
A discrete random variable X is said to have a
Poisson distribution with parameter   0 if
the pmf of X is
e   x
P  X  x   p  x;   
x!
for x satisfying x  0,1, 2,3,
2

If X has Poisson distribution
parameter  , then E  X   V ( X )   .
with
3

The assumption that   0 ensures that
p  x;    0 .

That the probabilities sum to 1 is a
consequence of the Maclaurin series for e  :
e  1   
2
2!

3
3!

  x 0

x
x!
4

Suppose that in the binomial pmf b(x;n,p) we let
in such a way that np
p0
n   and
approaches a value   0 . Then
b  x; n, p   p  x;  
5

According to this proposition, in any binomial
experiment in which n is large and p is small,
b  x; n, p   p  x; . 

As a rule of thumb, this approximation can
safely be applied if n > 50 and np < 5.
6

If a publisher of nontechnical books takes
great pains to ensure that its books are free
of typographical errors, so that the
probability of any given page containing at
least one such error is .005 and errors are
independent from page to page, what is the
probability that one of its 400-page novels
will contain exactly one page with errors?
7

Let S be a page with at least one error, let F
be a page with no errors, and let X be the
number of pages with at least one error.

Then X is binomial with n = 400, p = .005, and
np=2.
 400 
399
1
P  X  1  
.005 .995   .270669

 1 
8

The Poisson approximation is
e2 21
 .270671 ,
1!
which is very close to the true answer.

We used   np  2 in the approximation.
9

An important application of the Poisson
distribution arises in connection with the
occurrence of events over time.

The events might be visits to a website, email
messages to a particular address, or accidents
in an industrial facility.
10



There exists a parameter   0 such that for
any short interval of length t , the probability
that exactly one event occurs is   t  o  t 
The probability of more than one event
occurring during t is o  t 
The number of events occurring during the
time interval t
is independent of the
number that occur prior to this interval.
11

Under the assumptions above, the
probability Pk  t  of k events in a time interval
k
 t
of length t is Pk  t   e  t  / k ! , i.e. Poisson
with parameter    t .

Thus the expected number of events in an
interval of length t is t , and the expected
number in a unit interval is  .
12

Suppose pulses arrive at a counter at an
average rate of six per minute, so that   6 .

To find the probability that at least one pulse
is received in a ½-minute interval we use the
Poisson distribution with parameter  t  3 ,
and thus
P ( X  1)  1 
e
3
 3
0!
0
 .950
13