5.2The Poisson Random Variable 155
Now, given a total ofn+mevents, because each one of these events is independently
type 1 with probabilityp, it follows that the conditional probability that there are exactly
ntype 1 events (andmtype 2 events) is the probability that a binomial (n+m,p) random
variable is equal ton. Consequently,
P{N 1 =n,N 2 =m}=
(n+m)!
n!m!
pn(1−p)me−λ
λn+m
(n+m)!
=e−λp
(λp)n
n!
e−λ(1−p)
(λ(1−p))m
m!
(5.2.4)
The probability mass function ofN 1 is thus
P{N 1 =n}=
∑∞
m= 0
P{N 1 =n,N 2 =m}
=e−λp
(λp)n
n!
∑∞
m= 0
e−λ(1−p)
(λ(1−p))m
m!
=e−λp
(λp)n
n!
(5.2.5)
Similarly,
P{N 2 =m}=
∑∞
n= 0
P{N 1 =n,N 2 =m}=e−λ(1−p)
(λ(1−p))m
m!
(5.2.6)
It now follows from Equations 5.2.4, 5.2.5, and 5.2.6, thatN 1 andN 2 are independent
Poisson random variables with respective meansλpandλ(1−p).
The preceding result generalizes when each of the Poisson number of events can be
classified into any ofrcategories, to yield the following important property of the Poisson
distribution:If each of a Poisson number of events having meanλis independently classified as
being of one of the types1,...,r, with respective probabilities p 1 ,...,pr,
∑r
i= 1 pi=^1 , then
the numbers of type1,...,r events are independent Poisson random variables with respective
meansλp 1 ,...,λpr.
5.2.1 Computing the Poisson Distribution Function
IfXis Poisson with meanλ, then
P{X=i+ 1 }
P{X=i}
=
e−λλi+^1 /(i+1)!
e−λλi/i!
=
λ
i+ 1
(5.2.7)