SERVING THE QUANTITATIVE FINANCE COMMUNITY

Amin
Topic Author
Posts: 2578
Joined: July 14th, 2002, 3:00 am

### Re: Breakthrough in the theory of stochastic differential equations and their simulation

I have started working on backing out volatility from FPE equation that could be input to transition probabilities grid. Meanwhile, I worked a little bit more with optimization of coefficients for various terms of analytic solution of FPE equation. These new coefficients work excellent with mean-correction. I have not perfectly optimized for them but they capture the shape of the density in an excellent way.
Here is the value of coefficients that I have copied from my program

C22(wnStart:Nn)=-.2*(1-gamma)+.0875*sigma0^2./((1-gamma)*w(wnStart:Nn));
C33(wnStart:Nn)=0.0;
C44(wnStart:Nn)=0.1050;

I will do a definitive optimization with a new algorithm after completing the part to back out volatility from FPE. But with mean-correction, you can still use the program for everything.
Please note that C22 takes bessel drift term with a slight modification.

Amin
Topic Author
Posts: 2578
Joined: July 14th, 2002, 3:00 am

### Re: Breakthrough in the theory of stochastic differential equations and their simulation

This is a post I made in off-topic and I am posting it here since I consider it an important message for good Americans and good people of other countries and therefore I copied it here..

https://forum.wilmott.com/viewtopic.php?f=15&t=94796&p=863284#p863284

I also make these claim again that most of these corrupt people have absolutely no sincere intention of doing many things that are in true American interest. With so many corrupt Pakistani generals on their payroll, they still do not force them to stop negative activities in Afghanistan as long as Pakistani generals continue to follow their (based on their hardliner hatred) demands in damaging the interests of our own country, Pakistan. So it seems that it is more important for hardliner generals in American military to promote their agendas of personal hatred than promoting genuine US interests or establishing cherished American/human values and practices in our backward countries. Instead of stopping negative activities by our army that they easily can if they were earnest, these corrupt generals in US army continue to cite these negative activities by our army to increase their funding and destroy reputation of our good-natured public so that they can continue on the other front of pursuing their hateful agendas against the civilian public in our countries.
Can Americans believe that Shakeel Afridi who helped US military in reaching the greatest terrorist on earth still rots in jail despite Trump made tall claims of getting him back in US in a matter of hours. But later convinced by other hardliner conservatives, he totally gave up on the idea since a lot of people in US do not want Shakeel Afridi to be free. In a sophisticated US society, people like Shakeel Afridi would otherwise fail to earn much respect but once good people in US know that this was the guy who helped get the most dangerous terrorist, they would be full of praise and respect for him and this could by association help improve the image of many other good and peaceful Muslim Americans like him and many hardliner conservatives do not want anything like that to ever happen and therefore want Shakeel Afridi to instead rot in jail. I really want to ask good Americans to suggest to the new president Joe Biden to ask the corrupt crooks in US military to finally do one good thing by asking the Pakistani generals on their payroll to free Shakeel Afridi which is a genuine American demand that these corrupt crooks could have easily achieved several long years ago if they were following genuine American interests instead of the agendas based on their personal hatred.
And Diplomatic staff in many countries like mine are directly chosen by people in American army who have no other interest other than finding diplomats who would best help promote their agendas of hardliner conservative hatred by damaging the rightful interest of Pakistani people. I really want to request good Americans when diplomats are shuffled after new president takes oath, please find a nice gentleman/woman for our country who would not only promote genuine friendship between two countries, but also try to advance the good human/American values of democracy, freedom of speech, secularism and all other good things that modern humanity cherishes. I know good Americans are probably the nicest people on earth and I hope they will always be able to control bad elements in their society(that by the way exist in every society). Bad people are only able to prevail in many good and educated societies because good people do not know the complete truth about actions of bad people.
Tomorrow, my mother would be operated for angiography in AFIC by sister's family friend, I hope mind control agents would take it easy on their persecution and drugging of water while I would have to be with my mother.

