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Fixed Income Securities: Valuation, Risk, And Risk Management, 1st Edition Solution Manual

Fixed Income Securities: Valuation, Risk, And Risk Management, 1st Edition Solution Manual is your ultimate textbook solutions guide, providing answers to the most difficult questions.

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    Solution Manual to accompany the textbook Fixed Income Securities: Valuation, Risk, and Risk Management by Pietro Veronesi Chapters 2 - 8 Version 1 Date: October, 2009 Author: Anna Cieslak, Javier Francisco Madrid

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    Solutions to Chapter 2 Exercise 1. Compute the discount factors implied by the STRIPS: Z (0 , 3) = exp( − 3 × 0 . 1) = 0 . 741 (1) Z (0 , 5) = exp( − 5 × 0 . 05) = 0 . 779 . (2) Since Z (0 , 3) < Z (0 , 5), there is an arbitrage opportunity due to the violation of the positive time discount rate. To exploit it: buy the 3-year bond and sell the 5-year bond. Exercise 2. Compute the quoted price P of the T-bill as: P = 100 × [ 1 − n 360 × d ] , (3) using the discount rate given, d . The simple (bond equivalent) yield measures your annualized return as: BEY = 100 − P P × 365 n . (4) Let τ = n 365 be the time to maturity expressed as fraction of a year, and let T denote the maturity date of a given T-bill. The continuously compounded yield follows as: r ( t, T ) = − 1 τ ln P 100 . (5) Finally, to obtain the semi-annually compounded yield for the 1-year T-bill, use: r 2 (0 , 1) = 2 × ( 1 ( P/ 100) 1 / 2 − 1 ) (6) Cont. comp. Semi-annual Maturity n T − t Discount, d Price, P BEY yield comp. Date a. 4-week 28 0.083 3.48% 99.7293 3.5379% 3.53% – 12/12/2005 b. 4-week 28 0.083 0.13% 99.9899 0.13% 0.13% – 11/6/2008 c. 3-month 90 0.25 4.93% 98.7675 5.06% 5.03% – 7/10/2006 d. 3-month 90 0.25 4.76% 98.8100 4.88% 4.86% – 5/8/2007 e. 3-month 90 0.25 0.48% 99.8800 0.49% 0.49% – 11/4/2008 f. 6-month 180 0.5 4.72% 97.6400 4.90% 4.84% – 4/21/2006 g. 6-month 180 0.5 4.75% 97.6250 4.93% 4.87% – 6/6/2007 h. 6-month 180 0.5 0.89% 99.5550 0.91% 0.90% – 11/11/2008 i. 360-day 360 1 1.73% 98.2700 1.78% 1.77% 1.75% 9/30/2008 j. 360-day 360 1 1.19% 98.8100 1.22% 1.21% 1.20% 11/5/2008 1

