期权期货及其他衍生品章节习题
约翰.赫尔,期权期货和其他衍生品(third edition)习题答案
12.1 一个证券组合当前价值为＄1000万，β值为1.0，S&P100目前位于250，解释一个执行价格为240。
标的物为S&P100的看跌期权如何为该组合进行保险？当S&P100跌到480，这个组合的期望价值是10 ×（480/500）=＄9.6million.买看跌期权10，000，000/500=20，000可以防止这个组合下跌到＄9.6million下的损失。
因此总共需要200份合约12.2 “一旦我们知道了支付连续红利股票的期权的定价方法，我们便知道了股票指数期权、货币期权和期货期权的定价”。
请解释这句话。
一个股票指数类似一个连续支付红利的股票12.3 请说明日圆看涨期权与日圆期货看涨期权的不同之处一个日元的看涨期权给了持有者在未来某个时刻以确定的价格购买日圆的权利，一个日圆远期看涨期权给予持有者在未来时刻远期价格超过特定范围按原先价格购买日圆的权利。
如果远期齐权行使，持有者将获得一个日圆远期和约的多头。
12.4请说明货币期权是如何进行套期保值的？12.5 计算3个月期，处于平价状态的欧式看涨股票指数期权的价值。
指数为250。
无风险年利率为10%，指数年波动率为18%，指数的年红利收益率为3%。
一个日元的看涨期权给了持有者在未来某个时刻以确定的价格购买日圆的权利，一个日圆远期看涨期权给予持有者在未来时刻远期价格超过特定范围按原先价格购买日圆的权利。
如果远期齐权行使，持有者将获得一个日圆远期和约的多头。
12.6 有一美式看涨期货期权，期货合约和期权合约同时到期。
在任何情况下期货期权比相应的标的物资产的美式期权更值钱？当远期价格大于即期价格时，美式远期期权在远期和约到期前的价值大于相对应的美式期权/12.7 计算5个月有效期的欧式看跌期货期权的价值。
期货价格为＄19，执行价格为＄20，无风险年利率为12%。
期货价格的年波动率为20%。
本题中12.8 假设交易所构造了一个股票指数。
约翰.赫尔_期权期货和其他衍生品第八版部分习题答案汇总
计算题1.1一个投资者进入了一个远期合约的短头寸，在该合约中投资者能够以1.4000的汇率（美元/英镑）卖出1000 000英镑。
当远期合约到期时的汇率为以下数值时⑴ 1.3900，⑵1.4200，投资者的损益分别为多少？（1.4-1.39）*1000000=1000（1.4-1.42）*1000000=-20001.2 当前黄金市价为每盎司$1,000,一个一年期的远期合约的执行价格为$1,200,一个套利者能够以每年10%的利息借入资金，套利者应该如何做才能达到套利目的？假设黄金存储费为0，同时黄金不会带来任何利息收入借入1000美元买入黄金，一年后以1200美元卖出（1200-1000）- （1000*10%）= 1002.1 一家公司进入一个合约的短头寸，在合约中公司以每莆式耳450美分卖出5000蒲式耳小麦。
初始保证金为3000美元，维持保证金为2000美元，价格如何变化会导致保证金的催收？在什么情况下公司可以从保证金账户提取1500美元？如果合约亏损1000美元，则会导致保证金催收，假设价格为S时收到了保证金催收，则（S1-4.5）*5000=1000 S1=470美分，即如果未来价格高于470美分会收到保证呢过金催收通知假设价格为S2时可以从保证金账户提取1500美金，则（4.5-S2）*5000=1500 S2=420美分即价格跌至420美分时，可以从保证金账户中提取1500美元3.1 假定今天是7月16日，一家公司持有价值1亿美元的股票组合，此股票组合的beta系数是1.2，这家公司希望采用CME在12月份到期SP500股指期货在7月16日至11月16日之间将beta系数由1.2变为0.5，SP500股指当前价值为1000，而每一份期货合约面值为250美元与股指的乘积。
1）公司应做什么样的交易2）Beta由1.2变为1.5，公司应该持有什么样的头寸1)该公司需要空头（1.2-0.5）*100000000/1000*250=280份合约2)该公司需要多头（1.5-1.2）*100000000/1000*250=120份合约3.2 P52 3.294.1P71 4.304.2 P71 4.325.1 9.110.110.2假设你是一家杠杆比例很高的公司的经理及唯一所有者。
期权期货与其他衍生产品第九版课后习题与答案Chapter
CHAPTER 31Interest Rate Derivatives: Models of the Short Rate Practice QuestionsProblem 31.1.What is the difference between an equilibrium model and a no-arbitrage model?Equilibrium models usually start with assumptions about economic variables and derive the behavior of interest rates. The initial term structure is an output from the model. In ano-arbitrage model the initial term structure is an input. The behavior of interest rates in ano-arbitrage model is designed to be consistent with the initial term structure.Problem 31.2.Suppose that the short rate is currently 4% and its standard deviation is 1% per annum. What happens to the standard deviation when the short rate increases to 8% in (a) Vasicek’s model;(b) Rendleman and Bartter’s mod el; and (c) the Cox, Ingersoll, and Ross model?In Vasicek’s model the standard deviation stays at 1%. In the Rendleman and Bartter model the standard deviation is proportional to the level of the short rate. When the short rate increases from 4% to 8% the standard deviation increases from 1% to 2%. In the Cox, Ingersoll, and Ross model the standard deviation of the short rate is proportional to the square root of the short rate. When the short rate increases from 4% to 8% the standard deviation increases from 1% to 1.414%.Problem 31.3.If a stock price were mean reverting or followed a path-dependent process there would be market inefficiency. Why is there not market inefficiency when the short-term interest rate does so?If the price of a traded security followed a mean-reverting or path-dependent process there would be market inefficiency. The short-term interest rate is not the price of a traded security. In other words we cannot trade something whose price is always the short-term interest rate. There is therefore no market inefficiency when the short-term interest rate follows amean-reverting or path-dependent process. We can trade bonds and other instruments whose prices do depend on the short rate. The prices of these instruments do not followmean-reverting or path-dependent processes.Problem 31.4.Explain the difference between a one-factor and a two-factor interest rate model.In a one-factor model there is one source of uncertainty driving all rates. This usually means that in any short period of time all rates move in the same direction (but not necessarily by the same amount). In a two-factor model, there are two sources of uncertainty driving all rates. The first source of uncertainty usually gives rise to a roughly parallel shift in rates. The