An outside barrier option

This implies the existence and uniqueness of the risk-neutral measure.. IPfis the unique risk-neutral measure... If we have an option whose payoff depends only on S, then Y is superfluou

An outside barrier option

Barrier process:

dY ( t )

Y ( t ) =  dt +  1 dB 1 ( t ) :

Stock process:

dS ( t )

S ( t ) =  dt +  2 dB 1 ( t ) +q

1, 2  2 dB 2 ( t ) ;

where 1 > 0 ;  2 > 0 ; ,1 <  < 1, andB 1andB 2 are independent Brownian motions on some

;F;P) The option pays off:

( S ( T ),K ) + 1fY(T)<Lg

at timeT, where

0 < S (0) < K; 0 < Y (0) < L;

Y( T ) = max 0

tT Y ( t ) :

Remark 24.1 The option payoff depends on both theY andS processes In order to hedge it, we will need the money market and two other assets, which we take to beY andS The risk-neutral measure must make the discounted value of every traded asset be a martingale, which in this case means the discountedY andSprocesses

We want to find 1and 2and define

d Be1 =  1 dt + dB 1 ; d Be2 =  2 dt + dB 2 ;

239

so that

dY

Y = r dt +  1 d Be1

= r dt +  1  1 dt +  1 dB 1 ; dS

S = r dt +  2 d Be1 +q

1, 2  2 d Be2

= r dt +  2  1 dt +q

1, 2  2  2 dt

+  2 dB 1 +q

1, 2  2 dB 2 :

We must have

 = r +  2  1 +q

We solve to get

 1 = ,r

 1 ;

 2 = ,r, 2  1

p

1, 2  2 :

We shall see that the formulas for  1 and 2 do not matter What matters is that (0.1) and (0.2) uniquely determine 1and 2 This implies the existence and uniqueness of the risk-neutral measure

We define

Z ( T ) = expn

, 1 B 1 ( T ), 2 B 2 ( T ),

1

2 (  21 +  22 ) To

;

f

IP ( A ) =Z

A Z ( T ) dIP; 8A2 F:

UnderfIP,Be1 andBe2 are independent Brownian motions (Girsanov’s Theorem) IPfis the unique risk-neutral measure

Remark 24.2 Under bothIP andfIP,Y has volatility 1,Shas volatility 2and

dY dS

Y S =  1  2 dt;

i.e., the correlation between dY Y anddS S is

The value of the option at time zero is

v (0 ;S (0) ;Y (0)) = IEf h

e,rT ( S ( T ),K ) + 1fY(T)<Lg

i

:

We need to work out a density which permits us to compute the right-hand side

Recall that the barrier process is

dY

Y = r dt +  1 d Be1 ;

so

Y ( t ) = Y (0)expn

rt +  1 Be1 ( t ),1 2  21 to

:

Set

b

 = r= 1, 1 = 2 ;

b

B ( t ) = tb + Be1 ( t ) ;

c

M ( T ) = max 0

tT Bb( t ) :

Then

Y ( t ) = Y (0)expf 1 Bb( t )g;

Y( T ) = Y (0)expf 1 Mc( T )g:

The joint density ofBb( T )andMc( T ), appearing in Chapter 20, is

f

IPf b

B ( T )2d ^ b; Mc( T )2d m ^g

= 2(2^ m,^ b )

Tp

2 T exp

( ,

(2^ m,^ b ) 2

2 T + b^ b,

1

2 b2 T

)

d ^ b d m; ^

^

m > 0 ; ^ b < m: ^

The stock process.

dS

S = r dt +  2 d Be1 +q

1, 2  2 d Be2 ;

so

S ( T ) = S (0)expfrT +  2 Be1 ( T ),1 2  2  22 T +q

1, 2  2 Be2 ( T ),1 2 (1, 2 )  22 Tg

= S (0)expfrT, 1 2  22 T +  2 Be1 ( T ) +q

1, 2  2 Be2 ( T )g From the above paragraph we have

e

B 1 ( T ) =,

b

T + Bb( T ) ;

so

S ( T ) = S (0)expfrT +  2 Bb( T ),1 2  22 T, 2 Tb +q

1, 2  2 Be2 ( T )g

24.1 Computing the option value

v (0 ;S (0) ;Y (0)) =fIEh

e,rT ( S ( T ),K ) + 1fY(T)<Lg

i

= e,rT IEf  

S (0)exp

( r,1 2  22, 2 b) T +  2 Bb( T ) +q

1, 2  2 Be2 ( T )

,K+ : 1fY (0)exp[ 1

b

M(T)]<Lg



We know the joint density of( Bb( T ) ; Mc( T )) The density ofBe2 ( T )is

f

IPf e

B 2 ( T )2d ~ bg= 1p

2 T exp

( ,

~

b 2

2 T

)

d ~ b; ~ b2IR:

Furthermore, the pair of random variables( Bb( T ) ; Mc( T ))is independent ofBe2 ( T )becauseBe1and

e

B 2are independent underIPf Therefore, the joint density of the random vector( Be2 ( T ) ; Bb( T ) ; Mc( T ))

is

f

IPf

e

B 2 ( T )2d ~ b; Bb( T )2d ^ b; Mc( T )2d m; ^ g=fIPf

e

B 2 ( T )2d ~ bg: IPf f

b

B ( T )2d ^ b; Mc( T )2d m ^g The option value at time zero is

v (0 ;S (0) ;Y (0))

= e,rT

1

1

logYL(0)

Z

0

^ m

Z

,1

1 Z

,1



S (0)exp



( r,1 2  22, 2 b) T +  2 ^ b +q

1, 2  2 ~ b

,K+ :p1

2 T exp

( ,

~ b 2

2 T

)

