CHAPTER 8 RISK-AVERSION, CAPITAL
ASSET ALLOCATION, AND MARKOWITZ
PORTFOLIO-SELECTION MODEL
1.
RP  WA RA  WB RB
 .25(8)  .75(12)
 11%
2. The variance of a portfolio consists of two components:
a) A weighted average of the variances of the securities in the portfolio.
b) A component which measures the relationship between each pair of securities
in the portfolio. This relationship is measured by the covariance.
For a two security portfolio:
Var ( RP )  WA2Var ( RA )  (1  WA )2 Var ( RB )
 2WA (1  WA )Cov( RA , RB )
3. u(0)  0
u(0)  0
u(0)  0
implies risk averse
implies risk neutral
implies risk seeker
(a)
u (W )  AeBW
u(W )  BAe BW
u(W )  B 2 AeBW
u(0)  B 2 A  0
risk seeker
u (W ) W
5  W
2 3
4
u(W ) 5 W
6
u(W ) 2W
4 5
u( 0 ) 0
r i ks neutral
(b)
(c)
u (W )  W 2 / 3
u(W )  2 / 3W 5 / 3
u(W )  10 / 9W 8 / 3
u(0)  0
risk neutral
u (W )  3W  e2W
u(W )  3  2e2W
u(W )  4e2W
u(0)  4  0
risk averse
(d)
4. A fair game means that the expected returns given information set θ equal the
expected returns without the information set. In other words, the returns already
reflect all information.
Sometimes we define an investor’s risk preferences by looking at his preferences
towards a fair gamble. A fair gamble is defined as a gamble where the cost equals
the expected value.
Risk averse investors always reject a fair gamble.
Risk neutral investors are indifferent to a fair gamble.
Risk seeking investors will always accept a fair gamble.
5. a) RA  .30(10%)  .20(11%)  .50(12%)  11.2%
RB  .30(9%)  .20(7%)  .50(4%)  6.1%
Var ( RA )  .30(10  11.2)2  .20(11  11.2)2  .50(12  11.2)2  .76
Var ( RB )  .30(9  6.1)2  .20(7  6.1)2  .50(4  6.1)2  4.89
Cov( RA , RB )  .30(10  11.2)(9  6.1)
 .20(11  11.2)(7  6.1)
 .50(12  11.2)(4  6.1)
 1.92
b) RP  .60(11.2%)  .40(6.1%)  9.16%
Var ( RP )  (.60)2 (.76)  (.40)2 (4.89)
 2(.40)(.60)( 1.92)
 .1344
6. An efficient portfolio is one which has the highest expected return for a given
level of risk. Likewise, an inefficient portfolio is one which does not have the
highest expected return for a given level of risk.
We can use the dominance principle to show that inefficient portfolios will always
be dominated by other portfolios.
7. Short selling entails the selling of a security that you don’t own. In portfolio
theory, the selling short of one asset is often used to provide the funds to purchase
another asset.
The efficient frontier consists of all portfolios which satisfy the following two
conditions.
1) Have the highest expected return for a given level of risk,
and
2) Have the lowest risk for a given expected return.
Efficient Frontier
Without Short Selling
Efficient Frontier
With Short Selling
8. Margin requirements determine the amount of money an investor can borrow to
purchase shares of stock.
The Dyl Model, which incorporates margin requirements into the Markowitz
Model, causes the efficient frontier to shift to the northeast. [See Figure 8-10B
of text.]
9. a)
b)
c)
10. a) Minimum risk portfolio can be found by
dVar ( RP )
0
dWA
and solving for WA.
WA 
 B2   A B PAB
 A2   B2  2 A B PAB
(2)2  (3)(2)(.8)
(3)2  (2)2  (2)(3)(2)(.8)
.8

9  4  9.6
 .2352

So, sell short 23.52% of asset A and place 123.52% in asset B to minimize
the risk of the portfolio.
RP  .2352(16%)  1.2352(10%)  8.59%
Var(RP) = (–.2352)2(3)2 + (1.2352)2 (2)2 + 2(–.2352)(1.2352)(3)(2)(.8) =
3.81
b) By selling short the risk free asset, you can obtain the additional funds
necessary for investing an additional $5,000 in the risky portfolio.
WAB 
25, 000
 1.25
20, 000
RP  (1  WAB ) R f  WAB RAB
 (1  1.25)5%  (1.25)8.59%
 9.49%
2
Var ( RP )  WAB2  AB
 (1.25)2 (3.81)
 5.95
11. If we assume that Investors can borrow and lend at the risk free rate Rf and if
Investors have the same expectations, there will be only one best portfolio.
Both John Doe and Jane Roe will purchase the same portfolio, portfolio M,
because the combination of the risk free asset and portfolio M dominates all
other combinations. However, Jane Roe will probably place some of her money
in the riskless asset and some in portfolio M, i.e. she will be a lender. John Doe,
on the other hand, will most likely borrow at the risk free rate to buy more of
portfolio M, i.e. he will be a borrower.
12. To determine the optimal weights for a portfolio using the Markowitz approach,
the portfolio manager can either:
1) Set a target rate of return and find the portfolio which has the least risk.
or
2) Set a level of risk for the portfolio and maximize the expected return for this
given level of risk.
These two problems can be solved using calculus or linear programming
methods on a computer.
13. Based on Appendix 10C, we can use Microsoft Excel to calculate the optimal
weights of Markowitz model.
In Markowitz Model in equation (8.13), we know
2 12
2 13
 2 11
 2
2 22
2 23
 21
 2 31
2 32
2 33

