Download Random variable – Discrete Probability distribution – Continuous

UNIT II-PART II
Random variable – Discrete Probability distribution –
Continuous probability distributions – Expectation –
Moment generating function – probability generating
function - Probability mass and density functions.
Prepared by
Dr. V. Valliammal
1
Random Variables
Random variable
A real variable (X) whose value is determined by the
outcome of a random experiment is called a
random variable.
(e.g) A random experiment consists of two tosses
of a coin. Consider the random experiment which is
the number of heads (0, 1 or2)
Outcome: HH HT TH TT
Value of X: 2 1 1 0
2
Discrete Random Variable
A random variable x which takes a countable
number of real values is called a discrete
random variable.
(e.g) 1. number of telephone calls per unit time
2. marks obtained in a test
3. number of printing mistakes in each
page of a book
3
Probability Mass Function
If X is a discrete random variable taking atmost a
countably infinite number of values x1, x2, .., we
associate a number Pi = P(X = xi) = P(xi), called the
probability mass function of X. The function P(xi)
satisfies the following conditions:
(i) P(xi)  0  i = 1, 2, …, 

(ii)  P(x
i
) 1
i 1
4
Continuous Random Variable
A random variable X is said to be continuous if it
can take all possible values between certain
limits.
(e.g.)1. The length of a time during which a vacuum
tube installed is a continuous random variable .
2. number of scratches on a surface, proportion of
defective parts among 1000 tested,
3. number of transmitted in error.
5
Probability Density Function
Consider a small interval (x, x+dx) of length dx.
The function f(x)dx represents the probability that
X falls in the interval (x, x+dx)
i.e., P(x  X  x+dx) = f(x) dx.
The probability function of a continuous random
variable X is called as probability density function
and it satisfies the following conditions.
(i) f(x)  0  x

(ii)  f(x)dx  1

6
Distribution Function
The distribution function of a random variable X is
denoted as F(X) and is defined as F(x) = P(X  x).
The function is also called as the cumulative
probability function.
x
F(x)  P(X  x) 
 P(x) when X is discrete
x  
x

 F(x)dx when X is continuous

7
Properties on Cumulative Distribution
1. If x  b, F(a)  F(b), where a and b are real
quantities.
2. If F is the distribution function of a onedimensional random variable X, then 0  F(x)  1.
3. If F is the distribution function of a one
dimensional random variable X, then F() = 0 and
F() = 1.
8
Problems
1. If a random variable X takes the values 1, 2, 3, 4
such that 2P(X=1)=3P(X=2)=P(X=3)=5P(X=4). Find
the probability distribution of X.
Solution:
Assume P(X=3) = α By the given equation
P(X  1) 
α
α
α
P(X  2)  P(X  4) 
2
3
5
For a probability distribution (and mass function)
 P(x) = 1
P(1)+P(2)+P(3)+P(4) =1
9
 

    1
2 3
5
61
 1
30

 
30
61
15
10
30
6
P ( X  1) 
; P ( X  2)  ; P ( X  3) 
; P ( X  4) 
61
61
61
61
The probability distribution is given by
X
1
2
3
4
p ( x)
15
61
10
61
30
61
6
61
10
2. Let X be a continuous random variable having
the probability density function
 2
,
x 1
f ( x)   x3
 0 , otherwise
Find the distribution function of x.
Solution:
x
x
x
 1 
1
  1
F ( x)   f ( x) dx  
dx  
3
2
2

1
1x
x
 x 1
2
11
4. A continuous random variable X has the
probability density function f(x) given by
f ( x)  ce
x
,  x  
Find the value of c and CDF of X.

Solution:
f ( x) dx  1


 x
ce


2 ce
0

2 ce
x
x
dx  1
dx 1
dx 1
0

 x 
2c   e   1

0
2c1 1
c
1
2
12
Case(ii ) x  0
Case(i) x  0
F x  

x
F x  
 f ( x) dx

x x
ce

dx

c
x
 f ( x) dx

x x
ce

c
x x
 e dx
0 x
x x
 e dx  c  e dx


 x
 c e

x
 x
 c  ce
 ce 

 
1 x
 e
2
dx
0
0
x

 x

c

 e 

0
x
c
x 

 c 2  e 


x 
1
 2  e 
2

1 x
x0
 2 e ,
F ( x)  
x 
1
  2  e , x  0
 2 

13
5. A random variable X has the following
probability distribution.
X: 0
1
2
3
4
5
6
7
f(x): 0
k
2k 2k 3k k2 2k2 7k2+k
Find (i) the value of k (ii) p(1.5 < X < 4.5 | X >2) and
(iii) the smallest value of λ such that p(X≤λ) > 1/2.
Solution
 P ( x) 1
(i)
0  k  2k  2k  3k  k
10 k
2
 9k  1  0
1
k 
 0 .1
10
2
 2k
2
 k   1,
 7k
2
k 1
1
10
14
(ii)
A  1.5  X  4.5  2,3,4
B  X  2  3,4,5,6,7
A  B  3,4
p A  B 
p (3,4)

