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 (ru)).

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

$\mu =e(\lambda E_{x_0})=e^{-\lambda }(E_0+\lambda E_{x_0}+{\lambda }^2{E}_{2x_0}/2!+\dots +{\lambda }^n{E}_{n{x}_0}/n!+\dots ).$

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.