Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

3.2. The Poisson Distribution 169

Integrating both side with respect tot,wehaveforsomeconstantcthat

logg(0,t)=−λt+c or g(0,t)=e−λtec.

Finally, using the boundary conditiong(0,0) = 1, we haveec= 1. Hence,

g(0,t)=e−λt. (3.2.3)

So the result holds fork=0.
For the remainder of the proof, assume that, forka nonnegative integer,g(k, t)=
e−λt(λt)k/k!. By induction, the proof follows if we can show that the result holds
forg(k+1,t). Another reasonable boundary condition isg(k+1,0) = 0. Consider
g(k+1,t+h). In order to havek+ 1 occurrences in (0,t+h] either there arek+1
occurrences in (0,t] and no occurrences in (t, t+h]ortherearekoccurrences in
(0,t] and one occurrence in (t, t+h]. Because these events are disjoint we have by
the independence of Axiom 3 that

g(k+1,t+h)=g(k+1,t)[1−λh+o(h)] +g(k, t)[λh+o(h)],

that is,

g(k+1,t+h)−g(k+1,t)

h

=−λg(k+1,t)+g(k, t)λ+[g(k+1,t)+g(k, t)]

o(h)

h

.

Lettingh→0 and using the value ofg(k, t), we obtain the differential equation

d

dt

g(k+1,t)=−λg(k+1,t)+λe−λt[(λt)k/k!].

This is a linear differential equation of first order. Appealing to a theorem in
differential equations, its solution is

e

R

λdtg(k+1,t)=

∫

e

R

λdtλe−λt[(λt)k/k!]dt+c.

Using the boundary conditiong(k+1,0) = 0 and carrying out the integration, we

obtain

g(k+1,t)=e−λt[(λt)k+1/(k+1)!]

Therefore,Xthas a Poisson distribution with parameterλt.

LetXhave a Poisson distribution with parameterλ.ThemgfofXis given by

M(t)=

∑∞

x=0

etxp(x)=

∑∞

x=0

etx

λxe−λ

x!

= e−λ

∑∞

x=0

(λet)x

x!

= e−λeλe

t

=eλ(e

t−1)