FINANCE 1. DESS Gestion des Risques/Ingéniérie Mathématique Université d EVRY VAL D ESSONNE EXERCICES CORRIGES. Philippe PRIAULET

FINANCE 1 DESS Gestion des Risques/Ingéniérie Mathématique Université d EVRY VAL D ESSONNE EXERCICES CORRIGES Philippe PRIAULET 1

Exercise 1 On 12/04/01 consider a fixed coupon bond whose features are the following: face value: $1,000 coupon rate: 8% coupon frequency: semi-annual maturity: 05/06/04 What are the future cash-flows delivered by this bond? Solution 2 The coupon cash-flow is equal to $40 Coupon = 8% $1, 000 2 =$40 It is delivered on the following future dates: 05/06/02, 11/06/02, 05/06/03, 11/06/03 and 05/06/04. The redemption value is equal to the face value $1,000 and is delivered on maturity date 05/06/04. Exercise 3 Consider the same bond as in the previous exercise. We are still on 12/04/01. 1- Compute the accrued interest taking into account the Actual/Actual day-count basis. 2- Same question if we are now on 09/06/02. Solution 4 1- The last coupon has been delivered on 11/06/01. There are 28 days between 11/06/01 and 12/04/01, and 181 days between the last coupon date (11/06/01) and the next coupon date (05/06/02). Hence the accrued interest is equal to $6.188 Accrued Interest = 28 $40 = $6.188 181 2- The last coupon has been delivered on 05/06/02. There are 123 days between 05/06/02 and 09/06/02, and 184 days between the last coupon date (05/06/02) and the next coupon date (11/06/02). Hence the accrued interest is equal to $26.739 Accrued Interest = 123 $40 = $26.739 184 2

Exercise 5 An investor has a cash of $10,000,000 at disposal. He wants to invest in a bond with $1,000 nominal value and whose dirty price is equal to 107.457%. 1- What is the number of bonds he will buy? 2- Same question if the nominal value and the dirty price of the bond are respectively $100 and 98.453%. Solution 6 1- The number of bonds he will buy is given by the following formula Number of bonds bought = Here the number of bonds is equal to 9306 2- n is equal to 101,562 n = n = Cash Nomin al Value of the bond dirty price 10, 000, 000 1, 000 107.457% =9306.048 10, 000, 000 100 98.453% = 101562.024 Exercise 7 On 10/25/99 consider a fixed coupon bond whose features are the following: face value: Eur 100 coupon rate: 10% coupon frequency: annual maturity: 04/15/08 Compute the accrued interest taking into account the four different day-count bases: Actual/Actual, Actual/365, Actual/360 and 30/360. Solution 8 The last coupon has been delivered on 04/15/99. There are 193 days between 04/15/99 and 10/25/99, and 366 days between the last coupon date (04/15/99) and the next coupon date (04/15/00). the accrued interest with the Actual/Actual day-count basis is equal to Eur 5.273 193 10% Eur 100 = Eur 5.273 366 3

the accrued interest with the Actual/365 day-count basis is equal to Eur 5.288 193 10% Eur 100 = Eur 5.288 365 the accrued interest with the Actual/360 day-count basis is equal to Eur 5.361 193 10% Eur 100 = Eur 5.361 360 There are 15 days between 04/15/99 and 04/30/99, five months between May and September, and 25 days between 09/30/99 and 10/25/99, so that there are 190 days between 04/15/99 and 10/25/99 on the 30/360 day-count basis 15 + (5 30) + 25 = 190 finally the accrued interest with the 30/360 day-count basis is equal to Eur 5.278 190 10% Eur 100 = Eur 5.278 360 Exercise 9 Treasury bills are quoted using the yield on a discount basis or on a money market basis. 1- The yield on a discount basis denoted y d is computed as y d = F P F B n where F is the face value, P the price, B the year-basis (365 or 360) and n is the number of calendar days remaining to maturity. Prove in this case that the price of the T-bill is obtained using the following equation µ P = F 1 n y d B 4

2- The yield on a money market basis denoted y m is computed as y m = B y d B n y d Prove in this case that the price of the T-bill is obtained using the following equation P = ³ F 1+ n ym B Solution 10 1- From the equation y d = F P F B n we deduce n y d B 1= P F and finally we obtain µ P = F 1 n y d B 2- From the equation y m = B y d B n y d we deduce y m = B F F P B n F F P B n B n = ³ B n F F P 1 F P F Then we have Finally we obtain n y m B = F P F P F P = ³ = F P P F 1+ n ym B = F P 1 Exercise 11 1- What is the yield on a discount basis of a bill whose face value F is 1,000, price P is 975 and n the number of calendar days remaining to maturity is 126? We assume that the year-basis is 360. 2- What is the yield on a money market basis of the same bill? 5

Solution 12 1- The yield y d on a discount basis which satisfies the following equation is equal to 7.143% y d = 1, 000 975 1, 000 360 126 =7.143% 2- The yield y m on a money market basis which satisfies the following equation is equal to 7.326% y m = 360 7.143% 360 126 7.143% =7.326% Exercise 13 Suppose the interest rate is 12% per year compounded continuously. What is the effective annual interest rate? Solution 14 The effective annual interest rate is obtained as R = e 0.12 1=0.1274 = 12.74%. Exercise 15 If you deposit $2,500 in a bank account which earns 8% annually on a continuously compounded basis, what will be the account balance in 7.14 years. Solution 16 The account balance in 7.14 years will be $2500.e 8%.7.14 = $4425.98 Exercise 17 If an investment has a cumulative 63.45% rate of return over 3.78 years, what is the annual continuously compounded rate of return? Solution 18 The annual continuously compounded rate of return R C is such that 1.6345 = e 3.78RC We deduce R C =ln(1.6345)/3.78 = 13%. Exercise 19 How long does it take to double a $100 initial investment when investing at a 5% continuously compounded interest rate? 6

Solution 20 In general, the solution is given by xe RCT =2x or T = ln 2 R C Note that this does not depend on the principle x. In this example, we obtain T = ln 2 =13.86214 years 0.05 Exercise 21 A invests $1000 at 5% per annum continuously compounded. B invests $200 at 20% per annum continuously compounded. Does B ever catch up? How long does it take? Solution 22 B catches up if the difference between them becomes zero, Alternatively if the ratio of their amounts becomes 1. The equality condition can be stated as 1000e 0.05t =200e 0.2t equivalent to ln(5) + 0.05t =0.2t or t = ln(5) 0.15 =10.73 Exercise 23 1- What is the price of a 5-year bond with a nominal value of $100, a yield to maturity of 7%, a 10% coupon rate, and an annual coupon frequency? 2- Same question for a yield to maturity of 8%, 9% and 10%. Conclude Solution 24 1- The price P of a bond is given by the following formula P = nx i=1 N c N (1 + y) n + (1 + y) n 7

which simplifies into P = N c y 1 1 (1 + y) n + N (1 + y) n where N, c, y and n are respectively the nominal value, the coupon rate, the yield to maturity and the number of years to maturity of the bond. Here we obtain for P P = 10 " 1 7% # 1 (1 + 7%) 5 + 100 (1 + 7%) 5 P is then equal to 112.301% of the nominal value or $112.301. Note that we can also use the Excel function Price to obtain P. 2- Prices of the bond for different yields to maturity (YTM) are given in the following table YTM Price 8% $107.985 9% $103.890 10% $100 Bond prices decrease with an increase in rates. Exercise 25 1- What is the price of a 5-year bond with a nominal value of $100, a yield to maturity of 7%, a 10% coupon rate, and semi-annual coupon payments? 2- Same question for a yield to maturity of 8%, 9% and 10%. Solution 26 1- The price P of this bond is given by the following formula which simplifies into P = P = N c y 2nX i=1 " N c/2 (1 + y/2) i + N (1 + y/2) 2n 1 # 1 N (1 + y/2) 2n + (1 + y/2) 2n where N, c, y and n are respectively the nominal value, the coupon rate, the yield to maturity and the number of years to maturity of the bond. 8

