POISSON PROCESSES:

A Poisson process is a simple and widely used stochastic process for modeling the times at
which arrivals enter a system. It is a counting process; pure birth process; point process as
well as a renewal process.
We will focus on the process as a counting process.
Definition: A stochastic process {X(t), t ≥ 0} is said to be a counting process if X(t)
represents the total number of events that have occurred up to time t. X(t) must satisfy:

(i) X(t) ≥ 0

(ii) X(t) is integer valued.

(iii) If u < t, then X(u) ≤ X(t)

(iv) For u < t, X(t) − X(u) equals the number of events have occurred in the interval (u,t];
beyond time u up to and including time t.

Properties of counting process:

(a) A counting process is said to have stationary increments if the distribution of number
of events that occur in any interval of time depends only on the length of the time
interval; i.e the number of events X(t2 ) − X(t1 ) in (t1 , t2) has the same distribution as
the number of events X(t2 + u) − X(t1 + u) in (t1 + u, t2 + u).

(b) A counting process is said to have independent increments if the number of events that
occur in non overlapping (disjoint) time intervals are independent; i.e the number of
events that occur by time u, X(u) are independent of independent of number of events
that occur between times u and t, u < t; X(t) − X(u).

Definition:
A counting process {N (t); t ≥ 0} is sadi to be a Poisson process having rate λ, λ > 0; if

(i) N (0)=0

(ii) The process has independent and stationary increments

(iii) The number of events in any time interval of length t has distribution

e−λt (λt)n
Pr[N (t + u) − N (u)] = n = 0, 1, ...
n!

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 Inter-arrival and waiting times of Poisson process:
Let Xi denote the time period between the (i-1)-th and i-th events; the sequence {Xi ; i =
1, 2, ...} is called the sequence of inter-arrival times. These times are random variables and
hence we can define probability distributions for them.
Now for time X1 ; it is time it takes for the first event to occur; hence {X1 > t} iff no events
occur in time interval [0,t]. Thus

Pr[X1 > t] = Pr[N (t) = 0] = e−λt

F1 (t) = e−λt
f1 (t) = λe−λt

X1 is a random variable having exponential distribution with rate λ.

Next for X2 ;
X
∞
P (X2 > t) = P r(X2 > t|X1 = u)P (X1 = u)
u=0
and
(P r(X2 > t|X1 = u) = P [no events in (u, u + t]|X1 = u]
= P [no events in(u, u + t]] = e−λt
thus
X
∞
P (X2 > t) = e−λtP (X1 = u)
u=0
−λt
= e

X2 is a random variable having exponential distribution with rate λ and is independent of

X1 . Repeating the same argument with X3 , X4 , ... yields the following: Xi ; i = 1, 2, ... are
independent and identically distributed exponential random variables with rate λ.
Next let Sn be the waiting time until the n-th event occurs;
X
n
Sn = Xi ; n ≥ 1
i=1

. Since Sn is a sum of n i.i.d exponential random variables, it follows that it is a random

variable having gamma distribution with parameters λ and n.

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 Example 1:
The number of failures N (t), which occur in a computer network over the time interval [0, t)
can be described by a homogeneous Poisson process {N (t); t ≥ 0} with rate of one failure
every 4 hours.

(i) What is the probability of at most 1 failure in [0,8), at least 2 failures in [8,16) and at
most 1 failure in [16,24) (time unit:1 hr)?

(ii) What is the probability that the third failure occurs after 8 hours?