second gives rise to a twist where long and short rates moves in opposite directions.Problem 31.5.Can the approach described in Section 31.4 for decomposing an option on a coupon-bearing bond into a portfolio of options on zero-coupon bonds be used in conjunction with a two-factor model? Explain your answer.No. The approach in Section 31.4 relies on the argument that, at any given time, all bond prices are moving in the same direction. This is not true when there is more than one factor.Problem 31.6.Suppose that 01a =. and 01b =. in both the Vasicek and the Cox, Ingersoll, Ross model. In both models, the initial short rate is 10% and the initial standard deviation of the short rate change in a short time t ∆is 0.zero-coupon bond that matures in year 10.In Vasicek’s model, 01a =., 01b =., and 002σ=. so that01101(10)(1)63212101B t t e -.⨯,+=-=..22(63212110)(010100002)00004632121(10)exp 00104A t t ⎡⎤.-.⨯.-..⨯.,+=-⎢⎥..⎣⎦071587=. The bond price is therefore 63212101071587038046e -.⨯..=.In the Cox, Ingersoll, and Ross model, 01a =., 01b =.and 00200632σ=.=.. Also013416γ==. Define10()(1)2092992a e γβγγ=+-+=.102(1)(10)607650e B t t γβ-,+==.225()2(10)069746ab a e A t t σγγβ/+⎛⎫,+==. ⎪⎝⎭The bond price is therefore 60765001069746037986e -.⨯..=.Problem 31.7.Suppose that 01a =., 008b =., and 0015σ=. in Vasicek’s model with the initial value of the short rate being 5%. Calculate the price of a one-year European call option on azero-coupon bond with a principal of $100 that matures in three years when the strike price is $87.Using the notation in the text, 3s =, 1T =, 100L =, 87K =, and2010015(1002588601P e σ-⨯..=-=.. From equation (31.6), (01)094988P ,=., (03)085092P ,=., and 114277h =. so thatequation (31.20) gives the call price as call price is 100085092(114277)87094988(111688)259N N ⨯.⨯.-⨯.⨯.=. or $2.59.Problem 31.8.Repeat Problem 31.7 valuing a European put option with a strike of $87. What is the put –call parity relationship between the prices of European call and put options? Show that the put and call option prices satisfy put –call parity in this case.As mentioned in the text, equation (31.20) for a call option is essentially the same as Bl ack’s model. By analogy with Black’s formulas corresponding expression for a put option is (0)()(0)()P KP T N h LP s N h σ,-+-,- In this case the put price is 87094988(111688)100085092(114277)014N N ⨯.⨯-.-⨯.⨯-.=.Since the underlying bond pays no coupon, put –call parity states that the put price plus the bond price should equal the call price plus the present value of the strike price. The bond price is 85.09 and the present value of the strike price is 870949888264⨯.=.. Put –call parity is therefore satisfied:82642598509014.+.=.+.Problem 31.9.Suppose that 005a =., 008b =., and 0015σ=. in Vasicek’s model with the initialshort-term interest rate being 6%. Calculate the price of a 2.1-year European call option on a bond that will mature in three years. Suppose that the bond pays a coupon of 5% semiannually. The principal of the bond is 100 and the strike price of the option is 99. The strike price is the cash price (not the quoted price) that will be paid for the bond.As explained in Section 31.4, the first stage is to calculate the value of r at time 2.1 years which is such that the value of the bond at that time is 99. Denoting this value of r by r *, we must solve(2125)(2130)25(2125)1025(2130)99B r B r A e A e **-.,.-.,...,.+..,.=where the A and B functions are given by equations (31.7) and (31.8). In this case A (2.1, 2.5)=0.999685, A (2.1,3.0)=0.998432, B(2.1,2.5)=0.396027, and B (2.1, 3.0)= 0.88005. and Solver shows that 065989.0*=r . Since434745.2)5.2,1.2(5.2*)5.2,1.2(=⨯-r B e Aand56535.96)0.3,1.2(5.102*)0.3,1.2(=⨯-r B e Athe call option on the coupon-bearing bond can be decomposed into a call option with a strike price of 2.434745 on a bond that pays off 2.5 at time 2.5 years and a call option with a strike price of 96.56535 on a bond that pays off 102.5 at time 3.0 years. The options are valued using equation (31.20).For the first option L =2.5, K = 2.434745, T = 2.1, and s =2.5. Also, A (0,T )=0.991836, B (0,T ) = 1.99351, P (0,T )=0.880022 while A (0,s )=0.988604, B (0,s )=2.350062, andP (0,s )=0.858589. Furthermore σP = 0.008176 and h = 0.223351. so that the option price is 0.009084.For the second option L =102.5, K = 96.56535, T = 2.1, and s =3.0. Also, A (0,T )=0.991836, B (0,T ) = 1.99351, P (0,T )=0.880022 while A (0,s )=0.983904, B (0,s )=2.78584, andP (0,s )=0.832454. Furthermore σP = 0.018168 and h = 0.233343. so that the option price is0.806105.The total value of the option is therefore 0.0090084+0.806105=0.815189.Problem 31.10.Use the answer to Problem 31.9 and put –call parity arguments to calculate the price of a put option that has the same terms as the call option in Problem 31.9.Put-call parity shows that: 0()c I PV K p B ++=+ or 0()()p c PV K B I =+--where c is the call price, K is the strike price, I is the present value of the coupons, and 0B is the bond price. In this case 08152c =., ()99(021)871222PV K P =⨯,.