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Math6911 S08, HM Zhu

Chapter 5 Finite Difference

Methods

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Math6911, S08, HM ZHU

References

1. Chapters 5 and 9, Brandimarte

2. Section 17.8, Hull

3. Chapter 7, Numerical analysis, Burden and Faires

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Outline

Finite difference (FD) approximation to the derivatives Explicit FD method Numerical issues

Implicit FD method Crank-Nicolson method Dealing with American options Further comments

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5.1 Finite difference approximations


Methods

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Finite-difference mesh

Aim to approximate the values of the continuous functionf(t, S) on a set of discrete points in (t, S) plane

Divide the S-axis into equally spaced nodes at distance

S apart, and, the t-axis into equally spaced nodes adistance t apart (t, S) plane becomes a mesh with mesh points on

(i t, jS)We are interested in the values off(t, S) at mesh points

(i t, jS), denoted as

( )i , jf f i t , j S=

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The mesh for finite-difference

approximation

t

S

i t

j S

( )i , jf f i t , j S= Smax=MS

T=N t

?

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Black-Scholes Equation for a

European option with value V(S,t)

conditionsboundaryandfinalproperwith

Tt0andS0where

)1.5(021

2

222

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Finite difference approximations

The basic idea of FDM is to replace the partial derivativesby approximations obtained by Taylor expansions near thepoint of interests

( ) ( ) ( ) ( ) ( )

( )

( ) ( )( )

( )( )

0

2

For example,

for small using Taylor expansion at point

t

f S,t f S ,t t f S ,t f S ,t t f S ,tlim

t t t

t , S ,t

f S ,tf S ,t t f S ,t t O t

t

+ + =

+ = + +

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Forward-, Backward-, and Central-

difference approximation to 1

st

orderderivatives

( ) ( ) ( )

( )( ) ( ) ( )

( )

( ) ( ) ( ) ( )( )2

Forward:

Backward:

Central:2

f t ,S f t t ,S f t ,SO t

t t

f t ,S f t ,S f t t ,SO t

t t

f t ,S f t t ,S f t t ,S O tt t

+ +

+

+ +

t t+ tt t

centralbackward forward

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Symmetric Central-difference

approximations to 2nd

order derivatives

( ) ( ) ) ( )

( )( )( )

22

22

2f t ,S f t ,S S f t ,S f t ,S SO S

S S

+ + +

( ) ( )

( )

( )

( )

Use Taylor's expansions for and

around point

f t ,S S f t ,S S

t , S :

f t ,S S ?

f t ,S S ?

+

+ =+

=

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1 1

1 1

1 1 1 1

2 2

Forward Difference:

Backward Difference:

Central Difference:

As to the second

i , j i , j i , j i , j

i , j i , j i , j i , j

i , j i , j i , j i , j

f f f ff f,t t S S

f f f ff f,

t t S S

f f f ff f,

t t S S

+ +

+ +

( )

21 1

2

1 1

2

2

derivative, we have:

=

i , j i , j i , j i , j

i , j i , j i , j

f f f ffS

S SS

f f f

S

+

+

+

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Depending on which combination of schemes we use in

discretizing the equation, we will have explicit, implicit, orCrank-Nicolson methods

We also need to discretize the boundary and final

conditions accordingly. For example, for European Call,


( )

( )

0

Final Condition:

0 for 01

Boundary Conditions:

0for 01

where

N , j

i ,

r N i t

i ,M max

max

f max j S K , , j , ,...,M

f, i , ,...,N

f S Ke

S M S.

= =

==

= =

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5.2.1 Explicit Finite-Difference Method


Methods

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Explicit Finite Difference Methods

22 2

2

1

1 1

21 1

2 2

12

2

2

In , at point ( ), set

backward difference:

central difference: ,

and

i , j i , j

i , j i , j

i , j i , j i , j

i , j

f f frS S rf i t , j S t S S

f ff

t t

f ff

S S

f f ff

, r f rf , S j SS S

+

+

+ + =

+

= =

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( )

( )( )

1 1 1

2 2

2 2

2 2

1

2

11

2

Rewriting the equation, we get an explicit scheme:

(5.2)

where

* * *

i , j j i , j j i , j j i , j

*

j

*

j

*

j

f a f b f c f

a t j rj

b t j r

c t j rj

+= + +

=

= +

= +

1 2 1 0 1 2 1for andi N - , N - , ..., , j , , ..., M - .= =

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Numerical Computation Dependency

x

x

x

x

t

S

it

jS

Smax=MS

T=N t0

0

(i-1)t

(j+1)S

(j-1)S

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Implementation

1. Starting with the final values , we apply (5.2) to solveWe use the boundary condition

to determine

2. Repeat the process to determine and so on

N , jf1 for 1 1N , jf j M .

