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I have a process $Z_t$ that satisfies

$\mathrm{d} Z_t = \dfrac{a}{Z_t}\mathrm{d}t +\mathrm{d}W_t$,

Then I am given that $S=\min\{s>0: Z_s=\epsilon \text{ or } Z_s=\alpha\}$, then I need to find what is $\Pr\{Z_S=\epsilon\}$.

Any help will be appreciated, I have a feeling that I need to find a Martingale transform and go from there but I don't really what to do after that.

Edit: If I did it properly, the martingale transform should be $M_t=Z_t^{1-2a}$.

ActuariallyImpaired
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1 Answers1

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I think I found the solution:

Note that we need to find $\Pr\{Z_S=\epsilon\}$ where $S$ is a valid stopping time. Note that we have a martingale transform for $Z_t$, then we can rewrite $\Pr\{Z_S=\epsilon\}$ as, \begin{align} \Pr\{Z_S=\epsilon\} &= \Pr\{M_S^{\frac{1}{1-2a}}=\epsilon\} \\ &= \Pr\{M_S = \epsilon^{1-2a} \} \end{align} Now note that since $M_t$ is a martingale we have that $\mathrm{E} [M_S] = M_0 = Z_0^{1-2a} = 1$. Therefore we have that \begin{align} \epsilon^{1-2a} \Pr\{M_S = \epsilon^{1-2a} \} + \alpha^{1-2a} \left ( 1-\Pr\{M_S = \epsilon^{1-2a} \} \right ) &= 1 \\ \dfrac{1- \alpha^{1-2a} }{ \epsilon^{1-2a} - \alpha^{1-2a}} &= \Pr\{M_S = \epsilon^{1-2a} \} \\ \dfrac{1- \alpha^{1-2a} }{ \epsilon^{1-2a} - \alpha^{1-2a}} &= \Pr\{Z_S = \epsilon \} \end{align}

ActuariallyImpaired
  • 321