1
1
 1
 E ( R1 ) E ( R2 ) E ( R3 )
1 E ( R1 )  W1   0 
1 E ( R2 )  W2   0 
 
1 E ( R3 )  W3    0  . (8.13)
   
0
0  1   1 
0
0   2   E *
K
W
A
Where E* =10%, the required rate of return of the portfolio. From the matrix
calculation of Markowitz Model, we can get the optimal weights from multiplying
A1 and K .
2 12
2 13
W1   2 11
W   2
2 22
2 23
 2   21
W3    2 31
2 32
2 33
  
1
1
1   1
 2   E ( R1 ) E ( R2 ) E ( R3 )
1
1
1
0
0
E ( R1 ) 
E ( R2 ) 
E ( R3 ) 

0 
0 
1
A1
W
 0 
 0 
 
 0 .
 
 1 
 E *
K
Based on Excel function, the values of average and variance are calculated by
“AVERAGE” and “VARP” functions and the covariance matrix is calculated by
“Covariance” in “Data Analysis.”
JNJ
IBM
BA
E(R )
0.0082
0.0050
0.0113
Variance
0.0025
0.0070
0.0082
Variance Covariance Matrix
JNJ
IBM
BA
JNJ
0.0025
0.0007
0.0007
IBM
0.0007
0.0070
0.0006
BA
0.0007
0.0006
0.0082
Therefore, we can obtain matrix A and K as follows:
Matrix A
0.0050
0.0014
0.0015
1.0000
0.0082
0.0014
0.0141
0.0013
1.0000
0.0050
0.0015
0.0013
0.0164
1.0000
0.0113
1.0000
1.0000
1.0000
0.0000
0.0000
0.0082
0.0050
0.0113
0.0000
0.0000
 0 
 0 


 0 


 1 
10% 
K
Then, from the matrix calculation of Markowitz Model, we can get the optimal
weights from multiplying A1 and K .
2 12
2 13
W1   2 11
W   2
2 22
2 23
 2   21
W3    2 31
2 32
2 33
  
1
1
1   1
 2   E ( R1 ) E ( R2 ) E ( R3 )
1
1
1
0
0
E ( R1 ) 
E ( R2 ) 
E ( R3 ) 

0 
0 
1
 0 
 0 
 
 0 .
 
 1 
 E *
A1
W
K
Where matrix A1 is
96.4952
-47.8974
-48.5978
0.5160
17.4282
-47.8974
23.7749
24.1225
1.5313
-166.1896
-48.5978
24.1225
24.4752
-1.0473
148.7614
0.5160
1.5313
-1.0473
-0.0488
5.5748
17.4282
-166.1896
148.7614
5.5748
-690.3857
The optimal weights in W matrix as follows:
W1
W2
W3
λ1
λ2
2.2588
-15.0877
13.8289
0.5087
-63.4638
If short-selling is allowed, the optimal weights for JNJ, IBM and BA are 2.2588,
-15.0877, and 13.8289 respectively. If short-selling is not allowed, we can obtain the
adjusted weights from equation (8.19) as follows:
WJNJ 
WJNJ
WJNJ  WIBM  WBA
2.2588
 0.0725
2.2588  15.0877  13.8289
15.0877
WIBM 
 0.4840
2.2588  15.0877  13.8289
13.8289
WBA 
 0.4436
2.2588  15.0877  13.8289
WJNJ 
14. There are four companies in the portfolio, therefore the matrix to calculate the
optimal weights are shown as follows:
2 12
2 13
2 14
 2 11
 2
2 22
2 23
2 24
 21
 2 31
2 32
2 33
2 34

