Mixing
How to solve Black Scholes equation directly without using probability
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
Given constants $r,sigma, K>0$, considering the Black-Scholes PDE of a Europoean call:
begincases
fracpartial cpartial t+rsfracpartial cpartial s+frac12sigma^2s^2 fracpartial^2 cpartial s^2-rc = 0\
c(T,s) =(s-K)^+\
c(t,0) = 0
endcases
I learned that we can do the substitution like this:
begincases
tau=T-t\
u=ce^rtau\
x=lnfracsK+(r-frac12sigma^2)(T-t)
endcases
Set $u=u(tau,x)$, then $c(t,s)=e^-rtauu(tau,x)$. Then we have:
begingather*
fracpartial cpartial t
= re^-rtauu+e^-rtau(fracpartial upartial tau fracd taudt + fracpartial upartial x fracpartial xpartial t)
= e^-rtau[ru-fracpartial upartial tau- (r-frac12sigma^2)fracpartial upartial x]\
fracpartial cpartial s = e^-rtaufracpartial upartial x fracpartial xpartial s = e^-rtaufrac1sfracpartial upartial x\
fracpartial^2 cpartial s^2 = e^-rtau[-frac1s^2fracpartial upartial x+frac1s^2fracpartial^2 upartial x^2]=e^-rtaufrac1s^2(fracpartial^2 upartial x^2-fracpartial upartial x)
endgather*
Thus:
beginalign*
fracpartial cpartial t+rsfracpartial cpartial s+frac12sigma^2s^2 fracpartial^2 cpartial s^2-rc
&= e^-rtau[-fracpartial upartial tau- (r-frac12sigma^2)fracpartial upartial x + rfracpartial upartial x + frac12sigma^2(fracpartial^2 upartial x^2-fracpartial upartial x)]\
&= e^-rtau[-fracpartial upartial tau + frac12sigma^2fracpartial^2 upartial x^2]\
&=0
endalign*
Namely we get the heat equation with boundary condition:
begincases
fracpartial upartial tau = fracsigma^22fracpartial^2 upartial x^2\
u(0,x)=K(e^x - 1)^+\
u(tau, -infty) = 0
endcases
I am stuck here. Since the boundary condition is not integrable, I cannot use Fourier transformation with regard to $x$ here. I know there is a probabilistic way of doing it, by computing the conditional expectation. I want to solve the PDE directly. After getting the heat equation here, what should I do next to solve it?
Besides, is there any way to start directly from the PDE of $c$, without change of variable? And, how do we know that change of variable work here? It seems to me very coincidentally the PDE becomes a heat equation in the end.
Thank you so much!
probability pde finance heat-equation
asked Jul 23 at 19:38
Edward Wang
598311
add a comment |Â
up vote
3
down vote
favorite
Given constants $r,sigma, K>0$, considering the Black-Scholes PDE of a Europoean call:
begincases
fracpartial cpartial t+rsfracpartial cpartial s+frac12sigma^2s^2 fracpartial^2 cpartial s^2-rc = 0\
c(T,s) =(s-K)^+\
c(t,0) = 0
endcases
I learned that we can do the substitution like this:
begincases
tau=T-t\
u=ce^rtau\
x=lnfracsK+(r-frac12sigma^2)(T-t)
endcases
Set $u=u(tau,x)$, then $c(t,s)=e^-rtauu(tau,x)$. Then we have:
begingather*
fracpartial cpartial t
= re^-rtauu+e^-rtau(fracpartial upartial tau fracd taudt + fracpartial upartial x fracpartial xpartial t)
= e^-rtau[ru-fracpartial upartial tau- (r-frac12sigma^2)fracpartial upartial x]\
fracpartial cpartial s = e^-rtaufracpartial upartial x fracpartial xpartial s = e^-rtaufrac1sfracpartial upartial x\
fracpartial^2 cpartial s^2 = e^-rtau[-frac1s^2fracpartial upartial x+frac1s^2fracpartial^2 upartial x^2]=e^-rtaufrac1s^2(fracpartial^2 upartial x^2-fracpartial upartial x)
endgather*
Thus:
beginalign*
fracpartial cpartial t+rsfracpartial cpartial s+frac12sigma^2s^2 fracpartial^2 cpartial s^2-rc
&= e^-rtau[-fracpartial upartial tau- (r-frac12sigma^2)fracpartial upartial x + rfracpartial upartial x + frac12sigma^2(fracpartial^2 upartial x^2-fracpartial upartial x)]\
&= e^-rtau[-fracpartial upartial tau + frac12sigma^2fracpartial^2 upartial x^2]\
&=0
endalign*
Namely we get the heat equation with boundary condition:
begincases
fracpartial upartial tau = fracsigma^22fracpartial^2 upartial x^2\
u(0,x)=K(e^x - 1)^+\
u(tau, -infty) = 0
endcases
I am stuck here. Since the boundary condition is not integrable, I cannot use Fourier transformation with regard to $x$ here. I know there is a probabilistic way of doing it, by computing the conditional expectation. I want to solve the PDE directly. After getting the heat equation here, what should I do next to solve it?
