# The Arithmetic Average case of Asian Options. Reducing the PDE to the BS equation

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```					The Arithmetic Average Case of Asian Options: Reducing the PDE to the Black-Scholes Equation
Solomon M. Antoniou

SKEMSYS
Scientific Knowledge Engineering and Management Systems
37 oliatsou Street, Corinthos 20100, Greece
solomon_antoniou@yahoo.com

Abstract
We consider the path-dependent contingent claims where the underlying asset follows an arithmetic 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. We then transform the BS equation into the heat equation and we provide some general solutions to that equation. This procedure appears for the first time in the finance literature. Keywords: Path-Dependent Options, Asian Options, Arithmetic Average PathDependent Contingent Claims, Lie Symmetries, Exact Solutions.

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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 (refs. [1]-[8]). If the underlying asset price follows a lognormal stochastic process, then its geometric average has a lognormal probability density and in this case there is a closed-form solution (ref. [5]). This is not however the case by considering arithmetic average. In this case there is no closed-form solution. However in both cases there are some numerical procedures available (ref. [8]). It is the purpose of this article to provide a closed-form general solution for the arithmetic path-dependent options. 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 pathdependent arithmetic average options. Full details of the calculation are provided in Appendix A. In Section 4, considering two invariants of the PDE, we convert

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the equation into the Black-Scholes equation. In section 5, we transform the BS equation to the heat equation. We then find the general solution of our PDE (Theorem 3). In section 6 we provide some solutions to the heat equation. Because of the complexity of the calculations, we have included a great deal of calculation details, supplemented by many 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 0 ≤ ≤ :
T 0

( ) = ∫ f (S, ) d where f (S, t ) is a suitable function. Therefore, for every t, 0 ≤ t ≤ , we have ( t ) = ∫ f (S, ) d
0 t

(2.1)

(2.2)

or, in differential form, dA = f (S, t ) dt (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 V(S, A, t ) and set up a portfolio containing one of the path-dependent options and a number asset: = V(S, A, t ) − ⋅ S (2.4) of the underlying

The change in the value of this portfolio is given by, according to Itô’s formula, by  ∂ V 1 2 2 ∂ 2V   dt + ∂ V dA +  ∂ V −  + d = S  ∂S 2  ∂t 2 ∂A ∂S      dS   (2.5)

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Choosing = ∂V ∂S (2.6)

to hedge the risk and using (2.3), we find that  ∂ V 1 2 2 ∂ 2V ∂V  dt + + f (S, t ) d = S 2  ∂t 2 ∂A ∂S   This change is risk-free and thus earns the risk-free rate of interest r, d = r dt leading to the pricing equation ∂ V 1 2 2 ∂ 2V ∂V ∂V + + f (S, t ) + rS −rV = 0 S 2 ∂t 2 ∂A ∂S ∂S combining (2.7), (2.8) and (2.4), where for the value of . (2.9) (2.8) (2.7)

in (2.4) has been substituted by (2.6)

The previous PDE is solved by imposing the condition V(S, A, T ) = (S, A) where T is the expiration time. If A is considered to be an arithmetic average state variable, = ∫ S( ) d
0 t

(2.10)

(2.11)

then PDE (2.9) becomes ∂ V 1 2 2 ∂ 2V ∂V ∂V + +S + rS −rV = 0 S ∂t 2 ∂A ∂S ∂ S2 If A is considered to be a geometric average state variable, = ∫ ln S( ) d
0 t

(2.12)

(2.13)

then PDE (2.9) becomes

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∂ V 1 2 2 ∂ 2V ∂V ∂V + + (ln S) + rS −rV = 0 S ∂t 2 ∂A ∂S ∂ S2

(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 max (S − A, 0) For average strike put, we have the payoff max ( A − S, 0) For average rate call, we have the payoff max (A − E, 0) For average rate put, we have the payoff max (E − A, 0) where E is the strike price. This is an introduction to the path-dependent options. Details can be found in Wilmott’s 3-volume set on Quantitative Finance (ref. [3]). (2.18) (2.17) (2.16) (S, A) (2.15)

3. Lie Symmetries of the Partial Differential Equation
The technique of Lie Symmetries was introduced by S. Lie [11]-[12] and is best described in refs [13]-[28]. This technique was for the first time used in partial differential equations of Finance by Ibragimov and Gazizov [29]. The same technique was also used by the author in solving the Bensoussan-Crouhy-Galai equation [30] and the HJB equation for a portfolio selection problem [31]. There is however a growing lists of papers in applying this method to Finance. For a non complete set of references, see [29]-[40]. We consider the PDE (2.12) ∂ V 1 2 2 ∂ 2V ∂V ∂V + + rS +S −rV = 0 S ∂t 2 ∂S ∂A ∂ S2 (3.1)

