cf-1 bank hapoalim jun-2001 zvi wiener 02-588-3049 mswiener/zvi.html computational finance

CF-1 Bank Hapoalim Jun-2001

Zvi Wiener

02-588-3049http://pluto.mscc.huji.ac.il/~mswiener/zvi.html

Computational Finance

CF1 slide 2Zvi Wiener

Plan1. Introduction, deterministic methods.2. Stochastic methods.3. Monte Carlo I.4. Monte Carlo II.5. Advanced methods for derivatives.

Other topics:queuing theory, floaters, binomial trees, numeraire, ESPP, convertible bond, DAC, ML-CHKP.


Linear Algebra

-2 1

1

Vectors {1, 1}, {-2, 1}

rows or 1 -2columns 1 1


Linear Algebra


Basic Operations

1 2 3

2 + -1 = 1

-2 1 -1

1 3

3 2 = 6

2 6


Linear Algebra

vector

3

2

1

x

x

x

x Vectors form a linear space.

3

2

1

x

x

x

x

33

22

11

yx

yx

yx

yx Zero vector

Scalar multiplication


Linear Algebra

matrix

333231

232221

131211

aaa

aaa

aaa

AMatrices alsoform a linearspace.

Zero matrix

000

000

000

Unit matrix

100

010

001


Linear Algebra

Matrix can operate on a vector

3

2

1

333231

232221

131211

x

x

x

aaa

aaa

aaa

Ax

333232131

323222121

313212111

xaxaxa

xaxaxa

xaxaxa

How does zero matrix operate?

How does unit matrix operate?


Linear Algebra

Transposition of a matrix

A symmetric matrix is A=AT

for example a variance-covariance matrix.

333231

232221

131211

aaa

aaa

aaa

A

332313

322212

312111

aaa

aaa

aaa

AT


Linear Algebra

Matrix multiplication

333323321331323322321231313321321131

332323221321322322221221312321221121

331323121311321322121211311321121111

bababababababababa

bababababababababa

bababababababababa

333231

232221

131211

333231

232221

131211

bbb

bbb

bbb

aaa

aaa

aaa

AB


Scalar Product

i

iibaba

332211

3

2

1

321 ,, bababa

b

b

b

aaa

a is orthogonal to b if ab = 0


Linear Algebra

Scalar product of two vectors

332211

3

2

1

321 yxyxyx

y

y

y

xxxyxT

Euclidean norm

23

22

21

2xxxxxx T


Determinant

211222112221

1211 aaaaaa

aaDet

Determinant is 0 if the operator mapssome vectors to zero (and can not be inverted).


Linear Algebra

• Matrix multiplication corresponds to a

consecutive application of each operator.

• Note that it is not commutative! ABBA.

• Unit matrix does not change a vector.

• An inverse matrix is such that AA-1=I.


Linear Algebra

• Determinant of a matrix ...

• A matrix can be inverted if det(A)0

• Rank of a matrix

• Matrix as a system of linear equations Ax=b.

• Uniqueness and existence of a solution.

• Trace tr(A) – sum of diagonal elements.


Linear Algebra

• Change of coordinates C-1AC.

• Jordan decomposition.

• Matrix power Ak.

• Matrix as a quadratic form (metric) xTAx.

• Markov process.

• Eigenvectors, eigenvalues Ax=x,

optimization.


Problems

Check how the following matrices act on vectors:

00

01

10

01

10

01

cossin

sincos

01

10

01

10


Simple Exercises

• Show an example of ABBA.

• Construct a matrix that inverts each vector.

• Construct a matrix that rotates a two dimensional vector by an angle .

• Construct a covariance matrix, show that it is symmetric.

• What is mean and variance of a portfolio in matrix terms?


Examples

• Credit rating and credit dynamics.

• Variance-covariance model of VaR.

• Can the var-covar matrix be inverted

• VaR isolines (the ovals model).

• Prepayment model based on types of clients.

• Finding a minimum of a function.


