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

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Solution Manualto accompany the textbookFixed Income Securities:Valuation, Risk, and Risk Managementby Pietro VeronesiChapters 2 - 8Version 1Date: October, 2009Author: Anna Cieslak, Javier Francisco Madrid

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Solutions to Chapter 2Exercise 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)SinceZ(0,3)< Z(0,5), there is an arbitrage opportunity due to the violation of the positive time discountrate. To exploit it: buy the 3-year bond and sell the 5-year bond.Exercise 2.Compute the quoted pricePof the T-bill as:P= 100×[1−n360×d],(3)using the discount rate given,d. The simple (bond equivalent) yield measures your annualized return as:BEY= 100−PP×365n .(4)Letτ=n365be the time to maturity expressed as fraction of a year, and letTdenote the maturity date ofa given T-bill. The continuously compounded yield follows as:r(t, T) =−1τlnP100.(5)Finally, to obtain the semi-annually compounded yield for the 1-year T-bill, use:r2(0,1) = 2×(1(P/100)1/2−1)(6)Cont. comp.Semi-annualMaturitynT−tDiscount,dPrice,PBEYyieldcomp.Datea.4-week280.0833.48%99.72933.5379%3.53%–12/12/2005b.4-week280.0830.13%99.98990.13%0.13%–11/6/2008c.3-month900.254.93%98.76755.06%5.03%–7/10/2006d.3-month900.254.76%98.81004.88%4.86%–5/8/2007e.3-month900.250.48%99.88000.49%0.49%–11/4/2008f.6-month1800.54.72%97.64004.90%4.84%–4/21/2006g.6-month1800.54.75%97.62504.93%4.87%–6/6/2007h.6-month1800.50.89%99.55500.91%0.90%–11/11/2008i.360-day36011.73%98.27001.78%1.77%1.75%9/30/2008j.360-day36011.19%98.81001.22%1.21%1.20%11/5/20081

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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) = 0.970452.Z(t, t+ 1.5)=exp(−0.03) = 0.970453.Z(t, t+ 1.5)=1(1 + 0.021)1.5= 0.969314.Z(t, t+ 1.5)=1(1 + 0.0201/2)2×1.5= 0.97045Bond 3 is mispriced.Exercise 4.Using Table 2.4, obtain the discount factorZ(t, T) for each maturityT−tfrom 0.25 to 7.5 years:Z(t, T) =1(1 +r2(t,T)2)2(T−t).(7)UseZto price each bond:a.Pz(0,5) = 100×Z(0,5) = 72.80b.Pc=15%,n=2(0,7) =152×∑14i=1Z(0, i/2) + 100×Z(0,7) = 151.23c.Pc=7%,n=4(0,4) =74×∑16i=1Z(0, i/4) + 100×Z(0,4) = 101.28d.Pc=9%,n=2(0,3.25) =92×∑7i=1Z(0, i/2−0.25) + 100×Z(0,3.25) = 108.55e. 100 (see Fact 2.11)f.PF R,n=1,s=0=Z(0,0.5)×100×(1 +6.8%1) = 103.44, where we assume thatr1(0) = 6.8%g.PF R,n=4,s=0.35%(0,5.5) = 100 +0.354∑22i=1Z(0, i/4) = 101.6h.PF R,n=2,s=0.40%=Z(0,0.25)×100×(1 +6.4%2) +0.402∑15i=1Z(0, i/2−0.25) = 104, where we assumethatr2(0) = 6.4%Exercise 5.a. When couponcis equal to the yield to maturityythe bond trades at par; when coupon is below(above) the yield to maturity the bond trades above (below) par. Obtain bond prices given yield andthe coupon using:Pc(0, T) =20∑i=1c/2×100(1 +y/2)i+100(1 +y/2)20(8)It follows:2

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cyP5%6%107.796%6%1007%6%92.89b. Figure 1 plots bond prices implied by different yields to maturity.02468101214165060708090100110120130140150Bond price vs YTMBond priceYieldFig. 1.Bond price as function of yield to maturityExercise 6.a. To obtain bond prices use the expression:Pc(0, T) =c2×2T∑i=1Z(0, i/2) + 100×Z(0, T),(9)To compute the yield to maturity, solve equation (8) foryusing a numerical solver.cT−tPy15%7151.23065.9461%3%784.34825.7474%b. The yields to maturity are different since bonds have different coupons, despite having the same time tomaturity. Both bonds are priced using a no arbitrage discount curve. Therefore, their prices are fair.Exercise 7.a.Bootstrap the discount factorsZ(t, T) using the expression (9), and substituting recursively for the6-month, 1-year, 1.5-year, and 2-year bonds.E.g., givenZ(0,0.5) andZ(0,1), for the 1.5-year bond youhave:3

