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 [math]\displaystyle{ X }[/math] be a locally compact Abelian group. Denote by [math]\displaystyle{ M^1(X) }[/math] the convolution semigroup of probability distributions on [math]\displaystyle{ X }[/math], and by [math]\displaystyle{ E_x }[/math]the degenerate distribution concentrated at [math]\displaystyle{ x\in X }[/math]. Let [math]\displaystyle{ x_0\in X, \lambda\gt 0 }[/math].

The Poisson distribution generated by the measure [math]\displaystyle{ \lambda E_{x_0} }[/math] is defined as a shifted distribution of the form

[math]\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). }[/math]

One has the following

Raikov's theorem on locally compact Abelian groups

Let [math]\displaystyle{ \mu }[/math] be the Poisson distribution generated by the measure [math]\displaystyle{ \lambda E_{x_0} }[/math]. Suppose that [math]\displaystyle{ \mu=\mu_1*\mu_2 }[/math], with [math]\displaystyle{ \mu_j\in M^1(X) }[/math]. If [math]\displaystyle{ x_0 }[/math] is either an infinite order element, or has order 2, then [math]\displaystyle{ \mu_j }[/math] is also a Poisson's distribution. In the case of [math]\displaystyle{ x_0 }[/math] being an element of finite order [math]\displaystyle{ n\ne 2 }[/math], [math]\displaystyle{ \mu_j }[/math] can fail to be a Poisson's distribution.