=., 025(025)1025(03)874730B I P P -=.⨯,.+.⨯,=. so that the put price is0815287122287473004644.+.-.=.Problem 31.11.In the Hull –White model, 008a =. and 001σ=.. Calculate the price of a one-year European call option on a zero-coupon bond that will mature in five years when the term structure is flat at 10%, the principal of the bond is $100, and the strike price is $68.Using the notation in the text 011(0)09048P T e -.⨯,==. and 015(0)06065P s e -.⨯,==.. Also4008001(100329008P e σ-⨯..=-=.. and 04192h =-. so that the call price is10006065()6809048()0439P N h N h σ⨯.-⨯.-=.Problem 31.12.Suppose that 005a =. and 0015σ=. in the Hull –White model with the initial term structure being flat at 6% with semiannual compounding. Calculate the price of a 2.1-year European call option on a bond that will mature in three years. Suppose that the bond pays a coupon of 5% per annum semiannually. The principal of the bond is 100 and the strike price of the option is 99. The strike price is the cash price (not the quoted price) that will be paid for the bond.This problem is similar to Problem 31.9. The difference is that the Hull –White model, which fits an initial term structure, is used instead of Vasicek’s model where the initial term structure is determined by the model.The yield curve is flat with a continuously compounded rate of 5.9118%.As explained in Section 31.4, the first stage is to calculate the value of r at time 2.1 years which is such that the value of the bond at that time is 99. Denoting this value of r by r *, we must solve(2125)(2130)25(2125)1025(2130)99B r B r A e A e **-.,.-.,...,.+..,.=where the A and B functions are given by equations (31.16) and (31.17). In this case A (2.1, 2.5)=0.999732, A (2.1,3.0)=0.998656, B(2.1,2.5)=0.396027, and B (2.1, 3.0)= 0.88005. and Solver shows that 066244.0*=r . Since434614.2)5.2,1.2(5.2*)5.2,1.2(=⨯-r B e Aand56539.96)0.3,1.2(5.102*)0.3,1.2(=⨯-r B e Athe call option on the coupon-bearing bond can be decomposed into a call option with a strike price of 2.434614 on a bond that pays off 2.5 at time 2.5 years and a call option with a strike price of 96.56539 on a bond that pays off 102.5 at time 3.0 years. The options are valued using equation (31.20).For the first option L =2.5, K = 2.434614, T = 2.1, and s =2.5. Also, P (0,T )=exp(-0.059118×2.1)=0.88325 and P (0,s )= exp(-0.059118×2.5)=0.862609. Furthermore σP = 0.008176 and h = 0.353374. so that the option price is 0.010523. For the second option L =102.5, K = 96.56539, T = 2.1, and s =3.0. Also, P (0,T )=exp(-0.059118×2.1)=0.88325 and P (0,s )= exp(-0.059118×3.0)=0.837484. Furthermore σP = 0.018168 and h = 0.363366. so that the option price is 0.934074.The total value of the option is therefore 0.010523+0.934074=0.944596.Problem 31.13.Observations spaced at intervals ∆t are taken on the short rate. The ith observation is r i (1 ≤ i ≤ m). Show that the maximum likelihood estimates of a, b, and σ in Vasicek’s model are given by maximizing[]∑=--⎪⎪⎭⎫ ⎝⎛∆σ∆----∆σ-m i i i i t t r b a r r t 122112)()ln(What is the corresponding result for the CIR model?The change r i –r i -1 is normally distributed with mean a (b − r i -1) and variance σ2∆t. The probability density of the observation is⎪⎭⎫⎝⎛∆σ---∆πσ--t r b a r r ti i i 21122)(exp 21We wish to maximize∏=--⎪⎭⎫⎝⎛∆σ---∆πσmi i i i t r b a r r t121122)(exp 21Taking logarithms, this is the same as maximizing[]∑=--⎪⎪⎭⎫ ⎝⎛∆σ∆----∆σ-m i i i i t t r b a r r t 122112)()ln(In the case of the CIR model, the change r i –r i -1 is normally distributed with mean a (b − r i -1) and variance t r i ∆σ-12and the maximum likelihood function becomes[]∑=----⎪⎪⎭⎫ ⎝⎛∆σ∆----∆σ-mi i i i i i t r t r b a r r t r 11221112)()ln(Problem 31.14.Suppose 005a =., 0015σ=., and the term structure is flat at 10%. Construct a trinomial tree for the Hull –White model where there are two time steps, each one year in length.The time step, t ∆, is 1 so that 0002598r ∆=.