properties of Poisson random variables


Proposition 1.

If X1,X2 are independent Poisson random variables with parameters λ1,λ2, then X1+X2 is a Poisson random variable with parameter λ1+λ2.

Proof.

Let X:=X1+X2 and λ:=λ1+λ2, let us calculate the distribution functionMathworldPlanetmath of X:

FX(x) = P(Xx)=P(X1+X2x)=i=0xP(X1+X2=i)
= i=0xj=0iP(X1=j and X2=i-j)=i=0xj=0iP(X1=j)P(X2=i-j)
= i=0xj=0ie-λ1λ1jj!e-λ2λ2i-j(i-j)!=i=0xj=0ie-λi!(ij)λ1jλ2i-j
= i=0xe-λi!j=0i(ij)λ1jλ2i-j=i=0xe-λi!(λ1+λ2)i=i=0xe-λi!λi.

As a result, X is a Poisson random variable with parameter λ. Notice that in the fifth equation, we used the assumption that X1 and X2 are independent. ∎

As a corollary, any sum of independent Poisson random variables is Poisson, with parameter the sum of the parameters from the independent random variablesMathworldPlanetmath.

Proposition 2.

A Poisson random variable is infinitely divisible.

Proof.

Let X be a Poisson random variable with parameter λ. Let n be any positive integer. Let X1,,Xn be independent identically distributed Poisson random variables with parameter λn. Then the sum of these random variables is easily seen to be Poisson, with parameter λ, and is therefore identically distributed as X. ∎

Title properties of Poisson random variables
Canonical name PropertiesOfPoissonRandomVariables
Date of creation 2013-03-22 18:50:55
Last modified on 2013-03-22 18:50:55
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 4
Author CWoo (3771)
Entry type Derivation
Classification msc 62E15