36
Problems and Solutions
We then obtain the following results:
SR(n) (%)
5.20
5.36
5.49
5.61
5.71
5.75
5.79
5.82
5.84
n
2
3
4
5
6
7
8
9
10
R(0, n) (%)
5.207
5.374
5.512
5.642
5.753
5.795
5.839
5.872
5.893
2. We obtain the following interbank zero-coupon yield curve:
6.00
5.80
Zero-coupon rate (%)
5.60
5.40
5.20
5.00
4.80
4.60
4.40
4.20
4.00
0
1
2
3
4
5
6
7
8
9
10
Maturity
5
CHAPTER 5—Problems
Exercise 5.1
Calculate the percentage price change for 4 bonds with different annual coupon
rates (5% and 10%) and different maturities (3 years and 10 years), starting with
a common 7.5% YTM (with annual compounding frequency), and assuming successively a new yield of 5%, 7%, 7.49%, 7.51%, 8% and 10%.
Solution 5.1
Results are given in the following table:
37
Problems and Solutions
New Yield (%)
5.00
7.00
7.49
7.51
8.00
10.00
Change (bps)
−250
−50
−1
+1
+50
+250
5%/3yr
6.95
1.34
0.03
−0.03
−1.32
−6.35
10%/3yr
6.68
1.29
0.03
−0.03
−1.26
−6.10
5%/10yr
20.71
3.76
0.07
−0.07
−3.59
−16.37
10%/10yr
18.31
3.34
0.07
−0.07
−3.19
−14.65
Exercise 5.4
Show that the duration of a perpetual bond delivering annually a coupon c with a
YTM equal to y is 1+y
y .
Solution 5.4
The price P of the perpetual bond is given by the following formula:
∞
N ×c
N ×c
=
P =
(1 + y)i
y
i=1
where N is the face value of the perpetual bond.
The duration D of the perpetual bond is
− y2
P (y)
(1 + y)
= −(1 + y) c =
D = −(1 + y)
P (y)
y
y
c
Exercise 5.5
Show that the duration of a portfolio P invested in n bonds with weights wi ,
denominated in the same currency, is the weighted average of each bond’s duration:
Dp =
n
wi Di
i=1
Solution 5.5
Consider n bond prices denoted by Pi for i = 1, . . . n, and a bond portfolio that
is the sum of each of these n bonds. We denote by P , the price of this portfolio
and suppose that all the bonds have the same YTM equal to y. Then
P (y) =
n
Pi (y)
i=1
and
P (y) =
n
Pi (y)
i=1
Dividing the previous equation by P (y) and multiplying it by −(1 + y), we obtain
P (y)
P (y) =
−(1 + y) i
P (y)
P (y)
n
−(1 + y)
i=1
or
DP =
n
i=1
wi Di
38
Problems and Solutions
where DP is the portfolio duration, Di the duration of bond, i, and wi =
the weight of bond i in the portfolio P .
Exercise 5.7
is
Compute the dirty price, the duration, the modified duration, the $duration and the
BPV (basis point value) of the following bonds with $100 face value assuming
that coupon frequency and compounding frequency are (1) annual; (2) semiannual
and (3) quarterly.
Bond
Bond 1
Bond 2
Bond 3
Bond 4
Bond 5
Bond 6
Bond 7
Bond 8
Bond 9
Bond 10
Solution 5.7
Pi (y)
P (y)
Maturity (years)
1
1
5
5
5
5
20
20
20
20
Coupon Rate (%)
5
10
5
10
5
10
5
10
5
10
YTM (%)
5
6
5
6
7
8
5
6
7
8
We use the following Excel functions “Price”, “Duration” and “MDuration” to
obtain respectively the dirty price, the duration and the modified duration of each
bond. The $duration is simply given by the following formula:
$duration = −price × modified duration
The BPV is simply
−$duration
10,000
1. When coupon frequency and compounding frequency are assumed to be annual,
we obtain the following results:
BPV =
Bond 1
Bond 2
Bond 3
Bond 4
Bond 5
Bond 6
Bond 7
Bond 8
Bond 9
Bond 10
Price
Duration
100
103.77
100
116.85
91.8
107.99
100
145.88
78.81
119.64
1
1
4.55
4.24
4.52
4.2
13.09
11.04
12.15
10.18
Modified
Duration
0.95
0.94
4.33
4
4.23
3.89
12.46
10.42
11.35
9.43
$Duration
BPV
−95.24
−97.90
−432.95
−467.07
