Test bank fixed income securities valuation risk and risk management 1st edition solution

Page 1

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Solution Manual to accompany the textbook

Fixed Income Securities: Valuation, Risk, and Risk Management by Pietro Veronesi Chapters 2 - 8


Version 1 Date: October, 2009 Author: Anna Cieslak, Javier Francisco Madrid


Solutions to Chapter 2 Exercise 1. Compute the discount factors implied by the STRIPS: Z(0, 3)

= exp(−3 × 0.1) = 0.741

(1)

Z(0, 5)

= exp(−5 × 0.05) = 0.779.

(2)

Since Z(0, 3) < Z(0, 5), there is an arbitrage opportunity due to the violation of the positive time discount rate. To exploit it: buy the 3-year bond and sell the 5-year bond. Exercise 2. Compute the quoted price P of the T-bill as: n

h

i

P = 100 × 1 − 360 × d , using the discount rate given, d. The simple (bond equivalent) yield measures your annualized return as: BEY = 100 − P P Let τ =

(3)

×365 . n

(4)

be the time to maturity expressed as fraction of a year, and let T denote the maturity date of

N

365

a given T-bill. The continuously compounded yield follows as: P

1

r(t, T ) = −τ ln 100 .

(5)

Finally, to obtain the semi-annually compounded yield for the 1-year T-bill, use: (6)

r2 (0, 1) = 2 × (P /100)1/2 − 1

1

Cont. comp. Semi-annual Maturity

N

T−T

Discount, D

Price, P

BEY

yield

comp.

Date

a.

4-week

28

0.083

3.48%

99.7293

3.5379%

3.53%

12/12/2005

b.

4-week

28

0.083

0.13%

99.9899

0.13%

0.13%

11/6/2008

c.

3-month

90

0.25

4.93%

98.7675

5.06%

5.03%

7/10/2006

d.

3-month

90

0.25

4.76%

98.8100

4.88%

4.86%

5/8/2007

e.

3-month

90

0.25

0.48%

99.8800

0.49%

0.49%

11/4/2008

f.

6-month

180

0.5

4.72%

97.6400

4.90%

4.84%

4/21/2006

g.

6-month

180

0.5

4.75%

97.6250

4.93%

4.87%

6/6/2007

h.

6-month

180

0.5

0.89%

99.5550

0.91%

0.90%

11/11/2008

i.

360-day

360

1

1.73%

98.2700

1.78%

1.77%

1.75%

9/30/2008

j.

360-day

360

1

1.19%

98.8100

1.22%

1.21%

1.20%

11/5/2008


1


Exercise 3. Compute respective discount factors taking into account the convention on which the interest rate is given: 1. Z(t, t + 1.5)

= exp(−0.02 × 1.5) = 0.97045

2. Z(t, t + 1.5)

= exp(−0.03) = 0.97045 1 = 1.5 = 0.96931 (1 + 0.021) 1 = = 0.97045 2×1.5 (1 + 0.0201/2)

3. Z(t, t + 1.5) 4. Z(t, t + 1.5) Bond 3 is mispriced. Exercise 4.

Using Table 2.4, obtain the discount factor Z(t, T ) for each maturity T − t from 0.25 to 7.5 years: 1

Z(t, T ) =

1+R

2

T,T

.

2(T −T)

)

(2

(7)

Use Z to price each bond: a. PZ (0, 5) = 100 × Z(0, 5) = 72.80 b. PC=15%,N=2(0, 7) =

c. P

C=7%,N=4

(0, 4) =

15

2

×

P14

I=1

Z(0, i/2) + 100 × Z(0, 7) = 151.23

7 × P 16 4 9 I =17Z(0,

d. PC=9%,N=2(0, 3.25) = 2 × e. 100 (see Fact 2.11)

P

I=1

i/4) + 100 × Z(0, 4) = 101.28

Z(0, i/2 − 0.25) + 100 × Z(0, 3.25) = 108.55

f. PF R,N=1,S=0 = Z(0, 0.5) × 100 × (1 + (0, 5.5) = 100 +

g. P F R,N=4,S=0.35%

h. P

F R,N=2,S=0.40%

6.8%

0.35

4

1

) = 103.44, where we assume that r1 (0) = 6.8%

22

P

I=1

Z(0, i/4) = 101.6 6.4%

0.40

2

that r 2 (0) = 6 .4%

2

P

15 I=1

Z(0, i/2 − 0.25) = 104, where we assume

)+

= Z(0, 0.25) × 100 × (1 +

Exercise 5. a. When coupon c is equal to the yield to maturity y the bond trades at par; when coupon is below (above) the yield to maturity the bond trades above (below) par. Obtain bond prices given yield and the coupon using: 20 X

PC (0, T ) = I=1

It follows:

100

c/2 × 100 (1 + y/2)

I

+

20

(1 + y/2)

(8)