Solution

(i) Let λ be rate of failure per hour; λ = 0.25. Also let p be the required probability

p = P r[N (8) − N (0) ≤ 1; N (16) − N (8) ≥ 2; N (24) − N (16) ≤ 1]

= P r[N (8) ≤ 1; N (8) ≥ 2; N (8) ≤ 1] due to stationary increments
= P r[N (8) ≤ 1] × P r[N (8) ≥ 2] × P r[N (8) ≤ 1] due to independent increments

Now

P r[N (8) ≤ 1] = P r[N (8) = 0] + P r[N (8) = 1]

e−0.25×8(0.25 × 8)0 e−0.25×8(0.25 × 8)1

= +
0! 1!
= 0.1353 + 0.2707 = 0.406

and

P r[N (8) ≥ 2] = 1 − P r[N (8) ≤ 1] = 0.594

Thus
p = 0.406 × 0.594 × 0.406 = 0.098

(ii) Let S3 be the waiting until the third failure occurs; S3 ∼ gamma(3, 0.25).
Z ∞
0.253 2 −0.25s
P (S3 > 8) = s e ds = 0.6767
8 2!

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 Conditional distribution of arrival time of Poisson process:
Given N (t) = n; the n arrival times of the events, S1, S2 , .., Sn have the same distribution as
order statistics corresponding to n independent random variables uniformly distributed over
the interval(0,t); i.e
n!
f (s1 , s2 , .., sn|N (t) = n) =
tn
.
to obtain this conditional distribution, for 0 < s1 < s2 < ... < sn < t; the event S1 =
s1 ; S2 = s2 ... = Sn = sn ; N (t) = n is equivalent to the event that the first n+1 inter-arrival
times satisfy X1 = s1 , X2 = s2 − s1 ,...,Xn = sn − sn−1 , Xn+1 > t − sn ; hence

g(s1 , s2 , .., sn, n)

f (s1 , s2, .., sn|n) =
P r[N (t) = n]

λe−λs1 λe−λ(s2 −s1 ) .....λe−λ(sn−sn−1 )e−λ(t−sn )

= e−λt (λt)n
n!

n!
= ; 0 < s1 < s2 < ... < sn < t
tn
Further properties of Poisson process

(1) Splitting(Thinning) of Poisson process:

Let the events of a Poisson process {N (t); t ≥ 0} having rate λ be classified as being
either of type I or of type II. Further let an event of type I occur with probability p and
that of tpe II occur with probability (1-p). Also let N1 (t) and N2 (t) be the number of
type I and type II events occurring in [0,t); then N1 (t) and N2 (t) are Poisson proceesses
having rates λp and λ(1−p) respectively. These two Poisson processes are independent
of each other.
This concept can be extended to more than two types of events being observed; in this
case each process decomposed from overall process has rate intensity λpi , i = 1, 2, .., k
P
where ki=1 pi = 1

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 Example 2:
Customers demanding service at a central processing station arrive according to a Poisson
process of rate λ = 10 persons per hour. The demand of each customer, independent of
others, is classified as ”urgent” with probability pu = 0.1; ”high priority” with probability
ph = 0.2 and ”low priority” with probability pl = 0.7. What is the probability that 3
”urgent”, 6 ”high priority” and 12 ”low priority” demands arise in the first two units of
time?
Solution:
The rate intensity of process for customers with ”urgent” demand is λu = 10 × 0.1 = 1; the
rate intensity of process for customers with ”high priority” demand is λh = 10 × 0.2 = 2 and
the rate intensity of process for customers with ”low priority” demand is λl = 10 × 0.7 = 7
. Let N1 (t) be the number of customers with ”urgent” demand; N2 (t) be the number of
customers with ”high priority” demand and N3(t) be the number of customers with ”low
priority” demand.

Pr[N1(2) = 3, N2 (2) = 6, N3 (2) = 12] = Pr[N1(2) = 3]Pr[N2(2) = 6]Pr[N3 (2) = 12]

e−2×1 (2 × 1)3 e−2×2 (2 × 2)6

= ×
3! 6!

e−2×7 (2 × 7)12
× = 0.0019
12!