Amin
Topic Author
Posts: 2578
Joined: July 14th, 2002, 3:00 am

### Re: Breakthrough in the theory of stochastic differential equations and their simulation

Let us suppose we want to solve a general SDE given as
$dX(t)=\mu(X) dt + \sigma(X) dz(t)$          Equation (1)
The fokker planck equation of this SDE is given as
$\frac{\partial p(X,t)}{\partial t} = -\frac{\partial [\mu(X) p(X,t)]}{\partial X} + .5 \frac{\partial^2 [{\sigma(X)}^2 p(X,t)]}{\partial X^2}$ Equation(2)

First a very brief primer on derivatives of normal and derivatives of the SDE variable.
I denote the standard normal density as p(Z)
We want to convert derivatives of the SDE density of X(t) into derivatives of standard normal density. Here are the main equations
$p(Z)=\frac{1}{\sqrt{2 \pi}} exp[-.5 Z^2]$ Equation(3)
derivatives of standard normal density are given in terms of hermite polynomials.
$\frac{\partial p(Z)}{\partial Z}=-Z p(Z)$   Equation(4)
$\frac{\partial^2 p(Z)}{\partial Z^2}=(Z^2-1) p(Z)$  Equation (5)
now we want to convert the density of X(t) and its derivatives in terms of density of Z in which we also use above equations (4-5). In the density of X, each value of X(t) is associated with a particular value of Z through constant CDF points on corresponding densities. So we calculate the density and derivatives of X(t) in terms of Z.
$p(X)=p(Z) |\frac{\partial Z}{\partial X}|$  Equation(6)This follows from the standard change of variables in a density.
$\frac{\partial p(X)}{\partial X}=\frac{\partial p(Z)}{\partial Z} {(\frac{\partial Z}{\partial X})}^2+p(Z) \frac{\partial^2 Z}{\partial X^2} =Z p(Z){(\frac{\partial Z}{\partial X})}^2+p(Z) \frac{\partial^2 Z}{\partial X^2}$  Equation(7)
$\frac{\partial^2 p(X)}{\partial X^2}=\frac{\partial^2 p(Z)}{\partial Z^2} {(\frac{\partial Z}{\partial X})}^3+3 \frac{\partial p(Z)}{\partial Z} \frac{\partial Z}{\partial X} \frac{\partial^2 Z}{\partial X^2}+p(Z) \frac{\partial^3 Z}{\partial X^3}$
$=(Z^2-1) p(Z) {(\frac{\partial Z}{\partial X})}^3 -3 Z p(Z) \frac{\partial Z}{\partial X} \frac{\partial^2 Z}{\partial X^2} +p(Z) \frac{\partial^3 Z}{\partial X^3}$ Equation(8)

$\frac{\partial p(X)}{\partial t}=\frac{\partial [p(Z) \frac{\partial Z}{\partial X}]}{\partial t}= \frac{\partial p(Z)}{\partial Z} {(\frac{\partial Z}{\partial X})}^2 \frac{\partial X}{\partial t}+ p(Z) \frac{\partial^2 Z}{\partial X^2} \frac{\partial X}{\partial t}$
$=Zp(Z) {(\frac{\partial Z}{\partial X})}^2 \frac{\partial X}{\partial t}+ p(Z) \frac{\partial^2 Z}{\partial X^2} \frac{\partial X}{\partial t}$   Equation(9)

Now I write the fokker planck equation after applying the derivatives as
$\frac{\partial p(X,t)}{\partial t} = -\mu(X) \frac{\partial p(X,t)}{\partial X} - \frac{\partial \mu(X)}{\partial X} p(X,t)$
$+({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) p(X,t)$
$+2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} \frac{\partial p(X,t)}{\partial X}+ .5 {\sigma(X)}^2 \frac{\partial^2 p(X,t)}{\partial X^2}$  Equation(10)

Now we write the density of X(t) in terms of density of standard normal Z by substituting from above equations (6-9) in Fokker-Planck Equation (10)