Calculus

• Function of one and many variables.

• Continuity in one and many directions.

• Derivative and partial derivative.

• Gradient and Hessian.

• Singularities, optimization, ODE, PDE.


x

$

Linear and quadratic terms


Taylor series

22

2)(")(')()( xo

xxfxxfxfxxf

2222

22

),(),(

yxoy

gyxgx

g

ygxgyxgyyxxg

yyxyxx

yx

2)("

2

1)(')()( xoxxFxxxFxFxxF T


Variance-Covariance

),...,,()( 21 nxxxVxV

nnn

n

n

1

2221

112111

,


Variance-Covariance

nx

V

x

V

xV 1

)('Gradient vector:


Variance-Covariance

2)("2

1)(')()( xxVxxVxVxxV

2)("2

1)(')( xxVxxVxV


Variance-Covariance

22

)(")(' xE

xVxVxEVE

"2

1' VtrVVE T

'' VVV T


Variance-Covariance

For a short time period , the changes in the value are distributed approximately normal with the following mean and variance:

''

"2

1'

2 VVV

VtrVVE

T

T


Variance-Covariance

Then VaR can be found as:

)(33.2)(%,1 VVVVaR

''33.2"2

1' VVVtrV TT


Weighted Variance covariance

Volatility estimate on day i based on last M days.

i

Mijj RR

Mt 1

2)()1(

1

i

jj

ji RRt

2)(1


Weighted Variance covariance

Covariance on day i based on last M days.

i

Mijjj RRRR

Mt 1221112 ))((

)1(

1

It is important to check that the resulting

matrix is positive definite!


Positive Quadratic Form

For every vector x a we have x.A.x > 0

Only such a matrix can be used to define a

norm.

For example, this matrix can not have

negative diagonal elements. Any variance-

covariance matrix must be positive.


Positive Quadratic Form

Needs["LinearAlgebra`MatrixManipulation`"];

ClearAll[ positiveForm ];

positiveForm[ a_?MatrixQ ] := Module[{aa, i},

aa = Table[

Det[ TakeMatrix[ a, {1, 1}, {i, i}] ],

{i, Length[a]}];

{ aa, If[ Count[ aa, t_ /; t < 0] > 0, False, True]}

];


Stochastic (transition) Matrix

Used to define a Markov chain (only the last state matters).

A matrix P is stochastic if it is non-negative and sum of elements in each line is 1.

One can easily see that 1 is an eigenvalue of any stochastic matrix.

What is the eigenvector?


Markov chain

• credit migration

• prepayment and freezing of a program


Stochastic (transition) Matrix

Theorem: P0 is stochastic iff (1,1,…1) is an eigenvector with an eigenvalue 1 and this is the maximal eigenvalue.

If both P and PT are stochastic, then P is called double stochastic.


Cholesky decomposition

The Cholesky decomposition writes a symmetric positive definite matrix as the product of an upper triangular matrix and its transpose.

In MMA CholeskyDecomposition[m]


Generating Random Samples

We need to sample two normally distributed variables with correlation .

If we can sample two independent Gaussian variables x1 and x2 then the required variables can be expressed as

2212

11

1

xxy

xy


Generating Random Samples

We need to sample n normally distributed variables with correlation matrix ij, ( >0).

Sample n independent Gaussian variables x1…xn.

n

kkiki xay

1

xAy .

AAAanda Tn

kik ,,1

1

2


Function of a matrix

!3!2

32 AAAEeA

TJTA

TJTJTJTTTA

JTTA

nn 1

21112

1


ODE

Axdt

dx 00 )( xtx

0)( 0)( xetx ttA


ODE

)(tfAxdt

dx 00 )( xtx

t

t

tAttA dtfexetx0

0 )()( )(0

)(


Bisection method

If the function is monotonic, e.g. implied vol.

x

f


Newton’s method

x

f

x1x2x3


Solve and FindRoot

Solve[ 0 = = x2- 0.8x3- 0.3, {x}]

