Let $I(\cdot)$ be an indicator, and $B\_{t}$ be an 1-dim standard Brownian motion in a nice filtered probability space
$(\Omega, \mathcal{F}, P, \mathcal{F}\_{t})$. We consider a random process
$$Y\_{t} = I(t\ge 1) B\_{t-1}.$$
Obviously, $Y$ is an $\mathcal F\_{t}$-adapted process with finite quadratic variation 
$\langle Y \rangle\_{t} = (t-1) I(t\ge 1)$.


[Q1.] Is it a Ito process, i.e. there exists a representation of the form
$$Y\_{t} = \int\_{0}^{t} b\_{s}ds + \sigma\_{s}d B\_{s}, \quad \forall t\ge 0$$
for some $\mathcal F\_t$-adapted processes $b$ and $\sigma$?


[Q2.] If No for [Q1.], then is it semi-martingale?


[Q3.] If the answers for the above are both NO, then it gives an example of finite QV which is not semi-martingale. Is there any other such an example which belongs to QV but not in semi-martingale?



Is $B(t-1)$ an Ito process?
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](http://en.wikipedia.org/wiki/Fractional_Brownian_motion) with Hurst-parameter $h<1/2$. (It has even quadratic variation constant $0$).