$Zp(Z) {(\frac{\partial Z}{\partial X})}^2 \frac{\partial X}{\partial t}+ p(Z) \frac{\partial^2 Z}{\partial X^2} \frac{\partial X}{\partial t}$
$= -\mu(X) (Z p(Z){(\frac{\partial Z}{\partial X})}^2+p(Z) \frac{\partial^2 Z}{\partial X^2}) - \frac{\partial \mu(X)}{\partial X} p(Z) |\frac{\partial Z}{\partial X}|$
$+({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) p(Z) |\frac{\partial Z}{\partial X}|$
$+2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} (Z p(Z){(\frac{\partial Z}{\partial X})}^2+p(Z) \frac{\partial^2 Z}{\partial X^2})$
$+ .5 {\sigma(X)}^2 ((Z^2-1) p(Z) {(\frac{\partial Z}{\partial X})}^3 -3 Z p(Z) \frac{\partial Z}{\partial X} \frac{\partial^2 Z}{\partial X^2} +p(Z) \frac{\partial^3 Z}{\partial X^3})$  Equation(11)

We see that $p(Z)$ which is a static density is common throughout the equations. We eliminate it and write the new equation as

$Z{(\frac{\partial Z}{\partial X})}^2 \frac{\partial X}{\partial t}+ \frac{\partial^2 Z}{\partial X^2} \frac{\partial X}{\partial t}$
$=- \mu(X) (Z {(\frac{\partial Z}{\partial X})}^2+ \frac{\partial^2 Z}{\partial X^2}) - \frac{\partial \mu(X)}{\partial X} |\frac{\partial Z}{\partial X}|$
$+({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) |\frac{\partial Z}{\partial X}|$
$+2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} (Z {(\frac{\partial Z}{\partial X})}^2+ \frac{\partial^2 Z}{\partial X^2})$
$+ .5 {\sigma(X)}^2 ((Z^2-1) {(\frac{\partial Z}{\partial X})}^3 -3 Z \frac{\partial Z}{\partial X} \frac{\partial^2 Z}{\partial X^2} + \frac{\partial^3 Z}{\partial X^3})$ Equation(12)

We multiply both sides of equations with dt and rearrange the above equation(12) to write

$dX(Z{(\frac{\partial Z}{\partial X})}^2 + \frac{\partial^2 Z}{\partial X^2} )$
$= -\mu(X) (Z {(\frac{\partial Z}{\partial X})}^2+ \frac{\partial^2 Z}{\partial X^2}) dt$
$+2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} (Z {(\frac{\partial Z}{\partial X})}^2+ \frac{\partial^2 Z}{\partial X^2}) dt$
$+ .5 {\sigma(X)}^2 ((Z^2-1) {(\frac{\partial Z}{\partial X})}^3 -3 Z \frac{\partial Z}{\partial X} \frac{\partial^2 Z}{\partial X^2} + \frac{\partial^3 Z}{\partial X^3}) dt$
$- \frac{\partial \mu(X)}{\partial X} |\frac{\partial Z}{\partial X}| dt$
$+({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) |\frac{\partial Z}{\partial X}| dt$  Equation(13)

We multiply both sides of the above equation 13 with ${(\frac{\partial X}{\partial Z})}^3$ to get

$dX(Z (\frac{\partial X}{\partial Z}) + {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^2 Z}{\partial X^2} )$
$= -\mu(X) (Z (\frac{\partial X}{\partial Z})+ {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^2 Z}{\partial X^2}) dt$
$+2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} (Z (\frac{\partial X}{\partial Z})+ {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^2 Z}{\partial X^2}) dt$
$- .5 {\sigma(X)}^2 dt$
$+ .5 {\sigma(X)}^2 (Z^2 -3 Z {(\frac{\partial X}{\partial Z})}^2 \frac{\partial^2 Z}{\partial X^2} +{(\frac{\partial X}{\partial Z})}^3 \frac{\partial^3 Z}{\partial X^3}) dt$
$- \frac{\partial \mu(X)}{\partial X} {(\frac{\partial X}{\partial Z})}^2 dt$
$+({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) {(\frac{\partial X}{\partial Z})}^2 dt$  Equation(14)

We use the relationship  ${(\frac{\partial X}{\partial Z})}^3 \frac{\partial^2 Z}{\partial X^2} =- \frac{\partial^2 X}{\partial Z^2}$ in above equation (14) to get
the following equation

