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

Question

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.

Answer 1

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).