Elton Gruber 7e Solution Manual

Elton, Gruber, Brown, and Goetzmann 1-1 Modern Portfolio Theory and Investment Analysis, 7th Edition Solutions to Text Problems: Chapter 1

Elton, Gruber, Brown, and Goetzmann Modern Portfolio Theory and Investment Analysis, 7th Edition

Solutions to Text Problems: Chapter 1

Chapter 1: Problem 1 A. Opportunity Set With one dollar, you can buy 500 red hots and no rock candies (point A), or 100 rock candies and no red hots (point B), or any combination of red hots and rock candies (any point along the opportunity set line AB).

Algebraically, if X = quantity of red hots and Y = quantity of rock candies, then:

10012.0 =+ YX That is, the money spent on candies, where red hots sell for 0.2 cents a piece and rock candy sells for 1 cent a piece, cannot exceed 100 cents ($1.00). Solving the above equation for X gives:

YX 5500 −= which is the equation of a straight line, with an intercept of 500 and a slope of −5.


B. Indifference Map Below is one indifference map. The indifference curves up and to the right indicate greater happiness, since these curves indicate more consumption from both candies. Each curve is negatively sloped, indicating a preference of more to less, and each curve is convex, indicating that the rate of exchange of red hots for rock candies decreases as more and more rock candies are consumed. Note that the exact slopes of the indifference curves in the indifference map will depend an the individual’s utility function and may differ among students.

Chapter 1: Problem 2 A. Opportunity Set

The individual can consume everything at time 2 and nothing at time 1, which, assuming a riskless lending rate of 10%, gives the maximum time-2 consumption amount:

$20 + $20 × (1 + 0.1) = $42.

Instead, the individual can consume everything at time 1 and nothing at time 2, which, assuming a riskless borrowing rate of 10%, gives the maximum time-1 consumption amount: $20 + $20 ÷ (1 + 0.1) = $38.18


The individual can also choose any consumption pattern along the line AB

(the opportunity set) below.

The opportunity set line can be determined as follows. Consumption at

time 2 is equal to the amount of money available in time 2, which is the income earned at time 2, $20, plus the amount earned at time 2 from any money invested at time 1, ($20 − C1) × (1 + 0.1): C2 = $20 + ($20 − C1) × (1.1) or C2 = $42 − 1.1C1 which is the equation of a straight line with an intercept of $42 and a slope of −1.1. B. Indifference Map We are given that the utility function of the individual is:

501),( 21

2121CC

CCCCU +++=

A particular indifference curve can be traced by setting U(C1,C2) equal to a constant and then varying C1 and C2. By changing the constant, we can trace out other indifference curves. For example, by setting U(C1,C2) equal to 50 we get:

5050

1 2121 =+++

CCCC or 24505050 2121 =++ CCCC


This indifference curve appears in the graph of the indifference map below as the curve labeled “50” (the lowest curve shown). By setting U(C1,C2) equal to 60, we get the curve labeled “60,” etc.

C. Solution The optimum solution is where the opportunity set is tangent to the highest possible indifference curve (the point labeled “E” in the following graph).

This problem is meant to be solved graphically. Below, we show an analytical solution:

501),( 21

2121CC

CCCCU +++=

Substituting the equation of the opportunity set given in part A for C2 in the above equation gives:

501.142

1.1421),(2

111121

CCCCCCU

−+−++=

To maximize the utility function, we take the derivative of U with respect to C1 and set it equal to zero:

0502.2

50421.11 1

1=−+−=

CdCdU

which gives C1 = $16.82. Substituting $16.82 for C1 in the equation of the opportunity set given in part A gives C2 = $23.50.


Chapter 1: Problem 3 If you consume nothing at time 1 and instead invest all of your time-1 income at a riskless rate of 10%, then at time 2 you will be able to consume all of your time-2 income plus the proceeds earned from your investment: $5,000 + $5,000 (1.1) = $10,500. If you consume nothing at time 2 and instead borrow at time 1 the present value of your time-2 income at a riskless rate of 10%, then at time 1 you will be able to consume all of the borrowed proceeds plus your time-1 income: $5,000 + $5,000 ÷ (1.1) = $9,545.45 All other possible optimal consumption patterns of time 1 and time 2 consumption appear on a straight line (the opportunity set) with an intercept of $10,500 and a slope of −1.1: C2 = $5,000 + ($5,000 − C1) × (1.1) = $10,500 − 1.1C1

Chapter 1: Problem 4 If you consume nothing at time 1 and instead invest all of your wealth plus your time-1 income at a riskless rate of 5%, then at time 2 you will be able to consume all of your time-2 income plus the proceeds earned from your investment: $20,000 + ($20,000 + $50,000)(1.05) = $93,500.


If you consume nothing at time 2 and instead borrow at time 1 the present value of your time-2 income at a riskless rate of 5%, then at time 1 you will be able to consume all of the borrowed proceeds plus your time-1 income and your wealth: $20,000 + $50,000 + $20,000 ÷ (1.05) = $89,047.62 All other possible optimal consumption patterns of time-1 and time-2 consumption appear on a straight line (the opportunity set) with an intercept of $93,500 and a slope of −1.05: C2 = $20,000 + ($20,000 + $50,000 − C1) × (1.05) = $93,500 − 1.05C1

Chapter 1: Problem 5 With Job 1 you can consume $30 + $50 (1.05) = $82.50 at time 2 and nothing at time 1, $50 + $30 ÷ (1.05) = $78.60 at time 1 and nothing at time 2, or any consumption pattern of time 1 and time 2 consumption shown along the line AB: C2 = $82.50 − 1.05C1. With Job 2 you can consume $40 + $40 (1.05) = $82.00 at time 2 and nothing at time 1, $40 + $40 ÷ (1.05) = $78.10 at time 1 and nothing at time 2, or any consumption pattern of time 1 and time 2 consumption shown along the line CD: C2 = $82.00 − 1.05C1.

The individual should select Job 1, since the opportunity set associated with it (line AB) dominates the opportunity set of Job 2 (line CD).


Chapter 1: Problem 6 With an interest rate of 10% and income at both time 1 and time 2 of $5,000, the opportunity set is given by the line AB: C2 = $5,000 + ($5,000 − C1) × (1.1) = $10,500 − 1.1C1 With an interest rate of 20% and income at both time 1 and time 2 of $5,000, the opportunity set is given by the line CD: C2 = $5,000 + ($5,000 − C1) × (1.2) = $11,000 − 1.2C1

Lines AB and CD intersect at point E (where C2 = time-2 income = $5,000 and C1 = time-1 income = $5,000). Along either line above point E, the individual is lending (consuming less at time 1 than the income earned at time 1); along either line below point E, the individual is borrowing (consuming more at time 1 than the income earned at time 1). Since the individual can only lend at 10% and must borrow at 20%, the individual’s opportunity set is given by line segments AE and ED. Chapter 1: Problem 7

For P = 50, this is simply a plot of the function 1

12 1

50CC

C+−

= .

For P = 100, this is simply a plot of the function 1

12 1

100C

CC

+−

= .


Chapter 1: Problem 8 This problem is analogous to Problem 2. We present the analytical solution below. The problem could be solved graphically, as in Problem 2. From Problem 3, the opportunity set is C2 = $10,500 − 1.1C1. Substituting this equation into the preference function P = C1 + C2 + C1 C2 yields:

21111 1.1500,10$1.1500,10$ CCCCP −+−+=

02.2500,10$1.11 11

=−+−= CdCdP

C1 = $4,772.68 C2 = $5,250.05

Chapter 1: Problem 9

Let X = the number of pizza slices, and let Y = the number of hamburgers. Then, if pizza slices are $2 each, hamburgers are $2.50 each, and you have $10, your opportunity set is given algebraically by $2X + $2.50Y = $10 Solving the above equation for X gives X = 5 − 1.25Y, which is the equation for a straight line with an intercept of 5 and a slope of −1.25. Graphically, the opportunity set appears as follows:

Assuming you like both pizza and hamburgers, your indifference curves will be negatively sloped, and you will be better off on an indifference curve to the right of another indifference curve. Assuming diminishing marginal rate of substitution between pizza slices and hamburgers (the lower the number of hamburgers you have, the more pizza slices you need to give up one more burger without changing your level of satisfaction), your indifference curves will also be convex.


A typical family of indifference curves appears below. Although you would rather be on an indifference curve as far to the right as possible, you are constrained by your $10 budget to be on an indifference curve that is on or to the left of the opportunity set. Therefore, your optimal choice is the combination of pizza slices and hamburgers that is represented by the point where your indifference curve is just tangent to the opportunity set (point A below).

Chapter 1: Problem 10 If you consume C1 at time 1 and invest (lend) the rest of your time-1 income at 5%, your time-2 consumption (C2) will be $50 from your time-2 income plus ($50 − C1)(1.05) from your investment. Algebraically, the opportunity set is thus C2 = 50 + (50 - C1)(1.05) = 102.50 - 1.05C1 If C1 is 0 (no time-1 consumption), then from the above equation C2 will be $102.50. If C2 is 0, then C1 will be $97.62. Graphically, the opportunity set appears below, along with a typical family of indifference curves.


Chapter 1: Problem 11 If you consume C1 at time 1 and invest (lend) the rest of your time-1 income at 20%, your time-2 consumption (C2) will be $10,000 from your time-2 income plus $10,000 from your inheritance plus ($10,000 - C1)(1.20) from your investment. The opportunity set is thus C2 = $10,000 + $10,000 + ($10,000 - C1)(1.20) = $32,000 - 1.2C1 If C2 is 0 (no time-2 consumption), then you can borrow the present value of your time-2 income and your time-2 inheritance and spend that amount along with your time-1 income on time-1 consumption. Solving the above equation for C1 when C2 is 0 gives C1 = $26,666.67, which is the maximum that can be consumed at time 1. Similarly, if C1 is 0 (no time-1 consumption), then you can invest all of your time-1 income at 20% and spend the future value of your time-1 income plus your time-2 income and inheritance on time-2 consumption. From the above equation, C2 will be $32,000 when C1 is 0, which is the maximum that can be consumed at time 2. Chapter 1: Problem 12 If you consume nothing at time 2, then you can borrow the present value of your time-2 income for consumption at time 1. If the borrowing rate is 10% and your time-2 income is $100, then the present value (at time 1) of your time-2 income is $100/(1.1) = $90.91. You can borrow this amount and spend it along with your time- income of $100 on time-1 consumption. So the maximum you can consume at time 1 is $90.91 + $100 = $190.91. If you consume nothing at time 1 and instead invest all of your time-1 income of $100 at the lending rate of 5%, the future value (in period 2) of your period 1 income will be $100(1.05) = $105. You can then spend that amount along with your time 2 income of $100 on time-2 consumption. So the maximum you can consume at time 2 is $105 + $100 = $205. With two different interest rates, we have two separate equations for opportunity sets: one for borrowers and one for lenders. If you only consume some of your time 1 income at time 1 and invest the rest at 5%, you have the following opportunity set: C2 = 100 + (100 - C1)(1.05) = 205 - 1.05C1. If you only consume some of your time-2 income at time 2 and borrow the present value of the rest at 10% for consumption at time 1, your opportunity set is: C1 = 100 + (100 - C2)/(1.1) = 190.91 - C2/1.1,or, solving the equation for C2, C2 = 210 - 1.1C1


Graphically, the two lines appear as follows:

The lines intersect at point E (which represents your income endowment for times 1 and 2). Moving along the lines above point E represents lending (investing some time-1 income); moving along the lines below point E represents borrowing (spending more than your time-1 income on time-1 consumption). Since you can only lend at 5%, line segment AE represents your opportunity set if you choose to lend. Since you must borrow at 10%, line segment ED represents your opportunity set if you choose to borrow. So your total opportunity set is represented by AED.

Elton, Gruber, Brown, and Goetzmann 4-1 Modern Portfolio Theory and Investment Analysis, 7th Edition Solutions To Text Problems: Chapter 4

Elton, Gruber, Brown and Goetzmann Modern Portfolio Theory and Investment Analysis, 7th Edition


Chapter 4: Problem 1 A. Expected return is the sum of each outcome times its associated

probability.

Expected return of Asset 1 = =1R 16% × 0.25 + 12% × 0.5 + 8% × 0.25 = 12%

2R = 6%; 3R = 14%; 4R = 12%

Standard deviation of return is the square root of the sum of the squares of each outcome minus the mean times the associated probability. Standard deviation of Asset 1 =

1σ = [(16% − 12%)2 × 0.25 + (12% − 12%)2 × 0.5 + (8% − 12%)2 × 0.25]1/2 = 81/2

= 2.83%

2σ = 21/2 = 1.41%; 3σ = 181/2 = 4.24%; 4σ = 10.71/2 = 3.27% B. Covariance of return between Assets 1 and 2 = 12σ = (16 − 12) × (4 − 6) × 0.25 + (12 − 12) × (6 − 6) × 0.5 + (8 − 12) × (8 − 6) × 0.25

= − 4 The variance/covariance matrix for all pairs of assets is:

1 2 3 4 1 8 − 4 12 0 2 − 4 2 − 6 0 3 12 − 6 18 0 4 0 0 0 10.7

Correlation of return between Assets 1 and 2 = 141.183.2

412 −=

×−

=ρ .

The correlation matrix for all pairs of assets is:

1 2 3 4 1 1 − 1 1 0 2 − 1 1 − 1 0 3 1 − 1 1 0 4 0 0 0 1


C. Portfolio Expected Return A 1/2 × 12% + 1/2 × 6% = 9% B 13% C 12% D 10% E 13% F 1/3 × 12% + 1/3 × 6% + 1/3 × 14% = 10.67% G 10.67% H 12.67% I 1/4 × 12% + 1/4 × 6% + 1/4 × 14% + 1/4 × 12% = 11% Portfolio Variance A (1/2)2 × 8 + (1/2)2 × 2 + 2 × 1/2 × 1/2 × (− 4) = 0.5 B 12.5 C 4.6 D 2 E 7 F (1/3)2 × 8 + (1/3)2 × 2 + (1/3)2 × 18 + 2 × 1/3 × 1/3 × (− 4) + 2 × 1/3 × 1/3 × 12 + 2 × 1/3 × 1/3 × (− 6) = 3.6 G 2 H 6.7 I (1/4)2 × 8 + (1/4)2 × 2 + (1/4)2 × 18 + (1/4)2 × 10.7

+ 2 × 1/4 × 1/4 × (− 4) + 2 × 1/4 × 1/4 × 12 + 2 × 1/4 × 1/4 × 0 + 2 × 1/4 × 1/4 × (− 6) + 2 × 1/4 × 1/4 × 0 + 2 × 1/4 × 1/4 × 0 = 2.7


Chapter 4: Problem 2 A. Monthly Returns

Month Security 1 2 3 4 5 6 A 3.7% 0.4% -6.5% 1.4% 6.2% 2.1% B 10.5% 0.5% 3.7% 1.0% 3.4% -1.4% C 1.4% 14.9% -1.4% 10.8% 4.9% 16.9%

B. Sample Average (Mean) Monthly Returns

( ) %22.16

%1.2%2.6%4.1%5.6%4.0%7.3=

+++−+=AR

%95.2=BR

%92.7=CR C. Sample Standard Deviations of Monthly Returns

( ) ( ) ( ) ( ) ( ) ( )

%92.334.156

%22.1%1.2%22.1%2.6%22.1%4.1%22.1%5.6%22.1%4.0%22.1%7.3 222222

==

−+−+−+−−+−+−=Aσ

%8.342.14 ==Bσ %78.602.46 ==Cσ


D. Sample Covariances and Correlation Coefficients of Monthly Returns

( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )

17.26

%95.2%4.1%22.1%1.2%95.2%4.3%22.1%2.6%95.2%0.1%22.1%4.1%95.2%7.3%22.1%5.6%95.2%5.0%22.1%4.0%95.2%5.10%22.1%7.3

=

⎥⎦

⎤⎢⎣

⎡−−×−+−×−+−×−+−×−−+−×−+−×−

=ABσ

24.7=ACσ ; 89.19−=BCσ

15.08.392.3

17.2=

×=ABρ

27.0=ACρ ; 77.0−=BCρ E. Portfolio Returns and Standard Deviations Portfolio 1 (X1 = 1/2; X2= 1/2; X3= 0): %09.2%92.70%95.22/1%22.12/11 =×+×+×=PR

( ) ( ) ( )%92.253.8

89.1902/124.702/117.22/12/1202.46042.142/134.152/1 2221

==

××+××+×××+×+×+×=Pσ

Portfolio 2 (X1 = 1/2; X2= 0; X3= 1/2): %57.42 =PR %35.496.182 ==Pσ Portfolio 3 (X1 = 0; X2= 1/2; X3= 1/2): %44.53 =PR %27.217.53 ==Pσ Portfolio 4 (X1 = 1/3; X2= 1/3; X3= 1/3): %03.44 =PR %47.209.64 ==Pσ


Chapter 4: Problem 3 It is shown in the text below Table 4.8 that a formula for the variance of an equally weighted portfolio (where Xi = 1/N for i = 1, …, N securities) is

( ) σσσσ kjkj2j

2P 1/N = +−

where σ 2

j is the average variance across all securities, σ kj is the average covariance across all pairs of securities, and N is the number of securities. Using the above formula with σ 2

j = 50 and σ kj = 10 we have: Portfolio Size (N) σ 2

P 5 18 10 14 20 12 50 10.8 100 10.4 Chapter 4: Problem 4 As is shown in the text, as the number of securities (N) approaches infinity, an equally weighted portfolio’s variance (total risk) approaches a minimum equal to the average covariance of the pairs of securities in the portfolio, which in Problem 3 is given as 10. Therefore, 10% above the minimum risk level would result in the portfolio’s variance being equal to 11. Setting the formula shown in the above answer to Problem 3 equal to 11 and using σ 2

j = 50 and σ kj = 10 we have:

( ) 11101050/12 =+−×= NPσ Solving the above equation for N gives N = 40 securities.


Chapter 4: Problem 5 As shown in the text, if the portfolio contains only one security, then the portfolio’s average variance is equal to the average variance across all securities, σ 2

j . If instead an equally weighted portfolio contains a very large number of securities, then its variance will be approximately equal to the average covariance of the pairs of securities in the portfolio, σ kj . Therefore, the fraction of risk that of an individual security that can be eliminated by holding a large portfolio is expressed by the following ratio:

2

2

i

kji

σ

σσ −

From Table 4.9, the above ratio is equal to 0.6 (60%) for Italian securities and 0.8 (80%) for Belgian securities. Setting the above ratio equal to those values and solving for σ kj gives 24.0 ikj σσ = for Italian securities and 22.0 ikj σσ = for Belgian securities.

Thus, the ratio kj

kji

σ

σσ −2

equals 5.14.0

4.02

22

=−

i

ii

σ

σσ for Italian securities and

42.0

2.02

22

=−

i

ii

σ

σσ for Belgian securities.

If the average variance of a single security, σ 2

j , in each country equals 50, then

20504.04.0 2 =×== ikj σσ for Italian securities and 10502.02.0 2 =×== ikj σσ for Belgian securities. Using the formula shown in the preceding answer to Problem 3 with σ 2

j = 50 and

either σ kj = 20 for Italy or σ kj = 10 for Belgium we have: Portfolio Size (N securities) Italian σ 2

P Belgian σ 2P

5 26 18 20 21.5 12 100 20.3 10.4


Chapter 4: Problem 6 The formula for an equally weighted portfolio's variance that appears below Table 4.8 in the text is

( ) σσσσ kjkj2j

2P 1/N = +−

where σ 2

j is the average variance across all securities, σ kj is the average covariance across all securities, and N is the number of securities. The text below Table 4.8 states that the average variance for the securities in that table was 46.619 and that the average covariance was 7.058. Using the above equation with those two numbers, setting σ 2

P equal to 8, and solving for N gives: 8 = 1/N (46.619 - 7.058) + 7.058 .942N = 39.561 N = 41.997. Since the portfolio's variance decreases as N increases, holding 42 securities will provide a variance less than 8, so 42 is the minimum number of securities that will provide a portfolio variance less than 8.

