Chapter 30 HullandWhitemodel Consider drt= t,trt dt + t dW t; where t , t and t are nonrandom functions of t . We can solve the stochastic differential equation. Set K t= Z t 0 udu: Then d e K t rt = e K t trt dt + drt = e K t t dt + t dW t : Integrating, we get e K t rt=r0 + Z t 0 e K u u du + Z t 0 e K u u dW u; so rt= e ,Kt r0 + Z t 0 e K u u du + Z t 0 e K u u dW u : From Theorem 1.69 in Chapter 29, we see that rt is a Gaussian process with mean function m r t=e ,Kt r0 + Z t 0 e K u u du (0.1) and covariance function r s; t=e ,Ks,Kt Z s^t 0 e 2Ku 2 udu: (0.2) The process rt is also Markov. 293 294 We want to study R T 0 rt dt .Todothis,wedefine X t= Z t 0 e Ku udW u; Y T = Z T 0 e ,Kt Xtdt: Then rt= e ,Kt r0 + Z t 0 e K u u du + e ,K t X t; Z T 0 rt dt = Z T 0 e ,K t r0 + Z t 0 e K u u du dt + Y T : According to Theorem 1.70 in Chapter 29, R T 0 rt dt is normal. Its mean is IE Z T 0 rt dt = Z T 0 e ,K t r0 + Z t 0 e K u u du dt; (0.3) and its variance is var Z T 0 rt dt ! = IEY 2 T = Z T 0 e 2Kv 2 v Z T v e ,Ky dy ! 2 dv : The price at time 0 of a zero-coupon bond paying $1 at time T is B 0;T= IEexp , Z T 0 rt dt = exp ,1IE Z T 0 rt dt + 1 2 ,1 2 var Z T 0 rt dt ! = exp ,r0 Z T 0 e ,K t dt , Z T 0 Z t 0 e ,K t+K u u du dt + 1 2 Z T 0 e 2K v 2 v Z T v e ,K y dy ! 2 dv = expf,r 0C 0;T,A0;Tg; where C 0;T= Z T 0 e ,Kt dt; A0;T= Z T 0 Z t 0 e ,Kt+K u u du dt , 1 2 Z T 0 e 2K v 2 v Z T v e ,K y dy ! 2 dv : CHAPTER 30. HullandWhitemodel 295 u t u = t T Figure 30.1: Range of values of u; t for the integral. 30.1 Fiddling with the formulas Note that (see Fig 30.1) Z T 0 Z t 0 e ,K t+K u u du dt = Z T 0 Z T u e ,K t+K u u dt du y = t; v = u= Z T 0 e Kv v Z T v e ,Ky dy ! dv : Therefore, A0;T= Z T 0 2 4 e Kv v Z T v e ,Ky dy ! , 1 2 e 2K v 2 v Z T v e ,K y dy ! 2 3 5 dv ; C 0;T= Z T 0 e ,Ky dy ; B 0;T = exp f,r0C 0;T, A0;Tg: Consider the price at time t 2 0;T of the zero-coupon bond: B t; T =IE " exp , Z T t ru du F t : Because r is a Markov process, this should be random only through a dependence on rt . In fact, B t; T = exp f,rtC t; T , At; T g ; 296 where At; T = Z T t 2 4 e Kv v Z T v e ,Ky dy ! , 1 2 e 2K v 2 v Z T v e ,K y dy ! 2 3 5 dv ; C t; T = e Kt Z T t e ,Ky dy : The reason for these changes is the following. We are now taking the initial time to be t rather than zero, so it is plausible that R T 0 ::: dv should be replaced by R T t ::: dv: Recall that K v = Z v 0 udu; and this should be replaced by K v , K t= Z v t udu: Similarly, K y should be replaced by K y , K t . Making these replacements in A0;T ,we see that the K t terms cancel. In C 0;T ,however,the K t term does not cancel. 30.2 Dynamics of the bond price Let C t t; T and A t t; T denote the partial derivatives with respect to t . From the formula B t; T = exp f,rtC t; T , At; T g ; we have dB t; T =Bt; T h ,C t; T drt , 1 2 C 2 t; T drt drt , rtC t t; T dt , A t t; T dt i = B t; T , C t; T t , trt dt , C t; T t dW t , 1 2 C 2 t; T 2 t dt , rtC t t; T dt , A t t; T dt : Because we have used the risk-neutral pricing formula B t; T =IE " exp , Z T t ru du F t to obtain the bond price, its differential must be of the form dB t; T =rtBt; T dt +::: dW t: CHAPTER 30. HullandWhitemodel 297 Therefore, we must have ,C t; T t , trt , 1 2 C 2 t; T 2 t , rtC t t; T , A t t; T = rt: We leave the verification of this equation to the homework. After this verification, we have the formula dB t; T =rtBt; T dt , tC t; T B t; T dW t: In particular, the volatility of the bond price is tC t; T . 