Chapter 5 Finite Difference Methods - YorkU Math and...

Math6911 S08, HM Zhu

Chapter 5 Finite Difference Methods

2Math6911, S08, HM ZHU

References

1. Chapters 5 and 9, Brandimarte

2. Section 17.8, Hull

3. Chapter 7, “Numerical analysis”, Burden and Faires


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


5.1 Finite difference approximations



Finite-difference mesh

• Aim to approximate the values of the continuous function f(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 a distance ∆t apart

• (t, S) plane becomes a mesh with mesh points on (i ∆t, j∆S)

• We are interested in the values of f(t, S) at mesh points (i ∆t, j∆S), denoted as

( )i , jf f i t , j S= ∆ ∆


The mesh for finite-difference approximation

t

S

i ∆t

j ∆S

( )i , jf f i t , j S= ∆ ∆

Smax=M∆S

T=N ∆t

?

Math6911, S08, HM ZHU

Black-Scholes Equation for a European option with value V(S,t)

conditionsboundary and finalproper with Tt0 and S0 where

)1.5(021

2

222

<≤+∞<<

=−∂∂

++∂∂ rV

SVrS

SVS

tV

∂∂σ

Notes:This is a second-order hyperbolic, elliptic, or parabolic, forward or backward partial differential equationIts solution is sufficiently well behaved ,i.e. well-posed


Finite difference approximations

The basic idea of FDM is to replace the partial derivatives by approximations obtained by Taylor expansions near the point 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 tt , S ,t

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

t

∆ →

∂ + ∆ − + ∆ −= ≈

∂ ∆ ∆∆

∂+ ∆ = + ∆ + ∆

∂

Forward-, Backward-, and Central-difference approximation to 1st order derivatives

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )( )2

Forward:

Backward:

Central:2

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

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

O tt t

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

t t

∂ + ∆ −≈ + ∆

∂ ∆∂ − − ∆

≈ + ∆∂ ∆

∂ + ∆ − − ∆≈ + ∆

∂ ∆

t t+ ∆tt t− ∆

centralbackward forward


Symmetric Central-difference approximations to 2nd order derivatives

( ) ( ) ( ) ( )( )

( )( )2

222

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 ?

+ ∆ − ∆

+ ∆ =

+

− ∆ =



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 12

2

derivative, we have:

=

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

i , j i , j i , j

f f f ff SS SS

f f f

S

+ −

+ −

− −⎛ ⎞∂≈ − ∆⎜ ⎟∆ ∆∂ ⎝ ⎠

− +

∆


• Depending on which combination of schemes we use in discretizing the equation, we will have explicit, implicit, or Crank-Nicolson methods

• We also need to discretize the boundary and final conditions accordingly. For example, for European Call,


( )

( )0

Final Condition:0 for 0 1

Boundary Conditions:0

for 0 1

where

N , j

i ,

r N i ti ,M max

max

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

f, i , ,...,N

f S Ke

S M S.

− − ∆

= ∆ − =

=⎧⎪ =⎨= −⎪⎩

= ∆


5.2.1 Explicit Finite-Difference Method



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 , ji , j

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

f fft t

f ffS S

f f ff , r f rf , S j SS S

σ

−

+ −

+ −

∂ ∂ ∂+ + = ∆ ∆

∂ ∂ ∂−∂

≈∂ ∆

−∂≈

∂ ∆

+ −∂≈ = = ∆

∂ ∆



( )( )

( )

1 1 1

2 2

2 2

2 2

12

1

12

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

σ

σ

σ

− − += + +

= ∆ −

= − ∆ +

= ∆ +

1 2 1 0 1 2 1for and i N - , N - , ..., , j , , ..., M - .= =


Numerical Computation Dependency

x

x

x

x

t

S

i∆t

j∆S

Smax=M∆S

T=N ∆t00

(i-1)∆t

(j+1)∆S

(j-1)∆S


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 .− ≤ ≤ −

1 0 1 and N , N - ,Mf f .−

2N , jf −


Example

We compare explicit finite difference solution for a European put with the exact Black-Scholes formula, where T = 5/12 yr, S0=$50, K = $50, σ=30%, r = 10%.