Elton, Gruber, Brown and Goetzmann 5-1 Modern Portfolio Theory and Investment Analysis, 7th Edition Solutions To Text Problems: Chapter 5



Chapter 5: Problem 1 From Problem 1 of Chapter 4, we know that: R 1 = 12% R 2 = 6% R 3 = 14% R 4 = 12% σ21 = 8 σ22 = 2 σ23 = 18 σ24 = 10.7 σ 1 = 2.83% σ 2 = 1.41% σ 3 = 4.24% σ 4 = 3.27% σ 12 = − 4 σ 13 = 12 σ 14 = 0 σ 23 = − 6 σ 24 = 0 σ 34 = 0 ρ 12 = − 1 ρ 13 = 1 ρ 14 = 0 ρ 23 = − 1.0 ρ 24 = 0 ρ 34 = 0 In this problem, we will examine 2-asset portfolios consisting of the following pairs of securities: Pair Securities A 1 and 2 B 1 and 3 C 1 and 4 D 2 and 3 E 2 and 4 F 3 and 4 A. Short Selling Not Allowed (Note that the answers to part A.4 are integrated with the answers to parts A.1, A.2 and A.3 below.) A.1 We want to find the weights, the standard deviation and the expected return of the minimum-risk porfolio, also known as the global minimum variance (GMV) portfolio, of a pair of assets when short sales are not allowed.


We further know that the compostion of the GMV portfolio of any two assets i and j is:

ijji

ijjGMViX

σσσ

σσ

222

2

−+

−=

GMVi

GMVj XX −= 1

Pair A (assets 1 and 2): Applying the above GMV weight formula to Pair A yields the following weights:

31

186

)4)(2(28)4(2

2 1222

21

1222

1 ==−−+

−−=

−+

−=

σσσσσGMVX (or 33.33%)

32

3111 12 =−=−= GMVGMV XX (or 66.67%)

This in turn gives the following for the GMV portfolio of Pair A:

%8%632%12

31

=×+×=GMVR

( ) ( ) ( ) ( ) 0432

3122

328

31 22

2 =−⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛+⎟⎠

⎞⎜⎝

⎛+⎟⎠

⎞⎜⎝

⎛=GMVσ

0=GMVσ

Recalling that ρ 12 = − 1, the above result demonstrates the fact that, when two assets are perfectly negatively correlated, the minimum-risk portfolio of those two assets will have zero risk. Pair B (assets 1 and 3): Applying the above GMV weight formula to Pair B yields the following weights:

31 =GMVX (300%) and 23 −=GMVX (−200%)

This means that the GMV portfolio of assets 1 and 3 involves short selling asset 3. But if short sales are not allowed, as is the case in this part of Problem 1, then the GMV “portfolio” involves placing all of your funds in the lower risk security (asset 1) and none in the higher risk security (asset 3). This is obvious since, because the correlation between assets 1 and 3 is +1.0, portfolio risk is simply a linear combination of the risks of the two assets, and the lowest value that can be obtained is the risk of asset 1.


Thus, when short sales are not allowed, we have for Pair B:

11 =GMVX (100%) and 03 =GMVX (0%)

%121 == RRGMV ; 821

2 == σσ GMV ; %83.21 == σσ GMV

For the GMV portfolios of the remaining pairs above we have: Pair GMV

iX GMVjX GMVR GMVσ

C (i = 1, j = 4) 0.572 0.428 12% 2.14% D (i = 2, j = 3) 0.75 0.25 8% 0% E (i = 2, j = 4) 0.8425 0.1575 6.95% 1.3% F (i = 3, j = 4) 0.3728 0.6272 12.75% 2.59% A.2 and A.3 For each of the above pairs of securities, the graph of all possible combinations (portfolios) of the securities (the portfolio possibilties curves) and the efficient set of those portfolios appear as follows when short sales are not allowed: Pair A

The efficient set is the positively sloped line segment.


Pair B

The entire line is the efficient set. Pair C

Only the GMV portfolio is efficient.


Pair D

The efficient set is the positively sloped line segment. Pair E

The efficient set is the positively sloped part of the curve, starting at the GMV portfolio and ending at security 4.


Pair F

The efficient set is the positively sloped part of the curve, starting at the GMV portfolio and ending at security 3. B. Short Selling Allowed (Note that the answers to part B.4 are integrated with the answers to parts B.1, B.2 and B.3 below.) B.1 When short selling is allowed, all of the GMV portfolios shown in Part A.1 above are the same except the one for Pair B (assets 1 and 3). In the no-short-sales case in Part A.1, the GMV “portfolio” for Pair B was the lower risk asset 1 alone. However, applying the GMV weight formula to Pair B yielded the following weights:

31 =GMVX (300%) and 23 −=GMVX (−200%)

This means that the GMV portfolio of assets 1 and 3 involves short selling asset 3 in an amount equal to twice the investor’s original wealth and then placing the original wealth plus the proceeds from the short sale into asset 1. This yields the following for Pair B when short sales are allowed:

%8%142%123 =×−×=GMVR ( ) ( ) ( ) ( ) ( )( )( )( ) 01223218283 222 =−+−+=GMVσ

0=GMVσ Recalling that ρ 13 = +1, this demonstrates the fact that, when two assets are perfectly positively correlated and short sales are allowed, the GMV portfolio of those two assets will have zero risk.


B.2 and B.3 When short selling is allowed, the portfolio possibilities graphs are extended. Pair A

The efficient set is the positively sloped line segment through security 1 and out toward infinity. Pair B

The entire line out toward infinity is the efficient set.


Pair C

Only the GMV portfolio is efficient. Pair D

The efficient set is the positively sloped line segment through security 3 and out toward infinity.


Pair E

The efficient set is the positively sloped part of the curve, starting at the GMV portfolio and extending past security 4 toward infinity. Pair F

The efficient set is the positively sloped part of the curve, starting at the GMV portfolio and extending past security 3 toward infinity.


C. Pair A (assets 1 and 2): Since the GMV portfolio of assets 1 and 2 has an expected return of 8% and a risk of 0%, then, if riskless borrowing and lending at 5% existed, one would borrow an infinite amount of money at 5% and place it in the GMV portfolio. This would be pure arbitrage (zero risk, zero net investment and positive return of 3%). With an 8% riskless lending and borrowing rate, one would hold the same portfolio one would hold without riskless lending and borrowing. (The particular portfolio held would be on the efficient frontier and would depend on the investor’s degree of risk aversion.) Pair B (assets 1 and 3): Since short sales are allowed in Part C and since we saw in Part B that when short sales are allowed the GMV portfolio of assets 1 and 3 has an expected return of 8% and a risk of 0%, the answer is the same as that above for Pair A. Pair C (assets 1 and 4): We have seen that, regardless of the availability of short sales, the efficient frontier for this pair of assets was a single point representing the GMV portfolio, with a return of 12%. With riskless lending and borrowing at either 5% or 8%, the new efficient frontier (efficient set) will be a straight line extending from the vertical axis at the riskless rate and through the GMV portfolio and out to infinity. The amount that is invested in the GMV portfolio and the amount that is borrowed or lent will depend on the investor’s degree of risk aversion. Pair D (assets 2 and 3): Since assets 2 and 3 are perfectly negatively correlated and have a GMV portfolio with an expected return of 8% and a risk of 0%, the answer is identical to that above for Pair A. Pair E (assets 2 and 4): We arrived at the following answer graphically; the analytical solution to this problem is presented in the subsequent chapter (Chapter 6). With a riskless rate of 5%, the new efficient frontier (efficient set) will be a straight line extending from the vertical axis at the riskless rate, passing through the portfolio where the line is tangent to the upper half of the original portfolio possibilities curve, and then out to infinity. The amount that is invested in the tangent portfolio and the amount that is borrowed or lent will depend on the investor’s degree of risk aversion. The tangent portfolio has an expected return of 9.4% and a standard deviation of 1.95%. With a riskless rate of 8%, the point of tangency occurs at infinity.


Pair F (assets 3 and 4): We arrived at the following answer graphically; the analytical solution to this problem is presented in the subsequent chapter (Chapter 6). With a riskless rate of 5%, the new efficient frontier (efficient set) will be a straight line extending from the vertical axis at the riskless rate, passing through the portfolio where the line is tangent to the upper half of the original portfolio possibilities curve, and then out to infinity. The amount that is invested in the tangent portfolio and the amount that is borrowed or lent will depend on the investor’s degree of risk aversion. The tangent (optimal) portfolio has an expected return of 12.87% and a standard deviation of 2.61%. With a riskless rate of 8%, the new efficient frontier will be a straight line extending from the vertical axis at the riskless rate, passing through the portfolio where the line is tangent to the upper half of the original portfolio possibilities curve, and then out to infinity. The tangent (optimal) portfolio has an expected return of 12.94% and a standard deviation of 2.64%. Chapter 5: Problem 2 From Problem 2 of Chapter 4, we know that: R A = 1.22% R B = 2.95% R C = 7.92% σ2A = 15.34 σ2B = 14.42 σ2C = 46.02 σ A = 3.92% σ B = 3.8% σ C = 6.78% σ AB = 2.17 σ AC = 7.24 σ BC = −19.89 ρ AB = 0.15 ρ AC = 0.27 ρ BC = −0.77 In this problem, we will examine 2-asset portfolios consisting of the following pairs of securities: Pair Securities 1 A and B 2 A and C 3 B and C


A. Short Selling Not Allowed (Note that the answers to part A.4 are integrated with the answers to parts A.1, A.2 and A.3 below.) A.1 We want to find the weights, the standard deviation and the expected return of the minimum-risk porfolio, also known as the global minimum variance (GMV) portfolio, of a pair of assets when short sales are not allowed. We further know that the compostion of the GMV portfolio of any two assets i and j is:

ijji

ijjGMViX

σσσ

σσ

222

2

−+

−=

GMVi

GMVj XX −= 1

Pair 1 (assets A and B): Applying the above GMV weight formula to Pair 1 yields the following weights:

482.0)17.2)(2(42.1434.15

17.242.14222

2

=−+

−=

−+

−=

ABBA

ABBGMVAX

σσσσσ

(or 48.2%)

518.0482.011 =−=−= GMV

AGMVB XX (or 51.8%)

This in turn gives the following for the GMV portfolio of Pair 1:

%12.2%95.2518.0%22.1482.0 =×+×=GMVR

( ) ( ) ( ) ( ) ( )( )( )( ) 52.817.2518.0482.0242.14518.034.15482.0 222 =++=GMVσ

%92.2=GMVσ


For the GMV portfolios of the remaining pairs above we have: Pair GMV

iX GMVjX GMVR GMVσ

2 (i = A, j = C) 0.827 0.173 2.38% 3.73% 3 (i = B, j = C) 0.658 0.342 4.65% 1.63% A.2 and A.3 For each of the above pairs of securities, the graph of all possible combinations (portfolios) of the securities (the portfolio possibilties curves) and the efficient set of those portfolios appear as follows when short sales are not allowed: Pair 1

The efficient set is the positively sloped part of the curve, starting at the GMV portfolio and ending at security B.


Pair 2

The efficient set is the positively sloped part of the curve, starting at the GMV portfolio and ending at security C. Pair 3

The efficient set is the positively sloped part of the curve, starting at the GMV portfolio and ending at security C.


B. Short Selling Not Allowed (Note that the answers to part B.4 are integrated with the answers to parts B.1, B.2 and B.3 below.) B.1 When short selling is allowed, all of the GMV portfolios shown in Part A.1 above remain the same. B.2 and B.3 When short selling is allowed, the portfolio possibilities graphs are extended. Pair 1

The efficient set is the positively sloped part of the curve, starting at the GMV portfolio and extending past security B toward infinity.


Pair 2

The efficient set is the positively sloped part of the curve, starting at the GMV portfolio and extending past security C toward infinity. Pair 3

The efficient set is the positively sloped part of the curve, starting at the GMV portfolio and extending past security C toward infinity.


C. In all cases where the riskless rate of either 5% or 8% is higher than the returns on both of the individual securities, if short sales are not allowed, any rational investor would only invest in the riskless asset. Even if short selling is allowed, the point of tangency of a line connecting the riskless asset to the original portfolio possibilities curve occurs at infinity for all cases, since the original GMV portfolio’s return is lower than 5% in all cases. Chapter 5: Problem 3 The answers to this problem are given in the answers to part A.1 of Problem 2. Chapter 5: Problem 4 The locations, in expected return standard deviation space, of all portfolios composed entirely of two securities that are perfectly negatively correlated (say, security C and security S) are described by the equations for two straight lines, one with a positive slope and one with a negative slope. To derive those equations, start with the expressions for a two-asset portfolio's standard deviation when the two assets' correlation is −1 (the equations in (5.8) in the text), and solve for XC (the investment weight for security C). E.g., for the first equation:

( )

. + + = X

) + (X = + X + -X =

X X =

SC

SPC

SCCSP

SCSCCP

SCCCP

σσσσ

σσσσσσσσσσσ −− 1

Now plug the above expression for XC into the expression for a two-asset portfolio's expected return and simplify:

( )

. + RR

+ RR + R=

+ RRR + R + R =

R + + + R +

+ =

RX + RX = R

PSC

SCS

SC

SCS

SC

SSSPCSCPS

SSC

SPC

SC

SP

SCCCP

σσσ

σσσ

σσσσσσ

σσσσ

σσσσ

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −+

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −

−−

⎟⎟⎠

⎞⎜⎜⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛

−

1

1

The above equation is that of a straight line in expected return standard deviation space, with an intercept equal to the first term in brackets and a slope equal to the second term in brackets.


Solving for XC in the second equation in (5.8) gives:

( )

( )

. +

= X

+ X= X + X=

X+ X=

SC

PSC

SCCSP

SCSCCP

SCCCP

σσσσ

σσσσσσσσσσσ

−−−

−−−− 1

Substitute the above expression for XC into the equation for expected return and simplify:

( )

. + RR

+ RR + R=

+ R + RRR + R =

R + + R +

=

RX + RX = R

PSC

CSS

SC

SCS

SC

SPSSCPCSS

SSC

PSC

SC

PS

SCCCP

σσσ

σσσ

σσσσσσ

σσσσ

σσσσ

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −+

⎥⎥⎦

⎤

⎢⎢⎣

⎡ −

−−

⎟⎟⎠

⎞⎜⎜⎝

⎛ −−⎟⎟

⎠

⎞⎜⎜⎝

⎛ −

−

1

1

The above equation is also that of a straight line in expected return standard deviation space, with an intercept equal to the first term in brackets and a slope equal to the second term in brackets. The intercept term for the above equation is identical to the intercept term for the first derived equation. The slope term is equal to −1 times the slope term of the first derived equation. So when one equation has a positive slope, the other equation has a negative slope (when the expected returns of the two assets are equal, the two lines are coincident), and both lines meet at the same intercept. Chapter 5: Problem 5 When ρ equals 1, the least risky "combination" of securities 1 and 2 is security 2 held alone (assuming no short sales). This requires X1 = 0 and X2 = 1, where the X's are the investment weights. The standard deviation of this "combination" is equal to the standard deviation of security 2; σP = σ2 = 2. When ρ equals -1, we saw in Chapter 5 that we can always find a combination of the two securities that will completely eliminate risk, and we saw that this combination can be found by solving X1 = σ2/(σ1 + σ2). So, X1 = 2/(5 + 2) = 2/7, and since the investment weights must sum to 1, X2 = 1 - X1 = 1 - 2/7 = 5/7. So a combination of 2/7 invested in security 1 and 5/7 invested in security 2 will completely eliminate risk when ρ equals -1, and σP will equal 0.


When ρ equals 0, we saw in Chapter 5 that the minimum-risk combination of two assets can be found by solving X1 = σ22/(σ12 + σ22). So, X1 = 4/(25 + 4) = 4/29, and X2 = 1 - X1 = 1 - 4/29 = 25/29. When ρ equals 0, the expression for the standard deviation of a two-asset portfolio is

( ) 22

21

21

21 1 σσσ XX = P −+

Substituting 4/29 for X1 in the above equation, we have

%86.1841

2900841

2500841400

4292525

294 22

=

=

+=

×⎟⎠

⎞⎜⎝

⎛+×⎟⎠

⎞⎜⎝

⎛=Pσ

Chapter 5: Problem 6 If the riskless rate is 10%, then the risk-free asset dominates both risky assets in terms of risk and return, since it offers as much or higher expected return than either risky asset does, for zero risk. Assuming the investor prefers more to less and is risk averse, the optimal investment is the risk-free asset.



Solutions to Text Problems: Chapter 6 Chapter 6: Problem 1 The simultaneous equations necessary to solve this problem are: 5 = 16Z1 + 20Z2 + 40Z3 7 = 20Z1 + 100Z2 + 70Z3 13 = 40Z1 + 70Z2 + 196Z3 The solution to the above set of equations is: Z1 = 0.292831 Z2 = 0.009118 Z3 = 0.003309 This results in the following set of weights for the optimum (tangent) portfolio: X1 = .95929 (95.929%) X2 = .02987 (2.987%) X3 = .01084 (1.084% The optimum portfolio has a mean return of 10.146% and a standard deviation of 4.106%. Chapter 6: Problem 2 The simultaneous equations necessary to solve this problem are: 11 − RF = 4Z1 + 10Z2 + 4Z3 14 − RF = 10Z1 + 36Z2 + 30Z3

17 − RF = 4Z1 + 30Z2 + 81Z3


The optimum portfolio solutions using Lintner short sales and the given values for RF are: RF = 6% RF = 8% RF = 10%

Z1 3.510067 1.852348 0.194631 Z2 −1.043624 −0.526845 −0.010070 Z3 0.348993 0.214765 0.080537

X1 0.715950 0.714100 0.682350 X2 −0.212870 −0.203100 −0.035290 X3 0.711800 0.082790 0.282350

Tangent (Optimum) Portfolio Mean Return 6.105% 6.419% 11.812% Tangent (Optimum) Portfolio Standard Deviation 0.737% 0.802% 2.971% Chapter 6: Problem 3 Since short sales are not allowed, this problem must be solved as a quadratic programming problem. The formulation of the problem is:

P

FP

X

RRσ

θ −=max

subject to:

11

=∑=

N

iiX

0≥iX ∀ i


Chapter 6: Problem 4 This problem is most easily solved using The Investment Portfolio software that comes with the text, but, since all pairs of assets are assumed to have the same correlation coefficient of 0.5, the problem can also be solved manually using the constant correlation form of the Elton, Gruber and Padberg “Simple Techniques” described in a later chapter. To use the software, open up the Markowitz module, select “file” then “new” then “group constant correlation” to open up a constant correlation table. Enter the input data into the appropriate cells by first double clicking on the cell to make it active. Once the input data have been entered, click on “optimizer” and then “run optimizer” (or simply click on the optimizer icon). At that point, you can either select “full Markowitz” or “simple method.” If you select “full Markowitz,” you then select “short sales allowed/riskless lending and borrowing” and then enter 4 for both the lending and borrowing rate and click “OK.” A graph of the efficient frontier then appears. You may then hit the “Tab” key to jump to the tangent portfolio, then click on “optimizer” and then “show portfolio” (or simply click on the “show portfolio” icon) to view and print the composition (investment weights), mean return and standard deviation of the tangent (optimum) portfolio. If instead you select “simple method,” you then select “short sales allowed with riskless asset” and enter 4 for the riskless rate and click “OK.” A table showing the investment weights of the tangent portfolio then appears. Regardless of the method used, the resulting investment weights for the optimum portfolio are as follows:

Asset i Xi

1 −5.999% 2 −17.966% 3 21.676% 4 0.478% 5 −29.585% 6 12.693% 7 −59.170% 8 −14.793% 9 3.442%

