251v2outl 4/19/2006 (Open this document in 'Outline' view!)
K. Two Random Variables.
1. Regression (Summary).
2. Covariance (  xy and s xy )
a. Population Covariance
The population covariance is defined, using probability, as
Cov( x, y )   xy  E x   x  y   y  E xy    x  y . This



can be used to describe the relationship between x and y .
If the covariance is positive we can say that x and y tend to
move together, while if it is negative we can say that they
tend to move in opposite directions. In order to use this
formula we must realize that E xy    xyPx, y  .
This means that we must add together the product of x and y ,
together with their joint probability, for each possible pair of
values of x and y . For example, assume that x and y are
x
related by the following joint probability table:
400
y 600
800





400
.12
600
.15
.10
.05
.16
.07
We begin by taking the upper left hand probability, .12, which
is the probability that both x and y are 400, and multiplying
it by 400 twice. Then we take the next probability in the same
row, .15, which is the probability that x is 600 and y is 400,
and multiply it by both 600 and 400. If we continue in this way we get
E xy  
xyPxy 

 .12 400 400  .15600 400  .18800 400 
  .10 400 600  .05600 600  .08800 600 
  .16 400 800  .07 600 800  .09 800 800 
 19200 36000 57600 
 24000 18000 38400   335600 .
 51200 33600 57600 
We can now use the following tableau to compute the means
and variances of x and y .
800 .
.18 

.08 
.09 

x
400
y
600
800
Px 
xPx 





400
.12
600
.15
.10
.05
.16
.07
.38
.27
152 
162 
800
.18 

.08 
.09 

.35
280 
P y 
.45
yP y  y 2 P y 
180
72000
.23
138
82800
.32
256
204800
1.00
574
359600
594
x 2 Px  60800  97200  224000  382000
2
 Px   1 (a check),
  E x    xPx   594 , E x    x Px   382000 ,
 P y   1 ,   E y    yP y   574 and
E y    y P y   359600
To summarize
2
2
x
y
2
2
We will need the variances below. To complete what we have
done, write
 xy  Covxy  Exy   x  y  335600 594574  5356
b. The Sample Covariance
The sample covariance is much easier to compute, the formula
being
s xy 
 x  x  y  y    xy  nx y .
n 1
n 1
For example, assume that we have data on income ( x ) and
savings ( y )(in thousands) for 5 families.
Family
1
2
3
4
5
Sum
 x  34 .7,  y  2.8,  x
and  xy  24 .22 .
Then x 
s x2 
2
x
y
x2
y2
xy
1.9
12.4
6.4
7.0
7.0
34.7
0.0
0.9
0.4
1.2
0.3
2.8
3.61
153.76
40.96
49.00
49.00
296.33
0.00
0.81
0.16
1.44
0.09
2.50
0.00
11.16
2.56
8.40
2.10
24.22
 296 .33,
y
2
 2.50 ,
34 .7
2 .8
 6.94 and y 
 0.56 .
5
5
x
2
y
2
 nx 2
n 1

296 .33  56.94 2
 13 .878 ,
4
2.50  50.56 2
 0.2330 and since
n 1
4
24.22  56.94 0.56 
xy  24 .22 , s xy 
 1.197 .
5 1
The positive sign of s xy , the sample covariance, indicates
s 2y 
 ny 2


that x and y tend to move together.
3
3. The Correlation Coefficient (  xy and rxy )
The size of a covariance is relatively meaningless; to judge the strength
of the relationship between x and y we need to compute the
correlation, which is found by dividing the covariance by the standard
deviations of x and y .
a. Population Correlation.
For the population covariance, recall from above that
 
 x2  E x 2   x2  382000  594 2  29164 and
 
 y2  E y 2   y2  359600  5742  30124 . So that
 xy 
 xy
 x y

 5356

 5356
 0.181 .
170 .77 173 .56 
29164 30124
The correlation must always be between positive 1 and
negative 1  1.0    1.0 . A correlation close to zero is
called weak. A correlation that is close to one in absolute value
is called strong. (Actually statisticians prefer to look at the
value of the correlation squared.) A strong positive correlation
indicates that x and y have a relationship that is close to a
straight line with a positive slope. A strong negative
correlation means that the relationship approximates a straight
line with a negative slope. Unfortunately, the correlation only
indicates linear relationships; a nonlinear relationship that is
obvious on a graph may give a zero correlation.
b. Sample Correlation.
Recall that s xy  1.197 , s x2  13 .878 , and s 2y  0.2330 . If we
divide the correlation by the two standard deviations, we find
s xy
1.197
that rxy 