$dX(Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2} )$
$=- \mu(X) (Z (\frac{\partial X}{\partial Z})- \frac{\partial^2 X}{\partial Z^2}) dt$
$+2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} (Z (\frac{\partial X}{\partial Z})- \frac{\partial^2 X}{\partial Z^2}) dt$
$- .5 {\sigma(X)}^2 dt$
$+ .5 {\sigma(X)}^2 (Z^2 +3 Z {(\frac{\partial Z}{\partial X})} \frac{\partial^2 X}{\partial Z^2} +{(\frac{\partial X}{\partial Z})}^3 \frac{\partial^3 Z}{\partial X^3}) dt$
$- \frac{\partial \mu(X)}{\partial X} {(\frac{\partial X}{\partial Z})}^2 dt$
$+({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) {(\frac{\partial X}{\partial Z})}^2 dt$  Equation(15)

The first four terms of the equation can be solved to give a differential of following squared form

$dX(Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2} )$
$= -\mu(X) (Z (\frac{\partial X}{\partial Z})- \frac{\partial^2 X}{\partial Z^2}) dt$
$+2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} (Z (\frac{\partial X}{\partial Z})- \frac{\partial^2 X}{\partial Z^2}) dt$
$- .5 {\sigma(X)}^2 dt$
$=.5 d\Big[\int_{t_0}^{t_1} dw - \int_{t_0}^{t_1} \mu(X) ds - \int_{t_0}^{t_1} 2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} ds$
$+ ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]^2$  Equation(16)

I want to explain that the above term (equation 16)
$.5 d\Big[\int_{t_0}^{t_1} dX - \int_{t_0}^{t_1} \mu(X) ds - \int_{t_0}^{t_1} 2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} ds$
$+ ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]^2$
equals
$+.5 \Big[( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]^2$ at time $t_0$ since the first three integral terms vanish at start time.
and equation 16 equals at time $t_1$
$.5 \Big[X(t_1)-X(t_0) - \int_{t_0}^{t_1} \mu(X) ds - \int_{t_0}^{t_1} 2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} ds$
$+ ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]^2$

So we can write the above equation 15 after substituting equation 16 in it as
$.5 d\Big[\int_{t_0}^{t_1} dX - \int_{t_0}^{t_1} \mu(X) ds - \int_{t_0}^{t_1} 2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} ds$
$+ ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]^2$  Equation(16)
$=+ .5 {\sigma(X)}^2 (Z^2 +3 Z {(\frac{\partial Z}{\partial X})} \frac{\partial^2 X}{\partial Z^2} +{(\frac{\partial X}{\partial Z})}^3 \frac{\partial^3 Z}{\partial X^3}) dt$
$- \frac{\partial \mu(X)}{\partial X} {(\frac{\partial X}{\partial Z})}^2 dt$
$+({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) {(\frac{\partial X}{\partial Z})}^2 dt$  Equation(17)

We can write the above equation after noting that some terms are getting constant multiplier coefficients not immediately known. So we re-write equation (17) as

$.5 \Big[X(t_1)-X(t_0) - \int_{t_0}^{t_1} \mu(X) ds - \int_{t_0}^{t_1} 2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} ds$
$+ ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]^2$
$= +{\Big[ ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]}^2$
$+ .5 {\sigma(X)}^2 (Z^2 +3C_2 Z {(\frac{\partial Z}{\partial X})} \frac{\partial^2 X}{\partial Z^2} +C_4 {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^3 Z}{\partial X^3}) dt$
$- C_3\frac{\partial \mu(X)}{\partial X} {(\frac{\partial X}{\partial Z})}^2 dt$
$+C_5({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) {(\frac{\partial X}{\partial Z})}^2 dt$  Equation(18)

After taking square root on both sides and re-arranging, we get the evolution equation for X(t) as

$X(t_1)=X(t_0) + \int_{t_0}^{t_1} \mu(X) ds + \int_{t_0}^{t_1} 2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} ds$
$- Z (\frac{\partial X}{\partial Z}) + \frac{\partial^2 X}{\partial Z^2}$
$+\sqrt{\Big[ +{\big[ ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\big]}^2 + {\sigma(X)}^2 \big[ Z^2 +3C_2 Z {(\frac{\partial Z}{\partial X})} \frac{\partial^2 X}{\partial Z^2} +C_4 {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^3 Z}{\partial X^3} \big] dt - C_3\frac{\partial \mu(X)}{\partial X} {(\frac{\partial X}{\partial Z})}^2 dt+C_5({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) {(\frac{\partial X}{\partial Z})}^2 dt \Big]}$
Equation(19)

