Consider the differential system
\begin{cases}
dX_t &=& -\frac{1}{2}X_t dt - Y_tdB_t, \\
dY_t &=& -\frac{1}{2}Y_tdt + X_tdB_t,
\end{cases}
$X_0 = 1$, $Y_0 = 0$.
Let $X_t$ and $Y_t$ be a solution of the given system. For $m \in \mathbb{N}$, let us define the sequence of stopping time \begin{gather*} \tau_m = \begin{cases} \inf\{t \geq 0: X_t + Y_t \geq m\}, &\text{if}& \{t \geq 0: X_t + Y_t \geq m\} \neq \emptyset, \\ +\infty &\text{otherwise}&. \end{cases} \end{gather*} How to prove that $X_{t \wedge \tau_m}$ and $Y_{t\wedge \tau_m}$ are Itô processes ? (In my definition, we say that $X_t$ is an Itô process if $X_t = M_t + V_t$, where $(M_t)$ is a continuous martingale in $L^2(\Omega,\mathcal{F},P)$ and $(V_t)$ is supposed to be continuous, of bounded variation, adapted and equal to $0$ at $t=0$).
Thank you. Marcus
Suppose that $(X_t,Y_t)_t$ is a solution to the given system. Then in particular, we can write
$$X_t-X_0 = - \underbrace{\int_0^t Y_s \, dB_s}_{\text{local martingale}} - \frac{1}{2} \underbrace{\int_0^t X_s \, ds}_{\text{bounded variation, continuous}}.$$
Applying the optional stopping theorem yields that
$$(t,\omega) \mapsto \left( \int_0^{t \wedge \tau} Y_s \, dB_s \right)(\omega)$$
is a martingale for any stopping time $\tau$. Hence,
$$X_{t \wedge \tau}-X_0 = - \underbrace{\int_0^{t \wedge \tau} Y_s \, dB_s}_{\text{local martingale}} - \frac{1}{2} \underbrace{\int_0^{t \wedge \tau} X_s \, ds}_{\text{bounded variation, continuous}}.$$
Finally, note the "local martingale"-part is a martingale if
$$\sup_{s \in [0,\tau]} |Y_s| \leq c < \infty$$
for some constant $c$ which does not depend on $\omega$. Obviously, this assumption is satisfied for $\tau_m$ (see OP for the definition).
Clearly, the same argumentation applies to $(Y_t)_{t \geq 0}$.