# Chapter 2 Random Variables

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```					    Chapter 2. Random Variables

2.1 Discrete Random Variables
2.2 Continuous Random Variables
2.3 The Expectation of a Random Variable
2.4 The Variance of a Random Variable
2.5 Jointly Distributed Random Variables
2.6 Combinations and Functions of Random Variables

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2.1 Discrete Random Variable
2.1.1 Definition of a Random Variable (1/2)

• Random variable
– A numerical value to each outcome of a particular
experiment
S

R

-3   -2    -1     0    1     2    3

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2.1.1 Definition of a Random Variable (2/2)

• Example 1 : Machine Breakdowns
– Sample space : S  {electrical , mechanical , misuse}
– Each of these failures may be associated with a repair cost
– State space : {50, 200,350}
– Cost is a random variable : 50, 200, and 350

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2.1.2 Probability Mass Function (1/2)

• Probability Mass Function (p.m.f.)
– A set of probability value pi assigned to each of the values
taken by the discrete random variable xi
– 0  pi  1 and i pi  1
– Probability : P( X  xi )  pi

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2.1.2 Probability Mass Function (1/2)

• Example 1 : Machine Breakdowns
– P (cost=50)=0.3, P (cost=200)=0.2,
P (cost=350)=0.5
– 0.3 + 0.2 + 0.5 =1                   xi     50    200   350

0.3   0.2   0.5
pi
f ( x)
0.5

0.3
0.2

50     200       350   Cost(\$)
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2.1.3 Cumulative Distribution Function (1/2)

• Cumulative Distribution Function
– Function : F ( x)  P( X  x) F ( x)     P( X
y: y  x
 y)
– Abbreviation : c.d.f

F ( x)
1.0

0.5

0.3

0   50           200                  350    x(\$cost)
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2.1.3 Cumulative Distribution Function (2/2)

• Example 1 : Machine Breakdowns
  x  50  F ( x)  P(cost  x)  0
50  x  200  F ( x)  P(cost  x)  0.3
200  x  350  F ( x)  P(cost  x)  0.3  0.2  0.5
350  x    F ( x)  P(cost  x)  0.3  0.2  0.5  1.0

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2.2 Continuous Random Variables
2.2.1 Example of Continuous Random Variables (1/1)

• Example 14 : Metal Cylinder Production
– Suppose that the random variable X is the diameter of a
randomly chosen cylinder manufactured by the company.
Since this random variable can take any value between
49.5 and 50.5, it is a continuous random variable.

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2.2.2 Probability Density Function (1/4)

• Probability Density Function (p.d.f.)
– Probabilistic properties of a continuous random variable

f ( x)  0
statespace
f ( x)dx  1

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2.2.2 Probability Density Function (2/4)

• Example 14
– Suppose that the diameter of a metal cylinder has a p.d.f
f ( x)  1.5  6( x  50.2)2 for 49.5  x  50.5
f ( x)  0, elsewhere

f ( x)

49.5          50.5   x
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2.2.2 Probability Density Function (3/4)

• This is a valid p.d.f.
50.5

49.5
(1.5  6( x  50.0) 2 )dx  [1.5 x  2( x  50.0)3 ]50.5
49.5

 [1.5  50.5  2(50.5  50.0)3 ]
[1.5  49.5  2(49.5  50.0)3 ]
 75.5  74.5  1.0

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2.2.2 Probability Density Function (4/4)

• The probability that a metal cylinder has a diameter between
49.8 and 50.1 mm can be calculated to be
50.1
 49.8
(1.5  6( x  50.0) 2 )dx  [1.5 x  2( x  50.0)3 ]50.1
49.8

 [1.5  50.1  2(50.1  50.0)3 ]
[1.5  49.8  2(49.8  50.0)3 ]
f ( x)                                    75.148  74.716  0.432

49.5   49.8     50.1    50.5      x

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2.2.3 Cumulative Distribution Function (1/3)

• Cumulative Distribution Function
x
 F ( x)  P( X  x)          f ( y )dy


dF ( x )
 f ( x) 
dx

 P ( a  X  b)  P ( X  b)  P ( X  a )
 F (b)  F (a)

 P ( a  X  b)  P ( a  X  b)

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2.2.2 Probability Density Function (2/3)