 0.6657 .
sx s y
13 .878 0.2330
4. Functions of Two Random Variables.
Cov(ax  b, cy  d )  acCov( x, y)
and if w  ax  b and v  cy  d ,  wv  signac xy
or Corr (ax  b, cy  d )  (sign(ac))Corr ( x, y) , where
signac has the value 1 or 1 depending on whether the
product of a and c is negative or positive.
5. Sums of Random Variables.
4
a. Ex  y   Ex  E y  and
Var x  y    x2   y2  2 xy  Varx   Var y   2Covx, y 
b. Independence.
(i) Definition.
Px, y   Px P y 
(ii) Consequences
If x and y are independent,
E xy   Ex E y  , Covx, y   0 ,  xy  0
and Var x  y   Var x   Var  y  .
c. If a, c and d are constants, Var(ax  cy)  a 2Var( x)  c 2Var( y)  2acCov( x, y) .
This and a. imply that Eax  cy  d   aEx   cE y   d
and Var(ax  cy  d )  a 2Var( x)  c 2Var( y)  2acCov( x, y)
d. Application to portfolio analysis – Most of this is
from the document 251var2 in the supplement.
If R  P1 R1  P2 R2 and P1  P2  1 , then
E R   P1 E R1   P2 E R2  and VarR  P12VarR1   P22VarR2   2P1 P2CovR1 , R2 
is the variance of the return. Thus if P1 and P2 are both .50 ,
we can say
VarR =.25VarR1 +.25VarR2 +.50CovR1 R2  .
For example, assume that  R1  0.20 ,  R2  0.30 ,
but  R1R 2 is unknown. Then
CovR1 ,R2    R1R2  R1  R2 = R1R2 .20.30=.06 R1R2 .
If we use the formula for VarR  immediately above, VarR  .25.202 +.25.302 +.50.06 R1R2
= .0100 + .0225 + .0300 R1R 2  .0325 + .0300 R1R 2 .
Now we can see the effect various values of R1R 2 will have
on VarR  and  R  Var R  .
If  R1R2  1, VarR=.0325+.03001 =.0625 and  R  .2500.
If  R1R2  0, VarR=.0325+.03000 =.0325 and  R  .1803.
If  R1R2  1,VarR=.0325+.0300-1=.0025 and  R  .0500 .
The purpose of this section is to show how to find the minimum value
for Var R  . Since variance is a measure of risk, minimizing variance minimizes risk, though actually, the
best measure of risk is probably the coefficient of variation, the standard deviation divided by the mean, in
this case C 
R
E R 
.
Remember that VarR= R2 =P12VarR1 +P22VarR2 +2P1 P2 CovR1 R2  .
Also recall that, since P1 and P2 are shares of $1.00, P1+P2  1 , then P2  1  P1 .
5
Remember too, that CovR1 ,R2    R1R2  R1  R2 .
If we put all this together,
VarR  P12VarR1   1-P1 2 VarR2   2P1 1-P1  R1R2  R1 R2 .
Now let us assume some values for the standard deviations and the
correlation.
Let  R1  0.4 soVarR1   0.16 ,  R2  0.3 soVarR2   0.09 and
CorrR1 ,R2    R1R2  0.5 . Then
VarR   P12 0.16   1-P1 2 0.09   2P1 1-P1 0.50.40.6.
 0.16 P12  0.09 1-P1 2  20.06 P1 1  P1 



 0.16P12  0.09 1  2P1  P12  0.12 P1-P12
 0.16P12

 0.09  0.18P12
 0.09P12  0.12P1  0.12P12
If we collect terms in P1 and P12 , we get
Var R  0.16  0.09  0.12 P12   0.18  0.12 P1  0.09
or Var R  0.13P12  0.06P1  0.09 .
  
 



In order to minimize risk we pick our value of P1 to give
us a minimum variance.
If we know calculus, the way that we find this minimum
variance is by taking the first derivative of VarR  with
respect to P1 and setting it equal to zero.
Since
d
Var R   0.26 P1  0.06 , if we set the variance
dP1
equal to zero we get 0.26 P1  0.06  0 , which implies that
0.06
 0.2308 . Now since P1  P2  1 , we set
0.26
P2  1  P1  0.7692 . That is, to minimize risk, we put about
23% of our money in stock 1 and 77% in stock 2.
If we do not know calculus, we can still minimize VarR  0.13P12  0.06P1  0.09 . Try values of P1 at
intervals of 0.1 between zero and one. We will find that the smallest values of
VarR  occur at P1  0.2 and P1  0.3. Now we can try values
P1 
of P1 at intervals of 0.01 between 0.2 and 0.3. We will find
that the smallest value of VarR  occurs at P1  0.23 .
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