\displaystyle E_{x}}$the degenerate distribution concentrated at ${\displaystyle x\in X}$. Let ${\displaystyle x_{0}\in X,\lambda >0}$.

The Poisson distribution generated by the measure ${\displaystyle \lambda E_{x_{0}}}$ is defined as a distribution of the form

Theorem^[2] Let ${\displaystyle \mu }$ be the Poisson distribution generated by the measure ${\displaystyle \lambda E_{x_{0}}}$. Suppose that ${\displaystyle \mu =\mu _{1}*\mu _{2}}$, with ${\displaystyle \mu _{j}\in M^{1}(X)}$. Then each of ${\displaystyle \mu _{j}}$ is a shift of a Poisson distribution if and only if ${\displaystyle x_{0}}$ is either an infinite-order element or has order 2.