Hitting time of two dimensional continuous martingale

Let $(\Omega, \mathcal{F}, P)$ be a probability space, on which $\mathcal{F}_t$ is filtration satisfying general conditions. $W_{t}=\left(W_{t}^{1},W_{t}^{2}\right)^{T}$ is a two dimensional Brownian motion, where $x^T$ is the transpose of $x$. Let $Y_t=(Y_t^1,Y_t^2)^T$ be a two dimensional continuous martingale given by $$Y_t=\int_{0}^{t}\sigma\left(s,\omega\right)\thinspace dW_{s}$$ where $\sigma_t$ is a bounded $\mathcal{F}_t$ measurable process and the 2$\times$2 matrix-valued $\sigma(t,\omega)$ is non-singular for all $t\in[0,T]$ and $\omega\in\Omega$.

Define the stopping time $\tau$ of $Y_t$ hitting a point, say $(1,1)^T$ as $$\tau:=\inf\left\{ t>0:Y_{t}=(1,1)^{T}\right\}.$$

The question is, under what condition of $\sigma_t$, can we show that $\mathbb{P}\left(\tau\leq T\right)=0$?

• Suppose $\sigma(t,\omega)$ is diagonal for all $t\in[0,T]$ and $\omega\in\Omega$, then we can apply the time change results for multidimensional local continuous martingale (cf. 1.9 Theorem in Chap 5 of Revuz and Yor 1999) and non-attainability of the origin by the 2-d Brownian path (cf. Prop 3.22 in Chap 3 of Karatzas and Shreve 1991) to show the desired result. But when $\sigma(t,\omega)$ is not diagonal, one can not apply the time change results for multidimensional local continuous martingale, at least directly. Any comment or suggestion is deeply appreciated. Thank you. – Don Oct 21 '14 at 14:51
• P.S. Here is some study for one dimensional non-degenerate martingale. – Don Oct 21 '14 at 14:52