{{x -> -0.467297}, {x ->0.858648 -0.255363*I}, {x -> 0.858648 + 0.255363*I}}

FindRoot[ x2 + Sin[x] - 0.8x3 - 0.3, {x, 0,1}]

{x -> 0.251968}


Max, min of a multidimensional function

• Gradient method

• Solve a system of equations (both derivatives)

-1

-0.5

0

0.5

1-1

-0.5

0

0.5

1

0

0.5

1

-1

-0.5

0

0.5

1


-1 -0.5 0 0.5 1-1

-0.5

0

0.5

1

Gradient method


Level curve of a multivariate function

ContourPlot[ x^2+y^3, {x,-2,2}, {y,- 2,2}]

ContourPlot[ x^2+y^3, {x,-2,2}, {y,- 2,2}], Contours->{1 ,-0.5}, ContourShading->False];


ContourPlot[ x^2+y^3, {x,-2,2}, {y,- 2,2}]

-2 -1 0 1 2

-2

-1

0

1

2


ContourPlot[ x^2+y^3, {x,-2,2}, {y,- 2,2}], Contours->{1 ,-0.5}, ContourShading->False];

-2 -1 0 1 2

-2

-1

0

1

2


Example

Consider a portfolio with two risk factors and benchmark duration of 6M.

The VaR limit is 3 bp. and you have to make two decisions:

a – % of assets kept in spread products

q – duration mismatch

we assume that all instruments (both treasuries and spread) have the same duration T+q months.


q - durationmismatch

Contour Levels of VaR (static)

0 0.2 0.4 0.6 0.8 1

-6

-4

-2

0

2

4

6

a (% of spread)


0 0.2 0.4 0.6 0.8 1

-6

-4

-2

0

2

4

6


a (% of spread)

position

VaR=2 bp

VaR=3 bp


0 0.2 0.4 0.6 0.8 1

-6

-4

-2

0

2

4

6


a (% of spread)

In order to reducerisk one can increase duration(in this case).


0 0.2 0.4 0.6 0.8 1

-6

-4

-2

0

2

4

6

What we can do using limits

VaR = 6 bp


0 0.2 0.4 0.6 0.8 1

-0.2

-0.1

0

0.1

0.2

spread %

duration mismatch (yr)

Position 2M, and10% spread5% weekly VaR=2.2 bp

weekly VaR limit 3 bp


Splines

x1 x2 x3 … xn


Splines<<Graphics`Spline`

pts = {{0, 0}, {1, 2}, {2, 3}, {3, 1}, {4, 0}}

Show[

Graphics[

Spline[pts, Cubic, SplineDots -> Automatic]]]


Splinespts = Table[{i, i + i^2 + (Random[] - 0.5)}, {i, 0, 1, .05}];

Show[Graphics[Spline[pts,Cubic,SplineDots ->Automatic]]]


Fitting datadata = Table[7*x + 3 + 10*Random[], {x, 10}];

f[x_] := Evaluate[Fit[data, {1, x}, x]]

Needs["Graphics`Graphics`"]

DisplayTogether[

ListPlot[data, PlotStyle -> {AbsolutePointSize[3],

RGBColor[1, 0, 0]}],

Plot[f[x], {x, 0, 10}, PlotStyle -> RGBColor[0, 0, 1]]

];


Fitting data

2 4 6 8 10

20

40

60


Fitting data

data = {{1.0, 1.0, .126}, {2.0, 1.0, .219},

{1.0, 2.0, .076}, {2.0, 2.0, .126}, {.1, .0, .186}};

ff[x_, y_] = NonlinearFit[data,

a*c*x/(1 + a*x + b*y), {x, y}, {a, b, c}];

ff[x, y]

nonlinear, multidimensional


Orly - Nelson Siegel

50 100 150 200 250 300 350

0.125

0.13

0.135

0.14

0.145 MAKAM


Numerical Differentiation

22

2)(")(')()( xo

xxfxxfxfxxf

22

2)(")(')()( xo

xxfxxfxfxxf

2)('2)()( xoxxfxxfxxf

xoxfx

xxfxxf

)('2

)()(


Numerical Differentiation

22

2)(")(')()( xo

xxfxxfxfxxf

22

2)(")(')()( xo

xxfxxfxfxxf

22)(")()(2)( xoxxfxxfxfxxf

22

)(")()(2)(

xOxfx

xxfxfxxf


Finite Differences

Typically equal time and S (or logS) steps.