Black-Scholes Price: $2.8446EFD Method with Smax=$100, ∆S=2, ∆t=5/1200: $2.8288EFD Method with Smax=$100, ∆S=1, ∆t=5/4800: $2.8406


Example (Stability)

We compare explicit finite difference solution for a European put with the exact Black-Scholes formula, where T = 5/12 yr, S0=$50, K = $50, σ=30%, r = 10%.

Black-Scholes Price: $2.8446EFD Method with Smax=$100, ∆S=2, ∆t=5/1200: $2.8288EFD Method with Smax=$100, ∆S=1.5, ∆t=5/1200: $3.1414EFD Method with Smax=$100, ∆S=1, ∆t=5/1200: -$2.8271E22


5.2.2 Numerical Stability



Numerical Accuracy

• The problem itself• The discretization scheme used• The numerical algorithm used


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 xy x

x x x x.

f x f x ,

well - co

δ δ

=

= +

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


Conditioning Issue

• Conditioning issue is related to the problem itself, not to the specific numerical algorithm; Stability issue is related to the numerical algorithm

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

• But a bad numerical algorithm can produce poor solutions even to well-conditioned problems


Conditional Issue

( ) ( )( )

( )( )

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

δ−≤ ≠


Floating Point Number & Error

( )

( ) xxfled

xxxx.x xflx

xx.x x x

i

ed

ed

−

≤≤≠±=≈

±=

: or integer boundedan :

systempoint floating theofprecision integer,an : 9,0 0, where

10 :numberpoint floating Truncated

10 :expansion decimal Infinite number. realany be Let

1

21

21

errorroundoffpoint Floating


Error Propagation

( )

( )( )

00040011.0 :npropagatioError

0003.0 :errorpoint Floating3. where

10667.0 and 6667.0Let : errors.point floating theseofon accumulati

an istheredone, arenscalculatioadditionalWhen

22

0

=−

−=−=

−=−=

xxfl

xxfld

xfl x Example


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

− ∆ ∆

∆ ∆


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)


Numerical Accuracy

• These factors contribute the accuracy of a numerical solution. We can find a “good” estimate 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

⎫≈ ⎪ ≈⎬≈ ⎪⎭


5.2.3 Financial Interpretation of Numerical Instability



Financial Interpretation of instability(Hall, page 423-4)

2

1 1 1 1 1

1

11

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 , ji, j

ˆˆ ˆf a f b f c f

a

+ − + + +

∂ ∂ ∂ ∂ +

= + +

( )

2 2

2 20

2 2

11

1 111 1

1 11 1

1 2 1 0 1 2 1

1 12 2

1 1 2 2

for and

d

j

j u

t j t r jr t r t

b j tr t r t

c t j t r jr t r t

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

σ π

σ π

σ π

⎛ ⎞∆ − ∆ =⎜ ⎟+ ∆ + ∆⎝ ⎠

= − ∆ =+ ∆ + ∆

⎛ ⎞= ∆ + ∆ =⎜ ⎟+ ∆ + ∆⎝ ⎠= =



ƒi , j ƒi +1, j

ƒi +1, j –1

ƒi +1, j +1

These coefficients can be interpreted as probabilities times a discount factor. If one of these probability < 0, instability occurs.

πd

π0

πu


Explicit Finite Difference Method as Trinomial Tree

[ ]

( ) ( )

0

2 220

Check if the mean and variance of the Expected value of the increase in asset price during t: E 0Variance of the increment:

E 0

d u

d

S S rj S t rS t

S S

π π π

π π π

∆

∆ = −∆ + + ∆ = ∆ ∆ = ∆

⎡ ⎤∆ = − ∆ + + ∆⎣ ⎦ ( )

[ ] [ ] ( )

