solinst - Kennesaw State University

36Problems and Solutions

We then obtain the following results:

n SR(n) (%) R(0, n) (%)2 5.20 5.2073 5.36 5.3744 5.49 5.5125 5.61 5.6426 5.71 5.7537 5.75 5.7958 5.79 5.8399 5.82 5.87210 5.84 5.893

2. We obtain the following interbank zero-coupon yield curve:

4.00

4.20

4.40

4.60

4.80

5.00

5.20

5.40

5.60

5.80

6.00

0 1 2 3 4 5 6 7 8 9 10

Maturity

Zer

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ou

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n r

ate

(%)

5 CHAPTER 5—Problems

Exercise 5.1 Calculate the percentage price change for 4 bonds with different annual couponrates (5% and 10%) and different maturities (3 years and 10 years), starting witha common 7.5% YTM (with annual compounding frequency), and assuming suc-cessively a new yield of 5%, 7%, 7.49%, 7.51%, 8% and 10%.

Solution 5.1 Results are given in the following table:


New Yield (%) Change (bps) 5%/3yr 10%/3yr 5%/10yr 10%/10yr5.00 −250 6.95 6.68 20.71 18.317.00 −50 1.34 1.29 3.76 3.347.49 −1 0.03 0.03 0.07 0.077.51 +1 −0.03 −0.03 −0.07 −0.078.00 +50 −1.32 −1.26 −3.59 −3.19

10.00 +250 −6.35 −6.10 −16.37 −14.65

Exercise 5.4 Show that the duration of a perpetual bond delivering annually a coupon c with aYTM equal to y is 1+y

y.

Solution 5.4 The price P of the perpetual bond is given by the following formula:

P =∞∑i=1

N × c

(1 + y)i= N × c

y

where N is the face value of the perpetual bond.The duration D of the perpetual bond is

D = −(1 + y)P ′(y)

P (y)= −(1 + y)

− c

y2

cy

= (1 + y)

y

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∑

i=1

wiDi

Solution 5.5 Consider n bond prices denoted by Pi for i = 1, . . . n, and a bond portfolio thatis the sum of each of these n bonds. We denote by P , the price of this portfolioand suppose that all the bonds have the same YTM equal to y. Then

P (y) =n∑

i=1

Pi(y)

and

P ′(y) =n∑

i=1

P ′i (y)

Dividing the previous equation by P (y) and multiplying it by −(1+y), we obtain

−(1 + y)P ′(y)

P (y)=

n∑i=1

−(1 + y)P ′

i (y)

P (y)

or

DP =n∑

i=1

wiDi


where DP is the portfolio duration, Di the duration of bond, i, and wi = Pi(y)P (y)

isthe weight of bond i in the portfolio P .

Exercise 5.7 Compute the dirty price, the duration, the modified duration, the $duration and theBPV (basis point value) of the following bonds with $100 face value assumingthat coupon frequency and compounding frequency are (1) annual; (2) semiannualand (3) quarterly.

Bond Maturity (years) Coupon Rate (%) YTM (%)Bond 1 1 5 5Bond 2 1 10 6Bond 3 5 5 5Bond 4 5 10 6Bond 5 5 5 7Bond 6 5 10 8Bond 7 20 5 5Bond 8 20 10 6Bond 9 20 5 7Bond 10 20 10 8

Solution 5.7 We use the following Excel functions “Price”, “Duration” and “MDuration” toobtain respectively the dirty price, the duration and the modified duration of eachbond. The $duration is simply given by the following formula:

$duration = −price × modified duration

The BPV is simply

BPV = −$duration

10,000

1. When coupon frequency and compounding frequency are assumed to be annual,we obtain the following results:

Price Duration Modified $Duration BPVDuration

Bond 1 100 1 0.95 −95.24 0.00952Bond 2 103.77 1 0.94 −97.90 0.00979Bond 3 100 4.55 4.33 −432.95 0.04329Bond 4 116.85 4.24 4 −467.07 0.04671Bond 5 91.8 4.52 4.23 −388.06 0.03881Bond 6 107.99 4.2 3.89 −420.32 0.04203Bond 7 100 13.09 12.46 −1,246.22 0.12462Bond 8 145.88 11.04 10.42 −1,519.45 0.15194Bond 9 78.81 12.15 11.35 −894.72 0.08947