Here we obtain for P P = 10 7% 1 1 ³ 1+ 7% 2 10 + 100 ³ 1+ 7% 2 10 P is then equal to 112.475% of the nominal value or $112.475. Note that we can also use the Excel function Price to obtain P. 2- Prices of the bond for different yields to maturity (YTM) are given in the following table YTM Price 8% $108.111 9% $103.956 10% $100 Exercise 27 We consider the following zero-coupon curve Maturity Zero-Coupon Rate 1year 4.00% 2years 4.50% 3years 4.75% 4years 4.90% 5years 5.00% 1- What is the price of a 5-year bond with a $100 face value which delivers a 5% annual coupon rate? 2- What is the yield to maturity of this bond? 3- We suppose that the zero-coupon curve increases instantaneously and uniformly by 0.5%. What is the new price and the new yield to maturity of the bond? What is the impact of this rates increase for the bondholder? 4- We suppose now that the zero-coupon curve will remain stable over time. You hold the bond until maturity. What is the annual return rate of your investment? Why is this rate different from the yield to maturity? 9

Solution 28 1- The price P of the bond is equal to the sum of its discounted cash-flows and given by the following formula P = 5 1+4% + 5 (1 + 4.5%) 2 + 5 (1 + 4.75%) 3 + 5 (1 + 4.9%) 4 + 105 (1 + 5%) 5 = $100.136 2- The yield to maturity R of this bond verifies the following equation 100.136 = 4X i=1 5 (1 + R) i + 105 (1 + R) 5 Using the Excel function yield, we obtain 4.9686% for R. 3-ThenewpricePofthebondisgivenbythefollowingformula P = 5 1+4.5% + 5 (1 + 5%) 2 + 5 (1 + 5.25%) 3 + 5 (1 + 5.4%) 4 + 105 (1 + 5.5%) 5 =$97.999 The new yield to maturity R of this bond verifies the following equation 97.999 = 4X i=1 5 (1 + R) i + 105 (1 + R) 5 Using the Excel function yield, we obtain 5.4682% for R. The impact of this rates increase for the bondholder is an absolute capital loss of $2.137 Absolute Loss = 97.999 100.136 = $2.137 and a relative capital loss of 2.134% Re lative Loss = 2.137 100.136 = 2.134% 4-Before maturity, the bondholder receives intermediate coupons that he reinvests on the market: - after one year, he receives $5 that he reinvests for 4 years at the 4-year zero-coupon rate 10

to obtain at the maturity date of the bond 5 (1 + 4.9%) 4 =$6.0544 - after two years, he receives $5 that he reinvests for 3 years at the 3-year zero-coupon rate to obtain at the maturity date of the bond 5 (1 + 4.75%) 3 =$5.7469 - after three years, he receives $5 that he reinvests for 2 years at the 2-year zero-coupon rate to obtain at the maturity date of the bond 5 (1 + 4.5%) 2 =$5.4601 - after four years, he receives $5 that he reinvests for 1 year at the 1-year zero-coupon rate to obtain at the maturity date of the bond 5 (1 + 4%) = $5.2 - after five years, he receives the final cash-flow equal to $105. The bondholder finally obtains $127.4614 five years later 6.0544 + 5.7469 + 5.4601 + 5.2 + 105 = $127.4614 which corresponds to a 4.944% annual return rate µ 127.4614 1/5 1=4.944% 100.136 This return rate is different from the yield to maturity of this bond (4.9686%) because the curve is not flat at a 4.9686% level. With a flat curve at a 4.9686%, we obtain $127.6108 five years later 6.0703 + 5.7829 + 5.5092 + 5.2484 + 105 = $127.6108 11

which corresponds exactly to a 4.9686% annual return rate. µ 127.6108 1/5 1=4.9686% 100.136 Exercise 29 We consider the three zero-coupon bonds (strips) with the following features Bond Maturity Price Bond 1 1year 96.43 Bond 2 2years 92.47 Bond 3 3years 87.97 Each strip delivers $100 at maturity. 1- Deduce the zero-coupon yield curve from the bond prices 2- We anticipate a rates increase in one year so the prices of strips with residual maturity 1 year, 2 years and three years are respectively 95.89, 90.97 and 84.23. What is the zero-coupon yield curve anticipated in one year? Solution 30 1- The 1-year zero-coupon rate denoted R(0,1) is equal to 3.702% R(0, 1) = 100 96.43 1=3.702% The 2-year zero-coupon rate denoted R(0,2) is equal to 3.702% R(0, 2) = µ 100 1/2 1=3.992% 92.47 The 3-year zero-coupon rate denoted R(0,3) is equal to 3.702% R(0, 2) = µ 100 1/3 1=4.365% 87.97 2- The 1-year, 2-year and 3-year zero-coupon rates become respectively 4.286%, 4.846% and 5.887%. 12

Exercise 31 We consider the following increasing zero-coupon yield curve Maturity R(0,t) Maturity R(0,t) 1year 5.000% 6years 6.550% 2years 5.500% 7years 6.650% 3years 5.900% 8years 6.741% 4years 6.200% 9years 6.830% 5years 6.382% 10 years 6.900% where R(0,t) is the zero-coupon rate at date 0 with maturity t. 1- Compute the par curve. 2- Compute the forward rate curve in one year. 3- Draw the three curves in the same graph. Wwhat can you say about their relative position? Solution 32 1- Recall that the par yield c(n) for maturity n is given by the following formula c(n) = 1 1 (1+R(0,n)) n np i=1 1 (1+R(0,i)) i Using this equation we obtain the following par yields Maturity c(n) Maturity c(n) 1 5.000% 6 6.466% 2 5.487% 7 6.556% 3 5.867% 8 6.636% 4 6.147% 9 6.711% 5 6.315% 10 6.770% 2- Recall that F (0,x,y x), the forward rate as seen from date t =0,startingatdatet = x, andwithresidualmaturityy x is defined as (1 + R(0,y)) y F (0,x,y x) (1 + R(0,x)) x 1 y x 1 13

Using the previous equation,we obtain the forward rate curve in one year Maturity F(0,1,n) Maturity F(0,1,n) 1 6.002% 6 6.928% 2 6.353% 7 6.992% 3 6.603% 8 7.061% 4 6.730% 9 7.113% 5 6.863% 3- The graph of the three curves shows that the forward yield curve is above the zero-coupon yield curve, which is above the par yield curve. This is always the case when the par yield curve is increasing. 7,5% 7,0% 6,5% Yield 6,0% Par Yield Curve 5,5% Zero-Coupon Yield Curve Forward Yield Curve 5,0% 1 2 3 4 5 6 7 8 9 10 Maturity Exercise 33 At date t=0, we get in the market three bonds with the following features Coupon Maturity Price Bond 1 10 2years P0 1 =108.00 Bond 2 7.5 3years P0 2 =100.85 Bond 3 8.5 3years P0 3 =103.50 14

Derive the zero-coupon curve until the five-year maturity. Solution 34 Using the no-arbitrage relation, we obtain the following equations for the five bond prices 108 = 10B(0, 1) + 110B(0, 2) 100.85 = 7.5B(0, 1) + 7.5B(0, 2) + 107.5B(0, 3) 103.5 =9B(0, 1) + 9B(0, 2) + 109B(0, 3) which can be expressed in a matrix form 108 100.85 103.5 = 10 110 7.5 7.5 107.5 8.5 8.5 108.5 B(0, 1) B(0, 2) B(0, 3) Then we get the following discount factors B(0, 1) B(0, 2) B(0, 3) = 0.944275 0.895975 0.80975 we deduce the zero-coupon rates R(0, 1) = 5.901% R(0, 2) = 5.646% R(0, 3) = 7.288% Exercise 35 Suppose we know from market prices the following zero-coupon rates with matu- 15

rities inferior or equal to one year: Maturity Zero-Coupon Rate 1Day 3.20% 1Month 3.30% 2Months 3.40% 3Months 3.50% 6Months 3.60% 9Months 3.80% 1Year 4.00% Now we consider bonds priced by the market until the 4-year maturity: Maturity Coupon Gross Price 1Yearand3Months 4% 102.8 1Yearand6Months 4.5% 102.5 2Years 3.5% 98.3 3Years 4% 98.7 4Years 5% 101.6 1- Using the bootstrapping method, compute the zero-coupon rates for the following maturities1yearand3months,1yearand6months,2years,3yearsand4years. 2- Draw the zero-coupon yield curve using a linear interpolation Solution 36 1- We first extract the one-year-and-three-month maturity zero-coupon rate. In the absence of arbitrage opportunities, the price of this bond is the sum of its future discounted cash-flows: 102.8 = 4 (1 + 3.5%) 1/4 + 104 (1 + x) 1+1/4 where x is the one-year-and-three-month maturity zero-coupon rate to be determined. Solving this equation (for example with the Excel solver) we obtain 4.16% for x. Applying the same procedure with the one-year-and-six-month maturity and the two-year maturity bonds we obtain respectively 4.32% and 4.41% for x. Next we have to extract the 3-year maturity zero-coupon 16