Besides, is there any way to start directly from the PDE of $c$, without change of variable? And, how do we know that change of variable work here? It seems to me very coincidentally the PDE becomes a heat equation in the end.
Thank you so much!
probability pde finance heat-equation
asked Jul 23 at 19:38
Edward Wang
598311
1
It is well known that the Black-Scholes PDE can be transformed into the heat equation. Your change-of-variables looks OK, but to check here is something similar. The problem with the integrability of $u(0,x)$ is manageable as shown in the answer.
– RRL
Jul 24 at 4:05
@RRL Thank you! But I have another question. How do we know that way of change of variable?
– Edward Wang
Jul 25 at 2:19
Is the question how would you know to use these transformations? These are just standard tricks, like solving an ODE with an integrating factor: $x' +x = f implies (e^t x)' = e^tf$
– RRL
Jul 25 at 4:02
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Given constants $r,sigma, K>0$, considering the Black-Scholes PDE of a Europoean call:
begincases
fracpartial cpartial t+rsfracpartial cpartial s+frac12sigma^2s^2 fracpartial^2 cpartial s^2-rc = 0\
c(T,s) =(s-K)^+\
c(t,0) = 0
endcases
I learned that we can do the substitution like this:
begincases
tau=T-t\
u=ce^rtau\
x=lnfracsK+(r-frac12sigma^2)(T-t)
endcases
Set $u=u(tau,x)$, then $c(t,s)=e^-rtauu(tau,x)$. Then we have:
begingather*
fracpartial cpartial t
= re^-rtauu+e^-rtau(fracpartial upartial tau fracd taudt + fracpartial upartial x fracpartial xpartial t)
= e^-rtau[ru-fracpartial upartial tau- (r-frac12sigma^2)fracpartial upartial x]\
fracpartial cpartial s = e^-rtaufracpartial upartial x fracpartial xpartial s = e^-rtaufrac1sfracpartial upartial x\
fracpartial^2 cpartial s^2 = e^-rtau[-frac1s^2fracpartial upartial x+frac1s^2fracpartial^2 upartial x^2]=e^-rtaufrac1s^2(fracpartial^2 upartial x^2-fracpartial upartial x)
endgather*
Thus:
beginalign*
fracpartial cpartial t+rsfracpartial cpartial s+frac12sigma^2s^2 fracpartial^2 cpartial s^2-rc
&= e^-rtau[-fracpartial upartial tau- (r-frac12sigma^2)fracpartial upartial x + rfracpartial upartial x + frac12sigma^2(fracpartial^2 upartial x^2-fracpartial upartial x)]\
&= e^-rtau[-fracpartial upartial tau + frac12sigma^2fracpartial^2 upartial x^2]\
&=0
endalign*
Namely we get the heat equation with boundary condition:
begincases
fracpartial upartial tau = fracsigma^22fracpartial^2 upartial x^2\
u(0,x)=K(e^x - 1)^+\
u(tau, -infty) = 0
endcases
I am stuck here. Since the boundary condition is not integrable, I cannot use Fourier transformation with regard to $x$ here. I know there is a probabilistic way of doing it, by computing the conditional expectation. I want to solve the PDE directly. After getting the heat equation here, what should I do next to solve it?
Besides, is there any way to start directly from the PDE of $c$, without change of variable? And, how do we know that change of variable work here? It seems to me very coincidentally the PDE becomes a heat equation in the end.
