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K. Two Random Variables.
1. Regression (Summary).
2. Covariance (  xy and
s xy )
3. The Correlation Coefficient (  xy and rxy )
4. Functions of Two Random Variables.
5. Sums of Random Variables, Independence.
In the following problems (i) check for independence, (ii) Compute
Covx, y  , Corr x, y  and, (iii) Compute Ex  y  and


Var x  y  : D&C pg. 221 3, 4, 7,14. In problem 3, find the following: Px  y  4 , P x  y  4 x  0
Downing and Clark (formerly pg. 348 now posted at end of 251hwkadd) Old Computational Problem 1: For the sample data below b)
Compute Cov x, y and Corr x, y . c) Compute the mean of x  y and Var x  y .








x 34 26 9 30 47 10 34 34 45 10 47 32 47 8 45
y 6 57 89 60 95 42 31 28 90 25 45 23 52 95 48
Text 5.8 (Compute
Var x  y  ), 5.9, 5.11, 5.16, 5.17 [5.8, 5.9, 5.10, 5.13] (5.8, 5.10), K1-K4 .
This document includes Exercises 5.8, 5.9, 5.11, 5.16, 5.17 and Problems K1 to K4.
--------------------------------------------------------------------------------------------------------------------------------Exercise 5.8 (5.8 in 8th edition): For the joint probabilities below, compute: a) E x  and E x  ; b)  x and
 x ; c)  xy and d) Ex  y  and Var x  y  .
Joint Probability
.20
.40
.30
.10
X
-100
50
200
300
Y
50
30
20
20
Solution:
x
 100
0
50
0
200
.3
300
.1
y 30
0
.4
0
0
50
.2
0
0
0
20
We are given
x
y
20
 100
0
50
0
200
.3
300
.1
P y 
.4
yP y 
8
y 2 P y 
160
30
0
.4
0
0
.4
12
360
50
.2
0
0
0
.2
10
Px 
xPx 
.2
.4
.3
.1
 20
20
60
30
x Px  2000 1000 12000
2
and can expand this to the table below.
9000
1.0 E  y   30
90  E x 
 
24000  E x 2
500
   1020
E y
2
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 0 100 20   050 20   .3200 20   .1300 20 
E xy  
xyPxy     0 100 30   .450 30   0200 30   0300 30 
 .2 100 50   050 50   0200 50   0300 50 
0
 0  1200  600 


   0  600
0
 0  1400
0
0
 0
 1000

a)  x  E x  
 xPx  90
 
 E y  
 y  Ey  
 yP y   30
b)  x2  E x 2   x2  24000  90 2  15900 . So  x  15900  126 .095 .
 y2
2
2
y
 1020  302  120 . So  y  120  10.954 .
c)  xy  Covxy  Exy   x  y  1400  3090  1300 .
Note that  xy 
 xy

 1300
 0.9411 .
15900 120
e) From the outline Ex  y   Ex   E y  . So Ex  y   90  30  120 .
 x y
From the outline Var x  y    x2 y   x2   y2  2 xy  Var x   Var  y   2Covx, y  .
So  x2 y  15900  120  2 1300   13420 and  x y  13420  115.845 .
Exercise 5.9 (5.9 in 8th edition): If we assign the weight w  .4 to investment X (and .6 to investment Y in
a portfolio of two stocks), E x   50, E y   100 ,  x2  9000,  y2  15000 and  xy  7500 , find a) the
portfolio expected return and b) the portfolio risk.
Solution: a) If we use the formulas in the outline or the syllabus supplement, E ( R)  wE ( x)  1  wE ( y)
 .450   .6100   80 . b) If we measure risk by the standard deviation,
 R2  w 2 x2  1  w2  y2  2w1  w xy  .42 9000  .62 15000  2.4.67500
 1440  5400  3600  10440 , so  x  10440  102 .176 . It is really more appropriate to compute the
coefficient of variation. C 
std .deviation 102 .76

