Page 204 - Elements of Distribution Theory

P1: JZP
052184472Xc06 CUNY148/Severini May 24, 2005 2:41

190 Stochastic Processes

Proof. First suppose that conditions (PP1), (PP2), and (PP3) are satisfied so that the
process is a Poisson process. Then, for any t ≥ 0, N(t) has a Poisson process with mean
λt. Hence,

lim Pr[N(t) = 1]/t = lim λ exp(−λt) = λ
t→0 t→0
and
∞
j j−1
Pr[N(t) ≥ 2]/t = λ t exp(−λt)/j!
j=2
∞
j
≤ λ (λt) exp(−λt)/j!
j=1
≤ λ[1 − exp(−λt)].
It follows that

lim sup Pr[N(t) ≥ 2]/t ≤ lim sup λ[1 − exp(−λt)] = 0.
t→0 t→0
Hence, (PP5) holds. Condition (PP4) follows directly from (PP3).
Now assume that (PP1), (PP2), (PP4), and (PP5) hold. Note that (PP5) implies that

lim{Pr[N(t) = 0] − 1}/t =− lim{Pr[N(t) = 1] + Pr[N(t) ≥ 2]}/t =−λ. (6.7)
t→0 t→0
Using (PP2) and (PP4),

Pr[N(s + t) = 0] = Pr[N(s) = 0, N(s + t) − N(s) = 0] = Pr[N(t) = 0]Pr[N(s) = 0].
Hence,
Pr[N(s + t) = 0] − Pr[N(s) = 0] Pr[N(t) = 0] − 1
= Pr[N(s) = 0] .
t t
By (6.7),
d
Pr[N(s) = 0] =−λPr[N(s) = 0],
ds
that is,
d
log Pr[N(s) = 0] =−λ.
ds
Solving this differential equation yields
Pr[N(s) = 0] = exp{−λs}, s ≥ 0.

Now consider Pr[N(s + t) = 1]. Note that

Pr[N(s + t) = 1] = Pr[N(s) = 0]Pr[N(t) = 1] + Pr[N(s) = 1]Pr[N(t) = 0].
Hence,
Pr[N(s + t) = 1] Pr[N(t) = 1] Pr[N(s) = 1]
= + .
Pr[N(s + t) = 0] Pr[N(t) = 0] Pr[N(s) = 0]