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We shall determine the Lie Point Symmetries of the previous equation. We follow closely Olver (ref. [15]). Let (S, A, t , V ( 2) ) be defined by (S, A, t , V
( 2)

∂ V 1 2 2 ∂ 2V ∂V ∂V + + rS +S −rV )= S ∂t 2 ∂S ∂A ∂ S2 ∂ ∂ + 2 (S, A, t , V) + ∂S ∂A ∂ ∂ + (S, A, t , V) ∂t ∂V

(3.2)

We introduce the vector field X (the generator of the symmetries) by X = 1 (S, A, t , V)

+ 3 (S, A, t , V)

(3.3) and we find that

We calculate in Appendix A the coefficients 1 , 2 , 3 and

the Lie algebra of the infinitesimal transformations of the original equation (3.1) is spanned by the five vectors X1 = X2 = ∂ ∂t ∂ ∂A ∂ ∂ +A ∂S ∂A ∂ ∂ ∂ 2 + A2 + ( S − r A)V 2 ∂S ∂A ∂V (3.4)

(3.5)

X3 = S

(3.6)

X 4 = 2AS X5 = V

(3.7)

∂ ∂V

(3.8)

and the infinite dimensional sub-algebra ∂ ∂V where (S, A, t ) is an arbitrary solution of the original PDE (3.1). X = (S, A, t ) (3.9)

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4. Reduction of the equation to the BS equation
We shall now try to transform the pricing differential equation into another by reducing the number of independent variables. We shall retain the time variable and try to find a combination of the other two variables S and A. The best candidate for this job is the generator X 4 given by (3.7). We state and prove the following Theorem: Theorem 1. Any solution of the partial differential equation ∂ V 1 2 2 ∂ 2V ∂V ∂V + + rS +S −rV = 0 S ∂t 2 ∂S ∂A ∂ S2 can be expressed into the form  S V(S, A, t ) = A n u ( x , t ) × exp  b   A where the function u ( x , t ) satisfies the Black-Scholes equation ut + with x = S A2 1 2 2 x u xx + r x u x − r u = 0 2 2
2

(4.1)

(4.2)

, n = − r b and b =

.

Proof. Considering the generator X 4 , we can determine two invariants of the PDE. For this purpose we have to solve the following system
2 dS dA dV = = 2AS A 2 2 ( S − r A)V

(4.3)

The differential equation integrated to give S A
2

dS dA dS dA is equivalent to , which can be =2 = S A 2AS A 2

= C1 from which we can obtain our first invariant x: S A
2

x=

(4.4)

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The second differential equation dA A2 = dV 2 ( S − r A)V
2

can be written as (since S = x A 2 ) b or r dV  b  x −  dA = A V  where b = 2
2

x A2 − r A A2

dA =

dV , V

. By integration of this equation we find b ( x A − r ln A) + C 2 = ln V

from which there follows that V = C A − r b e b x A , where C is a constant. Therefore another invariant of the PDE is the function u ( x , t ) , related to V(S, A, t ) by  S V(S, A, t ) = A n u ( x , t ) exp  b   A with n = − r b = − 2r
2

(4.5)

.

Upon substitution of V(S, A, t ) given by (4.5) into equation (3.1), we find that the function u ( x , t ) satisfies the BS equation ut + 1 2 2 x u xx + r x u x − r u = 0 2 (4.6)

The proof is in Appendix B.

5. Transformation of the BS equation to the heat equation.
Under the substitution

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z = ln x and taking into account that x 2 u xx = u zz − u z x ux = uz the BS equation (4.6) takes the form ut + where =r− 1 2 2 1 2 u zz + u z − r u = 0 2

(5.1)

(5.2)

(5.3)

Under the substitution = −t (5.4)

where T is the expiry time, equation (5.2) becomes u = 1 2 u zz + u z − r u 2 (5.5)

At this stage we use the following Lemma. The equation u t = a 2 u xx + b u x + c u under the substitution = e − c t u(t, x = y − b t ) can be converted into the heat equation
t =a 2 yy

(5.6)

(5.7)

(5.8)

Proof. See Appendix C. Using the above Lemma, equation (5.5) under the substitution u ( , z) = e − r y=z+ ( , y) (5.9) (5.10)

