Poisson Process
Consider a random process representing the number of occurrences of an event up to time
t (over time interval (0, t]. Such a process is called a counting process and we shall
denote it by N (t ), t  0. Clearly N (t ), t  0 is a continuous-time discrete state
process and any of its realizations is non-decreasing function of time.
The counting process N (t ), t  0 is called Poisson’s process with the rate parameter  if
(i)
(ii)
N(0)=0
N (t) is an independent increment process.
Thus the increments N (t2 )  N (t1 ) and N (t4 )  N (t3 ), et. are independent.
(iii)
P N (t )  1  t  0(t )
where 0( t ) implies any function such that lim
t  0
(iv)
P N (t )  2  0  t 
0(t )
 0.
t
The assumptions are valid for many applications. Some typical examples are



Number of alpha particles emitted by a radio active substance.
Number of binary packets received at switching node of a communication
network.
Number of cars arriving at a petrol pump during a particular interval of time.
P  N (t  t )  n  = Probability of occurrence of n events up to time t  t
 P  N (t )  n, N (t )  0   P {N (t )  n  1, n(t )  1}  P  N (t )  n  1, n(t )  2 
 P  N (t )  n  (1  t  0(t ))  P  N (t )  n  1  (t  0(t ))  P  N (t )  (n  1)  (0(
P  N (t )  n, N (t )  0  P {N (t )  n  1, n(t )  1}  P N (t )  n  1, n(t )  2
 P  N (t )  n (1  t  0(t ))  P  N (t )  n  1 (t  0(t ))  P N (t )  (n  1) (0(t ))
lim
t  0
P  N (t  t )  n   P  N (t )  n 
t

   P {N (t )  n}  P {N (t )  n  1} 
d
P  N (t )  n     P  N (t )  n   P  N (t )  n  1  
dt
(1)
The above is a first-order linear differential equation with initial condition
P(N (0)  n)  0. . This differential equation can be solved recursively.
First consider the problem to find P N (t )  0
From (1)
d
P  N (t )  0    P  N (t )  0 
dt
 P  N (t )  0  e  t
Next to find P( N (t )  1)
d
P( N (t )  1   P  N (t )  1   P N (t )  0
dt
  P  N (t )  1   e  t
with initial condition P N (0)  1  0.
Solving the above first-order linear differential equation we get
P N (t )  1  tet
Now by mathematical indication we can show that
( t ) n e   t
P  N (t )  n 
n!
Remark
(1) The parameter  is called the rate or intensity of the Poisson process.
It can be shown that
( (t2  t1 )) n e  (t2 t1 )
P  N (t2 )  N (t1 )  n 
n!
Thus the probability of the increments depends on the length of the interval t2  t1 and not
on the absolute times t2and t1. Thus the Poisson process is a process with stationary
increments.
(2) The independent and stationary increment properties help us to compute the joint
probability mass function of N (t ). For example,
P  N (t1 )  n1 , N (t2 )  n2    P  N (t1 )  n1}) P({N (t2 )  n2  /{N (t1 )  n1}
 P  N (t1 )  n1}) P({N (t2 )  N (t1 )  n2  n1 
(t1 ) n1 e  t1 ( (t2  t1 )) n2  n1 e   (t2 t1 )