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    Exercise 3. Compute respective discount factors taking into account the convention on which the interest rate is given: 1. Z ( t, t + 1 . 5) = exp( − 0 . 02 × 1 . 5) = 0 . 97045 2. Z ( t, t + 1 . 5) = exp( − 0 . 03) = 0 . 97045 3. Z ( t, t + 1 . 5) = 1 (1 + 0 . 021) 1 . 5 = 0 . 96931 4. Z ( t, t + 1 . 5) = 1 (1 + 0 . 0201 / 2) 2 × 1 . 5 = 0 . 97045 Bond 3 is mispriced. Exercise 4. Using Table 2.4, obtain the discount factor Z ( t, T ) for each maturity T − t from 0.25 to 7.5 years: Z ( t, T ) = 1 ( 1 + r 2 ( t,T ) 2 ) 2( T − t ) . (7) Use Z to price each bond: a. P z (0 , 5) = 100 × Z (0 , 5) = 72 . 80 b. P c =15% ,n =2 (0 , 7) = 15 2 × ∑ 14 i =1 Z (0 , i/ 2) + 100 × Z (0 , 7) = 151 . 23 c. P c =7% ,n =4 (0 , 4) = 7 4 × ∑ 16 i =1 Z (0 , i/ 4) + 100 × Z (0 , 4) = 101 . 28 d. P c =9% ,n =2 (0 , 3 . 25) = 9 2 × ∑ 7 i =1 Z (0 , i/ 2 − 0 . 25) + 100 × Z (0 , 3 . 25) = 108 . 55 e. 100 (see Fact 2.11) f. P F R,n =1 ,s =0 = Z (0 , 0 . 5) × 100 × (1 + 6 . 8% 1 ) = 103 . 44, where we assume that r 1 (0) = 6 . 8% g. P F R,n =4 ,s =0 . 35% (0 , 5 . 5) = 100 + 0 . 35 4 ∑ 22 i =1 Z (0 , i/ 4) = 101 . 6 h. P F R,n =2 ,s =0 . 40% = Z (0 , 0 . 25) × 100 × (1 + 6 . 4% 2 ) + 0 . 40 2 ∑ 15 i =1 Z (0 , i/ 2 − 0 . 25) = 104, where we assume that r 2 (0) = 6 . 4% Exercise 5. a. When coupon c is equal to the yield to maturity y the bond trades at par; when coupon is below (above) the yield to maturity the bond trades above (below) par. Obtain bond prices given yield and the coupon using: P c (0 , T ) = 20 ∑ i =1 c/ 2 × 100 (1 + y/ 2) i + 100 (1 + y/ 2) 20 (8) It follows: 2

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    c y P 5% 6% 107.79 6% 6% 100 7% 6% 92.89 b. Figure 1 plots bond prices implied by different yields to maturity. 0 2 4 6 8 10 12 14 16 50 60 70 80 90 100 110 120 130 140 150 Bond price vs YTM Bond price Yield Fig. 1. Bond price as function of yield to maturity Exercise 6. a. To obtain bond prices use the expression: P c (0 , T ) = c 2 × 2 T ∑ i =1 Z (0 , i/ 2) + 100 × Z (0 , T ) , (9) To compute the yield to maturity, solve equation (8) for y using a numerical solver. c T − t P y 15% 7 151.2306 5.9461% 3% 7 84.3482 5.7474% b. The yields to maturity are different since bonds have different coupons, despite having the same time to maturity. Both bonds are priced using a no arbitrage discount curve. Therefore, their prices are fair. Exercise 7. a. Bootstrap the discount factors Z ( t, T ) using the expression (9), and substituting recursively for the 6-month, 1-year, 1.5-year, and 2-year bonds. E.g., given Z (0 , 0 . 5) and Z (0 , 1), for the 1.5-year bond you have: 3

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    Z (0 , 1 . 5) = 100 . 86 − 7 . 5 2 ( Z (0 , 0 . 5) + Z (0 , 1)) 100 + 7 . 5 2 . (10) This yields: T − t Coupon, c Price, P Issued Z ( t, T ) 0.5 0.00% $96.80 5/15/2000 0.9680 1 5.75% $99.56 5/15/1998 0.9407 1.5 7.50% $100.86 11/15/1991 0.9032 2 7.50% $101.22 5/15/1992 0.8740 b. Compute the no-arbitrage price of the two bonds given the discount function obtained above. The prices of the two bonds are: P c =8% = $101 . 71 (11) P c =13 . 13% = $106 . 60 , (12) i.e. both are higher than the market prices. There is an arbitrage opportunity. You could make riskless profit by buying the underpriced bond at the traded price and selling the corresponding portfolio of zeros that replicates the cash flows from the bond. Exercise 8. The quotes are obtained on May 15, 2000. Use the mid bid-ask quote to compute the price. You want to obtain the semi-annual curve. The provided maturities of bonds are spaced semi-annually. We can assume that the clean (quoted) price is equal to the dirty (invoice) price, i.e. the accrual is zero. For each maturity, bootstrap the discount factors Z ( t, T ) as in Exercise 7.a. The continuously compounded zero coupon yield is given as: r ( t, T ) = − 1 T − t ln Z ( t, T ) . (13) 4

