Consider, for all $x \in \mathbb R$, the process $\left( X_t^x\right)_{t\geq 0}$ unique solution of the following SDE:

$$X_t ^x =x + \int _0 ^t \sigma\left( X_s^x\right) ~dB_s + \int _0 ^t b\left( X_s^x\right) ~ds , \ t\geq 0$$

where $\sigma$ and $b$ are lipschitziennes from $\mathbb R$ into $\mathbb R$ and $\left( B_t\right)_{t\geq 0}$ is the real standard Brownian motion starting from zero.

Consider also the following stopping times:

$\forall x, y \in \mathbb R$, $$\tau_{\left(x,y\right)} = \inf \{ t\geq 0 : X_t ^x =y\}$$

$\forall \left] y,z \right[ \subset \mathbb R$ bounded, $$\tau_{\left(y,z\right)}(x) = \inf \{ t\geq 0 : X_t ^x \notin \left] y,z \right[\}$$

(H1): Suppose $b \equiv 0$and $x \in \left] y,z \right[$

1rst question:

I was asked to show that, $\forall \phi \in \mathcal C^2_c\left( \mathbb R, \mathbb R \right)$ we have

$$\mathbb E \left \{ \phi\left( X_{t\wedge \tau^x}^x\right)\right \} -\phi \left( x\right)= \frac{1}{2}\mathbb E \left \{\int _0^{t\wedge \tau^x}a \phi^"\left( X_s^x\right) ~ds\right \}$$ where $a(x) = \sigma ^2(x)$ and $\tau^x =\tau_{\left(y,z\right)}(x)$

However, I can't show that but I can show according to Itô's lemma that

$$\phi\left( X_{t\wedge \tau^x}^x\right)-\phi \left( x\right)= \frac{1}{2}\int _0^{t\wedge \tau^x}a \phi^"\left( X_s^x\right) ~ds +\int _0^{t\wedge \tau^x} b \phi'\left( X_s^x\right) ~ds +\int _0^{t\wedge \tau^x} \sigma \phi'\left( X_s^x\right) ~dBs$$ then, by taking expected value fo both sides, we have

$$\mathbb E \left \{ \phi\left( X_{t\wedge \tau^x}^x\right)\right \} -\phi \left( x\right)= \frac{1}{2}\mathbb E \left \{\int _0^{t\wedge \tau^x}a \phi^"\left( X_s^x\right) ~ds\right \} + \mathbb E \left \{\int _0^{t\wedge \tau^x}b \phi'\left( X_s^x\right) ~ds\right \}$$

Where am I making a mistake ? Or must it be a mistake in the exercise?

2nd question

After all, I was asked to deduce from this result that $\mathbb P \left\{ \tau^x < \infty\right\}=1$.

Following Ju'x's suggestion, we can choose $\phi \in \mathcal C^2_c\left( \mathbb R, \mathbb R \right)$ such that $\phi'' (x) = \frac {1}{a(x)}$, then by the first result , we have

$$\mathbb E \left \{ t\wedge \tau ^x \right \} = 2 \left( \mathbb E \left \{ \phi\left( X_{t\wedge \tau^x}^x\right)\right \} -\phi \left( x\right) \right )$$

How to conclude?

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1rst question: you wrote that $b \equiv 0$ under hypothesis (H1), this might solve you problem.

2nd question: I will assume that $\sigma$ does not vanish. Using a cut-off function we can find $\phi \in C^2_c(\mathbb{R})$ such that $\phi''=\frac{1}{a}$ on $[y,z]$. The result of the first question gives $$E[\phi(X_{t\wedge \tau^x}^x)] - \phi(x) = \frac{1}{2}E[t\wedge \tau^x].$$ Notice that the l.h.s. is bounded by $2\|\phi\|_\infty$ and that the r.h.s. tends to $\frac{1}{2}E[\tau^x]$ as $t\to\infty$ by Lebesgue's monotone conergence theorem. Hence we finally have $$E[\tau^x] \leq \|\phi\|_\infty < \infty.$$ In particular, $\tau^x < \infty$ a.s.

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Yes of course. I must have been working so much that I'm becoming tired and forgotting to read first of all. Thank you! – Paul Jan 18 '13 at 18:25
Still have no idea for the second question. I thinked to apply the first result for an specific choice of $\phi$. But I've not found anything yet. Does it inspire you? – Paul Jan 18 '13 at 18:27
Well, if $\sigma$ does not vanish, you can take $\phi$ such that $\phi''=a^{-1}$ on $[y,z]$. – Siméon Jan 18 '13 at 18:44
How to conclude? Please, see the edition I did. – Paul Jan 18 '13 at 21:04
Just to confirm. To show that $\tau^x < \infty$ a.s., we can suppose that $\mathbb P \left ( \tau^x =\infty\right)>0$ then $\mathbb E \left [ \tau^x\right ] = \mathbb E \left [ \tau^x\right(\mathbf 1_{\left \{ \tau^x < \infty\right\} } +\mathbf 1_{\left \{ \tau^x = \infty\right\} }\left) \right] = + \infty$. Then we have the result by contradiction. Correct? – Paul Jan 18 '13 at 22:20