Bond 10 119.64 10.18 9.43 −1,127.94 0.11279


2. When coupon frequency and compounding frequency are assumed to be semi-annual, we obtain the following results:

— Price Duration Modified Duration $Duration BPVBond 1 100 0.99 0.96 −96.37 0.009637Bond 2 103.83 0.98 0.95 −98.45 0.009845Bond 3 100 4.49 4.38 −437.60 0.04376Bond 4 117.06 4.14 4.02 −470.04 0.047004Bond 5 91.68 4.46 4.31 −394.87 0.039487Bond 6 108.11 4.1 3.94 −425.73 0.042573Bond 7 100 12.87 12.55 −1,255.14 0.125514Bond 8 146.23 10.77 10.46 −1,529.39 0.152939Bond 9 78.64 11.87 11.47 −902.13 0.090213Bond 10 119.79 9.87 9.49 −1,136.91 0.113691

3. When coupon frequency and compounding frequency are assumed to be quar-terly, we obtain the following results:

— Price Duration Modified Duration $Duration BPVBond 1 100 0.98 0.97 −96.95 0.009695Bond 2 103.85 0.96 0.95 −98.72 0.009872Bond 3 100 4.45 4.40 −439.98 0.043998Bond 4 117.17 4.08 4.02 −471.53 0.047153Bond 5 91.62 4.42 4.35 −398.40 0.03984Bond 6 108.18 4.04 3.96 −428.51 0.042851Bond 7 100 12.75 12.60 −1,259.67 0.125967Bond 8 146.41 10.64 10.48 −1,534.44 0.153444Bond 9 78.56 11.73 11.53 −905.89 0.090589Bond 10 119.87 9.71 9.52 −1,141.47 0.114147

Exercise 5.11 Zero-coupon Bonds

1. What is the price of a zero-coupon bond with $100 face value that matures inseven years and has a yield of 7%? We assume that the compounding frequencyis 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

P = $100(1 + 7%

2

)2×7= $61.77818

2. The modified duration MD is given by

MD = 1

P(

1 + 7%2

) ×∑

i

tiPV(CFi ) = 6.763


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 durationof 6.3. The clean price is 95.25. You have just received a coupon payment 12 daysago. 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

AI = 12

182× 3.5 = 0.23077

2. Since the price is so far below par, the yield must be higher than the couponrate.

3. The approximate change in price is given by the following equation

�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 investin a risk-free pure discount bond (nominal $100) that matures on 12/31/98 and ina risk-free coupon bond, nominal $100 that pays an annual interest (on 12/31) of8% and matures on 12/31/00. Assume a flat term structure of 7%. How many unitsof 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 amountwe will have to deposit

PV = 100

(1.07)1.5= 90.35

We also compute the price P1 of the 1-year pure discount bond

P1 = 100

1.07= 93.46

Similarly, the price P3 of the 3-year coupon bond is

P3 = 8

1.07+ 8

(1.07)2+ 108

(1.07)3= 102.6


The duration of the 1-year pure discount bond is obviously 1. The duration D3 ofthe 3-year coupon bond is

D3 = 1 ×8

1.07102.6

+ 2 ×8

(1.07)2

102.6+ 3 ×

8(1.07)3

102.6= 2.786

We now compute the number of units of the 1-year and the 3-year bonds (q1 andq3 respectively), so as to achieve a $duration equal to that of the debt, and also apresent value of the portfolio equal to that of the debt. We know that the durationof the debt we are trying to immunize is 1.5.

Therefore, q1 and q3 are given as the unique solution to the following systemof equations:{

93.46 × 1 × q1 + 102.6 × 2.786 × q3 = 90.35 × 1.593.46 × q1 + 102.6 × q3 = 90.35

�⇒{

q1 = 0.696117q3 = 0.2465

Exercise 5.19 An investor holds 100,000 units of a bond whose features are summarized in thefollowing table. He wishes to be hedged against a rise in interest rates.

Maturity Coupon Rate YTM Duration Price18 Years 9.5% 8% 9.5055 $114,181

Characteristics of the hedging instrument, which is here a bond are as fol-lows:

Maturity Coupon Rate YTM Duration Price20 Years 10% 8% 9.8703 $119.792

Coupon frequency and compounding frequency are assumed to be semian-nual. 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 tosell?

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 instanta-neously by 2%.

4. Conclude.

Solution 5.19 1. The quantity φ of the hedging instrument is obtained as follows:

φ = −11,418,100 × 9.5055

119.792 × 9.8703= −91,793

The investor has to sell 91,793 units of the hedging instrument.


