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Figure 1:
Counting process

Let Fig. 1 be a graph of the customers who are entering a bank.
Every time a customer comes, the counter is increased by one. The time of the
arrival of
th customer is . Since the customers are coming at
random, the sequence
, denoted shortly
by
, is a random sequence. Also, the number of customers
who came in the interval is a random variable
(process). Such a process is right continuous, as indicated by the graph in
Fig. 1.
As it is often case in the theory of stochastic processes, we assume that
the index set, i.e. the set where
is taking values
from, is
. Therefore, we have a sequence of nonnegative random
variables
as
WLOG^{1} let and
, then
is called a point process (counting process), and is denoted shortly by
.
Let
be interarrival time, then the sequence of
interarrival times
is another stochastic process.
Special case is when
is a sequence of
i.i.d.^{2} random
variables, then the
sequence is called a renewal process.
is the associated renewal point process, sometimes also
called renewal process. Also, keep in mind that
.
Definition (Poisson process) A point process
is
called a Poisson process if and satisfies the following
conditions
 its increments are stationary and its nonoverlapping increments are
independent


Remarks
Theorem Let
be a Poisson process, then

(1) 
The expression on the left hand side of (1) represents the
probability of arrivals in the interval .
Proof A generating function of a discrete random variable is
defined
via the following ztransform (recall that the moment generating function of a
continuous random variable is defined through Laplace transform):
where
. Let us assume that is a Poisson random variable with
parameter , then
and

(2) 
Going back to Poisson process, define the generating function as
Then we can write
Furthermore
Comparing this result to (2) we conclude that is a Poisson
random variable with parameter
.
Theorem If
is a Poisson process and
is the interarrival time between the
th and
th events, then
is a sequence of i.i.d. random variables with exponential
distribution, with parameter .
Proof
exponential
Figure 2:
Event description

Need to show that and are independent and is also
exponential.

(3) 
The event
is a subset of the event
described by Fig. 2, i.e.
no arrivals one arrival no arrivals
From (3)

(4) 
Similarly, the event described by Fig. 3 is a subset of
the event
, therefore
From (3)

(5) 
Figure 3:
Event description

From (4) and (5), using squeeze theorem
, it follows
Therefore, is independent of , and is exponentially
distributed random variable.
Theorem


Proof
Recall that
, then
Likewise
Theorem (Conditioning on the number of arrivals) Given that in the
interval the number of arrivals is , the arrival times
are independent and uniformly distributed on .
Proof
Figure 4:
Uniform bins

Independence of arrival times , etc. directly follows from
independence of nonoverlapping increments. In particular let and
be arrival times of first and second event, then
Suppose that we know exactly one event happened in the interval ,
and suppose the interval is partitioned into segments of length
, as shown in Fig. 1. Let be the probability of
event happening in the
th bin, then
. From the
definition of Poisson process it follows that
, say
. The
constant is determined from
Let be a random variable corresponding to the time of arrival, then
the probability density function (pdf) of can be defined as
where 

Therefore, is uniformly distributed on .
Let and be the arrival times of two events, and we know exactly two
events happened on . Also assume that and represent mere
labels of events, not necessarily their order. Given that happened in
th bin, the
probability of occurring in any bin of size
is proportional
to the size of that bin, i.e.
, except for
the
th bin, where
. By rendering the bin
size infinitesimal, we notice that the probability remains constant over
all but one bin, the bin in which occurred, where . But this set
is a set of measure zero, so the cumulative sum over again gives
rise to uniform distribution on .
Question
What is the probability of observing events at instances
on the interval ?
Since arrival times
are continuous random variables,
the answer is 0. However, we can calculate the associated pdf as
Question What is the power spectrum of Poisson process?
It does not make sense to talk about the power spectrum of Poisson process,
since it is not a stationary process.
In particular the mean of Poisson process is
and its autocorrelation
function is
Since
, we conclude that
is
not stationary (in weak sense), therefore it does not make sense to talk about
its power spectrum. Let us define the following stochastic process
(Fig. 5)
spike train 
(6) 
Figure 5:
Spike train

