Poisson Distribution
Poisson distribution is a discrete probability distribution, which is the limiting
case of the binomial distribution under certain conditions.
1. When n is very indefinitely very large
2. Probability of success is very small.
3. np =  is finite,   R 
Def: A discrete random variable X is said to be follow a Poisson distribution if the probability mass
function is given by
Siméon Denis Poisson
 x
(1781-1840)
e 
p( X  x)  P( x;  ) 
, x  0,1,2,3..........
France
x!
Where e = 2.7183 and   0
Here  is called the parameter of the Poisson distribution.

Examples where the Poisson distribution is used (or) Applications of Poisson distribution:
This distribution is used to describe the behavior of the rare events like
1.
The number of blind born per year in a large city.
2.
The number of printing mistakes per page in a large volume of a book.
3.
The number of air pockets in a glass sheet.
4.
The number of accidents occurred annually at a busy crossing of city.
5.
The number of defective articles produced by a quality machine.
6.
This is widely used in waiting lines or queuing problems in management studies.
7.
It has wide applications in industrial quality control.
8.
In determining the number of deaths in a given period by a rare disease.


The sum of the probabilities of the Poisson distribution is unity i.e.
 P( x;  )  1
x 0
Proof:
For a Poisson distribution the probability mass function is given by
p( X  x)  P( x;  ) 
e  x
, x  0,1,2,3..........
x!
e  x
Now,  P ( x;  ) = 
x!
x 0
x 0


= e 

x
 x!
x 0


1 2 3
 
 ...........  
 0! 1! 2! 3!

= e  
 
= e e
=1
0
 Mean and Variance of Poisson distribution:
For a Poisson distribution the probability mass function is given by
p( X  x)  P( x;  ) 
e  x
, x  0,1,2,3..........
x!
Now,
e  x
Mean = E(X)=  x P ( x;  ) =  x
x!
x 0
x 0


= e 

x
x 0
= e  

 x 1
x( x  1)!
 x 1
 ( x  1)!
x 1


1 2 3
 
 ...........  
 0! 1! 2! 3!

0
= e   
= e   e 
=
 Mean of the Poisson distribution is 
Variance = V(X) = E ( X 2 )  [ E ( X )] 2
= E ( X 2 )  [ ]2 ……… (1)[ Mean of the Poisson distribution is  ]
From (1)
e  x
E ( X )   x P( x;  ) =  x
x!
x 0
x 0
 x

e 
=  [ x( x  1)  x]
x!
x 0
 x


e 
e  x
=  x( x  1)
+ x
x!
x!
x 0
x 0


2
2
2
 x( x  1)
x 1

=e 

2
2  x  2

= e 
x( x  1)( x  2)!
+  [ Mean of the Poisson distribution is  ]
 x 2
 ( x  2)! + 
x2
 0 1 2 3

= e   

 ...........    
 0! 1! 2! 3!

 2 
= e  e 
2
 E ( X )  2   ……………… (2)

2
Putting (2) in (1) we get
V(X)
= E ( X 2 )  [ ]2
= 2   -  2
= 
 Variance of the Poisson distribution is 
Note: In Poisson distribution the mean and variance are equal i.e. 
 Moments of the Poisson distribution:
Non central moments (about zero):
11  Mean  E ( X )  
 21  E ( X 2 )  2  
31  E(X 3 )  3  32  
 41  E ( X 4 )  4  63  72  
Central moments (about mean):
1  0
 2  Variance  
3  
 4  32  
 Skew ness of Poisson Distribution:
Measure of Skew ness of Poisson distribution is given by
3 2
2
1
1 


3
3


2
 Kurtosis of Poisson Distribution:
Measure of Kurtosis of Poisson distribution is given by
2 
 4 32  
1

 3
2
2


2
Note:
1) Since Skew ness =
2) Since Kurtosis = 3 
1

1

>0, Poisson distribution is always positively skewed.
>3, Poisson distribution is always Leptokurtic
 Recursion formula:
If X is a Poisson variate with p.m.f P( X  x) 
e  x
, x  0,1,2,3..........
x!
e  x 1
, x  0,1,2,3..........
And P( X  x  1) 
( x  1)!
Now,
e  x
P( X  x)

 x!x 1 
P ( X  x  1) e 
x
( x  1)!

 P ( X  x)  P ( X  x  1)
x

Prove that the Poisson distribution is the limiting case of binomial distribution stating the
required conditions.
Sol:
The Poisson distribution can be limiting case of a binomial distribution under certain
conditions.
1. Number of trails i.e. n is indefinitely large i.e. n  
2. p , the probability of success in each trail is indefinitely small i.e p  0
3. np   is finite.
If X is a binomial variate then the probability mass function is given by
P(X=x) = n c x p x q n  x , x  0,1,2,........, n
Under the above conditions
lim B ( x; n, p )  lim n c x p x q n x
n 
n
n!
  
= lim
  1  
n  x!( n  x )! n
n
  
x
n x
[ np   ]
n(n  1)( n  2)...........(n  x  1)( n  x)!   
 
n 
x!(n  x)!
n
n x
 
1  
n

n
1  2 
 x  1    
x 
n
1
1

1

..........
........
1







1




n   n 
x
 n  n 


lim
=
x
x! n
nx
 
1



 n
= lim
 
1  
x

n
lim 1 . 
=
x
n


x!
 
1



 n
n
 
lim 1  
x n 

 x e  e  x
n

=
=
=
x
x!
x!
x! 1x
 
lim 1  
n 
n

n
x

Mode of the Poisson distribution:
Case: 1) When  is not an integer, the distribution is unimodal and the value of mode is the
integral part of the  .
Case: 2) When  is an integer, the distribution is bimodal and the value of modes are  and   1 .

Show that in a Poisson distribution with unit mean, mean deviation is 2/e times the standard
deviation.
Sol: The Poisson distribution with unit mean i.e.   1 is given by
P(x; 1) =
e 11x e 1

x!
x!
Now, we have to show that
Mean deviation about mean =
i.e. E X  E ( X ) 
2
1
e
2
i.e. E X  1 
e
i.e. E X  1 
2
Standard deviation
e
2
E ( X 2 )  [ E ( X )] 2
e
[ given mean = 1]
Now,
e 1
x!
x 0

1
x  1
= 1  
e  x 1 x! 

E X 1   x  1


1
1
1
 
= 1  
e  x 1 ( x  1)! x 1 x!
1
= 1  e  (e  1)
e
2
=
e
Hence in Poisson distribution with unit mean, mean deviation are 2/e times the standard deviation.

Properties of Poisson distribution:
The following are the some of the properties of the Poisson distribution.
1. Poisson distribution is a discrete probability distribution with single parameter  .
2. Both mean and variance of the Poisson distribution are equal to  .
3. The distribution is positively skewed and leptokurtic.
4. It is asymptotic form of binomial distribution when p is small, n is large and np is finite.
5. the normal distribution is a limiting form of a Poisson distribution as   0
6. The distributio0n of rare events generally approximates to a Poisson distribution.
7. If X 1 and X 2 are two independent Poisson variates with mean 1 and  2 respectively, then
X  X 1  X 2 is also a Poisson variate with mean 1 +  2 .