Thank you so much!
probability pde finance heat-equation
asked Jul 23 at 19:38
Edward Wang
598311
Given constants $r,sigma, K>0$, considering the Black-Scholes PDE of a Europoean call:
begincases
fracpartial cpartial t+rsfracpartial cpartial s+frac12sigma^2s^2 fracpartial^2 cpartial s^2-rc = 0\
c(T,s) =(s-K)^+\
c(t,0) = 0
endcases
I learned that we can do the substitution like this:
begincases
tau=T-t\
u=ce^rtau\
x=lnfracsK+(r-frac12sigma^2)(T-t)
endcases
Set $u=u(tau,x)$, then $c(t,s)=e^-rtauu(tau,x)$. Then we have:
begingather*
fracpartial cpartial t
= re^-rtauu+e^-rtau(fracpartial upartial tau fracd taudt + fracpartial upartial x fracpartial xpartial t)
= e^-rtau[ru-fracpartial upartial tau- (r-frac12sigma^2)fracpartial upartial x]\
fracpartial cpartial s = e^-rtaufracpartial upartial x fracpartial xpartial s = e^-rtaufrac1sfracpartial upartial x\
fracpartial^2 cpartial s^2 = e^-rtau[-frac1s^2fracpartial upartial x+frac1s^2fracpartial^2 upartial x^2]=e^-rtaufrac1s^2(fracpartial^2 upartial x^2-fracpartial upartial x)
endgather*
Thus:
beginalign*
fracpartial cpartial t+rsfracpartial cpartial s+frac12sigma^2s^2 fracpartial^2 cpartial s^2-rc
&= e^-rtau[-fracpartial upartial tau- (r-frac12sigma^2)fracpartial upartial x + rfracpartial upartial x + frac12sigma^2(fracpartial^2 upartial x^2-fracpartial upartial x)]\
&= e^-rtau[-fracpartial upartial tau + frac12sigma^2fracpartial^2 upartial x^2]\
&=0
endalign*
Namely we get the heat equation with boundary condition:
begincases
fracpartial upartial tau = fracsigma^22fracpartial^2 upartial x^2\
u(0,x)=K(e^x - 1)^+\
u(tau, -infty) = 0
endcases
I am stuck here. Since the boundary condition is not integrable, I cannot use Fourier transformation with regard to $x$ here. I know there is a probabilistic way of doing it, by computing the conditional expectation. I want to solve the PDE directly. After getting the heat equation here, what should I do next to solve it?
Besides, is there any way to start directly from the PDE of $c$, without change of variable? And, how do we know that change of variable work here? It seems to me very coincidentally the PDE becomes a heat equation in the end.
Thank you so much!
probability pde finance heat-equation
asked Jul 23 at 19:38
Edward Wang
598311
edited Jul 23 at 21:04
asked Jul 23 at 19:38
Edward Wang
598311
asked Jul 23 at 19:38
Edward Wang
598311
asked Jul 23 at 19:38
Edward Wang
598311
598311
1
It is well known that the Black-Scholes PDE can be transformed into the heat equation. Your change-of-variables looks OK, but to check here is something similar. The problem with the integrability of $u(0,x)$ is manageable as shown in the answer.
– RRL
Jul 24 at 4:05
@RRL Thank you! But I have another question. How do we know that way of change of variable?
– Edward Wang
Jul 25 at 2:19
Is the question how would you know to use these transformations? These are just standard tricks, like solving an ODE with an integrating factor: $x' +x = f implies (e^t x)' = e^tf$
– RRL
Jul 25 at 4:02
add a comment |Â
1
It is well known that the Black-Scholes PDE can be transformed into the heat equation. Your change-of-variables looks OK, but to check here is something similar. The problem with the integrability of $u(0,x)$ is manageable as shown in the answer.
– RRL
Jul 24 at 4:05
@RRL Thank you! But I have another question. How do we know that way of change of variable?
– Edward Wang
Jul 25 at 2:19
Is the question how would you know to use these transformations? These are just standard tricks, like solving an ODE with an integrating factor: $x' +x = f implies (e^t x)' = e^tf$
– RRL
Jul 25 at 4:02
1
1
It is well known that the Black-Scholes PDE can be transformed into the heat equation. Your change-of-variables looks OK, but to check here is something similar. The problem with the integrability of $u(0,x)$ is manageable as shown in the answer.
