asian geometric

38
1 Pricing Asian Options: the Geometric Average case Solomon M. Antoniou SKEMSYS Scientific Knowledge Engineering and Management Systems 37 Κoliatsou Street, Corinthos 20100, Greece [email protected] Dedicated to Gelly Abstract We consider the valuation of path-dependent contingent claims where the underlying asset follows a geometric average process. Considering the no- arbitrage PDE of these claims, we first determine the underlying Lie point symmetries. After determination of the invariants, we transform the PDE to the Black-Scholes (BS) equation with time-dependent coefficients. We then transform the BS equation into the heat equation which is solved using Poisson’s formula taking into account the payoff condition. We thus obtain a closed-form solution for the pricing of asian options in the geometric average case. This procedure appears for the first time in the finance literature. Keywords: Path-Dependent Options, Asian Options, Geometric Average Path- Dependent Contingent Claims, Lie Symmetries, Exact Solutions.

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Page 1: Asian Geometric

1

Pricing Asian Options:

the Geometric Average case

Solomon M. Antoniou

SKEMSYS Scientific Knowledge Engineering

and Management Systems

37 Κoliatsou Street, Corinthos 20100, Greece

[email protected]

Dedicated to Gelly

Abstract

We consider the valuation of path-dependent contingent claims where the

underlying asset follows a geometric average process. Considering the no-

arbitrage PDE of these claims, we first determine the underlying Lie point

symmetries. After determination of the invariants, we transform the PDE to the

Black-Scholes (BS) equation with time-dependent coefficients. We then transform

the BS equation into the heat equation which is solved using Poisson’s formula

taking into account the payoff condition. We thus obtain a closed-form solution

for the pricing of asian options in the geometric average case. This procedure

appears for the first time in the finance literature.

Keywords: Path-Dependent Options, Asian Options, Geometric Average Path-

Dependent Contingent Claims, Lie Symmetries, Exact Solutions.

Page 2: Asian Geometric

2

1. Introduction

A path-dependent option is an option whose payoff depends on the past history of

the underlying asset. In other words these options have payoffs that do not depend

on the asset’s value at expiry. Two very common examples of path-dependent

options are Asian options and lookback options.

The terminal payoff of an Asian option depends on the type of averaging of the

underlying asset price over the whole period of the option’s lifetime. According to

the way of taking average, we distinguish two classes of Asian options: arithmetic

and geometric options.

A lookback option is another type of path-dependent option, whose payoff

depends on the maximum or minimum of the asset price during the lifetime of the

option.

The valuation of path-dependent (Asian) European options is a difficult problem

in mathematical finance. There are only some simple cases where the price of

path-dependent contingent claims can be obtained in closed-form (Angus [3],

Barucci et al. [8], Bergman [9], Geman and Yor [23], Kemma and Vorst [33],

Rubinstein [42], Wilmott [49], Zhang [51]). However in both cases (geometric and

arithmetic average) there are some numerical procedures available (Barucci et al.

[8], Haug [25]).

In the case of arithmetic average there is no closed-form solution. However in one

of our previous paper (Antoniou [6]) we have been able to convert the PDE (2.12)

into the BS equation using symmetry methods.

If the underlying asset price follows a lognormal stochastic process, its geometric

average has a lognormal probability density and in this case there is a closed-form

solution (Angus [3]).

In this paper we provide a rigorous justification of the transformation which

converts the equation (2.14) into the BS equation, using symmetry methods. On

the other hand, we provide our own method of solution to the time-dependent BS

equation.

Page 3: Asian Geometric

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The paper is organized as follows: In Section 2 we consider the general problem of

pricing the path-dependent contingent claims. In Section 3 we consider the Lie

point symmetries of the partial differential equation (equation (3.1)) for the path-

dependent geometric average options. Full details of the calculation are provided

in Appendix A. In Section 4, considering one invariant of the PDE, we convert the

equation into the time-dependent Black-Scholes equation. In section 5, we

transform the BS equation to the heat equation. In section 6 we find a closed-form

solution, using Poisson’s formula and the payoff condition.

Because of the complexity of the calculations, we have included a great deal of

calculation details, supplemented by three Appendices.

2. Path-Dependent Contingent Claims

Suppose that an option pays off at expiration time T an amount that is a function

of the path taken by the asset between time zero and T. This path-dependent

quantity can be represented by an integral of some function of the asset over the

time period Ττ0 :

T

0

τd)τ,S(f)Τ(Α (2.1)

where )t,S(f is a suitable function. Therefore, for every t, Τt0 , we have

t

0

τd)τ,S(f)t(Α (2.2)

or, in differential form,

dt)t,S(fdA (2.3)

We have thus introduced a new state variable A, which will later appear in our

PDE.