−388.06
−420.32
−1,246.22
−1,519.45
−894.72
−1,127.94
0.00952
0.00979
0.04329
0.04671
0.03881
0.04203
0.12462
0.15194
0.08947
0.11279
39
Problems and Solutions
2. When coupon frequency and compounding frequency are assumed to be semiannual, we obtain the following results:
—
Bond 1
Bond 2
Bond 3
Bond 4
Bond 5
Bond 6
Bond 7
Bond 8
Bond 9
Bond 10
Price
100
103.83
100
117.06
91.68
108.11
100
146.23
78.64
119.79
Duration
0.99
0.98
4.49
4.14
4.46
4.1
12.87
10.77
11.87
9.87
Modified Duration
0.96
0.95
4.38
4.02
4.31
3.94
12.55
10.46
11.47
9.49
$Duration
−96.37
−98.45
−437.60
−470.04
−394.87
−425.73
−1,255.14
−1,529.39
−902.13
−1,136.91
BPV
0.009637
0.009845
0.04376
0.047004
0.039487
0.042573
0.125514
0.152939
0.090213
0.113691
3. When coupon frequency and compounding frequency are assumed to be quarterly, we obtain the following results:
—
Bond 1
Bond 2
Bond 3
Bond 4
Bond 5
Bond 6
Bond 7
Bond 8
Bond 9
Bond 10
Price
100
103.85
100
117.17
91.62
108.18
100
146.41
78.56
119.87
Duration
0.98
0.96
4.45
4.08
4.42
4.04
12.75
10.64
11.73
9.71
Modified Duration
0.97
0.95
4.40
4.02
4.35
3.96
12.60
10.48
11.53
9.52
$Duration
−96.95
−98.72
−439.98
−471.53
−398.40
−428.51
−1,259.67
−1,534.44
−905.89
−1,141.47
BPV
0.009695
0.009872
0.043998
0.047153
0.03984
0.042851
0.125967
0.153444
0.090589
0.114147
Exercise 5.11 Zero-coupon Bonds
1. What is the price of a zero-coupon bond with $100 face value that matures in
seven years and has a yield of 7%? We assume that the compounding frequency
is semiannual.
2. What is the bond’s modified duration?
3. Use the modified duration to find the approximate change in price if the bond
yield rises by 15 basis points.
Solution 5.11
1. The price P is given by
$100
P = 2×7 = $61.77818
1 + 7%
2
2. The modified duration MD is given by
MD =
1
P 1+
7%
2
×
i
ti PV(CFi ) = 6.763
40
Problems and Solutions
3. The approximate change in price is −$0.627
P −MD × y × P = −6.763 × 0.0015 × $61.77818
= −$0.627
Exercise 5.13 You own a 7% Treasury bond with $100 face value that has a modified duration
of 6.3. The clean price is 95.25. You have just received a coupon payment 12 days
ago. Coupons are received semiannually.
1. If there are 182 days in this coupon period, what is the accrued interest?
2. Is the yield greater than the coupon rate or less than the coupon rate? How do
you know?
3. Use the modified duration to find the approximate change in value if the yield
were to suddenly rise by 8 basis points.
4. Will the actual value change more or less than this amount? Why?
Solution 5.13
1. The accrued interest AI is given by
12
× 3.5 = 0.23077
182
2. Since the price is so far below par, the yield must be higher than the coupon
rate.
3. The approximate change in price is given by the following equation
AI =
P −MD × y × P = −6.3 × $(95.25 + 0.23077) × 0.0008
= −$0.48117
4. The actual loss will be less than this amount, due to convexity (see Chapter 6).
Exercise 5.15 Today is 01/01/98. On 06/30/99, we make a payment of $100. We can only invest
in a risk-free pure discount bond (nominal $100) that matures on 12/31/98 and in
a risk-free coupon bond, nominal $100 that pays an annual interest (on 12/31) of
8% and matures on 12/31/00. Assume a flat term structure of 7%. How many units
of each of the bonds should we buy in order to be perfectly immunized?
Solution 5.15 We first have to compute the present value PV of the debt, which is the amount
