Page 320 - A First Course In Stochastic Models

ASYMPTOTIC EXPANSIONS 315

must be required that the function a(x) is directly Riemann integrable. Direct Rie-
mann integrability can be characterized in several ways. A convenient definition is
the following one. A function a(x) defined on [0, ∞) is said to be directly Riemann
∞
integrable when a(x) is almost everywhere continuous and n=1 n < ∞, where
a
a n is the supremum of |a(x)| on the interval [n − 1, n). A sufficient condition
for a function a(x) to be directly Riemann integrable is that it can be written as
a finite sum of monotone, integrable functions. This condition suffices for most
applications.

Theorem 8.2.2 (key renewal theorem) Assume F(x) has a probability density
f (x). For a given function a(t) that is bounded on finite intervals, let the function
Z(t) be defined by the renewal equation

t
Z(t) = a(t) + Z(t − x)f (x) dx, t ≥ 0.
0
Suppose that a(t) is directly Riemann integrable. Then
1 ∞
lim Z(t) = a(x) dx.
t→∞ µ 1 0

The proof of this theorem is demanding and will not be given. The interested reader
is referred to Feller (1971). Next we derive a number of useful results from the
key renewal theorem.

2
Theorem 8.2.3 Suppose F(x) is non-arithmetic with µ 2 = E(X ) < ∞. Then
1
t µ 2

lim M(t) − = 2 − 1, (8.2.4)
t→∞ µ 1 2µ
1

t t 2 µ 2 µ 2 µ 3
lim M(x) dx − + − 1 t = 2 − , (8.2.5)
t→∞ 2µ 1 2µ 2 4µ 3 6µ 2
0 1 1 1
3
provided that µ 3 = E(X ) < ∞.
1
Proof The asymptotic result M(t)/t → 1/µ 1 as t → ∞ suggests that, for some
constant c, M(t) ≈ t/µ 1 + c for t large. Let us therefore define the function
Z 0 (t) by
t
Z 0 (t) = M(t) − , t ≥ 0.
µ 1
Assuming for ease that F(x) has a density f (x), we easily deduce from (8.1.3)
that

t
Z 0 (t) = a(t) + Z 0 (t − x)f (x) dx, t ≥ 0, (8.2.6)
0