Page 19 - A First Course In Stochastic Models

10 THE POISSON PROCESS AND RELATED PROCESSES

D, then the only customers present at time t are those customers who have arrived
in (t − D, t]. Hence the number of customers present at time t is Poisson dis-
tributed with mean λD proving (1.1.6) for the special case of deterministic service
times. Next consider the case that the service time takes on finitely many values
D 1 , . . . , D s with respective probabilities p 1 , . . . , p s . Mark the customers with the
same fixed service time D k as type k customers. Then, by Theorem 1.1.3, type k
customers arrive according to a Poisson process with rate λp k . Moreover the var-
ious Poisson arrival processes of the marked customers are independent of each
other. Fix now t with t > max k D k . By the above argument, the number of type k
customers present at time t is Poisson distributed with mean (λp k )D k . Thus, by the
independence property of the split Poisson process, the total number of customers
present at time t has a Poisson distribution with mean

s

λp k D k = λµ.
k=1
This proves (1.1.6) for the case that the service time has a discrete distribution
with finite support. Any service-time distribution can be arbitrarily closely approx-
imated by a discrete distribution with finite support. This makes plausible that the
insensitivity result (1.1.6) holds for any service-time distribution.

Rigorous derivation
The differential equation approach can be used to give a rigorous proof of (1.1.6).
Assuming that there are no customers present at epoch 0, define for any t > 0
p j (t) = P {there are j busy servers at time t}, j = 0, 1, . . . .

Consider now p j (t + t) for t small. The event that there are j servers busy at
time t + t can occur in the following mutually exclusive ways:

(a) no arrival occurs in (0, t) and there are j busy servers at time t + t due to
arrivals in ( t, t + t),

(b) one arrival occurs in (0, t), the service of the first arrival is completed before
time t + t and there are j busy servers at time t + t due to arrivals in
( t, t + t),
(c) one arrival occurs in (0, t), the service of the first arrival is not completed
before time t + t and there are j − 1 other busy servers at time t + t due
to arrivals in ( t, t + t),
(d) two or more arrivals occur in (0, t) and j servers are busy at time t + t.

Let B(t) denote the probability distribution of the service time of a customer.
Then, since a probability distribution function has at most a countable number of