Page 13 - A First Course In Stochastic Models

4 THE POISSON PROCESS AND RELATED PROCESSES

Proof Fix t ≥ 0. The event {γ t > x} occurs only if one of the mutually exclusive
events {X 1 > t + x}, {X 1 ≤ t, X 1 + X 2 > t + x}, {X 1 + X 2 ≤ t, X 1 + X 2 + X 3 >
t + x}, . . . occurs. This gives
∞

P {γ t > x} = P {X 1 > t + x} + P {S n ≤ t, S n+1 > t + x}.
n=1
By conditioning on S n , we find
t y n−1
P {S n ≤ t, S n+1 > t + x} = P {S n+1 > t + x | S n = y}λ n e −λy dy
0 (n − 1)!
t n−1
n y −λy
= P {X n+1 > t + x − y}λ e dy.
0 (n − 1)!
This gives

∞ t y n−1

λ
P {γ t > x} = e −λ(t+x) + e −λ(t+x−y) n e −λy dy
0 (n − 1)!
n=1
t
= e −λ(t+x) + e −λ(t+x−y) λ dy
0
λt
= e −λ(t+x) + e −λ(t+x) (e − 1) = e −λx ,
proving the desired result. The interchange of the sum and the integral in the second
equality is justified by the non-negativity of the terms involved.

The theorem states that at each point in time the waiting time until the next arrival
has the same exponential distribution as the original interarrival time, regardless
of how long ago the last arrival occurred. The Poisson process is the only renewal
process having this memoryless property. How much time is elapsed since the last
arrival gives no information about how long to wait until the next arrival. This
remarkable property does not hold for general arrival processes (e.g. consider the
case of constant interarrival times). The lack of memory of the Poisson process
explains the mathematical tractability of the process. In specific applications the
analysis does not require a state variable keeping track of the time elapsed since the
last arrival. The memoryless property of the Poisson process is of course closely
related to the lack of memory of the exponential distribution.
Theorem 1.1.1 states that the number of arrivals in the time interval (0, s) is
Poisson distributed with mean λs. More generally, the number of arrivals in any
time interval of length s has a Poisson distribution with mean λs. That is,

(λs) k
−λs
P {N(u + s) − N(u) = k} = e , k = 0, 1, . . . , (1.1.4)
k!
independently of u. To prove this result, note that by Theorem 1.1.2 the time
elapsed between a given epoch u and the epoch of the first arrival after u has the