赫尔《期权、期货及其他衍生产品》（第8版）复习笔记及课后习题详解 (50).docx

CHAPTER30Convexity,Timing,andQuantoAdjustmentsPracticeQuestionsProblem30.1.ExplainhowyouwouldvalueaderivativethatpaysoffIooRinfiveyearswhereRistheone-yearinterestrate(annuallycompounded)observedinfouryears.Whatdifferencewoulditmakeifthepayoffwerein(a)4yearsand(b)6years?Thevalueofthederivativeis100/?45P(0,5)whereP(V)isthevalueofat-yearzero-couponbondtodayandR1histheforwardratefortheperiodbetweenr1andt2,expressedwithannualcompounding.Ifthepayoffismadeinfouryearsthevalueis100(5+c)P(0,4)wherecistheconvexityadjustmentgivenbyequation(30.2).Theformulafortheconvexityadjustmentis:C=强生0+¾,5)wherefflyisthevolatilityoftheforwardratebetweentimesZ1andt2.Theexpression100(45+c)istheexpectedpayoffinaworldthatisforwardriskneutralwithrespecttoazero-couponbondmaturingattimefouryears.Ifthepayoffismadeinsixyears,thevalueisfromequation(30.4)givenbyIoO(R4$+C)P(0,6)exp_1+叫,6_wherepisthecorrelationbetweenthe(4,5)and(4,6)forwardrates.AsanapproximationwecanassumethatP=1,45=46,and/?45=/?46.Approximatingtheexponentialfunctionwethengetthevalueofthederivativeas100(R4一C)尸(0,6).Problem30.2.Explainwhetheranyconvexityortimingadjustmentsarenecessarywhen(a) Wewishtovalueaspreadoptionthatpaysoffeveryquartertheexcess(ifany)ofthefive-yearswaprateoverthethree-monthLlBORrateappliedtoaprincipalof$100.Thepayoffoccurs90daysaftertheratesareobserved.(b) Wewishtovalueaderivativethatpaysoffeveryquarterthethree-monthLIBORrateminusthethree-monthTreasuryhillrate.Thepayoffoccurs90daysaftertheratesareobserved.(a) Aconvexityadjustmentisnecessaryfortheswaprate(b) Noconvexityortimingadjustmentsarenecessary.Problem30.3.SupposethatinExample29.3ofSection29.2thepayoffoccursafteroneyear(i.e.,whentheinterestrateisobserved)ratherthanin15months.WhatdifferencedoesthismaketotheinputstoBlack,smodels?Therearetwodifferences.Thediscountingisdoneovera1.0-yearperiodinsteadofovera1.25-yearperiod.Alsoaconvexityadjustmenttotheforwardrateisnecessary.Fromequation(30.2)theconvexityadjustmentis:1 +0.25 × 0.070072×022×0-25×1=0.00005orabouthalfabasispoint.IntheformulaforthecapletWesetFk=0.07005insteadof0.07.Thismeansthat4=-0.5642andd2=-0.7642.Withcontinuouscompoundingthe15-monthrateis6.5%andtheforwardratebetween12and15monthsis6.94%.The12monthrateistherefore6.39%Thecapletpricebecomes0.25×10,000°69394x,00.07005V(-0.5642)-0.08N(-0.7642)=5.29or$5.29.Problem30.4.TheLIBORZswapyieldcurve(whichisusedfordiscounting)isflatat10%perannumwithannualcompounding.Calculatethevalueofaninstrumentwhere,infiveyears,time,thetwo-yearswaprate(Wifhannualcompounding)isreceivedandafixedrateof10%ispaid.Bothareappliedtoanotionalprincipalof$100.Assumethatthevolatilityoftheswaprateis20%perannum.Explainwhythevalueoftheinstrumentisdifferentfromzero.TheconvexityadjustmentdiscussedinSection30.1leadstotheinstrumentbeingworthanamountslightlydifferentfromzero.DefineG(y)asthevalueasseeninfiveyearsofatwo-yearbondwithacouponof10%asafunctionofitsyield.G(y) =0.11.117 + y (i + y)2G'(y) = 0.1(i+y)22.2(i + »Gy) =0.26.6r(l+y)3(l+y)4ItfollowsthatG'(0.1)=-1.7355andG*(0.1)=4.6582andtheconvexityadjustmentthatmustbemadeforthetwo-yearswap-rateis0.5×0.12×0.22×5×4,6582=0.002681.7355Wecanthereforevaluetheinstrumentontheassumptionthattheswapratewillbe10.268%infiveyears.Thevalueoftheinstrumentisor$0.167.0.2681.15= 0.167Problem30.5.WhatdifferencedoesitmakeinProblem30.4iftheswaprateisobservedinfiveyears,buttheexchangeofpaymentstakesplacein(a)sixyears,and(b)sevenyears?Assumethatthevolatilitiesofallforwardratesare20%.Assumealsothattheforwardswapratefortheperiodbetweenyearsfiveandsevenhasacorrelationof0.8withtheforwardinterestratebetweenyearsfiveandsixandacorrelationof0.95withtheforwardinterestratebetweenyearsfiveandseven.exp -0.8× 0.20 ×0.20×0.1×5=0.9856InthiscaseWehavetomakeatimingadjustmentaswellasaconvexityadjustmenttotheforwardswaprate.For(a)equation(30.4)showsthatthetimingadjustmentinvolvesmultiplyingtheswaprateby1+0.1sothatitbecomes10.268×0.9856=10.120.Thevalueoftheinstrumentis0.120C-=0.l)oo1.16or$0,068.For(b)equation(30.4)showsthatthetimingadjustmentinvolvesmultiplyingtheswapratebyexp -0.95 × 0.2 × 0.2 ×0.1×2×5=0.9660sothatitbecomes10.268×0.966=9.919.Thevalueoftheinstrumentisnow一”“or-$0,042.Problem30.6.ThepriceofabondattimeT,measuredintermsofitsyield,isG(y).AssumegeometricBrownianmotionfortheforwardbondyield,y,inaworldthatisforwardriskneutralwithrespecttoabondmaturingattimeT.Supposethatthegrowthrateoftheforwardbondyieldisaanditsvolatilityv.(a) UseIto,slemmatocalculatetheprocessfarthefatardbondpriceintermsofa,v,y,andG(y).(b) Theforwardbondpriceshouldfollowamartingaleintheworldconsidered.Usethisfacttocalculateanexpressionfora.(c) Showthattheexpressionforais,toafirstapproximation,consistentwithequation(30.J).(a) Theprocessforyisdy=aydt-yydzTheforwardbondpriceisG(y).FromIt,slemma,itsprocessisdG(y)=G,(y)ay+gGy)y1dt÷G,(y)yydz(b) Sincetheexpect