Page 66 - A First Course In Stochastic Models

POISSON ARRIVALS SEE TIME AVERAGES 57

(b) With probability 1, the long-run fraction of arrivals who find the system in the
set B of states equals the long-run fraction of time the system is in the set B of
states. That is, with probability 1,
n t
1 1
lim I k (B) = lim I B (u) du.
n→∞ n t→∞ t 0
k=1
Proof See Wolff (1982).
It is remarkable in Theorem 2.4.1 that E[number of arrivals in (0, t) finding the
system in the set B] is equal to λ × E[amount of time in (0, t) that the system is
in the set B], although there is dependency between the arrivals in (0, t) and the
evolution of the state of the system during (0, t). This result is characteristic for
the Poisson process.
The property ‘Poisson arrivals see time averages’ is usually abbreviated as
PASTA. Theorem 2.4.1 has a useful corollary when it is assumed that the continu-
ous-time process {X(t)} is a regenerative process whose cycle length has a finite
positive mean. Define the random variables T B and N B by
T B = amount of time the process {X(t)} is in the set B of states
during one cycle,
N B = number of arrivals during one cycle who find the process {X(t)}
in the set of B states.
The following corollary will be very useful in the algorithmic analysis of queueing
systems in Chapter 9.
Corollary 2.4.2 If the arrival process {N(t)} is a Poisson process with rate λ, then

E(N B ) = λE(T B ).

Proof Denote by the random variables T and N the length of one cycle and the
number of arrivals during one cycle. Then, by Theorem 2.2.3,
1 t E(T B )
lim I B (u) du = with probability 1
t→∞ t 0 E(T )
and
n
1 E(N B )
lim I k (B) = with probability 1.
n→∞ n E(N)
k=1
It now follows from part (b) of Theorem 2.4.1 that E(N B )/E(N) = E(T B )/E(T ).
Thus the corollary follows if we can verify that E(N)/E(T ) = λ. To do so, note
that the regeneration epochs for the process {X(t)} are also regeneration epochs
for the Poisson arrival process. Thus, by the renewal-reward theorem, the long-
run average number of arrivals per time unit equals E(N)/E(T ), showing that
E(N)/E(T ) = λ.