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. Theloss incurred is given by the following formula (exactly −$103,657 if wetake 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 anincrease (and a decrease) of the YTM curve. The P&L of the position isgiven 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 thefollowing 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 bythe 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 alarge move of the YTM curve, we see that the hedge is not perfect because ofthe 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 cheapbond and selling a rich bond. This is the rich and cheap bond strategy. Today, therich and cheap bonds have the following characteristics:

Bond Coupon (%) Maturity (years) YTM (%)Rich 5 10 7.50

Cheap 5.5 12 7.75

Coupon frequency and compounding frequency are assumed to be annual. Facevalue 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 Price MD BPVRich 82.840 7.34 0.06081

Cheap 82.822 8.136 0.06738


We take a long position of $100,000,000 in the 5.5%/12-year bond. Thehedge ratio HR is equal to

HR = 0.06738

0.06081= 1.10818

Then we have to take a short position of x in the 5%/10-year bond, where x

is given by

x = HR × $100,000,000 = $110,818,000


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 from1% 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

(1 + 6%)20= $31.18

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 bythe following formula:

New Price = 31.18 + $ Dur ×(New YTM − 6%)


Using the two-order Taylor expansion, the new price of the bond is given bythe following formula:

New Price = 31.18 + $ Dur ×(New YTM − 6%)

+$Conv

2× (New YTM − 6%)2

We finally obtain the following graphThe 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%.

0

10

20

30

40

50

60

70

80

90

1% 2% 3% 4% 5% 6% 7% 8% 9% 10% 11%

Actual priceOne-order taylor estimationTwo-order taylor estimation

Exercise 6.3 1. Compute the modified duration and convexity of a 6%, 25-year bond selling ata yield of 9%. Coupon frequency and compounding frequency are assumed tobe 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 to10.62, while convexity is equal to 182.92.

2. The estimated percentage price change, for a yield change from 9% to 11% isequal to

−10.62 × (0.02) + 0.5 × (182.92) × (0.02) 2 = −21.24 + 3.66 = −17.58%,

while the actual change is −18.03%.


3. If the yield decreases by 200 basis points, instead, then the estimated pricechange is +21.24% due to duration, and +3.66% due to convexity, that is24.90%; as a whole, the actual price change is 25.46%. The estimated pricechange is no longer symmetric around the current yield because the price func-tion has curvature.

Exercise 6.6 Assume a 2-year Euro-note, with a $100,000 face value, a coupon rate of 10% anda convexity of 4.53. If today’s YTM is 11.5% and term structure is flat. Couponfrequency 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? Whathappens to convexity when interest rates rise? Why?

3. What is the exact price change in dollars if interest rates increase by 10 basispoints (a uniform shift)?

4. Use the duration model to calculate the approximate price change in dollars ifinterest rates increase by 10 basis points.

5. Incorporate convexity to calculate the approximate price change in dollars ifinterest rates increase by 10 basis points.

Solution 6.6 1. Duration

D = 1 × 10,0001.115 + 2 × 10,000

1.1152 + 2 × 100,0001.1152

10,0001.115 + 10,000

1.1152 + 100,0001.1152

= 1 × 10,0001.115 + 2 × 10,000

1.1152 + 2 × 100,0001.1152

97,448.17= 1.908

2. Convexity measures the change in modified duration or the change in the slopeof the price-yield curve. Holding maturity constant, the higher the coupon, thesmaller the duration. Hence, for low duration levels the change in slope (con-vexity) is small. Alternatively, holding coupon constant, the higher the maturity,the higher the duration, and hence, the higher the convexity. When interest ratesrise, 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

P (11.6%) = 10,000

1.116+ 10,000

1.1162+ 100,000

1.1162= $97,281.64

Price has decreased by $166.53 from P (11.5%) = $97,448.17 to $97,281.644. We use

�P � −MD × �y × P (11.5%) = − D

1 + y× �y × P

= −1.908

1.115× 97,448.17 × 0.001 = −$166.754


5. We use

�P � −MD × �y × P + 1

2× RC × (�y)2 × P

= −1.908

1.115× 97,448.17 × 0.001 + 1

2×4.53 × (0.001)2 × 97,448.17 = −$166.531

Hedging error is smaller when we account for convexity.

