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).

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_xthe degenerate distribution concentrated at x\in X. Let. The Poisson distribution generated by the measure is defined as a shifted distribution of the form One has the following

Raikov's theorem on locally compact Abelian groups

Let \mu be the Poisson distribution generated by the measure. Suppose that, with. 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.