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.

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 [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 shifted distribution of the form

${\displaystyle \mu =e(\lambda E_{x_{0}})=e^{-\lambda }(E_{0}+\lambda E_{x_{0}}+\lambda ^{2}E_{2x_{0}}/2!+\ldots +\lambda ^{n}E_{nx_{0}}/n!+\ldots ).}$ 

One has the following

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 ${\displaystyle \mu _{j}\in M^{1}(X)}$ . If ${\displaystyle x_{0}}$  is either an infinite order element, or has order 2, then ${\displaystyle \mu _{j}}$  is also a Poisson's distribution. In the case of ${\displaystyle x_{0}}$  being an element of finite order ${\displaystyle n\neq 2}$ , ${\displaystyle \mu _{j}}$  can fail to be a Poisson's distribution.