=.. Also max 4j = showing that the branching method should change four steps from the center of the tree. With only three steps we never reach the point where the branching changes. The tree is shown in Figure S31.1.Figure S31.1: Tree for Problem 31.14Problem 31.15.Calculate the price of a two-year zero-coupon bond from the tree in Figure 31.6.A two-year zero-coupon bond pays off $100 at the ends of the final branches. At nodeB it is worth 01211008869e -.⨯=.. At nodeC it is worth 010********e -.⨯=.. At nodeD it is worth 00811009231e -.⨯=.. It follows that at node A the bond is worth 011(88690259048059231025)8188e -.⨯.⨯.+.⨯.+.⨯.=. or $81.88Problem 31.16.Calculate the price of a two-year zero-coupon bond from the tree in Figure 31.9 and verify that it agrees with the initial term structure.A two-year zero-coupon bond pays off $100 at time two years. At nodeB it is worth 0069311009330e -.⨯=.. At nodeC it is worth 0052011009493e -.⨯=.. At nodeD it is worth 0034711009659e -.⨯=.. It follows that at node A the bond is worth 003821(933001679493066696590167)9137e -.⨯.⨯.+.⨯.+.⨯.=.or $91.37. Because 00451229137100e -.⨯.=, the price of the two-year bond agrees with the initial term structure.Problem 31.17.Calculate the price of an 18-month zero-coupon bond from the tree in Figure 31.10 and verify that it agrees with the initial term structure.An 18-month zero-coupon bond pays off $100 at the final nodes of the tree. At node E it is worth 0088051009570e -.⨯.=.. At node F it is worth 00648051009681e -.⨯.=.. At node G it is worth 00477051009764e -.⨯.=.. At node H it is worth 00351051009826e -.⨯.=.. At node I it is worth 00259051009871e .⨯.=.. At node B it is worth 0056405(011895700654968102289764)9417e -.⨯..⨯.+.⨯.+.⨯.=.Similarly at nodes C and D it is worth 95.60 and 96.68. The value at node A is therefore 0034305(016794170666956001679668)9392e -.⨯..⨯.+.⨯.+.⨯.=.The 18-month zero rate is 0181500800500418e -.⨯..-.=.. This gives the price of the 18-month zero-coupon bond as 00418151009392e -.⨯.=. showing that the tree agrees with the initial term structure.Problem 31.18.What does the calibration of a one-factor term structure model involve?The calibration of a one-factor interest rate model involves determining its volatility parameters so that it matches the market prices of actively traded interest rate options as closely as possible.Problem 31.19.Use the DerivaGem software to value 14⨯, 23⨯, 32⨯, and 41⨯ European swap options to receive fixed and pay floating. Assume that the one, two, three, four, and five year interest rates are 6%, 5.5%, 6%, 6.5%, and 7%, respectively. The payment frequency on the swap is semiannual and the fixed rate is 6% per annum with semiannual compounding. Use the Hull –White model with 3a %= and 1%σ=. Calcula te the volatility implied by Black’s model for each option.The option prices are 0.1302, 0.0814, 0.0580, and 0.0274. The implied Black volatilities are 14.28%, 13.64%, 13.24%, and 12.81%Problem 31.20.Prove equations (31.25), (31.26), and (31.27).From equation (31.15) ()()()()r t B t t t P t t t A t t t e -,+∆,+∆=,+∆ Also ()()R t t P t t t e -∆,+∆= so that()()()()R t t r t B t t t e A t t t e -∆-,+∆=,+∆or()()()()()()()()R t B t T t B t t t r t B t T B t T B t t t e eA t t t -,∆/,+∆-,,/,+∆=,+∆ Hence equation (31.25) is true with()ˆ()(B t T t Bt T B t t t ,∆,=,+∆ and()()()ˆ()()B t T B t t t A t T At T A t t t ,/,+∆,,=,+∆ or()ˆln ()ln ()ln ()()B t T At T A t T A t t t B t t t ,,=,-,+∆,+∆Problem 31.21.(a) What is the second partial derivative of P(t,T) with respect to r in the Vasicek and CIR models?(b) In Section 31.2, ˆDis presented as an alternative to the standard duration measure D. What is a similar alternative ˆCto the convexity measure in Section 4.9? (c) What is ˆCfor P(t,T)? How would you calculate ˆC for a coupon-bearing bond? (d) Give a Taylor Series expansion for ∆P(t,T) in terms of ∆r and (∆r)2 for Vasicek and CIR.(a) ),(),(),(),(),(2),(222T t P T t B e T t A T t B rT t P r T t B ==∂∂- (b) A corresponding definition for Cˆis 221r QQ ∂∂(c) When Q =P (t ,T ), 2),(ˆT t B C=For a coupon-bearing bond C ˆis a weighted average of the Cˆ’s for the constituent zero -coupon bonds where weights are proportional to bond prices. (d)+∆+∆-=+∆∂∂+∆∂∂=∆22222),(),(21),(),(),(21),(),(r T t P T t B r T t P T t B r rT t P r r T t P T t PProblem 31.22.Suppose that the short rate r is 4% and its real-world process is0.1[0.05]0.01dr r dt dz =-+while the risk-neutral process is0.1[0.11]0.01dr r dt dz =-+(a) What is the market price of interest rate risk?