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]
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.
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 .
Let
X
M1(X)
X
Ex
x\inX
x0\inX,λ>0
The Poisson distribution generated by the measure
λ
E | |
x0 |
\mu=e(λ
E | |
x0 |
)=e-λ(E0+λ
E | |
x0 |
+λ2
E | |
2x0 |
/2!+\ldots+λn
E | |
nx0 |
/n!+\ldots).
One has the following
Let
\mu
λ
E | |
x0 |
\mu=\mu1*\mu2
\muj\inM1(X)
x0
\muj
x0
n\ne2
\muj