期权、期货及其他衍生产品第9版-赫尔】Ch(5)

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.。

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第五章确定远期和期货价格
5.3假定你签署了一个无股息股票上6个月期限的远期合约，股票当前价格是$30，无风险利率是12%, (连续复利)，远期价格为多少？
5.4一个股指的当前价格为350，无风险利率为每年8%(连续复利)，股指的股息收益为每年4%，4个月的期货价格是多少？
5.9在签署无股息股票1年期的远期合约时，股票的当前价格为$40,连续复利的无风险利率为每年10%，(1)远期合约的初始价值和期货价格分别为多少(2)6个月后，股票价格变为$45，无风险利率仍然为每年10%,这时的远期价格和远期合约的价值分别为多少？
(1)远期合约的初始价值是0，期货价格为
(2)远期合约价值是,远期价格是
5.14在瑞士和美国按连续复利的2个月期限利率分别是每年2%和每年5%，瑞士法郎的即期价格是$1.0500，在2个月后交割的期货价格也是$1.0500美元，这时会存在什么样的套利机会
期货的理论价格为
瑞士法郎期货的理论价格大于实际价格，说明存在套利机会，在现货市场上卖瑞士法郎换美元，在期货市场上用美元买瑞士法郎
5.15白银的现价为每盎司$25，每年储存费用是每盎司$0.24，储存费每季度预付一次，假定所有期限的利率均为每年5%(连续复利)，计算9个月后交割的期货价格
储存费用的即期价格是
期货价格为。

(NEW)赫尔《期权、期货及其他衍生产品》教材精讲讲义简介赫尔的《期权、期货及其他衍生产品》是一本经典的金融学教材，被广泛用于大学金融学课程的教学。

本文档将对该教材进行精讲，涵盖主要内容和关键概念，旨在帮助读者深入理解和掌握期权、期货及其他衍生产品领域的知识。

本文档采用Markdown格式，方便阅读和使用。

第一章：期权市场简介1.1 期权的定义和特点期权是一种金融衍生工具，它赋予持有者在未来某个时间以特定价格买入或卖出某一标的资产的权利。

期权的特点包括灵活性、杠杆作用、风险限定和多样性等。

1.2 期权市场的组织和参与者期权市场包括交易所市场和场外市场。

交易所市场由交易所组织和管理，参与者包括期权合约买方、卖方、证券公司和交易所监管机构等。

1.3 期权定价模型期权定价模型是评估期权价格的数学模型，常用的模型包括布莱克-斯科尔斯模型和基于风险中性定价的模型。

第二章：期权定价理论2.1 基本期权定价理论基本期权定价理论包括不含股息的欧式期权定价、含股息的欧式期权定价以及美式期权定价等。

2.2 期权市场交易策略期权市场交易策略包括买入期权、卖出期权、期权组合以及期权套利等。

2.3 隐含波动率与期权定价隐含波动率是指根据期权市场价格反推出的波动率水平，它对期权价格的波动具有重要影响。

第三章：期权交易策略3.1 期权买入策略期权买入策略包括买入认购期权、买入认沽期权和买入期权组合等，旨在获得价差和方向性收益。

3.2 期权卖出策略期权卖出策略包括卖出认购期权、卖出认沽期权和卖出期权组合等，旨在获取权利金收入和时间价值消耗。

3.3 期权组合策略期权组合策略包括多头组合和空头组合，以及各种组合的调整和套利策略。

第四章：期货市场简介4.1 期货合约的基本特点期货合约是一种标准化的合约，约定了在未来某个时间以特定价格交割特定数量的标的资产。

4.2 期货交易所和市场参与者期货交易所是组织和管理期货市场的机构，市场参与者包括期货合约买方、卖方、交易所监管机构和期货经纪人等。

第2章期货市场的运作机制第一部分本章要点1．期货市场的具体运作机制。

将讨论合约条款的约定、保证金账户的运作、交易所的组织结构、市场监管规则、期货报价方式及期货的财会和税务处理等内容。

我们还将讨论，并解释造成这两类合约所实现的收益不同的原因。

2．期货与远期合约的不同之处。

第二部分重难点导学2.1 背景知识假定现在是3月5日，一位纽约的投资者指示他的经纪人买入5000蒲式耳玉米，资产交割时间在7月份，经纪人会马上将这一指令通知芝加哥交易所。

同时，假定另一位在堪萨斯州的投资者指示她的经纪人卖出5000蒲式耳玉米，资产交割时间也是7月份，她的经纪人也会马上将客户的指令通知芝加哥交易所。

这时双方会同意某个交易价格，从而交易成交。

几个概念：期货的长头寸（long futures position）；期货的短头寸（short futures position）。

期货价格（futures price）；期货合约的平仓。

2.2 期货合约的规定·资产品种·合约的规模·交割地点·交割时间有时交割资产的备选方案也会被说明。

2.2.1 资产商品期货标的资产的质量可能有很大差别。

金融期货合约中的标的资产定义通常明确。

2.2.2 合约的规模合约的规模（contract size）：每一合约中交割资产的数量。

2.2.3 交割的安排交割地点必须由交易所指定，这对那些运输费用昂贵的商品尤其重要。

当选用交割地点时，期货的短头寸方所收入的价格会随着交割地点的不同而得以调整。

期货交割地点离商品生产地越远，交割价格越高。

2.2.4 交割月份期货合约通常以其交割月份来命名。

交易所必须指定交割月份内的准确交割时间。

对于许多期货而言，交割时间区间为整个月。

2.2.5 报价交易所定义报价的方式各不相同，例如，在美国，原油期货报价以美元和美分计量，而中长期国债期货报价是以1美元以及1/32美元来报价的。

2.2.6 价格和头寸的限额跌停（limit down）：价格下跌的金额等于每日价格限额。

第9章期权市场机制第一部分本章要点期权市场组织机构、市场专用语、市场的交易过程等。

第二部分重难点导学9.1 期权的类型看涨期权（call option）给期权持有者在将来一定时刻以一定的价格买入某种资产的权利；看跌期权（put option）给期权持有者在将来一定时刻以一定价格卖出某资产的权利。

美式期权可在到期之日前的任何时刻行使；欧式期权只能在到期日才能行使。

9.1.1 看涨期权例9-1：考虑某投资者买入当前价格为98美元，执行价格为100美元的100股股票的欧式看涨期权，期权的到期日为4个月，购买一股股票的期权价格为5美元。

图9-1 买入一股股票的看涨期权收益曲线9.1.2 看跌期权例9-2：考虑一个购买了执行价格为70美元的100股股票的欧式看跌期权。

假定当前股票价格为65美元，期权的到期日为3个月，卖出一股股票的期权价格为7美元。

图9-2 买入看跌期权的收益曲线9.2 期权头寸一方为取得期权的长头寸方（即买入期权），另一方为取得期权的短头寸方[即卖出期权或对期权进行承约（written the option）]。

期权交易共有4种头寸形式：·看涨期权长头寸；·看跌期权长头寸；·看涨期权短头寸；·看跌期权短头寸。

图9-3及图9-4显示了期权承约人的盈亏与股票价格之间的关系。

图9-3 卖出看涨期权的收益图（期权价格=5美元，执行价格=100美元）图9-4 卖出看跌期权的收益图（期权价格=7美元，执行价格=70美元）K——执行价格，S T——标的资产的到期价格，在S T>K时，行使期权；在S T≤K时，将不行使期权。

欧式看涨期权的长头寸的收益为max（S T-K，0）欧式看涨期权的短头寸的收益为-max（S T-K，0）=min（K-S T，O）欧式看跌期权的长头寸的收益为max（K-S T，0）欧式看跌期权的短头寸的收益为-max（K-S T，0）=min（S T-K，0）图9-5展示了期权的收益图形。

