Download 4.2 Variance and Covariance

E:
I:
A  B  ,
Mutually Exclusive(E) &
Statistically Independent(I)
P( A  B)  P( A) P( B);
1 Tossing a coin. Let A ={Head}, B={Tail}, then A and B
are ( ) but not ( );
2 Tossing two coins one by one. Let C ={First Head},
D={Second Head}, then C and D are ( ) but not ( );
Solution:
P(CD) = ¼= P(C) P(D)
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Chapter 4 Mathematical Expectation

4.1 Mean of Random Variables

4.2 Variance and Covariance

4.3 Means and Variances of Linear
Combinations of Random variables

4.4 Chebyshev’s Theorem
2
In general
If X and Y are two random variables, f(x, y) is the
joint density function, then:
 E(X)=   xf ( x, y ) =
x
y
 xg ( x)
x
 

 

 E(X) =   xf ( x, y )dxdy   xg ( x)dx
 E(Y) =   yf ( x, y ) =  yh( y )
y x
 
(discrete case)
y
(continuous case)
(discrete case)

 E(Y) =  yf ( x, y )dxdy  yh( y )dy
(continuous case)
g(x) and h(y) are marginal probability distributions of X and Y,
respectively.
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4.2 Variance and Covariance

The most important measure of variability of a
random variable X is obtained by letting
g(X) =
( X  )
2
then E[g(X)] gives a measure of the variability
of the distribution of X.
4
Definition 4.3 Let X be a random variable with
probability distribution f(x) and mean . The
variance of X is
 2  E[( X   ) 2 ]   ( x   ) 2 f ( x)
x
if X is discrete, and

 2  E[( X   ) 2 ]   ( x   ) 2 f ( x)dx
if X is continuous.



The positive square root of the variance, , is called the
standard deviation of X.
The quantity x -  is called the deviation of an
observation, x, from its mean.
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Note

2≥0

When the standard deviation of a random
variable is small, we expect most of the values
of X to be grouped around mean.

We often use standard deviations to compare
two or more distributions that have the same
unit measurements.
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Example 4.8 Page97
Let the random variable X represent the number of automobiles
that are used for official business purposes on any given workday.
The probability distributions of X for two companies are given
below:
Company A:
Company B:
x
1
2
3
f(x)
0.3
0.4
0.3
x
0
f(x)
0.2
1
2
0.1 0.3
3
4
0.3 0.1
Find the variances of X for the two companies.
Solution:
B  2
  (1)(0.3)  (2)(0.4)  (3)(0.3)  2
A
 A2  (1  2) 2 (0.3)  (2  2) 2 (0.4)  (3  2) 2 (0.3)  0.6
4
 B2   ( x  2) 2 f ( x)  1.6
x 0
 B2   A2
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Theorem 4.2
The variance of a random variable X is
 2  E( X 2 )   2 .
Proof: For the discrete case we can write
   ( x   ) f ( x)   ( x 2  2x   2 ) f ( x)
2
2
x
x
  x 2 f ( x)  2   x f ( x)   2  f ( x).
x
Since
x
   xf ( x)
by definition, and
x
 f ( x)  1
for any
x
x
discrete probability distribution, it follows that
 2   x 2 f ( x)   2  E ( X 2 )   2 .
x
For the continuous case the proof is step by step the same, with
summations replaced by integrations.
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Example 4.10
The weekly demand for Pepsi, in thousands of liters, from a
local chain of efficiency stores, is a continuous random
variable X having the probability density
1  x2
elsewhere
2( x-1) ,
f ( x)  
0,
Find the mean and variance of X.
Solution:
2
  E( X )  2 x( x  1)dx  5 3.
1
and
2
E ( X )  2 x 2 ( x  1)dx  17 6 .
2
1
Therefore,
 2  17 6  (5 3) 2  1 18.
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Theorem 4.3 Let X be a random variable with
probability distribution f(x). The variance of the
random variable g(X) is
 g ( X )  E[( g ( X )   g ( X ) ) 2 ]   ( g ( x)   g ( X ) ) 2 f ( x)
2
x
if X is discrete, and