10 189.224%


Given the above weights, the optimum (tangent) portfolio has a mean return of 18.907% and a standard deviation of 3.297%. The efficient frontier is a positively sloped straight line starting at the riskless rate of 4% and extending through the tangent portfolio (T) and out to infinity:

Chapter 6: Problem 5 Since the given portfolios, A and B, are on the efficient frontier, the GMV portfolio can be obtained by finding the minimum-risk combination of the two portfolios:

31

20216362016

222

2

−=×−+

−=

−+

−=

ABBA

ABBGMVAX

σσσσσ

3111 =−= GMV

AGMVB XX

This gives %33.7=GMVR and %83.3=GMVσ Also, since the two portfolios are on the efficient frontier, the entire efficient frontier can then be traced by using various combinations of the two portfolios, starting with the GMV portfolio and moving up along the efficient frontier (increasing the weight in portfolio A and decreasing the weight in portfolio B). Since XB = 1 − XA the efficient frontier equations are:

( ) ( )AABAAAP XXRXRXR −×+=−+= 18101

( ) ( )( ) ( )AAAA

ABAABAAAP

XXXX

XXXX

−+−+=

−+−+=

14011636

12122

2222 σσσσ


Since short sales are allowed, the efficient frontier will extend beyond portfolio A and out toward infinity. The efficient frontier appears as follows:



Solutions to Text Problems: Chapter 7 Chapter 7: Problem 1 We will illustrate the answers for stock A and the market portfolio (S&P 500); the answers for stocks B and C are found in an identical manner. The sample mean monthly return on stock A is:

%946.212

94.048.775.1207.118.197.879.216.357.112.427.1505.1212

12

1

=

−+++−−−++−+=

=∑=t

At

A

RR

The sample mean monthly return on the market portfolio (the answer to part 1.E) is:

%005.312

15.147.216.646.311.277.643.441.448.441.299.528.1212

12

1

=

−+++−−+++++=

=∑=t

mt

m

RR

Using data given in the problem and the above two sample mean monthly returns, we have the following: Month t AAt RR − ( )2AAt RR − mmt RR − ( )2mmt RR − ( )( )mmtAAt RRRR −−

1 9.104 82.883 9.275 86.026 84.44 2 12.324 151.881 2.985 8.910 36.79 3 -7.066 49.928 -0.595 0.354 4.2 4 -1.376 1.893 1.475 2.176 -2.03 5 0.214 0.046 1.405 1.974 0.3 6 -5.736 32.902 1.425 2.031 -8.17 7 -11.916 141.991 -9.775 95.551 116.48 8 -4.126 17.024 -5.115 26.163 21.1 9 -1.876 3.519 0.455 0.207 -0.85

10 9.804 96.118 3.155 9.954 30.93 11 4.534 20.557 -0.535 0.286 -2.43 12 -3.886 15.101 -4.155 17.264 16.15

Sum 0.00 613.84 0.00 250.90 296.91


The sample variance and standard deviation of the stock A’s monthly return are:

( )

15.5112

84.61312

12

1

2

2 ==−

=∑=t

AAt

A

RRσ

%15.715.51 ==Aσ The sample variance (the answer to part 1.F) and standard deviation of the market portfolio’s monthly return are:

( )

91.2012

90.25012

12

1

2

2 ==−

=∑=t

mmt

m

RRσ

%57.491.20 ==mσ The sample covariance of the returns on stock A and the market portfolio is:

( )( )[ ]

74.2412

91.29612

12

1 ==−−

=∑=t

mmtAAt

Am

RRRRσ

The sample correlation coefficient of the returns on stock A and the market portfolio (the answer to part 1.D) is:

757.057.415.7

74.24=

×==

mA

AmAm σσ

σρ

The sample beta of stock A (the answer to part 1.B) is:

183.191.2074.24

2 ===m

AmA σ

σβ

The sample alpha of stock A (the answer to part 1.A) is: %609.0%005.3183.1%946.2 −=×−=−= mAAA RR βα Each month’s sample residual is security A’s actual return that month minus the return that month predicted by the regression. The regression’s predicted monthly return is: mtAAedictedtA RR βα −=Pr,,


The sample residual for each month t is then: edictedtAAtAt RR Pr,,−=ε So we have the following: Month t AtR edictedtAR Pr,, Atε 2

Atε 1 12.05 13.92 -1.87 3.5 2 15.27 6.48 8.79 77.26 3 -4.12 2.24 -6.36 40.45 4 1.57 4.69 -3.12 9.73 5 3.16 4.61 -1.45 2.1 6 -2.79 4.63 -7.42 55.06 7 -8.97 -8.62 -0.35 0.12 8 -1.18 -3.11 1.93 3.72 9 1.07 3.48 -2.41 5.81

10 12.75 6.68 6.07 36.84 11 7.48 2.31 5.17 26.73 12 -0.94 -1.97 1.02 1.04

Sum: 0.00 262.36 Since the sample residuals sum to 0 (because of the way the sample alpha and beta are calculated), the sample mean of the sample residuals also equals 0 and the sample variance and standard deviation of the sample residuals (the answer to part 1.C) are:

( )

863.2112

36.2621212

12

1

12

12 ===−

=∑∑== t

Att

AAt

A

εεεσε

%676.4863.21 ==Aεσ


Repeating the above analysis for all the stocks in the problem yields: Stock A Stock B Stock C alpha −0.609% 2.964% −3.422% beta 1.183 1.021 2.322 correlation with market 0.757 0.684 0.652 standard deviation of sample residuals* 4.676% 4.983% 12.341% with %005.3=mR and 91.202 =mσ . *Note that most regression programs use N − 2 for the denominator in the sample residual variance formula and use N − 1 for the denominator in the other variance formulas (where N is the number of time series observations). As is explained in the text, we have instead used N for the denominator in all the variance formulas. To convert the variance from a regression program to our results, simply multiply the

variance by either N

N 2− or N

N 1− .

Chapter 7: Problem 2 A. A.1 The Sharpe single-index model's formula for a security's mean return is

R + = R miii βα

Using the alpha and beta for stock A along with the mean return on the market portfolio from Problem 1 we have: %946.2005.3183.1609.0 =×+−=AR Similarly: %032.6=BR ; %556.3=CR


The Sharpe single-index model's formula for a security's variance of return is: 2222

imii εσσβσ += Using the beta and residual standard deviation for stock A along with the variance of return on the market portfolio from Problem 1 we have: 14.51676.491.20183.1 222 =+×=Aσ Similarly:

62.462 =bσ ; 0.2652 =cσ A.2 From Problem 1 we have: %946.2=AR ; %031.6=BR ; %554.3=CR 15.512 =Aσ ; 61.462 =Bσ ; 0.2652 =Cσ B. B.1 According to the Sharpe single-index model, the covariance between the returns on a pair of assets is: 2

mjiijSIM σββσ = Using the betas for stocks A and B along with the variance of the market portfolio from Problem 1 we have: 254.2591.20021.1183.1 =××=ABSIMσ Similarly: 433.57=ACSIMσ ; 568.49=BCSIMσ


B.2 The formula for sample covariance from the historical time series of 12 pairs of returns on security i and security j is:

( )( )

12

12

1∑=

−−= t

jjtiit

ij

RRRRσ

Applying the above formula to the monthly data given in Problem 1 for securities A, B and C gives: 462.18=ABσ ; 618.61=ACσ ; 085.54=BCσ C. C.1 Using the earlier results from the Sharpe single-index model, the mean monthly return and standard deviation of an equally weighted portfolio of stocks A, B and C are:

%18.4%556.331%032.6

31%946.2

31

=×+×+×=PR

%348.8

57.493143.57

3125.25

3120.265

3162.46

3115.51

31 222222

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛×⎟

⎠

⎞⎜⎝

⎛+×⎟⎠

⎞⎜⎝

⎛+×⎟⎠

⎞⎜⎝

⎛×+×⎟⎠

⎞⎜⎝

⎛+×⎟⎠

⎞⎜⎝

⎛+×⎟⎠

⎞⎜⎝

⎛=Pσ

C.2 Using the earlier results from the historical data, the mean monthly return and standard deviation of an equally weighted portfolio of stocks A, B and C are:

%18.4%554.331%031.6

31%946.2

31

=×+×+×=PR

%374.8

08.543162.61

3146.18

3120.265

3162.46

3115.51

31 222222

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛×⎟

⎠

⎞⎜⎝

⎛+×⎟⎠

⎞⎜⎝

⎛+×⎟⎠

⎞⎜⎝

⎛×+×⎟⎠

⎞⎜⎝

⎛+×⎟⎠

⎞⎜⎝

⎛+×⎟⎠

⎞⎜⎝

⎛=Pσ


D. The slight differences between the answers to parts A.1 and A.2 are simply due to rounding errors. The results for sample mean return and variance from either the Sharpe single-index model formulas or the sample-statistics formulas are in fact identical. The answers to parts B.1 and B.2 differ for sample covariance because the Sharpe single-index model assumes the covariance between the residual returns of securities i and j is 0 (cov(εi εj ) = 0), and so the single-index form of sample covariance of total returns is calculated by setting the sample covariance of the sample residuals equal to 0. The sample-statistics form of sample covariance of total returns incorporates the actual sample covariance of the sample residuals. The answers in parts C.1 and C.2 for mean returns on an equally weighted portfolio of stocks A, B and C are identical because the Sharpe single-index model formula for the mean return on an individual stock yields a result identical to that of the sample-statistics formula for the mean return on the stock. The answers in parts C.1 and C.2 for standard deviations of return on an equally weighted portfolio of stocks A, B and C are different because the Sharpe single-index model formula for the sample covariance of returns on a pair of stocks yields a result different from that of the sample-statistics formula for the sample covariance of returns on a pair of stocks. Chapter 7: Problem 3 Recall from the text that the Vasicek technique’s forecast of security i’s beta ( 2iβ ) is:

121

21

21

121

21

21

2 iii

ii β

σσ

σβ

σσ

σβ

ββ

β

ββ

β ×+

+×+

=

where 1β is the average beta across all sample securities in the historical period (in this problem referred to as the “market beta”), 1iβ is the beta of security i in the historical period, 2

1βσ is the variance of all the sample securities’ betas in the

historical period and 21iβσ is the square of the standard error of the estimate of beta

for security i in the historical period. If the standard errors of the estimates of all the betas of the sample securities in the historical period are the same, then, for each security i, we have: ai =

21βσ

where a is a constant across all the sample securities.


Therefore, we have for any security i:

( ) 11121

21

121

2 1 iii XXaa

a βββσ

σβ

σβ

β

β

β

−+=×+

+×+

=

This shows that, under the assumption that the standard errors of all historical betas are the same, the forecasted beta for any security using the Vasicek technique is a simple weighted average (proportional weighting) of 1β (the “market beta”) and 1iβ (the security’s historical beta), where the weights are the same for each security. Chapter 7: Problem 4 Letting the historical period of the year of monthly returns given in Problem 1 equal 1 (t = 1), then the forecast period equals 2 and the Blume forecast equation is: 12 60.041.0 ii ββ += Using the earlier answer to Problem 1 for the estimate of beta from the historical period for stock A along with the above equation we obtain the stock’s forecasted beta: 120.1183.160.041.060.041.0 12 =×+=+= AA ββ Similarly: 023.12 =Bβ ; 803.12 =Cβ Chapter 7: Problem 5 A. The single-index model's formula for security i's mean return is

R + = R miii βα

Since Rm equals 8%, then, e.g., for security A we have:

%=+ =

x + =R + = R mAAA

14122

85.12βα


Similarly:

%4.13=BR ; %4.7=CR ; %2.11=DR B. The single-index model's formula for security i's own variance is:

. + = 2e

2m

2i

2i iσσβσ

Since σm = 5, then, e.g., for security A we have:

( ) ( ) ( )25.65

355.1 222

= =

+ = 2e

2m

2A

2A A

+×

σσβσ

Similarly: σ2B = 43.25; σ2C = 20; σ2D = 36.25 C. The single-index model's formula for the covariance of security i with security j is

σββσσ 2mjijiij = =

Since σ2m = 25, then, e.g., for securities A and B we have:

75.48

253.15.1== = 2

mBAAB

××σββσ

Similarly: σAC = 30; σAD = 33.75; σBC = 26; σBD = 29.25; σCD = 18 Chapter 7: Problem 6 A. Recall that the formula for a portfolio's beta is:

ββ ii

N

1 = iP X = ∑

The weight for each asset (Xi) in an equally weighted portfolio is simply 1/N, where N is the number of assets in the portfolio.


Since there are four assets in Problem 5, N = 4 and Xi equals 1/4 for each asset in an equally weighted portfolio of those assets. So:

( )

125.1

5.441

9.08.03.15.141

41

41

41

41

=

=

=

= DCBAP

×

+++

+++ βββββ

B. Recall that the definition of a portfolio's alpha is:

αα ii

N

1 = iP X = ∑

Using 1/4 as the weight for each asset, we have:

( )

5.2

1041

413241

41

41

41

41

=

=

=

= DCBAP

×

+++

+++ ααααα

C. Recall that a formula for a portfolio’s variance using the single-index model is:

∑=

+=N

ieimPP i

X1

22222 σσβσ

Using 1/4 as the weight for each asset, we have:

( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

52.33

1641916125125.1

4412

411

413

415125.1

2

22

22

22

22

222

=

++++×=

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛+=Pσ


D. Using the single-index model’s formula for a portfolio’s mean return we have:

%5.11

8125.15.2=

×+=

+= mPPP RR βα

Chapter 7: Problem 7 Using 12 677.0343.0 ii ββ += and the historical betas given in Problem 5 we can forecast, e.g., the beta for security A:

3585.10155.1343.0

5.1677.0343.0677.0343.0 12

=+=

×+=

+= AA ββ

Similarly: 2231.12 =Bβ ; 8846.02 =Cβ ; 9523.02 =Dβ Chapter 7: Problem 8 Using the historical betas given in Problem 5 and Vasicek’s formula, we can forecast, e.g., the beta of security A:

( )( ) ( )

( )( ) ( )

2932.18795.04137.0

5.15863.014137.0

5.10441.00625.0

0625.010441.00625.0

0441.0

5.121.025.0

25.0121.025.0

21.022

2

22

2

121

21

21

121

21

21

2

=+=

×+×=

×+

+×+

=

×+

+×+

=

×+

+×+

= AAA

AA β

σσ

σβ

σσσ

βββ

β

ββ

β

Similarly: 1137.12 =Bβ ; 8683.02 =Cβ ; 9390.02 =Dβ




Solutions to Text Problems: Chapter 8 Chapter 8: Problem 1 Given the correlation coefficient of the returns on a pair of securities i and j, the securities’ covariance can be expressed as the securities’ correlation coefficient times the product of their standard deviations:

jiijij σσρσ =

But if we assume that all pairs of securities have the same constant correlation, *ρ , then the constant-correlation expression for covariance is:

jiijCC σσρσ *= Given the assumptions of the Sharpe single-index model, the single-index model’s expression for the covariance between the returns on a pair of securities is:

jijmim

mm

mjjm

m

miim

mm

jm

m

im

mjiijSIM

σσρρ

σσ

σσρσσσρ

σσσ

σσ

σββσ

=

××=

××=

=

222

222

2

If the assumptions of both the constant correlation and single-index model hold, then we have ijij SIMCC σσ = :

jijmimji σσρρσσρ =* or jmimρρρ =* This must hold for all pairs of securities, including i and j, i and k and j and k. So we have:

jmimρρρ =*

kmimρρρ =*

kmjmρρρ =* The only solution to the above set of equations is:

*ρρρρ === kmjmim Therefore, for any security i we have:

imm

iim

m

miim

m

imi σ

σρ

σσρ

σσσρ

σσ

β ×====*

22

In other words, given that all pairs of securities have the same correlation coefficient and that the Sharpe single-index model holds, each security’s beta is proportional to its standard deviation, where the proportion is a constant across all

securities equal to mσρ *

.


Chapter 8: Problem 2 Start with a general 3-index model of the form:

iiiiii cIbIbIbaR +×+×+×+= *3

*3

*2

*2

*1

*1

* (1)

Set 1*1 II = and define an index I2 which is orthogonal to I1 as follows:

tdII +×+= 110

*2 γγ or ( )110

*22 IIdI t ×+−== γγ

which gives:

2110*2 III +×+= γγ

Substituting the above expression into equation (1) and rearranging we get:

( ) ( ) iiiiiiii cIbIbIbbbaR +×+×+××++×+= *3

*32

*211

*2

*10

*2

* γγ The first term in the above equation is a constant, which we can define as ′

1a . The coefficient in the second term of the above equation is also a constant, which we can define as ′

1ib . We can then rewrite the above equation as:

iiiiii cIbIbIbaR +×+×+×′+′= *3

*32

*211 (2)

Now define an index I3 which is orthogonal to I1 and I2 as follows:

teIII +×+×+= 22110*3 θθθ or ( )22110

*33 IIIeI t ×+×+−== θθθ

which gives:

322110*3 IIII +×+×+= θθθ

Substituting the above expression into equation (2) and rearranging we get:

( ) iiiiiiiii cIbIbbIbbbaR +×+××++×⎟⎠⎞⎜

⎝⎛ ×+′+⎟

⎠⎞⎜

⎝⎛ ×+′= 3

*322

*3

*2113103 θθθ

In the above equation, the first term and all the coefficients of the new orthogonal indices are constants, so we can rewrite the equation as:

iiiiii cIbIbIbaR +×+×+×+= 332211


Chapter 8: Problem 3 Recall from the earlier chapter on the single-index model that an expression for the covariance of returns on two securities i and j is:

( ) ( )[ ] ( )[ ] [ ]jimmjimmijmmjiij eeRReRReRR E E E E 2

+−+−+⎥⎦⎤

⎢⎣⎡ −= ββββσ

The first term contains the variance of the market portfolio, the second two terms contain the covariance of the market portfolio with the residuals and the last term is the covariance of the residuals. Given that one of the model’s assumptions is that the covariance of the market portfolio with the residuals is zero and that, from the problem, the covariance of the residuals equals a constant K, the derived covariance between the two securities is:

Kmjiij += 2σββσ One expression for the variance of a portfolio is:

∑∑∑=

≠==

+=N

j

N

jkk

jkkj

N

iiiP XXX

1 11

222 σσσ

Recalling that the single-index model’s expression for the variance of a security is

2222eimii σσβσ += and substituting that expression and the derived expression for

covariance into the above equation and rearranging gives:

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛++=

++⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

++=

+++=

∑∑∑

∑∑∑∑∑

∑∑∑∑∑

∑ ∑∑∑∑∑

=≠==

=≠====

=≠=== =

= =≠=

≠===

N

j

N

jkk

kj

N

ieiimP

N

j

N

jkk

kj

N

ieiim

N

iii

N

iii

N

j

N

jkk

kj

N

ieii

N

i

N

jmjiji

N

j

N

j

N

jkk

kj

N

jkk

mkjkj

N

ieii

N

imiiP

XXKX

KXXXXX

KXXXXX

KXXXXXX

1 11

2222

1 11

222

11

1 11

22

1 1

2

1 1 11

2

1

22

1

2222

σσβ

σσββ

σσββ

σββσσβσ


Chapter 8: Problem 4 Using the result from Problem 2, we have:

iiiiii cIbIbIbaR +×+×+×+= 332211 Since the residual ci always has a mean of zero (by construction if necessary), we have the following expression for expected return:

332211 IbIbIbaR iiiii ×+×+×+= The variance formula is:

( )( )( ) ( ) ( )( ) ⎥⎦

⎤⎢⎣⎡ +−+−+−=

⎥⎦⎤

⎢⎣⎡ ×+×+×+−+×+×+×+=

2333222111

2332211332211

2

E

E

iiii

iiiiiiiiii

cIIbIIbIIb

IbIbIbacIbIbIbaσ

Carrying out the squaring, noting that the indices are all orthogonal with each other and making the usual assumption that the residual is uncorrelated with any index gives us:

223

23

22

22

21

21

2ciIiIiIii bbb σσσσσ +++=

The covariance formula is:

( )( )( )( )

( ) ( ) ( )( ) ( ) ( ) ( )( )[ ]jjjjiiii

jjjjjjjjj

iiiiiiiiii

cIIbIIbIIbcIIbIIbIIb

IbIbIbacIbIbIba

IbIbIbacIbIbIba

+−+−+−×+−+−+−=

⎥⎥⎦

⎤

⎢⎢⎣

⎡

×+×+×+−+×+×+×+×

×+×+×+−+×+×+×+=

333222111333222111

332211332211

3322113322112

E

E σ

Carrying out the multiplication, noting that the indices are all orthogonal with each other, making the usual assumption that the residuals are uncorrelated with any index and assuming that the residuals are uncorrelated with each other gives us:

2333

2222

2111 IjiIjiIjiij bbbbbb σσσσ ++=


Chapter 8: Problem 5 The formula for a security's expected return using a general two-index model is:

2211 IbIbaR iiii ×+×+=

Using the above formula and data given in the problem, the expected return for, e.g., security A is:

%1249.088.02

2211

=×+×+=

×+×+= IbIbaR AAAA

Similarly:

%17=BR ; %6.12=CR The two-index model’s formula for a security’s own variance is:

222

22

21

21

2ciIiIii bb σσσσ ++=

Using the above formula, the variance for, e.g., security A is:

( ) ( ) ( ) ( ) ( )6225.1140625.556.2

25.29.028.0 22222

222

22

21

21

2

=++=++=

+= cAIAIAA bb σσσσ

Similarly, σ2B = 16.4025, and σ2C = 13.0525. C. The two-index model's formula for the covariance of security i with security j is:

2222

2111 IjiIjiijj bbbb σσσ +=

Using the above formula, the covariance of, e.g., security A with security B is:

( )( )( ) ( )( )( )8325.103125.752.3

5.23.19.021.18.0 22

2222

2111

=+=+=

+= IBAIBAAB bbbb σσσ

Similarly, σAC = 9.0675, and σBC = 12.8975.