– RRL
Jul 24 at 4:05
It is well known that the Black-Scholes PDE can be transformed into the heat equation. Your change-of-variables looks OK, but to check here is something similar. The problem with the integrability of $u(0,x)$ is manageable as shown in the answer.
– RRL
Jul 24 at 4:05
@RRL Thank you! But I have another question. How do we know that way of change of variable?
– Edward Wang
Jul 25 at 2:19
@RRL Thank you! But I have another question. How do we know that way of change of variable?
– Edward Wang
Jul 25 at 2:19
Is the question how would you know to use these transformations? These are just standard tricks, like solving an ODE with an integrating factor: $x' +x = f implies (e^t x)' = e^tf$
– RRL
Jul 25 at 4:02
Is the question how would you know to use these transformations? These are just standard tricks, like solving an ODE with an integrating factor: $x' +x = f implies (e^t x)' = e^tf$
– RRL
Jul 25 at 4:02
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
You can apply the Fourier transform with complex wavenumber, $k = k_r + ik_i$.
Since $u(0,x) = mathcalO(e^x)$ as $x to +infty$ and $u(0,x) = 0$ for $x leqslant 0$ the Fourier integral (applied to the initial condition) converges when $Im(k) > 1$,
$$hatu(0,k) = int_-infty^infty u(0,x) e ^ikx , dx = int_-infty^infty underbraceu(0,x) e^-k_i x_in L^1(mathbbR)e ^ik_rx , dx \ = int_0^infty K(e^x-1)e ^ikx,dx = -fracK1 + ik + fracKik \ = fracKik-k^2$$
Apply the transform to the PDE in the usual way and obtain an ODE for the transform $hatu(tau,k)$ of the form
$$fracpartial hatupartial tau = - fracsigma^2k^22 hatu,$$
with the solution
$$hatu(tau,k) = hatu(0,k) e^-sigma^2k^2tau/2 = fracKe^-sigma^2k^2tau/2ik - k^2$$
The inverse transform takes the form of a contour integral in the complex plane
$$u(tau,x) = frac12piint_ibeta-infty^ibeta+infty hatu(tau,x) e ^-ikx , dk, $$
for any $beta > 1$, leading ultimately to the Black-Scholes solution.
answered Jul 23 at 21:15
RRL
43.6k42260
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
You can apply the Fourier transform with complex wavenumber, $k = k_r + ik_i$.
Since $u(0,x) = mathcalO(e^x)$ as $x to +infty$ and $u(0,x) = 0$ for $x leqslant 0$ the Fourier integral (applied to the initial condition) converges when $Im(k) > 1$,
$$hatu(0,k) = int_-infty^infty u(0,x) e ^ikx , dx = int_-infty^infty underbraceu(0,x) e^-k_i x_in L^1(mathbbR)e ^ik_rx , dx \ = int_0^infty K(e^x-1)e ^ikx,dx = -fracK1 + ik + fracKik \ = fracKik-k^2$$
Apply the transform to the PDE in the usual way and obtain an ODE for the transform $hatu(tau,k)$ of the form
$$fracpartial hatupartial tau = - fracsigma^2k^22 hatu,$$
with the solution
$$hatu(tau,k) = hatu(0,k) e^-sigma^2k^2tau/2 = fracKe^-sigma^2k^2tau/2ik - k^2$$
The inverse transform takes the form of a contour integral in the complex plane
$$u(tau,x) = frac12piint_ibeta-infty^ibeta+infty hatu(tau,x) e ^-ikx , dk, $$
for any $beta > 1$, leading ultimately to the Black-Scholes solution.
answered Jul 23 at 21:15
RRL
43.6k42260
add a comment |Â
up vote
3
down vote
accepted
You can apply the Fourier transform with complex wavenumber, $k = k_r + ik_i$.