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(2) Merging of independent Poisson processes:

(a) Let {N1(t); t ≥ 0} and {N2 (t); t ≥ 0} be independent Poisson processes having
rates λ1 and λ2 respectively. Define N (t) = N1 (t) + N2 (t); i.e {N (t); t ≥ 0} is
a random process that combines the events that occur in the two independent
processes. It can easily be shown that N (t) is a Poisson process having rate
λ1 + λ2 .
(b) For two independent Poisson processes N1 (t) and N2 (t) with respective rates λ1
and λ2
P r[N1 (t) = j]P r[N2(t) = n − j]
P r[N1 (t) = j|N1 (t) + N2 (t) = n] =
P r[N1 (t) + N2 (t) = n]
e−λ1 t (λ1 t)j eλ2 t (λ2 t)n−j
j!
× (n−j)!
= e(λ1 +λ2 )t ((λ1 +λ2 )t)n
n!

n! e−λ1 t(λ1 t)j eλ2 t (λ2 t)n−j

=
j!(n − j)! e(λ1 +λ2 )t((λ1 + λ2 )t)n
 j  n−j
n! λ1 λ2
=
j!(n − j)! λ1 + λ2 λ1 + λ2

(c) Let {N1(t); t ≥ 0}, {N2(t); t ≥ 0},..., {Nk (t); t ≥ 0} be k independent Poisson
Pk
processes with respective rates λ1 ,λ2,...,λk . Define N (t) as N (t) = i=1 Ni (t);
P
then {N (t); t ≥ 0} is a Poisson process with rate ki=1 λi
P P
(d) Let N (t) = ki=1 Ni (t) with rate intensity λ = ki=1 λi ;

P r[N1(t) = n1, N2 (t) = n2 , .., Nk (t) = nk |N (t) = n]

P r[N1 (t) = n1 , N2(t) = n2, .., Nk (t) = nk ]

=
P r[N (t) = n]
 n1  n2  nk X
k
n! λ1 λ2 λk
= ... ni = n
n1 !n2!...nk ! λ λ λ i=1

the conditonal distribution of N1 (t), N2(t), ..., Nk(t) given N (t) is multinomial
λi
distribution with probabilities λ

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 (3) Order of events in independent Poisson processes:
Let {N1 (t); t ≥ 0} and {N2 (t); t ≥ 0} be independent Poisson processes having rates
λ1 and λ2 . The probability that n events occur in the first process before m events
occur in the second process is

X 
n+m−1
n+m−1

λ1
k 
λ2
n+m−1−k

k=n
k λ1 + λ2 λ1 + λ2

The probability we are interested in is the probability that among the first n + m − 1
events in the combined process, n or more events belong to the first process.

(i) Since the two processes are independent, all events that occur are independent of
each other.
(ii) There are only two possible outcomes: event of first process or event of second
process
(iii) The probability that an event from first process occurs before the second process
λ1
is λ1 +λ2
; which is equivalent to probability of ”success”
(iv) The number of events that occur is fixed; n + m − 1

The characteristics above defined a binomial process hence the probability is derived
using binomial distribution.
Alternatively, Let Sn1 denote the time of the n − th event of the first process and Sm
2

denote the time of the m − th event of the second process;

X
n+m−1   k  n+m−1−k
n+m−1 λ1 λ2
P r[Sn1 < 2
Sm ] =
k=n
k λ1 + λ2 λ1 + λ2

Example 3: A Poisson process {N (t); t ≥ 0} has a rate of λ = 6 per hour. The events that
occur are classified as type 1 and type 2. When an event occurs, there is a 0.3 probability
that it is a type 1 event and a probability 0.7 that it is a type 2 event. Assuming the event
types are independent of each other;

(i) Find the probability that the second event of type 1 occurs before the third event of
type 2.

(ii) Find the probability that not more than three events of type 1 and not fewer than five
events of type 2 occur in any gien three hour period.

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(iii) Find the probability that the total number of events that occur in any any seven hour
period are more than fifty-seven.

