Hull and White model

We can solve the stochastic differential equation... From assumption 3, we have the initial condition 0.. We can solve the equation numerically to determine the function t ; 0tT.. Rem

Hull and White model

Consider

dr ( t ) = ( ( t ), ( t ) r ( t )) dt +  ( t ) dW ( t ) ;

where ( t ), ( t )and ( t )are nonrandom functions oft

We can solve the stochastic differential equation Set

K ( t ) =Z t

0 ( u ) du:

Then

d

e K(t) r ( t )

= e K(t)

( t ) r ( t ) dt + dr ( t )

= e K(t) ( ( t ) dt +  ( t ) dW ( t )) :

Integrating, we get

e K(t) r ( t ) = r (0)+Z t

0 e K(u) ( u ) du +Z t

0 e K(u)  ( u ) dW ( u ) ;

so

r ( t ) = e,K(t)

r (0) +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 thatr ( t )is a Gaussian process with mean function

m r ( t ) = e,K(t)

r (0) +Z t

0 e K(u) ( u ) du

(0.1) and covariance function

 r ( s;t ) = e,K(s),K(t)Z s^t

The processr ( t )is also Markov

293

We want to study 0 T r ( t ) dt To do this, we define

X ( t ) =Z t

0 e K(u)  ( u ) dW ( u ) ; Y ( T ) =Z T

0 e,K(t) X ( t ) dt:

Then

r ( t ) = e,K(t)

r (0) +Z t

0 e K(u) ( u ) du

+ e,K(t) X ( t ) ;

Z T

0 r ( t ) dt =Z T

0 e,K(t)

r (0) +Z t

0 e K(u) ( u ) du

dt + Y ( T ) :

According to Theorem 1.70 in Chapter 29,

R0 T r ( t ) dtis normal Its mean is

IEZ T

0 r ( t ) dt =Z T

0 e,K(t)

r (0) +Z t

0 e K(u) ( u ) du

and its variance is

var Z T

0 r ( t ) dt

!

= IEY 2 ( T )

=Z T

0 e 2K(v)  2 ( v ) Z T

v e,K(y) dy

!2 dv:

The price at time 0 of a zero-coupon bond paying $1 at timeT is

B (0 ;T ) = IE exp

( ,

Z T

0 r ( t ) dt

)

= exp

(

(,1) IEZ T

0 r ( t ) dt + 1 2 (,1) 2 var Z T

0 r ( t ) dt

!)

= exp

,r (0)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 (0) C (0 ;T ),A (0 ;T )g;

where

C (0 ;T ) =Z T

0 e,K(t) dt;

A (0 ;T ) =Z T

0

Z t

0 e,K(t)+K(u) ( u ) du dt,1 2Z T

0 e 2K(v)  2 ( v ) Z T

v e,K(y) dy

!2 dv:

t

u = t

T

Figure 30.1: Range of values ofu;tfor the integral.

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 K(v) ( v ) Z T

v e,K(y) dy

!

dv:

Therefore,

A (0 ;T ) =Z T

0

2

4e K(v) ( v ) Z T

v e,K(y) dy

! ,1 2 e 2K(v)  2 ( v ) Z T

v e,K(y) dy

!23

5 dv;

C (0 ;T ) =Z T

0 e,K(y) dy;

B (0 ;T ) = expf,r (0) C (0 ;T ),A (0 ;T )g:

Consider the price at timet2[0 ;T ]of the zero-coupon bond:

B ( t;T ) = IE

"

exp

( ,

Z T

t r ( u ) du

)

F( t )

#

:

Becauseris a Markov process, this should be random only through a dependence onr ( t ) In fact,

B ( t;T ) = exp r ( t ) C ( t;T ) A ( t;T ) ;

A ( t;T ) =Z T

t

2

4e K(v) ( v ) Z T

v e,K(y) dy

! ,1 2 e 2K(v)  2 ( v ) Z T

v e,K(y) dy

!23

5 dv;

C ( t;T ) = e K(t)Z T

t e,K(y) dy:

The reason for these changes is the following We are now taking the initial time to betrather than zero, so it is plausible that

R0 T ::: dvshould be replaced by

RT

t ::: dv:Recall that

K ( v ) =Z v

0 ( u ) du;

and this should be replaced by

K ( v ),K ( t ) =Z v

t ( u ) du:

Similarly,K ( y ) should be replaced byK ( y ),K ( t ) Making these replacements inA (0 ;T ), we see that theK ( t )terms cancel InC (0 ;T ), however, theK ( t )term does not cancel