we will have to deposit
100
= 90.35
(1.07)1.5
We also compute the price P1 of the 1-year pure discount bond
PV =
100
= 93.46
1.07
Similarly, the price P3 of the 3-year coupon bond is
P1 =
P3 =
8
8
108
+
+
= 102.6
1.07 (1.07)2
(1.07)3
41
Problems and Solutions
The duration of the 1-year pure discount bond is obviously 1. The duration D3 of
the 3-year coupon bond is
8
8
8
(1.07)2
(1.07)3
+3×
= 2.786
D3 = 1 × 1.07 + 2 ×
102.6
102.6
102.6
We now compute the number of units of the 1-year and the 3-year bonds (q1 and
q3 respectively), so as to achieve a $duration equal to that of the debt, and also a
present value of the portfolio equal to that of the debt. We know that the duration
of the debt we are trying to immunize is 1.5.
Therefore, q1 and q3 are given as the unique solution to the following system
of equations:
q1 = 0.696117
93.46 × 1 × q1 + 102.6 × 2.786 × q3 = 90.35 × 1.5
⇒
93.46 × q1 + 102.6 × q3 = 90.35
q3 = 0.2465
Exercise 5.19 An investor holds 100,000 units of a bond whose features are summarized in the
following table. He wishes to be hedged against a rise in interest rates.
Maturity
18 Years
Coupon Rate
9.5%
YTM
8%
Duration
9.5055
Price
$114,181
Characteristics of the hedging instrument, which is here a bond are as follows:
Maturity
20 Years
Coupon Rate
10%
YTM
8%
Duration
9.8703
Price
$119.792
Coupon frequency and compounding frequency are assumed to be semiannual. YTM stands for yield to maturity. The YTM curve is flat at an 8% level.
1. What is the quantity φ of the hedging instrument that the investor has to
sell?
2. We suppose that the YTM curve increases instantaneously by 0.1%.
(a) What happens if the bond portfolio has not been hedged?
(b) And if it has been hedged?
3. Same question as the previous one when the YTM curve increases instantaneously by 2%.
4. Conclude.
Solution 5.19
1. The quantity φ of the hedging instrument is obtained as follows:
11,418,100 × 9.5055
= −91,793
119.792 × 9.8703
The investor has to sell 91,793 units of the hedging instrument.
φ=−
42
Problems and Solutions
2. Prices of bonds with maturity 18 years and 20 years become respectively
$113.145 and $118.664.
(a) If the bond portfolio has not been hedged, the investor loses money. The
loss incurred is given by the following formula (exactly −$103,657 if we
take all the decimals into account):
Loss = $100,000 × (113.145 − 114.181) = −$103,600
(b) If the bond portfolio has been hedged, the investor is quasi-neutral to an
increase (and a decrease) of the YTM curve. The P&L of the position is
given by the following formula:
P&L = −$103,600 + $91,793 × (119.792 − 118.664) = −$57
3. Prices of bonds with maturity 18 years and 20 years become respectively
$95.863 and $100.
(a) If the bond portfolio has not been hedged, the loss incurred is given by the
following formula:
Loss = $100,000 × (95.863 − 114.181) = −$1,831,800
(b) If the bond portfolio has been hedged, the P&L of the position is given by
the following formula:
P&L = −$1,831,800 + $91,793 × (119.792 − 100) = −$15,032
4. For a small move of the YTM curve, the quality of the hedge is good. For a
large move of the YTM curve, we see that the hedge is not perfect because of
the convexity term that is no more negligible (see Chapter 6).
Exercise 5.22 A trader implements a duration-neutral strategy, which consists in buying a cheap
bond and selling a rich bond. This is the rich and cheap bond strategy. Today, the
