Page 314 - A First Course In Stochastic Models

THE RENEWAL FUNCTION 309

Theorem 8.1.1 Assume that the probability distribution function F(x) of the inte-
roccurrence times has a probability density f (x). Then the renewal function M(t)
satisfies the integral equation
t
M(t) = F(t) + M(t − x)f (x) dx, t ≥ 0. (8.1.3)
0
This integral equation has a unique solution that is bounded on finite intervals.
Proof The proof of (8.1.3) is instructive. Fix t > 0. To compute E[N(t)], we
condition on the time of the first renewal and use that the process probabilistically
starts over after each renewal. Under the condition that X 1 = x, the random variable
N(t) is distributed as 1+N(t −x) when 0 ≤ x ≤ t and N(t) is 0 otherwise. Hence,
by conditioning upon X 1 , we find

∞
t
E[N(t)] = E[N(t) | X 1 = x]f (x) dx = E[1 + N(t − x)]f (x) dx,
0 0
which gives (8.1.3). To prove that the equation (8.1.3) has a unique solution,
t
suppose that H(t) = F(t) + H(t − x)f (x) dx, t ≥ 0 for a function H(t)
0
that is bounded on finite intervals. We substitute this equation repeatedly into itself
and use the convolution formula
t
F n (t) = F(t − x)f n−1 (x) dx,
0
where f k (x) denotes the probability density of F k (x). This gives
n t

H(t) = F k (t) + H(t − x)f n (x) dx, n = 1, 2, . . . . (8.1.4)
0
k=1
Fix now t > 0. Since H(x) is bounded on [0, t], the second term on the right-hand
side of (8.1.4) is bounded by cF n (t) for some c > 0. Since M(t) < ∞, we have
∞

F n (t) → 0 as n → ∞. By letting n → ∞ in (8.1.4), we find H(t) = k=1 F k (t)
showing that H(t) = M(t).
Theorem 8.1.1 allows for the following important generalization.

Theorem 8.1.2 Assume that F(x) has a probability density f (x). Let a(x) be a
given, integrable function that is bounded on finite intervals. Suppose the function
Z(t), t ≥ 0, is defined by the integral equation
t
Z(t) = a(t) + Z(t − x)f (x) dx, t ≥ 0. (8.1.5)
0
Then this equation has a unique solution that is bounded on finite intervals. The
solution is given by