10 1andN , N - ,Mf f .

2N , jf

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Example

We compare explicit finite difference solution for aEuropean put with the exact Black-Scholes formula, where

T = 5/12 yr, S0=$50, K = $50, =30%, r = 10%.

Black-Scholes Price: $2.8446

EFD Method with Smax=$100, S=2, t=5/1200: $2.8288


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Example (Stability)

We compare explicit finite difference solution for a Europeanput with the exact Black-Scholes formula, where T = 5/12 yr,

S0=$50, K = $50, =30%, r = 10%.



EFD Method with Smax=$100, S=1.5, t=5/1200: $3.1414EFD Method with Smax=$100, S=1, t=5/1200: -$2.8271E22

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5.2.2 Numerical Stabili ty


Methods

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Numerical Accuracy

The problem itselfThe discretization scheme used

The numerical algorithm used

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Conditioning Issue

( )

( ) ( )

Suppose we have mathematically posed problem:

where is to evaluated given an input .

Let for small change

If hen is near then we call the problem is

*

*

y f x

y x

x x x x.

f x f x ,

well - co

=

= +

. Otherwise, it is ill-posed/ill-conditioned.nditioned

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Conditioning Issue

Conditioning issue is related to the problem itself, not to thespecific numerical algorithm; Stability issue is related to thenumerical algorithm

One can not expect a good numerical algorithm to solve an ill-conditioned problem any more accurately than the datawarrant

But a bad numerical algorithm can produce poor solutions

even to well-conditioned problems

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Conditional Issue

( ) ( )( )

( )( )

The concept "near" can be measured by further information about

the particular problem:

0

where C is called of this problem. If C is large,the problem is ill-conditioned.

*f x f x xC f x

xf x

condition number

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Floating Point Number & Error

( )

( ) xxfled

xx

xx.xxflx

xx.xx

x

i

ed

e

d

=

=

:or

integerboundedan:systempointfloatingtheofprecisioninteger,an:

9,00,where

10:numberpointfloatingTruncated

10:expansiondecimalInfinite

number.realanybeLet

1

21

21

errorroundoffpointFloating

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Error Propagation

( )

( )( )

00040011.0:npropagatioError

0003.0:errorpointFloating

3.where10667.0and6667.0Let:

errors.pointfloatingtheseofonaccumulati

anistheredone,arenscalculatioadditionalWhen

22

0

=

=

===

xxfl

xxfl

dxflxExample

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Numerical Stability or Instability

( )

Stability ensures if the error between the numerical

soltuion and the exact solution remains bounded

as numerical computation progresses.

That is, (the solution of a slightly perturbed problem)

is n

*

f x

( )( )

( )

ear the computed solution

Stability concerns about the behavior of

as numerical computation progresses for fixed discretization

steps and S.

*

i , j

f x .

f f i t , j S

t

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Convergence issue

( )

( )

Convergence of the numerical algorithm concerns about

the behavior of as S 0 for fixed

values .

For well-posed linear initial value problem,

Stability Convergence

(Lax's eq

i , jf f i t , j S t ,

i t , j S

uivalence theorem, Richtmyer and Morton, "Difference

Methods for Initial Value Problems" (2nd) 1967)

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Numerical Accuracy

These factors contribute the accuracy of a numerical solution.

We can find a goodestimate if our problem is well-conditioned and the algorithm is stable

( ) ( )( ) ( )

( ) ( )Stable:Well-conditioned:

* *

*

*

f x f xf x f x

f x f x

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5.2.3 Financial Interpretation of Numerical Instability


Methods

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Financial Interpretation of instability

(Hall, page 423-4)

2

1 1 1 1 1

1

1

1

2If and are assumed to be the same at ( )

as they are at ( ), we obtain equations of the form:

(5.3)

where

=

i , j j i , j j i , j j i , j

j

f S f S i , j

i, j

f a f b f c f

a

+ + + +

+

= + +

( )

2 2

2 20

2 2

1

1

1 111 1

1 1

1 11 2 1 0 1 2 1

1 1

2 2

1 1

2 2for and

d

j

j u

t j t r jr t r t

b j tr t r t

c t j t r j

r t r t i N - , N - , ..., , j , , ..., M - .

