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1、CHAPTER 31Interest Rate Derivatives: Models of the Short RatePractice 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 t
2、erm structure is an output from the model. In ano-arbitrage model the initial term structure is an input. The behavior of interest rates in a no-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 devi
3、ation is 1% per annum. Whathappens to the standard deviation when the short rate increases to 8% in (a) Vasiceks model;(b) Rendleman and Bartters model; and (c) the Cox, Ingersoll, and Ross model?In Vasiceks model the standard deviation stays at 1%. In the Rendleman and Bartter modestandard deviatio
4、n 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 incre
5、ases 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 sec
6、urity 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
7、 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 on
8、e-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
9、 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 cou
10、pon-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
11、 than one factor.Problem 31.6.Suppose that a = 0.1 and b = 0.1 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 Dt is 0.02 Dt . Compare the prices given by the models for
12、azero-coupon bond that matures in year 10.1In Vasiceks modeal=, 0.1, b = 0.1, and s = 0.02 so thatB(t, t +10) =(1- e-0.110 ) = 6.321210.1A(t, t +10) = exp (6.32121-10)(0.12 0.1- 0.0002) - 0.0004 6.321212 0.010.4= 0.71587The bond price is therefore 0.71587e-6.321210.1 = 0.38046a2 + 2s 2In the Cox, In
13、gersoll, and Ross model, a = 0.1, b = 0.1 and s = 0.02 /0.1 = 0.0632 . AlsoDefineg = 0.13416b = (g + a)(e10g -1) + 2g = 0.92992B(t, t +10) = 2(e10g -1) = 6.07650b 2g e5( a+g ) 2ab/s 2A(t, t +10) = b= 0.69746The bond price is therefore 0.69746e-6.076500.1 = 0.37986Problem 31.7.Suppose that a = 0.1, b
14、 = 0.08 , and s = 0.015 in Vasiceks model with the initial value ofthe 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$87.1- e-20.112 0.1Using the notation in the text, s =
15、3 , T = 1 , L = 100 , K = 87 , and 0.015o=(1- e-20.1 )= 0.025886P0.1From equation (31.6), P(0,1) = 0.94988 , P(0, 3) = 0.85092 , and h = 1.14277 so thatequation (31.20) gives the call price as call price is100 0.85092 N (1.14277) - 87 0.94988 N (1.11688) = 2.59or $2.59.Problem 31.8.Repeat Problem 31
16、.7 valuing a European put option with a strike of $87. What is the putcall parity relationship between the prices of European call and put options? Show that the put and call option prices satisfy putcall parity in this case.As mentioned in the text, equation (31.20) for a call option is essentially
17、 the same as Blacks model. By analogy with Blacks formulas corresponding expression for a put option isIn this case the put price isKP(0,T )N (-h +s) - LP(0, s)N (-h)P87 0.94988 N (-1.11688) -100 0.85092 N (-1.14277) = 0.14Since the underlying bond pays no coupon, putcall parity states that the put
18、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 87 0.94988 = 82.64 . Putcall parity is therefore satisfied:82.64 + 2.59 = 85.09 + 0.14Problem 31.9.Suppose that a = 0.05 , b = 0.08 ,
19、 and s = 0.015 in Vasiceks 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 o
20、ption 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
21、solve2.5A(2.1, 2.5)e- B(2.1,2.5) r* +102.5A(2.1,3.0)e- B(2.1,3.0) r* = 99where the A and B functions are given by equations (31.7) and (31.8). In this caseA(2.1, 2.5)=0.999685, A(2.1,3.0)=0.998432, B(2.1,2.5)=0.396027, andB(2.1, 3.0)= 0.88005.and Solver shows that r* = 0.065989 . Since2.5A(2.1,2.5)e
22、- B(2.1,2.5)r* = 2.434745and102.5A(2.1,3.0)e- B (2.1,3.0)r* = 96.56535the 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 of
23、f 102.5 at time 3.0 years.The options are valued using equation (31.20).For the first optionL=2.5, K= 2.434745, T = 2.1, ands=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 sP= 0.008176 and h= 0.223351. so that the
24、option price is 0.009084.For the second optionL=102.5, K= 96.56535, T = 2.1, ands=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 sP= 0.018168 andh= 0.233343. so that the option price is0.806105.The total value of the
25、 option is therefore 0.0090084+0.806105=0.815189.Problem 31.10.Use the answer to Problem 31.9 and putcall 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:orc + I + PV (K ) = p + B0p = c + PV (K ) - (B0- I )
26、where c is the call price, Kis the strike price, Iis the present value of the coupons, andBis the bond price. In this case c = 0.8152 , PV (K ) = 99 P(0, 2.1) = 87.1222 ,00B - I = 2.5 P(0, 2.5) +102.5 P(0, 3) = 87.4730 so that the put price is0.8152 + 87.1222 - 87.4730 = 0.4644Problem 31.11.In the H