HullOFOD9eSolutionsCh13第九版期权、期货及其他衍生品课后答案CHAPTER 13 Binomial TreesPractice QuestionsProblem 13.1.A stock price is currently $40. It is known that at the end of one month it will be either $42 or $38. The risk-free interest rate is 8% per annum with continuous compounding. What is the value of a one-month European call option with a strike price of $39?Consider a portfolio consisting of 1-: Call option +?: Shares If the stock price rises to $42, the portfolio is worth 423?-. If the stock price falls to $38, it is worth 38?. These are the same when42338?-=? or 075?=.. The value of the portfolio in one month is 28.5 for both stock prices. Its value today must be the present value of 28.5, or 0080083332852831e -.?..=.. This means that 402831f -+?=.where f is the call price. Because 075?=., the call price is 400752831$169?.-.=.. As an alternative approach, we can calculate the probability, p , of an up movement in a risk-neutral world. This must satisfy: 0080083334238(1)40p p e .?.+-= so that 00800833344038p e .?.=-or 05669p =.. The value of the option is then its expected payoff discounted at the risk-free rate: 008008333[305669004331]169e -.?.?.+?.=. or $1.69. This agrees with the previous calculation.Problem 13.2.Explain the no-arbitrage and risk-neutral valuationapproaches to valuing a European option using a one-step binomial tree.In the no-arbitrage approach, we set up a riskless portfolio consisting of a position in the option and a position in the stock. By setting the return on the portfolio equal to the risk-free interest rate, we are able to value the option. When we use risk-neutral valuation, we first choose probabilities for the branches of the tree so that the expected return on the stock equals the risk-free interest rate. We then value the option by calculating its expected payoff and discounting this expected payoff at the risk-free interest rate.Problem 13.3.What is meant by the delta of a stock option?The delta of a stock option measures the sensitivity of the option price to the price of the stock when small changes are considered. Specifically, it is the ratio of the change in the price of the stock option to the change in the price of the underlying stock.Problem 13.4.A stock price is currently $50. It is known that at the end of six months it will be either $45 or $55. The risk-free interest rate is 10% per annum with continuous compounding. What is the value of a six-month European put option with a strike price of $50?Consider a portfolio consisting of 1-: Put option +?: Shares If the stock price rises to $55, this is worth 55?. If the stock price falls to $45, the portfolio is worth 455?-. These are the same when 45555?-=?or 050?=-.. The value of the portfolio in six months is 275-. for both stock prices. Its value today must be the present valueof 275-., or 010********e -.?.-.=-.. This means that 502616f -+?=-.where f is the put price. Because 050?=-., the put price is $1.16. As an alternative approach we can calculate the probability, p , of an up movement in a risk-neutral world. This must satisfy: 01055545(1)50p p e .?.+-= so that 010*******p e .?.=- or 07564p =.. The value of the option is then its expected payoff discounted at the risk-free rate: 0105[007564502436]116e -.?.?.+?.=. or $1.16. This agrees with the previous calculation.Problem 13.5.A stock price is currently $100. Over each of the next two six-month periods it is expected to go up by 10% or down by 10%. The risk-free interest rate is 8% per annum with continuous compounding. What is the value of a one-year European call option with a strike price of $100?In this case 110u =., 090d =., 05t ?=., and 008r =., so that00805090***********e p .?.-.==..-.The tree for stock price movements is shown in Figure S13.1. We can work back from the end of the tree to the beginning, as indicated in the diagram, to give the value of the option as $9.61. The option value can also be calculated directly from equation (13.10): 22200805[0704121207041029590029590]961e -?.?..?+?.?.?+.?=. or $9.61.Figure S13.1: Tree for Problem 13.5Problem 13.6.For the situation considered in Problem 13.5, what is the value of a one-year European put option with a strike price of $100? Verify that the European call and European put prices satisfy put –call parity.Figure S13.2 shows how we can value the put option using the same tree as in Problem 13.5. The value of the option is $1.92. The option value can also be calculated directly from equation (13.10): 20080522[0704102070410295910295919]192e -?.?..?+?.?.?+.?=.or $1.92. The stock price plus the put price is 10019210192$+.=.. The present value of the strike price plus the call price is 008110096110192e $-.?+.=.. These are the same, verifyingthat put –call parity holds.Figure S13.2: Tree for Problem 13.6Problem 13.7.What are the formulas for u and d in terms of volatility?u e =and d e -=Problem 13.8.Consider the situation in which stock price movements during the life of a European option are governed by a two-step binomial tree. Explain why it is not possible to set up a position in the stock and the option that remains riskless for the whole of the life of the option.The riskless portfolio consists of a short position in the option and a long position in ? shares. Because ? changes during the life of the option, this riskless portfolio must also change.Problem 13.9.A stock price is currently $50. It is known that at the end of two months it will be either $53 or $48. The risk-free interest rate is 10% per annum with continuous compounding. What is thevalue of a two-month European call option with a strikeprice of $49? Use no-arbitrage arguments.At the end of two months the value of the option will be either $4 (if the stock price is $53) or $0 (if the stock price is $48). Consider a portfolio consisting of:shares1option+?:-:The value of the portfolio is either 48? or 534?- in two months. If48534?=?- i.e.,08?=. the value of the portfolio is certain to be 38.4. For this value of ? the portfolio is therefore riskless. The current value of the portfolio is: 0850f .?-where f is the value of the option. Since the portfolio must earn the risk-free rate of interest010212(0850)384f e .?/.?-=.i.e.,223f =.The value of the option is therefore $2.23.This can also be calculated directly from equations (13.2) and (13.3). 106u =., 096d =. so that01021209605681106096e p .?/-.==..-. and010212056814223f e -.?/=?.?=.Problem 13.10.A stock price is currently $80. It is known that at the end of four months it will be either $75or $85. The risk-free interest rate is 5% per annum with continuous compounding. What is the value of a four-monthEuropean put option with a strike price of $80? Use no-arbitrage arguments.At the end of four months the value of the option will be either $5 (if the stock price is $75) or $0 (if the stock price is $85). Consider a portfolio consisting of:shares1option-?:+:(Note: The delta, ? of a put option is negative. We have constructed the portfolio so that it is +1 option and -? shares rather than 1- option and +? shares so that the initial investment is positive.)The value of the portfolio is either 85-? or 755-?+ in four months. If 85755-?=-?+ i.e.,05?=-. the value of the portfolio is certain to be 42.5. For this value of ? the portfolio is therefore riskless. The current value of the portfolio is: 0580f .?+where f is the value of the option. Since the portfolio is riskless005412(0580)425f e .?/.?+=.i.e.,180f =.The value of the option is therefore $1.80.This can also be calculated directly from equations (13.2) and (13.3). 10625u =., 09375d =. so that00541209375063451062509375e p .?/-.==..-. 103655p -=. and005412036555180f e -.?/=?.?=.Problem 13.11.A stock price is currently $40. It is known that at the end ofthree months it will be either $45 or $35. The risk-free rate of interest with quarterly compounding is 8% per annum. Calculate the value of a three-month European put option on the stock with an exercise price of $40. Verify that no-arbitrage arguments and risk-neutral valuation arguments give the same answers.At the end of three months the value of the option is either $5 (if the stock price is $35) or $0 (if the stock price is $45).Consider a portfolio consisting of:shares1option-?:+:(Note: The delta, ?, of a put option is negative. We have constructed the portfolio so that it is +1 option and -? shares rather than 1- option and +? shares so that the initial investment is positive.)The value of the portfolio is either 355-?+ or 45-?. If:35545-?+=-?i.e.,05?=-.the value of the portfolio is certain to be 22.5. For this value of ? the portfolio is therefore riskless. The current value of the portfolio is 40f -?+where f is the value of the option. Since the portfolio must earn the risk-free rate of interest (4005)102225f ?.+?.=. Hence 206f =. i.e., the value of the option is $2.06.This can also be calculated using risk-neutral valuation. Suppose that p is the probability of an upward stock price movement in a risk-neutral world. We must have 4535(1)40102p p +-=?. i.e., 1058p =. or: 058p =.The expected value of the option in a risk-neutral world is:00585042210?.+?.=. This has a present value of210206102.=..This is consistent with the no-arbitrage answer.Problem 13.12.A stock price is currently $50. Over each of the next two three-month periods it is expected to go up by 6% or down by 5%. The risk-free interest rate is 5% per annum with continuous compounding. What is the value of a six-month European call option with a strike price of $51?A tree describing the behavior of the stock price is shown in Figure S13.3. The risk-neutral probability of an up move, p , is given by00531209505689106095e p .?/-.==..-. There is a payoff from the option of 561851518.-=. for the highest final node (which corresponds to two up moves) zero in all other cases. The value of the option is therefore 2005612518056891635e -.?/.?.?=.This can also be calculated by working back through the tree as indicated in Figure S13.3. The value of the call option is the lower number at each node in the figure.Figure S13.3:Tree for Problem 13.12Problem 13.13.For the situation considered in Problem 13.12, what is the value of a six-month European put option with a strike price of $51? Verify that the European call and European put prices satisfy put–call parity. If the put option were American, would it ever be optimal to exercise it early at any of the nodes on the tree?The tree for valuing the put option is shown in Figure S13.4. We get a payoff of-.=.if -.=.if the middle final node is reached and a payoff of 51451255875 515035065the lowest final node is reached. The value of the option is therefore2005612.??.?.+.?.=.(06520568904311587504311)1376e-.?/This can also be calculated by working back through the tree as indicated in Figure S13.4. The value of the put plus the stock price is.+=.137********The value of the call plus the present value of the strike price is005612e-.?/.+=.16355151376This verifies that put–call parity holdsTo test whether it worth exercising the option early wecompare the value calculated for the option at each node with the payoff from immediate exercise. At node C the payoff from -.=.. Because this is greater than 2.8664, the option should immediate exercise is 5147535be exercised at this node. The option should not be exercised at either node A or node B.Figure S13.4:Tree for Problem 13.13Problem 13.14.A stock price is currently $25. It is known that at the end of two months it will be either $23 or $27. The risk-free interest rate is 10% per annum with continuous compounding. Suppose T S is the stock price at the end of two months. What is the value of a derivative that pays off2T S at this time?At the end of two months the value of the derivative will be either 529 (if the stock price is 23) or 729 (if the stock price is 27). Consider a portfolio consisting of:shares1derivative+?:-:The value of the portfolio is either 27729?- or 23529?- in twomonths. If2772923529?-=?- i.e.,50?= the value of the portfolio is certain to be 621. For this value of ? the portfolio is therefore riskless. The current value of the portfolio is: 5025f ?-where f is the value of the derivative. Since the portfolio must earn the risk-free rate of interest 010212(5025)621f e .?/?-= i.e., 6393f =. The value of the option is therefore $639.3.This can also be calculated directly from equations (13.2) and (13.3). 108u =., 092d =. so that01021209206050108092e p .?/-.==..-. and 010212(0605072903950529)6393f e -.?/=.?+.?=.Problem 13.15.Calculate u , d , and p when a binomial tree is constructed to value an option on a foreign currency. The tree step size is one month, the domestic interest rate is 5% per annum, the foreign interest rate is 8% per annum, and the volatility is 12% per annum.In this case (005008)11209975a e .-.?/==.010352u e .==.109660d u =/=.0997509660045531035209660p .-.==..