Chapter 8: Problem 6 For an industry-index model, the text gives two formulas for the covariance between securities i and k. If firms i and k are both in industry j, the covariance between their securities' returns is given by:

22Ijkjijmkmimik bbbb σσσ +=

Otherwise, if the firms are in different industries, the covariance of their securities' returns is given by:

2mkmimik bb σσ =

If only firms A and B are in the same industry, then:

( )( )( ) ( )( )( )8325.103125.752.3

5.23.19.021.18.0 22

2222

2

=+=+=

+= IBAmBmAmAB bbbb σσσ

The second formula should be used for the other pairs of firms:

( )( )( ) 88.229.08.0 2

2

==

= mCmAmAC bb σσ

( )( )( ) 96.329.01.1 2

2

==

= mCmBmBC bb σσ

Chapter 8: Problem 7

The answers for this problem are found in the same way as the answers for problem 6, except that now only firms B and C are in the same industry. So for firms B and C, the covariance between their securities' returns is:

( )( )( ) ( )( )( )8975.129375.896.3

5.21.13.129.01.1 22

2222

2

=+=+=

+= ICBmCmBmBC bbbb σσσ


The other formula should be used for the other pairs of firms:

( )( )( ) 52.321.18.0 2

2

==

= mBmAmAB bb σσ

( )( )( ) 88.229.08.0 2

2

==

= mCmAmAC bb σσ

Chapter 8: Problem 8 To answer this problem, use the procedure described in Appendix A of the text. First, I1 is defined as being equal to I*1 , then I*2 is regressed on I1 to obtain the given regression equation. Since dt is uncorrelated with I1 by the techniques of regression analysis, dt is an orthogonal index to I1. So, define I2 = dt. Then express the given regression equation as: I*2 = 1 + 1.3 I1 + I2. Now, substitute the above equation for I*2 into the given multi-index model and simplify: iR = 2 + 1.1 I*1 + 1.2 I*2 + ci = 2 + 1.1 I1 + 1.2 (1 + 1.3 I1 + I2) + ci = 2 + 1.1 I1 + 1.2 + 1.56 I1 + 1.2 I2 + ci = 3.2 + 2.66 I1 + 1.2 I2 + ci The two-index model has now been transformed into one with orthogonal indices I1 and I2, where I1 = I*1, and I2 = dt = I*2 - 1 - 1.3 I1.




Chapter 9: Problem 1 In the table below, given that the riskless rate equals 5%, the securities are ranked in descending order by their excess return over beta.

Security Rank i Fi RR − i

Fi RRβ− ( )

2ei

iFi RRσ

β− 2

2

ei

i

σβ

( )∑=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −i

j ej

jFj RR

12σ

β ∑

=⎟⎟

⎠

⎞

⎜⎜

⎝

⎛i

j ej

j

12

2

σ

β Ci

1 1 10 10.0000 0.3333 0.0333 0.3333 0.0333 2.5000 6 2 9 6.0000 1.3500 0.2250 1.6833 0.2583 4.6980 2 3 7 4.6667 0.5250 0.1125 2.2083 0.3708 4.6910 5 4 4 4.0000 0.2000 0.0500 2.4083 0.4208 4.6242 4 5 3 3.7500 0.2400 0.0640 2.6483 0.4848 4.5286 3 6 6 3.0000 0.3000 0.1000 2.9483 0.5848 4.3053

The numbers in the column above labeled Ci were obtained by recalling from the text that, if the Sharpe single-index model holds:

( )

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛+

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −

=

∑

∑

=

=

i

j ej

jm

i

j ej

jFjm

i

RR

C

12

22

12

2

1σ

βσ

σ

βσ

Thus, given that 2mσ = 10:

500.2333.1333.3

0333.01013333.010

1 ==×+

×=C

698.4583.3833.16

2583.01016833.110

2 ==×+

×=C

etc.

With no short sales, we only include those securities for which ii

Fi CRR>

−β

. Thus,

only securities 1 and 6 (the highest and second highest ranked securities in the above table) are in the optimal (tangent) portfolio. We could have stopped our

calculations after the first time we found a ranked security for which ii

Fi CRR<

−β

,

(in this case the third highest ranked security, security 2), but we did not so that we

could demonstrate that ii

Fi CRR<

−β

for all of the remaining lower ranked securities

as well.


Since security 6 (the second highest ranked security, where i = 2) is the last ranked

security in descending order for which ii

Fi CRR>

−β

, we set C* = C2 = 4.698 and

solve for the optimum portfolio’s weights using the following formulas:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⎟⎟⎠

⎞⎜⎜⎝

⎛= *

2 CRRZi

Fi

ei

ii βσ

β

∑=

= 2

1ii

ii

Z

ZX

This gives us:

( ) 1767.0698.410301

1 =−⎟⎠

⎞⎜⎝

⎛=Z

( ) 1953.0698.4610

5.12 =−⎟

⎠

⎞⎜⎝

⎛=Z

3720.01953.01767.021 =+=+ ZZ

475.03720.01767.0

1 ==X

525.03720.01953.0

2 ==X

Since i = 1 for security 1 and i = 2 for security 6, the optimum (tangent) portfolio when short sales are not allowed consists of 47.5% invested in security 1 and 52.5% invested in security 6. Chapter 9: Problem 2 This problem uses the same input data as Problem 1. When short sales are allowed, all securities are included and C* is equal to the value of Ci for the lowest ranked security. Referring back to the table given in the answer to Problem 1, we see that the lowest ranked security is security 3, where i = 6. Therefore, we have C* = C6 = 4.3053.


To solve for the optimum portfolio’s weights, we use the following formulas:

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⎟⎟⎠

⎞⎜⎜⎝

⎛= *

2 CRRZi

Fi

ei

ii βσ

β

and

∑=

= 6

1ii

ii

Z

ZX (for the standard definition of short sales)

or

∑=

= 6

1ii

ii

Z

ZX (for the Lintner definition of short sales)

So we have:

( ) 1898.03053.410301

1 =−⎟⎠

⎞⎜⎝

⎛=Z

( ) 2542.03053.4610

5.12 =−⎟

⎠

⎞⎜⎝

⎛=Z

( ) 0271.03053.4667.420

5.13 =−⎟

⎠

⎞⎜⎝

⎛=Z

( ) 0153.03053.44201

4 −=−⎟⎠

⎞⎜⎝

⎛=Z

( ) 0444.03053.475.310

8.05 −=−⎟

⎠

⎞⎜⎝

⎛=Z

( ) 0653.03053.4340

0.26 −=−⎟

⎠

⎞⎜⎝

⎛=Z

3461.00653.00444.00153.00271.02542.01898.06

1=−−−++=∑

=iiZ

5961.00653.00444.00153.00271.02542.01898.06

1=+++++=∑

=iiZ


This gives us the following weights (by rank order) for the optimum portfolios under either the standard definition of short sales or the Lintner definition of short sales: Standard Definition Lintner Definition

Security 1 (i = 1) 5484.03461.01898.0

1 ==X 3184.05961.01898.0

1 ==X

Security 6 (i = 2) 7345.03461.02542.0

2 ==X 4264.05961.02542.0

2 ==X

Security 2 (i = 3) 0783.03461.00271.0

3 ==X 0455.05961.00271.0

3 ==X

Security 5 (i = 4) 0442.03461.00153.0

4 −=−

=X 0257.05961.00153.0

4 −=−

=X

Security 4 (i = 5) 1283.03461.00444.0

5 −=−

=X 0745.05961.00444.0

5 −=−

=X

Security 3 (i = 6) 1887.03461.00653.0

6 −=−

=X 1095.05961.00653.0

6 −=−

=X

Chapter 9: Problem 3 With short sales allowed but no riskless lending or borrowing, the optimum portfolio depends on the investor’s utility function and will be found at a point along the upper half of the minimum-variance frontier of risky assets, which is the efficient frontier when riskless lending and borrowing do not exist. As is described in the text, the entire efficient frontier of risky assets can be delineated with various combinations of any two efficient portfolios on the frontier. One such efficient portfolio was found in Problem 2. By simply solving Problem 2 using a different value for RF , another portfolio on the efficient frontier can be found and then the entire efficient frontier can be traced using combinations of those two efficient portfolios.


Chapter 9: Problem 4 In the table below, given that the riskless rate equals 5%, the securities are ranked in descending order by their excess return over standard deviation.

Security Rank i Fi RR − i

Fi RRσ−

( )∑=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −i

j j

Fj RR

1 σ

ρρρ

i+−1 Ci

1 1 10 1.00 1.00 0.5000 0.5000 2 2 15 1.00 2.00 0.3333 0.6667 5 3 5 1.00 3.00 0.2500 0.7500 6 4 9 0.90 3.90 0.2000 0.7800 4 5 7 0.70 4.60 0.1667 0.7668 3 6 13 0.65 5.25 0.1429 0.7502 7 7 11 0.55 5.80 0.1250 0.7250

The numbers in the column above labeled Ci were obtained by recalling from the text that, if the constant-correlation model holds:

( )⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ −×⎟⎟⎠

⎞⎜⎜⎝

⎛+−

= ∑=

i

j j

Fji

RRi

C11 σρρ

ρ

Thus, given that ρ = 0.5 for all pairs of securities: 5000.00.15.01 =×=C 6667.00.23333.02 =×=C etc.

With no short sales, we only include those securities for which ii

Fi CRR>

−σ

. Thus,

only securities 1, 2, 5 and 6 (the four highest ranked securities in the above table) are in the optimal (tangent) portfolio. We could have stopped our calculations

after the first time we found a ranked security for which ii

Fi CRR<

−σ

, (in this case

the fifth highest ranked security, security 4), but we did not so that we could

demonstrate that ii

Fi CRR<

−σ

for all of the remaining lower ranked securities as

well.


Since security 6 (the fourth highest ranked security, where i = 4) is the last ranked

security in descending order for which ii

Fi CRR>

−σ

, we set C* = C4 = 0.78 and solve

for the optimum portfolio’s weights using the following formulas:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

= *

11 CRRZ

i

Fi

ii σσρ

∑=

= 4

1ii

ii

Z

ZX

This gives us:

( )( ) ( ) 0440.078.01105.0

11 =−⎟⎟

⎠

⎞⎜⎜⎝

⎛=Z

( )( ) ( ) 0293.078.01155.0

12 =−⎟⎟

⎠

⎞⎜⎜⎝

⎛=Z

( )( ) ( ) 0880.078.0155.0

13 =−⎟⎟

⎠

⎞⎜⎜⎝

⎛=Z

( )( ) ( ) 0240.078.09.0105.0

14 =−⎟⎟

⎠

⎞⎜⎜⎝

⎛=Z

1853.00240.00880.00293.00440.04321 =+++=+++ ZZZZ

2375.01853.00440.0

1 ==X

1581.01853.00293.0

2 ==X

4749.01853.00880.0

3 ==X

1295.01853.00240.0

4 ==X

Since i = 1 for security 1, i = 2 for security 2, i = 3 for security 5 and i = 4 for security 6, the optimum (tangent) portfolio when short sales are not allowed consists of 23.75% invested in security 1, 15.81% % invested in security 2, 47.49% % invested in security 5 and 12.95% invested in security 6.


Chapter 9: Problem 5 This problem uses the same input data as Problem 4. When short sales are allowed, all securities are included and C* is equal to the value of Ci for the lowest ranked security. Referring back to the table given in the answer to Problem 4, we see that the lowest ranked security is security 7, where i = 7. Therefore, we have C* = C7 = 0.725. To solve for the optimum portfolio’s weights, we use the following formulas:

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−

−⎟⎟⎠

⎞⎜⎜⎝

⎛−

= *

11 CRRZ

i

Fi

ii σσρ

and

∑=

= 7

1ii

ii

Z

ZX (for the standard definition of short sales)

or

∑=

= 7

1ii

ii

Z

ZX (for the Lintner definition of short sales)

So we have:

( )( ) ( ) 0550.0725.01105.0

11 =−⎟⎟

⎠

⎞⎜⎜⎝

⎛=Z

( )( ) ( ) 0367.0725.01155.0

12 =−⎟⎟

⎠

⎞⎜⎜⎝

⎛=Z

( )( ) ( ) 1100.0725.0155.0

13 =−⎟⎟

⎠

⎞⎜⎜⎝

⎛=Z

( )( ) ( ) 0350.0725.09.0105.0

14 =−⎟⎟

⎠

⎞⎜⎜⎝

⎛=Z

( )( ) ( ) 0050.0725.07.0105.0

15 −=−⎟⎟

⎠

⎞⎜⎜⎝

⎛=Z

( )( ) ( ) 0075.0725.065.0205.0

16 −=−⎟⎟

⎠

⎞⎜⎜⎝

⎛=Z

( )( ) ( ) 0175.0725.055.0205.0

17 −=−⎟⎟

⎠

⎞⎜⎜⎝

⎛=Z

2067.00175.00075.00050.00350.01100.00367.00550.07

1=−−−+++=∑

=iiZ

2667.00175.00075.00050.00350.01100.00367.00550.07

1=++++++=∑

=iiZ


This gives us the following weights (by rank order) for the optimum portfolios under either the standard definition of short sales or the Lintner definition of short sales: Standard Definition Lintner Definition

Security 1 (i = 1) 2661.02067.00550.0

1 ==X 2062.02667.00550.0

1 ==X

Security 2 (i = 2) 1776.02067.00367.0

2 ==X 1376.02667.00367.0

2 ==X

Security 5 (i = 3) 5322.02067.01100.0

3 ==X 4124.02667.01100.0

3 ==X

Security 6 (i = 4) 1703.02067.00350.0

4 ==X 1312.02667.00350.0

4 ==X

Security 4 (i = 5) 0242.02067.00050.0

5 −==X 0187.02667.00050.0

5 −==X

Security 3 (i = 6) 0363.02067.00075.0

6 −==X 0281.02667.00075.0

6 −==X

Security 7 (i = 7) 0847.02067.00175.0

7 −==X 0656.02667.00175.0

7 −==X

Chapter 9: Problem 6 With short sales allowed but no riskless lending or borrowing, the optimum portfolio depends on the investor’s utility function and will be found at a point along the upper half of the minimum-variance frontier of risky assets, which is the efficient frontier when riskless lending and borrowing do not exist. As is described in the text, the entire efficient frontier of risky assets can be delineated with various combinations of any two efficient portfolios on the frontier. One such efficient portfolio was found in Problem 5. By simply solving Problem 5 using a different value for RF , another portfolio on the efficient frontier can be found and then the entire efficient frontier can be traced using combinations of those two efficient portfolios.




Chapter 11: Problem 1 Expected utility of investment A = 1/3 × −7.5 + 1/3 × −12.5 + 1/3 × −31.5 = −17.0 Expected utility of investment B = 1/4 × −4.0 + 1/2 × −17.5 + 1/4 × −40.0 = −19.75 Expected utility of investment C = 1/5 × 0.5 + 3/5 × −31.5 + 1/5 × −144.0 = −47.8 Investment A is preferred because it has the highest level of expected utility. Chapter 11: Problem 2 Expected utility of investment A = 1/3 × −0.45 + 1/3 × −0.41 + 1/3 × −0.33 = −0.40 Expected utility of investment B = 1/4 × −0.50 + 1/2 × −0.38 + 1/4 × −0.32 = −0.39 Expected utility of investment C = 1/5 × −1 + 3/5 × −0.33 + 1/5 × −0.24 = −0.45 Investment B is preferred because it has the highest level of expected utility. Chapter 11: Problem 3 Expected utility of investment A = 2/5 × −8.96 + 1/5 × −14 + 2/5 × −21.84 = −15.12 Expected utility of investment B = 1/2 × −6 + 1/4 × −17.76 + 1/4 × −36 = −16.44 Investment A is preferred because it has the highest level of expected utility. Chapter 11: Problem 4 For an investor to be indifferent, the expected utility of investment B must be set equal to that of investment A. Referring back to Problem 3, we see that the given probabilities are for the first two of the three outcomes in investment B. So we need to solve for the probabilities of those two outcomes that make investment B’s expected utility level equal to that of A’s. Since the last outcome in investment B has a probability of 1/4, the first two probabilities must sum to 3/4. Therefore we have: X × −6 + (3/4 −X) × −17.76 + 1/4 × −36 = −15.12


Solving for X: X = 0.61 Therefore, the first outcome’s probability of 0.5 would have to be increased by 0.11 to 0.61, and the second outcome’s probability of 0.25 would have to be reduced by 0.11 to 0.14. Chapter 11: Problem 5 The investor will prefer the investment that maximizes expected utility of terminal wealth. Recall that the formula for expected utility of wealth (E[U(W)]) is: ( )[ ] ( ) ( )∑ ×=

WWPWUWUE

where each P(W) is the probability associated with each particular outcome of wealth (W). Since ( ) 205.0 WWWU −= , we have: Investment A:

( )[ ] ( ) ( ) ( )

525.43.055.055.42.075.3

3.01005.0105.0705.072.0505.05E 222

=×+×+×=

××−+××−+××−=WU

Investment B:

( )[ ] ( ) ( ) ( )

635.41.095.46.08.43.02.4

1.0905.098.0805.083.0605.06E 222

=×+×+×=

××−+××−+××−=WU

Investment B is preferred over investment A since B provides higher expected utility. Chapter 11: Problem 6 To solve this problem, set the expected utility of investment A in Problem 5 equal to 4.635 (the expected utility of investment B) and solve for the value of the first outcome in investment A: ( ) ( ) ( ) 635.43.01005.0105.0705.072.005.0 222 =××−+××−+×− XX 635.45.1275.201.2.0 2 =++− XX 086202 =+− XX


The equation above is a quadratic equation with two roots. Using the quadratic formula, the roots are found to be 6.26 and 13.74. So, the minimum amount that the first outcome of investment A would have to change by for the investor to be indifferent between investments A and B would be $6.26 − $5 = $1.26 (an increase), since both investments would then provide the same level of expected utility. Chapter 11: Problem 7 Roy’s safety-first criterion is to minimize Prob(RP < RL). If RL = 5%, then (assuming an initial investment of $100) for the outcomes in Problem 1 we have: Prob(RA < 5%) = 0.0 Prob(RB < 5%) = 0.25 Prob(RC < 5%) = 0.20 Thus, using Roy’s safety-first criterion, investment in A is preferred over investments in B and C, and investment in C is preferred over investment in B. Chapter 11: Problem 8 Kataoka's safety-first criterion is to maximize RL subject to Prob(RP < RL) ≤ α. If α = 10%, then (assuming an initial investment of $100) for the outcomes in Problem 1 the maximum RL is: 4.99% for A 3.99% for B 0.99% for C Thus, A is preferred to B and C, and B is preferred to C. Chapter 11: Problem 9 Employing Telser's criterion, we see that (assuming an initial investment of $100) Projects A, B and C in Problem 1 do not satisfy the constraint Prob(Rp ≤ 5%) ≤ 10%. So, investments A, B, and C are indistinguishable using Telser’s criterion with RL = 5%.


