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Fractiles
Given a probability distribution F(x) and a number p, we define
a p-fractile xp by the following formulas
lim F  x   p
x x p 
lim F  x   p
x x p 
For a continuous increasing F(x) a fractile is actually the
inverse of F(x)
F x p   p
x p  F 1  p 
For a distribution F(x) with jumps, we can determine fractiles as
follows
F  x   p1 , F  x   p2  x p  
x  
for any p1  p  p2
x  
F(x)
p
x
xp
In intervals where the distribution is constant, the corresponding
fractile is not determined uniquely
Special fractiles
x0.01 - percentile: x0.35 - 35-th percentile
x0.25 - first quartile
x0.50 - second quartile = median
x0.75 - third quartile
Let, for a discrete random variable, the probability distribution be
given by the probability function
xi
x1
x2
x3
...
xn
pi
p1
p2
p3
...
pn
k
If
p
i 1
i
k 1
  and
p
i 1
i
  we take
f  xk   xk 1  xk 
k
   pi
i 1
pk 1
Example
Calculate the median of a discrete random variable whose
probability function is given by the following table
xi
pi
1.2
1.5
1.8
2.1
2.4
2.7
3.0
3.3
0.12 0.25 0.18 0.01 0.15 0.2 0.04 0.05
We have 0.12 + 0.25 = 0.37 and 0.12 + 0.25 + 0.18 = 0.55
so that
0.5  0.37
x0.5  1.5  1.8  1.5
 1.71667
0.18
Important distributions of discrete random variables
 binomial distribution
 Poisson distribution
Important distributions of continuous random variables
 uniform distribution
 normal distribution
 chi-square distribution
 t-distribution
 F distribution
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
n
n
i 0
i 1
E X 2    i 2Cni p i q n i np  iCni 11 p i 1q n i 
 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  t2  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
0
t1 t 2 t  t
2
1
lim
The Poisson distribution Po(λ) is given by the probability
function
p x   e

e
x 0
x


x


x!
2
3

  




 e  1       e e  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   
D X   

E X    e
x


x!
x 0

E X 2    e
x


x!
x 0

 e   
x 0

x  e
x


x 1

x 2  e
x 1
x!
x 1

x e   
x


x!

x
 e    
x 1  x  1!
x  0 x!

x 2 e   
x 1
x 1
 x  1!
x
x
x
x



  



 x  1  e    x    e  e   e  
x!
x  0 x! 
 x 0 x!
 2  
D X   E X 2   E  X   2    2  
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)
Poisson
18
15
12
9
6
Binomial
3
0
0,25
0,2
0,15
0,1
0,05
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
2 
 x   2
2 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
2
 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.