rate solving the following equation 98.7 = 4 (1 + 4%) + 4 (1 + 4.41%) 2 + 104 (1 + y%) 3 y is equal to 4.48% and finally we extract the 4-year maturity zero-coupon rate denoted z by solving the following equation z is equal to 4.57%. 101.6 = 5 (1 + 4%) + 5 (1 + 4.41%) 2 + 5 (1 + 4.48%) 3 + 105 (1 + z%) 4 2- Using the linear graph option in Excel we draw the zero-coupon yield curve 4,80% 4,60% 4,40% 4,20% Zero-Coupon Rate 4,00% 3,80% 3,60% 3,40% 3,20% 3,00% 0 1 2 3 4 Maturity Exercise 37 From the prices of zero-coupon bonds quoted in the market, we obtain the following 17

zero-coupon curve Maturity Zero-Coupon Rate R(0,t) Discount Factor B(0,t) 1year 5.000% 0.95238 2years 5.500% 0.89845 3years 5.900% 0.84200 4years 6.200% 0.78614 5years?? 6years 6.550% 0.68341 7years 6.650% 0.63720 8years?? 9years 6.830% 0.55177 10 years 6.900% 0.51312 where R(0,t) is the zero-coupon rate at date 0 for maturity t, and B(0,t) is the discount factor at date 0 for maturity t. We need to know the value for the 5-year and the 8-year zero-coupon rates. We have to estimate them, and test four different methods. 1- We use a linear interpolation with the zero-coupon rates. Deduce R(0,5), R(0,8) and the corresponding values for B(0,5) and B(0,8). 2- We use a linear interpolation with the discount factors. Deduce B(0,5), B(0,8) and the corresponding values for R(0,5) and R(0,8). 3- We postulate the following form for the zero-coupon rate function R _ (0,t): _ R (0,t)=a + bt + ct 2 + dt 3 Estimate the coefficients a, b, c and d which best approximate the given zero-coupon rates using the following optimization program Min a,b,c,d X i ³ R(0,i) R _ 2 (0,i) 18

where R(0,i) are the zero-coupon rates given by the market. Deduce the value for R(0, 5) = R _ (0, 5), R(0, 8) = R _ (0, 8), and the corresponding values for B(0,5) and B(0,8). 4- We postulate the following form for the discount factor function B _ (0,t): _ B (0,t)=a + bt + ct 2 + dt 3 Estimate the coefficients a, b, c and d which best approximate the given discount factors using the following optimization program Min a,b,c,d X i ³ B(0,i) B _ 2 (0,i) where B(0,i) are the discount factors given by the market. Deduce the value for B(0, 5) = _ B (0, 5), B(0, 8) = _ B (0, 8), and the corresponding values for R(0,5) and R(0,8). 5- Conclude Solution 38 1- Consider that we know R(0,x) and R(0,z), respectively, the x-year and the z-year maturity zero-coupon rates and that we need R(0,y) the y-years maturity zero-coupon rate with y [x; z]. Using the linear interpolation, R(0,y) is given by the following formula R(0,y)= (z y)r(0,x)+(y x)r(0,z) z x From this equation, we deduce the value for R(0,5) and R(0,8) R(0, 5) = R(0, 8) = (6 5)R(0, 4) + (5 4)R(0, 6) 6 4 (9 8)R(0, 7) + (8 7)R(0, 9) 9 7 = = R(0, 4) + R(0, 6) 2 R(0, 7) + R(0, 9) 2 =6.375% =6.740% Using the standard following equation which lies the zero-coupon rate R(0,t) and the discount factor B(0,t) B(0,t)= 1 (1 + R(0,t)) t 19

we obtain 0.73418 for B(0,5) and 0.59345 for B(0,8). 2- We use the same formula as in question 1 but adapted to discount factors B(0,y)= (z y)b(0,x)+(y x)b(0,z) z x we obtain 0.73478 for B(0,5) and 0.59449 for B(0,8). Using the standard following equation which lies the zero-coupon rate R(0,t) and the discount factor B(0,t) R(0,t)= µ 1 1/t 1 B(0,t) we obtain 6.358% for R(0,5) and 6.717% for R(0,8). 3- Using the Excel function DroiteReg, we obtain the following values for the parameters Parameters Value a 0.04351367 b 0.00720757 c -0.000776521 d 3.11234E-05 which provides us with the following values for the zero-coupon rates and associated discount factors Maturity R(0,t) _ R (0,t) B(0,t) _ B (0,t) 1 5.000% 4.998% 0.95238 0.95240 2 5.500% 5.507% 0.89845 0.89833 3 5.900% 5.899% 0.84200 0.84203 4 6.200% 6.191% 0.78614 0.78641 5? 6.403%? 0.73322 6 6.550% 6.553% 0.68341 0.68330 7 6.650% 6.659% 0.63720 0.63681 8? 6.741%? 0.59339 9 6.830% 6.817% 0.55177 0.55237 10 6.900% 6.906% 0.51312 0.51283 20

4- We first note that there is a constraint in the minimization because we must have B(0, 0) = 1 So the value for a is necessarily equal to 1. Using the Excel function DroiteReg, we obtain the following values for the parameters Parameters Value a 1 b -0.04945479 c -0.001445358 d 0.000153698 which provides us with the following values for the discount factors and associated zero-coupon rates Maturity B(0,t) _ B (0,t) R(0,t) _ R (0,t) 1 0.95238 0.94925 5.000% 5.346% 2 0.89845 0.89654 5.500% 5.613% 3 0.84200 0.84278 5.900% 5.867% 4 0.78614 0.78889 6.200% 6.107% 5? 0.73580? 6.328% 6 0.68341 0.68444 6.550% 6.523% 7 0.63720 0.63571 6.650% 6.686% 8? 0.59055? 6.805% 9 0.55177 0.54988 6.830% 6.871% 10 0.51312 0.51461 6.900% 6.869% 5- The table below provides the results obtained using the four different methods of interpo- 21

lation and minimization Rates Interpol. DF Interpol. Rates Min. DF Min. R(0,5) 6.375% 6.358% 6.403% 6.328% R(0,8) 6.740% 6.717% 6.741% 6.805% B(0,5) 0.73418 0.73478 0.73322 0.73580 B(0,8) 0.59345 0.59449 0.59339 0.59055 Rates Interpol. is for interpolation on rates (question 1). DF Interpol. is for interpolation on discount factors (question 2). Rates Min is for minimization with rates (question 3). DF Min. is for minimization with discount factors (question 4). The table shows that results are quite similar according to the two methods based on rates. Differences appear when we compare the four methods. In particular, we can obtain a spread of 7.5 bps for the estimation of R(0,5) between Rates Min. and DF Min., and a spread of 8.8 bps for the estimation of R(0,8) between the two methods based on discount factors. We conclude that the zero-coupon rates and discount factors estimations are sensitive to the method of interpolation or minimization used. Exercise 39 We want to derive the current zero-coupon yield curve for maturities inferior to 10 years. For that goal, we use a basket of bonds quoted by the market and a discount function modelled as a three-order polynomial spline. The features of the bonds used to derive this curve 22

are summarized in the following table Bond Coupon Rate Maturity Market Price Bond 1 0% 7/365 year 99.92 Bond 2 0% 1/12 year 99.65 Bond 3 0% 0.25 year 98.92 Bond 4 0% 0.5 year 97.77 Bond 5 5% 1year 100.02 Bond 6 6% 2years 101.56 Bond 7 5% 2.5 years 101.72 Bond 8 7% 3.25 years 109.72 Bond 9 8% 4years 108.65 Bond 10 5% 4.5 years 100.26 Bond 11 7% 5.5 years 109.89 Bond 12 7% 7years 107.55 Bond 13 6% 8.75 years 102.75 Bond 14 7% 10 years 108.21 The coupon frequency of these bonds is annual, and the face value is Eur 100. We model the discount function B(0,s) as a standard polynomial spline with two splines B 0 (s) =1+c 0 s + b 0 s 2 + a 0 s 3 for s [0, 3] B (0,s)= B 10 (s) =1+c 0 s + b 0 s 2 + a 0 hs 3 (s 3) 3i + a 1 (s 3) 3 for s [3, 10] 1- Write the theoretical price of bond 1. 2- Calculate the coefficients behind each parameter and the constant number for bond 1 3- Do the same job for bond 2 to bond 14. 4- Estimate the coefficients c 0,b 0,a 0 and a 1 which best approximate the market prices of given bonds using the following optimization program Min c 0,b 0,a 0,a 1 14X i=1 ³ P i P _ 2 i 23