Please note that all the terms on RHS appear under square-root. The last matlab program I posted has two terms outside the square-root and are being linearly added and that is an error. I will post a new program with this solution in a day and it works far better closer to zero.
Here I mention that Lamperti/Bessel form SDE is given as
$dX(t)=\mu(X) dt + \sigma dz(t)$  Equation(20)
and it has a solution given as

$X(t_1)=X(t_0) + \int_{t_0}^{t_1} \mu(X) ds - Z (\frac{\partial X}{\partial Z}) + \frac{\partial^2 X}{\partial Z^2}$
$+\sqrt{\Big[ +{\big[ ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\big]}^2+ {\sigma(X)}^2 \big[Z^2 +3C_2 Z {(\frac{\partial Z}{\partial X})} \frac{\partial^2 X}{\partial Z^2} +C_4 {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^3 Z}{\partial X^3}\big] dt- C_3\frac{\partial \mu(X)}{\partial X} {(\frac{\partial X}{\partial Z})}^2) dt \Big]}$ Equation (21)

Please pardon any error with signs. I quickly wrote the post but I will carefully check it again and fix any errors in a day or two.

Friends, I redid some of the analysis of analytic solution of Fokker-Planck equation. I would further test it on computer and see how the existing algorithm needs to be improved. There might be some minor errors but I am presenting it in the hope that it would be useful. I will come with improvements in FPE solution algorithm code in a few days and post it here. Here is the analytic analysis I have redone.

I am starting with equation(13) of copied post.
$dX(Z{(\frac{\partial Z}{\partial X})}^2 + \frac{\partial^2 Z}{\partial X^2} )$
$= -\mu(X) (Z {(\frac{\partial Z}{\partial X})}^2+ \frac{\partial^2 Z}{\partial X^2}) dt$
$+2 \sigma(X) \frac{\partial \sigma(X)}{\partial X} (Z {(\frac{\partial Z}{\partial X})}^2+ \frac{\partial^2 Z}{\partial X^2}) dt$
$+ .5 {\sigma(X)}^2 ((Z^2-1) {(\frac{\partial Z}{\partial X})}^3 -3 Z \frac{\partial Z}{\partial X} \frac{\partial^2 Z}{\partial X^2} + \frac{\partial^3 Z}{\partial X^3}) dt$
$- \frac{\partial \mu(X)}{\partial X} |\frac{\partial Z}{\partial X}| dt$
$+({(\frac{\partial \sigma(X)}{\partial X})}^2+\sigma(X) \frac{\partial^2 \sigma(X)}{\partial X^2}) |\frac{\partial Z}{\partial X}| dt$  Equation(13)

For simplicity, I just consider the FPE of SDEs in Bessel coordinates with constant volatility. I am sure once this works for SDEs in Bessel coordinates, we would be able to extrapolate it to original coordinates. So we write the Above equation(13) of copied post for SDEs in Bessel coordinates as

$dX(Z{(\frac{\partial Z}{\partial X})}^2 + \frac{\partial^2 Z}{\partial X^2} )$
$= -\mu(X) (Z {(\frac{\partial Z}{\partial X})}^2+ \frac{\partial^2 Z}{\partial X^2}) dt$
$+ .5 {\sigma(X)}^2 ((Z^2-1) {(\frac{\partial Z}{\partial X})}^3 -3 Z \frac{\partial Z}{\partial X} \frac{\partial^2 Z}{\partial X^2} + \frac{\partial^3 Z}{\partial X^3}) dt$
$- \frac{\partial \mu(X)}{\partial X} |\frac{\partial Z}{\partial X}| dt$    Equation(13 A)

I re-write the equation as (I have changed the sign on second line below. It was probably wrong but I will fix this later in previous equations)

$dX(Z{(\frac{\partial Z}{\partial X})}^2 + \frac{\partial^2 Z}{\partial X^2} )$
$-\mu(X) (Z {(\frac{\partial Z}{\partial X})}^2+ \frac{\partial^2 Z}{\partial X^2}) dt$
$=+ .5 {\sigma(X)}^2 ((Z^2-1) {(\frac{\partial Z}{\partial X})}^3 -3 Z \frac{\partial Z}{\partial X} \frac{\partial^2 Z}{\partial X^2} + \frac{\partial^3 Z}{\partial X^3}) dt$
$- \frac{\partial \mu(X)}{\partial X} |\frac{\partial Z}{\partial X}| dt$    Equation(13 B)