= + +

= =+ +

= + = + + = =

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i , j i +1, j

i +1, j 1

i +1, j +1

These coefficients can be interpreted as probabilities timesa discount factor. If one of these probability < 0, instabilityoccurs.

d

0

u

E li i Fi i Diff M h d

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Explicit Finite Difference Method asTrinomial Tree

[ ]

( ) ( )

0

2 22

0

Check if the mean and variance of the

Expected value of the increase in asset price during t:

E 0

Variance of the increment:

E 0

d u

d

S S rj S t rS t

S S

= + + = =

= + + ( )

[ ] [ ] ( )

2 2 2 2 2

22 2 2 2 2 2 2 2Var E E

which is coherent with geometric Brownian motion in a risk-neutral world

u j S t S t

S t r S t S t

= =

= =

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Change of Variable

2 2 2

2

2 21 1 1 1 1 1 1 1 1 1 1

2

2 2

2

2 2 2

Define The B-S equation becomes

The corresponding difference equation is

or

i , j i , j i , j i , j i , j i , j i , j

i , j

Z ln S.

f f fr rf

t Z Z

f f f f f f fr rf

t Z Z

f

+ + + + + + + + +

=

+ + =

+ + + =

1 1 1 1 1 54( )* * *

i , j j i , j j i , j j i , jf f f . + + + += + +

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Change of Variable

2 2

2

2

2

2 2

2

1

1 2 2 2

11

1

11 2 2 2

3

where

It can be shown that it is numerically most efficient if

*

j

*

j

*j

t tr

r t Z Z

t

r t Z

t trr t Z Z

Z t .

= + +

= +

= + +

=

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Reduced to Heat Equation

Get rid of the varying coefficients S and S by usingchange of variables:

Equation (5.1) becomes heat equation (5.5):

( )( ) ( )

( )

=

===+

2

14

11

2

12

21

,,,2,2

rk

xueEtSVTtEeSkxk

x

2

2(5.5)

for and 0

u u

x

x

=

< < + >

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( )

( )( )

( )( )

( ) ( )( )

( )

( )

===

++=

++

=+

=

+

++

++

T

,...M,mNn-N

uuuu

xOO

xOx

uuuO

uu

x,mnuu

m

n

m

n

m

n

m

n

m

n

m

n

m

n

m

n

m

n

m

n

2

2

111

2

2

211

1

2

1

10andfor,x

where

21

bythis

eapproximatcanwe,andoftermsIgnoring

2

:formtheofequationsdifferencefinite

ofsystemasolvinginvolvesthis,With

Explicit Finite Difference Method

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Stability and Convergence

(P. Wilmott, et al, Option Pricing)

( )

2

1,

2

:

The solution of Eqn (5.5) is

1 1i) Stable if 0 ; ii) Unstable if

2 2x

:

If 0 then the explicit finite-difference approximation

converges to the exact solution as ,

< = >

<

Stability

Convergence

( )

( )

,mn

x 0

(in the sense that as , x 0)

is O

u u n x m

Rate of Convergence

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5.3.1 Implicit Finite-Difference Method


Methods

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Implicit Finite Difference Methods

2

2 22

1

1 1

21 1

2 2

12

2

2

,j ,

, ,

, , ,

In , we use

difference:

central difference: ,

and

i i j

i j i j

i j i j i j

i , j

f f frS S r f t S S

f ff

t t

f ff

S S

f f ff, rf rf

S S

+

+

+

+ + =

+ =

forward

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( )

( )( )

1 1 1

2 2

2 2

2 2

1

, , ,

Rewriting the equation, we get an implicit scheme:

= (5.6)

where

1=2

1

2

for

j i j j i j j i j i , j

j

j

j

a f b f c f f

a t j rj

b t j r

c t j rj

+ ++ +

+

= + += +

1 2 1 0 1 2 1andi N - , N - , ..., , j , , ..., M - .= =

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x

x

x

x

t

S

it

jS

Smax=MS

T=N t0

0

(i-1)t

(j+1)S

(j-1)S

(i+1)t

Implementation

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Implementation

( )1

1 2 3 1 1 0 1

Equation (5.6) can be rewritten in matrix form:

57

where and are ( 1) dimensional vectors

0 0 0

and is ( 1) ( 1) symmetric

i i i

i i

T T

i i , i , i , i ,M i i , M i ,M

.