 1.2845 or 128.45%.
mean
80
Exercise 5.11 [5.10 in 9th] (5.10 in 8th edition): In the problem in the text, Dow-Jones Fund has
E x   105 ,  x2  14725, the Weak Economy Fund has E  y   35,  y2  11025 and their covariance is
 xy  12675. Compute the portfolio expected return and risk if 50% is invested in the Dow-Jones Fund
(and the rest in the Weak Economy Fund), a) 30% is invested in the Dow-Jones Fund, b) 70% is invested in
the Dow-Jones Fund and c) Explain which of these strategies you would recommend.
Solution: From the outline and syllabus supplement, we know that E (ax  cy)  aE( x)  cE( y) and
Var(ax  cy)  a 2Var( x)  c 2Var( y)  2acCov( x, y) . If R represents the total return of a portfolio and x
is the return on asset X and y is the return on asset Y, w is the proportion of our portfolio in asset X and
1  w is the proportion of our portfolio in Asset Y, our return on the portfolio is R  wx  1  wy. Note
that because w is a proportion, it must be between 1 and 0. If we let w replace a and 1  w replace c ,
we find that E ( R)  wE ( x)  1  wE ( y) and Var ( R)  w 2Var ( x)  1  w2 Var ( y)  2w1  wCov( x, y) or
 R2  w 2 x2  1  w2  y2  2w1  w xy .
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Originally w  .5 , E x   105 , E  y   35,  x2  14725,  y2  11025 and  xy  12675. So
E ( R)  wE ( x)  1  wE ( y)  .5105   .535   70 . If we measure risk by the standard deviation,
 R2  w 2 x2  1  w2  y2  2w1  w xy  .52 14725 .52 11025 2.5.5 12675  100 , so
 R  100  10 .
The coefficient of variation is C 
a) w  .30 , 1  w  .70 . E ( R)  wE ( x)  1  wE ( y)
std .deviation 10

 0.1429 or 14.29%
mean
70
 .3105   .735   31.50  24.50  56.00.  R2  w 2 x2  1  w2  y2  2w1  w xy
 .32 14725  .72 11025  2.3.7 12675   1325 .25  5402 .25  5323 .5  1404 , so
 R  1404  37 .47 .
The coefficient of variation is C 
 37 .47

 0.6691 or 66.91%
 56 .00
b) w  .70 , 1  w  .30 . E ( R)  wE ( x)  1  wE ( y)  .7105   .335   73.50  10.50  84.00.
 R2  w 2 x2  1  w2  y2  2w1  w xy  .72 14725  .32 11025  2.7.3 12675 
 7215 .25  992 .25  5323 .5  2884 , so  x  2884  53 .70 C 
 53 .70

