Page 24 - A First Course In Stochastic Models

THE POISSON PROCESS 15

mean λ j β j for j = 1, . . . , N. In particular,

the long-run average number of back orders outstanding at base j
∞ k
(λ j β j )
≈ (k − S j )e −λ j β j , j = 1, . . . , N.
k!
k=S j
This expression and the expression for W 0 enables us to calculate the total average
number of outstanding back orders at the bases for a given assignment (S 0 , S 1 , . . . ,
S N ). Next, by some search procedure, the optimal values of S 0 , S 1 , . . . , S N can be
calculated.

1.1.4 The Poisson Process and the Uniform Distribution

In any small time interval of the same length the occurrence of a Poisson arrival is
equally likely. In other words, Poisson arrivals occur completely randomly in time.
To make this statement more precise, we relate the Poisson process to the uniform
distribution.

Lemma 1.1.4 For any t > 0 and n = 1, 2, . . . ,

x j
n

n
x n−j
P {S k ≤ x | N(t) = n} = 1 − (1.1.7)
j t t
j=k
for 0 ≤ x ≤ t and 1 ≤ k ≤ n. In particular, for any 1 ≤ k ≤ n,
kt t
E(S k | N(t) = n) = and E(S k − S k−1 | N(t) = n) = . (1.1.8)
n + 1 n + 1
Proof Since the Poisson process has independent and stationary increments,

P {S k ≤ x, N(t) = n}
P {S k ≤ x | N(t) = n} =
P {N(t) = n}
P {N(x) ≥ k, N(t) = n}
=
P {N(t) = n}
n
1
= P {N(x) = j, N(t) − N(x) = n − j}
P {N(t) = n}
j=k
n j n−j
1 −λx (λx) −λ(t−x) [λ(t − x)]
= −λt n e e
e (λt) /n! j! (n − j)!
j=k
n
x j
n

x n−j
= 1 − ,
j t t
j=k