• Example 14
x
F ( x)  P( X  x)   (1.5  6( y  50.0) 2 )dy
49.5

 [1.5 y  2( y  50.0)3 ]49.5
x

 [1.5 x  2( x  50.0)3 ]  [1.5  49.5  2(49.5  50.0) 3 ]
1.5 x  2( x  50.0)3  74.5
P(49.7  X  50.0)  F (50.0)  F (49.7)
 (1.5  50.0  2(50.0  50.0)3  74.5)
(1.5  49.7  2(49.7  50.0)3  74.5)
 0.5  0.104  0.396

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2.2.2 Probability Density Function (3/3)

P(49.7  X  50.0)  0.396
1

P( X  50.0)  0.5
F ( x)

P( X  49.7)  0.104

49.5      49.7       50.0              50.5   x

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2.3 The Expectation of a Random Variable
2.3.1 Expectations of Discrete Random Variables (1/2)

• Expectation of a discrete random variable with p.m.f
P( X  xi )  pi

E( X )       px
i
i    i

• Expectation of a continuous random variable with p.d.f f(x)

E( X )     state space
xf ( x)dx

• The expected value of a random variable is also called the
mean of the random variable

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2.3.1 Expectations of Discrete Random Variables (2/2)

• Example 1 (discrete random variable)
– The expected repair cost is
E (cost)  (\$50  0.3)  (\$200  0.2)  (\$350  0.5)  \$230

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2.3.2 Expectations of Continuous Random Variables (1/2)

• Example 14 (continuous random variable)
– The expected diameter of a metal cylinder is
50.5
E( X )             x(1.5  6( x  50.0)2 )dx
49.5
– Change of variable: y=x-50
0.5
E ( x)           ( y  50)(1.5  6 y 2 )dy
0.5
0.5
           (6 y 3  300 y 2  1.5 y  75)dy
0.5

 [3 y 4 / 2  100 y 3  0.75 y 2  75 y ]0.5
0.5

 [25.09375]  [24.90625]  50.0

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2.3.2 Expectations of Continuous Random Variables (2/2)

• Symmetric Random Variables
– If x has a p.d.f f ( x) that is
f ( x)   E( X )  
so that
f (   x)  f (   x)

– Then, E ( X )     (why?)

– So that the expectation of
the random variable is equal
to the point of symmetry
         x

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E ( X )   xf ( x)dx
              
    xf ( x)dx +  xf ( x)dx
-            

 y  2  x
                   
    xf ( x)dx +   
-                  -
yf ( y )dy



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2.3.3 Medians of Random Variables (1/2)

• Median
– Information about the “middle” value of the random variable
F ( x)  0.5
• Symmetric Random Variable
– If a continuous random variable is symmetric about a point    ,
then both the median and the expectation of the random
variable are equal to 

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2.3.3 Medians of Random Variables (2/2)

• Example 14

F ( x)  1.5 x  2( x  50.0)3  74.5  0.5

x  50.0

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2.4 The variance of a Random Variable
2.4.1 Definition and Interpretation of Variance (1/2)

• Variance(  )
2

– A positive quantity that measures the spread of the
distribution of the random variable about its mean value
– Larger values of the variance indicate that the distribution is

– Definition:       Var( X )  E (( X  E ( X ))2 )
 E ( X 2 )  ( E ( X ))2

• Standard Deviation
– The positive square root of the variance
– Denoted by 

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2.4.1 Definition and Interpretation of Variance (2/2)

Var( X )  E (( X  E ( X )) 2 )
 E ( X 2  2 XE ( X )  ( E ( X )) 2 )
 E ( X 2 )  2 E ( X ) E ( X )  ( E ( X )) 2
 E ( X 2 )  ( E ( X )) 2

f ( x)
Two distribution with
identical mean values but
different variances

x
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2.4.2 Examples of Variance Calculations (1/1)

• Example 1

Var( X )  E (( X  E ( X )) 2 )   pi ( xi  E ( X )) 2
i

 0.3(50  230) 2  0.2(200  230) 2  0.5(350  230) 2
17,100   2

  17,100  130.77

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2.4.3 Chebyshev’s Inequality (1/1)

• Chebyshev’s Inequality
– If a random variable has a mean  and a variance  2 then
,
1
P (   c  X    c )  1 
c2
for c  1
– For example, taking c  2 gives

1
P(   2  X    2 )  1  2  0.75
2

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• Proof

2    

( x   )2 f ( x)dx       

| x  | c
( x   )2 f ( x)dx  c 2 2      

| x  | c
f ( x)dx.