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Z(0,1.5) = 100.86−7.52(Z(0,0.5) +Z(0,1))100 +7.52.(10)This yields:T−tCoupon,cPrice,PIssuedZ(t, T)0.50.00%$96.805/15/20000.968015.75%$99.565/15/19980.94071.57.50%$100.8611/15/19910.903227.50%$101.225/15/19920.8740b. Compute the no-arbitrage price of the two bonds given the discount function obtained above. The pricesof the two bonds are:Pc=8%=$101.71(11)Pc=13.13%=$106.60,(12)i.e.both are higher than the market prices.There is an arbitrage opportunity.You could make risklessprofit by buying the underpriced bond at the traded price and selling the corresponding portfolio of zerosthat 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 toobtain the semi-annual curve. The provided maturities of bonds are spaced semi-annually. We can assumethat the clean (quoted) price is equal to the dirty (invoice) price, i.e. the accrual is zero. For each maturity,bootstrap the discount factorsZ(t, T) as in Exercise 7.a. The continuously compounded zero coupon yieldis given as:r(t, T) =−1T−tlnZ(t, T).(13)4

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CusipT−tMid bid-ask quoteCoupon,cAccrualZ(t, T)Yield,r(t, T)912827ZN0.5100.90638.500%00.967936.520%912810CU1105.996113.125%00.935086.713%912810CX1.5112.410215.750%00.903126.793%912827F42101.21887.500%00.874186.724%912810DA2.5110.683611.625%00.843876.790%912810DD3110.343810.750%00.816386.762%912810DG3.5115.324211.875%00.789286.761%912810DH4118.914112.375%00.762676.773%912810DM4.5118.312511.625%00.739516.706%912810DQ5121.628912.000%00.715446.697%912827V85.596.00005.875%00.694406.631%912827X86100.62116.875%00.672296.618%912827Z66.598.76566.500%00.650806.609%9128272U799.57816.625%00.631526.566%9128273X7.593.14845.500%00.612256.542%9128274F893.80085.625%00.594786.494%9128274V8.587.99224.750%00.576406.482%9128275G992.83985.500%00.561516.413%5

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Solutions to Chapter 3Exercise 1.a. 3; equal to the maturity of the zero bondb. 2.9542; the duration of the coupon bond is the weighted average of the coupon payment timesc. 0.9850d. 0.5; equal to the time left to the next coupon paymente. 0.5111; obtain the pricePF Rof the floating rate bond (see Chapter 2, equation (2.39)). In analogy toa coupon bond, the duration is computed as:DF R= 100PF R×0.5 + 0.5s×∑3t=0.5Z(0, t)×tPF R.(14)f. 0.2855; proceed as in point e. above but recognize that the valuation is outside the reset date. Assumethat the coupon applying to the next reset date has been set atr2(0) = 6.4%.Exercise 2.PortfolioA:SecurityDuration,DWeight,wD×w4.5yr @ 5% semi3.866040%1.557yr @ 2.5% semi6.404925%1.601.75 fl + 30bps semi0.254020%0.051yr zero1.000010%0.102yr @ 3% quart1.95305%0.10Port.D3.40PortfolioB:SecurityDuration,DWeight,wD×w7yr @ 10% semi5.426240%2.174.25yr @ 3% quart3.983825%1.0090 day zero0.250020%0.052yr fl semi0.500010%0.051.5yr @ 6% semi1.45645%0.07Port.D3.34The investors would select the shorter duration portfolioB.6