M

f , f , f , f , a f , , , , , c f

M M

+

= +

= =

Cf f b

f b

f b

C

1 1

2 2 2

3 3

2

1 1

matrices0 0

0

0

0 0

M

M M

b c

a b c

a b

c

a b

=

C

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Implementation

1. Starting with the final values , we need to solve alinear system (5.7) to obtain

using LU factorization or iterative methods. We use theboundary condition to determine

2. Repeat the process to determine and so on

N , jf1 for 1 1N , jf j M

10 1and

N , N - ,Mf f .

2N , jf

E l

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Example

We compare implicit finite difference solution for aEuropean put with the exact Black-Scholes formula, whereT = 5/12 yr, S0=$50, K = $50, =30%, r = 10%.


IFD Method with Smax=$100, S=2, t=5/1200: $2.8194


E l (St bilit )

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Example (Stability)

We compare implicit finite difference solution for a Europeanput with the exact Black-Scholes formula, where T = 5/12 yr,

S0=$50, K = $50, =30%, r = 10%.



IFD Method with Smax=$100, S=1.5, t=5/1200: $3.1325IFD Method with Smax=$100, S=1, t=5/1200: $2.8348

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Implicit vs Explicit Finite Difference Methods

i +1, ji , j

i , j 1

i , j +1

Implicit Method

(always stable)

i , j i +1, j

i +1, j 1

i +1, j +1

Explicit Method

Implicit vs Explicit Finite Difference

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p pMethod

The explicit finite difference method isequivalent to the trinomial tree approach:

Truncation error: O(t) Stability: not always

The implicit finite difference method is

equivalent to a multinomial tree approach: Truncation error: O(t) Stability: always

Other Points on Finite Difference

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Methods

It is better to have ln Srather than Sasthe underlying variable in general

Improvements over the basic implicit and

explicit methods: Crank-Nicolson method, average of

explicit and implicit FD methods, trying

to achieve Truncation error: O((t)2 ) Stability: always

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5.3.2 Solving a linear system using direct methods


Methods

Solve Ax=b

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Solve Ax=b

A x=b

Various shapes of matrix A

Lower triangular Upper triangular General

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5.3.2.A Triangular Solvers

Example: 3 x 3 upper triangular system

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Example: 3 x 3 upper triangular system

=

20

10

100

400

110

164

3

2

1

x

x

x

465

65*61004

510

54/20

1

231

32

3

=

====

==

x

xxx

xx

x

Solve an upper triangular system Ax=b

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Solve an upper triangular system Ax=b

( )

1,1,

,,1,002

1

=

=

=

==+=

>

>

niaxabx

abx

nibxaxa

x

x

x

aa

ii

ij

jijii

nnnn

i

ij

jijiii

n

inii

Implementation

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Implementation

Function x = UpperTriSolve(A, b)n = length(b);

x(n) = b(n)/A(n,n);

for i = n-1:-1:1

sum = b(i);

for j = i+1:n

sum = sum - A(i,j)*x(j);

endx(i) = sum/A(i,i);

end

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5.3.2.B Gauss Elimination

To solve Ax=b for general formA

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To solve Ax b for general form A

To solve Ax = b :

Suppose A = LU. Then

Ax = LUx = b

z

= *

Solve two triangular systems:

1) solve z from Lz = b

2) solve x from Ux = z

Gauss Elimination = *

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Gauss Elimination =

Goal: Make A an upper triangular matrix through using

fundamental row operations

For example, to zero the elements in the lower triangular part of A

1) Zero a21

=

333231

11

132123

11

122122

131211

333231

232221

131211

11

21 0

100

01

001

aaa

aaaa

aaaa

aaa

aaa

aaa

aaa

aa

E 21A


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Gauss Elimination

2) Zero a31

11 12 13 11 12 13

21 13 21 1321 12 21 1222 23 22 23

11 11 11 11

31

31 32 33 31 12 31 1311 32 33

11 11

1 0 0

0 1 0 0 0

0 10

a a a a a a

a a a aa a a aa a a aa a a a

aa a a a a a a

a a aa a

=

11 12 13

22 23

32 33

0

0

=

a a a

a a

a a

E 31 E 21A


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Gauss Elimination

3) Zero

11 12 13 11 12 13

22 23 22 23

32 32 33 32 2333

22 22

1 0 0

0 1 0 0 00

0 1 0 0

a a a a a a

a a a aa a a a a

aa a

=

U

32

a

E 32 E 31 E 21A = U

lower triangular


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Claim 1:21

32 31 21

11

31 32

11 22

1 0 0

1 0

1

a

a

a aa a

=

E E E

Claim 2:

( )1 21

32 31 21

11

31 32

11 22

1 0 0

1 0

1

a

a

a aa a

=

E E E

LU Factorization = *

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Therefore, through Gauss Elimination, we have

E 32 E 31 E 21A = U

A = E 32 E 31 E 21( )1

U

A =LU

It is called LU factorization. When A is symmetric,

which is called Cholesky DecompositionT=A LL

To solve Ax=b for general form A

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g

To solve Ax = b becomes1) Use Gauss elimination to make A upper triangular, i.e.

