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.^[1][2][3]

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 [ru]).

An extension to locally compact Abelian groups

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

Raikov's theorem on locally compact Abelian groups

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.

References