notes on solving the black-scholes equation

8/2/2019 Notes on Solving the Black-Scholes Equation

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Notes on Solving the Black-Scholes Equation

Eola Investments, LLC

[email protected]

October 2009

All Rights Reserved

Quantitative Discove

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ABSTRACT

Solving the Black-Scholes equation is a good exercise for oneself in that it involves several very

useful techniques such as asymptotic analysis, change of variables, Green's function, Fourier transform,

and complex integration that can be applied elsewhere. Presented here is a systematic and self-

contained exercise to derive a closed-form solution of the Black-Scholes equation for a European call

option for the purpose of fun.


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STEP 1: TRANSFORM BLACK-SCHOLES EQUATION TO DIFFUSION EQUATION

It should not be surprised that the Black-Scholes equation can be reduced to a diffusion

equation, given that the Black-Scholes equation comes from a geometrical Brownian motion. However,

the transformation is not obvious. The following derivations can be regarded as side notes to

Silverman's work [1] .

With common notations, the Black-Scholes equation is

V t

rSV S

1

22 S

2 2V

S2rV=0 (1)

It is easy to spot that the coefficients of the second and third terms can be changed to constants by a

transformation of variables S lnS . However, strictly speaking, S has an unit and the log operator

can only apply to unitless quantities. We need to normalize S by some constant with the same unit as S.

A natural choice is the strike price K. Let x=ln S/K , we have

V S

=Vx

dx

dS=Vx

1

S

2V

S2= S

V

S= S

1

S

V

x=1

S2

V

x

1

S

x

V

x

dx

dS=1

S2

V

x

1

S2

2V

x2

and Eq. (1) becomes

V

tr

1

22

V

x

1

2 2

2V

x2rV=0 (2)

In solving a partial differential equation we typically specify initial conditions, where the values

of the function at the start are known. That is, we would like to work on time to maturity, Tt , to

solve the equation, since at the time to maturity the value of a call option is exactly known. Using time

to maturity as the time variable, we have




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V

Ttr

1

22

Vx

1

222V

x2rV=0 (3)

Now let us consider a special case that r=2 / 2 , (which may not be so remote if the current

account deficits of US continue skyrocketing and Fed continues printing papers to finance it), and a

region thatx is large enough that the curvature can be neglected. In this case the second and third terms

are dropped from Eq. (3) and the equation asymptotically becomes

VTt

~rV , r=2/ 2 and x (4)

The solution to Eq. (4) is straightforward; it takes the form

V~CerTt (5)

where Cis not a function of t. This relationship strongly suggests that if we construct a new function

W=VerTt (6)

this function can absorb the rV term. Let us check:

VTt = We

r Tt

Tt =er Tt WTtWe

rTt

V

x=WerTt

x=e

rTt W

x

2V

x2=2 WerTt

x2

=erTt

2W

x2

Inserting above terms into Eq. (3) we have a more simplified Black-Scholes equation

W

Ttr

1

22

Wx

1

2 2

2W

x2=0 (7)

Again let us consider another special case where the time to maturity is so long that a few days

change in time will not affect the value of an option, given the same stock price. In this case, the time




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derivative term is neglected, resulting in a second order ordinary differential equation.

r1

2 2

Wx

1

222W

x2~0 , Tt (8)

Rearranging Eq. (8), we get

2W

x2

r2/2

2/2

Wx

~0 (9)

This equation is very easy to solve. Its solution takes the form of

W~ABe

r2/2

2/2

x

whereA andB are not functions ofx. Note the coefficient ofx. If we group it intox to construct a new

variabley, we may be able to eliminate the coefficients in Eq (8), thus in Eq (7). Let us check.

y=r2/2

2/2

x (10)

W

x=W

y

dy

dx=W

y

r 2/2

2/ 2

2W

x2=

x

W

x=

xW

y

r2/2

2/2=

r 2/2

2/ 2

yW

y

dy

dx=

r 2/2

2 /22 2W

y2

Combining the above equations with Eq. (7), we have

W

Ttr

2/2

2

2/ 2

W y

r

2/2

2

2/2

2W

y2=0

W

[ r2/2

2

2 /2Tt]

W

y

2W

y2=0

Let




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=r2/22

2/2

Tt (11)

we get a cleaner version of the equation

W

=Wy

2

W y

2 (12)

Once again let us consider another special case where the curvature effect can be neglected

(deep out of money options with short time to maturity) . In this case Eq. (12) is reduced to

W

~Wy

, y and 0 (13)

Its solution is

W~D y (14)

whereD is a constant. This solution prompts us to replacey with a new variable,

z=y (15)