(b) What is the expected return and volatility for a 5-year zero-coupon bond in the risk-neutral world?(c) What is the expected return and volatility for a 5-year zero-coupon bond in the real world?(a) The risk neutral process for r has a drift rate which is 0.006/r higher than the real world process. The volatility is 0.01/r . This means that the market price of interest rate risk is −0.006/0.01 or −0.6.(b) The expected return on the bond in the risk-neutral world is the risk free rate of 4%. The volatility is 0.01×B (0,5) where935.31.01)5,0(51.0=-=⨯-e Bi.e., the volatility is 3.935%.(c) The process followed by the bond price in a risk-neutral world isPdz Pdt dP 03935.004.0-=Note that the coefficient of dz is negative because bond prices are negatively correlated with interest rates. When we move to the real world the return increases by the product of the market price of dz risk and −0.03935. The bond price process becomes:Pdz Pdt dP 03935.0)]03935.06.0(04.0[--⨯-+=orPdz Pdt dP 03935.006361.0-=The expected return on the bond increases from 4% to 6.361% as we move from the risk-neutral world to the real world.Further QuestionsProblem 31.23.Construct a trinomial tree for the Ho and Lee model where 002σ=.. Suppose that the initial zero-coupon interest rate for a maturities of 0.5, 1.0, and 1.5 years are 7.5%, 8%, and 8.5%. Use two time steps, each six months long. Calculate the value of a zero-coupon bond with a face value of $100 and a remaining life of six months at the ends of the final nodes of the tree. Use the tree to value a one-year European put option with a strike price of 95 on the bond. Compare the price given by your tree with the analytic price given by DerivaGem.The tree is shown in Figure S31.2. The probability on each upper branch is 1/6; the probability on each middle branch is 2/3; the probability on each lower branch is 1/6. The six month bond prices nodes E, F, G, H, I are 0144205100e -.⨯., 0119705100e -.⨯., 0095205100e -.⨯., 0070705100e -.⨯., and 0046205100e -.⨯., respectively. These are 93.04, 94.19, 95.35, 96.53, and 97.72. The payoffs from the option at nodes E, F, G, H, and I are therefore 1.96, 0.81, 0, 0, and 0. The value at node B is 0109505(0166719606667081)08192e -.⨯..⨯.+.⨯.=.. The value at node C is00851050166708101292e -.⨯..⨯.⨯=.. The value at node D is zero. The value at node A is 0075005(01667081920666701292)0215e -.⨯..⨯.+.⨯.=. The answer given by DerivaGem using the analytic approach is 0.209.Figure S31.2: Tree for Problem 31.23Problem 31.24.A trader wishes to compute the price of a one-year American call option on a five-year bond with a face value of 100. The bond pays a coupon of 6% semiannually and the (quoted) strike price of the option is $100. The continuously compounded zero rates for maturities of sixmonths, one year, two years, three years, four years, and five years are 4.5%, 5%, 5.5%, 5.8%, 6.1%, and 6.3%. The best fit reversion rate for either the normal or the lognormal model has been estimated as 5%.A one year European call option with a (quoted) strike price of 100 on the bond is actively traded. Its market price is $0.50. The trader decides to use this option for calibration. Use the DerivaGem software with ten time steps to answer the following questions.(a) Assuming a normal model, imply the σ parameter from the price of the European option.(b) Use the σ parameter to calculate the price of the option when it is American. (c) Repeat (a) and (b) for the lognormal model. Show that the model used does notsignificantly affect the price obtained providing it is calibrated to the known European price.(d) Display the tree for the normal model and calculate the probability of a negative interest rate occurring.(e) Display the tree for the lognormal model and verify that the option price is correctly calculated at the node where, with the notation of Section 31.7, 9i = and 1j =-.Using 10 time steps:(a) The implied value of σ is 1.12%.(b) The value of the American option is 0.595(c) The implied value of σ is 18.45% and the value of the American option is 0.595. Thetwo models give the same answer providing they are both calibrated to the same European price.