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    Cusip T − t Mid bid-ask quote Coupon, c Accrual Z ( t, T ) Yield, r ( t, T ) 912827ZN 0.5 100.9063 8.500% 0 0.96793 6.520% 912810CU 1 105.9961 13.125% 0 0.93508 6.713% 912810CX 1.5 112.4102 15.750% 0 0.90312 6.793% 912827F4 2 101.2188 7.500% 0 0.87418 6.724% 912810DA 2.5 110.6836 11.625% 0 0.84387 6.790% 912810DD 3 110.3438 10.750% 0 0.81638 6.762% 912810DG 3.5 115.3242 11.875% 0 0.78928 6.761% 912810DH 4 118.9141 12.375% 0 0.76267 6.773% 912810DM 4.5 118.3125 11.625% 0 0.73951 6.706% 912810DQ 5 121.6289 12.000% 0 0.71544 6.697% 912827V8 5.5 96.0000 5.875% 0 0.69440 6.631% 912827X8 6 100.6211 6.875% 0 0.67229 6.618% 912827Z6 6.5 98.7656 6.500% 0 0.65080 6.609% 9128272U 7 99.5781 6.625% 0 0.63152 6.566% 9128273X 7.5 93.1484 5.500% 0 0.61225 6.542% 9128274F 8 93.8008 5.625% 0 0.59478 6.494% 9128274V 8.5 87.9922 4.750% 0 0.57640 6.482% 9128275G 9 92.8398 5.500% 0 0.56151 6.413% 5

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    Solutions to Chapter 3 Exercise 1. a. 3; equal to the maturity of the zero bond b. 2.9542; the duration of the coupon bond is the weighted average of the coupon payment times c. 0.9850 d. 0.5; equal to the time left to the next coupon payment e. 0.5111; obtain the price P F R of the floating rate bond (see Chapter 2, equation (2.39)). In analogy to a coupon bond, the duration is computed as: D F R = 100 P F R × 0 . 5 + 0 . 5 s × ∑ 3 t =0 . 5 Z (0 , t ) × t P F R . (14) f. 0.2855; proceed as in point e. above but recognize that the valuation is outside the reset date. Assume that the coupon applying to the next reset date has been set at r 2 (0) = 6 . 4%. Exercise 2. Portfolio A : Security Duration, D Weight, w D × w 4.5yr @ 5% semi 3.8660 40% 1.55 7yr @ 2.5% semi 6.4049 25% 1.60 1.75 fl + 30bps semi 0.2540 20% 0.05 1yr zero 1.0000 10% 0.10 2yr @ 3% quart 1.9530 5% 0.10 Port. D 3.40 Portfolio B : Security Duration, D Weight, w D × w 7yr @ 10% semi 5.4262 40% 2.17 4.25yr @ 3% quart 3.9838 25% 1.00 90 day zero 0.2500 20% 0.05 2yr fl semi 0.5000 10% 0.05 1.5yr @ 6% semi 1.4564 5% 0.07 Port. D 3.34 The investors would select the shorter duration portfolio B . 6