We are now going to derive the PDE of pricing of the path-dependent option. To

value the contract, we consider the function )t,A,S(V and set up a portfolio

Page 4: Asian Geometric

4

containing one of the path-dependent options and a number Δ of the underlying

asset:

SΔ)t,A,S(VΠ (2.4)

The change in the value of this portfolio is given by, according to Itô’s formula, by

dSΔS

VdA

A

Vdt

t

VSσ

2

1

t

VΠd

2

222

(2.5)

Choosing

S

(2.6)

to hedge the risk and using (2.3), we find that

dtA

V)t,S(f

t

VSσ

2

1

t

VΠd

2

222

(2.7)

This change is risk-free and thus earns the risk-free rate of interest r,

dtΠrΠd (2.8)

leading to the pricing equation

0VrS

VSr

A

V)t,S(f

t

VSσ

2

1

t

V

2

222

(2.9)

combining (2.7), (2.8) and (2.4), where Δ in (2.4) has been substituted by (2.6)

for the value of Δ.

The previous PDE is solved by imposing the condition

)A,S(Ω)T,A,S(V (2.10)

where T is the expiration time.

If A is considered to be an arithmetic average state variable,

t

0

τd)τ(SΑ (2.11)

then PDE (2.9) becomes

Page 5: Asian Geometric

5

0VrS

VSr

A

VS

t

VSσ

2

1

t

V

2

222

(2.12)

If A is considered to be a geometric average state variable,

t

0

τd)τ(SlnΑ (2.13)

then PDE (2.9) becomes

0VrS

VSr

A

V)S(ln

t

VSσ

2

1

t

V

2

222

(2.14)

We list below the payoff types we have to consider along with PDEs (2.12) and

(2.14).

For average strike call, we have the payoff )T,A,S(V

)0,AS(max (2.15)

For average strike put, we have the payoff

)0,SA(max (2.16)

For average rate call, we have the payoff

)0,EA(max (2.17)

For average rate put, we have the payoff

)0,AE(max (2.18)

where E is the strike price.

This is an introduction to the path-dependent options. Details can be found in

Wilmott’s treatise (Wilmott [49], Vol.2 and Vol.3) on Quantitative Finance.

3. Lie Symmetries of the Partial Differential Equation

The technique of Lie Symmetries was introduced by S. Lie (Lie [36] and [37]) and

is best described in Anderson [2], Bluman and Anco [11], Bluman and Cole [12],

Bluman and Kumei [13], Campbell [17], Cantwell [18], Cohen [19], Dickson [20],

Dresner [21], Emanuel [22], Hydon [27], Ibragimov [28] and [29], Ince [31],

Page 6: Asian Geometric

6

Olver [38], Ovsiannikov [39], Page [40], Schwarz [43], Steeb [46], Stephani [47]

and Wulfman [50]. This technique was for the first time used in partial

differential equations of Finance by Ibragimov and Gazizov [30]. The same

technique was also used by the author in solving the Bensoussan-Crouhy-Galai

equation (Antoniou [4]) , the HJB equation for a portfolio optimization problem

(Antoniou [5]) and the PDE reduction to the BS equation in the case of arithmetic

average asian options (Antoniou [6]).

There is however a growing lists of papers in applying this method to Finance. For

a non complete set of references, see Antoniou [7], Bordag [14], Bordag and

Chmakova [15], Bordag and Frey [16], Goard [24], Leach et al. [34], Pooe et al.

[41], Silberberg [42], Sinkala et al. [45].

Equation (2.12) of the arithmetic average case has been considered before

(Antoniou [6]) where it was reduced to the Black-Scholes equation using again

group-theoretic methods.

We now consider the PDE (2.14)

0VrS

VSr

A

V)S(ln

S

VS

2

1

t

V2

222

(3.1)

We shall determine the Lie Point Symmetries of the previous equation.

We follow closely Olver [37].

Let )V,t,A,S(Δ )2( be defined by

0VrS

VSr

A

V)S(ln

t

VSσ

2

1

t

V

2

222

(3.2)

We introduce the vector field X (the generator of the symmetries)

by

A)V,t,A,S(ξ

S)V,t,A,S(ξX 21

V

)V,t,A,S(φt

)V,t,A,S(ξ3

(3.3)

Page 7: Asian Geometric

7

We calculate in Appendix A the coefficients 321 ξ,ξ,ξ and φ and we find that

the Lie algebra of the infinitesimal transformations of the original equation (3.1)

is spanned by the eight vectors

t

X1

(3.4)

AA

2

3

tt

S)SlnS(

2

1X2

V

Vtrtσ8

)r2σ()S(ln

σ4

)r2σ(

2

22

2

2

(3.5)

tt

AAt3

S]AS3t)SlnS[(X 2

3

2

2

222

22

2

tσ8

)r2σ()S(ln

σ

2)S(lnt2

σ4

)r2σ(

VVA

σ2

)r2σ(3t2tr

2

22

(3.6)

AX4

(3.7)

At

SSX5

(3.8)

VVt

σ

)r2σ()S(ln

σ

2

At

SSt2X

2

2

2

26

(3.9)

At

SSt3X 32

7

VVA

σ

6t

σ2

)r2σ(3)S(lnt

σ

6

2

2

2

2

2

(3.10)

VVX8

(3.11)

Page 8: Asian Geometric

8

and the infinite dimensional sub-algebra

V

)t,A,S(βXβ

(3.12)

where )t,A,S(β is an arbitrary solution of the original PDE (3.1).

