Page 466 - A First Course In Stochastic Models

E. LAPLACE TRANSFORM THEORY 461

obtained by numerical inversion of its Laplace transform f (s). Numerical inver-
∗
sion methods that perform well for probability functions f (x) are discussed in
Appendix F.

Example E.1 The Erlang distribution
Suppose that X 1 , . . . , X n are independent random variables having a common
exponential distribution with mean 1/µ. Then X 1 + · · · + X n has the probability
density
n n−1 −µx
µ x e
, x ≥ 0,
(n − 1)!
that is, X 1 + · · · + X n is Erlang (n, µ) distributed. To prove this, note that the
Laplace transform of the probability density f n (x) of X 1 + · · · + X n is given by

f (s) = E[e −s(X 1 +···+X n ) ]
∗
n
= E(e −sX 1 ) · · · E(e −sX n ).
∞ −sx −µx
Noting that E(e −sX i ) = e µe dx = µ/(s+µ) for all s with Re(s) > −µ,
0
it follows that
µ n
∗
f (s) = n .
n
(s + µ)
∗
Using (E.10), the inversion of f (s) shows that f n (x) is indeed given by the Erlang
n
(n, µ) density.
Example E.2 The renewal function
Consider a renewal process for which the probability distribution function B(x) of
the interoccurrence times of the events has a probability density b(x). The renewal
function M(x) is defined by

∞

M(x) = B (x), (E.11)
n
n=1
where B n (x) is the probability distribution function of X 1 +· · ·+X n . That is, B n (x)
is the n-fold convolution of B(x) with itself. The distribution function B n (x) has
a probability density b n (x). Since b n (x) is the density of X 1 + · · · + X n ,

∞
−sx −s(X 1 + ··· +X n ) n
∗
e b n (x) dx = E e = b (s) ,
0
∞ −sx
∗
where b (s) = e b(x) dx. By (E.4),
0
n
∞ b (s)
∗
e −sx B n (x) dx = . (E.12)
0 s