Page 318 - A First Course In Stochastic Models

ASYMPTOTIC EXPANSIONS 313

n
n+1

∗
∗
∞ f (s) − f (s)
e −st P {N(t) = n} dt = , n = 0, 1, . . .
0 s
Hence for fixed n the probability P {N(t) = n} can be calculated by numerical
Laplace inversion. Another interesting question is how to compute the probability
lim P {N(t + D) − N(t) = n}, n = 0, 1, . . .
t→∞
for a given constant D. Denote this probability by a n (D). Using the limiting dis-
tribution P {γ t ≤ x} from Theorem 8.2.5 in the next subsection and the relations
(E.4) and (E.6) in Appendix E, it is not difficult for the reader to verify that

∗
∞ 1 1 − f (s)
e −sx a 0 (x) dx = − 2 (8.1.8)
0 s µ 1 s
∞ 1 − f (s) [f (s)] − [f (s)]
∗
∗
∗
n−1 n
e −sx a n (x) dx = , n ≥ 1. (8.1.9)
0 µ 1 s s
This is a useful result. For example, the probability distribution {a n (D)} gives
the limiting distribution of the number of busy servers in the infinite-server queue
with renewal input and deterministic service times (GI /D/∞ queue). This result
is easily proved. Since each customer gets immediately assigned a free server upon
arrival and the service time of each customer equals the constant D, the only
customers present at time t + D are those who have arrived in (t, t + D].

8.2 ASYMPTOTIC EXPANSIONS

In Section 2.2 we proved a law of large numbers for the process {N(t)}:
N(t) 1
lim = with probability 1. (8.2.1)
t→∞ t µ 1

The proof was elementary. It is tempting to conclude from (8.2.1) that M(t)/t →
1/µ 1 as t → ∞. Although this result is correct, it cannot be directly concluded
from (8.2.1). The reason is that the random variable N(t)/t need not be bounded in
t. For a sequence of unbounded random variables Y n it is not necessarily true that
lim n→∞ E(Y n ) = E(Y) when Y n converges to Y with probability 1 as n → ∞.
Consider the counterexample in which Y n = 0 with probability 1−1/n and Y n = n
with probability 1/n. Then E(Y n ) = 1 for all n, whereas Y n converges to 0 with
probability 1.

Theorem 8.2.1 (elementary renewal theorem)
M(t) 1
lim = .
t→∞ t µ 1