rich and cheap bonds have the following characteristics:
Bond
Rich
Cheap
Coupon (%)
5
5.5
Maturity (years)
10
12
YTM (%)
7.50
7.75
Coupon frequency and compounding frequency are assumed to be annual. Face
value are $100 for the two bonds.
Compute the BPV of the two bonds and find the hedge position.
Solution 5.22 We first calculate the price, modified duration (MD) and BPV of each bond.
Bond
Rich
Cheap
Price
82.840
82.822
MD
7.34
8.136
BPV
0.06081
0.06738
43
Problems and Solutions
We take a long position of $100,000,000 in the 5.5%/12-year bond. The
hedge ratio HR is equal to
0.06738
= 1.10818
0.06081
Then we have to take a short position of x in the 5%/10-year bond, where x
is given by
HR =
x = HR × $100,000,000 = $110,818,000
6
CHAPTER 6—Problems
Exercise 6.1
We consider a 20-year zero-coupon bond with a 6% YTM and $100 face value.
Compounding frequency is assumed to be annual.
1. Compute its price, modified duration, $duration, convexity and $convexity?
2. On the same graph, draw the price change of the bond when YTM goes from
1% to 11%
(a) by using the exact pricing formula;
(b) by using the one-order Taylor estimation;
(c) by using the second-order Taylor estimation.
Solution 6.1
1. The price P of the zero-coupon bond is simply
P =
$100
= $31.18
(1 + 6%)20
Its modified duration is equal to 20/(1 + 6%) = 18.87
Its $duration, denoted by $Dur, is equal to
$ Dur = −18.87 × 31.18 = −588.31
Its convexity, denoted by RC, is equal to
RC = 20 × 21 ×
100
(1 + 6%)22
= 373.80
Its $convexity, denoted by $Conv, is equal to
$Conv = 373.80 × 31.18
= 11,655.20
2. Using the one-order Taylor expansion, the new price of the bond is given by
the following formula:
New Price = 31.18 + $ Dur ×(New YTM − 6%)
44
Problems and Solutions
Using the two-order Taylor expansion, the new price of the bond is given by
the following formula:
New Price = 31.18 + $ Dur ×(New YTM − 6%)
$Conv
× (New YTM − 6%)2
2
We finally obtain the following graph
The straight line is the one-order Taylor estimation. Using the two-order
Taylor estimation, we underestimate the actual price for YTM inferior to 6%,
and we overestimate it for YTM superior to 6%.
+
90
80
70
Actual price
One-order taylor estimation
Two-order taylor estimation
60
50
40
30
20
10
0
1%
2%
3%
4%
5%
6%
7%
8%
9%
10%
11%
Exercise 6.3
1. Compute the modified duration and convexity of a 6%, 25-year bond selling at
a yield of 9%. Coupon frequency and compounding frequency are assumed to
be semiannual.
2. What is its estimated percentage price change for a yield change from 9% to 11%
using the one-order Taylor expansion? Using the two-order Taylor expansion?
Compare both of them with the actual change?
3. Same question when the yield decreases by 200 basis points. Conclude.
Solution 6.3
1. For a 6%, 25-year bond selling at a yield of 9%, modified duration amounts to
10.62, while convexity is equal to 182.92.
2. The estimated percentage price change, for a yield change from 9% to 11% is
equal to
−10.62 × (0.02) + 0.5 × (182.92) × (0.02)
while the actual change is −18.03%.
2
= −21.24 + 3.66 = −17.58%,
45
Problems and Solutions
3. If the yield decreases by 200 basis points, instead, then the estimated price
change is +21.24% due to duration, and +3.66% due to convexity, that is
24.90%; as a whole, the actual price change is 25.46%. The estimated price
