Download The Poisson Process

The Poisson Process1
Let us assume that we are counting the occurrence of rare events over a period of time (or
distance, or volume, etc.). We will assume that for some constant  > 0, the following
statements hold true:
1) The probability that exactly one event occurs in an interval of length h is the same for all
such intervals and is equal to h + o(h), where o(h) stands for any function for which
 oh 
lim 
 0.
h 0
 h 
2) The probability that 2 or more events occur in an interval of length h is the same for all
such intervals and is o(h).
3) For any integers n, j1, j2, …, jn and any set of n non-overlapping intervals, if we define Ei
to be the event that exactly ji of the events under consideration occur in the ith interval,
then events E1, E2, …, En are independent.
Loosely put, Assumptions 1 and 2 state that for small values of h, the probability of exactly 1
event occurring in an interval of length h equals h plus something that is small relative to h,
whereas the probability of 2 or more events occurring is small relative to h. Assumption 3 states
that whatever occurs in one interval has no (probability) effect on what will occur in other, nonoverlapping intervals. These three assumptions are known as the Poisson Postulates.
Under Assumptions 1, 2, and 3, we shall now show that the number of events occurring in any
interval of length t is a Poisson random variable with parameter t. To be precise, let us call the
interval [0, t] and denote by N(t) the number of events occurring in that interval. To obtain an
expression for P(N(t) = k), we start by breaking the interval [0, t] into n non-overlapping
subintervals each of length t/n.
Now,
P(N(t) = k) =
P(k of the n subintervals contain exactly 1 event and the other n – k contain 0 events)
+ P(N(t) = k and at least 1 subinterval contains 2 or more events)
This follows because the event on the left side of the equation, that is {N(t) = k}, is the disjoint
union of the two events appearing on the right-hand side of the equation. Letting A and B denote
these two events, we have
P(B)  P(at least one subinterval contains 2 or more events)
 n

 P {i th sub int erval contains 2 or more events}
 i 1

n


  P {i th sub int erval contains 2 or more events}
(Boole’s Inequality)
i 1
n
t
  o 
i 1  n 
t
 no 
n
(Assumption 2)
  t 
 o n  
 t   .
t 
  n  


Now, for any t, t/n  0 as n  +, and so lim PB   0 .
n  
On the other hand, since Assumptions 1 and 2 imply that
P({0 events occur in an interval of length h}) = 1  h  oh  oh  1  h  oh , we see from
the independence assumption, number 3, that
P(A) = P({k of the subintervals contain exactly 1 event and the other n-k contain 0 events})
 n   t
 t 
     o 
 n 
 k  n
k
  t   t  
1   n   o n 
    
nk
.
However, since
  t 
 o n  
 t
 t 
n   o   t  t      t as n  +, it follows, by the same argument that verified
t 
 n 
n
  n  


the Poisson approximation to the binomial distribution, that P A  e
Thus, we obtain, by letting n  +,
P N (t )  k   e
 t
 t
t k
k!
as n  +.
t k
, for k = 0, 1, 2, ….
k!
Hence, if the Poisson Postulates are satisfied, the number of events occurring an any fixed
interval of length t is a Poisson random variable with parameter t; and we say that the events
occur in accordance with a Poisson process having rate .
1 This derivation is from A First Course in Probability, 2nd Edition, by Sheldon Ross.