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$$dX_t = -\frac{1}{2}e^{-2X_t}\ \ dt+e^{-X_t}dB_t, X_0=x_0$$

Hint: solve this equation using the substitution $X_t=u(B_t)$, show that the solution blows up at a finite random time.

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Can one have less input in a question? –  Did Jun 9 '12 at 16:51
    
I checked it again. It's the same as that on the exam paper. –  Bonnie Jun 10 '12 at 9:15
    
So there was a hint, after all. By the way: what is blocking you here? What similar questions can you solve? What did you try? What are your thoughts? –  Did Jun 10 '12 at 10:14
    
I want to use Ito's lemma and then to apply the coefficient matching. But I get a strange result. I think there may be something wrong, so I just keep the oringinal problem here. –  Bonnie Jun 10 '12 at 15:18
    
You might want to show your strange result. –  Did Jun 10 '12 at 15:39

1 Answer 1

Suppose $X_t=u(B_t)$.

Using Ito's lemma, $dX_t=u'dB_t+\frac{1}{2}u''dt$

By coefficient matching with the given fomula $dX_t=e^{-X_t}dB_t-\frac{1}{2}e^{-2X_t} dt$, we get $u'=e^{-u}, u''=-e^{-2u}$.

let $f=u'$, then $f^2=-f'$.

Solve above ODE, $f(t)=\frac{1}{t+C}$, C is constant.

$u'(t)=f(t)=\frac{1}{t+C}$ $\Rightarrow u(t)=\ln(t+C)$ $\Rightarrow X_t=u(B_t)=\ln(B_t+C)$

$X_0=x_0 \Rightarrow \ln(B_0+C)=x_0 \Rightarrow C=e^{x_0}$.

$C=e^{x_0}$ is bounded. After finite random time, $B_t$ hits $-e^{x_0}$ with probability one, and $X_t$ goes to $\infty$.

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