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 ().

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 $\lambda E_$ is defined as a shifted distribution of the form

$\mu=e(\lambda E_)=e^(E_0+\lambda E_+\lambda^2 E_/2!+\ldots+\lambda^n E_/n!+\ldots).$

One has the following

Raikov's theorem on locally compact Abelian groups

Let $\mu$ be the Poisson distribution generated by the measure $\lambda E_$. 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.

References

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Characterization of probability distributions
Probability theorems
Theorems in statistics