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is transformed into the equation = 1 2 yy 2 by (5.12) (5.11)

Introducing further a new independent variable y= 2

equation (5.11) is transformed to the heat equation w =w where w = w ( , ) = ( , y) We thus arrive at the following Theorem 2. The solution to the BS equation (4.6) is given by u (x, t ) = e − r w ( , ) where = = 2 y= 2 (ln x + ) (5.16) (5.17) (5.15) (5.14) (5.13)

−t

and w ( , ) satisfies the heat equation w = w . Using Theorems 1 and 2, we get the following Theorem 3. The differential equation (3.1) has the general solution  S  V(S, A, t ) = A − m w ( , ) × exp  b − r   A  where the function w ( , ) satisfies the heat equation w = w . The variables m = rb, b = 2
2

(5.18)

and ,

are defined by (5.16) and (5.17) respectively, while is given by (5.3) and x by (4.4).Therefore equation (5.18)

can be written in full notation as

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1 V(S, A, t ) =   A where =

2r
2

 2 S w ( , ) exp   2 A −r 

   

2   S   1 2    ln 2  +  r −    A   2  

What we need now is some solutions to the heat equation, which are not singular at = 0 ( t = T ). We come to this point next.

6. General solutions of the heat equation
The Lie algebra of the infinitesimal symmetries of the heat equation w = w produced by the six vector fields X1 = X2 = ∂ ∂ ∂ ∂ ∂ ∂w X4 = X5 = 2 X6 = 4 ∂ ∂ +2 ∂ ∂ ∂ ∂ − w ∂ ∂w ∂ ∂ ∂ +4 2 −( 2 + 2 )w ∂ ∂ ∂w is

X3 = w

and the infinite dimensional sub-algebra X = ( , )∂w where ( , ) is an arbitrary solution of the heat equation (refs. [15], [37]).

6.1. The solution that is invariant with respect to the generator X 2 + X6 + a X3 = 4 ∂ + ( 4 2 + 1) ∂ − ( 2 + 2 − a ) w ∂ w

(a is a constant) is given by ([15], [37])  a    2   ,  + K ⋅ W  a , − 2  × w( , ) = K1 ⋅ W  2 4 2  2 4 2 +1  2  4 2 + 1  4 +1      1
2  a × exp  − + arctan(2  4 2 +1 2 

 )  

(6.1)

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where W (a , z) is the parabolic cylindrical function of imaginary argument (Appendix D). Note. The solution given by Olver [15] (Example 3.17, p.208) contains some misprints. We have repeated the calculations [28] and we have found the solution given by (6.1). 6.2. The solution that is invariant with respect to the generator X 2 − X5 = ∂ − 2 ∂ + w ∂ w is given by ([15], [37])  w ( , ) = { K1 ⋅ Ai ( + 2 ) + K 2 ⋅ Bi ( + 2 ) } exp    + 2 3  3   =0 (6.2)

where Ai( x ) and Bi( x ) are the Airy functions (Appendix E). Both solutions (6.1) and (6.2) we have considered, are not singular for (2.18) valid for t = T ( = 0 ). ( t = T ). This is because we want to take into account the payoff conditions (2.15)-

Appendix A. Lie Symmetry Analysis of the pricing PDE.
We consider the PDE (3.1) : ∂ V 1 2 2 ∂ 2V ∂V ∂V + + rS +S −rV = 0 S ∂t 2 ∂S ∂A ∂ S2 We shall determine the Lie Point Symmetries of the previous equation. We follow closely Olver (ref. [15]). Let (S, A, t , V ( 2) ) be defined by (S, A, t , V
( 2)

(A.1)

∂ V 1 2 2 ∂ 2V ∂V ∂V + + rS +S −rV )= S ∂t 2 ∂S ∂A ∂ S2

(A.2)

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

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X = 1 (S, A, t , V)

∂ ∂ ∂ + 2 (S, A, t , V) + 3 (S, A, t , V) + ∂S ∂A ∂t ∂ ∂V (A.3)

+ (S, A, t , V)

The second prolongation is defined in our case by the equation ∂ ∂ ∂ ∂ + A + t + SS pr ( 2) X = X + S ∂ VS ∂ VA ∂ Vt ∂ VSS The Lie point symmetries of the equation are determined by pr ( 2) X [ (S, A, t , V ( 2) )] = 0 as long as (S, A, t , V ( 2) ) = 0 . (A.4)

