Raikov's theorem

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] Contents 1 Statement of the theorem 2 Comment 3 An extension to locally compact Abelian groups 4 Raikov's theorem on locally compact Abelian groups 5 References 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 xin 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 nneq 2} , {displaystyle mu _{j}} can fail to be a Poisson's distribution.

References ^ D. Raikov (1937). "On the decomposition of Poisson laws". Dokl. Acad. Sci. URSS. 14: 9–11. ^ Rukhin A. L. (1970). "Certain statistical and probability problems on groups". Trudy Mat. Inst. Steklov. 111: 52–109. ^ Linnik, Yu. V., Ostrovskii, I. V. (1977). Decomposition of random variables and vectors. Providence, R. I.: Translations of Mathematical Monographs, 48. American Mathematical Society. Categories: Characterization of probability distributionsProbability theoremsTheorems in statistics

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