Exercise 6.8 Modified Duration/Convexity Bond Portfolio HedgeAt date t , the portfolio P to be hedged is a portfolio of Treasury bonds with

various possible maturities. Its characteristics are as follows:

Price YTM MD Convexity$28,296,919 7.511% 5.906 67.578

We consider Treasury bonds as hedging assets, with the following features:

Bond Price ($) Coupon Rate (%) Maturity dateBond 1 108.039 7 3 yearsBond 2 118.786 8 7 yearsBond 3 97.962 5 12 years

Coupon frequency and compounding frequency are assumed to be annual. At datet , we force the hedging portfolio to have the opposite value of the portfolio to behedged.

1. What is the number of hedging instruments necessary to implement a modifiedduration/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 YTM (%) MD ConvexityBond 1 6.831 2.629 9.622Bond 2 7.286 5.267 36.329Bond 3 7.610 8.307 90.212

3. We then are looking for the quantities φ1, φ2 and φ3 of each hedging instrument1, 2, 3 as solutions to the following linear system:

φ1φ2φ3

= 100.445 103.808 79.929

−264.057 −546.791 −663.947966.460 3,771.257 7,210.58

−1 −28,296,919167,143,615

−1,912,260,201

=−279,536

290,043−379,432

Exercise 6.10 Computing the Level, Slope and Curvature $Durations of a Bond Portfolio using

the Nelson and Siegel Extended Model


On 09/02/02, the values of the Nelson and Siegel Extended parameters are asfollows:

β0 β1 β2 τ1 β3 τ2

5.9% −1.6% −0.5% 5 1% 0.5

Recall from Chapter 4 that the continuously compounded zero-coupon rateRc(0, θ) is given by the following formula:

Rc(0, θ) = β0 + β1

1 − exp(− θ

τ1

)θτ1

+ β2

1 − exp(− θ

τ1

)θτ1

− exp

(− θ

τ1

)+ β3

1 − exp(− θ

τ2

)θτ2

− exp

(− θ

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.

Maturity Coupon(years) (%)

Bond 1 3 4Bond 2 7 5Bond 3 15 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, 200units of Bond 2 and 100 units of Bond 3.

3. The parameters of the Nelson and Siegel Extended model change instanta-neously to become

β0 β1 β2 τ1 β3 τ2

5.5% −1% 0.1% 5 2% 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 $durationof the bond portfolio associated with parameter βi .

4. Same questions when the coupon frequency is semiannual.


Solution 6.10 1. We obtain the following zero-coupon yield curve

4.20

4.40

4.60

4.80

5.00

5.20

5.40

5.60

0 5 10 15 20 25 30

Maturity

Zer

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(%)

2. For each bond, the price and level, slope and curvature $durations denotedrespectively by D0, D1, D2 and D3 are given in the following table:

— Price ($) D0 D1 D2 D3

Bond 1 97.61 −281.60 −212.97 −56.49 −47.08Bond 2 99.58 −604.87 −337.83 −171.36 −48.57Bond 3 106.58 −1,109.21 −418.35 −299.10 −51.82

Portfolio 40,333.43 −260,074.82 −130,697.65 −69,831.80 −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:


4.00

4.20

4.40

4.60

4.80

5.00

5.20

5.40

5.60

5.80

6.00

0 5 10 15 20 25 30

Maturity

Zer

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ate

(%)

Curve at the originNew curve

(b) The new price is equal to

New Price = 39,977

whereas the new estimated price obtained by using the equation given in theexercise 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 byNelson–Siegel $durations multiplied by the change in value of the differentparameters.

4. When the coupon frequency is semiannual, we obtain the following results:

— Price ($) D0 D1 D2 D3Bond 1 97.74 −279.25 −211.16 −55.89 −46.69Bond 2 99.94 −598.81 −334.78 −169.50 −48.17Bond 3 107.40 −1,099.32 −415.52 −296.38 −51.54

Portfolio 40,502.52 −257,618.44 −129,623.23 −69,126.35 −19,457.59

Exercise 6.12 Bond Portfolio Hedge using the Nelson–Siegel Extended ModelWe consider the Nelson–Siegel Extended zero-coupon rate function

Rc(t, θ) = β0 + β1

1 − exp(− θ

τ1

)θτ1

+ β2

1 − exp(− θ

τ1

)θτ1

− exp

(− θ

τ1

)+β3

1 − exp(− θ

τ2

)θτ2

− exp

(− θ

τ2

)


where Rc(t, θ) is the continuously compounded zero-coupon rate at date t withmaturity θ .