: 2(2^ m,^ b )

Tp

2 T exp

( ,

(2 ^ m,^ b ) 2

2 T + b^ b,1 2 b2 T

)

:d ~ b d ^ b d m: ^

The answer depends onT;S (0)andY (0) It also depends on 1 ; 2 ;;r;K andL It does not

depend on;; 1 ;nor 2 The parameterbappearing in the answer isb=  r1

,1

2 :

Remark 24.3 If we had not regardedY as a traded asset, then we would not have tried to set its

mean return equal tor We would have had only one equation (see Eqs (0.1),(0.2))

 = r +  2  1 +q

to determine 1 and 2 The nonuniqueness of the solution alerts us that some options cannot be

hedged Indeed, any option whose payoff depends onY cannot be hedged when we are allowed to

trade only in the stock

If we have an option whose payoff depends only on S, then Y is superfluous Returning to the original equation forS,

dS

S =  dt +  2 dB 1 +

q

1, 2  2 dB 2 ;

we should set

dW =  dB 1 +q

1, 2 dB 2 ;

soW is a Brownian motion underIP (Levy’s theorem), and

dS

S =  dt +  2 dW:

Now we have only Brownian motion, there will be only one, namely,

 = ,r

 2 ;

so withd Wf=  dt + dW;we have

dS

S = r dt +  2 d W;f and we are on our way

24.2 The PDE for the outside barrier option

Returning to the case of the option with payoff

( S ( T ),K ) + 1fY(T)<Lg;

we obtain a formula for

v ( t;x;y ) = e,r(T,t) IEft;x;yh

( S ( T ),K ) + 1fmaxtuTY (u) < Lg;i

by replacingT,S (0)andY (0)byT ,t,xandyrespectively in the formula forv (0 ;S (0) ;Y (0)) Now start at time 0 atS (0)andY (0) Using the Markov property, we can show that the stochastic process

e,rt v ( t;S ( t ) ;Y ( t ))

is a martingale underfIP We compute

dh

e,rt v ( t;S ( t ) ;Y ( t ))i

= e,rt

 ,rv + v t + rSv x + rY v y + 1 2  22 S 2 v xx +  1  2 SY v xy + 1 2  21 Y 2 v yy



dt

+  2 Sv x d Be1 +q

1  2  2 Sv x d Be2 +  1 Y v y d Be1



v(t, 0, 0) = 0

x

y

v(t, x, L) = 0, x >= 0

Figure 24.1: Boundary conditions for barrier option Note thatt2[0 ;T ]is fixed.

Setting thedtterm equal to 0, we obtain the PDE

,rv + v t + rxv x + ryv y + 1

2  22 x 2 v xx

+  1  2 xyv xy + 1 2  21 y 2 v yy = 0 ;

0t < T; x0 ; 0yL:

The terminal condition is

v ( T;x;y ) = ( x,K ) + ; x0 ; 0y < L;

and the boundary conditions are

v ( t; 0 ; 0) = 0 ; 0tT;

v ( t;x;L ) = 0 ; 0 t T; x 0 :

x = 0 y = 0

,rv + v t + ryv y + 1 2  21 y 2 v yy = 0 ,rv + v t + rxv x + 1 2  22 x 2 v xx = 0

This is the usual Black-Scholes formula

iny

This is the usual Black-Scholes formula

inx The boundary conditions are The boundary condition is

v ( t; 0 ;L ) = 0 ; v ( t; 0 ; 0) = 0; v ( t; 0 ; 0) = e,r(T,t) (0,K ) + = 0;

the terminal condition is the terminal condition is

v ( T; 0 ;y ) = (0,K ) + = 0 ; y0 : v ( T;x; 0) = ( x,K ) + ; x0 :

On thex = 0boundary, the option value

isv ( t; 0 ;y ) = 0 ; 0yL: On therelevant, and the option value is given byy = 0boundary, the barrier is

ir-the usual Black-Scholes formula for a Eu-ropean call

24.3 The hedge

After setting thedtterm to 0, we have the equation

dh

e,rt v ( t;S ( t ) ;Y ( t ))i

= e,rt

 2 Sv x d Be1 +q

1, 2  2 Sv x d Be2 +  1 Y v y d Be1



;

wherev x = v x ( t;S ( t ) ;Y ( t )),v y = v y ( t;S ( t ) ;Y ( t )), and Be1 ; Be2 ;S;Y are functions of t Note that

dh

e,rt S ( t )i

= e,rt [,rS ( t ) dt + dS ( t )]

= e,rt

 2 S ( t ) d Be1 ( t ) +q

1, 2  2 S ( t ) d Be2 ( t )



:

dh

e,rt Y ( t )i

= e,rt [,rY ( t ) dt + dY ( t )]

= e,rt  1 Y ( t ) d Be1 ( t ) :

Therefore,

dh

e,rt v ( t;S ( t ) ;Y ( t ))i

= v x d [ e,rt S ] + v y d [ e,rt Y ] :

Let
2 ( t )denote the number of shares of stock held at timet, and let
1 ( t )denote the number of

“shares” of the barrier processY The valueX ( t )of the portfolio has the differential

dX =
2 dS +
1 dY + r [ X
2 S
1 Y ] dt:

This is equivalent to

d [ e,rt X ( t )] =
2 ( t ) d [ e,rt S ( t )] +
1 ( t ) d [ e,rt Y ( t )] :

To getX ( t ) = v ( t;S ( t ) ;Y ( t ))for allt, we must have

X (0) = v (0 ;S (0) ;Y (0))

and

2 ( t ) = v x ( t;S ( t ) ;Y ( t )) ;

1 ( t ) = v y ( t;S ( t ) ;Y ( t )) :