We implement next the equation pr ( 2) X [ (S, A, t , V ( 2) )] = 0 .We have pr ( 2) X [ (S, A, t , V ( 2) )] = 0 1 2 2   ⇔ pr ( 2) X Vt + S VSS + r S VS + S VA − r V  = 0 2   ⇔ 1[ 2 S VSS + r VS + VA ] + (− r ) + 1 2 2 SS + S (r S) + A (S) + t + =0 S 2 We have the following expressions for the coefficients S , and SS
S 2 = S + ( V − 1S ) VS − 1V VS − 3S Vt − A

(A.5) ,
t

− 2S VA − 3V VS Vt − 2V VS VA
A 2 = A + ( V − 2A ) VA − 2V VA − 3A Vt −

− 1A VS − 1V VA VS − 3V VA Vt
t

= t + ( V − 3t ) Vt − 3V Vt2 − 1t VS − − 2 t VA − 2V VA Vt − 1V VS Vt

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SS

2 = SS + (2 SV − 1SS ) VS − 3SS Vt + ( VV − 2 1SV ) VS − 3 2 − 2 3SV VS Vt − 1VV VS − 3VV VS Vt + ( V − 2 1S ) VSS −

− 2 2S VSA − 3 1V VS VSS − 2V VA VSS − 2 3V VS VSt − − 2 2SV VA VS − 2 3S VSt − 2 2V VS VSA − 3V Vt VSS −
2 − 2VV VS VA − 2SS VA

Using the previous expressions for S , the equation (A.5) that
1[ 2

A

,

t

and SS we get from

S VSS + r VS + VA ] − r +

2 + r S{ S + ( V − 1S ) VS − 1V VS − 3S Vt − 2S VA −

− 3V VS Vt − 2V VS VA } +
2 + S{ A + ( V − 2A ) VA − 2V VA − 3A Vt − 1A VS −

− 1V VA VS − 3V VA Vt } + + t + ( V − 3t ) Vt − 3V Vt2 − 1t VS − 2 t VA − − 2V VA Vt − 1V VS Vt + + 1 2 2 2 S { SS + (2 SV − 1SS ) VS − 3SS Vt + ( VV − 2 1SV ) VS − 2
3 2 − 2 3SV VS Vt − 1VV VS − 3VV VS Vt + ( V − 2 1S ) VSS −

− 2 2S VSA − 3 1V VS VSS − 2V VA VSS − 2 3V VS VSt − − 2 2SV VA VS − 2 3S VSt − 2 2V VS VSA − 3V Vt VSS −
2 − 2VV VS VA − 2SS VA } = 0

(A.6) (S, A, t , V ( 2) ) = 0 .

We have now to take into account condition For this purpose we substitute u t by

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−

1 2 2 S VSS − r S VS − S VA + r V 2

into (A.6). We thus derive the equation
1[ 2

S VSS + r VS + VA ] − r +

2 + r S{ S + ( V − 1S ) VS − 1V VS − 2S VA − 2V VS VA } +

 1 2 2  + r S{− 3S − 3V VS } − S VSS − r S VS − S VA + r V  +  2 
2 + S{ A + ( V − 2A ) VA − 2V VA − 1A VS − 1V VA VS } +

 1 2 2  + S{ − 3A − 3V VA } − S VSS − r S VS − S VA + r V  +  2   1 2 2  + t + ( V − 3t ) − S VSS − r S VS − S VA + r V  −  2   1 2 2  − 3V  − S VSS − r S VS − S VA + r V  − 1t VS − 2 t VA +  2   1 2 2  + {− 2V VA − 1V VS }− S VSS − r S VS − S VA + r V  +  2  1 2 2 2 + S { SS + ( 2 SV − 1SS ) VS + ( VV − 2 1SV ) VS } + 2 1 2 2 2 + S {− 3SS − 2 3SV VS − 3VV VS } × 2  1 2 2  × − S VSS − r S VS − S VA + r V  +  2  1 2 2 3 + S {− 1VV VS + ( V − 2 1S ) VSS − 2 2S VSA − 2 − 3 1V VS VSS − 2V VA VSS − 2 3V VS VSt − − 2 2SV VA VS − 2 3S VSt − 2 2V VS VSA } + 1 2 2  1 2 2  S {− 3V VSS } − S VSS − r S VS − S VA + r V  + 2  2  1 2 2 2 + S {− 2VV VS VA − 2SS VA } = 0 2 +
2

(A.7)

