Page 327 - A First Course In Stochastic Models

322 ADVANCED RENEWAL THEORY

mutually independent. For any s > 0, let

P on (s) = P {the process is in the on-state at time s}

and

U(s) = P {the amount of time the process is in the on-state during [0, s]}.

Theorem 8.3.1 Suppose that the on-times and off-times have exponential distri-
butions with respective means 1/α and 1/β. Then, assuming that an on-time starts
at epoch 0,

β α −(α+β)s
P on (s) = + e , s ≥ 0 (8.3.1)
α + β α + β
and
∞ n n k
−β(s−x) β(s − x) −αx (αx)
P {U(s) ≤ x} = e 1 − e , 0 ≤ x < s.
n! k!
n=0 k=0
(8.3.2)
The distribution function P {U(s) ≤ x} has a mass of e −αs at x = s.

Proof Let P off (s) = P {the process is in the off-state at time s}. By considering
what may happen in the time interval (s, s+ s] with s small, it is straightforward
to derive the linear differential equation

P (s) = βP off (s) − αP on (s), s > 0.
′
on
′
Since P off (s) = 1−P on (s), we find P (s) = β−(α+β)P on (s), s > 0. The solution
on
of this equation is given by (8.3.1). The proof of (8.3.2) is more complicated. The
random variable U(s) is equal to s only if the first on-time exceeds s. Hence
P {U(s) ≤ x} has mass e −αs at x = s. Now fix 0 ≤ x < s. By conditioning on the
lengths of the first on-time and the first off-time, we obtain

x ∞
−αy −βu
P {U(s) ≤ x} = αe dy P {U(s − y − u) ≤ x − y}βe du.
0 0
Noting that P {U(s − y − u) ≤ x − y} = 1 if s − y − u ≤ x − y, we next obtain

P {U(s) ≤ x} = e −β(s−x) (1 − e −αx )
x s−x
−αy −βu
+ αe dy P {U(s − y − u) ≤ x − y}βe du.
0 0
Substituting this equation repeatedly into itself leads to the desired result (8.3.2).