 0.6393 or 66.93%
 84 .00
c) The Instructor’s Solutions Manual says “Investing 50% in the Dow Jones index fund will yield the
lowest risk per unit average return ….”. In the 8th edition, where the coefficients of variation were not
reported someone screwed up the numbers for this problem in the Instructor’s Solutions Manual . The
person who wrote the answer book had E y   70 ,  y2  37900 and  xy  23300 , which gave a much
more interesting problem. The problem here is a no-brainer. The first stock has a higher expected return
and lower variance. We would not take the 30% strategy, because it has lower returns and higher variance
than the others. The question is whether the higher risk with the 70% strategy rather than the 50% strategy
is offset by the higher return.
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Exercise 5.16 [5.13a-f in 9th]: We are setting up a portfolio that consists of a corporate bond fund and a
common stock fund. The numbers below represent the annual return per thousand dollars of the
two funds under various economic conditions and the probability that these conditions will occur.
Probability
State of the
Corporate Bond
Common Stock
Economy
Fund
Fund
.10
-$30
-$150
Recession
.15
$50
-$20
Stagnation
.35
$90
$120
Slow Growth
.30
$100
$160
Moderate Growth
.10
$110
$250
High Growth
Find a) Expected returns for both funds, b) the standard deviation of the return for both funds and c) the
covariance between the two funds. d) In which of the two funds would you invest? Explain.
Solution: Let x = return of the corporate bond fund and y = return of the common stock fund. I will use
the same format that I have used for previous problems, though the absence of off-diagonal elements means
that a more concise format is possible.
x
We are given
y
 150
 30
.10
50
0
90
0
100
0
 20
0
.15
0
0
0
120
0
0
.35
0
0
160
0
0
0
.30
0
250
0
0
0
0
.10
110
0
and can expand this to the table
below.
x
y
 150
 20
 30
.10
0
50
0
.15
90
0
0
100
0
0
110
0
0
P y 
.10
.15
yP y 
 15
3
y 2 P y 
2250
60
120
160
250
0
0
0
0
0
0
.35
0
0
0
.30
0
0
0
.10
.35
.30
.10
42
48
25
5040
7680
6250
Px 
xPx 
x 2 Px 
.10
 3.0
90
.15
.35 .30 .10
1.00 E  y   97
7.5 31 .5 30 .0 11 .0
77  E x 
375 2835 3000 1210 7510  E x 2
 
 
E y 2  21280
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E xy  

0
450
  0  150

  0
0

0
 0
 0
0

a)  x  E x  
 
0
0
0
0
.10  30  250 

0 .1550  20 
0
0
0

xyPxy   
0
0 .3590 120 
0
0


0
0
0 .30 100 160 
0


0
0
0
0  .10 110 250 

0
0
 0
0
0
 0
 3780
0
 0  11630

 0  4800
 0
0
 0  2750 
xPx   77
 y  Ey  
yP y   97


b)  x2  E x 2   x2  7510  77   1581 . So  x  1581  39 .76 and CV x 
39 .76
 51 .63 % .
77
108 .95
 112 .32 % .
 y2  E y 2   y2  21280 972  11871. So  y  111871 108.95 and CV y 
97
c)  xy  Covxy  Exy   x  y  11630 7797  4161. The text answer book fails to note that the
2
 
 xy
4161 2
 .9225 , which means that the securities move so similarly,
1581 11871 
 x y
that there may not be much advantage in putting them in the same portfolio.
d) The Instructor’s Solutions Manual says that the ‘Common stock fund gives the investor a higher
expected return than corporate bond fund, but also has a standard deviation better than 2.5 times higher
than that for corporate bond fund. An investor should carefully weigh the increased risk.’
correlation is  xy 

Exercise 5.17 [5.13g-k in 9th]: Compute the portfolio expected return and portfolio risk for the following
percentages invested in the corporate bond fund as in Problem 5.17: a) 30%; b) 50% and c) 70%. d) On the
basis of results in a)-c), which portfolio would you recommend?
Solution: Recall that  x  77 ,  y  97,  x2  1581,  y2  11871 and  xy  Covxy  4161.
We know from a previous problem that, if R is our total return and w is the fraction of our portfolio in
security x, E ( R)  wE ( x)  1  wE ( y) and Var ( R)  w 2Var ( x)  1  w2 Var ( y)  2w1  wCov( x, y) or
 R2  w 2 x2  1  w2  y2  2w1  w xy . So we can state the following.
a) If w  .30 , E ( R)  .30(77 )  .70 (97 )  23.1  67.9  91 and
Var ( R)  .30 2 1580   .70 2 (11871 )  2.30 .70 4161   142 .20  5816 .79  1747 .62  7706 .61
 R  7706 .61  87 .79 .
The coefficient of variation is C 
b) If w  .50 , E ( R)  .50(77 )  .50 (97 )  38.5  48.5  87 and
 87.79

 .9647 or 96.47%

91
Var ( R)  .50 2 1580   .50 2 (11871 )  2.50 .50 4161   395 .00  2967 .75  2080 .50  5443 .25
 R  5443 .25  73.78 .
The coefficient of variation is C 
c) If w  .70 , E ( R)  .70(77 )  .30 (97 )  53.9  29.1  83 and
 73.78

 .8480 or 84.80%.