4. Reduction of the equation to the time-dependent

Black-Scholes equation

The invariant associated to the generator 5X , given by (3.8) is

SlntAy (4.1)

This transformation, in a slightly different form, was used by Angus [3]. In our

case however we derived this transformation rigorously, using group-theoretic

arguments.

Considering the substitution

)t,y(u)t,A,S(V (4.2)

we can transform equation (3.1) to the equation

0urutσ2

1utrσ

2

1u yy

22y

2t

(4.3)

Under the substitution

zlny (4.4)

we can transform (4.3) to the time-dependent BS equation

0uruzt2

1uzt

2

1tr

2

1u zz

222z

222t

(4.5)

Details are given in Appendix B.

5. Solution of the time-dependent Black-Scholes equation

The Black-Scholes equation

0V)t(rS

VS)t(r

S

VS)t(

2

1

t

V2

222

(5.1)

Page 9: Asian Geometric

9

with time-dependent coefficients, can, in complete analogy to the reasoning

developed in Appendix C, be converted into the heat equation. The reader is

advised to look at Appendix C before start reading this section.

Introducing the transformation

)S,t(Ue)S,t(V )t(f (5.1)

where

T

t

ds)s(r)t(f (5.2)

equation (5.1) takes the form

0S

US)t(r

S

US)t(

2

1

t

U2

222

(5.3)

We now consider the transformation

)t(

)t(Slnx (5.4)

where )t( and )t( are functions to be determined next.

Under this transformation, since

)t(x

U)t(

U

t

U

and

x

U

S

US

,

x

U

x

U

S

US

2

2

2

22

equation (5.3) transforms into

0x

U)t(r

x

U

x

U)t(

2

1)t(

x

U)t(

U2

22

which is equivalent to

0x

U)t(r)t(

2

1)t(

x

U)t(

2

1)t(

U 2

2

22

(5.5)

The choices

Page 10: Asian Geometric

10

)t(2

1)t( 2 and 0)t(r)t(

2

1)t( 2 (5.6)

convert equation (5.3) into the heat equation

2

2

x

UU

(5.7)

Therefore the required transformation, coming from (5.6), reads

T

t

2 ds)s(2

1)t( and

T

t

2 ds)s(2

1)s(r)t( (5.8)

where again T is the strike time.

Conclusion. The Black-Scholes equation

0V)t(rS

VS)t(r

S

VS)t(

2

1

t

V2

222

with time-dependent coefficients, under the transformations

)S,t(Ue)S,t(V )t(f , T

t

ds)s(r)t(f

and

)t(

)t(Slnx,

T

t

2 ds)s(2

1)t( ,

T

t

2 ds)s(2

1)s(r)t(

is converted into the heat equation 2

2

x

UU

.

The above argument of solving the time-dependent Black-Scholes equation

appears for the first time in literature. For another method see Wilmott [49] (Vol.

2, section ).

6. The Option Pricing Formula.

We shall now derive the option pricing formula for the geometric average case of

Asian Options. We shall follow the following steps:

Step 1. We consider equation (4.5). We shall transform this equation into the Heat

Equation, along the lines of the previous section.

Page 11: Asian Geometric

11

Under the transformation

)t,z(Ue)t,z(u )tT(r (6.1)

since

t)tT(r)tT(r

t UeUeru

z)tT(r

z Ueu

zz)tT(r

zz Ueu

equation (4.5) transforms into the equation

0Uzt2

1Uzt

2

1tr

2

1U zz

222z

222t

(6.2)

Under the new transformation (see equation (C.8), Appendix C)

)t(

)t(zlnx (6.3)

since

)t(U)t(UU xt

xz UUz

xxxzz2 UUUz

equation (6.2) takes on the form

0U)t(tr2

1Ut

2

1U)t( x

2xx

22

(6.4)

The choice

22 t

2

1)t( and tr

2

1)t( 2

(6.5)

converts equation (6.4) into the heat equation

xxUU (6.6)

Transformations (6.5) give us by integration in the interval ]T,t[ :

Page 12: Asian Geometric

12

)tT(6

1s

2

1)t( 332

T

t

22 (6.7)

and

)tT(r2

1

2

1dssr

2

1)t( 222

T

t

2

(6.8)

Step 2. We shall derive the option pricing formula under the boundary condition

K

T

Aexp)A,S(f (6.9)

From the formula (4.1), we obtain

SlnT

t

T

y

T

A

and then

T

Slntyexp

T

Aexp

Since zlny and )t(xzln , with )t( given by (6.8), we obtain

K

T

Slnt)t(xexp)A,S(f (6.10)

Using Poisson’s formula (see for example Tikhonov and Samarskii [48] or

Lebedev [35]) we have

dwe)0,w(U2

1),x(U 4

)wx( 2

(6.11)

where )0,x(U can be derived from (6.10) for 0 , i.e. Tt :

0,K

T

SlnTxexpmax)0,x(U (6.12)

Therefore

dwe0,KT

SlnTwexpmax

2

1),x(U 4

)wx( 2

(6.13)

Page 13: Asian Geometric

13

Under the substitution

2

xwv

i.e.