Now we regard Was a function of andz, i.e., W=W, z, y . The consequences of this

change of variable are

W, y

=W, z, y

W, z, y

zz

=W, z, y

W, z, y

z

W, y

y=W, z, y

zzy

=W, z, y

z

2W, y

y2 =y

W, yy =

y

W, z, yz =

2W , z, y

z2zy=

2 W, z, y

z2

Substituting the above relationships into the Eq. 12, we finally have the diffusion equation

W

=

2W

z2 , 0 and z [16]




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Let us summarize all changes of variables here:

W=VerTt (6)

=r2/22

2

/2

Tt (11)

z=r2 /22

2/2Tt

r2/2

2/2ln

S

K (17)




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STEP 2: GREEN'S FUNCTION FOR DIFFUSION EQUATION

Let us use the diffusion equation to review the Green's function method for solving differential

equations. A general explicit one dimensional diffusion equation using our notations here is like this:

W

2W

z2=Q, z , 0 and z (17)

which is subject to certain initial condition W0,z and boundary conditions. A very lucky

characteristics of the above equation is its linearity. For example, if

W

2W

z2=Q1, z has a solution W= f1, z , and

W

2W

z2=Q2, z has a solution W= f2, z

under the same initial and boundary conditions, we can immediately know that

W

2W

z2=Q1, zQ2, z has a solution W= f1, zf2, z .

Another aspect of linearity is that if

W

2W

z2=Q, z has a solution W= f, z

then we know that

W

2W

z2=cQ , z has a solution W=cf, z

where c is a constant.

Having these two extremely handy properties in mind, we can solve Eq. (17) by first dividing

the function Q, z into tiny pieces and solve them individually, then add the solutions together to

obtain the global solution. Actually we do not need to solve numerous equations as one might think so




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since we have divided the function Q, z numerous times. We only need to solve it once with one

single piece of Q, z which can be representative to any other tiny pieces. Such a representative

piece is described by a unit impulse and the representative diffusion equation is

W, z

2W, zz

2=' zz ' (18)

where is the Dirac delta function, which is an extension of the Kronecker delta function in discrete

space

Kronecker ij={1, i=j0, i j

(19)

to continuous space. The Dirac delta function has the following two properties:

xx '=0, xx ' (20)

xx 'dx '=1 (21)

The above properties leads to several very useful results

fx xx 'dx '= fx (22)

xx '=d

dxh xx ' , h xx '=

xx 'dx (23)

where h(x-x') is the Heaviside step function

h xx '={0, xx '

1/2, x=x '1, xx '

(24)

Note the notations in Eq. (18). andx are the variables spanning the domain of interest

while ' andx'are the location of the impulse. We call the solution of Eq. (18) the Green's function,

G , ' , z ,z ' , i.e.,




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G , ' , z , z '

2G , ' , z , z '

z2

= 'zz ' (25)

By moving the impulse at ' , z ' to cover the whole surface of Q, z , we can use Eq. (18) to

reconstruct Eq. (17). And the solution of Eq. (17) is just the integration of G, ' , z ,z ' with

respect to ' , z ' over , z domain and with amplitude corrections defined by Q , z .

(Remember the impulse we have studied is an unit impulse. And the amplitude of a real tiny piece of

Q, z is, of course, defined by the amplitude ofQ at ' , z ' ). Mathematically, the solution of

Eq. (17) is

W, z=

0

G,' , z, z ' Q ' , z'dz' d ' (26)

Let us check this claim:

W, z

2W , z

z2

=0

G, ' , z ,z ' Q ' , z ' dz' d ' 2

z2

0

G,' , z , z ' Q ' , z ' dz' d '

= 0

G , ' , z , z ' Q ' , z ' dz ' d '

0

2

z2

G , ' , z , z ' Q ' , z ' dz ' d '

= 0

Q ' , z ' [ G, ' , z , z ' 2

z2

G, ' , z , z ' ]dz' d '

= 0

Q ' , z ' 'zz 'dz' d'

= Q , z

Wait! Not so quick. In the above discussion we have intentionally omitted the initial conditions.

In fact, we can consider the initial conditions are related to a special type of Q, z located at




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=0 . That is, for a homogeneous diffusion equation (Q = 0), we can write

W

2W

z2=0=Q , z , 0 and z (27)

Let us define an effective half width of the Dirac delta function, , so that

d1

Integrating Eq. (27) in the domain [, ] , we have

W, z

d

2W , z

z2

d=

Q , z d

lim0

W, z== lim

0

2 W, z

z2d=Q 0,z

Now, W, z is out of bound so we set it to be zero, and we obtain an astonishing result: the

source function at =0 is the initial condition itself!

lim 0

W , z=Q 0,z (28)

And the solution to Eq. (27) is then

W, z=0

G, ' , z , z ' '0Q ' , z ' dz ' d'

=

G, z , z 'Q 0,z 'dz '

=

G , z , z 'W0,z 'dz ' (29)

The Green's function of a homogeneous diffusion equation is reduced from Eq. (25) by setting '=0

to


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G , z , z '

2

G , z , z '

z2

=zz ' (30)

And using the same argument that we derived Eq. (28) we know that the Green's function at =0 is

a delta function

G0,z , z '=zz ' (31)

Now let us check that the Eq. (29) is indeed the solution of (27).