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Equate all the coefficients of the partial derivatives of the function V to zero. We are then going to obtain a system of partial differential equations of the unknown coefficients 1 , 2 , 3 and of the vector field (A.3). Before that, and just for the

economy of the calculations, we observe that we can get some simple expressions, by first looking at the coefficients of the mixed partial derivatives. In fact the coefficient of the derivative VSA is
1 2 2 S ( −2 2S ) and that of 2

1 2 2 S ( −2 2 V ) which means that 2S = 0 and 2 V = 0 . 2 Therefore the coefficient 2 does not depend either on S or V:

VS VSA is

2 = 2 (A, t )

(A.8)
1 2 2 S ( −2 3V ) and that of VSt is 2

The coefficient of the derivative VS VSt is

1 2 2 S ( −2 3S ) which means that 3V = 0 and 3S = 0 . Therefore the 2 coefficient 3 does not depend neither on V nor on S:
3 = 3 (A, t )

(A.9)

Because of (A.8) and (A.9), equation (A.7) becomes
1[ 2

S VSS + r VS + VA ] − r +

2 + r S{ S + ( V − 1S ) VS − 1V VS } +

+ S{ A + ( V − 2A ) VA − 1A VS − 1V VA VS } +  1 2 2  + S ( − 3A )  − S VSS − r S VS − S VA + r V  +  2   1 2 2  + t + ( V − 3t ) − S VSS − r S VS − S VA + r V  −  2   1 2 2  − 1t VS − 2 t VA + ( − 1V VS ) − S VSS − r S VS − S VA + r V  +  2 

16

1 2 2 2 S { SS + (2 SV − 1SS ) VS + ( VV − 2 1SV ) VS } + 2 1 2 2 3 + S {− 1VV VS + ( V − 2 1S ) VSS − 3 1V VS VSS } = 0 2 The coefficient of VS VSS is the sum of the terms +
1 2 2  1 2 2 (− 1V )  − S  and S ( −3 1V ) 2  2 

(A.10)

and therefore 1V = 0 , which means that
1 = 1 (S, A, t )

(A.11)

Therefore (A.10) becomes
1[ 2

S VSS + r VS + VA ] − r +

+ r S{ S + ( V − 1S ) VS } + S{ A + ( V − 2A ) VA − 1A VS } +  1 2 2  + S ( − 3A )  − S VSS − r S VS − S VA + r V  +  2   1 2 2  + t + ( V − 3t ) − S VSS − r S VS − S VA + r V  −  2  − 1t VS − 2 t VA +
+

1 2 2 2 S { SS + (2 SV − 1SS ) VS + VV VS } + 2 (A.12)

1 2 2 S ( V − 2 1S ) VSS = 0 2 From (A.12) we get the following coefficients, which have to be zero:

Coefficient of V 2 : S 1 2 2 S VV = 0 2 Coefficient of VA :
S 2 3 + S ( 3t − 2 ) + 1 − 2 t = 0

(A.13)

(A.14)

Coefficient of VSS :

17

2

S 1+

1 2 2 1 2 3 S ( 3t − 2 1S ) + S 3A = 0 2 2

(A.15)

Coefficient of VS : r 1 + r S ( 3t − 1S ) − S 1A + r S2 3A + + 1 2 2 S (2 SV − 1SS ) − 1t = 0 2 (A.16)

Zero-th order coefficient − r + rS S + S A − rS 3 V + t + + r ( V − 3t ) V + 1 2 2 S SS = 0 2 (A.17)

We have now to solve the system of PDEs (A.13)-(A.17). From (A.13) we get V: = (S, A, t ) ⋅ V + (S, A, t ) From (A.15) we get
1S − VV = 0 and therefore

is a linear function with respect to

(A.18)

1 1 1 1= 3 t + S 3A S 2 2

(A.19)

which is a linear differential equation with unknown function 1 . The solution of the previous differential equation is
1=

1 1 (S ⋅ ln S) 3t + S 2 3A + S ⋅ ( , t ) 2 2

(A.20)

where ( , t ) is a function to be determined. From (A.14), using the expression (A.20) for the function equation 3 2 1 3 S + 3t (S ⋅ ln S) + 2 2 + [ 3t − 2 + ( , t )]S − 2 t = 0 (A.21)
1 , we derive the

18

Since the previous equation should hold for every S, each one of the coefficients has to be zero. Therefore we obtain the following four equations:
3

=0

(A.22) (A.23) + ( , t) = 0 (A.24) (A.25)

3t = 0 3t − 2 2t = 0

From the equations (A.22) and (A.23) and taking into account (A.9),we get that 3 is a constant:
3 = a1

(A.26)

