1期权、期货和其他衍生品 1-3习题解答zhaoyang
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。
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。
赫尔《期权、期货及其他衍生产品》(第7版)课后习题详解(期权市场的运作过程)
8.2 课后习题详解一、问答题1. 某投资者以3美元的价格买入欧式看跌期权，股票价格为42美元，执行价格为40美元，在什么情况下投资者会盈利?在什么情况下期权会被行使?画出在到期时投资者盈利与股票价格的关系图。
An investor buys a European put on a share for $3. The stock price is $42 and the strike price is $40. Under what circumstances does the investor make a profit? Under what circumstances will the option be exercised? Draw a diagram showing the variation of the investor's profit with the stock price at the maturity of the option.答：如果到期日股票价格低于37美元，投资者执行将获得利润。
在这种情况下执行期权获得的收益高于3美元。
如果到期日股票价格低于40美元，期权就会被执行。
图8-1显示了投资者的利润随股票价格而变化的情况。
图8-1 投资者的利润2. 某投资者以4美元的价格卖出一欧式看涨期权，股票价格为47美元，执行价格为50美元，在什么情况下投资者会盈利?在什么情况下期权会被行使?画出在到期时投资者盈利与股票价格的关系图。
An investor sells a European call on a sbare for $4. The stock price is $47 and the strike price is $50. Under what circumstances does the investor make a profit? Under what circumstances will the option be exercised? Draw a diagram showing the variation of tlle investor's profit with tlle stock price at the maturity of the option.答：如果到期日股票价格低于54美元，投资者将获得利润；如果到期日股票价格低于50美元，期权将不被执行，投资者将获得利润4美元；如果到期日股票价格介于50美元与54美元之间，期权将被执行，投资者的利润介于0到4美元之间。
期权、期货课后题答案
第1章引言1．3远期合约多头与远期合约空头的区别是什么？答：持有远期合约多头的交易者同意在未来某一确定的时间以某一确定的价格购买一定数量的标的资产；而持有远期合约空头的交易者则同意在未来某一确定的时间以某一确定的价格出售一定数量的标的资产。
1.6某交易员进入期货价格每磅50美分的棉花远期合约空头方。
合约的规模是50000磅棉花。
当合约结束时棉花的价格分别为（a）每磅48.20美分，（b）每磅51.30美分，对应以上价格交易员的盈亏为多少？答：（a）此时交易员将价值48.20美分/磅的棉花以50美分/磅的价格出售，收益=（0.50 00-0.482）×50000=900（美元）。
（b）此时交易员将价值51.30美分/磅的棉花以50美分/磅的价格出售，损失=（0.513 -0.500）×50000=650（美元）。
1.9你认为某股票价格将要上升，股票的当前价格为29美元，而3个月期限，执行价格为30美元的看涨期权价格为2.90美元，你总共有5800美元的资金。
说明两种投资方式：一种是利用股票，另一种是利用期权。
每种方式的潜在盈亏是什么？答：在目前的资金规模条件下，一种方式为买入200只股票，另一种方式是买入2000个期权（即20份合约）。
如果股票价格走势良好，第二种方式将带来更多收益。