I multiply the whole equation with ${(\frac{\partial X}{\partial Z})}^3$ and also re-arrange to get

$dX(Z (\frac{\partial X}{\partial Z}) + {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^2 Z}{\partial X^2} )$
$-\mu(X) (Z (\frac{\partial X}{\partial Z})+ {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^2 Z}{\partial X^2}) dt$
$- .5 {\sigma(X)}^2 dt$
$=+ .5 {\sigma(X)}^2 (Z^2 -3 Z {(\frac{\partial X}{\partial Z})}^2 \frac{\partial^2 Z}{\partial X^2} +{(\frac{\partial X}{\partial Z})}^3 \frac{\partial^3 Z}{\partial X^3}) dt$
$- \frac{\partial \mu(X)}{\partial X} {(\frac{\partial X}{\partial Z})}^2 dt$   Equation(14A)

We want to convert the first three lines of the above Equation(14 A) into a complete differential of square with following terms and we want to calculate the squared differential below so that we would know what terms have to be added/subtracted to first three lines of Equation(14 A) in order to make it a complete squared differential
$.5 d\Big[\int_{t_0}^{t} dX(s) - \int_{t_0}^{t} \mu(X) ds + ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]^2$  Equation(16A)

We apply the square inside the differential and get the following

$.5 d\Big[ {[\int_{t_0}^{t} dX]}^2 -2 [\int_{t_0}^{t} dX] [ \int_{t_0}^{t} \mu(X) ds] +2 [\int_{t_0}^{t} dX] ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})$
$+ {[\int_{t_0}^{t} \mu(X) ds]}^2 -2 [\int_{t_0}^{t} \mu(X) ds] ( Z (\frac{\partial X}{\partial Z})- \frac{\partial^2 X}{\partial Z^2})$
$+ {( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2}^2)}^2 \Big]$  Equation(16A)

We apply the differential on the above equation to get
$[\int_{t_0}^{t} dX(s)] dX(s)+ .5 {\sigma(x)}^2 ds- dX(s) [ \int_{t_0}^{t} \mu(X) ds]- [\int_{t_0}^{t} dX(s)] \mu(X) ds +dX(s) ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})$
$+ [\int_{t_0}^{t} \mu(X) ds] \mu(X) ds - \mu(X) ds ( Z (\frac{\partial X}{\partial Z})- \frac{\partial^2 X}{\partial Z^2})$  Equation(17A)

We only have second,  fifth and sixth term present in the equation coming from FPE equation. We will have to add rest of the terms on both sides and then complete the squared differential on LHS.
Considering first term, $][\int_{t_0}^{t} dX(s)] dX(s)$ we have
$[\int_{t_0}^{t} dX(s)] = \mu(X) (t-t_0) + \sigma(X) (Z(t)-Z(t_0))$
and
$dX(s)] = \mu(X) ds + \sigma(X) dZ(s)$
so we have the first terms in Equation(17A) as
$[\int_{t_0}^{t} dX(s)] dX(s)= {\mu(X)}^2 (t-t_0) dt + {\sigma(X)}^2 (Z(t)-Z(t_0))dZ(t) +\mu(X) \sigma(X) (t-t_0)dZ(t) +\mu(X) \sigma(X) (Z(t)-Z(t_0))d(t)$
and the third term in Equation(17A) as
$dX(s) [ \int_{t_0}^{t} \mu(X) ds]= {\mu(X)}^2 (t-t_0) dt + \mu(X) \sigma(X) (t-t_0) dZ(t)$\
and fourth term in Equation(17A) as
$[\int_{t_0}^{t} dX(s)] \mu(X) ds={\mu(X)}^2 (t-t_0) dt + +\mu(X) \sigma(X) (Z(t)-Z(t_0))d(t)$
and sixth term in Equation(17A) as
$[\int_{t_0}^{t} \mu(X) ds] \mu(X) ds={\mu(X)}^2 (t-t_0) dt$

