Download 2.1 Random Variables, Expected Values and Variance

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2. Probability Distributions
2.1. Random Variables, Expected Values and Variance
Definition (random variable): A random variable is a numerical description
of the outcome of an experiment.
There are two common types of random variables. They are:
Discrete random variable: a quantity assumes either a finite number of
values or an infinite sequence of values, such as 0, 1, 2, 
Continuous random variable: a quantity assumes any numerical value in an
interval or collection of intervals, such as time, weight, distance, and
temperature.
Definition (probability distribution (or density) function (p.d.f)): a
function describes how probabilities are distributed over the values of the
random variable.
Required conditions for a discrete probability distribution function:
Let a1 , a 2 ,, a n , be all the possible values of the discrete random variable X.
Then, the required conditions for
f (x)
to be the discrete probability
distribution for X are
(a) P X  ai   f (ai )  0, for every i.

(b)
 f a   f a   f a     f a     1
i 1
i
1
2
n
Required conditions for a continuous probability density:
Let the continuous random variable Z taking values in subsets of  ,  .
Then, the required conditions for f (x) to be the continuous probability density
function for Z are
(a)
f ( x)  0,    x  .

(b)

f ( x )dx  1

Note: As X is discrete,
P(c  X  d ) 
 f a  .
cai d
15
i
16
As X is continuous,
d
P (c  X  d ) 
 f ( x)dx.
c
Expected Value and Variance:
As X is discrete,

E ( X )     ai f ai   a1 f ( a1 )  a2 f (a2 )    an f (an )  
i 1
and
Var ( X )   2  E X  E ( X )   (ai   ) 2 f (ai )
2
i
 (a1   ) f (a1 )  (a2   ) f (a2 )    (an   ) 2 f (an )  
2
2
As X is continuous,
E( X )   

 xf ( x) dx

and
Var ( X )   2  E  X  E ( X ) 
2

 ( x  u)
2
f ( x) dx

Important Properties of Expected Value and Variance:
a, b are constants and X is discrete or continuous.
1.
 
Var X   E X 2   2 .
2. EaX  b  aE X   b .
3. Var aX  b   a 2Var  X 
Example 1:
The probability distribution functions (discrete random variable) or probability
density functions (continuous random variable) for a random variable X are
(a)
c exp  6 x , x  0

f ( x)    cx,  1  x  0

0, otherwise

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(b)


f x   cx 2 exp  x 3 , x  0
(c)
x
1
f  x   c  , x  0, 1, 2, 
3
Find c .
[solution:]
(a)
  cx 2 
  c exp  6 x 



cxdx

c
exp

6
x
dx

1




1
0

 1
2
6

0

 1
c c
3c  c
3
   1
 1 c 
2 6
6
2

0

0
(b)


 

  c exp  x 3 
1
3
3
cx
exp

x
dx

1

c
exp

x
dx

1


 1
0
0 3
3

0
c
  1 c  3
3
 
2
3
 
(c)
x

1
c   1 

x 0  3 
c 
 1 
 1  1 2

  1  3c  1
c 1        1  c
2
 1  1 
 3  3 

3

2
3
Example 2:
The probability distribution function for a discrete random variable X is
f ( x )  2k , x  10
k , x  20
k  0.2, x  30
0, otherwise
where k is some constant. Please find
(a) k (b) P( X  15 or X  40 ) (c) E X  and Var X  (d) E5 X  2
(e) Var 3 X  7
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[solution:]
(a)
 f ( x)  f (10)  f (20)  f (30)  2k  k  k  0.2  1
x
 k  0.3 .
(b) P( X  15 or X  40)  P( X  10 or X  20 or X  30)  1 .
(c)
u  E ( X )   xf ( x)  10  f (10)  20  f (20)  30  f (30)
x
 10  0.6  20  0.3  30  0.1  15
and
E ( X 2 )   x 2 f ( x)  10 2  f 10  20 2  f 20  30 2  f 30
x
 100  0.6  400  0.3  900  0.1  270
 
 Var  X   E X 2   2  270  225  45
(d) E5 X  2  5E X   2  5 15  2  77.
(e) Var (3 X  7)  32 Var X   9  45  405 .
Example 3:
Let X be a discrete random variable representing the number of hours
a college student spending on reading novels per week. The following
probability distribution has been proposed.
i3
f i  
, i  1, 2, 3 ,
9c
where k is some constant.
(a) Compute c. (b) Compute P X  1.2 and P X  2.2 .
(c) Compute E X  a nd Var X  .
[solution:]
(a)
 f (i)  f (1)  f (2)  f (3) 
i
13 23 33 36 4