-.Problem 13.16.The volatility of a non-dividend-paying stock whose price is $78, is 30%. The risk-free rate is 3% per annum (continuously compounded) for all maturities. Calculate values for u, d, and pwhen a two-month time step is used. What is the value of a four-month European call option with a strike price of $80 given by a two-step binomial tree. Suppose a trader sells 1,000 options (10 contracts). What position in the stock is necessary to hedge the trader’s position at the time of the trade?4898.08847.01303.18847.08847.0/11303.112/230.01667.030.0=--=====??e p u d e uThe tree is given in Figure S13.5. The value of the option is $4.67. The initial delta is 9.58/(88.16 –69.01) which is almost exactly 0.5 so that 500 shares should be purchased.Figure S13.5: Tree for Problem 13.16Problem 13.17.A stock index is currently 1,500. Its volatility is 18%. The risk-free rate is 4% per annum (continuously compounded) for all maturities and the dividend yield on the index is 2.5%. Calculate values for u, d, and p when a six-month time step is used. What is the value a 12-month American put option with a strike price of 1,480 given by a two-step binomial tree.4977.08805.01357.18805.08805.0/11357.15.0)025.004.0(5.018.0=--=====?-?e p u d e uThe tree is shown in Figure S13.6. The option is exercised at the lower node at the six-month point. It is worth 78.41.Figure S13.6: Tree for Problem 13.17Problem 13.18.The futures price of a commodity is $90. Use a three-step tree to value (a) a nine-month American call option with strike price $93 and (b) a nine-month American put option with strike price $93. The volatility is 28% and the risk-free rate (all maturities) is 3% with continuous compounding.4651.08694.01503.18694.018694.0/11503.125.028.0=--=====?u u d e u The tree for valuing the call is in Figure S13.7a and that for valuing the put is in Figure S13.7b. The values are 7.94 and 10.88, respectively.824637Figure S13.7a : CallFigure S13.7b : PutFurther QuestionsProblem 13.19.The current price of a non-dividend-paying biotech stock is $140 with a volatility of 25%. The risk-free rate is 4%. For a three-month time step: (a) What is the percentage up movement? (b) What is the percentage down movement?(c) What is the probability of an up movement in a risk-neutral world? (d) What is the probability of a down movementin a risk-neutral world?Use a two-step tree to value a six-month European call option and a six-month European put option. In both cases the strike price is $150.(a) 25.025.0?=e u = 1.1331. The percentage up movement is13.31% (b) d = 1/u = 0.8825. The percentage down movement is11.75%(c) The probability of an up movement is 5089.0)8825.1331.1/()8825.()25.004.0=--?e (d) The probability of a down movement is0.4911.The tree for valuing the call is in Figure S13.8a and that for valuing the put is in Figure S13.8b. The values are 7.56 and 14.58, respectively.Figure S13.8a : CallFigure S13.8b : PutProblem 13.20.In Problem 13.19, suppose that a trader sells 10,000 European call options. How many shares of the stock are needed to hedge the position for the first and second three-month period? For the second time period, consider both the case where the stock price moves up during the first period and the case where it moves down during the first period.The delta for the first period is 15/(158.64 – 123.55) = 0.4273. The trader should take a long position in 4,273 shares. If there is an up movement the delta for the second period is 29.76/(179.76 – 140) = 0.7485. The trader should increase the holding to 7,485 shares. If there is a down movement the trader should decrease the holding to zero.Problem 13.21.A stock price is currently $50. It is known that at the end of six months it will be either $60 or $42. The risk-free rate of interest with continuous compounding is 12% per annum. Calculate the value of a six-month European call option on the stock with an exercise price of $48. Verify that no-arbitrage arguments and risk-neutral valuation arguments give the same answers.At the end of six months the value of the option will be either $12 (if the stock price is $60) or $0 (if the stock price is $42). Consider a portfolio consisting of:shares1option+?:-:The value of the portfolio is either 42? or 6012?- in sixmonths. If426012?=?- i.e.,06667?=. the value of the portfolio is certain to be 28. For this value of ? the portfolio is therefore riskless. The current value of the portfolio is: 0666750f .?-where f is the value of the option. Since the portfolio must earn the risk-free rate of interest01205(0666750)28f e .?..?-=i.e.,696f =.The value of the option is therefore $6.96.This can also be calculated using risk-neutral valuation. Suppose that p is the probability of an upward stock price movement in a risk-neutral world. We must have 0066042(1)50p p e .+-=? i.e., 181109p =. or: 06161p =.The expected value of the option in a risk-neutral world is: 120616100383973932?.+?.=. This has a present value of 00673932696e -..=.Hence the above answer is consistent with risk-neutral valuation.Problem 13.22.A stock price is currently $40. Over each of the next two three-month periods it is expected to go up by 10% or down by 10%. The risk-free interest rate is 12% per annum with continuous compounding.a. What is the value of a six-month European put option witha strike price of $42?b. What is the value of a six-month American put option with a strike price of $42?a. A tree describing the behavior of the stock price is shownin Figure S13.9. The risk-neutral probability of an up move, p , is given by012312090065231109e p .?/-.==..-.Calculating the expected payoff and discounting, we obtain the value of the option as 2012612[24206523034779603477]2118e -.?/.??.?.+.?.=.The value of the European option is 2.118. This can also be calculated by working back through the tree as shown in Figure S13.9. The second number at each node is the value of the European option.b. The value of the American option is shown as the third number at each node on the tree. It is 2.537. This is greater than the value of the European option because it is optimal to exercise early at node C.40.0002.1182.53744.000 0.8100.81036.0004.7596.00048.4000.0000.00039.6002.4002.40032.4009.6009.600ABCFigure S13.9: Tree to evaluate European and American put options in Problem 13.22. At each node, upper number is the stock price, the next number is the European put price, and the final number is the American put priceProblem 13.23.Using a “trial -and-error” approach, estimate how high the strike price has to be in Problem 13.17 for it to be optimal toexercise the option immediately.Trial and error shows that immediate early exercise is optimal when the strike price is above 43.2. This can be also shown to be true algebraically. Suppose the strike price increases by a relatively small amount q . This increases the value of being at node C by q and the value of being at node B by 0030347703374e q q -..=.. It therefore increases the value of being at node A by 003(065230337403477)0551q q e q -..?.+.=.For early exercise at node A we require 253705512q q .+.<+ or 1196q >.. This corresponds to the strike price being greater than 43.196.Problem 13.24.A stock price is currently $30. During each two-month period for the next four months it is expected to increase by 8% or reduce by 10%. The risk-free interest rate is 5%. Use a two-step tree to calculate the value of a derivative that pays off 2[max(300)]T S -, whereT S is the stock price in four months? If the derivative is American-style, should it be exercised early?This type of option is known as a power option. A tree describing the behavior of the stock price is shown in Figure S13.10. The risk-neutral probability of an up move, p , is given by 005212090602010809e p .?/-.==..-. Calculating the expected payoff and discounting, we obtain the value of the option as393.5]3980.049.323980.06020.027056.0[12/405.02=?+-eThe value of the European option is 5.393. This can also be calculated by working back through the tree as shown in Figure S13.10. The second number at each node is the value of theEuropean option. Early exercise at node C would give 9.0 which is less than 13.2435. The option should therefore not be exercised early if it is American.Figure S13.10: Tree to evaluate European power option in Problem 13.24. At each node, upper number is the stock price and the next number is the option priceProblem 13.25.Consider a European call option on a non-dividend-paying stock where the stock price is $40, the strike price is $40, the risk-free rate is 4% per annum, the volatility is 30% per annum, and the time to maturity is six months.a. Calculate u , d , and p for a two step treeb. Value the option using a two step tree.c. Verify that DerivaGem gives the same answerd. Use DerivaGem to value the option with 5, 50, 100, and 500 time steps.0.0000 29.1600 0.705624.3000 32.4900 F(a) This problem is based on the material in Section 13.8. In this case 025t ?=.so that03011618u e .==., 108607d u =/=., and00402508607049591161808607e p .?.-.==..-.(b) and (c) The value of the option using a two-step tree as given by DerivaGem is shown in Figure S13.11 to be 3.3739. To use DerivaGem choose the first worksheet, select Equity as the underlying type, and select Binomial European as the Option Type. After carrying out the calculations select Display Tree.(d) With 5, 50, 100, and 500 time steps the value of the option is 3.9229, 3.7394, 3.7478, and 3.7545, respectively.Figure S13.11: Tree produced by DerivaGem to evaluate European option in Problem 13.25Problem 13.26.Repeat Problem 13.25 for an American put option on a futures contract. The strike price and the futures price are $50, the risk-free rate is 10%, the time to maturity is six months, and the volatility is 40% per annum.(a) In this case 025t ?=.and 04012214u e .==., 108187d u =/=., and0102508187045021221408187e p .?.-.==..-.(b) and (c) The value of the option using a two-step tree is4.8604.(d) With 5, 50, 100, and 500 time steps the value of the option is 5.6858, 5.3869, 5.3981, and 5.4072, respectively.Problem 13.27.Footnote 1 shows that the correct discount rate to use for the real world expected payoff inAt each node:Upper v alue = Underlying Asset PriceLower v alue = Option Price Values in red are a result of early exercise.Strike price = 40Discount factor per step = 0.9900Time step, dt = 0.2500 years, 91.25 daysGrowth factor per step, a = 1.0101Probability of up mov e, p = 0.4959Up step size, u = 1.16180.00000.25000.5000the case of the call option considered in Figure 13.1 is 42.6%. Show that if the option is a put rather than a call the discount rate is –52.5%. Explain why the two real-world discount rates are so different.The value of the put option is012312.?+.?=.*************)10123e-.?/The expected payoff in the real world is.?+.?=.*************)08877The discount rate R that should be used in the real world is therefore given by solving025.=.1012308877Re-.-.or 52.5%.The solution to this is 0525The underlying stock has positive systematic risk because it expected return is higher than the risk free rate. This means that the stock will tend to do well when the market does well. The call option has a high positive systematic risk because it tends to do very well when the market does well. As a result a high discount rate is appropriate for its expected payoff. The put option is in the opposite position. It tends to provide a high return when the market does badly. As a result it is appropriate to use a highly negative discount rate for its expected payoff.Problem 13.28.A stock index is currently 990, the risk-free rate is 5%, and the dividend yield on theindex is 2%. Use a three-step tree to value an 18-month American put option with astrike price of 1,000 when the volatility is 20% per annum. How much does the option holder gain by being able to exercise early? When is the gain made?The tree is shown in Figure S13.12. The value of the option is 87.51. It is optimal to exercise at the lowest node at time one year. If early exercise were not possible the value at this node would be 236.63. The gain made at the one year point is therefore 253.90 – 236.63= 17.27.Figure 13.12: Tree for Problem 13.28Problem 13.29.Calculate the value of nine-month American call option on a foreign currency using athree-step binomial tree. The current exchange rate is 0.79 and the strike price is 0.80 (both expressed as dollars per unit of the foreign currency). The volatility of the exchange rate is 12% per annum. The domestic and foreign risk-free rates are 2% and 5%, respectively. Suppose a company has bought options on 1 million units of the foreign currency. What position in the foreign currency is initially necessary to hedge its risk?The tree is shown in Figure S13.13. The cost of an American option to buy one million units of the foreign currency is $18,100. The delta initially is (0.0346 ?0.0051)/(0.8261 – 0.7554) = 0.4176. The company should sell 417,600 units of the foreign currencyFigure S13.13: Tree for Problem 13.29。