Chapter 11: Problem 10 The geometric mean returns of the outcomes shown in Problem 1 (assuming an initial investment of $100) are:

AGR = (1.05)1/3 (1.06)1/3 (1.09)1/3 − 1 = 0.0665 (6.65%)

BGR = (1.04)1/4 (1.07)1/2 (1.1)1/4 − 1 = 0.0698 (6.98%)

CGR = (1.01)1/5 (1.09)3/5 (1.18)1/5 − 1 = 0.0907 (9.07%). Thus, C > B > A. Chapter 11: Problem 11 Roy's criterion is to minimize Prob(RP < RL). When RL = 3%, Prob(RA < 3%) = 0, Prob(RB < 3%) = 0, and Prob(RC < 3%) = 0. So, investments A, B, and C are indistinguishable using Roy's criterion with RL = 3%. Chapter 11: Problem 12 The geometric mean returns of the investments shown in Problem 11 are:

AGR = (1.03).4 (1.04).3 (1.06).1 (1.07).1 (1.09).1 − 1 = 0.0458 (4.58%)

BGR = (1.05).1 (1.06).2 (1.08).1 (1.09).2 (1.1).4 − 1 = 0.0828 (8.28%)

CGR = (1.05).1 (1.07).1 (1.08).2 (1.09).2 (1.11).4 − 1 = 0.0898 (8.98%). Thus, C > B > A.



Solutions to Text Problems: Chapter 12 Chapter 12: Problem 1 Equation (12.1) in the text can be used to answer this question:

USNUS

FUS

N

FN RRRR,ρ

σσ×

−>

−

As is explained in the text, if the above inequality holds, then the foreign investment will be attractive to a U.S. investor. USR and NR for the foreign countries are given in the problem's table. From the tables in the text, we have: σN ρN,US

Austria 24.50 0.281 France 17.76 0.534 Japan 25.70 0.348 U.K. 15.59 0.646 Also, from the text tables, σUS = 13.59. Given that RF = 6%, we have:

N

FN RRσ− USN

US

FUS RR,ρ

σ×

−

Austria 0.327 0.289 France 0.563 0.550 Japan 0.311 0.358 U.K. 0.577 0.665 For Austria and France, the above inequality holds, so a U.S. investor should consider those foreign markets as attractive investments; for Japan and the U.K., the above inequality does not hold, so a U.S. investor should not consider those foreign markets as attractive investments.


Chapter 12: Problem 2 To answer this question, use the formula introduced in Chapter 5 for finding the minimum-risk portfolio of two assets:

1221

22

21

122122

1 2 ρσσσσρσσσ

−+

−=GMVX

where X1 is the investment weight for asset 1 and X2 = 1 - X1. For equities, σUS = 13.59, σN = 16.70 and ρN,US = 0.423. So the minimum-risk portfolio is:

( ) ( )( )( )

( ) ( ) ( )( )( )( )( )%34.67 6734.0

423.07.1659.1327.1659.13423.07.1659.137.16

22

2

=−+

−=GMV

USX

( )%66.32 3266.01 =−= GMV

USGMVN XX

For bonds, σUS = 7.90, σN = 9.45 and ρN,US = 0.527. So the minimum-risk portfolio is:

( ) ( )( )( )

( ) ( ) ( )( )( )( )( )%41.68 6841.0

527.045.99.7245.99.7527.045.99.745.9

22

2

=−+

−=GMV

USX

( )%59.31 3159.01 =−= GMV

USGMVN XX

For T-bills, σUS = 0.35, σN = 6.77 and ρN,US = −0.220. So the minimum-risk portfolio is:

( ) ( )( )( )

( ) ( ) ( )( )( )( )( )%63.98 9863.0

22.077.635.0277.635.022.077.635.077.6

22

2

=−−+

−−=GMV

USX

( )%37.1 0137.01 =−= GMV

USGMVN XX


Chapter 12: Problem 3 In the text, the return due to exchange-rate changes (RX) is shown to be equal to fxt/fxt-1 - 1, where fxt is the foreign exchange rate at time t expressed in terms of the investor's home currency per unit of foreign currency. Let fxt be the exchange rate expressed in terms of dollars and fx*t be the exchange rate expressed in terms of pounds. These two rates are simply reciprocals, i.e., fx*t = 1/fxt. So from the table in the problem we have:

Period (1 + RX)

(for US investor) (1 + R*X)

(for UK investor) 1 2.5/3 = 0.833 3/2.5 = 1.200 2 2.5/2.5 = 1.000 2.5/2.5 = 1.000 3 2/2.5 = 0.800 2.5/2 = 1.250 4 1.5/2 = 0.750 2/1.5 = 1.333 5 2.5/1.5 = 1.667 1.5/2.5 = 0.600

The total return to a U.S. investor from a U.K. investment is (1 + RX)(1 + RUK) − 1; the total return to a U.K. investor from a U.S. investment is(1 + R*X)(1 + RUS) − 1. So: Return to U.S. Investor

Period From U.S.

Investment From U.K. Investment 1

10% (0.833)(1.05) − 1 = −12.5%

2 15% (1)(0.95) − 1 = − 5.0% 3 −5% (0.8)(1.15) − 1 = − 8.0% 4 12% (0.75)(1.08) − 1 = −19.0% 5 6% (1.667)(1.1) − 1 = 83.3%

Average 7.6% 7.76%

Return to U.K. Investor

Period From U.K.

Investment From U.S. Investment 1

5% (1.2)(1.1) − 1 = 32.0%

2 −5% (1)(1.15) − 1 = 15.0% 3 15% (1.25)(0.95) − 1 = 18.75% 4 8% (1.333)(1.12) − 1 = 49.3% 5 10% (0.6)(1.06) − 1 = −36.4%

Average 6.6% 15.73%


Chapter 12: Problem 4 Using the data and averages from Problem 3 we have: For U.S. Investor

( ) ( ) ( ) ( ) ( )

%95.65

6.766.7126.756.7156.710 22222

=

−+−+−−+−+−=USσ

( ) ( ) ( ) ( ) ( )

%06.385

76.73.8376.71976.7876.7576.75.12 22222

=

−+−−+−−+−−+−−=UKσ

For U.K. Investor

( ) ( ) ( ) ( ) ( )

%65.65

6.6106.686.6156.656.65 22222

=

−+−+−+−−+−=UKσ

( ) ( ) ( ) ( ) ( )

%70.285

73.154.3673.153.4973.1575.1873.151573.1532 22222

=

−−+−+−+−+−=USσ

Chapter 12: Problem 5 This problem is essentially the same as Problem 3, except that the exchange rate is given in indirect (yen/$) terms rather than direct ($/yen) terms. From the table in the problem we have:

Period (1 + RX)

(for US investor) (1 + R*X)

(for Japanese investor) 1 200/180 = 1.111 180/200 = 0.900 2 180/190 = 0.947 190/180 = 1.056 3 190/150 = 1.267 150/190 = 0.789 4 150/170 = 0.882 170/150 = 1.133 5 170/180 = 0.944 180/170 = 1.059


The total return to a U.S. investor from a Japan investment is (1 + RX)(1 + RJ) − 1; the total return to a Japanese investor from a U.S. investment is(1 + R*X)(1 + RUS) − 1. So: Return to U.S. Investor

Period From U.S.

Investment From Japan Investment 1

12% (1.111)(1.18) − 1 = 31.10%

2 15% (0.947)(1.12) − 1 = 6.06% 3 5% (1.267)(1.1) − 1 = 39.37% 4 10% (0.882)(1.12) − 1 = −1.22% 5 6% (0.944)(1.07) − 1 = 1.01%

Average 9.6% 15.26%

Return to Japanese Investor

Period From Japan Investment From U.S. Investment

1

18% (0.9)(1.12) − 1 = 0.80%

2 12% (1.056)(1.15) − 1 = 21.44% 3 10% (0.789)(1.05) − 1 = −17.16% 4 12% (1.133)(1.1) − 1 = 24.63% 5 7% (1.059)(1.06) − 1 = 12.25%

Average 11.8% 8.39%

Chapter 12: Problem 6 The answers to this problem are found in the same way as those to Problem 4. For the U.S. investor: σUS = 3.72%; σJ = 16.68% For the Japanese investor: σJ = 3.6%; σUS = 15.227%


Chapter 12: Problem 7 Use the formula for the sample correlation coefficient ρ with five observations:

( )( )

( ) ( )∑ ∑

∑

= =

=

−−

−−=

5

1

5

1

22

5

1

t tJJUSUS

tJJUSUS

RRRR

RRRR

tt

tt

ρ

For the U.S. investor, ρ = −0.251. For the Japanese investor, ρ = −0.050.




Chapter 13: Problem 1 The equation for the security market line is: ( ) iFmFi RRRR β−+= Thus, from the data in the problem we have: ( ) 5.06 ×−+= FmF RRR for asset 1 ( ) 5.112 ×−+= FmF RRR for asset 2 Solving the above two equations simultaneously, we find RF = 3% and mR = 9%. Using those values, an asset with a beta of 2 would have an expected return of: 3 + (9 − 3) × 2 = 15% Chapter 13: Problem 2 Given the security market line in this problem, for the two stocks to be fairly priced their expected returns must be: ( )8% 08.05.008.004.0 =×+=XR ( )20% 20.0208.004.0 =×+=YR If the expected return on either stock is higher than its return given above, the stock is a good buy. Chapter 13: Problem 3 Given the security market line in this problem, the two funds’ expected returns would be: ( )21.2% 212.08.019.006.0 =×+=AR ( )28.8% 288.02.119.006.0 =×+=BR


Comparing the above returns to the funds’ actual returns, we see that both funds performed poorly, since their actual returns were below those expected given their beta risk. Chapter 13: Problem 4 Given the security market line in this problem, the riskless rate equals 0.04 (4%), the intercept of the line, and the excess return of the market above the riskless rate (also called the “market risk premium”) equals 0.10 (10%), the slope of the line. (The return on the market portfolio must therefore be 0.04 + 0.10 = 0.14, or 14%.) Chapter 13: Problem 5 The price form of the CAPM’s security market line equation is:

( ) ( )( ) ⎥

⎦

⎤⎢⎣

⎡××−−=

m

mimFmi

Fi Y

YYPrYYr

P var cov1

where ( )FF Rr += 1 and m

mmm P

PYR −= .

From Problem 4, we have 04.0=FR and 14.0=mR . Therefore m

mm

PPY −

=14.0

which gives mm YP =14.1 . Substituting these vales into the above security market line equation, we have:

( ) ( )

( )( )( ) ⎥

⎦

⎤⎢⎣

⎡××−=

⎥⎦

⎤⎢⎣

⎡××−×−=

m

mimi

m

mimmii

YYYPY

YYYPPYP

var cov10.0

04.11

var cov04.114.1

04.11


Chapter 13: Problem 6 To be rigorous, one should use the four Kuhn-Tucker conditions shown in Appendix E of Chapter 6. To find the optimum portfolio when short sales are not allowed, we have, for each asset i, the following Kuhn-Tucker conditions:

0=+ ii

UdXdθ (1)

0=iiUX (2) 0≥iX (3) 0≥iU (4) We have already seen that, given the assumptions of the standard CAPM, setting

0=idX

dθ gives the equilibrium first order condition for asset i, which is the standard

CAPM’s security market line: ( ) iFmFi RRRR β−+= or equivalently ( ) 0=−−− iFmFi RRRR β When short sales are not allowed, Kuhn-Tucker condition (1) implies that: ( ) 0=+−−− iiFmFi URRRR β But, since all assets are held long in the market portfolio, Xi > 0 for each asset and therefore, given Kuhn-Tucker condition (2), Ui = 0 for each asset. Thus, the standard CAPM holds even if short sales are not allowed.


Chapter 13: Problem 7 Using the two assets in Problem 1, a portfolio with a beta of 1.2 can be constructed as follows: 0.5X1 + (1.5)(1 – X1) = 1.2 X1 = 0.3; X2 = 0.7 The return on this combination would be: 0.3(6%) + 0.7(12%) = 10.2% Asset 3 has a higher expected return than the portfolio of assets 1 and 2, even though asset 1 and the portfolio have the same beta. Thus, buying asset 3 and financing it by shorting the portfolio would produce a positive (arbitrage) return of 15% − 10.2% = 4.8% with zero net investment and zero beta risk. Chapter 13: Problem 8 The security market line is: ( ) iFmFi RRRR β−+= Substituting the given values for assets 1 and 2 gives two equations with two unknowns: ( ) 8.04.9 ×−+= FmF RRR ( ) 3.14.13 ×−+= FmF RRR Solving simultaneously gives: %3=FR ; %11=mR Chapter 13: Problem 9 Substituting the given betas in the given equation yields: ( )%8.17 178.01 =R ; ( )%1.15 151.02 =R




Chapter 14: Problem 1 Given the zero-beta security market line in this problem, the return on the zero-beta portfolio equals 0.04 (4%), the intercept of the line, and the excess return of the market above the zero-beta portfolio’s return (also called the “market risk premium”) equals 0.10 (10%), the slope of the line. The return on the market portfolio must therefore be 0.04 + 0.10 = 0.14, or 14%. Chapter 14: Problem 2

ZR has the same role in the zero-beta model as RF does in the standard model. So, referring back to the answer to Problem 5 in Chapter 13, simply replace RF with ZR to obtain:

( ) ( )( ) ⎥

⎦

⎤⎢⎣

⎡××−−=

m

mimZmi

Zi Y

YYPrYYr

P var cov1

where ( )ZZ Rr += 1 . Chapter 14: Problem 3 As is shown in the text, the post-tax form of the CAPM’s equilibrium pricing equation is: ( )( ) ( )FiiFmFmFi RRRRRR −×+×−×−−+= δτβδτ Rearranging the above equation to isolate δi we have: ( ) ( )( ) iiFmFmFi RRRRR τδβδττ +×−×−−+−= 1 Comparing the above general equation to the specific one given in the problem,

we see that ( ) 05.01 =−τFR , or ( )τ−=1

05.0FR , and that 24.0=τ . Therefore:

( ) ( )%58.6 0658.024.01

05.0=

−=FR


Chapter 14: Problem 4 Since we are given ZR and only one RF , and since ZR > RF , this situation is where there is riskless lending at RF and no riskless borrowing. The efficient frontier will therefore be a ray in expected return-standard deviation space tangent to the minimum-variance curve of risky assets and intersecting the expected return axis at the riskless rate of 3% plus that part of the minimum-variance curve of risky assets to the right of the tangency point. This is depicted in the graph below, where the efficient frontier extends along the ray from RF to the tangent portfolio L, then to the right of L along the curve through the market portfolio M and out toward infinity (assuming unlimited short sales). Note that, unless all investors in the economy choose to lend or invest solely in portfolio L, the market portfolio M will always be on the minimum-variance curve to the right of portfolio L.

Since both M and Z are on the minimum-variance curve, the entire minimum-variance curve of risky assets can be traced out by using combinations (portfolios) of M and Z. Letting X be the investment weight for the market portfolio, the expected return on any combination portfolio P of M and Z is: ( ) ZmP RXRXR −+= 1 (1) Recognizing that M and Z are uncorrelated, the standard deviation of any combination portfolio P of M and Z is:

( ) 2222 1 ZmP XX σσσ −+= (2)


Substituting the given values for mR and ZR into equation (1) gives:

( )510

1515+=

−+=X

XXRP (3)

Substituting the given values for mσ and Zσ into equation (2) gives:

( )

64128548

6412864484

8122

2

22

2222

+−=

+−+=

×−+×=

XX

XXX

XXPσ

(4)

Using equations (3) and (4) and varying X (the fraction invested in the market portfolio M) gives various coordinates for the minimum-variance curve; some of them are given below: X 0 0.2 0.4 0.6 0.8 1.0 1.5 2.0 PR 5 7 9 11 13 15 20 25 Pσ 8 7.77 10.02 13.58 17.67 22 33.24 44.72 The zero-beta form of the security market line describes equilibrium beta risk and expected return relationship for all securities and portfolios (including portfolio L) except those combination portfolios composed of the riskless asset and tangent portfolio L along the ray RF - L in the above graph:

( )i

iZmZi RRRRβ

β105 +=

−+=

The equilibrium beta risk and expected return relationship for any combination portfolio C composed of the riskless asset and tangent portfolio L along the ray RF - L in the above graph is described by the following line:

( )C

L

FLFC

RRRR ββ

×−

+=


Combining the two lines yields the following graph:

Chapter 14: Problem 5 If the post-tax form of the equilibrium pricing model holds, then: ( ) ( )( ) ( )τδβτδ FiiFmFmFi RRRRRR −+−−−+= If the standard CAPM model holds, then: ( ) iFmFi RRRR β−+= Assume that the post-tax model holds instead of the standard model, and

Fm R=δ . For a stock with ( ) 0>− τδ Fi R , the institution that uses the post-tax model would correctly believe that the stock has a higher expected return than the stock’s return expected by the institution using the standard model. Similarly, for a stock with ( ) 0<− τδ Fi R , the institution that uses the post-tax model would correctly believe the stock has a lower expected return than the stock’s return expected by the institution using the standard model.


Now consider a specific example using the following data for stocks A and B, the market portfolio and the riskless asset:

0.1=Aβ ; %8=Aδ ; 0.1=Bβ ; %0=Bδ ; %14=mR ; %4=mδ ; %4=FR ; 25.0=τ If the post-tax model holds, then the institution using that model would correctly believe that the equilibrium expected returns for the two stocks are:

( ) ( )( ) ( )

%151104

25.0480.125.0444144

=++=

×−+××−−−+=AR

( ) ( )( ) ( )

%131104

25.0400.125.0444144

=−+=

×−+××−−−+=BR

The institution using the standard model would incorrectly believe that the stocks’ equilibrium expected returns are:

( )%14104

0.14144=+=

×−+=AR

( )%14104

0.14144=+=

×−+=BR

The institution using the post-tax model would tend to buy stock A and sell stock B short. Of course, residual risk puts a limit to the amount of unbalancing the institution would do. But by some unbalancing, the institution earns an excess return. The institution using the standard model would be indifferent between the two stocks. However, by buying stock B, the institution loses excess return.


