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PROBAILITY AND STOCHASTIC PROCESSES
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PROBAILITY AND STOCHASTIC PROCESSES
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PROBABILITY AND STOCHASTIC PROCESS 818. On Poisson process
We now discuss one of the most important process, called Poisson process.
Definition 8.1 (Counting process). A counting process is a stochastic process {X(t), t ≥ 0}
with values that are non-negative, integer, and non-decreasing.
There are two ways to define the Poisson process. One way is given as follows: A non-negative,
integer-valued, non-decreasing stochastic process {X(t) : t ≥ 0} is said to be Poisson process
with intensity λ > 0, if the following hypotheses hold:
a) X(0) = 0 P-a.s.
b) It has independent and stationary increments.
c) It has unit jumps i.e., for any infinitesimal h > 0,
P(∆X(h) := X(t + h) − X(t) = 1) = λh + o(h)
P(∆X(h) ≥ 2) = o(h)
where o(h) is the function of h such that o(h)
h → 0 as h → 0.
Another way (constructive way) to define the Poisson process is as follows:
Definition 8.2 (Poison Process). Let (τi )i≥1 be a sequence of independent exponential ran-
dom variables with parameter λ i.e.,
(
λe−λy , y ≥ 0
fτi (y) =
0, otherwise
Pn
and Tn = i=1 τi . The process N (t) : t ≥ 0 defined by
X
N (t) = 1{t≥Tn }
n≥1
is called a Poisson process with intensity λ.
• The Poisson process is therefore a counting process: it counts the number of random
times (Tn ) which occur between 0 and t, where Tn − Tn−1 n≥1 is an i.i.d. sequence of
exponential random variables.
• The arrival times T1 , T2 , · · · have Gamma(n, λ) distribution, i.e., the probability density
function of Tn is given by
λn tn−1 −λt
fTn (t) = e , t > 0.
(n − 1)!
In particular, for n = 1, 2, · · · , we have
n n
E Tn = , Var(Tn ) = 2 .
λ λ
The following properties of the Poisson process can be easily deduced:
a) For any t > 0, N (t) is almost surely finite. To show it, we proceed as follows. Let
Ω1 = {ω ∈ Ω : Tnn → λ1 }. Then by strong law of large number, Tnn → λ1 with probability
1. So, P(Ω1 ) = 1. Thus, for any ω ∈ Ω1 , Tn (ω) → ∞ and so, for all ω ∈ Ω1 , there exists
n0 (ω) ≥ 1 such that for all n ≥ n0 (ω), Tn (ω) > t. So,
P N (t) < ∞ = P(Ω1 ) = 1.
b) For any ω, the sample path t 7→ N (t, ω) is piecewise constant and increases by
jumps of size 1.
c) For a.s. ω, the sample path is right continuous with left limit (cádlág) such
that N (t−) = N (t) for any t > 0.
, 82 A. K. MAJEE
d) (N (t)) is continuous in probability:
P
∀t > 0, N (s) → N (t) (s → t) .
e) For any t > 0, N (t) follows a Poisson distribution with parameter λt:
(λt)n
P N (t) = n = e−λt
(n ∈ N) .
n!
Indeed, P(N (t) = 0) = P(T1 > t) = e−λt . For n ≥ 1,
P(N (t) = n) = P(Tn ≤ t < Tn+1 = Tn + τn+1 )
Z t t
λn sn−1 −λs −λ(t−s)
Z
= P(Tn = s)P(τn+1 > t − s) ds = e e ds
0 0 (n − 1)!
(λt)n
= e−λt .
n!
One can easily check that
E[N (t)] = λt = Var(N (t)).
f) The characteristic function of N (t) is given by
φN (t) (u) = exp λt(eiu − 1)
∀u ∈ R.
To see this,
∞ ∞ ∞
X X (λt)n X (λteiu )n
E[eiuN (t) ] = eiun P(N (t) = n) = eiun e−λt = e−λt
n! n!
n=0 n=0 n=0
−λt λteiu iu
=e e = exp λt(e − 1) .
g) N (t) has independent increments: for any t1 < t2 < · · · < tn ,
N (tn ) − N (tn−1 ), · · · , N (t2 ) − N (t1 ), N (t1 )
are independent random variables. Moreover, the increments of N are ho-
mogeneous: for any t > s, N (t) − N (s) has the same distribution as N (t − s).
Remark 8.1. Since N (t) has independent increments, it is a Markov process.
Example 8.1. Suppose that incoming calls in a call centre arrive according to a Poisson process
with intensity of 30 calls per hour.
a) What is the probability that no call received in a 5-minute period?
b) What is the probability that more than 12 calls are received in a 30-minute intervals?
Solution: Let N (t) denotes the no. of incoming calls in t minutes. Then N (t) is a Poisson
process with intensity λ = 12 .
a) Now the probability that no call received in a 5-minute period is equal to
5
P(N (5) = 0) = e− 2 .
b) Let p be the probability that more than 12 calls are received in a 30-minute intervals.
Then
∞ 1 k ∞
X
−30× 12 30 × 2 −15
X (15)k
p = P(N (30) ≥ 13) = e =e .
k! k!
k=13 k=13
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