CHAPTER 3Hedging Strategies Using FuturesPractice QuestionsProblem 3.1.Under what circumstances are (a) a short hedge and (b) a long hedge appropriate?A short hedge is appropriate when a company owns an asset and expects to sell that asset in the future. It can also be used when the company does not currently own the asset but expects to do so at some time in the future. A long hedge is appropriate when a company knows it will have to purchase an asset in the future. It can also be used to offset the risk from an existing short position.Problem 3.2.Explain what is meant by basis risk when futures contracts are used for hedging.Basis risk arises from the hedger’s uncertainty as to the difference between the spot price and futures price at the expiration of the hedge.Problem 3.3.Explain what is meant by a perfect hedge. Does a perfect hedge always lead to a better outcome than an imperfect hedge? Explain your answer.A perfect hedge is one that completely eliminates the hedger’s risk. A p erfect hedge does not always lead to a better outcome than an imperfect hedge. It just leads to a more certain outcome. Consider a company that hedges its exposure to the price of an asset. Suppose the asset’s price movements prove to be favorable to the company. A perfect hedge totally neutralizes the company’s gain from these favorable price movements. An imperfect hedge, which only partially neutralizes the gains, might well give a better outcome.Problem 3.4.Under what circumstances does a minimum-variance hedge portfolio lead to no hedging at all?A minimum variance hedge leads to no hedging when the coefficient of correlation between the futures price changes and changes in the price of the asset being hedged is zero.Problem 3.5.Give three reasons why the treasurer of a company might not hedge the company’s exposure to a particular risk.(a) If the company’s competitors are not hedging, the treasurer might feel that the company will experience less risk if it does not hedge. (See Table 3.1.) (b) The shareholders might not want the company to hedge because the risks are hedged within their portfolios. (c) If there is a loss on the hedge and a gain from the company’s exposure to the underlying asset, the treasurer might feel that he or she will have difficulty justifying the hedging to other executives within the organization.Problem 3.6.Suppose that the standard deviation of quarterly changes in the prices of a commodity is $0.65, the standard deviation of quarterly changes in a futures price on the commodity is $0.81, and the coefficient of correlation between the two changes is 0.8. What is the optimal hedge ratio for a three-month contract? What does it mean?The optimal hedge ratio is 065080642081..⨯=.. This means that the s ize of the futures position should be 64.2% of the size of the company’s exposure in a three-month hedge.Problem 3.7.A company has a $20 million portfolio with a beta of 1.2. It would like to use futures contracts on a stock index to hedge its risk. The index futures is currently standing at 1080, and each contract is for delivery of $250 times the index. What is the hedge that minimizes risk? What should the company do if it wants to reduce the beta of the portfolio to 0.6?The formula for the number of contracts that should be shorted gives 20000000128891080250,,.⨯=.⨯ Rounding to the nearest whole number, 89 contracts should be shorted. To reduce the beta to 0.6, half of this position, or a short position in 44 contracts, is required.Problem 3.8.In the corn futures contract, the following delivery months are available: March, May, July, September, and December. State the contract that should be used for hedging when the expiration of the hedge is in a) June, b) July, and c) JanuaryA good rule of thumb is to choose a futures contract that has a delivery month as close aspossible to, but later than, the month containing the expiration of the hedge. The contracts that should be used are therefore(a) July(b)September(c)MarchProblem 3.9.Does a perfect hedge always succeed in locking in the current spot price of an asset for a future transaction? Explain your answer.No. Consider, for example, the use of a forward contract to hedge a known cash inflow in a foreign currency. The forward contract locks in the forward exchange rate — which is in general different from the spot exchange rate.Problem 3.10.Explain why a short hedger’s position improves when the basis strengthens unexpectedly and worsens when the basis weakens unexpectedly.The basis is the amount by which the spot price exceeds the futures price. A short hedger is long the asset and short futures contracts. The value of his or her position therefore improves as the basis increases. Similarly, it worsens as the basis decreases.Problem 3.11.Imagine you are the treasurer of a Japanese company exporting electronic equipment to the United States. Discuss how you would design a foreign exchange hedging strategy and the arguments you would use to sell the strategy to your fellow executives.The simple answer to this question is that the treasurer should1.Estimate the company’s future cash flows in Japanese yen and U.S. dollars2.Enter into forward and futures contracts to lock in the exchange rate for the U.S. dollarcash flows.However, this is not the whole story. As the gold jewelry example in Table 3.1 shows, the company should examine whether the magnitudes of the foreign cash flows depend on the exchange rate. For example, will the company be able to raise the price of its product in U.S. dollars if the yen appreciates? If the company can do so, its foreign exchange exposure may be quite low. The key estimates required are those showing the overall effect on the company’s profitability of changes in the exchange rate at various times in the future. Once these estimates have been produced the company can choose between using futures and options to hedge its risk. The results of the analysis should be presented carefully to other executives. It should be explained that a hedge does not ensure that profits will be higher. It means that profit will be more certain. When futures/forwards are used both the downside and upside are eliminated. With options a premium is paid to eliminate only the downside.Problem 3.12.Suppose that in Example 3.2 of Section 3.3 the company decides to use a hedge ratio of 0.8. How does the decision affect the way in which the hedge is implemented and the result?If the hedge ratio is 0.8, the company takes a long position in 16 December oil futures contracts on June 8 when the futures price is $88.00. It closes out its position on November 10. The spot price and futures price at this time are $90.00 and $89.10. The gain on the futures position is(89.10 − 88.00) × 16,000 = 17,600The effective cost of the oil is therefore20,000 × 90 – 17,600 = 1,782, 400or $89.12 per barrel. (This compares with $88.90 per barrel when the company is fully hedged.)Problem 3.13.“If the minimum -variance hedge ratio is calculated as 1.0, the hedge must be perfect." Is this statement true? Explain your answer.The statement is not true. The minimum variance hedge ratio is S Fσρσ It is 1.0 when 05=.ρ and 2S F =σσ. Since 10<.ρ the hedge is clearly not perfect.Problem 3.14.“If there is no basis risk, the minimum variance hedge ratio is always 1.0." Is this statement true? Explain your answer.The statement is true. Using the notation in the text, if the hedge ratio is 1.0, the hedger locks in a price of 12F b +. Since both 1F and 2b are known this has a variance of zero and must be the best hedge.Problem 3.15“For an asset where futures prices for contracts on the asset are usually less than spot prices, long hedges are likely to be particularly attractive." Explain this statement.A company that knows it will purchase a commodity in the future is able to lock in a price close to the futures price. This is likely to be particularly attractive when the futures price is less than the spot price.Problem 3.16.The standard deviation of monthly changes in the spot price of live cattle is (in cents per pound)1.2. The standard deviation of monthly changes in the futures price of live cattle for the closestcontract is 1.4. The correlation between the futures price changes and the spot price changes is 0.7. It is now October 15. A beef producer is committed to purchasing 200,000 pounds of live cattle on November 15. The producer wants to use the December live-cattle futures contracts to hedge its risk. Each contract is for the delivery of 40,000 pounds of cattle. What strategy should the beef producer follow?The optimal hedge ratio is 12070614..⨯=.. The beef producer requires a long position in 20000006120000⨯.=, lbs of cattle. The beef producer should therefore take a long position in 3 December contracts closing out the position on November 15.Problem 3.17.A corn farmer argues “I do not use futures co ntracts for hedging. My real risk is not the price of corn. It is that my whole crop gets wiped out by the weather.”Discuss this viewpoint. Should the farmer estimate his or her expected production of corn and hedge to try to lock in a price for expected production?If weather creates a significant uncertainty about the volume of corn that will be harvested, the farmer should not enter into short forward contracts to hedge the price risk on his or her expected production. The reason is as follows. Suppose that the weather is bad and the farmer’sproduction is lower than expected. Other farmers are likely to have been affected similarly. Corn production overall will be low and as a consequence the price of corn will be relatively high. The farmer’s problems arising from the bad harvest will be made worse by losses on the short futures position. This problem emphasizes the importance of looking at the big picture when hedging. The farmer is correct to question whether hedging price risk while ignoring other risks is a good strategy.Problem 3.18.On July 1, an investor holds 50,000 shares of a certain stock. The market price is $30 per share. The investor is interested in hedging against movements in the market over the next month and decides to use the September Mini S&P 500 futures contract. The index is currently 1,500 and one contract is for delivery of $50 times the index. The beta of the stock is 1.3. What strategy should the investor follow? Under what circumstances will it be profitable?A short position in 50000301326501500,⨯.⨯=⨯,contracts is required. It will be profitable if the stock outperforms the market in the sense that its return is greater than that predicted by the capital asset pricing model.Problem 3.19.Suppose that in Table 3.5 the company decides to use a hedge ratio of 1.5. How does the decision affect the way the hedge is implemented and the result?If the company uses a hedge ratio of 1.5 in Table 3.5 it would at each stage short 150 contracts. The gain from the futures contracts would be.1=⨯50.2$5570.1per barrel and the company would be $0.85 per barrel better off than with a hedge ratio of 1.Problem 3.20.A futures contract is used for hedging. Explain why the daily settlement of the contract can give rise to cash flow problems.Suppose that you enter into a short futures contract to hedge the sale of an asset in six months. If the price of the asset rises sharply during the six months, the futures price will also rise and you may get margin calls. The margin calls will lead to cash outflows. Eventually the cash outflows will be offset by the extra amount you get when you sell the asset, but there is a mismatch in the timing of the cash outflows and inflows. Your cash outflows occur earlier than your cash inflows.A similar situation could arise if you used a long position in a futures contract to hedge the purchase of an asset at a future time and the asset’s price fe ll sharply. An extreme example of what we are talking about here is provided by Metallgesellschaft (see Business Snapshot 3.2).Problem 3.21.An airline executive has argued: “There is no point in our using oil futures. There is just as much chance that the price of oil in the future will be less than the futures price as there is that it will be greater than this price.” Discuss the executive’s viewpoint.It may well be true that there is just as much chance that the price of oil in the future will be above the futures price as that it will be below the futures price. This means that the use of a futures contract for speculation would be like betting on whether a coin comes up heads or tails. But it might make sense for the airline to use futures for hedging rather than speculation. The futures contract then has the effect of reducing risks. It can be argued that an airline should not expose its shareholders to risks associated with the future price of oil when there are contracts available to hedge the risks.Problem 3.22.Suppose the one-year gold lease rate is 1.5% and the one-year risk-free rate is 5.0%. Both rates are compounded annually. Use the discussion in Business Snapshot 3.1 to calculate the maximum one-year forward price Goldman Sachs should quote for gold when the spot price is $1,200.Goldman Sachs can borrow 1 ounce of gold and sell it for $1200. It invests the $1,200 at 5% so that it becomes $1,260 at the end of the year. It must pay the lease rate of 1.5% on $1,200. This is $18 and leaves it with $1,242. It follows that if it agrees to buy the gold for less than $1,242 in one year it will make a profit.Problem 3.23.The expected return on the S&P 500 is 12% and the risk-free rate is 5%. What is the expected return on the investment with a beta of (a) 0.2, (b) 0.5, and (c) 1.4?a)00502(012005)0064.