Solved-Computational Finance Lab II -Solution

Description

Consider the following Black-Scholes PDE for European call:

@V 1 2S2 @2V @V rV = 0; (0; 1) (0; T ]; T > 0
+ + (r )S
@t 2 @S2 @S
V (S; t) = 0; for S = 0;

V (S; t) = S Ke r(T t) ; for S
! 1

with suitable initial condition V (S; 0):

With the following transformation

S = Kex; t = T 2 ;

2
x 2

V (s; t) = V Ke ; T
2

(

v(x; ) =: K exp 1 (q
2
{

q := 2r ; q := 2(r ) ;

2 2
) =: v(x; ); and ] }
[
1)x 1 (q 1)2 + qy(x; )

4

the above Black-Scholes PDE becomes the following 1-D heat conduction parabolic PDE:

@y @2y
x 0; x
@ = @x2 ; x
2 R;
2

{ }

y(x; 0) = max exp( (q + 1)) exp( (q 1)); 0 ; x R ;
2 2

y(x; ) = 0; for1x !
; 1 2 ! 1

( )

y(x; ) = exp (q + 1)x + (q + 1) for x :
2 4

Solve the transformed PDE by the following schemes:
(i) Forward-Euler for time & central difference for space (FTCS) scheme.

(ii) Backward-Euler for time & central difference for space (BTCS) scheme.

(iii) Crank-Nicolson finite difference scheme

The values of the parameters are T = 1; K = 10; r = 0:06; = 0:3 and = 0.

Continued on the next page

1

2. Consider the following Black-Scholes PDE for European put:
@V 1 @2V @V
+ 2S2 + (r )S rV = 0; (0; ) (0; T ]; T > 0
@S2 @S
@t 2 t) 1
r(T

S; for S = 0;
V (S; t) = Ke
V (S; t) = 0; for S ! 1

with suitable initial condition V (S; 0):

Using the transformation given above the Black-Scholes PDE becomes the following problem:

@y @2y
@ = @x2 ; x 2 R;0;

x x

y(x; 0) = max exp( (q 1)) exp( (q + 1)); 0 ; x R ;
2 2

2

1 1 2
}
{ for x !
y(x; ) = exp 2 (q 1)x + 4 (q 1) ; ;
( )

y(x; ) = 0 for x ! 1;

Solve the transformed PDE by the following schemes:

(i) Forward-Euler for time & central difference for space (FTCS) scheme.

(ii) Backward-Euler for time & central difference for space (BTCS) scheme.

(iii) Crank-Nicolson finite difference scheme

The values of the parameters are T = 1; K = 10; r = 0:06; = 0:3 and = 0.

2