where P i are the market prices and _ P ithe theoretical prices. We suppose that residual are homoscedastic so that each bond has the same weight in the minimization program. 5- Calculate P 14 i=1 ³P i P _ 2 i 6- For each bond, calculate the spread between the market price and the theoretical price. 7- Draw the graph of the zero-coupon yield curve. 8- Draw the graph of the forward yield curve in one, two and three months. Solution 40 1- The theoretical price _ P 1ofbond1is µ " _ 7 P i= 100.B 0, =100. 1+c 0. 365 7 365 + b 0. µ 7 2 µ # 7 3 + a 0. 365 365 2- From question 1, we deduce the coefficients behind each parameter and the constant number for bond 1 constant number c 0 b 0 a 0 100 700 365 100. ³ 7 365 2 100. ³ 7 365 3 3- For each bond, we obtain the coefficients behind each parameter and the constant number 24

in the following table Bond constant number c 0 b 0 a 0 a 1 Bond 1 100 1.9178 0.0368 0.0007 0 Bond 2 100 8.3333 0.6944 0.0579 0 Bond 3 100 25 6.25 1.5625 0 Bond 4 100 50 25 12.5 0 Bond 5 105 105 105 105 0 Bond 6 112 218 430 854 0 Bond 7 115 272.5 668.75 1658.125 0 Bond 8 128 374 1177 3764.9531 1.6719 Bond 9 132 480 1840 7092 108 Bond 10 125 512.5 2231.25 9523.125 355 Bond 11 142 676 3525.5 17177.625 1696.375 Bond 12 149 896 5880 32688 7100 Bond 13 154 1131.5 9234.625 56638.9688 21270.5 Bond 14 170 1385 12695 81387 39788 Note that Z, what we call the coefficients matrix; is the matrix with dimension 14 4 which is represented by the four last columns of the table. We give below the details for Bond 8: constant number n n =7 4 + 100 = 128 coefficient behind c 0 7 [(3.25 3) + (3.25 2) + (3.25 1)] + 107 3.25 = 374 coefficient behind b 0 h 7 (3.25 3) 2 +(3.25 2) 2 +(3.25 1) 2i +107 3.25 2 = 1177 25

coefficient behind a 0 7 h (3.25 3) 3 +(3.25 2) 3 +(3.25 1) 3i +107 h 3.25 3 (3.25 3) 3i = 3764.9531 coefficient behind a 1 107 (3.25 3) 3 =1.6719 4- β =(c 0,b 0,a 0,a 1 ) T, the vector of parameters is the solution of the standard OLS (Ordinary Least Squared) procedure. It is the result of the following matricial calculation β = ³ Z T Z 1 Z T P where Z T is the transposed matrix of Z, X 1 is the inverse matrix of X and P is the following vector resulting for each bond of the difference between the constant number and the market price of the bond. P = 99.92 100 99.65 100 98.92 100 97.77 100 100.02 105 101.56 112 101.72 115 109.72 128 108.65 132 100.26 125 109.89 142 107.55 149 102.75 154 108.21 170 = 0.08 0.35 1.08 2.23 4.98 10.44 13.28 18.28 23.35 24.74 32.11 41.45 51.25 61.79 26

We finally obtain for β β T = 0.043700313 0.003444022 0.000566232 9.21747E 05 β. Note that we can also use the Excel function Droitereg to obtain the vector of parameters 5- The sum of squared spreads P 14 i=1 ³P i P _ i 2is equal to 0.043885606. 6- In the following table we examine for each bond the spread between the market price and the theoretical price Bond Market Price Theoretical Price Spread Bond 1 99.92 99.9161 0.0039 Bond 2 99.65 99.6335 0.0165 Bond 3 98.92 98.8869 0.0331 Bond 4 97.77 97.7360 0.0340 Bond 5 100.02 100.1093-0.0893 Bond 6 101.56 101.4760 0.0840 Bond 7 101.72 101.7274-0.0074 Bond 8 109.72 109.7342-0.0142 Bond 9 108.65 108.6926-0.0426 Bond 10 100.26 100.2787-0.0187 Bond 11 109.89 109.8869 0.0031 Bond 12 107.55 107.4482 0.1018 Bond 13 102.75 102.8591-0.1091 Bond 14 108.21 108.1697 0.0403 27

7- We draw below the graph of the zero-coupon yield curve 6,25% 5,75% Zero-Coupon Rate 5,25% 4,75% 4,25% 0 1 2 3 4 5 6 7 8 9 10 Maturity We draw below the graph of the three forward yields curves beginining in one, two and three months. Recall first that F (0,x,y x), the forward rate as seen from date t =0,startingat date t = x, and with residual maturity y x is defined as (1 + R(0,y)) y F (0,x,y x) (1 + R(0,x)) x 1 y x 1 28

We give successively the values 1/12, 2/12 and 3/12 to x. 6,10% 5,90% 5,70% Forward Yield 5,50% 5,30% 5,10% 4,90% Forward Yield Curve in 3 Months Forward Yield Curve in 2 Months Forward Yield Curve in 1 Month 4,70% 4,50% 0 20 40 60 80 100 120 Maturity in Months Because the zero-coupon curve is increasing, the forward yield curve in three months is above the forward yield curve in two months, which is above the forward yield curve in one month. Exercise 41 We consider three bonds with the following features: Bond Maturity Coupon Rate YTM Bond 1 2years 5% 5% Bond 2 10 years 6% 5.5% Bond 3 30 years 7% 6% YTM is for yield to maturity. Coupon frequency is annual. 1- Compute the dirty price and the modified duration of each bond 2- a) The YTM of each of these bonds decreases instantaneously by 0.2%. For each bond, compute the new exact price given by discounting its future cash-flows, the price approximation given using the first order Taylor expansion, and the difference between these two prices. b) Same question if the YTM of each of these bonds decreases instantaneously by 1%. Conclude 29

c) For bond 3, draw the difference between the two prices (the new exact price given by discounting its future cash-flowsandthepricegivenusingthefirst order Taylor expansion) depending on the YTM change. 3- Compute the convexity of each bond 4- We suppose that the YTM of each of these bonds decreases instantaneously by 1%. Compute the price approximation given using the second order Taylor expansion. Compare it to the exact price given by discounting its future cash-flows. Solution 42 1-Thedirtypriceandthemodified duration of each of these bonds is given in the following table: Bond Price Modified Duration Bond 1 100 1.859 Bond 2 103.769 7.438 Bond 3 113.765 13.394 2- a) When the YTM of each of these bonds decreases by 0.2%, we obtain the following results Bond New Exact Price FOTE Price Spread Bond 1 100.373 100.372 0.001 Bond 2 105.327 105.312 0.015 Bond 3 116.877 116.812 0.065 Recall that the FOTE price is given by the following formula FOTE Pr ice = P + MD y P where P is the original price, MD the modified duration and y, the YTM change. b) When the YTM of each of these bonds decreases by 1%, we obtain the following results Bond New Exact Price FOTE Price Spread Bond 1 101.886 101.859 0.027 Bond 2 111.869 111.487 0.382 Bond 3 130.745 129.003 1.742 30

When the YTM change is high, the spread between the two prices is not negligible. We have to use the second order approximation. c) For bond 3, we draw below the difference between the new exact price given by discounting its future cash-flows and the price given using the first order Taylor expansion depending on the YTM change. 20 Difference between the Two Prices 15 10 5 0-5 3.0% 3.3% 3.6% 3.9% 4.2% 4.5% 4.8% 5.1% 5.4% 5.7% 6.0% YTM Level 3- The convexity of each of these bonds is given in the following table 6.3% 6.6% 6.9% 7.2% 7.5% 7.8% 8.1% 8.4% 8.7% 9.0% Bond Convexity Bond 1 5.27 Bond 2 70.95 Bond 3 280.97 4- Recall that the SOTE (second order Taylor expansion) price is given by the following formula SOTE Pr ice = P + MD P y + RC P ( y) 2 /2 where RC is the (relative) convexity. 31