+.⨯.-.=. or 6.4%b)00505(012005)0085.+.⨯.-.=. or 8.5%c)00514(012005)0148.+.⨯.-.=. or 14.8%Further QuestionsProblem 3.24.It is now June. A company knows that it will sell 5,000 barrels of crude oil in September.It uses the October CME Group futures contract to hedge the price it will receive. Each contract is on 1,000 barrels of ‘‘light sweet crude.’’ What position should it take? What price risks is it still exposed to after taking the position?It should short five contracts. It has basis risk. It is exposed to the difference between the October futures price and the spot price of light sweet crude at the time it closes out its position in September. It is also possibly exposed to the difference between the spot price of light sweet crude and the spot price of the type of oil it is selling.Problem 3.25.Sixty futures contracts are used to hedge an exposure to the price of silver. Each futures contract is on 5,000 ounces of silver. At the time the hedge is closed out, the basis is $0.20per ounce. What is the effect of the bas is on the hedger’s financial position if (a) the traderis hedging the purchase of silver and (b) the trader is hedging the sale of silver?The excess of the spot over the futures at the time the hedge is closed out is $0.20 per ounce. If the trader is hedging the purchase of silver, the price paid is the futures price plus the basis. Thetrader therefore loses 60×5,000×$0.20=$60,000. If the trader is hedging the sales of silver, the price received is the futures price plus the basis. The trader therefore gains $60,000.Problem 3.26.A trader owns 55,000 units of a particular asset and decides to hedge the value of her position with futures contracts on another related asset. Each futures contract is on 5,000 units. The spot price of the asset that is owned is $28 and the standard deviation of the change in this price over the life of the hedge is estimated to be $0.43. The futures price of the related asset is $27 and the standard deviation of the change in this over the life of the hedge is $0.40. The coefficient of correlation between the spot price change and futures price change is 0.95.(a) What is the minimum variance hedge ratio?(b) Should the hedger take a long or short futures position?(c) What is the optimal number of futures contracts with no tailing of the hedge?(d) What is the optimal number of futures contracts with tailing of the hedge?(a) The minimum variance hedge ratio is 0.95×0.43/0.40=1.02125.(b) The hedger should take a short position.(c) The optimal number of contracts with no tailing is 1.02125×55,000/5,000=11.23 (or 11 when rounded to the nearest whole number)(d) The optimal number of contracts with tailing is 1.012125×(55,000×28)/(5,000×27)=11.65 (or 12 when rounded to the nearest whole number).Problem 3.27.A company wishes to hedge its exposure to a new fuel whose price changes have a 0.6correlation with gasoline futures price changes. The company will lose $1 million for each 1 cent increase in the price per gallon of the new fuel over the next three months. The new fuel's price change has a standard deviation that is 50% greater than price changes in gasoline futures prices. If gasoline futures are used to hedge the exposure what should the hedge ratio be? What is the company's exposure measured in gallons of the new fuel? What position measured in gallons should the company take in gasoline futures? How many gasoline futures contracts should be traded? Each contract is on 42,000 gallons.Equation (3.1) shows that the hedge ratio should be 0.6 × 1.5 = 0.9. The company has anexposure to the price of 100 million gallons of the new fuel. It should therefore take a position of 90 million gallons in gasoline futures. Each futures contract is on 42,000 gallons. The number of contracts required is therefore9.2142000,42000,000,90 or, rounding to the nearest whole number, 2143.Problem 3.28.A portfolio manager has maintained an actively managed portfolio with a beta of 0.2. During the last year the risk-free rate was 5% and equities performed very badly providing a return of−30%. The portfolio manage produced a return of −10% and claims that in the circumstances it was good. Discuss this claim.When the expected return on the market is −30% the expected return on a portfolio with a beta of 0.2 is0.05 + 0.2 × (−0.30 − 0.05) = −0.02or –2%. The actual return of –10% is worse than the expected return. The portfolio manager done 8% worse than a simple strategy of forming a portfolio that is 20% invested in an equity index and 80% invested in risk-free investments. (The manager has achieved an alpha of –8%!)Problem 3.29. (Excel file)The following table gives data on monthly changes in the spot price and the futures price for a certain commodity. Use the data to calculate a minimum variance hedge ratio.Denote i x and i y by the i -th observation on the change in the futures price and the change in the spot price respectively.096130i i x y =.=.∑∑222447423594i i x y =.=.∑∑2352i i x y =.∑ An estimate of F σ is05116=. An estimate of S σ is04933=.An estimate of ρ is0981=.The minimum variance hedge ratio is049330981094605116SF.=.⨯=..σρσProblem 3.30.It is July 16. A company has a portfolio of stocks worth $100 million. The beta of the portfolio is 1.2. The company would like to use the December futures contract on a stock index to change beta of the portfolio to 0.5 during the period July 16 to November 16. The index is currently1,000, and each contract is on $250 times the index.a)What position should the company take?b)Suppose that the company changes its mind and decides to increase the beta of theportfolio from 1.2 to 1.5. What position in futures contracts should it take?a)The company should short(1205)1000000001000250.-.⨯,,⨯or 280 contracts.b)The company should take a long position in(1512)1000000001000250.-.⨯,,⨯or 120 contracts.Problem 3.31. (Excel file)A fund manager has a portfolio worth $50 million with a beta of 0.87. The manager is concerned about the performance of the market over the next two months and plans to use three-month futures contracts on the S&P 500 to hedge the risk. The current level of the index is 1250, one contract is on 250 times the index, the risk-free rate is 6% per annum, and the dividend yield on the index is 3% per annum. The current 3 month futures price is 1259.a)What position should the fund manager take to eliminate all exposure to the market overthe next two months?b)Calculate the effect of your strategy on the fund manager’s returns if the level of themarket in two months is 1,000, 1,100, 1,200, 1,300, and 1,400. Assume that the one-month futures price is 0.25% higher than the index level at this time.a) The number of contracts the fund manager should short is 50000000087138201259250,,.⨯=.⨯ Rounding to the nearest whole number, 138 contracts should be shorted.b) The following table shows that the impact of the strategy. To illustrate the calculations in the table consider the first column. If the index in two months is 1,000, the futures price is 1000×1.0025. The gain on the short futures position is therefore(1259100250)2501388849250$-.⨯⨯=,,The return on the index is 3212⨯/=0.5% in the form of dividend and250125020%-/=- in the form of capital gains. The total return on the index istherefore 195%-.. The risk-free rate is 1% per two months. The return is therefore205%-. in excess of the risk-free rate. From the capital asset pricing model we expect the return on the portfolio to be 0872*******%%.⨯-.=-. in excess of the risk-free rate. The portfolio return is therefore 16835%-.. The loss on the portfolio is01683550000000.⨯,, or $8,417,500. When this is combined with the gain on the futures the total gain is $431,750.Problem 3.32.It is now October 2014. A company anticipates that it will purchase 1 million pounds of copper in each of February 2015, August 2015, February 2016, and August 2016. The company has decided to use the futures contracts traded in the COMEX division of the CME Group to hedge its risk. One contract is for the delivery of 25,000 pounds of copper. The initial margin is $2,000per contract a nd the maintenance margin is $1,500 per contract. The company’s policy is to hedge 80% of its exposure. Contracts with maturities up to 13 months into the future are considered to have sufficient liquidity to meet the company’s needs. Devise a hedging stra tegy for the company. (Do not make the “tailing” adjustment described in Section 3.4.)Assume the market prices (in cents per pound) today and at future dates are as follows. What is the impact of the strategy you propose on the price the company pays for copper? What is the initial margin requirement in October 2014? Is the company subject to any margin calls?To hedge the February 2015 purchase the company should take a long position in March 2015 contracts for the delivery of 800,000 pounds of copper. The total number of contracts required is 8000002500032,/,=. Similarly a long position in 32 September 2015 contracts is required to hedge the August 2015 purchase. For the February 2016 purchase the company could take a long position in 32 September 2015 contracts and roll them into March 2016 contracts during August 2015. (As an alternative, the company could hedge the February 2016 purchase by taking a long position in 32 March 2015 contracts and rolling them into March 2016 contracts.) For the August 2016 purchase the company could take a long position in 32 September 2015 and roll them into September 2016 contracts during August 2015.The strategy is therefore as followsOct. 2014: Enter into long position in 96 Sept. 2015 contractsEnter into a long position in 32 Mar. 2015 contractsFeb 2015: Close out 32 Mar. 2015 contractsAug 2015: Close out 96 Sept. 2015 contractsEnter into long position in 32 Mar. 2016 contractsEnter into long position in 32 Sept. 2016 contractsFeb 2016: Close out 32 Mar. 2016 contractsAug 2016: Close out 32 Sept. 2016 contractsWith the market prices shown the company pays.+.⨯.-.=.3690008(3723036910)37156for copper in February, 2015. It pays.+.⨯.-.=.3650008(3728036480)37140for copper in August 2015. As far as the February 2016 purchase is concerned, it loses 3728036480800.-.=. on the .-.=. on the September 2015 futures and gains 37670364301240 February 2016 futures. The net price paid is therefore.+.⨯.-.⨯.=.377000880008124037348.-.=. on theAs far as the August 2016 purchase is concerned, it loses 3728036480800.-.=. on the September 2016 futures. The September 2015 futures and gains 38820364202400net price paid is therefore388000880008240037520.+.⨯.-.⨯.=.The hedging strategy succeeds in keeping the price paid in the range 371.40 to 375.20.In October 2014 the initial margin requirement on the 128 contracts is 1282000$⨯, or $256,000. There is a margin call when the futures price drops by more than 2 cents. This happens to the March 2015 contract between October 2014 and February 2015, to the September 2015 contract between October 2014 and February 2015, and to the September 2015 contract between February 2015 and August 2015. (Under the plan above the March 2016 contract is not held between February 2015 and August 2015, but if it were there would be a margin call during this period.)。