W, z

2W , z

z2

=

G, z , z 'W0,z 'dz '

2

z2

W, z , z 'Q0,z 'dz '

=

G , z , z 'W0,z ' dz '

2

z2

G , z , z ' W0,z ' dz '

=

W0,z '[ G , z , z ' 2

z2

G , z , z ']dz '

=

W0,z 'zz 'dz '

= W0,z=Q 0,z=Q , z

And check that the initial condition is fulfilled:

W0,z=

G 0,z , z 'W0,z 'dz '=

z , z 'W0,z 'dz '=W0,z

Combining Eq. (26) and Eq. (29), we obtain the general solution of a diffusion equation:

W, z=0

G, ' , z , z'Q ' , z 'dz' d'

G , z , z 'W 0,z 'dz ' (32)

Next, we need to find out what exactly the Green's function is by solving Eq. (24). Denote the




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spacial Fourier transform of the Green's function as g, ' , k , k ' , we have

G, ' , z ,z ' =1

2

g,' ,k , k ' eikz

dk (33)

G , ' , z , z '

= 12

g, ' ,k , k 'eikzdk

2G , ' , z , z '

z2

=1

2

2

z2

g,' ,k , k ' eikz

dk=k2

2

g, ' ,k ,k ' eikz

dk

The Fourier transform pair of the delta function are

k=1

2

zz 'eikzdz=eikz'

2

zz '=1

2

keikz

dk=1

2

eikz'

eikz

dk

Therefore, the Fourier transform of Eq. (24) is

g, ' , k ,k '

k

2g, ' ,k , k ' =

1

2 'e

ikz'(34)

We just transformed a partial differential equation to a first order ordinary differential equation. When

' Eq. (34) becomes

g, ' , k , k '

k2g , ' , k , k '=0

Now, we are concerning with the causal Green's function which means when ' the Green's

function is zero. This is because the impulse at ' can never give a response before ' . The arrow

of time has only one direction. when ' the solution to the above equation is

g, ' , k , k ' =Eek2

, '

where E is a constant to be determined by the continuity of g, ' ,k , k ' at ' below. The value




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of g, ' ,k , k ' at ' is determined by integrating Eq. (34) with respect to [ ', ' ]

and take limit 0 :

'

'g, ' ,k , k '

d

'

'

k2g, ' , k ,k ' d=

1

2'

'

'eikz'd

lim0

g, ' , k ,k ' = '= '

=lim 0

g ',' , k , k '=1

2eikz'

Therefore

lim 0

Eek2 ' =

1

2e

ikz'

E=1

2eikz'

ek2 '

and the Green's function in the k domain is

g, ' ,k ,k ' ={0, '

1

2ek2 'ikz'

, ' (35a)

Or we use the Heaviside step function

g, ' ,k ,k ' =h '

2ek

2'ikz'

(35b)

Substituting Eq. (35b) into Eq. (33)

G, ' , z , z '=1

2

h'ek2 'ikzz '

dk

Now the integration as-is is not readily solvable. Here is a trick. It looks like a error function

erfx =2

0

x

eu

2

du (36)

and we are going to transform it to a error function. Let




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a kb2c=k2 'ikzz '

it is easy to find out that

a=', b=i zz '

2 '

, c=zz '2

4 '

Therefore we can transform the integral to

G, ' , z , z'=h '

2ezz '

2

4 '

e'k izz '2 '

2

dk

Let

u=k i zz '2 '

then

G, ' , z , z'=h '

2ezz '

2

4 ' i zz '/ 2 '

izz '/2 '

e'u

2

du

Now the integration becomes along a complex line and we need to construct a closed integration path

to evaluate this integral, as shown below.

