Convergence of exponential Brownian martingale to zero almost surely

Solution 1:

As @Did already pointed out, it is much easier to show that

$$a W_t - \frac{1}{2} a^2 t \to - \infty \qquad \text{almost surely as $t \to \infty$.}$$

Recall that the process $$B_t := \begin{cases} t W_{\frac{1}{t}}, & t>0, \\ 0, & t=0 \end{cases}$$ defines a Brownian motion; in particular

$$\lim_{t \to 0} t W_{\frac{1}{t}} = \lim_{t \to 0} B_t = 0.$$

This implies

$$\lim_{t \to \infty} \frac{W_t}{t}= \lim_{s \to 0} s W_{\frac{1}{s}} = 0.$$

Hence,

$$a W_t - \frac{1}{2} a^2 t = t \underbrace{\left( a \frac{W_t}{t} -\frac{a^2}{2} \right)}_{\to - \frac{a^2}{2}} \to - \infty$$

almost surely as $t \to \infty$.