Download MTH/STA 561 POISSON DISTRIBUTION Many important

MTH/STA 561
POISSON DISTRIBUTION
Many important applications of probability theory are concerned with modeling the time
instants at which events occur. For example, in planning the number of telephone lines
and the types of equipment that might best serve the needs of a given organization, one
must have some idea of the number of incoming and outgoing telephone calls that might be
expected during various periods of time. The e¢ cient design of a roadway system depends
upon the expected number and times of interval of vehicles that will use the system. A
hospital, department store, or any other group that maintains a sta¤ to serve individuals is
very interested in the number of, and times at which, demands for service may occur.
Experiments yielding the number of outcomes occurring during a given time interval or
in a speci…ed region are often called Poisson experiments. The given time interval may be
of any length, such as a minute, a day, a week, a month, or even a year. Hence, a Poisson
experiment might generate observations for the random variable representing the number of
telephone calls per hour received by an o¢ ce, the number of days school is closed due to
snow during the winter, or the number of postponed games due to rain during a baseball
season. The speci…ed region could be a line segment, an area, a volume, or perhaps a piece
of material. In this case, the random variable might represent the number of …eld mice per
acre, the number of bacteria in a given culture, or the number of typing errors per page.
De…nition 1. A Poisson experiment is one that possesses the following properties:
1. In a su¢ cient short length of time, say of length t (or in a su¢ cient small region),
only 0 or 1 event might occur. The probability that more than one event will occur in a
short time interval of length t (or fall in such a small region) is negligible.
2. The probability of exactly one event occurring in this short time interval of length t
or falling in such a small region is equal to
t, proportional to the length of the interval.
3. Any nonoverlapping intervals of length t (or any nonoverlapping regions) are independent Bernoulli trials.
4. The random variable of interest, Y , is the number of outcomes occurring during a
given time interval or in a speci…ed region. This random variable is called a Poisson random
variable.
Example 1. The following are typical Poisson random variables:
1. The number of radioactive particles passing through a counter during a millisecond in
a laboratory experiment.
2. The number of telephone calls per hour received by an o¢ ce.
3. The number of days school is closed due to snow storm during the winter.
4. The number of postponed games due to rain during a baseball season.
5. The number of deaths from a certain noncontagious disease per year in a certain city.
Let the interval of t be divided into n = t= t nonoverlapping, equal length pieces. These
small subintervals of time then are independent Bernoulli trials, each with probability of
1
success (an event occurs) equal to p =
t. The probability of no event occurring on each
trial is q = 1
t. Then Y; the number of events in the interval of length t, is binomial
distributed with parameters n and p =
t = t=n; that is,
n
y
P (Y = y) =
t
n
y
t
n
1
n y
If we now take the limit of this probability distribution as t ! 0 (and thus n ! 1), we
arrive at the Poisson distribution, which gives the probability of occurrence for any number
y of events in the continuous interval of length t, as long as the …rst three properties of
De…nition 1 hold regarding the experiment generating the events. We can write
y
n
t
( t)y
t
n!
1
1
y! (n y)! ny
n
n
y
n
t
t
n (n 1) (n 2) (n y + 1) ( t)y
=
1
1
ny
y!
n
n
y
n
y
n n 1 n 2
n y + 1 ( t)
t
t
=
1
1
n
n
n
n
y!
n
n
y
y
1
2
y 1 ( t)
t
t
= 1 1
1
1
1
1
n
n
n
y!
n
n
P (Y = y) =
Since
lim
n!1
t
n
1
n
= lim
n!1
lim
n!1
lim 1
n!1
1
1
#
t
=e
t
y
=1
2
n
y
1
=1
y
1
n
= 1;
y
P (Y = y) =
where
1
n= t
1
1+
( n= t)
t
n
1
n
we have
"
n
e
y!
for y = 0; 1; 2;
= t. We have proved the following result.
Poisson Distribution. The probability distribution of the Poisson random variable
Y , representing the number of outcomes occurring during a time interval or in a speci…ed
region, is given by
y
e
p (y) =
for y = 0; 1; 2;
y!
where > 0 is the average number of outcomes occurring during the time interval or the
speci…ed region and e is the base of the natural logarithm.
Example 2. The average number of radioactive particles passing through a counter
during 1 millisecond in a laboratory is 4. What is the probability that 6 particles enter the
counter in a given millisecond?
2
Solution. Using the Poisson distribution with parameter
p (6) =
46 e
6!