L1Ax = L1b Ux = z

2) Solve x from Ux = z

This suggests that when doing Gauss elimination, we can do it

to the augmented matrix A b[ ] associated with the linear system.

An exercise

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An exercise

=

4

0

0

21

121

12

3

2

1

x

x

x

% build the matrix A

A = [2, -1, 0; -1, 2, -1; 0, -1, 2]

% build the vector b

x_true = [1:3]';

b = A * x_true;

% lu decomposition of A

[l, u] = lu(A)

% solve z from lz = b where z = uxz = l\b;

% solve x from ux = z

x = u\z

General Gauss Elimination

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65


row i

row j -aji

aiirow i

What if we have zero or very smalldiagonal elements?

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66


diagonal elements?

LUPA

LUA

PA

A

=

=

=

=

=

=

=

.9999011

10.000101

1110.0001

0110

9999-0

10.0001

110,000

01

11

10.0001=

10

32

32

10

01

10

3210=

exampleFor

pivoting.partialcalledisThisLU.=PAi.e.,stably,computedorcomputedbecan

neliminatioGaussthatsoAofrowspermutetoneedweSomtimes,

Can we always have A = LU?

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( )( )1 1 0 1 1nxn

No!

If : , : for =,..., - ,

then A has an LU factorization.Proof:

See Golub and Loan, Matrix Computation, 3rd edition

det k k k n

A

R

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Methods

5.4.1 Crank-Nicolson Method


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( )

( )

( )( )

22 2

2

1

21 1 1 1 1 1 1 1 12 2

2

2

1

1

2

21

2 2

With ,

we can obtain an explicite form:

i , j i , j

i , j i , j i , j i , j i , j

i , j

f f frS S rf t S S

f f O tt

f f f f f

rj S j S S S

rf O S

+

+ + + + + + +

+

+ + =

+ =

+

+ +


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( )

( )

( )

( )

2

2 22

1

21 1 1 12 22

2

12

21

2 2

With ,

we can obtain an implicit form :

i , j i , j

i , j i , j i , j i , j i , j

i , j

f f frS S r f t S S

f fO t

t

f f f f frj S j S

S S

rf O S

+

+ +

+ + =

+ =

+

+ +

Crank-Nicolson Methods: average of explicitand implicit methods

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(Brandmarte, p485)

( )( )

( )

( ) ( )( )

21

1 1 1 1 1 1

2 1 1 1 1 1 1 12 2

2 2

2

1

2 2 2

2 21

4

2

i , j i , j

i , j i , j i , j i , j

i , j i , j i , j i , j i , j i , j

i , j i , j

f f O tt

f f f frj S

S S

f f f f f fj S

S S

rf f O S

+

+ + + +

+ + + + +

+

+ =

+ + +

+

+ + +

Crank-Nicolson Methods

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Crank Nicolson Methods

( )( )

( )

1 1

1 1 1 1 1

2 2

1

1

R ewriting the equation, we get an C rank-N icolson scheme:

(5.5)

where

=4

j i , j j i , j j i , j

j i , j j i , j j i , j

j

j

f f f

f f f

tj rj

+

+ + + +

+ =

+ + +

= ( )

( )

2 2

2 2

2

1 2 1 0 1 2 1

4

for and

j

tj r

tj rj

i N - , N - , ..., , j , , ..., M - .

+

= +

= =


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x

x

x

x

t

S

it

jS

Smax=MS

T=N t00

(i-1)t

(j+1)S

(j-1)S

(i+1)t

x

x

x

x

Implementation

Equation (5 5) can be rewritten in matrix form:

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( ) ( )

1 2 1

1 2 3 1

1 0 10 1 1

1 2

1

0

Equation (5.5) can be rewritten in matrix form:

where and are ( ) dimensional vectors

and and are (

i i

i i

Ti i , i , i , i ,M

T

i , i , M i ,M i ,M

M

f , f , f , f ,

f f , , f f

M

+

+ +

= +

=

= + +

M f M f b

f b

f

b

M M

1 1

2 2 2

1 3

2

1 1

1 1

1 0 0

1 00

0 0 1

) ( ) symmetric matrices

M

M M

M

=

M

Implementation

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75


p

1 1

2 2 2

2 3

2

1 1

and1 0 0

1 0

0

0 0 1

M

M M

+ +

=

+

M

Example

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76


We compare Crank-Nicolson Methods for a European putwith the exact Black-Scholes formula, where T = 5/12 yr,S0=$50, K = $50, =30%, r = 10%.


