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Question: Solve the stochastic differential equation:

$$ dX_t=X^3_t\,dt-X^2_t\,dW_t $$ where: $$ X_0=1 $$

My Attempt:

Using Ito's with: $$ f(x)=\log(x) $$ I get that: $$ d\log(X_t)=dt\left(0+\left(\frac 1{X_T}\right)X_T^3+\left(\frac {-1}{X_T^2}\right)X^4_T\right)+\left(\frac 1{X_T}\right)\left(-X_T^2\right)\,dW_t $$ which gives me that: $$ d\log X_t=-X_t\,dW_t $$ and integrating from 0 to t: $$ X_t=\exp\left(-\int_0^tX_s\,dW_s\right) $$ Is this correct? I don't have too much experience with SDEs so would appreciate any hints/advice.

Cheers

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  • $\begingroup$ Note that your result gives $(X_t)$ as a function of itself. Also, $\log$ is not $C^2$ on $\mathbb R_+$ so you have to be careful when applying Itô's formula to $\log$ when you don't know if $(X_t)$ hits $0$ or not. $\endgroup$ – Ian May 21 '14 at 0:53
  • $\begingroup$ So, could I use another function instead? $\endgroup$ – dimebucker May 21 '14 at 0:57
  • $\begingroup$ Usually, you either consider the process stopped at $T_0$ the hitting time of zero, and then check that $T_0=+\infty$ almost surely, or you solve "as if it were $C^2$", and then check that the derived solution indeed solves the SDE. Aside from this technicality, you have proved that considering $\log$ does not help you derive a solution in this case. $\endgroup$ – Ian May 21 '14 at 0:59
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If $X_t=F(W_t)$, one knows that $\mathrm dX_t=F'(W_t)\mathrm dW_t+\frac12F''(W_t)\mathrm dt$. If ever there exists some function $F$ such that $$F(0)=1,\qquad F'(w)=-F(w)^2,\qquad F''(w)=2F(w)^3,$$ the proof is complete. Can you identify such a function $F$? Be aware though that there might be no solution $(X_t)$ defined for every nonnegative $t$.

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  • $\begingroup$ $$F(W_t)=\frac 1{W_t}$$but why those conditions? $\endgroup$ – dimebucker May 21 '14 at 7:34
  • $\begingroup$ You want $X$ to solve the SDE $dX=X^3dt-X^2dW$. Hence, if $dX=F'(W)dW+\frac12F''(W)dt$, you need that... // Note that $1/W$ is not defined at time $0$ and that you need $F(0)=1$ (as written in my answer) hence no, $F(w)=1/w$ is not a solution. $\endgroup$ – Did May 21 '14 at 7:37
  • $\begingroup$ thanks, this makes sense now, I get that $F(W_t)=\frac {1}{W_t+1}$ $\endgroup$ – dimebucker May 29 '14 at 13:39

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