Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

168 Some Special Distributions

that is,p(x) satisfies the conditions of being a pmf of a discrete type of random
variable. A random variable that has a pmf of the formp(x)issaidtohavea
Poisson distributionwith parameterλ, and any suchp(x) is called aPoisson
pmfwith parameterλ.
As the following remark shows, Poisson distributions occur in many areas of
applications.

Remark 3.2.1.Consider a process that counts the number of certain events occur-
ring over an interval of time; for example, the number of tornados that touch down
in Michigan per year, the number of cars entering a parking lot between 8:00 and
12:00 on a weekday, the number of car accidents at a busy intersection per week,
the number of typographical errors per page of a manuscript, and the number of
blemishes on a manufactured car door. As in the third and fourth examples, the
occurrences need not be over time. It is convenient, though, to use the time rep-
resentation in the following derivation. LetXtdenote the number of occurrences
of such a process over the interval (0,t]. The range ofXtis the set of nonnegative
integers{ 0 , 1 , 2 ,...}. For a nonnegative integerkand a real numbert>0, denote
the pmf ofXtbyP(Xt=k)=g(k, t). Under the following three axioms, we next
show thatXthas a Poisson distribution.

1.g(1,h)=λh+o(h), for a constantλ>0.

2.

∑∞

t=2g(t, h)=o(h).

3. The number of occurrences in nonoverlapping intervals are independent of one
   another.

Here theo(h) notation means thato(h)/h→0ash→0. For instance,h^2 =o(h)
ando(h)+o(h)=o(h). Note that the first two axioms imply that in a small
interval of timeh, either one or no events occur and that the probability of one
event occurring is proportional toh.
By the method of induction, we now show that the distribution ofXtis Poisson
with parameterλt.First,weobtaing(k, t)fork= 0. Note that the boundary
conditiong(0,0) = 1 is reasonable. No events occur in time (0,t+h] if and only if
no events occur in (0,t] and no events occur in (t, t+h]. By Axioms (1) and (2),
the probability that no events occur in the interval (0,h]is1−λh+o(h). Further,
the intervals (0,t]and(t, t+h] do not overlap. Hence, by Axiom (3) we have

g(0,t+h)=g(0,t)[1−λh+o(h)]. (3.2.2)

That is,

g(0,t+h)−g(0,t)

h

=−λg(0,t)+

g(0,t)o(h)

h

→−λg(0,t), ash→ 0.

Thus,g(0,t) satisfies the differential equation

dtg(0,t)

g(0,t)

=−λ