(d) We get a negative interest rate if there are 10 down moves. The probability of this is0.16667×0.16418×0.16172×0.15928×0.15687×0.15448×0.15212×0.14978×0.14747 ×0.14518=8.3×10-9 (e) The calculation is0052880101641791707502789e -.⨯..⨯.⨯=.Problem 31.25.Use the DerivaGem software to value 14⨯, 23⨯, 32⨯, and 41⨯ European swap options to receive floating and pay fixed. Assume that the one, two, three, four, and five year interest rates are 3%, 3.5%, 3.8%, 4.0%, and 4.1%, respectively. The payment frequency on the swap is semiannual and the fixed rate is 4% per annum with semiannual compounding. Use thelognormal model with 5a %=, 15%σ=, and 50 time steps. Calculate the volatility implied by Black’s model for each option.The values of the four European swap options are 1.72, 1.73, 1.30, and 0.65, respectively. The implied Black volatilities are 13.37%, 13.41%, 13.43%, and 13.42%, respectively.Problem 31.26.Verify that the DerivaGem software gives Figure 31.11 for the example considered. Use the software to calculate the price of the American bond option for the lognormal and normalmodels when the strike price is 95, 100, and 105. In the case of the normal model, assume that a = 5% and σ = 1%. Discuss the results in the context of the heavy-tails arguments of Chapter 20.With 100 time steps the lognormal model gives prices of 5.585, 2.443, and 0.703 for strike prices of 95, 100, and 105. With 100 time steps the normal model gives prices of 5.508, 2.522, and 0.895 for the three strike prices respectively. The normal model gives a heavier left tail and a less heavy right tail than the lognormal model for interest rates. This translates into a less heavy left tail and a heavier right tail for bond prices. The arguments in Chapter 20 show that we expect the normal model to give higher option prices for high strike prices and lower option prices for low strike. This is indeed what we find.Problem 31.27.Modify Sample Application G in the DerivaGem Application Builder software to test theconvergence of the price of the trinomial tree when it is used to price a two-year call option on a five-year bond with a face value of 100. Suppose that the strike price (quoted) is 100, the coupon rate is 7% with coupons being paid twice a year. Assume that the zero curve is as in Table 31.2. Compare results for the following cases:(a) Option is European; normal model with 001σ=. and 005a =..(b) Option is European; lognormal model with 015σ=. and 005a =..(c) Option is American; normal model with 001σ=. and 005a =..(d) Option is American; lognormal model with 015σ=. and 005a =..The results are shown in Figure S31.3.Figure S31.3: Tree for Problem 31.27Problem 31.28.Suppose that the (CIR) process for short-rate movements in the (traditional) risk-neutral world is()dr a b r dt =-+and the market price of interest rate risk is λ(a) What is the real world process for r?(b) What is the expected return and volatility for a 10-year bond in the risk-neutral world? (c) What is the expected return and volatility for a 10-year bond in the real world?(a) The volatility of r (i.e., the coefficient of rdz in the process for r ) is real world process for r is therefore increased by r r λσ⨯ so that the process isdz r dt r r b a dr σ+λσ+-=])([(b) The expected return is r and the volatility is (,B t T σin the risk-neutral world.(c) The expected return is r T t B r ),(λσ+ and the volatility is as in (b) in the real world.。
期权期货和其它衍生产品约翰赫尔答案
期权期货和其它衍生产品约翰赫尔答案1.1请说明远期多头与远期空头的区别。
答：远期多头指交易者协定今后以某一确定价格购入某种资产；远期空头指交易者协定今后以某一确定价格售出某种资产。
1.2请详细说明套期保值、投机与套利的区别。
答：套期保值指交易者采取一定的措施补偿资产的风险暴露；投机不对风险暴露进行补偿，是一种〝赌博行为〞；套利是采取两种或更多方式锁定利润。
1.3请说明签订购买远期价格为$50的远期合同与持有执行价格为$50的看涨期权的区别。
答：第一种情形下交易者有义务以50$购买某项资产〔交易者没有选择〕，第二种情形下有权益以50$购买某项资产〔交易者能够不执行该权益〕。
1.4一位投资者出售了一个棉花期货合约，期货价格为每磅50美分，每个合约交易量为50，000磅。
请问期货合约终止时，当合约到期时棉花价格分别为〔a〕每磅48.20美分；〔b〕每磅51.30美分时，这位投资者的收益或缺失为多少？答：(a)合约到期时棉花价格为每磅$0.4820时，交易者收入：〔$0.5000-$0.4820〕×50，000＝$900;(b)合约到期时棉花价格为每磅$0.5130时，交易者缺失:($0.5130-$0.5000) ×50，000=$6501.5假设你出售了一个看跌期权，以$120执行价格出售100股IBM的股票，有效期为3个月。
IBM股票的当前价格为$121。
你是如何考虑的？你的收益或缺失如何？答：当股票价格低于$120时，该期权将不被执行。
当股票价格高于$120美元时，该期权买主执行该期权，我将缺失100(st-x)。
1.6你认为某种股票的价格将要上升。
现在该股票价格为$29,3个月期的执行价格为$30的看跌期权的价格为$2.90.你有$5,800资金能够投资。
现有两种策略：直截了当购买股票或投资于期权，请问各自潜在的收益或缺失为多少？答：股票价格低于$29时，购买股票和期权都将缺失，前者缺失为$5,800$29×(29-p),后者缺失为$5，800;当股票价格为〔29，30〕，购买股票收益为$5,800$29×(p-29),购买期权缺失为$5,800;当股票价格高于$30时，购买股票收益为$5,800$29×(p-29)，购买期权收益为$$5,800$29×(p-30)-5,800。
期权期货与其他衍生产品第九版课后习题与答案Chapter(.
