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).

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 $n\neq 2$, $\mu _{j}$ can fail to be a Poisson's distribution.