LetC t ( t;T )andA t ( t;T )denote the partial derivatives with respect tot From the formula

B ( t;T ) = expf,r ( t ) C ( t;T ),A ( t;T )g;

we have

dB ( t;T ) = B ( t;T )h

,C ( t;T ) dr ( t ),1 2 C 2 ( t;T ) dr ( t ) dr ( t ),r ( t ) C t ( t;T ) dt,A t ( t;T ) dti

= B ( t;T )

,C ( t;T )( ( t ), ( t ) r ( t )) dt

,C ( t;T )  ( t ) dW ( t ),1 2 C 2 ( t;T )  2 ( t ) dt

,r ( t ) C 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 r ( u ) du

)

F( t )

#

to obtain the bond price, its differential must be of the form

dB ( t;T ) = r ( t ) B ( t;T ) dt + ( ::: ) dW ( t ) :

Therefore, we must have

,C ( t;T )( ( t ), ( t ) r ( t )),1 2 C 2 ( t;T )  2 ( t ),r ( t ) C t ( t;T ),A t ( t;T ) = r ( t ) :

We leave the verification of this equation to the homework After this verification, we have the formula

dB ( t;T ) = r ( t ) B ( t;T ) dt, ( t ) C ( t;T ) B ( t;T ) dW ( t ) :

In particular, the volatility of the bond price is ( t ) C ( t;T )

Recall:

dr ( t ) = ( ( t ), ( t ) r ( t )) dt +  ( t ) dB ( t ) ;

K ( t ) =Z t

0 ( u ) du;

A ( t;T ) =Z T

t

2

4e K(v) ( v ) Z T

v e,K(y) dy

! ,

1

2 e 2K(v)  2 ( v ) Z T

v e,K(y) dy

!23

5 dv;

C ( t;T ) = e K(t)Z T

t e,K(y) dy;

B ( t;T ) = expf,r ( t ) C ( t;T ),A ( t;T )g:

Suppose we obtainB (0 ;T )for allT 2[0 ;T]from market data (with some interpolation) Can we determine the functions ( t ), ( t ), and ( t )for allt2[0 ;T]? Not quite Here is what we can do

We take the following input data for the calibration:

1 B (0 ;T ) ; 0T T

;

2 r (0);

3 (0);

4  ( t ) ; 0tT

(usually assumed to be constant);

5  (0) C (0 ;T ) ; 0T T

, i.e., the volatility at time zero of bonds of all maturities

Step 1 From 4 and 5 we solve for

C (0 ;T ) =Z T

0 e,K(y) dy:

We can then compute

@

@T C (0 ;T ) = e,K(T)

=) K ( T ) =,log @

@T C (0 ;T ) ;

@

@T K ( T ) = @T @

Z T

0 ( u ) du = ( T ) :

We now have ( T )for allT 2[0 ;T]

Step 2 From the formula

B (0 ;T ) = expf,r (0) C (0 ;T ),A (0 ;T )g;

we can solve forA (0 ;T )for allT 2[0 ;T] Recall that

A (0 ;T ) =Z T

0

2

4e K(v) ( v ) Z T

v e,K(y) dy

! ,1 2 e 2K(v)  2 ( v ) Z T

v e,K(y) dy

!23

5 dv:

We can use this formula to determine ( T ) ; 0T T

as follows:

@

@T A (0 ;T ) =

Z T 0

"

e K(v) ( v ) e,K(T),e 2K(v)  2 ( v ) e,K(T) Z T

v e,K(y) dy

!#

dv;

e K(T) @

@T A (0 ;T ) =

Z T 0

"

e K(v) ( v ),e 2K(v)  2 ( v ) Z T

v e,K(y) dy

!#

dv;

@

@T



e K(T) @

@T A (0 ;T )



= e K(T) ( T ),

Z T

0 e 2K(v)  2 ( v ) e,K(T) dv;

e K(T) @

@T



e K(T) @

@T A (0 ;T )



= e 2K(T) ( T ),

Z T

0 e 2K(v)  2 ( v ) dv;

@

@T



e K(T) @

@T



e K(T) @

@T A (0 ;T )



= 0( T ) e 2K(T) + 2 ( T ) ( T ) e 2K(T),e 2K(T)  2 ( T ) ; 0T T:

This gives us an ordinary differential equation for, i.e.,

0( t ) e 2K(t) + 2 ( t ) ( t ) e 2K(t),e 2K(t)  2 ( t ) =known function oft:

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 ) ; 0tT

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 8x2IR;

g ( x ) = sin(1000 x )

