Page 338 - A First Course In Stochastic Models

RUIN PROBABILITIES 333

case, the asymptotic expansion (8.4.12) for the heavy-tailed case is typically bad for
x-values of interest. It takes very large x before the asymptotic expansion (8.4.12)
applies. In practice one has to use numerical Laplace inversion to calculate the tail
probabilities Q(x) in the heavy-tailed case; see Appendix F.
We give no rigorous proof for the result (8.4.12), but we do make it plausible.
To do so, we first establish the relation

∞
n
Q(x) = (1 − ρ)ρ [1 − B n,e (x)], x ≥ 0, (8.4.13)
n=0
where B 0,e (x) = 1 for all x ≥ 0 and B n,e (x) is the n-fold convolution of B e (x)
with itself for n ≥ 1. The formula (8.4.13) does not require any condition on the
distribution function B(x). To prove (8.4.13), denote the Laplace transform of Q(x)
∞ −sx
∗ Q(x) dx. Taking Laplace transforms of both sides of (8.4.5),
by Q (s) = e
0
we find
ρ ρb (s)
∗
∗ e ∗ ∗
Q (s) = − + ρQ (s)b (s),
e
s s
where b (s) is the Laplace transform of the derivative b e (x) = (1/µ)[1 − B(x)]
∗
e
of the equilibrium excess distribution function B e (x). This gives
∞
∗
ρ − ρb (s) ρ − ρb (s)
∗
∗ e e n ∗ n
Q (s) = = ρ [b (s)]
e
s[1 − ρb (s)] s
∗
e n=0
∞
1 n [b (s)] n
∗
e
= − (1 − ρ) ρ . (8.4.14)
s s
n=0
n
It is left to the reader to verify that [b (s)] /s is the Laplace transform of B n,e (x);
∗
e
see also relation (E.12) in Appendix E. Inversion of (8.4.14) yields
∞ ∞

n n
Q(x) = 1 − (1 − ρ) ρ B n,e (x) = (1 − ρ) ρ [1 − B n,e (x)],
n=0 n=0
proving (8.4.13). Next the expansion (8.4.12) can be made plausible. Assume that
B(x) is subexponential. If in addition an integrability condition is imposed on
B(x) to exclude pathological cases, it can be shown that the equilibrium excess
distribution function B e (x) is subexponential as well. Then 1 − B n,e (x) ∼ n[1 −
B e (x)] as x → ∞ for all n and thus
∞
n
Q(x) ∼ (1 − ρ)ρ n[1 − B e (x)] as x → ∞,
n=0
∞ n
which yields (8.4.12) by noting that n=0 (1 − ρ)ρ n = ρ/(1 − ρ).