2 42
2 43
2 44
 2 41
 1
1
1
1

 E ( R1 ) E ( R2 ) E ( R3 ) E ( R4 )
1 E ( R1 )  W1   0 
1 E ( R2 )  W2   0 
 
1 E ( R3 )  W3   0 
    
1 E ( R4 )  W4   0 
0
0   1   1 
   
0
0   2   E *
W
A
K
Where E* =12%, the required rate of return of the portfolio.
The information about the mean and variance of rate of returns for four companies are
as follows:
JNJ
IBM
BA
CAT
E(R )
0.0082
0.0050
0.0113
0.0169
Variance
0.0025
0.0070
0.0082
0.0097
Variance Covariance Matrix
JNJ
IBM
BA
CAT
JNJ
0.0025
0.0007
0.0007 0.0011
IBM
0.0007
0.0070
0.0006 0.0025
BA
0.0007
0.0006
0.0082 0.0031
CAT
0.0011
0.0025
0.0031 0.0097
Therefore, we can obtain matrix A and K as follows:
0.0050
0.0014
0.0015
0.0021
1.0000
0.0082
0.0014
0.0141
0.0013
0.0049
1.0000
0.0050
0.0015
0.0013
0.0164
0.0063
1.0000
0.0113
0.0021
0.0049
0.0063
0.0194
1.0000
0.0169
1.0000
1.0000
1.0000
1.0000
0.0000
0.0000
0.0082
0.0050
0.0113
0.0169
0.0000
0.0000
 0 
 0 


 0 


 0 
 1 


12% 
K
By excel function, the inverse of matrix A can be calculated as:
99.9015
-55.8418
-36.9166
-7.1432
0.8552
-24.9153
-55.8418
42.3581
-3.2389
16.7226
0.7365
-66.8810
-36.9166
-3.2389
64.7865
-24.6310
0.1239
2.3676
-7.1432
16.7226
-24.6310
15.0516
-0.7156
89.4287
0.8552
0.7365
0.1239
-0.7156
-0.0148
1.3243
-24.9153
-66.8810
2.3676
89.4287
1.3243
-159.0383
Then, from the matrix calculation of Markowitz Model, we can get the optimal
weights from multiplying A1 and K .
2 12
2 13
2 14
W1   2 11
W   2
2 22
2 23
2 24
 2   21
W3   2 31
2 32
2 33
2 34
 
2 42
2 43
2 44
W4   2 41
 1   1
1
1
1
  
 2   E ( R1 ) E ( R2 ) E ( R3 ) E ( R4 )
W
A1
1 E ( R1 ) 
1 E ( R2 ) 
1 E ( R3 ) 

1 E ( R4 ) 
0
0 

0
0 
1
 0 
 0 
 
 0 
 
 0 
 1 
 
 E *
K
The optimal weights in W matrix as follows:
W1
W2
W3
W4
λ1
λ2
-2.1346
-7.2892
0.4080
10.0159
0.1442
-17.7603
If short-selling is allowed, the optimal weights for JNJ, IBM, BA, and CAT are
-2.1346, -7.2892, 0.4080 and 10.0159 respectively. If short-selling is not allowed, we
can obtain the adjusted weights from equation (8.19) as follows:
WJNJ 
WJNJ
WJNJ  WIBM  WBA  WCAT
2.1346
 0.1075
2.1346  7.2892  0.4080  10.0159
7.2892
WIBM 
 0.3673
2.1346  7.2892  0.4080  10.0159
0.4080
WBA 
 0.0206
2.1346  7.2892  0.4080  10.0159
10.0159
WCAT 
 0.5046
2.1346  7.2892  0.4080  10.0159
WJNJ 
15. Use the same way and same information in question13 to calculate the optimal
weights when E* =0.5% and inverse matrix of A is the same as in question13.
2 12
2 13
W1   2 11
W   2
2 22
2 23
 2   21
W3    2 31
2 32
2 33
  
1
1
1   1
 2   E ( R1 ) E ( R2 ) E ( R3 )
A1
W
1
1
1
0
0
E ( R1 ) 
E ( R2 ) 
E ( R3 ) 

0 
0 
1
 0 
 0 
 
 0 
 
 1 
 E *
K
Then we can obtain optimal weight matrix W as follows:
W1
W2
W3
λ1
λ2
0.6031
0.7003
-0.3035
-0.0209
2.1228
If short-selling is allowed, the optimal weights for JNJ, IBM and BA are 0.6031,
0.7003, and -0.3035 respectively.
16. Use the same way and same information in question14 to calculate the optimal
weights when E* =1% and inverse matrix of A is the same as in question14.
2 12
2 13
2 14
W1   2 11
W   2
2 22
2 23
2 24
 2   21
W3   2 31
2 32
2 33
2 34
 
2 42
2 43
2 44
W4   2 41
 1   1
1
1
1
  
 2   E ( R1 ) E ( R2 ) E ( R3 ) E ( R4 )
1 E ( R1 ) 
1 E ( R2 ) 
1 E ( R3 ) 

1 E ( R4 ) 
0
0 

0
0 
1
 0 
 0 
 
 0 
 
 0 
 1 
 
 E *
A1
W
K
Then we can obtain optimal weight matrix W as follows:
W1
W2
W3
W4
λ1
0.6061
0.0677
0.1476
0.1787
-0.0015
λ2
-0.2661
Whether short-selling is allowed or not, the optimal weights for JNJ, IBM, BA, and
CAT are 0.6061, 0.0677, 0.1476 and 0.1787 respectively.