CHAPTER 15The Black-Scholes-Merton ModelPractice QuestionsProblem 15.1.What does the Black –Scholes –Merton stock option pricing model assume about the probability distribution of the stock price in one year? What does it assume about theprobability distribution of the continuously compounded rate of return on the stock during the year?The Black –Scholes –Merton option pricing model assumes that the probability distribution of the stock price in 1 year (or at any other future time) is lognormal. It assumes that thecontinuously compounded rate of return on the stock during the year is normally distributed.Problem 15.2.The volatility of a stock price is 30% per annum. What is the standard deviation of the percentage price change in one trading day?The standard deviation of the percentage price change in time t ∆is where σ is the volatility. In this problem 03=.and, assuming 252 trading days in one year, 12520004t ∆=/=. so that 00019=.=. or 1.9%.Problem 15.3.Explain the principle of risk-neutral valuation.The price of an option or other derivative when expressed in terms of the price of theunderlying stock is independent of risk preferences. Options therefore have the same value in a risk-neutral world as they do in the real world. We may therefore assume that the world is risk neutral for the purposes of valuing options. This simplifies the analysis. In a risk-neutral world all securities have an expected return equal to risk-free interest rate. Also, in arisk-neutral world, the appropriate discount rate to use for expected future cash flows is the risk-free interest rate.Problem 15.4.Calculate the price of a three-month European put option on a non-dividend-paying stock with a strike price of $50 when the current stock price is $50, the risk-free interest rate is 10% per annum, and the volatility is 30% per annum.In this case 050S =, 50K =, 01r=., 03σ=., 025T =., and12102417000917d d d ==.=-.=.The European put price is 0102550(00917)50(02417)N e N -.⨯.-.--.0102550046345004045237e -.⨯.=⨯.-⨯.=.or $2.37.Problem 15.5.What difference does it make to your calculations in Problem 15.4 if a dividend of $1.50 is expected in two months?In this case we must subtract the present value of the dividend from the stock price before using Black –Scholes-Merton. Hence the appropriate value of 0S is01667010501504852S e -.⨯.=-.=.As before 50K =, 01r =., 03σ=., and 025T =.. In this case12100414001086d d d ==.=-.=-.The European put price is 0102550(01086)4852(00414)N e N -.⨯..-.-. 010255005432485204835303e -.⨯.=⨯.-.⨯.=. or $3.03.Problem 15.6.What is implied volatility? How can it be calculated?The implied volatility is the volatility that makes the Black –Scholes-Merton price of an option equal to its market price. The implied volatility is calculated using an iterativeprocedure. A simple approach is the following. Suppose we have two volatilities one too high (i.e., giving an option price greater than the market price) and the other too low (i.e., giving an option price lower than the market price). By testing the volatility that is half way between the two, we get a new too-high volatility or a new too-low volatility. If we search initially for two volatilities, one too high and the other too low we can use this procedure repeatedly to bisect the range and converge on the correct implied volatility. Other more sophisticated approaches (e.g., involving the Newton-Raphson procedure) are used in practice.Problem 15.7.A stock price is currently $40. Assume that the expected return from the stock is 15% and its volatility is 25%. What is the probability distribution for the rate of return (with continuous compounding) earned over a two-year period?In this case 015=.μ and 025=.σ. From equation (15.7) the probability distribution for the rate of return over a two-year period with continuous compounding is:⎪⎪⎭⎫⎝⎛-ϕ225.0,225.015.022i.e.,)03125.0,11875.0(ϕThe expected value of the return is 11.875% per annum and the standard deviation is 17.7%per annum.Problem 15.8.A stock price follows geometric Brownian motion with an expected return of 16% and a volatility of 35%. The current price is $38.a) What is the probability that a European call option on the stock with an exercise price of $40 and a maturity date in six months will be exercised?b) What is the probability that a European put option on the stock with the same exercise price and maturity will be exercised?a) The required probability is the probability of the stock price being above $40 in six months time. Suppose that the stock price in six months is T S⎥⎦⎤⎢⎣⎡⨯⎪⎪⎭⎫ ⎝⎛-+5.035.0,5.0235.016.038ln ~ln 22ϕT Si.e.,()2247.0,687.3~ln ϕT SSince ln 403689=., we require the probability of ln(S T )>3.689. This is3689368711(0008)0247N N .-.⎛⎫-=-. ⎪.⎝⎭Since N(0.008) = 0.5032, the required probability is 0.4968.b) In this case the required probability is the probability of the stock price being less than $40 in six months time. It is10496805032-.=.Problem 15.9.Using the notation in the chapter, prove that a 95% confidence interval for T S is between22(2)196(2)19600andT T S e S e -/-.-/+.μσμσFrom equation (15.3):⎥⎦⎤⎢⎣⎡⎪⎪⎭⎫ ⎝⎛-+T T S S T 220,2ln ~ln σσμϕ95% confidence intervals for ln T S are therefore20ln ()1962S T +--.σμand20ln ()1962S T +-+.σμσ95% confidence intervals for T S are therefore2200ln (2)196ln (2)196and S T S T e e +-/-.+-/+.μσμσi.e.22(2)196(2)19600andT T S e S e -/-.-/+.μσμσProblem 15.10.A portfolio manager announces that the average of the returns realized in each of the last 10 years is 20% per annum. In what respect is this statement misleading?This problem relates to the material in Section 15.3. The statement is misleading in that a certain sum of money, say $1000, when invested for 10 years in the fund would have realized a return (with annual compounding) of less than 20% per annum.The average of the returns realized in each year is always greater than the return per annum (with annual compounding) realized over 10 years. The first is an arithmetic average of the returns in each year; the second is a geometric average of these returns.Problem 15.11.Assume that a non-dividend-paying stock has an expected return of μ and a volatility of σ. An innovative financial institution has just announced that it will trade a derivative that pays off a dollar amount equal to ln S T at time T where T S denotes the values of the stock price at time T .a) Use risk-neutral valuation to calculate the price of the derivative at time t in term of the stock price, S, at time tb) Confirm that your price satisfies the differential equation (15.16)a) At time t , the expected value of ln S T is from equation (15.3) 2ln (/2)()S T t μσ+--In a risk-neutral world the expected value of ln S T is therefore2ln (/2)()S r T t σ+--Using risk-neutral valuation the value of the derivative at time t is()2[ln (/2)()]r T t e S r T t σ--+--b) If()2[ln (/2)()]r T t f e S r T t σ--=+--then()()2()2()2()22[ln (/2)()]/2r T t r T t r T t r T t fre S r T t e r tf e S S f e S S σσ--------∂=+----∂∂=∂∂=-∂The left-hand side of the Black-Scholes-Merton differential equation is()222()2ln (/2)()(/2)/2ln (/2)()r T t r T t e r S r r T t r r e r S r r T t rfσσσσ----⎡⎤+----+-⎣⎦⎡⎤=+--⎣⎦=Hence the differential equation is satisfied.Problem 15.12.Consider a derivative that pays off n T S at time T where T S is the stock price at that time. When the stock pays no dividends and its price follows geometric Brownian motion, it can be shown that its price at time t ()t T ≤ has the form()n h t T S ,where S is the stock price at time t and h is a function only of t and T .(a) By substituting into the Black –Scholes –Merton partial differential equation derive an ordinary differential equation satisfied by ()h t T ,.(b) What is the boundary condition for the differential equation for ()h t T ,? (c) Show that2[05(1)(1)]()()n n r n T t h t T eσ.-+--,= where r is the risk-free interest rate and σ is the stock price volatility.If ()()n G S t h t T S ,=, then n t G t h S ∂/∂=, 1n G S hnS -∂/∂=, and 222(1)n G S hn n S -∂/∂=- where t h h t =∂/∂. Substituting into the Black –Scholes –Merton differential equation we obtain21(1)2t h rhn hn n rh σ++-=The derivative is worth n S when t T =. The boundary condition for this differential equation is therefore ()1h T T ,= The equation2[05(1)(1)]()()n n r n T t h t T e σ.-+--,=satisfies the boundary condition since it collapses to 1h = when t T =. It can also be shown that it satisfies the differential equation in (a). Alternatively we can solve the differential equation in (a) directly. The differential equation can be written21(1)(1)2t h r n n n h σ=----The solution to this is21ln [(1)(1)]()2h r n n n T t σ=-+--or2[05(1)(1)]()()n n r n T t h t T e σ.-+--,=Problem 15.13.What is the price of a European call option on a non-dividend-paying stock when the stock price is $52, the strike price is $50, the risk-free interest rate is 12% per annum, the volatility is 30% per annum, and the time to maturity is three months?In this case 052S =, 50K =, 012r =., 030=.