2


c

y

P

5%

6% 107.79

6%

6%

7%

6% 92.89

100

b. Figure 1 plots bond prices implied by different yields to maturity. Bond price vs YTM

150 140 130

Bond price

120 110 100 90 80 70 60 50 0

2

4

6

8 Yield

10

12

14

16

Fig. 1. Bond price as function of yield to maturity

Exercise 6. a. To obtain bond prices use the expression: c

2T

X

PC (0, T ) = 2 ×

I=1

Z(0, i/2) + 100 × Z(0, T ),

(9)

To compute the yield to maturity, solve equation (8) for y using a numerical solver. c

T−t

P

y

15%

7

151.2306

5.9461%

3%

7

84.3482

5.7474%

b. The yields to maturity are different since bonds have different coupons, despite having the same time to maturity. Both bonds are priced using a no arbitrage discount curve. Therefore, their prices are fair. Exercise 7. a. Bootstrap the discount factors Z(t, T ) using the expression (9), and substituting recursively for the 6-month, 1year, 1.5-year, and 2-year bonds. E.g., given Z(0, 0.5) and Z(0, 1), for the 1.5-year bond you have:


3


100.86 −

Z(0, 1.5) =

7.5

(Z(0, 0.5) + Z(0, 1))

2

100 +

7.5

.

(10)

2

This yields: T−t

Coupon, c

0.5

0.00%

1

Price, P

Issued

Z(t, T )

$96.80

5/15/2000

0.9680

5.75%

$99.56

5/15/1998

0.9407

1.5

7.50%

$100.86

11/15/1991

0.9032

2

7.50%

$101.22

5/15/1992

0.8740

b. Compute the no-arbitrage price of the two bonds given the discount function obtained above. The prices of the two bonds are:

P

C=8%

P

C=13.13%

= $101.71

(11)

= $106.60,

(12)

i.e. both are higher than the market prices. There is an arbitrage opportunity. You could make riskless profit by buying the underpriced bond at the traded price and selling the corresponding portfolio of zeros that replicates the cash flows from the bond. Exercise 8. The quotes are obtained on May 15, 2000. Use the mid bid-ask quote to compute the price. You want to obtain the semi-annual curve. The provided maturities of bonds are spaced semi-annually. We can assume that the clean (quoted) price is equal to the dirty (invoice) price, i.e. the accrual is zero. For each maturity, bootstrap the discount factors Z(t, T ) as in Exercise 7.a. The continuously compounded zero coupon yield is given as: r(t, T ) = −

1 ln Z(t, T ). T−t

(13)


4 Cusip 912827ZN

T − T Mid bid-ask quote

Coupon, C

Accrual

Z(T, T )

0.5

100.9063

8.500%

0

0.96793

912810CU

1

105.9961

13.125%

0

0.93508

912810CX

1.5

112.4102

15.750%

0

0.90312

912827F4

2

101.2188

7.500%

0

0.87418

912810DA

2.5

110.6836

11.625%

0

0.84387

912810DD

3

110.3438

10.750%

0

0.81638

912810DG

3.5

115.3242

11.875%

0

0.78928

912810DH

4

118.9141

12.375%

0

0.76267

912810DM

4.5

118.3125

11.625%

0

0.73951

912810DQ

5

121.6289

12.000%

0

0.71544

912827V8

5.5

96.0000

5.875%

0

0.69440

912827X8

6

100.6211

6.875%

0

0.67229

912827Z6

6.5

98.7656

6.500%

0

0.65080

9128272U

7

99.5781

6.625%

0

0.63152

9128273X

7.5

93.1484

5.500%

0

0.61225

9128274F

8

93.8008

5.625%

0

0.59478

9128274V

8.5

87.9922

4.750%

0

0.57640

9128275G

9

92.8398

5.500%

0

0.56151


5


Solutions to Chapter 3 Exercise 1. a. 3; equal to the maturity of the zero bond b. 2.9542; the duration of the coupon bond is the weighted average of the coupon payment times c. 0.9850 d. 0.5; equal to the time left to the next coupon payment e. 0.5111; obtain the price PF R of the floating rate bond (see Chapter 2, equation (2.39)). In analogy to a coupon bond, the duration is computed as: DF R = 100 PF R

× 0.5 +0.5s ×

3 P

Z(0, t) × t

.

PF R

T=0.5

(14)

f. 0.2855; proceed as in point e. above but recognize that the valuation is outside the reset date. Assume that the coupon applying to the next reset date has been set at r2(0) = 6.4%. Exercise 2. Portfolio A: Security

Duration, D

Weight, w

D×w

4.5yr @ 5% semi

3.8660

40%

1.55

7yr @ 2.5% semi

6.4049

25%

1.60

1.75 fl + 30bps semi

0.2540

20%

0.05

1yr zero

1.0000

10%

0.10

2yr @ 3% quart

1.9530

5%

0.10

Port. D

3.40

Portfolio B: Security

Duration, D

Weight, w

D×w

7yr @ 10% semi

5.4262

40%

2.17

4.25yr @ 3% quart

3.9838

25%

1.00

90 day zero

0.2500

20%

0.05

2yr fl semi

0.5000

10%

0.05

1.5yr @ 6% semi

1.4564

5%

0.07

Port. D

3.34

The investors would select the shorter duration portfolio B.