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Exercise 3.Obtain yield to maturityyfor each security. Compute modified and Macaulay duration accodring to equation(3.19) and (3.20) in the book.YieldDurationModifiedMacaulaya.6.95%332.8993b.6.28%2.95422.99742.9061c.6.66%0.98500.98500.9689d.0.00%0.50.50.5e.6.82%0.51110.51110.4943f.6.76%0.28550.28550.2761Exercise 4.Compute the duration of each asset and use the fact that the dollar duration is the bond price times itsduration.PriceDuration$ Durationa.$89.564.55$407.88b.$67.63-7.00($473.39)c.$79.463.50$277.74d.$100.000.5$50.00e.$100.00-0.25($25.00)f.$102.70-0.2763($28.38)Exercise 5.a. Compute number of units of each security (N) in the portfolio and apply Fact 3.5.PortfolioA:SecurityPriceDurationWeightND×P×N4.5yr @ 5% semi94.033.866040%0.43154.647yr @ 2.5% semi81.566.404925%0.307160.121.75 fl + 30bps semi102.090.254020%0.205.081yr zero93.611.000010%0.1110.002yr @ 3% quart92.541.95305%0.059.76$D339.617

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PortfolioB:SecurityPriceDurationWeightND×P×N7yr @ 10% semi123.365.426240%0.32217.054.25yr @ 3% quart86.833.983825%0.2999.5990 day zero98.450.250020%0.205.002yr fl semi100.000.500010%0.105.001.5yr @ 6% semi98.831.45645%0.057.28$D333.92b. For 1 bps increase, we have (see Definition 3.5):PortfolioA: 339.61×0.01/100 =−$0.0340PortfolioB: 333.92×0.01/100 =−$0.0334c. Yes.Exercise 6.After the reshuffling of the portfolio, its value becomes $50 mn.a. Short -0.307 units of long bond in portfolioA, and -0.081 units in portfolioB.b. New dollar durations are: 19.36 and 62.61 for portfolioAandB, respectively.c. The conclusion reverses.NLT bondNew weight LT bondNew $DPort. A-0.307-25%19.36Port. B-0.081-10%62.61Exercise 7.a. $10 mnb. Compute the dollar duration of the cash flows in each bond, and then the dollar duration of theportfolio:SecurityPosition$ (mn)PriceN$D$D×N6yr IF @ 20% - fl quartLong20.00146.480.1371,140.28155.694yr fl 45bps semiLong20.00101.620.19753.5410.545yr zeroShort(30.00)76.41-0.393382.052-150.00Port. value$10.00 mnPort. $D16.23Exercise 8.a. The price of the 3yr @ 5% semi bond is $97.82. You want the duration of the hedged portfolio to bezero. 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.OriginalNow∆ valueUnhedged port.$10.00$8.97($1.03)Hedge($5.69)($5.44)$0.25Total$4.31$3.53($0.78)a. $8.97 mnb. $3.53 mnc. 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. Thechange in value is due to the change in interest rates only.OriginalNow∆ valueUnhedged port.$10.00$9.97($0.03)Hedge($5.69)($5.43)$0.26Total$4.31$4.54$0.23Exercise 11.Use the curve given on February 15, 1994, but change the times to maturity to those on May 15, 1994. Thechange in value is due to coupon only.OriginalNow∆ valueUnhedged port.$10.00$9.12($0.88)Hedge($5.69)($5.68)$0.01Total$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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OriginalNow∆C∆rTotalEx.8Ex.9Ex.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$0.25Total$4.31$3.53($0.87)$0.08($0.78)10

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Solutions to Chapter 4Exercise 1.a. 16b. 4.7 (see Fact 4.3)c. 3.85d. 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). Assumer2(0) = 6.40%.Exercise 2.PortfolioA:SecurityPriceWeight,wDuration,DConvexity,Cw×Dw×C5yr @ 4% quart89.2030%4.551021.97031.376.594.25yr @ 6% semi99.1525%3.753515.28070.943.8290-day zero98.4520%0.250.06250.050.012.5 yr float quart100.0015%0.250.06250.040.016yr zero69.5410%6.0360.603.60Port.100%2.9914.03PortfolioB:SecurityPriceWeight,wDuration,DConvexity,Cw×Dw×C7yr @ 2% semi78.7740%6.507144.39232.6017.763.25yr float 50 bps103.1630%0.27170.11940.080.044yr @ 3.5% semi89.0020%3.750614.64060.752.9390-day zero98.4510%0.250.06250.030.01Port.100%3.4620.73Assume the interest rates move up by 1% in a parallel fashion, i.e.dr= 1%. The change in the value of theportfolios is:dPAPA=−2.99×0.01 + 12×14.03×(0.01)2=−0.0292(15)dPBPB=−3.46×0.01 + 12×20.73×(0.01)2=−0.0336(16)∆ due toD∆ due toCTotal ($ mn)Port.A-0.02990.0007-0.0292Port.B-0.03460.0010-0.033611