87
Var ( R)  .70 2 1580   .30 2 (11871 )  2.70 .30 4161   774 .20  1068 .39  1747 .62  3590 .21
 R  3590 .21  59 .92 .
The coefficient of variation is C 
 59.92

 .7219 or 72.19%.

83
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d) Here is what the Instructor’s Solutions Manual says:
Based on the results of (a)-(c), you should recommend a portfolio with 70% of
corporate bonds and 30% of common stocks because it has the lowest risk per unit average return.
I say Nonsense! If the lowest risk per unit average return is our goal, clearly we should put all our money in
the bond fund. Other than that since higher risk seems to give higher returns, the choice is a personal one based
on the age and preferences of the investor.
Note that the 9th edition version is slightly longer and the answers given in the Instructor’s Solutions Manual
are:
(a)
E(X) = $77
(b)
E(Y) = $97
(c)
(d)
 X = 39.76
 Y = 108.95
 XY = 4161
(e)
(f)
Stock Y gives the investor a higher expected return than stock X, but also has a standard
deviation better than 2.5 times higher than that for stock X. An investor should carefully
weigh the increased risk.
I have added coefficients of variation to the answers given.
w  .10 , E(R) = $95,  R = 101.88 and CV  107 .24% .
(g)
w  .30 , E(R) = $91,  R = 87.79 and CV  96 .47 % .
(h)
(i)
w  .50 , E(R) = $87,  R = 73.78 and CV  84 .80 % .
(j)
w  .70 , E(R) = $83,  R = 59.92 and CV  72 .19 % .
(k)
(l)
w  .90 , E(R) = $79,  R = 46.35 and CV  58 .67 % .
Based on the results of (g)-(k), an investor should recognize that as the expected return
increases, so does the portfolio risk.
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Problem K1: Let us assume that x has a mean of 2 and a standard deviation of 4, and that y has a mean of
3 and a standard deviation of 6. Also assume that the correlation between x and y is .9. If
w  7 x  3 and v  4 y  5 , find the covariance and correlation between w and v .
Solution: The problem says  x  4 ,  y  6 and  xy  .9 , so that  xy   xy x y  .946  21.60 . The
Outline says that Cov(ax  b, cy  d )  acCov( x, y) and if w  ax  b and v  cy  d , wv  signac xy .
If we use the first formula from the outline with a  7 , b  3 , c  4 and d  5 , we get
 wv  Cov(w, v)  Cov(ax  b, cy  d )  acCov( x, y)  ac xy  7421.60  604.8 . Now we know that
 w2  Var ax  b  a 2Varx   72 42  784 and that  v2  Varcy  d   c 2Var y    42 62  576 .

 604 .8
So  wv  wv 
 .9 .
 w v
784 576
Much better, use the second formula from the outline,
 wv  signac xy  sign7 4.9  sign 28.9  1.9  .9 .
Problem K2: The table below shows average Fahrenheit temperature and yield in lbs./acre for an industrial
crop.
F
Y
a. Find the covariance and correlation between Fahrenheit
70
15
temperature and yield.
75
17
b. If the conversion formula for Celsius temperature is
79
16
C = 5/9(F-32), find the covariance and correlation between
80
20
Celsius temperature and yield.
Solution: a) These are sample data, so we first compute the sample variances and covariance.
obs
x
y
x2
y2
xy
TF 
70 15 lb A
75 17 lb A
74 16 lb A
1
2
3
4
sum

Y 
4900
5625
6241
225 1050
289 1275
256 1264
 x  304 ,  x
2
 23166 ,
 y  68 ,  y
2
 1170 and
80 20 lb A 6400 400 1600
304 68
23166 1170 5189
xy  5189 , x 
 x  304  76 , s   x
n
2
x
4
 y  68  17 , s   y
y
n
2
y
4
covariance is sTF Y  s xy 
rxy  rTF Y 
s xy
sx s y