v2xw , dv2dw (6.14)

equation (6.13) takes on the form

dv2e0,KT

SlnTv2xexpmax

2

1),x(U

2v

(6.15)

We now have that

0KT

SlnTv2xexp

if

2

xS

KlnT

v (6.16)

We thus get from (6.15) and (6.16)

dveKT

SlnTv2xexp

1),x(U

2v

or

dveK1

dve1

),x(U2

2

vv

T

SlnTv2x

(6.17)

Using the identity

SlnTT

x

Tvv

T

SlnTv2x2

22

equation (6.17) becomes

dveK1

dveT

TxexpS

1),x(U

2

2

vTv

2

(6.18)

Page 14: Asian Geometric

14

The substitution

T

vs

converts equation (6.18) into the equation

dveK1

dseT

TxexpS

1),x(U

22 vs

2

(6.19)

where

T2

xS

KlnT

(6.20)

At this stage we consider the complementary error function (for the special

functions used in the text see for example Abramowitz and Stegun [1], or Lebedev

[35]) defined by

de2

)x(erfc

x

2

Therefore equation (6.19) gives us

)(erfcK2

1)(erfc

T

TxexpS

2

1),x(U

2

(6.21)

There is a known identity

)x2()x(erfc2

1 (6.22)

where )x( is the cumulative function of the normal distribution )1,0(N with

mean 0 and variance 1, given by (see for example Abramowitz and Stegun [1])

due2

1)x(

x2

u2

Equation (6.21) then gives

)2(K)2(T

TxexpS),x(U

2

(6.23)

Page 15: Asian Geometric

15

Let us now express everything in terms of the original variables.

Since )t(zlnx and SlntAyzln , we finally have

)t(SlntAx (6.24)

where )t( is given by (6.8).

We also have that )t( (see equations (6.3)) where )t( is given by (6.7).

Introducing the notation

2d1 and 2d2 (6.25)

we find

T

)t(2

)t(2

)t(S

KlnTSlntA

d1

(6.26)

)t(2

)t(S

KlnTSlntA

d2

(6.27)

Therefore (6.23) becomes

)d(K)d(TT

)t(SlntAexpS),x(U 212

(6.28)

Collecting everything together, we obtain the following option pricing formula for

the Asian Options, for the geometric average case

)d(K)d(

T

)t(

T

)t(SlntAexpSe)t,A,S(V 212

)tT(r (6.29)

Appendix A. Lie Symmetries of the Pricing Equation.

We consider the partial differential equation

0VrA

V)S(ln

S

VSr

S

VS

2

1

t

V2

222

The Lie Symmetries of the previous equation are to be determined next.

Step 1. Let )V,t,A,S(Δ )2( be defined by

Page 16: Asian Geometric

16

VrA

V)S(ln

S

VSr

S

VSσ

2

1

t

V)V,t,A,S(Δ

2

222)2(

Step 2. Introduce the vector field X (the generator of the symmetries) by

t)V,t,A,S(

A)V,t,A,S(

S)V,t,A,S(X 321

V

)V,t,A,S(φ

(A.1)

Step 3. The second prolongation is defined in our case by the equation

SS

SS

t

t

A

A

S

S)2(

VVVVXXpr

Step 4. The Lie Symmetries of the equation are determined by the condition

0)]V,t,A,S([Xpr )2()2( as long as 0V,t,A,S( )2( .

Implementation of the equation 0)]V,t,A,S([Xpr )2()2( :

We have

0)]V,t,A,S([Xpr )2()2(

0VrV)S(lnVSrVS2

1VXpr ASSS

22t

)2(

)r(φ]VS

1VrVSσ[ξ ASSS

21

0φSσ2

1φ)S(lnφ)Sr(φ SS22tAS

Step 5. Calculation of Sφ , Aφ , tφ and SSφ

tS32SV1SS1VS

S VξVξV)ξφ(φφ

ASV2tSV3AS2 VVξVVξVξ

tA32AV2AA2VA

A VξVξV)ξφ(φφ

tAV3SAV1SA1 VVξVVξVξ

Page 17: Asian Geometric

17

St12tV3tt3Vt

t VξVξV)ξφ(φφ

tSV1tAV2At2 VVξVVξVξ

2SSV1VVtSS3SSS1SVSS

SS V)ξ2φ(VξV)ξφ2(φφ

t2SVV3

3SVV1tSSV3 VVξVξVVξ2

SSSV1SAS2SSS1V VVξ3Vξ2V)ξ2φ(

SASV2StSV3SSAV2 VVξ2VVξ2VVξ

SStV3SASV2StS3 VVξVVξ2Vξ2

ASS2A2SVV2 VξVVξ

Step 6. Thus, using the previous expressions for Sφ , Aφ , tφ and SSφ from the

last equation of Step 4:

φr]VS

1VrVSσ[ξ ASSS

21

AS2tS3

2SV1SS1VS VξVξVξV)ξφ(φ{Sr

}VVξVVξ ASV2tSV3

tA32AV2AA2VA VξVξV)ξφ(φ{)S(ln

}VVξVVξVξ tAV3SAV1SA1

St12tV3tt3Vt VξVξV)ξφ(φ

tSV1tAV2At2 VVξVVξVξ

2SSV1VVtSS3SSS1SVSS

22 V)ξ2φ(VξV)ξφ2(φ{Sσ2

1

SSS1Vt2SVV3

3SVV1tSSV3 V)ξ2φ(VVξVξVVξ2

StSV3SSAV2SSSV1SAS2 VVξ2VVξVVξ3Vξ2

SStV3SASV2StS3SASV2 VVξVVξ2Vξ2VVξ2

}VξVVξ ASS2A2SVV2 (A.2)

Page 18: Asian Geometric

18

Step 7. We now have to take into account the condition 0V,t,A,S( )2( .

