Page 321 - A First Course In Stochastic Models

316 ADVANCED RENEWAL THEORY

where
t 1 t
a(t) = F(t) − + (t − x)f (x) dx, t ≥ 0.
µ 1 µ 1 0
t ∞ ∞
Writing (t − x)f (x) dx = (t − x)f (x) dx − (t − x)f (x) dx, we find
0 0 t
1 ∞
a(t) = −[1 − F(t)] + (x − t)f (x) dx.
µ 1 t
This shows that a(t) is the sum of two monotone, integrable functions. We have
1
∞ ∞ ∞ ∞

a(t) dt = − [1 − F(t)] dt + dt (x − t)f (x) dx
0 0 µ 1 0 t
1 ∞ x
= −µ 1 + f (x) dx (x − t) dt
µ 1 0 0
1 ∞ 1 2
= −µ 1 + x f (x) dx
µ 1 0 2
µ 2
= −µ 1 + .
2µ 1
By applying the key renewal theorem to (8.2.6), the result (8.2.4) follows. The
proof of (8.2.5) proceeds along the same lines. The relation (8.2.4) suggests that,
for some constant c,

t t 2 µ 2
M(x) dx ≈ + t − 1 + c for t large.
0 2µ 1 2µ 2 1
To determine the constant c, define the function

t 2
t µ 2
Z 1 (t) = M(x) dx − + t 2 − 1 , t ≥ 0.
0 2µ 1 2µ 1
By integrating both sides of the equation (8.1.3) over t and interchanging the order
of integration, we get the following renewal equation for the function U(x) =
x
0 M(t) dt:
t t
U(t) = F(x) dx + U(t − x)f (x) dx, t ≥ 0.
0 0
From this renewal equation, we obtain after some algebra

t
Z 1 (t) = a(t) + Z 1 (t − x)f (x) dx, t ≥ 0, (8.2.7)
0