MS03 Support Material
Proofs of mean and variance of binomial and Poisson distributions
For use with AQA A-level Mathematics Specification (6360)
1
Binomial
1.1
Mean
Mean = E(X) =  =
x p
i
n
=

x 0
where
i
 n
n x
x  p x 1  p  =
 x
n
=
p
i
1
n

x 1
Definition
 n
n x
x  p x 1  p 
 x
 x x! n  x ! p 1  p 
n!
 n
n x
x
Definition of  
x
 
x 1
 n  1 !
n
= np
  x  1 !  n  x  !
p x 1 1  p 
x 1
m
= np
m!
 y ! m  y  !
p y 1  p 
y 0
= np 
p
i
= np  1 = np
Value is zero at x = 0
m y
n x
Take out factor of np
Cancel x as x! = x(x – 1)!
Substitution of y = x  1
Upper limit of summation is
then (n  1)
Substitution of m = n  1
Noting (n – x) = (m – y)
Sum of all binomial terms =
sum of all probabilities = 1
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1.2
Variance
x
Variance = E(X ) – (E(X)) =  =
2
(a)
2
2
2
i
pi  2 where
p
i
1
Firstly consider E(X(X – 1)) = E(X ) – E(X) = E(X )  
2
2
(b)
Add & subtract  = (E(X))
= 2 + 2  
Substitution
2
2
2 2
2
Rearranging & substitution
of  = np
Secondly reconsider E(X(X – 1))

 n
n x
xx  1  p x 1  p 
 x
Definition
n
 n
n x
xx  1  p x 1  p 
 x
Value is zero at x = 0 and 1
n
=
x 0
=

x2
n
=
 n
 x x 1 x! n  x ! p 1  p
n!
x
n x
Definition of  
x
 
x 2
 n  2!
x2  x  2  !  n  x  !
n
= n  n  1 p 2 
m
= n  n  1 p 2 
y 0
= n(n – 1)p2 
(c)
Expansion
= [E(X2)  2] + 2  
Thus  = E(X(X – 1))   +  = E(X(X – 1)) – n p + np
2
Definition
p x 2 1  p 
n x
m!
m y
p y 1  p 
y ! m  y  !
p
i
= n(n – 1)p2  1 = n(n – 1)p2
Take out factor of
n(n – 1)p2
Cancel x(x – 1) as
x! = x(x – 1)(x – 2)!
Substitution of y = x  2
Upper limit of summation is
then (n  2)
Substitution of m = n  2
Noting (n – x) = (m – y)
Sum of all binomial terms =
sum of all probabilities = 1
Finally, using parts (a) and (b)
2 = E(X(X – 1)) – n2p2 + np = n(n – 1)p2 – n2p2 + np
= n2p2 – np2 – n2p2 + np
Expanding
= np – np2 = np(1 – p)
Cancelling & factorising
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2
Poisson
2.1
Mean
x p
Mean = E(X) =  =
i


=
x
x 0

= 

x 1
=
2.2
p
i
i
where
e   x
=
x!

1
i

Definition
e   x
x!
x
x 1
e   x1
= 
x  1!


y 0
Value is zero at x = 0
Take out factor of 
Cancel x as x! = x(x – 1)!
Substitution of y = x  1
e   y
y!
=1=
Sum of all Poisson terms =
sum of all probabilities = 1
Variance
Variance = E(X ) – (E(X)) =  =
2
(a)
p
2
2
x
2
i
pi  2 where
p
i
1
Firstly consider E(X(X – 1)) = E(X ) – E(X) = E(X )  
2
2
(b)
Add & subtract  = (E(X))
= 2 + 2  
Substitution
2
2
2
2
Rearranging & substitution
of  = 
Secondly reconsider E(X(X – 1))

=

x x  1
x 0

= 2

x2
= 2 
(c)
Expansion
= [E(X2)  2] + 2  
Thus  = E(X(X – 1))   +  = E(X(X – 1)) –  + 
2
Definition
e   x
=
x!
e   x2
= 2
x  2!
p
i


x x  1
x2


y 0
e y
y!
= 2  1 = 2
e   x
x!
Definition
Value is zero at x = 0 and 1
Take out factor of 
Cancel x(x – 1) as
2
x! = x(x – 1)(x – 2)!
Substitution of y = x  2
Sum of all Poisson terms =
sum of all probabilities = 1
Finally, using parts (a) and (b)
2 = E(X(X – 1)) – 2 +  = 2 – 2 + 
=
Expanding
Cancelling
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3
Some alternative approaches
3.1
Binomial
The following results may be quoted/used in the proofs.
Probability generating function
G(t) = E(tX) = (q + pt)n
where q = 1 – p
Moment generating function
M(t) = E(etX) = (q + pet )n
where q = 1 – p
The following properties may then be used to find the mean and variance.

d G t 
d t t 1
2 
3.2
d2 G t 
   2
2
d t t 1
d M t 
d t t 0
or

or
2 
d2 M t 
 2
2
d t t 0
Poisson
The following results may be quoted/used in the proofs.
Probability generating function
G(t) = E(tX) = et 
Moment generating function
t
M(t) = E(etX) = e  e  
The following properties may then be used to find the mean and variance.

d G t 
d t t 1
2 
d2 G t 
   2
2
d t t 1
d M t 
d t t 0
or

or
2 
d2 M t 
 2
2
d t t 0
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