Substitute tu by VrV)S(lnVSrVSσ2

1ASSS

22 into (A.2):

φr]VS

1VrVSσ[ξ ASSS

21

}VVξVξVξV)ξφ(φ{Sr ASV2AS22SV1SS1VS

VrV)S(lnVSrVSσ2

1}Vξξ{Sr ASSS

22SV3S3

}VVξVξVξV)ξφ(φ{)S(ln SAV1SA12AV2AA2VA

VrV)S(lnVSrVSσ2

1}Vξξ{)S(ln ASSS

22AV3A3

VrV)S(lnVSrVSσ2

1)ξφ(φ ASSS

22t3Vt

At2St1

2

ASSS22

V3 VξVξVrV)S(lnVSrVSσ2

VrV)S(lnVSrVSσ2

1}VξVξ{ ASSS

22SV1AV2

}V)ξ2φ(V)ξφ2(φ{Sσ2

1 2SSV1VVSSS1SVSS

22

}VξVξ2ξ{Sσ2

1 2SVV3SSV3SS3

22

VrV)S(lnVSrVSσ2

1ASSS

22

SAS2SSS1V3SVV1

22 Vξ2V)ξ2φ(Vξ{Sσ2

1

StSV3SSAV2SSSV1 VVξ2VVξVVξ3

}VVξ2Vξ2VVξ2 SASV2StS3SASV2

VrV)S(lnVSrVSσ2

1}Vξ{Sσ

2

1ASSS

22SSV3

22

Page 19: Asian Geometric

19

0}VξVVξ{Sσ2

1ASS2A

2SVV2

22 (A.3)

Step 8. Equate all the coefficients of the partial derivatives of the function V to

zero. We are then going to have a system of partial differential equations. Before

that, and just for the economy of the calculations, we observe that we get some

simple expressions looking at the coefficients of the mixed partial derivatives.

In fact the coefficient of the derivative SAV is )ξ2(Sσ2

1S2

22 and that of

SAS VV is )ξ2(Sσ2

1V2

22 which means that 0ξ S2 and 0ξ V2 . Therefore

the coefficient 2ξ does not depend either on S or V:

)t,A(ξξ 22 (A.4)

The coefficient of the derivative StS VV is )ξ2(Sσ2

1V3

22 and that of StV is

)ξ2(Sσ2

1S3

22 which means that 0ξ V3 and 0ξ S3 . Therefore the

coefficient 3ξ does not depend either on V or S:

)t,A(ξξ 33 (A.5)

Because of (A.4) and (A.5), equation (A.3) becomes

φr]VS

1VrVSσ[ξ ASSS

21

}VξV)ξφ(φ{Sr 2SV1SS1VS

}VVξVξV)ξφ(φ{)S(ln SAV1SA1AA2VA

VrV)S(lnVSrVSσ2

1)ξ()S(ln ASSS

22A3

VrV)S(lnVSrVSσ2

1)ξφ(φ ASSS

22t3Vt

VrV)S(lnVSrVSσ2

1)Vξ(VξVξ ASSS

22SV1At2St1

Page 20: Asian Geometric

20

}V)ξ2φ(V)ξφ2(φ{Sσ2

1 2SSV1VVSSS1SVSS

22

0}VVξ3V)ξ2φ(Vξ{Sσ2

1SSSV1SSS1V

3SVV1

22 (A.6)

The coefficient of SSSVV is the sum of the terms

22

V1 S2

1)( and )ξ3(Sσ

2

1V1

22

and therefore 0ξ V1 , which means that

)t,A,S(ξξ 11 (A.7)

Therefore (A.6) becomes

φr]VS

1VrVSσ[ξ ASSS

21

}VξV)ξφ(φ{)S(ln}V)ξφ(φ{Sr SA1AA2VASS1VS

VrV)S(lnVSrVSσ2

1)ξ()S(ln ASSS

22A3

VrV)S(lnVSrVSσ2

1)ξφ(φ ASSS

22t3Vt

}VφV)ξφ2(φ{Sσ2

1VξVξ 2

SVVSSS1SVSS22

At2St1

0V)ξ2φ(Sσ2

1SSS1V

22 (A.8)

From (A.8) we get the following coefficients (which are equated to zero):

Coefficient of 2SV :

0φSσ2

1VV

22 (A.9)

Coefficient of AV :

0ξ)ξξ()S(lnξ)S(lnξS

1t2Α2t3Α3

21 (A.10)

Coefficient of SSV :

0)ξ()S(lnSσ2

1)ξ2ξ(Sσ

2

1ξSσ A3

22S1t3

221

2 (A.11)

Page 21: Asian Geometric

21

Coefficient of SV :

A3A1S1t31 ξ)S(lnSrξ)S(ln)ξξ(Srξr

0ξ)ξφ2(Sσ2

1t1SS1SV

22 (A.12)

Zero-th order coefficient

tΑ3AS φV)ξ()S(lnrφ)S(lnφSrφr

0φSσ2

1V)ξφ(r SS

22t3V (A.13)

Step 9. We have now to solve the system of PDEs (A.9)-(A.13).