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    Exercise 3. Obtain yield to maturity y for each security. Compute modified and Macaulay duration accodring to equation (3.19) and (3.20) in the book. Yield Duration Modified Macaulay a. 6.95% 3 3 2.8993 b. 6.28% 2.9542 2.9974 2.9061 c. 6.66% 0.9850 0.9850 0.9689 d. 0.00% 0.5 0.5 0.5 e. 6.82% 0.5111 0.5111 0.4943 f. 6.76% 0.2855 0.2855 0.2761 Exercise 4. Compute the duration of each asset and use the fact that the dollar duration is the bond price times its duration. Price Duration $ Duration a. $89.56 4.55 $407.88 b. $67.63 -7.00 ($473.39) c. $79.46 3.50 $277.74 d. $100.00 0.5 $50.00 e. $100.00 -0.25 ($25.00) f. $102.70 -0.2763 ($28.38) Exercise 5. a. Compute number of units of each security ( N ) in the portfolio and apply Fact 3.5. Portfolio A : Security Price Duration Weight N D × P × N 4.5yr @ 5% semi 94.03 3.8660 40% 0.43 154.64 7yr @ 2.5% semi 81.56 6.4049 25% 0.307 160.12 1.75 fl + 30bps semi 102.09 0.2540 20% 0.20 5.08 1yr zero 93.61 1.0000 10% 0.11 10.00 2yr @ 3% quart 92.54 1.9530 5% 0.05 9.76 $ D 339.61 7

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    Portfolio B : Security Price Duration Weight N D × P × N 7yr @ 10% semi 123.36 5.4262 40% 0.32 217.05 4.25yr @ 3% quart 86.83 3.9838 25% 0.29 99.59 90 day zero 98.45 0.2500 20% 0.20 5.00 2yr fl semi 100.00 0.5000 10% 0.10 5.00 1.5yr @ 6% semi 98.83 1.4564 5% 0.05 7.28 $ D 333.92 b. For 1 bps increase, we have (see Definition 3.5): Portfolio A : 339 . 61 × 0 . 01 / 100 = − $0 . 0340 Portfolio B : 333 . 92 × 0 . 01 / 100 = − $0 . 0334 c. Yes. Exercise 6. After the reshuffling of the portfolio, its value becomes $50 mn. a. Short -0.307 units of long bond in portfolio A , and -0.081 units in portfolio B . b. New dollar durations are: 19.36 and 62.61 for portfolio A and B , respectively. c. The conclusion reverses. N LT bond New weight LT bond New $ D Port. A -0.307 -25% 19.36 Port. B -0.081 -10% 62.61 Exercise 7. a. $10 mn b. Compute the dollar duration of the cash flows in each bond, and then the dollar duration of the portfolio: Security Position $ (mn) Price N $ D $ D × N 6yr IF @ 20% - fl quart Long 20.00 146.48 0.137 1,140.28 155.69 4yr fl 45bps semi Long 20.00 101.62 0.197 53.54 10.54 5yr zero Short (30.00) 76.41 -0.393 382.052 -150.00 Port. value $10.00 mn Port. $ D 16.23 Exercise 8. a. The price of the 3yr @ 5% semi bond is $97.82. You want the duration of the hedged portfolio to be zero. You need to short 0.058 units of the 3-year bond, i.e. the short position is -$5 . 69. 8

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    b. The total value of the portfolio is: $4 . 31 mn. Exerxise 9. Compute the new value of the portfolio assuming the term structure of interest rates as of May 15, 1994. Original Now ∆ value Unhedged port. $10.00 $8.97 ($1.03) Hedge ($5.69) ($5.44) $0.25 Total $4.31 $3.53 ($0.78) a. $8.97 mn b. $3.53 mn c. The immunization covered part of the loss. The change in the value of the portfolio is both due to ( i ) the passage of time (coupon) and ( ii ) the increase in interest rates. Exercise 10. Use the curve given on May 15, 1994, but keep the times to maturity unchanged from the initial ones. The change in value is due to the change in interest rates only. Original Now ∆ value Unhedged port. $10.00 $9.97 ($0.03) Hedge ($5.69) ($5.43) $0.26 Total $4.31 $4.54 $0.23 Exercise 11. Use the curve given on February 15, 1994, but change the times to maturity to those on May 15, 1994. The change in value is due to coupon only. Original Now ∆ value Unhedged port. $10.00 $9.12 ($0.88) Hedge ($5.69) ($5.68) $0.01 Total $4.31 $3.45 ($0.87) Exercise 12. a.,b. Loss of $0.87 mn. c. Gain of $0.08 mn. 9