2 2 2 2 2

22 2 2 2 2 2 2 2 Var 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

σ σ

σ σ

= ∆ ∆ = ∆

⎡ ⎤∆ = ∆ − ∆ = ∆ − ∆ ≈ ∆⎣ ⎦


Change of Variable

2 2 2

2

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

2

2 2

22 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 , ji , j

Z ln S.

f f fr rft Z Z

f f f f f f fr rf

t Z Z

f

σ σ

σ σ+ + + + − + + + − + +

=

⎛ ⎞∂ ∂ ∂+ − + =⎜ ⎟∂ ∂ ∂⎝ ⎠

− − − +⎛ ⎞+ − + =⎜ ⎟∆ ∆ ∆⎝ ⎠

1 1 1 1 1 5 4( )* * *i , j j i , j j i , j j i , jf f f .α β γ+ − + + += + +


Change of Variable

2 2

2

22

2 2

2

11 2 2 2

1 11

11 2 2 2

3

where

It can be shown that it is numerically most efficient if

*j

*j

*j

t trr t Z Z

tr t Z

t trr t Z Z

Z t .

σ σα

β σ

σ σγ

σ

⎡ ⎤⎛ ⎞ ∆ ∆= − − +⎢ ⎥⎜ ⎟+ ∆ ∆ ∆⎝ ⎠⎣ ⎦

∆⎛ ⎞= −⎜ ⎟+ ∆ ∆⎝ ⎠⎡ ⎤⎛ ⎞ ∆ ∆

= − +⎢ ⎥⎜ ⎟+ ∆ ∆ ∆⎝ ⎠⎣ ⎦

∆ = ∆


Reduced to Heat Equation

Get rid of the varying coefficients S and S² by using change of variables:

Equation (5.1) becomes heat equation (5.5):

( ) ( ) ( ) ( )

⎟⎠⎞

⎜⎝⎛=

=−==+−−−

2

1411

21

2

21

,,,2,2

σ

τσττ

rk

xueEtSVTtEeSkxkx

2

2 (5.5)

for and 0

u ux

x

∂ ∂∂τ ∂

τ

=

− ∞ < < +∞ >


( )

( )( )

( )( )( ) ( )( )

( )

( ) ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

==≤≤=

+−+=

++−

=+−

=

+−

−++

−++

δτ

T,...M,mNn-N

uuuu

δxOδτO

δxOδx

uuuδτO δτ

uu

x,mδnuu

mn

mn

mn

mn

mn

mn

mn

mn

mn

mn

2

2

111

2

22

111

21

10 and for ,x

where

21by this

eapproximatcan we, and of termsIgnoring

2

:form theof equations difference finite of system a solving involves this,With

σ

δδτα

ααα

τδ

Explicit Finite Difference Method


Stability and Convergence(P. Wilmott, et al, Option Pricing)

( )2

1 ,2

: The solution of Eqn (5.5) is

1 1 i) Stable if 0 ; ii) Unstable if 2 2x

:

If 0 then the explicit finite-difference approximation

converges to the exact solution as ,

δτα αδ

α

δτ δ

< = ≤ >

< ≤

Stability

Convergence

( )( )

,mn

x 0 (in the sense that as , x 0)

is O

u u n x mδ δτ δτ δ

δτ

→

→ →

Rate of Convergence


5.3.1 Implicit Finite-Difference Method



Implicit Finite Difference Methods

22 2

2

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 ji , j

f f frS S r ft S S

f fft t

f ffS S

f f ff , rf rfS S

σ

∂∂

∂∂

∂∂

+

+ −

+ −

∂ ∂ ∂+ + =

∂ ∂ ∂−

≈∆

−≈

∆

+ −≈ =

∆

forward



( )( )( )