30.3 Calibration of the Hull & Whitemodel Recall: drt= t,trt dt + t dB t; K t= Z t 0 udu; At; T = Z T t 2 4 e Kv v Z T v e ,Ky dy ! , 1 2 e 2K v 2 v Z T v e ,K y dy ! 2 3 5 dv ; C t; T =e Kt Z T t e ,Ky dy ; B t; T = exp f,rtC t; T , At; T g : Suppose we obtain B 0;T for all T 2 0;T from market data (with some interpolation). Can we determine the functions t , t ,and t for all t 2 0;T ? Not quite. Here is what we can do. We take the following input data for the calibration: 1. B 0;T; 0 T T ; 2. r0 ; 3. 0 ; 4. t; 0 t T (usually assumed to be constant); 5. 0C 0;T; 0 T T , i.e., the volatility at time zero of bonds of all maturities. Step 1. From4and5wesolvefor C 0;T= Z T 0 e ,Ky dy : 298 We can then compute @ @T C0;T= e ,KT = KT=,log @ @T C0;T; @ @T KT = @ @T Z T 0 u du = T : We now have T for all T 2 0;T . Step 2. From the formula B 0;T = expf,r0C 0;T, A0;Tg; we can solve for A0;T for all T 2 0;T . Recall that A0;T= Z T 0 2 4 e Kv v Z T v e ,Ky dy ! , 1 2 e 2K v 2 v Z T v e ,K y dy ! 2 3 5 dv : We can use this formula to determine T ; 0 T T as follows: @ @T A0;T= Z T 0 " e Kv ve ,KT ,e 2Kv 2 ve ,KT Z T v e ,Ky dy ! dv ; e K T @ @T A0;T= Z T 0 " e Kv v,e 2Kv 2 v Z T v e ,Ky dy ! dv ; @ @T e KT @ @T A0;T = e KT T, Z T 0 e 2Kv 2 v e ,KT dv ; e K T @ @T e KT @ @T A0;T = e 2KT T , Z T 0 e 2Kv 2 v dv ; @ @T e KT @ @T e KT @ @T A0;T = 0 Te 2KT +2TTe 2KT ,e 2KT 2 T; 0 T T : This gives us an ordinary differential equation for , i.e., 0 te 2K t +2tte 2Kt ,e 2Kt 2 t= known function of t: From assumption 4 and step 1, we know all the coefficients in this equation. From assumption 3, we have the initial condition 0 . We can solve the equation numerically to determine the function t; 0 t T . Remark 30.1 The derivation of the ordinary differential equation for t requires three differ- entiations. Differentiation is an unstable procedure, i.e., functions which are close can have very different derivatives. Consider, for example, f x=0 8x2 IR; g x= sin1000x 100 8x 2 IR: CHAPTER 30. HullandWhitemodel 299 Then jf x , g xj 1 100 8x 2 IR; but because g 0 x = 10 cos1000x; we have jf 0 x , g 0 xj =10 for many values of x . Assumption5 for the calibration was that we know the volatility at time zero of bonds of all maturi- ties. These volatilities can be implied by the prices of options on bonds. We consider now how the model prices options. 30.4 Option on a bond Consider a European call option on a zero-coupon bond with strike price K and expiration time T 1 . The bond matures at time T 2 T 1 . The price of the option at time 0 is IE e , R T 1 0 ru du BT 1 ;T 2 , K + = IEe , R T 1 0 ru du expf,rT 1 C T 1 ;T 2 , AT 1 ;T 2 g, K + : = Z 1 ,1 Z 1 ,1 e ,x expf,yCT 1 ;T 2 , AT 1 ;T 2 g,K + fx; y dx dy ; where f x; y is the joint density of R T 1 0 ru du; rT 1 . We observed at the beginning of this Chapter (equation (0.3)) that R T 1 0 ru du is normal with 1 4 = IE " Z T 1 0 ru du = Z T 1 0 IEru du = Z T 1 0 r0e ,K v + e ,K v Z v 0 e K u u du dv ; 2 1 4 =var " Z T 1 0 ru du = Z T 1 0 e 2K v 2 v Z T 1 v e ,K y dy ! 