2
 ny 2
n 1

2
 nx 2
n 1

23166  476 2 62

 20 .67 ,
3
3
1170  417 2 14

 4.67 The formula for the sample
3
3
 xy  nx y  5189  476 17   7 so
n 1
7
20 .67 4.67
 0.71 .
3
251solnK2 3/07/07 (Open this document in 'Page Layout' view!)
b) The formula relating Celsius and Fahrenheit is TC  5 9 TF  5 9 32  . The Outline says that if
w  ax  b and v  cy  d , Cov(w, v)  Cov(ax  b, cy  d )  acCov( x, y) and rwv  signacrxy . If we let
TC  w , TC  x and Y  y  v , use the first formula from the outline with a  5 9 , b   5 9 32  , c  1 and
d  0 , we get Cov(w, v)  Cov(ax  b, cy  d )  acCov( x, y)  acs xy  5 9 17  3.89 . Now we know that
 w2  Var ax  b  a 2Var x   5 9 2 20.67   6.38 and that
 v2  Varcy  d   c 2Var y   12 4.67   4.67 . So rTCY  rwv 
rvz

s w sv
3.89
 .71 .
6.38 4.67
Much better, use the second formula from the outline,
rTCY  rwv  signacrxy  sign5 9 1.71  sign5 9 .71   1.71  .71 .
Problem K3:
a. What is Covx, x  ?
b. What is (i) Cov5x  3,7 x  9 and (ii) Corr 5x  3,7 x  9 ?
c. So what is  xx ?
d. If m represents the per item markup and FC is fixed cost, and we sell x units of an
item, the profit is w  mx  FC . Assume that we have two goods, good X and Good
Y, and that the quantity of good X sold is x and the quantity of good Y sold is y . Let
w be the profits on good X and v be the profits on good Y. Assume that
Covx, y   20 , and that data concerning
Good X Good Y sales are as in the table at left. Find the
Markup
5
2
variance of profits on each of the two goods.
Fixed Cost
100
50
Then find the covariance and correlation between
Mean
90
90
profits on the two goods.
Variance
50
50
Solution: a) What is Covx, x  ? The formula far the population variance is  xy  Covxy  Exy   x  y ,
 
so if we substitute x for y , we get  xx  Exx   x  x  E x 2   x2   x2 . If, instead, we want to work
with the sample variance s xy 
conclude Covx, x   Varx  .
 xy  nx y , we find that s
n 1
xx

 xx  nx x   x
n 1
2
 nx 2
n 1
 s x2 . So we
b) What is (i) Cov5 x  3, 7 x  9 ? The formula we have been using is
Cov(w, v)  Cov(ax  b, cy  d )  acCov( x, y) . If we substitute x for y , we get
Cov(w, v)  Cov(ax  b, cx  d )  acCov( x, x)  acVarx  . Let a  5 , b  3 , c  7 and d  9 , then
Cov(5x  3,7 x  9)  57Covx   35Covx  )
What is (ii) Corr 5x  3, 7 x  9 ? The formula we have been using is
Corr w, v   Corrax  b, cy  d   signacCorr x, y  . If we substitute x for y , we get
Corr ax  b, cx  d   signacCorrx, x   sign57Corr x, x   Corrx, x  .
251solnK2 3/07/07 (Open this document in 'Page Layout' view!)
c) So what is  xx ? The formula for the population correlation is  xy 
y , we get  xy 
 xy
 x y
. If we substitute x for
s xy
 xx
2
. If we substitute
 x2  1 . The formula for the sample correlation is rxy 
 x x  x
sx s y
s xx
s2
 x2  1 , so if we go back to the previous part of the problem,
sx sx sx
Corr5x  3,7 x  9  sign57Corrx, x   Corrx, x   1 .
x for y , we get rxy 
d) If m represents the per item markup and FC is fixed cost, and we sell x units of an item, the profit is
w  mx  FC . Assume that we have two goods, good X and Good Y, and that the quantity of good X sold
is x and the quantity of good Y sold is y . Let w be the profits on good X and v be the profits on good
Y. Assume that Covx, y   20 , and that data concerning sales is as in the table above. Find the
variance of profits on each of the two goods. Then find the covariance and correlation between the profits
on the two goods.
If we look at the table above we find that the profit on X is w  5x  100 and the profit on Y is
2
v  2 y  50 . We know  w2  Varax  b  a 2Varx   5 50   1250 and that
 v2  Varcy  d   c 2Var y   22 50 2  200 .
Cov(w, v)  Cov(ax  b, cy  d )  acCov( x, y)  ac xy  5220  200 . So
 wv 
 xy
 vz
200
20

