1
Pricing Asian Options:
the Geometric Average case
Solomon M. Antoniou
SKEMSYS Scientific Knowledge Engineering
and Management Systems
37 Κoliatsou Street, Corinthos 20100, Greece
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.
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.
3
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
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
VΔ
(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
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],
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)
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)
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)
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
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.
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[ :
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)
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)
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)
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
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
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)
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
1ξ
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
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
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)
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
1ξ
2
1ξ
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)
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
23
)t,Α(θ)SlnS()t,Α(θξ)Sln1(2
1Sr At3
0)t,Α(θSξ)SlnS(2
1ξ
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
4ξ
σ
r2σ
4
1)t,A(F t2t32
2
)t,Α(θσ
4ξ
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)
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:
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)
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.
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
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
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:
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
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)
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
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
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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