The fundamental lemma says that if
, where is a linear
operator, then
Since differentiation is a linear operator we have
Also, it can be shown using theory of linear operators that
Thus, is WWS^{3}
stochastic process, and it makes sense to define the power spectrum of such a
process as a Fourier transform of its autocorrelation function i.e.
Therefore, the spike train
of independent times
behaves almost as a white noise, since its power spectrum is flat
for all frequencies, except for the spike at . The process
defined by (6) is
a simple version of what is in engineering literature known as a
shot noise.
Definition (Inhomogeneous Poisson process) A Poisson process with a
nonconstant rate
is called inhomogeneous Poisson process.
In this case we have
 nonoverlapping increments are independent (the stationarity is lost though).


Theorem If
is a Poisson process with the rate
, then is a Poisson random variable with parameter
i.e.

(7) 
Proof The proof of this theorem is identical to that of
homogeneous case except that
is replaced by
. In particular,
one can easily get

(8) 
from which (7) readily follows.
Theorem Let
be an inhomogeneous Poisson process with the
rate
and let , then

(9) 
The application of this theorem stems from the fact that we cannot use
, since the increments are no longer stationary.
Proof
Thus, is a Poisson random variable with parameter
, and (9) easily follows.
Theorem


Proof Recall that
and 

From (8) we have
, and the two results
follow after immediate calculations.
Theorem (Conditioning on the number of arrivals) Given that in the
interval the number of arrivals is , the arrival times
are independently distributed on with the pdf
.
Proof The proof of this theorem is analogous to that of the homogeneous
case. The probability of a single event happening at any of bins (Fig.
1) is given by
, where is the bin index. Given that exactly one event
occurred in the interval , we have
The argument for independence of two or more arrival times is identical to that
of the homogeneous case.
Question
What is the probability of observing events at instances
on the interval ?
Since arrival times
are continuous random variables,
the answer is 0. However, we can calculate the associated pdf as
Question How to generate a sample path of a point (Poisson)
process on ? It can be done in many different ways. For a homogeneous
process we use three methods:
 Conditioning on the number of arrivals. Draw the number of arrivals
from a Poisson distribution with parameter
, and then
draw numbers uniformly distributed on .
 Method of infinitesimal increments. Partition the segment into
sufficiently small subsegments of length
(say
). For each subsegment draw a number
from a uniform distribution on . If
, an event occurres in
that subsegment. We repeat the procedure for all subsegments. We can use the
approximation
for a less expensive procedure.
 Method of interarrival times. Using the fact that interarrival times are
independent and exponentially distributed, we draw random variables
from exponential distribution with parameter , where
is the smallest integer that satisfies the criterion
.
Then the sequence
generates the required point process.
For an inhomogeneous process, the procedures 1 and 2 can be modified using
the appropriate theory of inhomegeneous Poisson process.
The illustration of the three procedures for a homogeneous case is given below.
We use and
. For each different procedure
we generate sample paths, and calculate the statistics by averaging over
ensemble. Raster plots of out of samples for the three methods
are shown in Fig. 6, Fig. 7 and Fig. 8,
respectively. Since the generated processes are homogeneous, the interarrival
time distribution is exponential with parameter . The histograms
of interarrival times corresponding to the three methods are also given below.
They clearly exhibit an exponenital trend. The sample mean and sample standard
deviation (unbiased) are also shown. Keep in mind that for exponentially
distributed random variables, both mean and standard deviation are given by
. Sample statistics indicates that is in the
vicinity of .
Figure 6:
Realization of a point process using conditioning on the number of
arrivals. (Top) Ten different sample paths of the same point process
shown as raster plots. (Bottom) The histogram of interarrival times, showing
the exponential trend

Figure 7:
Realization of a point process using method of infinitesimal
increments. (Top) Ten different sample paths of the same point process
shown as raster plots. (Bottom) The histogram of interarrival times, showing
the exponential trend

Figure 8:
Realization of a point process using method of independent
interarrival times. (Top) Ten different sample paths of the same point
process
shown as raster plots. (Bottom) The histogram of interarrival times, showing
the exponential trend

Next: About this document ...
Up: Poisson Process, Spike Train
Previous: Poisson Process, Spike Train
Zoran Nenadic
20021213