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

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1、CHAPTER30Convexity,Timing,andQuantoAdjustmentsPracticeQuestionsProblem30.1.ExplainhowyouwouldvalueaderivativethatpaysoffIooRinfiveyearswhereRistheone-yearinterestrate(annuallycompounded)observedinfouryears.Whatdifferencewoulditmakeifthepayoffwerein(a)4yearsand(b)6years?Thevalueofthederivativeis100/?

2、45P(0,5)whereP(V)isthevalueofat-yearzero-couponbondtodayandR1histheforwardratefortheperiodbetweenr1andt2,expressedwithannualcompounding.Ifthepayoffismadeinfouryearsthevalueis100(5+c)P(0,4)wherecistheconvexityadjustmentgivenbyequation(30.2).Theformulafortheconvexityadjustmentis:C=强生0+,5)wherefflyisth

3、evolatilityoftheforwardratebetweentimesZ1andt2.Theexpression100(45+c)istheexpectedpayoffinaworldthatisforwardriskneutralwithrespecttoazero-couponbondmaturingattimefouryears.Ifthepayoffismadeinsixyears,thevalueisfromequation(30.4)givenbyIoO(R4$+C)P(0,6)exp_1+叫,6_wherepisthecorrelationbetweenthe(4,5)a

4、nd(4,6)forwardrates.AsanapproximationwecanassumethatP=1,45=46,and/?45=/?46.Approximatingtheexponentialfunctionwethengetthevalueofthederivativeas100(R4一C)尸(0,6).Problem30.2.Explainwhetheranyconvexityortimingadjustmentsarenecessarywhen(a) Wewishtovalueaspreadoptionthatpaysoffeveryquartertheexcess(ifan

5、y)ofthefive-yearswaprateoverthethree-monthLlBORrateappliedtoaprincipalof$100.Thepayoffoccurs90daysaftertheratesareobserved.(b) Wewishtovalueaderivativethatpaysoffeveryquarterthethree-monthLIBORrateminusthethree-monthTreasuryhillrate.Thepayoffoccurs90daysaftertheratesareobserved.(a) Aconvexityadjustm

6、entisnecessaryfortheswaprate(b) Noconvexityortimingadjustmentsarenecessary.Problem30.3.SupposethatinExample29.3ofSection29.2thepayoffoccursafteroneyear(i.e.,whentheinterestrateisobserved)ratherthanin15months.WhatdifferencedoesthismaketotheinputstoBlack,smodels?Therearetwodifferences.Thediscountingis

7、doneovera1.0-yearperiodinsteadofovera1.25-yearperiod.Alsoaconvexityadjustmenttotheforwardrateisnecessary.Fromequation(30.2)theconvexityadjustmentis:1 +0.25 0.0700720220-251=0.00005orabouthalfabasispoint.IntheformulaforthecapletWesetFk=0.07005insteadof0.07.Thismeansthat4=-0.5642andd2=-0.7642.Withcont

8、inuouscompoundingthe15-monthrateis6.5%andtheforwardratebetween12and15monthsis6.94%.The12monthrateistherefore6.39%Thecapletpricebecomes0.2510,00069394x,00.07005V(-0.5642)-0.08N(-0.7642)=5.29or$5.29.Problem30.4.TheLIBORZswapyieldcurve(whichisusedfordiscounting)isflatat10%perannumwithannualcompounding.

9、Calculatethevalueofaninstrumentwhere,infiveyears,time,thetwo-yearswaprate(Wifhannualcompounding)isreceivedandafixedrateof10%ispaid.Bothareappliedtoanotionalprincipalof$100.Assumethatthevolatilityoftheswaprateis20%perannum.Explainwhythevalueoftheinstrumentisdifferentfromzero.Theconvexityadjustmentdis

10、cussedinSection30.1leadstotheinstrumentbeingworthanamountslightlydifferentfromzero.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)=4.6582andtheco

11、nvexityadjustmentthatmustbemadeforthetwo-yearswap-rateis0.50.120.2254,6582=0.002681.7355Wecanthereforevaluetheinstrumentontheassumptionthattheswapratewillbe10.268%infiveyears.Thevalueoftheinstrumentisor$0.167.0.2681.15= 0.167Problem30.5.WhatdifferencedoesitmakeinProblem30.4iftheswaprateisobservedinf

12、iveyears,buttheexchangeofpaymentstakesplacein(a)sixyears,and(b)sevenyears?Assumethatthevolatilitiesofallforwardratesare20%.Assumealsothattheforwardswapratefortheperiodbetweenyearsfiveandsevenhasacorrelationof0.8withtheforwardinterestratebetweenyearsfiveandsixandacorrelationof0.95withtheforwardintere

13、stratebetweenyearsfiveandseven.exp -0.8 0.20 0.200.15=0.9856InthiscaseWehavetomakeatimingadjustmentaswellasaconvexityadjustmenttotheforwardswaprate.For(a)equation(30.4)showsthatthetimingadjustmentinvolvesmultiplyingtheswaprateby1+0.1sothatitbecomes10.2680.9856=10.120.Thevalueoftheinstrumentis0.120C-

14、=0.l)oo1.16or$0,068.For(b)equation(30.4)showsthatthetimingadjustmentinvolvesmultiplyingtheswapratebyexp -0.95 0.2 0.2 0.125=0.9660sothatitbecomes10.2680.966=9.919.Thevalueoftheinstrumentisnow一”“or-$0,042.Problem30.6.ThepriceofabondattimeT,measuredintermsofitsyield,isG(y).AssumegeometricBrownianmotio

15、nfortheforwardbondyield,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)y1dtG,(y)yydz(b) Sincetheexpect