CN Method with Smax=$100, S=2, t=5/1200: $2.8241


Example (Stability)

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77


We compare Crank-Nicolson Method for a European put with

the exact Black-Scholes formula, where T = 5/12 yr, S0=$50,K = $50, =30%, r = 10%.


CN Method with Smax=$100, S=1.5, t=5/1200: $3.1370


Example: Barrier Options

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Barrier options are options where the payoff depends on whether

the underlying assets price reaches a certain level during a certainperiod of time.

Types:

knock-out: option ceases to exist when the asset price reaches a barrier

Knock-in: option comes into existence when the asset price reaches a barrier

Example: Barrier Option

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79


We compare Crank-Nicolson Method for a European down-

and-out put, where T = 5/12 yr, S0=$50, K = $50, Sb=$50=40%, r = 10%.

What are the boundary conditions for S?

Example: Barrier Option

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80


We compare Crank-Nicolson Method for a European down-

and-out put, where T = 5/12 yr, S0=$50, K = $50, Sb=$50=40%, r = 10%.

Boundary conditions are

Exact Price (Hall, p533-535): $0.5424

CN Method with Smax=$100, S=0.5, t=1/1200: $0.5414

( ) ( )0 0andmax bf t ,S f t ,S= =

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Appendix A.

Matrix Norms

Vector Norms

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82


( )

??,?,3].1-4[2example,For

max

normEuclideantheis2p

normadefinetowaysvariousareThere-

anyfor

anyfor

0iff0;anyfor0s.t.numberrealatomappingfunctionaisnormA vector-

matrixaorvectoraoflengththemeasureway toaasserveNorms-

21

1

1

1

n

====

=

++

=

==>

=

vvvv

x

x

yxyxyx

xx

xx0xxxx

ini

pn

i

p

ip

n

x

x

,

ccc

Matrix Norms

nm

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( )( ) { }BB

AAA

AA

Ax

AxA

BABABA

AA

0A0A0AA

AA

x

ofeigenvalueanis:max

wherenorm,spectralthe,

maxmax

sup

normsmatrixusedcommonlyVarious-

anyfor

anyfor

iff;anyfor0

s.t.numberrealatomappingfunctionaisnormmatrixaSimilarly,-

2

11111

21

1 1

2

0

nm

kk

T

n

j ijmi

m

i ijnj

m

i

n

j

ijF

p

p

p

nm

aa

a

,

ccc

++

=

==>

==

= =

An Example

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84


????

132

513142

1

2

====

=

FAAAA

A

Basic Properties of Norms

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85


.5.4

numberrealaiswhere.3.2

0and;0.1Then.andLet

BABAxAAx

xxyxyx

0xxxyx,BA,

=++

==

nnn

Condition number of a square matrix

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86


(n m n

n n

1 2

1 2

1

All norms in are equivalent. That is, if and are norms

on then 0 such that for all we have

: , where A

The

n n

, c ,c ,

c c

C .

>

x

x x x

Condition Number of A Matrix A A

condition number gives a measure of how close a matrix is

close to singular. The bigger the C, the harder it is to solve .Ax=b

Convergence

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87


- vectors xk

converges to x xk

x converges to 0

- matrix A

k

0 Ak

0 0

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Appendix B.

Basic Row Operations

Basic row operations= *

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Three kinds of basic row operations:

1) Interchange the order of two rows or (equations)

0 1 0

1 0 0

0 0 1

a11 a12 a13

a21 a22 a23

a31 a32 a33

=

a21 a22 a23

a11 a12 a13

a31 a32 a33

Basic row operations = *

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90Math6911, S08, HM ZHU

2) Multiply a row by a nonzero constant

3) Add or subtract rows

c 0 0

0 1 0

0 0 1

a11 a12 a13

a21 a22 a23

a31 a32 a33

=

ca11 ca12 ca13

a21 a22 a23

a31 a32 a33

1 0 0

1 1 0

0 0 1

a11 a12 a13

a21 a22 a23

a31 a32 a33

=

a11 a12 a13

a21 a11 a22 a12 a23 a13a31 a32 a33

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