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

σ

σ

σ

− + ++ +

∆ − +

= + ∆ +

= − ∆ +

1 2 1 0 1 2 1 and i N - , N - , ..., , j , , ..., M - .= =



x

x

x

x

t

S

i∆t

j∆S

Smax=M∆S

T=N ∆t00

(i-1)∆t

(j+1)∆S

(j-1)∆S

(i+1)∆t

Implementation

( )1

1 2 3 1 1 0 1

Equation (5.6) can be rewritten in matrix form:5 7

where and are ( 1) dimensional vectors

0 0 0

and is ( 1) ( 1) symmetric

i i i

i iT 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 bf b

f b

C

1 1

2 2 2

3 3

2

1 1

matrices0 0

0 0

0 0M

M M

b ca b c

a bc

a b−

− −

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

C


Implementation

1. Starting with the final values , we need to solve a linear system (5.7) to obtain using LU factorization or iterative methods. We use the boundary condition to determine

2. Repeat the process to determine and so on

N , jf1 for 1 1N , jf j M− ≤ ≤ −

1 0 1 and N , N - ,Mf f .−

2N , jf −


Example

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

Black-Scholes Price: $2.8446IFD Method with Smax=$100, ∆S=2, ∆t=5/1200: $2.8194IFD Method with Smax=$100, ∆S=1, ∆t=5/4800: $2.8383


Example (Stability)

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

Black-Scholes Price: $2.8846IFD Method with Smax=$100, ∆S=2, ∆t=5/1200: $2.8288IFD Method with Smax=$100, ∆S=1.5, ∆t=5/1200: $3.1325IFD Method with Smax=$100, ∆S=1, ∆t=5/1200: $2.8348


Implicit vs Explicit Finite Difference Methods

ƒi +1, jƒi , j

ƒi , j –1

ƒi , j +1

Implicit Method

(always stable)

ƒi , j ƒi +1, j

ƒi +1, j –1

ƒi +1, j +1

Explicit Method


Implicit vs Explicit Finite Difference Method

• The explicit finite difference method is equivalent 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 Methods

• It is better to have ln S rather than S as the 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


5.3.2 Solving a linear system using direct methods



Solve Ax=b

A x=b

Various shapes of matrix A

Lower triangular Upper triangular General


5.3.2.A Triangular Solvers


Example: 3 x 3 upper triangular system

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2010

100

400110164

3

2

1

xxx

46565*61004

51054/20

1

231

32

3

=∴=−−=⇒

=−=⇒==⇒

xxxx

xxx


Solve an upper triangular system Ax=b

( )

1,1,

,,1,00 2

1

−=⎟⎟⎠

⎞⎜⎜⎝

⎛−=

=⇒

==+=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

∑

∑

>

>

niaxabx

abx

nibxaxa

x

xx

aa

iiij

jijii

nnnn

iij

jijiii

n

inii …


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


5.3.2.B Gauss Elimination


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 = *

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

01001

aaaa

aaaa

aaa

aaa

aaaaaaaaa

aa

⇓E 21A



2) Zero a31

11 12 13 11 12 13

21 13 21 1321 12 21 1222 23 22 23

11 11 11 1131

31 32 33 31 12 31 1311 32 33

11 11

1 0 00 1 0 0 0

0 10

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

a a a aa

a a a a a a aa a aa a

⎡ ⎤⎡ ⎤ ⎢ ⎥⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥− − = − −⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥− ⎢ ⎥⎣ ⎦⎢ ⎥ − −⎢ ⎥⎣ ⎦

⎣ ⎦

11 12 13

22 23

32 33

00

= a a a

a aa a

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

⇓E 31 E 21A



3) Zero

11 12 13 11 12 13

22 23 22 23

32 32 33 32 2333

22 22

1 0 00 1 0 0 0

00 1 0 0

a a a a a aa a a a

a a a a aaa a

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ = ≡⎢ ⎥ ⎢ ⎥⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦⎢ ⎥ ⎢ ⎥− −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

U

32a

⇓E 32 E 31 E 21A = U

⇓lower triangular



Claim 1: 21

32 31 2111

31 32

11 22

1 0 0

1 0

1

aaa aa a

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= −⎢ ⎥⎢ ⎥⎢ ⎥

− −⎢ ⎥⎣ ⎦

E E E

Claim 2:

( ) 1 2132 31 21

11

31 32

11 22

1 0 0

1 0

1

aaa aa a

−

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥

= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

E E E


LU Factorization = *

Therefore, through Gauss Elimination, we have

E 32 E 31 E 21A = U

A = E 32 E 31 E 21( )−1U

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

To solve Ax = b becomes

1) Use Gauss elimination to make A upper triangular, i.e.

