What is the probability that the arithmetic OU process $dx_t= \theta(\mu-x_t)dt+\sigma dW_t$ hits barrier $U$ before hitting barrier $L$ when $L<x_0<U$ ?


Assuming $\theta>0$ (take $\tilde{X}=\mu-X$ if it is not the case)

Let us denote $\text{erfi}(x)$ the imaginary error function Let us denote $\tau_L$,resp.$\tau_U$ the hitting time of $L$resp.$U$ where $L<U$

1) Using Ito's lemma, prove that : $$Y_t = \text{erfi}\left(\sqrt{\frac{\theta}{\sigma^2}}\left(X_t-\mu\right)\right) \text{ is a martingale}$$

2) Using optimal stopping theorem, prove that : $$\mathbb{P}(\tau_L\leq \tau_U) = \frac{\text{erfi}\left(\sqrt{\frac{\theta}{\sigma^2}}\left(x_0-\mu\right)\right)-\text{erfi}\left(\sqrt{\frac{\theta}{\sigma^2}}\left(U-\mu\right)\right)}{\text{erfi}\left(\sqrt{\frac{\theta}{\sigma^2}}\left(L-\mu\right)\right)-\text{erfi}\left(\sqrt{\frac{\theta}{\sigma^2}}\left(U-\mu\right)\right)}$$

  • $\begingroup$ Thank you, I'll have a look. But is $\beta$ meant to be $\sigma$ ? $\endgroup$
    – OldSchool
    Jul 15 '16 at 10:14
  • $\begingroup$ you're right, I correct it $\endgroup$
    – MJ73550
    Jul 15 '16 at 12:30
  • $\begingroup$ Thanks. Very nice technique. If you can put more details, more people are able to appreciate it. $\endgroup$
    – Gordon
    Jul 16 '16 at 15:21
  • $\begingroup$ I appreciate your taking the time to respond to this question. Having looked at your answer, a plot of the given $P( T_L ≤ T_U )$ over the range of possible $x_0$'s between $L$ and $U$, seems wrong (the convexity seems to be the wrong way). If I change the signs of $(x_0-μ), (U-μ)$ and $(L-μ)$ to $(x_0+μ), (U+μ)$ and $(L+μ)$ I can get a plausible plot. However, simulations consistently show much greater probability of touching U than this formula suggests, when $x_0$ is near $L$ and μ is set near or beyond U,. I may be doing something wrong, but is it worth checking your answer? $\endgroup$
    – OldSchool
    Jul 17 '16 at 7:49
  • $\begingroup$ $\mathbb{P}(\tau_L\leq \tau_U)=1-\mathbb{P}(\tau_U <\tau_L) $ I.e the probability to hit L before U is equal to 1 minus the probability to hit U before L. I admit I answer not exactly to your question. Sorry. $\endgroup$
    – MJ73550
    Jul 17 '16 at 8:38

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