Solution:

(i) Let Sn1 denote the time of the n − th event of type 1 process and Sm
2
denote the time of
the m − th event of type 2 process; lambda1 = 6 × 0.3 = 1.8, λ2 = 6 × 0.7 = 4.2,n = 2
and m = 3 hence

4  
X k  4−k
4 1.8 4.2
P r[S21 < S32 ] =
k 6 6
k=2
     
4 2 2 4 3 1 4
= (0.3) (0.7) + (0.3) (0.7) + (0.3)4 (0.7)0
2 3 4
= 0.3483

(ii) Let N1 (t) and N2(t) be independent Poisson processes of type 1 and type 2 respectively.
For three hour period lambda1 = 6 × 0.3 × 3 = 5.4, λ2 = 6 × 0.7 × 3 = 12.6

P r[N1 (3) ≤ 3, N2 (3) ≥ 5] = P r[N1 (3) ≤ 3] × P r[N2 (3) ≥ 5]

(5.4)2 −5.4 (5.4)3 −5.4
P r[N1 (3) ≤ 3] = e−5.4 + 5.4e−5.4 + e + e
2 6
= 0.21329
X 4
(12.6)n
P r[N2 (3) ≥ 5] = 1 − e−12.6
n=0
n!
= 1 − 0.004979 = 0.99502
P r[N1 (3) ≤ 3] × P r[N2 (3) ≥ 5] = 0.21329 × 0.99502 = 0.21223

(iii) N (7) ∼ P o(6 × 7 = 42). We can approximate Poisson distribution using normal
distribution; hence N (7) ∼ N (42, 42). Hence
 
N (7) − 42 57 − 42
P r[N (7) > 57] = P r √ > √
42 42

= P r[Z > 2.31] = 0.0104

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 Compound Poisson process:
In a compound Poisson process, each arrival in an ordinary Poisson process comes with an
associated real-valued random variable that represents the value of the arrival in a sense.
These variables are independent and identically distributed, and are independent of the
underlying Poisson process. Our interest centers on the sum of the random variables for all
the arrivals up to a fixed time t which thus is a Poisson-distributed random sum of random
variables. Some examples include;

(i) The arrivals are customers at a store and each customer spends a random amount of
money.

(ii) The arrivals are visits to a website and each visitor spends a random amount of time
at the site.

(iii) The arrivals are failure times of a complex system and each failure requires a random
repair time.

Defintion:
Let {N (t); t ≥ 0} be a Poisson process with rate intensity λ. Further let {Yn ; n = 1, 2, ...} be
a sequence of real-valued random variables that are i.i.d. with p.d.f f (y) and are independent
of N (t). Then
N (t)
X
Z(t) = Yi
i=1

is a stochastic process and it is referred to as compound Poisson process. {Z(t); t ≥ 0} has

stationary and independent increments.
Mean and variance of Z(t):
Let E[Y ] = µ and Var[Y ] = σ 2 ;

E[Z(t)] = E [E[Z(t)|N (t)]] = E[N (t)]E[Y ]

= µλt
V ar[Z(t)] = E [V ar(Z(t)|N (t))] + V ar [[E[Z(t)|N (t)]] = E[N (t)]E[Y 2 ]
= (µ2 + σ 2)λt

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 Asymptotic distribution of Z(t):
As t increases; the mean of Poisson distribution increases. Thus by Central limit Theorem,
the distribution of Z(t) converges to the normal distribution.

lim Z(t) − E[Z(t)]

p ∼ N (0, 1)
t→∞ V ar[Z(t)]

Example 4:
Suppose that health claims are filed with a health insurer according to a Poisson process
with rate λ = 5 per day, and that the independent amounts claimed X are Exponential
random variables with mean µ= $500. What is the probability that the aggregate amount
of claims S(t) in the first ten days exceeeds $ 33,650?
Solution:
E[S(10)] = E[N (10)]E[X] = (5 × 10 × 500) = 25, 000 and V ar[S(10)] = E[N (10)]E[X 2 ] =
5 × 10 × (5002 + 5002 ) = 25, 000, 000. Hence
 
S(t) − 25000 33650 − 25000
P r[S(10) > 33650] = P r √ > √
25000000 25000000

= P r[Z > 1.73] = 0.042

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