Sums of independent Poisson random variables are Poisson random variables. Let X and
Y be independent Poisson random variables with parameters λ1 and λ2 , respectively.
Define λ = λ1 + λ2 and Z = X + Y . Claim that Z is a Poisson random variable with
parameter λ. Why?
pZ (z) = P (Z = z)
z
X
=
P (X = j & Y = z − j)
=
=
=
j=0
z
X
j=0
z
X
j=0
z
X
j=0
z
X
so X + Y = z
P (X = j)P (Y = z − j)
since X and Y are independent
e−λ1 λj1 e−λ2 λz−j
2
j! (z − j)!
1
e−λ1 λj1 e−λ2 λz−j
2
j!(z − j)!
z!
e−λ1 λj1 e−λ2 λz−j
2
multiply and divide by z!
j!(z
−
j)!
z!
j=0
z −λ1 j −λ2 z−j
X
z e λ1 e λ2
=
using the form of binominal coefficients
j
z!
j=0
z e−λ X z j z−j
=
λλ
factoring out z! and e−λ1 e−λ2 = e−λ1 −λ2 = e−λ
z! j=0 j 1 2
=
e−λ
(λ1 + λ2 )z
z!
e−λ λz
=
z!
=
using binomial expansion (in reverse)
−λ z
So altogether we showed that pZ (z) = e z!λ . So Z = X + Y is Poisson, and we just sum the
parameters.
What about a sum of more than two independent Poisson random variables? Say X1 ,
X2 , X3 are independent Poissons? Then (X1 + X2 ) is Poisson, and then we can add on X3
and still have a Poisson random variable. So X1 + X2 + X3 is a Poisson random variable.
Works in general. Once we know it for two, we can keep adding more and more of them. So
if X1 , X2 , . . . , Xn are independent Poisson random variables with parameters λ1 , λ2 , . . . , λn ,
then X1 + X2 + . . . + Xn is a Poisson random variable too, with parameter λ1 + λ2 + . . . + λn .
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Lesson 20

solutions 1