 g ( X ) 2  E[( g ( X )   g ( X ) ) 2 ]   ( g ( x)   g ( X ) ) 2 f ( x)dx

if X is continuous.
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Definition 4.4 Covariance of two random variables
Let X and Y be random variables with joint probability
distribution f(x, y). The covariance of X and Y is
 XY  E[( X   X )(Y  Y )]   ( x   X )( y  Y ) f ( x, y )
x
y
if X and Y are discrete, and
 XY  E[( X   X )(Y  Y )] 
 
  (x  
X
)( y  Y ) f ( x, y )dxdy
  
if X and Y are continuous.
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Remark




If large values of X often result in large values of Y or small
values of X result in small values of Y, then positive X–  X
will often result in positive Y – Y .Thus the product (X –  X)(Y
– Y) will tend to be positive (positive association between X
and Y).
Similarly, if small values of X often result in large values of Y
or large values of X result in small values of Y, then positive X
–  X will often result in negative Y – Y. Thus the product (X
–  X )(Y – Y ) will tend to be negative (negative association
between X and Y).
The sign of the covariance indicates whether the relationship
between two variables is positive or negative.
It can be shown that when two variables are independent, then
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the covariance of the two variables is zero.
Theorem 4.4 The covariance of two random
variables X and Y with means  X and Y ,
respectively, is given by
 XY  E( XY )   X Y .
Proof: For the discrete case we can write
 XY  
 (x  
x

)( y  Y ) f ( x, y )
y
 ( xy  
x

X
X
y  Y x   X  Y ) f ( x , y )
y
 xyf ( x, y)     yf ( x, y)
Y
X
x
y
 Y 
x
x
 xf ( x, y)  
y
X
X
y
Y 
x
 f ( x, y )
y
1
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Example 4.14, page 101
The fraction X of male runners and the fraction Y of female
runners who compete in marathon races is described by the joint
density function:
0  x  1, 0  y  x
elsewhere
8 xy,
f ( x, y )  
0,
Find the covariance of X and Y.
Solution: We first compute the marginal density function. They are
4 y (1  y 2 ), 0  y  1
h( y )  
elsewhere
0
4 x 3 , 0  x  1
g ( x)  
0 elsewhere
1
1
 X  E ( X )   4 x 4 dx  54
Y  E(Y )   4 y 2 (1  y 2 )dy  158
0
0
1
E ( XY )  
0
1

y
8 x 2 y 2 dxdy 
4
9
 XY  E ( XY )   X Y  94  ( 54 )( 158 ) 
4
225
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Correlation Coefficient
The measure of the strength of the relationship
Definition 4.5 Let X and Y be random variables
with covariance  XY and standard deviation
 X and  Y , respectively. The correlation
coefficient of X and Y is
 XY
 XY

 XY
15
Example 4.14, page 101
The fraction X of male runners and the fraction Y of female
runners who compete in marathon races is described by the joint
density function:
8 xy,
f ( x, y )  
0,
0  x  1, 0  y  x
elsewhere
Find the correlation coefficient of X and Y.
Solution: We first compute the marginal density function. They are
4 x 3 , 0  x  1
g ( x)  
0 elsewhere
1
 X  E ( X )   4 x 4 dx  54
0
 XY

 XY 
 XY
4 y (1  y 2 ), 0  y  1
h( y )  
elsewhere
0
1
Y  E(Y )   4 y 2 (1  y 2 )dy  158
0
4
225
2 11
75 225

4
1

66 2
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Remark

 XY
is free of units.
Satisfies the inequality
  XY  0

 XY  1
if
1   XY  1
 XY  0
(X and Y are independent)
if Y = a + bX
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