From equation (A.25) there follows that 2 is independent of the time t and so
2 = f (A)

(A.27)

Therefore equation (A.24) gives us ( , t ) = f ′(A) The function 1 becomes, as follows from (A.20),
1 = S ⋅ f ′( )

(A.28)

(A.29)

From the above equation we get the following expressions to be used later on.
1S = f ′( ) 1SS = 0 1A = S ⋅ f ′′( A ) 1t = 0

(A.30a) (A.30b) (A.30c) (A.30d)

Equation (A.16), using the expressions (A.18)-(A.20), takes the form
r S f ′( A ) − r S f ′( A ) − S 2 ⋅ f ′′( A ) + 2 S 2 S = 0

Solving the previous equation with respect to S , we find
2 S = f ′′( A )

(A.31)

and therefore

19

2

= f ′′( A ) S + ( A , t )

(A.32)

where ( A, t ) is a function to be determined. We also find from (A.31) that S 2 SS = 0 an expression we shall need later on. From (A.32) we also get
2 t = t (A, t )

(A.33)

(A.34)

and
2 A = f ′′′( A ) S + A ( A , t )

(A.35) )+ (A.36)

From (A.17), using (A.18) we obtain the equation − r ( ⋅ V + ) + r S( S ⋅ V + S ) + S( A ⋅ V + + ( t ⋅ V + t ) + r ( − 3t ) V +

1 2 2 S ( SS ⋅ V + SS ) = 0 2

Equating the coefficients of V to zero, we get from the previous equation the following two equations: − r + rS S + S A + + t + r ( − 3t ) + and − r + rS S + S + t+ 1 2 2 S SS = 0 2 (A.38) introduced in (A.14) is a 1 2 2 S SS = 0 2 (A.37)

Equation (A.38) expresses the fact that the function solution of the original PDE. Equation (A.37) can be expressed as rS 1
2

f ′′(A) + S

1
2

{ f ′′′(A) S + A (A, t )} +

20

+

1
2

t ( A, t ) = 0

(A.39)

The previous equation can be written in equivalent form
t ( A , t ) + { A ( A , t ) + r ⋅ f ′′( A )} S + f ′′′( A ) S 2

=0

(A.40)

Since the previous equation should hold for any value of S, we get the following equations:
f ′′′( A ) = 0
A (A, t ) + r ⋅ f ′′(A) = 0 t ( A, t ) = 0

(A.41) (A.42) (A.43)

From (A.41) we conclude that f (A) is a second degree polynomial with respect to A: f ( A) = a 2 + a 3 A + a 4 A 2 From (A.43), we obtain that ( A , t ) does not depend on t. Therefore
(A, t ) = g(A )

(A.44)

(A.45)

and so equation (A.42) becomes, because of (A.45) and (A.44) g′(A) + r ⋅ 2 a 4 = 0 from which we determine the function g ( A) : g ( A) = −2 r a 4 A + a 5 Therefore (A, t ) = −2 r a 4 A + a 5 Le us collect now everything together. We find from (A.29) and (A.44) that
1 = S ⋅ (a 3 + 2 a 4 A )

(A.46)

(A.47)

(A.48)

From (A.27) and (A.44) we find that
2 = a 2 + a3 A + a4 A 2

(A.49)

We also have from (A.26)

21

3 = a1

(A.50)

From (A.32), using (A.44) and (A.47) we find (S, A, t ) = and so using (A.18), (S, A, t ) = 1
2

1
2

( 2a 4 S − 2 r a 4 A + a5)

(A.51)

( 2 a 4 S − 2 r a 4 A + a 5 ) V + (S, A, t )

(A.52)

Therefore the vector X, the generator of the symmetries, can be expressed as X = (a 3 S + 2 a 4 A S) ∂ + ∂S ∂ ∂ + a1 + ∂A ∂t

+ (a 2 + a 3 A + a 4 A 2 )

1  ∂ +  ( 2 a 4 S − 2 r a 4 A + a 5 ) V + (S, A, t )  2 ∂V The previous expression can be written as X = a1  ∂ ∂ ∂ ∂  + a2 + a3 S  ∂S + A ∂ A  +  ∂t ∂A    ∂ ∂ ∂  2 + + a 4  2AS + A 2 + ( S − r A)V  2 ∂S ∂A ∂V    1 ∂  ∂ + a5   2 V ∂ V  + (S, A, t ) ∂ V    (A.53)

Therefore, we conclude from (A.53) that the Lie algebra of the infinitesimal transformations of the original equation (3.1) is spanned by the five vectors given by (3.4)-(3.8) and the infinite dimensional Lie sub-algebra with generator given by (3.9).