cHAPTra KI betUrM M«4duxn«x<^ ar ••••• f«g ・t plM OM<a to MB ll ・—TL —▼rMdV — ■ M 9•&・ Mtw«aH 叩■厂 1—、M»A-l S*2*0«M ■■ ^7521 1*4SeM 0|«tc4WB* ap*w ^«K « <* o az «•«!io •e ・fOMHVu ・w« Aft. xW*・9«v MBMk t» W*»|w «WM !•«•••— %-V> K ・9. ■・•( •*©> T-»» ■.空竺J 马A 空兰.…计•A SK<•4 «>A7・・1・•・R ・08*・5ZE 11*.F<<O<MI4WV ■❷•waw JttM <«•<«■■ J ・I9H74・ MrtMyiiPH—W ・(VAlw«・・，•■■■>•・" M «M 4 MI « E f R —W •字•字）vfViurutMjnr^vMtai 111 A —i"・^«hW —WI4»••一 y wt 卯 v >•*♦ *» <wr* 戸 x« tv•I Mt «wM ■*■<> «W ■・vfl !• mtvwvF k» VKM »> t»l • p< c» IW Mk vtt»•••• MM• tl»>■斗・4r.•丄防| n*kw !•*J M **vWatAM)' H MB*・朴・卜:卞”・ 4tJJ• Mm K -W ・ r ・Q.・ 9-QS.M > T-«» ・4CMP»c»to« A ■a0>■!«<•(■・•《 Mkwtaa 血”•>« «WE>y«»«vP • IF —・••—■—■科M4_ <tl ・oaMalWMi .11创n 11MM ■— 一A""・v«r■n«^»a w'«•!•<.•..»■ •> i- a-k>―初|・(« _F0R »*i!<■<■« i ・9” ■*!•«••«・》MM E ・》*k>e« —«E.r ■MIC ■一* »» ■•• ”・$«s» •• ts. no ¥ wr ■5— wK ・・«■•*・•••■•••・•••• *♦"■・••••・”•・•・~•l*« f*ASK"*""b QU ，,•・"♦(■* rI・・t ・K •WR XVIM •一刪・0|«・・・<||"• 8K kfk •■>•••<• ■…m»«*w<WH ” IK* lMW«IM iewr mu ■・—A« 巧 ki j«"U >• >^*10* t••<!•.»<»• ••••(■a * 、・r R if 。
赫尔《期权、期货及其他衍生产品》（第7版）课后习题详解（曲率、时间与Quanto调整）
赫尔《期权、期货及其他衍⽣产品》（第7版）课后习题详解（曲率、时间与Quanto调整）29.2 课后习题详解⼀、问答题1. 解释你如何去对⼀个在5年后付出100R 的衍⽣产品定价，其中R 是在4年后所观察到的1年期利率(按年复利)。
当⽀付时间在第4年时，会有什么区别?当⽀付时间在第6年时，会有什么区别?Explain how you would value a derivative that pays off 100R in five years where R is the one-year interest rate (annually compounded) observed in four years. What difference would it make if the payoff were in four years? What difference would it make if tile payoff were in six years?答：衍⽣产品的价值是，其中P(0，t)是⼀个t 期零息债券的价格，为期限在和之间的远期利率，以年复利计息。
当⽀付时间在第4年时，价值为，其中c 为由教材中⽅程（29-2）得到的曲率调整。
曲率调整公式为：其中，是远期利率在时间和之间的波动率。
表达式100(R4,5 + c)为在⼀个远期风险中性的世界中，⼀个4年后到期的零息债券的预期收益。
如果在6年后进⾏⽀付，由教材中的⽅程（29-4）得到其价值为：其中，ρ为(4,5)和(4,6)远期利率之间的相关系数。
作为估计，假定，近似计算其指数函数，得到衍⽣产品的价值为：。
2. 解释在下⾯情况下，有没有必要做出任何曲率或时间调整?(a)要对⼀种期权定价，期权每个季度⽀付⼀次，数量等于5年的互换利率超出3个⽉LIBOR利率的部分(假如超出的话)，本⾦为100美元，收益发⽣在利率被观察到后的90天。
(b)要对⼀种差价期权定价，期权每季度⽀付⼀次，数量等于3个⽉的LIBOR利率减去3个⽉的短期国库券利率，收益发⽣在利率被观察后的90天。
约翰.赫尔,期权期货和其他衍生品(third edition)习题答案
CH99.1 股票现价为$40。
已知在一个月后股价为$42或$38。
无风险年利率为8％（连续复利）。
执行价格为$39的1个月期欧式看涨期权的价值为多少？ 解：考虑一资产组合：卖空1份看涨期权；买入Δ份股票。
若股价为$42，组合价值则为42Δ－3；若股价为$38，组合价值则为38Δ 当42Δ－3＝38Δ，即Δ＝0.75时，组合价值在任何情况下均为$28.5，其现值为：，0.08*0.0833328.528.31e −=即：－f ＋40Δ＝28.31 其中f 为看涨期权价格。