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    Original Now ∆ C ∆ r Total Ex.8 Ex.9 Ex.11 (9)-(8)-(11) (9)-(8) Unhedged port. $10.00 $8.97 ($0.88) ($0.15) ($1.03) Hedge ($5.69) ($5.44) $0.01 $0.24 $0.25 Total $4.31 $3.53 ($0.87) $0.08 ($0.78) 10

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    Solutions to Chapter 4 Exercise 1. a. 16 b. 4.7 (see Fact 4.3) c. 3.85 d. 0.28 = 0.25 (due to floater with zero spread) + 0.03 (due to the spread) e. 0.14 = 0.06 (due to floater with zero spread) + 0.08 (due to the spread). Assume r 2 (0) = 6 . 40%. Exercise 2. Portfolio A : Security Price Weight, w Duration, D Convexity, C w × D w × C 5yr @ 4% quart 89.20 30% 4.5510 21.9703 1.37 6.59 4.25yr @ 6% semi 99.15 25% 3.7535 15.2807 0.94 3.82 90-day zero 98.45 20% 0.25 0.0625 0.05 0.01 2.5 yr float quart 100.00 15% 0.25 0.0625 0.04 0.01 6yr zero 69.54 10% 6.0 36 0.60 3.60 Port. 100% 2.99 14.03 Portfolio B : Security Price Weight, w Duration, D Convexity, C w × D w × C 7yr @ 2% semi 78.77 40% 6.5071 44.3923 2.60 17.76 3.25yr float 50 bps 103.16 30% 0.2717 0.1194 0.08 0.04 4yr @ 3.5% semi 89.00 20% 3.7506 14.6406 0.75 2.93 90-day zero 98.45 10% 0.25 0.0625 0.03 0.01 Port. 100% 3.46 20.73 Assume the interest rates move up by 1% in a parallel fashion, i.e. dr = 1%. The change in the value of the portfolios is: dP A P A = − 2 . 99 × 0 . 01 + 1 2 × 14 . 03 × (0 . 01) 2 = − 0 . 0292 (15) dP B P B = − 3 . 46 × 0 . 01 + 1 2 × 20 . 73 × (0 . 01) 2 = − 0 . 0336 (16) ∆ due to D ∆ due to C Total ($ mn) Port. A -0.0299 0.0007 -0.0292 Port. B -0.0346 0.0010 -0.0336 11

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    The portfolio A is less sensitive to the parallel shifts in interest rates. Exercise 3. It follows that: V ar ( dr ) = E ( dr 2 ) = 3 . 451 e − 6 . Therefore: E ( dP A P A ) = 1 2 × C A × E ( dr 2 ) = 2 . 42 e − 5 × 252 = 0 . 61% annualized (17) E ( dP B P B ) = 1 2 × C B × E ( dr 2 ) = 3 . 58 e − 5 × 252 = 0 . 90% annualized (18) Exercise 4. Follow the steps in Example 4.3 using a 2-year bond instead of a 10-year bond. Duration Hedging Spot Curve Change in Shift P c (0 , 10) P z (0 , 2) Position Portfolio Value Initial Values 103.58 91.39 -4.5507 dr = . 1% 102.75 91.21 -4.5507 0.0030 dr = 1% 95.63 89.58 -4.5507 0.2880 dr = 2% 88.38 87.81 -4.5507 1.1087 dr = − . 1% 104.41 91.58 -4.5507 0.0030 dr = − 1% 112.29 93.24 -4.5507 0.3113 dr = − 2% 121.84 95.12 -4.5507 1.2957 The hedge performs better in that for any scenario the change in the value of the portfolio is positive. Exercise 5. Take into account the fact that the time to maturity of both securities is changing: a. by 1/252, b. by 1/52 and c. by 1/12 fraction of a year. Price the 10-year coupon bond and the 2-year zero bond for each of those cases and for the different levels of interest rates. Finally, compute the change in the value of the portfolio relative to the initial one. a. Day Trade Spot Curve Change in Shift P c (0 , 10) P z (0 , 2) Position Portfolio Value Initial Values 103.58 91.39 -4.5507 dr = . 1% 102.77 91.23 -4.5507 -0.0540 dr = 1% 95.65 89.60 -4.5507 0.2199 dr = 2% 88.40 87.83 -4.5507 1.0284 dr = − . 1% 104.43 91.59 -4.5507 -0.0515 dr = − 1% 112.30 93.25 -4.5507 0.2680 dr = − 2% 121.86 95.13 -4.5507 1.2648 12