100 8x2IR:

jf ( x ),g ( x )j 

1

100 8x2IR;

but because

g0( x ) = 10cos(1000 x ) ;

we have

jf0

( x ),g0

( x )j= 10

for many values ofx

Assumption 5 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

Consider a European call option on a zero-coupon bond with strike priceKand expiration timeT 1 The bond matures at timeT 2 > T 1 The price of the option at time 0 is

IE

e,

RT1

0 r(u) du ( B ( T 1 ;T 2 ),K ) +

= IEe,

RT1

0 r(u) du (expf,r ( T 1 ) C ( T 1 ;T 2 ),A ( T 1 ;T 2 )g ,K ) + :

=Z

1

,1

Z 1

,1

e,x

expf,yC ( T 1 ;T 2 ),A ( T 1 ;T 2 )g ,K+

f ( x;y ) dx dy;

wheref ( x;y )is the joint density of



RT1

0 r ( u ) du; r ( T 1 )

We observed at the beginning of this Chapter (equation (0.3)) that

RT1

0 r ( u ) duis normal with

 1 4

= IE

"

Z T1

0 r ( u ) du

#

=Z T1

0 IEr ( u ) du

=Z T1

0



r (0) e,K(v) + e,K(v)Z v

0 e K(u) ( u ) du

dv;

 21 4

= var

"

Z T1

0 r ( u ) du

#

=Z T1

0 e 2K(v)  2 ( v ) Z T1

v e,K(y) dy

!2 dv:

We also observed (equation (0.1)) thatr ( T 1 )is normal with

 2 4

= IEr ( T 1 ) = r (0) e,K(T1) + e,K(T1)Z T1

0 e K(u) ( u ) du;

 22 4

= var( r ( T 1 )) = e,2K(T1)Z T1

0 e 2K(u)  2 ( u ) du:

In fact, 0 T r ( u ) du; r ( T 1 ) is jointly normal, and the covariance is

 1  2 = IE

"

Z T1

0 ( r ( u ),IEr ( u )) du: ( r ( T 1 ),IEr ( T 1 ))

#

=Z T1

0 IE [( r ( u ),IEr ( u )) ( r ( T 1 ),IEr ( T 1 ))] du

=Z T1

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,yC ( T 1 ;T 2 ),A ( T 1 ;T 2 )g ,K+

1

2  1  2p

1, 2 exp

( ,

1 2(1, 2 )

"

x 2

 21 + 2  xy 1  2 + y 2

 22

#)

dx dy: (4.1)

The price of the option at timet2[0 ;T 1 ]is

IE

e,

RT1

t r(u) du ( B ( T 1 ;T 2 ),K ) +

F( t )

= IE

e,

RT1

t r(u) du (expf,r ( T 1 ) C ( T 1 ;T 2 ),A ( T 1 ;T 2 )g ,K ) +

F( t )

 (4.2)

Because of the Markov property, this is random only through a dependence onr ( t ) To compute this option price, we need the joint distribution of



RT1

t r ( u ) du; r ( T 1 )

conditioned onr ( t ) This

pair of random variables has a jointly normal conditional distribution, and

 1 ( t ) = IE

"

Z T1

t r ( u ) du

F( t )

#

=Z T1

t



r ( t ) e,K(v)+K(t) + e,K(v)Z v

t e K(u) ( u ) du

dv;

 21 ( t ) = IE

2 4

Z T1

t r ( u ) du, 1 ( t )

!2

F( t )

3 5

=Z T1

t e 2K(v)  2 ( v ) Z T1

v e,K(y) dy

!2 dv;

 2 ( t ) = IE

r ( T 1 )

r ( t )

= r ( t ) e,K(T1)+K(t) + e,K(T1)Z T1

t e K(u) ( u ) du;

 22 ( t ) = IE

( r ( T 1 ), 2 ( t )) 2

F( t )

= e,2K(T1)Z T1

t e 2K(u)  2 ( u ) du;

 ( t )  1 ( t )  2 ( t ) = IE

"

Z T1

t r ( u ) du, 1 ( t )

!

( r ( T 1 ), 2 ( t ))

F( t )

#

=Z T1

t e,K(u),K(T1)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

r ( t )

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 ) = expf,r ( t ) C ( t;T ),A ( t;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 r ( u ) du

)

F( t )

#

can exceed 1

The variances and covariances are not random The means are random through a dependence on

r ( t )

Advantages of the Hull & White model:

1 Leads to...

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... class="text_page_counter">Trang 9

pair of random variables has a jointly normal conditional distribution, and< /p>

 ( t ) = IE

"

Z