Page 32 - A First Course In Stochastic Models

NON-STATIONARY POISSON PROCESSES 23

as s → 0. Subtracting p k (s) from both sides of this equation and dividing by
s, we obtain

′
p (s) = −λ(t + s)[p k (s) − p k−1 (s)], k = 1, 2, . . . .
k
For k = 0, we have p (s) = −λ(t + s)p 0 (s). The boundary conditions p 0 (0) = 1
′
0
and p k (0) = 0 for k ≥ 1 apply. It is well known from the theory of differential
equations that the solution of the first-order differential equation
y (s) + a(s)y(s) = b(s), s ≥ 0
′
is given by

s
−A(s) A(x) −A(s)
y(s) = e b(x)e dx + ce
0
s
for some constant c, where A(s) = a(x) dx. The constant c is determined by a
0
boundary condition on y(0). This gives after some algebra
−[M(s+t)−M(t)]
p 0 (s) = e , s ≥ 0.
′
By induction the expression for p k (s) next follows from p (s) + λ(t + s)p k (s) =
k
λ(t + s)p k−1 (s). We omit the details.
Note that M(t) represents the expected number of arrivals up to time t.

Example 1.3.1 A canal touring problem

A canal touring boat departs for a tour through the canals of Amsterdam every T
minutes with T fixed. Potential customers pass the point of departure according to
a Poisson process with rate λ. A potential customer who sees that the boat leaves
t minutes from now joins the boat with probability e −µt for 0 ≤ t ≤ T . Which
stochastic process describes the arrival of customers who actually join the boat
(assume that the boat has ample capacity)? The answer is that this process is a
non-stationary Poisson process with arrival rate function λ(t), where

λ(t) = λe −µ(T −t) for 0 ≤ t < T and λ(t) = λ(t − T ) for t ≥ T.
This follows directly from the observation that for t small

P {a customer joins the boat in (t, t + t)}
= (λ t) × e −µ(T −t) + o( t), 0 ≤ t < T.

Thus, by Theorem 1.3.1, the number of passengers joining a given tour is Poisson
T −µT
distributed with mean λ(t) dt = (λ/µ)(1 − e ).
0