Appendix B.
In this Appendix we transform the original PDE (3.1) to the BS equation.

22

Using the transformation (4.5), we can express the various partial derivatives of the function V(S, A, t ) in terms of the partial derivatives of the function u ( x , t ) . We find that  S Vt = u t A n exp  b   A  S VS = (u x A n − 2 + b u A n −1 ) exp  b   A  S VSS = (u xx A n − 4 + 2b u x A n − 3 + b 2 u A n − 2 ) exp  b   A  S VA = ( −2 u x S A n − 3 + n u A n −1 − b u S A n − 2 ) exp  b   A Substituting the above expressions into the original equation (3.1), we obtain An u t + 1 2 2 S ( u xx A n − 4 + 2b u x A n − 3 + b 2 u A n − 2 ) + 2

+ r S (u x A n − 2 + b u A n −1 ) + + S ( −2 u x S A n − 3 + n u A n −1 − b u S A n − 2 ) − − r u An = 0 The previous equation can be put into the form An u t + 1 2 2 n −4 (S A ) u xx + 2

1 2  + 2 b (S2 A n − 3 ) + r (S A n − 2 ) − 2(S2 A n − 3 ) u x + 2  1 +  2 b 2 (S2 A n − 2 ) + r b (S A n −1 ) + n (S A n −1 ) − 2 − b (S 2 A n − 2 ) − r A n u = 0

]

(B.1)

We now have to simplify the expressions appearing within the brackets of the previous equation. We have

23

(I) S 2 A n − 4 = ( x 2 A 4 ) A n − 4 = x 2 A n (II) 1 2 2b (S2 A n − 3 ) + r (S A n − 2 ) − 2(S2 A n − 3 ) = 2 = 1 2 2 2 4 n −3 + r ( xA 2 ) A n − 2 − 2( x 2 A 4 ) A n − 3 = 2⋅ (x A ) A 2 2

= 2 x 2 A n +1 + r x A n − 2 x 2 A n +1 = r x A n (III) 1 2 2 2 n −2 b (S A ) + r b (S A n −1 ) + n (S A n −1 ) − b (S 2 A n − 2 ) − r A n = 2 = 1 2 4 2 4 n −2 2 +r (x A ) A ( x A 2 ) A n −1 + 4 2 2  2r  2 2 4 n −2  ( xA 2 ) A n −1 − + − − r An = ( x A )A   2 2   = 2
2

x2 An+2 + − 2r
2

2r
2

x A n +1 − 2
2

x A n +1 −

x 2 A n + 2 − r A n = −r A n

Substituting the previous expressions (I)-(III) into equation (B.1), we obtain the equation 1 2 2   An  u t + x u xx + r x u x − r u  = 0 2   from which we get ut + 1 2 2 x u xx + r x u x − r u = 0 2

which is the well-known Black-Scholes (BS) equation.

Appendix C. In this Appendix we shall transform the equation
u t = a 2 u xx + b u x + c u to the heat equation, where a, b, c are constants.

24

C1. Lie Symmetry Analysis of the equation
After performing a Lie symmetry analysis, we have found that the above equation has the six symmetry generators (six vector fields) X1 = ∂ ∂t (C.1) (C.2)

 b2   ∂ 1 ∂ ∂   b − +t + − c t  u X2 =  x x − 2 ∂x  4a 2   ∂u ∂ t   4a 2       ∂ ∂  b 1 2 + t2 + − X3 = x t x 2t − x ⋅2−  2 ∂x ∂ t  4a 8a 2  b2  1  ∂ − − c t 2 − t  u 2  ∂u  4a 2    

(C.3) (C.4)

X4 =

∂ ∂x

X5 = t

∂  1 b + − x−  ∂ x  2a 2 2a 2

 ∂ tu  ∂u 

(C.5) (C.6)

∂ ∂u and the infinite dimensional sub-algebra ∂ X = ( x, t ) ∂u where ( x , t ) is an arbitrary solution of the original equation (ref. [37]) X6 = u

(C.7)

C.2. Transformation of the equation to the heat equation
We shall consider a linear combination of generators such as to convert the BS equation to the heat equation. For this reason, we consider X= where and ∂ 1 ∂ 1 ∂ + +u ∂t ∂x ∂u (C.8)

are constants to be determined.