 0.4 . Better, observe that  xy 

 0.4
 w v


1250 200
50 50
x y
Corr w, v   Corr ax  b, cy  d   signacCorr x, y   sign52Corr x, y   0.4 .
x
Problem K4: Find  xy for a)
x
1
4
.10
5
0
6
0
y 2
0
.40
0
3
0
0
.50
P y 
.10
yP y  y 2 P y 
0.1
0.1
and b)
1
0
4
0
5
6
.10
y 2
0
.40
0
3 .50
0
0
Solution: a)
x
4
 .10

 0
 0

0.1
1
y
2
3
Px 
xPx 
0.4 
5
0
0.4
6
0

0
.50 

0.5
2.0 
3.0 
.40
0
x 2 Px  1.6  10 .0  18 .0 
.40
0.8
1.6
.50
1.5
4.5
1.00
2.4
6.2
5.4
29 .6
 Px   1 (a check),   E x   xPx  5.4 , E x    x
 P y   1 ,   E  y    yP y   2.4 and E y    y P y   6.2 .
2
To summarize
x
2
2
Px   29 .6 ,
2
y
E xy  

051
061  0.4 0
0 
 .10 41
xyPxy    042 .40 52 062   0 4.0 0   13 .4
  043 .053 .50 63  0 0 9.0
251solnK2 3/07/07 (Open this document in 'Page Layout' view!)
 
To complete what we have done, write  x2  E x 2   x2  29 .6  5.42  0.44 ,
 y2
 
E y
2
  y2
 6.2  2.4  0.44 ,  xy  Covxy  Exy   x  y  13.4  5.42.4  0.44 .
2
 xy
So that  xy 
 x y
0.44

 1.00 .
0.44 0.44
b)
x
4
0
1
y
2
3
Px 
xPx 
x Px 


 0
 .50

0.5
`2.0 
2
5
0
.40
0
0.4
6
.10 

0
.0

0.1
P y 
.10
yP y  y 2 P y 
0.1
0.1
.40
0.8
1.6
.50
1.5
4.5
1.00
2.4
6.2
2.0  0.6 
4.6
8.0  10 .0  3.6 
21 .6
 Px   1 (a check),   E x   xPx  4.6 , E x    x
 P y   1 ,   E  y    yP y   2.4 and E y    y P y   6.2 .
2
To summarize
x
2
2
Px   21 .6 ,
2
y
E xy  

051 .10 61  0
0 0.6
 041



xyPxy     042 .40 52 062   0 4.0 0   10 .6
 50 43 .053 063 6.0 0
0 
 
To complete what we have done, write  x2  E x 2   x2  21 .6  4.62  0.44 ,
 
 y2  E y 2   y2  6.2  2.42  0.44 ,  xy  Covxy  Exy   x  y  10.6  4.62.4  0.44 .
So that  xy 
 xy

 0.44
 1.00 . In general, joint probability tables with only the diagonals
0.44 0.44
filled produce correlations close to +1 or -1.
 x y