= 4, we …nd
4
= 0:1042:
Theorem 1. For the Poisson random variable Y ,
1
P
y
e
y!
y=0
= 1:
Proof. From the Maclaurin series expansion for ex ,
ex =
1 xk
P
k=0 k!
we can easily show that the sum of the probability distribution for the Poisson random
variable over its range is equal to 1; that is,
1
P
p (y) =
y=0
because the in…nite sum
1
P
y
y!
1
P
y
e
y!
y=0
y=0
The partial sums
1
P
=e
y
y!
y=0
=e
e =1
is a series expansion of e .
a
P
p (y) =
y=0
a
P
y
e
y!
y=0
for the Poisson distribution for many values of
3 of Appendix III.
between 0:02 and 25 are provided in Table
Example 3. A certain type of tree has seedlings randomly dispersed in a large area,
with the mean density of seedlings being approximately 5 per square yard, that is = 5. If
a forester randomly locates a square yard in the area, then
(1) the probability that the it contains less than or equal to 4 seedlings is
P (Y
4) =
4
P
y=0
y
e
y!
= 0:440
(2) the probability that the it contains more than 7 seedlings is
P (Y > 7) = 1
P (Y
7
P
7) = 1
y=0
y
e
y!
=1
0:867 = 0:133
(3) the probability that the it contains exactly 6 seedlings is
P (Y = 6) = P (Y
6)
P (Y
5) =
6
P
y=0
= 0:762
0:616 = 0:146
3
y
e
y!
5
P
y=0
y
e
y!
Theorem 2. The mean and variance of the Poisson distribution are
and
E (Y ) =
Proof.
To …nd the mean, we write
E (Y ) =
1
P
=
y
1
P
e
=e
y!
1)!
y=1
y=1 (y
x
1
1
P
= e
= e e =
1)!
x=0 x!
y
e
y!
y
y=0
e
1
P
y=1
with x = y
V ar (Y ) =
=
y
(y
1
P
y
y
1. The variance is obtained by …rst …nding
E [Y (Y
1)] =
1
P
y
e
y!
y (y
1)
1
P
y
y=0
= e
y=2
=
2
e
(y
1
P
w=0
with w = y
2. Hence,
2)!
=
1
P
=
2
=
e
1
P
y=2
2
e
y!
1)
y=2
w
w!
y
y (y
e
e =
y 2
(y
2)!
2
[E (Y )]2
V ar (Y ) = E Y 2
= E [Y (Y
= 2+
1)] + E (Y )
2
=
[E (Y )]2
Example 4. The number, Y , of seedlings randomly dispersed in a large area as given
in Example 3 is a Poisson random variable with parameter = 5. So
E (Y ) = 5
and
V ar (Y ) = 5
Now consider the time interval as being split up into n subintervals, each of which is
so small that at most one outcome could occur in it with probability p. For all practical
purposes, we assume that
P (one outcome occur in a subinterval) = p
P (no outcomes occur in a subinterval) = 1
p
Then the total number of outcomes occurring in the time interval is just the total number
of subintervals that contains one outcome. Also, assume that the occurrence of outcomes
can be regarded as independent from subinterval to subinterval. Then the total number of
outcomes has a binomial distribution. Letting = np, we have the following theorem.
4
Theorem 3. Let Y be a binomial random variable with n independent identical trials
and success probability p. When n ! 1, p ! 0 and = np remains constant. Then
n y
p (1
y
lim
n!1
p)n
y
y
e
:
y!
=
Proof. Write
n y
p (1
y
p)n
y
n!
py (1 p)n y
y! (n y)!
n (n 1) (n 2) (n y + 1) y
p (1 p)n y
y!
y
n y
n (n 1) (n 2) (n y + 1)
1
y!
n
n
y
n n (n
y
1) (n 2) (n y + 1)
1
1
y!
n
ny
n
y
n
y n n
1 n 2
n y+1
1
1
y!
n
n
n
n
n
n
y
n
y
1
2
y 1
1
1
1 1
1
1
y!
n
n
n
n
n
=
=
=
=
=
=
While y and
remain constant, we see that
1
n
lim 1 1
n!1
2
n
1
y
lim
n!1
1
n
lim
n!1
1
n
= lim
n!1
"
=1
n
y
1
y
1
n
=1
=1
n=
1
1+
( n= )
#
=e
Hence,
lim
n!1
n y
p (1
y
n y
p)
y
=
e
y!