L−1Ax = L−1b ⇒ 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

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

−

400

21121

12

3

2

1

xxx

% 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 = ux z = l\b;

% solve x from ux = z x = u\z


General Gauss Elimination

row i

row j -a ji

aii

∗row i


What if we have zero or very small diagonal elements?

LUPA

LUA

PA

A

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=

⎥⎦

⎤⎢⎣

⎡

.9999011

10.000101

1110.0001

0110

9999-010.0001

110,00001

1110.0001

=

1032

3210

0110

3210

=

exampleFor pivoting. partial

called is This LU.=PA i.e., stably, computedor computed becan neliminatio Gauss that soA of rows permute toneed weSomtimes,


Can we always have A = LU?

( )( )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



5.4.1 Crank-Nicolson Method



( )

( )

( )( )

22 2

2

1

21 1 1 1 1 1 1 1 12 22

21

12

212 2

W ith ,

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 rft S S

f fO t

tf f f f f

rj S j SS S

rf O S

σ

σ

+

+ + + − + + + − +

+

∂ ∂ ∂+ + =

∂ ∂ ∂

−+ ∆ =

∆− + −

− ∆ − ∆∆ ∆

+ + ∆



( )

( )

( )( )

22 2

2

1

21 1 1 12 22

2

12

212 2

W ith ,

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 ft S S

f fO t

tf f f f f

rj S j SS S

rf O S

σ

σ

+

+ − + −

∂ ∂ ∂+ + =

∂ ∂ ∂

−+ ∆ =

∆− + −

− ∆ − ∆∆ ∆

+ + ∆


Crank-Nicolson Methods: average of explicit and implicit methods(Brandmarte, p485)

( )( )

( )

( ) ( )( )

21

1 1 1 1 1 1

2 1 1 1 1 1 1 12 22 2

21

2 2 2

2 214

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 fO t

tf f f frj S

S S

f f f f f fj S

S Sr f f O S

σ

+

+ + + − + −

+ + + − + + −

+

−+ ∆ =

∆− −⎛ ⎞∆

− +⎜ ⎟∆ ∆⎝ ⎠+ − + −⎛ ⎞

− ∆ +⎜ ⎟∆ ∆⎝ ⎠

+ + + ∆


Crank-Nicolson Methods

( )( )

( )

1 1

1 1 1 1 1

2 2

1

1

R ew riting the equa tion , w e ge t an C rank-N ico lson schem e:

(5 .5 )

w here

=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

t j rj

α β γ

α β γ

α σ

β

− +

+ − + + +

− + − − =

+ + +

∆−

= − ( )

( )

2 2

2 2

2

1 2 1 0 1 2 1

4

fo r and

j

t j r

t j rj

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

σ

γ σ

∆+

∆= +

= =



x

x

x

x

t

S

i∆t

j∆S

Smax=M∆S

T=N ∆t00

(i-1)∆t

(j+1)∆S

(j-1)∆S

(i+1)∆t

x

x

x

x

Implementation

( ) ( )

1 2 1

1 2 3 1

1 0 1 0 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 iT

i i , i , i , i ,M

T

i , i , M i ,M i ,M

M

f , f , f , f ,

f f , , f f

M

α γ

+

−

+ − +

= +

−

= ⎡ ⎤⎣ ⎦

⎡ ⎤= + +⎣ ⎦−

M f M f bf b

f

b

M M

1 1

2 2 2

1 3

2

1 1

1 11 0 0

1 00

0 0 1

) ( ) symmetric matrices

M

M M

Mβ γ

α β γα

γα β

−

− −

× −

− −⎡ ⎤⎢ ⎥− −⎢ ⎥⎢ ⎥=⎢ ⎥−⎢ ⎥⎢ ⎥− −⎣ ⎦

M


Implementation

1 1

2 2 2

2 3

2

1 1

and 1 0 0

1 00

0 0 1M

M M

β γα β γ

αγ

α β−

− −

+⎡ ⎤⎢ ⎥+⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎢ ⎥+⎣ ⎦

M


Example

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

Black-Scholes Price: $2.8446CN Method with Smax=$100, ∆S=2, ∆t=5/1200: $2.8241CN Method with Smax=$100, ∆S=1, ∆t=5/4800: $2.8395