6


Exercise 3. Obtain yield to maturity y for each security. Compute modified and Macaulay duration accodring to equation (3.19) and (3.20) in the book. Yield

Duration

Modified

Macaulay

a.

6.95%

3

3

2.8993

b.

6.28%

2.9542

2.9974

2.9061

c.

6.66%

0.9850

0.9850

0.9689

d.

0.00%

0.5

0.5

0.5

e.

6.82%

0.5111

0.5111

0.4943

f.

6.76%

0.2855

0.2855

0.2761

Exercise 4. Compute the duration of each asset and use the fact that the dollar duration is the bond price times its duration.

Price

Duration

$ Duration

a. $89.56

4.55

$407.88

b. $67.63

-7.00

($473.39)

c. $79.46

3.50

$277.74

d. $100.00

0.5

$50.00

e. $100.00

-0.25

($25.00)

f. $102.70

-0.2763

($28.38)

Exercise 5. a. Compute number of units of each security (N ) in the portfolio and apply Fact 3.5. Portfolio A: Security

Price

Duration

Weight

N

DĂ—PĂ—N

4.5yr @ 5% semi

94.03

3.8660

40%

0.43

154.64

7yr @ 2.5% semi

81.56

6.4049

25%

0.307

160.12

1.75 fl + 30bps semi

102.09

0.2540

20%

0.20

5.08

1yr zero

93.61

1.0000

10%

0.11

10.00

2yr @ 3% quart

92.54

1.9530

5%

0.05

9.76

$D

339.61


7


Portfolio B : Security

Price

Duration

Weight

N

D×P×N

7yr @ 10% semi

123.36

5.4262

40%

0.32

217.05

4.25yr @ 3% quart

86.83

3.9838

25%

0.29

99.59

90 day zero

98.45

0.2500

20%

0.20

5.00

2yr fl semi

100.00

0.5000

10%

0.10

5.00

1.5yr @ 6% semi

98.83

1.4564

5%

0.05

7.28

$D

333.92

b. For 1 bps increase, we have (see Definition 3.5): Portfolio A: 339.61 × 0.01/100 = −$0.0340 Portfolio B: 333.92 × 0.01/100 = −$0.0334 c. Yes. Exercise 6. After the reshuffling of the portfolio, its value becomes $50 mn. a. Short -0.307 units of long bond in portfolio A, and -0.081 units in portfolio B. b. New dollar durations are: 19.36 and 62.61 for portfolio A and B, respectively. c. The conclusion reverses. N LT bond

New weight LT bond

New $D

Port. A

-0.307

-25%

19.36

Port. B

-0.081

-10%

62.61

Exercise 7. a. $10 mn b. Compute the dollar duration of the cash flows in each bond, and then the dollar duration of the portfolio:

Security

Position

$ (mn)

Price

N

$D

$D × N

6yr IF @ 20% - fl quart

Long

20.00

146.48

0.137

1,140.28

155.69

4yr fl 45bps semi

Long

20.00

101.62

0.197

53.54

10.54

5yr zero

Short

(30.00)

76.41

-0.393

382.052

-150.00

Port. $D

16.23

Port. value

$10.00 mn

Exercise 8. a. The price of the 3yr @ 5% semi bond is $97.82. You want the duration of the hedged portfolio to be zero. You need to short 0.058 units of the 3-year bond, i.e. the short position is -$5.69.


8


b. The total value of the portfolio is: $4.31 mn. Exerxise 9. Compute the new value of the portfolio assuming the term structure of interest rates as of May 15, 1994. Original

Now

value

Unhedged port.

$10.00

$8.97

($1.03)

Hedge

($5.69)

($5.44)

$0.25

Total

$4.31

$3.53

($0.78)

a. $8.97 mn b. $3.53 mn c. The immunization covered part of the loss. The change in the value of the portfolio is both due to ( i) the passage of time (coupon) and (ii) the increase in interest rates. Exercise 10. Use the curve given on May 15, 1994, but keep the times to maturity unchanged from the initial ones. The change in value is due to the change in interest rates only. Original

Now

value

Unhedged port.

$10.00

$9.97

($0.03)

Hedge

($5.69)

($5.43)

$0.26

Total

$4.31

$4.54

$0.23

Exercise 11. Use the curve given on February 15, 1994, but change the times to maturity to those on May 15, 1994. The change in value is due to coupon only. Original

Now

value

Unhedged port.

$10.00

$9.12

($0.88)

Hedge

($5.69)

($5.68)

$0.01

$4.31

$3.45

($0.87)

Total Exercise 12. a.,b. Loss of $0.87 mn. c. Gain of $0.08 mn.


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