Example 5. The production line in a large manufacturing plant produces items of
which 1% are defective. In a random sample of 1; 000 items selected from the production
line, what is the probability we could …nd exactly 9 defective items?
Solution. The number of defective items in this sample is a binomial random variable
with parameters n = 1; 000 and p = 0:01. Then the desired probability is
p (9) =
1; 000
(0:01)9 (0:99)991 :
9
The calculation of this probability is an enormous task. However, since p is close to zero and
n is large, we can approximate with the Poisson distribution using = 1; 000 0:01 = 10;
that is,
109 e 10
p (9)
= 0:1251
9!
5
APPENDIX.
An experiment of chance that continues in time (or in space) and is observed as it unfolds
is called a stochastic process, or a random process, or simply process. A process that leads
to a Poisson distribution is called Poisson process. The symbol o (h) represents any function
satisfying
o (h)
lim
= 0:
h!0 h
For example, h2 = o (h) because
h2
= lim h = 0;
h!0 h
h!0
lim
and o (h) + o (h) = o (h) because
o (h) + o (h)
o (h)
= 2 lim
= 0:
h!0
h!0 h
h
lim
Let us also denote by Pn (h) the probability that n events occur in an interval of length h;
that is,
Pn (h) = P fn events occur in (0; h)g
De…nition 2. The following four properties regarding a Poisson experiment are known
as the Poisson postulates:
(1) Events de…ned according to the numbers of events in nonoverlapping intervals of time
are independent.
(2) The probability structure of the process is time invariant.
(3) The probability of exactly one event in a small interval of time is approximately
proportional to the size of the interval; that is,
P1 (h) = h + o (h)
for
>0
(4) The probability of more than one event in a small interval is negligible in comparison
with the probability of one event in that interval; that is,
1
X
Pn (h) = o (h) :
n=2
Theorem 3. Under the Poisson postulates,
Pn (t) =
( t)n e
n!
t
for n = 0; 1; 2;
6
Proof.
Write
Pn (t + h) = P fn events occur in (0; t + h)g
= P fn events occur in (0; t) and none in (t; t + h)g
+P fn 1 events occur in (0; t) and 1 in (t; t + h)g
+
+P fnone in (0; t) and n in (t; t + h)g
By Postulate (1),
Pn (t + h) = P fn events occur in (0; t)g P fnone in (t; t + h)g
+P fn 1 events occur in (0; t)g P f1 in (t; t + h)g
+
+P fnone in (0; t)g P fn in (t; t + h)g
Then it follows from Postulates (2), (3), and (4) that
Pn (t + h) = Pn (t) [1
h o (h)] + Pn
+
+ P0 (t) o (h)
1
(t) [ h + o (h)] + Pn
2
(t) o (h)
for n > 0 and
P0 (t + h) = P0 (t) [1
h
o (h)]
for n = 0. Transposing the Pn (t), dividing by h, and passing to the limit as h ! 0, we
obtain the derivative of Pn (t):
Pn0 (t) =
Pn (t) + Pn
1
for n = 1; 2; 3; : : :
(t)
P00 (t) =
P0 (t)
The appropriate initial conditions are P0 (0) = 1 and Pn (0) = 0 for n = 1; 2; 3;
, it follows that
P00 (t) =P0 (t) =
d
ln P0 (t) =
dt
which implies that
ln P0 (t) =
t+A
and so
P0 (t) = Ce
t
where C = eA
The condition P0 (0) = 1 implies that C = 1; so
P0 (t) = e
t
Substitution of this expression in the equation for P10 (t) yields
P10 (t) + P1 (t) = e
7
t
. Since
Multiplying both sides by e
t
gives
e t P10 (t) + e t P1 (t) =
that is,
d t
e P1 (t) =
dt
This becomes integrable with the result
e t P1 (t) = t + K
The condition P1 (0) = 0 implies that K = 0 and thus
t
P1 (t) = te
Now substitution of the induction assumption
Pn
1 (t) =
( t)n 1 e
(n 1)!
t
in the equation for Pn (t) yields
Pn0 (t) + Pn (t) =
Multiplying both sides by e
t
( t)n 1 e
(n 1)!
t
gives
n
e
t
Pn0
t
(t) + e Pn (t) =
that is,
d t
e Pn (t) =
dt
(n
(n
1)!
tn
n
1)!
Thus,
tn
1
( t)n
e Pn (t) =
+M
n!
Then the condition Pn (0) = 0 implies that M = 0 and hence
t
Pn (t) =
( t)n e
n!
8
t
1