change is no longer symmetric around the current yield because the price function has curvature.
Exercise 6.6
Assume a 2-year Euro-note, with a $100,000 face value, a coupon rate of 10% and
a convexity of 4.53. If today’s YTM is 11.5% and term structure is flat. Coupon
frequency and compounding frequency are assumed to be annual.
1. What is the Macaulay duration of this bond?
2. What does convexity measure? Why does convexity differ among bonds? What
happens to convexity when interest rates rise? Why?
3. What is the exact price change in dollars if interest rates increase by 10 basis
points (a uniform shift)?
4. Use the duration model to calculate the approximate price change in dollars if
interest rates increase by 10 basis points.
5. Incorporate convexity to calculate the approximate price change in dollars if
interest rates increase by 10 basis points.
Solution 6.6
1. Duration
D=
=
1×
10,000
10,000
100,000
1.115 + 2 × 1.1152 + 2 × 1.1152
10,000
10,000
100,000
1.115 + 1.1152 + 1.1152
1×
10,000
1.115
+2×
10,000
1.1152
+2×
100,000
1.1152
= 1.908
97,448.17
2. Convexity measures the change in modified duration or the change in the slope
of the price-yield curve. Holding maturity constant, the higher the coupon, the
smaller the duration. Hence, for low duration levels the change in slope (convexity) is small. Alternatively, holding coupon constant, the higher the maturity,
the higher the duration, and hence, the higher the convexity. When interest rates
rise, duration (sensitivity of prices to changes in interest rates) becomes smaller.
Hence, we move toward the flatter region of the price-yield curve. Therefore,
convexity will decrease parallel to duration.
3. Price for a 11.6% YTM is
10,000 10,000 100,000
+
+
= $97,281.64
P (11.6%) =
1.116
1.1162
1.1162
Price has decreased by $166.53 from P (11.5%) = $97,448.17 to $97,281.64
4. We use
D
× y × P
P −MD × y × P (11.5%) = −
1+y
1.908
× 97,448.17 × 0.001 = −$166.754
=−
1.115
46
Problems and Solutions
5. We use
1
× RC × (y)2 × P
2
1
1.908
× 97,448.17 × 0.001 +
=−
1.115
2
×4.53 × (0.001)2 × 97,448.17 = −$166.531
P −MD × y × P +
Hedging error is smaller when we account for convexity.
Exercise 6.8
Modified Duration/Convexity Bond Portfolio Hedge
At date t, the portfolio P to be hedged is a portfolio of Treasury bonds with
various possible maturities. Its characteristics are as follows:
Price
$28,296,919
YTM
7.511%
MD
5.906
Convexity
67.578
We consider Treasury bonds as hedging assets, with the following features:
Bond
Bond 1
Bond 2
Bond 3
Price ($)
108.039
118.786
97.962
Coupon Rate (%)
7
8
5
Maturity date
3 years
7 years
12 years
Coupon frequency and compounding frequency are assumed to be annual. At date
t, we force the hedging portfolio to have the opposite value of the portfolio to be
hedged.
1. What is the number of hedging instruments necessary to implement a modified
duration/convexity hedge?
2. Compute the YTM, modified duration and convexity of the three hedging assets.
3. Which quantities φ1 , φ2 and φ3 of each of the hedging asset 1, 2, 3 do we have
to consider to hedge the portfolio P ?
Solution 6.8
1. We need three hedging instruments.
2. We obtain the following results:
Bond
Bond 1
Bond 2
Bond 3
YTM (%)
6.831
7.286
7.610
MD
2.629
5.267
8.307
Convexity
9.622
36.329
90.212
3. We then are looking for the quantities φ1 , φ2 and φ3 of each hedging instrument
1, 2, 3 as solutions to the following linear system:

−1 
 
 
−279,536
φ1
100.445 103.808 79.929
−28,296,919
 φ2  =  −264.057 −546.791 −663.947   167,143,615  =  290,043 
−379,432
966.460 3,771.257 7,210.58
−1,912,260,201
φ3

Exercise 6.10 Computing the Level, Slope and Curvature $Durations of a Bond Portfolio using
the Nelson and Siegel Extended Model
47
Problems and Solutions
On 09/02/02, the values of the Nelson and Siegel Extended parameters are as
follows:
β0
β1
β2
τ1 β 3
τ2
5.9% −1.6% −0.5% 5 1% 0.5
Recall from Chapter 4 that the continuously compounded zero-coupon rate
R c (0, θ ) is given by the following formula:


1 − exp − τθ1

R c (0, θ ) = β0 + β1 
θ
τ1

+ β2 

+ β3 
1 − exp − τθ1
θ
τ1
1 − exp − τθ2
θ
τ2

θ 
− exp −
τ1

θ 
− exp −
τ2
1. Draw the zero-coupon yield curve associated with this set of parameters.
2. We consider three bonds with the following features. Coupon frequency is
annual.
Bond 1
Bond 2
Bond 3
Maturity
(years)
3
7
15
Coupon
(%)
4
5
6
Compute the price and the level, slope and curvature $durations of each bond.
Compute also the same $durations for a portfolio with 100 units of Bond 1, 200
units of Bond 2 and 100 units of Bond 3.
3. The parameters of the Nelson and Siegel Extended model change instantaneously to become
β0
5.5%
β1
−1%
β2
0.1%
τ1
5
β3
2%
τ2
0.5
(a) Draw the new zero-coupon yield curve.
(b) Compute the new price of the bond portfolio and compare it with the value
given by the following equation:
New Estimated Price = Former Price + β0 .D0,P + β1 .D1,P
+ β2 .D2,P + β3 .D3,P
where βi is the change in value of parameter βi , and Di,P is the $duration
of the bond portfolio associated with parameter βi .
4. Same questions when the coupon frequency is semiannual.
48
Problems and Solutions
Solution 6.10
1. We obtain the following zero-coupon yield curve
5.60
Zero-coupon rate (%)
5.40
5.20
5.00
4.80
4.60
4.40
4.20
0
5
10
15
20
25
30
Maturity
2. For each bond, the price and level, slope and curvature $durations denoted
respectively by D0 , D1 , D2 and D3 are given in the following table:
—
Bond 1
Bond 2
Bond 3
Portfolio
Price ($)
97.61
99.58
106.58
40,333.43
D0
−281.60
−604.87
−1,109.21
−260,074.82
D1
−212.97
−337.83
−418.35
−130,697.65
D2
−56.49
−171.36
−299.10
−69,831.80
D3
−47.08
−48.57
−51.82
−19,604.17
The level, slope and curvature $durations of the bond portfolio denoted by D0,P ,
D1,P , D2,P and D3,P are simply obtained by using the following formulas:
D0,P = 100 × (−281.60) + 200 × (−604.87) + 100 × (−1,109.21)
= −260,074.82
D1,P = 100 × (−212.97) + 200 × (−337.83) + 100 × (−418.35)
= −130,697.65
D2,P = 100 × (−56.49) + 200 × (−171.36) + 100 × (−299.10)
= −69,831.80
D3,P = 100 × (−47.08) + 200 × (−48.57) + 100 × (−51.82)
= −19,604.17
3. (a) We draw the new curve on the following graph:
49
Problems and Solutions
6.00
5.80
Zero-coupon rate (%)
5.60
5.40
5.20
5.00
Curve at the origin
New curve
4.80
4.60
4.40
4.20
4.00
0
5
10
15
20
25
30
Maturity
(b) The new price is equal to
New Price = 39,977
whereas the new estimated price obtained by using the equation given in the
exercise is
New Estimated Price = 40,333 + (−0.4%).(−260,075) + 0.6%.(−130,698)
+0.6%.(−69,832) + 1%.(−19,604) = 39,975
We conclude that the price change of the bond portfolio is well explained by
Nelson–Siegel $durations multiplied by the change in value of the different
parameters.
4. When the coupon frequency is semiannual, we obtain the following results:
—
Bond 1
Bond 2
Bond 3
Portfolio
Price ($)
97.74
99.94
107.40
40,502.52
D0
−279.25
−598.81
−1,099.32
−257,618.44
D1
−211.16
−334.78
−415.52
−129,623.23
D2
−55.89
−169.50
−296.38
−69,126.35
D3
−46.69
−48.17
−51.54
−19,457.59
Exercise 6.12 Bond Portfolio Hedge using the Nelson–Siegel Extended Model
We consider the Nelson–Siegel Extended zero-coupon rate function