 1
9c 9c 9c 9c c
 c  4.
(b) P ( X  1.2)  P ( X  1)  f 1 
1
1
.and

9  4 36
P( X  2.2)  P( X  3)  f 3 
(c)
18
33
27

.
9  4 36
19
u  E ( X )   if (i )  1  f (1)  2  f (2)  3  f (3)
i
 1
1
8
27 98 49
 2
 3


36
36
36 36 18
and
E ( X 2 )   i 2 f (i )  12  f 1  2 2  f 2   32  f 3
i
 1
1
8
27 276 46
 4
 9


36
36
36
36
6
 
 Var  X   E X 2   2 
2
46  49 
    0.2561
6  18 
Example 4:
The probability density function for a continuous random variable X is
f ( x )  a  bx 2 , 0  x  1
0, otherwise.
where a, b are some constants. Please find
(a) a, b if E ( X ) 
3
(b) Var( X ) .
5
[solution:]
(a)
 a  bx dx  1 
1

1
f ( x)dx  1 
2
0
ax 
0
 a
b 3 1
x |0  1
3
b
1
3
and
1
1
0
0


E ( X )   xf ( x)dx   x a  bx 2 dx 
Solve for the two equations, we have a 
a 2 b 4 1 a b 3
x  x |0   
2
4
2 4 5
3
6
.
, b 
5
5
(b)
f ( x) 
3
6

x2 , 0  x  1
5
5
0, otherwise.
Thus,
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Var ( X )  E  X  E ( X )  E ( X )  E ( X )
2
2
2
 3
 E( X )   
5
2
9
3 6 2
  x dx 
25
5 5

0
1
6 5 1
9
1
6
9
2
 x3 
x |0 
 


5
25
25 5 25 25 25
1
1
9
  x f ( x ) dx 

25
0
x
2
2
Example 5:
The probability density function for a continuous random variable X is
x  2

,2 x  4
f ( x )   18

0, otherwise



Please find (a) P X  1 (b) P X 2  9 (c) E X  and Var X 


(d) E 9 X 2  8 X  2 (e) Var6 X  8 .
[solution:]
1
 x2 x
x2
 1 1  1 1 2
P X  1  P 1  X  1  
dx             
18
 36 9  1  36 9   36 9  9
1
1
(a)
(b)


P X 2  9  P 3  X  3 
x2
3 18 dx 
3
2
 0dx 
3
3
 x2 x 
x2
25
2 18 dx   36  9   36
2
3
(c)
4
EX    

2
x  2 dx 
x
18
4
 x2 x 
 x3 x2 
  18  9 dx   54  18   2 .
 2
2
4
Since
   x
4
E X
2
2
2
x  2 dx 

18
4
 x3 x2 
 x4 x3 
  18  9 dx   72  27   6 ,
 2
2
4
 
Var  X   E  X     E X 2   2  6  2 2  2 .
2


 
(d) E 9 X 2  8 X  2  9E X 2  8E  X   2  9  6  8  2  2  72 .
(e) Var 6 X  8  6 2 Var  X   36  2  72
20
2
21
Definition (cumulative distribution function (c.d.f)): F x  P X  x.
Important Properties of CDF:
1. CDF is right continuous.
2. As X is discrete, F x  
 f a  ;
ai  x
i
As X is continuous and f is continuous, f x   F ' x  .
Example 2 (continuous):
The probability distribution function for a discrete random variable X is
f ( x )  0.6, x  10
0.3, x  20
0.1, x  30
0, otherwise
Thus, the CDF is
F ( x )  0, x  10
0.6, 10  x  20
0.3, 20  x  30
1, x  30
Example 1 (a):
The probability distribution functions (discrete random variable) or probability
density functions (continuous random variable) for a random variable X are
3
 2 exp  6 x , x  0

3

f ( x)   
x,  1  x  0
2

0, otherwise



Thus, as  1  x  0 ,
x
  3x 2 
 3x
3
2
F x   
dx  
  1 x
2
 4  1 4
1
x


as 0  x   ,
 3x
3
3   3 exp  6 x 
3 1
F x   
dx   exp  6 x dx   

  1  exp  6 x  .

2
2
4  2
6
0 4 4
1
0
x
x
x
21
22
Thus, the CDF is
F ( x )  0, x  1


3
1 x2 , 1  x  0
4
3
1
1  exp  6 x , 0  x  

4
4
22