27、ullWhite model, a = 0.08 and s = 0.01. 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 P(0,T ) = e-0.11 =
28、0.9048 and P(0, s) = e-0.15 = 0.6065 . Also1- e-20.0812 0.080.01o=(1- e-40.08 )= 0.0329P0.08and h = -0.4192 so that the call price is100 0.6065N (h) - 68 0.9048N (h -sProblem 31.12.) = 0.439PSuppose that a = 0.05 and s = 0.015 in the HullWhite model with the initial term structure being flat at 6% w
29、ith 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 pric
30、e (not the quoted price) that will be paid for the bond.This problem is similar to Problem 31.9. The difference is that the HullWhite model, which fits an initial term structure, is used instead of Vasiceks model where the initial term structure is determined by the model.The yield curve is flat wit
31、h 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 solve2.5A(2.1, 2.5)e- B(2.1,2.5) r* +102.5A(2.1,3.0)e- B(2
32、.1,3.0) r* = 99where the A and B functions are given by equations (31.16) and (31.17). In this caseA(2.1, 2.5)=0.999732, A(2.1,3.0)=0.998656,B(2.1,2.5)=0.396027, andB(2.1, 3.0)= 0.88005.and Solver shows that r* = 0.066244 . Sinceand2.5A(2.1,2.5)e- B (2.1,2.5)r* = 2.434614102.5A(2.1,3.0)e- B (2.1,3.0
33、)r* = 96.56539the 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 e
34、quation (31.20).For the first optionL=2.5, K= 2.434614, T = 2.1, ands=2.5. Also,P(0,T)=exp(-0.0591182.1)=0.88325and P(0,s)= exp(-0.0591182.5)=0.862609.Furthermore sP= 0.008176 and h= 0.353374. so that the option price is 0.010523. For the second optionL=102.5, K= 96.56539, T = 2.1, ands=3.0. Also,P(
35、0,T)=exp(-0.0591182.1)=0.88325and P(0,s)= exp(-0.0591183.0)=0.837484.Furthermore sP= 0.018168and 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 Dt are taken on the short rate. The
36、 ith observation is ri(1 i m). Show that the maximum likelihood estimates of a, b, and s in Vasiceksmodel are given by maximizingm r - r- a(b - r )Dt 2 - ln(s2 Dt) - ii-1s2 Dti-1i=1 What is the corresponding result for the CIR model?The change ri ri-1is normally distributed with mean a(b ri-1) and v
37、ariance s2Dt. The probability density of the observation is1 r - r- a(b - r ) 2ps2 Dtexp ii-1i-1 2s2 DtWe wish to maximize2ps2 Dtm1 r - r- a(b - r ) i=1exp ii-12s2 Dti-1 Taking logarithms, this is the same as maximizingm r - r- a(b - r )Dt 2 - ln(s2 Dt) - ii-1s2 Dti-1i=1 In the case of the CIR model
38、, the change ri ri-1is normally distributed with mean a(b ri-1)and variance s2 ri-1Dt and the maximum likelihood function becomesm - ln(s2 r Dt) -r - rii-1- a(b - ri-1)Dt 2 i=1 i-1s2 r Dti-1Problem 31.14.Suppose a = 0.05 , s = 0.015 , and the term structure is flat at 10%. Construct a trinomial tree
39、 for the HullWhite model where there are two time steps, each one year in length.3The time step, Dt , is 1 so that Dr = 0.015= 0.02598 . Alsojmax= 4 showing that thebranching method should change four steps from the center of the tree. With only three steps we never reach the point where the branchi
40、ng changes. The tree is shown in Figure S31.1.EBFACGDHINodeABCDEFGHIr10.00%12.61%10.01%7.41%15.24%12.64%10.04%7.44%4.84%p0.16670.14290.16670.19290.12170.14290.16670.19290.2217up0.66660.66420.66660.66420.65670.66420.66660.66420.6567mp0.16670.19290.16670.14290.22170.19290.16670.14290.1217dFigure S31.1
41、: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 node B it is worth 100e-0.121 = 88.69 . At node C it is worth 100e-0.101 = 90.48 . At node D it is wort
42、h 100e-0.081 = 92.31 . It follows that at node A the bond is worth(88.69 0.25 + 90.48 0.5 + 92.31 0.25)e-0.11 = 81.88or $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
43、 bond pays off $100 at time two years. At node B it is worth 100e-0.06931 = 93.30 . At node C it is worth 100e-0.05201 = 94.93 . At node D it is worth 100e-0.03471 = 96.59 . It follows that at node A the bond is worth(93.30 0.167 + 94.93 0.666 + 96.59 0.167)e-0.03821 = 91.37or $91.37. Because 91.37
44、= 100e-0.045122 , 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 no
45、des of the tree. At node E it is worth 100e-0.0880.5 = 95.70 . At node F it is worth 100e-0.06480.5 = 96.81. At node G it is worth 100e-0.04770.5 = 97.64 . At node H it is worth 100e-0.03510.5 = 98.26 . At node I it is worth 100e0.02590.5 = 98.71. At node B it is worth(0.118 95.70 + 0.654 96.81+ 0.2
46、28 97.64)e-0.05640.5 = 94.17 Similarly at nodes C and D it is worth 95.60 and 96.68. The value at node A is therefore(0.167 94.17 + 0.666 95.60 + 0.167 96.68)e-0.03430.5 = 93.92The 18-month zero rate is 0.08 - 0.05e-0.181.5 = 0.0418 . This gives the price of the 18-month zero-coupon bond as 100e-0.04181.5 = 93.92 showing that the tree agrees with the initial term structure.Problem 31.18.