So we could have from Equation() and Equation()
$.5 d\Big[\int_{t_0}^{t} dX(s) - \int_{t_0}^{t} \mu(X) ds + ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]^2$
$= +{\Big[ ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]}^2$
$+ .5 {\sigma(X)}^2 (Z^2 +3C_2 Z {(\frac{\partial Z}{\partial X})} \frac{\partial^2 X}{\partial Z^2} +C_4 {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^3 Z}{\partial X^3}) dt$
$- C_3\frac{\partial \mu(X)}{\partial X} {(\frac{\partial X}{\partial Z})}^2 dt$
$+{\mu(X)}^2 (t-t_0) dt + {\sigma(X)}^2 (Z(t)-Z(t_0))dZ(t) +\mu(X) \sigma(X) (t-t_0)dZ(t) +\mu(X) \sigma(X) (Z(t)-Z(t_0))d(t)$
$-{\mu(X)}^2 (t-t_0) dt - \mu(X) \sigma(X) (t-t_0) dZ(t)$
$-{\mu(X)}^2 (t-t_0) dt - \mu(X) \sigma(X) (Z(t)-Z(t_0))d(t)$
$+{\mu(X)}^2 (t-t_0) dt$

After simplification, we have
$.5 d\Big[\int_{t_0}^{t} dX(s) - \int_{t_0}^{t} \mu(X) ds + ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]^2$
$= +{\Big[ ( Z (\frac{\partial X}{\partial Z}) - \frac{\partial^2 X}{\partial Z^2})\Big]}^2$
$+ .5 {\sigma(X)}^2 (Z^2 +3C_2 Z {(\frac{\partial Z}{\partial X})} \frac{\partial^2 X}{\partial Z^2} +C_4 {(\frac{\partial X}{\partial Z})}^3 \frac{\partial^3 Z}{\partial X^3}) dt$
$- C_3\frac{\partial \mu(X)}{\partial X} {(\frac{\partial X}{\partial Z})}^2 dt$
$+{\sigma(X)}^2 (.5 dH_2(t)-Z(t_0)dZ(t) )$

We can apply time integral on both sides and then take a square-root to solve the equation. The above analysis is very preliminary and only suggestive and may have some algebra errors. I will come with a detailed program in a few days.

Amin
Topic Author
Posts: 2578
Joined: July 14th, 2002, 3:00 am

### Re: Breakthrough in the theory of stochastic differential equations and their simulation

I want to bring the attention of friends to this slightly old post that is very relevant especially when next president takes office.

viewtopic.php?f=15&t=94796&start=1155#p861398

Amin
Topic Author
Posts: 2578
Joined: July 14th, 2002, 3:00 am

### Re: Breakthrough in the theory of stochastic differential equations and their simulation

Eleven days after receiving the antipsychotic injection(mixed with mind control drugs), I was feeling slightly better today after a very bad patch of more than ten days and had very little bouts of anxiety and uneasiness today.
My mother's heart surgery operation went well on Monday and she had to receive intensive care from hospital staff and our family and it will take her several months to recover properly.
In my research, I tried to back out volatility from the solution of FP equation so that the volatility could be used in the solution of SDE with transition probabilities. I made some progress but could not fully solve the problem. I also found a way to solve for covariance/correlation between SDE solution at two different stages(t1 and t2) in time. I will share it with friends after verifying it further. I also think that it would help us find the transition probabilities volatility analytically for diffusions with mean-reverting and more complicated drifts in the SDE and get us to solve for analytically beyond simple noises. I hope to do more experiments tomorrow and then come back to friends in a few days with the worked out programs.
I also want to request friends to approach the new president and his staff to end mind control of good, nice and talented individuals both in America and abroad. He has far more pressing things to take care at the start but I hope that plight of mind control victims also improves. I, for one, is on brutal mind control torture for past twenty two years and there are many others like me who have been on mind control for more than a decade. And I want to thank all those good and nice humans who have tried for my mind control to end in the past and I have been able to do some creative work in mathematics because my mind control decreased due to efforts from those great human beings. I will be indebted to all those good people for the rest of my life who tried to end my mind control. It is not possible for me to repay this debt other than to have regard and good wishes for those people who helped decrease my mind control.