Chapter 14: Problem 6 Using Ross’s APT model, we can create an arbitrage portfolio as follows: 01=×∑

i

ARBiX (1)

0== ∑ i

i

ARBiARB aXa (2)

0== ∑ i

i

ARBiARB bXb (3)

Since the above portfolio has zero net investment and zero risk with respect to the given two-factor model, by the force of arbitrage its expected return must also be zero: 0== ∑

ii

ARBiARB RXR (4)

From a theorem of linear algebra, since the above orthogonality conditions (1), (2) and (3) with respect to the ARB

iX result in orthogonality condition (4) with respect to the ARB

iX , iR can be expressed as a linear combination of 1, ai and bi: iii baR 210 1 λλλ ++×= (5) We can create a zero-risk investment portfolio as follows: 1=∑

i

ZiX

0== ∑ i

i

ZiZ aXa

0== ∑ i

i

ZiZ bXb

Substituting the above equations into equation (5) gives:

0

210

λ

λλλ

=

++== ∑∑∑∑ ii

Zii

i

Zi

i

Zii

i

ZiZ bXaXXRXR


We can create a strictly market-risk investment portfolio as follows: 1=∑

i

MiX

1== ∑ i

i

MiM aXa

0== ∑ i

i

MiM bXb


10

210

λλ

λλλ

+=

++== ∑∑∑∑ ii

Mii

i

Mi

i

Mii

i

MiM bXaXXRXR

or ZMM RRR −=−= 01 λλ We can create a strictly interest rate-risk investment portfolio as follows: 1=∑

i

CiX

0== ∑ i

i

CiC aXa

1== ∑ i

i

CiC bXb


20

210

λλ

λλλ

+=

++== ∑∑∑∑ ii

Cii

i

Ci

i

Cii

i

CiC bXaXXRXR

or ZCC RRR −=−= 02 λλ Substituting the derived values for λ0, λ1 and λ2 into equation (5), we have: ( ) ( ) iZCiZMZi bRRaRRRR ×−+×−+=


Chapter 14: Problem 7 In the graph below, the efficient frontier with riskless lending but no riskless borrowing is the ray extending from RF to the tangent portfolio L and then along the minimum-variance curve through the market portfolio M and out toward infinity (assuming unlimited short sales). All investors who wish to lend will hold tangent portfolio L in some combination with the riskless asset, since no other portfolio offers a higher slope. Furthermore, unless all investors lend or invest solely in portfolio L, the market portfolio M will be along the minimum-variance curve to the right of portfolio L, since the market portfolio is a wealth-weighted average of all the efficient risky-asset portfolios held by investors, and no rational investor would hold a risky-asset portfolio along the curve to the left of L. The expected return on a zero-beta asset is the intercept of a line tangent to the market portfolio, and the zero-beta portfolio on the minimum-variance frontier must be below the global minimum variance portfolio of risky assets by the geometry of the graph. Furthermore, by the geometry of the graph, since the risk-free lending rate is the intercept of the line tangent to portfolio L, and since L is to the left of M on the minimum-variance curve, the risk-free lending rate must be below the expected return on a zero-beta asset.


The zero-beta security market line is the line in the graph below extend from the expected return on a zero-beta asset through the market portfolio and out toward infinity (assuming unlimited short sales). The expected return-beta relationships of all risky securities risky-asset portfolios (including the market portfolio M and portfolio L) are described by that line. The other line from the risk-free lending rate to portfolio L only describes the expected return-beta relationships of combination portfolios of the risk-free asset and portfolio L; those combination portfolios are not described by the zero-beta security market line.

Chapter 14: Problem 8 Assume the same situation as in Problem 5. The investor who believes in the standard (pre-tax) CAPM expects a return of 14% on either security. You expect a return before taxes of 15% on stock A and 13% on stock B. If your tax factor was below the aggregate tax factor (τ lower than 0.25) then you should buy stock B from the other investor and sell that investor stock A. The fact that this will lead to higher after-tax cash flows for you is straightforward.


Chapter 14: Problem 9 This problem can be answered directly by using the equation developed for non-marketable assets. The equation also holds for deleted assets, with the subscript H now standing for those assets that were left out:

( )

( ) ( )⎟⎟⎠

⎞⎜⎜⎝

⎛+×

+

−+= Hi

m

Hmi

Hmm

Hm

FmFi RR

PPRR

RRPP

RRRR cov covcov 2σ

The effect of leaving out bonds depends on two factors: 1.) Whether or not the returns on the aggregate of all bonds are negatively or positively correlated with the returns on the aggregate of all stocks; 2.) The correlation between the returns on a particular stock and the returns on the aggregate of all bonds. From the above equation, if returns on stocks and bonds are generally positively correlated (as empirical evidence shows), then the denominator in the second term of the equation will tend to lower the expected return on any stock. If the return on a particular stock is negatively correlated with bonds, that will further lower the stock’s expected return. However, if the stock is positively correlated with bonds, this will offset the effect of positive correlation between all stocks and bonds and may actually result in a higher expected return for the stock.




Chapter 15: Problem 1 That is NOT a valid test of the theory, and the empirical evidence IS consistent with the theory. If high-beta stocks always gave higher returns, then they would be less risky than low-beta stocks. It is precisely because the returns on high-beta stocks are more risky, and hence sometimes below and sometimes above the returns on low-beta stocks, that high-beta stocks have higher expected (and over long periods of time higher actual) returns. Chapter 15: Problem 2 Let:

AiR = the expected percentage change in alcoholism in city i; GR = the expected percentage change in the price of gold; PR = the expected percentage change in professors’ salaries.

Then we have:

( ) ( )( )P

PAiGPGAi

RRRRRRR

var cov

×−+=

The above equation is exactly parallel to the zero-beta CAPM equation, with expected percentage change in alcoholism in a city playing the role of the expected return on a security. The analogy between variables is seen from:

( ) ( )( )m

miZmZi

RRRRRRR

var cov

×−+=

Therefore, tests exactly parallel to those employed in the text can be used.


Chapter 15: Problem 3 Since the equality shown in equation (15.7) in the text only holds for an efficient portfolio, if the market portfolio is inefficient the equality will not hold and instead we have: Fkkm RR −≠λσ The remaining proof follows the proof shown in the text below equation (15.7), but with the not-equal sign replacing the equal sign in all the remaining equations in the proof. Chapter 15: Problem 4 One way to use general equilibrium theory to evaluate a stock portfolio manager’s performance would be to estimate the equilibrium security market line using historical time series of returns over a period of time along with the portfolio’s average return and beta. If, given the portfolio’s beta, the portfolio had an average return above the equilibrium return predicted by the estimated security market line, it would indicate superior performance. (This performance measure is known as “Jensen’s alpha” and is discussed at length in Chapter 24.) Chapter 15: Problem 5 If the post-tax form of the CAPM holds, then the real relationship as a cross-sectional regression model is: ( ) iFiiFi RRR εδγβγγ +−++=− 210 If the standard CAPM security market line is tested, the cross-sectional regression model is: iiFi RR εβγγ ++=− 10 If δ was uncorrelated with β across securities, then the regression estimates of γ0 and γ1 in the standard model would be unaffected. However, empirical evidence shows that δ and β are negatively correlated across securities (high-dividend securities tend to have low betas and low-dividend securities tend to have high betas) and that δ is positively correlated with R across securities, so this is a classic case of missing-variable bias. The effect of the bias is to raise the estimate of the intercept (γ0) and lower the estimate of the slope (γ1).

Elton, Gruber, Brown and Goetzmann 1 Modern Portfolio Theory and Investment Analysis, 6th Edition Solutions To Text Problems: Chapter 16

Elton, Gruber, Brown and Goetzmann Modern Portfolio Theory and Investment Analysis, 6th Edition

Solutions to Text Problems: Chapter 16 Chapter 16: Problem 1 From the text we know that three points determine a plane. The APT equation for a plane is: 22110 iii bbR Assuming that the three portfolios given in the problem are in equilibrium (on the plane), then their expected returns are determined by: 210 5.012 (a) 210 2.034.13 (b) 210 5.0312 (c) The above set of linear equations can be solved simultaneously for the three unknown values of 0, 1 and 2. There are many ways to solve a set of simultaneous linear equations. One method is shown below. Subtract equation (a) from equation (b): 21 3.024.1 (d) Subtract equation (a) from equation (c): 2120 (e) Subtract equation (e) from equation (d): 27.04.1 or 22 Substitute 22 into equation (d): 6.024.1 1 or 11 Substitute 11 and 22 into equation (a): 1112 0 or 100 Thus, the equation of the equilibrium APT plane is: 21 210 iii bbR


Chapter 16: Problem 2 According to the equilibrium APT plane derived in Problem 1, any security with b1 = 2 and b2 = 0 should have an equilibrium expected return of 12%: %1202210210 21 iii bbR Assuming the derived equilibrium APT plane holds, since portfolio D has bD1 = 2 and bD2 = 0 with an expected return of 10%, the portfolio is not in equilibrium and an arbitrage opportunity exists. The first step is to use portfolios in equilibrium to create a replicating equilibrium investment portfolio, call it portfolio E, that has the same factor loadings (risk) as portfolio D. Using the equilibrium portfolios A, B and C in Problem 1 and recalling that an investment portfolio’s weights sum to 1 and that a portfolio’s factor loadings are weighted averages of the individual factor loadings we have:

21331 1 11111 DBABACBABBAAE bXXXXbXXbXbXb

015.02.05.0 1 22222 DBABACBABBAAE bXXXXbXXbXbXb Simplifying the above two equations, we have:

12 AX or 21

AX

217.0 BA XX

Since 21

AX , 0BX and 211 BAC XXX .

Since portfolio E was constructed from equilibrium portfolios, portfolio E is also on the equilibrium plane. We have seen above that any security with portfolio E’s factor loadings has an equilibrium expected return of 12%, and that is the expected return of portfolio E:

%1212214.13012

21

CCBBAAE RXRXRXR

So now we have two portfolios with exactly the same risk: the target portfolio D and the equilibrium replicating portfolio E. Since they have the same risk (factor loadings), we can create an arbitrage portfolio, combining the two portfolios by going long in one and shorting the other. This will create a self-financing (zero net investment) portfolio with zero risk: an arbitrage portfolio.


In equilibrium, an arbitrage portfolio has an expected return of zero, but since portfolio D is not in equilibrium, neither is the arbitrage portfolio containing D and E, and an arbitrage profit may be made. We need to short sell either portfolio D or E and go long in the other. The question is: which portfolio do we short and which do we go long in? Since both portfolios have the same risk and since portfolio E has a higher expected return than portfolio D, we want to go long in E and short D; in other words, we want 1ARB

EX and 1ARBDX . This gives us:

011 ARB

DARBE

i

ARBi XXX (zero net investment)

But since portfolio E consists of a weighted average of portfolios A, B and C,

1ARBEX is the same thing as

21

ARBAX , 0ARB

BX and 21

ARBCX , so we have:

01210

21

ARBD

ARBC

ARBB

ARBA

i

ARBi XXXXX (zero net investment)

0

21321301

21

1111

11

DARBDC

ARBCB

ARBBA

ARBA

ii

ARBiARB

bXbXbXbX

bXb

(zero factor 1 risk)

0

015.0212.005.0

21

2222

22

DARBDC

ARBCB

ARBBA

ARBA

ii

ARBiARB

bXbXbXbX

bXb


%2

10112214.13012

21

D

ARBDC

ARBCB

ARBBA

ARBA

ii

ARBiARB

RXRXRXRX

RXR

(positive arbitrage return)

As arbitrageurs exploit the opportunity by short selling portfolio D, the price of portfolio D will drop, thereby pushing portfolio D’s expected return up until it reaches its equilibrium level of 12%, at which point the expected return on the arbitrage portfolio will equal 0. There is no reason to expect any price effects on portfolios A, B and C, since the arbitrage with portfolio D can be accomplished using other assets on the equilibrium APT plane.


Chapter 16: Problem 3 From the text we know that three points determine a plane. The APT equation for a plane is: 22110 iii bbR Assuming that the three portfolios given in the problem are in equilibrium (on the plane), then their expected returns are determined by: 21012 (a) 210 25.113 (b) 210 35.017 (c) Solving for the three unknowns in the same way as in Problem 1, we obtain the following solution to the above set of simultaneous linear equations: 80 ; 61 ; 22 ; Thus, the equation of the equilibrium APT plane is: 21 268 iii bbR Chapter 16: Problem 4 According to the equilibrium APT plane derived in Problem 3, any security with b1 = 1 and b2 = 0 should have an equilibrium expected return of 12%: %140268268 21 iii bbR Assuming the derived equilibrium APT plane holds, since portfolio D has bD1 = 1 and bD2 = 0 with an expected return of 15%, the portfolio is not in equilibrium and an arbitrage opportunity exists. The first step is to use portfolios in equilibrium to create a replicating equilibrium investment portfolio, call it portfolio E, that has the same factor loadings (risk) as portfolio D. Using the equilibrium portfolios A, B and C in Problem 3 and recalling that an investment portfolio’s weights sum to 1 and that a portfolio’s factor loadings are weighted averages of the individual factor loadings we have:

115.05.11 1 11111 DBABACBABBAAE bXXXXbXXbXbXb

0132 1 22222 DBABACBABBAAE bXXXXbXXbXbXb


Simplifying the above two equations, we have:

21

21

BA XX

354 BA XX Solving the above two simultaneous equations we have:

31

AX , 31

BX and 311 BAC XXX .

Since portfolio E was constructed from equilibrium portfolios, portfolio E is also on the equilibrium plane. We have seen above that any security with portfolio E’s factor loadings has an equilibrium expected return of 14%, and that is the expected return of portfolio E:

%14173113

3112

31

CCBBAAE RXRXRXR

So now we have two portfolios with exactly the same risk: the target portfolio D and the equilibrium replicating portfolio E. Since they have the same risk (factor loadings), we can create an arbitrage portfolio, combining the two portfolios by going long in one and shorting the other. This will create a self-financing (zero net investment) portfolio with zero risk: an arbitrage portfolio. In equilibrium, an arbitrage portfolio has an expected return of zero, but since portfolio D is not in equilibrium, neither is the arbitrage portfolio containing D and E, and an arbitrage profit may be made. We need to short sell either portfolio D or E and go long in the other. The question is: which portfolio do we short and which do we go long in? Since both portfolios have the same risk and since portfolio D has a higher expected return than portfolio E, we want to go long in D and short E; in other words, we want 1ARB

DX and 1ARB

EX . This gives us: 011 ARB

EARBD

i

ARBi XXX (zero net investment)


But since portfolio E consists of a weighted average of portfolios A, B and C,

1ARBEX is the same thing as

31

ARBAX ,

31

ARBBX and

31

ARBCX , so we have:

0131

31

31

ARBD

ARBC

ARBB

ARBA

i

ARBi XXXXX (zero net investment)

0

115.0315.1

311

31

1111

11

D

ARBDC

ARBCB

ARBBA

ARBA

ii

ARBiARB

bXbXbXbX

bXb


0

013312

311

31

2222

22

D

ARBDC

ARBCB

ARBBA

ARBA

ii

ARBiARB

bXbXbXbX

bXb


%1

151173113

3112

31

D

ARBDC

ARBCB

ARBBA

ARBA

ii

ARBiARB

RXRXRXRX

RXR

(positive arbitrage return)

As arbitrageurs exploit the opportunity by buying portfolio D, the price of portfolio D will rise, thereby pushing portfolio D’s expected return down until it reaches its equilibrium level of 14%, at which point the expected return on the arbitrage portfolio will equal 0. There is no reason to expect any price effects on portfolios A, B and C, since the arbitrage with portfolio D can be accomplished using other assets on the equilibrium APT plane. Chapter 16: Problem 5 The general K-factor APT equation for expected return is:

K

kikki bR

10

where 0 is the return on the riskless asset, if it exists.


Given the data in the problem and in Table 16.1 in the text, along with a riskless rate of 8%, the Sharpe multi-factor model for the expected return on a stock in the construction industry is:

%034.1059.1122.012.04.056.5624.02.136.58

iR

The last number, 1.59, enters because the stock is a construction stock. Chapter 16: Problem 6 A. From the text we know that, for a 2-factor APT model to be consistent with the standard CAPM, jFmj RR . Given that 4 Fm RR and using results from Problem 1, we have: 141 or 25.01 ; 242 or 5.02 . B. From the text we know that 2211 iii bb . So we have: 5.05.05.025.01 A 85.05.02.025.03 B 5.05.05.025.03 C C. Assuming all three portfolios in Problem 1 are in equilibrium, then we can use any one of them to find the risk-free rate. For example, using portfolio A gives: AFmfA RRRR or AFmAF RRRR Given that %12AR , 5.0A and %4 Fm RR , we have: %105.0412 FR



Solutions to Text Problems: Chapter 17 Chapter 17: Problem 1 The simplest trading strategy would be to buy a stock at the opening price on the day that the “heard on the street” column indicates analysts have reported positive recommendations and to short sell it if analysts have reported negative recommendations. Since any stock price effect occurs very shortly after the news is released, the stock position could be unwound after five days. Naturally, in examining returns from this strategy, purchases and sales would have to be adjusted for transactions costs. The results of Davies and Canes suggest that round-trip transactions costs must be less than two percent and perhaps less than one percent for this rule to produce excess returns. In testing this strategy, we would have to be sure to adjust the returns for risk. Following this strategy will lead to a changing portfolio of stocks being held over time. Either the beta or standard deviation of this portfolio could be used as a risk measure. Chapter 17: Problem 2 See the section in the text entitled “Relative Strength” for the answer to this question. Chapter 17: Problem 3 There are several ways this rule could be tested. One way would be to rank all stocks by their P/E ratios, select the X percent (e.g., 20%) of the stocks with the lowest P/E ratios, then select from that group the Y percent (e.g., 20%) with the largest five-year growth rates. After then making sure that transactions costs are included, risk-adjusted excess returns for the final group could be obtained and examined using one of the methodologies outlined in the text. Chapter 17: Problem 4 If a market is semi-strong-form efficient, the efficient market hypothesis says that prices should reflect all publicly available information. If you have access to a “good” and significant piece of information that you believe is not yet public information, you could examine the residuals from a model such as the “market model” to see if there were recently positive excess returns, indicating whether or not the market had already incorporated that information.


Chapter 17: Problem 5 If a market is semi-strong-form efficient, the efficient market hypothesis says that prices should reflect all publicly available information. If publicly available information is already fully reflected in market prices, one would strongly suspect the market to be weak-form efficient as well. The only rational explanation for weak-form inefficiency is if information is incorporated into prices slowly over time, thus causing returns to be positively autocorrelated. The only exception to this might be if the market is strong-form inefficient and monopoly access to information disseminates through widening circles of investors over time. Chapter 17: Problem 6 You could test that by following any of the test methodologies outlined in the text for semi-strong-form efficiency, where day zero (the “event day”) is defined as the day at which the block of stocks becomes available for trading. Chapter 17: Problem 7 Recall that the zero-beta CAPM leads to lower expected returns for high-beta (above 1) stocks and higher expected returns for low-beta stocks than does the standard CAPM. If we were testing a phenomenon that tended to occur for low-beta stocks and not for high-beta stocks, then the zero-beta CAPM could show inefficiency while the standard CAPM showed efficiency. Chapter 17: Problem 8 The betting market at roulette is in general an efficient market. Though betting on the roulette wheel has a negative expected return, there is no way that that information can be used to change the expected return. The only exception to this might be if the roulette wheel was not perfectly balanced. Since the house does not change the odds (prices) to reflect an unbalanced roulette wheel, an unbalanced wheel would make the betting market at roulette inefficient. Chapter 17: Problem 9 As in Problem 6, you could test that by following any of the test methodologies outlined in the text for semi-strong-form efficiency, where day zero (the “event day”) could be defined either as the day of retirement or as the day the retirement is first announced to the public.