When the YTM of each of these bonds decreases by 1%, we obtain the following results Bond Exact Price SOTE Price Spread Bond 1 101.886 101.886 0.000 Bond 2 111.869 111.855 0.014 Bond 3 130.745 130.601 0.144 Exercise 43 Today is 1/1/98. On 6/30/99 we will have to make a payment of $100. We can only invest in a riskfree pure discount bond (nominal $100) that matures on 12/31/98 and in a riskfree 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 44 We first have to compute the present value PV of the debt, which is the amount we will have to deposit PV = 100 1.5 =90.35 (1.07) We also compute the price P 1 of the one-year pure discount bond P 1 = 100 1.07 =93.46 Similarly, the price P 3 of the three year coupon bond is P 3 = 8 1.07 + 8 (1.07) 2 + 108 (1.07) 3 =102.6 The duration of the one-year pure discount bond is obviously one. three-year coupon bond is The duration D 3 of the D 3 =1 8 1.07 102.6 +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 one-year and the tree-year bonds (q 1 and q 3 respectively), so as to achieve a dollar duration equal to that of debt, and also a present value of the portfolio equal to that of debt. We know that the duration of the debt we are trying to 32

immnunize is 1.5, therefore q 1 and q 3 are given as the unique solution to the following system of equations 93.46 1 q 1 +102.6 2.786 q 3 =90.35 1.5 93.46 q 1 +102.6 q 3 =90.35 q 1 =0.696117 = q 3 =0.2465 In terms of dollar amount 93.46 q 1 =$65.0591invested in 1 year maturity bond and 102.6 q 3 = 25.2939 invested in 3 years maturity bond. Exercise 45 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 Coupon Rate YTM Duration Price 18 Years 9.5% 8% 9.5055 $114,181 Characteristics of the hedging instrument which is here a bond are the following: Maturity Coupon Rate YTM Duration Price 20 Years 10% 8% 9.8703 $119.792 Coupon frequency is semi-annual. YTM is for yield to maturity. The YTM curve is flat at a 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 46 1- The quantity φ of the hedging instrument is obtained by resolving the following equation (??) 11, 418, 100 9.5055 φ = = 91, 793 119.792 9.8703 The investor has to sell 91,793 units of the hedging instrument. 33

2- Prices of bonds with maturity 18 years and 20 years becomes 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 becomes 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 which is no more negligible. Exercise 47 Same exercise as the previous one except that we now consider a non flat YTM curve. The YTM of the bond with maturity 18 years is 7.5%, and the YTM of the bond with maturity 20 years is 8%. Solution 48 We compute the modified duration and the price of the bond to hedge. Maturity Coupon Rate YTM Modified Duration Price 18 Years 9.5% 7.5% 9.6928 $119.581 34

The modified duration of the hedging instrument is 9.4906. 1- The quantity φ of the hedging instrument is obtained by resolving the following equation (??) 11, 958, 100 9.6928 φ = = 98, 264 119.792 9.4906 The investor has to sell 98,264 units of the hedging instrument. With a non flat curve, note that we use the modified duration instead of the duration in the previous equation. 2- Prices of bonds with maturity 18 years and 20 years becomes respectively $118.471 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 Loss =$100, 000 (118.471 119.581) = $111, 000 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 = $111, 000 + $98, 264 (119.792 118.664) = $158 3- Prices of bonds with maturity 18 years and 20 years becomes respectively $100 and $100. a) If the bond portfolio has not been hedged, the loss incurred is given by the following formula Loss = $100, 000 (100 119.581) = $1, 958, 100 b) If the bond portfolio has been hedged, the P&L of the position is given by the following formula P &L = $1, 958, 100 + $98, 264 (119.792 100) = $13, 258 Exercise 49 Riding the Yield Curve At date t =0, we consider the following zero-coupon curve for the short-term segment which 35

is upward sloping. Maturity Zero-Coupon Rate Zero-Coupon Bond Price 92 days 4.00% $99.016 183 days 4.50% $97.817 274 days 4.80% $96.542 An investor has funds to invest for six months. He has two different opportunities: buying a 183-days T-Bill and holding it until the maturity. or riding down the yield curve by buying a 274-day T-bill and selling it six months after. 1- Calculate the total return rate of these two strategies assuming that the zero-coupon curve remains stable. 2- If the zero-coupon yield curve rises by 2% after date t=0 and stay stable at this level, whatwouldbethetotalreturnrateofridingdowntheyieldcurve? Solution 50 1- From the table above, the price of the 183-days T-Bill is 99.016$ so that buying and holding it will provide a total return rate of 100 99.016 99.016 =2.231% From the table above, the price of the 274-day T-Bill is $96.542. The 274-day T-Bill becomes a 92-day T-Bill at our horizon date. If we assume no change in the zero-coupon yield curve, ridingdowntheyieldcurvewouldbringatotalreturnrateof 99.016 96.542 96.542 =2.563% that is to say a 0.332% surplus of total return rate compared to the buy and hold strategy. 2- But if the zero-coupon yield curve rise by 2%, the price of the 92-day T-Bill is $98.542 so riding down the yield curve only brings a total return rate of 98.542 96.542 96.542 =2.072% In this case riding down the yield curve is worse than the buy-and-hold strategy. 36

Exercise 51 Rollover Strategy An investor has funds to invest for one year. He anticipates an 1% increase of the curve in six months. Six-month and one-year zero-coupon rates are respectively 3% and 3.2%. He has two different opportunities: he can buy the 1-year zero-coupon T-bond and hold it until maturity. or he can applicate the rollover strategy by buying the 6-month T-bill, holding it until maturity, and buying a new six-month T-bill in six month and holding it until maturity 1- Calculate the annualized total return rate of these two strategies assuming that the investor s anticipation is correct. 2- Same question when rates decrease by 1% in six months? Solution 52 1- The annualized total return rate of the first strategy is of course 3.2% 100 100 (1+3.2%) 100 (1+3.2%) = 100 96.899 96.899 =3.2% By doing a rollover, the investor will invest at date t =0$98.533 to obtain $100 six months later. Note that $98.533 is obtained as follows $100 (1 + 3%) 6 12 =$98.533 Six months later he will then buy a quantity α =100/98.058 of a six-month T-Bill which pays $100 at maturity so that the annualized total return rate of the second strategy is 3.5% 100 α 98.533 98.533 =3.5% 2- The annualized total return rate of the first strategy is still 3.2% as it is only 2.5% for the rollover strategy where β =100/99.015. 100 β 98.533 98.533 =2.5% In this case the rollover strategy is worse than the simple buy-and-hold strategy. Exercise 53 Bond versus Strips Arbitrage 37

On 05/15/02, we suppose that the price of strips with maturity 05/15/03, 05/15/04, 05/15/05 and 05/15/06 are respectively 96.05, 91.23, 86.5 and 81.1. The principal amount of strips is $100. At the same time the price of the bond with maturity 05/15/06, coupon rate 5% and principal amount $1,000 is 98.75. 1- Compute the price of the reconstructed bond 2- A trader wants to buy 10,000 bonds. Is there an arbitrage he can benefit? Solution 54 1- The price of the reconstructed bond using strips is 98.844 5 0.9605 + 5 0.9123 + 5 0.865 + 105 0.811 = 98.844 2- For the trader, the arbitrage consists in buying the bond and selling the strips. Then he buys a quantity of 10,000 bonds, sells a quantity of 5,000 strips with maturity 05/15/03, 5,000 strips with maturity 05/15/04, 5,000 strips with maturity 05/15/05 and 105,000 strips with maturity 05/15/06. The gain of the trader is $1, 000 10, 000 (98.844 98.75) % = $9, 400 Exercise 55 Butterfly We consider three bonds with short, medium and long maturities whose features are summarized in the following table Maturity Coupon Rate YTM Bond Price $Duration Quantity 2Years 6% 6% 100 183.34 q s 10 Years 6% 6% 100 736.01 1, 000 30 Years 6% 6% 100 1, 376.48 q l Face value of bonds is $100, YTM stands for yield-to-maturity, bond prices are dirty prices and we assume a flat yield-to-maturity curve in the exercise. We structure a butterfly inthe following way: we sell 10, 000 10-year maturity bonds we buy q s 2-year maturity bonds and q l 30-year maturity bonds 38

