Suppose $B(t),t\in[0,\infty)$ is Standard Brownian Motion. I know that $\exp(\alpha B_t-\dfrac{\alpha^2t}{2})$ is a martingale. I want to use this to show that almost surely, $B(t)/t\to0$ as $t\to\infty$.
I understand I need to use a proper maximal inequality.
Usually what we do is we first fix $\theta>1$ and then $\rho<1$ with $\rho\theta>1$. Then we show $\sum_n P[\sup_{\theta^n\leq t\leq \theta^{n+1}}|\dfrac{B_t}{t}|>\rho^n]<\infty$ which implies by Borel Cantelli that $B(t)/t\to 0$ a.s.
Here we used the Doob Maximal Inequality.
But suppose I want to show using the exponential martingale. How do I show this?
Write $M^{\alpha}_t = \exp\left( \alpha B_t - \frac{\alpha^2}{2}t \right)$ and notice that, if $\alpha > 0$, then
$$ \limsup_{t\to\infty} B_t/t > \alpha \quad \Rightarrow \quad \limsup_{t\to\infty} M^{\alpha}_t = \infty. $$
But since $M^{\alpha}_t$ is a positive martingale, Doob's martingale convergence theorem tells that $M^{\alpha}_t$ converges a.s., and so, $\mathbb{P}\left( \limsup_{t\to\infty} B_t/t > \alpha \right) = 0$. Then letting $\alpha \downarrow 0$ shows that $\limsup_{t\to\infty} B_t/t \leq 0$ a.s. Since the law of $B_t$ is symmetric, this also tells that $\liminf_{t\to\infty} B_t/t \geq 0$ a.s., and so, the desired convergence follows.
But if we are not forced to use $M^{\alpha}_t$, then my favorite proof is to utilize $(B_t/t)_{t > 0} \stackrel{d}= (B_{1/t})_{t > 0}$. Then the solution is as trivial as the statement $\lim_{s\to0} B_s = 0$ a.s.