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−λλn+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= 0P{N 1 =n,N 2 =m}=e−λp(λp)n
n!∑∞m= 0e−λ(1−p)(λ(1−p))m
m!=e−λp(λp)n
n!(5.2.5)Similarly,
P{N 2 =m}=∑∞n= 0P{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)