1- Determine the quantities q s and q l so that the butterfly is cash and $duration neutral. 2- What is the P&L of the butterfly if the yield to maturity curve still flat goes up to a 7% level? And down to a 5% level? 3- Draw the P&L of the butterfly depending on the value of the yield to maturity Solution 56 1- The quantities q s and q l, which are determined so that the butterfly iscashand $duration neutral, satisfy the following system (q s 183.34) + (q l 1, 376.48) = 10, 000 736.01 (q s 100) + (q l 100) = 10, 000 100 which solution is q s q l = 183.34 1, 376.48 100 100 1 7, 360, 100 1, 000, 000 = 5, 368 4, 632 2- If the yield to maturity curve goes up to a 7% level or goes down to a 5% level, bond prices become Maturity Price if YTM = 7% Price if YTM = 5% 2Years 98.19 101.86 10 Years 92.98 107.72 30 Years 87.59 115.37 P&Ls are respectively $3,051 and $3,969 when the yield to maturity curve goes up to a 7% level or goes down to a 5% level. 3- We draw below the profile of the P&L s butterfly depending on the value of the yield to 39

maturity 100000 90000 80000 70000 Total Return in $ 60000 50000 40000 30000 20000 10000 0 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 Yield to Maturity The butterfly has a positive convexity. Whatever the value of the yield to maturity the strategy always generates a gain. This gain is all the more substantial as the yield-to-maturity reaches a level further away from 6%. Exercise 57 We consider two firms A and B which have the same financial needs in terms of maturity and principal. The two firms can borrow money in the market at the following conditions: Firm A: 11% at a fixed rate or Libor + 2% for a $10 million loan and a five-year maturity. Firm B: 9% at a fixed rate or Libor + 0.25% for a $10 million loan and a five-year maturity. 1- We suppose that firm B prefers a floating rate debt as firm A prefers a fixed rate debt. What is the swap they will structure to optimize their financial conditions? 2- If firm B prefers a fixed rate debt as firm A prefers a floating rate debt, is there a swap to structure so that the two firms optimize their financial conditions? Conclude Solution 58 1- Firm B has 2% better conditions at a fixed rate and 1.75% better conditions at a floating rate than firm A. The spread between the conditions obtained by firm A and firm B at 40

a fixedrateandthespreadbetweentheconditionsobtainedbyfirm A and firm B at a floating rate is different from 0.25%. To optimize their financial conditions: Firm B borrows money at a 9% fixed rate, firm A contracts a loan at a Libor + 2% floating rate and they structure the following swap. Firm B pays Libor + 0.75% and receives the fixed 9% as firm A receives Libor + 0.75% and pays the fixed 9%. The financing operation is summarized in the following table Firm A Firm B Initial Financing (Libor + 2%) (9%) Swap A to B (9%) 9% Swap B to A Libor + 0.125% (Libor + 0.125%) Financing Cost (10.875%) (Libor + 0.125%) Financing Cost without Swap (11%) (Libor + 0.25%) Gain 0.125% 0.125% By structuring a swap, firm A and firm B have optimized their financial conditions and each firm has gained 0.125%. 2- There is no swap to structure between the two firms so that they ameliorate their financial conditions. Exercise 59 We consider at date T 0, a 6-month LIBOR standard swap contract with maturity 6 years with the following cash-flow schedule F F F F F F T 0 T 1 T 2 T 3 T 4 T 5 T 6 T 7 T 8 T 9 T 10 T 11 T 12 V 1 V 2 V 3 V 4 V 5 V 6 V 7 V 8 V 9 V 10 V 11 V 12 Note that T i+1 T i =6months, i {0, 1, 2,..., 11}. Wesupposethattheswapfacevalueis $1 million, and that the rate F of the fixed leg is 6%. At date T 0, zero-coupon rates and discount factors for maturities T 1,T 2,...T 12 are given in 41

the following table Maturity ZC rates Maturity ZC rates T 1 4.005% T 7 5.785% T 2 4.575% T 8 5.896% T 3 4.925% T 9 6.001% T 4 5.134% T 10 6.069% T 5 5.412% T 11 6.121% T 6 5.599% T 12 6.148% 1- What is the pricing formula for this plain vanilla swap using the zero-coupon method? 2- Compute the discount factors for maturities T 1,T 2,...T 12. 3- Give the price of this swap? 4- What is the swap rate such that the price of this swap is zero? 5- An investor has bought 100,000 5-year bond with a 7.2% annual coupon rate and a nominal amount of $1,000. What is the price, the yield to maturity and the modified duration of this bond. 6- The investor fears a rates increase. How many swaps must he sell to protect its bond portfolio? 7- The yield to maturity curve increases uniformly by 0.3%. What is his new position with and without the hedge? Solution 60 1- The price of this swap at date T 0 is given by SWAP T0 =$1, 000, 000 Ã i=6! X F.B(T 0,T 2i ) 1+B(T 0,T 12 ) 2- Discount factors are given in the following table using the following formula B(0,t)= i=1 1 (1 + R(0,t)) t where, at date 0, B(0,t) is the discount factor for maturity t, and R(0,t) the zero-coupon rate 42

for maturity t. Maturity ZC rates Discount Factors T 1 4.005% 0.98056 T 2 4.575% 0.95625 T 3 4.925% 0.93043 T 4 5.134% 0.90472 T 5 5.412% 0.87655 T 6 5.599% 0.84922 T 7 5.785% 0.82133 T 8 5.896% 0.79521 T 9 6.001% 0.76932 T 10 6.069% 0.74483 T 11 6.121% 0.72126 T 12 6.148% 0.69908 3-ThepriceoftheswapatdateT 0 is simply $3957.5. SWAP T0 =$1, 000, 000 6% 0.95625 + 0.90472 + 0.84922 +0.79521 + 0.74483 + 0.69908 1+0.69908 = $3, 957.5 4- In the previous formula, we have to replace 6% with 6.07996% to obtain a swap price equal to zero. 5- The price P of the bond is obtained by discounting its future cash-flows according to P =7.2 (0.95625 + 0.90472 + 0.84922 + 0.79521) + 107.2 0.74483 = 105.0847 The yield to maturity R verifies the following formula 105.0847 = 4X i=1 7 (1 + R) i + 107 (1 + R) 5 Using for example the Excel function yield, we obtain 5.9931% for R. The modified duration of this bond is given again by using the Excel function MDuration. 43

We obtain 4.14094. 6- The solution of the third question shows that the swap is the sum of two assets: a 6-year bond with a 6% annual coupon rate and $1,000,000 face value minus a $1,000,000 cash value. The price of the bond contained in the swap is given by the following formula 6 0.95625 + 0.90472 + 0.84922 + 0.79521 + 0.74483) + 106 0.69908 = 99.6043 In the same way as in question 5, we deduce the yield to maturity and the modified duration of this bond which are respectively equal to 6.08068% and 4.9119. Assuming that the yield to maturity move of the bond to hedge is equal to the yield to maturity of the bond contained in the swap, we deduce the number n of swaps that the investor has to sell according to the following equation Q B N B $Duration B = n N S $Duration S where Q B is the quantity of bonds bought by the investor, N B the nominal amount of the bond, N S the nominal amount of the swap. $Duration B is the $Duration of the bond to hedge as $Duration S is the $Duration of the bond contained in the swap. We finally obtain n = 100, 000 1, 000 105.0847 4.14094 1, 000, 000 99.6043 4.9119 =88.94 so that the investor has to sell 89 swaps. 7- The new price of the bond to hedge is 103.78982. the new price of the bond contained in the swap is 98.1502 so that the new swap price is SWAP =$1, 000, 000 (0.981502 1) = $18, 498 Without the hedge, the investor s loss is Loss =100, 000 $1, 000 (103.78982 105.0847) % = $1, 294, 880 44