time

S Following P. Wilmott, “Derivatives”


Finite Differences

Time step tasset step S(i,k) node of the grid is t = T - kt, iS0 i I, 0 k K

assets value at each node is

note the direction of time!

),( tkTSiVV ki


The Black-Scholes equation

Linear parabolic PDE

0)(2

12

222

rVS

VSrr

S

VS

t

Vf

Final conditions )(),( SPayoffTSV

Boundary conditions ...


Transformation of BS

U

x

U2

2

.2

,

,12

4

1

,12

2

1

),(),(

2

2

2

2

rTt

eS

r

r

xUetSV

x

x


Approximating

h

tSVhtSV

t

Vh

),(),(lim

0

tOt

VVtS

t

V ki

ki

1

),(


Approximating

211

2),( SO

S

VVtS

S

V ki

ki


Approximating

22

112

2 2),( SO

S

VVVtS

S

V ki

ki

ki


Bilinear Interpolation

A3A4

A2A1

V3V4

V2V1

4

1

4

1

jj

jjj

A

VA

V

Area of the rectangle


Final conditions and payoffs

)(),( SPayoffTSV

)(0 SiPayoffVi

For example a European Call option

)0,(0 ESiMaxVi


Boundary conditions Call

00 kV

For example a Call option

trkkI EeSIV

For large S the Call value asymptotes toS-Ee-r(T-t)


Boundary conditions Put

trkk EeV 0

For example a Put option

0kIV

For large S


Boundary conditions S=0

0),0(),0(

trVtt

V

100 )1( kk VtrV

10

01

0

k

kk

rVt

VV


Explicit scheme

0),(),(),(2

2

VtScS

VtSb

S

VtSa

t

V

),(2

2

211

211

1

StOVcS

VVb

S

VVVa

t

VV

ki

ki

ki

kik

i

ki

ki

kik

i

ki

ki


Explicit scheme

),(2

2

211

211

1

StOVcS

VVb

S

VVVa

t

VV

ki

ki

ki

kik

i

ki

ki

kik

i

ki

ki

ki

ki

ki

ki

ki

ki

ki VCVBVAV 11

1 )1(

Local truncation error 22 , SttO


Explicit scheme

ki

ki

ki

ki

ki

ki

ki VCVBVAV 11

1 )1(

time

S

Value here is calculated

These values arealready known


Explicit scheme

ki

ki

ki

ki

ki

ki

ki VCVBVAV 11

1 )1(

This equation is defined for 1i I-1,

for i=1 and i=I we use boundary conditions.


Explicit scheme

ki

ki

ki

ki

ki

ki

ki VCVBVAV 11

1 )1(

For the BS equation (with dividends)

,2/))((

,)(

,2/))((

22

22

22

tiDriC

triB

tiDriA

ki

ki

ki


Explicit scheme

Stability problems related to step sizes.

These relationships should guarantee stability.

Note that reducing asset step by half we must reduce the time step by a factor of four.

a

St

b

aS

2,

2 2


Explicit scheme

Advantages

easy to program, hard to make a mistake

when unstable it is obvious

coefficients can be S and t dependent

Disadvantages

there are restrictions on time step, which may cause slowness.


