Page 335 - A First Course In Stochastic Models

330 ADVANCED RENEWAL THEORY

Assumption 8.4.1 There are positive numbers a and b such that the complemen-
tary distribution function 1 − B(y) ≤ ae −by for all y sufficiently large.

This assumption excludes probability distributions with long tails like the log-
normal distribution. The assumption implies that the number s 0 defined by

∞
!

sy
s 0 = sup s | e {1 − B(y)} dy < ∞
0
exists and is positive (possibly s 0 = ∞). In addition to Assumption 8.4.1 we make
the technical assumption
λ ∞ sy
lim e {1 − B(y)} dy > 1.
s→s 0 σ
0
Then it is readily verified that the equation
λ ∞ δy
e {1 − B(y)} dy = 1 (8.4.7)
σ 0
has a unique solution δ on the interval (0, s 0 ). Next we convert the defective
renewal equation (8.4.6) into a standard renewal equation. This enables us to apply
the key renewal theorem to obtain the asymptotic behaviour of Q(x). Let

λ
∗ δx
h (x) = e {1 − B(x)}, x ≥ 0.
σ
∗
Then h (x), x ≥ 0 is a probability density with finite mean. Multiplying both sides
of equation (8.4.6) by e δx and defining the functions
δx
δx
∗
∗
Q (x) = e Q(x) and a (x) = e a(x), x ≥ 0,
we find that the defective renewal function (8.4.6) is equivalent to
x

∗ ∗ ∗ ∗
Q (x) = a (x) + Q (x − y)h (y) dy, x ≥ 0. (8.4.8)
0
This is a standard renewal equation to which we can apply the key renewal theorem.
The function a (x) is directly Riemann integrable as can be shown by verifying
∗
that |a (x)| ≤ ce −(a−δ)x as x → ∞ for finite constants c > 0 and a > δ. Using
∗
∞
definition (8.4.7) for δ and the relation 0 {1 − B(y)} dy = µ, we find
∞ ∞ λ ∞

∗ δx
a (x) dx = e {1 − B(y)} dy dx
0 0 σ x
λ ∞ y δx
= {1 − B(y)} e dx dy
σ 0 0
λ ∞ " δy # 1 − ρ
= e − 1 {1 − B(y)} dy = ,
δσ 0 δ