n1 !
(n2  n1 )!
Mean, Variance and Covariance of the Poisson process
We observe that at any time t  0, N (t ) is a Poisson random variable with the parameter
t. Therefore,
EN (t )   t
and var N (t )  t
Thus both the mean and variance of a Poisson process varies linearly with time.
As N (t ) is a random process with independent increment, we can readily show that
C N (t1 , t2 )  var( N (min(t1 , t2 )))
  min(t1 , t2 )
 RN (t1 , t2 )  C N (t1 , t2 )  EN (t1 ) EN (t 2 )
  min(t1 , t2 )   2t1t2
A typical realization of a Poisson process is shown below:
Example: A petrol pump serves on the average 30cars per hour. Find the probability that
during a period of 5 minutes (i) no car comes to the station, (ii) exactly 3 cars come to the
station and (iii) more than 3 cars come to the station.
Average arrival = 30 cars/hr
= ½ car/min
Probability of no car in 5 minutes
(i) P  N (5)  0  e
1
 5
2
 e 2.5  0.0821
3
1 
  5
2  2.5
(ii) P  N (5)  3  
e
3!
(iii) P{N (5)  3}  1  0.21  0.79
Binomial model:
P = probability of car coming in 1 minute=1/2
n=5
 P( X  0)  (1  P)5  0.55
Inter-arrival time and Waiting time of Poisson Process:
o
T1
T2
tn
tn 1
t2
t1
Tn 1
Tn
Let Tn = time elapsed between (n-1) st event and nth event. The random process
{Tn , n  1, 2,..} represent the arrival time of the Poisson process.
T1 = time elapsed before the first event take place. Clearly T1 is a continuous random
variable.
Let us find out the probability P({T1  t})
P({T1  t})  P({0 event upto time t})
 e  t
 FT1 (t )  1  P{T1  t}  1  et
 fT1 (t )  et t  0
Similarly,
P({Tn  t})  P( {0 event occurs in the interval (tn 1 , tn 1  t )/(n-1)th event occurs at (0, tn 1 ] })
 P( {0 event occurs in the interval (tn 1 , tn 1  t ]})
 e  t
 fTn (t )  et
Thus the inter-arrival times of a Poisson process with the parameter  are exponentially
distributed. with
fTn (t )  et
n0
Remark
 We have seen that the inter-arrival times are identically distributed with the
exponential pdf. Further, we can easily prove that the inter-arrival times are
independent also. For example
FT1 ,T2 (t1 , t2 )  P({T1  t1 , T2  t2 })
 P({T1  t1}) P({T2  t2 }/{T1  t1})
 P({T1  t1}) P({zero event occurs in (T1 , T1 +t2 ) }/{one event occurs in (01 , T1 ]})
 P({T1  t1}) P ({zero event occurs in (T1 , T1 +t2 ) })
 P({T1  t1}) P({T2  t2 })
 FT1 (t1 ) FT2 (t2 )

It is interesting to note that the converse of the above result is also true. If the
inter-arrival times between two events of a discrete state {N (t ), t  0} process are
1
exponentially distributed with mean , then {N (t ), t  0} is a Poisson process

with the parameter .
 The exponential distribution of the inter-arrival process indicates that the arrival
process has no memory. Thus
P({Tn  t0  t1 / Tn  t1})  P({Tn  t0 }) t0 , t1
Another important quantity is the waiting time Wn . This is the time that elapses before the
nth event occurs. Thus
n
Wn   Ti
i 1
How to find the first order pdf of Wn is left as an exercise. Note that Wn is the sum of n
independent and identically distributed random variables.
n
n
n
i 1
i 1

 EWn   ETi   ETi 
and
n
n
n
i 1
i 1
2
var (Wn )  var  (Ti )   var(Ti ) 
Example
The number of customers arriving at a service station is a Poisson process with a rate
10 customers per minute.
(a) What is the mean arrival time of the customers?
(b) What is the probability that the second customer will arrive 5 minutes after the
first customer has arrived?
(c) What is the average waiting time before the 10th customer arrives?
Semi-random Telegraph signal
Consider a two-state random process { X (t )} with the states X (t )  1 and X (t )  1.
Suppose X (t )  1 if the number of events of the Poisson process N (t ) in the interval
(0, t ] is even and X (t )  1 if the number of events of the Poisson process N (t ) in the
interval (0, t ] is odd. Such a process { X (t )} is called the semi-random telegraph signal
because the initial value X (0) is always 1.
p X ( t ) (1)  P ({ X (t )  1})
 P({ N (t )  0}  P({ N (t )  2})  ...
( t ) 2
 ....)
2!
 e  t cosh  t
 e  t {1 
and
p X ( t ) (1)  P({ X (t )  1})
 P({ N (t )  1}  P({ N (t )  3})  ...
( t ) 3
 ....)
1!
3!
 e  t sinh  t
 e  t {
t