From (A.9) we get 0φVV and therefore φ is a linear function with respect to

V:

)t,A,S(βV)t,A,S(αφ (A.14)

From (A.11) we get

)ξ()S(ln2

2

S

1ξ A3t31S1

which is a linear differential equation with unknown function 1ξ .

The solution of the previous differential equation is

)t,Α(θS)ξ()S(lnS4

1ξ)SlnS(

2

1ξ A3

2t31 (A.15)

where )t,Α(θ is a function to be determined.

From (A.10), using (A.15), we get

0)t,Α(θξ)S(lnξξ2

3)S(lnξ

4

5t2A2t3

2A3

(A.16)

Since the previous equation should hold for any value of S, we get from the

previous equation the following three equations:

0ξ A3 (A.17)

0ξξ2

3A2t3 (A.18)

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22

0)t,Α(θξ t2 (A.19)

Equation (A.17) expresses the fact that the function 0ξ3 does not depend on

the variable A. Therefore, using also (A.5), we have that

)t(fξ3 (A.20)

Using (A.18) and (A.20) we get

)t(f2

3ξ A2

and therefore

)t(gA)t(f2

3ξ2 (A.21)

From (A.19), using (A.21) we get

)t(gA)t(f2

3)t,Α(θ (A.22)

Because of (A.17), we get another expression for the function 1ξ :

)t,Α(θSξ)SlnS(2

1ξ t31 (A.23)

From (A.23) we also get the following expressions for the partial derivatives of the

function 1ξ to be used later on.

)t,Α(θξ)Sln1(2

1ξ t3S1 (A.24a)

t3SS1 ξS

1

2

1ξ (A.24b)

)t,Α(θSξ)SlnS(2

1ξ ttt3t1 (A.24c)

)t,Α(θSξ AA1 (A.24d)

Equation (A.12), using the expressions (A.14), (A.23) and (A.24), takes the form

t3t3 ξSr)t,Α(θSξ)SlnS(2

1r

Page 23: Asian Geometric

23

)t,Α(θ)SlnS()t,Α(θξ)Sln1(2

1Sr At3

0)t,Α(θSξ)SlnS(2

S

1

2

1α2Sσ

2

1ttt3t3S

22

Solving the previous equation with respect to Sα , we find

S

Sln)t,A(G

S

1)t,A(FαS (A.25)

where we have put

)t,Α(θ

σ

σ

r2σ

4

1)t,A(F t2t32

2

)t,Α(θσ

σ

2

4

1)t,A(G Α2tt32

Integrating (A.25) with respect to S, we find that

)t,Α(ρ)S)(lnt,A(G2

1)S)(lnt,A(F)t,A,S(α 2 (A.26)

We also find from (A.25) that

)Sln1)(t,A(G)t,A(FαS SS2 (A.27)

an expression we shall need later on.

From (A.26) we also get

)t,Α(ρ)S)(lnt,A(G2

1)S)(lnt,A(Fα t

2ttt (A.28)

and

)t,Α(ρ)S)(lnt,A(G2

1)S)(lnt,A(Fα A

2AAA (A.29)

From (A.13), using (A.14) we obtain the equation

V)ξ()S(lnr)βVα()S(ln)βVα(Sr)βVα(r Α3ΑASS

0)βVα(Sσ2

1V)ξα(r)βVα( SSSS

22t3tt (A.30)

Page 24: Asian Geometric

24

Equating the coefficients of V to zero, we get from the previous equation the

following two equations:

)(r)()S(lnr)S(lnSrr t3t3AS

0S2

1SS

22 (A.31)

and

0βSσ2

1ββ)S(lnβSrβr SS

22tΑS (A.32)

Equation (A.32) expresses the fact that the function β introduced in (A.14) is a

solution of the original PDE.

