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Binomial distribution
A binomial experiment has four properties:
1) it consists of a sequence of n identical trials;
2) two outcomes, success or failure, are possible on each trial;
3) the probability of success on any trial, denoted p, does not
change from trial to trial;
4) the trials are independent.
Let us conduct a binomial experiment with n trials, a probability
p of success and let X be a random variable giving the number
of successes in the binomial experiment. Then X is a discrete
random variable with the range {0, 1, 2, ..., n} and the
probability function
p(i ) = Cni p i (1 − p )
n −i
We say that X has a binomial distribution.
Binomial distribution with n=10, p=0.5
0,3
0,25
0,2
0,15
p(x)
0,1
0,05
0
0
2
4
6
8
10
12
Binomial distribution with n=10, p=0.25
0,3
0,25
0,2
0,15
p(x)
0,1
0,05
0
0
2
4
6
8
10
12
Expectancy and variance of a binomial distribution
For a random variable X with a binomial distribution Bi(n,p) we
have
E ( X ) = pn
D( X ) = np(1 − p )
Using the identity iCni =
i n!
n (n − 1)!
=
= nCni −−11
i !(n − i )! (i − 1)!(n − i )!
we can calculate
n
E ( X ) = ∑ iC p q
i
n
i =0
i
n −i
n
=∑ iC p q
i =1
i
n
i
n −i
n
n −1
i =1
i =0
n
=∑ nCni −−11 p i q n −i =
i =1
= p ∑ nCni −−11 p i −1q n −i = pn∑ Cni −1 p i q n −i = pn( p + q )
The same identity can be used to calculate D(X)
n −1
= pn
E ( X 2 ) = ∑ i 2Cni p i q n −i =np ∑ iCni −−11 p i −1q n −i =
n
n
i =0
i =1
 n i −1 i −1 n −i n

= np ∑ Cn −1 p q + ∑ (i − 1)Cni −−11 p i −1q n −i  =
 i =1

i =1
n
 n −1 i i n −i
i − 2 i − 2 n −i 
= np ∑ Cn −1 p q + (n − 1) p ∑ Cn − 2 p q  =
 i =0

i =2
n−2


n −1
= np ( p + q ) + (n − 1) p ∑ Cni − 2 p i q n −i − 2  =


i =0
{
= np 1 + (n − 1) p( p + q )
n−2
}= np{1 + (n − 1) p} =
= np(1 − p + np ) = npq + (np )
2
D( X ) = E (X 2 ) − [E ( X )] ⇒ D( X ) = npq
2
Poisson distribution
Poisson distribution is defined for a discrete random variable X
with a range {0, 1, 2, 3, ... } denoting the number of occurrences of an event A during a time interval (T1 ,T2 ) if the following
conditions are fulfilled:
1) given any two occurrences A1 and A2, we have
P( A2 | A1 ) = P( A2 )
2) the probability of A occurring during an interval (t1 ,t2 )
with T1 ≤ t1 < t 2 ≤ T2 equals to c(t2 − t1 ) where c is fixed
3) denoting by P12 the probability that E occurs at least
twice between t1 and t2, we have
P12
lim
=0
t1 →t 2 t − t
2
1
The Poisson distribution Po(λ) is given by the probability
function
p( x ) = e
∞
∑e
x =0
x
λ
−λ
x
λ
−λ
x!
2
3


λ
λ
λ
= e − λ 1 + + + + L  = e − λ e λ = 1
x!
 1! 2! 3!

Poisson law with E(X) = 5
0,2
0,15
0,1
p(x)
0,05
0
0
5
10
15
The expectancy and variance of a random variable X with Poisson
distribution are given by the following formulas
E( X ) = λ
D( X ) = λ
∞
E( X ) = ∑ e
−λ
λx
x!
x =0
∞
E (X 2 ) = ∑ e
x
λ
−λ
x!
x =0
∞
= e −λ λ ∑
x =0
∞
x =∑ e
−λ
x =1
∞
x 2 =∑ e
x =1
λx
x!
x
λ
−λ
x!
λx −1
∞
x =e λ ∑
−λ
x =1
( x − 1)!
∞
x 2 =e −λ λ ∑
x =1
∞
λx
x =0
x!
== e λ ∑
λx −1
( x − 1)!
−λ
=λ
x=
λx
x
x
∞
∞


λ
λ
( x + 1) = e −λ λ  ∑ x + ∑  = e −λ λ (eλ λ + e λ ) =
x!
x = 0 x! 
 x =0 x!
= λ2 − λ
D( X ) = E (X 2 ) − [E ( X )] = λ2 + λ − λ2 = λ
2
Relationship between Bi(n,p) and Po(λ)
For large n, we can write
Bi (n, p ) ≈ Po (np )
Comparing Bi(20,0.25) with Po(5)
0,25
0,2
0,15
0,1
0,05
Poisson
Binomial
18
15
12
9
6
3
0
0
Normal distribution law
A continuous random variable X has a normal distribution law
with E(X) = µ and D(X) = σ if its probability density is given
by the formula
1
f (x) =
e
2π σ
− ( x − µ )2
2σ 2
Standardized normal distribution
A continuous random variable X has a standardized normal
distribution if it has a normal distribution with E(X) = 0 and
D(X) = 1 so that its probability density is given by the formula
1
f (x) =
e
2π
− x2
2
Standardized normal distribution law (Gaussian curve)
There are several different ways of obtaining a random variable X
with a normal distribution law.
For large n, binomial and Poisson distributions asymptotically
approximate normal distribution law
If an activity is undertaken with the aim to achieve a value µ, but
a large number of independent factors are influencing the
performance of the activity, the random variable describing the
outcome of the activity will have a normal distribution with the
expectancy µ.
Of all the random variables with a given expectancy µ and
variance σ2, a random variable with the normal distribution
N(µ,σ2) has the largest entropy.
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