gogo Posted July 1, 2007 Share Posted July 1, 2007 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 Link to comment Share on other sites More sharing options...

cosine Posted July 1, 2007 Share Posted July 1, 2007 Where's the F? Link to comment Share on other sites More sharing options...

gogo Posted July 1, 2007 Author Share Posted July 1, 2007 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) ) Link to comment Share on other sites More sharing options...

Tom Mattson Posted July 1, 2007 Share Posted July 1, 2007 What does the "o" represent? Surely it isn't zero. And what is "dWt"? Link to comment Share on other sites More sharing options...

gogo Posted July 1, 2007 Author Share Posted July 1, 2007 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 Link to comment Share on other sites More sharing options...

Dave Posted July 1, 2007 Share Posted July 1, 2007 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]. Link to comment Share on other sites More sharing options...

the tree Posted July 1, 2007 Share Posted July 1, 2007 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. Link to comment Share on other sites More sharing options...

Tartaglia Posted July 1, 2007 Share Posted July 1, 2007 [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] Link to comment Share on other sites More sharing options...

Tartaglia Posted July 1, 2007 Share Posted July 1, 2007 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 Link to comment Share on other sites More sharing options...

Tartaglia Posted July 1, 2007 Share Posted July 1, 2007 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) Link to comment Share on other sites More sharing options...

timo Posted July 1, 2007 Share Posted July 1, 2007 [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. Link to comment Share on other sites More sharing options...

Tartaglia Posted July 1, 2007 Share Posted July 1, 2007 Thanks Atheist Link to comment Share on other sites More sharing options...

Dave Posted July 1, 2007 Share Posted July 1, 2007 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. Link to comment Share on other sites More sharing options...

the tree Posted July 1, 2007 Share Posted July 1, 2007 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. Link to comment Share on other sites More sharing options...

gogo Posted July 2, 2007 Author Share Posted July 2, 2007 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) ) Link to comment Share on other sites More sharing options...

Tartaglia Posted July 2, 2007 Share Posted July 2, 2007 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 Link to comment Share on other sites More sharing options...

river_rat Posted July 2, 2007 Share Posted July 2, 2007 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. Link to comment Share on other sites More sharing options...

gogo Posted July 3, 2007 Author Share Posted July 3, 2007 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) Link to comment Share on other sites More sharing options...

river_rat Posted July 4, 2007 Share Posted July 4, 2007 Gogo, can you show me how you got that SDE from my suggested substitution? Link to comment Share on other sites More sharing options...

gogo Posted July 5, 2007 Author Share Posted July 5, 2007 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 Link to comment Share on other sites More sharing options...

river_rat Posted July 9, 2007 Share Posted July 9, 2007 Why didn't you use Ito's formula? Link to comment Share on other sites More sharing options...

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