On 09/02/02, the model is calibrated, parameters being as follows:

β0 β1 β2 τ1 β3 τ2

5.9% −1.6% −0.5% 5 1% 0.5

At the same date, a manager wants to hedge its bond portfolio P against interest-rate risk. The portfolio contains the following Treasury bonds (delivering annualcoupons, with a $100 face value):

Bond Maturity Coupon QuantityBond 1 01/12/05 4 10,000Bond 2 04/12/06 7.75 10,000Bond 3 07/12/07 4 10,000Bond 4 10/12/08 7 10,000Bond 5 03/12/09 5.75 10,000Bond 6 10/12/10 5.5 10,000Bond 7 01/12/12 4 10,000Bond 8 03/12/15 4.75 10,000Bond 9 07/12/20 4.5 10,000

Bond 10 01/12/25 5 10,000Bond 11 07/12/30 4.5 10,000Bond 12 01/12/31 4 10,000Bond 13 07/12/32 5 10,000

We consider Treasury bonds as hedging instruments with the following fea-tures:

Hedging Asset Coupon MaturityHedging Asset 1 4.5 04/15/06Hedging Asset 2 5 12/28/12Hedging Asset 3 6 10/05/15Hedging Asset 4 6 10/10/20Hedging Asset 5 6.5 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 portfolioto 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 instanta-

neously to become


β0 β1 β2 τ1 β3 τ2

6.5% −1% 0.1% 5 2% 0.5

(a) What is the price of the bond portfolio after this change? If the manager hasnot hedged its portfolio, how much money has he lost?

(b) What is the variation in price of the global portfolio (where the global port-folio 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 aregiven in the following table:

Price D0 D1 D2 D3

$12,723,603 −110,075,273 −42,504,124 −25,924,204 −6,021,090

where Di = ∂P∂βi

for i = 0, 1, 2 and 3 are the level, slope and curvature $dura-tions of P in the Nelson–Siegel Extended model.

2. The level, slope and curvature $durations of the five hedging instruments aregiven in the following table:

Hedging Asset Price ($) D0 D1 D2 D3

Hedging Asset 1 100.41 −337.70 −242.65 −75.01 −48.11Hedging Asset 2 101.71 −813.99 −373.12 −235.42 −47.97Hedging Asset 3 111.97 −1013.91 −412.53 −282.76 −51.75Hedging Asset 4 112.05 −1242.56 −424.00 −315.54 −51.84Hedging Asset 5 120.22 −1799.54 −457.82 −354.97 −55.57

3. We are looking for the quantities φ1, φ2, φ3, φ4 and φ5 of each hedging assetas the solutions to the following linear system:

φ1

φ2

φ3

φ4

φ5

=

−337.70 −813.99 −1013.91 −1242.56 −1799.54−242.65 −373.12 −412.53 −424.00 −457.82−75.01 −235.42 −315.54 −354.97 −354.97−48.11 −47.97 −51.75 −51.84 −55.57100.41 101.71 111.97 112.05 120.22

−1

×

108,004,51642,733,84426,093,9546,044,277

−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


(b) The prices of the five hedging assets, after the change in parameters, aregiven in the following table

Hedging Asset Price ($)Hedging Asset 1 96.08Hedging Asset 2 93.10Hedging Asset 3 101.76Hedging Asset 4 100.42Hedging Asset 5 105.62

With the hedging portfolio (which contains the five hedging assets in ade-quate 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 + hedginginstruments) is then equal to −$5,546 ($1,054,624–$1,060,170).

(c) The hedge is efficient.


Exercise 7.1 Would you say it is easier to track a bond index or a stock index. Why or whynot?

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 hugenumber of bonds. Other difficulties include that many of the bonds in the indicesare 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 per-form than statistical replication of stock indices, in the sense that a significantlylower tracking error can usually be achieved for a given number of instrumentsin the replicating portfolio. This is because bonds with different maturities tend toexhibit a fair amount of cross-sectional correlation so that a very limited numberof factors account for a very large fraction of changes in bond returns. Typically, 2or 3 factors (level, slope, curvature) account for more than 80% of these variations.Stocks typically exhibit much more idiosyncratic risk, so that one typically needsto use a large number of factors to account for not much more than 50% of thechanges in stock prices.

Exercise 7.2 What are the pros and cons of popular indexing methodologies in the fixed-incomeuniverse?

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