σ and 025T =..212105365003865d d d ==.=-.=. The price of the European call is 01202552(05365)50(03865)N e N -.⨯..-. 00352070425006504e -.=⨯.-⨯. 506=. or $5.06.Problem 15.14.What is the price of a European put option on a non-dividend-paying stock when the stock price is $69, the strike price is $70, the risk-free interest rate is 5% per annum, the volatility is 35% per annum, and the time to maturity is six months?In this case 069S =, 70K =, 005r =., 035=.σ and 05T =..212101666000809d d d ==.=-.=-. The price of the European put is 0050570(00809)69(01666)e N N -.⨯..--. 002570053236904338e -.=⨯.-⨯.640=. or $6.40.Problem 15.15.Consider an American call option on a stock. The stock price is $70, the time to maturity is eight months, the risk-free rate of interest is 10% per annum, the exercise price is $65, and the volatility is 32%. A dividend of $1 is expected after three months and again after sixmonths. Show that it can never be optimal to exercise the option on either of the two dividend dates. Use DerivaGem to calculate the price of the option.Using the notation of Section 15.12, 121D D ==, 2()0101667(1)65(1)107r T t K e e ---.⨯.-=-=., and 21()01025(1)65(1)160r t t K e e ---.⨯.-=-=.. Since 2()1(1)r T t D K e --<- and 21()2(1)r t t D K e --<-It is never optimal to exercise the call option early. DerivaGem shows that the value of the option is 1094..Problem 15.16.A call option on a non-dividend-paying stock has a market price of $2.50. The stock price is $15, the exercise price is $13, the time to maturity is three months, and the risk-free interest rate is 5% per annum. What is the implied volatility?In the case 25c =., 015S =, 13K =, 025T =., 005r =.. The implied volatility must becalculated using an iterative procedure.A volatility of 0.2 (or 20% per annum) gives 220c =.. A volatility of 0.3 gives 232c =.. A volatility of 0.4 gives 2507c =.. A volatility of 0.39 gives 2487c =.. By interpolation the implied volatility is about 0.396 or 39.6% per annum.The implied volatility can also be calculated using DerivaGem. Select equity as theUnderlying Type in the first worksheet. Select Black-Scholes European as the Option Type. Input stock price as 15, the risk-free rate as 5%, time to exercise as 0.25, and exercise price as 13. Leave the dividend table blank because we are assuming no dividends. Select the button corresponding to call. Select the implied volatility button. Input the Price as 2.5 in the second half of the option data table. Hit the Enter key and click on calculate. DerivaGem will show the volatility of the option as 39.64%.Problem 15.17.With the notation used in this chapter (a) What is ()N x '?(b) Show that ()12()()r T t SN d Ke N d --''=, where S is the stock price at time t21d =22d =(c) Calculate 1d S ∂/∂ and 2d S ∂/∂. (d) Show that when()12()()r T t c SN d Ke N d --=-()21()(r T t c rKe N d SN d t --∂'=--∂ where c is the price of a call option on a non-dividend-paying stock. (e) Show that 1()c S N d ∂/∂=.(f) Show that the c satisfies the Black –Scholes –Merton differential equation.(g) Show that c satisfies the boundary condition for a European call option, i.e., thatmax(0)c S K =-, as t tends to T.(a) Since ()N x is the cumulative probability that a variable with a standardized normal distribution will be less than x , ()N x ' is the probability density function for a standardized normal distribution, that is,22()x N x -'=(b)12()(N d N d ''=+2221()22d d T t σσ⎡⎤=---⎢⎥⎣⎦221()exp ()2N d d T t σσ⎡⎤'=--⎢⎥⎣⎦Because22d =it follows that()21exp ()2r T t Ked T t S σσ--⎡⎤--=⎢⎥⎣⎦As a result()12()()r T t SN d Ke N d --''=which is the required result.(c)22212S K d σσ==Hence1d S ∂=∂ Similarly222d σ=and2d S ∂=∂ Therefore:12d d S S∂∂=∂∂(d)()12()()12122()()()()()r T t r T t r T t c SN d Ke N d d d c SN d rKe N d Ke N d t t t------=-∂∂∂''=--∂∂∂ From (b):()12()()r T t SN d Ke N d --''=Hence()1221()()r T t d d c rKe N d SN d t tt --∂∂∂⎛⎫'=-+- ⎪∂∂∂⎝⎭ Since12d d -=12(d d t t t∂∂∂-=∂∂∂=Hence()21()(r T t c rKe N d SN d t --∂'=--∂(e) From differentiating the Black –Scholes –Merton formula for a call price weobtain()12112()()()r T t d d cN d SN d Ke N d S S dS--∂∂∂''=+-∂∂ From the results in (b) and (c) it follows that1()cN d S ∂=∂(f) Differentiating the result in (e) and using the result in (c), we obtain21121()(d cN d S SN d ∂∂'=∂∂'= From the results in d) and e)222()2122211()121()(21()(2[()()]r T t r T t c c c rS S rKe N d SN d t S S rSN d S N d r SN d Ke N d rcσσ----∂∂∂'++=--∂∂∂'++=-=This shows that the Black –Scholes –Merton formula for a call option does indeed satisfy the Black –Scholes –Merton differential equation(g) Consider what happens in the formula for c in part (d) as t approaches T . IfS K >, 1d and 2d tend to infinity and 1()N d and 2()N d tend to 1. If S K <, 1d and 2d tend to zero. It follows that the formula for c tends to max(0)S K -,.Problem 15.18.Show that the Black –Scholes –Merton formulas for call and put options satisfy put –call parity.The Black –Scholes –Merton formula for a European call option is 012()()rT c S N d Ke N d -=- so that 012()()rT rT rT c Ke S N d Ke N d Ke ---+=-+ or 012()[1()]rT rT c Ke S N d Ke N d --+=+- or 012()()rT rT c Ke S N d Ke N d --+=+-The Black –Scholes –Merton formula for a European put option is 201()()rT p Ke N d S N d -=--- so that 02010()()rT p S Ke N d S N d S -+=---+ or 0201()[1()]rT p S Ke N d S N d -+=-+-- or 0201()()rT p S Ke N d S N d -+=-+ This shows that the put –call parity result 0rT c Ke p S -+=+ holds.Problem 15.19.A stock price is currently $50 and the risk-free interest rate is 5%. Use the DerivaGemsoftware to translate the following table of European call options on the stock into a table of implied volatilities, assuming no dividends. Are the option prices consistent with the assumptions underlying Black –Scholes –Merton?Using DerivaGem we obtain the following table of implied volatilitiesBlack-Scholes European as the Option Type. Input stock price as 50, the risk-free rate as 5%, time to exercise as 0.25, and exercise price as 45. Leave the dividend table blank because we are assuming no dividends. Select the button corresponding to call. Select the implied volatility button. Input the Price as 7.0 in the second half of the option data table. Hit theEnter key and click on calculate. DerivaGem will show the volatility of the option as 37.78%. Change the strike price and time to exercise and recompute to calculate the rest of the numbers in the table.The option prices are not exactly consistent with Black –Scholes –Merton. If they were, the implied volatilities would be all the same. We usually find in practice that low strike price options on a stock have significantly higher implied volatilities than high strike price options on the same stock. This phenomenon is discussed in Chapter 20.Problem 15.20.Explain carefully why Black’s approach to evaluating an American call option on a dividend-paying stock may give an approximate answer even when only one dividend is anticipated. Does the answer given by Black’s approach understate or overstate the true option value? Explain your answer.Black’s approach in effect assumes that the holder of option must decide at time zero whether it is a European option maturing at time n t (the final ex-dividend date) or a European optionmaturing at time T . In fact the holder of the option has more flexibility than this. The holder can choose to exercise at time n t if the stock price at that time is above some level but nototherwise. Furthermore, if the option is not exercised at time n t , it can still be exercised attime T .It appears that Black ’s approach should understate the true option value. This is because the holder of the option has more alternative strategies for deciding when to exercise the option than the two strategies implicitly assumed by the approach. These alternative strategies add value to the option.However, this is not the whole story! The standard approach to valuing either an American or a European option on a stock paying a single dividend applies the volatility to the stock price less the present value of the dividend. (The procedure for valuing an American option is explained in Chapter 21.) Black’s approach when considering exercise just prior to the dividend date applies the volatility to the stock price itself. Black’s approach thereforeassumes more stock price variability than the standard approach in some of its calculations. In some circumstances it can give a higher price than the standard approach.Problem 15.21.Consider an American call option on a stock. The stock price is $50, the time to maturity is 15 months, the risk-free rate of interest is 8% per annum, the exercise price is $55, and the volatility is 25%. Dividends of $1.50 are expected in 4 months and 10 months. Show that it can never be optimal to exercise the option on either of the two dividend dates. Calculate the price of the option.With the notation in the text12121500333308333125008and 55D D t t T r K ==.,=.,=.,=.,=.