 P(| x   | c )  1/ c 2

 P(| x   | c )  1  P(| x   | c )  1  1/ c 2

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2.4.4 Quantiles of Random Variables (1/2)

• Quantiles of Random variables
– The pth quantile of a random variable X
F ( x)  p
– A probability of p that the random variable takes a value
less than the pth quantile
• Upper quartile
– The 75th percentile of the distribution
• Lower quartile
– The 25th percentile of the distribution
• Interquartile range
– The distance between the two quartiles

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2.4.4 Quantiles of Random Variables (2/2)

• Example 14
F ( x)  1.5 x  2( x  50.0)3  74.5 for 49.5  x  50.5
– Upper quartile : F ( x)  0.75 x  50.17

– Lower quartile : F ( x)  0.25    x  49.83

– Interquartile range :   50.17  49.83  0.34

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2.5 Jointly Distributed Random Variables
2.5.1 Jointly Distributed Random Variables (1/4)

• Joint Probability Distributions
– Discrete
P ( X  xi , Y  y j )  pij  0
satisfying           p
i    j
ij   1
– Continuous

f ( x, y)  0 satisfying   
state space
f ( x, y)dxdx  1

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2.5.1 Jointly Distributed Random Variables (2/4)

• Joint Cumulative Distribution Function
– Discrete

F ( x, y )  P( X  xi , Y  y j )
– Continuous

F ( x, y )      
i:xi  x j: y j  y
pij

x            y
F ( x, y )                           f ( w, z )dzdw
w        z 

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2.5.1 Jointly Distributed Random Variables (3/4)

• Example 19 : Air Conditioner Maintenance
– A company that services air conditioner units in residences
and office blocks is interested in how to schedule its
technicians in the most efficient manner
– The random variable X, taking the values 1,2,3 and 4, is the
service time in hours
– The random variable Y, taking the values 1,2 and 3, is the
number of air conditioner units

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2.5.1 Jointly Distributed Random Variables (4/4)

• Joint p.m.f
X=service time
Y=
number                                        p
i   j
ij   0.12  0.18
of units    1      2        3       4
     0.07  1.00

1       0.12   0.08    0.07     0.05
• Joint cumulative
distribution function
2       0.08   0.15    0.21     0.13
F (2,2)  p11  p12  p21  p22
 0.12  0.18  0.08  0.15
3       0.01   0.01    0.02     0.07
 0.43

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2.5.2 Marginal Probability Distributions (1/2)

• Marginal probability distribution
– Obtained by summing or integrating the joint probability
distribution over the values of the other random variable
– Discrete
P( X  i )  pi    pij
j

– Continuous

f X ( x)         f ( x, y ) dy


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2.5.2 Marginal Probability Distributions (2/2)

• Example 19
– Marginal p.m.f of X
3
P ( X  1)   p1 j  0.12  0.08  0.01  0.21
j 1

– Marginal p.m.f of Y
4
P(Y  1)   pi1  0.12  0.08  0.07  0.05  0.32
i 1

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• Example 20: (a jointly continuous case)

• Joint pdf:    f ( x, y )
• Marginal pdf’s of X and Y:

f X ( x)   f ( x, y)dy
fY ( y)   f ( x, y)dx

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2.5.3 Conditional Probability Distributions (1/2)

• Conditional probability distributions
– The probabilistic properties of the random variable X under
the knowledge provided by the value of Y
– Discrete
P( X  i, Y  j ) pij
pi| j  P( X  i | Y  j )                   
P(Y  j )       p j
– Continuous

f ( x, y )
f X |Y  y ( x) 
fY ( y )
– The conditional probability distribution is a probability
distribution.

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2.5.3 Conditional Probability Distributions (2/2)

• Example 19
– Marginal probability distribution of Y

P(Y  3)  p3  0.01  0.01  0.02  0.07  0.11
– Conditional distribution of X
p13 0.01
p1|Y 3    P ( X  1| Y  3)           0.091
p3 0.11

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2.5.4 Independence and Covariance (1/5)

• Two random variables X and Y are said to be independent if
– Discrete
pij  pi  p j for all values iof  X and jof Y
– Continuous

f ( x, y)  f X ( x) fY ( y) for all x and y

– How is this independency different from the independence
among events?

