solving this SDE

[gogo]

Hello

 

Could someone help me solving this SDE

 

dX(t) = -0.5*o^2*exp( -2*X(t) )dt + 2*o*exp( -X(t) )dWt

 

by using Itos formula on function F(t,x)=exp(x)

 

It is quite urgent and I would apriciete Your solution

 

Thank you

[cosine]

Where's the F?

[gogo]

the problem just states that

that by using Ito's formula on function F(t,x)=exp(x)

you must find an explicit term(or formula,i'm not sure which word to us) for proces (X(t) )

[Tom Mattson]

What does the "o" represent? Surely it isn't zero. And what is "dWt"?

[gogo]

o is actualy greek simbol "sigma" but i cant write it on my computer

 

W represents Brown's movement

 

SDE is a stohastic differential equation

 

I was hoping that someone with experiance in solving this could help me

[Dave]

o is actualy greek simbol "sigma" but i cant write it on my computer

 

You may wish to consider using the LaTeX feature to typeset the SDE properly: [math]\sigma[/math].

[the tree]

So the whole thing was meant to say

[math]dX(t) = -0.5\sigma^2 e^{ -2X(t) }dt + 2\sigma e^{-X(t)} dWt[/math]

presumably?

There seems to be too many d's, I don't think I could have read that right.

 

Also, Brown's Movement? Neither Mathworld nor Wikipedia have anything to say on the subject, which is unusual.

[Tartaglia]

[math]

\frac{\partial F}{\partial X} = F(X,t)

\frac{\partial^2 F}{\partial X^2} = F(X,t)

 

dF(X, t) = \frac{\partial F}{\partial X}*dX(t) +0.5*\frac{\partial^2 F}{\partial X^2}* (dX(t))^2

 

(dX(t))^2 = 4\sigma^2 e^{-2X(t)} dt

 

 

[/math]

[Tartaglia]

Latex is giving me a little agravation here but here are the first few steps

 

Wt is a Wiener process or Brownian motion not Browns motion

 

let F(X) = exp(X)

 

F'(X) = exp(X) = F(X)

F''(X) = exp(X) = F(X)

 

dF(X) = F'(X) dX +0.5 F''(X) (dX)^2

 

Bearing in mind dWt - N(0,dt)

 

(dX)^2 = 4*sigma^2*exp(-2X)dt

 

substituting and simplifying gives

 

dF(x) = {1.5*sigma^2dt}/F(X) + 2*sigma*dWt

[Tartaglia]

The effect here is to turn X(t) which has a large negative drift and a large volatility when X(t) becomes negative into a process F(X) which has constant volatility and positive drift which depends inversely on F(X)

[timo]

[math]\frac{\partial F}{\partial X} = F(X,t) [/math]

[math]\frac{\partial^2 F}{\partial X^2} = F(X,t)[/math]

[math]dF(X, t) = \frac{\partial F}{\partial X}*dX(t) +0.5*\frac{\partial^2 F}{\partial X^2}* (dX(t))^2[/math]

[math](dX(t))^2 = 4\sigma^2 e^{-2X(t)} dt[/math]

Try skipping the carriage returns inbetween the latex tags and wrap each line with [ math] ... [ /math] seperately. The multiplication dot is "\cdot", btw.

[Tartaglia]

Thanks Atheist

[Dave]

Also, Brown's Movement? Neither Mathworld nor Wikipedia have anything to say on the subject, which is unusual.

 

MathWorld and Wikipedia are not exhaustive, and should be considered tools at best. I have recently completed a course covering Ecalle's alien derivative which most certainly is a topic in mathematics, but extremely unlikely to appear on Wikipedia.

[the tree]

I wasn't denying that there was such thing as "Brown's movement", I was just saying that I didn't have the slightest inkling as to what he meant. But thankfully, Tartaglia cleared things up anyway.

[gogo]

Latex is giving me a little agravation here but here are the first few steps

 

Wt is a Wiener process or Brownian motion not Browns motion

 

let F(X) = exp(X)

 

F'(X) = exp(X) = F(X)

F''(X) = exp(X) = F(X)

 

dF(X) = F'(X) dX +0.5 F''(X) (dX)^2

 

Bearing in mind dWt - N(0,dt)

 

(dX)^2 = 4*sigma^2*exp(-2X)dt

 

substituting and simplifying gives

 

dF(x) = {1.5*sigma^2dt}/F(X) + 2*sigma*dWt

 

 

 

could you give me the exact solution for X(t) from that??

With some supstitutions i've came to the ekvivalent SDE to the first one I posted:

 

dY(t) = -(0.5*sigma^2)*(1/Y(t))dt + 2*sigma*dW(t)

 

(suptitution is: Y(t) = exp( X(t) )

[Tartaglia]

The discrepency between yours and mine is due to you missing out the second term in the Taylor's expansion, which is really the whole basis of Ito calculus.

 

I've been looking for a second substitution, but I haven't found an appropriate one yet.

 

Substituting exp(0.25x) instead of exp(x) gives you a Martingale (ie no drift) and this may be useful.

 

My advice is to take this to a financial economics or options forum

[river_rat]

I wasn't denying that there was such thing as "Brown's movement", I was just saying that I didn't have the slightest inkling as to what he meant. But thankfully, Tartaglia cleared things up anyway.

 

What you were looking for was "brownian motion" and not "brown's movement" which sounds like the results of a rather strong curry if you ask me.

 

Have you tried the substitution [math]F(X) = e^{2 X(t)}[/math] ? I get a linear SDE (assuming no silly errors that is) which you can solve using the standard methods.

[gogo]

What you were looking for was "brownian motion" and not "brown's movement" which sounds like the results of a rather strong curry if you ask me.

 

Have you tried the substitution [math]F(X) = e^{2 X(t)}[/math] ? I get a linear SDE (assuming no silly errors that is) which you can solve using the standard methods.

 

the result of that is:

 

dy = - (sigma^2)dt + 4*sigma*y^(1/2)dWt

 

i dont know, i dont think there's an explicit solution for "my" SDE

(My apologies for not writing in Latex)

[river_rat]

Gogo, can you show me how you got that SDE from my suggested substitution?

[gogo]

y(X) = exp(2X) => dy=(exp(2X))*2*dx=y*2*dx=2ydx

 

=> dy/2y=-(sigma^2)*(1/y) + 2*sigma*y^(-1/2) * dWt

dy = -2(sigma^2) + 4*sigma*y^(1/2)dWt

[river_rat]

Why didn't you use Ito's formula?