pB 
p (3,4,5,6,7)
5
2k  3k
5k
5


 10 
2
2
2
2
7
7
2k  3k  k  2k  7k  k 10k  6k
10
p(X)
F(X)
p (1.5  X  4.5 | X  2)  p  A | B  
(iii)
X
0
0
0
2
2k = 0.2
0.3
3
2k = 0.2
0.5
4
3k = 0.3
0.8
5
k2=0.01 0.81
6
2k2 = 0.02
0.83
7
7k2+k = 0.17 1.00
From the table for X = 4,5,6,7 p(X) > and the smallest value is 4
Therefore λ = 4.
15
Expectation of a Random Variable
The expectation of a random variable X is denoted
as E(X). It returns a representative value for a
probability distribution.
For a discrete probability distribution
E(X) =  x p(x).
For a continuous random variable X which
assumes values in (a, b)
b
E(X)   xf(x)dx
a
16
Properties on Expectation
1. Expectation of a constant is a constant.
2. E[aX] = aE(X), where a is a constant.
3. E(aX + b) = aE(X) + b, where a and b are
constants.
4. |E(X)|  E|X|, for any random variable X.
5. If X  Y, E(X)  E(Y).
17
Variance of a Random Variable
The variance of a Random variable X, which is
represented as V(X) is defined as the expectation
of squares of the derivations from the expected
value.
V(X) = E(X2) – (E(X))2
Properties On Variance
1. Variance of a constant is 0.
2. V(aX) = a V(X), where a is a constant.
18
Moments and Other Statistical Constants
Raw Moments
Raw moments about origin
b
μr   x r f(x)dx
a
Raw moments about any arbitrary value A
b
μr   (x  A) r f(x)dx
a
Central moments
b
r
μ r  E[X  E(X)]   (X  E(X))r f(x)dx
a
19
Relationship between Raw Moments and Central
Moments
μ1  0 (always)
μ 2  μ2  μ12
μ 3  μ 3  3μ 2μ1  2μ13
μ 4  μ4  4μ3μ1  6μ2μ1 2  3μ14
20
Moment Generating Function (M.G.F)
It is a function which automatically generates the
raw moments. For a random variable X, the
moment generating function is denoted as MX(t)
and is derived as MX(t) = E(etX).
Reason for the name M.G.F
M X (t )  E (e tx )


t 2 X 2 t 3X 2
 E 1  tX 

 
2!
3!


 t 2X2 
  
 E(1)  E(tx)  E
 2! 
21
t2
 1  tE(X)  E(X 2 )  
2!
t2
 1  tμ1  μ 2  
2!
Here
μ1
μ 2
= coefficient of t in MX(t)
= coefficient of
t2
2!
in MX(t)
In general μr = coefficient of
t2
in
2!
MX(t).
22
Problems
1. The p.m.f of a RV X, is given by Find MGF, mean
and variance.
Solution
  e
M X t   E e tX 


 etx
x 0
p ( x)
1
2x
 et 
   
x 0  2 

tx
x
2
3
4
 e t   e t   et   et 
             ....
2 2 2 2
2
3
4

e t   e t   e t   e t   e t 

1              ..

2 2 2 2 2


et 1
et


t
2
e
2  et
1
2
23
Differentiating twice with respect to t
t  t  t  t 

t
 2  e  e   e   e 
2
e
 


M X t   
2
2
t
t


2e 
2e 




2
M X t  
t  t
t 
t 
t

 2  e   2e   2e 2 2  e   e 

 




t 4

2t
t

2  e 



2  e 


put t = 0 above
t
4e  2e
3
E ( X )  M X 0  2
 
E X 2  M X 0   6
 
2
Variance  E X 2  E  X   6  4  2
24
2. Find MGF of the RV X, whose pdf is given by and
hence find the first four central moments.
Solution
 
M X t   E e tX


tx
e
 f x dx


  e tx e x dx
0

   e   t x dx
0
  t  x

e

 





t


0


  t 
25
Expanding in powers of t

M X t  

  t 
2
3
1
t t t
 1           ...
t
  
1  

Taking the coefficient we get the raw moments
about origin
E  X   coefficient of t 1! 
1

2
2
 2 
E  X    coefficient of t 2! 
2

 


6
3
24
 coefficient of t 4 4!  4

  

  

E X 3  coefficient of t 3 3! 
E X4
26
and the central moments are
1  0
2
 2   2  2C11 1  1

2

2
2
1

2

1

2

1

2
2
3
 3   3  3C1  2 1  3C 2 1 1  1
6
2 1
1 1
1
2

3
3


3
2 
3
3
 2





2
4
4
 4   4  4C1 3 1  4C 2  2 1  4C 3 1  1
24
6 1
2 1
1
1
9

4
6
4


4
3 
2 2
4
4
4


 



27
3. If the MGF of a (discrete) RV X is
distribution of X and p ( X = 5 or 6).
Solution
M X t  
1
1

5  4e t
 4e t
51 
5

1   4e t
 1  
5  5

1
5  4e
t
find the



  4e t
  
  5
2
3

  4e t 
  
  ...

  5 
By definition
 tX 
M X t   E  e  


tx
 e p ( x)
1  e
t0
t
p (0)  e p (1)  e
t2
p (2)  ...
28