所以，f ＝40×0.75－28.31＝$1.69另解：（计算风险中性概率p ） 42p －38（1－p ）＝，p ＝0.56690.08*0.0833340e期权价值是其期望收益以无风险利率贴现的现值，即： f ＝（3×0.5669＋0×0.4331）＝$1.690.08*0.08333e−9.2 用单步二叉树图说明无套利和风险中性估值方法如何为欧式期权估值。
解：在无套利方法中，我们通过期权及股票建立无风险资产组合，使组合收益率等价于无风险利率，从而对期权估值。
在风险中性估值方法中，我们选取二叉树概率，以使股票的期望收益率等价于无风险利率，而后通过计算期权的期望收益并以无风险利率贴现得到期权价值。
9.3什么是股票期权的Delta ？解：股票期权的Delta 是度量期权价格对股价的小幅度变化的敏感度。
即是股票期权价格变化与其标的股票价格变化的比率。
9.4某个股票现价为$50。
已知6个月后将为$45或$55。
无风险年利率为10％（连续复利）。
执行价格为$50，6个月后到期的欧式看跌期权的价值为多少？ 解：考虑如下资产组合，卖1份看跌期权，买Δ份股票。
若股价上升为$55，则组合价值为55Δ；若股价下降为$45，则组合价值为：45Δ－5 当55Δ＝45Δ－5，即Δ＝－0.50时，6个月后组合价值在两种情况下将相等，均为$－27.5，其现值为：，即：0.10*0.5027.5$26.16e −−=− －P ＋50Δ＝－26.16所以，P ＝－50×0.5＋26.16＝$1.16 另解：求风险中性概率p0.10*0.505545(1)50p p e+−= 所以，p ＝0.7564看跌期权的价值P ＝0.10*0.50(0*0.75645*0.2436)$1.16e −+=9.5 某个股票现价为$100。
约翰.赫尔,期权期货和其他衍生品(third edition)习题答案
8.14 执行价格为$60 的看涨期权成本为$6，相同执行价格和到期日的看跌期权成
本为$4，制表说明跨式期权损益状况。请问：股票价格在什么范围内时，
跨式期权将导致损失呢？
解：可通过同时购买看涨看跌期权构造跨式期权：max（ ST －60，0）+max(60
－ ST )－（6＋4），其损益状况为：
股价 ST
解：(a)该组合等价于一份固定收益债券多头，其损益V = C ，不随股票价格变化。 （V 为组合损益，C 为期权费，下同）如图 8.2： (b)该组合等价于一份股票多头与一份固定收益债券多头，其损益V = ST + C ， 与股价同向同幅度变动。（ ST 为最终股票价格，下同）如图 8.3 (c)该组合等价于一份固定收益债券多头与一份看涨期权空头，其损益为
8.18 盒式价差期权是执行价格为 X 1 和 X 2 的牛市价差期权和相同执行价格的熊 市看跌价差期权的组合。所有期权的到期日相同。盒式价差期权有什么样的 特征？
解：牛市价差期权由 1 份执行价格为 X 1 欧式看涨期权多头与 1 份执行价格为 X 2 的欧式看涨期权空头构成（ X 1 < X 2 ），熊市价差期权由 1 份执行价格为 X 2 的 欧式看跌期权多头与 1 份执行价格为 X 1 的看跌期权空头构成，则盒式价差
8.17 运用期权如何构造出具有确定交割价格和交割日期的股票远期合约？ 解：假定交割价格为 K，交割日期为 T。远期合约可由买入 1 份欧式看涨期权，
同时卖空 1 份欧式看跌期权，要求两份期权有相同执行价格 K 及到期日 T。 可见，该组合的损益为 ST －K，在任何情形下，其中 ST 为 T 时股票价格。 假定 F 为远期合约价格，若 K=F，则远期合约价值为 0。这表明，当执行价 格为 K 时，看涨期权与看跌期权价格相等。
- 1、下载文档前请自行甄别文档内容的完整性，平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益，请联系客服反馈,我们会尽快为您处理(人工客服工作时间：9:00-18:30)。
期权期货及其他衍生品章节习题
第一章：引言
1. 远期合约多头，是指在将来某一时间以约定价格买入的资产。
空头则指以约定价格卖出的资产。
2. 对冲：投资人用来把控风险，将风险控制在一定范围内的行为。
期货对冲比较死一点，使对于当前时间来说风险变动范围较小。
而期权对冲更加灵活，为风险提供了保护下限，同时收益与不买期权斜率一致，但是会带来期权购买费的损失。
投机：对资产的走势进行赌注，预测，单纯的投资盈利行为。
可以认为是从0开始。
套利者：寻找不同市场间的价格差异进行无风险套利，但是这个套利点很难抓住。
3. (a) 签署一份期货合约，执行价格为50美元，在将来某一个时间按照执行价格交易卖出。
需要有一定的资产进行保证，但保证金不属于额外费用。
(b) 购买一份看涨期权合约，执行价格为50美元，在到期期间内可以行使以50美元卖出资产的权利，也可以不行使。
需要缴纳期权费不归还。
4. 卖出一个看涨期权(一般是具有期货合约的多头方，同时看跌期货价格，此时一般是对冲损失)，在这期间卖方必须履行以执行价格卖出资产的义务，并不是完全看跌，而是卖方认为期权中标的资产不可能涨到某一个价格以上。
当然如果跌了最好。
卖出的时候会立刻拿到期权费，如果买方不行使权利，则赚了期权费，如果买方行使了权利，那么说明标的资产的确涨价了，但是如果没有涨的很厉害，仍然不亏。
所以，卖出看涨期权，实则看跌，属于履行义务方，赚取期权费，亏损履行义务带来的损失。
买入一个看跌期权（一般也是具有期货合约的多头方，同时看跌期货价格，此时一般也是用来对冲），在这期间买方可以行使从卖方那里以执行价格卖出资产给卖方（空头），这是行使的权利。
买入看跌期权，确实是看跌，属于权利的行使，亏损的是期权费，赚取行使权利带来的利润。