例如，如果股票价格上升到40美元，将从第二种方式获得2000×（40-30）-5800=14200（美元），而从第一种方式中仅能获得200×（40-29）=2200（美元）。
然而，当股票价格下跌时，第二种方式将导致更大的损失。
例如，如果股票价格下跌至25美元，第一种方式的损失为200×（29-25）=800（美元），而第二种方式的损失为全部5800美元的投资。
这个例子说明了期权交易的杠杆作用。
1.12解释为什么期货合约既可以用于投机也可以用于对冲。
答：如果一个交易员对一资产的价格变动有风险敞口，他可以用一个期货合约来进行对冲。
赫尔《期权、期货及其他衍生产品》复习笔记及课后习题详解（利率期货）【圣才出品】
赫尔《期权、期货及其他衍⽣产品》复习笔记及课后习题详解（利率期货）【圣才出品】第6章利率期货6.1 复习笔记1．天数计算和报价惯例天数计算常表⽰为X/Y，计算两个⽇期间获得的利息时，X定义了两个⽇期间天数计算的⽅式，Y定义了参照期内总天数计算的⽅式。
两个⽇期间获得的利息为：（两个⽇期之间的天数/参考期限的总天数）×参考期限内所得利息在美国常⽤的三种天数计算惯例为：①实际天数/实际天数；②30/360；③实际天数/360。
（1）美国短期债券的报价货币市场的产品报价采⽤贴现率⽅式，该贴现率对应于所得利息作为最终⾯值的百分⽐⽽不是最初所付出价格的百分⽐。
⼀般来讲，美国短期国债的现⾦价格与报价的关系式为：P＝360（100－Y）/n其中，P为报价，Y为现⾦价格，n为短期债券期限内以⽇历天数所计算的剩余天数。
（2）美国长期国债美国长期国债是以美元和美元的1/32为单位报出的。
所报价格是相对于⾯值100美元的债券。
报价被交易员称为纯净价，它与现⾦价有所不同，交易员将现⾦价称为带息价格。
⼀般来讲，有以下关系式：现⾦价格＝报价（即纯净价）＋从上⼀个付息⽇以来的累计利息2．美国国债期货（1）报价超级国债和超级国债期货合约的报价与长期国债本⾝在即期市场的报价⽅式相同。
（2）转换因⼦当交割某⼀特定债券时，⼀个名为转换因⼦的参数定义了空头⽅的债券交割价格。
债券的报价等于转换因⼦与最新成交期货价格的乘积。
将累计利息考虑在内，对应于交割100美元⾯值的债券收⼊的现⾦价格为：最新的期货成交价格×转换因⼦＋累计利息（3）最便宜可交割债券在交割⽉份的任意时刻，许多债券可以⽤于长期国债期货合约的交割，这些可交割债券有各式各样的券息率及期限。
空头⽅可以从这些债券中选出最便宜的可交割债券⽤于交割。
因为空头⽅收到的现⾦量为：最新成交价格×转换因⼦＋累计利息买⼊债券费⽤为：债券报价＋累计利息因此最便宜交割债券是使得：债券报价－期货的最新报价×转换因⼦达到最⼩的债券。
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)。
第一章1.1请解释远期多头与远期空头的区别。
答：远期多头指交易者协定将来以某一确定价格购入某种资产；远期空头指交易者协定将来以某一确定价格售出某种资产。
1.2请详细解释套期保值、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请解释签订购买远期价格为$50的远期合同与持有执行价格为$50的看涨期权的区别。
答：第一种情况下交易者有义务以50$购买某项资产（交易者没有选择），第二种情况下有权利以50$购买某项资产（交易者可以不执行该权利）。
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。
请问期货合约结束时，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损失为多少？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，交易者损失:($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。
你是怎么考虑的？你的收益或损失如何？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，我将损失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资金可以投资。