1 − exp − τθ1
1 − exp − τθ1
θ 
 + β2 
− exp −
R c (t, θ ) = β0 + β1 
θ
θ
τ
1
τ
τ
1

+β3 
1 − exp − τθ2
θ
τ2
1

θ 
− exp −
τ2
50
Problems and Solutions
where R c (t, θ ) is the continuously compounded zero-coupon rate at date t with
maturity θ .
On 09/02/02, the model is calibrated, parameters being as follows:
β0
5.9%
β1
−1.6%
β2
−0.5%
τ1
5
β3
1%
τ2
0.5
At the same date, a manager wants to hedge its bond portfolio P against interestrate risk. The portfolio contains the following Treasury bonds (delivering annual
coupons, with a $100 face value):
Bond
Bond 1
Bond 2
Bond 3
Bond 4
Bond 5
Bond 6
Bond 7
Bond 8
Bond 9
Bond 10
Bond 11
Bond 12
Bond 13
Maturity
01/12/05
04/12/06
07/12/07
10/12/08
03/12/09
10/12/10
01/12/12
03/12/15
07/12/20
01/12/25
07/12/30
01/12/31
07/12/32
Coupon
4
7.75
4
7
5.75
5.5
4
4.75
4.5
5
4.5
4
5
Quantity
10,000
10,000
10,000
10,000
10,000
10,000
10,000
10,000
10,000
10,000
10,000
10,000
10,000
We consider Treasury bonds as hedging instruments with the following features:
Hedging Asset
Hedging Asset 1
Hedging Asset 2
Hedging Asset 3
Hedging Asset 4
Hedging Asset 5
Coupon
4.5
5
6
6
6.5
Maturity
04/15/06
12/28/12
10/05/15
10/10/20
10/10/31
Coupons are assumed to be paid annually, and the face value of each bond is $100.
At date t, we force the hedging portfolio to have the opposite value of the portfolio
to be hedged.
1. Compute the price and level, slope and curvature $durations of portfolio P .
2. Compute the price and level, slope and curvature $durations of the five hedging
assets.
3. Which quantities φ1 , φ2 , φ3 , φ4 and φ5 of each hedging asset 1, 2, 3, 4 and 5 do
we have to consider to hedge the portfolio P ?
4. The parameters of the Nelson and Siegel Extended model change instantaneously to become
51
Problems and Solutions
β0
6.5%
β1
−1%
β2
0.1%
τ1
5
β3
2%
τ2
0.5
(a) What is the price of the bond portfolio after this change? If the manager has
not hedged its portfolio, how much money has he lost?
(b) What is the variation in price of the global portfolio (where the global portfolio is the bond portfolio plus the hedging instruments)?
(c) Conclusion.
Solution 6.12
1. The price and level, slope and curvature $durations of bond portfolio P are
given in the following table:
Price
$12,723,603
D0
−110,075,273
D1
−42,504,124
D2
−25,924,204
D3
−6,021,090
∂P
where Di = ∂β
for i = 0, 1, 2 and 3 are the level, slope and curvature $durai
tions of P in the Nelson–Siegel Extended model.
2. The level, slope and curvature $durations of the five hedging instruments are
given in the following table:
Hedging Asset
Hedging Asset 1
Hedging Asset 2
Hedging Asset 3
Hedging Asset 4
Hedging Asset 5
Price ($)
100.41
101.71
111.97
112.05
120.22
D0
−337.70
−813.99
−1013.91
−1242.56
−1799.54
D1
−242.65
−373.12
−412.53
−424.00
−457.82
D2
−75.01
−235.42
−282.76
−315.54
−354.97
D3
−48.11
−47.97
−51.75
−51.84
−55.57
3. We are looking for the quantities φ1 , φ2 , φ3 , φ4 and φ5 of each hedging asset
as the solutions to the following linear system:
−1
  