Since there is no singularity in the function, the integration along the closed line I1I2I3I4 is

zero. Therefore I1I2I3=I4 . Furthermore, at infinite the function become zero so I1 and I3

are all zero and I2=I4 . That means the integral along the complex line is the same as the integral

along the real line, i.e.,

k=izz '/2 '

k=0

I1

I2

I3

I4




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G, ' , z , z ' =h '

2e

zz '

2

4 '

e' u

2

du

=h '

2'ezz '

2

4'

e'u

2

d' u

=h '

'e

zz '

2

4 '0

e ' u

2

d ' u

=h '

2 'ezz '

2

4 ' [ 20

e'u

2

d' u]=

h '

2 'ezz '

2

4 ' erf

Therefore

G, ' , z , z ' =h '

4'e

zz '

2

4 ' (37)




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STEP 3: SOLVE BLACK-SCHOLES EQUATION USING GREEN'S FUNCTION

The procedure we derived in Step 2 is more than enough to solve the Black-Scholes equation

for vanilla European options. In fact, the procedure is more reserved for solving other exotic options

That we will demonstrate in other places.

To solve the Black-Scholes equation for a European call option, the general Green's function for

the diffusion equation is reduced to

G, z , z '=1

4ezz '

2

4 (38)

The initial condition in (t,S) domain is

V0,S=h SKSK (39)

Again here we use a Heaviside step function. In the , z domain, the initial condition is

W0,z=h z0 Ke

2/2

r2/2

z1 (40)

The solution is, according to Eq. (29)

W, z=

1

4ezz '

2

4 h z '0Ke

2/2

r2/2

z'1dz '

=K

4

0

ezz '

2

4

2/2

r 2 /2z'

dz 'K

40

ezz '

2

4 dz '

The second integral is easier to solve

K

4

0

ezz '

2

4 dz '=K

z

4

ez 'z

2

4 dz 'z4

=K

[ z40

eu

2

d u0

eu

2

d u]= K

2 [ 2 0z

4

eu2

d u2

0

eu 2

d u ]




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=K

2 [erf z41]Noticing that the error function relates to the normal cumulative distribution function by

Nx =

1

2

[erf

x

21

](41)

we obtain

K

4

0

ezz '

2

4 dz '=K N z2 We use the same trick (besides using Matlab, Mathematica, Mathcad, or any software stating with M)

presented in Step 2 to carry out the first integration

K

4

0

ezz '

2

4

2/2

r 2 /2z'

dz '

= K

40

ez 'z2

2/2

r2/2

2

4

2

/2

r 2/2z 2/2

r 2 /2dz '

= K

4 e

2/2

r2/2 z

2 /2

r2/20

e

z 'z2

2/2

r2/ 2

2

4

dz '

=K

2e

2/2

r2

/2z2 /2

r2

/2[ 2 z2 2 /2r 2/24

ez 'z2

2/2

r 2 /22

4 d

z 'z2

2/2

r2/2

4]

= K2

e

2/2

r2

/2

z

2 /2

r2

/2

[erf

z2

2/2

r2

/ 24

1

]= K e

2/2

r 2/2z 2/2

r2 /2N z22

2/2

r2/2

Therefore, the solution to Eq. (16) for a European call option is




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W, z=K e

2

/2

r2 /2z 2

/2

r 2/2N z2 2

2/2

r2/ 2 K N

z

2 (42)

Recall that

W=VerTt (6)

=r2/22

2/2

Tt (11)

z=r2 /2

2

2/2Tt

r2/2

2/2ln

S

K (17)

we can easily show

z

2=d2=

ln SKr2

2 TtTt

(43)

z

22

2/ 2

r2/2

=d1=z

2 Tt=d2 Tt (44)

and

V t ,S=S Nd1K erTt

N d2 (45)

which is the well known Black-Scholes formula. And some people prefer to call it Bachelier-Thorp

formula. [2]

Happy typing with OpenOffice Writer!




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REFERENCES

[1] Dennis Silverman, Solution of the Black Scholes Equation Using the Green's Function of the

Diffusion Equation (1999) http://www.physics.uci.edu/~silverma/bseqn/bs/bs.html. (Last access

October 16 2009). Note: there is a typo in Eq. (21) of the paper by Silverman.

[2] Espen Gaarder Haug & Nassim Nicholas Taleb, Why we have never used the Black-Scholes-

Merton Option Pricing Formula Fifth Version (2009), Social Science Research Network,

http://papers.ssrn.com/sol3/papers.cfm?abstract_id=1012075&rec=1&srcabs=8156 (Last access

October 22, 2009)

http://www.physics.uci.edu/~silverma/bseqn/bs/bs.htmlhttp://papers.ssrn.com/sol3/papers.cfm?abstract_id=1012075&rec=1&srcabs=8156http://www.eolainvestments.com/http://www.physics.uci.edu/~silverma/bseqn/bs/bs.htmlhttp://papers.ssrn.com/sol3/papers.cfm?abstract_id=1012075&rec=1&srcabs=8156

notes on solving the black-scholes equation

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