Implicit scheme

0),(),(),(2

2

VtScS

VtSb

S

VtSa

t

V

),(2

2

2111

11

11

2

11

1111

1

StOVcS

VVb

S

VVVa

t

VV

ki

ki

ki

kik

i

ki

ki

kik

i

ki

ki


Implicit scheme

),(2

2

2111

11

11

2

11

1111

1

StOVcS

VVb

S

VVVa

t

VV

ki

ki

ki

kik

i

ki

ki

kik

i

ki

ki

11

11111

1 )1(

ki

ki

ki

ki

ki

ki

ki VCVBVAV


Implicit scheme

time

S

Values here are calculated

This value is used

11

11111

1 )1(

ki

ki

ki

ki

ki

ki

ki VCVBVAV


Crank-Nicolson scheme

),(2222

22

2

2

2

2

211

11

11

11

1

211

2

11

111

11

StOVc

Vc

S

VVb

S

VVb

S

VVVa

S

VVVa

t

VV

ki

kik

i

ki

ki

ki

ki

ki

ki

ki

ki

ki

ki

ki

ki

ki

ki

ki

ki

ki



ki

ki

ki

ki

ki

ki

ki

ki

ki

ki

ki

ki

VCVBVA

VCVBVA

11

11

11111

1

)1(

)1(



time

S

Values here are calculated

These values are used



0

0

0

10

10

000

1

01 10

11

11

12

12

11

11

11

12

12

12

12

11

11

kk

kI

kI

k

k

kI

kI

kI

kI

kk

kk VA

V

V

V

V

BA

CB

BA

CB

kkkR

kkL rvMvM 11

A general form of the linear equation is:

Note that M are tridiagonal!



kkkR

kL

k rvMMv 111

Theoretically this equation can be solved as

In practice this is inefficient!


LU decomposition

M is tridiagonal, thus M=LU, where L is

lower triangular, and U is upper triangular.

In fact L has 1 on the diagonal and one

subdiagonal only, U has a diagonal and one

superdiagonal.


LU decomposition

Then in order to solve Mv=q

orLUv=q

We will solve

Lw=q

first, and then

Uv=w.


LU decomposition

Very fast, especially when M is time

independent.

Disadvantages:

Needs a big modification for American options


Other methods

• SOR successive over relaxation

• Douglas scheme

• Three time-level scheme

• Alternating direction method

• Richardson Extrapolation

• Hopscotch method

• Multigrid methods


Multidimensional case

r

S

Fixed t layer

These values are usedto calculate space derivatives

Note additional boundary conditions.


Geometrical Brownian Motion

SdBrSdtdS

)()( SpayoffEeoption neutralrisktTr


Lognormal process

SdBrSdtdS

dBdtrSd

2)(log

2

tZtr N

eStS)1,0(

2

2)0()(


Euler Scheme

SdBrSdtdS

)()1,0( tOtSZtrSS N


Milstein Scheme

SdBrSdtdS

)()1(2

1 22)1,0(

22

)1,0(

tOtZS

tSZtrSS

N

N


Stochastic Calculus

Standard Normal

Diffusion process

x z

dzexN 2

2

2

1)(

tdBtXdttXdX ),(),(

ABM tdBdtdX

GBM tXdBXdtdX


Ito’s Lemma

tdBtXdttXdX ),(),(

22

2

),( dXX

FdX

X

Fdt

t

FtXdF

dt dXdt 0 0dX 0 dt


)1,0(

2

2)0()(

Ztt

eXtX

)ln(XY

22

)(11

0 dXX

dXX

dtdY

tdBdtdY

2

2

)1,0(

2

2)0()( ZttYtY

tXdBXdtdX


Siegel’s paradoxConsider two currencies X and Y. Define S an exchange rate (the number of units of currency Y for a unit of X).

The risk-neutral process for S is

tSXY SdBSdtrrdS )(

By Ito’s lemma the process for 1/S is

tSSXY dBS

dtS

rrS

d111 2


Siegel’s paradoxThe paradox is that the expected growth rate of 1/S is not ry - rX, but has a correction term.

cf-1 bank hapoalim jun-2001 zvi wiener 02-588-3049 mswiener/zvi.html computational finance

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