Consider the following SDE: $dX_t=\frac{1}{1+X_t^2}dW_t$, $X_0=1$. For some $a<1<b$ define stopping times $\tau_a=\inf\{t\geqslant 0 : X_t\leqslant a\}$ and $\tau_b=\inf\{t\geqslant 0 : X_t\geqslant b\}$. Find $P(\tau_a<\tau_b)$.

My attempts:

Maybe it is possible to find a solution of this equation. My idea was to use Ito formula for $Z_t=X_t+\frac{1}{3}X_t^3$, but it didn't help.

Thank you in advance.



  1. $(X_t)_{ \geq 0}$ is a martingale.
  2. By the optional stopping theorem, $(X_{t \wedge \tau})_{t \geq 0}$ is a martingale for any stopping time $\tau$. For $\tau := \tau_a \wedge \tau_b$ this implies $$\mathbb{E}(X_{t \wedge \tau})=0. \tag{1}$$
  3. Show that $a \leq X_{t \wedge \tau} \leq b$ and $X_{\tau} \in \{a,b\}$. Conclude from the dominated convergence theorem and Step 2 that $$\mathbb{E}(X_{\tau}) = 0$$ i.e. $$a \mathbb{P}(X_{\tau}=a) + b \mathbb{P}(X_{\tau}=b). \tag{2}$$
  4. We have $$\mathbb{P}(X_{\tau}=a) + \mathbb{P}(X_{\tau}=b) = 1. \tag{3}$$
  5. $(2)$ and $(3)$ is a system of linear equations for $\mathbb{P}(X_{\tau}=a) = \mathbb{P}(\tau_a<\tau_b)$ and $\mathbb{P}(X_{\tau}=b) = \mathbb{P}(\tau_b<\tau_a)$. Solve it.
  • $\begingroup$ How do we know that X is a (true) martingale and not just a local martingale? How does it follow from the SDE? It strikes me as odd at you haven't used it in any way. $\endgroup$ – Rory Nov 11 '17 at 11:43
  • $\begingroup$ @Rory Well, that's a well-known statement. If $H_t$ is such that $$\int_0^t \mathbb{E}(H_s^2) \, ds < \infty \quad \text{for all $t \geq 0$}$$ then $$M_t := \int_0^t H_s \, dW_s$$ is a martingale. For this particular example we are interested in, we have $H_t := 1/(1+X_t^2)$ which clearly satisfies the integrability condition as $|H_t| \leq 1$. $\endgroup$ – saz Nov 11 '17 at 13:37
  • $\begingroup$ Is this equivalent to the statement that if a local martingale is square integrable at every time t, it is a true martingale? Where we have used Itô isometry to calculate the second moment? $\endgroup$ – Rory Nov 11 '17 at 20:22
  • $\begingroup$ @Rory Why would you expect this to be equivalent to the statement which I stated in my comment? Here, we are just considering one particular form of (local) martingales. Anyway, using Ito's formula you can indeed compute/estimate the second moments of $X_t$. $\endgroup$ – saz Nov 11 '17 at 20:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.