现有两种策略：直接购买股票或投资于期权，请问各自潜在的收益或损失为多少？答：股票价格低于$29时，购买股票和期权都将损失，前者损失为$5,800×(29-p),后者损失为$5，800;当股票价格为（29，30），购买$29×(p-29),购买期权损失为$5,800;当股票价格高股票收益为$5,800$29于$30时，购买股票收益为$5,800×(p-29)，购买期权收益为$29×(p-30)-5,800。
$$5,800$291.7假设你拥有5,000股每股价值$25的股票，如何运用看跌期权来确保你的股票价值在未来的四个月中不会受到股价下跌的影响。
答：通过购买5，000份价格为$25，期限为4个月的看跌期权来保值。
1.8一种股票在首次发行时会为公司提供资金。
请说明CBOE股票期权是否有同样的作用。
答：股票期权不为公司提供资金，它只是交易者之间相互买卖的一种证券，公司并不参与交易。
1.9请解释为什么远期合同既可用来投机又可用来套期保值？答：如果投资者预期价格将会上涨，可以通过远期多头来降低风险暴露，反之，预期价格下跌，通过远期空头化解风险。
如果投资者资产无潜在的风险暴露，远期合约交易就成为投机行为。
1.10假设一个执行价格为$50的欧式看涨期权价值$2.50,并持有到期。
在何种情况下期权的持有者会有盈利？在何种情况下，期权会被执行？请画图说明期权的多头方的收益是如何随期权到期日的股价的变化而变化的。
答：由欧式看涨期权多头的损益计算公式：max(,0)T S X -2.5=T S -52.5,该欧式看涨期权的持有者在标的资产的市场价格大于$52.5时，会有盈利；当标的资产的市场价格高于$50时，期权就会被执行。
图形如下：损益T S 0 52.51.11假设一欧式看跌期权执行价格为$60，价值为$4.00并持有到期。
在何种情况下，期权持有者（即空头方）会有盈利？在何种情况下，期权会被执行？请画图说明期权的空头方的收益是如何随期权到期日的股价的变化而变化的。
答：由欧式看跌期权多头的损益计算公式：max(,0)T X S --4=56-T S ,该欧式看跌期权的持有者在标的资产的市场价格低于$56时，会有盈利；当标的资产的市场价格低于$60时，期权就会被执行。
图形如下： 损益T S 0 561.12一位投资者出售了一个欧式9月份到期的看涨期权，执行价格为$20。
现在是5月，股票价格为18,期权价格为$20,现在是5月，股票价格为$18,期权价格为$2如果期权持有到期，并且到期时的股票价格为$25,请描述投资者的现金流状况。
答：由欧式看涨期权空头的损益计算公式：max(,0)T X S -+2=20-25+2=-3，投资者到期时将损失$3。
1.13一位投资者出售了一个欧式12月份到期的看跌期权，执行价格为$30,期权价值为$4。
在什么情况下，投资者会有盈利？答：当市场价格高于$20时，该看跌期权不被执行，投资者盈利为$4,当市场价格为（30，34）时，投资者盈利为4-(30-T S )。
1.14请说明在1.4节中描述的标准石油公司的债券是一种普通债券、一个执行价格为$25的基于油价的看涨期权的多头和一个执行价格为$40的基于油价的看涨期权的空头的组合。
max(25,0)T S -+min(40,0)T T S S -+,若市场价格高于$25,低于$40，则投资者损失为$25；若市场价格高于或等于$40,投资者收入为T S -25+40-T S +T S =T S +15;因此，该组合为一种普通债券，一个执行价格为$25的看涨期权多头和一个执行价格为$40的看涨期权空头。
1.15一家公司将在4个月后收到一笔外币。
选用哪种期权合约，可以进行套期保值？答：通过购入四月期的看跌期权进行套期保值。
1.16黄金的现价为每盎司$500。
一年后交割的远期价格为每盎司$700。
一位套期保值者可以10％的年利率借到钱。
套利者应当如何操作才能获利？假设储存黄金费用不计。
答：套利者以10%的年利率借入货币，购买黄金现货，卖出黄金远期，一年后交割收益为700-（1＋10％）1.17芝加哥交易所提供标的物为长期国债的期货合约。
请描述什么样的投资者会使用这种合约。
答：投资者预期长期利率下降的套期保值者；长期利率的投机者以及在现货和期货市场套利者，可购买该期货合约。