Example (Stability)

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

Black-Scholes Price: $2.8446CN Method with Smax=$100, ∆S=1.5, ∆t=5/1200: $3.1370CN Method with Smax=$100, ∆S=1, ∆t=5/1200: $2.8395


Example: Barrier Options

Barrier options are options where the payoff depends on whether the underlying asset’s price reaches a certain level during a certain period of time.

Types: knock-out: option ceases to exist when the asset price reaches a barrierKnock-in: option comes into existence when the asset price reaches a barrier


Example: Barrier Option

We compare Crank-Nicolson Method for a European down-and-out put, where T = 5/12 yr, S0=$50, K = $50, Sb=$50 σ=40%, r = 10%.

What are the boundary conditions for S?


Example: Barrier Option

We compare Crank-Nicolson Method for a European down-and-out put, where T = 5/12 yr, S0=$50, K = $50, Sb=$50 σ=40%, r = 10%.

Boundary conditions are

Exact Price (Hall, p533-535): $0.5424CN Method with Smax=$100, ∆S=0.5, ∆t=1/1200: $0.5414

( ) ( )0 0 and max bf t ,S f t ,S= =


Appendix A.

Matrix Norms


Vector Norms

( )

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

max

normEuclidean theis 2p

norm a define to ways variousare There -any for

any for

0 iff 0 ;any for 0

s.t. number real a to mappingfunction a is normA vector -matrix aor vector a oflength themeasure way toa as serve Norms -

21

1

1

1

n

====−

≡

=⎟⎠

⎞⎜⎝

⎛≡

ℜ∈+≤+•

ℜ∈=•

==≠>•

ℜ∈

∞

≤≤∞

=∑

vvvv

x

x

yxyxyx

xx

xx0xx

xx

ini

pn

i

pip

n

x

x

,

ccc

Matrix Norms

( )( ) { }BB

AAA

AA

Ax

AxA

BABABA

AA

0A0A0AA

AA

x

of eigenvaluean is :max

wherenorm, spectral the,

maxmax

sup

normsmatrix usedcommonly Various -any for

any for

iff ;any for 0

s.t. number real a to mappingfunction a is normmatrix a Similarly, -

2

11111

21

1 1

2

0

nm

kk

T

n

jij

mi

m

iij

nj

m

i

n

jijF

p

pp

nm

aa

a

,

ccc

λλρ

ρ

≡

≡

≡≡

⎟⎟⎠

⎞⎜⎜⎝

⎛≡≡

ℜ∈+≤+•

ℜ∈=•

==≠>•

ℜ∈

∑∑

∑∑

=≤≤∞

=≤≤

= =≠

×

×


An Example

??

??

132513142

1

2

==

==

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−=

∞

FAA

AA

A


Basic Properties of Norms

.5

.4

number real a is where .3

.2

0 and ;0 .1Then . and Let

BABA

xAAx

xx

yxyx

0xxxyx,B A,

≤

≤

=

+≤+

=⇔=≥

ℜ∈ℜ∈ ×

ααα

nnn


Condition number of a square matrix

( )n m n

n n1 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 .

α β

α β α

×

− ×

ℜ ℜ • •

ℜ ∃ > ∈ℜ

≤ ≤

≡ ∈ℜ

xx 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


- vectors xk converges to x ⇔ xk − x converges to 0

- matrix Ak → 0 ⇔ A k − 0 → 0

Convergence


Appendix B.

Basic Row Operations


Basic row operations = *

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 = *

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 −a13

a31 a32 a33

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

Chapter 5 Finite Difference Methods - YorkU Math and...

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