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The portfolioAis less sensitive to the parallel shifts in interest rates.Exercise 3.It follows that:V ar(dr) =E(dr2) = 3.451e−6. Therefore:E(dPAPA)= 12×CA×E(dr2) = 2.42e−5×252 = 0.61% annualized(17)E(dPBPB)= 12×CB×E(dr2) = 3.58e−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 HedgingSpot CurveChange inShiftPc(0,10)Pz(0,2)PositionPortfolio ValueInitial Values103.5891.39-4.5507dr=.1%102.7591.21-4.55070.0030dr= 1%95.6389.58-4.55070.2880dr= 2%88.3887.81-4.55071.1087dr=−.1%104.4191.58-4.55070.0030dr=−1%112.2993.24-4.55070.3113dr=−2%121.8495.12-4.55071.2957The 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 1/52and c. by 1/12 fraction of a year. Price the 10-year coupon bond and the 2-year zero bond for each of thosecases and for the different levels of interest rates. Finally, compute the change in the value of the portfoliorelative to the initial one.a. Day TradeSpot CurveChange inShiftPc(0,10)Pz(0,2)PositionPortfolio ValueInitial Values103.5891.39-4.5507dr=.1%102.7791.23-4.5507-0.0540dr= 1%95.6589.60-4.55070.2199dr= 2%88.4087.83-4.55071.0284dr=−.1%104.4391.59-4.5507-0.0515dr=−1%112.3093.25-4.55070.2680dr=−2%121.8695.13-4.55071.264812

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b. Week TradeSpot CurveChange inShiftPc(0,10)Pz(0,2)PositionPortfolio ValueInitial Values103.5891.39-4.5507dr=.1%102.8491.29-4.5507-0.2734dr= 1%95.7389.68-4.5507-0.0422dr= 2%88.4987.92-4.55070.7194dr=−.1%104.5091.65-4.5507-0.2614dr=−1%112.3693.30-4.55070.1013dr=−2%121.9095.17-4.55071.1461c. Month TradeSpot CurveChange inShiftPc(0,10)Pz(0,2)PositionPortfolio ValueInitial Values103.5891.39-4.5507dr=.1%103.1491.56-4.5507-1.1966dr= 1%96.0789.99-4.5507-1.1455dr= 2%88.8688.29-4.5507-0.5817dr=−.1%104.7991.91-4.5507-1.1443dr=−1%112.6293.51-4.5507-0.6000dr=−2%122.1095.32-4.55070.6470Exercise 6.Level is computed as the average term structure, slope is the difference between the 10-year and the 1-monthrate, and the butterfly (curvature) is obtained as:Curvature =−1M yield + 2×5Y yield−10Y yield(19)DateLevelSlopeCurvature∆ Slope∆ Curvature9/26/20082.14%3.64%2.04%1.57%0.20%9/10/20082.40%2.07%0.59%-0.06%-0.14%8/25/20082.51%2.13%0.63%-0.09%-0.20%8/11/20082.69%2.22%0.78%-0.19%-0.14%7/25/20082.75%2.41%1.05%0.06%0.30%7/10/20082.54%2.35%0.89%-0.28%-0.29%6/25/20082.82%2.63%1.47%0.52%0.01%6/10/20082.92%2.11%0.97%--a. 9/26/2008b. 9/26/2008c. See Figure 213

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d. 9/10/2008–9/26/2008e. 7/10/2008–7/25/200801234567891000.0050.010.0150.020.0250.030.0350.040.045MaturityYieldYield curve on different days09/26/0809/10/0808/25/0808/11/0807/25/0807/10/0806/25/0806/10/08Fig. 2.Term structures 6/10/2008–9/26/2008 from CRSPExercise 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 theβ’s between available maturities.Tocompute factor durations, use Fact 4.5.SecurityPriceLevelDSlopeDCurvatureDa.4 yr zero81.454.100.200.99b.2.5 yr @ 3% semi96.242.50-0.610.77c.3.25 yr float 0 bps100.710.25-0.06-0.08d.4.25 yr float 35 bps102.120.29-0.07-0.07Exercise 8.Use the assumptions of Chapter 3 (Exercise 7) for the pricing of the inverse floater.a. i. $40 mnii. Refer to Exercise 7 in Chapter 3 for the pricing of the respective securities.14

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