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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.

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Hints:

  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.
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  • $\begingroup$ 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

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