We can also find the conditional and joint probability mass functions. For example, for
t1  t2 ,
p X ( t1 ), X ( t2 ) (1,1)  P ({ X (t1 )  1}) P ({ X (t2 )  1)}/{ X (t1 )  1)}
 e  t1 cosh t1 P({ N (t2 ) is even )}/{N (t1 ) is even)}
 e  t1 cosh t1 P({ N (t2 )  N (t1 ) is even )}/{N (t1 ) is even)}
 e  t1 cosh t1 P({ N (t2 )  N (t1 ) is even )}
 e  t1 cosh t1e   ( t2 t1 ) cosh  (t2  t1 )
 e  t2 cosh  t1 cosh  (t2  t1 )
Similarly
p X ( t1 ), X ( t2 ) (1, 1)  e  t2 cosh t1 sinh  (t2  t1 ),
p X ( t1 ), X ( t2 ) (1,1)  e  t2 sinh t1 sinh  (t2  t1 )
p X ( t1 ), X ( t2 ) (1, 1)  e  t2 sinh t1 cosh  (t2  t1 )
Mean, autocorrelation and autocovariance function of X (t )
EX (t )  1 e  t cosh t  1 e  t sinh t
 e t (cosh t  sinh t )
 e2 t
EX 2 (t )  1 e  t cosh t  1 e  t sinh t
 e  t (cosh t  sinh t )
 e  t et
1
 var( X (t ))  1  e 4 t
For t1  t2
RX (t1 , t2 )  EX (t1 ) X (t2 )
 11 p X ( t1 ), X ( t2 ) (1,1)  1 ( 1)  p X ( t1 ), X (t2 ) (1, 1)
 ( 1) 1 p X ( t1 ), X ( t2 ) ( 1,1)  ( 1)  ( 1)  p X ( t1 ), X ( t2 ) ( 1, 1)
 e  t2 cosh t1 cosh  (t2  t1 )  e  t2 cosh t1 sinh  (t2  t1 )
 e  t2 sinh t1 sinh  (t2  t1 )  e  t2 sinh t1 cosh  (t2  t1 )
 e  t2 cosh t1 (cosh  (t2  t1 )  sinh  (t2  t1 ))  e  t2 sinh t1 (cosh  (t2  t1 ) 1 sinh  (t2  t1 ))
 e  t2 e   ( t2 t1 ) (cosh t1  sinh t1 )
 e  t2 e   ( t2 t1 ) et1
 e 2  ( t2 t1 )
Similarlrly for t1  t2
RX (t1 , t2 )  e 2  ( t1 t2 )
 RX (t1 , t2 )  e 2  t1 t2
Random Telegraph signal
Consider a two-state random process {Y (t )} with the states Y (t )  1 and Y (t )  1.
1
1
Suppose P({Y (0)  1})  and P({Y (0)  1}) 
and Y (t ) changes polarity with an
2
2
equal probability with each occurrence of an event in a Poisson process of parameter .
Such a random process {Y (t )} is called a random telegraph signal and can be expressed
as
Y (t )  AX (t )
where { X (t )} is the semirandom telegraph signal and A is a random variable
1
1
independent of X (t ) with P ({ A  1})  and P({ A  1})  .
2
2
Clearly,
1
1
EA  (1)   1  0
2
2
and
1
1
EA2  (1) 2   12   1
2
2
Therefore,
EY (t )  EAX (t )
 EAEX (t )
A and X (t ) are independent
0
RY (t1 , t2 )  EAX (t1 ) AX (t2 )
 EA2 EX (t1 ) X (t2 )
e
2  t1 t2