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johnmate1122@gmail.com Bank Fit and Well 4th Canadian Edition
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 â&#x2C6;&#x2019; 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&#x2014;PĂ&#x2014;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.
9