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2.5.4 Independence and Covariance (2/5)

• Covariance
Cov( X , Y )  E (( X  E ( X ))(Y  E (Y )))
 E ( XY )  E ( X ) E (Y )

Cov( X , Y )  E (( X  E ( X ))(Y  E (Y )))
 E ( XY  XE (Y )  E ( X )Y  E ( X ) E (Y ))
 E ( XY )  E ( X ) E (Y )  E ( X ) E (Y )  E ( X ) E (Y )
 E ( XY )  E ( X ) E (Y )

– May take any positive or negative numbers.
– Independent random variables have a covariance of zero
– What if the covariance is zero?

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2.5.4 Independence and Covariance (3/5)

• Example 19 (Air conditioner maintenance)

E ( X )  2.59,           E (Y )  1.79
4    3
E ( XY )   ijpij
i 1 j 1

 (11 0.12)  (1 2  0.08)
    (4  3  0.07)  4.86
Cov( X , Y )  E ( XY )  E ( X ) E (Y )
 4.86  (2.59 1.79)  0.224

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2.5.4 Independence and Covariance (4/5)

• Correlation:

Cov( X , Y )
Corr( X , Y ) 
Var( X )Var(Y )

– Values between -1 and 1, and independent random
variables have a correlation of zero

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2.5.4 Independence and Covariance (5/5)

• Example 19: (Air conditioner maintenance)

Var( X )  1.162, Var(Y )  0.384
Cov( X , Y )
Corr( X , Y ) 
Var( X )Var(Y )
0.224
                0.34
1.162  0.384

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• What if random variable X and Y have linear relationship, that is,
Y  aX  b
where     a0
Cov( X , Y )  E[ XY ]  E[ X ]E[Y ]
 E[ X (aX  b)]  E[ X ]E[aX  b]
 aE[ X 2 ]  bE[ X ]  aE 2 [ X ]  bE[ X ]
 a ( E[ X 2 ]  E 2 [ X ])  aVar ( X )
Cov( X , Y )          aVar ( X )
Corr ( X , Y )                    
Var ( X )Var (Y )   Var ( X )a 2Var ( X )
That is, Corr(X,Y)=1 if a>0; -1 if a<0.

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2.6 Combinations and Functions of Random
Variables
2.6.1 Linear Functions of Random Variables (1/4)
• Linear Functions of a Random Variable
– If X is a random variable and Y  aX  b
for some numbers a, b  R then E (Y )  aE( X )  b
and Var(Y )  a 2 Var( X )

• Standardization
-If a random variable X has an expectation of  and a
variance of  ,      X  1       
2

Y             X  
            
has an expectation of zero and a variance of one.

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2.6.1 Linear Functions of Random Variables (2/4)

• Example 21:Test Score Standardization
– Suppose that the raw score X from a particular testing
procedure are distributed between -5 and 20 with an
expected value of 10 and a variance 7. In order to
standardize the scores so that they lie between 0 and 100,
the linear transformation Y  4 X  20is applied to the
scores.

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2.6.1 Linear Functions of Random Variables (3/4)

– For example, x=12 corresponds to a standardized score of
y=(4ⅹ12)+20=68

E (Y )  4E( X )  20  (4 10)  20  60

Var(Y )  42 Var( X )  42  7  112

 Y  112  10.58

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2.6.1 Linear Functions of Random Variables (4/4)

• Sums of Random Variables
– If X 1 and X 2 are two random variables, then
E ( X1  X 2 )  E ( X1 )  E ( X 2 )  (why ?)
and
Var( X1  X 2 )  Var( X1 )  Var( X 2 )  2Cov( X1, X 2 )
– If X 1 and X 2 are independent, so that Cov( X 1 , X 2 )  0
then
Var( X1  X 2 )  Var( X1 )  Var( X 2 )

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• Properties of       Cov( X 1 , X 2 )
Cov( X 1 , X 2 )  E[ X 1 X 2 ]  E[ X 1 ]E[ X 2 ]

 Cov( X 1 , X 2 )  Cov( X 2 , X 1 )
Cov( X 1 , X 1 )  Var ( X 1 ) Cov( X 2 , X 2 )  Var ( X 2 )
Cov( X 1  X 2 , X 1 )  Cov( X 1 , X 1 )  Cov( X 2 , X 1 )
Cov( X1  X 2 , X1  X 2 )  Cov( X1 , X1 )  Cov( X1 , X 2 ) 
Cov( X 2 , X1 )  Cov( X 2 , X 2 )