In order to determine the invariants corresponding to the symmetry (C.8), we consider the system

25

dt dx du = = 1 1 u From the equation dt dx we get by integration = 1 1 t + C1 = x

(C.9)

′ or x = t + C1 and hence one invariant of the equation is
y=x− t

(C.10) t = ln(C 2 u )

From the equation

dt du we get by integration = 1 u

or e t = C 2 u and hence another invariant is = u e− t Therefore u= e t Using (C.12) and (C.10) we find that ut = ( t − ux = y e t u xx = yy e t Therefore the equation u t = a 2 u xx + b u x + c u transforms to the equation (divided by the factor e t )
t− y+ y+

(C.11)

(C.12)

)e t

= a 2 yy + b y + c

from which we get
t =a 2 yy + ( + b) y + (c − )

(C.13)

We determine now of
y and

and

such as to become zero the coefficients

: +b=0

26

c− =0 Therefore we should have invariants will be y = x + b t and Lemma The substitution = e − c t ⋅ u ( t , x = y − bt ) converts the equation u t = a 2 u xx + b u x + c u to the heat equation
t =a 2 yy

= − b and = e−c t ⋅ u

= c , and hence the two

Therefore we arrive at the following

Appendix D. Weber’s Differential Equation
Abramowitz and Stegun ([41], §19.17) define as standard solutions of Weber’s equation 1   y′′ =  a − x 2  y 4   the functions W (a , ± x ) where (cosh a )1 / 4 W (a , ± x ) = (G1y1 2 with G1 = 1 1   + ia 4 2  G3 = 3 1   + ia 4 2  2G 3 y 2 )

and y1 , y 2 are defined by x2  2 1  x4  3 7  x6 + a −  + a − a + y1 = 1 + a 2!  2  4!  2  6! 15  x 211  x   +  a 4 − 11a 2 +  +  a 5 − 25 a 3 + + a 4  8!  4  8! 
8 8

27

and y2 = x + a x 3  2 3  x 5  3 13  x 7 + a −  + a − a  + 3!  2  5!  2  7!

9 11  4  5 2 63  x 3 531  x +  a − 17 a +  +  a − 35 a + + a 4  9!  4  11! 

respectively. xn In the previous expressions, the non-zero coefficients a n of n! are connected by 1 a n + 2 = a ⋅ a n − n (n − 1) a n − 2 4 It is also worth noting that W (a , ± x ) can be expressed in terms of the Confluent Hypergeometric functions ([41], §19.25)  W (a , ± x ) = 2 − 3 / 4   G1  3 1 1 2  H  − , a, x  ± G3  4 2 4  ± where H( m, n , x ) = e − i x 1F1 (m + 1 − i n; 2m + 2; 2i x ) = = e − i x M ( m + 1 − i n; 2m + 2; 2i x ) Note. Weber’s equation can be obtained by separation of variables for the Laplace equation in parabolic cylindrical coordinates (ref [42], Chapter 10). 2G 3  1 1 1  x H  − , a , x 2  G1  4 2 4 

Appendix E. The Airy equation
The differential equation of Airy w ′′ − z w = 0

28

has as solutions the linearly independent solutions Ai(z) and Bi(z)

(Abramowitz and Stegun [41], §10.4). These functions are defined by Ai(z) = c1 f (z) − c 2 g (z) Bi(z) = 3 [ c1 f (z) + c 2 g (z)] where f (z) = 1 + and 2 2 ⋅ 5 7 2 ⋅ 5 ⋅ 8 10 g(z) = z + z 4 + z + z + 4! 7! 10! respectively. The constants c1 and c 2 are given by c1 = Ai(0) = and c 2 = − Ai′(0) = Bi′(0) 1 = 3 31 / 3 (1 / 3) Bi(0) 1 = 3 32 / 3 (2 / 3) z 3k +1 = ∑ 3    3  k (3k + 1)! k =0
∞ k 2 ∞

1 3 1⋅ 4 6 1⋅ 4 ⋅ 7 9 z + z + z + 3! 6! 9!

=

3k 1 z 3k   ∑  3  k (3k ) ! k =0

They also can be expressed in terms of Bessel’s functions of fractional order: Ai(z) = Bi(z) = where 2 = z z 3 1 z K1 / 3 ( ) 3 ([41], §10.4.14)

z { I −1 / 3 ( ) + I1 / 3 ( )} ([41], §10.4.18) 3

References.

29

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[2]

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Description: We consider the path-dependent contingent claims where the underlying asset follows an arithmetic 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. We then transform the BS equation into the heat equation and we provide some general solutions to that equation. This procedure appears for the first time in the finance literature.