With the hedge, the investor loses -$1,294,880 in the bond market and gains $1,294,105 in the swap market Gain =89 ( 3, 957.5 18, 498) = $1, 294, 105 so that the net position is very near from zero (exactly -$775). Exercise 61 Deriving the current interbank zero-coupon yield curve with the Nelson and Siegel model we obtain the following parameter values β 0 β 1 β 2 τ 6.9% 3.5% 1% 3 The goal of this exercise is to obtain the Nelson and Siegel level, slope and curvature $Durations of some plain vanilla swaps, and to hedge a bond portfolio against a change of these parameters. 1- Compute the price and the Nelson and Siegel level, slope and curvature $durations of the three following swaps: a- the 6-month Libor plain vanilla swap with a nominal amount of $1,000,000, a maturity of 2 years, semi-annual payments on the fixedlegandafixed rate equal to 4%. b- the 6-month Libor plain vanilla swap with a nominal amount of $1,000,000, a maturity of 5 years, semi-annual payments on the fixedlegandafixed rate equal to 5%. c- the 6-month Libor plain vanilla swap with a nominal amount of $1,000,000, a maturity of 10 years, semi-annual payments on the fixedlegandafixed rate equal to 5.50%. 2-Aninvestorwantstohedgeabondportfoliowhoselevel,slopeandcurvature$durations are respectively -789,456,145, -142,256,548 and -97,897,254. He decides to use the three swaps to hedge his position. How many of these three swaps must he buy or sell? 3- Compute the financing cost of the hedge by assuming that the investor hedges his position during 14 days with no rebalancing. The Libor 2 weeks is equal to 3.5%. 4- Explain why hedging with swaps is better than hedging with bonds. Solution 62 1- Recall that the discount factor B(0, θ) is given by the following formula using 45

the Nelson and Siegel model B(0, θ) =e θ.rc (0,θ) θ. = e µβ 0 +β 1 1 exp( τ θ ) 1 exp( τ θ +β 2 θ ) θ exp( θ τ ) τ τ and that the level, slope and curvature $durations of a bond delivering cash-flows C i at dates θ i, which are denoted respectively S 0, S 1 and S 2, are given by S 0 = P θ i C i e θ ir C (0,θ i ) i ³ S 1 = P θ i 1 exp θ i τ 1 θ i C i e θ ir C (0,θ i ) i τ ³ 1 S 2 = P θ i 1 exp θ i ³ τ 1 θ i exp θ i τ 1 C i e θ ir C (0,θ i ) i τ 1 a-thepricingformulaforsuchaswapusingthezero-couponmethodis à 4X! SWAP =10, 000, 000. 2%.B(0,i/2) 1+B(0, 2) = $3, 103 i=1 As we can see in the previous formula, a standard swap is equivalent to: abondwiththenominalamountandthefixed rate of the swap, and the coupon frequency of the fixed leg minus the nominal amount of the swap. We deduce that the level, slope and curvature $durations of a swap are equivalent of those of the bond contained in the swap. For this swap we obtain S 1 = 10 6 µ 4P S 2 = 10 6 µ 4P S 0 = 10 6 µ i=1 4P i=1 i 2 i 2 i=1 1 e i/2 τ i/2 τ 1 e i/2 τ i/2 τ i e i/2 τ 2 2%B(0,i/2) + 2B(0, 2) = 1, 935, 651 1 e 2%B(0,i/2) + 2 i/2 τ 2 B(0, 2) = 1, 418, 594 τ 2%B(0,i/2) + 2 e 2 τ B(0, 2) = 414, 983 1 e 2 τ 2 τ 46

b- Using the same method we obtain for the second swap SWAP =$4, 072 S 0 = 4, 499, 361 S 1 = 2, 261, 944 S 2 = 1, 317, 652 c- Using the same method we obtain for the third swap SWAP = $5, 291 S 0 = 7, 723, 248 S 1 = 2, 537, 714 S 2 = 1, 997, 192 2- This conducts to search for the quantity q 1,q 2 and q 3 to invest respectively in the three swaps with maturity 2, 5 and 10 years that verify the following linear system q 1 1, 935, 651 + q 2 1, 418, 594 + q 3 414, 983 = 789, 456, 145 q 1 4, 499, 361 + q 2 2, 261, 944 + q 3 1, 317, 652 = 142, 256, 548 q 1 7, 723, 248 + q 2 2, 537, 714 + q 3 1, 997, 192 = 97, 897, 254 which provides The investor will: q 1 =527.42 q 2 =667.11 q 3 = 2, 838.21 buy 527 swaps with 2-year maturity buy 667 swaps with 5-year maturity sell 2,838 swaps with 10-year maturity to protect his bond portfolio against a change of the level, slope and curvature factors. 3- The financing cost of the hedge is equal to [(527 3, 103) + (667 4, 072) (2, 838 5, 291)] 14 3.5% = $21, 909 360 47

4- It is preferable to use swaps instead of bonds to hedge a position because the hedging cost is usually lower with these instruments. Exercise 63 The Cheapest to Deliver on the Repartition Date We are now on the repartition date. Suppose a futures contract with Eur100,000 contract size whose price is 95 and three bonds denoted A,B and C with the following features. What is the bond that the seller of the futures contract will choose to deliver. Clean Price Conversion Factor Bond A 112.67 119.96% Bond B 111.54 118.66% Bond C 111.47 119.78% Solution 64 The seller of the futures contract chooses to deliver the bond that maximises the difference between the invoice price IP and the cost of purchasing the bond CP, whichiscalled the cheapest to deliver. The quantity IP -CP is given by the following formula IP CP = futures price CF clean price where CF is the conversion factor. Here we obtain the following results IP-CP Bond A 1.292 Bond B 1.187 Bond C 2.321 The seller of the futures contract will choose to deliver Bond C. Exercise 65 A treasurer of a firm knows today that he will have to invest Eur 50,000,000 in a particular bond A in one month. Today, this bond quotes 116.414 (the accrued interest is included in the price). He fears a decrease in rates and then uses the futures market to hedge its interest rate risk. The price of the ten year futures contract which expires in one month is 98.55. Its nominal amount is Eur 100,000. The conversion factor for bond A is 1.18125. 48

1- Has the treasurer to buy or sell futures contracts to hedge its interest rate risk? 2- What is the position he has to take on the futures market? 3- One month later, the price of bond A is 121.137 as the price of the futures contract is 102.55. What is the result of the hedge transaction for the treasurer? Solution 66 1- The tresurer fears a decrease in rates and then an increase in price of bond A. To hedge this interest rate risk, he has to buy futures contract because prices of futures contracts move in the same way as bond prices when interest rates decrease or increase. 2- The number n of futures contract is given by the following hedging ratio n = MD A.P A.N A 50, 000, 000.CF = 1.18125 = 590.775 MD A.P A.N F 100, 000 where MD A is the modified duration of bond A, P A its gross price, N A the amount the treasurer will invest in bond A, and N F the nominal amount of the futures contract. Note here that bond A is the cheapest to deliver which simplifies the previous formula. Here, the treasurer buys 591 futures contracts. 3- The result is the following for the treasurer: gain on the futures market: 591 (102.55 98.55) % Eur 100, 000 = Eur 2,364,000 opportunity loss on the bond market: (116.414 121.137) % Eur 50, 000, 000 = Eur 2.361,500 which represents a net gain of Eur 2,500. The net position is not exactly equal to zero because the treasurer buys 591 futures contract and not 590.775 as given by the hedge ratio. Exercise 67 Speculation with Futures Today, the gross price of a 10-year maturity bond with $1,000 principal amount is 116.277. At the same moment the price of the ten year futures contract which expires in two month is 98.03. Its nominal amount is $100,000, and the deposit margin is $1,000. One month later the price of the bond is 120.815 as the price of the futures is 102.24. 49

1- What is the leverage effect on this futures contract? 2- An investor anticipates that rates will decrease in a short-term period. His cash at disposal is $100,000. a) What is the position he can take on the market using the bond? What is his absolute gain after one month? What is the return rate of its investment? b) Same question as the previous one using the futures contract? 3- Conclude Solution 68 1- The leverage effect of the futures contract is 100 Leverage Effect = No min al Amount Deposit Margin = 100, 000 1, 000 =100 2- a) The investor anticipates a decrease in rates so he will buy bonds. His cash at disposal is $100,000. then he buys 86 bonds Number of bonds bought =. His absolute gain over the period is $4,723 $100, 000 $1, 000 116.277% =86 Absolute Gain =86 $1, 000 [120.815 116.277] % = $3, 902.68 The return rate of its investment over the period is 3.903% Re turn Rate = $3, 902.68 $100, 000 =3.903% b) Futures contracts move in the same way as bonds when interest rates change, so the investor will buy futures contracts. His cash at disposal is $100,000. then he buys 102 futures contracts Number of futures contracts bought = $100, 000 deposit margin futures price = $100, 000 $1, 000 98.03% =102 50