1.18一种股票的现价为$94,执行价格为$95的3个月期的看涨期权价格为$4.70。
一位投资者预计股票价格将要上升，正在犹豫是购买100股股票，还是购买20份看涨期权（每份合约为100股）。
两种策略都须投资$9,400。
你会给他什么建议？股票价格上升到多少时，购买期权会盈利更大？答：购买股票盈利更大些。
由：max(,0)T S X C --=20max(95,0)9,400T S --因此，当股票价格高于$94009520+=$565时，期权会盈利更大。
1.19“期权和期货是零合游戏”你是怎样理解这句话的？答：这句话是说期权和期货的一方损失程度等于另一方的盈利程度，总的收入为零。
1.20请描述下述组合的损益：同时签订一项资产的远期多头合约和有同样到期日的基于该项资产的欧式看跌期权的多头，并且在构造该组合时远期价格等于看跌期权的执行价格。
答：T S -X +max(X -T S ,0),当T S >X 时，收入为T S -X ,当T S <X 时，收入为0。
1.21说明在1.4节中描述的ICON 是由一种普通股票和两种期权组合而成。
答：假设ICON 中外汇汇率为T S ，则ICON 的收益为1000，若T S <X1000-a(T S -X),当1000T X S X a+>> 0，当1000T X S a+< 因此，ICON 的收益来自：（a ） 普通债券（b ） 执行价格为X 的欧式空头看涨期权（c ） 执行价格为1000X a+的欧式多头看涨期权 如下图所示：普通债券 空头看涨期权 多头看涨期权 总收益T S <X 1000 0 0 10001000X a+>T S >X 1000 -a(T S -X ) 0 1000-a(T S -X )T S >X +1000a 1000 -a(T S -X ) -a(T S -X -1000a) 01.22说明在1.4节中描述的范围远期合约可由两种期权组合而成。
如何构造价值为零的范围远期合约？答:假设用范围远期合约去购买一单位的外汇，T S 为汇率，则（a ） 若T S <1X ，支付1X（b ） 若T S >1X ,支付2X（c ） 若1X ≤T S ≤2X ，支付即期利率范围远期合约可以看作由一个执行价格为1X 的空头看跌期权和一个执行价格为2X 的多头看涨期权组成。
如下表所示：外汇成本 看跌期权价值 看涨期权价值 净成本T S <1X －T S －（1X －T S ） 0－1X1X <T S <2X －T S 0 0－T S2X <T S －T S 0 T S －2X －2X由于范围远期合约看跌期权与看涨期权头寸在建立初相等，因此构建范围远期合约不需要成本。
1.23某公司在1996年7月1日签订了一份远期合约，在1997年1月1日，购买1000万日元。
1996年9月1日，又签订了在1997年1月1日出售1000万日元的远期合约。
请描述这项策略的损益。
答：第一份远期合约的收益为T S -1F ,第二份远期合约的收益为T S -2F ,因此总收益为2F -1F 。
1.24假设英镑兑美元的即期和远期汇率如表1.1所示。
在下列情况中，投资者会有何获利机会？（A ） 一个180天的欧式看涨期权执行价格为1英镑兑1.5700美元，成本2美分。
（B ） 一个90天的欧式看跌期权执行价格为1英镑兑1.6400美元，成本2美分。
答：交易者通过卖出（A ），90天后买入（B ）来套利。
则（A ）合约的损失为min(,0)T X S -+0.02＝0.0118,(B)合约盈利为max(,0)T X S --0.02=0.0144,净收益为0.0026。
1.25请解释下面这句话：“一个远期合约的多头等价于一个欧式看涨期权的多头和一个欧式看跌期权的空头。
”答：由欧式看涨期权和看跌期权的损益公式得，一个欧式看涨期权的多头和一个欧式看跌期权的空头组合的损益为：max(,0)T S X -+min(,0)T X S -,当T S >X 时，总收入为T S -X +X -T S =0;当T S <X 时，总收入亦为0。
与远期合约多头相一致。
第二章2.1请说明未平仓合约数与交易量的区别。
答：未平仓合约数既可以指某一特定时间里多头合约总数，也可以指空头合约总数，而交易量是指在某一特定时间里交易的总和约数。
2.2请说明自营经纪人与佣金经纪人的区别。