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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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    Thanks very much. But I'm still confused about the meaning of "the solution blows up at a finite random time" mean? Does it mean that after finite random time, $B_t$ hits $−C$ , and $X_t$ goes to infinity?2012-06-10

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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 $\infty$.