 Var ( X 1  X 2 )  Var ( X 1 )  Var ( X 2 )  2Cov( X 1 , X 2 )

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2.6.2 Linear Combinations of Random Variables (1/5)

• Linear Combinations of Random Variables
– If X 1 , , X n is a sequence of random variables and a1 ,                , an
and b are constants, then
E (a1 X 1     an X n  b)  a1E ( X 1 )      an E ( X n )  b

– If, in addition, the random variables are independent, then

Var(a1 X 1     an X n  b)  a1 Var( X1 )       an Var( X n )
2                    2

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2.6.2 Linear Combinations of Random Variables (2/5)

• Averaging Independent Random Variables
– Suppose that X 1 , , X n is a sequence of independent
random variables with an expectation  and a variance  2.

– Let         X1          Xn
X
n
– Then
E( X )  
and
2
Var( X ) 
n
– What happened to the variance?

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2.6.2 Linear Combinations of Random Variables (3/5)

1      1       1                     1
E( X )  E  X1     X n   E( X1)               E( X n )
n      n       n                     n
1      1
        
n      n

2                      2
1          1     1                    1
Var( X )  Var  X 1      X n     Var( X 1 )        Var( X n )
n          n     n                    n
1 2 
2               2
1
2
  2            
n              n     n

NIPRL
2.6.2 Linear Combinations of Random Variables (4/5)

• Example 21
– The standardized scores of the two tests are
10             5      50
Y1  X1 and Y2  X 2 
3             3       3
– The final score is
2    1     20     5      50
Z  Y1  Y2     X1  X 2 
3    3     9      9       9

NIPRL
2.6.2 Linear Combinations of Random Variables (5/5)

– The expected value of the final score is
20          5            50
E (Z )      E( X1 )  E ( X 2 ) 
9          9            9
 20      5         50
  18     30  
 9       9         9
 62.22

– The variance of the final score is
 20     5     50 
Var( Z )  Var     X1  X 2  
 9      9     9 
2                 2
 20              5
    Var( X 1 )     Var( X 2 )
 9               9
2           2
 20      5
    24     60  137.04
 9       9

NIPRL
2.6.3 Nonlinear Functions of a Random Variable (1/3)

• Nonlinear function of a Random Variable
– A nonlinear function of a random variable X is another
random variable Y=g(X) for some nonlinear function g.
Y  X 2, Y  X , Y  eX

– There are no general results that relate the expectation and
variance of the random variable Y to the expectation and
variance of the random variable X

NIPRL
2.6.3 Nonlinear Functions of a Random Variable (2/3)
– For example,                       f(x)=1
E(x)=0.5
f X ( x)  1 for 0  x  1
f ( x)  0 elsewhere

0              1   x
FX ( x)  x for 0  x  1
f(y)=1/y
E(y)=1.718
– Consider

Y  eX
where 1  Y  2.718

– What is the pdf of Y?
1.0                  2.718   y

NIPRL
2.6.3 Nonlinear Functions of a Random Variable (3/3)

• CDF methd

FY ( y )  P(Y  y )  P(e X  y )  P( X  ln( y ))  Fx (ln( y ))  ln( y )

– The p.d.f of Y is obtained by differentiation to be
dFY ( y ) 1
fY ( y )                                 for 1  y  2.718
dy       y
• Notice that
2.718                     2.718
E (Y )             zfY ( z )dz             1dz  2.718  1  1.718
z 1                     z 1

E (Y )  e E ( X )  e0.5  1.649

NIPRL
• Determining the p.d.f. of nonlinear relationship between r.v.s:
Given f X ( x) and Y  g ( X ) ,what is fY ( y ) ?

If    x1 , x2 ,    , xn    are all its real roots, that is,

y  g ( x1 )  g ( x2 )          g ( xn )
then                    n
f X ( xi )
fY ( y )  
i 1 | g '( xi ) |

where
dg ( x)
g '( x) 
dx

NIPRL
•   Example: determine   fY ( y )
f X ( x)  1 for 0  x  1      Y  eX
f ( x)  0 elsewhere                   where 1  Y  2.718

y  g ( x)  e x
 ln y  x             -> one root is possible in   0  x 1
dg
 ex  y
dx
1   1
fY ( y )      
| y| y

NIPRL

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