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 function of X:
FX(x) | = | P(X≤x)=P(X1+X2≤x)=x∑i=0P(X1+X2=i) | ||
= | x∑i=0i∑j=0P(X1=j and X2=i-j)=x∑i=0i∑j=0P(X1=j)P(X2=i-j) | |||
= | x∑i=0i∑j=0e-λ1λj1j!e-λ2λi-j2(i-j)!=x∑i=0i∑j=0e-λi!(ij)λj1λi-j2 | |||
= | x∑i=0e-λi!i∑j=0(ij)λj1λi-j2=x∑i=0e-λi!(λ1+λ2)i=x∑i=0e-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 variables.
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 |