Page 12 - A First Course In Stochastic Models

THE POISSON PROCESS 3

k−1 j
−λt (λt)
P {S k ≤ t} = 1 − e , t ≥ 0. (1.1.1)
j!
j=0
k k−1 −λt
The Erlang (k, λ) distribution has the probability density λ t e /(k − 1)!.
Theorem 1.1.1 For any t > 0,

(λt) k
−λt
P {N(t) = k} = e , k = 0, 1, . . . . (1.1.2)
k!
That is, N(t) is Poisson distributed with mean λt.

Proof The proof is based on the simple but useful observation that the number
of arrivals up to time t is k or more if and only if the kth arrival occurs before or
at time t. Hence

P {N(t) ≥ k} = P {S k ≤ t}
k−1 j
−λt (λt)
= 1 − e .
j!
j=0
The result next follows from P {N(t) = k} = P {N(t) ≥ k} − P {N(t) ≥ k + 1}.

The following remark is made. To memorize the expression (1.1.1) for the dis-
tribution function of the Erlang (k, λ) distribution it is easiest to reason in reverse
order: since the number of arrivals in (0, t) is Poisson distributed with mean λt
and the kth arrival time S k is at or before t only if k or more arrivals occur in
−λt j
∞
(0, t), it follows that P {S k ≤ t} = e (λt) /j!.
j=k
The memoryless property of the Poisson process
Next we discuss the memoryless property that is characteristic for the Poisson
process. For any t ≥ 0, define the random variable γ t as
γ t = the waiting time from epoch t until the next arrival.

The following theorem is of utmost importance.

Theorem 1.1.2 For any t ≥ 0, the random variable γ t has the same exponential
distribution with mean 1/λ. That is,
−λx
P {γ t ≤ x} = 1 − e , x ≥ 0, (1.1.3)
independently of t.