Since $u(0,x) = mathcalO(e^x)$ as $x to +infty$ and $u(0,x) = 0$ for $x leqslant 0$ the Fourier integral (applied to the initial condition) converges when $Im(k) > 1$,
$$hatu(0,k) = int_-infty^infty u(0,x) e ^ikx , dx = int_-infty^infty underbraceu(0,x) e^-k_i x_in L^1(mathbbR)e ^ik_rx , dx \ = int_0^infty K(e^x-1)e ^ikx,dx = -fracK1 + ik + fracKik \ = fracKik-k^2$$
Apply the transform to the PDE in the usual way and obtain an ODE for the transform $hatu(tau,k)$ of the form
$$fracpartial hatupartial tau = - fracsigma^2k^22 hatu,$$
with the solution
$$hatu(tau,k) = hatu(0,k) e^-sigma^2k^2tau/2 = fracKe^-sigma^2k^2tau/2ik - k^2$$
The inverse transform takes the form of a contour integral in the complex plane
$$u(tau,x) = frac12piint_ibeta-infty^ibeta+infty hatu(tau,x) e ^-ikx , dk, $$
for any $beta > 1$, leading ultimately to the Black-Scholes solution.
answered Jul 23 at 21:15
RRL
43.6k42260
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
You can apply the Fourier transform with complex wavenumber, $k = k_r + ik_i$.
Since $u(0,x) = mathcalO(e^x)$ as $x to +infty$ and $u(0,x) = 0$ for $x leqslant 0$ the Fourier integral (applied to the initial condition) converges when $Im(k) > 1$,
$$hatu(0,k) = int_-infty^infty u(0,x) e ^ikx , dx = int_-infty^infty underbraceu(0,x) e^-k_i x_in L^1(mathbbR)e ^ik_rx , dx \ = int_0^infty K(e^x-1)e ^ikx,dx = -fracK1 + ik + fracKik \ = fracKik-k^2$$
Apply the transform to the PDE in the usual way and obtain an ODE for the transform $hatu(tau,k)$ of the form
$$fracpartial hatupartial tau = - fracsigma^2k^22 hatu,$$
with the solution
$$hatu(tau,k) = hatu(0,k) e^-sigma^2k^2tau/2 = fracKe^-sigma^2k^2tau/2ik - k^2$$
The inverse transform takes the form of a contour integral in the complex plane
$$u(tau,x) = frac12piint_ibeta-infty^ibeta+infty hatu(tau,x) e ^-ikx , dk, $$
for any $beta > 1$, leading ultimately to the Black-Scholes solution.
answered Jul 23 at 21:15
RRL
43.6k42260
You can apply the Fourier transform with complex wavenumber, $k = k_r + ik_i$.
Since $u(0,x) = mathcalO(e^x)$ as $x to +infty$ and $u(0,x) = 0$ for $x leqslant 0$ the Fourier integral (applied to the initial condition) converges when $Im(k) > 1$,
$$hatu(0,k) = int_-infty^infty u(0,x) e ^ikx , dx = int_-infty^infty underbraceu(0,x) e^-k_i x_in L^1(mathbbR)e ^ik_rx , dx \ = int_0^infty K(e^x-1)e ^ikx,dx = -fracK1 + ik + fracKik \ = fracKik-k^2$$
Apply the transform to the PDE in the usual way and obtain an ODE for the transform $hatu(tau,k)$ of the form
$$fracpartial hatupartial tau = - fracsigma^2k^22 hatu,$$
with the solution
$$hatu(tau,k) = hatu(0,k) e^-sigma^2k^2tau/2 = fracKe^-sigma^2k^2tau/2ik - k^2$$
The inverse transform takes the form of a contour integral in the complex plane
$$u(tau,x) = frac12piint_ibeta-infty^ibeta+infty hatu(tau,x) e ^-ikx , dk, $$
for any $beta > 1$, leading ultimately to the Black-Scholes solution.
answered Jul 23 at 21:15
RRL
43.6k42260
edited Jul 24 at 4:01
answered Jul 23 at 21:15
RRL
43.6k42260
answered Jul 23 at 21:15
RRL
43.6k42260
answered Jul 23 at 21:15
RRL
43.6k42260
43.6k42260
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2860715%2fhow-to-solve-black-scholes-equation-directly-without-using-probability%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
It is well known that the Black-Scholes PDE can be transformed into the heat equation. Your change-of-variables looks OK, but to check here is something similar. The problem with the integrability of $u(0,x)$ is manageable as shown in the answer.
– RRL
Jul 24 at 4:05
@RRL Thank you! But I have another question. How do we know that way of change of variable?
– Edward Wang
Jul 25 at 2:19
Is the question how would you know to use these transformations? These are just standard tricks, like solving an ODE with an integrating factor: $x' +x = f implies (e^t x)' = e^tf$
– RRL
Jul 25 at 4:02