2 dv : We also observed (equation (0.1)) that rT 1 is normal with 2 4 = IErT 1 =r0e ,K T 1 + e ,K T 1 Z T 1 0 e K u u du; 2 2 4 =var rT 1 = e ,2K T 1 Z T 1 0 e 2K u 2 u du: 300 In fact, R T 1 0 ru du; rT 1 is jointly normal, and the covariance is 1 2 = IE " Z T 1 0 ru , IEru du: rT 1 , IErT 1 = Z T 1 0 IE ru , IEru rT 1 , IErT 1 du = Z T 1 0 r u; T 1 du; where r u; T 1 is defined in Equation 0.2. The option on the bond has price at time zero of Z 1 ,1 Z 1 ,1 e ,x expf,yCT 1 ;T 2 , AT 1 ;T 2 g,K + 1 2 1 2 p 1 , 2 exp , 1 21 , 2 " x 2 2 1 + 2xy 1 2 + y 2 2 2 dx dy : (4.1) The price of the option at time t 2 0;T 1 is IE e , R T 1 t ru du B T 1 ;T 2 , K + Ft = IE e , R T 1 t ru du expf,rT 1 C T 1 ;T 2 , AT 1 ;T 2 g,K + Ft (4.2) Because of the Markov property, this is random only through a dependence on rt . To compute this option price, we need the joint distribution of R T 1 t ru du; rT 1 conditioned on rt .This CHAPTER 30. HullandWhitemodel 301 pair of random variables has a jointly normal conditional distribution, and 1 t=IE " Z T 1 t rudu F t = Z T 1 t rte ,K v+K t + e ,K v Z v t e K u u du dv ; 2 1 t=IE 2 4 Z T 1 t rudu , 1 t ! 2 F t 3 5 = Z T 1 t e 2K v 2 v Z T 1 v e ,K y dy ! 2 dv ; 2 t=IE rT 1 rt =rte ,KT 1 +K t + e ,K T 1 Z T 1 t e K u u du; 2 2 t=IE rT 1 , 2 t 2 F t = e ,2K T 1 Z T 1 t e 2K u 2 u du; t 1 t 2 t=IE " Z T 1 t ru du , 1 t ! rT 1 , 2 t F t = Z T 1 t e ,K u,K T 1 Z u t e 2K v 2 v dv du: The variances and covariances are not random. The means are random through a dependence on rt . Advantages of the Hull & White model: 1. Leads to closed-form pricing formulas. 2. Allows calibration to fit initial yield curve exactly. Short-comings of the Hull & White model: 1. One-factor, so only allows parallel shifts of the yield curve, i.e., B t; T = exp f,rtC t; T , At; T g ; so bond prices of all maturities are perfectly correlated. 2. Interest rate is normally distributed, and hence can take negative values. Consequently, the bond price B t; T =IE " exp , Z T t ru du F t can exceed 1. 302 [...]... t F t t Z T1 Zu = e,K u,KT1 e2K v 2v dv du: t t The variances and covariances are not random The means are random through a dependence on rt Advantages of the Hull & White model: 1 Leads to closed-form pricing formulas 2 Allows calibration to fit initial yield curve exactly Short-comings of the Hull & White model: 1 One-factor, so only allows parallel shifts of the yield curve, i.e.,... AT1; T2g , K + F t (4.2) Because of the Markov property, this is random only through a dependence on rt To compute R this option price, we need the joint distribution of tT1 ru du; rT1 conditioned on rt This CHAPTER 30 Hull and White model 301 pair of random variables has a jointly normal conditional distribution, and 1 t = IE = "Z T 1 t ZT 1 t ru du F t Zv rte,Kv+K t... so only allows parallel shifts of the yield curve, i.e., B t; T = exp f,rtC t; T , At; T g ; so bond prices of all maturities are perfectly correlated 2 Interest rate is normally distributed, and hence can take negative values Consequently, the bond price " B t; T = IE exp , can exceed 1 ZT t ru du F t 302 . Chapter 30 Hull and White model Consider drt= t,trt dt + t dW t; where t , t and t are nonrandom functions of t. on rt .This CHAPTER 30. Hull and White model 301 pair of random variables has a jointly normal conditional distribution, and 1 t=IE " Z T 1