Page 205 - Elements of Distribution Theory

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

6.5 Counting Processes 191

Extending this argument shows that, for any m = 1, 2,...,
Pr[N(s) = 1] Pr[N(s/m) = 1]
= m .
Pr[N(s) = 0] Pr[N(s/m) = 0]
Since this holds for any m,
Pr[N(s) = 1] Pr[N(s/m) = 1]
= lim m .
Pr[N(s) = 0] m→∞ Pr[N(s/m) = 0]
Note that, from (6.7),
lim Pr[N(s/m) = 0] = 1
m→0
and, from (PP5), writing t for s/m,

lim mPr[N(s/m) = 1] = s lim Pr[N(t) = 1]/t = sλ.
m→∞ t→0
It follows that

Pr[N(s) = 1] = λPr[N(s) = 0] = sλ exp{−λs}.
In a similar manner, we may write

m 2 m−2
Pr[N(s) = 2] = Pr[N(s/m) = 1] Pr[N(s/m) = 0]
2
+ mPr[N(s/m) = 2]Pr[N(s/m) = 0] m−1
for any m = 1, 2,.... Using the expressions for Pr[N(s) = 0] and Pr[N(s) = 1] derived
above, we have that, for any m = 1, 2,...,
2
Pr[N(s) = 2] = (1 − 1/m)(sλ) exp{−sλ}/2 + mPr[N(s/m) = 2] exp{−(1 − 1/m)sλ}.
Letting m →∞ it follows that
2
(sλ) exp{−sλ}
Pr[N(s) = 2] = .
2!
The general case follows along similar lines:

m n m−n
Pr[N(s) = n] = Pr[N(s/m) = 1] Pr[N(s/m) = 0] + R
n
m(m − 1) ··· (m − n + 1) n
= (λs) exp{−λs}+ R
m n
where R is a finite sum, each term of which includes a factor of the form Pr[N(s/m) = j]
for some j ≥ 2. Letting m →∞, and using (PP5), yields the result.

Distribution of the interarrival times
Consider a Poisson process {N(t): t ≥ 0} and let T 1 denote the time until the first arrival
occurs. Since, for any t > 0,
X 1 > t if and only if N(t) = 0,

it follows that
Pr(X 1 ≤ t) = 1 − Pr(N(t) = 0) = 1 − exp{−λt}.