His absolute gain over the period is Absolute Gain = 102 $100, 000 [102.24 98.03] % = $429, 420 The return rate of its investment over the period is 429.42% Re turn Rate = $429, 420 $100, 000 =429.42% 3- The difference of performance between the two investments is explained by the leverage effect of the futures contract. Exercise 69 Let us consider a caplet contracted at date t =05/13/02 with ³ nominal amount $10, 000, 000, exerciseratee =5%, based upon the six-month Libor R L t, 1 2 and with the following schedule t =05/13/02 T 0 =06/03/02 T 1 =12/03/02 When the caplet is contracted When the caplet starts Pay-off payment At date T 1 =12/03/02, the cap holder receives the cash-flow C µ C =$10, 000, 000 Max 0; δ µr L T 0, 1 5% 2 where δ = T 1 T 0 (expressed in fraction of years using the Exact/360 day-count basis). 1- Draw the P&L of this caplet considering that the premium paid by the buyer is equal to 0.1% of the nominal amount. 2- What is the price formula of this caplet using the Black model? 3- Assuming that the 6-month Libor forward is 5.17% at date t, its volatility 15% and that the zero-coupon rate R C (t, T 1 t) is equal to 5.25%, give the price of this caplet. 4- Deduce the margin taken by the seller of this caplet. We know want to compute the sensitivities of the caplet 5- Delta a- Compute the delta of the caplet. b- Recalculate the price of the caplet if the 6-month forward goes from 5.17% to 5.18%. 51

c- Compare the price difference with the quantity delta change of the forward rate. 6- Gamma a- Compute the gamma of the caplet. b- The 6-month forward goes from 5.17% to 5.18%. Compare the price difference of the caplet with the quantity delta change of the forward rate+gamma (change of the forward rate) 2 /2. 5.35%. 7- Vega a- Compute the vega of the caplet. b- Recalculate the price of the caplet if the volatility goes from 15% to 16%. c- Compare the price difference with the quantity vega change of volatility. 8- Rho a- Compute the rho of the caplet. b- Recalculate the price of the caplet if the interest rate R C (t, T 1 t) goes from 5.25% to c- Compare the price difference with the quantity rho change of rate. 9- Theta a- Compute the theta of the caplet. b- Recalculate the price of the caplet one day later on 05/14/02.. c- Compare the price difference with the quantity theta 365 1. Solution 70 1- The P&L of the caplet depends on the value of the 6-month Libor at date T 0, and is given by the following formula (considering the buyer position; of course the seller position is the opposite one) ½ µ P &L =$10, 000, 000 Max 0; δ µr L T 0, 1 ¾ 5% 0.1% 2 52

It appears on the followng graph 100000 90000 80000 70000 60000 P&L in $ 50000 40000 30000 20000 10000 0 3,00% -10000 3,50% 4,00% 4,50% 5,00% 5,50% 6,00% 6,50% 7,00% -20000 Value of the 6-month Libor on 06/03/02 2- The caplet price at date t in Black (1976) model is given by h ³ Caplet t = N δ B(t, T 1 ) F (t, T 0,T 1 )Φ(d) EΦ d σ p i T 0 t where: N is the nominal amount. B(t, T 1 ) is the discount factor equal to B(t, T 1 )=e (T 1 t).r C (t,t 1 t) Φ is the cumulative distribution function of the standard Gaussian distribution. F (t, T 0,T 1 ) is the 6-month Libor forward as seen from date t, starting at date T 0 and finishing at date T 1. Note in particular that µ F (T 0,T 0,T 1 )=R L T 0, 1 2 53

d is given by ³ d = ln F (t,t0,t 1 ) E +0.5σ 2 (T 0 t) σ T 0 t σ is the volatility of the underlying rate F (t, T 0,T 1 ), which is usually referred to as the caplet volatility. 3- N =$10, 000, 000 and δ = 183 360. The discount factor is equal to B(t, T 1 )=e 204 365.5.25% =0.97108384 We obtain the following values for d and d σ T 0 t d = 0.94100172 d σ p T 0 t = 0.90477328 so that Φ(d) and Φ(d σ T 0 t) are equal to Φ(d) = 0.82664803 Φ(d σ p T 0 t) = 0.81720728 Note that Φ the cumulative distribution function of the standard Gaussian law is already preprogrammed in Excel (see Excel functions). We finally obtain the caplet price Caplet =$9, 267 = 0.09267% of the nominal amount 4- The seller of this caplet takes a margin equal to 7.91% 5- Delta 10, 000 9, 267 1=7.91% a- The delta, which is the first derivative of the caplet price with respect to the underlying 54

rate F (t, T j 1,T j ), is given by = N δ B(t, T 1 )Φ(d) =4, 080, 618 b- If the 6-month Libor forward goes from 5.17% to 5.18%, the caplet price becomes $9,678. c- The price difference is equal to $411 Pr ice Difference =$9, 678 $9, 267 = $411 very near from the quantity delta change of the forward rate equal to $408 delta change of the forward rate =4, 080, 618 0.01% = $408 6- Gamma a- The gamma γ, which is the second derivative of the caplet price with respect to the underlying rate F (t, T j 1,T j ), is given by γ = N δ B(t, T 1 ) σ T 0 tf (t, T 0,T 1 ) Φ0 (d 1 )=675, 296, 853 where Φ 0 is the density function of the standard Gaussian law Φ 0 (x) = 1 2π e x2 /2 b- If the 6-month Libor forward goes from 5.17% to 5.18%, the caplet price becomes $9,678 so that the price difference of the caplet is equal to $411 (see 5-c) and very well explained by the quantity delta change of the forward rate+gamma (change of the forward rate) 2 /2 delta change of the forward rate+gamma (change of the forward rate) 2 /2 = $408+$3 = $411 7- Vega a- The vega ν, which is the first derivative of the caplet price with respect to the volatility 55

parameter σ, is given by υ = N δ B(t, T 1 ) p T 0 tf (t, T 0,T 1 )Φ 0 (d 1 )=15, 794 b- If the volatility goes from 15% to 16%, the caplet price becomes $9,429. c- The price difference is equal to $162 Pr ice Difference =$9, 429 $9, 267 = $162 very near from the quantity vega change of volatility equal to $157.94 delta change of rate =15, 794 1% = $157.94 8- Rho a- The rho ρ, whichisthefirst derivative of the caplet price with respect to the interest rate R C (t, T 1 t), is given by ρ = (T 1 t) Caplet = 5, 251 b- If the interest rate R C (t, T 1 t) goes from 5.25% to 5.35%, the caplet price becomes $9,262. c- The price difference is equal to -$7 Pr ice Difference =$9, 262 $9, 267 = $5 very near from the quantity rho change of rate equal to -$5.25 delta change of rate = 5, 251 0.1% = $5.25 9- Theta a- The theta θ, whichisthefirst derivative of the caplet price with respect to time, is given 56

by µ θ = N δ R C (t, T 1 t) Caplet where r is the short term rate. N δ B(t, T 1)σF (t, T 0,T 1 ) 2 T 0 t b- One day later on 05/14/02, the caplet price becomes $9,212. c- The price difference is equal to -$55 Φ 0 (d 1 ) = 19, 820 Pr ice Difference =$9, 212 $9, 267 = $55 very near from the quantity theta 1 365 equalto-$54.3 theta 1 1 = 19, 820 365 365 = $54.3 Exercise 71 On 05/13/02 a firm buys a barrier caplet up and in whose features are the following: notional amount: $ 10,000,000 reference rate: 3-month Libor strike rate: 5% starting date: 06/03/02 barrier: 6% day-count: Actual/360 1- What is the pay-off of this option for the buyer? 2- Draw the P&L of this caplet considering that the premium paid by the buyer is equal to 0.08% of the nominal amount. 3- What is the advantage and the drawback of this option compared to a classical caplet? Solution 72 1- The pay-off of this option for the buyer on 09/03/02 is Pay-Off =$10, 000, 000 92 Max µ 360 0; R 06/03/02, 1 5% 1 4 R(06/03/02, 1 4) 6% ³ where R 06/03/02, 1 4 is the 3-month Libor rate observed on 06/03/02, 92 is the number of days between the 06/03/02 and the 09/03/02, and 1 A =1if event A occurs and 0 otherwise. 2- The P&L is given by the following formula P &L = Pay-Off $10, 000, 000 0.08% 57

It appears in the following graph. 80000 70000 60000 50000 40000 P&L in $ 30000 20000 10000 0 3,0% 3,5% 4,0% 4,5% 5,0% 5,5% 6,0% 6,5% 7,0% 7,5% 8,0% -10000-20000 Value of the 3-month Libor on 06/03/02 3- The advantage of this option compared to a classical caplet is that the buyer will pay a lower premium. The drawback is that he will gain only if the reference rate is equal or above the barrier. 58