Return to Answer

Note that for fixed $t$, we have by the martingale representation theorem that

$$Y_t = \int_0^t I\bigl( s \in [0,t-1) \bigr) \, dB_s.$$

Note in particular that for $t\leq 1$ the indicator yields $0$ trivially. However, the integrand is not progressively measurable in the Brownian filtration, thus it is not an Ito process (and by uniqueness of the martingale represenation theorem we can also find no other representation - it is thus neither an Ito process nor a seminmartingale).

However, if we define the filtration $\tilde{\mathcal{F}}_t = \mathcal{B}(\mathbb{R}_{\geq0}) \otimes \mathcal{F}_{(t-1) \vee 0}$, then it is a martingale with respect to this filtration.

Note that for fixed $t_0$, we have by the martingale representation theorem that

$$Y_{t_0} = \int_0^{t_0} I\bigl( s \in [0,t_0-1) \bigr) \, dB_s.$$

In particular that for $t_0\leq 1$ the indicator yields $0$ trivially. However, if we want to consider the process $(Y_t)$, then the integrand $I\bigl( s \in [0,t-1) \bigr)$ is not progressively measurable in the Brownian filtration (i.e. not measurable w.r.t. the $\sigma$-field $\mathcal{B}([0,s]) \otimes \mathcal{F}_s$). Thus it is not an Ito process and also not a semi-martingale as the martingale representation is unique. This answers [Q1] and [Q2].

As for [Q3], one classical example for a process with finite quadratic variation that is not a semimartingale is fractional Brownian motion with Hurst-parameter $h<1/2$. (It has even quadratic variation constant $0$).

Note that for fixed $t$, we have by the martingale representation theorem that

$$Y_t = \int_0^t I\bigl( s \in [0,t-1) \bigr) \, dB_s.$$

Note in particular that for $t\leq 1$ the indicator yields $0$ trivially. However, the integrand is not progressively measurable in the Brownian filtration, thus it is not an Ito process (and by uniqueness of the martingale represenation theorem we can also find no other representation - it is thus neither an Ito process nor a seminmartingale).

However, if we enlarge the filtration to $\tilde{\mathcal{F}}_t = \mathcal{B}(\mathbb{R}_{\geq0}) \otimes \mathcal{F}_t$, then it is a martingale with respect to this filtration.

Note that for fixed $t$, we have by the martingale representation theorem that

$$Y_t = \int_0^t I\bigl( s \in [0,t-1) \bigr) \, dB_s.$$

Note in particular that for $t\leq 1$ the indicator yields $0$ trivially. However, the integrand is not progressively measurable in the Brownian filtration, thus it is not an Ito process (and by uniqueness of the martingale represenation theorem we can also find no other representation - it is thus neither an Ito process nor a seminmartingale).

However, if we define the filtration $\tilde{\mathcal{F}}_t = \mathcal{B}(\mathbb{R}_{\geq0}) \otimes \mathcal{F}_{(t-1) \vee 0}$, then it is a martingale with respect to this filtration.

Yes, it is both, an Ito process and a semimartingale, if you are ready to enlarge the filtration. We have

$$Y_t = \int_0^t I\bigl( s \in [0,t-1) \bigr) \, dB_s.$$

Note in particular that for $t\leq 1$ the indicator yields $0$ trivially. However, the integrand is not measurable in the Brownian filtration. However, if we enlarge the filtration to encompass also all indicator functions, we get a martingale with respect to this filtration which is an "Ito proceess" (or maybe better called "Skorohod process" as this is a Skorohod integral).

Note that for fixed $t$, we have by the martingale representation theorem that

$$Y_t = \int_0^t I\bigl( s \in [0,t-1) \bigr) \, dB_s.$$

Note in particular that for $t\leq 1$ the indicator yields $0$ trivially. However, the integrand is not progressively measurable in the Brownian filtration, thus it is not an Ito process (and by uniqueness of the martingale represenation theorem we can also find no other representation - it is thus neither an Ito process nor a seminmartingale). 

However, if we enlarge the filtration to $\tilde{\mathcal{F}}_t = \mathcal{B}(\mathbb{R}_{\geq0}) \otimes \mathcal{F}_t$, then it is a martingale with respect to this filtration.