Solutions to Text Problems: Chapter 18 Chapter 18: Problem 1 Since the company’s growth rate of 10% extends into the future indefinitely, use the constant-growth model to value its stock:

( )13.15$

10.014.01.155.0101

0 =−×

=−+

=−

=gkgD

gkDP

Chapter 18: Problem 2 Using equation (18.5b) in the text, we have:

00.20$5.014.012.0

110 =

×−=

−=

rbkDP

Chapter 18: Problem 3 Solving equation (18.5b) in the text for k (the required rate of return) we have:

( )%3.10 103.05.014.0301

0

1 =×+=+= rbPDk

Chapter 18: Problem 4 Solving equation (18.5b) in the text for r (the rate of return on new investment) we have:

( )%7.20 207.05.0

160112.01

0

1 =×⎟⎠

⎞⎜⎝

⎛ −=×⎟⎟⎠

⎞⎜⎜⎝

⎛−=

bPDkr

So the rate of return on new investment would have to change from 14% to 20.7%, an increase of 6.7 percentage points.


Chapter 18: Problem 5 This problem can be solved using the two-period growth model shown in the text, where the first growth period is 5 years with a growth rate of 10% (g1) followed by a growth rate of 6% (g2) indefinitely:

( )( ) ( )

( )( ) ( )

( )

( )( )5

2

6

1

51

10

52

6

1

51

1

52

65

1

111

55

5

1

111

0

1111

1

1111

111

111

k

gkD

gkkg

gD

k

gkD

gkkg

D

k

gkD

kgD

kP

kgDP

tt

t

tt

t

+

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟⎠

⎞⎜⎜⎝

⎛++

−+=

+

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟⎠

⎞⎜⎜⎝

⎛++

−=

+

⎟⎟⎠

⎞⎜⎜⎝

⎛−

++

+=

++

+

+=

∑

∑

=

−

=

−

Recognizing that the dividend at the end of period 6 is equal to the dividend at the end of period 5 compounded 1 period at g2 and then adjusted by a factor of 0.5/0.3 to reflect the increased dividend payout rate, we have:

( )

( ) ( )

( )565.1$

3.05.006.11.155.0

3.05.011

3.05.01

5

25

10

256

=

×××=

×++=

×+=

ggD

gDD


So we have:

( )

64.12$163.10474.2

925.1563.19

04.016355.0605.0

14.106.014.0

565.1

10.014.014.11.11

1.155.0 5

5

0

=+=

+×=

⎟⎠

⎞⎜⎝

⎛−

+

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎠

⎞⎜⎝

⎛−××=P

Chapter 18: Problem 6 This problem can be solved using the three-period growth model shown in the text, where the first growth period is 5 years with a growth rate of 10% (g1) followed each year by linearly declining growth rates (g2, g3, g4 and g5) over a second period of 4 years down to a 6% steady-state growth rate (gs) indefinitely thereafter. Since the growth rate is declining linearly over the 4-year period, the annual

decline is 8.05

610=

− percentage points per year. So we have g1 = 10% (first 5

years), g2 = 9.2% (year 6), g3 = 8.4% (year 7), g4 = 7.6% (year 8), g5 = 6.8% (year 9) and gS = 6% (year 10 and thereafter), and the model is:

( ) ( )

( )

( ) ( )

( )( ) ( )

( ) ( )9

109

6

4

2

510

1

51

10

9

109

6

4

25

1

51

1

99

9

61

51

10

11

11111

1

11

1111

11111

k

gkD

k

ggD

gkkg

gD

k

gkD

k

gD

gkkg

D

kP

kD

gkkg

DP

S

tt

t

jj

S

tt

t

jj

tt

t

+

⎟⎟⎠

⎞⎜⎜⎝

⎛−

++

++

+

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟⎠

⎞⎜⎜⎝

⎛++

−+=

+

⎟⎟⎠

⎞⎜⎜⎝

⎛−

++

+

+

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟⎠

⎞⎜⎜⎝

⎛++

−=

++

++

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟⎠

⎞⎜⎜⎝

⎛++

−=

∑∏

∑∏

∑

=

−

=

=

−

=

=


Recognizing that the dividend at the end of period 10 is equal to the dividend at the end of period 9 compounded 1 period at gS and then adjusted by a factor of 0.5/0.3 to reflect the increased dividend payout rate, we have:

( )

( ) ( ) ( )

( )129.2$

3.05.006.1068.1076.1084.1092.11.155.0

3.05.0111

3.05.01

5

5

2

510

910

=

×××××××=

×+×+×+=

×+=

∏=

Sj

j

S

gggD

gDD

So we have:

( )( ) ( )

( ) ( )

( )( )

( )( )

( )( )

( )( )

( )

29.12$184.8371.0396.0419.0441.0474.2

14.106.014.0

129.214.1

068.1076.1084.1092.11.155.014.1

076.1084.1092.11.155.0

14.1084.1092.11.155.0

14.1092.11.155.0

10.014.014.11.11

1.155.0

11

11111

1

9

9

5

8

5

7

5

6

5

5

96

109

6

4

2

510

1

51

100

=+++++=

⎟⎠

⎞⎜⎝

⎛−

+

×××××+

××××+

×××+

××+

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎠

⎞⎜⎝

⎛−××=

+

⎟⎟⎠

⎞⎜⎜⎝

⎛−

++

++

+

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟⎠

⎞⎜⎜⎝

⎛++

−+= ∑

∏=

−

=

k

gkD

k

ggD

gkkg

gDPt

t

t

jj

Chapter 18: Problem 7 Solving equation (18.5b) in the text for k (the expected rate of return) we have:

( )%1.18 181.05.014.091

0

1 =×+=+= rbPDk


Chapter 18: Problem 8 Since the company’s growth rate of 10% extends into the future indefinitely, use the equation (18.6) in the text from the constant-growth model:

( )

( )%7.16 167.0

1.09

1.155.0

1

0

0

0

1

=

+×

=

++

=+= gP

gDgPDk

Chapter 18: Problem 9 This problem can be solved using the two-period growth model shown in the text, where the first growth period is 10 years with a growth rate of rb = g1 followed by a growth rate of 5% (g2) indefinitely. The model is:

( )10

2

11

1

101

101

111

k

gkD

gkkg

DP+

⎟⎟⎠

⎞⎜⎜⎝

⎛−

+

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟⎠

⎞⎜⎜⎝

⎛++

−×=

Given r = 0.14 and b = 0.5, g1 = 0.14 × 0.5 = 0.07 (7%). Also,

( )( ) ( )( )93.1$

05.107.11

11

1

92

911

21011

=××=

++=

+=

ggD

gDD

So we have:

( )

21.16$88.833.7

12.105.012.0

93.1

07.012.012.107.11

1 10

10

0

=+=

⎟⎠

⎞⎜⎝

⎛−

+

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎠

⎞⎜⎝

⎛−×=P


Chapter 18: Problem 10 This problem can be solved iteratively by substituting various values for k into the first formula shown in the answer for Problem 9. By trial and error the solution is k = 9.6%. Chapter 18: Problem 11 As with Problem 10, this problem can be solved iteratively by substituting various values for the length of the first growth period into the first formula shown in the answer for Problem 9. By trial and error the solution is 24 years. Chapter 18: Problem 12 The solution to this problem is a general form of the model shown in the answer to Problem 6:

( )( ) ( )

( )( ) ( )

( )( ) ( ) 21

21

21

1

21

1

21

1

1

11

111

111

1

11111

1

1

1

1 2

1110

1

1

10

2121

11

1

100

NNS

NNNN

Ntt

N

j

SNN

NNNN

NN

Ntt

t

N

k

gkD

k

NggjggD

gkkg

gD

kP

kD

gkkg

gDP

+

++

+

+=

=

++

+

+=

+

⎟⎟⎠

⎞⎜⎜⎝

⎛

−+

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−×−++

+

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟⎠

⎞⎜⎜⎝

⎛++

−×+=

++⎟

⎟⎠

⎞⎜⎜⎝

⎛

++

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟⎠

⎞⎜⎜⎝

⎛++

−×+=

∑∏

∑

where D0 = the just-paid dividend g1 = the annual growth rate during the first period of years N1 = the number of years in the first growth period N2 = the number of years in the second growth period of linearly changing growth rates gS = the annual steady-state growth rate after the second period of linearly changing growth rates Note that the step value for linearly changing rates from g1 to gS is (g1 − gS) / (N2 + 1), not (g1 − gS) / N2.



Solutions to Text Problems: Chapter 19 Chapter 19: Problem 1 If earnings follow a mean-reverting process, then it is appropriate to use historical data to forecast future earnings. There are many appropriate techniques. Three specific ones are presented below. A. If earnings follow a mean-reverting process with no trend or cycle, the following exponential smoothing model could be used to forecast future earnings: ( )11

ˆˆ−− −+= tttt EEaEE

where

tE = the time-t forecast for earnings at time t + 1; Et = the actual earnings at time t; a = a constant less than 1.0. B. If earnings follow a mean-reverting process with a trend but no cycle, either smoothed earnings plus the trend or smoothed earnings times the trend could be used, depending on whether the trend was additive or multiplicative. For example, with an additive trend the forecast would be: tt gE ˆˆ + where

( ) ( )[ ]1111 ˆˆˆˆˆ−−−− +−++= tttttt gEEagEE ;

tg is the time-t estimate of the trend.

See footnote 7 in the text for further details on this technique. C. If earnings follow a mean-reverting process with a trend and a cycle, then the forecast is smoothed earnings adjusted for the trend and the cycle. For example, with an additive trend and a multiplicative cycle the forecast would be: ( ) ttt fgE ˆˆˆ ×+

where tf is the time-t estimate of the cycle.


Chapter 19: Problem 2 If there was a strong relationship between a firm’s earnings and the overall industry’s and economy’s earnings, then, for example, a linear model could be estimated:

EIi cEbEaE ++= where Ei = the firm i’s earnings; EI = the industry’s earnings; EE = the economy’s earnings. Such an equation switches the forecasting task from forecasting Ei directly to forecasting it indirectly by first forecasting EI and EE and estimating the parameters a, b and c. Chapter 19: Problem 3 YES. Mean reversion could be present in the industry’s and economy’s earnings, too. Chapter 19: Problem 4 YES. The economy could also exhibit independence in earnings changes. Chapter 19: Problem 5 If earnings expectations are important in determining share prices, then a valuable analyst is one who can forecast changes in investors’ expectations. If forecasts in general become more accurate over time, a valuable analyst is one who at any point in time can forecast more accurately than the average analyst can.



Solutions to Text Problems: Chapter 21 Chapter 21: Problem 1 We can use the cash flows bonds A and B to replicate the cash flows of bond C. Let YA be the fraction of bond A purchased and YB be the fraction of bond B purchased. (Note that these are not investment weights that sum to 1.) Then we have: t = 1: $100 YA + $80 YB = $90 t = 2: $1,100 YA + $1,080 YB = $1,090 Solving the above two equations simultaneously gives YA= YB = 1/2. So buying 1/2 of bond A and 1/2 of bond B gives the same cash flows as buying 1 bond C (or, equivalently, buying 1 bond A and 1 bond B gives the same cash flows as buying 2 of bond C). Therefore, if the Law of One Price held, the bonds’ current prices would be related as follows: 1/2 PA + 1/2 PB = PC But, since we are given that PA = $970, PB = $936 and PC = $980, we have instead: 1/2 × $970 + 1/2 × $936 = $953 < $980 The Law of One Price does not hold. Given that the future cash flows of the portfolio of bonds A and B are identical in timing and amount to those of bond C, and assuming that all three bonds are in the same risk class, an investor should purchase 1 bond A and 1 bond B rather than 2 of bond C. Chapter 21: Problem 2 A. A bond’s current yield is simply its annual interest payment divided by its current price, so we have: Current Yield = $100 ÷ $960 = 0.1042 (10.42%)


B. A bond’s yield to maturity is the discount rate that makes the sum of the present values of the bond’s future cash flows equal to the bond’s current price. Since this bond has annual cash flows, we need to find the rate, y, that solves the following equation:

( ) ( )5

5

1 11000$

1100$960$

yytt +

+⎟⎟⎠

⎞⎜⎜⎝

⎛

+= ∑

=

We can find y iteratively by trial and error, but the easiest way is to use a financial calculator and input the following: PV = − 960 PMT = 100 FV = 1000 N = 5 After entering the above data, compute I to get I = y = 11.08%. Chapter 21: Problem 3 In general, the nominally annualized spot rate for period t (S0t) is the yield to maturity for a t-period zero-coupon (pure discount) instrument:

ttS

FP⎟⎠

⎞⎜⎝

⎛ +

=

21 0

0

where P0 is the zero’s current market price, F is the zero’s face (par) value, and t is the number of semi-annual periods left until the zero matures. The zero-coupon bonds in this problem all have face values equal to $1,000.


If semi-annual periods are assumed, then bond A is a one-period zero, bond B is a two-period (one-year) zero, bond C is a three-period zero, and bond D is a four-period (two-year) zero. So we have:

(7.99%) 0799.0

21

1000=855

(8.31%) 0831.0

21

1000=885

(8.51%) 0851.0

21

1000=920

(8.33%) 0833.0

21

1000960

04404

03303

02202

01101

=⇒

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=⇒

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=⇒

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=⇒

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

SS

SS

SS

SS

The nominally annualized implied forward rates (ft,t+j) can be obtained from the above spot rates. A general expression for the relationship between current spot rates and implied forward rates is:

21

21

21

1

,0

,0

, ×

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

++

+

j

tt

jtjt

jttS

S

f

where t is the semi-annual period at the end of which the forward rate begins, j is the number of semi-annual periods spanned by the forward rate, and both t and j are integers greater than 0.


We can an obtain a set of one-period forward rates by setting j equal to 1 and varying t from 1 to 3 in the preceding equation:

( )( )

( )( )

( )( )

(7.04%) 0704.0210416.10400.121

21

21

(7.92%) 0792.0210426.10416.121

21

21

(8.70%) 0870.0210417.10426.121

21

21

3

4

303

404

34

2

3

202

303

23

1

2

101

202

12

=×⎟⎟⎠

⎞⎜⎜⎝

⎛−=×

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

=×⎟⎟⎠

⎞⎜⎜⎝

⎛−=×

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

=×⎟⎟⎠

⎞⎜⎜⎝

⎛−=×

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=

S

S

f

S

S

f

S

S

f

If instead we wanted the expected spot yield curve one period from now under the pure expectations theory, we can set t equal to 1 and vary j from 1 to 3 in the preceding equation:

( )( )

( )( )

( )( )

(7.89%) 0789.0210417.10400.121

21

21

(8.31%) 0831.0210417.1

1.041621

21

21

(8.70%) 0870.0210417.10426.121

21

21

31

1

4

31

101

404

1414

21

1

3

21

101

303

1313

1

2

101

202

1212

=×⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−⎟⎟⎠

⎞⎜⎜⎝

⎛=×

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

==

=×⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−⎟⎟⎠

⎞⎜⎜⎝

⎛=×

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

==

=×⎟⎟⎠

⎞⎜⎜⎝

⎛−=×

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

−

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+

⎟⎟⎠

⎞⎜⎜⎝

⎛+

==

S

S

fS

S

S

fS

S

S

fS


Chapter 21: Problem 4 We can use the cash flows bonds A and B to replicate the cash flows of bond C. Let YA be the fraction of bond A purchased and YB be the fraction of bond B purchased. Then we have: t = 1: $80 YA + $1,100 YB = $120 t = 2: $1,080 YA + $0 YB = $1,120 Solving the above two equations simultaneously gives:

297308

2728

080,1120,1

===AY

29710

700,29000,1

700,2927

27000,1

27700,29

27240,2

27240,3

100,1272880120

==×=⎟⎠

⎞⎜⎝

⎛ −=

⎟⎠

⎞⎜⎝

⎛ ×−=BY

So buying 308/297 of bond A and 10/297 of bond B gives the same cash flows as buying 1 bond C (or, equivalently, buying 308 of bond A and 10 of bond B gives the same cash flows as buying 297 of bond C). Therefore, if the Law of One Price held, the bonds’ current prices would be related as follows: 308/297 PA + 10/297 PB = PC But, since we are given that PA = $982, PB = $880 and PC = $1,010, we have instead: 308/297 × $982 + 10/297 × $880 = $1,048 > $1,010 The Law of One Price does not hold. For the Law of One Price to hold, bond C would have to sell for $1,048.


Chapter 21: Problem 5 If the Law of One Price holds, then the same discount rate (which is a spot rate) applies for the cash flows in a particular period for all three bonds. Also, in the presence of taxes, the price of each bond must equal the sum of the present values of its future after-tax cash flows, where the present values are calculated using the spot rates. Each bond’s capital gain or loss is simply its principal (par) value minus its price. Given that each bond has a par value of $1,000, bond A has a capital gain of $1,000 − $985 = $15, bond B has a capital gain of $1,000 − $900 = $100, and bond C has a capital loss of $1,000 − $1,040 = − $40. Given that the periods shown are annual, that taxes must be paid on capital gains and can be deducted on capital losses, and that the capital gain or loss tax rate is one-half of the ordinary income tax rate, we need to find the discount factors and ordinary income tax rate that makes the following set of equations hold simultaneously:

( ) ( ) 985$000,1$2

15$180$180$ 4442 =×+××−×−×+×−× ddTdTdT

( ) 900$000,1$2

1001100$ 222 =×+××−×−× ddTdT

( ) ( ) 040,1$000,1$2

40$1120$1120$ 4442 =×+××+×−×+×−× ddTdTdT

where T = the ordinary income tax rate;

202

2

21

1

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=S

d = the two-semi-annual-period (one-year) discount factor;

404

4

21

1

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=S

d = the four-semi-annual-period (two-year) discount factor.

The solution to the above set of simultaneous equations is: T = 0.3303; d2 = 0.8568; d4 = 0.8934



Solutions to Text Problems: Chapter 22 Chapter 22: Problem 1 The duration formula shown in the text for annual payments can easily be modified to reflect semi-annual payments as follows:

0

1

22

1

P

i

tCF

D

T

tt

t

×

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎠

⎞⎜⎝

⎛ +

×

=

∑=

where T is the number of semi-annual periods remaining to maturity. Given P0 = $1,000, semi-annual interest payments of $50, a principal of $1,000 paid at the end of 5 years and a flat yield curve at 10%, we have:

05.421.8

10002210.01

101000

210.01

5010

10

1

==×

⎟⎠

⎞⎜⎝

⎛ +

×+

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎠

⎞⎜⎝

⎛ +

×

=

∑=t

tt

D years.

Chapter 22: Problem 2 The duration formula for annual payments annual payments is:

( )

0

1 1P

itCF

D

T

tt

t∑=

⎟⎟⎠

⎞⎜⎜⎝

⎛

+

×

=

where T is the number of years remaining to maturity.