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    b. Week Trade Spot Curve Change in Shift P c (0 , 10) P z (0 , 2) Position Portfolio Value Initial Values 103.58 91.39 -4.5507 dr = . 1% 102.84 91.29 -4.5507 -0.2734 dr = 1% 95.73 89.68 -4.5507 -0.0422 dr = 2% 88.49 87.92 -4.5507 0.7194 dr = − . 1% 104.50 91.65 -4.5507 -0.2614 dr = − 1% 112.36 93.30 -4.5507 0.1013 dr = − 2% 121.90 95.17 -4.5507 1.1461 c. Month Trade Spot Curve Change in Shift P c (0 , 10) P z (0 , 2) Position Portfolio Value Initial Values 103.58 91.39 -4.5507 dr = . 1% 103.14 91.56 -4.5507 -1.1966 dr = 1% 96.07 89.99 -4.5507 -1.1455 dr = 2% 88.86 88.29 -4.5507 -0.5817 dr = − . 1% 104.79 91.91 -4.5507 -1.1443 dr = − 1% 112.62 93.51 -4.5507 -0.6000 dr = − 2% 122.10 95.32 -4.5507 0.6470 Exercise 6. Level is computed as the average term structure, slope is the difference between the 10-year and the 1-month rate, and the butterfly (curvature) is obtained as: Curvature = − 1M yield + 2 × 5Y yield − 10Y yield (19) Date Level Slope Curvature ∆ Slope ∆ Curvature 9/26/2008 2.14% 3.64% 2.04% 1.57% 0.20% 9/10/2008 2.40% 2.07% 0.59% -0.06% -0.14% 8/25/2008 2.51% 2.13% 0.63% -0.09% -0.20% 8/11/2008 2.69% 2.22% 0.78% -0.19% -0.14% 7/25/2008 2.75% 2.41% 1.05% 0.06% 0.30% 7/10/2008 2.54% 2.35% 0.89% -0.28% -0.29% 6/25/2008 2.82% 2.63% 1.47% 0.52% 0.01% 6/10/2008 2.92% 2.11% 0.97% - - a. 9/26/2008 b. 9/26/2008 c. See Figure 2 13

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    d. 9/10/2008–9/26/2008 e. 7/10/2008–7/25/2008 0 1 2 3 4 5 6 7 8 9 10 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 Maturity Yield Yield curve on different days 09/26/08 09/10/08 08/25/08 08/11/08 07/25/08 07/10/08 06/25/08 06/10/08 Fig. 2. Term structures 6/10/2008–9/26/2008 from CRSP Exercise 7. You need factor sensitivities for yields with maturities from 3 months to 4.25 years with a 0.25-year spacing. To obtain the 0.25-year grid of sensitivities, interpolate linearly the β ’s between available maturities. To compute factor durations, use Fact 4.5. Security Price Level D Slope D Curvature D a. 4 yr zero 81.45 4.10 0.20 0.99 b. 2.5 yr @ 3% semi 96.24 2.50 -0.61 0.77 c. 3.25 yr float 0 bps 100.71 0.25 -0.06 -0.08 d. 4.25 yr float 35 bps 102.12 0.29 -0.07 -0.07 Exercise 8. Use the assumptions of Chapter 3 (Exercise 7) for the pricing of the inverse floater. a. i. $40 mn ii. Refer to Exercise 7 in Chapter 3 for the pricing of the respective securities. 14

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