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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?

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  • $\begingroup$ I have not thought of this issue. Just let the filtration be any larger one which runs at least 1-d BM. $\endgroup$ – kenneth May 9 '15 at 8:09
  • $\begingroup$ If it were an Ito process, wouldn't we have to have $b_s = \sigma_s = 0$ for $0 \le s < 1$? Then it ought to be possible to show that $Y$ is $\sigma(B_t - B_1 : t \ge 1)$-measurable, hence independent of $\sigma(B_t : t \le 1\}$. That is absurd since $Y_{3/2} = B_{1/2}$. $\endgroup$ – Nate Eldredge May 10 '15 at 0:55
  • $\begingroup$ @NateEldredge Probably the original question was not precise. In fact, $b$ and $\sigma$ may not be deterministic, and I modified it now. For instance $\sigma_s = B(1/2)$ for all $s>1/2$, then the process $Y_t$ shall not be $\sigma(B_t-B_1: t\ge 1)$-measurable. $\endgroup$ – kenneth May 10 '15 at 5:25
  • $\begingroup$ @StephanSturm Yes, I agree. $\endgroup$ – kenneth May 10 '15 at 13:53
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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 $0$).

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  • $\begingroup$ Thanks. However, I need one $\sigma$ for all $t\ge 0$, that is, $Y_t = \int_0^t \sigma(s, \omega) d B_s$. However, your representation is $\int_0^t \sigma(s, t, \omega) d B_s$. $\endgroup$ – kenneth May 10 '15 at 5:30
  • $\begingroup$ I got your point. Uniqueness of MRT actually tells us it's not a semi-martingale, and I agree. However, I do not quite understand, what do you mean by $I(s\in [0,t-1))$ being not progressively measurable? And how progressive measurability plays a role here? $\endgroup$ – kenneth May 10 '15 at 14:00
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Let $\sigma_t=I(t\ge 1)$ and $b_t=0$ for all $t$. So $\sigma_t$ and $b_t$ are deterministic and certainly predictable, adapted, and integrable. Then the Ito process $$ \int_0^t b_d\,ds + \sigma_s\,dB_s = \int_0^t I(s\ge 1)\,dB_s = I(t\ge 1)(B_t-B_1) $$ is identically distributed (but not identical) with your process $I(t\ge 1)B_{t-1}$.

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    $\begingroup$ I think the question was for identical, not for identical law (that is, the OP really wants to rewrite $Y_t$ itself as a stochastic integral against $B$). Clearly, one must take $b_t=0$ for equality in law (to preserve the martingale property), and I suspect that no $\sigma_t$ will do the job, but at the moment do not see a convincing argument. $\endgroup$ – ofer zeitouni May 9 '15 at 20:17
  • $\begingroup$ I wants to know if $B(t-1)$ is Ito process or not. I agree that your representation is giving an identical distribution. But I am not sure it is sufficient to answer it is Ito process? $\endgroup$ – kenneth May 10 '15 at 5:35
  • $\begingroup$ @kenneth I guess it's not quite $\endgroup$ – Bjørn Kjos-Hanssen May 10 '15 at 5:54

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