CHAPTER 10 INDEX MODELS FOR
PORTFOLIO SELECTION
1.
RD
5.3
6.25
-2.12
RE
7.3
8.2
-10.2
RNYSE
4.5
5.2
1. 1
12 .75
-3.4
8. 6
15 .4
-2.0
9.5
10. 3
-1.3
10.5
9.2
-2.1
-8.3
10.2
5.4
6.2
8.2
-1.0
-10.2
11.5
5.9
7.2
7.6
0
-7.4
9.3
4.2
8.5
STDD = 6.17
STDE = 7.86
STDNYSE = 5.21
VAR NYSE = 27.14
Regression Output for D return:
Regression Output for E return:
Constant
–0.99432
Constant
–1.73399
Std Err of Y Est
1.823686
Std Err of Y Est
3.889611
R Squared
0.927299
R Squared
0.795688
No. of Observations
12
No. of Observations
12
Degrees of Freedom
10
Degrees of Freedom
10
X Coefficient(s)
1.141178
X Coefficient(s)
188.0751
1.344912
Std Err of Coef.
0.101044
Std Err of Coef.
12.86892
0.215510
a.
αD = –.99
βD = 1.14
b.
Var(ei)D = (1.823686)2 = 3.325
c.
αE = –1.79
βE = 1.34
Var(ei)E = (3.889611)2 = 15.13
D 
1.14 
Cov( D, NYSE )

 D , NYSE
2
NYSE

 DNYSE D NYSE
2
 NYSE

 ENYSE E NYSE
2
 NYSE
( 6.17 )( 5.21)
27.14
 D , NYSE  .96
E 
1.34 
Cov( E , NYSE )

 E , NYSE
2
NYSE
( 7 . 8 6 ) ( 5 . 2 1 )
27.14
 E , NYSE  .89
d. If we consider the minimum variance portfolio to be optimal, then we can find the
optimal proportions by solving
dVAr ( RP )
0
dWD
 E2   D , E D E
WD  2
 D   E2  2 D , E D E
Cov( RD , RE )   D  E M2
 (1.14)(1.34)(27.14)
 41.459
 D,E 
Cov( RD , RE )
 D E
41.459
(6.17)(7.86)
 .855

(7.86)2  (.855)(6.17)(7.86)
(6.17)2  (7.86)2  2(.855)(6.17)(7.86)
20.32

16.92
 1.20
WD 
Sell short 20% of stock E and use proceeds to buy more of stock D.
RP  1.20(3.998)  (.20)(4.15)  3.9676%
 P2  (1.20)2 (6.17)2  (.20)2 (7.86)2
 2(1.20)( .20)(.855)(6.17)(7.86)
 1.44(38.07)  (.04)(61.78)  19.90
 37.392
2. Yes, to assure that the Indexes are unrelated to each other, the index variables can
be orthogondized (made uncorrelated) by completing inter-regressions on the
indexes themselves.
3. The expected return
E(Ri) = E(ai + b1I1 + b2I2 + ei)
= E(ai) + E(b1I1) + E(b2I2)+ E(ei)
E(Ri) = ai + b1I1 + b2I2
since a1, b1, b2 are constants and E(ei) = 0
the variance
 i2  E ( Ri  R1 )2
 E[(a1  b1 I1  b2 I 2  ei )  (a1  b1 I1  b2 I 2 )]2
 E[(a1  ai )  bi ( I1  I1 )  b1 ( I 2  I )  ei ]2
 bi2 E[( I1  I1 )2  b1b2 E ( I1  I 2 ) E ( I 2  I 2 )]
 b12 E ( I 2  I 2 )  b1 E[( I1  I1 )ei ]
 b2 E[( I 2  I 2 )ei ]  E (ei2 )
if
E[( E1  I1 )( I 2  I 2 )]  0
E[( I1  I1 )ei ]  0
E[( I 2  I 2 )ei ]  0
E ( I1  I1 )2   12
E ( I 2  I 21 )2   22
E (ei2 )   ei2
 i2  b1212  b22 22   ei2
Covariance
 ij  E ( Ri  Ri )( R j  R j )
 E{[(a1  b1 I1  b2 I 2  e1 )  (ai  b1 I1  b2 I 2 )] 
[(a j  bj1 I1  bj 2 I 2  e j )  (a j  b j1 I1  b j 2 I 2 )]}
 E[bi1bj1 ( I  I1 )2  bi 2bj 2 ( I 2  I 2 )2 ]
 bi1bj1 12  b12bj 2 22
4.
a. ai is the intercept term. It shows the relationship between R. and all the
indexes assuming that all the indexes are zero.
bi1 is the change in Ri caused by a change in index 1.
•
•
•
biL is the change in Ri caused by a change in index 1.
b. 12  bi2112 
 biL2  22  eei2
5. The Markowitz model of portfolio theory requires information about the variances
of each individual security and the covariance between every pair of securities. If
we assume that each individual security is related to some Index I and is unrelated
to every other security, we have the single index model.
For n securities, the Markowitz model requires
n+ (n2 – n)/2
n2  n
estimates of covariance
2
For n securities, the single-index model requires
where n estimates of variances and
3n+2
Where n estimates of α of each security, n estimates of β of each security, n
estimates of variance of each security, one estimate of variance of the market and
one estimate of the expected return of the market.
6.
 x1  2 x2  4 x3  15
2 x1  8 x2  2 x3  8
4 x1  2 x2  5 x3  7
(1)
(2)
(3)
Substitution method:
Eq(1) X 2+Eq(2), we can obtain
4 x2  10 x3  22
x3  2.2  0.4 x2
(4)
Eq(2) X 5+Eq(3) X 2, we can obtain
18x1  36 x2  54
x1  3  2 x2
(5)
Substitute Eq(4) and Eq(5) into Eq(1), then we can get
3  2 x2  2 x2  1.6 x2  8.8  15
x2  2
Substitute x2=2 into Eq(4) and Eq(5), we can get x1=1 and x3=3
Cramer’s rule:
15 2 4
x1  8 8 2 /
7 2 5
1 15 4
x2  2 8 2 /
4 7 5
1 2 15
x3  2 8 8 /
4 2 7
1
2
4
1
2
4
1
2
4
2
8
2
2
8
2
2
8
2
4
2 1
5
4
2 2
5
4
2 3
5
Matrix Inversion:
A
x =k
 1 2 4   x1  15 
 2 8 2   x2    8 
 4 2 5  x3   7 
1
0.25 
 1 2 4 
 0.25 0.125
A1   2 8 2   0.125 0.0764 0.0694 
 4 2 5
 0.25 0.0694 0.0278
0.25  15   1 
 0.25 0.125
x =A1 k= 0.125 0.0764 0.0694   8    2 
 0.25 0.0694 0.0278  7   3 
Then we can obtain x1=1, x2=2 and x3=3