# A local martingale subtract half its quadratic variation tends to negative infinity

Let $$M$$ be a continuous local martingale with $$M_0=0$$ and $$[M]_\infty=\infty$$ almost surely. I am required to show that $$M_t-\frac{1}{2}[M]_t\to-\infty$$ almost surely as $$t\to\infty$$.

Unfortunately I don't really have any idea how to do this so any advice would be greatly appreciated. I suspect it could possibly be related to the Dubins-Schwarz theorem (that is, $$M$$ can be represented as a stochastic time change of brownian motion) as the hint given is that $$W_t/t\to0$$ almost surely, where $$W$$ is a brownian motion, but I don't see how to obtain the result.

Following your train of thought, we can represent $$M_t$$ as a time-changed Brownian motion $$\beta$$ such that $$\beta_{\langle M \rangle_t} = M_t$$. Thus, $$M_t - \frac{1}{2} \langle M \rangle_t = -\langle M \rangle_t \left( \frac{\beta_{\langle M \rangle_t}}{\langle M \rangle_t} + \frac{1}{2} \right) .$$ As $$t \to \infty$$, the second bracketed term tends to $$1/2$$ almost surely (since $$\langle M \rangle_\infty = \infty$$ a.s., we can treat this as our new time).