We consider now equation (A.31). Using (A.26)-(A.29), this equation becomes

)}S)(lnt,A(G)t,A(F{r

)t,Α(ρ)S)(lnt,A(G2

1)S)(lnt,A(F)S(ln A

2AA

)t,Α(ρ)S)(lnt,A(G2

1)S)(lnt,A(Fξr t

2ttt3

0)}Sln1)(t,A(G)t,A(F{σ2

1 2

The previous equation can be expressed as

)t,Α(Gσ2

1)t,A(ρξr)t,Α(F)r2σ(

2

1 2tt3

2

)S(ln)t,Α(F)t,A(ρ)t,Α(G)r2σ(2

1tA

2

2tA )S(ln)t,Α(G

2

1)t,A(F

0)S(ln)t,Α(G2

1 3A (A.33)

Since this equation should hold for any value of S, we obtain the following three

equations:

Page 25: Asian Geometric

25

0)t,Α(Gσ2

1)t,A(ρξr)t,Α(F)r2σ(

2

1 2tt3

2 (A.34)

0)t,Α(F)t,A(ρ)t,Α(G)r2σ(2

1tA

2 (A.35)

0)t,Α(G2

1)t,A(F tA (A.36)

0)t,Α(GA (A.37)

Before proceeding to the solution of the above system of equations, it is instructive

to find some equivalent expressions for the functions )t,A(F and )t,A(G :

In fact, using (A.20) and (A.22), we find that

)t(gA)t(f

2

3

σ

4)t(f

σ

r2σ

4

1)t,A(F

22

2

(A.38)

and

)t(fσ

2)t,A(G

2 (A.39)

Equation (A.39) is compatible with (A.37). Equation (A.36), because of (A.38)

and (A.39), is equivalent to

0)t(fσ

2

2

1)t(f

2

3

σ

4

4

1

22

and therefore 0)t(f , which means that )t(f is a second degree polynomial

with respect to t:

2

321 tataa)t(f (A.40)

and using (A.20),

2

3213 tataaξ (A.41)

Because of (A.40), the functions )t,A(F and )t,A(G become

)t(gσ

1)ta2a(

σ4

r2σ)t,A(F

2322

2

(A.42)

Page 26: Asian Geometric

26

and

32a

σ

4)t,A(G (A.43)

respectively.

Equation (A.35), because of (A.42) and (A.43), becomes

)t(gσ

1a

σ

r2σ

2

3)t,A(ρ

232

2

A

(A.44)

Similarly, equation (A.34) gives us

)t(gσ

1)ta2a(

σ4

r2σ)r2σ(

2

1)t,A(ρ

2322

22

t

332 a2)ta2a(r (A.45)

Equations (A.44) and (A.45) is a system of DEs with unknown functions )t(g and

)t,A(ρ . The compatibility condition

)t,A(ρ)t,A(ρ tAAt

taking into account (A.44) and (A.45), is equivalent to

0)t(gσ

1 )iv(

2

from which there follows that )t(g is a third degree polynomial with respect to t:

3

72

654 tatataa)t(g (A.46)

Therefore (A.44) and (A.45) become

}a4a)r2σ{(σ2

3)t,A(ρ 73

2

2A (A.47)

and

)ta6a2(σ

1)ta2a(

σ4

r2σ)r2σ(

2

1)t,A(ρ 762322

22

t

332 a2)ta2a(r (A.48)

respectively.

Page 27: Asian Geometric

27

Integrating them with respect to A and t respectively, we obtain

)t(hA}a4a)r2σ{(σ2

3)t,A(ρ 73

2

2

)ta3ta2(σ2

r2σ)tata(

σ8

)r2σ()t,A(ρ 2

762

22

322

22

)A(kta2)tata(r 32

32

Comparing the two previous expressions we get the function )t,A(ρ :

)ta3ta2(σ2

r2σ)tata(

σ8

)r2σ()t,A(ρ 2

762

22

322

22

ta2)tata(r 32

32

8732

2aA}a4a)r2σ{(

σ2

3 (A.49)

Using (A.46), the function )t,A(F becomes

)ta3a(σ

2)ta2a(

σ4

r2σ)t,A(F 762322

2

(A.50)

From (A.21) we get

37

2654322 tatataaA)ta2a(

2

3ξ (A.51)

From (A.22) we get the function )t,Α(θ :

2

7653 ta3ta2aAa3)t,Α(θ (A.52)

From (A.23), using (A.41) and (A.52) we find

)ta3ta2aAa3(S)ta2a()SlnS(2

1ξ 2

7653321 (A.53)

From (A.26), using (A.50), (A.43) and (A.49) we find

)S(ln)ta3a(

σ

2)ta2a(

σ4

r2σ)t,A,S(α 762322

2

Page 28: Asian Geometric

28

)tata(σ8

)r2σ()S(lna

σ

2 2322

222

32

ta2)tata(r)ta3ta2(

2

r23

232

2762

2

8732

2aA}a4a)r2σ{(

σ2

3 (A.54)

Therefore we can now determine the function )V,t,A,S(φ introduced in (A.14):

)S(ln)ta3a(

σ

2)ta2a(

σ4

r2σ)V,t,A,S(φ 762322

2

)tata(σ8

)r2σ()S(lna

σ

2 2322

222

32

ta2)tata(r)ta3ta2(σ2

r2σ3

232

2762

2

)t,A,S(βVaA}a4a)r2σ{(σ2

3873

2

2

(A.55)

