Amin wrote:Instead of Black Scholes, I was a little more bold and tried solution of CEV noise fokker planck pde. I have given initial steps and I will complete full numerical analysis tomorrow.

We try to find the solution of following fokker planck equation

[$]\frac{\partial p}{\partial t}=.5 \sigma ^2\frac{\partial ^2\left[x^{2\gamma }p\right]}{\partial x^2}=.5 \sigma ^2x^{2\gamma }\frac{\partial ^2p}{\partial x^2}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial p}{\partial x}+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}p[$] Equation(1)

We change coordinates from p(x,t) to [$]\text{w($\zeta $,t)}[$] such that [$]\text{p(x,t)=w($\zeta $,t)}[$] We have after the change of coordinates

[$]\frac{\partial p}{\partial x}=\frac{\partial w}{\partial \zeta }\frac{\partial \zeta }{\partial x}[$]

[$]\frac{\partial p}{\partial t}=\frac{\partial w}{\partial \zeta }\frac{\partial \zeta }{\partial t}+\frac{\partial w}{\partial t}[$]

[$]\frac{\partial ^2p}{\partial x^2}=\frac{\partial ^2w}{\partial \zeta ^2}\left(\frac{\partial \zeta }{\partial x}\right)^2+\frac{\partial w}{\partial \zeta }\frac{\partial ^2\zeta }{\partial x^2}[$] Equations(2)

Substituting Equations(2) in Equation(1), we get

[$]\frac{\partial w}{\partial t}=.5 \sigma ^2x^{2\gamma }(\frac{\partial \zeta }{\partial x})^2\frac{\partial ^2w}{\partial \zeta ^2}+\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)\frac{\partial w}{\partial \zeta }[$]

[$]+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}w[$] Equation(3)

We want to convert the transformed equation(3) into standard heat equation in new coordinates. Assuming we can successfully do that we will get a standard brownian motion as the solution to heat equation in new coordinates.

For that to happen, we will have to impose the conditions

[$].5 \sigma ^2x^{2\gamma }(\frac{\partial \zeta }{\partial x})^2=1[$] Equation(4)

[$]\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)\frac{\partial w}{\partial \zeta }+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}w=0[$] Equation(5)

But we also know once we have satisfied the above conditions, as we stated earlier we will get a standard brownian motion as the solution to heat equation in new coordinates.

[$]w=\frac{1}{\sqrt{4\text{pi} t}}\text{Exp}\left[-\frac{\zeta ^2}{4t}\right][$] Equation(6)

[$]\frac{\partial w}{\partial \zeta }=\frac{1}{\sqrt{4\text{pi} t}}\text{Exp}\left[-\frac{\zeta ^2}{4t}\right]\left(-\frac{\zeta }{2t}\right)[$] Equation(7)

From Equation (4) , we get

[$]\frac{\partial \zeta }{\partial x}=\frac{1}{\sqrt{.5 }\text{$\sigma $x}^{\gamma }}[$] Equation(8)

From above equation

[$]\zeta (x)=\frac{x^{1-\gamma }}{\sqrt{.5 }\sigma (1-\gamma )}-\frac{x_0{}^{1-\gamma }}{\sqrt{.5 }\sigma (1-\gamma )}[$] Equation(9)

[$]\frac{\partial ^2\zeta }{\partial x^2}=\frac{(-\gamma )}{\sqrt{.5 }\text{$\sigma $x}^{\gamma +1}}[$] Equation(10)

Putting Equations (6) and (7) in Equaiton(5) and simplifying to cancel the exponential, we get

[$]\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}-\frac{\partial \zeta }{\partial t}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)\left(-\frac{\zeta }{2t}\right)+\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}=0[$] Equation(11)

We try to solve the following ODE which can be solved for both zeta or x.(I will attempt to give the complete solution tomorrow but we will have to convert the eqution into zeta or x because a map between both variables exists as given by equation(9) for example.

[$]\frac{\partial \zeta }{\partial t}=\left.(.5 \sigma ^2x^{2\gamma }\frac{\partial ^2\zeta }{\partial x^2}+2\gamma \sigma ^2x^{2\gamma -1}\frac{\partial \zeta }{\partial x}\right)-\left(\frac{2t}{\zeta }\right)\gamma \sigma ^2(2\gamma -1)x^{2\gamma -2}[$]

I will try to give complete numerically worked out solution tomorrow. The PDE can then easily be solved in the light of example I cited in the previous post. Here is the reference again. Please look at equation 4.24 in example 4.7 on page 48 of the book, "Essential partial differential equations: Analytical and computational aspects" by David F. Griffiths, John W. Dold, and David J. Silvester. I followed the reference in the above derivation.

Sorry for late update. I will do numerics tomorrow but quickly, equation(11) is not an ODE, it is again a PDE. zeta is a function of both x and t and the map in equation(9) should be valid for t=0 though equation(8) for d/dx_zeta(x,t) should be valid everywhere in t. We can expand zeta(x,t) by method of iterated integrals as

zeta(x,t)=zeta(x,0)+ integral(0 to t) d/dt_(zeta(x,0)) dt + double-integral(0,t) d2/dt2_(zeta(x,0)) dt dt + triple-integral(0,t) d3/dt3_(zeta(x,0)) dt dt dt +....higher order terms

all of the d/dt_(zeta(x,0), d2/dt2_(zeta(x,0)) and d3/dt3_(zeta(x,0)) can be found from the equations 8,9,10 for the initial data combined with the equation 11. Sorry for the quick, non-standard rough notation but will put everything in latex tomorrow and try to add numerics.

Paul wrote:There's a pattern here: list and Amin both ignore the advice of the experts and go merrily their own way, learning nothing in the process.

Do the Black-Scholes PDE with call payoff.

Yes, I will give numerics for Black Scholes lognormal model first and will try to do that tomorrow. Sorry for going the wrong way first.