=2()00804167155(1)180r T t K e e ---.⨯.⎡⎤⎢⎥⎣⎦-=-=. Hence2()21r T t D K e --⎡⎤⎢⎥⎣⎦<- Also:21()00805155(1)216r t t K e e ---.⨯.⎡⎤⎢⎥⎣⎦-=-=. Hence:21()11r t t D K e --⎡⎤⎢⎥⎣⎦<- It follows from the conditions established in Section 15.12 that the option should never be exercised early.The present value of the dividends is033330080833300815152864e e -.⨯.-.⨯..+.=.The option can be valued using the European pricing formula with:05028644713655025008125S K r T σ=-.=.,=,=.,=.,=.212100545003340d d d ==-.=-.=-.12()04783()03692N d N d =.,=.and the call price is00812547136047835503692417e -.⨯..⨯.-⨯.=. or $4.17.Problem 15.22.Show that the probability that a European call option will be exercised in a risk-neutral world is, with the notation introduced in this chapter, 2()N d . What is an expression for the value of a derivative that pays off $100 if the price of a stock at time T is greater than K ?The probability that the call option will be exercised is the probability that T S K > where T S is the stock price at time T . In a risk neutral world[]T T r S S T 220,)2/(ln ~ln σσϕ-+The probability that T S K > is the same as the probability that ln ln T S K >. This is21N ⎡⎤-2N =2()N d =The expected value at time T in a risk neutral world of a derivative security which pays off $100 when T S K > is therefore2100()N dFrom risk neutral valuation the value of the security at time zero is2100()rT e N d -Problem 15.23.Use the result in equation (15.17) to determine the value of a perpetual American put option on a non-dividend-paying stock with strike price K if it is exercised when the stock price equals H where H<K. Assume that the current stock price S is greater than H. What is the value of H that maximizes the option value? Deduce the value of a perpetual American put option with strike price K.If the perpetual American put is exercised when S=H , it provides a payoff of (K −H ). We obtain its value, by setting Q=K−H in equation (15.17), as 22/2/2)()(σσ-⎪⎭⎫ ⎝⎛-=⎪⎭⎫ ⎝⎛-=r r S H H K H S H K VNow1/222/222222/21/22/2222222)(212)(212-σσσ-σσ⎪⎭⎫ ⎝⎛σ⎪⎭⎫ ⎝⎛σ-+-+⎪⎭⎫ ⎝⎛σ-=⎪⎭⎫ ⎝⎛σ-+-⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛σ-+⎪⎭⎫ ⎝⎛-=r r r r r S H S r H H K r S H H rK dH V d H H K r S H S H r S H K S H dH dVdV/dH is zero when222σ+=r rK H and, for this value of H , d 2V /dH 2 is negative indicating that it gives the maximum value of V .The value of the perpetual American put is maximized if it is exercised when S equals this value of H . Hence the value of the perpetual American put is 2/2)(σ-⎪⎭⎫ ⎝⎛-r H S H Kwhen H =2rK /(2r +σ2). The value is 2/22222)2(2σ-⎪⎪⎭⎫ ⎝⎛σ+σ+σr rK r S r K This is consistent with the more general result produced in Chapter 26 for the case where the stock provides a dividend yield.Problem 15.24.A company has an issue of executive stock options outstanding. Should dilution be taken into account when the options are valued? Explain you answer.The answer is no. If markets are efficient they have already taken potential dilution intoaccount in determining the stock price. This argument is explained in Business Snapshot 15.3.Problem 15.25.A company’s stock price is $50 and 10 million shares are outstanding. The company is considering giving its employees three million at-the-money five-year call options. Option exercises will be handled by issuing more shares. The stock price volatility is 25%, the five-year risk-free rate is 5% and the company does not pay dividends. Estimate the cost to the company of the employee stock option issue.The Black-Scholes-Merton price of the option is given by setting 050S =, 50K =, 005r =., 025σ=., and 5T =. It is 16.252. From an analysis similar to that in Section 15.10 the cost to the company of the options is 1016252125103⨯.=.+ or about $12.5 per option. The total cost is therefore 3 million times this or $37.5 million. Ifthe market perceives no benefits from the options the stock price will fall by $3.75.Further QuestionsProblem 15.26.If the volatility of a stock is 18% per annum, estimate the standard deviation of the percentage price change in (a) one day, (b) one week, and (c) one month.(a) %13.125218=(b) %50.25218=(c) %20.51218=Problem 15.27.A stock price is currently $50. Assume that the expected return from the stock is 18% per annum and its volatility is 30% per annum. What is the probability distribution for the stock price in two years? Calculate the mean and standard deviation of the distribution. Determine the 95% confidence interval.In this case 050S =, 018=.μ and 030=.σ. The probability distribution of the stock price in two years, T S , is lognormal and is, from equation (15.3), given by:⎥⎦⎤⎢⎣⎡⨯⎪⎭⎫ ⎝⎛-+23.0,2209.018.050ln ~ln 2ϕT Si.e.,)42.0,18.4(~ln 2ϕT SThe mean stock price is from equation (15.4)018203650507167e e .⨯.==.and the standard deviation is from equation (15.5)018503183e .⨯=.95% confidence intervals for ln T S are418196042and 418196042.-.⨯..+.⨯.i.e.,335and 501..These correspond to 95% confidence limits for T S of 335501and e e ..i.e.,2852and 15044..Problem 15.28. (Excel file)Suppose that observations on a stock price (in dollars) at the end of each of 15 consecutive weeks are as follows:30.2, 32.0, 31.1, 30.1, 30.2, 30.3, 30.6, 33.0,32.9, 33.0, 33.5, 33.5, 33.7, 33.5, 33.2Estimate the stock price volatility. What is the standard error of your estimate?The calculations are shown in the table below2009471001145i i uu =.=.∑∑and an estimate of standard deviation of weekly returns is:002884=.The volatility per annum is therefore 002079.=. or 20.79%. The standard error of this estimate is 00393=. or 3.9% per annum.Problem 15.29.A financial institution plans to offer a security that pays off a dollar amount equal to 2T S at time T .(a) Use risk-neutral valuation to calculate the price of the security at time t in terms of the stock price, S , at time t . (Hint: The expected value of 2T S can be calculatedfrom the mean and variance of T S given in section 15.1.)(b) Confirm that your price satisfies the differential equation (15.16).(a) The expected value of the security is 2[()]T E S From equations (15.4) and (15.5), at time t :2()2()()2()var()[1]T t T T t T t T E S Se S S e e μμσ---==-Since 22var()[()][()]T T T S E S E S =-, it follows that 22[()]var()[()]T T T E S S E S =+ so that 22222()()22()(2)()2[()][1]T t T t T t T T t E S S e e S e S e μσμμσ---+-=-+=In a risk-neutral world r μ= so that222(2)()ˆ[()]r T t T E S S e σ+-= Using risk-neutral valuation, the value of the derivative security at time t is()2ˆ[()]r T t Te E S --22(2)()()r T t r T t S e e σ+---=22()()r T t S e σ+-=(b) If: 22222()()()()22()()2()()2()22r T t r T t r T t r T t f S e f S r e tf Se Sf e S σσσσσ+-+-+-+-=∂=-+∂∂=∂∂=∂ The left-hand side of the Black-Scholes –Merton differential equation is:222222()()2()()22()()()()2()2r T t r T t r T t r T t S r e rS e S e rS e rfσσσσσσ+-+-+-+--+++==Hence the Black-Scholes equation is satisfied.Problem 15.30.Consider an option on a non-dividend-paying stock when the stock price is $30, the exercise price is $29, the risk-free interest rate is 5% per annum, the volatility is 25% per annum, and the time to maturity is four months.a. What is the price of the option if it is a European call?b. What is the price of the option if it is an American call?c. What is the price of the option if it is a European put?d. Verify that put –call parity holds.In this case 030S =, 29K =, 005r =., 025=.σ and 412T =/2104225d ==.2202782d ==.(04225)06637(02782)06096N N .=.,.=.(04225)03363(02782)03904N N -.=.,-.=.a. The European call price is00541230066372906096252e -.⨯/⨯.-⨯.=.or $2.52.b. The American call price is the same as the European call price. It is $2.52.c. The European put price is00541229039043003363105e -.⨯/⨯.-⨯.=.or $1.05.d. Put-call parity states that:rT p S c Ke -+=+In this case 252c =., 030S =, 29K =, 105p =. and 09835rT e -=. and it is easy to verify that the relationship is satisfied,Problem 15.31.Assume that the stock in Problem 15.30 is due to go ex-dividend in 1.5 months. The expected dividend is 50 cents.a. What is the price of the option if it is a European call?b. What is the price of the option if it is a European put?c. If the option is an American call, are there any circumstances when it will beexercised early?a. The present value of the dividend must be subtracted from the stock price. This givesa new stock price of:01250053005295031e -.⨯.-.=.and 2103068d ==. 2201625d ==.12()06205()05645N d N d =.;=. The price of the option is therefore005412295031062052905645221e -.⨯/.⨯.-⨯.=.or $2.21.b. Because12()03795()04355N d N d -=.,-=. the value of the option when it is a European put is005412290435529503103795122e -.⨯/⨯.-.⨯.=.。