The vector X, the generator of the symmetries, can be expressed as

S

)ta3ta2aAa3(S)ta2a()SlnS(2

1X 2

765332

A

tatataaA)ta2a(2

3 37

265432

t)tataa( 2

321

)S(ln)ta3a(

σ

2)ta2a(

σ4

r2σ762322

2

)tata(σ8

)r2σ()S(lna

σ

2 2322

222

32

Page 29: Asian Geometric

29

ta2)tata(r)ta3ta2(σ2

r2σ3

232

2762

2

V

)t,A,S(βVaA}a4a)r2σ{(σ2

3873

2

2

The previous expression can thus be written as

taX 1

AA

2

3

tt

S)SlnS(

2

1a2

VVtrt

8

)r2()S(ln

4

)r2(2

22

2

2

tt

AAt3

S]AS3t)SlnS[(a 2

3

2

2

222

22

2

t8

)r2()S(ln

2)S(lnt2

4

)r2(

VVA

2

)r2(3t2tr

2

22

At

SSa

Aa 54

VVt

σ

)r2σ()S(ln

σ

2

At

SSt2a

2

2

2

26

VVA

σ

6t

σ2

)r2σ(3)S(lnt

σ

6

At

SSt3a

2

2

2

2

2

327

V)t,A,S(β

VVa8

Therefore, the generators of the symmetries are:

Page 30: Asian Geometric

30

tX1

AA

2

3

tt

S)SlnS(

2

1X2

V

Vtrtσ8

)r2σ()S(ln

σ4

)r2σ(

2

22

2

2

tt

AAt3

S]AS3t)SlnS[(X 2

3

2

2

222

22

2

tσ8

)r2σ()S(ln

σ

2)S(lnt2

σ4

)r2σ(

V

VAσ2

)r2σ(3t2tr

2

22

AX4

At

SSX5

VVt

σ

)r2σ()S(ln

σ

2

At

SSt2X

2

2

2

26

At

SSt3X 32

7

VVA

σ

6t

σ2

)r2σ(3)S(lnt

σ

6

2

2

2

2

2

VVX8

V)t,A,S(βXβ

Page 31: Asian Geometric

31

Note: The generators listed above can also provide some invariants which in turn

can be used to determine general solutions to the nonlinear equation (3.1).

Therefore the Lie symmetry analysis has a value on its own.

Appendix B. Reduction of equation (3.1) to the

time-dependent BS-equation.

We consider the generator 5X , given by (3.8)

A

tS

SX5

In order to find the invariant corresponding to this generator, we have to solve the

equation

t

dA

S

dS

The general solution of the above equation is given by

1CASlnt

Therefore the invariant associated to the generator 5X , is

SlntAy (B.1)

Considering the substitution

)t,y(u)t,A,S(V (B.2)

we can transform the partial derivatives:

ytt u)S(lnuV

tVS S

yyy2

SS2 ututVS

Using the previous substitutions, equation (3.1) is transformed to

0urutσ2

1utrσ

2

1u yy

22y

2t

(B.3)

Under the substitution

zlny (B.4)

Page 32: Asian Geometric

32

we find for the partial derivatives

zy uzu and zzz2

yy uzuzu

Therefore equation (B.3) takes the form

0uruztσ2

1uztσ

2

1trσ

2

1u zz

222z

222t

(B.5)

which is a Black-Scholes equation (Black and Scholes [10], Wilmott [49]) with

time-dependent coefficients.

Appendix C.

The Lie Algebra of the infinitesimal transformations of the constant-coefficient

Black-Scholes equation

0FrS

FSr

S

FS

2

1

t

F2

222

(C.1)

contains the two generators

F

F

and

t

(C.2)

of its underlying symmetries.

We consider the linear combination of the above generators ( is a constant to be

determined)

tF

F

(C.3)

and try to determine the corresponding invariant to the (C.3). Solving the equation

dtF

dF (C.4)

we find as the general solution CtFln , from which we determine one

invariant G:

)S,t(Ge)S,t(F t (C.5)

We then substitute (C.5) into (C.1) and we obtain the equation

Page 33: Asian Geometric

33

0G)r(S

GSr

S

GS

2

1

t

G2

222

The choice r removes the last term of this equation, leading to

0S

GSr

S

GS

2

1

t

G2

222

(C.6)

which is a simplified version of the original equation (C.1).

Therefore the transformation

)S,t(Ge)S,t(F )tT(r (C.7)

converts equation (C.1) into (C.7), where T is the strike time.

We can further transform equation (C.6) into the Heat Equation. The crucial point

is that equation (C.6) contains S

S

as generator of its infinitesimal Lie

symmetries. Therefore we introduce the transformation

)t(b

)t(aSln (C.8)

where )t(a and )t(b are functions to be determined next.

Under this transformation, since

)t(aG

)t(bG

t

G

and

G

S

GS ,

GG

S

GS

2

2

2

22

equation (C.6) transforms into

0G

rGG

2

1)t(a

G)t(b

G2

22

which is equivalent to

Page 34: Asian Geometric

34

0G

r2

1)t(a

G

2

1)t(b

G 2

2

22

(C.9)

The choices

2

2

1)t(b and 0r

2

1)t(a 2 (C.10)

convert equation (C.9) into the heat equation

2

2 GG

(C.11)

Therefore the required transformation, coming from (C.10), reads

)tT(2

1)t(a 2 and )tT(

2

1r)t(b 2

(C.12)

where again T is the strike time.

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