−337.70 −813.99 −1013.91 −1242.56 −1799.54
φ1
 φ2   −242.65 −373.12 −412.53 −424.00 −457.82 

  
 φ3  =  −75.01 −235.42 −315.54 −354.97 −354.97 

  
 φ4   −48.11 −47.97 −51.75 −51.84 −55.57 
100.41 101.71 111.97
112.05
120.22
φ5


108,004,516
 42,733,844 



×
 26,093,954 
 6,044,277 
−12,769,376
4. (a) The price of the bond portfolio P , after the change in parameters, is equal
to $11,663,433. With no hedge, the manager has lost $1,060,170
Loss = $11,715,756 − $12,769,376 = $1,053,620
52
Problems and Solutions
(b) The prices of the five hedging assets, after the change in parameters, are
given in the following table
Hedging Asset
Hedging Asset 1
Hedging Asset 2
Hedging Asset 3
Hedging Asset 4
Hedging Asset 5
Price ($)
96.08
93.10
101.76
100.42
105.62
With the hedging portfolio (which contains the five hedging assets in adequate quantity), the manager gains $1,054,624.
Gain = −39,507.(96.10 − 100.41) − 74,586.(93.14 − 101.71)
+ 52,726.(101.80 − 111.97) − 37,764.(100.47 − 112.05)
− 23,649.(105.68 − 120.22)
= $1,054,624
The variation in price of the global portfolio (bond portfolio + hedging
instruments) is then equal to −$5,546 ($1,054,624–$1,060,170).
(c) The hedge is efficient.
7 CHAPTER 7—Problems
Exercise 7.1
Would you say it is easier to track a bond index or a stock index. Why or why
not?
Solution 7.1
As is often the case, the answer is yes and no.
On the one hand, it is harder to perform perfect replication of a bond index
compared to a stock index. This is because bond indices typically include a huge
number of bonds. Other difficulties include that many of the bonds in the indices
are thinly traded and the fact that the composition of the index changes regularly,
as they mature.
On the other hand, statistical replication on bond indices is easier to perform than statistical replication of stock indices, in the sense that a significantly
lower tracking error can usually be achieved for a given number of instruments
in the replicating portfolio. This is because bonds with different maturities tend to
exhibit a fair amount of cross-sectional correlation so that a very limited number
of factors account for a very large fraction of changes in bond returns. Typically, 2
or 3 factors (level, slope, curvature) account for more than 80% of these variations.
Stocks typically exhibit much more idiosyncratic risk, so that one typically needs
to use a large number of factors to account for not much more than 50% of the
changes in stock prices.
Exercise 7.2
What are the pros and cons of popular indexing methodologies in the fixed-income
universe?