Given P0 = $1,000, annual interest payments of $100, a principal of $1,000 paid at maturity and a flat yield curve at 10%, we have:

( ) ( )

100010.01

100010.01

1001∑= +

×+⎟

⎟⎠

⎞⎜⎜⎝

⎛

+

×

=

T

tTt

Tt

D

where T has values of 10, 8, 5 and 3 years. Using the above equation, we have:

T D

10

6.76

8 5.87 5 4.17 3 2.74

Chapter 22: Problem 3 Let XA be the portfolio’s investment weight for bond A, XB be the portfolio’s investment weight for bond B, and, since an investment portfolio’s weights sum to 1, XC = (1 − XA − XB) be the portfolio’s investment weight for bond C. Given the individual bonds’ durations, the duration of a portfolio of those bonds is: DP = 5XA + 10XB + 12(1 − XA − XB) Setting the portfolio’s duration equal to the target duration of 9, we have: 5XA + 10XB + 12(1 − XA − XB) = 9 Since there is just one equation with two unknowns, there are an infinite number of solutions (portfolios) that will satisfy the equation. Either XA or XB can be arbitrarily set and then the remaining weights solved for. Three of the infinite number of solutions are: 1.) XA = 22/56; XB = 7/56; XC = 27/56 2.) XA = 4/14; XB = 7/14; XC = 3/14 3.) XA = 10/28; XB = 7/28; XC = 11/28


Chapter 22: Problem 4 Since in this problem there are three bonds with three sets of cash flows to meet the three liabilities, we have three equations with three unknowns and therefore one unique solution. In a more realistic situation, there would be many more bonds than the number of liabilities (many more unknowns than the number of equations) and thus there would be an infinite number of solutions. In that case, the linear programming procedure shown in the text’s Appendix B would be required to find the least-cost solution. Let YA be the fraction of A bonds to buy, YB be the fraction of B bonds to buy, and YC be the fraction of C bonds to buy. (Note that these are not investment weights that sum to 1.) We want to form a portfolio of these three bonds that replicates the timing and amounts of the liabilities. At t = 1: $50 YA + $100 YB + $1,000 YC = $250 At t = 2: $1,050 YA + $100 YB + $0 YC = $500 At t = 3: $0 YA + $1,100 YB + $0 YC = $550 The solution to the above set of simultaneous linear equations is: YA = 9/21; YB = 1/2; YC = 15/84 Assuming fractional purchases may be made, the cost of the bond portfolio is then: YA PA + YB PB + YC PC = 9/21 × $950 + 1/2 × $1,000 + 15/84 × $920 = $1,071.43 Chapter 22: Problem 5 Equation (21.6) in the text is a form of a single-index model for bonds:

( ) immm

iii eRR

DDRR +−−=

If the yield curve is flat at 10%, then the first period’s expected return is 10% for each of the three bonds. Since the market portfolio is a weighted average of the three bonds, the market portfolio also has an expected return of 10%. The duration of the market portfolio is a weighted average of the three bonds’ durations. Since the three bonds are assumed to be of equal value, the value-weighted market portfolio is also an equally weighted portfolio. Therefore, the duration of the market portfolio is: Dm = 1/3 × 5 + 1/3 × 10 +1/3 × 12 = 9 years


Therefore we have:

( ) imA eRR +−×−= %1095%10

( ) imB eRR +−×−= %109

10%10

( ) imC eRR +−×−= %109

12%10

We have seen in an earlier chapter that, under the assumptions of the Sharpe single-index model, the covariance between the returns on any pair of securities i and j is: 2

mjiijj σββσ = Making the same assumptions as those for the Sharpe single-index model and

recognizing that m

i

DD

− in the bond single-index model (equation (21.6)) is

analogous to βi in the Sharpe single-index model, the covariance between the returns on any pair of bonds i and j is:

2m

m

j

m

iijj D

DDD

σσ ××=

Therefore we have:

22

910

95

mmm

B

m

AAB D

DDD

σσσ ××=××=

22

912

95

mmm

C

m

AAC D

DDD

σσσ ××=××=

22

912

910

mmm

C

m

BBC D

DDD

σσσ ××=××=



Solutions to Text Problems: Chapter 23 Chapter 23: Problem 1 Although selling calls today would generate a positive cash flow for the client now, the client would lose the potential profit he would make if the stock were to appreciate in value, because the stock would be called away from the client. Thus, there is a potential opportunity cost of engaging in that strategy. Chapter 23: Problem 2 The profit diagram of buying the two puts appears as follows:


The profit diagram of buying the call appears as follows:

Combining the two profit diagrams we have:

The thicker line in the above diagram represents the profit from the combination. If the options finish at the money, where the stock price at their expiration equals their strike prices of $50, the profit would be − $17 (a $17 loss). For the combination to have a positive profit, the stock must either be below $41.50 or above $67 on the day the options expire. Algebraically, letting the stock price on the expiration date = P, the profit is:

50$≤P : Profit = − $17 + 2 × ($50 − P) = $83 − 2P (since only the two put options would be exercised) $50 < P: Profit = − $17 + P − $50 = P − $67 (since only the call option would be exercised)


Chapter 23: Problem 3 The profit diagram of writing the two $45 calls appears as follows:

The profit diagram of buying the $40 call appears as follows:


Combining the two profit diagrams we have:

The thicker line in the above diagram represents the profit from the combination. If the $40 call option finishes out of or at the money, where the stock price at its expiration is below or equal to that option’s strike price of $40, the profit would be $2, because none of the options would be exercised and therefore the profit is simply the net profit from buying the $40 call option (− $8) and selling the two $45 call options ($10). If the two $45 call options finish at the money, where the stock price on their expiration equals their strike price of $45, the profit would be $7, equal to the net profit of $2 from buying and writing the options plus the $5 gain from exercising the $40 call option. $7 is the maximum profit because, at stock prices higher than $45, although exercising the $40 call option continues to contribute a gain, the two $45 call options that were sold will be exercised against the seller and therefore contribute twice the loss, so the profit declines, reaching zero at a stock price of $52. (At a stock price of $52, the profit from the $40 call will be $12 − $8 = $4 and the profit from the two $45 calls will be −$14 + $10 = − $4, giving a total profit of 0.) If the stock price is greater than $52 on the expiration date, the profit will be negative (a loss). Algebraically, letting the stock price on the expiration date = P, the profit is:

40$≤P : Profit = $2 (since no options would be exercised)

45$40$ << P : Profit = $2 + P − $40 (since only the $40 call option would be exercised)

P≤45$ : Profit = $2 + P − $40 − 2 × (P − $45) = $52 − P (since all options would be exercised)


Chapter 23: Problem 4 From the text, we know that a is the lowest number of upward moves in the stock price at which the call takes on a positive value at expiration (finishes in the money), u is the size of each up movement, d is the size of each down movement and n is the number of periods remaining to the option’s expiration. Given that the option’s exercise price (E) is $60 and the current stock price (S0) is $50, we need to solve for the minimum integer a such that: ( ) EduS aa >×× −1

0 So we have: ( ) 60$9.02.150$ 1 >×× −aa and the solution is a = 5. To value the call option, we use the binomial formula: [ ] [ ]PnaBErPnaBSC n ,,,,0

−−′= where

67.09.02.19.01.1=

−−

=−−

=dudrP

73.067.01.12.1

=×=×=′ PruP

[ ] [ ] 972.073.0,10,5,, ==′ BPnaB [ ] [ ] 926.067.0,10,5,, == BPnaB So we have: ( ) 18.27$926.01.160$972.050$ 10 =××−×= −C


Chapter 23: Problem 5 The Black-Scholes option-pricing formula for valuing a call option is:

( ) ( )210 dNeEdNSC rt−=

We are given: S0 = $95; E = $105; t = 2/3 years (8 months); σ = 0.60; r = 0.08 (8%) Solving for d1 and d2 we have:

149.0490.0073.0

3260.0

3236.0

2108.0

10595ln

21ln 20

1 ==

×

×⎟⎠

⎞⎜⎝

⎛ ×++⎟⎠

⎞⎜⎝

⎛

=×⎟⎠

⎞⎜⎝

⎛ ++⎟⎟⎠

⎞⎜⎜⎝

⎛

=t

trES

dσ

σ

341.0490.0167.0

3260.0

3236.0

2108.0

10595ln

21ln 20

2 −=−

=

×

×⎟⎠

⎞⎜⎝

⎛ ×−+⎟⎠

⎞⎜⎝

⎛

=×⎟⎠

⎞⎜⎝

⎛ −+⎟⎟⎠

⎞⎜⎜⎝

⎛

=t

trES

dσ

σ

From the normal distribution we have: ( ) ( ) 560.0149.01 == NdN ( ) ( ) 367.0341.02 =−= NdN So the value of the call option is:

67.16$367.0105$560.095$3208.0

=×−×=×

eC



Solutions to Text Problems: Chapter 24 Chapter 24: Problem 1 The no-arbitrage condition for stock-index futures appears in the text as:

( ) ( )DPVPR

F−=

+1

Given that F = $200, P = $190, R = 6%, and PV(D) = $4, we have:

186$4$190$68.188$06.1

200$=−>=

so the futures are overpriced relative to the underlying index. Therefore, the arbitrage would involve selling the futures, borrowing the present value of the futures price and the present value of the dividends at 6% for six months, using some of the borrowed funds to buy the index today (t = 0), and keeping the remainder as arbitrage profit. Six months from now (t = 1), receive the futures price for the index, receive the future value of the dividends, and use the proceeds to pay off the loan. The cash flows are as follows: t = 0 t = 1 sell futures 0 $200 borrow $200/(1.06) $188.68 + $4 −$192.68(1.06) plus $4 at 6% for = $192.68 = −$204.24 six months buy index for −$190 0 delivery against futures contract in six months receive six months 0 $4(1.06)=$4.24 of dividends and invest them at 6% ________ _______________ total cash flow $2.68 0 So the arbitrage profit is $2.68 per futures contract. If the present value (at t = 0) of transactions costs is $2.68 or greater then the arbitrage opportunity is negated.


Chapter 24: Problem 2 Yes. Farmers need to be assured that they can sell their crops at harvest time, regardless of market conditions, so that they can make planting and farm equipment decisions in advance of the harvest. Even if both farmers and General Mills believe that the spot price at the expiration of the futures contract will be higher than the futures contract price (so that the farmers would get more money selling their crops later on at the spot price than by selling futures), futures contracts make sense economically to the farmers, since selling futures now eliminates the uncertainty of selling their crops later. General Mills needs to ensure a steady supply of wheat for their products regardless of market conditions, and knowing the price of wheat in advance helps in making pricing and working capital decisions. So a futures contract makes economic sense from their point of view as well, even if they share the same distributional assumptions as the farmers that spot wheat prices will be lower at the expiration of the futures contract than the futures contract price, since buying futures now eliminates the uncertainty of the cost and availability of wheat later. Chapter 24: Problem3 One equation for interest rate parity that appears in the text is:

( ) 11 −+= FD RSFR

where RD is the domestic interest rate, RF is the foreign interest rate, F is the domestic futures price for one unit of foreign currency, and S is the spot exchange rate expressed as domestic currency per unit foreign currency; i.e., both F and S are expressed in direct terms. From a U.S. viewpoint, the quotes given in the problem are in indirect terms, so, if RD is the U.S. rate and RF is the rate for Japan, then, from the problem, F = 1/115 and S = 1/120. So solving the above equation for the U.S. rate gives:

( )

( )( )%52.8 0852.0

104.1115120

104.1

1201

1151

=

−×=

−×=DR


Chapter 24: Problem 4 Assume you match the durations (interest rate sensitivities) of long-term and short-term bond futures by holding them long or short in the necessary proportion. Assuming a normal yield curve, you believe that long-term rates will fall relative to short-term rates. If the market does not share your belief today, and if long-term rates fall and short-term rates rise, then the prices of long-term bonds and long-term bond futures will rise and the prices of short-term bonds and short-term bond futures will fall. Therefore, you want to be long in long-term bond futures and short in short-term bond futures. If instead the entire yield curve shifted up, short-term rates would have to rise more than long-term rates for the spread to narrow, so the above position would still be profitable. If the entire yield curve shifted down, long-term rates would have to fall more than short-term rates for the spread to narrow, so the above position would still be profitable. Chapter 24: Problem 5 Assuming that a futures market exists for corporate bonds, sell futures contracts to deliver $100 million of 19-year corporates one year from today. In one year, close out your futures position by delivering your 19-year corporate bonds; from your viewpoint today, your 20-year corporates have thus been shortened to 1-year corporates. A strategy that uses futures that are in fact traded would require selling futures today on 20-year government bonds. In one year, sell your corporate bonds and use the proceeds to purchase an offsetting futures contract on 19-year government bonds to close out your futures position. The additional risk with this strategy is basis risk, which is the risk that the prices of government bonds and government bond futures will not move in exactly the same way as corporate bonds of the same maturity. Chapter 24: Problem 6 To lock in today's rates, sell $40 million of 10-year government bond futures. If interest rates rise, the value of the futures will fall, which means a profit for you since you are short the futures. At the end of three months, when your own bond issue is floated, close out your futures position by buying an offsetting futures contract. If interest rates have in fact risen, use the profits from your futures position to finance the increased interest payments on your bond issue. If interest rates have fallen, use some of the proceeds from your bond issue to cover your loss from your futures position. Either way, ignoring basis risk, the effective interest rate on your bond issue is locked in at today's rates by selling futures.



Solutions to Text Problems: Chapter 25 Chapter 25: Problem 1 Using standard deviation as the measure for variability, the reward-to-variability ratio for a fund is the fund’s excess return (average return over the riskless rate) divided by the standard deviation of return, i.e., the fund’s Sharpe ratio. E.g., for fund A we have:

833.16

314=

−=

−

A

FA RRσ

See the table in the answers to Problem 5 for the remaining funds’ Sharpe ratios. Chapter 25: Problem 2 The Treynor ratio is similar to the Sharpe ratio, except the fund’s beta is used in the denominator instead of the standard deviation. E.g., for fund A we have:

833.75.1

314=

−=

−

A

FA RRβ

See the table in the answers to Problem 5 for the remaining funds’ Treynor ratios. Chapter 25: Problem 3 A fund’s differential return, using standard deviation as the measure of risk, is the fund’s average return minus the return on a naïve portfolio, consisting of the market portfolio and the riskless asset, with the same standard deviation of return as the fund’s. E.g., for fund A we have:

%165

313314 −=⎟⎠

⎞⎜⎝

⎛ ×−

+−=⎟⎟⎠

⎞⎜⎜⎝

⎛×

−+− A

m

FmFA

RRRR σσ

See the table in the answers to Problem 5 for the remaining funds’ differential returns based on standard deviation.


Chapter 25: Problem 4 A fund’s differential return, using beta as the measure of risk, is the fund’s average return minus the return on a naïve portfolio, consisting of the market portfolio and the riskless asset, with the same beta as the fund’s. This measure is often called “Jensen’s alpha.” E.g., for fund A we have: ( )( ) ( )( ) %45.1313314 −=×−+−=×−+− AFmFA RRRR β See the table in the answers to Problem 5 for the remaining funds’ Jensen alphas. Chapter 25: Problem 5 This differential return measure is the same as the one used in Problem 4, except that the riskless rate is replaced with the average return on a zero-beta asset. E.g., for fund A we have: ( )( ) ( )( ) %5.35.1413414 −=×−+−=×−+− AZmZA RRRR β The answers to Problems 1 through 5 for all five funds are as follows:

Fund Sharpe Ratio

Treynor Ratio

Differential Return

Based On Standard Deviation

Differential

Return Based On

Beta and RF

Differential

Return Based On

Beta and ZR

A

1.833

7.333

−1%

−4%

−3.5%

B 2.250 18.000 1% 4% 3.5% C 1.625 13.000 −3% 3% 3.0% D 1.063 14.000 −5% 2% 1.5% E 1.700 8.500 −3% −3% −2.0%


Chapter 25: Problem 6 Looking at the table in the answers to Problem 5, we see that Fund B is ranked higher than Fund A by their Sharpe ratios. Solving for the average return that would make Fund B’s Sharpe ratio equal to Fund A’s we have:

833.14

3=

−=

− B

B

fB RRRσ

or %33.10=BR So, for the ranking to be reversed, Fund B’s average return would have to be lower than 10.33%.



Solutions to Text Problems: Chapter 26 Chapter 26: Problem 1 A. The points on a Predictive Realization Diagram would have the following coordinates (where Pi = predicted change in earnings and Ri = realized change in earnings): Industry/Firm Pi Ri A1

0.05

0.00

A2 0.05 0.03 A3 0.75 −0.25 B4 0.04 0.06 B5 0.05 0.04 B6 0.65 0.20 B7 −0.01 −0.01 C8 .070 0.40 C9 −0.03 −0.01 C10 −0.02 0.02 While there are only ten points on the Prediction Realization Diagram, certain tendencies can be detected. It is very clear from the diagram that analysts in this brokerage firm systematically overestimate earnings. Their forecasts have a strong upward bias. The second marked tendency is fro the degree of overestimation to grow as positive increases in earnings become larger. Similarly, there is a slight (based on one observation) tendency for analysts to overestimate the size of a decrease in earnings when a decrease takes place. The analysts misestimated the direction of a change in earnings in only two out of the ten cases. B. Recall from the text that, for the computation of mean square forecast error (MSFE), the results are the same whether we use predicted levels or predicted changes in earnings. We will do the MSFE analysis using levels and the following formula:

( )∑=

−=N

iii AF

N 1

21MSFE

where Fi is the forecasted level of earnings for firm i per share Ai is the actual earnings per share for firm i.


Industry/Firm Fi Ai ( )2ii AF − A1

$1.10

$1.05

0.0025

A2 $1.37 $1.35 0.0004 A3 $4.25 $3.25 1.0000 B4 $2.10 $2.12 0.0004 B5 $2.13 $2.12 0.0001 B6 $3.25 $2.80 0.2025 B7 $1.06 $1.06 0.0000 C8 $2.70 $2.40 0.0900 C9 $0.52 $0.54 0.0004 C10 $1.16 $1.20 0.0016

Sum

1.2979

Therefore:

1298.0102979.1MSFE ==

C. From the text, we know that the MSFE can be decomposed by level of aggregation as follows:

( ) ( ) ( )[ ] ( ) ( )[ ]∑∑==

−−−+−−−+−=N

iaiai

N

iaa RRPP

NRRPP

NRP

1

2

1

22 11MSFE

where the first term measures the forecast error due to all analysts misestimating the average earnings in the economy, the second term measures the error due to individual analysts misestimating the differential earnings for particular industries from the average for the economy, and the third term measures the error due to individual analysts misestimating the differential earnings for particular companies within an industry from the average for that industry. So we have: Error due forecasting sector of economy: ( ) ( ) 0306.0048.0223.0 22

=−=− RP


Error due forecasting each industry:

( ) ( )[ ]

( ) ( )[ ]( ) ( )[ ]( ) ( )[ ]

( ) 0143.00271.00169.00989.0101

048.01367.0223.02167.03

048.00725.0223.01825.04

048.00733.0223.02833.03

1011

2

2

2

1

2

=++×=

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

−−−×+

−−−×+

−−−−×

×=−−−∑=

N

iaa RRPP

N

Error due forecasting each firm:

( ) ( )[ ] 0849.011

2=−−−∑

=

N

iaiai RRPP

N

Notice that the sum of the three components equals 0.1298, which is the total MSFE we calculated earlier. To express each component as a percentage of the total MSFE, simply divide each component by 0.1298 and multiply by 100: Percent of forecast error due to forecasting sector of economy = 23.57% Percent of forecast error due to forecasting each industry = 11.02% Percent of forecast error due to forecasting each firm = 65.41% D. 1. MSFE for each analyst:

( ) 3343.00029.131

31MSFE(A)

3

1

2 =×=−×= ∑=i

ii RP

MSFE(B) = 0.0508 MSFE(C) = 0.0307


2. MSFE decomposition for each analyst: For analyst A, Industry Error = ( )2AA RP − = 0.1272

Company Error = ( ) ( )[ ]∑=

−−−3

1

2

31

iAiAi RRPP

= ( ) ( )( )[ ]∑=

−−−−3

1

20733.02833.031

iii RP

= 0.2071

% Industry Error = 1003343.01272.0

× = 38.05%

% Company Error = 1003343.02071.0

× = 61.95%

For analyst B, Industry Error = ( )20725.01825.0 − = 0.0121


−−−4

1

20725.01825.041

iii RP

= 0.0387 % Industry Error = 23.8% % Company Error = 76.2% For analyst C, Industry Error = ( )21367.02167.0 − = 0.0064


−−−3

1

21367.02167.031

iii RP

= 0.0243 % Industry Error = 20.8% % Company Error = 79.2%


E. The calculations in this part use N, not N − 1, in the denominator for variances. Error Due To Bias = ( )2RP − = ( )2048.0223.0 − = 0.0306 Error Due To Variance = ( )2RP σσ − = ( )21569.03144.0 − = 0.0248 Error Due To Covariance = ( ) RPPR σσρ−× 12 = ( ) 1569.03144.02461.012 ××−× = 0.0744

% Error Due To Bias = 1001298.00306.0

× = 23.57%

% Error Due To Variance = 1001298.00248.0

× = 19.11%

% Error Due To Covariance = 1001298.00744.0

× = 57.32%

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