Page 25 - A First Course In Stochastic Models

16 THE POISSON PROCESS AND RELATED PROCESSES

∞

proving the first assertion. Since E(U) = P {U > u} du for any non-negative
0
random variable U, the second assertion follows from (1.1.7) and the identity
(p + q + 1)! 1 p q
y (1 − y) dy = 1, p, q = 0, 1, . . . .
p!q! 0
The right-hand side of (1.1.7) can be given the following interpretation. Let
U 1 , . . . , U n be n independent random variables that are uniformly distributed on
the interval (0, t). Then the right-hand side of (1.1.7) also represents the probability
that the smallest kth among U 1 , . . . , U n is less than or equal to x. This is expressed
more generally in Theorem 1.1.5.

Theorem 1.1.5 For any t > 0 and n = 1, 2, . . . ,

P {S 1 ≤ x 1 , . . . , S n ≤ x n | N(t) = n} = P {U (1) ≤ x 1 , . . . , U (n) ≤ x n },

where U (k) denotes the smallest kth among n independent random variables
U 1 , . . . , U n that are uniformly distributed over the interval (0, t).

The proof of this theorem proceeds along the same lines as that of Lemma 1.1.4.
In other words, given the occurrence of n arrivals in (0, t), the n arrival epochs
are statistically indistinguishable from n independent observations taken from the
uniform distribution on (0, t). Thus Poisson arrivals occur completely randomly
in time.

Example 1.1.5 A waiting-time problem

In the harbour of Amsterdam a ferry leaves every T minutes to cross the North
Sea canal, where T is fixed. Passengers arrive according to a Poisson process with
rate λ. The ferry has ample capacity. What is the expected total waiting time of all
passengers joining a given crossing? The answer is

1 2
E(total waiting time) = λT . (1.1.9)
2
To prove this, consider the first crossing of the ferry. The random variable N(T )
denotes the number of passengers joining this crossing and the random variable S k
represents the arrival epoch of the kth passenger. By conditioning, we find

E(total waiting time)
∞

= E(total waiting time | N(T ) = n)P {N(T ) = n}
n=0